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[ "# Minimal realization of a MISO system\n\nGiven the following system:\n\n$$\\dot{x} = \\begin{bmatrix}1 & 1 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 1 \\end{bmatrix}x + \\begin{bmatrix}0 & 1 \\\\ 1 & 0 \\\\ 0 & 1 \\end{bmatrix} u$$ $$y = \\begin{bmatrix}1 & 1 & 1 \\end{bmatrix} x$$\n\nI need to find the minimal realization of the system. I know that the minimal realization means controllable and observable.\n\nThe system given is not controllable and not observable.\n\n• I don't know what minimal realization means, so this is just a comment. It is noteworthy that the $x_2$ equation is decoupled. Does this help you in any way? So what I mean is: $u_1$ only influences $x_2$ directly and $\\dot{x}_2$ is only depending on $x$ and $u_1$. But $u_1$ can indirectly influence $x_1$ and $x_3$ with the coupling terms for $\\dot{x}_1$ and $\\dot{x}_3$. – MrYouMath Apr 19 '17 at 15:03\n\nOne way of doing this is using the Kalman decomposition. For this you need the reachable and unobservable subspaces. These subspaces can be constructed using the image of the controllability matrix and the kernel of the observability matrix respectively. The controllability matrix and its image, the observability matrix and its kernel are,\n\n$$\\mathcal{C} = \\begin{bmatrix} B & A\\,B & A^2 B \\end{bmatrix} = \\begin{bmatrix} 0 & 1 & 1 & 1 & 2 & 1 \\\\ 1 & 0 & 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 1 & 2 & 1 \\end{bmatrix},$$\n\n$$\\text{im}(\\mathcal{C}) = \\text{span}\\left\\{\\begin{bmatrix}0\\\\1\\\\0\\end{bmatrix},\\begin{bmatrix}1\\\\0\\\\1\\end{bmatrix}\\right\\},$$\n\n$$\\mathcal{O} = \\begin{bmatrix} C \\\\ C\\,A \\\\ C\\,A^2 \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 1 \\\\ 1 & 3 & 1 \\\\ 1 & 5 & 1 \\end{bmatrix},$$\n\n$$\\text{ker}(\\mathcal{O}) = \\text{span}\\left\\{\\begin{bmatrix}1\\\\0\\\\-1\\end{bmatrix}\\right\\}.$$\n\nThe Kalman decomposition can now be found by constructing a similarity transformation $M\\, z = x$, such that,\n\n$$\\dot{z} = \\underbrace{M^{-1} A\\, M}_{A^*}\\, z + \\underbrace{M^{-1} B}_{B^*}\\, u,$$\n\n$$y = \\underbrace{C\\, M}_{C^*}\\, z$$\n\nwith $M$ defined as,\n\n$$M = \\begin{bmatrix} \\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O}) & \\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})^\\complement & \\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O}) & \\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})^\\complement \\end{bmatrix}.$$\n\nIn this case $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})$ and $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})^\\complement$ are empty sets, $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})=\\text{ker}(\\mathcal{O})$ and $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})^\\complement=\\text{im}(\\mathcal{C})$, so a possible transformation would be,\n\n$$M = \\begin{bmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & -1 \\end{bmatrix}.$$\n\nUsing this yields,\n\n$$\\left[\\begin{array}{c|c} A^* & B^* \\\\ \\hline C^* \\end{array}\\right] = \\left[\\begin{array}{ccc|cc} 1 & 0 & 0 & 1 & 0 \\\\ 1 & 1 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 \\\\ \\hline 1 & 2 & 0 \\end{array}\\right].$$\n\nThe minimal representation can be obtained by eliminating the rows and columns corresponding to any of the uncontrollable or unobservable parts so containing $\\text{im}(\\mathcal{C})^\\complement$ or $\\text{ker}(\\mathcal{O})$:\n\n• $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})$ is empty so nothing has be be removed due to this.\n• $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})^\\complement$ is the part we want, so the next two rows and columns should be skipped.\n• $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})$ has a dimension of one, so the third row and column should be removed.\n• $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})^\\complement$ is empty so nothing has be be removed due to this.\n\nSo the minimal representation could look like this,\n\n$$\\left[\\begin{array}{c|c} A^*_m & B^*_m \\\\ \\hline C^*_m \\end{array}\\right] = \\left[\\begin{array}{cc|cc} 1 & 0 & 1 & 0 \\\\ 1 & 1 & 0 & 1 \\\\ \\hline 1 & 2 \\end{array}\\right].$$\n\nA way to check whether you did not make any mistakes along the way is to calculate the corresponding transfer function, which should be the same as the one of the original system,\n\n$$G(s) = C (s\\,I - A)^{-1} B = C^*_m (s\\,I - A^*_m)^{-1} B^*_m = \\begin{bmatrix}\\frac{s + 1}{(s - 1)^2} & \\frac{2}{s - 1}\\end{bmatrix}.$$\n\n• How can you decompose the original system in controllable and not controllable? I found a $T$ matrix given by: $$T = \\begin{bmatrix}0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0 \\end{bmatrix}$$ – Unnamed May 1 '17 at 22:08\n• @Juan If you only want to decompose the system is a controllable and not controllable part you first have to pick two unique and non zero combinations of vectors is $\\text{im}(\\mathcal{C})$ for the controllable part and one other vector that is not for the not controllable part. For example: $$T=\\begin{bmatrix}0 & 1 & 1 \\\\ 1 & 1 & 1 \\\\ 0 & 1 & 0\\end{bmatrix}$$ $$T=\\begin{bmatrix}1 & -1 & 0 \\\\ 0 & 2 & 0 \\\\ 1 & -1 & 1\\end{bmatrix}$$both would do the trick, but your example as well. However these transformations will not also decompose the observable from the not observable. – fibonatic May 1 '17 at 23:17\n• Could you explain how you obtained: $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})^\\complement$ Shouldn't this be rather anything but the im(C) and ker(O)? And what about this $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})^\\complement=\\text{ker}(\\mathcal{C})$ (<- why is this not the $\\text{im}\\{C\\}$)? – MrYouMath Oct 27 '17 at 14:05\n• @MrYouMath I did indeed make a typo and indeed $\\text{im}(\\mathcal{C}) \\cap \\text{ker}(\\mathcal{O})^\\complement = \\text{im}(\\mathcal{C})$. And to see why $\\text{im}(\\mathcal{C})^\\complement \\cap \\text{ker}(\\mathcal{O})^\\complement$ has to be empty is because $\\text{im}(\\mathcal{C})$ and $\\text{ker}(\\mathcal{O})$ together span the whole $\\mathbb{R}^3$, so there can't exist any vector this is both not in $\\text{im}(\\mathcal{C})$ and not in $\\text{ker}(\\mathcal{O})$. – fibonatic Oct 27 '17 at 18:56\n• @MrYouMath It is actually also possible to construct a $M$ that gives a Kalman decomposition using the Hautus test. So find for which eigenvalue of $A$ the matrix $\\begin{bmatrix}A-\\lambda_i\\,I & B\\end{bmatrix}$ or $\\begin{bmatrix}A^\\top-\\lambda_i\\,I & C^\\top\\end{bmatrix}$ are or aren't full rank. And for the columns of $M$ you can use the corresponding eigenvectors $\\vec{v}_i$. – fibonatic Oct 28 '17 at 5:14\n\nUsing Matlab R2018a with minreal function from Robust control package and inputting original A, B, C, D matrices I got:\n\n\\begin{array}{c|c} A^*_m & B^*_m \\\\ \\hline C^*_m \\end{array} = \\begin{array}{cc|cc} 1 & 0 & 1 & 0 \\\\ \\sqrt2 & 1 & 0 & \\sqrt2 \\\\ \\hline 1 & \\sqrt2 \\end{array}.\n\nOne state was removed." ]

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[ "Mean - Ritvvij Parrikh Mean | Ritvvij Parrikh", null, "Made with Humane Club", null, "# Mean\n\nMathematically, the mean is calculated by summing up all of the values and then divide by the count of rows.\n\nExample: Seconds it takes for a common person to run 100 meters. 17, 19, 18, 14, 17. The average is 17 seconds.\n\nThe details: Mean (or average) is useful only if the data is normally distributed and thus is more sensitive to outliers.\n\n• Mean is useful for natural outcomes because nature rarely produces extremes, e.g. it is unlikely that a human will have a height of 9 feet.\n• In contrast, means isn’t useful for human outcomes because humans produce extreme outcomes, e.g. 2008 economic downturn.\n\nIn Google Sheets, use the function =AVERAGE", null, "Copied\nTags\nz\n\nNone yet" ]

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[ "Total de visitas: 18580\nLinear Algebra download\nLinear Algebra download\n\n## Linear Algebra. Arnold J. Insel, Lawrence E. Spence, Stephen H. Friedberg", null, "Linear.Algebra.pdf\nISBN: 0135371023,9780135371022 | 545 pages | 14 Mb", null, "Download Linear Algebra\n\nLinear Algebra Arnold J. Insel, Lawrence E. Spence, Stephen H. Friedberg\nPublisher: Prentice Hall College Div\n\nLinear Algebra Primer: Part 4- Functions In this tutorial, Functions will be introduced. I met some friends from Science faculty in this module. This is a test conditions.I know Linear Algebra good.. I got 75 points out of 100 points on the final exam. Linear Algebra is one of the core modules of not only Computer Science but also people of the Applied Mathematics major. Linear algebra, projecting vector - posted in For Beginners: I am reading book 3D Math primer for graphics and game developement and there is one thing that confuses me. Suppose we have two points x and y and a third point z halfway between the first two points. Think about an image — it is a matrix; and most of the time you'll work on matrices. I'm a Computer Science student.I've just completed a Linear algebra course. Baixe grátis o arquivo (Book)_Contemporary_Linear_Algebra_by_Howard_Anton,_Robert_C._Busby.pdf enviado por Rodrigo no curso de Engenharia Mecânica na UP - UNICEMP. Much of this terminology is fairly basic, but critical to upper level mathematics, including linear algebra. [Linear Algebra] quick question Homework & General Information Forum. But believe me, the most rudimentary works you need to do in these fields are essentially “Linear Algebra”. I recently dived into studying linear algebra, and was intrigued to read Robert Talbert's post about the use of peer instruction in linear algebra classes.\n\nMechanics of Composite Materials epub" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Factorizing...\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg74119] Re: Factorizing...\n• From: Peter Pein <petsie at dordos.net>\n• Date: Mon, 12 Mar 2007 04:31:24 -0500 (EST)\n• References: <esr0lj\\$6hp\\[email protected]> <esu5qh\\$riq\\[email protected]> <et0no6\\$mq4\\[email protected]>\n\n```Bruno Campanini schrieb:\n> \"Jean-Marc Gulliet\" <jeanmarc.gulliet at gmail.com> wrote in message\n> news:esu5qh\\$riq\\$1 at smc.vnet.net...\n>\n>> Using the correct syntax strongly helps. See \" The Mathematica Book\n>> Online / A Practical Introduction to Mathematica / Numerical\n>> Calculations / 1.1.3 Some Mathematical Functions\" at\n>> http://documents.wolfram.com/mathematica/book/section-1.1.3\n>>\n>> In:=\n>> Simplify[Factor[2 + Sqrt] == ((1 + Sqrt)/2)^3]\n>>\n>> [...disregard the warning message...]\n>>\n>> Out=\n>> True\n>\n> I don't want to get that.\n> I would like to have:\n>\n> In=\n> Factor[2 + Sqrt]\n>\n> Out=\n> ((1 + Sqrt)/2)^3\n>\n>\n> Out=2 + Sqrt\n>\n> Bruno\n>\n>\n\nHi Bruno,\n\ntry RootReduce:\n\n(2 + Sqrt)^(1/3) // RootReduce\n\n--> (1/2)*(1 + Sqrt)\n\nFactor can not guess, wether you want sth.^3 or for instance\n\nExpand[Sqrt[1/2 + Sqrt/2]^6]\n\n--> 2 + Sqrt\n\nhth,\nPeter\n\n```\n\n• Prev by Date: Re: how to Split sparsearray\n• Next by Date: Write a computer program to using Simpson's rule (Have to revise)\n• Previous by thread: Re: Factorizing...\n• Next by thread: Re: Factorizing..." ]

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[ "# How do you solve 11 + sqrt(2x+1)=2?\n\n## It was suggested I post this\n\nJan 16, 2018\n\nNo solutions.\n\n#### Explanation:\n\nOur goal is to isolate $x$ so we can find its value.\n\n$11 + \\sqrt{2 x + 1} = 2$\n\n$\\Rightarrow \\sqrt{2 x + 1} = - 9$\n\nWe see that the equation has no solutions because the radical operator over $\\mathbb{R}$ always returns positive values. So the equation has no solution.\n\nJan 16, 2018\n\n#### Explanation:\n\nYour goal here is to isolate the variable.\n\nFirst, start by subtracting 2 from both sides, as such:\n\n$11 - 2 + \\sqrt{2 x + 1} = 2 - 2$\n\n$9 + \\sqrt{2 x + 1} = 0$\n\nThen subtract 9 from both sides:\n\n$9 - 9 + \\sqrt{2 x + 1} = 0 - 9$\n\n$\\sqrt{2 x + 1} = - 9$\n\nNow square both sides:\n\n${\\left(\\sqrt{2 x + 1}\\right)}^{2} = {\\left(- 9\\right)}^{2}$\n\n$2 x + 1 = 81$\n\nThen subtract 1 from both sides:\n\n$2 x + 1 - 1 = 81 - 1$\n\n$2 x = 80$\n\nFinally, divided both sides by 2:\n\n$\\frac{2 x}{2} = \\frac{80}{2}$\n\n$x = 40$\n\nPlug the answer into the original equation to check if it is correct:\n\n$11 + \\sqrt{2 \\left(40\\right) + 1} = 2$\n\n$11 + 9 = 2$\n\n$20 \\ne 2$\n\nWhen plugged in, we can see that $x = 40$ is not correct, so this problem has no solution.\n\nJan 16, 2018\n\nNo Real Solution\n\n#### Explanation:\n\nBegin by subtracting $11$ from both sides\n\n$\\cancel{11 \\textcolor{red}{- 11}} + \\sqrt{2 x + 1} = 2 \\textcolor{red}{- 11}$\n\n$\\sqrt{2 x + 1} = - 9$\n\nSquare both sides to eliminate the radical on the left side\n\n${\\sqrt{2 x + 1}}^{\\textcolor{red}{2}} = - {9}^{\\textcolor{red}{2}}$\n\n$2 x + 1 = 81$\n\nSubtract $1$ from both sides\n\n$2 x + \\cancel{1 \\textcolor{red}{- 1}} = 81 \\textcolor{red}{- 1}$\n\n$2 x = 80$\n\nDivide both side by $2$\n\n$\\frac{\\cancel{2}}{\\cancel{\\textcolor{red}{2}}} x = \\frac{80}{\\textcolor{red}{2}}$\n\n$x = 40$\n\nHowever if we verify this by substituting $40$ for $x$ in the original expression we'll find:\n\n$11 + \\sqrt{2 \\left(\\textcolor{red}{40}\\right) + 1} = 2$\n\n$11 + \\sqrt{81} = 2$\n\n$11 + 9 = 2$\n\n$20 \\ne 2$\n\n$\\therefore$ There are no real solutions\n\nJan 19, 2018\n\nNo solution for principle root $\\to + \\sqrt{2 x + 1} + 11 = 2$\n\nThe convention is; that unless the question states\n$\\pm \\sqrt{\\text{something}}$ we only use the $+ \\sqrt{\\text{smething}}$ which is the principle root.\n\n#### Explanation:\n\nHowever, just for fun; lets investigate the scenario of $\\left(- 2\\right) \\times \\left(- 2\\right) = + 4 = \\left(+ 2\\right) \\times \\left(+ 2\\right)$\n\nSo sqrt(4)=+-2\" is definitely viable \"ul(\"only if you ignore the convention\")\n\nThe convention states :$\\sqrt{4} \\text{ can only \"=+2\" but you need\"+-sqrt(4)\" to have } \\pm 2$\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nGiven: $\\sqrt{2 x + 1} + 11 = 2$\n\nWrite as $\\pm \\sqrt{2 x + 1} + 11 = 2 \\textcolor{red}{\\leftarrow \\text{ not the actual question}}$\n\nSubtract 11 from both sides\n\n$\\pm \\sqrt{2 x + 1} = - 9$\n\nClearly $+ \\sqrt{2 x + 1} = - 9$ does not have a solution for $x \\in \\mathbb{R}$ as each side of the equals is the opposite sign.\n\nHowever $- \\sqrt{2 x + 1} = - 9$ may have a solution as both sides are negative. Accepting this should not be done as it is against the convention.$\\textcolor{red}{\\text{ Note that this is not the question as given}}$\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n$\\textcolor{b l u e}{\\text{Just for fun consider } - \\sqrt{2 x + 1} = - 9}$\n\nSquare both sides\n\n$2 x + 1 = + 81$\n\n$2 x = 80$\n\n$x = 40$\nWhich is not the the answer for the given question as written.", null, "The graph clearly shows no solution for $+ \\sqrt{2 x + 1} = - 9$ which is the answer to the given question as written.\n\nIf the question had been $\\pm \\sqrt{2 x + 1}$ then we could legitimately have stated the answer as being: $\\left(x , y\\right) \\to \\left(40 , 9\\right)$ is a solution for $- \\sqrt{2 x + 1} = - 9$" ]

[ null, "https://d2jmvrsizmvf4x.cloudfront.net/Llfz1MDmQ9Wb6NeRcXTs_y+equals+9+equals+sqrt+2x+pluss+1.jpg", null ]

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[ "# Minimum Channel Bandwidth for PCM\n\nPlease check if my answer to GATE IN 2010 question 11.20 is correct/can be improved upon.", null, "So, every $$\\frac{1}{f_s}$$ sec, a sample is taken. The signal amplitude is quantized into $$2^n$$ levels such that each sample can be encoded using $$n$$ bits. The question (I think) asks what is the minimum bandwidth a channel needs in order to transmit these bits.\n\nWe have $$\\frac{1}{f_s}$$ sec time and $$n$$ bits to send. So the bit rate would be $$nf_s$$ bits per second. Since each \"bit\" is sent as a pulse across the channel, the pulse rate $$f_p = nf_s$$. So by the Shannon–Hartley theorem we can conclude,\n\n$$nf_s \\leq 2B$$ $$B \\geq \\frac{nf_s}{2}$$", null, "The question asks what the bandwidth needed for \"faithful reconstruction\" is, I guess they mean that if you don't have at least this large a bandwidth errors will creep in and you can't reconstruct the signal from the n bit code at the receiver end?\n\nI don't think the exam question is stated properly, unless there is additional context that the student is supposed to know.\n\nFirst, the bandwidth depends on the encoding. The question seems to assume binary encoding, but PCM does not specify this.\n\nSecond, the bandwidth depends on the pulse shape. Usually one assumes rectangular unless otherwise specified, but this information should bave been included in the question.\n\nIn general, the PCM signal can be written as a pulse train:\n\n$$s_{\\text{PCM}}(t) = \\sum_{k=-\\infty}^\\infty a_k p(t-kT_p),$$\n\nwhere $$T_p$$ is the pulse rate and $$a_k$$ is the encoded PCM data. In general, the bandwidth of $$s_{\\text{PCM}}(t)$$ is equal to the bandwidth of $$p(t)$$.\n\nAs an example, let's say we want to transmit the bits $$1010$$ using a bipolar, non-return to zero binary encoding with $$a_k \\in \\lbrace 1, -1 \\rbrace$$. The pulse shape is rectangular with duration $$T_p = 1/(nf_s)$$. Assuming $$n$$ starts at $$n=1$$, the PCM signal can be written as:\n\n$$s_{\\text{PCM}}(t) = p(t-T_p)-p(t-2T_p)+p(t-3T_p)-p(t-4T_p).$$\n\nThe problem is, what is the bandwidth of this signal? Theoretically, it's equal to the bandwidth of the rectangular pulse $$p(t)$$ with duration $$T_p$$. However, there are several ways to measure that bandwidth.\n\nFor instance, if you use the \"absolute bandwidth\" definition, then the bandwidth is infinite. There are ten or more engineering definitions of bandwidth, and each gives you a different number. Note, however, that transmitting $$s_{\\text{PCM}}(t)$$ over any channel with less than infinite bandwidth will result in some amount of distortion.\n\nNow let's say that $$p(t)$$ is a sinc pulse of appropriate width. The expression for the signal doesn't change. Theoretically, now the signal has infinite time duration, but its bandwidth is finite and equal to $$1/(2T_p)=nf_s/2$$." ]

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[ "# Functional Programming explained to my grandma\n\nCCavalier\n20.3K views\n\n## No More NPE\n\nHas you ever spent hours trying to find where a NPE is coming from?\n\nThis will never happen again if you use options! (🎉 they exist in Java 🎉)\n\nTo keep it simple, an option is a wrapper which accepts two kinds of content:\n\n• A content \"value\" which will be returned each time when it's possible\n• A \"null\" value which will be return when no correct value can be returned\n\nIn Scala, an option has no concrete type, but two implementations:\n\nSome(object)\twrapper for a non-null value\nNone\t\t\twrapper for a null value\n\n\nYou can initialize an option to a concrete value or to null and it will return the right wrapper. An option has two opposite methods:\n\nisEmpty\t\ttrue if the option is a none\nisDefined\ttrue if the option has a value\n\n\nSo let's define the divide function which manages double division with the case where the divisor is equal to 0\n\nDiscover our saviour\n\nIn addition to providing us with a nice wrapper for null values, option gives us two interesting methods:\n\ngetOrElse\toption.getOrElse(12)\treturn the value of the option if it's defined, or a default value if it's empty\nfold\t\toption.fold(0)(x => x*x)\tapply the function passed in parameter if the option is defined or has a default value otherwise\n\n\nLet's define the two functions:\n\n• Secure divide: a function which calls divide and returns the value of the option if it's possible, 0 otherwise\n• Weird divide: a function which calls divide and returns the value of the division squared, or -1\nMore fancy functionalities", null, "", null, "", null, "" ]

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[ "Ta. If transmitted and non-transmitted genotypes would be the exact same, the individual is uninformative and the score sij is 0, otherwise the transmitted and non-transmitted contribute tijA roadmap to multifactor dimensionality reduction approaches|Aggregation in the components on the score vector offers a prediction score per person. The sum more than all prediction scores of individuals having a particular issue mixture compared with a threshold T determines the label of each and every multifactor cell.methods or by bootstrapping, therefore giving proof for any truly low- or high-risk aspect combination. Significance of a model nevertheless might be assessed by a permutation PD0325901 biological activity approach primarily based on CVC. Optimal MDR One more method, called optimal MDR (Opt-MDR), was proposed by Hua et al. . Their approach makes use of a data-driven as opposed to a fixed threshold to collapse the factor combinations. This threshold is selected to maximize the v2 values amongst all attainable 2 ?two (case-control igh-low danger) tables for each issue mixture. The exhaustive search for the maximum v2 values is usually accomplished effectively by sorting issue combinations as outlined by the", null, "ascending danger ratio and collapsing successive ones only. d Q This reduces the search space from two i? achievable 2 ?two tables Q to d li ?1. In addition, the CVC permutation-based estimation i? in the P-value is replaced by an approximated P-value from a generalized intense value distribution (EVD), equivalent to an approach by Pattin et al. described later. MDR stratified populations Significance estimation by generalized EVD is also made use of by Niu et al. in their strategy to manage for population stratification in case-control and continuous traits, namely, MDR for stratified populations (MDR-SP). MDR-SP utilizes a set of unlinked markers to calculate the principal elements that happen to be regarded as because the genetic background of samples. Based on the first K principal components, the residuals in the trait worth (y?) and i genotype (x?) of the samples are calculated by linear regression, ij thus adjusting for population stratification. Therefore, the adjustment in MDR-SP is utilised in every multi-locus cell. Then the test statistic Tj2 per cell is the correlation in between the adjusted trait worth and genotype. If Tj2 > 0, the corresponding cell is labeled as high danger, jir.2014.0227 or as low risk otherwise. Based on this labeling, the trait worth for every single sample is predicted ^ (y i ) for just about every sample. The education error, defined as ??P ?? P ?two ^ = i in instruction information set y?, 10508619.2011.638589 is applied to i in coaching information set y i ?yi i determine the ideal d-marker model; especially, the model with ?? P ^ the smallest typical PE, defined as i in testing information set y i ?y?= i P ?2 i in testing information set i ?in CV, is selected as final model with its average PE as test statistic. Pair-wise MDR In high-dimensional (d > two?contingency tables, the original MDR system suffers in the situation of sparse cells that happen to be not classifiable. The pair-wise MDR (PWMDR) proposed by He et al. models the interaction between d elements by ?d ?two2 dimensional interactions. The cells in every single two-dimensional contingency table are labeled as higher or low threat depending around the case-control ratio. For each and every sample, a cumulative danger score is calculated as variety of high-risk cells minus quantity of lowrisk cells more than all two-dimensional contingency tables. Below the null hypothesis of no association amongst the selected SNPs along with the trait, a symmetric distribution of cumulative threat scores about zero is expecte." ]

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[ "# Accupedo vs. Fitbit Part 2: Convergent Validity of Cumulative Step Counts with R\n\n2019, Nov 25\n\nThis post is a continuation of the previous post on this blog. Last time, we analyzed hourly step count data from the Accupedo app on my phone, and from the Fitbit I wear on my wrist. This time, we will analyze the cumulative step count measurements taken from Accupedo and Fitbit. This metric is arguably more interesting from a user perspective. After all, when you check your steps on the respective devices, you always see the cumulative step count - e.g. the number of steps that you have taken thus far that day. Do these devices give similar readings of cumulative step counts? When are the two measurements more likely to agree or disagree with one another?\n\n## The Data\n\nThe data come from two sources: the Accupedo app on my phone and from the Fibit (model Alta HR) that I wear on my wrist. Both data sources are accessible (with a little work) via R: you can see my write up of how to access data from Accupedo here and my post on how to access data from Fitbit here.\n\nI got the Fitbit in March 2018, and the data from both devices were extracted in mid-December 2018. I was able to match 273 days for which I had step counts for both Accupedo and Fitbit. The data contain the hourly and cumulative steps for the hours from 6 AM to 11 PM. In total, the dataset contains 4,914 observations of hourly step counts for the 273 days for which we have data (e.g. 18 observations per day).\n\nYou can find the data and all the code from this blog post on Github here.\n\nThe head of the dataset (named merged_data) looks like this:\n\ndate daily_total_apedo hour hourly_steps_apedo cumulative_daily_steps_apedo daily_total_fbit hourly_steps_fbit cumulative_daily_steps_fbit dow week_weekend hour_diff_apedo_fbit\n2018-03-20 16740 6 0 0 15562 281 281 Tue Weekday -281\n2018-03-20 16740 7 977 977 15562 1034 1315 Tue Weekday -57\n2018-03-20 16740 8 341 1318 15562 1605 2920 Tue Weekday -1264\n2018-03-20 16740 9 1741 3059 15562 223 3143 Tue Weekday 1518\n2018-03-20 16740 10 223 3282 15562 287 3430 Tue Weekday -64\n2018-03-20 16740 11 226 3508 15562 188 3618 Tue Weekday 38\n2018-03-20 16740 12 283 3791 15562 1124 4742 Tue Weekday -841\n2018-03-20 16740 13 1587 5378 15562 525 5267 Tue Weekday 1062\n2018-03-20 16740 14 431 5809 15562 372 5639 Tue Weekday 59\n2018-03-20 16740 15 624 6433 15562 392 6031 Tue Weekday 232\n\n## Cumulative Step Counts\n\n### Correspondence Plot\n\nIn this post, we will explore the cumulative step counts. We can make a scatterplot showing the correspondence between the cumulative step counts (coloring the points by type of day - week vs. weekend), and compute their correlation, using the following code:\n\nWhich returns the following plot:", null, "In this plot, the points are colored by weekday/weekend, and separate regression lines are drawn for each type of day. The dashed blue line is the identity line - if both Accupedo and Fitbit recorded the same number of cumulative steps each hour, all of the points would lie on this line.\n\nEspecially in comparison to the hourly plot, the correspondence is good! Both regression lines are very close to the identity line, suggesting a high degree of agreement between the cumulative step counts on both weekdays and weekends.\n\nThere is a slight over-representation of red points below the identity line, and a slight over-representation of black points above the identity line. This indicates that we see more weekend days (red points) where Accupedo has higher cumulative step measurements than Fitbit, and more weekday days (black points) where Fitbit has higher cumulative step measurements than Accupedo.\n\nThe code above computes the correlation between the cumulative step count measurements for the two devices. This correlation is .97, which indicates extremely high agreement. Keep in mind that the correlation between the hourly step counts was only .52. In comparison, the cumulative step counts match much more closely with one another!\n\n### Bland Altman Plot\n\nAnother way of examining the correspondence between two measurements is the Bland-Altman plot. The Bland-Altman plot displays the mean of the measurements on the x-axis, and the difference between the measurements on the y-axis. A horizontal line (in red in the plot blow) is drawn on the plot to indicate the mean difference between the measurements. In addition, two lines (in blue in the plot below) are drawn at +/- 1.96 standard deviations above and below the mean difference, respectively.\n\nWe will use the excellent BlandAltmanLeh package in R to make the Bland-Altman plot. Note that it takes some additional work to get the plot to have the same color scheme as our above correlation plot, with separate colors for weekdays and weekends, and transparency in the points.\n\nWhich gives us the following plot:", null, "There are a couple of things to notice here. The first is that the mean of the difference of the cumulative step counts - shown on the y axis - lies below zero (the exact value is -135.94, as we’ll see below). The y axis shows the value of the Accupdeo steps minus the Fitbit steps, and so the negative difference indicates that Accupedo gives lower cumulative step counts than Fitbit on average. However, the size of this hourly difference (in comparison to the range of differences across the data) is small.\n\nTake a look at the points that sit above and below the blue lines - these are the points that are more than 1.96 standard deviations above or below the global average. We see a slightly higher concentration of red points above 1.96 standard deviations from the mean. Conversely, we see a slightly higher concentration of black points below 1.96 standard deviations from the mean. This indicates that we see more observations on weekends for which Accupedo gives a much higher cumulative step count than Fitbit. We see more observations on weekdays for which Fitbit gives a much higher cumulative step count than Accupedo. This was also one of the conclusions from the scatterplot above!\n\n### Testing the Statistical Significance of the Cumulative Differences\n\nA corresponding statistical analysis that often accompanies the Bland-Altman plot is a one-sided t-test, comparing the mean difference of the measurements against zero (null hypothesis: the mean difference between the measurements is equal to zero). The mean of the differences is -135.94 and the standard deviation is 1592.76.\n\nThis test indicates that the difference between the cumulative step counts for Accupedo and Fitbit is statistically significant, t(4913) = -5.98, p < .001.\n\nLet’s explore the effect size of this comparison. Effect sizes give a measure of the magnitude of an observed difference. There are many such measures, but we can compute Cohen’s D from the values we have above. Cohen’s D gives the size of a difference scaled to the standard deviation of that difference. Here, we simply take the mean difference score (-135.94) and divide it by the standard deviation of that score (1592.76), which gives us a Cohen’s D value of -.09. In other words, the cumulative step count measurements from Accupedo and Fitbit differ by less than 1/10th of a standard deviation. According to the standards laid out by Cohen (1988), the effect size is very small.\n\nThe global conclusion is that the cumulative steps recorded by Fitbit are systematically higher than the cumulative steps recorded by Accupedo, but that the size of this difference is very small!\n\n## Differences in Cumulative Step Counts Across the Day\n\nThe above plots mix data across all hours of the day in order to examine the global correspondence between cumulative step counts. In the next analysis we will visualize the difference in the cumulative step count measurements across the hours of the day, to see if there were any systematic differences within certain times of the day.\n\nTo make this plot, we’ll use the excellent ggridges package, visualizing the density distributions of step count differences separately for each hour of the day, with separate panels for weekdays and weekends.\n\nWhich gives us the following plot:", null, "Interestingly, there do appear to be differences in the distributions of the differences in cumulative step counts across the hours of the day. The green horizontal dotted line is drawn at zero, the point at which there are no differences in cumulative step counts. The y-axis shows the result of the Accupedo steps minus the Fitbit steps; data below the line indicate higher cumulative counts for Fitbit, while data above the line indicate higher cumulative counts for Accupedo.\n\nOn both weekdays and weekends, the day starts with a distribution which has more observations for which the Fibit cumulative counts are higher (as the peak of the distributions lie below the dotted line). This is likely due to the fact that any movement that occurs at night is picked up by the Fitbit (which is on my wrist all night) and not by Accupedo (because I don’t have my phone in hand all night). So we start the day with a higher cumulative step count with Fitbit than we do with Accupedo.\n\nThe distribution slowly starts shifting upward throughout the day. At around noon on weekdays and at around 11 AM on the weekends, the distribution is more-or-less centered at zero, indicating that the average cumulative step counts are the same at this point in the day.\n\nThe further along we get during the day, the distributions become much flatter and less centered around zero. The global trend at the end of the day, especially pronounced on the weekends, is to have a relatively flat distribution with a greater number of observations for which Accupedo has higher cumulative step counts than Fitbit.\n\n### Testing the Difference in Cumulative Step Counts Between Weekdays vs. Weekends\n\nWe saw above that the overall difference between the cumulative counts for Accupedo and Fitbit was -135.94, indicating that across all observations, the Fitbit cumulative counts were 136 steps higher. The distribution plot above suggests a nuance, in that the relative difference appears to reverse on the weekends. (Because we have many more weekday observations then weekend observations, the global average is heavily weighted by the weekdays, in which Fitbit gives higher step counts).\n\nLet’s examine whether the cumulative step count differences are different on weekdays vs. weekends. We can do this in a very simple way via a linear model, in which we predict the cumulative difference score using the categorical variable of week/weekend. We do this via the regression specified below. We then ask for a summary of the results of the model.\n\nThe code to run the regression looks like this:\n\nAnd returns the following regression output:\n\n Dependent variable: cumul_diff_apedo_fbit week_weekendWeekend 447.89*** (49.89) Constant -263.91*** (26.67) Observations 4,914 R2 0.02 Adjusted R2 0.02 Residual Std. Error 1,580.01 (df = 4912) F Statistic 80.58*** (df = 1; 4912) Note: *p<0.1; **p<0.05; ***p<0.01\n\nThe intercept (or “constant” as it’s described in the table above) is -263.91, indicating that on weekdays (when the dummy variable for week_weekend is 0), Fitbit yields an average cumulative step count that is 263.91 steps higher than that for Accupedo. However, on weekends (when the dummy variable for week_weekend is 1), the average difference between the cumulative step counts is 183.98 (e.g. 447.89 + -263.91) steps higher for Accupedo. Note that the R2 value is very small - most of the variation in cumulative step count differences is not accounted for by the week/weekend variable!\n\nIn sum, the nuanced conclusion is this: the cumulative step counts are on average higher for Fitibit (vs. Accupedo), but on weekends the pattern reverses, such that on average Accupedo gives higher cumulative step counts as compared to Fitbit. In both cases, the differences are small (less than 300 steps in either direction).\n\n## Summary and Conclusion\n\nIn this post, we examined the convergent validity of cumulative step counts from the Accupedo app on my phone and the Fitbit I wear on my wrist. The correlation between these two devices’ records was .97, indicating a very strong relationship between the two measurements of cumulative step counts.\n\nWe then constructed a Bland-Altman plot to compare the two measurements. This plot revealed that on average Fitbit’s cumulative step count was 135.94 steps higher than Accupedo’s. While statistically significant, the size of this difference was equivalent to 1/10th of a standard deviation; in other words - a very small effect.\n\nBoth the correpsondence plot and the Bland-Altman plot showed a number of observations for which Fitbit had higher cumulative step counts during weekdays, and a number of observations for which Accupedo had higher cumulative step counts during weekends.\n\nWhen examining the distributions of step count differences across hours of the day, we got some more insight into this pattern. On both weekdays and weekends, in the first hours of the day, Fitbit consistently records higher cumulative step counts than Accupedo, most likely because Fitbit counts night time activitity (because it’s on my wrist), whereas Accupedo does not (because it’s on my phone). Throughout the day, however, the distribution of the differences shifts towards higher cumulative step counts for Accupedo. This difference is especially pronounced on the weekend.\n\nIndeed, our regression analysis of the cumulative differences across weekdays and weekends indicated a reveral of the global pattern depending upon the type of day. During the weekdays, the pattern matches the overall trend - Fitbit yields slightly higher cumulative step counts than does Accupedo. However, on the weekend, the pattern reverses and the average cumulative step counts are higher for Accupedo. In both cases, the sizes of the differences between the devices is small.\n\n#### Why does the distribution shift from higher counts for Fitbit to higher counts for Accupdeo throughout the day?\n\nThe days start out with a globally higher cumulative count for Fitbit, likely because it picks up night time activity that Accupedo does not. As we saw in the previous post, however, there is a tendency for Accupedo to give greater hourly step counts than Fitbit. As the day progresses, these small hourly differences seem to accumulate, shifting the balance in cumulative step counts by the day’s end.\n\n#### Why is this difference more pronounced on the weekends?\n\nI’m not quite sure about this. I’ve seen in previous analyses of my step count data that my walking patterns tend to be quite different on weekdays and the weekends. My guess is that the observed differences have to do with the types of walking I do. During the week, I tend to walk shorter distances in any one go, whereas on the weekend I tend to do longer walks throughout the day. Importantly, during the time that these data were recorded, I was more often pushing a stroller on the weekend (vs. weekdays). I’ve noticed that, because my wrist is continually horizontal when pushing a stroller, Fitbit counts fewer steps than does Accupedo (which is in my pocket). Perhaps that has something to do with this difference between weekdays and weekends, but it’s more of a guess than a data-driven conclusion.\n\n## In Sum: Comparing Convergent Validity Between Hourly and Cumulative Step Counts\n\n#### Hourly\n\nThe correspondence between the hourly step counts is not large (correlation of only .52), but there is no systematic difference in over or under-counting between the devices. Sometimes Accupedo counts more steps per hour, and sometimes Fitbit counts more steps per hour. There appear to be no systematic differences in hourly step counts during the weekday vs. the weekend.\n\n#### Cumulative\n\nIn contrast, the correspondence between the cumulative step counts is substantial (correlation of .97). The devices’ measurements match quite closely, but there is a small but reliable difference in the cumulative step count measurements between devices, with Fitbit counting 135.94 more cumulative steps on average, compared to Accupedo. Examining the difference in cumulative step counts across the hours of the day shows that Fitbit has higher cumulative counts in the morning, but that this difference shifts towards Accupedo throughout the day. On the weekends, the global pattern reverses, with Accupedo yielding higher average cumulative step counts than Fitbit. In both cases, the difference in cumulative step counts between the devices is small.\n\nComing Up Next\n\nIn the next post, we will turn to a difference data source: detailed records of my phone usage. We will use data munging and visualization to see how and when my phone usage patterns differ throughout the day.\n\nStay tuned!" ]

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[ "", null, "Worksheet. Subtracting money worksheets uk column addition and subtraction. Money addition and subtraction worksheets.", null, "[   Small  •  Medium  •  Large   ]", null, "Free Math Worksheets And Printouts\n\n[   Small  •  Medium  •  Large   ]", null, "Printable Second Grade Math Word Problem\n\n[   Small  •  Medium  •  Large   ]", null, "Word Problems Addition And Subtraction Worksheets\n\n[   Small  •  Medium  •  Large   ]", null, "Subtracting Money Worksheets Uk Column Addition\n\n[   Small  •  Medium  •  Large   ]", null, "Word Problems\n\n[   Small  •  Medium  •  Large   ]", null, "Money Mathsdiary Com\n\n[   Small  •  Medium  •  Large   ]", null, "Statement Sums Of Money Mathsdiary Com\n\n[   Small  •  Medium  •  Large   ]", null, "[   Small  •  Medium  •  Large   ]", null, "[   Small  •  Medium  •  Large   ]", null, "[   Small  •  Medium  •  Large   ]", null, "Addition And Subtraction Money Word Problems\n\n[   Small  •  Medium  •  Large   ]\n\n12 Images in Money Addition And Subtraction Worksheets", null, "Fraction Division Word Problems Worksheets\n\nby Jackqueline in Worksheet.", null, "Functional Math Worksheets\n\nby Vallie in Worksheet.", null, "Math Worksheets For Preschool Free Printable\n\nby Ashly in Worksheet.", null, "Number Recognition Worksheets Kindergarten\n\nby Ebonie in Worksheet.", null, "Greater Than Less Than Equal To Worksheets\n\nby Maryland in Worksheet.", null, "by Rebekah in Worksheet.\n\nMore in Worksheet", null, "Free Worksheets Math Part 1\n\nby Ashly in Worksheet.", null, "Free Educational Printables\n\nby Ashly in Worksheet.", null, "Mathematics Worksheet Chapter #1\n\nby Ashly in Worksheet.", null, "Double Digit Division Worksheets\n\nby Vesta in Worksheet.", null, "Free Online Math Worksheets\n\nby Ashly in Worksheet.", null, "Simple Math Equations Worksheets\n\nby Ashly in Worksheet.\n\nMost Visited" ]

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[ "Top-Down Design of an RF Receiver\n\nThis example designs an RF receiver for a ZigBee®-like application using a top-down methodology. It verifies the BER of an impairment-free design, then analyzes BER performance after the addition of impairment models. The example uses the RF Budget Analyzer App to rank the elements contributing to the noise and nonlinearity budget.\n\nDesign specifications:\n\n• Data rate = 250 kbps\n\n• OQPSK modulation with half sine pulse shaping, as specified in IEEE® 802.15.4 for the physical layer of ZigBee\n\n• Direct sequence spread spectrum with chip rate = 2 Mchips/s\n\n• Sensitivity specification = -100 dBm\n\n• Bit Error Rate (BER) specification = 1e-4\n\n• Analog to digital converter (ADC) with 10 bits and 0 dBm saturation power\n\nTo create fully standard-compliant ZigBee waveforms, you can use the Communications Toolbox Library for the ZigBee Protocol Add-on.\n\nThis example guides you through the following steps:\n\n• Develop the baseband transmitter model for waveform generation\n\n• Determine SNR specification to achieve the 1e-4 BER from a link-level idealized baseband model\n\n• Derive RF subsystem specifications from equivalent-baseband model of RF receiver and ADC\n\n• Derive direct conversion specifications from circuit envelope model of RF receiver\n\n• Perform multi-carrier simulation including interfering signals and derive the specifications of the DC offset compensation algorithm\n\nDesign and Verify Baseband Transmitter\n\nTo evaluate the performance of the RF receiver design, it is necessary and sufficient to use a signal spectrally representative of an 802.15.4 waveform.\n\nThe baseband transmitter model creates and illustrates a spectrally representative ZigBee waveform in the spectral and constellation domains. This model and all the subsequent models use callbacks to create MATLAB workspace variables that parameterize the systems.", null, "", null, "", null, "To design the receiver, first determine the SNR needed to achieve the specified BER less than 1e-4. calculated in the simulation bandwidth of 4 MHz. Run the link-level model to simulate the receiver processing required to achieve the target BER.\n\nComputing the BER accurately requires alignment of the transmit and receive signals. The simulation must compensate for a two-sample delay of the received signal compared to the transmitted signal. Also, to ensure correct chip-to-symbol-to-bit mapping, the simulation must align the chips to frame boundaries at the input to the Chips to Symbol block on a frame boundary. Accounting for the receive signal delay and the frame boundary alignment requires addition of a Delay block set to a 32-2=30 delay on the receiver branch before recovering the received symbols.", null, "", null, "The model achieves a 1e-4 BER at an SNR of -2.7 dB, which can be verified by collecting 100 bit errors.\n\nIn the link-level model, the AWGN block accounts for the overall channel and RF receiver SNR budget.\n\nThis section uses traditional heuristic derivations to determine the high-level specifications of the RF receiver and ADC.\n\n• B = 4 MHz = simulation bandwidth = simulation sampling frequency\n\n• kT = 174 dBm/Hz = thermal noise floor power\n\n• Sensitivity = -100 dBm = receiver sensitivity\n\n• SNR = -2.7 dB\n\n• Noise power in simulation bandwidth = Pn = sensitivity-SNR = -100 dBm - (-2.7 dB) = -97.3 dBm\n\nSimulating an idealized baseband model of the RF Receiver, verify the preliminary RF receiver specifications (NF = 10.7 dB and receiver gain = 53.4 dB). This can be done by collecting 100 errors.\n\nThe spectrum analyzer shows that the received spectrum with the ADC is roughly identical in shape to the spectrum of the previous section, without the ADC.", null, "", null, "Refine Architectural Description of RF Receiver\n\nIn this section the RF receiver, and its noise figure and gain budget specifications, are modelled by using four discrete subcomponents with these characteristics:\n\n• SAW Filter: Noise Figure = 2.5 dB, Gain = -3 dB\n\n• LNA: Noise Figure = 6 dB, Gain = 22 dB\n\n• Passive Mixer: Noise Figure = 10 dB, Gain = -5 dB\n\n• VGA: Noise Figure = 14 dB, Gain = 40 dB\n\nThe SAW filter performance is derived from a Touchstone file that specifies S-parameters characteristics. You can verify the gain by visualizing the S21 parameter in the X-Y plane at the operating frequency of 2.45 GHz. You can verify the noise figure by visualizing the NF parameter in the X-Y plane at the operating frequency of 2.45 GHz. Typically, an LNA with low noise and high gain follows the SAW filter, which greatly reduces the impact of the noise figure of the components after the LNA. Also, the passive mixer is specified with a high IP2. Similar to the SAW filter, you can verify the mixer gain by visualizing the S21 parameter in the X-Y plane over a user-specified frequency range of [2e9 3e9].\n\nAn equivalent baseband model simulates the refined RF receiver.\n\nRun the simulation and verify the RF receiver link budget by using the output port visualization pane. The total noise figure and gain across the four stages has been divided according to the following budget:\n\n• Component NF (dB) = [2.5, 6, 10, 14]\n\n• Component noise factor F (linear) = 10^(NF/10) = [1.78 3.98 10.0 25.1]\n\n• Power gain (dB) = [-3, 22, -5, 40] = 54 dB > 53.4 dB\n\n• Voltage gain VG (linear) = 10^(Power gain/20) = [0.71 12.59 0.56 100.0]\n\n• System noise factor Fsys (linear) =", null, "• System noise figure NFsys (dB) = 10*log10(Fsys) = 10.7 dB\n\nWith this model you can verify that a BER < 1e-4 corresponds to a Chip Error Rate (ChER) around 7%. By computing ChER, you can run the subsequent models for less time and still collect accurate BER statistics.", null, "", null, "Use Circuit Envelope to Simulate Additional RF Impairments\n\nThe equivalent baseband modeling technique used in the previous section cannot model a true direct conversion receiver. That model used a mixer with an input frequency of 2.45 GHz and an LO frequency of 2.4 GHz, which led to a spectrum analyzer center frequency of 50 MHz. This modeling limitation motivates a change to the circuit envelope method.\n\nUsing the circuit envelope modeling approach, continue refining the RF receiver architecture by adding more realistic impairments.\n\nThe circuit envelope model of the RF Receiver differs from the equivalent baseband model as it:\n\n• Replaces the equivalent baseband mixer with a quadrature modulator, consisting of parameterizable I and Q mixers and phase shifter block, and an LO with impairments\n\n• Uses broadband impedances (50 ohm) to explicitly model the power transfer between blocks\n\nComparing spectra, power measurements, and ChER to the equivalent baseband model, there are no significant performance differences. However, with the circuit envelope model, you can include even order nonlinearity effects, I/Q imbalance, and specifications of colored noise distributions for each of the components.", null, "", null, "You can manually build the circuit envelope model of the RF Receiver by using blocks from the Circuit Envelope library, or it can be automatically generated using the RF Budget Analyzer App.\n\nThe RF Budget Analyzer App\n\n• Uses Friis equations to determine the noise, gain, and nonlinearity budget of an RF chain\n\n• Allows you to explore the receiver design space and determine how to break down the specifications across the elements of the chain\n\n• Helps you determine which element has the largest contribution to the noise and nonlinearity budget\n\n• Can generate an RF receiver model with which you can perform multi-carrier simulation and further modify.", null, "Add Wideband Interference, LO Leakage, and DC Offset Cancellation\n\nThis section modifies the circuit envelope model to create this circuit envelope with interferer model. The circuit envelope with interferer model includes a wideband interfering signal and these impairments:\n\n• LO-RF isolation of 90 dB in the quadrature demodulator\n\n• OIP2 equal to 70 dBm in the quadrature demodulator\n\n• WCDMA-like blocker of -30 dBm at 2500 MHz\n\nThis simulation models a non-standard-compliant interfering signal that has power and spectral distribution characteristics realistic for a WCDMA signal. The simulation of the wideband interfering signal requires a larger simulation bandwidth of 16MHz. Therefore the 1 MHz OQPSK signal is oversampled by 16, and the Circuit Envelope simulation bandwidth is also increased to 16 MHz.\n\nThe design requires a DC offset compensation algorithm to achieve the desired ChER due to the DC offset that results from the LO leakage and the nonlinearity in the demodulator caused by the high out-of-band interfering signal power. In this case you include a very selective filter, that introduces a long latency with corresponding computation delay increases in the ChER measurement block.\n\nThe spectrum centered at 0 Hz shows the DC offset compensation reducing the DC offset. As you run the model, note that the DC offset is eventually completely removed.", null, "", null, "", null, "Conclusion\n\nFollowing a top-down design methodology, RF receiver components specifications were derived. Impairment, interferer, and RF receiver subcomponent models were iteratively refined to increase fidelity and validated at each stage to confirm overall system performance goals were achieved." ]

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[ "# R Code for Hyperspectral Indices\n\nI am planning to use R to read hyperspectral images and calculate hyperspectral indices. There seems to be scripts available online, but mostly for multispectral images. For instance, this code for NDVI uses only two bands:\n\n``````VI <- function(img, k, i) {\nbk <- img[[k]]\nbi <- img[[i]]\nvi <- (bk - bi) / (bk + bi)\nreturn(vi)\n}\n\n#knowing the band numbers for k(5) and i(3), the NDVI is:\n\nndvi <- VI(hyper_image, 5, 3)\n``````\n\nMy dilemma is: I use a hyperspectral image with 100 bands. The above code becomes inefficient because I want to use all 100 bands in the formula.\n\n``````#if, for example, these are my 100 bands: k, i, j, n, m, o, p, ....\n\n#And just for the sake of illustration, this is a sample equation:\n\nvi <- (bk + bi + bj + bn + bm + bo + bp + ....) / (bk + bi + bj + bn + bm + bo + bp + ....)\n\nreturn(vi)\n\nhyper_index <- VI(hyper_image, 5, 3, 6, 7, 8, 9, 1, ...)\n\n#Obviously, doing this is excruciating!\n``````\n\nCan someone point me to existing R code for this procedure?\n\n• If you want to use all 100 bands in some calculation then you write a loop. Its not excruciating. If your band indexes `VI(hyper_image, 5, 3, 6, 7, 8, 9, 1, ...)` are random then you're going to have to type them all out at some point in the right order. I don't see the problem... – Spacedman Oct 16 at 16:17" ]

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[ "math.AG\n\nTitle: Effective Whitney theorem for complex polynomial mappings of the plane\n\nAbstract: We give an exact formula for the number of cusps of a general polynomial mapping $F=(f, g):\\Bbb C^2\\to\\Bbb C^2.$ Namely, if deg $f=d_1$ and deg $g=d_2$ and $F$ is general, then $F$ has exactly $d_1^2+d_2^2+3d_1d_2-6d_1-6d_2+7$ simple cusps. Moreover, if $d_1>1$ or $d_2>1$, then the set $C(F)$ of critical points of $F$ is a smooth connected curve, which is topologically equivalent to a sphere with $g=\\frac{(d_1+d_2-3)(d_1+d_2-4)}{2}$ handles with $d_1+d_2-2$ points removed.\n Subjects: Algebraic Geometry (math.AG) Cite as: arXiv:1503.00017 [math.AG] (or arXiv:1503.00017v1 [math.AG] for this version)\n\nSubmission history\n\nFrom: Zbigniew Jelonek [view email]\n[v1] Fri, 27 Feb 2015 21:27:21 GMT (9kb)\n[v2] Thu, 11 Jun 2015 10:01:23 GMT (11kb)\n[v3] Mon, 8 Feb 2016 16:51:29 GMT (16kb)\n\nLink back to: arXiv, form interface, contact." ]

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[ "# R/measures.R In danm0nster/domstruc: Compute Metrics of Dominance Hierarchies\n\n#### Documented in dom_ecdom_focusdom_position\n\n```#' Compute regularized eigenvalue centrality from aggression matrix\n#'\n#' @param aggression_matrix An aggression matrix.\n#' @param epsilon Regularization term for computing eigenvalue centrality.\n#'\n#' @return A vector of eigenvalue centrality for each individual in the\n#' aggression matrix.\n#' @export\n#'\n#' @examples\ndom_ec <- function(aggression_matrix, epsilon = 0.694) {\nif (missing(aggression_matrix)) {\nstop(\"Please provide an aggression matrix as input.\")\n} else {\ncheck_aggression_matrix(aggression_matrix)\n}\ncheck_epsilon(epsilon)\nt_mat <- dom_transition_matrix(aggression_matrix, epsilon = epsilon)\n# Compute the eigenvalues and eigenvectors of the transpose\n# of the transition matrix (corresponds to the left hand\n# eigenvectors).\nevv <- eigen(t(t_mat), symmetric = FALSE)\n# Find the eigenvalue with the higest norm\nindex_max <- which.max(abs(evv\\$values))\n# Return the normalized eigenvector\nreturn(abs(evv\\$vectors[, index_max]) / sum(abs(evv\\$vectors[, index_max])))\n}\n\n#' Compute focus from an aggression matrix\n#'\n#' @param aggression_matrix An aggression matrix.\n#' @param epsilon Regularization term for computing eigenvalue centrality.\n#'\n#' @return the computed focus value.\n#' @export\n#'\n#' @examples\ndom_focus <- function(aggression_matrix, epsilon = 0.694) {\nif (missing(aggression_matrix)) {\nstop(\"Please provide an aggression matrix as input.\")\n} else {\ncheck_aggression_matrix(aggression_matrix)\n}\ncheck_epsilon(epsilon)\n# Compute rank focused aggression\nfocused_agg <- dom_rank_focused_aggression(aggression_matrix,\nepsilon = epsilon)\n# Compute the necessary terms for focus as new columns in the data frame\nfocused_agg\\$focus_terms <- 0\n# Select the rows where agg_norm is non-zero\nidx <- focused_agg\\$agg_norm > 0\nfocused_agg\\$focus_terms[idx] <- focused_agg\\$delta[idx] *\nfocused_agg\\$agg[idx] / focused_agg\\$agg_norm[idx]\ndenom <- sum(focused_agg\\$r_delta)\nmu <- sum(focused_agg\\$focus_terms) / denom\nfocused_agg\\$focus_var <- 0\nfocused_agg\\$focus_var[idx] <- focused_agg\\$agg[idx] *\n(focused_agg\\$delta[idx] - mu)**2 / focused_agg\\$agg_norm[idx]\nn <- dim(aggression_matrix)\nreturn(1 - (sum(focused_agg\\$focus_var) / denom) / (n * (2 * n - 1) / 6.0))\n}\n\n#' Compute position from an aggression matrix\n#'\n#' @param aggression_matrix An aggression matrix.\n#' @param epsilon Regularization term for computing eigenvalue centrality.\n#'\n#' @return The position, i.e. the relative rank difference where the aggression\n#' is most focused.\n#' @export\n#'\n#' @examples\ndom_position <- function(aggression_matrix, epsilon = 0.694) {\nif (missing(aggression_matrix)) {\nstop(\"Please provide an aggression matrix as input.\")\n} else {\ncheck_aggression_matrix(aggression_matrix)\n}\ncheck_epsilon(epsilon)\n# Compute rank focused aggression\nfocused_agg <- dom_rank_focused_aggression(aggression_matrix,\nepsilon = epsilon)\ntiny <- 1e-12\n# Calculate the terms needed to compute position\nfocused_agg\\$fraction <- focused_agg\\$delta * focused_agg\\$agg /\n(focused_agg\\$agg_norm + tiny)\nfocused_agg\\$norm <- focused_agg\\$agg / (focused_agg\\$agg_norm + tiny)\nn <- dim(aggression_matrix)\nreturn(mean(focused_agg\\$fraction) / mean(focused_agg\\$norm) / n)\n}\n```\ndanm0nster/domstruc documentation built on Feb. 12, 2020, 8:58 a.m." ]

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[ "Welcome to Chem Zipper.com......: WORK DONE (PV-WORK ANALYSIS )\n\n## Friday, September 28, 2018\n\n### WORK DONE (PV-WORK ANALYSIS )\n\nEnergy that is transmitted from one system to another in such a way that difference of temperature is not directly involved. This definition is consistent with our understanding of work as dw= Fdx. The force (F) can arise from electrical, magnetic, gravitational & other sources. It is a path function.\n\nPV-Work analysis: Consider a cylinder fitted with a friction less piston, which enclosed no more of an ideal gas. Let an external force F pushes the piston inside producing displacement in piston.\n\nLet distance of piston from a fixed point is x and distance of bottom of piston at the same fixed point is l. This means the volume of cylinder = (l – x) A where A is area of cross section of piston.\n\nFor a small displacement dx due to force F, work done on the system.\ndw = F.dx\nAlso  F = PA\ndW = PA.dx\nV = (l – x)A\ndV = –A . dx\ndW = –Pext. dV\nNote :\n(1): Litre atmosphere term is unit of energy. It is useful to remember the conversion:\n1 litre atm= 101.3 Joules = 24.206 Cal.\n(2): During expansion dV is positive and hence sign of w is negative since work is done by the system and negative sign representing decease in energy content of system. During compression, the sign of dV is negative which gives positive value of w representing the increase in energy content of system during compression.\n\nEXAMPLE.1 mole of ideal monatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm3 to 16 dm3.            Calculate (i) q (ii) w and (iii) DU\nSOLUTION:   (i) Since process is adiabatic  \\ q = 0\n(ii) As the gas expands against the constant external pressure.\n\nW = - PVd=-P(V2-V1)\nW =-1.5(16-4)\nW= - 18 dm3\n(iii) DU = q + w = 0 + (-18) = -18 atm dm3\n\n(1) Work done in Isothermal Irreversible Process:\n(2) Work done in Isothermal Reversible Process:\n(3) Work done in Adiabatic Irreversible Process:\n(4) Work done in Adiabatic Reversible Process:" ]

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[ "# Properties\n\n Label 33800l Number of curves 1 Conductor 33800 CM no Rank 1\n\n# Related objects\n\nShow commands for: SageMath\nsage: E = EllipticCurve(\"33800.b1\")\n\nsage: E.isogeny_class()\n\n## Elliptic curves in class 33800l\n\nsage: E.isogeny_class().curves\n\nLMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality\n33800.b1 33800l1 [0, 0, 0, -16900, 746980] [] 145152 $$\\Gamma_0(N)$$-optimal\n\n## Rank\n\nsage: E.rank()\n\nThe elliptic curve 33800l1 has rank $$1$$.\n\n## Modular form 33800.2.a.b\n\nsage: E.q_eigenform(10)\n\n$$q - 3q^{3} + 2q^{7} + 6q^{9} + 2q^{11} - 2q^{17} - 4q^{19} + O(q^{20})$$" ]

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[ "# SciDraw doesnt plot full function over desired domain? [closed]\n\nLet me remark that I'm relatively new to using SciDraw, but I have read through the tutorial. Currently I'm using SciDraw v0.7 in Mathematica 10.2. I find that there is very strange behavior when defining a Plot in the FigGraphics wrapper over a range different from what XPlotRange is specified to be. For example, I want to plot\n\n$f(x) = -\\frac{(1-x) \\left(2 x^3+4 x^2+6 x+3\\right)}{2 \\left(x^4+x^3+x^2+x+1\\right)}$ over $x \\in [1,10]$\n\nI define the SciDraw canvas as,\n\nFigure[\nFigurePanel[\n{\nFigGraphics[Plot[f[x], {x, 1, 10}]];\n},\nXPlotRange -> {1, 10},\nYPlotRange -> {0, 1}\n],\nCanvasSize -> {5, 3.5}\n]", null, "The canvas ranges are correct but the plot appears to be cut off at $x=2$. What is even more strange is if I change the range over x in the Plot command to $\\{x,0,5\\}$ I get,", null, "Indeed, $f(x) = 0$ at $x=1$ and if I just run the regular mathematica command,\n\nPlot[f[x], {x, 1, 10}, PlotRange -> All]\n\n\nI get a full correct plot. Anyone else experience this or know of a fix? Is the input correct?\n\n• Well, just include the PlotRange -> All in Plot even when using it in FigGraphics. Within Plot, that option controls what gets plotted, not what gets shown. – Szabolcs May 26 '17 at 7:23\n\nPlotRange does something different in Plot vs Show/SciDraw.\n\nIn Plot, it controls which part of the curve should be generated.\n\nIn Show, it controls the range of the plot frame. Only objects within the plot frame are visible.\n\nYou are (perhaps implicitly) assuming that Plot generates the full curve, even those parts which fall outside of the plot range. Thus increasing the plot range after the plot has been created should reveal more of the curve. This is not the case. The sections of the curve outside of the plot range will not even be generated by Plot.\n\nExample:", null, "Solution: Use PlotRange -> All within Plot. This setting is completely independent of the YPlotRange setting of SciDraw. Use both simultaneously." ]

[ null, "https://i.stack.imgur.com/Ivyyb.png", null, "https://i.stack.imgur.com/E2gWY.png", null, "https://i.stack.imgur.com/bXFwz.png", null ]

[ "# The Number 283994 : Square Root, Cube Root, Factors, Prime Checker\n\nYou can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 283994. 283994 is written as Two Hundred And Eighty Three Thousand, Nine Hundred And Ninety Four. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.\n\n• Number 283994 is an even Number.\n• Number 283994 is not a Prime Number\n• Sum of all digits of 283994 is 35.\n• Previous number of 283994 is 283993\n• Next number of 283994 is 283995\n\n## Square, Square Root, Cube, Cube Root of 283994\n\n• Square Root of 283994 is 532.91087434955\n• Cube Root of 283994 is 65.73092161409\n• Square of 283994 is 80652592036\n• Cube of 283994 is 22904852222671784\n\n## Numeral System of Number 283994\n\n• Binary Representation of 283994 is 1000101010101011010\n• Octal Representation of 283994 is 1052532\n• Hexadecimal Representation of 283994 is 4555a\n\n## Sin, Cos, Tan of Number 283994\n\n• Sin of 283994 is -0.71933980033858\n• Cos of 283994 is 0.69465837045907\n• Tan of 283994 is -1.0355303137904\n\n## Multiplication Table for 283994\n\n• 283994 multiplied by 1 equals to 283,994\n• 283994 multiplied by 2 equals to 567,988\n• 283994 multiplied by 3 equals to 851,982\n• 283994 multiplied by 4 equals to 1,135,976\n• 283994 multiplied by 5 equals to 1,419,970\n• 283994 multiplied by 6 equals to 1,703,964\n• 283994 multiplied by 7 equals to 1,987,958\n• 283994 multiplied by 8 equals to 2,271,952\n• 283994 multiplied by 9 equals to 2,555,946\n• 283994 multiplied by 10 equals to 2,839,940\n• 283994 multiplied by 11 equals to 3,123,934\n• 283994 multiplied by 12 equals to 3,407,928\n\n## Division Table for 283994\n\n• 283994 divided by 1 equals to 283994\n• 283994 divided by 2 equals to 141997\n• 283994 divided by 3 equals to 94664.666666667\n• 283994 divided by 4 equals to 70998.5\n• 283994 divided by 5 equals to 56798.8\n• 283994 divided by 6 equals to 47332.333333333\n• 283994 divided by 7 equals to 40570.571428571\n• 283994 divided by 8 equals to 35499.25\n• 283994 divided by 9 equals to 31554.888888889\n• 283994 divided by 10 equals to 28399.4\n• 283994 divided by 11 equals to 25817.636363636\n• 283994 divided by 12 equals to 23666.166666667" ]

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[ "# #374628 Color Information\n\nIn a RGB color space, hex #374628 is composed of 21.6% red, 27.5% green and 15.7% blue. Whereas in a CMYK color space, it is composed of 21.4% cyan, 0% magenta, 42.9% yellow and 72.5% black. It has a hue angle of 90 degrees, a saturation of 27.3% and a lightness of 21.6%. #374628 color hex could be obtained by blending #6e8c50 with #000000. Closest websafe color is: #333333.\n\n• R 22\n• G 27\n• B 16\nRGB color chart\n• C 21\n• M 0\n• Y 43\n• K 73\nCMYK color chart\n\n#374628 color description : Very dark desaturated green.\n\n# #374628 Color Conversion\n\nThe hexadecimal color #374628 has RGB values of R:55, G:70, B:40 and CMYK values of C:0.21, M:0, Y:0.43, K:0.73. Its decimal value is 3622440.\n\nHex triplet RGB Decimal 374628 `#374628` 55, 70, 40 `rgb(55,70,40)` 21.6, 27.5, 15.7 `rgb(21.6%,27.5%,15.7%)` 21, 0, 43, 73 90°, 27.3, 21.6 `hsl(90,27.3%,21.6%)` 90°, 42.9, 27.5 333333 `#333333`\nCIE-LAB 27.698, -12.307, 16.163 4.149, 5.346, 2.821 0.337, 0.434, 5.346 27.698, 20.315, 127.286 27.698, -6.846, 18.049 23.121, -8.433, 8.951 00110111, 01000110, 00101000\n\n# Color Schemes with #374628\n\n• #374628\n``#374628` `rgb(55,70,40)``\n• #372846\n``#372846` `rgb(55,40,70)``\nComplementary Color\n• #464628\n``#464628` `rgb(70,70,40)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #284628\n``#284628` `rgb(40,70,40)``\nAnalogous Color\n• #462846\n``#462846` `rgb(70,40,70)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #282846\n``#282846` `rgb(40,40,70)``\nSplit Complementary Color\n• #462837\n``#462837` `rgb(70,40,55)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #283746\n``#283746` `rgb(40,55,70)``\n• #463728\n``#463728` `rgb(70,55,40)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #283746\n``#283746` `rgb(40,55,70)``\n• #372846\n``#372846` `rgb(55,40,70)``\n• #11150c\n``#11150c` `rgb(17,21,12)``\n• #1e2615\n``#1e2615` `rgb(30,38,21)``\n• #2a361f\n``#2a361f` `rgb(42,54,31)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #445631\n``#445631` `rgb(68,86,49)``\n• #51663b\n``#51663b` `rgb(81,102,59)``\n• #5d7744\n``#5d7744` `rgb(93,119,68)``\nMonochromatic Color\n\n# Alternatives to #374628\n\nBelow, you can see some colors close to #374628. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #3f4628\n``#3f4628` `rgb(63,70,40)``\n• #3c4628\n``#3c4628` `rgb(60,70,40)``\n• #3a4628\n``#3a4628` `rgb(58,70,40)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #354628\n``#354628` `rgb(53,70,40)``\n• #324628\n``#324628` `rgb(50,70,40)``\n• #304628\n``#304628` `rgb(48,70,40)``\nSimilar Colors\n\n# #374628 Preview\n\nThis text has a font color of #374628.\n\n``<span style=\"color:#374628;\">Text here</span>``\n#374628 background color\n\nThis paragraph has a background color of #374628.\n\n``<p style=\"background-color:#374628;\">Content here</p>``\n#374628 border color\n\nThis element has a border color of #374628.\n\n``<div style=\"border:1px solid #374628;\">Content here</div>``\nCSS codes\n``.text {color:#374628;}``\n``.background {background-color:#374628;}``\n``.border {border:1px solid #374628;}``\n\n# Shades and Tints of #374628\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, #060804 is the darkest color, while #fbfcfa is the lightest one.\n\n• #060804\n``#060804` `rgb(6,8,4)``\n• #10140b\n``#10140b` `rgb(16,20,11)``\n• #1a2113\n``#1a2113` `rgb(26,33,19)``\n• #232d1a\n``#232d1a` `rgb(35,45,26)``\n• #2d3a21\n``#2d3a21` `rgb(45,58,33)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #41522f\n``#41522f` `rgb(65,82,47)``\n• #4b5f36\n``#4b5f36` `rgb(75,95,54)``\n• #546b3d\n``#546b3d` `rgb(84,107,61)``\n• #5e7845\n``#5e7845` `rgb(94,120,69)``\n• #68844c\n``#68844c` `rgb(104,132,76)``\n• #729153\n``#729153` `rgb(114,145,83)``\n• #7c9d5a\n``#7c9d5a` `rgb(124,157,90)``\n• #85a764\n``#85a764` `rgb(133,167,100)``\n• #8fae71\n``#8fae71` `rgb(143,174,113)``\n• #99b57d\n``#99b57d` `rgb(153,181,125)``\n• #a3bc8a\n``#a3bc8a` `rgb(163,188,138)``\n``#adc396` `rgb(173,195,150)``\n• #b7caa3\n``#b7caa3` `rgb(183,202,163)``\n• #c0d1af\n``#c0d1af` `rgb(192,209,175)``\n``#cad9bc` `rgb(202,217,188)``\n• #d4e0c8\n``#d4e0c8` `rgb(212,224,200)``\n• #dee7d5\n``#dee7d5` `rgb(222,231,213)``\n• #e8eee1\n``#e8eee1` `rgb(232,238,225)``\n• #f1f5ee\n``#f1f5ee` `rgb(241,245,238)``\n• #fbfcfa\n``#fbfcfa` `rgb(251,252,250)``\nTint Color Variation\n\n# Tones of #374628\n\nA tone is produced by adding gray to any pure hue. In this case, #373935 is the less saturated color, while #376c02 is the most saturated one.\n\n• #373935\n``#373935` `rgb(55,57,53)``\n• #373e30\n``#373e30` `rgb(55,62,48)``\n• #37422c\n``#37422c` `rgb(55,66,44)``\n• #374628\n``#374628` `rgb(55,70,40)``\n• #374a24\n``#374a24` `rgb(55,74,36)``\n• #374e20\n``#374e20` `rgb(55,78,32)``\n• #37531b\n``#37531b` `rgb(55,83,27)``\n• #375717\n``#375717` `rgb(55,87,23)``\n• #375b13\n``#375b13` `rgb(55,91,19)``\n• #375f0f\n``#375f0f` `rgb(55,95,15)``\n• #37640a\n``#37640a` `rgb(55,100,10)``\n• #376806\n``#376806` `rgb(55,104,6)``\n• #376c02\n``#376c02` `rgb(55,108,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #374628 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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[ "# 10 Awesome Python One-Liners Look Like A Pro 😎", null, "", null, "Subscribe to my newsletter and never miss my upcoming articles\n\nHello, buddies! Python programs generally are smaller than other programming languages like Java. Programmers have to type relatively less and indentation requirements of the language make them readable all the time. However, Python programs can be made more concise using some one-liner codes. These can save time by having fewer codes to type 😉", null, "So today we're going to see some awesome Python one-liners. Scroll down!\n\n## 1. For loop in One line\n\nFor loop is a multi-line statement, But in Python, we can write for loop in one line using the List Comprehension method. Let take an example of filtering values that are lesser than 250. Check out the Below code example.\n\n``````\nmylist = [100, 200, 300, 400, 500]\n#Orignal way\nresult = []\nfor x in mylist:\nif x > 250:\nresult.append(x)\nprint(result) # [300, 400, 500]\n\n#One Line Way\nresult = [x for x in mylist if x > 250]\nprint(result) # [300, 400, 500]\n``````\n\n## 2. While Loop in One Line\n\nThis One-Liner snippet will show you how to use the While loop code in One Line, there are two methods of doing this.\n\n``````#method 1 Single Statement\nwhile True: print(1) # infinite 1\n\n#method 2 Multiple Statement\nx = 0\nwhile x < 5: print(x); x= x + 1 # 0 1 2 3 4 5\n``````\n\n## 3. If Else Statement in One Line\n\nWell, to write the IF Else statement in One Line we will use the ternary operator. Syntax of Ternary is `[on true] if [expression] else [on false]`\n\nThere are 3 examples in the below example code to make your understanding clear that how to use the ternary operator for one line if-else statement. To use the `Elif` statement we had to use multiple Ternary operators. 👇\n\n``````\n#Example 1 if else\nprint(\"Yes\") if 8 > 9 else print(\"No\") # No\n\n#Example 2 if elif else\nE = 2\nprint(\"High\") if E == 5 else print(\"Meidum\") if E == 2 else print(\"Low\") # Medium\n\n#Example 3 only if\nif 3 > 2: print(\"Exactly\") # Exactly\n``````\n\n## 4. Function in One Line\n\nWe have two methods to write Functions in one line, In the first method, we will use the same function definition with the ternary operator or one-line loop methods. The second method is to define functions with lambda. Check out the example code below for more clear understanding 👇\n\n``````#method 1\ndef fun(x): return True if x % 2 == 0 else False\nprint(fun(2)) # False\n#method 2\nfun = lambda x : x % 2 == 0\nprint(fun(2)) # True\nprint(fun(3)) # False\n``````\n\n## 5. Recursion in One Line\n\nThis One-Liner Snippet will show how to use Recursion in one line. we will use one line function definition with one line if-else statement. Below is an example of finding the Fibonacci numbers.\n\n``````#Fibonaci example with one line Recursion\ndef Fib(x): return 1 if x in {0, 1} else Fib(x-1) + Fib(x-2)\nprint(Fib(5)) # 8\nprint(Fib(15)) # 987\n``````\n\n## 6. Read File in One Line\n\nIt is possible to read the file in one line correctly without using the statement or normal reading method.\n\n``````#Normal Way\nwith open(\"data.txt\", \"r\") as file:\nprint(data) # Hello world\n#One Line Way\ndata = [line.strip() for line in open(\"chat.txt\",\"r\")]\nprint(data) # ['hello buddies', 'Are you doing well?']\n``````\n\n## 7. Class in One Line\n\nClass is always multi-line work. But in Python, there are some ways in which you can use class-like features in One line of codes.\n\n``````#Normal way\nclass Emp:\ndef __init__(self, name, age):\nself.name = name\nself.age = age\nemp1 = Emp(\"Unity Buddy\", 1000)\nprint(emp1.name, emp1.age) # Unity Buddy 1000\n#One Line Way\n#method 1 Lambda with Dynamic Artibutes\nEmp = lambda: None; Emp.name = \"Unity Buddy\"; Emp.age = 1000\nprint(Emp.name, Emp.age) # Unity Buddy 1000\n#method 2\n\nfrom collections import namedtuple\nEmp = namedtuple('Emp', [\"name\", \"age\"]) (\"Unity Buddy\", 1000)\nprint(Emp.name, Emp.age) # Unity Buddy 1000\n``````\n\n## 8. Map Function in One Line\n\nThe Map function is a higher-order function that applies. That applies a function to every element. Below is the Example that how we can use the map function in one line of Code.\n\n``````print(list(map(lambda a: a + 2, [5, 6, 7, 8, 9, 10])))\n#output\n# [7, 8, 9, 10, 11, 12]\n``````\n\n## 9. Find Prime Number in One Line\n\nThis snippet will show you how to write a one-liner code to find the Prime number within the range.\n\n``````print(list(filter(lambda a: all(a % b != 0 for b in range(2, a)), range(2,20))))\n#Output\n# [2, 3, 5, 7, 11, 13, 17, 19]\n``````\n\n## 10. Printing any data in an easy-to-read format\n\nDo you sometimes have a lot of data or data which varies in type, and you want to print it all out to the terminal for further inspection but it all prints in one huge line and you can’t see anything? Well for that you can use the built-in `pprint` module in Python, this will automatically format your data and print it nicely to your terminal. Here I have a dictionary with random values and use the `pprint` function from the `pprint` module to print it out!\n\n``````from pprint import pprint\n\ndata = {\"ahead\": {\"between\": \"boy\", \"blew\": \"allow\", \"path\": 1972198692, \"inside\": False, \"fly\": \"throat\", \"toward\": -1359495784.298975}}\n\npprint(data)\n``````\n\nBONUS: Import all the libraries in one go\n\n`pyforest` is a python module that can help us import all the libraries in a single line of python. You can expect all kinds of general use python libraries to get imported. This also includes most of the libraries used for machine learning tasks. This library also has some helper modules such as os, tqdm, re, and many more.\n\n``````pip install pyforest\n``````\n\nTo check the list of libraries, run this `dir(pyforest)`. After importing pyforest, you can directly run `pd.read_csv()` , `sns.barplot()` , `os.chdir()` and many more!\n\nSo buddies, that's it! Thanks for reading and happy Coding!" ]

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[ "How do you write the sum of the number 55+66 as the product of their GCF and another sum?\n\nOct 19, 2016\n\n$55 + 66 = 11 \\times \\left(5 + 6\\right)$\n\nExplanation:\n\nThe prime factors of $55$ are $5 \\times 11$\n\nand those of $66$ are $2 \\times 3 \\times 11$\n\nAs only $11$ is common, GCF is $11$\n\nand as such $55 + 66$ can be written as\n\n$55 + 66 = 11 \\times 5 + 11 \\times 6 = 11 \\times \\left(5 + 6\\right)$" ]

[ null ]

[ "# Entire function with prescribed values\n\nI am trying to solve the following problem from Ahlfors' Complex Analysis Chapter 5, Section 2.3: Suppose that $\\{a_n\\}$ is a sequence of distinct complex numbers such that $a_n\\to \\infty$ and let $\\{c_n\\}$ be a sequence of arbitrary complex numbers. Show that there exists an entire function $f(z)$ satisfying $f(a_n)=c_n$.\n\nThe hint that is in Ahlfors' book is to let $g(z)$ be a function with simple zeros at each $a_n$. Such a function exists by Weierstrass' Theorem. Then, the hint says to look for appropriate $\\gamma_n$ such that the following series converges $$\\sum_1^\\infty g(z)\\frac{e^{\\gamma_n(z-a_n)}}{z-a_n}\\frac{c_n}{g'(a_n)}.$$\n\nI have been playing around with this for a while. I know that I need to find the $\\gamma_n$ so that on any compact ball, $|z|\\leq R$, the values\n\n$$\\left|g(z)\\frac{e^{\\gamma_n(z-a_n)}}{z-a_n}\\frac{c_n}{g'(a_n)}\\right|$$ are bounded by some values $M_n(R)$ so that for each $R>0$, the sum $\\sum M_n(R)$ converges. If I can do this then I know that the sum will converge uniformly on compact subsets, and so I know that the sum will be an analytic function, and then it will clearly have the right properties. However, I am having difficulty figuring out what I am supposed to choose for $\\gamma_n$. I tried expressing $g$ as a Taylor series around each $a_n$ and then finding some upper bound of the terms in the sum which depended only on $R$, but have had no luck so far.\n\nCan anyone provide a hint about how to go about finding such $\\gamma_n$?\n\n• I cannot add anything constructive, but let me just remark that I am always absolutely amazed how the theory of complex functions of one variable is so much deeper than what is even hinted at in an introductory course. (Where it maybe seems that one has gained total control over holomorphic functions...) – Piotr Pstrągowski Feb 11 '13 at 1:27\n\n## 1 Answer\n\nUnfortunately, Ahlfors' hint is very misleading, and there is in fact a simpler way to solve this problem, especially since at this point of the book Ahlfors has proven both Mittag-Leffler and Weierstrass Theorems.\n\nLet $g$ be an entire function with simple zeros at $a_n$. Recall that Mittag-Leffler's Theorem not only asserts the existence of meromorphic functions with poles at $a_n$, but allows us to control the singular part of the function at each $a_n$. So let $h$ be a meromorphic function on $\\mathbb{C}$ with simple poles at each $a_n$ with singuar part $c_n/(z-a_n)$. Then $f:=gh$ has the desired properties.\n\n• Slight correction: the residues of $h$ should be $c_n/g'(a_n)$. – Unit Apr 16 '16 at 23:09" ]

[ null ]

[ "", null, "", null, "", null, "ISSN 1088-6850(online) ISSN 0002-9947(print)\n\nThe Albanese mapping for a punctual Hilbert scheme. II. Symmetrized differentials and singularities\n\nAuthor: Mark E. Huibregtse\nJournal: Trans. Amer. Math. Soc. 274 (1982), 109-140\nMSC: Primary 14C05; Secondary 14C25, 14F07, 14J99, 14K99\nDOI: https://doi.org/10.1090/S0002-9947-1982-0670923-0\nMathSciNet review: 670923\nFull-text PDF Free Access\n\nAbstract: Let", null, "be the canonical mapping from the irreducible and nonsingular surface", null, "to its Albanese variety", null, ",", null, "the", null, "-fold symmetric product of", null, ", and", null, "the punctual Hilbert scheme parameterizing 0-dimensional closed subschemes of length", null, "on", null, ". The latter is an irreducible and nonsingular variety of dimension", null, ", and the \"Hilbert-Chow\" morphism", null, "is a birational map which desingularizes", null, ". Let", null, "denote the map induced by", null, "by addition on", null, ". This paper studies the singularities of the composite morphism", null, "which is a natural generalization of the mapping", null, ", where", null, "is an irreducible and nonsingular curve and", null, "is its Jacobian. Unlike the latter, however,", null, "need not be smooth for", null, ". We prove that", null, "is smooth for", null, "only if", null, "is smooth (Theorem 3), and over", null, "we prove the converse (Theorem 4). In case", null, "is an abelian surface, we show", null, "is smooth for", null, "prime to the characteristic (Theorem 5), and give a counterexample to smoothness for all", null, "(Theorem 6). We exhibit a connection (over", null, ") between singularities of", null, "and generalized Weierstrass points of", null, "(Theorem 9).\n\nOur method is as follows: We first show that the singularities of", null, "are the zeros of certain holomorphic", null, "-forms on", null, "which are the \"symmetrizations\" of holomorphic", null, "-forms on", null, ". We then study \"symmetrized differentials\" and their zeros on", null, "(Theorems 1,2). Our method works for curves", null, "as well; we give an alternative proof of a result of Mattuck and Mayer [10, p. 226] which shows that the zeros of symmetrized differentials on", null, "represent (for", null, "complete nonsingular) the special divisors of degree", null, "on", null, ".\n\n[Enhancements On Off] (What's this?)\n\nRetrieve articles in Transactions of the American Mathematical Society with MSC: 14C05, 14C25, 14F07, 14J99, 14K99\n\nRetrieve articles in all journals with MSC: 14C05, 14C25, 14F07, 14J99, 14K99" ]

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[ "Here, we use molecular dynamics simulations as a tool to investigate vacancy clustering in pure aluminum single crystals. A 1% superconcentration of single vacancies are randomly introduced into an otherwise perfect lattice, and the system is allowed to evolve for 500 ns at an elevated temperature of 728 K. Under these conditions, the individual vacancies rapidly agglomerate into larger clusters to reduce their overall energy. The systems are then subject to mechanical deformation to failure. The results of a total of 35 molecular dynamics simulations are reported. The mechanical behavior of these systems is found to be highly sensitive to the vacancy cluster microstructure, with the largest cluster size being most closely correlated with the cavitation strength. Since the largest cluster size evolves, an interesting time–structure–property coupling governs the behavior of these supersaturated metals. Despite the idealizations of the microstructure and loading conditions, we find a remarkably favorable agreement with laser-driven spall experiments.\n\nThe equilibrium vacancy concentration in metals at ambient temperature and pressure is extremely low, e.g., $O(10−10)$ for lead (low melting temperature) and $O(10−22)$ for copper (high melting temperature). At such low concentrations, their direct effects on the mechanical behavior are fairly negligible. However, supersaturated vacancy concentrations $O(10−4)−O(10−2)$ can be metastable under extreme conditions, e.g., high temperature (Berger et al., 1979; Adams and Wolfer, 1993), high tensile pressure (Gavini, 2009; Reina et al., 2011), radiation environments (Song et al., 2011; Zinkle and Busby, 2009; Zinkle and Was, 2013; Zhang et al., 2018; Zinkle and Farrell, 1989), or severe plastic deformation (Gray and Huang, 1991; Rose and Berger, 1968; Nancheva and Saarinen, 1986; Kanel, 1998; Nancheva et al., 1986). Such high concentrations of point defects can have significant deleterious effects on the mechanical behavior of metals, e.g., accelerated creep deformation rates, embrittlement, loss of ductility, and some positive effects like improved hardness. Furthermore, under extreme conditions, the vacancies may diffuse and rearrange themselves into lower energy states through the formation of vacancy clusters (Ho et al., 2007; Gavini, 2009). Given that size effects are ubiquitous in the strength of materials, it is reasonable to expect that the mechanical behavior of metals with supersaturated concentrations of vacancies will be somehow correlated with the evolving size and/or spacing distribution of these vacancy clusters.\n\nThe vacancy concentration in aluminum subject to a tensile pressure of $4.5$ GPa at room temperature is $C∼0.06%$ and increases even further at higher tensile pressures (Reina et al., 2011). This is approximately the same vacancy concentration achieved in aluminum near its melting temperature at ambient pressures. Moshe et al. (2000) subjected aluminum foils to short pulses of tensile pressures as large as $8$ GPa via dynamic loading, which will generate extremely large equilibrium vacancy concentrations in excess of 5%, even at room temperature. One of our primary interests here include studying the maximum tensile pressure a material can sustain before unstable failure as a function of vacancy microstructure. This property is termed the cavitation strength (or the spall strength) and is an important property in a range of applications from ballistic performance (cf. Antoun et al., 2003) to tree transpiration (cf. Cochard, 2006; Wang et al., 2017). Contrary to conventional wisdom, Wilkerson and Ramesh (2016a) and earlier Reina et al. (2011) analyzed the experiments of Moshe et al. (2000) and concluded that vacancy-driven processes are likely to play an important role in failure at such high tensile pressures, even on these extremely short timescales. This claim is further supported by observations of spall fracture surfaces of high-purity aluminum reported in Dalton et al. (2011) and Pedrazas et al. (2012) and reprinted here in Fig. 1. For samples subjected to tensile pressures in excess of $∼3.5$ GPa, Pedrazas et al. (2012) found that the majority of fracture sites have no obvious initiation sites and instead fracture seems to spontaneously appear. This is contrary to fracture samples subjected to lower tensile pressures, e.g., $≲3$ GPa, in which all fracture sites have an obvious associated second-phase particle as their fracture initiation site. This transition was captured in idealized models, assuming pre-exisiting vacancy clusters (Wilkerson, 2014, 2017). Similar observations of seemingly spontaneous void nucleation have been observed ahead of blunt crack tips in high-purity FCC metals (Lyles and Wilsdorf, 1975; Wilsdorf, 1982; Wang and Anderson, 1991) and austenitic stainless steel (Bauer and Wilsdorf, 1973), even at room temperature.\n\nFIG. 1.\n\nSEM images of fracture surfaces reprinted from Pedrazas et al. (2012) are shown here for (a) an Al high-purity single crystal with a cavitation strength of 3.9 GPa and (b) an AA1100 single crystal with a cavitation strength of 3.0 GPa. Most of the voids in the commercially pure aluminum shown in (b) nucleate from second-phase particles, e.g., Al$3$Fe, whereas for the high-purity aluminum, only a very small fraction of the dimples $(∼5%)$ contained second-phase particles, implying that the bulk of these voids nucleates by an alternative mechanism, e.g., vacancy clustering and subsequent growth by plastic deformation.\n\nFIG. 1.\n\nSEM images of fracture surfaces reprinted from Pedrazas et al. (2012) are shown here for (a) an Al high-purity single crystal with a cavitation strength of 3.9 GPa and (b) an AA1100 single crystal with a cavitation strength of 3.0 GPa. Most of the voids in the commercially pure aluminum shown in (b) nucleate from second-phase particles, e.g., Al$3$Fe, whereas for the high-purity aluminum, only a very small fraction of the dimples $(∼5%)$ contained second-phase particles, implying that the bulk of these voids nucleates by an alternative mechanism, e.g., vacancy clustering and subsequent growth by plastic deformation.\n\nClose modal\n\nSize effects are ubiquitous in structure–property relationships. Orowan strengthening, (Orowan, 1948; Embury et al., 1989; Nie and Muddle, 1998; Dixit et al., 2008) e.g., in over-aged 2000-, 6000-, and 7000-series aluminum alloys, is governed by the mean spacing of precipitates, which serve as obstacles to dislocation motion. The Taylor hardening law assumes that strain hardening is dictated by the evolving dislocation density (i.e., the material hardens as the mean free-path between forest dislocations decreases) (Taylor, 1938; Bishop and Hill, 1951; Kocks and Mecking, 2003; Ma and Roters, 2004). The Hall–Petch relationship established an empirical relationship between grain size and yield strength (cf. Hall, 1951; Petch, 1953). A similar relationship between grain size and cavitation strength has been recently derived (Wilkerson and Ramesh, 2016b). Solid solution strengthening of the yield strength, e.g., in 3000-series Al-Mn alloys and 5000-series Al–Mg alloys, is typically found to scale with roughly the concentration of point defects raised to the $n$th power, where $n∼0.75$ for binary Al–Mg alloys (Cahn and Haasen, 1996) and $n∼1$ for commercial 3000-series and 5000-series aluminum alloys (Ryen et al., 2006). Yield strength of nanoporous materials generally scales with both ligament size (Dou and Derby, 2009; El-Awady, 2015; Wilkerson, 2019) and relative density (Ashby and Gibson, 1997). In fracture mechanics, the fracture strength of brittle materials scales roughly with the inverse square root of the largest flaw size (Griffith, 1921). In all these examples, the smaller size of volume defects and/or smaller spacing between defects generally induces a higher strength.\n\nHere, we report a systematic study to demonstrate that the mechanical behavior of aluminum is primarily sensitive to the size of these vacancy clusters. This diffusion-driven vacancy clustering process is time-dependent, and for aluminum with $C=1%$ at $T=728$ K is shown to be fast enough for significant degradation of the cavitation strength over nanoseconds of clustering time. This paper is organized as follows. Section II provides an overview of the modeling approach, which makes use of the Large-Scale Atomic/Molecular Massively Parallel Simulator (LAMMPS) molecular dynamics (MD) package. Section III discusses results and analysis of the molecular dynamics simulations. Finally, Sec. IV provides a brief summary and conclusions.\n\nIn the present work, molecular dynamics (MD) simulations are performed using the open source package LAMMPS (Plimpton, 1995). The model system under consideration is a face centered cubic (FCC) aluminum (Al) single crystal containing a $C=1%$ concentration of monovacancies. Our MD system is held fixed at a constant temperature of $T=728$ K chosen to be consistent with related lattice kinetic Monte Carlo (LKMC) simulations on a similar system (Reina et al., 2011). The system is subjected to either a prescribed pressure $(P)$ or volume $(V)$. One particularly powerful advantage of studying these systems via MD rather than LKMC is that mechanical behavior may be analyzed with ease in MD, whereas LKMC is limited to defect evolution. That said, LKMC has significant advantages over MD in the study of large systems over long timescales. In order to study the mechanical behavior of these materials as a function of evolved vacancy cluster structure, we deform the system to failure following a desired clustering time ($Δtc$). The effect of clustering time on the mechanical behavior and cavitation strength is reported. A total of 35 MD simulations are carried out. To our knowledge, this is the first study of its kind to systematically study nonequilibrium structure–property relationships in metals with nonequilibrium concentrations of vacancies.\n\nThe atomic interactions are calculated using the Finnis/Sinclair embedded atom method (EAM) (Mendelev et al., 2008). It is a well developed potential for studying the behavior of Al and accurately captures experimentally measured elastic constants and vacancy formation energy of Al. The initial material defect structure is prepared according to the schematic shown in Fig. 2. First, a perfect Al system consisting of 108 000 atoms in a cubic periodic cell (with a side length of 12.3 nm) is equilibrated at the desired temperature with the pressure held constant at ambient conditions ($P=0$ GPa) by a Nosé–Hoover isothermal-isobaric (NPT) barostat. Once the system comes to equilibrium, a prescribed number of atoms are randomly removed from the system to produce a structure with the desired supersaturated concentration of vacancies. After removing the atoms, the ensemble is switched to an isothermal-isochroic (NVT) canonical ensemble in order to maintain a constant volume of the periodic cell. A Laplace (tensile) pressure is generated by the creation of vacancies. To study the kinetics of the system in driving toward its lowest energy state, we allow the system to evolve under NVT conditions for a desired clustering time. Following clustering for a prescribed clustering time, triaxial extension deformation is carried out at a constant strain rate of $108$ s$−1$ via an NVT ensemble. This particular strain rate is motivated by a suite of laser-driven spall experiments on aluminum carried out at similar strain rates (Moshe et al., 2000). These spall experiments identified an abrupt transition in strain rate-sensitivity at a strain rate of $∼108$ s$−1$, cf. Fig. 3, which could be indicative of a transition in the underlying failure mechanism.\n\nFIG. 2.\n\nSchematic of the preparation of the initial microstructure consisting of cubic periodic cells of FCC aluminum with $1%$ of atoms randomly removed to generate a $C=1%$ vacancy concentration. The microstructure is then allowed to evolve over a prescribed clustering time. The final microstructure is composed of vacancy clusters of various sizes, which are assigned different colors by size for the purpose of visualization. Subsequently, the systems are deformed to failure to understand the evolving structure–property relationships.\n\nFIG. 2.\n\nSchematic of the preparation of the initial microstructure consisting of cubic periodic cells of FCC aluminum with $1%$ of atoms randomly removed to generate a $C=1%$ vacancy concentration. The microstructure is then allowed to evolve over a prescribed clustering time. The final microstructure is composed of vacancy clusters of various sizes, which are assigned different colors by size for the purpose of visualization. Subsequently, the systems are deformed to failure to understand the evolving structure–property relationships.\n\nClose modal\nFIG. 3.\n\nComparison of our MD predicted cavitation strengths with experimentally measured spall strength of aluminum across a range of tensile strain rates (Fortov et al., 1991; Moshe et al., 1998, 2000). The MD markers are for aluminum with $C=1%$ at $T=728$ K with a cavitation strength of $∼4.2$ GPa, which relaxes to a cavitation strength of $∼3.5$ GPa after $Δtc=500$ ns of vacancy clustering time. The shock-induced temperature rise and shock-induced vacancy concentrations in the experiments are unknown.\n\nFIG. 3.\n\nComparison of our MD predicted cavitation strengths with experimentally measured spall strength of aluminum across a range of tensile strain rates (Fortov et al., 1991; Moshe et al., 1998, 2000). The MD markers are for aluminum with $C=1%$ at $T=728$ K with a cavitation strength of $∼4.2$ GPa, which relaxes to a cavitation strength of $∼3.5$ GPa after $Δtc=500$ ns of vacancy clustering time. The shock-induced temperature rise and shock-induced vacancy concentrations in the experiments are unknown.\n\nClose modal\n\nThe random distribution of vacancies is a nonequilibrium state that is far from the lowest energy state for the system. A lower system energy is achievable through the formation of vacancy clusters. Here, a vacancy cluster is defined as a single volume defect composed of a closed set of $ℓ$ vacancies with nearest neighbor connectivity. We denote a vacancy cluster of size $ℓ$ as a cluster containing $ℓ$ vacancies, and we utilize $ℓ^$ to denote the largest cluster size at a given time. Note that this particular definition of clusters does not take into account the geometry of the clusters.\n\nTo identify vacancies in simulation cells, the geometric method of Wigner–Seitz cells (Nordlund et al., 1998) is used. In this method, the atom positions in a simulation cell are analyzed with respect to a lattice defined by undisturbed simulation cell regions. A lattice site which includes an empty Wigner–Seitz cell is labeled a vacancy. Likewise, a Wigner–Seitz cell with multiple atoms is labeled an interstitial. After identifying specific vacancies, the system can be decomposed into disconnected sets of vacancies (clusters of vacancies) based on a distance criterion. The cut-off distance for vacancy clustering here is taken to be 2.86 Å, which is the first neighbor distance of pure aluminum in the simulations. The samples are compared with a reference configuration, which is taken here to be the state of the system after NPT ensemble at the particular temperature and pressure being analyzed. To calculate the vacancy evolution, the state of the system is taken to be characterized by the spatial distribution of vacancies on a frozen lattice. The visualization of MD simulation results is performed with OVITO (Stukowski, 2010).\n\nThe present section provides results of the microstructure evolution of aluminum with a supersaturated concentration of vacancies as predicted by molecular dynamics calculations. Since this particular material system is in a nonequilibrium state, the kinetics of microstructure, e.g., vacancy cluster evolution is of particular interest and is the focus of Subsection III A. Additionally, the effect of these evolving nonequilibrium microstructures on the mechanical response of the system is investigated in Subsection III B. Finally, Subsection III C provides some implications of this diffusion-failure coupled process for the structure–property modeling of the mechanical behavior and cavitation strength under triaxial loading.\n\nOnce the material system is seeded with the prescribed number of vacancies, the system begins to evolve toward a more energetically favorable state. In a real material system with a vast hierarchy of points, lines, and surface defects, the material may lower its overall energy by (i) reducing the overall vacancy concentration via recombination with interstitials and vacancy annihilation at vacancy sinks, e.g., dislocations and grain boundaries; (ii) aggregating vacancies to form vacancy clusters via diffusion; or (iii) a combination of both mechanisms. However, our particular simulations contain no vacancy sinks or interstitials; as such, mechanism (i) is inoperative in our molecular dynamics simulations, and the system is forced to evolve toward lower energy states via mechanism (ii) exclusively. Again, although this is not necessarily the physical situation for most applications, it is very useful for understanding the influence of clustering on the system in the absence of other competing mechanisms.\n\nFigure 4 shows an example of the kinetic evolution for an aluminum system with $1%$ vacancy concentration subject to ambient pressure and a constant temperature of 728 K. The evolution of the microstructure is remarkably fast. In this particular simulation, relatively large vacancy clusters composed of as many as 12 aggregated single vacancies are formed within about 5 ns of clustering time. The clusters continue to grow with time. For example, the largest clusters reach a size of 58 in 100 ns and 109 in 500 ns. Even after 500 ns, the clusters have not yet reach an equilibrium state. This is not particularly surprising given that the LKMC simulations of Reina et al. (2011) revealed that similar systems continued to evolve even after 100 $μ$s. The clustering kinetics are driven by the energetics, and in this particular simulation shown, the overall energy of the system reduces by 500 eV in evolving from the randomly seeded structure of monovacancies ($Δtc=0$) to the defect structure shown in Fig. 4 at $Δtc=500$ ns. This energy relaxation is primarily associated with a reduction in the elastic strain energy (or bond energies), which reduces by approximately 0.14% in the first 500 ns after vacancy generation.\n\nFIG. 4.\n\nKinetics of microstructural evolution of Al with 1% vacancy concentration at constant temperature of $T=728$ K, in the absence of applied external stresses. Individual vacancies agglomerate into vacancy clusters that further ripen with clustering time $(Δtc)$. Vacancy clusters are colored by their size, i.e., the number of single vacancies composing them.\n\nFIG. 4.\n\nKinetics of microstructural evolution of Al with 1% vacancy concentration at constant temperature of $T=728$ K, in the absence of applied external stresses. Individual vacancies agglomerate into vacancy clusters that further ripen with clustering time $(Δtc)$. Vacancy clusters are colored by their size, i.e., the number of single vacancies composing them.\n\nClose modal\n\nIn studying Fig. 4, some interesting questions arise pertaining to the spatial correlation between the initial conditions and the evolving structure. One might assume that the spatial location of the randomly generated trivacancies (the largest clusters at $t=0$) will be the likely location of the largest clusters at later times. However, this does not necessarily seem to be the case. Perhaps, because the trivacancies themselves are somewhat mobile. In addition, it may be seen that even rather large clusters $(>20)$ are absent from spatial locations that they previously occupied. In this case, it is less likely that such large clusters are mobile and more likely that these clusters are being cannibalized by even larger ones.\n\nFurther insight may be gleaned from statistically quantifying the kinetic evolution of each cluster size as shown by the histogram in Fig. 5. The histogram shows the fraction of vacancies belonging to each cluster size as a function of time. Initially, nearly $100%$ of the vacancies are single vacancies; however, this fraction very rapidly drops down to only about $15%$ of the vacancies with the remaining $85%$ of vacancies belonging to clusters. Another interesting observation from this analysis is a relative dearth of clusters in the size range of $ℓ=$3$−$10 for all time after about 10 ns. After this time, very few $(<5%)$ of vacancies belong to this intermediate size range, with about $80%$ of the vacancies belonging to clusters larger than 10. In addition, after about 500 ns, most of the clusters in the size of $ℓ=$3$−$40 are cannibalized by the larger clusters. This seems to indicate that the intermediate size clusters are the least energetically favorable. Perhaps, this phenomenon is associated with the large Laplace pressure (and associated strain energy penalty) generated by such intermediate clusters. Due to the constant volume constraint of the NVT ensemble, a tensile Laplace pressure is induced by the creation of the vacancies. The induced Laplace (tensile) pressure is $∼0.25$ GPa in magnitude.\n\nFIG. 5.\n\nHistogram representation of the temporal evolution of the fraction of vacancies belonging to vacancy clusters of various sizes for Al with 1% vacancy concentration at a constant temperature of $T=728$ K. Color indicates $Δtc$. The inset shows the temporal evolution of the largest vacancy cluster.\n\nFIG. 5.\n\nHistogram representation of the temporal evolution of the fraction of vacancies belonging to vacancy clusters of various sizes for Al with 1% vacancy concentration at a constant temperature of $T=728$ K. Color indicates $Δtc$. The inset shows the temporal evolution of the largest vacancy cluster.\n\nClose modal\n\nCloser inspection of Fig. 5 shows that the percentage of monovacancies rapidly drops in the first 10 ns of clustering time, from around 100% of total vacancies to around 50% and this decrease continues rapidly to about 15% in 100 ns of clustering time. This is also reflected in the evolution of the number of discrete clusters (including monovacancies) as shown in Fig. 6. In Fig. 5, the percentage of monovacancies continues to slowly decrease to around 8% after 580 ns of clustering time. It is interesting to note that although a high percentage of monovacancies (around 500) cluster together in the first 10 ns of clustering time, it takes about 400 ns for a cluster as large as 100 to be grown. A possible explanation for this might be that the mobility of vacancy clusters likely decreases with their size.\n\nFIG. 6.\n\nTemporal evolution of the discrete number of vacancy clusters in Al with 1% vacancy concentration at a constant temperature of $T=728$ K. The number of clusters is normalized by the total number of lattices in a simulation box. Note that FCC systems have four atoms per lattice.\n\nFIG. 6.\n\nTemporal evolution of the discrete number of vacancy clusters in Al with 1% vacancy concentration at a constant temperature of $T=728$ K. The number of clusters is normalized by the total number of lattices in a simulation box. Note that FCC systems have four atoms per lattice.\n\nClose modal\n\nAs a practical matter, for vacancy clustering to play any role in real applications involving a supersaturated vacancy concentration, the timescales for vacancy clustering must be faster than the time required for vacancies to diffuse to vacancy sinks, e.g., dislocation cores and grain boundaries, or to recombine with interstitials. Otherwise, the supersaturated vacancy concentration will drop back down toward the equilibrium concentration before any significant clustering takes place. Moreover, some applications in which vacancy clustering is speculated to play a role, e.g., shock and spall failure (Reina et al., 2011; Wilkerson and Ramesh, 2016a), involve very short timescales $O$(1 ns–1 $μ$s). As such, the kinetics of vacancy clustering must be relatively fast for clustering to play any meaningful role in such applications. At least for this case of extremely high vacancy concentrations and fairly high temperatures, the results shown in Figs. 4 and 5 seem to suggest that vacancy clustering is indeed fast enough to play a role in shock loading applications.\n\nAs noted, these particular simulation results (shown in Figs. 4 and 5) represent a fairly high temperature relative to the melting temperature of perfect aluminum (933 K). Additionally, the vacancy concentration is likely one- to two-orders of magnitude higher than what is expected to be achievable in any of the motivating applications discussed earlier in Sec. I. For comparison, the equilibrium vacancy concentration of aluminum at 728 K is $∼10−4$. Nevertheless, we argue that the insights that may be gleaned from such molecular dynamics calculations are quite valuable in the development of continuum-level constitutive models, as will be discussed further in Sec. III C.\n\nThus far, this paper has focused on the kinetics of vacancy clustering in otherwise defect-free pure Al with a supersaturated concentration of vacancies. We have reported that the clustering kinetics are quite fast, even on nanosecond timescales. This section will discuss how vacancy clusters of various sizes and spacing distributions affect the mechanical behavior of these systems. In particular, it will be shown that the rearrangement of randomly seeded monovacancies (and divacancies, trivacancies, etc.) into lower energy states composed of large vacancy clusters weakens the cavitation strength of these systems. Thus, since this vacancy clustering process is time-dependent, it must follow that the strength is also time-dependent, provided the microstructure and strength are correlated. To study this time evolving microstructure–strength relationship, after a prescribed amount of clustering time to allow a particular degree of microstructural evolution, the MD samples are then deformed under triaxial (volumetric) tensile loading at a constant engineering strain rate of $ε˙=108$ s$−1$.\n\nFigure 7 shows the true mean stress ($Σm≜−P$) vs longitudinal engineering strain ($ε$) with response of the FCC Al samples for an initial vacancy concentration of $C=1%$ at a constant temperature of $T=728$ K as a function of clustering time $Δtc$. In all cases, the stress starts at a value higher than zero (an initial tensile stress), then increases nonlinearly with deformation to a peak tensile (cavitation) strength, followed by a complex softening response. The nonlinear response is essentially pure elastic until right before the peak stress is achieved. This general nonlinear elastic response may be captured by an appropriate equation of state that accounts for the nonlinear relationship between pressure and volume at high elastic strains, e.g., the equation of state (Mie, 1903; Grüneisen, 1912). The stiffness (nonlinear bulk modulus) is independent of the clustering time, i.e., independent of the size distribution of the vacancy clusters at a particular vacancy concentration.\n\nFIG. 7.\n\nStress–strain response of Al with a vacancy concentration of 1% at a constant temperature of $T=728$ K. Each line indicates the mean stress vs longitudinal strain response after a particular prescribed clustering time $Δtc$ over which the microstructure evolves. The figure inset shows the peak tensile (cavitation) strength as a function of clustering time. The samples are deformed under triaxial tensile loading at a strain rate of $ε˙=108$ s$−1$.\n\nFIG. 7.\n\nStress–strain response of Al with a vacancy concentration of 1% at a constant temperature of $T=728$ K. Each line indicates the mean stress vs longitudinal strain response after a particular prescribed clustering time $Δtc$ over which the microstructure evolves. The figure inset shows the peak tensile (cavitation) strength as a function of clustering time. The samples are deformed under triaxial tensile loading at a strain rate of $ε˙=108$ s$−1$.\n\nClose modal\n\nOn the other hand, the peak tensile (cavitation) strength, denoted here as $Σm∗$, is quite sensitive to the clustering time, as shown in the inset of Fig. 7. For example, for $Δtc=0$ ns (immediately after vacancy generation and without any clustering time), the tensile cavitation strength is about 4.2 GPa at an engineering strain of about 2.8%. After $Δtc=500$ ns of clustering time, the tensile cavitation strength is about 3.5 GPa at 2.4% engineering strain. It is noteworthy that this cavitation strength agrees remarkably well with the laser-driven spall strength experiments carried out by Moshe et al. (2000) on aluminum at a similar strain rate. It is quite impressive that a diffusion-driven $∼17%$ (700 MPa) weakening of the tensile cavitation strength is achieved in less than a microsecond for aluminum with $C=1%$ at $T=728$ K. It should be noted that this $∼17%$ (700 MPa) weakening of the tensile cavitation strength cannot be explained by the initial (zero strain) tensile (Laplace) pressure alone, i.e., $Σm(ε=0)$, which only reduces by $150$ MPa from $Δtc=0$ ns to $Δtc=100$ ns. In other words, even if the initial tensile stresses in Fig. 7 are zeroed out, there will still be a 550 MPa weakening of the cavitation strength.\n\nThis observed drop in the strength of samples by increasing the clustering time is obviously associated with the change in microstructure, i.e., creation of large vacancy clusters at the expense of monovacancy populations. In examining the dislocation nucleation process (via OVITO), we find that dislocations tend to most favorably nucleate at the largest vacancy clusters, i.e., the largest flaw theory as gleaned by Griffith (1921) nearly 100 years ago. More recently, Lubarda et al. (2004) derived a theory for dislocations nucleating from a nanovoid surface, which was later expanded upon by Wilkerson and Ramesh (2016a). Their models predict that the macroscopic stress required for heterogeneous dislocation nucleation at the surface of a nanovoid is a function of the size of said nanovoid. Generally speaking, the nucleation stress reduces nonlinearly with increasing size of the nanovoid and is generally easier than homogeneously nucleating dislocations within the perfect lattice. This size effect is primarily driven by size effects associated with surface tension, dislocation image stresses, length scales associated with stress field perturbations around the nanovoid, and the hop distance of dislocation nuclei under thermal excitation, cf. Lubarda et al. (2004) and Wilkerson and Ramesh (2016a) for details. Once dislocations are nucleated, the nanovoids rapidly grow and drive a sudden (nearly brittle) relaxation of the macroscopic stress. From our MD simulations here, it seems that this process is (to first order) essentially the same for a material with a complex distribution of vacancy clusters that have agglomerated via diffusion processes. Section III C will quantify these relations in greater detail.\n\nFollowing the above discussion, it is now apparent that the clustering time–strength relationship is essentially a size effect that would be analogous to the Hall–Petch size effects for a microstructure experiencing dynamic grain growth. As such, the correlation of interest is between the shape of the largest cluster size evolution vs clustering time (which can be seen in Fig. 5 by tracing out the extreme value of the histogram for each time, which is shown in the inset of Fig. 5) with the general shape of the cavitation strength vs clustering time shown in the inset of Fig. 7. In comparing these two curves, it is interesting to note that most of the diffusion-driven weakening is achieved in the first 100 ns of clustering, with a fairly small change over the next 400 ns of clustering. That said, Fig. 5 demonstrates that there is still substantial evolution of vacancy cluster size distribution during this interval from $Δtc=100$ ns to $Δtc=500$ ns. If we assume, as argued above, that this evolving structure–strength relationship is fundamentally governed by a vacancy cluster size effect, then we can conclude that the size effects are strongest for very small cluster sizes, e.g., $ℓ∼1−$30, and the size effect is not as strong as the clusters continue to grow, which comports with the models of Lubarda et al. (2004) and Wilkerson and Ramesh (2016a).\n\nIt should be noted that our simulations of clustering kinetics are not truly subject to ambient pressure conditions due to the tensile Laplace pressures generated by the creation of vacancies following NPT equilibration. As the vacancies cluster into larger sizes, this tensile stress relaxes. For example, the Laplace pressure prior to any vacancy clustering ($Δtc=0$) is about 0.25 GPa. After $Δtc=500$ ns of clustering time, the Laplace pressure has relaxed to about 0.1 GPa. This is to be expected as the Laplace pressure scales with the inverse of the void radius, i.e., $γ/a,$ with $γ$ denoting the surface energy. Recall from Fig. 5 that the largest vacancy cluster (denoted here as $ℓ^$) grows from an initial size of $ℓ^=3$ to $ℓ^∼100$ after $Δtc=500$ ns of clustering time for $C=1%$ at $T=728$ K. Hence, the initial Laplace pressure is fairly sensitive to the size distribution of vacancy clusters, as expected. Since vacancy mobility increases with tensile pressure (Ho et al., 2007), it is anticipated that this Laplace pressure plays some role in slightly accelerating the vacancy clustering process from what it would otherwise be under truly ambient conditions. That said, the Laplace pressure is a physical phenomena that would occur in real systems, so it should not be neglected.\n\nOur overall assessment thus far is that the rearrangement of monovacancies into larger vacancy clusters generally weakens the cavitation strength (and associated failure strain), relaxes the Laplace pressure, but does not affect the general elastic response (as expected). Any trends in the post-peak stress softening response are fairly difficult to discern in Fig. 7.\n\nHaving discussed how the evolving vacancy microstructure affects mechanical behavior, the final technical piece of this paper addresses implications of our findings for reduced-order continuum models. The fundamental questions we aim to address here:\n\n• Does the largest cluster size dominate the cavitation strength of these supersaturated FCC metals?\n\n• Is it possible to adequately predict the cavitation strength of these systems by tracking the largest cluster size?\n\n• Or is it necessary to track the full statistics of the vacancy cluster size distributions?\n\nTo help answer these questions, we study an idealized set of microstructures with a single (periodic) spherical void as shown in Fig. 8 (on the right). These idealized microstructures are constructed such that the volume of the single void is identical to the volume of the largest vacancy cluster, i.e., $ℓ^ΔΩv$, in the random clustering simulation under comparison, shown in Fig. 8 (on the left). Next, the simulation box size of the idealized microstructure is chosen such that the void volume fraction is identical to the vacancy concentration in the random clustering simulation, i.e., $C=1%=ℓ^ΔΩv/w′3$. We may deduce that $ℓ^$ is the most important parameter governing the cavitation strength, if the cavitation strength of the random clustering microstructure agrees reasonably well with that of the idealized microstructure. In other words, that the details of the vacancy cluster size distribution are secondary in comparison to $ℓ^$, since the idealized microstructure has a Dirac delta size distribution, and the random vacancy cluster has a complex (evolving) size distribution as shown in Fig. 5. To compare the cavitation strength of the random clustering microstructure and the idealized microstructure, we generated several idealized microstructures and subjected them to the same loading conditions, i.e., triaxial expansion loading at a constant strain rate of $108$ s$−1$. Representative stress–strain curves for the idealized microstructure are shown in Fig. 9.\n\nFIG. 8.\n\nSnapshot (on left) of a random microstructure (with a complex vacancy size distribution whose largest size is denoted as $ℓ^$) generated via our vacancy clustering approach with a box size of $w3$. Snapshot (on right) of an idealized microstructure with a single (periodic) spherical void of size $ℓ^$, i.e., a Dirac delta size distribution with a box size of $w′3$.\n\nFIG. 8.\n\nSnapshot (on left) of a random microstructure (with a complex vacancy size distribution whose largest size is denoted as $ℓ^$) generated via our vacancy clustering approach with a box size of $w3$. Snapshot (on right) of an idealized microstructure with a single (periodic) spherical void of size $ℓ^$, i.e., a Dirac delta size distribution with a box size of $w′3$.\n\nClose modal\nFIG. 9.\n\nRepresentative stress–strain response of nanoporous Al at a constant temperature of $T=728$ K for various nanovoid sizes. The idealized microstructure consists of a repeating periodic array of nanovoids of radius $a$ on a square lattice of spacing $w$. The figure inset shows the peak tensile (cavitation) strength as a function of clustering time. The samples are deformed under triaxial tensile loading at a strain rate of $ε˙=108$ s$−1$.\n\nFIG. 9.\n\nRepresentative stress–strain response of nanoporous Al at a constant temperature of $T=728$ K for various nanovoid sizes. The idealized microstructure consists of a repeating periodic array of nanovoids of radius $a$ on a square lattice of spacing $w$. The figure inset shows the peak tensile (cavitation) strength as a function of clustering time. The samples are deformed under triaxial tensile loading at a strain rate of $ε˙=108$ s$−1$.\n\nClose modal\n\nIn Fig. 10, the cavitation strength is plotted as a function of $ℓ^$, equivalent to the volume of the largest vacancy cluster. The crosses correspond to the cavitation strength of the random vacancy clustering microstructures (the peak stresses in Fig. 7). The closed circles correspond to the cavitation strength of the idealized (single periodic void) microstructures (the peak tensile stresses in Fig. 9). Since the largest vacancy clusters may grow during the mechanical loading phase, the cluster sizes are extracted at a tensile longitudinal strain of $ε=0.02.$\n\nFIG. 10.\n\nCavitation strength of aluminum with $C=1%$ at $T=728$ K for various vacancy cluster size distributions. Crosses correspond to the cavitation strength of the random vacancy clustering microstructures at various clustering times, e.g., snapshot on left in Fig. 8. The closed circles correspond to the cavitation strength of the idealized (single periodic void) microstructures, e.g., snapshot on right in Fig. 8.\n\nFIG. 10.\n\nCavitation strength of aluminum with $C=1%$ at $T=728$ K for various vacancy cluster size distributions. Crosses correspond to the cavitation strength of the random vacancy clustering microstructures at various clustering times, e.g., snapshot on left in Fig. 8. The closed circles correspond to the cavitation strength of the idealized (single periodic void) microstructures, e.g., snapshot on right in Fig. 8.\n\nClose modal\n\nBy comparing the crosses and closed circles in regions of overlap ($20≲ℓ^≲110$) in Fig. 10, we see that the idealized microstructures are only slightly weaker than the random vacancy clustering microstructure. Overall, the trends are in strong agreement with a similar size dependence in both microstructures. Therefore, we conclude that this is strong evidence that $ℓ^$ is adequate to capture the dominate trends in the cavitation strength. Thus, we can safely state that the details of the vacancy cluster size distribution are secondary when it comes to the cavitation strength. This finding has important implications for developing atomistically informed constitutive models of the mechanical behavior of these supersaturated materials, given that tracking full size distribution statistics is typically fairly computationally expensive.\n\nHere, we have utilized molecular dynamics simulations to study vacancy clustering processes in pure aluminum single crystals. A superconcentration of single vacancies are randomly introduced into an otherwise perfect lattice, and the system is allowed to evolve for 500 ns under (nearly) adiabatic pressure. The individual vacancies will rapidly agglomerate into larger clusters to reduce the system’s overall energy.\n\nFurthermore, the mechanical behavior of these systems is found to be strongly dependent on the microstructure, e.g., the manner in which the randomly seeded monovacancies rearrange themselves into lower energy states composed of distributions of vacancy clusters. Thus, since this vacancy clustering process is time-dependent, it must follow that the mechanical behavior is also time-dependent. To study these nonequilibrium microstructure–property relationships, after a prescribed amount of clustering time to allow a particular degree of microstructural evolution, the MD samples were then deformed under triaxial (volumetric) tensile loading at a constant engineering strain rate of $ε˙=108$ s$−1$. Some of the key findings are summarized below:\n\n• The timescales of clustering are sufficiently fast to be applicable even under shock loading conditions.\n\n• An initial Laplace tensile pressure is generated by the creation of vacancies, which relaxes as clusters grow.\n\n• The nonlinear elastic response is fairly insensitive to the arrangement of vacancies into clusters.\n\n• The cavitation strength is found to be strongly sensitive to the arrangement of vacancies into clusters.\n\n• The relationship between microstructure and the post-peak stress softening response is indiscernible.\n\n• Cavitation strength is most sensitive to the largest cluster size, with the details of the size distribution playing a secondary role.\n\n• Our MD predicted cavitation strength after $Δtc≳100$ ns of clustering time agrees remarkably well with the laser-driven spall strength measurements reported in Moshe et al. (2000).\n\nThis material is based upon work supported by the Army Research Laboratory under the MEDE Collaborative Research Alliance through Cooperative Agreement No. W911NF-12-2-0022. The views and conclusions contained in this document are those of the authors and should not be interpreted as representing the official policies, either expressed or implied, of the Army Research Laboratory or the U.S. Government. The U.S. Government is authorized to reproduce and distribute reprints for Government purposes notwithstanding any copyright notation herein. This work was supported by the TACC Computational Resource Center through the use of its high performance computing facilities. Portions of this research were conducted with high performance research computing resources provided by the Texas A&M University (https://hprc.tamu.edu).\n\n1\n,\nJ.\nand\nWolfer\n,\nW.\n, “\nVoid formation in rapidly-solidified metals\n,”\nActa Metall. Mater.\n41\n(\n9\n),\n2625\n2632\n(\n1993\n).\n2\nAntoun\n,\nT.\n,\nCurran\n,\nD. R.\n,\nSeaman\n,\nL.\n,\nKanel\n,\nG. I.\n,\nRazorenov\n,\nS. V.\n, and\nUtkin\n,\nA. V.\n,\nSpall Fracture\n(\n,\n2003\n).\n3\nAshby\n,\nM.\nand\nGibson\n,\nL.\n,\nCellular Solids: Structure and Properties\n(\nPress Syndicate of the University of Cambridge\n,\nCambridge\n,\n1997\n), pp.\n175\n231\n.\n4\nBauer\n,\nR.\nand\nWilsdorf\n,\nH.\n, “\nVoid initiation in ductile fracture\n,”\nScr. Metall.\n7\n(\n11\n),\n1213\n1220\n(\n1973\n).\n5\nBerger\n,\nA. S.\n,\nOckers\n,\nS. T.\n, and\nSiegel\n,\nR. W.\n, “\nMeasurement of the monovacancy formation enthalpy in copper\n,”\nJ. Phys. F Metal Phys.\n9\n(\n6\n),\n1023\n1033\n(\n1979\n).\n6\nBishop\n,\nJ.\nand\nHill\n,\nR.\n, “\nCXXVIII. A theoretical derivation of the plastic properties of a polycrystalline face-centred metal\n,”\nPhilos. Mag.\n42\n(\n334\n),\n1298\n1307\n(\n1951\n).\n7\nCahn\n,\nR. W.\nand\nHaasen\n,\nP.\n,\nPhysical Metallurgy\n(\nNorth-Holland\n,\nAmsterdam\n,\n1996\n), Vol. 70.\n8\nCochard\n,\nH.\n, “\nCavitation in trees\n,”\nC. R. Phys.\n7\n(\n9–10\n),\n1018\n1026\n(\n2006\n).\n9\nDalton\n,\nD. A.\n,\nWorthington\n,\nD. L.\n,\nSherek\n,\nP. A.\n,\nPedrazas\n,\nN. A.\n,\nQuevedo\n,\nH. J.\n,\nBernstein\n,\nA. C.\n,\nRambo\n,\nP.\n,\nSchwarz\n,\nJ.\n,\nEdens\n,\nA.\n,\nGeissel\n,\nM.\n,\nSmith\n,\nI. C.\n,\nTaleff\n,\nE. M.\n, and\nDitmire\n,\nT.\n, “\nMicrostructure dependence of dynamic fracture and yielding in aluminum and an aluminum alloy at strain rates of $2×106s−1$ and faster\n,”\nJ. Appl. Phys.\n110\n(\n10\n),\n103509\n(\n2011\n).\n10\nDixit\n,\nM.\n,\nMishra\n,\nR.\n, and\nSankaran\n,\nK.\n, “\nStructure–property correlations in Al 7050 and Al 7055 high-strength aluminum alloys\n,”\nMater. Sci. Eng. A\n478\n(\n1–2\n),\n163\n172\n(\n2008\n).\n11\nDou\n,\nR.\nand\nDerby\n,\nB.\n, “\nA universal scaling law for the strength of metal micropillars and nanowires\n,”\nScr. Mater.\n61\n(\n5\n),\n524\n527\n(\n2009\n).\n12\n,\nJ. A.\n, “\nUnravelling the physics of size-dependent dislocation-mediated plasticity\n,”\nNat. Commun.\n6\n,\n5926\n(\n2015\n).\n13\nEmbury\n,\nJ.\n,\nLloyd\n,\nD.\n, and\nRamachandran\n,\nT.\n, “Strengthening mechanisms in aluminum alloys,” in Treatise on Materials Science & Technology (Elsevier, 1989), Vol. 31, pp. 579–601.\n14\nFortov\n,\nV.\n,\nKostin\n,\nV.\n, and\nEliezer\n,\nS.\n, “\nSpallation of metals under laser irradiation\n,”\nJ. Appl. Phys.\n70\n(\n8\n),\n4524\n4531\n(\n1991\n).\n15\nGavini\n,\nV.\n, “\nRole of the defect core in energetics of vacancies\n,”\nProc. R. Soc. A\n465\n,\n3239\n3266\n(\n2009\n).\n16\nGray\n,\nG. T.\nand\nHuang\n,\nJ.\n, “\nInfluence of repeated shock loading on the substructure evolution of 99.99 wt.% aluminum\n,”\nMater. Sci. Eng. A\n145\n(\n1\n),\n21\n35\n(\n1991\n).\n17\nGriffith\n,\nA. A.\n, “\nThe phenomena of rupture and flow in solids\n,”\nPhilos. Trans. R. Soc. London\n221\n(\n582–593\n),\n163\n198\n(\n1921\n).\n18\nGrüneisen\n,\nE.\n, “\nTheorie des festen zustandes einatomiger elemente\n,”\nAnn. Phys.\n344\n(\n12\n),\n257\n306\n(\n1912\n).\n19\nHall\n,\nE.\n, “\nThe deformation and ageing of mild steel\n,”\nProc. Phys. Soc. London\n64\n(\n747–753\n),\n495\n(\n1951\n).\n20\nHo\n,\nG.\n,\nOng\n,\nM. T.\n,\nCaspersen\n,\nK. J.\n, and\nCarter\n,\nE. A.\n, “\nEnergetics and kinetics of vacancy diffusion and aggregation in shocked aluminium via orbital-free density functional theory\n,”\nPhys. Chem. Chem. Phys.\n9\n(\n36\n),\n4951\n4966\n(\n2007\n).\n21\nKanel\n,\nG.\n, “\nSome new data on deformation and fracture of solids under shock-wave loading\n,”\nJ. Mech. Phys. Solids\n46\n(\n10\n),\n1869\n1886\n(\n1998\n).\n22\nKocks\n,\nU.\nand\nMecking\n,\nH.\n, “\nPhysics and phenomenology of strain hardening: The fcc case\n,”\nProg. Mater. Sci.\n48\n(\n3\n),\n171\n273\n(\n2003\n).\n23\nLubarda\n,\nV.\n,\nSchneider\n,\nM.\n,\nKalantar\n,\nD.\n,\nRemington\n,\nB.\n, and\nMeyers\n,\nM.\n, “\nVoid growth by dislocation emission\n,”\nActa Mater.\n52\n(\n6\n),\n1397\n1408\n(\n2004\n).\n24\nLyles, Jr.\n,\nR. L.\nand\nWilsdorf\n,\nH.\n, “\nMicrocrack nucleation and fracture in silver crystals\n,”\nActa Metall.\n23\n(\n2\n),\n269\n277\n(\n1975\n).\n25\nMa\n,\nA.\nand\nRoters\n,\nF.\n, “\nA constitutive model for fcc single crystals based on dislocation densities and its application to uniaxial compression of aluminium single crystals\n,”\nActa Mater.\n52\n(\n12\n),\n3603\n3612\n(\n2004\n).\n26\nMendelev\n,\nM.\n,\nKramer\n,\nM.\n,\nBecker\n,\nC.\n, and\nAsta\n,\nM.\n, “\nAnalysis of semi-empirical interatomic potentials appropriate for simulation of crystalline and liquid Al and Cu\n,”\nPhilos. Mag.\n88\n(\n12\n),\n1723\n1750\n(\n2008\n).\n27\nMie\n,\nG.\n, “\nZur kinetischen theorie der einatomigen körper\n,”\nAnn. Phys.\n316\n(\n8\n),\n657\n697\n(\n1903\n).\n28\nMoshe\n,\nE.\n,\nEliezer\n,\nS.\n,\nDekel\n,\nE.\n,\nLudmirsky\n,\nA.\n,\nHenis\n,\nZ.\n,\nWerdiger\n,\nM.\n,\nGoldberg\n,\nI.\n,\nEliaz\n,\nN.\n, and\nEliezer\n,\nD.\n, “\nAn increase of the spall strength in aluminum, copper, and Metglas at strain rates larger than $107s−1$\n,”\nJ. Appl. Phys.\n83\n(\n8\n),\n4004\n4011\n(\n1998\n).\n29\nMoshe\n,\nE.\n,\nEliezer\n,\nS.\n,\nHenis\n,\nZ.\n,\nWerdiger\n,\nM.\n,\nDekel\n,\nE.\n,\nHorovitz\n,\nY.\n,\nMaman\n,\nS.\n,\nGoldberg\n,\nI.\n, and\nEliezer\n,\nD.\n, “\nExperimental measurements of the strength of metals approaching the theoretical limit predicted by the equation of state\n,”\nAppl. Phys. Lett.\n76\n(\n12\n),\n1555\n1557\n(\n2000\n).\n30\nNancheva\n,\nN.\nand\nSaarinen\n,\nK.\n, “\n,”\nScr. Metall.\n20\n(\n8\n),\n1085\n1088\n(\n1986\n).\n31\nNancheva\n,\nN. M.\n,\nSaarinen\n,\nK.\n, and\nPopov\n,\nG. S.\n, “\nPositron annihilation in shock loaded titanium and titanium alloy BT14\n,”\nPhys. Status Solidi A\n95\n(\n2\n),\n531\n536\n(\n1986\n).\n32\nNie\n,\nJ. F.\nand\nMuddle\n,\nB. C.\n, “\nMicrostructural design of high-strength aluminum alloys\n,”\nJ. Phase Equilib.\n19\n(\n6\n),\n543\n(\n1998\n).\n33\nNordlund\n,\nK.\n,\nGhaly\n,\nM.\n,\nAverback\n,\nR. S.\n,\nCaturla\n,\nM.\n,\nDiaz De La Rubia\n,\nT.\n, and\nTarus\n,\nJ.\n, “\nDefect production in collision cascades in elemental semiconductors and fcc metals\n,”\nPhys. Rev. B\n57\n(\n13\n),\n7556\n7571\n(\n1998\n).\n34\nOrowan\n,\nE.\n, Symposium on Internal Stresses in Metals and Alloys (Institute of Metals, London, 1948), p. 451.\n35\nPedrazas\n,\nN. A.\n,\nWorthington\n,\nD. L.\n,\nDalton\n,\nD. A.\n,\nSherek\n,\nP. A.\n,\nSteuck\n,\nS. P.\n,\nQuevedo\n,\nH. J.\n,\nBernstein\n,\nA. C.\n,\nTaleff\n,\nE. M.\n, and\nDitmire\n,\nT.\n, “\nEffects of microstructure and composition on spall fracture in aluminum\n,”\nMater. Sci. Eng. A\n536\n,\n117\n123\n(\n2012\n).\n36\nPetch\n,\nN. J.\n, “\nThe cleavage strength of polycrystals\n,”\nJ. Iron Steel Inst.\n174\n,\n25\n28\n(\n1953\n), available at https://ci.nii.ac.jp/naid/10005741398/en/.\n37\nPlimpton\n,\nS.\n, “\nFast parallel algorithms for short-range molecular dynamics\n,”\nJ. Comput. Phys.\n117\n(\n1\n),\n1\n19\n(\n1995\n).\n38\nReina\n,\nC.\n,\nMarian\n,\nJ.\n, and\nOrtiz\n,\nM.\n, “\nNanovoid nucleation by vacancy aggregation and vacancy-cluster coarsening in high-purity metallic single crystals\n,”\nPhys. Rev. B\n84\n(\n10\n),\n104117\n(\n2011\n).\n39\nRose\n,\nM. F.\nand\nBerger\n,\nT. L.\n, “\nShock deformation of polycrystalline aluminium\n,”\nPhilos. Mag.\n17\n(\n150\n),\n1121\n1133\n(\n1968\n).\n40\nRyen\n,\nØ.\n,\nHolmedal\n,\nB.\n,\nNijs\n,\nO.\n,\nNes\n,\nE.\n,\nSjölander\n,\nE.\n, and\nEkström\n,\nH.-E.\n, “\nStrengthening mechanisms in solid solution aluminum alloys\n,”\nMetall. Mater. Trans. A\n37\n(\n6\n),\n1999\n2006\n(\n2006\n).\n41\nSong\n,\nK.\n,\nPauly\n,\nS.\n,\nZhang\n,\nY.\n,\nScudino\n,\nS.\n,\nGargarella\n,\nP.\n,\nSurreddi\n,\nK.\n,\nKühn\n,\nU.\n, and\nEckert\n,\nJ.\n, “\nSignificant tensile ductility induced by cold rolling in $Cu47.5Zr47.5Al5$ bulk metallic glass\n,”\nIntermetallics\n19\n(\n10\n),\n1394\n1398\n(\n2011\n).\n42\nStukowski\n,\nA.\n, “\nVisualization and analysis of atomistic simulation data with OVITO–the Open Visualization Tool\n,”\nModell. Simul. Mater. Sci. Eng.\n18\n(\n1\n),\n015012\n(\n2010\n).\n43\nTaylor\n,\nG. I.\n, “\nPlastic strain in metals\n,”\nJ. Inst. Metals\n62\n,\n307\n324\n(\n1938\n).\n44\nWang\n,\nJ.-S.\nand\nAnderson\n,\nP.\n, “\nFracture behavior of embrittled fcc metal bicrystals\n,”\nActa Metall. Mater.\n39\n(\n5\n),\n779\n792\n(\n1991\n).\n45\nWang\n,\nP.\n,\nGao\n,\nW.\n,\nWilkerson\n,\nJ.\n,\nLiechti\n,\nK. M.\n, and\nHuang\n,\nR.\n, “\nCavitation of water by volume-controlled stretching\n,”\nExtreme Mech. Lett.\n11\n,\n59\n67\n(\n2017\n).\n46\nWilkerson\n,\nJ.\n, “\nOn the micromechanics of void dynamics at extreme rates\n,”\nInt. J. Plast.\n95\n,\n21\n(\n2017\n).\n47\nWilkerson\n,\nJ.\nand\nRamesh\n,\nK.\n, “\nA closed-form criterion for dislocation emission in nano-porous materials under arbitrary thermomechanical loading\n,”\nJ. Mech. Phys. Solids\n86\n,\n94\n116\n(\n2016a\n).\n48\nWilkerson\n,\nJ.\nand\nRamesh\n,\nK.\n, “\nUnraveling the anomalous grain size dependence of cavitation\n,”\nPhys. Rev. Lett.\n117\n(\n21\n),\n215503\n(\n2016b\n).\n49\nWilkerson\n,\nJ. W.\n, “Multiscale mechanics of failure in extreme environments,” Ph.D. thesis (Johns Hopkins University, 2014).\n50\nWilkerson\n,\nJ. W.\n, “\nAnomalous size effects in nanoporous materials induced by high surface energies\n,”\nJ. Mater. Res.\n34\n,\n1\n10\n(\n2019\n).\n51\nWilsdorf\n,\nH. G.\n, “\nThe role of glide and twinning in the final separation of ruptured gold crystals\n,”\nActa Metall.\n30\n(\n6\n),\n1247\n1258\n(\n1982\n).\n52\nZhang\n,\nX.\n,\nHattar\n,\nK.\n,\nChen\n,\nY.\n,\nShao\n,\nL.\n,\nLi\n,\nJ.\n,\nSun\n,\nC.\n,\nYu\n,\nK.\n,\nLi\n,\nN.\n,\nTaheri\n,\nM. L.\n,\nWang\n,\nH.\n,\nWang\n,\nJ.\n, and\nNastasi\n,\nM.\n, “\n,”\nProg. Mater. Sci.\n96\n,\n217\n321\n(\n2018\n).\n53\nZinkle\n,\nS.\nand\nFarrell\n,\nK.\n, “\nVoid swelling and defect cluster formation in reactor-irradiated copper\n,”\nJ. Nucl. Mater.\n168\n(\n3\n),\n262\n267\n(\n1989\n).\n54\nZinkle\n,\nS. J.\nand\nBusby\n,\nJ. T.\n, “\nStructural materials for fission & fusion energy\n,”\nMater. Today\n12\n(\n11\n),\n12\n19\n(\n2009\n).\n55\nZinkle\n,\nS. J.\nand\nWas\n,\nG. S.\n, “\nMaterials challenges in nuclear energy\n,”\nActa Mater.\n61\n,\n735\n(\n2013\n)." ]

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[ "Multiplying complex numbers worksheet. About. By using this website, you agree to our Cookie Policy. Complex Numbers; Trigonometry; ELA Worksheets; Multiplying 2-digit by 2-digit Numbers Worksheets . Here's an example: Example One Multiply (3 + 2i)(2 - i). Multiply Polar Complex - Displaying top 8 worksheets found for this concept.. 28 scaffolded questions that start relatively easy and end with some real … Free worksheetpdf and answer key on multiplying complex numbers. Adding and Subtracting Complex Numbers Worksheet and Multiplying and Dividing Rational Expressions. Answers to Dividing Complex Numbers 1) i 2) i 2 3) 2i 4) − 7i 4 5) 1 8 − i 2 6) 1 10 − i 2 7) − 1 7 + 9i 7 8) 3 2 + 3i 2 9) − 1 5 + i 15 10) − 3 13 + 2i 13 11) 2 5 + 3i 10 12) 4 5 − 2i 5 13) − 27 113 − 47i 113 14) − 59 53 + 32i 53 15) 3 29 + 22i 29 16) − 17 25 − 4i 25 17) 57 89 − 69i 89 18) 41 145 − 28i 145 19) 36 + 11i 109 20) −2 − i 2. F i2 n0o12f ekunt la i zs3onf mtmwtaqruec 0lwlocxo f ha jl jln … Imaginary Numbers Lesson Plans & Worksheets from multiplying complex numbers worksheet , image source: www.lessonplanet.com. Student answer sheet provided to allow students to record their work and answers. Multiplying integers is no different from multiplying whole numbers, except taking care of following the rules of determining the sign of the answer. Complex multiplication is a more difficult operation to understand from either an algebraic or a geometric point of view. Multiplying complex numbers is much like multiplying binomials. A Complex Number is a combination of a Real Number and an Imaginary Number: A Real Number is the type of number we use every day. Types: Worksheets, Assessment, Homework. Some of the worksheets for this concept are Multiplying complex numbers, Multiplication and division in polar form, Multiplication and division in polar form, Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Appendix e complex numbers e1 e complex numbers, Complex numbers. 9. 2. Who said multiplying 98 by 67 is a tall order? Multiply using the FOIL method. You will discover a number of the templates are free to use and others call for a premium account. For example, multiply (1+2i)⋅(3+i). Khan Academy is a 501(c)(3) nonprofit organization. Learn how to multiply two complex numbers. Displaying top 8 worksheets found for - Multiplying And Dividing Imaginary And Complex Numbers. Show Ads. They will have 4 problems multiplying complex numbers in polar form written in degrees, 3 more problems in radians, then 4 problems where they divide complex numbers written in polar form in degrees, then 4 mor . This website uses cookies to ensure you get the best experience. Multiplying Complex Numbers Worksheet – If you find a template that you would like to use, begin customizing it and you could also to open it in your document window! Hide Ads About Ads. Simple subtraction worksheets showing negative answers introduced with the number line. The two numbers are already in standard form. Answers to multiplying complex numbers 1 64i 2 14i 3 18 6i 4 8i 5 24 6 64 7 20 46i 8 25 49i 9 20 50i 10 18 66i 11 2 18i 12 30 20i 13 21 18i 14 24 36i 15 126 210i 16 7 35i 17 7 199i 18 568 144i 19 252 84i 20 224 288i. Learn how to multiply two complex numbers. Multiplying a Complex Number by a Real Number. The materials are organized by chapter and lesson, with one Word Problem Practice D: Find the training resources you need for all your activities. 3. 1 true or false. Learn about imaginary numbers, complex numbers, a + bi forms, and negative radicals. Complex numbers name multiple choice. Complex Number Calculator. Complex Number Multiplication . 1) i 2) i 3) (cos isin ) 4) (cos isin ) Plot each point in the complex plane. I say \"almost\" because after we multiply the complex numbers, we have a little bit of simplifying work. Show more details Add to cart. Includes-- --Set of 24 Cards (12 Question 12 5) i Real Imaginary .. Donate or volunteer today! Precalculus Eureka Math EngageNY Math from adding and subtracting complex numbers worksheet , source:khanacademy.org. Multiplying complex numbers matching practice. This topic covers: - Adding, subtracting, multiplying, & dividing complex numbers - Complex plane - Absolute value & angle of complex numbers - Polar coordinates of complex numbers. Courses. ©f i2 N0O12F EKunt la i ZS3onf MtMwtaQrUeC 0LWLoCX.o F hA jl jln DrDiag ght sc fr 1ersve1r2vte od P.a G XMXaCdde 9 9waiht5hB 1I2nAfUizn ZibtMeV fA Sl Agesb 7rfa G G2D.Z Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Operations with Complex Numbers Date_____ Period____ Simplify. You should always be looking for ways to simplify your life and be a more efficient person. Worksheet November 24, 2018 130 views. For example, multiply (1+2i)⋅(3+i). Subjects: Math, PreCalculus. The major difference is that we work with the real and imaginary parts separately. Our mission is to provide a free, world-class education to anyone, anywhere. ©V f2 20P1 64o gK 6u tKaf TSZoMfAt4w Kalr 6eg RLmLJC N.S 9 jA dl VlV cr idgTh LtPsK TrFetsSeSrJvexd e.2 a zM Ta 4d9e 1 2wFintfhL BIhn mfYiwn ViDtqe o rA el1g qeYbsrBab 12K.1 Worksheet by Kuta Software LLC Answers to Multiplying and Dividing Complex Numbers (ID: 1) 1) 15 + 112 i 2) 2 − 19 i 3) −35 − 12 i 4) −46 − 43 i We can convert the numbers in the sequence into a list by using “multiply by zero” as our conversion formula. Let’s examine a few simple examples. We distribute the real number just as we would with a binomial. How to Multiply Powers of I Example 1. Multiply Complex Numbers Worksheet Pdf And Answer Key 28 Scaffolded Questions On Simplify Complex Numbers Simplifying Rational Expressions Number Worksheets . Let’s do it algebraically first, and let’s take specific complex numbers to multiply, say 3 + 2i and 1 + 4i. Solution Use the distributive property to write this as. Multiplying complex numbers is almost as easy as multiplying two binomials together. First up, let’s multiply (2 – 3i) with (-4 – 2i) 1. We are going to prove them wrong, and this set of worksheets helps. Answers to Multiplying Complex Numbers 1) 64i 2) 14i 3) −18 − 6i 4) 8i 5) 24 6) −64 7) −20 − 46i 8) −25 + 49i 9) 20 − 50i 10) 18 + 66i 11) 2 − 18i 12) 30 + 20i 13) −21 + 18i 14) −24 − 36i 15) 126 + 210i 16) 7 − 35i 17) 7 − 199i 18) 568 + 144i 19) 252 + 84i 20) 224 + 288i . Complex numbers worksheet pdf. whole numbers B. Multiplying complex numbers is much like multiplying binomials. The Multiplying Complex Numbers Worksheet can be found on the web and you should be able to find one of these books that explains the use of it. Simplify by adding like terms, and then put the answer into standard form. 3(2 - i) + 2i(2 - i) 6 - 3i + 4i - 2i 2. Our imaginary numbers worksheets come with an answer key for every worksheet and a free video tutorial BEFORE you buy! Try the variations with all facts, or print the worksheets that focus only on specific families of multiplication facts that need more practice! 4. Tomorrow's lesson is on multiplying and dividing rational numbers. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Learn more Accept. Adding and subtracting complex numbers might seem complicated, but it really is not. Related Posts of \"Adding and Subtracting Complex Numbers Worksheet\" Waves Worksheet Answer Key Physics. Dividing by a fraction is the same as multiplying by its solution. Replace i 2 with -1. Let’s begin by multiplying a complex number by a real number. Worksheet with answer keys complex numbers. 4 8 B. b 7. Learning to multiply and divide does just that. Whether its addition or subtraction, you can follow these steps. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step . You can use it with software that is designed for this and it can help you make the most of it. Or, we can use formulas and functions that allow us to perform the conversions automatically. Before referring to Multiplying Complex Numbers Worksheet, remember to understand that Education and learning is actually the crucial for a better next week, as well as studying does not only halt after a institution bell rings.This currently being claimed, we all offer you a variety of uncomplicated nonetheless helpful articles and also web themes built suited to almost any informative purpose. The general form of complex numbers is a + bi a is the real part of the number, and b is the imaginary part. Put the like terms together. Simplify the Imaginary Number $$i^9 \\\\ i ^1 \\\\ \\boxed{i}$$ Example 2. 28 Subtraction Worksheets. SHARE ON Twitter Facebook WhatsApp Pinterest. Students will simplify 18 algebraic expressions with complex numbers imaginary numbers including adding subtracting multiplying and dividing complex numbers includes rationalizing the denominator by multiplying by the conjugate algebra 2 curriculum this resource works well as independent prac. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Gallery of 50 Multiplying Complex Numbers Worksheet how to use the relation i squared = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.worksheets, games and activities that are suitable for Common Core High School: Number & Quantity, HSN-CN.A.2, imaginary numbers, examples and step by … If you're seeing this message, it means we're having trouble loading external resources on our website. Grades: 11 th, 12 th. For K-12 kids, teachers and parents. Advanced. Products & Quotients of Complex Numbers in … If you are trying to understand the Waves Worksheet, you should first … Video Tutorial on Multiplying Imaginary Numbers. Dividing complex numbers simplify. Worksheet by kuta … This collection of free worksheets ensures students take home relevant, adequate practice on multiplication of two-digit numbers with and without regrouping. Wish List. Use the rules of exponents (in other words add 6 + 3) $$i^{\\red{6 + 3}} = i ^9$$ Step 2. Complex number any … This is your starting point for negative numbers. Introduction to Negative Numbers Subtraction with Negative Results. Count on our printable multiplying mixed numbers worksheets for all the practice you need to perfect your skills in multiplying a mixed number with another mixed number, or multiplying three mixed numbers or completing the fraction multiplication equations. I 1 1 a true b false write the number as a product of a real number and i. Some of the worksheets for this concept are Multiplying complex numbers, Dividing complex numbers, Infinite algebra 2, Chapter 5 complex numbers, Operations with complex numbers, Plainfield north high school, Introduction to complex numbers, Complex numbers and powers of i. Site Navigation. Examples: 12.38, ½, 0, −2000. Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. This activity allows students to use the foil method to multiply two complex numbers to get an answer in the form of a+bi. Free worksheet pdf and answer key on complex numbers. Nov 8, 2016 - Free worksheet(pdf) and answer key on multiplying complex numbers. Name: _____Math Worksheets Date: _____ www.EffortlessMath.com 12 Answers Multiplying and Dividing Complex Numbers 1) 5 2) 20 3) 7 4) 12 5) −7−6 6) −8 7) 6−42 − 8) 9+40 9) 8−20 10) −34−34 11) −60+2 65 12) 25+77 13) 25+46 14) 2 15) 3 4 + 16) 9−5 17) 2 5 −6 5 These printable multiplying integers worksheets do the needful in building multiplication skills. Worksheet by Kuta Software LLC Kuta Software - Infinite Precalculus Complex Numbers and Polar Form Name_____ Date_____ Period____-1-Find the absolute value. Simplify the following product: $$i^6 \\cdot i^3$$ Step 1. Add like terms, simplify and put the answer into standard form (a+bi) Right, time for some examples. On specific families of multiplication facts that need more practice make the most of.. Numbers Worksheet '' Waves Worksheet answer key 28 scaffolded questions that start relatively easy and with... Education to anyone, anywhere games, quizzes, worksheets and a free video BEFORE. Sure that the domains *.kastatic.org and *.kasandbox.org are unblocked specific families of multiplication facts that need more!! Answer in the form of a+bi \\\\ i ^1 \\\\ \\boxed { i } i^6 i^3. Worksheets showing negative answers introduced with the number is given in rectangular form and polar coordinates the. Record their work and answers we have a little bit of Simplifying.! Addition or subtraction, you agree to our Cookie Policy into standard form sign of templates... -4 – 2i ) ( 3 ) nonprofit organization you 're seeing this,. A web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked relevant, practice... About imaginary numbers worksheets 0, −2000 activity allows students to use the distributive property to write this as a... Work and answers 12.38, ½, 0, −2000 and be a more efficient person Trigonometry ; worksheets! Numbers to get an answer in the form of a+bi do the needful in building multiplication.....Kasandbox.Org are unblocked ways to simplify your life and be a more efficient.. 12.38, ½, 0, −2000 the best experience *.kasandbox.org are unblocked number by a fraction is same! Worksheet '' Waves Worksheet answer key 28 scaffolded questions that start relatively easy end. Scaffolded questions on simplify complex Expressions using algebraic rules step-by-step this concept & Quotients of complex numbers in these! An answer in the form of a+bi precalculus Eureka Math EngageNY Math from adding and subtracting complex Calculator! The foil method to multiply two complex numbers Worksheet and multiplying and Dividing Rational numbers multiply polar -..., multiply ( 1+2i ) ⋅ ( 3+i ) of free worksheets ensures students take home relevant, adequate on! Tutorial BEFORE you buy 1 a true b false write the number is given in rectangular and., quizzes, worksheets and a free, world-class education to anyone,.... Adding like terms, and then put the answer into standard form functions that allow us to the... These printable multiplying integers is no different from multiplying whole numbers, except taking care of the! A premium account this set of worksheets helps free, world-class education to anyone, anywhere of following rules! And polar coordinates when the number line, anywhere or subtraction, you can use formulas and multiplying complex numbers worksheet answers... That we work with the real and imaginary parts separately using algebraic rules step-by-step 3i + 4i - 2. From multiplying whole numbers B. multiplying complex numbers Worksheet '' Waves Worksheet answer key on multiplying numbers. 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Looking for ways to simplify your life and be a more efficient person students to record their work and.! Answer into standard form worksheets helps write this as multiply polar complex - displaying top worksheets. The templates are free to use and others call for a premium.. Free worksheetpdf and answer key on multiplying complex numbers Worksheet pdf and answer key Physics is much multiplying... B. multiplying complex numbers Calculator - simplify complex numbers Worksheet and multiplying and Rational. These steps s multiply ( 1+2i ) ⋅ ( 3+i ) with ( -4 – 2i ) 1 always... Agree to our Cookie Policy quizzes, worksheets and a forum, you to., let ’ s multiply ( 2 - i ) it means 're! Efficient person and a forum to allow students to record their work and answers two complex might... I ^1 \\\\ \\boxed { i } multiplying complex numbers worksheet answers 2 and subtracting complex numbers, we use! Try the variations with all facts, or print the worksheets that focus only specific. 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Or, we can use formulas and functions that allow us to perform the conversions.... It really is not start relatively easy and end with some real … Worksheet. Number line means we 're having trouble loading external resources on our website two complex numbers Trigonometry ; worksheets. Top 8 worksheets found for this and it can help you make the most of it Worksheet answer key complex. Binomials together One multiply ( 3 + 2i ) 1 worksheets helps example One multiply ( )!, −2000 who said multiplying 98 by 67 is a 501 ( c ) ( +..., but it really is not a more efficient person \\$ Step 1 as a product of a real and. Lesson is on multiplying complex numbers in … these printable multiplying integers do! A free video tutorial BEFORE you buy language, plus puzzles, games, quizzes, multiplying complex numbers worksheet answers and a video... It really is not 're behind a web filter, please make sure that the domains.kastatic.org. By multiplying a complex number by a real number and i facts that need more practice please make sure the! Tomorrow 's lesson is on multiplying complex numbers, except taking care of the... Eureka Math EngageNY Math from adding and subtracting complex numbers Calculator - simplify complex numbers try the variations with facts...\n\nUnrequited Love 2020, Cavapoo Rescue Ny, Höfchen Frankfurt Speisekarte, Back Office Jobs In Gvk Airport, Is Silver Fullbuster Evil, Canadian Leftist Podcast, Assam News Today, Spray Adhesive Canada, No Leadership Positions In High School, Love, Rosie Summary, Cost Cutters Printable Coupons," ]

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[ "", null, "Download", null, "Download Presentation", null, "4.1 Coordinates Objective : To plot points and name points in the coordinate plane.\n\n# 4.1 Coordinates Objective : To plot points and name points in the coordinate plane.\n\nDownload Presentation", null, "## 4.1 Coordinates Objective : To plot points and name points in the coordinate plane.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. 4.1 CoordinatesObjective: To plot points and name points in the coordinate plane. is formed by two real number lines that intersect at the origin. (x-axis and y-axis) A coordinate plane is a point in the coordinate plane represented by real numbers. The x-coordinate is the first number. The y-coordinate is the second number. Ex. (3,6) An ordered pair (x, y)  (right or left, up or down) How do you tell?\n\n2. Coordinate plane (x, y) y-axis Quadrant II (-, +) Quadrant I (+, +) Origin (0,0) x- axis Quadrant III (-, -) Quadrant IV (+, -)\n\n3. Plotting points To plot a point: (3,4) Start at (0,0) Move 3 to the right (positive) Then 4 up (positive) make a point • Plot these points: • (-2, -4) • (0, 3) • (-1,0) • (6,-2) • (-4, 5)\n\n4. Practice Name the following points and give the quadrant or axis where they lie. A: B: C: D: A D C B Plot the following points and label each!! E (-3, 4) F(-5, 0) G(-3, -1) H(0, 0)\n\n5. The Coordinate Plane Steps to Make a Scatter Plot: • Determine what will be x and y. • x – is in charge, it changes automatically • y – depends on x, is not automatic • Determine units of each axis and label. • Find range of variable • Divide range by number of squares • Always round up to “nice” unit • Plot points.\n\n6. 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Example Make a Scatter Plot The age (in years) of seven used cars and the price (in thousands of dollars) paid for the cars are recorded in the table. Make a scatter plot and explain what it indicates. price in \\$1,000 age of car How much would a 2-year old car cost?\n\n7. Make a Scatter Plot Example The amount (in millions of dollars) spent in the United States on snowmobiles is shown in the table. Make a scatter plot and explain what it indicates.\n\n8. Reminders • Math Lab Tomorrow - Stocks • 4.1-4.3 Quiz on Monday, Oct. 7th • Homework: • P. 206-207 #’s 10-26 EVEN, 35-37 • EXTRA CREDIT – Halloween “Goblin’ Goblin” Grid\n\n9. 4.2 Graphing Linear Equations Objective: To graph linear equations using a table of values. Note (1) All the Eqs. in Chap 4 refer 2 variable linear Eqs. (2) The graph of each linear eq. is a LINE The solution to a linear equation -- is an ordered pair (x, y). There are many solutions to a linear equation and all of the solutions together form a straight line, \n\n10. Find out if the ordered pairs are solutions. HOW? A) -5x – 8y = 15 (-3, 0) B) -2x – 9y = 7 (-1, -1)\n\n11. -1 0 1 Graph a line 5 Steps to graph a line 3 1. Pick three values for x 1 • Plug in values for x then solve for y • Solve for y, then evaluate y for all input x 3. Graph the ordered pairs 4. Connect these order pairs. This should form a straight line! y = -2x + 3\n\n12. -1 0 1 Function form  When an equation is solved for y = What are the advantages of putting the equation into function form? Graph the given linear equations. A) 3x – y = 2 Solve for y: We select three x values, and evaluate the corresponding y values.\n\n13. Graph the given linear equations. Solve for y, B) 2x – 2y = 10 We select three x values, and evaluate the corresponding y values. What are the “good” x values we should select? The x values should make “nice” y values. (or, no fraction values for y)\n\n14. Make a table of values and graph the following line: 6x – 3y = 12 2y = 4x + 1\n\n15. Ex 4) y = -3 Ex 5) x = 4 Ex 6) x = -2 Ex 7) y = 0 Special Linear Equations always a vertical line x = #  always a horizontal line y = #  MEMORIZE THESE!!!! It’s easy! x = # , label the # on x-axis, then “cut” there y = # , label the # on y-axis, then “cut” there\n\n16. Make a table of values and graph the following line: y = -3 x = 4\n\n17. Summary • A two variables linear equation represents a line in x-y coordinate plan. • An ordered pair is a solution to a two variable linear equation, then the point represented by the ordered pair is on the line represented by the linear equation, and vice versa. • Remember the two types of special line by an easy way: • x = #  no y  cut x-axis at that #  parallel to y-axis • y = #  no x  cut y-axis at that #  parallel to x-axis\n\n18. Summary • When graphing a linear equation, remember the 4 steps: • Pick a few x values • Solve for y, then evaluate y for all input x • Graph ordered pairs • Connect ordered pairs with a line\n\n19. Lesson 4.2 DHQ • Decide whether the given ordered pair is a solution of 2x – 3y = 8. • (-2, -4) b. (7, -2) • Rewrite 4x – 2y = 18 in function form.\n\n20. Reminders • 4.1-4.3 Quiz on Monday, Oct. 7th • Homework: • P. 214-215 #’s 15-20, 30-32, 36-37, 60\n\n21. y-intercept?( , ) x-intercept? ( , ) 4.3 Quick Graphs Using Intercepts Objective: To graph lines using x and y-intercepts. Where the line or curve crosses the x-axis. This should be written as the point (x, 0) . (WHY?) What is an x-intercept? What is a y-intercept? Where the line or curve crosses the y-axis. This should be written as the point (0, y) . (WHY?) 0 3 2 0 Note x or y intercept is a point!!!\n\n22. 6x + 3y = 12 Remember – TWO POINTS CAN MAKE A LINE! Ex 1) Find the x and y-intercepts To find the y-intercept; let x = 0 and solve for y. To find the x-intercept; let y = 0 and solve for x. 6( 0 ) +3y = 12 6x + 3( 0 ) = 12 3y = 12 6x = 12 y = 4 x = 2 The y intercept is (0, 4) The x intercept is (2, 0)\n\n23. You try these: Calculate the x and y-intercept. Then graph each line. 1. 2x – y = 4 2. 3y – 2x = -6\n\n24. What happens with horizontal and vertical lines? Find the x and y-intercepts (if possible). Graph each line. Ex 2) y = 4 Ex 3) x = -1 You can set x = 0 but end with “No solution”  can not find x, or no x-intercept. You can set y = 0 but end with “No solution”  can not find y, or no y-intercept. y-int. (0, 4) x-int. (-1, 0) Variable x does not show up  no x-intercept  line is parallel to x-axis Variable y does not show up  no y-intercept  line is parallel to y-axis\n\n25. You try these: Calculate the x and/or y-intercept. Then graph each line. 4. y = 2x + 4 5. y = -3\n\n26. Graph and Write the equation of the special line 7. Vertical line passing through (-2, 3) 6. Horizontal line passing (-3, 4) and (4, 4).\n\n27. Summary • x-intercept (y-intercept) is a point where the line or curve crosses the x(y)-axis. • To find x-intercept (y-intercept), just setting • y = 0 (x = 0) in an equation. • 3. When graph a line, just find x and/or y-intercept and then connect two intercepts with a line.\n\n28. Find the intercepts and graph. 3) 2x – 4y = –8 4) x – 4y = 2 The point where line(curve) crosses the x-axis. Graphing with Intercepts What is an x-intercept? How do you find an x-intercept? (Why does this work?) How should you write the x-intercept? set y = 0 and then solve for x. In an order pair (x, y). The point where line(curve) crosses the y-axis. What is a y-intercept? How do you find a y-intercept? (Why does this work?) How should you write the y-intercept? set x = 0 and then solve for y. In an order pair (x, y). – 4y = –8 – 4y = 2 y = 2 y = –1/2 2x = –8 x = 2 x = –4 So – If I gave you a quiz over this material, how would you do?\n\n29. Lesson 4.3 DHQ • Give the x- and y- intercepts of the graph of 2x – y = -4. • 2. Graph 2x – 3y = 6 using x and y intercepts.\n\n30. Reminders • 4.1-4.3 Quiz on Monday, Oct. 7th • Homework: • P. 221-222 #’s 35-37, 44-49, 56-57\n\n31. What is represented by this BrainBat? C H I M A D E N A\n\n32. (ORDER MATTERS!)  Slope: m = m = m = To read the slope from a graph, choose 2 lattice points and write a ratio of the vertical to horizontal change. Explain.   4.4 The Slope of a Line Objective: To calculate the slope of a line using 2 points, to read slope from a given line and to understand some applications of slope. 1) The measurement of the steepness and direction of a line (m) Slope is:\n\n33. The slope formula – finding slopes from ordered pairs If you are given 2 points  you have 2 x-values and 2 y-values. You need memorize it!!!\n\n34. 6) Describe how to move a slope of…….  a) -2  b) c) d) 5a) Sketch a positive slope. c) Sketch a 0 slope. d) Sketch an undefined slope. b) Sketch a negative slope.\n\n35. Ex 4) The store plane descends 100 feet for every 2000 feet it travels. Find the slope of decent. Ex 5) The road rises 2 feet for every 50 yards. Find the slope of the road. * REMEMBER TO CONVERT YDs to FT\n\n36. (order matters!!) • Summary • There are 3 different formulas to calculate the slope:\n\n37. Lesson 4.4 DHQ • Find the slope of each line. Use formula to find the slope. • (-2, 2), (0, 4) • (1, 1), (4, 2) • 2. Find the slope of the line.\n\n38. Reminders • 4.4+4.6 Quiz on Wednesday, Oct. 16th • Homework: • P. 230-231 #’s 23-28, 35-36 ***38\n\n39. 4.6 Graph Using Slope-Intercept Form Objective: 1. To use slope intercept form to graph a line. 2. Tell the slope of two parallel lines. Slope-Intercept Form is :Memorize this form!! y = mx + b (m = slope & b = y-intercept.) • The equation must be y = to use this short-cut • m is always the coefficient of • y-intercept (b) is always x Constant – (no variable)\n\n40. Solve the following equations for y = then pick out the m and b for each. -2x -2x y = –2x - 5 -2 -5\n\n41. Use the slope (m) and y-intercept (b) to graph each line. 2x + y = 5 m = b = x - 2y = 4 m = b = You need to graph a) the y-intercept first and then b) use the slope to get the second (third) point (move the slope) c) connect the two (three) points by a line.\n\n42. One more for you to try.  3x – 2y = –6 m = b = Can you solve for y? Can you locate the m and b? Can you put this on the graph paper? 3x – 2y = –6 -3x -3x\n\n43. Summary • The slope-intercept form is the form in which y is solved. • In the slope-intercept, the number in front of x is the slope m and the number right after x is the y-intercept. • When you graph a line of slope-intercept form, you have to graph • a) the y-intercept first and then • b) use the slope to get the second (third) point • c) connect the two (three) points by a line. • 4. Parallel lines have equal slope.\n\n44. Lesson 4.6 DHQ • Write x + y + 3 = 0 in slope-intercept form. Then graph the equation.\n\n45. Reminders • 4.4+4.6 Quiz on Wednesday, Oct. 16th • Homework: • P. 244 #’s 13-18, 29-31\n\n46. Slope-Intercept Form Organization Check: We can now graph an equation in three different ways: Using an x, y table Using x-intercepts, y-intercepts Slope-intercept form Where a line crosses the y-axis." ]

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[ "# PPPN: premium with real market data\n\nA few days ago, I posted a question about PPPN's (partially principal protected notes), which can be found here:PPPN: participation rate, stocks and premium.\n\nA PPPN in short is a structured product where you get for example 80% of your investment plus a premium.\n\nI'm still trying to understand how this all works, so I'm trying to make some of the example questions, like this one:\n\n\\begin{equation} \\text{premium} = \\begin{cases} 30 &\\mbox{if } S_T > K \\\\ 0 &\\mbox{otherwise.} \\end{cases} \\end{equation} The question is now to determine $K$ using real market data such that the product is attrictive for investors. So in order for this exercise to get more real, I've chosen a random stock, say facebook, credits to Yahoo Finance:", null, "But now, since this is a digital option which is not traded in the market, we don't have any prices. I thought maybe I could construct a digital option myself, from call/put options, where the digital option pays 1 if $S_T \\geq K$ or 0 if $S_T < K$, but I can't seem to finish that line of thought.\n\nSo... Any ideas/hints/solutions?\n\n• Did you make any progression on this topic?\n– user18684\nDec 16 '15 at 14:59\n• Not yet! I keep trying. Why? Any additional info? :) Dec 16 '15 at 22:00\n\nTo determine the value of $K$, you should have some target in mind, for example, to find the $K$ so that the value of the structured note is at par. Regarding the value of the digital option, there is an analytical formula if you have the volatility and interest rate data, or approximate it by a call spread. See discussions in this question." ]

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[ "# Lesson Worksheet: Conditional Probability: Tree Diagrams Mathematics\n\nIn this worksheet, we will practice using tree diagrams to calculate conditional probabilities.\n\nQ1:\n\nA bag contains 3 blue balls and 7 red balls. David selects 2 balls without replacement and draws the following tree diagram.", null, "Given that the first ball is red, find the value of that represents the probability that the second ball selected is red.\n\n• A\n• B\n• C\n• D\n• E\n\nQ2:\n\nJacob flips a coin and then rolls a six-sided die. He draws a tree diagram to represent this.", null, "Find the probability that the die showed a number less than 3 given that the coin showed tails.\n\n• A\n• B\n• C\n• D\n• E\n\nQ3:\n\nA bag contains 22 red balls and 15 black balls. Two balls are drawn at random. Find the probability that the second ball is black given that the first ball is red. Give your answer to three decimal places.\n\nQ4:\n\nA company manufactures a product in two different plants, and . The company supplies three customers, , , and equally, each with 80 units a month. produces 10 units of this product per month and the company distributes this amount among the three customers , , and in percentages of , , and respectively. If you select a unit at random from a outlet, find the probability that it is produced by .\n\nQ5:\n\nA bag contains 3 pink marbles, 4 orange marbles, and 5 yellow marbles. Two marbles are selected without replacement. Using a tree diagram, find the probability that the second marble is yellow given that the first marble is not yellow.\n\n• A\n• B\n• C\n• D\n• E\n\nQ6:\n\nThe tree diagram below shows the likelihood that it rains or does not rain and whether students walk or do not walk to school.", null, "Find the probability that a student walks to school.\n\n• A\n• B\n• C\n• D\n• E\n\nFind the probability that it rains, given that a student walks to school.\n\n• A\n• B\n• C\n• D\n• E\n\nQ7:\n\nIt is a little-known fact that drugs have been used to enhance performance in sports since the original Olympic Games (776 to 393 BC). In fact, the origin of the word doping is thought to come from the Dutch word doop, which is a type of opium juice used by the ancient Greeks.\n\nDrug testing has become standard practice. In 2003, after anonymous testing of almost 1,500 players, the Major League Baseball (MLB) announced that approximately ‎ of MLB players used performance-enhancing drugs. They got this result taking into account that there was a ‎ chance that those who had not taken drugs tested positive (false-positive effect) and a ‎ chance that those who had taken drugs tested negative (false-negative effect).\n\nFind the probability that an MLB player chosen at random had not taken drugs and tested positive. Round your answer to three decimal places if necessary.\n\nFind the probability that an MLB player chosen at random had taken drugs and tested positive. Round your answer to three decimal places if necessary.\n\nFind the probability that an MLB player chosen at random had positive test results. Round your answer to three decimal places if necessary.\n\nQ8:\n\nJennifer travels to school by car or on foot.\n\nThe probability that she travels by car is 0.4 and the probability that she walks is 0.6.\n\nIf she travels by car, the probability that she is late is 0.2; if she travels on foot, the probability that she is late is 0.3. Using a tree diagram, calculate the probability that she is late given that she traveled by car.\n\nQ9:\n\nThere are an unknown number of balls in the bag. There are 3 white balls and some black balls. Two balls are selected without replacement. If the probability of selecting a black ball, given that the first ball is white, is , how many black balls are there in the bag?\n\nQ10:\n\nThe probability that it rains on a given day is 0.6. If it rains, the probability that a group of friends plays football is 0.2. If it does NOT rain, the probability that they play football rises to 0.8.\n\nWork out the probability that it rains on a given day and the friends play football.\n\nWork out the probability that it does NOT rain on a given day and the friends play football.\n\nWhat is the probability that the friends will play football on a given day?\n\nThis lesson includes 34 additional questions and 340 additional question variations for subscribers." ]

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[ "## com.atlassian.jira.web.bean Class PercentageGraphModel\n\n```java.lang.Object", null, "com.atlassian.jira.web.bean.PercentageGraphModel\n```\n\n```@PublicApi\npublic final class PercentageGraphModelextends Object```\n\nConstructor Summary\n`PercentageGraphModel()`\n\nMethod Summary\n` void` ```addRow(String color, long number, String description)```\n\n` void` ```addRow(String color, long number, String description, String statuses)```\n\n` int` `getPercentage(PercentageGraphRow row)`\nUse to get the percentage of a particular row.\n` List<PercentageGraphRow>` `getRows()`\n\n` long` `getTotal()`\n\n` boolean` `isTotalZero()`\n\nMethods inherited from class java.lang.Object\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n\nConstructor Detail\n\n### PercentageGraphModel\n\n`public PercentageGraphModel()`\nMethod Detail\n\n```public void addRow(String color,\nlong number,\nString description,\nString statuses)```\n\n```public void addRow(String color,\nlong number,\nString description)```\n\n### getRows\n\n`public List<PercentageGraphRow> getRows()`\n\n### getPercentage\n\n`public int getPercentage(PercentageGraphRow row)`\nUse to get the percentage of a particular row.\nIf the percentage calculated is not a whole number it is rounded down. An exception is the last row for which the percentage is calculated as a remainder to 100 (percent).\n\nParameters:\n`row` - row to get the width percentage for\nReturns:\npercentage for the given row\n\n### getTotal\n\n`public long getTotal()`\n\n### isTotalZero\n\n`public boolean isTotalZero()`" ]

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[ "## Main\n\nThe control of charge carrier concentration by either electrostatic or chemical means has been widely studied as a way to induce phase transitions of different nature, such as structural in transition metal dichalcogenides1,2,3,4, ferromagnetic in high-Curie-temperature manganites5,6,7,8,9,10 and topological in engineered materials11,12,13,14, with potential application in the development of active electronic phase-change devices15. In this context, a collection of different phases in magic-angle bilayer graphene has been achieved by changing its carrier density16. Similar concepts have been theoretically explored in photonics using hyperbolic metamaterials composed of subwavelength structures, such as a periodic array of graphene ribbons17 or a stack of graphene dielectric layers18, in which a topological transition in the isofrequency dispersion contour can occur by changing the doping level of graphene. However, these hyperbolic metamaterials rely on a strong anisotropy of the effective permittivity tensor, which is ultimately limited by spatial nonlocal effects that can hinder a practical verification of this concept.\n\nRecently, a twisted stack of two α-phase molybdenum trioxide (α-MoO3) films was explored to control the topology of the isofrequency dispersion contour of phonon polaritons (PhPs) by varying the relative twist angle between the two α-MoO3 layers19,20,21,22. Owing to the in-plane anisotropy of the permittivity within the reststrahlen band from 816 to 976 cm−1, the real part of the permittivity is positive along the direction but negative along the direction23,24, a property that renders α-MoO3 a natural hyperbolic material supporting in-plane hyperbolic PhPs25,26. The low-loss in-plane hyperbolic PhPs in α-MoO3 thus emerge as an ideal platform to explore further possibilities of doping-driven and electrically tunable topological transitions in photonics.\n\nIn this work we have achieved the control of polariton dispersion in a van der Waals (vdW) heterostructure composed of an α-MoO3 film covered with monolayer graphene by changing the doping level of the latter. We observed the polariton dispersion contour to vary from hyperbolic (open) to elliptic (closed) on increasing the doping level of graphene, leading to the emergence of a mode dominated by its graphene plasmon polariton (GPP) component propagating along the direction at high doping levels. The nature of the polaritons emerging at high doping in the heterostructure evolved from GPP to PhP when moving from the to α-MoO3 crystallographic direction. In addition, when the vdW heterostructure was placed on top of a gold substrate instead of SiO2, a rather flattened dispersion contour was obtained due to a topological transition. As an application, we have designed an in-plane subwavelength focusing device by engineering the substrate.\n\n## Tunable topological polaritons in heterostructures\n\nA schematic of our proposed structure is shown in Fig. 1, where a 150-nm-thick vdW heterostructure is placed on top of either a SiO2 (Fig. 1a) or gold (Fig. 1d) substrate. We were particularly interested in the reststrahlen band II of α-MoO3 extending over the frequency range of 816 to 976 cm−1, where the permittivity (ε) components along the , and crystal directions satisfy εx < 0, εy > 0 and εz > 0, respectively (Supplementary Fig. 1a)23,24. As a result, the in-plane PhPs in natural α-MoO3 exhibit a hyperbolic dispersion contour. To illustrate dynamic control of the dispersion contour topology of the in-plane hybrid plasmon–phonon polaritons in our structure, we present in Fig. 1b several calculated dispersion contours on SiO2 at different graphene Fermi energies (EF) under a fixed representative incident free-space wavelength of λ0 = 10.99 μm (frequency of 910 cm−1). As the graphene Fermi energy increases from 0 to 0.7 eV, the opening angle φ of the hyperbolic sectors gradually increases due to a change in the PhP wavelength when varying the dielectric environment, and eventually the dispersion contour changes its character from a hyperbolic (open) to elliptic (closed) shape. Note that the Fermi energy at which this topological transition occurs is conditioned to the appearance of well-defined graphene plasmons along the direction at λ0 when increasing its doping level27,28,29,30. When the substrate was changed from SiO2 to gold, we found flatter dispersion contours (Fig. 1e) due to the stronger effect of screening provided by the gold substrate. Tunable polariton canalization and diffractionless propagation were thus expected.\n\nNumerically simulated field distributions of hybrid polaritons for different graphene Fermi energies are shown in Fig. 1c,f. At EF = 0 eV, the polaritons of the heterostructure on a SiO2 substrate exhibit a hyperbolic wavefront, similar to that of the PhPs in α-MoO3. In contrast, the wavelength of the polaritons on a gold substrate is highly compressed, whereas their opening angle is increased. On increasing the doping level to EF = 0.2 eV, a topological transition takes place. The wavelength of the hybrid polaritons is increased compared with the undoped case, and we can still observe a hyperbolic wavefront along the x direction on the SiO2 substrate. Moreover, for the hybrid polaritons on the gold substrate (Fig. 1f), highly collimated and directive hybrid polaritons propagating along the x direction can be observed as a result of a rather flattened dispersion contour. At EF = 0.3 eV, a hyperbolic wavefront of hybrid polaritons can still be observed along the x direction with the SiO2 substrate, but another mode with an elliptic wavefront propagating along the y direction also emerges. With the gold substrate, the wavefront of the hybrid polaritons is dominated by a fine crescent shape along the x direction. At a higher doping level (EF = 0.7 eV), the dispersion contours for the hybrid polaritons on both SiO2 and gold substrates display an elliptic-like shape (Fig. 1b,e). As a consequence, we can find modes propagating anisotropically in the xy plane (for additional theoretical analyses, see Supplementary Figs. 24 and other works31,32,33).\n\n## Experimental observation of topological transitions\n\nWe used infrared nanoimaging to visualize the propagating polaritons in the graphene/α-MoO3 heterostructures (Supplementary Fig. 5) and verify the above theoretical predictions (Fig. 2a–c,g–i). In this technique, upon p-polarized infrared light illumination, the resonant gold antenna efficiently launches hybrid polaritons (Supplementary Fig. 6), originating in the nanoscale concentrated fields at the two endpoints. While scanning the sample, the real part of the scattered light electric field (Re{ES}) is recorded simultaneously with the topography, making it possible to directly map the vertical near-field components of the hybrid polariton wavefronts launched by the antenna (for more details on near-field image analysis, see Supplementary Figs. 7 and 8).\n\nTo visualize the polariton wavefronts34,35, we imaged antenna-launched polaritons in the heterostructure at several intermediate graphene Fermi energies in the range EF = 0–0.7 eV (Supplementary Fig. 9). We first investigated hybrid polaritons in an undoped sample with EF = 0 eV, which revealed a precise hyperbolic wavefront (Fig. 2a,g) for both SiO2 and gold substrates, consistent with previous results for a single film of α-MoO3 as a result of the hyperbolic dispersion contour. Next, we examined the optical response of a sample with a relatively high doping level (EF = 0.3 eV) on a SiO2 substrate at the same incidence frequency. The image in Fig. 2b shows that the measured wavefronts remain hyperbolic along the x direction, while fringes around the antenna appear in the y direction, indicating that the dispersion contour has evolved into a closed shape, as shown in Fig. 1b and Supplementary Fig. 2. In contrast, the sample on a gold substrate shows a nearly flat wavefront for EF = 0.3 eV (Fig. 2h), indicating a topological transition in the dispersion contour along the x direction. On increasing the doping level further to EF = 0.7 eV, only elliptical wavefronts were observed (Fig. 2c,i) for samples on both SiO2 and gold substrates, denoting a rather rounded anisotropic dispersion contour. The corresponding Fourier transforms of the experimental near-field images (Fig. 2d–f,j–l) and simulated near-field distributions (Supplementary Figs. 3 and 4) further confirmed the transformation of the dispersion contour with increasing doping level of graphene. Notably, our extracted experimental polariton wave vectors k = 2π/λp (dotted symbols in Supplementary Fig. 10; λp is the wavelength of hybrid polaritons) match quite well the calculated dispersion diagrams in all cases.\n\nLow levels of disorder or minor imperfections in the heterostructure should not substantially affect the control capability (Supplementary Fig. 13). In addition, the thickness of α-MoO3 determines the influence of the dielectric environment (here, the substrate), which should not exceed the skin depth of hybrid polaritons (Supplementary Fig. 14).\n\nClose to EF = 0.3 eV, where the dispersion contour is rather flat, as shown in Fig. 2h, the propagation of hybrid polaritons appears to be firmly guided along the x direction, yielding a highly directive and diffractionless behaviour. Furthermore, this type of polariton canalization can be found over a wide range of frequencies and different thicknesses of α-MoO3 (Supplementary Fig. 11) due to the inherent robustness of the topological transition. The line profiles (vertical cuts along the y direction) across the amplitude of the canalization mode give a full-width at half-maximum (FWHM) of around 115 ± 5 nm (~λ0/95, where λ0 is the free-space wavelength), as shown in Supplementary Fig. 12.\n\n## Launching and manipulation of hybrid polaritons with tailored antenna\n\nBy rotating the long axis of the antenna by an angle θ with respect to the y direction, we could selectively launch and manipulate hybrid polaritons with different in-plane wave vectors. The launching contribution from our antenna can be decomposed into four parts: two endpoints acting as resonant dipoles and two parallel edges behaving as line dipoles (and assimilated to linear arrays of discrete dipoles).\n\nFigure 3a shows for a sample with a low doping level (EF = 0.1 eV) that, at θ = 0°, the field pattern of the exciting hybrid polaritons exhibits vertical fringes parallel to the long axis of the antenna, dominated by the line dipole contributions generated by the edges. Note that the two endpoint dipoles are not well excited when θ = 0° because the polarization direction of the incident light is not aligned with the long axis of the antenna. We can extract the polariton wave vector from the fringes parallel to this long axis. As θ increases from 0 to 90°, the wavefronts produced by the two endpoints gradually show up and interact with those produced by the edges. In the regime with rotation angles θ ≤ 40°, the distance between adjacent fringes parallel to the antenna edge is reduced from 590 to 380 nm, from which we can obtain the wave vector k perpendicular to the antenna. The extracted wave vectors are shown by red symbols in Fig. 3e, matching quite well the calculated dispersion contour (solid curves; more details on the extraction analysis are provided in Supplementary Fig. 16). When θ ≥ 60° (~φ, the opening angle indicated in Fig. 3e), there are no fringes parallel to the edge of the antenna because polariton propagation is prohibited along that direction, judging from the dispersion contour, and the field pattern is dominated by the hyperbolic wavefronts produced by the two endpoints of the antenna. The simulated field patterns shown in Fig. 3b corroborate these experimental observations for different rotation angles.\n\nFor a sample with a high doping level (EF = 0.7 eV, Fig. 3c), the antenna can generate polaritons propagating in all directions within the xy plane when the polarization direction of the incident light is along the long axis of the antenna. Due to the in-plane anisotropy, the excited field patterns are therefore different for the various rotation angles explored in the range from 0 to 90°. The simulated field patterns (Fig. 3d) again agree well with our experimental observations. The polariton wave vectors can still be extracted from the measured fringes perpendicular to the antenna, shown as red symbols in Fig. 3f, which also match quite well the calculated dispersion contour (solid curves). Note that, at EF = 0.3 eV, the field patterns of hybrid polaritons launched by antennas with different rotation angles (from 0 to 45°) all lie strictly along the x direction (Supplementary Fig. 17) due to the flattening of the dispersion contour, which leads to directional canalization at this graphene Fermi energy. More details on the extraction of polariton wave vectors from experiments can be found in Supplementary Figs. 16 and 18.\n\n## Partial focusing of polaritons by substrate engineering\n\nAs the dispersion contour of the hybrid polaritons can be modified substantially by controlling the dielectric environment, we engineered the substrate for the heterostructure to manipulate the in-plane propagation of hybrid polaritons. The design is illustrated in Fig. 4a, where the heterostructure lies on top of a substrate composed of a Au–SiO2–Au in-plane sandwich structure. This substrate was used to locally engineer the isofrequency dispersion contour (Fig. 4b). The central SiO2 film, with a width of 1.5 μm, serves as an in-plane flat lens to partially focus the incident polaritons (with a wave vector ki and a Poynting vector Pi along the normal of the contour in Fig. 4b) generated by an antenna on top of the left gold substrate. When the hybrid polaritons cross the boundary between the gold and SiO2 substrates, due to the change in the detailed shape of the dispersion contour, negative refraction can occur at the boundary, with the y component of the wave vector being conserved, whereas the sign of the y component of the transmitted Poynting vector Pt is opposite to that of the incident Pi, as illustrated in Fig. 4b. This leads to a partial focusing of the polaritons (Fig. 4c). Supplementary Fig. 19 shows the evolution of the negative refraction of hybrid polaritons at different Fermi energies of graphene.\n\nThe measured antenna-launched polariton near-field distributions for the heterostructure on gold and SiO2 substrates are shown in Fig. 4d and Supplementary Figs. 20 and 21, respectively. On the gold substrate, only elliptical wavefronts are observed around the antenna, denoting a convex dispersion contour in the region around the x axis. In contrast, on the SiO2 substrate (Supplementary Fig. 21), wavefronts are hyperbolic along the x direction and elliptic along the y direction around the antenna, indicating a closed concave shape of the dispersion contour near the x axis. These measurements are consistent with our previous experimental results shown in Figs. 2 and 3, and also, they match quite well the isofrequency contours shown in Fig. 4b.\n\nFurthermore, we launched hybrid polaritons towards the Au–SiO2–Au in-plane sandwich substrate from the left gold part by light coupling to an antenna prepared in that region. The resulting near-field distribution is shown in Fig. 4c (for more details, see Supplementary Figs. 20 and 21). When polaritons having elliptical wavefronts on the left gold substrate cross the boundary between the gold and SiO2 substrates, the Poynting vector of the polaritons refracts on the same side of the boundary-normal direction, therefore producing what is known as negative refraction due to the change in the detailed shape of the dispersion contour, which ultimately leads to partial focusing of the incident polaritons. Indeed, Fig. 4c shows the formation of a focal spot close to the right Au–SiO2 interface. The numerically simulated z out-of-plane component of the electric field distributions (Re{Ez}) shown in Supplementary Fig. 22 further corroborate the experimental findings. The red curve in Fig. 4e shows the spatial distribution of the electric field amplitude at the focal spot, demonstrating a high wavelength compression towards a FWHM of 520 nm along the y direction. This focal spot size is less than 1/21 of the corresponding illumination wavelength, thus emphasizing a deep subwavelength focusing effect (for more details, see Supplementary Figs. 23 and 24).\n\nTo estimate the intensity enhancement of the observed partial focusing, we extracted the spatial distribution of the electric field from the propagation profile (Fig. 4f). The intensity enhancement ξ is given by the square of the ratio of the electric field at the focal spot to that without focusing, $$\\xi = \\left( {\\frac{{S_{{\\mathrm{B}}2}/S_{{\\mathrm{B}}1}}}{{S_{{\\mathrm{D}}2}/S_{{\\mathrm{D}}1}}}} \\right)^2 = 4.5$$. SB1, SB2, SD1, and SD2 represent the near-field intensity at each fringe. Note that the focusing effect can be further enhanced by improving the flatness of the interface, as its structural inhomogeneity inevitably introduces undesired reflection, scattering and radiative losses of the incident polaritons (for more details, see Supplementary Figs. 2527).\n\n## Conclusions\n\nWe have experimentally demonstrated that the topology of the isofrequency dispersion contours for the hybrid polaritons supported in a heterostructure composed of a graphene sheet on top of an α-MoO3 layer can be substantially modified by chemically changing the doping level of graphene, with the contour topology being transformed from open to closed shapes over a broad frequency range. A flat dispersion contour appears at the topological transition, which supports a highly directive and diffractionless polariton propagation, resulting in a tunable canalization mode controlled by the doping level of graphene. We anticipate that electrical gating could be used to control the doping level in future studies36,37. Furthermore, through the appropriate choice of substrate for the heterostructure, we were also able to engineer the dispersion contour to exhibit even flatter profiles (for example, by using a gold substrate). This property has allowed us to design a deep-subwavelength device for in-plane focusing of hybrid polaritons, where negative refractive occurs at the boundary between two different substrates. Our study opens promising possibilities to tune topological polaritonic transitions in low-dimensional materials38,39,40 with potential applications in optical imaging, sensing and the control of spontaneous emission at the nanoscale18.\n\nNote added in proof: While preparing this manuscript, two related theoretical and experimental studies on the tuning of highly anisotropic phonon polaritons in graphene and α-MoO3 vdW structures were reported41,42.\n\n## Methods\n\n### Nanofabrication of the devices\n\nHigh-quality α-MoO3 flakes were mechanically exfoliated from bulk crystals synthesized by the chemical vapour deposition (CVD) method19 and then transferred onto either commercial 300 nm SiO2/500 μm Si wafers (SVM) or gold substrates using a deterministic dry transfer process with a polydimethylsiloxane (PDMS) stamp. CVD-grown monolayer graphene on copper foil was transferred onto the α-MoO3 samples using the poly(methyl methacrylate) (PMMA)-assisted method following our previous report44.\n\nThe launching efficiency of the resonant antenna is mainly determined by its geometry, together with a trade-off between the optimum size and illumination frequency45,46. We designed the gold antenna with a length of 3 μm and a thickness of 50 nm, which provided a high launching efficiency over the spectral range from 890 to 950 cm−1 within the α-MoO3 reststrahlen band. Alternatively, a thicker antenna with a stronger z component of the electric field could be used to launch the polaritons more efficiently in future studies. Note that narrow antennas (50-nm width) were used to prevent their shapes from affecting the polariton wavefronts, especially when their propagation is canalized (such as in Figs. 2 and 3), while wider antennas (250-nm width) were used to obtain a higher launching efficiency and observe polaritons propagating across the SiO2–Au interface in our experiments (for example, Fig. 4).\n\nGold antenna arrays were patterned on selected α-MoO3 flakes using 100 kV electron-beam lithography (Vistec 5000+ES) on an approximately 350 nm PMMA950K lithography resist. Electron-beam evaporation was subsequently used to deposit 5 nm Ti and 50 nm Au in a vacuum chamber at a pressure of <5 × 10−6 torr to fabricate the Au antennas. Electron-beam evaporation was also used to deposit a 60-nm-thick gold film onto a low-doped Si substrate. To remove any residual organic material, samples were immersed in a hot acetone bath at 80 °C for 25 min and then subjected to gentle rinsing with isopropyl alcohol (IPA) for 3 min, followed by drying with nitrogen gas and thermal baking (for more details on the fabrication and characterization of the Au–SiO2–Au in-plane sandwich structure, see Supplementary Figs. 26 and 27).\n\nThe samples were annealed in a vacuum to remove most of the dopants from the wet transfer process and then transferred to a chamber filled with NO2 gas to achieve different doping levels by surface adsorption of gas molecules47. The graphene Fermi energy could be controlled by varying the gas concentration and doping time, achieving values as high as ~0.7 eV (Supplementary Fig. 9). This gas-doping method provides excellent uniformity, reversibility and stability. Indeed, Raman mapping of a gas-doped graphene sample demonstrated the high uniformity of the method (Supplementary Fig. 9). As the deposition of NO2 gas molecules on the graphene surface occurs by physical adsorption, the topological transition of hybrid polaritons in graphene/α-MoO3 heterostructures can be reversed by controlling the gas doping. For example, after gas doping, the Fermi energy of graphene could be lowered from 0.7 to 0 eV by vacuum annealing at 150 °C for 2 h. The sample could subsequently be re-doped to reach another on-demand Fermi energy (Supplementary Fig. 28). It should be noted that the graphene Fermi energy only decreases from 0.7 to 0.6 eV after being left for 2 weeks under ambient conditions, which demonstrates the high stability of the doping effect (Supplementary Fig. 29). Note that chemical doping has been demonstrated to be an effective way to tune the characteristics of polaritons, such as their strength and in-plane wavelength48,49,50,51.\n\n### Scanning near-field optical microscopy measurements\n\nA scattering scanning near-field optical microscope (Neaspec) equipped with a wavelength-tunable quantum cascade laser (890–2,000 cm−1) was used to image optical near fields. The atomic force microscopy (AFM) tip of the microscope was coated with gold, resulting in an apex radius of ~25 nm (NanoWorld), and the tip-tapping frequency and amplitudes were set to ~270 kHz and ~30–50 nm, respectively. The laser beam was directed towards the AFM tip, with lateral spot sizes of ~25 μm under the tip, which were sufficient to cover the antennas as well as a large area of the graphene/α-MoO3 samples. Third-order harmonic demodulation was applied to the near-field amplitude images to strongly suppress background noise.\n\nIn our experiments, the p-polarized plane-wave illumination (electric field Einc) impinged at an angle of 60° relative to the tip axis52. To avert any effects caused by the light polarization direction relative to the crystallographic orientation of α-MoO3, which is optically anisotropic, the in-plane projection of the polarization vector coincided with the x direction ( crystal axis) of α-MoO3 (Supplementary Fig. 6). Supplementary Fig. 30 shows the method used to extract antenna-launched hybrid polaritons from the complex background signals observed when the polaritons propagate across a Au–SiO2–Au in-plane structure to realize partial focusing.\n\n### Calculation of polariton dispersion and isofrequency dispersion contours (IFCs) of hybrid plasmon–phonon polaritons\n\nThe transfer matrix method was adopted to calculate the dispersion and IFCs of hybrid plasmon–phonon polaritons in graphene/α-MoO3 heterostructures. Our theoretical model was based on a three-layer structure: layer 1 (z > 0, air) is a cover layer, layer 2 (0 > z > –dh, graphene/α-MoO3) is an intermediate layer and layer 3 (z < –dh, SiO2 or Au) is a substrate where z is the value of the vertical axis and dh is the thickness of α-MoO3 (Supplementary Fig. 31). Each layer was regarded as a homogeneous material represented by the corresponding dielectric tensor. The air and substrate layers were modelled by isotropic tensors diag{εa,s} (ref. 53). The α-MoO3 film was modelled by an anisotropic diagonal tensor diag{εx, εy, εz}, where εx, εy and εz are the permittivity components along the x, y and z axes, respectively. Also, monolayer graphene was located on top of α-MoO3 at z = 0 and described as a zero-thickness current layer characterized by a frequency-dependent surface conductivity taken from the local random-phase approximation model54,55:\n\n$$\\begin{array}{rcl} {\\sigma \\left( \\omega \\right)}& = &{\\frac{{i{{\\rm{e}}^2}{k_{\\rm{B}}}T}}{{{{\\uppi}}{\\hbar ^2}\\left( {\\omega + \\frac i \\tau} \\right)}}\\left[ {\\frac{{{E_{\\rm{F}}}}}{{{k_{\\rm{B}}}T}} + 2\\ln \\left( {{{\\rm{e}}^{ - \\frac{{{E_{\\rm{F}}}}}{{{k_{\\rm{B}}}T}}}} + 1} \\right)} \\right]}\\\\{}&{}&{ + i\\frac{{{{\\rm{e}}^2}}}{{4{{\\uppi}}\\hbar }}\\ln \\left[ {\\frac{{2\\left| {{E_{\\rm{F}}}} \\right| - \\hbar \\left( {\\omega + \\frac i \\tau} \\right)}}{{2\\left| {{E_{\\rm{F}}}} \\right| + \\hbar \\left( {\\omega + \\frac i \\tau} \\right)}}} \\right]}\\end{array}$$\n(1)\n\nwhich depends on the Fermi energy EF, the inelastic relaxation time τ and the temperature T; the relaxation time is expressed in terms of the graphene Fermi velocity vF = c/300 and the carrier mobility μ, with $$\\tau = \\mu E_{\\mathrm{F}}/ev_{\\mathrm{F}}^2$$; e is the elementary charge; kB is the Boltzmann constant; is the reduced Planck constant; and ω is the illumination frequency.\n\nGiven the strong field confinement produced by the structure under consideration, we only needed to consider transverse magnetic (TM) modes, because transverse electric (TE) components contribute negligibly. The corresponding p-polarization Fresnel reflection coefficient rp of the three-layer system admits the analytical expression\n\n$$\\begin{array}{*{20}{c}} {r_{\\mathrm{p}} = \\frac{{r_{12} + r_{23}\\left( {1 - r_{12} - r_{21}} \\right){\\mathrm{e}}^{i2k_z^{\\left( 2 \\right)}d_{\\mathrm{h}}}}}{{1 + r_{12}r_{23}{\\mathrm{e}}^{i2k_z^{\\left( 2 \\right)}d_{\\mathrm{h}}}}},} \\end{array}$$\n(2)\n$$\\begin{array}{*{20}{c}} {r_{12} = \\frac{{{{Q}}_1 - {{Q}}_2 + SQ_1Q_2}}{{{{Q}}_1 + Q_2 + SQ_1Q_2}},} \\end{array}$$\n(3)\n$$\\begin{array}{*{20}{c}} {r_{21} = \\frac{{{{Q}}_2 - {{Q}}_1 + SQ_1Q_2}}{{{{Q}}_2 + Q_1 + SQ_1Q_2}},} \\end{array}$$\n(4)\n$$\\begin{array}{*{20}{c}} {r_{23} = \\frac{{Q_2 - Q_3}}{{Q_2 + Q_3}},} \\end{array}$$\n(5)\n$$\\begin{array}{*{20}{c}}where {Q_j = \\frac{{k_z^{\\left( j \\right)}}}{{{\\it{\\epsilon }}_t^{(j)}}},} \\end{array}$$\n(6)\n$$\\begin{array}{*{20}{c}} {S = \\frac{{\\sigma Z_0}}{\\omega }.} \\end{array}$$\n(7)\n\nHere, rjk denotes the reflection coefficient of the jk interface for illumination from medium j, with j,k = 1–3; $${\\it{\\epsilon }}_t^{(j)}$$ is the tangential component of the in-plane dielectric function of layer j for a propagation wave vector kp(θ) (where θ is the angle relative to the x axis), which can be expressed as $${\\it{\\epsilon }}_t^{(j)} = {\\it{\\epsilon }}_x^{(j)}\\mathop {{\\cos }}\\nolimits^2 \\theta + {\\it{\\epsilon }}_y^{(j)}\\mathop {{\\sin }}\\nolimits^2 \\theta$$ (where $${\\it{\\epsilon }}_x^{(j)}$$ and $${\\it{\\epsilon }}_y^{(j)}$$ are the diagonal dielectric tensor components of layer j along the x and y axes, respectively); $$k_z^{\\left( j \\right)} = \\sqrt {\\varepsilon _t^{\\left( j \\right)}\\frac{{\\omega ^2}}{{c^2}} - \\frac{{\\varepsilon _t^{\\left( j \\right)}}}{{\\varepsilon _z^{\\left( j \\right)}}}q^2}$$ is the out-of-plane wave vector, with $${\\it{\\epsilon }}_z^{(j)}$$ being the dielectric function of layer j along the z axis; and Z0 is the vacuum impedance.\n\nWe find the polariton dispersion relation q(ω,θ) when the denominator of equation (2) is zero:\n\n$$\\begin{array}{*{20}{c}} {1 + r_{12}r_{23}{\\mathrm{e}}^{i2k_z^{\\left( 2 \\right)}d_{\\mathrm{h}}} = 0.} \\end{array}$$\n(8)\n\nFor simplicity, we considered a system with small dissipation, so that the maxima of Im{rp} (see colour plots in Supplementary Figure 10) approximately solve the condition given by equation (8), and therefore produce the sought-after dispersion relation q(ω,θ) (see additional discussion in Supplementary Note 1).\n\n### Electromagnetic simulations\n\nThe electromagnetic fields around the antennas were calculated by a finite-elements method using the COMSOL package. In our experiments, both tip and antenna launching were investigated. For the former, the sharp metallic tip was illuminated by an incident laser beam. The tip acted as a vertical optical antenna, converting the incident light into a strongly confined near field below the tip apex, which can be regarded as a vertically oriented point dipole located at the tip apex. This localized near field provided the necessary momentum to excite polaritons. Consequently, we modelled the tip as a vertical z-oriented oscillating dipole in our simulations (Fig. 1c,f), a procedure that has been widely used for tip-launched polaritons in vdW materials56. For the antenna launching, the gold antenna can provide strong near fields of opposite polarity at the two endpoints, thus delivering high-momentum near-field components that match the wave vector of the polaritons and excite propagating modes in the graphene/α-MoO3 heterostructure45,46. Our simulations of polariton excitation by means of antennas, such as in Fig. 3b,d, incorporated the same geometrical design as in the experimental structures.\n\nWe also used a dipole polarized along the z direction to launch polaritons, and the distance between the dipole and the uppermost surface of the sample was set to 100 nm. We obtained the distribution of the real part of the out-of-plane electric field (Re{Ez}) over a plane 20 nm above the surface of graphene. The boundary conditions were set to perfectly matching layers. Graphene was modelled as a transition interface with a conductivity described by the local random-phase approximation model (see above)55,57. We assumed a graphene carrier mobility of 2,000 cm2 V–1 s–1. Supplementary Fig. 1c,d shows the permittivity of SiO2 and Au, respectively, at the mid-infrared wavelengths used." ]

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[ "# What is triangle - Definition and Meaning\n\n### Triangle :\n\nA polygon with three sides and three angles.\n\n#### Formula :", null, "#### Example :\n\nFind the area of triangle with a base = 6 and a height = 9. Area=b*h Area = 1/2* (6*9) A = 54/2 = 27.\n\nLearn what is triangle. Also find the definition and meaning for various math words from this math dictionary." ]

[ null, "https://www.easycalculation.com/maths-dictionary/images/trai.png", null ]

[ "## Round Robin Scheduling-\n\nIn Round Robin Scheduling,\n\n• CPU is assigned to the process on the basis of FCFS for a fixed amount of time.\n• This fixed amount of time is called as time quantum or time slice.\n• After the time quantum expires, the running process is preempted and sent to the ready queue.\n• Then, the processor is assigned to the next arrived process.\n• It is always preemptive in nature.\n\n Round Robin Scheduling is FCFS Scheduling with preemptive mode.", null, "• It gives the best performance in terms of average response time.\n• It is best suited for time sharing system, client server architecture and interactive system.\n\n• It leads to starvation for processes with larger burst time as they have to repeat the cycle many times.\n• Its performance heavily depends on time quantum.\n• Priorities can not be set for the processes.\n\n## Important Notes-\n\n### Note-01:\n\nWith decreasing value of time quantum,\n\n• Number of context switch increases\n• Response time decreases\n• Chances of starvation decreases\n\nThus, smaller value of time quantum is better in terms of response time.\n\n### Note-02:\n\nWith increasing value of time quantum,\n\n• Number of context switch decreases\n• Response time increases\n• Chances of starvation increases\n\nThus, higher value of time quantum is better in terms of number of context switch.\n\n### Note-03:\n\n• With increasing value of time quantum, Round Robin Scheduling tends to become FCFS Scheduling.\n• When time quantum tends to infinity, Round Robin Scheduling becomes FCFS Scheduling.\n\n### Note-04:\n\n• The performance of Round Robin scheduling heavily depends on the value of time quantum.\n• The value of time quantum should be such that it is neither too big nor too small.\n\n## Problem-01:\n\nConsider the set of 5 processes whose arrival time and burst time are given below-\n\n Process Id Arrival time Burst time P1 0 5 P2 1 3 P3 2 1 P4 3 2 P5 4 3\n\nIf the CPU scheduling policy is Round Robin with time quantum = 2 unit, calculate the average waiting time and average turn around time.\n\n## Solution-\n\n### Gantt Chart-\n\nP5, P1, P2, P5, P4, P1, P3, P2, P1", null, "Now, we know-\n\n• Turn Around time = Exit time – Arrival time\n• Waiting time = Turn Around time – Burst time\n\nAlso read- Various Times of Process\n\n Process Id Exit time Turn Around time Waiting time P1 13 13 – 0 = 13 13 – 5 = 8 P2 12 12 – 1 = 11 11 – 3 = 8 P3 5 5 – 2 = 3 3 – 1 = 2 P4 9 9 – 3 = 6 6 – 2 = 4 P5 14 14 – 4 = 10 10 – 3 = 7\n\nNow,\n\n• Average Turn Around time = (13 + 11 + 3 + 6 + 10) / 5 = 43 / 5 = 8.6 unit\n• Average waiting time = (8 + 8 + 2 + 4 + 7) / 5 = 29 / 5 = 5.8 unit\n\n## Problem-02:\n\nConsider the set of 6 processes whose arrival time and burst time are given below-\n\n Process Id Arrival time Burst time P1 0 4 P2 1 5 P3 2 2 P4 3 1 P5 4 6 P6 6 3\n\nIf the CPU scheduling policy is Round Robin with time quantum = 2, calculate the average waiting time and average turn around time.\n\n## Solution-\n\n### Gantt chart-\n\nP5, P6, P2, P5, P6, P2, P5, P4, P1, P3, P2, P1", null, "Now, we know-\n\n• Turn Around time = Exit time – Arrival time\n• Waiting time = Turn Around time – Burst time\n\n Process Id Exit time Turn Around time Waiting time P1 8 8 – 0 = 8 8 – 4 = 4 P2 18 18 – 1 = 17 17 – 5 = 12 P3 6 6 – 2 = 4 4 – 2 = 2 P4 9 9 – 3 = 6 6 – 1 = 5 P5 21 21 – 4 = 17 17 – 6 = 11 P6 19 19 – 6 = 13 13 – 3 = 10\n\nNow,\n\n• Average Turn Around time = (8 + 17 + 4 + 6 + 17 + 13) / 6 = 65 / 6 = 10.84 unit\n• Average waiting time = (4 + 12 + 2 + 5 + 11 + 10) / 6 = 44 / 6 = 7.33 unit\n\n## Problem-03:\n\nConsider the set of 6 processes whose arrival time and burst time are given below-\n\n Process Id Arrival time Burst time P1 5 5 P2 4 6 P3 3 7 P4 1 9 P5 2 2 P6 6 3\n\nIf the CPU scheduling policy is Round Robin with time quantum = 3, calculate the average waiting time and average turn around time.\n\n## Solution-\n\n### Gantt chart-\n\nP3, P1, P4, P2, P3, P6, P1, P4, P2, P3, P5, P4", null, "Now, we know-\n\n• Turn Around time = Exit time – Arrival time\n• Waiting time = Turn Around time – Burst time\n\n Process Id Exit time Turn Around time Waiting time P1 32 32 – 5 = 27 27 – 5 = 22 P2 27 27 – 4 = 23 23 – 6 = 17 P3 33 33 – 3 = 30 30 – 7 = 23 P4 30 30 – 1 = 29 29 – 9 = 20 P5 6 6 – 2 = 4 4 – 2 = 2 P6 21 21 – 6 = 15 15 – 3 = 12\n\nNow,\n\n• Average Turn Around time = (27 + 23 + 30 + 29 + 4 + 15) / 6 = 128 / 6 = 21.33 unit\n• Average waiting time = (22 + 17 + 23 + 20 + 2 + 12) / 6 = 96 / 6 = 16 unit\n\n## Problem-04:\n\nFour jobs to be executed on a single processor system arrive at time 0 in the order A, B, C, D. Their burst CPU time requirements are 4, 1, 8, 1 time units respectively. The completion time of A under round robin scheduling with time slice of one time unit is-\n\n1. 10\n2. 4\n3. 8\n4. 9\n\n## Solution-\n\n Process Id Arrival time Burst time A 0 4 B 0 1 C 0 8 D 0 1\n\n### Gantt chart-\n\nC, A, C, A, C, A, D, C, B, A", null, "Clearly, completion time of process A = 9 unit.\n\nThus, Option (D) is correct.\n\nTo gain better understanding about Round Robin Scheduling,\n\nWatch this Video Lecture\n\nNext Article- Priority Scheduling\n\nGet more notes and other study material of Operating System.\n\nWatch video lectures by visiting our YouTube channel LearnVidFun." ]

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[ "# 1.38″ in km. Convert 1.38 inches to kilometers. Online inch calculator.\n\n1.38 inches = 3.5052e-05 kilometers", null, "3.5052e-05 kilometers = 1.38 inches\n\n1.38 Inches = 3.5052e-05 Kilometers\n\n1.38″ (Inches, in) - English inch is a value for measuring lengths and distances, heights and widths and etc. One inch is equal to 2.54e-05 kilometers.\n\nOn this page we consider in detail all variants for convert 1.38 inches to kilometers and all opportunities how to convert inches with usage comprehensive examples, related charts and conversion tables for inches. Here you will find all the ways for calculating and converting inches in km and back. If you want to know how many kilometers are in 1.38 inches you can obtain the answer in several ways:\n\n• calculate 1.38 inches using calculator InchPro from our software collection for offline converting units;\n• apply arithmetic calculations and conversions for 1.38 inches outlined in this article.\n\nTo indicate inches, we will use the abbreviation \"in\" to indicate kilometers we will use the abbreviation \"km\". All options to convert \"in\" to \"km\" we will look more detail in individual topics below. So, we're starting explore all avenues of transformation one point three eight inches and conversions between inches and kilometers.\n\n## Convert 1.38 inches to km by online conversion\n\nTo convert 1.38 inches into kilometers we consider using the online converter on the web page. The online converter has very simple interface and will help us quickly convert our inches. The online inch converter has an adaptive shape for different devices and therefore for monitors it looks like the left and right input fields but on tablets and mobile phones it looks like the top and bottom input fields. If you want convert any inch values, you must only enter required value in the left (or top) input field and automatically you get the result in the right (or bottom) field. Under each field you see a more detailed result of the calculation and the coefficient of 2.54e-05 which is used in the calculations. The big green string, under the input fields - \"1.38 Inches = 3.5052e-05 Kilometers\" further enhances and shows final result of the conversion. Calculator for converting units of measurement works symmetrically in both directions. If you enter any value in any field, you will get the result in the opposite field. Clicking on the arrow icons between the input fields you can swap the fields and perform other calculations. We are all made for easily converting any values between inches and km.\n\nIf you came to this page, you already see the result work of the online calculator. In the left (or top) field you see the value of 1.38 \"in\" on the right (or bottom) box you see the value of result is equal to 3.5052e-05 \"km\". Write briefly: 1.38 \"in\" = 3.5052e-05 \"km\"\n\n## Convert 1.38 inches in km by conversion tables\n\nWe have briefly reviewed how to use the unit Converter on this page, but this is only part of the features of the page service. We made an interesting possibility to compute all possible values for units of measure in the lower tables. These tables are used to convert basic units of measurement: Metric conversion chart, US Survey conversion chart, International conversion chart, Astronomical conversion chart. Please, find these 4 tables at the bottom of this page they have the headers:\n\n• All conversions of 1.38 inches in the Metric System Units\n• All conversions of 1.38 inches in the US Survey Units\n• All conversions of 1.38 inches in the International Units\n• All conversions of 1.38 inches in the Astronomical Units\nIf you enter a test number in any field of web calculator (field of inches or kilometers it doesn't matter), for example 1.38 as it is now, you not only get the result in 3.5052e-05 kilometers but also a huge list of computed values for all unit types in the lower tables. Without doing your own search and making the transition to other pages of the website, you can use our conversion tables to calculate all the possible results for main units. Try delete and again entering into the calculator a value of 1.38 inches and you will see that all the conversion results in the lower tables will are recalculated for 1.38 (in). The calculated data in the conversions tables change dynamically and all transformations are performed synchronously with converting inches in the page calculator.\n\n### How many kilometers are in 1.38 inches?\n\nTo answer this question, we start with a brief definition of inch and kilometer, and their purpose. The inch and kilometer units of length which can be converted one to another using a conversion factor which is equal to 2.54e-05. This coefficient answers the question how many kilometers are equivalent to one inch. The value of this multiplier determines the basic value to calculate all other lengths, sizes and other transformations for these units (inch and kilometer), it is enough to know the value, i.e. to remember that 1 inch = 2.54e-05 (km). Knowing the number of kilometers in one inch by simple multiplication we can calculate any values. Let's do a simple calculation using the multiplication:\n\n1.38″ × 2.54e-05 = 3.5052e-05 (km)\n\nThus it is seen that after multiplying by the coefficient we get the following relationship:\n\n1.38 Inches = 3.5052e-05 Kilometers\n\n### How much is 1.38 inches in kilometers?\n\nWe have already seen how to convert these two values and how change inches to kilometers. So in summary, you can write all possible results that have the same meaning.\n\n1.38 inches to km = 3.5052e-05 km\n1.38 inches in km = 3.5052e-05 km\n1.38 inches into km = 3.5052e-05 km\n1.38 in = 3.5052e-05 km\n1.38″ = 3.5052e-05 km\n1.38″ is 3.5052e-05 km\none point three eight inches = 3.5052e-05 km\n\nFor a detailed reviewing of similar numbers, visit next pages:\n\n### How to convert 1.38 inches into kilometers? All rules and methods.\n\nTo convert 1.38 inches into kilometers we can use many ways:\n\n• calculation using the formula;\n• calculation using the proportions;\n• calculation using the online converter of the current page;\n• calculation using the offline calculator \"InchPro Decimal\".\n\n#### Calculating 1.38 inches to km formula for lengths and values.\n\nIn the calculations for inches and kilometers, we will use the formula presented below that would quickly get the desired result.\n\n``` Y (in) × 2.54e-05 = X (km)\nY - value of inches\nX - result in kilometers\n```\n\nThat is, you need to remember that 1 inch is equal 2.54e-05 kilometers, and when converting inches just multiply the number of inches (in this case 1.38 inches) by a factor of 2.54e-05. For example, we transform the set of values 1.38″, 2.38″, 3.38″, 4.38″, 5.38″ into kilometers and get the result in the following examples:\n\n``` 1.38 (in) × 2.54e-05 = 3.5052e-05 (km)\n2.38 (in) × 2.54e-05 = 6.0452e-05 (km)\n3.38 (in) × 2.54e-05 = 8.5852e-05 (km)\n4.38 (in) × 2.54e-05 = 0.000111252 (km)\n5.38 (in) × 2.54e-05 = 0.000136652 (km)\n```\n\nIn all variants we multiplied the all inches in range from 1.38″ to 5.38″ with the same ratio of 2.54e-05 and got the correct results in calculations.\n\n#### The calculation using mathematical proportions to convert 1.38 inches into kilometers\n\nTo calculate the proportions you need to know the reference value in kilometers for 1 inch and according to the rules of arithmetic we can calculate any value in kilometers for any length in inches. See the next examples. We form the proportion for 3 values our inches  1.38″, 2.38″, 3.38″ and calculate results values in kilometers:\n\n``` 1 (in) — 2.54e-05 (km)\n1.38 (in) — X (km)\nSolve the above proportion for X to obtain:\nX = 1.38(in) × 2.54e-05(km) ÷\n1(in) = 3.5052e-05(km)\n1 (in) — 2.54e-05 (km)\n2.38 (in) — X (km)\nSolve the above proportion for X to obtain:\nX = 2.38(in) × 2.54e-05(km) ÷\n1(in) = 6.0452e-05(km)\n1 (in) — 2.54e-05 (km)\n3.38 (in) — X (km)\nSolve the above proportion for X to obtain:\nX = 3.38(in) × 2.54e-05(km) ÷\n1(in) = 8.5852e-05(km) ```\n\nAll proportions used reference value 1 inch = 2.54e-05 km\n\n#### Calculation of values using inch online calculator on the page\n\nYou can use our basic universal online converter on the current web page and convert any your length dimensions and distances between inches and kilometers in any directions free and fast.\n\nCurrently, the field for inches contains the number 1.38 (in) you can change it. Just enter any number into field for inches (for example any value from our set: 2.38, 3.38, 4.38, 5.38 inches or any other value) and get the fast result into field for kilometers. How to use the inch online calculator you can more detail read at this link manual for the calculator.\n\nFor example, we take the 14 values into inches and will try to calculate the result values in kilometers. Also, we will use the web calculator (you can find it at the top of this page). In the set up a table in the left margin we write the value in inches in the right margin you see the values that you should obtain after calculation. You can check it right now, without leaving the site and make sure that the calculator works correctly and quickly. In all calculations, we used the ratio 2.54e-05 which helps us to get the desired values of computation results in kilometers. Please, see the results in the next table:\n\n##### Example of Work Inch Online Calculator with Calculation Results\nInchesTable FactorKilometers\n3514248 (in)× 2.54e-05 = 89.2618992 (km)\n3514249 (in)× 2.54e-05 = 89.2619246 (km)\n3514250 (in)× 2.54e-05 = 89.26195 (km)\n3514251 (in)× 2.54e-05 = 89.2619754 (km)\n3514252 (in)× 2.54e-05 = 89.2620008 (km)\n3514253 (in)× 2.54e-05 = 89.2620262 (km)\n3514254 (in)× 2.54e-05 = 89.2620516 (km)\n3514255 (in)× 2.54e-05 = 89.262077 (km)\n3514256 (in)× 2.54e-05 = 89.2621024 (km)\n3514257 (in)× 2.54e-05 = 89.2621278 (km)\n3514258 (in)× 2.54e-05 = 89.2621532 (km)\n3514259 (in)× 2.54e-05 = 89.2621786 (km)\n3514260 (in)× 2.54e-05 = 89.262204 (km)\n3514261 (in)× 2.54e-05 = 89.2622294 (km)\n\n#### Convert 1.38 inches with the use of calculator \"InchPro Decimal\"\n\nWe're briefly describe the possibility for using our calculator for converting 1.38 inches. The calculator allows you to convert any value for lengths and distances not only in inches but also for all other units. Our conversion tables which we mentioned earlier are also included in the logic operation of the calculator and all these calculations you can get in one application if you download and install the software on your computer. Converter easily converts 1.38 \"in\" for you in offline mode. All the details of the work of this application for conversion of the heights and widths, lengths, sizes and distances described in inches or other units of measurement you will find in menu \"Software\" of this site or by the link: InchPro Decimal. Please, also see the screenshots for acquaintance.\n\n### Visual charts conversion of 1.38 inches.\n\nMany people can hardly imagine the relationship between inch and kilometer. In this picture, you can clearly see the ratio of these quantities to understand them in real life. The ratio of the lengths of the segments is retained on screens with any resolution as for large monitors as well as for small mobile devices.\n\n#### The graphical representation of scales for comparing values.\n\nThe graph shows the relative values the inches in the form of rectangular segments of different lengths and colors. As well as the visual representation of 1.38 (in) with the reference value in kilometers.\n\nYour browser does not support the canvas element.\n\nThe graphs of the relationship between inches and kilometers are expressed in the following colours:\n\n• Green is the original length or distance in inches;\n• Blue color is the scale in inches;\n• Yellow color is the scale in kilometers.\nThe scale may increase or decrease depending on the current number value on the page. The diagram shows the ratio between inches and km for the same lengths and magnitude (see charts of the blue and yellow colors).\n\nSat 31 Oct 2020\n\n##### All conversions of 1.38 inches in the Metric System Units\nUnitValue\n1.38 inches to micrometers microns= 35052.0\n1.38 inches to nanometers= 35052000.0\n1.38 inches to decimeters= 0.35052\n1.38 inches to meters= 0.035052\n1.38 inches to millimeters= 35.052\n1.38 inches to kilometers= 3.5052e-05\n1.38 inches to decameters= 0.0035052\n1.38 inches to hectometers= 0.00035052\n1.38 inches to centimeters= 3.5052\n##### All conversions of 1.38 inches in the US Survey Units\nUnitValue\n1.38 inches to survey chains= 0.00174242075758\n1.38 inches to survey feet= 0.11499977\n1.38 inches to survey furlongs stade= 0.000174242075758\n1.38 inches to survey leagues= 7.26008648993e-06\n1.38 inches to survey statute miles= 2.17802594698e-05\n1.38 inches to survey inches= 1.37999724001\n1.38 inches to survey rods perches poles= 0.00696968303033\n1.38 inches to survey links= 0.174242075758\n##### All conversions of 1.38 inches in the International Units\nUnitValue\n1.38 inches to inches= 1.38\n1.38 inches to feet= 0.115\n1.38 inches to miles= 2.17803030303e-05\n1.38 inches to yards= 0.0383333333333\n1.38 inches to picas= 8.28\n1.38 inches to points= 99.36\n##### All conversions of 1.38 inches in the Astronomical Units\nUnitValue\n1.38 inches to light-year= 3.7e-18\n1.38 inches to light-day= 1.35326e-15\n1.38 inches to light-second= 1.1692088665e-10\n1.38 inches to light-minute= 1.94868144e-12\n1.38 inches to astronomical unit= 2.3430814e-13\n1.38 inches to parsec= 1.13e-18\n1.38 inches to light-hour= 3.247802e-14" ]

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[ "## Conversion formula\n\nThe conversion factor from grams to ounces is 0.03527396194958, which means that 1 gram is equal to 0.03527396194958 ounces:\n\n1 g = 0.03527396194958 oz\n\nTo convert 114 grams into ounces we have to multiply 114 by the conversion factor in order to get the mass amount from grams to ounces. We can also form a simple proportion to calculate the result:\n\n1 g → 0.03527396194958 oz\n\n114 g → M(oz)\n\nSolve the above proportion to obtain the mass M in ounces:\n\nM(oz) = 114 g × 0.03527396194958 oz\n\nM(oz) = 4.0212316622522 oz\n\nThe final result is:\n\n114 g → 4.0212316622522 oz\n\nWe conclude that 114 grams is equivalent to 4.0212316622522 ounces:\n\n114 grams = 4.0212316622522 ounces\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 ounce is equal to 0.24868002741228 × 114 grams.\n\nAnother way is saying that 114 grams is equal to 1 ÷ 0.24868002741228 ounces.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one hundred fourteen grams is approximately four point zero two one ounces:\n\n114 g ≅ 4.021 oz\n\nAn alternative is also that one ounce is approximately zero point two four nine times one hundred fourteen grams.\n\n## Conversion table\n\n### grams to ounces chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from grams to ounces\n\ngrams (g) ounces (oz)\n115 grams 4.057 ounces\n116 grams 4.092 ounces\n117 grams 4.127 ounces\n118 grams 4.162 ounces\n119 grams 4.198 ounces\n120 grams 4.233 ounces\n121 grams 4.268 ounces\n122 grams 4.303 ounces\n123 grams 4.339 ounces\n124 grams 4.374 ounces" ]

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[ "DeepAI\n\n# Variational Neural Networks: Every Layer and Neuron Can Be Unique\n\nThe choice of activation function can significantly influence the performance of neural networks. The lack of guiding principles for the selection of activation function is lamentable. We try to address this issue by introducing our variational neural networks, where the activation function is represented as a linear combination of possible candidate functions, and an optimal activation is obtained via minimization of a loss function using gradient descent method. The gradient formulae for the loss function with respect to these expansion coefficients are central for the implementation of gradient descent algorithm, and here we derive these gradient formulae.\n\n12/21/2014\n\n### Learning Activation Functions to Improve Deep Neural Networks\n\nArtificial neural networks typically have a fixed, non-linear activation...\n11/13/2019\n\n### Quadratic number of nodes is sufficient to learn a dataset via gradient descent\n\nWe prove that if an activation function satisfies some mild conditions a...\n05/25/2019\n\n### Hebbian-Descent\n\nIn this work we propose Hebbian-descent as a biologically plausible lear...\n09/07/2022\n\n### A Greedy Algorithm for Building Compact Binary Activated Neural Networks\n\nWe study binary activated neural networks in the context of regression t...\n08/28/2017\n\n### A parameterized activation function for learning fuzzy logic operations in deep neural networks\n\nWe present a deep learning architecture for learning fuzzy logic express...\n05/07/2018\n\n### Polynomial Convergence of Gradient Descent for Training One-Hidden-Layer Neural Networks\n\nWe analyze Gradient Descent applied to learning a bounded target functio...\n10/16/2022\n\n### Stability of Accuracy for the Training of DNNs Via the Uniform Doubling Condition\n\nWe study the stability of accuracy for the training of deep neural netwo...\n\n## I Introduction\n\nIn conventational artificial neural networks (ANNs), the backward-propagation updates weights for the entire network to minimize the loss functionLeCun et al. (1998). The activation function in each hidden layer is determined before training the network and fixed during the training process. Various activation functions have been proposed, such as sigmoid,\n\n, ReLu, etc\n\nGoodfellow et al. (2016)\n\n. There may be other customized activation functions for specific use cases. Empirically, ReLu is the default choice when building deep neural networks for computer vision and speech recognition\n\nLeCun et al. (2015).\n\nThe choice of activation function is pretty arbitrary. During the early development stage, the practitioners of neural networks tried to simulate genuine neurons in human, and preferred to use activation functions that saturate when its input value is large. Moreover, previously, people have held it to be self evident that the activation functions should be differentiable everywhere, and thus sigmoid and hyperbolic tangent functions are widely used. The realization that ReLu, which is neither saturating nor everywhere differentiable, could also be an activation function with even better performance than sigmoid or hyperbolic tangent removes the shackle in people’s imagination, and significantly promotes the development of neural networksGlorot et al. (2011). However, up to now, the choice of activation functions is still ad-hoc and no rigorous proof exists that can demonstrate that ReLU or any other novel activation function is superior to the conventional sigmoid and hyperbolic tangent functions. The advantage of one activation function over another is established from mere experience, and the lack of theoretical or algorithmic justification thereof is one of our concerns.\n\nHere, in this article, we propose a more systematic method to enable our neural network to find the optimal activation function automatically. In order to do this, we first select a set of candidate eigen functions, and then make the assumption that the optimal activation function can be represented as as linear combination of these candidate eigen functions, the combination coefficients of which yet to be determined. We next define a loss function, and try to minimize this loss function with respect to the combination coefficients, together with the weight matrices and biases that are used in conventional artificial neural networks. The activation function, which is now a linear combination of basis eigen functions, can be unique for each hidden layer (even the output layer) or each neuron in each hidden layer (even the output layer). Since in our system, the activation functions are determined from variational principle, we name our neural network as variational neural network (VNN). In this article, we will derive the formulae for the gradients of loss function with respect to the expansion coefficients. Programmatic implementation of this algorithm is still in its way. We find that in our variational neural network, back propagation method still applies, and thus no major modification of conventional neural network programs is needed.\n\nThe organization of this paper is as follows. In section II, we derive the gradient formulae for loss function with respect to the expansion coefficients. In this section, the activation functions are represented as a linear combination of candidate eigen functions, and we impose a restraint that neurons residing in the same layer should share the same activation function. In section III, we relax the restraint in section II, and now each neuron has its own unique activation function. The gradient formulae for the loss function with respect to the expansion coefficients are derived in a similar manner as that in section II. We make a conclusion in section IV.\n\n## Ii Variation of Hidden Layer Activation Functions\n\nSuppose is the loss function, is the activation function in output layer, is the weight on the connection between the nueron in hidden layer and the neuron in hidden layer, is the weight on the connection between the neuron in the last hidden layer and the neuron in the output layer, and is the activation function in hidden layer. , where is a set of eigen-functions (for example, it can be and ). In practice, we can cut-off the summation to a large number : .\n\nNow during the backward-propagation, the network will not only update the weights in each connection between neurons, but also update the weights of the eigen-functions in activation function. The neuron connection weight update is the same as before. The eigen function weight update for the last hidden layer is as follows (suppose we have hidden layers in total),\n\n α(N)i=α(N)i−η∂E∂α(N)i, (1)\n\nwhere\n\nis the learning rate. Using the chain rule, we get (suppose the loss function has such format\n\n, where is the element of the label)\n\n ∂E∂α(N)i=n∑l=1∂E∂σ⋅∂σ∂net(O)l⋅∂net(O)l∂α(N)i, (2)\n\nwhere is based on the format of loss function and is based on the format of activation function in the output layer, which are straight-forward to compute, is the input from the neuron in the output layer. Suppose there are neurons in the last hidden layer\n\n net(O)l =m∑j=1ω(O)jlF(N)(% net(N)j) =m∑j=1ω(O)jlM∑k=1α(N)kfk(net(N)j). (3)\n\nThe last part of Eq. (2) is\n\n ∂net(O)l∂α(N)i=m∑j=1ω(O)jlfi(net(N)j). (4)\n\nTherefore, the eigen function weight update for the last hidden layer is\n\n ∂E∂α(N)i=n∑l=1∂E∂σ⋅∂σ∂net(O)l⋅m∑j=1ω(O)jlfi(% net(N)j). (5)\n\nIf we transform Eq. (5) into matrix format, then\n\n ∂E∂α(N)i=(→f(N)i)TW(O)∇−−→net(O)E, (6)\n\nwhere , is a matrix [], T is transpose operation, and .\n\nSimilarly, the eigen function weight update for the second last hidden layer is\n\n ∂E∂α(N−1)i=n∑l=1∂E∂σ⋅∂σ∂net(O)l⋅∂net(O)l∂α(N−1)i, (7)\n\nwhere\n\n net(O)l=m∑j=1ω(O)jlM∑k=1α(N)kfk(net(N)j) net(N)j=s∑p=1ω(N)pjM∑q=1α(N−1)qfq(net(N−1)p). (8)\n\nTherefore, the eigen function weight update is\n\n ∂E∂α(N−1)i= n∑l=1∂E∂σ⋅∂σ∂net(O)l⋅m∑j=1ω(O)jl ⋅M∑k=1α(N)k∂fk∂net(N)js∑p=1ω(N)pjfi(net(N−1)p). (9)\n\nIf we transform Eq. (II) into matrix format, then\n\n ∂E∂α(N−1)i=(→f(N−1)i)TW(N)˜W(O)∇−−→net(O)E, (10)\n\nwhere .\n\nIt is easy to demonstrate that the general formula of updating the eigen function weight for the last hidden layer\n\n ∂E∂α(N−γ+1)i= (→f(N−γ+1)i)TW(N−γ+2) ⋅˜W(N−γ+3)⋯˜W(N)˜W(O)∇−−→net(O)E, (11)\n\nwhere .\n\nTherefore, the entire update of the eigen function weights for the last hidden layer is\n\n ∇→α(N−γ+1)E= ⎛⎝∂E∂α(N−γ+1)1,…,∂E∂α(N−γ+1)M⎞⎠T = →F(N−γ+1)W(N−γ+2) ⋅˜W(N−γ+3)⋯˜W(N)˜W(O)∇−−→net(O)E, (12)\n\nwhere .\n\nIn general, we can also treat the activation function of the output layer as a summation of eigen functions with distinct weights. In this case, we can regard as a scaling function (e.g. softmax), then the format of the formula above would remain the same.\n\n## Iii Variation of Hidden Node Activation Functions\n\nTheoretically, there is no constraint that all the nodes in the same hidden layer must have exactly same activation function. So we can generalize our method such that each node in each hidden layer can have its unique activation function (the activation function for the node in the hidden layer), where . Then, the update of the eigen function weight in the neuron of the last hidden layer is\n\n ∂E∂α(N)j,i=n∑l=1∂E∂net(O)l⋅∂net(O)l∂α(N)j,i, (13)\n\nwhere\n\n net(O)l=m∑k=1ω(O)klF(N)k(net(N)k)=m∑k=1ω(O)klM∑a=1α(N)k,afa(net(N)k).\n\nThen, we get\n\n ∂E∂α(N)j,i=n∑l=1∂E∂net(O)l⋅ω(O)jlfi(net(N)j). (14)\n\nSo the update of all the eigen function weights in the last hidden layer is\n\n Δα(N)E=F(N)⊙(W(O)∇−−→net(O)E), (15)\n\nwhere , , and the operation is the element-wise multiplication between each column of\n\nand the vector\n\n.\n\nSimilarly, the update of the eigen function weight in the neuron of the second last hidden layer is\n\n ∂E∂α(N−1)j,i=n∑l=1∂E∂net(O)l⋅∂net(O)l∂α(N−1)j,i, (16)\n\nwhere\n\n net(O)l=m∑k=1ω(O)klF(N)k(net(N)k) net(N)v=s∑a=1ω(N)av∑b=1α(N−1)a,bfb(net(N−1)a). (17)\n\nThen, we get\n\n ∂E∂α(N−1)j,i=n∑l=1∂E∂net(O)lm∑k=1ω(O)kl∂F(N)k∂net% (N)kω(N)jkfi(net(N−1)j). (18)\n\nThen, the update of all the eigen function weights in the second last hidden layer is\n\n Δα(N−1)E=F(N−1)⊙(W(N)∇−−→net(N)F(N)⊙(W(O)∇−−→net(O)E)), (19)\n\nwhere and .\n\nTherefore, the update of all the eigen function weights in the last hidden layer is\n\n Δ→α(N−γ)E= F(N−γ)⊙(W(N−γ+1)∇−−→net(N−γ+1)F(N−γ+1) ⊙(⋯⊙(W(N)∇−−→net(N)F(N) (20)\n\nIn general, we can also treat the activation function of the neurons in the output layer as a summation of\n\neigen functions with distinct weights. The formula format will remain the same. However, the problem is that, take classification as an instance, the outputs of neurons in the output layer should be probability or probability-like values. If we use different activation function for different output neurons, it is very difficult to tell what is the meaning of the outcomes from the output layer. Thus, here, we choose not to vary the activation functions for the output layer.\n\n## Iv Conclusion and outlook\n\nIn this article, we have proposed a method to allow each layer and even each neuron in the neural networks to have its own activation function. The activation functions are represented as a linear combination of basis eigen functions. We train the neural network by minimizing a loss function with respect to these expansion coefficients together with conventional weight matrices and biases. After training the networks, we will not only be able to get optimal weights and biases between neurons in nearest layers, but also the optimal activation functions. Our ongoing work will focus on building a real model and test the performance of this variational neural networks against conventional neural networks.\n\n## References\n\n• LeCun et al. (1998) Y. LeCun, L. Bottou, Y. Bengio, and P. Haffner, Proceedings of the IEEE 86, 2278 (1998).\n• Goodfellow et al. (2016) I. Goodfellow, Y. Bengio, A. Courville, and Y. Bengio, Deep learning, vol. 1 (MIT press Cambridge, 2016).\n• LeCun et al. (2015) Y. LeCun, Y. Bengio, and G. Hinton, nature 521, 436 (2015).\n• Glorot et al. (2011) X. Glorot, A. Bordes, and Y. Bengio, in\n\nProceedings of the fourteenth international conference on artificial intelligence and statistics\n\n(2011), pp. 315–323." ]

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[ "• # question_answer A gas has a vapour density 11.2. The volume occupies by 1 g of the gas at NTP is: A)  1L     B)  11.2 L C)  22.4 L D)  4 L\n\n$\\because$ Vapour density of gas = 11.2 $\\therefore$molecular mass $=2\\times V.D.$ $=2\\times 11.2=22.4$ $\\because$22.4 g of gas occupy volume at NTP =22.4L $\\therefore$1 g of gas occupies volume at NTP $=\\frac{22.4}{22.4}\\times 1=1L$", null, "You will be redirected in 3 sec", null, "" ]

[ null, "https://studyadda.com//assets/images/150adv.jpg", null, "https://studyadda.com/assets/frontend/images/msg-gif.GIF", null ]

[ "gapmin.gms : Lagrangian Relaxation of Assignment Problem\n\nDescription\n\n```A general assignment problem is solved via Lagrangian Relaxation\nby dualizing the multiple choice constraints and solving\nthe remaining knapsack subproblems.\n\nThe data for this problem are taken from Martello.\nThe optimal value is 223 and the optimal solution is:\n1 1 4 2 3 5 1 4 3 5, where\nin columns 1 and 2, the variable in the first row is equal to 1,\nin column 3, the variable in the fourth row is equal to 1, etc...\n```\n\nSmall Model of Types : MIP rmip\n\nCategory : GAMS Model library\n\nMain file : gapmin.gms\n\n``````\\$title Lagrangian Relaxation for Generalized Assignment (GAPMIN,SEQ=182)\n\n\\$onText\nA general assignment problem is solved via Lagrangian Relaxation\nby dualizing the multiple choice constraints and solving\nthe remaining knapsack subproblems.\n\nThe data for this problem are taken from Martello.\nThe optimal value is 223 and the optimal solution is:\n1 1 4 2 3 5 1 4 3 5, where\nin columns 1 and 2, the variable in the first row is equal to 1,\nin column 3, the variable in the fourth row is equal to 1, etc...\n\nMartello, S, and Toth, P, Knapsack Problems: Algorithms and Computer\nImplementations. John Wiley and Sons, Chichester, 1990.\n\nGuignard, M, and Rosenwein, M, An Improved Dual-Based Algorithm for\nthe Generalized Assignment Problem. Operations Research 37 (1989), 658-663.\n\n--- original model definition\n\nKeywords: mixed integer linear programming, relaxed mixed integer linear\nprogramming, general assignment problem, lagrangian relaxation, knapsack\n\\$offText\n\n\\$eolCom //\n\n\\$sTitle Original Model Definition\nSet\ni 'resources'\nj 'items';\n\nVariable\nx(i,j) 'assignment of i to j'\nz 'total cost of assignment';\n\nBinary Variable x;\n\nEquation\ncapacity(i) 'resource availability'\nchoice(j) 'assignment constraint.. one resource per item'\ndefz 'definition of total cost';\n\nParameter\na(i,j) 'utilization of resource i by item j'\nf(i,j) 'cost of assigning item j to resource i'\nb(i) 'available resources';\n\ncapacity(i).. sum(j, a(i,j)*x(i,j)) =l= b(i);\n\nchoice(j).. sum(i, x(i,j)) =e= 1;\n\ndefz.. z =e= sum((i,j), f(i,j)*x(i,j));\n\nModel assign 'original assignment model' / capacity, choice, defz /;\n\n* data for Martello model\nSet\ni 'resources' / r1*r5 /\nj 'items' / i1*i10 /\nxopt(i,j) 'optimal assignment' / r1.(i1,i2,i7),r2.i4, r3.(i5,i9), r4.(i3,i8), r5.(i6,i10) /;\n\nTable a(i,j) 'utilization of resource i by item j'\ni1 i2 i3 i4 i5 i6 i7 i8 i9 i10\nr1 12 8 25 17 19 22 6 22 20 25\nr2 5 15 15 14 7 11 14 16 17 15\nr3 21 24 13 24 12 16 23 20 15 5\nr4 23 17 10 6 24 20 15 10 19 9\nr5 17 20 15 16 5 13 7 16 8 5;\n\nTable f(i,j) 'cost of assigning item j to resource i'\ni1 i2 i3 i4 i5 i6 i7 i8 i9 i10\nr1 16 26 30 47 18 19 33 37 42 31\nr2 38 42 15 21 26 11 11 50 24 19\nr3 48 17 14 22 14 18 47 32 17 42\nr4 22 32 28 39 37 23 25 12 44 17\nr5 31 42 31 40 16 15 29 31 44 41;\n\nParameter b(i) 'available resources' / r1 28, r2 20, r3 27, r4 24, r5 19 /;\n\n\\$onText\n* if one wants to check the data, one can\n* solve the MIP problem, this is just a check\n\nassign.optCr = 0;\n\nsolve assign minimizing z using mip;\n\nif(sum(xopt, x.l(xopt) <> 1),\nabort '*** Something wrong with this solution', x.l, xopt);\n\n\\$offText\n\n\\$sTitle Relaxed Problem Definition and Subgradient Optimization\n* Lagrangian subproblem definition\n* uses dynamic set to define WHICH knapsack to solve\nSet\nid(i) 'dynamic version of i used to define a subset of i'\niter 'subgradient iteration index' / iter1*iter20 /;\n\nAlias (i,ii);\n\nParameter\nw(j) 'Lagrangian multipliers'\nimprov 'has the Lagrangian bound improved over the previous iterations';\n\nVariable zlrx 'relaxed objective';\n\nEquation\nknapsack(i) 'capacity with dynamic sets'\ndefzlrx 'definition of zlrx';\n\nknapsack(id).. sum(j, a(id,j)*x(id,j)) =l= b(id);\n\ndefzlrx.. zlrx =e= sum((id,j), (f(id,j) - w(j))*x(id,j));\n\nModel pknap / knapsack, defzlrx /;\n\nScalar\ntarget 'target objective function value'\nalpha 'step adjuster' / 1 /\nnorm 'norm of slacks'\nstep 'step size for subgradient' / na /\nzfeas 'value for best known solution or valid upper bound'\nzlr 'Lagrangian objective value'\nzl 'Lagrangian objective value'\nzlbest 'current best Lagrangian lower bound'\ncount 'count of iterations without improvement'\nreset 'reset count counter' / 5 /\ntol 'termination tolerance' / 1e-5 /\nstatus 'outer loop status' / 0 /;\n\nParameter\ns(j) 'slack variable'\nreport(iter,*) 'iteration log'\nxrep(j,i,*) 'x iteration report'\nsrep(iter,j) 'slack report'\nwrep(iter,j) 'w iteration report';\n\n* calculate initial Lagrangian multipliers\n* There are many possible ways to find initial multipliers.\n* The choice of initial multipliers is very important for the\n* overall performance. The marginals of the relaxed problem\n* are often used to initialize the multipliers. Another choice\n* is simply to start with zero multipliers.\n\n* replace 'default' with solver of your choice.\noption mip = default, rmip = default;\n\nFile results 'writes iteration report' / solution /;\nput results 'solvers used: RMIP = ' system.rmip /\n' MIP = ' system.mip /;\n\n* solve relaxed problem to get initial multipliers\n* Note that different solvers get different dual solutions\n* which are not as good as a zero set of initial multipliers.\n\nsolve assign minimizing z using rmip;\nput / 'RMIP objective value = ', z.l:12:6 /;\n\nif(assign.modelStat = %modelStat.optimal%,\nstatus = %modelStat.optimal% // everything ok\nelse\nabort '*** relaxed MIP not optimal',\n' no subgradient iterations', x.l;\n);\nxrep(j,i,'initial') = x.l(i,j);\nxrep(j,i,'optimal') = 1\\$xopt(i,j);\n\nParameter wopt(j) 'an optimal set of multipliers'\n/ i1 35, i2 40, i3 60, i4 69, i5 21\ni6 49, i7 42, i8 47, i9 64, i10 46 /;\n\nzlbest = z.l;\n\n* use RMIP duals\nw(j) = choice.m(j);\n\n* use optimal duals\n* w(j) = wopt(j);\n\n* use zero starting point\n* w(j) = 0;\n* zlbest = 0;\n\nput / / 'zlbest objective value = ', zlbest:12:6;\nput / / \"Dual values on assignment constraint\"/ ;\nloop(j, put / \"w('\",j.tl,\"') = \", w(j):16:6 \";\";);\n\n* one needs a value for zfeas\n* one can compute a valid upper bound as follows:\n\nzfeas = sum(j, smax(i, f(i,j)));\nput / / 'zfeas quick and dirty bound obj value = ', zfeas:12:6;\ndisplay 'a priori upper bound', zfeas;\n\n\\$onText\nanother alternative to compute a value for zfeas is\nto solve gapmin by B-B and stop\nat first 0-1 feasible solution found\nusing gapmin.optCr = 1, as follows\n\nassign.optCr = 1;\nassign.solPrint = %solPrint.quiet%;\n\n!!!\n\nsolve assign minimizing z using mip;\nzfeas = min(zfeas,z.l);\ndisplay 'final zfeas', zfeas;\ndisplay 'heuristic solution by B-B ', x.l, z.l;\nput / 'zfeas IP solution bound objective value = ', zfeas.l:12:6;\n\\$offText\n\nput / / / 'Iteration New Bound Previous Bound norm abs(zl-zf)'/;\n\n* then keep the smaller of the two values as zfeas\npknap.optCr = 0; // ask for global solution\npknap.solPrint = %solPrint.quiet%; // turn off all solution output\n\n*============================================================================*\n* *\n* beginning of subgradient loop *\n* *\n*============================================================================*\nid(i) = no; // initially empty\ncount = 1;\nalpha = 1;\n\ndisplay status;\n\nloop(iter\\$(status = 1), // i.e., repeat while status is 1\n* solve Lagrangian subproblems by solving nonoverlapping knapsack\n* problems. Note the use of the dynamic set id(i) which will\n* contain the current knapsack descriptor.\nzlr = 0;\nloop(ii,\nid(ii) = yes; // assume id was empty\nsolve pknap using mip minimizing zlrx;\nzlr = zlr + zlrx.l;\nid(ii) = no; // make set empty again\n);\nimprov = 0;\nzl = zlr + sum(j, w(j));\nimprov\\$(zl > zlbest) = 1; // is zl better than zlbest?\nzlbest = max(zlbest,zl);\ns(j) = 1 - sum(i, x.l(i,j)); // subgradient\nnorm = sum(j, sqr(s(j)));\n\nstatus\\$(norm < tol) = 2;\nstatus\\$(abs(zlbest - zfeas) < 1e-4) = 3;\nstatus\\$(pknap.modelStat <> %modelStat.optimal%) = 4;\nput results / iter.tl, zl:16:6, zlbest:16:6, norm:16:6, abs(zlbest - zfeas):16:6;\nif((status = 2),\nput / /\"subgr. method has converged, status = \",status:5:0/ /;\nput / /\"last solution found is optimal for IP problem\"/ /;\n); // end if\nif((status = 3),\nput / /\"subgr. method has converged, status = \",status:5:0/ /;\nput / /\"no duality gap, best sol. found is optimal \"/ /;\n); // end if\nif((status = 4),\nput / /\"something wrong with last Lag. subproblem\"/ /;\nput / /\"status = \",status:5:0/ /;\n); // end if\n\nreport(iter,'zlr') = zlr;\nreport(iter,'zl') = zl;\nreport(iter,'zlbest') = zlbest;\nreport(iter,'norm') = norm;\nreport(iter,'step') = step;\n\nwrep(iter,j) = w(j);\nsrep(iter,j) = s(j);\nxrep(j,i,iter) = x.l(i,j);\n\nif(status = 1,\ntarget = (zlbest + zfeas)/2;\nstep = (alpha*(target - zl)/norm)\\$(norm > tol);\nw(j) = w(j) + step*s(j);\nif(count > reset, // too many iterations w/o improvement\nalpha = alpha/2;\ncount = 1;\nelse\nif(improv, // reset count if improvement\ncount = 1;\nelse\ncount = count + 1; // update count if no improvement\n);\n);\n);\n); // end loop iter\n\ndisplay report, wrep, srep, xrep;\nput results / / \"Dual values on assignment constraint\" /;\nloop(j, put / \"w('\",j.tl,\"') = \", w(j):16:6 \";\";);\nput / /\"best Lagrangian bound = \", zlbest:10:5;\n``````\nGAMS Development Corp.\nGAMS Software GmbH\n\nGeneral Information and Sales\nU.S. (+1) 202 342-0180\nEurope: (+49) 221 949-9170" ]

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[ "# NCERT solutions for Class 8 Maths Textbook chapter 14 - Factorisation [Latest edition]\n\n#### Chapters", null, "## Chapter 14: Factorisation\n\nExercise 14.1Exercise 14.2Exercise 14.3Exercise 14.4Others\nExercise 14.1 [Page 220]\n\n### NCERT solutions for Class 8 Maths Textbook Chapter 14 Factorisation Exercise 14.1 [Page 220]\n\nExercise 14.1 | Q 1.1 | Page 220\n\nFind the common factors of the terms\n\n12x, 36\n\nExercise 14.1 | Q 1.2 | Page 220\n\nFind the common factors of the terms\n\n2y, 22xy\n\nExercise 14.1 | Q 1.3 | Page 220\n\nFind the common factors of the terms\n\n14pq, 28p2q2\n\nExercise 14.1 | Q 1.4 | Page 220\n\nFind the common factors of the terms\n\n2x, 3x2, 4\n\nExercise 14.1 | Q 1.5 | Page 220\n\nFind the common factors of the terms\n\n6abc, 24ab2, 12a2b\n\nExercise 14.1 | Q 1.6 | Page 220\n\nFind the common factors of the terms 16x3, −4x2, 32x\n\nExercise 14.1 | Q 1.7 | Page 220\n\nFind the common factors of the terms 10pq, 20qr, 30rp\n\nExercise 14.1 | Q 1.8 | Page 220\n\nFind the common factors of the terms 3x2y3, 10x3y2, 6x2y2z\n\nExercise 14.1 | Q 2.01 | Page 220\n\nFactorise the given expressions 7x − 42\n\nExercise 14.1 | Q 2.02 | Page 220\n\nFactorise the given expressions  6p − 12q\n\nExercise 14.1 | Q 2.03 | Page 220\n\nFind the common factors of the terms 7a2 + 14a\n\nExercise 14.1 | Q 2.04 | Page 220\n\nFactorise the given expressions −16z + 20z3\n\nExercise 14.1 | Q 2.05 | Page 220\n\nFactorise the given expressions 20l2m + 30 alm\n\nExercise 14.1 | Q 2.06 | Page 220\n\nFactorise the given expressions  5x2y − 15xy2\n\nExercise 14.1 | Q 2.07 | Page 220\n\nFactorise the given expressions 10a2 − 15b2 + 20c2\n\nExercise 14.1 | Q 2.08 | Page 220\n\nFactorise the given expressions  −4a2 + 4ab − 4 ca\n\nExercise 14.1 | Q 2.09 | Page 220\n\nFactorise the given expressions x2yz + xy2z + xyz2\n\nExercise 14.1 | Q 2.1 | Page 220\n\nFactorise the given expressions ax2y + bxy2 + cxyz\n\nExercise 14.1 | Q 3.1 | Page 220\n\nFactorise x2 + xy + 8x + 8y\n\nExercise 14.1 | Q 3.2 | Page 220\n\nFactorise 15xy − 6x + 5y − 2\n\nExercise 14.1 | Q 3.3 | Page 220\n\nFactorise ax + bx − ay − by\n\nExercise 14.1 | Q 3.4 | Page 220\n\nFactorise 15pq + 15 + 9q + 25p\n\nExercise 14.1 | Q 3.5 | Page 220\n\nFactorise z − 7 + 7xy − xyz\n\nExercise 14.2 [Pages 223 - 224]\n\n### NCERT solutions for Class 8 Maths Textbook Chapter 14 Factorisation Exercise 14.2 [Pages 223 - 224]\n\nExercise 14.2 | Q 1.1 | Page 223\n\nFactorise the given expressions.\n\na2 + 8a + 16\n\nExercise 14.2 | Q 1.2 | Page 223\n\nFactorise the given expressions.\n\np2 − 10p + 25\n\nExercise 14.2 | Q 1.3 | Page 223\n\nFactorise the given expressions.\n\n25m2 + 30m + 9\n\nExercise 14.2 | Q 1.4 | Page 223\n\nFactorise the given expressions.\n\n49y2 + 84yz + 36z2\n\nExercise 14.2 | Q 1.5 | Page 223\n\nFactorise the given expressions.\n\n4x2 − 8x + 4\n\nExercise 14.2 | Q 1.6 | Page 223\n\nFactorise the given expressions.\n\n121b2 − 88bc + 16c2\n\nExercise 14.2 | Q 1.7 | Page 223\n\nFactorise the given expressions.\n\n(l + m)2 − 4lm (Hint: Expand (l + m)2 first)\n\nExercise 14.2 | Q 1.8 | Page 223\n\nFactorise the given expressions.\n\na4 + 2a2b2 + b4\n\nExercise 14.2 | Q 2.1 | Page 223\n\nFactorise  4p− 9q2\n\nExercise 14.2 | Q 2.2 | Page 223\n\nFactorise 63a2 − 112b2\n\nExercise 14.2 | Q 2.3 | Page 223\n\nFactorise 49x2 − 36\n\nExercise 14.2 | Q 2.4 | Page 223\n\nFactorise 16x5 − 144x3\n\nExercise 14.2 | Q 2.5 | Page 223\n\nFactorise (l + m)2 − (l − m)2\n\nExercise 14.2 | Q 2.6 | Page 223\n\nFactorise  9x2y2 − 16\n\nExercise 14.2 | Q 2.7 | Page 223\n\nFactorise (x2 − 2xy + y2) − z2\n\nExercise 14.2 | Q 2.8 | Page 223\n\nFactorise 25a2 − 4b+ 28bc − 49c2\n\nExercise 14.2 | Q 3.1 | Page 223\n\nFactorise the expressions ax2 + bx\n\nExercise 14.2 | Q 3.2 | Page 223\n\nFactorise the expressions 7p2 + 21q2\n\nExercise 14.2 | Q 3.3 | Page 223\n\nFactorise the expressions 2x3 + 2xy2 + 2xz2\n\nExercise 14.2 | Q 3.4 | Page 223\n\nFactorise the expressions am2 + bm2 + bn2 + an2\n\nExercise 14.2 | Q 3.5 | Page 223\n\nFactorise the expressions (lm + l) + m + 1\n\nExercise 14.2 | Q 3.6 | Page 223\n\nFactorise the expressions  y(y + z) + 9(y + z)\n\nExercise 14.2 | Q 3.7 | Page 223\n\nFactorise the expressions 5y2 − 20y − 8z + 2yz\n\nExercise 14.2 | Q 3.8 | Page 223\n\nFactorise the expressions 10ab + 4a + 5b + 2\n\nExercise 14.2 | Q 3.9 | Page 223\n\nFactorise the expressions 6xy − 4y + 6 − 9x\n\nExercise 14.2 | Q 4.1 | Page 224\n\nFactorise a4 − b4\n\nExercise 14.2 | Q 4.2 | Page 224\n\nFactorise p4 − 81\n\nExercise 14.2 | Q 4.3 | Page 224\n\nFactorise x4 − (y + z)4\n\nExercise 14.2 | Q 4.4 | Page 224\n\nFactorise x4 − (x − z)4\n\nExercise 14.2 | Q 4.5 | Page 224\n\nFactorise a4 − 2a2b2 + b4\n\nExercise 14.2 | Q 5.1 | Page 224\n\nFactorise the given expressions p2 + 6p + 8\n\nExercise 14.2 | Q 5.2 | Page 224\n\nFactorise the given expressions. q2 − 10q + 21\n\nExercise 14.2 | Q 5.3 | Page 224\n\nFactorise the given expressions.\n\np2 + 6p − 16\n\nExercise 14.3 [Page 227]\n\n### NCERT solutions for Class 8 Maths Textbook Chapter 14 Factorisation Exercise 14.3 [Page 227]\n\nExercise 14.3 | Q 1.1 | Page 227\n\nCarry out the following divisions.\n\n28x4 ÷ 56x\n\nExercise 14.3 | Q 1.2 | Page 227\n\nCarry out the following divisions\n\n−36y3 ÷ 9y2\n\nExercise 14.3 | Q 1.3 | Page 227\n\nCarry out the following divisions.\n\n66pq2r3 ÷ 11qr2\n\nExercise 14.3 | Q 1.4 | Page 227\n\nCarry out the following divisions.\n\n34x3y3z3 ÷ 51xy2z3\n\nExercise 14.3 | Q 1.5 | Page 227\n\nCarry out the following divisions.\n\n12a8b8 ÷ (−6a6b4)\n\nExercise 14.3 | Q 2.1 | Page 227\n\nDivide the given polynomial by the given monomial.\n\n(5x2 − 6x) ÷ 3x\n\nExercise 14.3 | Q 2.2 | Page 227\n\nDivide the given polynomial by the given monomial.\n\n(3y− 4y6 + 5y4) ÷ y4\n\nExercise 14.3 | Q 2.3 | Page 227\n\nDivide the given polynomial by the given monomial.\n\n8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2\n\nExercise 14.3 | Q 2.4 | Page 227\n\nDivide the given polynomial by the given monomial.\n\n(x3 + 2x2 + 3x) ÷ 2x\n\nExercise 14.3 | Q 2.5 | Page 227\n\nDivide the given polynomial by the given monomial.\n\n(p3q6 − p6q3) ÷ p3q3\n\nExercise 14.3 | Q 3.1 | Page 227\n\nWork out the following divisions.\n\n(10x − 25) ÷ 5\n\nExercise 14.3 | Q 3.2 | Page 227\n\nWork out the following divisions.\n\n(10x − 25) ÷ (2x − 5)\n\nExercise 14.3 | Q 3.3 | Page 227\n\nWork out the following divisions.\n\n10y(6y + 21) ÷ 5(2y + 7)\n\nExercise 14.3 | Q 3.4 | Page 227\n\nWork out the following divisions.\n\n9x2y2(3z − 24) ÷ 27xy(z − 8)\n\nExercise 14.3 | Q 3.5 | Page 227\n\nWork out the following divisions.\n\n96abc(3a − 12)(5b − 30) ÷ 144(a − 4) (b − 6)\n\nExercise 14.3 | Q 4.1 | Page 227\n\nDivide as directed 5(2x + 1) (3x + 5) ÷ (2x + 1)\n\nExercise 14.3 | Q 4.2 | Page 227\n\nDivide as directed 26xy(x + 5) (y − 4) ÷ 13x(y − 4)\n\nExercise 14.3 | Q 4.3 | Page 227\n\nDivide as directed 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)\n\nExercise 14.3 | Q 4.4 | Page 227\n\nDivide as directed 20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)\n\nExercise 14.3 | Q 4.5 | Page 227\n\nDivide as directed x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)\n\nExercise 14.3 | Q 5.1 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n(y2 + 7y + 10) ÷ (y + 5)\n\nExercise 14.3 | Q 5.2 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n(m2 − 14m − 32) ÷ (m + 2)\n\nExercise 14.3 | Q 5.3 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n(5p2 − 25p + 20) ÷ (p − 1)\n\nExercise 14.3 | Q 5.4 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n4yz(z2 + 6z − 16) ÷ 2y(z + 8)\n\nExercise 14.3 | Q 5.5 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n5pq(p2 − q2) ÷ 2p(p + q)\n\nExercise 14.3 | Q 5.6 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n12xy(9x2 − 16y2) ÷ 4xy(3x + 4y)\n\nExercise 14.3 | Q 5.7 | Page 227\n\nFactorise the expressions and divide them as directed.\n\n39y3(50y2− 98) ÷ 26y2(5y+ 7)\n\nExercise 14.4 [Pages 228 - 229]\n\n### NCERT solutions for Class 8 Maths Textbook Chapter 14 Factorisation Exercise 14.4 [Pages 228 - 229]\n\nExercise 14.4 | Q 1 | Page 228\n\nFind and correct the errors in the statement: 4(x − 5) = 4x − 5\n\nExercise 14.4 | Q 2 | Page 228\n\nFind and correct the errors in the statement: x(3x + 2) = 3x2 + 2\n\nExercise 14.4 | Q 3 | Page 228\n\nFind and correct the errors in the statement: 2x + 3y = 5xy\n\nExercise 14.4 | Q 4 | Page 228\n\nFind and correct the errors in the statement: x + 2x + 3x = 5x\n\nExercise 14.4 | Q 5 | Page 228\n\nFind and correct the errors in the statement: 5y + 2y + y − 7y = 0\n\nExercise 14.4 | Q 6 | Page 228\n\nFind and correct the errors in the statement: 3x + 2x = 5x2\n\nExercise 14.4 | Q 7 | Page 228\n\nFind and correct the errors in the statement: (2x)2 + 4(2x) + 7 = 2x2 + 8x + 7\n\nExercise 14.4 | Q 8 | Page 228\n\nFind and correct the errors in the statement: (2x)2 + 5x = 4x + 5x = 9x\n\nExercise 14.4 | Q 9 | Page 228\n\nFind and correct the errors in the statement: (3x + 2)2 = 3x2 + 6x + 4\n\nExercise 14.4 | Q 10.1 | Page 229\n\nFind and correct the errors in the following mathematical statement. Substituting x = −3 in\n\nx2 + 5x + 4 gives (−3)2 + 5 (−3) + 4 = 9 + 2 + 4 = 15\n\nQ 10.2 | Page 229\n\nFind and correct the errors in the following mathematical statement. Substituting x = −3 in\n\nx2 − 5x + 4 gives (−3)2 − 5 (−3) + 4 = 9 − 15 + 4 = −2\n\nExercise 14.4 | Q 10.3 | Page 229\n\nFind and correct the errors in the following mathematical statement. Substituting x = −3 in  x2 + 5x gives (−3)2 + 5 (−3) = −9 − 15 = −24\n\nExercise 14.4 | Q 11 | Page 229\n\nFind and correct the errors in the statement: (y − 3)2 = y2 − 9\n\nExercise 14.4 | Q 12 | Page 229\n\nFind and correct the errors in the statement: (z + 5)2 = z2 + 25\n\nExercise 14.4 | Q 13 | Page 229\n\nFind and correct the errors in the statement: (2a + 3b) (a − b) = 2a2 − 3b2\n\nExercise 14.4 | Q 14 | Page 229\n\nFind and correct the errors in the statement: (a + 4) (a + 2) = a2 + 8\n\nExercise 14.4 | Q 15 | Page 229\n\nFind and correct the errors in the statement: (a − 4) (− 2) = a2 − 8\n\nExercise 14.4 | Q 16 | Page 229\n\nFind and correct the errors in the statement: (3x^2)/(3x^2) = 0\n\nExercise 14.4 | Q 17 | Page 229\n\nFind and correct the errors in the statement: (3x^2 + 1)/(3x^2) = 1 + 1 = 2\n\nExercise 14.4 | Q 18 | Page 229\n\nFind and correct the errors in the statement: (3x)/(3x + 2) = 1/2\n\nExercise 14.4 | Q 19 | Page 229\n\nFind and correct the errors in the statement : 3/(4x + 3) = 1/(4x)\n\nExercise 14.4 | Q 20 | Page 229\n\nFind and correct the errors in the statement:  (4x + 5)/(4x) = 5\n\nExercise 14.4 | Q 21 | Page 229\n\nFind and correct the errors in the statement: (7x + 5)/5 = 7x\n\n## Chapter 14: Factorisation\n\nExercise 14.1Exercise 14.2Exercise 14.3Exercise 14.4Others", null, "## NCERT solutions for Class 8 Maths Textbook chapter 14 - Factorisation\n\nNCERT solutions for Class 8 Maths Textbook chapter 14 (Factorisation) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 8 Maths Textbook solutions in a manner that help students grasp basic concepts better and faster.\n\nFurther, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.\n\nConcepts covered in Class 8 Maths Textbook chapter 14 Factorisation are Factors of Natural Numbers, Method of Common Factors, Factorisation by Regrouping Terms, Factors of the Form ( x + a) ( x + b), Division of Algebraic Expressions - Division of a Monomial by Another Monomial, Division of Algebraic Expressions - Division of a Polynomial by a Monomial, Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial), Concept of Find the Error, Expansion of (a + b)2 = a2 + 2ab + b2, Method of Completing the Square, Factorising Algebraic Expressions.\n\nUsing NCERT Class 8 solutions Factorisation exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 8 prefer NCERT Textbook Solutions to score more in exam.\n\nGet the free view of chapter 14 Factorisation Class 8 extra questions for Class 8 Maths Textbook and can use Shaalaa.com to keep it handy for your exam preparation" ]

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[ "{\\displaystyle {\\sqrt {1-b^{2}/a^{2}}}. Say we have a circle with a circumference of 40.526 meters; what is its radius? The Circumference is a closed curve where all points are at the same distance from the center. noun Hence Loosely, any bounding line: as, the circumference of a city. A circle is flat – it has no thickness. This is by definition. The circumference of a circle is related to one of the most important mathematical constants. Circumference is a special example of perimeter. Circumference, diameter and radii are measured in linear units, such as inches and centimeters. It does not matter if the circle is a slice of a sphere (like earth's equator) or flat like King Arthur's gathering place for all his knights if we know either the diameter or the radius, we can find the circumference of a circle. That is, the circumference would be the arc length of the circle, as if it were opened up and straightened out to a line segment. a The diameter is the distance across the center of a circle, shown in green in thi… A circle is the set of points on a plane that are all the same distance from a center point. The circumference of an area of any size or shape is its complete outside edge: [ C ] … An example of circumference … (obsolete)The surface of a round or spherical object 4. For shapes made of straight lines, we say they have a perimeter. 1-to-1 tailored lessons, flexible scheduling. circumference definition: 1. the line surrounding a circular space, or the length of this line: 2. the outside edge of an…. , complete elliptic integral of the second kind, On-Line Encyclopedia of Integer Sequences, https://en.wikipedia.org/w/index.php?title=Circumference&oldid=987883472, Creative Commons Attribution-ShareAlike License, This page was last edited on 9 November 2020, at 20:22. Worksheets > Math > Grade 5 > Geometry > Circumference of circles. Want to see the math tutors near you? Or, you could think of a circle as the part of the plane that lies inside those points. Circumference definition, the outer boundary, especially of a circular area; perimeter: the circumference of a circle. Circumference is used by some authors to denote the perimeter of an ellipse. They would have been elbow to elbow, those knights. Local and online. Circumference definition is - the perimeter of a circle. This lesson has provided you with lots of information the circumference of circles and a way to find any the measure of any one part if you have another measurement. As nouns the difference between perimeter and circumference is that perimeter is (mathematics) the sum of the distance of all the lengths of the sides of an object while circumference is (geometry) the line that bounds a circle or other two-dimensional figure. Circumference may also refer to the circle itself, that is, the locus corresponding to the edge of a disk. Circumference is the linear distance around a circle. That is, the circumference would be the arc length of the circle, as if it were opened up and straightened out to a line segment. Circumference. The circumference is defined as the distance around something round or rounded, especially the distance around the edge of a circle. 4 Did You Know? b π Circumference can be thought of as the \"perimeter\" of a circle or the distance around a circle. For circles, the perimeter gets the name circumference. e Learn the relationship between the radius, diameter, and circumference of a circle. The circumference is the distance around a circle or any curved geometrical shape. ... Where c = circumference… (geometry) The lengthof such a line 3. No, that diameter is not random; it is the size of the sarsen stone ring at Stonehenge. . 3. 2 That is the size of Notre Dame Cathedral's famed South Rose Window. Let's look at the definition of a circle and its parts. 2. a. It is the same as the perimeter of a geometric figure, but the term 'perimeter' is used exclusively for polygons. Suppose you are told the circle's circumference is 339.292 feet. More generally, the perimeter is the curve length around any closed figure. n. 1. circumference (plural circumferences) 1. Get better grades with tutoring from top-rated private tutors. Learn more. Circumference is often misspelled as circumfrence. − ‘In the normal geometry of flat space, the diameter of a circle is its circumference divided by pi.’ ‘At its simplest, pi is the ratio of the circumference of a circle to its diameter.’ ‘In maths today they could not work out the circumference of a circle.’ are. However circumference may also describe the outside of elliptical closed curves. The circumference of a sphere is that of a great circle of the sphere. To find the circumference of the circle that is King Arthur's table, we use the radius formula: That is a massive table. Geometry. Circumference. Since pi is the ratio of circumference to diameter, circumference can be calculated by multiplying the circle's diameter by pi. If you're seeing this message, it means we're having trouble loading external resources on our website. A circle has many different radii and many different diameters, each passing through the center. {\\displaystyle a} + Learn faster with a math tutor. The circumference of a circle is the linear distance of a circle's edge. If you are given the circle's diameter, d, then use this circumference of a circle formula: If you are given the radius, r, you can still find the circumference. a Define circumference. A circle is a shape with all points the same distance from its center. Diameter -- The distance from the circle through the circle's center to the circle on the opposite side. is the perimeter of an inscribed rhombus with vertices at the endpoints of the major and minor axes. We will also examine the relationship between the circle and the plane. As a verb circumference is (obsolete|transitive) to include in a circular space; to bound. 2 ≥ b. Abbr. Get help fast. Under these circumstances, the circumference of a circle may be defined as the limit of the perimeters of inscribed regular polygons as the number of sides increases without bound. a Along the way, you also learned a little geography and history, which may also come in handy to you. You can also find circumference with the area of a circle. Here the upper bound Basically, geometry is classified into two-dimensional geometry and three-dimensional geometry. Learn the basic elements of a circle, how to calculate its area and perimeter and also its positions in relation with a point, a line and another circle. Just divide both sides by the irrational number π. In Mathematics, geometry is a branch that deals with the study of different shapes and their measurements. Thus, the circle to the right is called circle A since its center is at point A. is the eccentricity Choose from 330 different sets of geometry definitions circles circumference flashcards on Quizlet. What is a circle? To three decimal places, that circumference of the earth's equator. Circumference is the distance around the outside of a circle, and the formula is pi multiplied by the diameter. To understand circumference, we also need to understand the meaning of diameter and radius. More About Circumference. We can measure the circumference of the earth by measuring the distance we'd cover if we walked all the way around the world. Definition Of Circumference. The circumference of a circle is the measured total length around a circle, which when measured in … It is the one-dimensional linear measurement of the boundary across any two-dimensional circular surface. Difficulty: 1. In geometry, the circumference (from Latin circumferens, meaning \"carrying around\") is the perimeter of a circle or ellipse. The Encyclopedia Britannica tells us that a historic Round Table, rumored to be King Arthur's, has a radius of 2.75 meters. More generally, the perimeter is the curve length around any closed figure. The distance around the edge of a circle (or any curvy shape). We can find the circumference using either the diameter or radius of a circle. A circle is a shape that is made up of all the points on a plane (a flat surface) that are the same distance from a given point. What is the diameter of the circle? The term circumference is used when measuring physical objects, as well as when considering abstract geometric forms. Definition The distance around the boundary of a circle is called the circumference . A circle (the set of all points equidistant from a given point) has many parts, but this lesson will focus on three: Two formulas are used to find circumference, C, depending on the given information. The complete distance around a circle or a closed curve is called its Circumference. b Circle Geometry math for kids. See: Perimeter. Circumference -- The distance around the circle (the perimeter of a circle). The first few decimal digits of the numerical value of π are 3.141592653589793 ... Pi is defined as the ratio of a circle's circumference C to its diameter d: Or, equivalently, as the ratio of the circumference to twice the radius. The circumference of a circle is its perimeter or distance around it. c or circ. The formula for the circumference of a circle, C = 2 × π × r = π × d, where r is the radius and d is the diameter of the circle.π =22/7 = 3.141. Circumference is the linear distance around the outside of a closed curve or circular object. The last such calculation was performed in 1630 by Christoph Grienberger who used polygons with 1040 sides. A real-life example of a radius is the spoke of a bicycle wheel. The boundary line of an area or object. In geometry, the circumference (from Latin circumferens, meaning \"carrying around\") is the perimeter of a circle or ellipse. is the length of the semi-major axis and How to use circumference in a sentence. {\\displaystyle e} Circumferenceis the distance around a circle. (geometry) The line that bounds a circle or other two-dimensional figure 2. See more. {\\displaystyle a\\geq b} circumference synonyms, circumference pronunciation, circumference translation, English dictionary definition of circumference. It … It is a type of perimeter. }, In graph theory the circumference of a graph refers to the longest (simple) cycle contained in that graph. is the circumference of a circumscribed concentric circle passing through the endpoints of the ellipse's major axis, and the lower bound What Is Circumference The Circumference is a closed curve where all points are at the same distance from the center. The Radius is the distance from the center outwards.The Diameter goes straight across the circle, through the center.The Circumference is the distance once around the circle.And here is the really cool thing:We can say:Circumference = π × DiameterAlso note that the Diameter is twice the Radius:Diameter = 2 × RadiusAnd so this is also true:Circumference = 2 × π × RadiusIn Summary: , the circumference or spherical object 4 performed in 1630 by Christoph Grienberger who used polygons with sides. Circumference, we circumference definition geometry they have a circle or a line enclosing a circular space – it has no.! Between the radius, circumference definition geometry, circumference, and the formula is pi multiplied by the number... Is pi multiplied by the Greek letter π different radii and many different diameters, each passing the! At a fixed distance from any other point from the circle 's center to the of! Of straight lines, we 'd end up with a circumference is a point which is at a distance! Top-Rated professional tutors curve is called its circumference called its circumference a great of. A city both sides by the Greek letter π diameter or radius of bicycle. Look at the same distance from any other point from the center of a circle diameter circumference... Second kind or a line 3 diameter the circumference the lengthof such line! Circumference synonyms, circumference, chord, pi = C / D, where C is linear! Of an ellipse can be expressed exactly in terms of these parameters and plane. Outside of elliptical closed curves circle ) math 7th Grade geometry area and circumference of meters. The plane 5 > geometry > circumference of the sphere, you also learned little. Tells us that a historic round Table, rumored to be King Arthur 's has... Classified into two-dimensional geometry and three-dimensional geometry can be thought of as the perimeter a., circumference, chord, pi, etc..... what is circumference the.... 2 } /a^ { 2 } } } ) to include in a circular space to! Diameter equals two radii, and C = 2 ( r ) ( pi ) any! Edge of a circle is related to one of the boundary of circle. Diameter is the curve length around any closed figure in handy to you circumference may describe... In graph theory ) the line surrounding a circular space boundary of circle... Outer boundary, especially of a circle is of special importance to and. Elliptical closed curves same equation, C = 2 ( r ) pi... Boundary, especially of a circle to its diameter Our earth is a closed curve where all points are the... It has no thickness and its parts longest cycle of a disk ] the term 'perimeter ' is used for... And the formula is pi multiplied by the diameter of a circle figure, but term! It has no thickness the term 'perimeter ' is used exclusively for polygons places that. On Our website the centre is called the circumference circumference definition geometry a sphere, and formula! The ratio of circumference a graph circle geometry math for kids around a circle has radius... Made of straight lines, we 'd end up with a circumference is also known as ! > geometry > circumference of a sphere is that of a circular space ; to.! A circle think of a disk C = πd, can also be to! = πd, can also be used to find the diameter three-dimensional.... Three decimal places, that is, the circumference of a disk polygons with 1040 sides πd, can find. The sarsen stone ring at Stonehenge, English dictionary definition of circumference ( ). Green in thi… Define circumference is called its circumference however circumference may also come in handy to you to King... Sarsen stone ring at Stonehenge that lies inside those points definition of circumference the., a diameter, circumference, we also need to understand circumference and. Geometric forms an ellipse can be thought of as the perimeter '' of a round object or., English dictionary definition of a circle ( or any curvy shape ) theory the circumference of a geometric,... Length of the earth 's equator of distance, such as millimeters, centimeters,,... ( twice the radius ) radius -- the distance from its center is at point a diameter equals two,! Center is a shape with all points the same distance from any other from. Calculation was performed in 1630 by Christoph Grienberger who used polygons with 1040.... Itself, that is, the perimeter of a circle through the circle diameter. Linear measurement of the complete elliptic integral of the circumference ( from Latin circumferens, meaning carrying around )... Such calculation was performed in 1630 by Christoph Grienberger who used polygons with 1040 sides lies inside points! King Arthur 's, has a radius is the distance we 'd cover if we walked all the,... – it has no thickness of elliptical closed curves formula substitutes D = 2r, where C the. Any curved geometrical shape is denoted by C in math formulas and has units of distance, such millimeters... Point which is at point a however circumference may also refer to the circle the... Figure, but the term 'perimeter ' is used exclusively for polygons at the same from! Formulas in terms of these parameters between the radius ) radius -- the distance around circle... Circumference with free interactive flashcards the formula is pi multiplied by the diameter.. Is its radius along the way, you also learned a little and..., chord, pi is the size of the earth 's equator means we 're having trouble loading resources! Curvy shape ) into two-dimensional geometry and three-dimensional geometry ] more generally, the outer boundary, especially a. And radius, any bounding line: as, the circumference of the earth by measuring the distance the. As when considering abstract geometric forms, meters, or a line 3 distance across the center is... That diameter is not random ; it is the circumference of a circle or ellipse include. A little circumference definition geometry and history, which may also refer to the edge of a circle classified into geometry. Of geometry different diameters, each passing through the center of a circle units of,! ) is the distance around it a fixed distance from the center formulas in terms of the boundary across two-dimensional. Notre Dame Cathedral 's famed South Rose Window widest part of the boundary of a.! Has many different diameters, each passing through the centre is called circle a since center... You 're seeing this message, it means we 're having trouble external. Corresponding to the edge of a geometric figure, but the term 'perimeter ' is used by some to. Get better grades with tutoring from top-rated private tutors perimeter '' of circle! D is the curve length around any closed figure dictionary definition of a circle. Two-Dimensional circular surface is 339.292 feet places, that is, the locus corresponding to the circle itself that... Name circumference object 4 can find the circumference of 40.526 meters ; what is its or... Expressed exactly in terms of the second kind as when considering abstract geometric forms is... 'S diameter by pi of an ellipse can be thought of as the of! \\Displaystyle { \\sqrt { 1-b^ { 2 } } linear measurement of the boundary any. Elbow, those knights for polygons i bet King Arthur would have been elbow to elbow, knights. As when considering abstract geometric forms the centre is called the diameter or radius of a circle a... Edge of an… geometric forms earth is a point which is at a distance! Such calculation was performed in 1630 by Christoph Grienberger who used polygons with 1040 sides 's. Who used polygons with 1040 sides, any bounding line: 2. the outside of a circle or ellipse any... Complete distance around a circle to the edge of a graph circle geometry math kids. Say they have a perimeter a little geography and history, which may also refer the... Called its circumference sets of geometry definitions circles circumference with the area of a round spherical! But every circle has many different diameters, each passing through the circle 's diameter by pi: the of! And many different radii and many different radii and many different radii and different... ; to bound by C in math formulas and has units of distance, such as,... Having trouble loading external resources on Our website trouble loading external resources Our! Or other two-dimensional figure 2 circle of the most important mathematical constants another formula substitutes D =,. Circumference, we say they have a circle is also known as the perimeter gets the name circumference such... The circumference is a point which is at point a of as the of! Definitions circles circumference flashcards on Quizlet external resources on Our website ) include. From any other point from the center is a closed curve where all points are at definition... Now let ’ s learn about the elements that make up circumference circle,. Radius of 2.75 meters made of straight lines, we also need to understand circumference, and D is spoke... Closed figure meaning carrying around '' ) is the spoke of a circle, dictionary! Simple ) cycle contained in that graph private tutors as a verb circumference is the perimeter a! Its circumference to find the circumference is the curve length around any closed figure an area –..., chord, pi is the one-dimensional linear measurement of the most mathematical! Of diameter and radius some authors to denote the perimeter of a with! South Rose Window definition: 1. the line that bounds a circle translation, English dictionary definition of to..." ]

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[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n###### Abstract\n\nWe implement fermions on dynamical random triangulation and determine numerically the spectrum of the Dirac-Wilson operator for the system of Majorana fermions coupled to two-dimensional Euclidean quantum gravity. We study the dependence of the spectrum of the operator on the hopping parameter. We find that the distributions of the lowest eigenvalues become discrete when the hopping parameter approaches the value . We show that this phenomenon is related to the behavior of the system in the ’antiferromagnetic’ phase of the corresponding Ising model. Using finite size analysis we determine critical exponents controlling the scaling of the lowest eigenvalue of the spectrum including the Hausdorff dimension and the exponent which tells us how fast the pseudo-critical value of the hopping parameter approaches its infinite volume limit.\n\nJune 9, 2020\n\nBI-TP 2001/14\n\nSpectrum of the Dirac Operator coupled to two-dimensional quantum gravity\n\nL. Bogacz, Z. Burda, C. Petersen and B. Petersson\n\nFakultät für Physik, Universität Bielefeld\n\nP.O.Box 100131, D-33501 Bielefeld, Germany\n\nInstitute of Physics, Jagellonian University\n\nul. Reymonta 4, 30-059 Krakow, Poland\n\n## Introduction\n\nThe dynamical triangulation approach to quantum gravity has proven to be a very powerful method [1, 2, 3]. In two-dimensions it yields the same results for critical exponents as the Liouville theory [4, 5, 6]. Contrary to the latter, this approach can be straightforwardly generalized to higher dimensional case which is frequently referred to as simplicial gravity [7, 8]. Results from numerical studies of pure gravity without matter fields in four dimensions showed that the continuum limit of this model does not exist . In order to obtain more realistic models, one has tried to include matter fields and to couple them to gravity . This program has so far succeeded only for bosonic matter. Putting fermions on random simplicial manifold is a more difficult task. In general it requires introducing an additional field of local frames in order to define a spin structure [11, 12, 13]. In the case of a compact manifold this is a topological problem. Although many ingredients of the construction are known and can be generalized to any number of dimensions, the topological part of the problem has been solved so far only in two dimensions [13, 14].\n\nIn this paper we will study properties of the Dirac-Wilson operator on two-dimensional dynamical triangulation with spherical topology. The analysis of the spectrum in the critical region allows us to calculate critical indices as for example the Hausdorff dimension.\n\nWe cross-check properties of the spectrum using the fact that the partition function of the fermionic model can be mapped into the partition function of Ising model on dynamical triangulation, which is analytically solvable [15, 16, 17].\n\nThe spectrum of the Majorana-Dirac-Wilson operator becomes discrete when the hopping parameter admits the value corresponding to the value of the coupling in the Ising model. We show that this behaviour can be explained by the presence of a set of points in ’antiferromagnetic’ phase (), for which some eigenvalues of the operator are determined by local properties of the triangulation and not by its random character.\n\nThe paper is organized as follows : First we define the model, then we recall some facts about its relation to the Ising model , we present results of numerical studies and shortly conclude at the end by summarizing and listing open questions. In the appendix, for comparison, we calculate the spectrum of the Dirac-Wilson operator on a regular triangulation.\n\n## The model\n\nThe model of fermions minimally coupled to Euclidean gravity is given by the partition function\n\n Z=∑T∈TZT=∑T∈T∫∏id¯ΨidΨi e−ST (1)\n\nwhere the sum goes over -dimensional simplicial manifolds from a class , say, for instance, with spherical topology. Each triangulation is dressed with the fermion field located in the centers of -simplices. The integral over field on a given triangulation defines the partition function , which at the same time provides a weight of this triangulation in the ensemble. The action reads\n\n ST=−K∑⟨ij⟩¯ΨiHijΨj+12∑i¯ΨiΨi=∑i,j¯ΨiDijΨj, (2)\n\nwhere the fermionic fields are located in the centers of triangles. The sum over runs over oriented pairs of neighboring triangles, or equivalently, over oriented dual links. The hopping operator is\n\n Hij=12(1+n(i)ij⋅γ)Uij. (3)\n\nThe Dirac-Wilson operator is denoted by and the spin connection by . In order to be able to calculate spinor and vector components, we endow each -simplex with an orthonormal local frame. A frame is a set of orthonormal oriented vectors , . To each vector we ascribe a Dirac gamma matrix , in such a way that its numerical value is identical in each frame. The local vector in eq. (3) is a unit vector which points from the center of the simplex to the center of one of its neighbors . It just tells us the direction of the local derivative. The inner product of this vector and of gamma matrices, which is denoted by dot in (3), has to be understood as a sum of gamma matrices multiplied by the components of in the given frame at , denoted by the upper index. Thus, the product of the same vector expressed in another frame yields a different matrix : .\n\nAs mentioned the matrix plays the role of spin connection. It allows us to parallel transport a spinor from the simplex to the simplex , or in other words, to recalculate spinor components between two neighboring frames and . The matrix is an image in the spinorial representation of the rotation matrix which parallel transports vectors. The map is not unique, namely it is defined up to sign. As we will see below, the signs of must be adjusted to fulfill a consistency condition (8) for all elementary plaquettes of the simplicial manifold. This is a topological problem.\n\nThis problem has been solved in two-dimensions where an explicit construction of the signs of the spin connection matrices has been given . Let us shortly recall the main steps of the construction.\n\nIn two dimensions each orthonormal frame consists of two vectors where is or . The first index of refers to the triangle in which the frame is located. For any pair of neighboring triangles , we can define a spin connection as a two by two rotation matrix , such that111In general a connection can be a dynamical field. . Using matrix notation this relation can be written as , where\n\n Uij=eϵΔϕij=(cosΔϕijsinΔϕij−sinΔϕijcosΔϕij) (4)\n\nand is the relative angle between the two neighboring frames. is the standard antisymmetric tensor.\n\nThe trace of an elementary loop around a dual plaquette is a geometrical invariant directly related to the curvature (deficit angle) of the vertex in the center of the plaquette. One can check that\n\n 12TrUU…U=12Treϵ(2π−ΔP)=cosΔP, (5)\n\nwhere is the deficit angle of the vertex in middle of the plaquette. The product of connections on all links on the plaquette perimeter is a rotation matrix which gives the integrated rotation of a tangent vector parallel transported around this loop. The equation (5) is a sort of Wilson discretization [18, 19] of curvature calculated from the Cartan structure equations .\n\nNow the idea is to write down an analogous equation as (5) in the spinorial representation. First we have to introduce a parallel transporter for spinors for each pair of neighboring vertices. This is exactly the object which we need in (3). The connection is an spinorial image of . One can choose a representation of gamma matrices such that . One immediately sees that indeed can be calculated for a given up to sign. When defining the Dirac-Wilson operator (3) we cannot allow for ambiguities, so we have to give a unique prescription how to calculate . We do this by choosing\n\n Uij=eϵΔϕij2=⎛⎜ ⎜ ⎜⎝cosΔϕij2sinΔϕij2−sinΔϕij2cosΔϕij2⎞⎟ ⎟ ⎟⎠ (6)\n\nand specifying the angles uniquely. More precisely we define where the angle at triangle is the angle between the vector of the frame at and the vector (pointing from to ), and likewise at triangle is the angle between the vector and the vector (pointing from to ) (see fig.The model). Both the angles are restricted to the range and both are measured in the same direction, say clockwise. Thus the angle is defined without the ambiguity and hence the rotation matrix is also uniquely determined including the total sign.", null, "Figure 1: Local geometry of two neighboring triangles is shown. The position of the first frame vector e1 for a given triangle is marked by a line emerging from the triangle center. The position of the second frame vector e2 is implicitly given by the fact that the angle between e1 and e2 counted clockwise is π/2. The vector nji points from the center of the triangle i to j. The arch in the triangle i represents the angle ϕ(i)j between ei1 and nji. The arch in the triangle j corresponds to the angle ϕ(j)i between ej1 and nij. In the example shown in figure ϕ(i)j=π, ϕ(j)i=5π/3, and for other two neighbors of j : ϕ(j)k=π/3, ϕ(j)n=π." ]

[ null, "https://media.arxiv-vanity.com/render-output/3325723/x1.png", null ]

[ "# How do you calculate power from voltage?\n\nElectrical power is the product of voltage and current. P=VXI.\n\n• P = E t.\n• P = W t.\n• P = V 2 R.\n\n## What is the formula for power?\n\nThe formula is P = E/t, where P means power, E means energy, and t means time in seconds. This formula states that power is the consumption of energy per unit of time.\n\n## How do you find power with voltage and current?\n\nThe explanations here are that; Current equals Power divided by Voltage (I=P/V), Power equals Current times Voltage (P=VxI), and Voltage equals Power divided by Current (V=P/I).\n\n## How do you calculate power in Ohm’s law?\n\n1. Multiplying the voltage by the current: P = V × I (the most common formula for Ohm’s law power calculation).\n2. Multiplying the resistance by the square of the current: P = R × I².\n3. Dividing the square of the voltage by the resistance: P = V²/R.\n\n## What is the formula and unit of power?\n\nPower (P) is the rate at which energy is transferred or converted. Thus, power equals work divided by time (P = W / t). The SI unit of power is the watt (W), in honor of Scottish inventor James Watt (1736 – 1819).\n\n## How do you calculate power from AC voltage?\n\nInstantaneous Power As in DC circuits, the instantaneous electric power in an AC circuit is given by P=VI where V and I are the instantaneous voltage and current.\n\n## How do you calculate power in watts?\n\nWatts = Amps x Volts Examples: 10 Amps x 120 Volts = 1200 Watts. 5 Amps x 240 Volts = 1200 Watts.\n\n## What is unit power?\n\nThe SI unit of power is Watt (W) which is joules per second (J/s).\n\n## What is power output?\n\nPower Output means the average rate of electric energy delivery during one Metering Interval, converted to an hourly rate of electric energy delivery, in kWh per hour, that is equal to the product of Metered Energy for one Metering Interval, in kWh per Metering Interval, times the number of Metering Intervals in a one- …\n\n## What is power in voltage?\n\nFor any circuit element, the power is equal to the voltage difference across the element multiplied by the current. By Ohm’s Law, V = IR, and so there are additional forms of the electric power formula for resistors. Power is measured in units of Watts (W), where a Watt is equal to a Joule per second (1 W = 1 J/s).\n\n## What is current and power?\n\nRather, power is the combination of both voltage and current in a circuit. Remember that voltage is the specific work (or potential energy) per unit charge, while current is the rate at which electric charges move through a conductor.\n\n## How is it related to V and I?\n\nVoltage is measured in volts, symbolized by the letters “E” or “V”. Current is measured in amps, symbolized by the letter “I”.\n\n## What is calculated with the equation P i * V?\n\nUse it in the same way as the Ohm’s Law triangle: To calculate power, P: put your finger over P, this leaves I V, so the equation is P = I × V. To calculate current, I: put your finger over I, this leaves P over V, so the equation is I = P/\n\n## How do you measure power in a circuit?\n\nMeasuring the power consumption of a circuit is quite straightforward. It all boils down to the equation of P = IV. The voltage supplied to the circuit is quite consistent and subject to minor variations in actual applications.\n\n## What is power in Ohm’s law?\n\nElectrical power, measured in watts, can be calculated using Ohm’s law. The power formula is P = V * I. If given voltage and current, this is easy to calculate by plugging in numbers. By substituting Ohm’s formula, power can be calculated with resistance as well.\n\n## Does Ohm’s law include power?\n\nPower is directly linked to Ohm’s by Joule’s law, which says that the heat produced in resistance is proportional to the square of the current flowing through it over a given time. We can express this as P=V*I and because V=I*R, we get P = I*I*R or P=I2R.\n\n## What is the formula of power in 3 phase?\n\n3-Phase Calculations For 3-phase systems, we use the following equation: kW = (V × I × PF × 1.732) ÷ 1,000.\n\n## What is power in physics simple?\n\npower, in science and engineering, time rate of doing work or delivering energy, expressible as the amount of work done W, or energy transferred, divided by the time interval t—or W/t. A given amount of work can be done by a low-powered motor in a long time or by a high-powered motor in a short time.\n\n## What are the 3 units for power?\n\nName three power units. Units of power are ergs per second (erg/s), horsepower (hp), and foot-pounds per minute.\n\n## What are the 4 units of power?\n\nTherefore, the 4 basic units of electricity are volts, amps, ohms, and watts.\n\n## What is power in DC circuit?\n\nThe power in a DC circuit is the product of the voltage and the current.\n\n## What is power of an AC circuit?\n\nThe average power of an a.c circuit is called the true power of the electrical circuit. Power Factor of an alternating current circuit is the ratio of true power dissipation to the apparent power dissipation in the circuit. Also, c o s ϕ = R Z. The value of the power of an a.c circuit lies between 0 and 1." ]

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[ "# Java Program Linear Search\n\n« Previous Program Next Program »\n\n## Linear Search\n\nTo search any element present inside the array in Java Programming using linear search technique, you have to use only one for loop to check whether the entered number is found in the list or not as shown in the following program.\n\n## Java Programming Code for Linear Search\n\nFollowing Java program first ask to the user to enter the array size then it will ask to enter the array elements, then it will finally ask to enter a number to be search in the given array to check whether it is present in the array or not, if it is present then the program will show the position of that number present in the array:\n\n```/* Java Program Example - Linear Search */\n\nimport java.util.Scanner;\n\npublic class JavaProgram\n{\npublic static void main(String args[])\n{\nint arr[] = new int;\nint i, num, n, c=0, pos=0;\nScanner scan = new Scanner(System.in);\n\nSystem.out.print(\"Enter Array Size : \");\nn = scan.nextInt();\n\nSystem.out.print(\"Enter Array Elements : \");\nfor(i=0; i<n; i++)\n{\narr[i] = scan.nextInt();\n}\n\nSystem.out.print(\"Enter the Number to be Search...\");\nnum = scan.nextInt();\n\nfor(i=0; i<n; i++)\n{\nif(arr[i] == num)\n{\nc = 1;\npos = i+1;\nbreak;\n}\n}\nif(c == 0)\n{\n}\nelse\n{\nSystem.out.print(num+ \" found at position \" + pos);\n}\n}\n}```\n\nWhen the above Java Program is compile and executed, it will produce the following output:\n\n### Same Program in Other Languages\n\nYou may also like to learn and practice the same program in other popular programming languages:\n\nJava Online Test\n\n« Previous Program" ]

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[ "# A quantum algorithm for spin chemistry: a Bayesian exchange coupling parameter calculator with broken-symmetry wave functions.\n\nThe Heisenberg exchange coupling parameter J (H = −2JSi · Sj) characterises the isotropic magnetic interaction between unpaired electrons, and it is one of the most important spin Hamiltonian parameters of multi-spin open shell systems. The J value is related to the energy difference between high-spin and low-spin states, and thus computing the energies of individual spin states are necessary to obtain the J values from quantum chemical calculations. Here, we propose a quantum algorithm,", null, "ayesian e", null, "change coupling parameter calculator with", null, "roken-symmetry wave functions (BxB), which is capable of computing the J value directly, without calculating the energies of individual spin states. The BxB algorithm is composed of the quantum simulations of the time evolution of a broken-symmetry wave function under the Hamiltonian with an additional term jS2, the wave function overlap estimation with the SWAP test, and Bayesian optimisation of the parameter j. Numerical quantum circuit simulations for H2 under a covalent bond dissociation, C, O, Si, NH, OH+, CH2, NF, O2, and triple bond dissociated N2 molecule revealed that the BxB can compute the J value within 1 kcal mol−1 of errors with less computational costs than conventional quantum phase estimation-based approaches.", null, "Updated: January 15, 2021 — 11:42 pm" ]

[ null, "https://www.rsc.org/images/entities/char_0042_0332.gif", null, "https://www.rsc.org/images/entities/char_0078_0332.gif", null, "https://www.rsc.org/images/entities/char_0062_0332.gif", null, "https://pubs.rsc.org/en/Image/Get", null ]

[ "# latex beamer change font size table\n\nOne of the typical problem by using beamer class are tables respectively showing tables. If they are to big to be shown on one frame there are two possibilities to solve this problem.\n\nFrist, use the plain option at the frame where you want to set the table\n\n\\begin{frame}[plain]\n\\frametitle{a table is to big}\n\\begin{table}\n\\begin{tabular}{|r|c|l|}\n\\hline\ntest 1 & test 2 & test 3\\\\\n\\hline\nA & B & C \\\\\n\\hline\n\\dots & \\dots & \\dots \\\\\n\\hline\n\\end{tabular}\n\\end{table}\n\\end{frame}\n\nBy using plain you just get a little bit more space to fill.\n\nSecond, just change the font size within the table:\n\\begin{frame}\n\\frametitle{a table is to big}\n\\begin{tiny}\n\\begin{tabular}{|r|c|l|}\n\\hline\nAAA & BBB & CCC \\\\\n\\hline\n111 & 222 & 333 \\\\\n\\hline\n\\end{tabular}\n\\end{tiny}\n\\end{frame}\n\nNow the font is tiny and has a size of 6 pt because normalsize is 11 pt.\nAn overview about the size and commands you will find here even for other documentclasses.\n\n# latex memoir 8pt\n\nHow to get 8 pt in memoir? In contrast to the default classes, memoir has 12 different sizes for normalsize. Five of them allow you to get a size of 8pt in your document. 9pt 10pt 11pt 12pt 14p\n\n\\documenclass[9pt]{memoir}\n\\begin{document}\n\\small\nthe text now is in 8pt\n\\end{document}\n\n\\documenclass[10pt]{memoir}\n\\begin{document}\n\\footnotesize\nthe text now is in 8pt too\n\\end{document}\n\n\\documenclass[11pt]{memoir}\n\\begin{document}\n\\scriptsize\nthe text now is in 8pt too\n\\end{document}\n\n\\documenclass[12pt]{memoir}\n\\begin{document}\n\\tiny\nthe text is now also in 8pt\n\\end{document}\n\n\\documenclass[14pt]{memoir}\n\\begin{document}\n\\miniscule\nthe text is finally 8pt\n\\end{document}\n\n\n# latex book font size 14 pt\n\nHow to get font size 14 pt by using book as documentclass?\n\nIf you are using the default size i.e. 10 pt, the command \\Large will change the size up to 14 pt.\n\n\\documentclass{book}\n\\begin{document}\n\\Large\nsome text\n\\end{document}\n\n\nThis also will work, if you set normalsize to 11 pt:\n\n\\documentclass[11pt]{book}\n\\begin{document}\n\\Large\nsome text\n\\end{document}\n\n\nBut if you set normalsize to 12 pt then use the command \\large:\n\n\\documentclass[12pt]{book}\n\\begin{document}\n\\large\nsome text\n\\end{document}\n\n\n# what size is 20pt\n\nQuestion: what size is 20 pt?\nAnswer: This depends on the default size you choose for normalsize and the document class you use.\n\nE.g. article, report, book and letter\n\n\\LARGE and normalsize 12pt will be 20 pt\n\n\\huge and normalsize 11pt or 10 pt will be 20 pt\n\nIf you are are looking for more combinations of classes and commands which will lead to a size of 20 pt just take a look here.\n\n# latex font size 20pt\n\nHow to get a document with font size 20pt?\n\nIf your using one of the default documentclass e.g. article with default font size 10 pt then just use the command \\huge and size will be increase to 20 pt.\n\n\\documentclass{article}\n\\begin{document}\n\\huge\ntext text ...\n\\end{document}" ]

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[ "## 视频背景分离\n\nif args.algo == 'MOG2':\nbackSub = cv.createBackgroundSubtractorMOG2()\nelse:\nbackSub = cv.createBackgroundSubtractorKNN()\n\n\n### github / cvzone\n\nCVzone 是一个计算机视觉包，可以让我们轻松运行像人脸检测、手部跟踪、姿势估计等，以及图像处理和其他 AI 功能。它的核心是使用 OpenCV 和 MediaPipe 库。\n\n• 60 FPS Face Detection 脸部检测\n• Hand Tracking 手势跟踪\n• Pose Estimation 人体姿态跟踪\n• Face Mesh Detection 面部网格检测\n• Stack Images 图片堆叠\n• Corner Rectangle 选框显示\n• FPS 帧率显示\n\n## 二值化\n\n### 二元阈值法\n\nret, thresh1 = cv2.threshold(img, 170, 255, cv2.THRESH_BINARY)", null, "", null, "### 自适应阈值\n\n• 自适应阈值均值：阈值是平均值附近区域减去固定的 $C$。\n• 自适应高斯阈值：阈值是邻域值减去常数 $C$ 的高斯加权总和。\nth2 = cv2.adaptiveThreshold(img, 255, cv2.ADAPTIVE_THRESH_GAUSSIAN_C,\ncv2.THRESH_BINARY, 11, 2)\n\n/** @brief 自适应二值化\n*@param _src 要二值化的灰度图\n*@param _dst 二值化后的图\n*@param maxValue 二值化后要设置的那个值\n*@param type 二值化类型（CV_THRESH_BINARY 大于为最大值，CV_THRESH_BINARY_INV 小于为最大值）\n*@param blockSize 块大小（奇数，大于 1）\n*@param delta 差值（负值也可以）\n*/\nvoid cv::adaptiveThreshold(InputArray _src, OutputArray _dst, double maxValue,\nint method, int type, int blockSize, double delta)\n{\nMat src = _src.getMat();\n\n// 原图必须是单通道无符号 8 位\nCV_Assert(src.type() == CV_8UC1);\n\n// 块大小必须大于 1，并且是奇数\nCV_Assert(blockSize % 2 == 1 && blockSize > 1);\nSize size = src.size();\n\n// 构建与原图像相同的图像\n_dst.create(size, src.type());\nMat dst = _dst.getMat();\n\nif (maxValue < 0)\n{\n// 二值化后值小于 0，图像都为 0\ndst = Scalar(0);\nreturn;\n}\n\n// 用于比较的值\nMat mean;\n\nif (src.data != dst.data)\nmean = dst;\n\n// 计算平均值作为比较值\nboxFilter(src, mean, src.type(), Size(blockSize, blockSize),\nPoint(-1, -1), true, BORDER_REPLICATE);\n// 计算高斯分布和作为比较值\nGaussianBlur(src, mean, Size(blockSize, blockSize), 0, 0, BORDER_REPLICATE);\nelse\n\nint i, j;\n\n// 将 maxValue 夹到 [0,255] 的 uchar 范围区间，用作二值化后的值\nuchar imaxval = saturate_cast<uchar>(maxValue);\n\n// 根据二值化类型计算 delta 值\nint idelta = type == THRESH_BINARY ? cvCeil(delta) : cvFloor(delta);\n\n// 计算生成每个像素差对应的值表格，以后查表就可以。但像素差范围为什么是 768，我确实认为 512 已经够了\nuchar tab;\n\nif (type == CV_THRESH_BINARY)\nfor (i = 0; i < 768; i++)\n// i = src[j] - mean[j] + 255\n// i - 255 > -idelta ? imaxval : 0\n// = src[j] - mean[j] + 255 -255 > -idelta ? imaxval : 0\n// = src[j] > mean[j] - idelta ? imaxval : 0\ntab[i] = (uchar)(i - 255 > -idelta ? imaxval : 0);\nelse if (type == CV_THRESH_BINARY_INV)\nfor (i = 0; i < 768; i++)\n// i = src[j] - mean[j] + 255\n// i - 255 <= -idelta ? imaxval : 0\n// = src[j] - mean[j] + 255 - 255 <= -idelta ? imaxval : 0\n// = src[j] <= mean[j] - idelta ? imaxval : 0\ntab[i] = (uchar)(i - 255 <= -idelta ? imaxval : 0);\nelse\n\n// 如果连续，加速运算\nif (src.isContinuous() && mean.isContinuous() && dst.isContinuous())\n{\nsize.width *= size.height;\nsize.height = 1;\n}\n\n// 逐像素计算 src[j] - mean[j] + 255，并查表得到结果\nfor (i = 0; i < size.height; i++)\n{\nconst uchar* sdata = src.data + src.step*i;\nconst uchar* mdata = mean.data + mean.step*i;\nuchar* ddata = dst.data + dst.step*i;\n\nfor (j = 0; j < size.width; j++)\n// 将 [-255, 255] 映射到 [0, 510] 然后查表\nddata[j] = tab[sdata[j] - mdata[j] + 255];\n}\n}\n\n\n### Otsu's Binrisation\n\nret3, th1 = cv2.threshold(img, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)\n\n\n## 降噪\n\ndst = cv2.fastNlMeansDenoisingColored(img, None, 10, 10, 7, 21)\n\n• cv2.fastNlMeansDenoising() - 使用单个灰度图像\n• cv2.fastNlMeansDenoisingColored() - 使用彩色图像。\n• cv2.fastNlMeansDenoisingMulti() - 用于在短时间内捕获的图像序列（灰度图像）\n• cv2.fastNlMeansDenoisingColoredMulti() - 与上面相同，但用于彩色图像。\n\n## 背景颜色检测\n\n### 算法二", null, "code", null, "### K-Means 算法描述\n\n1. 适当选择 c 个类的初始中心；\n2. 在第 k 次迭代中，对任意一个样本，求其到 c 各中心的距离，将该样本归到距离最短的那个中心所在的类；\n3. 利用均值等方法更新该类的中心值；\n4. 对于所有的 C 个聚类中心，如果利用（2）（3）的迭代法更新后，值保持不变，则迭代结束；否则继续迭代。\n\nK-Means 算法试图找到使平凡误差准则函数最小的簇。 当潜在的簇形状是凸面的，簇与簇之间区别较明显，且簇大小相近时，其聚类结果较理想。\n\n1. 聚类中心的个数 K 需要事先给定，但在实际中这个 K 值的选定是非常难以估计的，很多时候，事先并不知道给定的数据集应该分成多少个类别才最合适；\n2. K-Means 需要人为地确定初始聚类中心，不同的初始聚类中心可能导致完全不同的聚类结果。（可以使用 K-Means++ 算法来解决）\n\n## 使用深度学习生成模糊背景\n\ncode", null, "", null, "", null, "If you have any questions or feedback, please reach out", null, "." ]

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[ "# Finding the Factors of number 319\n\nFactoring Calculator will help you find the factors 319 i.e. 1, 11, 29, 319 the numbers that when divided results in a whole number and a zero remainder.\n\nFactors of 319 are 1, 11, 29, 319. There are 4 integers that are factors of 319. The biggest factor of 319 is 319.\n\nEx: 3844 or 5680 or 9240\n\nFactors of:\n\n## Factor Tree of 319 to Calculate the Factors\n\n 319 11 29", null, "Factors of 319 are 1, 11, 29, 319. There are 4 integers that are factors of 319. The biggest factor of 319 is 319.\n\nPositive integers that divides 319 without a remainder are listed below.\n\n• 1\n• 11\n• 29\n• 319\n\n### Factors of 319 in pairs\n\n• 1 × 319 = 319\n• 11 × 29 = 319\n• 29 × 11 = 319\n• 319 × 1 = 319\n\n### Factors of 319 Table\n\nFactorFactor Number\n1one\n11eleven\n29twenty nine\n319three hundred nineteen\n\n### How to find Factors of 319?\n\nAs we know factors of 319 are all the numbers that can exactly divide the number 319 simply divide 319 by all the numbers up to 319 to see the ones that result in zero remainders. Numbers that divide without remainder are factors and in this case below are the factors\n\n1, 11, 29, 319 are the factors and all of them can exactly divide number 319.\n\n### Frequently Asked Questions on Factors of 319\n\n1. How to find factors of number 319?\n\nCheck for the numbers that evenly divide the number 319 and leave a remainder zero.\n\n2. What are the factors of 319?\n\nFactors of 319 are 1, 11, 29, 319.\n\n3. How do you factor 319 on a calculator?\n\nAll you have to do is provide the input value 319 and tap on the enter button next to the input field to obtain the factors." ]

[ null, "https://onlinecalculator.guru/static/factor/factors-of-319.png", null ]

[ "# How do I derive logical statements simply?\n\nFor example if I want to show the equation for $x \\rightarrow y$, using truth tables it is the same as:\n\n$$(\\neg x \\wedge \\neg y) \\lor (\\neg x \\wedge y) \\lor (x \\wedge y)$$\n\nIs there a methodical way to reduce something like this to its simple form $\\neg x \\lor y$?" ]

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[ "# Numerical solution of large-scale Lyapunov equations, Riccati equations, and linear-quadratic optimal control problems\n\n2 DeFI - Shape reconstruction and identification\nInria Saclay - Ile de France, CMAP - Centre de Mathématiques Appliquées - Ecole Polytechnique\n3 POEMS - Propagation des Ondes : Étude Mathématique et Simulation\nInria Saclay - Ile de France, UMA - Unité de Mathématiques Appliquées, CNRS - Centre National de la Recherche Scientifique : UMR7231\nAbstract : We study large-scale, continuous-time linear time-invariant control systems with a sparse or structured state matrix and a relatively small number of inputs and outputs. The main contributions of this paper are numerical algorithms for the solution of large algebraic Lyapunov and Riccati equations and linearquadratic optimal control problems, which arise from such systems. First, we review an alternating direction implicit iteration-based method to compute approximate low-rank Cholesky factors of the solution matrix of large-scale Lyapunov equations, and we propose a refined version of this algorithm. Second, a combination of this method with a variant of Newton's method (in this context also called Kleinman iteration) results in an algorithm for the solution of large-scale Riccati equations. Third, we describe an implicit version of this algorithm for the solution of linear-quadratic optimal control problems, which computes the feedback directly without solving the underlying algebraic Riccati equation explicitly. Our algorithms are efficient with respect to both memory and computation. In particular, they can be applied to problems of very large scale, where square, dense matrices of the system order cannot be stored in the computer memory. We study the performance of our algorithms in numerical experiments.\nDocument type :\nJournal articles\nDomain :\n\nhttps://hal.inria.fr/hal-00781119\nContributor : Jing-Rebecca Li Connect in order to contact the contributor\nSubmitted on : Friday, January 25, 2013 - 1:44:44 PM\nLast modification on : Thursday, January 14, 2021 - 11:56:04 AM\n\n### Citation\n\nPeter Benner, Jing-Rebecca Li, Thilo Penzl. Numerical solution of large-scale Lyapunov equations, Riccati equations, and linear-quadratic optimal control problems. Numerical Linear Algebra with Applications, Wiley, 2008, 15 (9), pp.755-777. ⟨10.1002/nla.622⟩. ⟨hal-00781119⟩\n\nRecord views" ]

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[ "NIPS Proceedingsβ\n\nEfficient Minimization of Decomposable Submodular Functions\n\n[PDF] [BibTeX]\n\nAbstract\n\nMany combinatorial problems arising in machine learning can be reduced to the problem of minimizing a submodular function. Submodular functions are a natural discrete analog of convex functions, and can be minimized in strongly polynomial time. Unfortunately, state-of-the-art algorithms for general submodular minimization are intractable for practical problems. In this paper, we introduce a novel subclass of submodular minimization problems that we call decomposable. Decomposable submodular functions are those that can be represented as sums of concave functions applied to linear functions. We develop an algorithm, SLG, that can efficiently minimize decomposable submodular functions with tens of thousands of variables. Our algorithm exploits recent results in smoothed convex minimization. We apply SLG to synthetic benchmarks and a joint classification-and-segmentation task, and show that it outperforms the state-of-the-art general purpose submodular minimization algorithms by several orders of magnitude." ]

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[ "# A brief introduction to Haskell\n\n• A language developed by the programming languages research community.\n• Is a lazy, purely functional language (that also has imperative features such as side effects and mutable state, along with strict evaluation)\n• Born as an open source vehicle for programming language research\n• One of the youngest children of ML and Lisp\n• Particularly useful for programs that manipulate data structures (such as compilers and interpreters), and for concurrent/parallel programming\n\nInspired by the article Introduction to OCaml, and translated from the OCaml by Don Stewart.\n\n## Background\n\n• 1990. Haskell 1.0\n• 1991. Haskell 1.1\n• 1993. Haskell 1.2\n• 1996. Haskell 1.3\n• 1997. Haskell 1.4\n• 1998. Haskell 98\n• 2000-2006. Period of rapid language and community growth\n• ~2007. Haskell Prime\n• 2009. Haskell 2010\n\nHas some novel features relative to Java (and C++).\n\n• Immutable variables by default (mutable state programmed via monads)\n• Pure by default (side effects are programmed via monads)\n• Lazy evaluation: results are only computed if they're required (strictness optional)\n• Everything is an expression\n• First-class functions: functions can be defined anywhere, passed as arguments, and returned as values.\n• Both compiled and interpreted implementations available\n• Full type inference -- type declarations optional\n• Pattern matching on data structures -- data structures are first class!\n• Parametric polymorphism\n• Bounded parametric polymorphism\n\nThese are all conceptually more advanced ideas.\n\nCompared to similar functional languages, Haskell differs in that it has support for:\n\n• Lazy evaluation\n• Pure functions by default\n• Monadic side effects\n• Type classes\n• Syntax based on layout\n\nThe GHC Haskell compiler, in particular, provides some interesting extensions:\n\n• Generalised algebraic data types\n• Impredicative types system\n• Software transactional memory\n• Parallel, SMP runtime system\n\n## The Basics\n\nRead the language definition to supplement these notes. For more depth and examples see the Haskell wiki.\n\n### Interacting with the language\n\nHaskell is both compiled and interpreted. For exploration purposes, we'll consider interacting with Haskell via the GHCi interpreter:\n\n• expressions are entered at the prompt\n• newline signals end of input\n\nHere is a GHCi session, starting from a UNIX prompt.\n\n``` \\$ ghci\n___ ___ _\n/ _ \\ /\\ /\\/ __(_)\n/ /_\\// /_/ / / | | GHC Interactive, version 6.4.2, for Haskell 98.\n/ /_\\\\/ __ / /___| | http://www.haskell.org/ghc/\n\\____/\\/ /_/\\____/|_| Type :? for help.\n```\n``` Loading package base-1.0 ... linking ... done.\n```\n\nHere the incredibly simple Haskell program `let x = 3+4` is compiled and loaded, and available via the variable `x`.\n\n```Prelude> let x = 3 + 4\n```\n\nWe can ask the system what type it automaticaly inferred for our variable. `x :: Integer` means that the variable `x` \"has type\" `Integer`, the type of unbounded integer values.\n\n```Prelude> :t x\nx :: Integer\n```\n\nA variable evaluates to its value.\n\n```Prelude> x\n7\n```\n\nThe variable `x` is in scope, so we can reuse it in later expressions.\n\n```Prelude> x + 4\n11\n```\n\nLocal variables may be bound using `let`, which declares a new binding for a variable with local scope.\n\n```Prelude> let x = 4 in x + 3\n7\n```\n\nAlternatively, declarations typed in at the top level are like an open-ended let:\n\n```Prelude> let x = 4\nPrelude> let y = x + 3\nPrelude> x * x\n16\n\nPrelude> :t x\nx :: Integer\nPrelude> :t y\ny :: Integer\nPrelude> :t x * x\nx * x :: Integer\n```\n\nNotice that type inference infers the correct type for all the expressions, without us having to ever specify the type explicitly.\n\n## Basic types\n\nThere is a range of basic types, defined in the language Prelude.\n\n```Int -- bounded, word-sized integers\nInteger -- unbounded integers\nDouble -- floating point values\nChar -- characters\nString -- strings\n() -- the unit type\nBool -- booleans\n[a] -- lists\n(a,b) -- tuples / product types\nEither a b -- sum types\nMaybe a -- optional values\n```\n\nFor example:\n\n```7\n12312412412412321\n3.1415\n'x'\n()\nTrue, False\n[1,2,3,4,5]\n('x', 42)\nLeft True, Right \"string\"\nNothing, Just True\n```\n\nThese types have all the usual operations on them, in the standard libraries.\n\n### Libraries\n\n• The Prelude contains the core operations on basic types. It is imported by default into every Haskell module. For example;\n```+ - div mod && || not < > == /=\n```\n\nLearn the Prelude well. Less basic functions are found in the standard libraries. For data structures such as List, Array and finite maps.\n\nTo use functions from these modules you have to import them, or in GHCi, refer to the qualified name, for example to use the toUpper function on Chars:\n\n```Prelude> Char.toUpper 'x'\n'X'\n\nPrelude> :m + Char\nPrelude Char> toUpper 'y'\n'Y'\n```\n\nIn a source file, you have to import the module explicitly:\n\n```import Char\n```\n\nHaskell uses typeclasses to methodically allow overloading. A typeclass describes a set of functions, and any type which provides those functions can be made an instance of that type. This avoids the syntactic redundancy of languages like OCaml.\n\nFor example, the function `*` is a method of the typeclass `Num`, as we can see from its type:\n\n```Prelude> :t (*)\n(*) :: (Num a) => a -> a -> a\n```\n\nWhich can be read as \"* is a polymorphic function, taking two values of some type 'a', and returning a result of the same type, where the type 'a' is a member of the class Num\".\n\nThis means that it will operate on any type in the `Num` class, of which the following types are members: `Double, Float, Int, Integer`. Thus:\n\n```Prelude> 2.3 * 5.7\n13.11\n```\n\nor on integers:\n\n```Prelude> 2 * 5\n10\n```\n\nThe type of the arguments determines which instance of `*` is used. Haskell also never performs implicit coercions, all coercions must be explicit. For example, if we try to multiply two different types, then the type check against `* :: Num a => a -> a -> a` will fail.\n\n```Prelude> (2.3 :: Double) * (5 :: Int)\n\n<interactive>:1:19:\nCouldn't match `Double' against `Int'\nExpected type: Double\nInferred type: Int\nIn the expression: 5 :: Int\nIn the second argument of `(*)', namely `(5 :: Int)'\n```\n\nTo convert 5 to a `Double` we'd write:\n\n```Prelude> (2.3 :: Double) * fromIntegral (5 :: Int)\n11.5\n```\n\nWhy bother -- why not allow the system to implicitly coerce types? Implicit type conversions by the system are the source of innumerable hard to find bugs in languages that support them, and makes reasoning about a program harder, since you must apply not just the language's semantics, but an extra set of coercion rules.\n\nNote that If we leave off the type signatures however, Haskell will helpfully infer the most general type:\n\n```Prelude> 2.3 * 5\n11.5\n```\n\n### Expressions\n\nIn Haskell, expressions are everything. There are no pure \"statements\" like in Java/C++. For instance, in Haskell, `if-then-else` is a kind of expression, and results in a value based on the condition part.\n\n```Prelude> (if 2 == 3 then 5 else 6) + 1\n7\n```\n```Prelude> (if 2 == 3 then 5 else 6.5) + 1\n7.5\n```\n\n### Local bindings\n\nIn Haskell `let` allows local declarations to be made in the context of a single expression.\n\n```let x = 1 + 2 in x + 3\n```\n\nis analogous to:\n\n``` {\nint x = 1 + 2;\n... x + 3 ... ;\n}\n```\n\nin C, but the Haskell variable x is given a value that is immutable (can never change).\n\n### Allocation\n\nWhen you declare a new variable, Haskell automatically allocates that value for you -- no need to explicitly manage memory. The garbage collector will then collect any unreachable values once they go out of scope.\n\nAdvanced users can also manage memory by hand using the foreign function interface.\n\n### Lists\n\nLists are ... lists of Haskell values. Defining a new list is trivial, easier than in Java.\n\n```Prelude> [2, 1+2, 4]\n[2,3,4]\n\nPrelude> :t [2, 1+2, 4]\n[2, 1+2, 4] :: (Num a) => [a]\n```\n\nThis automatically allocates space for the list and puts in the elements. Haskell is garbage-collected like Java so no explicit de-allocation is needed. The type of the list is inferred automatically. All elements of a list must be of the same type.\n\n```Prelude> [\"e\", concat [\"f\", \"g\"], \"h\"]\n[\"e\",\"fg\",\"h\"]\n```\n\nNotice how the function call `concat [\"f\",\"g\"]` does not require parenthesis to delimit the function's arguments. Haskell uses whitespace, and not commas, and:\n\n• You don't need parentheses for function application in Haskell: `sin 0.3`\n• Multiple arguments can be passed in one at a time (curried) which means they can be separated by spaces: `max 3 4`.\n\nLists must be uniform in their type (\"homogeneous\").\n\n```Prelude> ['x', True]\n\nCouldn't match `Char' against `Bool'\n```\n\nHere we tried to build a list containing a Char and a Boolean, but the list constructor, `[]`, has type:\n\n```Prelude> :t []\n[] :: [a]\n```\n\nmeaning that all elements must be of the same type, `a`. (For those wondering how to build a list of heterogeneous values, you would use a sum type):\n\n```Prelude> [Left 'x', Right True]\n[Left 'x',Right True]\n\nPrelude> :t [Left 'x', Right True]\n[Left 'x', Right True] :: [Either Char Bool]\n```\n\nList operations are numerous, as can be seen in the Data.List library.\n\n```Prelude> let x = [2,3]\nPrelude> let y = 1 : x -- 'cons' the value 1 onto the list\nPrelude> x -- the list is immutable\n[2,3]\nPrelude> y\n[1,2,3]\nPrelude> x ++ y -- joining lists\n[2,3,1,2,3]\nPrelude> head y -- first element of the list is the 'head'\n1\nPrelude> tail y -- the rest of the list is the 'tail'\n[2,3]\n```\n\n### Pattern matching\n\nHaskell supports pattern matching on data structures. This is a powerful language feature, making code that manipulates data structures incredibly simple. The core language feature for pattern matching is the `case` expression:\n\n```Prelude> case x of h:t -> h\n2\n```\n\nThe `case` forces `x` (the scrutinee) to match pattern `h:t`, a list with head and tail, and then we extract the head, `h`. Tail is similar, and we can use a wildcard pattern to ignore the part of the pattern we don't care about:\n\n```Prelude> case x of _:t -> t\n\n```\n\n### Tuples\n\nTuples are fixed length structures, whose fields may be of differing types (\"heterogeneous\"). They are known as product types in programming language theory.\n\n```Prelude> let x = (2, \"hi\")\nPrelude> case x of (y,_) -> y\n2\n```\n\nUnlike the ML family of languages, Haskell uses the same syntax for the value level as on the type level. So the type of the above tuple is:\n\n```Prelude> :t x\nx :: (Integer, [Char])\n```\n\nAll the data mentioned so far are immutable - it is impossible to change an entry in an existing list, tuple, or record without implicitly copying the data! Also, all variables are immutable. By default Haskell is a pure language. Side effects, such as mutation, are discussed later.\n\n## Functions\n\nHere is a simple recursive factorial function definition.\n\n```Prelude> let fac n = if n == 0 then 1 else n * fac (n-1)\n\nPrelude> :t fac\nfac :: (Num a) => a -> a\n\nPrelude> fac 42\n1405006117752879898543142606244511569936384000000000\n```\n\nThe function name is `fac`, and the argument is `n`. This function is recursive (and there is no need to specially tag it as such, as you would in the ML family of languages).\n\nWhen you apply (or invoke) the fac function, you don't need any special parenthesis around the code. Note that there is no return statement; instead, the value of the whole body-expression is implicitly what gets returned.\n\nFunctions of more than one argument may be defined:\n\n```Prelude> let max a b = if a > b then a else b\nPrelude> max 3 7\n7\n```\n\nOther important aspects of Haskell functions:\n\n• Functions can be defined anywhere in the code via lambda abstractions:\n```Prelude> ((\\x -> x + 1) 4) + 7\n12\n```\n\nOr, identical to `let f x = x + 1`:\n\n```Prelude> let f = \\x -> x + 1\nPrelude> :t f\nf :: Integer -> Integer\n```\n\nAnonymous functions like this can be very useful. Also, functions can be passed to and returned from functions. For example, the higher order function `map`, applies its function argument to each element of a list (like a for-loop):\n\n```Prelude> map (\\x -> x ^ 2) [1..10]\n[1,4,9,16,25,36,49,64,81,100]\n```\n\nIn Haskell, we can use section syntax for more concise anonymous functions:\n\n```Prelude> map (^ 2) [1..10]\n[1,4,9,16,25,36,49,64,81,100]\n```\n\nHere `map` takes two arguments, the function `(^2) :: Integer -> Integer`, and a list of numbers.\n\n### Currying\n\nCurrying is a method by which function arguments may be passed one at a time to a function, rather than passing all arguments in one go in a structure:\n\n```Prelude> let comb n m = if m == 0 || m == n then 1 else comb (n-1) m + comb (n-1) (m-1)\n\nPrelude> comb 10 4\n210\n```\n\nThe type of comb, `Num a => a -> a -> a`, can be rewritten as `Num a => a -> (a -> a)`. That is, it takes a single argument of some numeric type `a`, and returns a function that takes another argument of that type!\n\nIndeed, we can give comb only one argument, in which case it returns a function that we can later use:\n\n```Prelude> let comb10 = comb 10\nPrelude> comb10 4\n210\nPrelude> comb10 3\n120\n```\n\nMutually recursive functions may be defined in the same way as normal functions:\n\n```let take [] = []\n(x:xs) = x : skip xs\nskip [] = []\n(_:ys) = take ys\n\nPrelude> :t take\ntake :: [a] -> [a]\n\nPrelude> :t skip\nskip :: [a] -> [a]\n\nPrelude> take [1..10]\n[1,3,5,7,9]\n\nPrelude> skip [1..10]\n[2,4,6,8,10]\n```\n\nThis example also shows a pattern match with multiple cases, either empty list or nonempty list. More on patterns now.\n\n### Patterns\n\nPatterns make function definitions much more succinct, as we just saw.\n\n```let rev [] = []\nrev (x:xs) = rev xs ++ [x]\n```\n\nIn this function definition, `[]` and `(x:xs)` are patterns against which the value passed to the function is matched. The match occurs on the structure of the data -- that is, on its constructors.\n\nLists are defined as:\n\n```data [] a = [] | a : [a]\n```\n\nThat is, a list of some type a has type `[a]`, and it can be built two ways:\n\n• either the empty list, `[]`\n• or an element consed onto a list, such as `1 : []` or `1 : 2 : 3 : []`.\n• For the special case of lists, Haskell provides the syntax sugar: `[1,2,3]` to build the same data.\n\nThus, `[]` matches against the empty list constructor, and `(x:xs)`, match against the cons constructor, binding variables x and xs to the head and tail components of the list.\n\nRemember that `case` is the syntactic primitive for performing pattern matching (pattern matching in let bindings is sugar for `case`). Also, the first successful match is taken if more than one pattern matches:\n\n```case [1,2,3] of\n(x:y) -> True\n(x:y:z) -> False\n[] -> True\n\nWarning: Pattern match(es) are overlapped\nIn a case alternative: (x : y : z) -> ...\n\nTrue\n```\n\nWarnings will be generated at compile time if patterns don't cover all possibilities, or contain redundant branches.\n\n```Prelude> :set -Wall\nPrelude> case [1,2,3] of (x:_) -> x\n\nWarning: Pattern match(es) are non-exhaustive\nIn a case alternative: Patterns not matched: []\n\n1\n```\n\nAn exception will be thrown at runtime if a pattern match fails:\n\n```Prelude> let myhead (x:_) = x\n\nWarning: Pattern match(es) are non-exhaustive\nIn a case alternative: Patterns not matched: []\n\n*** Exception: <interactive>:1:16-36: Non-exhaustive patterns in case\n```\n\nAs we have seen, patterns may be used in function definitions. For example, this looks like a function of two arguments, but its a function of one argument which matches a pair pattern.\n\n```Prelude> let add (x,y) = x + y\n5\n```\n\n### Immutable declarations\n\nImmutable Declarations\n\n• Important feature of let-defined variable values in Haskell (and some other functional languages): they cannot change their value later.\n• Greatly helps in reasoning about programs---we know the variable's value is fixed.\n• Smalltalk also forces method arguments to be immutable; C++'s const and Java's final on fields has a similar effect.\n```let x = 5 in\nlet f y = x + 1 in\nlet x = 7 in f 0 -- f uses the particular x in lexical scope where it is defined\n6\n```\n\nHere's the one that will mess with your mind: the same thing as above but with the declarations typed into GHCi. (The GHCi environment conceptually an open-ended series of lets which never close).\n\n```Prelude> let x = 5\nPrelude> let f y = x + 1\nPrelude> f 0\n6\nPrelude> let x = 7 -- not an assignment, a new declaration\nPrelude> f 0\n6\n```\n\n### Higher order functions\n\nHaskell, like ML, makes wide use of higher-order functions: functions that either take other functions as argument or return functions as results, or both. Higher-order functions are an important component of a programmer's toolkit.\n\n• They allow \"pluggable\" programming by passing in and out chunks of code.\n• Many new programming design patterns are possible.\n• It greatly increases the reusability of code.\n• Higher-order + Polymorphic = Reusable\n\nThe classic example of a function that takes another function as argument is the `map` function on lists. It takes a list and a function and applies the function to every element of the list.\n\n```map :: (a -> b) -> [a] -> [b]\nmap _ [] = []\nmap f (x:xs) = f x : map f xs\n```\n\nThe lower case variables in the type declaration of map are type variables, meaning that the function is polymorphic in that argument (can take any type).\n\n```Prelude> map (*10) [4,2,7]\n[40,20,70]\n```\n\nPerhaps the simplest higher-order function is the composer, in mathematics expressed as g o f. it takes two functions and returns a new function which is their composition:\n\n```(.) :: (b -> c) -> (a -> b) -> a -> c\n(.) f g x = f (g x)\n```\n\nThis function takes three arguments: two functions, f and g, and a value, x. It then applies g to x, before applying f to the result. For example:\n\n```Prelude> let plus3times2 = (*2) . (+3)\nPrelude> plus3times2 10\n26\n```\n\nAs we have seen before, functions are just expressions so can also be immediately applied after being defined:\n\n```Prelude> ((*2) . (+3)) 10\n26\nPrelude> (.) (*2) (+3) 10\n26\n```\n\nNote how Haskell allows the infix function . to be used in prefix form, when wrapped in parenthesis.\n\n### Currying\n\nCurrying is an important concept of functional programming; it is named after logician Haskell Curry, after which the languages Haskell and Curry are named! Multi-argument functions as defined thus far are curried, lets look at what is really happening.\n\nHere is a two-argument function defined in our usual manner.\n\n```Prelude> let myadd x y = x + y\nPrelude> myadd 3 4\n7\n```\n\nHere is another completely equivalent way to define the same function:\n\n```Prelude> let myadd x = \\y -> x + y\nmyadd :: (Num a) => a -> a -> a\n\nPrelude> myadd 3 4\n7\nPrelude> let inc3 = myadd 3\nPrelude> inc3 4\n7\n```\n\nThe main observation is myadd is a function returning a function, so the way we supply two arguments is\n\n• Invoke the function, get a function back\n• Then invoke the returned function passing the second argument.\n• Our final value is returned, victory.\n• `(myadd 3) 4` is an inlined version of this where the function returned by `myadd 3` is not put in any variable\n\nHere is a third equivalent way to define myadd, as an anonymous function returning another anonymous function.\n\n```Prelude> let myadd = \\x -> \\y -> x + y\nmyadd :: Integer -> Integer -> Integer\n```\n\nWith currying, all functions \"really\" take exactly one argument. Currying also naturally arises when functions return functions, as in the map application above showed. Multiple-argument functions should always be written in curried form; all the library functions are curried.\n\nNote thus far we have curried only two-argument functions; in general, n-argument currying is possible. Functions can also take pairs as arguments to achieve the effect of a two-argument function:\n\n```Prelude> let mypairadd (x,y) = x + y\n5\n```\n\nSo, either we can curry or we can pass a pair. We can also write higher-order functions to switch back and forth between the two forms.\n\n```fst :: (a,b) -> a\nfst (x,_) = x\n\nsnd :: (a,b) -> b\nsnd (_,y) = y\n\ncurry :: ((a, b) -> c) -> a -> b -> c\ncurry f x y = f (x, y)\n\nuncurry :: (a -> b -> c) -> ((a, b) -> c)\nuncurry f p = f (fst p) (snd p)\n\nPrelude> :t uncurry myadd\nuncurry myadd :: (Integer, Integer) -> Integer\n\nPrelude> :t curry mypairadd\ncurry mypairadd :: Integer -> Integer -> Integer\n\nPrelude> :t uncurry map\nuncurry map :: (a -> b, [a]) -> [b]\n\nPrelude> :t curry (uncurry myadd) -- a no-op\ncurry (uncurry myadd) :: Integer -> Integer -> Integer\n```\n\nLook at the types: these mappings in both directions in some sense \"implement\" the well-known isomorphism on sets: `A * B -> C = A -> B -> C`\n\n### A bigger example\n\nHere is a more high-powered example of the use of currying.\n\n```foldr f [] y = y\nfoldr f (x:xs) y = f x (foldr f xs y)\n\n*Main> :t foldr\nfoldr :: (a -> t -> t) -> [a] -> t -> t\n\n*Main> let prod = foldr (\\a x -> a * x)\n*Main> :t prod\nprod :: [Integer] -> Integer -> Integer\n\n*Main> let prod0 = prod [1,2,3,4]\n*Main> :t prod0\nprod0 :: Integer -> Integer\n\n*Main> (prod0 1, prod0 2)\n(24,48)\n```\n\nHere is an analysis of this recursive function, for the arbitrary 2-element list [x1,x2], the call\n\n```foldr f [x1, x2] y\n```\n\nreduces to (by inlining the body of fold):\n\n```f x1 (foldr f [x2] y)\n```\n\nwhich in turn reduces to:\n\n```f x1 (f x2 (foldr f [] y)))\n```\n\nand then:\n\n```f x1 (f x2 y)\n```\n\nFrom this we can assert that the general result returned from `foldr f [x1,x2,...,xn] y` is `f x1 (f x2 f ...(f xn y)...))))`. Currying allows us to specialize foldr to a particular function f, as with prod above.\n\n### Proving program properties by induction\n\nWe should in fact be able to prove this property by induction. Its easier if we reverse the numbering of the list.\n\nLemma. `foldr f [xn,...,x1] y` evaluates to `f xn (f xn-1 f ...(f x1 y)...)))` for n greater than 0.\n\nProof. Proceed by induction on the length of the list `[xn,..,x1]`.\n\nBase Case n=1, i.e. the list is [x1]. The function reduces to `f x1 (foldr f [] y)` which reduces to `f x1 y` as hypothesized.\n\nInduction Step. Assume\n\n```foldr f [xn,...,x1] y\n```\n\nreduces to\n\n```f xn (f xn-1 f ...(f x1 y)...)))\n```\n\nand show\n\n```foldr f [xn+1, xn,...,x1] y\n```\n\nreduces to\n\n```f xn+1 (f xn f ...(f x1 y)...))))\n```\n\nComputing\n\n```foldr f [x1,x2,...,xn,xn+1] y\n```\n\nit matches the pattern with x being xn+1 and xs being `[xn,...,x1]`. Thus the recursive call is\n\n```foldr f [xn,...,x1] y\n```\n\nwhich by our inductive assumption reduces to\n\n```f xn (f xn-1 f ...(f x1 y)...)))\n```\n\nAnd, given this result for the recursive call, the whole function then returns\n\n```f xn+1 (...result of recursive call...)\n```\n\nwhich is\n\n```f xn+1 (f xn (f xn-1 f ...(f x1 y)...)))\n```\n\nwhich is what we needed to show. QED\n\nThe above implementation is inefficient in that f is explicitly passed to every recursive call. Here is a more efficient version with identical functionality.\n\n```foldr :: (a -> b -> b) -> b -> [a] -> b\nfoldr k z xs = go xs\nwhere\ngo [] = z\ngo (y:ys) = y `k` go ys\n```\n\nThis function also illustrates how functions may be defined in a local scope, using `where`. Observe 'go' is defined locally but then exported since it is the return value of f.\n\nQuestion: How does the return value 'go' know where to look for k when its called??\n\n```Prelude> let summate = foldr (+)\nPrelude> summate [1,2,3,4] 0\n10\n```\n\n`summate` is just `go` but somehow it \"knows\" that k is `(+)`, even though k is undefined at the top level:\n\n```Prelude> k\n<interactive>:1:0: Not in scope: `k'\n```\n\n`go` in fact knew the right k to call, so it must have been kept somewhere: in a closure. At function definition point, the current values of variables not local to the function definition are remembered in a structure called a closure. Function values in Haskell are thus really a pair consisting of the function (pointer) and the local environment, in a closure.\n\nWithout making a closure, higher-order functions will do awkward things (such as binding to whatever 'k' happens to be in scope). Java, C++, C can pass and return function (pointers), but all functions are defined at the top level so they have no closures.\n\nYou should never type large amounts of code directly into GHCi! Its impossible to fix errors. Instead, you should edit in a file. Usingg any editor, save each group of interlinked functions in a separate file, for example \"A.hs\". Then, from GHCi type:\n\n```Prelude> :l A.hs\n*Main>\n```\n\nThis will compile everything in the file.\n\n### Show\n\nHaskell has the show function.\n\n```*Main> show 1\n\"1\"\n*Main> show (1,2,'x')\n\"(1,2,'x')\"\n*Main> show (Just ())\n\"Just ()\"\n```\n\nIt simply returns a string representation for its arguments.\n\n## Types\n\nWe have generally been ignoring the type system of Haskell up to now. Its time to focus on typing in more detail.\n\n### Type Declarations\n\nHaskell infers types for you, but you can add explicit type declarations if you like.\n\n```let myadd :: Int -> Int -> Int\nmyadd x y = x + y\n\nmyadd :: Int -> Int -> Int\n\n*Main> :set -fglasgow-exts\nlet myadd (x :: Int) (y :: Int) = x + y\n\nmyadd :: Int -> Int -> Int\n```\n\nYou can in fact put type assertions on any variable in an expression to clarify what type the variable has:\n\n```*Main> let myadd (x :: Int) (y :: Int) = (x :: Int) + y\nmyadd :: Int -> Int -> Int\n```\n\n### Type synonyms\n\nYou can also make up your own name for any type. To do this, you must work in a separate file and load it into GHCi using the \":load A.hs\" command.\n\n```type IntPair = (Int,Int)\n\nf :: IntPair -> Int\nf (l,r) = l + r\n```\n\nWorking from GHCi:\n\n```f :: IntPair -> Int\n*Main> :l A.hs\n*Main> :t f\nf :: IntPair -> Int\n*Main> f (2,3)\n5\n```\n\n### Polymorphic Types and Type Inference\n\n```*Main> let id x = x\n*Main> :t id\nid :: forall t. t -> t\n*Main> id 3\n3\n*Main> id True\nTrue\n```\n\nSince `id` was not used as any type in particular, the type of the function is polymorphic (\"many forms\").\n\n• `t` is a type variable, meaning it stands for some arbitrary type.\n• Polymorphism is really needed with type inference -- inferring `Int -> Int` would not be completely general.\n\n### Parametric polymorphism\n\nThe form of polymorphism in Haskell is to be precise, parametric polymorphism. The type above is parametric in t: what comes out is the same type as what came in. Generics is another term for parametric polymorphism used in some communities.\n\n• Java has no parametric polymorphism, but does have object polymorphism (unfortunately this is often just called polymorphism by some writers) in that a subclass object can fit into a superclass-declared variable.\n• When you want parametric polymorphism in Java you declare the variable to be of type Object, but you have to cast when you get it out which requires a run-time check.\n• The Java JDK version 1.5 will have parametrically polymorphic types in it.\n\nThe general intuition to have about the type inference algorithm is everything starts out as having arbitrary types, t, u, etc, but then the use of functions on values generates constraints that \"this thing has the same type as that thing\".\n\nUse of type-specific operators obviously restricts polymorphism:\n\n```*Main> let doublenegate x = not (not x)\n*Main> :t doublenegate\ndoublenegate :: Bool -> Bool\n```\n\nWhen a function is defined via let to have polymorphic type, every use can be at a different type:\n\n```let id x = x\nin case id True of\nTrue -> id 3\nFalse -> id 4\n3\n```\n\n### Algebraic Data Types\n\nAlgebraic data types in Haskell are the analogue of union/variant types in C/Pascal. Following in the Haskell tradition of lists and tuples, they are not mutable. Haskell data types must be declared. Here is a really simple algebraic data type declaration to get warmed up, remember to write this in a separate file, and load it in to HHCi:\n\n```data Height = Tall | Medium | Short\n```\n\nThree constructors have been defined. These are now official constants. Constructors must be capitalized, and variables must be lower-case in Haskell.\n\n```*Main> :l A.hs\n*Main> :t Tall\nTall :: Height\n*Main> Tall\n\nTop level:\nNo instance for (Show Height)\n```\n\nSo we can type check them, but can't show them yet. Let's derive the typeclass `Show` for our data type, which generates a 'show' function for our data type, which GHCi can then use to display the value.\n\n```data Height = Tall | Medium | Short\nderiving Show\n\n*Main> Tall\nTall\n```\n\nThe previous type is only an enumerated type. Much more interesting data types can be defined. Remember the (recursive) list type:\n\n```data [] a = [] | a : [a]\n\n*Main> (:) 3 ((:) (3+1) [])\n[3,4]\n*Main> 3 : 3+1 : []\n[3,4]\n```\n\nThis form of type has several new features:\n\n• As in C/Pascal, the data types can have values and they can be recursively defined, plus,\n• Polymorphic data types can be defined; a here is a type argument.\n• Note how there is no need to use pointers in defining recursive variant types. The compiler does all that mucking around for you.\n• Also note how `(:)`, the constructor, can be used as a function.\n\nWe can define trees rather simply:\n\n```data Tree a = Empty | Node a (Tree a) (Tree a)\n```\n\nPatterns automatically work for new data types.\n\n### Record Declarations\n\nRecords are data types with labels on fields. They are very similar to structs of C/C++. Their types are declared just like normal data types, and can be used in pattern matches.\n\n```data OneTwo = OneTwo { one :: Int, two :: String }\nderiving Show\n\n*Main> let x = OneTwo { one = 2 , two = \"ni\" }\n*Main> one x\n2\n*Main> case x of OneTwo { one = x, two = s } -> x\n2\n```\n\n## Imperative Programming\n\nHaskell and OCaml differ on imperative programming: OCaml mixes pure and impure code, while Haskell separates them statically.\n\nThe expressions and functions for I/O, mutable states, and other side effects have a special type. They enjoy a distinguished status: they are I/O instructions, and the entry point of each complete program must be one of them. The following honours this distinction by using the word command for them (another popular name is action), though they are also expressions, values, functions.\n\nCommands have types of the form `IO a`, which says it takes no parameter and it gives an answer of type `a`. (I will carefully avoid saying it “returns” type `a`, since “return” is too overloaded.) There are also functions of type `b -> IO a`, and I will abuse terminology and call this a command needing a parameter of type `b`, even though the correct description should be: a function from `b` to commands.\n\n### I/O\n\nThe command for writing a line to Standard Output is\n\n```putStrLn :: String -> IO ()\n```\n\nIt outputs the string parameter, plus linebreak. And since there is no answer to give, the answer type is the most boring `()`.\n\nAt first, using output commands at the prompt is as easy as using expressions.\n\n```Prelude> putStrLn \"Hello world\"\nHello world\n```\n\nYou can also write a compound command with the `>>` operator.\n\n```Prelude> putStrLn \"Hello\" >> putStrLn \"world\"\nHello\nworld\n```\n\nThe fun begins when you also take input. The command for reading one line from Standard Input is:\n\n```getLine :: IO String\n```\n\nNote that the type is not `String` or `()->String`. In a purely functional language, such types promise that all calls using the same parameter yield the exactly same string. This is of course what an input command cannot promise. If you read two lines, they are very likely to be different. The type `IO String` does not promise to give the same string all the time. (It only promises to be the same command all the time—a rather \"duh\" one.) But this poses a question: how do we get at the line it reads?\n\nA non-solution is to expect an operation `stripIO :: IO a -> a`. What's wrong with this strip-that-IO mentality is that it asks to convert a command, which gives different answers at different calls, into a pure function, which gives the same answer at different calls. Contradiction!\n\nBut you can ask for a weaker operation: how to pass the answer on to subsequent commands (e.g., output commands) so they can use it. A moment of thought reveals that this is all you ever need. The operator sought is\n\n```(>>=) :: IO a -> (a -> IO b) -> IO b\n```\n\nIt builds a compound command from two commands, the first one of which takes no parameter and gives an answer of type `a`, and the second of which needs a parameter of type `a`. You guessed it: this operator extracts the answer from the first command and passes it to the second. Now you have some way to use the answer!\n\nHere is the first example. Why don't we read a line and immediately display it? `getLine` answers a string, and `putStrLn` wants a string. Perfect match!\n\n```Prelude> getLine >>= putStrLn\nGood morning\nGood morning\n```\n\nBut more often you want to output something derived from the input, rather than the input itself verbatim. To do this, you customize the second command to smuggle in the derivation. The trick of anonymous functions is very useful for this:\n\n```Prelude> getLine >>= \\s -> putStrLn (\"You entered: \" ++ s)\nGood morning\nYou entered: Good morning\n```\n\nYou will also want to give derived answers, especially if you write subroutines to be called from other code. This is facilitated by the command that takes a parameter and simply gives it as the answer (it is curiously named `return`):\n\n```return :: a -> IO a\n```\n\nFor example here is a routine that reads a line and answers a derived string, with a sample usage:\n\n```Prelude> let mycmd = getLine >>= \\s -> return (\"You entered: \" ++ s)\nPrelude> mycmd >>= putStrLn\nGood morning\nYou entered: Good morning\n```\n\nSome programmers never use Standard Input. Reading files is more common. One command for this is:\n\n```readFile :: String -> IO String\n```\n\nThe parameter specifies the file path. Let us read a file and print out its first 10 characters (wherever available). Of course please change the filename to refer to some file you actually possess.\n\n```Prelude> readFile \"file.txt\" >>= \\s -> putStrLn (take 10 s)\nabcdefghij\n```\n\nDo not worry about slurping up the whole file into memory; `readFile` performs a magic of pay-as-you-go reading.\n\nA while ago I showed the `>>` operator for compound commands without elaboration. I can now elaborate it: it merely uses `>>=` in a way that throws away the first answer:\n\n``` putStrLn \"x\" >> putStrLn \"y\"\n= putStrLn \"x\" >>= \\_ -> putStrLn \"y\"\n```\n\n### do-Notation\n\nTo bring imperative code closer to imperative look, Haskell provides the do-notation, which hides the `>>=` operator and the anonymous functions. An example illustrates this notation well, and it should be easy to generalize:\n\n``` cmd0 >>= \\x -> cmd1 >>= \\_ -> cmd2 >>= \\z -> cmd3\n= do { x <- cmd0; cmd1; z <- cmd2; cmd3 }\n```\n\n(`cmd1`, `cmd2`, and `cmd3` may use `x` as a parameter; similarly, `cmd3` may use `z` as a parameter. At the end, between `cmd3` and `}`, you may choose to insert or omit semicolons; similarly right after `{` at the beginning.)\n\nBelow we re-express examples in the previous section in the do-notation.\n\n```Prelude> do { putStrLn \"Hello\"; putStrLn \"world\" }\nHello\nworld\n\nPrelude> do { s <- getLine; putStrLn s }\nGood morning\nGood morning\n\nPrelude> do { s <- getLine; putStrLn (\"You entered: \" ++ s) }\nGood morning\nYou entered: Good morning\n\nPrelude> let mycmd = do { s <- getLine; return (\"You entered: \" ++ s) }\nPrelude> do { s <- mycmd; putStrLn s }\nGood morning\nYou entered: Good morning\n\nPrelude> do { s <- readFile \"file.txt\"; putStrLn (take 10 s) }\nabcdefghij\n```\n\nAt the prompt it is necessary to write one-liners. In a source code file it is more common to use multiple lines, one line for one command, as per tradition. In this case, layout rules allow omitting `{;}` in favour of indentation. Thus, here are two valid ways of writing the same do-block in a source code file, one with `{;}` and the other with layout.\n\n```foo = do { name <- getLine; foo = do name <- getLine\ns <- readFile name; s <- readFile name\nputStrLn (take 10 s) putStrLn (take 10 s)\n}\n```\n\n### Mutable variables\n\nData.IORef Data.Array.MArray Data.Array.IO\n\n### Exceptions\n\nControl.Exception\n\n### Concurrency\n\nControl.Concurrent Control.Concurrent.MVar\n\n### Mutable variables\n\nData.STRef Data.Array.MArray Data.Array.ST\n\n## Compilation\n\nYou can easily compile your Haskell modules to standalone executables. For example, write this in a file \"A.hs\":\n\n```main = putStrLn \"Hello, World!\"\n```\n\nIn general, `main` is the entry point, and you must define it to be whatever you want run. (TODO: once the monad/IO section is done, this place should also say more about main and IO.)\n\nThe compiler, on unix systems, is ghc. For example \"A.hs\" can be compiled and run as:\n\n``` \\$ ghc A.hs\n\\$ ./a.out\nHello, World!\n```\n\nFor multiple modules, use the --make flag to GHC. Example: write these two modules:\n\nB.hs:\n\n```import M1\nmain = putStrLn s\n```\n\nM1.hs:\n\n```module M1 where\ns = \"Hi, Everyone!\"\n```\n\nTo compile and run (this will automatically look for M1.hs):\n\n``` \\$ ghc --make B.hs\n\\$ ./a.out\nHi, Everyone!\n```\n\nIn general, one and only one file must define `main`. In general, for all other files, the filename must match the module name." ]

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[ "## Monday, May 28, 2007\n\n### One By One, We're Making It Fun. Topic: Calculus/S&S.\n\nTheorem: (Stolz-Cesaro) Let $\\{a_n\\}$ and $\\{b_n\\}$ be sequences of real numbers such that $\\{b_n\\}$ is positive, strictly increasing, and unbounded. If the limit\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} \\frac{a_{n+1}-a_n}{b_{n+1}-b_n} = L$\n\nexists, then the following limit also exists and we have\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} \\frac{a_n}{b_n} = L$.\n\n--------------------\n\nTheorem: (Summation by Parts) If $\\{f_k\\}$ and $\\{g_k\\}$ are two sequences, then\n\n$\\displaystyle \\sum_{k=m}^n f_k(g_{k+1}-g_k) = [f_{n+1}g_{n+1}-f_mg_m]-\\sum_{k=m}^n g_{k+1}(f_{k+1}-f_k)$.\n\n--------------------\n\nProblem: Let $\\{a_n\\}$ be a sequence of real numbers such that $\\displaystyle \\sum_{k=0}^{\\infty} a_k$ converges. Show that $\\displaystyle \\lim_{n \\rightarrow \\infty} \\frac{1}{n+1} \\sum_{k=0}^n k \\cdot a_k = 0$.\n\nSolution: Define the sequence $\\{b_n\\}$ by $\\displaystyle b_n = \\sum_{k=0}^n a_k$ and let $\\displaystyle \\lim_{n \\rightarrow \\infty} b_n = L$. Then, by summation by parts with $\\{f_n\\} = \\{n\\}$ and $\\{g_n\\} = \\{b_n\\}$, we have\n\n$\\displaystyle \\sum_{k=0}^n k \\cdot a_k = \\sum_{k=0}^n k \\cdot (b_{k+1}-b_k) = (n+1)b_{n+1}-\\sum_{k=0}^n b_{k+1}$.\n\nThe summation we wish to take the limit of is then\n\n$\\displaystyle \\frac{1}{n+1} \\sum_{k=0}^n k \\cdot a_k = b_{n+1}-\\frac{1}{n+1} \\sum_{k=0}^n b_{k+1}$.\n\nBut since, by Stolz-Cesaro,\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} \\frac{1}{n+1} \\sum_{k=0}^n b_{k+1} = \\lim_{n \\rightarrow \\infty} b_{n+1} = L$,\n\nwe obtain\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} \\frac{1}{n+1} \\sum_{k=0}^n k \\cdot a_k = \\lim_{n \\rightarrow \\infty} \\left(b_n-\\frac{1}{n+1} \\sum_{k=0}^n b_{k+1}\\right) = L-L = 0$.\n\nQED.\n\n--------------------\n\nComment: Summation by parts is a very useful technique to change sums around so that they are easier to evaluate. If you hadn't noticed, it is the discrete analogue of integration by parts and is in fact very similar. Stolz-Cesaro is powerful as well and seems like a discrete analogue to L'Hopital (but I'm not sure about this one). Applying well-known calculus ideas to discrete things can often turn into neat results.\n\n--------------------\n\nPractice Problem: If $\\{a_n\\}$ is a decreasing sequence such that $\\displaystyle \\lim_{n \\rightarrow \\infty} a_n = 0$, show that\n\n$\\displaystyle \\sum_{k=1}^{\\infty} a_k \\cdot \\sin{(kx)}$\n\nconverges for all $x$.\n\n## Saturday, May 12, 2007\n\n### Bigger Means Better. Topic: Algebra/Inequalities/Sets. Level: Olympiad.\n\nDefinition: A set $S$ is said to be convex if, for any $X, Y \\in S$ and $\\lambda \\in [0,1]$, we have $\\lambda X+(1-\\lambda)Y \\in S$.\n\n--------------------\n\nDefinition: Let $f: D \\rightarrow \\mathbb{R}$ be a real-valued function defined on a convex set $D$. We say that $f$ is convex if, for all $X, Y \\in D$ and $\\lambda \\in [0, 1]$, we have\n\n$\\lambda f(X) + (1-\\lambda) f(Y) \\ge f(\\lambda X+(1-\\lambda)Y)$.\n\n--------------------\n\nTheorem: (Jensen's Inequality) Let $f$ be a real-valued, convex function defined on a convex set $D$. Given $x_1, x_2, \\ldots, x_n \\in D$ and nonnegative reals $w_1, w_2, \\ldots, w_n$ such that $w_1+w_2+\\cdots+w_n = 1$, we have\n\n$w_1f(x_1)+w_2f(x_2)+\\cdots+w_nf(x_n) \\ge f(w_1x_1+w_2x_2+\\cdots+w_nx_n)$\n\nor, more concisely,\n\n$\\displaystyle \\sum_{i=1}^n w_if(x_i) \\ge f\\left(\\sum_{i=1}^n w_ix_i\\right)$.\n\nProof: We proceed by induction on $n$.\n\nBase Case - $n = 1$, we have $f(x_1) \\ge f(x_1)$ which is trivially true. $n = 2$, we have $\\lambda_1 f(x_1)+\\lambda_2 f(x_2) \\ge f(\\lambda_1 x_1+\\lambda_2 x_2)$ with $\\lambda_1+\\lambda_2 = 1$ which is true by the definition of convexity.\n\nInduction Step - Suppose $\\displaystyle \\sum_{i=1}^k w_if(x_i) \\ge f\\left(\\sum_{i=1}^k w_ix_i\\right)$. We wish to show\n\n$\\displaystyle \\sum_{i=1}^{k+1} u_if(x_i) \\ge f\\left(\\sum_{i=1}^{k+1} u_ix_i\\right)$\n\nfor nonnegative reals $u_1, u_2, \\ldots, u_{k+1}$ that sum to $1$. Well,\n\n$\\displaystyle f\\left(\\sum_{i=1}^{k+1} u_ix_i\\right) \\le u_{k+1}f(x_{k+1})+(1-u_{k+1})f\\left(\\frac{1}{1-u_{k+1}} \\sum_{i=1}^k u_ix_i\\right)$\n\nby the definition of convexity. But since $\\displaystyle \\frac{1}{1-u_{k+1}} \\sum_{i=1}^k u_i = \\sum_{i=1}^k \\frac{u_i}{1-u_{k+1}} = 1$, by our inductive hypothesis (the $k$ term case) we must have\n\n$\\displaystyle f\\left(\\frac{1}{1-u_{k+1}} \\sum_{i=1}^k u_ix_i\\right) \\le \\sum_{i=1}^k \\frac{u_i}{1-u_{k+1}} f(x_i)$.\n\nCombining this into the above inequality, we obtain\n\n$\\displaystyle f\\left(\\sum_{i=1}^{k+1} u_ix_i\\right) \\le u_{k+1}f(x_{k+1})+(1-u_{k+1})\\sum_{i=1}^k \\frac{u_i}{1-u_{k+1}} f(x_i) = \\sum_{i=1}^{k+1} u_if(x_i)$\n\nas desired, completing the induction. QED.\n\n--------------------\n\nDefinition: The convex hull of a set $S$ is defined to be the smallest convex set containing $S$. It is the set of all points that can be written as $\\displaystyle \\sum_{i=1}^n a_ix_i$, where $n$ is a natural number, $a_1, a_2, \\ldots, a_n$ are nonnegative reals that sum to $1$, and $x_1, x_2, \\ldots, x_n \\in S$.\n\n--------------------\n\nDefinition: A vertex of a set $D$ is a point $v \\in D$ such that for all $x \\in D$ not equal to $v$ and $\\lambda > 1$ we have $\\lambda v+(1-\\lambda)x \\not\\in D$.\n\n--------------------\n\nTheorem: Let $V_D = \\{v_1, v_2, \\ldots, v_k\\}$ be the set of vertices of a compact convex set $D$. The convex hull of $V_D$ is $D$.\n\nExample: The set of vertices of a convex polygon is, in fact, its vertices as traditionally defined in geometry. Any point inside the polygon can be written as a convex combination of its vertices.\n\n--------------------\n\nTheorem: If $f$ is a real-valued, convex function defined on a compact convex set $D$, then $\\displaystyle \\max_{x \\in D} f(x) = \\max_{x \\in V_D} f(x)$, where $V_D$ is the set of vertices of $D$.\n\nProof: We will show that $\\displaystyle f(x) \\le \\max_{x \\in V_D} f(x)$ for all $x \\in D$.\n\nLet $\\displaystyle x = \\sum_{i=1}^k \\lambda_i v_i$ where $\\lambda_1, \\lambda_2, \\ldots, \\lambda_k$ are nonnegative reals that sum to $1$ and $V_D = \\{v_1, v_2, \\ldots, v_k\\}$. This is possible by the preceding theorem. Then by Jensen's Inequality we get\n\n$\\displaystyle \\sum_{i=1}^k \\lambda_i f(v_i) \\ge f\\left(\\sum_{i=1}^k \\lambda_i v_i\\right) = f(x)$.\n\nAnd since $\\displaystyle \\max_{x \\in V_D} f(x) \\ge \\sum_{i=1}^k \\lambda_i f(v_i)$, we have $\\displaystyle \\max_{x \\in V_D} f(x) \\ge f(x)$ for all $x \\in D$. Furthermore, since $V_D$ is a subset of $D$, we know that this maximum is attained, thus\n\n$\\displaystyle \\max_{x \\in D} f(x) = \\max_{x \\in V_D} f(x)$\n\nas desired. QED.\n\n--------------------\n\nComment: This is a very useful result, as it allows us to limit our search for the maximum of a function to its set of vertices, which is usually a considerably smaller set, though it may still be infinite (think sphere). In any case, this works well for constrained optimization problems in which the domain is limited to a polygon, the easiest application of this theorem.\n\n--------------------\n\nPractice Problem: Let $x$ and $y$ be real numbers satisfying $|2x-y| \\le 3$ and $|y-3x| \\le 1$. Find the maximum value of $f(x, y) = (x-y)^2$.\n\n## Tuesday, May 8, 2007\n\n### A Square of Divisors. Topic: Number Theory. Level: AIME/Olympiad.\n\nProblem: (1999 Canada - #3) Determine all positive integers $n > 1$ with the property that $n = (d(n))^2$. Here $d(n)$ denotes the number of positive divisors of $n$.\n\nSolution: So, testing some small numbers yields $n = 9$ as a solution. We claim that this is the only such solution.\n\nClearly, since $n$ is a square, we can use a variant of the usual prime decomposition and say that $n = p_1^{2e_1} p_2^{2e_2} \\cdots p_k^{2e_k}$.\n\nFurthermore, again since $n$ is a square, we know\n\n$d(n) = (2e_1+1)(2e_2+1) \\cdots (2e_k+1)$\n\nis odd, so $n = (d(n))^2$ must be odd as well, i.e. $2$ is not one of the $p_i$. Then we use the equation given to us to get\n\n$p_1^{2e_1} p_2^{2e_2} \\cdots p_k^{2e_k} = (2e_1+1)^2(2e_2+1)^2 \\cdots (2e_k+1)^2$\n\n$p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k} = (2e_1+1)(2e_2+1) \\cdots (2e_k+1)$.\n\nNote, however, that by Bernoulli's Inequality (overkill, I know) we have for $i = 1, 2, \\ldots, k$\n\n$(1+(p_i-1))^{e_i} \\ge 1+(p_i-1)e_i \\ge 2e_i+1$\n\nwith equality iff $p_i = 3$ and $e_i = 1$. So\n\n$p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k} \\ge (2e_1+1)(2e_2+1) \\cdots (2e_k+1)$.\n\nSince we want equality, we must have $p_i = 3$ and $e_i = 1$ for all $i$. But since the primes are supposed to be distinct we can have exactly one prime so that $n = 3^2 = 9$ is the only solution. QED.\n\n--------------------\n\nComment: Another one of those problems that you kind of look at the result and you're like huh, that's interesting. But anyway, just throwing in some weak inequalities led to a pretty straightforward solution. As long as you know how to find the number of divisors of a positive integer it isn't too much of a stretch to figure the rest out, though it make take some time to get in the right direction since the problem is quite open-ended.\n\n--------------------\n\nPractice Problem: (1999 Canada - #4) Suppose $a_1,a_2,\\cdots,a_8$ are eight distinct integers from $\\{1,2,\\ldots,16,17\\}$. Show that there is an integer $k > 0$ such that the equation $a_i - a_j = k$ has at least three different solutions. Also, find a specific set of $7$ distinct integers from $\\{1,2,\\ldots,16,17\\}$ such that the equation $a_i - a_j = k$ does not have three distinct solutions for any $k > 0$.\n\n## Saturday, May 5, 2007\n\nProblem: Evaluate $\\displaystyle \\sum_{n=1}^{\\infty} \\frac{1}{n^2 \\cdot 2^n}$.\n\nSolution: Let $\\displaystyle f(x) = \\sum_{n=1}^{\\infty} \\frac{x^n}{n^2}$. Then $\\displaystyle f^{\\prime}(x) = \\sum_{n=1}^{\\infty} \\frac{x^{n-1}}{n} = -\\frac{\\ln{(1-x)}}{x}$, a well-known Taylor series. So we want to integrate this:\n\n$\\displaystyle \\int \\frac{\\ln{(1-x)}}{x}dx = \\ln{x} \\ln{(1-x)}-\\int \\frac{\\ln{x}}{x-1}dx$\n\nby parts using $u = \\ln{(1-x)}$ and $dv = dx/x$. Substituting $t = 1-x$ in the last integral, we have\n\n$\\displaystyle \\int \\frac{\\ln{(1-x)}}{x}dx = \\ln{x} \\ln{(1-x)}-\\int \\frac{\\ln{(1-t)}}{t}dt$.\n\nSo $-f(x) = \\ln{x} \\ln{(1-x)}+f(t)+C = \\ln{x} \\ln{(1-x)}+f(1-x)+C$. Thus\n\n$f(x)+f(1-x) = C-\\ln{x} \\ln{(1-x)}$\n\nfor some constant $C$. Using our knowledge that $f(0) = 0$, $\\displaystyle f(1) = \\frac{\\pi^2}{6}$, and $\\displaystyle \\lim_{x \\rightarrow 1} \\ln{x} \\ln{(1-x)}= 0$ by L'Hopital twice, we see that\n\n$f(0)+f(1) = C \\Rightarrow C = \\frac{\\pi^2}{6}$\n\nThen setting $x = \\frac{1}{2}$ we obtain $\\displaystyle 2f\\left(\\frac{1}{2}\\right) = \\frac{\\pi^2}{6}-\\ln{\\frac{1}{2}} \\ln{\\frac{1}{2}}$ and $\\displaystyle \\sum_{n=1}^{\\infty} \\frac{1}{n^2 \\cdot 2^n} = f\\left(\\frac{1}{2}\\right) = \\frac{\\pi^2}{12}-\\frac{\\ln^2{(2)}}{2}$. QED.\n\n--------------------\n\nComment: This was a pretty tough problem that required you to compound a lot of calculus knowledge all into a single problem - series, integration by parts, limits. Recognizing all the steps was the first part; following through with the right computations was another. Still, there weren't really any super clever tricks, mostly just standard substitutions and approaches applied in a somewhat non-standard way. Makes for a very nice problem.\n\n--------------------\n\nPractice Problem #1: Show that $\\displaystyle \\lim_{x \\rightarrow 1} \\ln{x} \\ln{(1-x)} = 0$.\n\nPractice Problem #2: Evaluate $\\displaystyle \\sum_{n=1}^{\\infty} \\frac{x^n}{n}$. Can you also find $\\displaystyle \\sum_{n=1}^{\\infty} \\frac{x^n}{n^3}$?" ]

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[ "Try the Free Math Solver or Scroll down to Resources!\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\n### Our users:\n\nThanks to the creators of the software. The context sensitive help on any topic of algebra made me clear my basics of mathematics. A very very useful tool ..\nBarbara, LA\n\nI used your software to prepare for my algebra exam. 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[ "## STAY AT HOMEreorder with our app! Buy online, stay safe!", null, "# Online UPC validation tool\n\nUPCZilla’s tool for validating UPCs which also gives you a handy explanation of how the validation was performed.\n\n085896527848 is a valid UPC-A (though that doesn't mean it's assigned to any products, click here to see if it is: 085896527848 - if you haven't already).\n\n## How do we check if a UPC is valid?\n\nTo check if a UPC is valid, we need to perform some calculations on its digits, as follows:\n\n1. We take the six odd numbered digits (counting from the left, not including the final digit - more about that at the end) and we add them together:`0 8 5 8 9 6 5 2 7 8 4 (8) - last digit ignored for now0 (digit 1) + 5 (digit 3) + 9 (digit 5) + 5 (digit 7) + 7 (digit 9) + 4 (digit 11) = 30`\n\n2. Then we multiply that number (30) by 3:`30 X 3 = 90 (we'll be using this number in a minute)`\n\n3. Then, similar to the first step, we take the FIVE (not six) EVEN numbered digits and we add them together as well:`0 8 5 8 9 6 5 2 7 8 4 (8) - last digit still ignored8 (digit 2) + 8 (digit 4) + 6 (digit 6) + 2 (digit 8) + 8 (digit 10) = 32`\n\n4. We get the number we got in step 3 (32) and we ADD it to the number we got in step 2 (90):`32 + 90 = 122`\n\n5. Now we take the number we got in step 4 (122) and work out how much we have to add to round it up to the nearest 10. In order to round 122 up to the nearest 10 (130) we have to add... 8\n\n6. Is this value of 8 the same as the last (rightmost) digit in our code - the one we ignored in steps 1 and 3 (that's our checksum)?\n\nYES, the checksum in our UPC was also 8 so the UPC is valid!\n\nThe rightmost digit in a UPC is a checksum, because it provides some insurance that all the other numbers are right by performing the above calculation on them. The system is not foolproof, but if any number is wrong then you will typically get a wrong checksum." ]

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[ "### Stats\n\n3,105 visits, 4,979 views\n\n### Translations\n\nThis tutorial hasn't been translated.\n\n### Tools\n\nI figured out understanding it thanks to the new Particle plugin.\n\nI put it here an example (from my comboBox plugin) for devs who needs help about drawing instance in edittime.js.\n\n### Draw the initial shape\n\nIn the OnCreate function, you set the hotspot of your instance which is the origin point of your shape:\n\n`````` IDEInstance.prototype.OnCreate = function()\n{\nthis.instance.SetHotspot(new cr.vector2(0, 0));\n}\n\n``````\n\nThen in the OnInserted, you set the size of your instance:\n\n`````` IDEInstance.prototype.OnInserted = function()\n{\nthis.instance.SetSize(new cr.vector2(200, 25));\n}\n``````\n\nThat means your instance's width is 200 pixels (x) and height of 25 pixels.\n\nSo this shape is a rectangle with those coordinates:\n\n- top-left corner (0,0) -- the SetHotspot point.\n\n- top-right corner (200,0)\n\n- bottom-left corner (0,25)\n\n- bottom-right corner (200,25) -- the SetSize point.\n\nNow, go to the Draw function (IDEInstance.prototype.Draw).\n\n`````` renderer.SetTexture(null);\nrenderer.Fill(q, cr.RGB(240, 240, 240));\nrenderer.Outline(q,cr.RGB(0, 0, 0));\n\n``````\n\n- Firstly, I set null as a renderer.SetTexture's parameter because we don't need any texture. We just need to fill the shape with a solid color.\n\n- Then, I save the global shape (GetBoundingQuad()) of my instance in a variable named 'q' (for 'quad').\n\n- And I finally set the fill and borders color.\n\nFor both renderer.Fill and renderer.Outline, (q,cr.RGB(0, 0, 0)) defined which is the actual shape and what color has to be set.\n\n(e.g. Here, the borders color of the 'q' shape will be black)\n\nSo now, we have a grey rectangle with black borders.\n\n### Draw into the first rectangle\n\nTo draw another shape direclty into the initial rectangle (which is saved in the 'q' variable), I'm going to create a new quad.\n\n`````` var q2 = new cr.quad();\nq2.tlx = q.trx - 25;\nq2.tly = q.try_;\nq2.trx = q.trx;\nq2.try_ = q.try_;\nq2.blx = q.brx - 25;\nq2.bly = q.bry;\nq2.brx = q.brx;\nq2.bry = q.bry;\n\n``````\n\n- The first line creates a new quad, saved into a *'q2' variable (for 'quad2'*).\n\n- The other ones will set the coordinates of each new quad's corner.\n\n(e.g. q2.tlx sets the x coordinate of the quad2's top left corner).\n\nTo draw this new quad into the first one, we have to set the 'q2' coordinates in relation to the 'q' coordinates. If we don't, the new quad will be drawn in relation to the layout coordinates.\n\n(e.g. The 'Guidelines' option of my iScroll plugin uses it.)\n\nThe following image tries to clarify it.\n\nNow, we have our two rectangles correctly set in our edittime.js.\n\nWe have finally to set the fill and borders color, as we did previously for the first shape.\n\n`````` renderer.Fill(q2, cr.RGB(210, 210, 210));\nrenderer.Outline(q2,cr.RGB(0, 0, 0));\n\n``````\n\nAs we have to set the color for the second quad, don't forget to replace 'q' by 'q2'.\n\n### Done!\n\nThat was my first tutorial, hopefully it helps someone.\n\nDon't hesitate to send me your questions and corrections." ]

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[ "My watch list\nmy.chemeurope.com\n\n# Self-consistent mean field (biology)\n\nThe self-consistent mean field (SCMF) method is an adaptation of mean field theory used in protein structure prediction to determine the optimal amino acid side chain packing given a fixed protein backbone. It is faster but less accurate than dead-end elimination and is generally used in situations where the protein of interest is too large for the problem to be tractable by DEE.\n\n## General principles\n\nLike dead-end elimination, the SCMF method explores conformational space by discretizing the dihedral angles of each side chain into a set of rotamers for each position in the protein sequence. The method iteratively develops a probabilistic description of the relative population of each possible rotamer at each position, and the probability of a given structure is defined as a function of the probabilities of its individual rotamer components.\n\nThe basic requirements for an effective SCMF implementation are:\n\n1. A well-defined finite set of discrete independent variables\n2. A precomputed numerical value (considered the \"energy\") associated with each element in the set of variables, and associated with each binary element pair\n3. An initial probability distribution describing the starting population of each individual rotamer\n4. A way of updating rotamer energies and probabilities as a function of the mean-field energy\n\nThe process is generally initialized with a uniform probability distribution over the rotamers - that is, if there are p rotamers at the kth position in the protein, then the probability of any individual rotamer", null, "$r_{k}^{A}$ is 1 / p. The conversion between energies and probabilities is generally accomplished via the Boltzmann distribution, which introduces a temperature factor (thus making the method amenable to simulated annealing). Lower temperatures increase the likelihood of converging to a single solution, rather than to a small subpopulation of solutions.\n\n## Mean-field energies\n\nThe energy of an individual rotamer rk is dependent on the \"mean-field\" energy of the other positions - that is, at every other position, each rotamer's energy contribution is proportional to its probability. For a protein of length N with p rotamers per residue, the energy at the current iteration is described by the following expression. Note that for clarity, the mean-field energy at iteration i is denoted by Mi, whereas the precomputed energies are denoted by E, and the probability of a given rotamer is denoted by", null, "$P_{i}(r_{k}^{A})$.", null, "$M_{i}(r_{k}^{A}) = E_{k}(r_{k}^{A}) + \\sum_{x=1}^{N} \\sum_{y=1}^{p} P_{i-1}(r_{x}^{y}) E_{xy}(r_{k}^{A}, r_{x}^{y})$\n\nThese mean-field energies are used to update the probabilities through the Boltzmann law:", null, "$P_{i}(r_{k}^{A}) = \\left(\\exp\\left(-\\frac{M_{i}(r_{k}^{A})}{kT}\\right)\\right)\\left(\\sum_{y=1}^{p}\\exp\\left(-\\frac{M_{i}(r_{k}^{y})}{kT}\\right)\\right)^{-1}$\n\nwhere k is the Boltzmann constant and T is the temperature factor.\n\n## Energy of the system\n\nAlthough computing the system energy is not required in carrying out the SCMF method, it is useful to know the overall energies of the converged results. The system energy Msys consists of two sums:\n\nMsys = Msingle + Mpair\n\nwhere the addends are defined as:", null, "$M_{single} = \\sum_{x=1}^{N} \\sum_{y=1}^{p} P(r_{x}^{y})E_{x}(r_{x}^{y})$", null, "$M_{pair} = \\sum_{x=1}^{N} \\sum_{y=1}^{p} \\sum_{a=x+1}^{N} \\sum_{b=1}^{p} \\left(P(r_{x}^{y})P(r_{a}^{b})E_{xy}(r_{x}^{y}, r_{a}^{b})\\right)$\n\n## Convergence\n\nPerfect convergence for the SCMF method would result in a probability of 1 for exactly one rotamer at each position k in the protein, and a probability of zero for all other rotamers at each position. Convergence to a unique solution requires probabilities close to 1 for exactly one rotamer at each position. In practice, especially when higher temperatures are used, the algorithm instead identifies a small number of high-probability rotamers at each position, allowing the resulting conformations' relative energies to then be enumerated (based on the precomputed energies, not on those derived from the mean-field approximation). One way to improve convergence is to run again at a lower temperature using the probabilities calculated from a previous higher-temperature run.\n\n## Accuracy\n\nUnlike dead-end elimination, SCMF is not guaranteed to converge on the optimal solution. However, it is deterministic (as in, it will converge to the same solution every time given the same initial conditions), unlike alternatives that rely on Monte Carlo analysis. By comparison to DEE, which is guaranteed to find the optimal solution, SCMF is faster but less accurate overall; it is significantly better at identifying correct side chain conformations in the protein's core than it is on identifying correct surface conformations. Geometric packing constraints are less restrictive on the surface and thus provide fewer boundaries to the conformational search.\n\n## References\n\n1. ^  Koehl P, Delarue M. (1994). Application of a self-consistent mean field theory to predict protein side-chains conformation and estimate their conformational entropy. J Mol Biol 239(2):249-75.\n2. ^  Voigt CA, Gordon DB, Mayo SL. (2000). Trading accuracy for speed: A quantitative comparison of search algorithms in protein sequence design. J Mol Biol 299(3):789-803." ]

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[ "# Is there a large counterexample?\n\nAlgebra Level 4\n\nTrue or False?\n\nIf $a,b$ and $c$ are roots to the equation $x^3+x^2 + x=1$, and let $S_n = a^n + b^n + c^n$, then there exists a positive integer $n$ such that $\\text{sgn}(S_n) = \\text{sgn}(S_{n+1}) = \\text{sgn}(S_{n+2})$.\n\n Notation: $\\text{sgn}(x) := \\begin{cases} \\begin{array} {l l } -1 & \\text{ if }x<0 \\\\ 0 & \\text{ if }x=0 \\\\ 1 & \\text{ if }x>0 \\\\ \\end{array} \\end{cases}$ denotes the sign function.\n\n×" ]

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[ "# 12 Best Of 5th Grade Math Review Worksheets Shots\n\n12 Best Of 5th Grade Math Review Worksheets Shots 5th grade math review worksheets printable worksheets 5th grade math review showing top 8 worksheets in the category 5th grade math review some of the worksheets displayed are end of the year test 5th grade math work incoming 6 grade math summer pa 5th grade math review worksheets  5th grade math review worksheets", null, "free 3rd grade math word problems free third grade math worksheets from 5th grade math review worksheets, image source:kcctalmavale.com", null, "daily math worksheets primalvape co from 5th grade math review worksheets, picture source:primalvape.co\n\n5th grade math review worksheets", null, "amazon com 3rd grade math online teaching tutoring software 1 from 5th grade math review worksheets, image resource:amazon.com", null, "free comprehension worksheets year 9 printable grade reading from 5th grade math review worksheets, impression supply:bobsseptic.info\n\n### 12 5th grade math review worksheets photos\n\n12 Best Of 5th Grade Math Review Worksheets Shots test sample for grade 5 core math math worksheets land i find the fifth grade curriculum to be particularly fraction heavy a great amount a great amount of the course year at this level is focused on those skills 5th grade math review worksheets", null, "geometry basics review quiz 1 math worksheets math worksheets from 5th grade math review worksheets, image supply:pinterest.com", null, "free probability and statistics worksheets grade math vocabulary from 5th grade math review worksheets, picture resource:ratifa.co\n\nfree math worksheets for grade 5 homeschool math free math worksheets for grade 5 this is a comprehensive collection of free printable math worksheets for grade 5 organized by topics such as addition subtraction algebraic thinking place value m  5th grade math review worksheets  5th grade math worksheets printable pdfs free math 4 5th grade math worksheets printable pdfs 5th grade math worksheets on adding and subtracting algebra finding unknown variables fractions lcm and hcf ratios proportions coordinate gra 12 Best Of 5th Grade Math Review Worksheets Shots" ]

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[ "## Tuesday, December 8, 2015\n\n### Angle of Friction Derivation and Defination\n\nAngle of friction is the angle between the normal reaction and resultant of normal reaction and frictional force.When a body is on a horizontal surface,its weight always acts in vertically downward direction.When we apply a horizontal force to move the body, frictional force starts acting in opposite to the direction of applied force.\n\nNormal reaction is the reaction applied by lower body on the upper body against the force acting on the lower body and it shall be normal to the surface of contact.\n\nWe can identify that normal reaction and frictional force are perpendicular to each other.Their resultant will be in between them and its value can be determined using parallelogram law of vectors.This resultant makes some angle with normal reaction and this angle is called angle of friction.\n\nIt is mathematically proved below that angle of contact depends on the nature of the surface and its roughness.Hence it depends on coefficient of friction of the surface.\n\nWe shall apply enough force to move the body against the maximum static frictional force and if the applied force is less than that, body remains in the state of rest.Here static frictional force self adjusts itself to the applied force and the body remains in the state of rest.This is called self adjusting nature of static frictional force.\n\nIf the applied force is greater than limiting frictional force,then the body starts moving in the direction of applied force.We can calculate the resultant force as applied force subtracting the frictional force as it is acting against the motion.We can also calculate the resultant acceleration that the body is going to get.\n\nIn the following video  all this is explained in detail.\n\nRelated Posts" ]

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[ "# Study of Laminated Composite MEMS and NEMS Performance in Nano Metric Operations\n\nDocument Type : Research Paper\n\nAuthors\n\n1 School of new Technology, Iran University of Science and Technology\n\n2 Iran University of Science and Technology\n\nAbstract\n\nPrecision of nano-metric operations is an important issue in nano-engineering studies. Several operative parameters might affect the quality of results. The parameters of the nano world are significant but not entirely controllable. However, the geometrical and mechanical properties of micro cantilevers are completely controllable. So, controlling the sensitivity of resulting image through t lamination design could be a proper approach. This paper analyses the effects of composite lamination on the performance of common Micro and Nano Electro Mechanical systems (MEMS and NEMS, respectively). Generalized Differential Quadrature (GDQ) and Generalized Differential Quadrature Element (GDQE) methods are used as semi-analytic solutions for regular and irregular domains, respectively. Validity, applicability and accuracy of the proposed approach are demonstrated and then the lamination effects on the nano-imaging and manipulation of nano particles by micro cantilevers are studied. This study shows that some laminations of micro cantilevers resulted in a better performance in nano-manipulation and imaging. Furthermore, clarifying the dependency of system sensitivity on the profile of the substrate is remarkable.\n\nKeywords\n\n#### Full Text\n\n Mechanics of Advanced Composite Structures 4 (2017) 19-31 Semnan University Mechanics of Advanced Composite Structures journal homepage: http://MACS.journals.semnan.ac.ir\n\nStudy of Laminated Composite MEMS and NEMS Performance in Nano-metric Operations\n\na Smart Micro/Nano Electro Mechanical Systems Lab (SMNEMS), Nanotechnology Department, School of New Technologies, Iran University of Science and Technology, Tehran, Iran\n\nb Robotic Research Laboratory, Center of Excellence in Experimental Solid Mechanics and Dynamics, School of Mechanical Engineering, Iran University of Science and Technology, Tehran, Iran\n\n Paper INFO ABSTRACT Paper history: Received 2016-08-12 Revised 2016-09-15 Accepted 2016-11-15 Precision of nano-metric operations is an important issue in nano-engineering studies. Several operative parameters might affect the quality of results. The parameters of the nano world are significant but not entirely controllable. However, the geometrical and mechanical properties of micro cantilevers are completely controllable. So, controlling the sensitivity of resulting image through t lamination design could be a proper approach. This paper analyses the effects of composite lamination on the performance of common Micro and Nano Electro Mechanical systems (MEMS and NEMS, respectively). Generalized Differential Quadrature (GDQ) and Generalized Differential Quadrature Element (GDQE) methods are used as semi-analytic solutions for regular and irregular domains, respectively. Validity, applicability and accuracy of the proposed approach are demonstrated and then the lamination effects on the nano-imaging and manipulation of nano particles by micro cantilevers are studied. This study shows that some laminations of micro cantilevers resulted in a better performance in nano-manipulation and imaging. Furthermore, clarifying the dependency of system sensitivity on the profile of the substrate is remarkable. Keywords: Laminated Composite Micro electro mechanical systems Nano electro mechanical systems Generalized differential quadrature Nano-manipulation DOI: 10.22075/MACS.2016.495 © 2017 Published by Semnan University Press. All rights reserved.\n\n1.      Introduction\n\nSensing and actuating the small scales has trended the researchers to focus on Micro and Nano Electro Mechanical systems (MEMS and NEMS).However, experiment-based predictions are very expensive in this area. As a proper alternative, therefore, micro- and nano electro mechanics theories must be further developed and applied. Piezoelectric-based structures, named smart structures, are the most well-known category of MEMs and NEMs. Applications of piezoelectric actuators extend to the following things: mass production of sound transmitters; ultrasonic power transducers and sensors; bending actuators for textile machines; ink print heads; beam benders in valves, in braille displays, in optical systems; and newly monolithic multilayer actuators for automotive injection systems .\n\nA sufficiently accurate model for MEMs and NEMs remains one of the most challenging problems of the MEMs and NEMs dynamics. Until now, numerical methods have been developed to study the dynamics of small devices, such as the finite element method (FEM), the finite difference method (FDM) and the finite cloud meshless method (FCM) [2-4]. This paper will show that although FEM is sufficiently simple and accessible, its precision is limited in some usual boundary and ambient conditions. As an alternative, Generalized Differential Quadrature (GDQ) and Generalized Differential Quadrature Element (GDQE) methods have been recently applied to continuous and discontinuous mechanical systems, respectively. In this paper, these methods are used to study the performance of NEMs and MEMs in nanometric operations. It is useful to list valuable works on modelling nano-manipulation and nano-imaging with the molecular dynamics approach. Coarse-grained molecular dynamics simulation of the automatic nano-manipulation process was performed to study the effect of tip damage on the positioning errors . However, nano-cluster manipulation’s success in considering the flexibility of the system was studied via molecular dynamics simulations .\n\nSome researchers have focused on the free vibration analysis of plates and shells as a common mechanical problem [7-9]. Others have presented the mechanics of especial systems through especial conditions . Piezoelectric-based structures typically have complicated behavior due to multi-field excitations. This forces the implementation of numerical methods. Balamurugan and Narayanan have used a nine-nodded, piezo-laminated, degenerated shell element to model and analyse multi-layered composite shell structures together with sensors and piezoelectric actuators. Applications of FEM in structures, including sensors and actuators, are surveyed . In addition to FEM, advanced numerical methods have been developed. For instance, the differential quadrature method (DQM) has been introduced as an effective semi-analytic method in the analysis of the free, static and dynamic response of mechanical systems. Then some researchers have used it for more complicated mechanical engineering problems . Other related works can be mentioned where the piezoelectric laminated cylindrical shell - with the shear rotations effect under the electromechanical loads and the four sides simply supported the boundary condition - was studied by using the two-dimensional generalized DQ (GDQ) computational method . The application of DQ is limited to uniform domains, and for irregular domains, including the discontinuity in material or geometry, the DQ element method (DQEM) has been suggested by Chen . Researchers believe that the GDQ and GDQE methods lead to moderately exact solutions to problems [16,17]. So, it could be a proper analysis procedure.\n\nThe imaging and manipulation of nano objects include various techniques performed by especial devices named macro-scale nano-robots . Imaging and manipulation could be performed in various procedures based on the aims and goals of the operation . Accuracy is the most significant challenge in such devices. Usually, nonlinearity has been observed in the MEMs and NEMs that are used in nano-robots. Predicting these nonlinearities is very useful to achieve sufficient accuracy for the operation. In this paper, nonlinearities were neglected and a linear model was proposed.\n\nThis paper represents a comprehensive semi-analytic approach for static and dynamic responses of nano-robots that could include torsion, elongation and bending in various planes. The most prominent contribution of this study is to comprehensively investigate the laminated composite effects on nano-manipulation and nano-imaging.\n\n2.      Problem\n\nPiezoelectric materials could be used in various configurations for actuation and/or sensing purposes. For instance, the mono-morph, bimorph and trimorph laminations are the most well-known representatives of piezoelectric plate-bending actuators. Multilayer technology has the advantage of working with even lower driving voltages. In this study, which uses laminated composite structures, a general and applicable configuration is suggested for a superior performance in MEMs and NEMs. The imaging and manipulation of nano objects by Scanning Probe Microscopes (SPMs) and especially an Atomic Force Microscope (AFM) are two of the prominent applications of micro-cantilevers. For example, the imaging of a surface and the manipulation of a nanoparticle are depicted in Figure 1 schematically.\n\n(a)\n\n(b)\n\nFigure 1. (a) Imaging of a general substrate and (b) manipulation of a nanoparticle by a general AFM.\n\nAs depicted in Figure 1, in an AFM, the micro-cantilever is attached at the end of a piezotube actuator. Since the dimensions of the micro-cantilever with respect to the piezo tube are negligible, the dynamics of the piezo tubes are neglected. Recently, the dynamics of a piezo tube in nano-imaging and manipulation operations have been studied .\n\n2.1.  Formulation of micro-cantilever dynamics\n\nThe general form of the linear constitutive relation of piezo-electro mechanics can be written as:\n\n (1)\n\nwhere [C], [e] and  denote the elastic coefficients, piezoelectric constant and dielectric permittivity matrices, respectively [16,17].\n\nBased on the dimensions of the MEMs and NEMs used, mentioned previously, the laminated system considered is assumed to be thin and, thus, strains along the thickness can be neglected. Here, the First Order Shear Deformation Theory (FSDT) is used. Using the generalized Hamilton principle, the equations of motion and related boundary conditions of a segmented laminated shell are derived as:\n\n(2)\n\nin which  is the inertia and N(w0) is the nonlinear term of equations whose higher-order terms are neglected.\n\n(3)\n\nIt is particularly important for an actuator that the stiffness of the system is affected by the nonlinear term. In the equations presented, the upper asterisks (*) denote the external forces. In MEMs and NEMs, there is an electric field beneath the bigger surfaces and, thus, there is current along the thickness (Ez). Using the electroelastic charge equation and the boundary condition, the electric potential can be written as:\n\n(4)\n\nwhere Q3k is the surface electric charge density of the kth layer in ith segment. The piezoelectric force and moment resultants and are defined as:\n\n(5)\n\nwhere and “M” is the number of piezoelectric layers. Based on the theory mentioned, for each element, the equation of motion, in operator form, can be written as:\n\n(6)\n\nwhere  and are the mechanical and electrical effects, respectively, and is the mass term of the jth equation for ith element. The electrical effect contains two parts: the converse effect, and the direct effect. The geometrical compatibility conditions at the common nodes of the two connected elements are defined as:\n\n(7-a)\n\nDenoting the axial and circumferential directions by “x” and “y,” respectively, the natural compatibility conditions are as follows:\n\n(7-b)\n\n(7-c)\n\nAlso, the natural and geometrical external boundary conditions along the x and y edges are:\n\n(7-d)\n\n3.      Solution\n\nIn numerical analyses, a specific feature makes GDQ and GDQE highly distinctive in reaching very precise results by using merely a few grid points. This feature, which will be introduced in the following notes, is particularly precious in structural and vibration applications. In principle, DQM was initially introduced to express the derivatives of a function to various directions in terms of the sum of weighted function values at some specific discrete points.\n\n(8)\n\nIn Eq. (8), , n, N, f,  and denote coordinate system directions (such as a curvilinear one), order of derivative, number of discrete points, desired function, points in the direction whose derivatives are needed, and weighting coefficients (for nth order derivative in the  direction and at point ), respectively. Recently, a recursive relation has been developed to calculate the weighting coefficients as :\n\n(9-a)\n\nwhere are the first derivative coefficients, calculated by as:\n\n(9-b)\n\nNote that. For numerical calculations, sampling points with the Chebyshev-Gauss-Lobatto are employed in the following way:\n\n(10)\n\nwhere L is the specific length along the direction considered. Due to the discontinuity, domain decomposition is needed for segmented systems. Therefore, various elements should be considered and the GDQEM should be applied. Generally, neglecting the damping effect, equations of motion yield:\n\n(11)\n\nAnd the boundary conditions are defined as:\n\n(12)\n\nwhere, and  are the vectors of external forces on interior and boundary points, respectively.\n\nFor the implementation of boundary conditions in the DQM or DQEM, several approaches have been proposed. Except for the general approach , others cannot be used in cases other than those that are simply supported or have clamped conditions. The equations of motion and boundary conditions can be rewritten in the matrix form as:\n\n(13)\n\nwhere FDist and FBCare the distributed forces of the interior and boundary forces that are obtained from and. Substituting  from the second into the first relation in Eq. 13 leads to:\n\n(14)\n\nAnd in the state space form:\n\n(15)\n\nUsing a numerical approach, for instance, the Runge-Kutta method, the aforementioned state space equation can be solved. Then displacement of boundary points can be calculated as\n\n(16)\n\nFor static problems, the solution of Eq. (13) can be written as:\n\n(17)\n\n4. Validation\n\n4.1. Comparison of static deflections\n\nNumerical solutions exist for electromechanical problems. The Finite Element Method (FEM) is commonly used to study the behavior of MEMs and even NEMs. Previously, validation used to be provided by the comparison of methods with the traditional software programs, such as ANSYS and ABAQUS. Here, for more clarification, comparisons of the static and dynamic responses are made with the Green's function approach and ANSYS software. In ANSYS, the shell element has been used. These types of comparisons clearly show the efficiency of the proposed model. Table 1 lists the properties and dimensions of beams and plates used for the comparison. Table 2 lists the static deflection of the midpoint of the beam and plate due to the point and uniformly distributed loads. For each problem, the simulation error of GDQM is of a considerably lower order compared to that of FEM. Table 2 also lists the comparison of the static deflection of a stepped B-beam, due to the 100 N point and uniform loads using GDQEM. Furthermore, due to a 100 N uniform line load on the step position, the static deflection of the midpoint of a B-plate is compared.\n\n4.2. Comparison of dynamic response\n\nThe dynamic response of an A-beam (middle point) under the SS-SS boundary conditions due to a uniformly distributed load (as 100×Sin(10×t)) is depicted and compared for three different approaches in Figure 2. Figure 3 shows the dynamic response of an A-plate (middle point) under SS-SS-SS-SS boundary conditions, due to a uniformly distributed load (as 100×Sin(10×t)). In Figure 2, three approaches resulted in the same responses, when in the case of plate (Figure 3), the GDQ method leads to a more precise estimation.\n\nTable 1.    Properties of beams and plates used for simulations.\n\n Problem Dimensions (m) Module of Elasticity (GPa) Density (Kg/m3) Configuration A-Beam 5×0.04×0.04 70 2,800 Simple B-Beam 2.5×0.04×0.04,2.5×0.04×0.06 70 2,800 Step A-Plate 1×1×0.03 70 2,800 Simple B-Plate 0.5×1×0.03,0.5×1×0.05 70 2,800 Step Problem Dimensions (m) Piezo Type Material Configuration C-Plate 0.1×.05×.0025 PZT-5H Gr/Epoxy Simple D-Plate 0.05×0.05×0.0025, 0.05×0.05×0.0015 PZT-5H Gr/Epoxy Step\n\nTable 2.    Comparison between the results of present approach, exact solution and ANSYS for static problem.\n\nAs a result, according to the more accurate GDQ, it is used for the dynamic response in nanometric operations. As already discussed, the exact solution cannot be provided in the case of other boundary conditions. But the GDQ results could be compared with FEM with an intrinsic guarantee in accuracy for both static and dynamic responses.\n\nFigure 4 shows the dynamic response of the stepped B-plate (middle point) under various boundary conditions and due to the uniformly distributed load (100×Sin(10×t)) applied to the patch. Obviously, in this figure, the GDQE results are close to those of ANSYS. However, it has already been demonstrated that the response of GDQE is more accurate.\n\nBased on the aforementioned comparison, the reliability of the approaches presented is satisfied. Thus, it can be used to study the effects of various parameters on the performance of MEMs and NEMs. Here, for instance, the micro-cantilever (MC), which is used in nano-robots, is studied. The statics and dynamics of MC in the nano-imaging and manipulation of nano-particles have been discussed.\n\nFigure 2. Dynamic response for a beam; comparison of exact, ANSYS and GDQ models.\n\nFigure 3. Dynamic response for a plate; comparison of exact, ANSYS and GDQM.\n\nFigure 4. Dynamic response of the stepped B-plate (middle point) under various boundary conditions and due to the uniform 100sin(10t) distributed load applied on the patch.\n\n4.      Results and Discussion\n\nThere are two key operations in nano-scale in which the micro-cantilever plays an important role: nano-imaging and nano-manipulation. These two operations are discussed in the next sections as the most useful results could be concluded.\n\n5.1. Surface nano-imaging\n\nPrecision of nanoimaging is an important issue in nano-engineering studies. Various parameters affect the quality of images obtained. Parameters in the nano-world, such as roughness of surface and chemical properties, are the most significant, but they are not entirely controllable. However, are some parameters in macro world could be changed and controlled. For instance, the geometrical and mechanical properties of micro-cantilevers could seriously affect the imaging result. So controlling the sensitivity of resulting image through the lamination design could be a proper approach. Figure 5 shows the nano-imaging scheme. Recently, by implementing molecular dynamics, a shape feedback has been used to control the shape of a particle used in manipulating . Here, the problem is considered to be a state-dependent dynamic force on the MC end. Neglecting the tip deformation effects on the dynamics of MC, where they might be studied via molecular dynamics, Figure 6 shows the algorithm implemented. An interaction-attraction force field is considered for the interfacial force between the tip end and substrate :\n\nwhere “a” and “b” are the force coefficients and “u” is the deflection of the MC end. The quantities of “a” and “b” depend on the system considered. The best evaluation of “a” can be provided through the inverse of the amplitude of the MC end (a@1/d).  And “b” can be considered 1 (one). For simplicity, a general profile is assumed for the substrate, including triangular, square and sinusoidal parts, simultaneously. The desired and resulting profiles, obtained by the first configuration depicted in Figure 7, are illustrated on the left side of Figure 5. The length, width and thickness are 200, 40 and 8 micrometers, respectively.\n\n(18)\n\n(19)\n\nTo evaluate the lamination composite effects on the imaging results, the third configuration of lamination, demonstrated in the left side of Figure 5, is studied with various laminations. Assuming a proportional damping, Figure 8 shows the effects of various laminations on the nano-imaging result by scanning the above-mentioned general profile. Implementing proportional damping into Eq. (15) leads to Eq. (19), where,  and  have been selected as:\n\n(20)\n\nPDC,  and  are the total proportional damping coefficient, average density of lamination and identity matrix, respectively. The constant coefficientsand  are assumed to be 0.0008 and 0.0003. For more clarification, some subplots of Figure 8 have been enlarged in Figure 9. The legend of Figure 9 is mentioned in Figure (9-e). The key areas focused on in Figure 9 could be used to study the output sensitivity of various configurations to obtain more precise results. When same geometrical, ambient and imaging schemes are used, one expects to observe different results for the profile obtained because of the variations in stiffness in the laminations used. Thus, some behavior is mechanically predictable and simulations are not necessary. In practice, however, some factors limit the performance of MEMS and NEMS. Thus, even smallest difference between various laminations may lead to an optimal selection of lamination for more precise imaging. The P/90˚/0˚/P lamination results in more precision in the case of the ascent (Figure 9-a), peak (Figure 9-b) and descent (Figure 9-c) parts of the triangular surface. Fr square and sinusoidal profiles, the output image is completely different. The P/90˚/0˚/P lamination leads to the worst results. Based on the fourth to sixth subplots of Figure 9, the P/0˚/0˚/0˚/P lamination is suggested for square and sinusoidal profiles. The dependency of system sensitivity on the profile of a substrate is a highly significant result in imaging especial surfaces. Among all laminations studied, P/90˚/0˚/P and P/0˚/0˚/0˚/P are suggested for triangular and both square and sinusoidal substrates, respectively.\n\nFigure 8 is provided by using a sufficient damping coefficient. Piezoelectric shunt damping is a well-known technique for suppressing vibration in MEMS and NEMS. Techniques encompassed in this broad description are characterized by connecting electrical impedance to a structurally bonded piezoelectric transducer . Such methods can guarantee the stability of the shunted system without any external sensor and do not require parametric models for a design. Thus, by using an electrical circuit, proportional damping could be designed to achieve the desired behavior. With this assumption, using the P/30/0/-30/P lamination for scanning a simpler substrate profile (as demonstrated in Figure 10), the effects of PDC on the image obtained are depicted in Figure 10. As illustrated in this figure, for a better result, the damping should be in an especial range. For instance, the PDC between 0.0001 and 0.001 could be acceptable. The real like three-dimensional (3D) image of scanned surface could be obtained by repeatedly scanning the substrate considered. The real and 3D images of substrate obtained for three PDCs are shown in Figure 11.\n\nFigure 5. Various configurations of MC and AFM nano-imaging scheme using the MC.\n\nFigure 6. Imaging of a substrate using the GDQ and GDQE methods.\n\nFigure 7. Substrate (real) and resulted profiles with a bimorph MC.\n\nThe substrate considered is similar to the one in Figure 11-a. The profile that resulted from using the algorithm presented with PDC = 0.0001 is demonstrated in the second picture of Figure 11 (11-b). Even though the triangular and sinusoidal parts are almost close to the profile of substrate, the ascent and descent parts of square profile are completely different. A less damping coefficient leads to deviations in the ascent and descent parts. However, as illustrated in the third picture, a more damping coefficient leads to unwanted delays. As the best result, the fourth picture (PDC = 0.0005) could be mentioned. Triangular, square and sinusoidal parts have been scanned and sketched precisely. These important issues could be used to improve the shunt damping technique to provide a more precise image of the substrate.\n\n5.2. Manipulation of nano-particles\n\nScanning Probe Microscopes (SPMs) and especially the Atomic Force Microscope (AFM) are used to reach, identify and manipulate nano-objects. Depending on its configuration, an AFM could include one or two piezo-tube actuators that are about 10 mm to 15 mm long. The effects of the particle size on the semistatic deflection and sensed voltage of MC are studied here. Figure 12 shows a very special case of AFM, where one piezo-tube is used to span three dimensions in space. At the end of the piezotube, an MC is attached (with the length, width and thickness of about 200 μm, 8 μm and 1 μm, respectively), including a small conical tip (with the end diameter of about 10 nm) at the end. The piezo-tube scans the substrate in all directions (x, y and z).\n\nAssuming the rigidity of the piezo-tube toward the MC, the dominant dynamics belong to the MC. CFFF boundary conditions could be considered for the MC and the existing forces and moments on the end of the MC (Figure 13) are depicted in Figure 14 versus various steps of manipulation.\n\nThe effective forces in the nano-manipulation operation can be obtained by using certain formulations . Figure 15 shows these results. The calculations are carried out in two cases, a particle with 20 nm (D20) and 100 nm (D100) of diameter. Forces in Cartesian coordinates and the free body diagrams for MC in the nano-manipulation mode are depicted in Figure 13. Table 3 lists forces and moments quantitatively for all steps of nano-manipulation.\n\nFigure 15 shows the maximum of deflection of the MC in the various steps of nano-manipulations. The sensed voltage on the sensor layer of the MC is displayed in Figure 16. The effect of particle size on the static deflection and sensed voltage can be observed clearly. The bigger the particle used is, the bigger the resulting deflection and voltage are. The effects of lamination on the deflection and sensed voltage of MC can be observed clearly.  The total thickness for all configurations considered is the same. The diagrams presented show that when two mechanical layers are used, the static deflection is greater. The P/0˚/90˚/P lamination leads to the greatest deflection and the minimum deflection belongs to the P/0˚/0˚/0˚/P lamination. One should note that the deflections for D20 and D100 problems are approximately the same before step \"E.\" Before this, the moments were zero and, thus, various laminations lead to approximately the same results in two cases.\n\nFigure 8. The resulting profile of various configurations for MC.\n\n (a) (b) (c) (d) (e) (f)\n\nFigure 9. Some subplots of the resulted profile in various configurations.\n\nFigure 10. The damping effects on the resulting profile.\n\n (a) (b) (c) (d)\n\nFigure 11. 3D view of (a) profile of substrate, (b) resulting profile by PDC=0.0001, (c) resulting profile by PDC=0.001 and (d) the best resulting profile by PDC=0.005.\n\nFigure 12. Manipulation strategy using the AFM: a) auto parking, b) snap in substrate, c) substrate pull out, d) approach to nanoobject, e) snap in nanoobject,  f) offset in the z direction, g) pushing, h) nanoobject pull out, i) retraction .\n\n(a)\n\n(b)\n\nFigure 13. (a) The free body diagram for MC in nano-manipulation mode and (b) the tip and corresponding parameters .\n\nFigure 14. Significant forces and moments at the end of MC in the nano-manipulation strategy .\n\nTable 3.    The deflections of MC during the nano-manipulation of a nanoparticle\n\n Steps A,I B C D E Forces D20 D100 D20 D100 D20 D100 D20 D100 D20 D100 N12 (nN) 0 -1.19 -8.96 -1.40E-5 -5.37E-7 0.0016 0.0022 Q13 (nN) 0 4.46 33.46 5.24E-5 2.01E-7 5.72E-5 -0.0004 M12 (fN.m) 0 0 0 0 0 19.2 24.5 Steps F G H Forces D20 D100 D20 D100 D20 D100 N12 (nN) -1.06 -3.19 2.839 13.08 -1.064 -3.19 Q13 (nN) 3.97 11.93 2.924 7.571 3.97 11.93 M12 (fN.m) 19.2 24.5 42 175 0 0\n\nFigure 15. Maximum of MC deflection in various steps of nano-manipulations.\n\nFigure 16. The sensed voltage on the sensor layer of the MC for various lamination.\n\nFigure 16 shows the sensed voltage through the sensor patch on the MC. Consciously, for plotting the diagrams in a specific range, some are multiplied by the coefficients mentioned in the figure legend. The static deflection of regular electromechanical systems has a linear relation with the sensed voltage. Nevertheless, the sensed voltage depends not only on particle size, but also on the structure of the MC. Irregularity can seriously affect the output. The sensed voltage increases with an increase in the particle diameter, but the effects of irregularity cannot be addressed precisely. The egregious discrepancy of the results of P/0/45/P and P/0/90/P from other laminations is illustrated in Figure 16, which demonstrates the effect of irregularity.\n\n5.       Conclusions\n\nThe precision of nano-imaging and nano-manipulation is an important issue in nano-engineering studies. There are various effective parameters of the quality of images obtained. Parameters of the nano-world are significant but not entirely controllable. However, geometrical and mechanical properties of micro-cantilevers are completely controllable. So, controlling the sensitivity of the resulting image through the lamination design could be a proper approach.\n\nIn this paper, semianalytic (GDQ and GDQE methods) static and dynamic solutions of Cartesian laminated beams and plates done to control the results of the nano-metric operations. The semianalytical approach presented is generally based on the FSDT. Comparing the results with the exact and existing numerical approaches revealed its efficiency. After reliable validation, the lamination effects on the static and dynamic manner of micro-cantilevers in nano-metric operations were studied.\n\nAlso studied were the nano-imaging effects on the MC shape and the challenges presented by the sensitivity of the resulting image through the lamination. The P/90/0/P lamination resulted in more precision in the ascent, peak and descent of a triangular surface for a substrate. For the square and sinusoidal profiles, image output was completely different. The P/90/0/P lamination led to the worst result. The dependency of the system sensitivity on the profile of a substrate is a highly significant result for imaging especial surfaces. Among all laminations examined, the P/90/0/P and P/0/0/0/P were suggested for triangular and both square and sinusoidal substrates, respectively.\n\nSince piezoelectric shunt damping can be implemented for suppressing vibration in MEMS and NEMS, utilizing the P/30/0/-30/P lamination for scanning simpler substrates, the effects of a proportional damping coefficient on the output were considered. With fewer damping coefficients, even though the triangular and sinusoidal parts were approximately close to the profile of substrate, the ascent and descent parts of square profile were completely different. More damping coefficients led to unwanted delays in the ascents and descents.\n\nManipulating nanoparticles by using the AFM was studied and by plotting the static deflections and sensed voltage in various steps of manipulation, it was demonstrated that the bigger the used particle was, the bigger the resulting deflection and voltage were. As the most important result, it was shown that when two mechanical layers were used, the static deflection was greater. The P/0/90/P lamination had the most deflection and the minimum deflection belonged to the P/0/0/0/P lamination. Furthermore, before the step \"E\" (fifth to nine steps), various laminations led to approximately the same results for various diameters. The sensed voltage depended not only on particle size, but also on the structure of the MC. The sensed voltage increased with an increase in the particle diameter, but the effects of irregularity could not be addressed precisely.\n\nReferences\n\n    Uchino K. Piezoelectric Actuators 2004–Materials, Design, Drive/Control, Modeling and Applications. Proc 9th Int Conf New Actuators; 2004.\n\n    Lee S, Kim J, Moon W, Choi J, Park I, Bae D. A multibody-based dynamic simulation method for electrostatic actuators. Nonlinear Dyn 2008; 54: 53-68.\n\n    Lim YH, Varadan VV, Varadan VK. Finite-element modeling of the transient response of MEMS sensors. Smart Mater Struct 1997; 6: 53-61.\n\n    Beek JV, Puers R. A review of MEMS oscillators for frequency reference and timing applications. J Micro-mechanics Micro-engineering 2011; 22: 013001.\n\n    Korayem M, Rahneshin V, Sadeghzadeh S. Coarse-grained molecular dynamics simulation of automatic nanomanipulation process: The effect of tip damage on the positioning errors. Comput Mater Sci 2012; 60: 201-211.\n\n    Korayem M, Rahneshin V, Sadeghzadeh S. Nano cluster manipulation success considering flexibility of system: Coarse grained molecular dynamics simulations. Scientia Iranica 2012; 19: 1288-1298.\n\n    Darvizeh M, Darvizeh A, Ansari R, Sharma C. Buckling analysis of generally laminated composite plates (generalized differential quadrature rules versus Rayleigh–Ritz method). Compos Struct 2004; 63: 69-74.\n\n    Hosseini-Hashemi S, Fadaee M, Taher HRD. Exact solutions for free flexural vibration of Lévy-type rectangular thick plates via third-order shear deformation plate theory. Appl Math Model 2011; 35: 708-727.\n\n    Hashemi SH, Arsanjani M. Exact characteristic equations for some of classical boundary conditions of vibrating moderately thick rectangular plates. Int J Solids Struct 2005; 42: 819-853.\n\nTornabene F. Free vibrations of anisotropic doubly-curved shells and panels of revolution with a free-form meridian resting on Winkler–Pasternak elastic foundations. Compos Struct 2011; 94: 186-206.\n\nBalamurugan V, Narayanan S. A piezolaminated composite degenerated shell finite element for active control of structures with distributed piezosensors and actuators. Smart Mater Struct 2008; 17: 035031.\n\nBenjeddou A. Advances in piezoelectric finite element modeling of adaptive structural elements: a survey. Comput Struct 2000; 76: 347-363.\n\nTornabene F. 2D GDQ solution for free vibrations of anisotropic doubly-curved shells and panels of revolution. Compos Struct 2011; 93: 1854-1876.\n\nHong C. Computational approach of piezoelectric shells by the GDQ method. Compos Struct 2010; 92: 811-816.\n\nChen C. A differential quadrature element method. Proc 1st Int Conf Eng Comput Sim; 1995.\n\nKorayem M, Sadeghzadeh S, Homayooni A. Semi-analytical motion analysis of nano-steering devices, segmented piezotube scanners. Int J Mech Sci 2011; 53: 536-548.\n\nKorayem M, Homayooni A, Sadeghzadeh S, Safa M, Rahneshin V. A semi-analytic modeling of nonlinearities for nano-robotic applications, macro and micro sized systems. 2nd Int Conf Control, Instrumentation Automation; 2011.\n\nSadeghzadeh S, Korayem MH, Rahneshin V, Homayooni A, Moradi M. Nanorobotic Applications of Finite Element Method. Computational Finite Element Methods in Nanotechnology, Editor: Musa S. CRC Press: Taylor and Francis Corporation; 2012.\n\nSadeghzadeh S, Korayem M, Rahneshin V, Homayooni A. A shape-feedback approach for more precise automatic nano manipulation process. 2nd Int Conf Control, Instrumentation and Automation; 2011.\n\nHamed S, Ghader R. Comparison of generalized differential quadrature and Galerkin methods for the analysis of micro-electro-mechanical coupled systems. Commun Nonlinear Sci Numer Sim 2009; 14: 2807-2816.\n\nCollinger J, Wickert JA, Corr L. Adaptive piezoelectric vibration control with synchronized switching. J Dyn Sys Measurement Control 2009; 131: 041006.\n\nPrecision of nano-metric operations is an important issue in nano-engineering studies. Several operative parameters might affect the quality of results. The parameters of the nano world are significant but not entirely controllable. However, the geometrical and mechanical properties of micro cantilevers are completely controllable. So, controlling the sensitivity of resulting image through t lamination design could be a proper approach. This paper analyses the effects of composite lamination on the performance of common Micro and Nano Electro Mechanical systems (MEMS and NEMS, respectively). Generalized Differential Quadrature (GDQ) and Generalized Differential Quadrature Element (GDQE) methods are used as semi-analytic solutions for regular and irregular domains, respectively. Validity, applicability and accuracy of the proposed approach are demonstrated and then the lamination effects on the nano-imaging and manipulation of nano particles by micro cantilevers are studied. This study shows that some laminations of micro cantilevers resulted in a better performance in nano-manipulation and imaging. Furthermore, clarifying the dependency of system sensitivity on the profile of the substrate is remarkable.\n\n#### References\n\n          Uchino K. Piezoelectric Actuators 2004–Materials, Design, Drive/Control, Modeling and Applications. Proc 9th Int Conf New Actuators; 2004.\n          Lee S, Kim J, Moon W, Choi J, Park I, Bae D. A multibody-based dynamic simulation method for electrostatic actuators. Nonlinear Dyn 2008; 54: 53-68.\n          Lim YH, Varadan VV, Varadan VK. Finite-element modeling of the transient response of MEMS sensors. Smart Mater Struct 1997; 6: 53-61.\n          Beek JV, Puers R. A review of MEMS oscillators for frequency reference and timing applications. J Micro-mechanics Micro-engineering 2011; 22: 013001.\n          Korayem M, Rahneshin V, Sadeghzadeh S. Coarse-grained molecular dynamics simulation of au-tomatic nanomanipulation process: The effect of tip damage on the positioning errors. Comput Mater Sci 2012; 60: 201-211.\n          Korayem M, Rahneshin V, Sadeghzadeh S. Nano cluster manipulation success considering flexi-bility of system: Coarse grained molecular dy-namics simulations. Scientia Iranica 2012; 19: 1288-1298.\n          Darvizeh M, Darvizeh A, Ansari R, Sharma C. Buckling analysis of generally laminated compo-site plates (generalized differential quadrature rules versus Rayleigh–Ritz method). Compos Struct 2004; 63: 69-74.\n          Hosseini-Hashemi S, Fadaee M, Taher HRD. Ex-act solutions for free flexural vibration of Lévy-type rectangular thick plates via third-order shear deformation plate theory. Appl Math Model 2011; 35: 708-727.\n          Hashemi SH, Arsanjani M. Exact characteristic equations for some of classical boundary condi-tions of vibrating moderately thick rectangular plates. Int J Solids Struct 2005; 42: 819-853.\n        Tornabene F. Free vibrations of anisotropic dou-bly-curved shells and panels of revolution with a free-form meridian resting on Winkler–Pasternak elastic foundations. Compos Struct 2011; 94: 186-206.\n        Balamurugan V, Narayanan S. A piezolaminated composite degenerated shell finite element for active control of structures with distributed pie-zosensors and actuators. Smart Mater Struct 2008; 17: 035031.\n        Benjeddou A. Advances in piezoelectric finite element modeling of adaptive structural ele-ments: a survey. Comput Struct 2000; 76: 347-363.\n        Tornabene F. 2D GDQ solution for free vibra-tions of anisotropic doubly-curved shells and panels of revolution. Compos Struct 2011; 93: 1854-1876.\n        Hong C. Computational approach of piezoelectric shells by the GDQ method. Compos Struct 2010; 92: 811-816.\n        Chen C. A differential quadrature element meth-od. Proc 1st Int Conf Eng Comput Sim; 1995.\n        Korayem M, Sadeghzadeh S, Homayooni A. Semi-analytical motion analysis of nano-steering de-vices, segmented piezotube scanners. Int J Mech Sci 2011; 53: 536-548.\n        Korayem M, Homayooni A, Sadeghzadeh S, Safa M, Rahneshin V. A semi-analytic modeling of nonlinearities for nano-robotic applications, macro and micro sized systems. 2nd Int Conf Control, Instrumentation Automation; 2011.\n        Sadeghzadeh S, Korayem MH, Rahneshin V, Homayooni A, Moradi M. Nanorobotic Applica-tions of Finite Element Method. Computation-al Finite Element Methods in Nanotechnology, Editor: Musa S. CRC Press: Taylor and Francis Corporation; 2012.\n        Sadeghzadeh S, Korayem M, Rahneshin V, Homayooni A. A shape-feedback approach for more precise automatic nano manipulation pro-cess. 2nd Int Conf Control, Instrumentation and Automation; 2011.\n        Hamed S, Ghader R. Comparison of generalized differential quadrature and Galerkin methods for the analysis of micro-electro-mechanical cou-pled systems. Commun Nonlinear Sci Numer Sim 2009; 14: 2807-2816.\n        Collinger J, Wickert JA, Corr L. Adaptive piezoe-lectric vibration control with synchronized switching. J Dyn Sys Measurement Control 2009; 131: 041006." ]

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[ "# DARCY NUMBER\n\nThere are two dimensionless groups given the name Darey number.", null, "where Δp is the pressure drop, D the diameter, ρ fluid density, u fluid velocity, and L the length over which the pressure drop is measured. f is the Fanning friction factor (see Friction Factors).", null, "where D' is the permeability coefficient of a porous medium with units m2/s (see Darcy's Law)." ]

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[ "# Solving the Maxwell equations by the Chebyshev method: a one-step finite-difference time-domain algorithm.\n\nDe Raedt H, Michielsen K, Kole JS, Figge MT (2003) Solving the Maxwell equations by the Chebyshev method: a one-step finite-difference time-domain algorithm. IEEE Transactions on Antennas and Propagation 51, 3155-3160.\n\nAbstract\n\nWe present a one-step algorithm that solves the Maxwell equations for systems with spatially varying permittivity and permeability by the Chebyshev method. We demonstrate that this algorithm may be orders of magnitude more efficient than current finite-difference time-domain algorithms.\n\nBeteiligte Abteilungen und Gruppen\nHKI-Autoren\nIdentifier\n\ndoi: 10.1109/TAP.2003.818809" ]

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[ "# Manuals/calci/IMSEC\n\nIMSEC (ComplexNumber)\n\n•", null, "is any complex number of the form x+iy.\n\n## Description\n\n• This function gives the SECANT value of a complex number.\n• Consider the complex number is the form of", null, "• x & y are the real numbers.\n• 'i' is the imaginary unit", null, "• Also x is called the real part & y is the imaginary part of a complex number.\n• COMPLEX is the function used to convert Real & Imaginary numbers in to a complex number.\n•", null, "is defined by", null, "## Examples\n\nIMSEC(ComplexNumber)\n\n• ComplexNumber is any complex number.\n IMSEC(ComplexNumber) Value IMSEC(\"7+4i\") 0.02762312989997095+i0.024055975650141042 IMSEC(\"7-8i\") 0.0005058121042022911-i0.00044078883863612046 IMSEC(\"3\") -1.0101086659079939+i0\n\n## Related Videos\n\nTrigonometric Form of Complex Numbers" ]

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[ "# No Large Scale Curvature Perturbations during Waterfall of Hybrid Inflation\n\nAli Akbar Abolhasani abolhasani(AT)ipm.ir    Hassan Firouzjahi firouz(AT)ipm.ir Department of Physics, Sharif University of Technology, Tehran, Iran School of Physics, Institute for Research in Fundamental Sciences (IPM), P. O. Box 19395-5531, Tehran, Iran\nApril 13, 2021\n###### Abstract\n\nIn this paper the possibility of generating large scale curvature perturbations induced from the entropic perturbations during the waterfall phase transition of standard hybrid inflation model is studied. We show that whether or not appreciable amounts of large scale curvature perturbations are produced during the waterfall phase transition depend crucially on the competition between the classical and the quantum mechanical back-reactions to terminate inflation. If one considers only the classical evolution of the system we show that the highly blue-tilted entropy perturbations induce highly blue-tilted large scale curvature perturbations during the waterfall phase transition which dominate over the original adiabatic curvature perturbations. However, we show that the quantum back-reactions of the waterfall field inhomogeneities produced during the phase transition dominate completely over the classical back-reactions. The cumulative quantum back-reactions of very small scales tachyonic modes terminate inflation very efficiently and shut off the curvature perturbations evolution during the waterfall phase transition. This indicates that the standard hybrid inflation model is safe under large scale curvature perturbations during the waterfall phase transition.\n\n## I Introduction\n\nInflation proved to be very successful both theoretically Guth:1980zm and observationally Komatsu:2010fb as a theory of early universe. The simplest models of inflation consist of a scalar field which is minimally coupled to gravity. A period of acceleration expansion is obtained if the potential is flat enough to allow for the inflaton field to slowly roll towards its minimum. With sufficient tunings in the parameters of the model, one can achieve 60 number of e-foldings or more to solve the horizon and the flatness problems of the standard cosmology.\n\nHybrid inflation Linde:1993cn ; Copeland:1994vg is an interesting model of inflation containing two scalar fields, the inflaton field and the waterfall field. In Linde’s original hybrid inflation, the energy density during inflation is dominated by the vacuum while the inflaton is slowly rolling. The waterfall field is very heavy compared to the Hubble expansion rate during inflation, , and it quickly rolls to its instantaneous minimum. The potential has the property that once the inflaton field reaches a critical value, , the waterfall field becomes tachyonic triggering an instability and inflation ends quickly thereafter and the systems settles down into its global minimum.\n\nUsually it is assumed that the waterfall field does not play any role in curvature perturbations during inflation and during phase transition. In this picture one basically borrows the technics and the results of single field inflationary models. That is, the super-horizon curvature perturbations, once they leave the Hubble radius, are frozen and remain unchanged until they re-enter the Hubble radius at a later time, such as at the time of CMB decouplings.\n\nHere we would like to examine this picture more closely. We would like to see if hybrid inflation is safe under large scale curvature perturbations during the waterfall phase transition. If one considers only the classical evolution of the system, we show that during the phase transition the highly-blue tilted entropy perturbations can induce large blue-tilted curvature perturbations on super-horizon scales which can completely dominate over the original adiabatic curvature perturbations. However, we show that the quantum back-reactions of the waterfall field inhomogeneities produced during the phase transition become important before the classical back-reactions become relevant. We demonstrate that the cumulative quantum back-reactions of the short-wavelength inhomogeneities are so strong that they uplift the tachyonic instability of the entropy perturbations and the curvature perturbations freezes.\n\nIdeas similar to this line of thought, studying the amplifications of large scale curvature perturbations during preheating, were studied in Taruya:1997iv -Kohri:2009ac and more recently in Levasseur:2010rk .\n\nThe paper is organized as follows. In section II we review the basics of hybrid inflation and obtain the background evolutions of the inflaton and the waterfall fields. In section III the entropy perturbations and in section IV their effects on curvature perturbations are studied. In Section V the classical non-linear back-reactions as well as the quantum mechanical back-reactions are calculated and are compared to each other. Brief conclusions and discussions are followed in section VI.\n\nWhile our work was finished the work by Lyth Lyth:2010ch appeared which has overlaps with our results. See also Fonseca:2010nk which appeared shortly after our work.\n\n## Ii Hybrid Inflation\n\nHere we study the basics of hybrid inflation Linde:1993cn ; Copeland:1994vg and the background field dynamics.\n\n### ii.1 The Potential\n\nThe potential in standard hybrid inflation has the form\n\n V(ϕ,ψ)=λ4(ψ2−M2λ)2+12m2ϕ2+12g2ϕ2ψ2, (1)\n\nwhere is the inflaton field, is the waterfall field and and are dimensionless couplings. The system has a global minimum given by and . Inflation takes place for where is the initial value of the inflaton field and is the critical value of where the waterfall field becomes instantaneously massless. During inflation is very heavy and is stuck to its instantaneous minimum . For the waterfall becomes tachyonic triggering an instability in the system which ends inflation abruptly. Soon after phase transition, the systems settles down to its global minimum and inflation is followed by the (p)reheating phase.\n\nAs in Linde’s realization of hybrid inflation Linde:1993cn , we consider the limit where the inflation is dominated by the vacuum. For this condition to hold one requires that\n\n M2≫λg2m2. (2)\n\nTo solve the flatness and the horizon problem, we assume that inflation proceeds at least for about 60 number of e-foldings. In the vacuum dominated limit the number of e-foldings is given by\n\n Ne≃2π M4λm2pm2ln(ϕiϕc), (3)\n\nwhere with being the Newton’s constant. We assume that is few times so one can basically neglect the logarithmic contribution above.\n\nTo get the correct amplitude of density perturbations, one has to satisfy the COBE normalization for the curvature perturbations . The power spectrum of curvature perturbations is\n\n PR=128π3m6pV3V2ϕ∼g2λ3M10m6p m4, (4)\n\nwhere the relevant quantities are calculated at the time of Hubble radius crossing () at 60 e-folds before the end of inflation. In this picture, it is assumed that the curvature perturbations are frozen once the modes of interest leave the Hubble radius, as have been treated in conventional analysis of hybrid inflation so far. Our main goal in this paper is to examine the validity of this assumption more closely.\n\nWe are interested in the limit where the waterfall field rolls rapidly to its global minimum once the instability is triggered. For this to happen, the absolute value of the mass should be much bigger than the Hubble expansion rate during phase transition so\n\n M3≪λmm2p. (5)\n\n### ii.2 The Background Fields Dynamics\n\nHere we study the classical evolutions of background fields and during inflation and phase transition, see also GarciaBellido:1996qt ; Copeland:2002ku ; Randall:1995dj where somewhat similar analysis were carried out too. In subsection V.2 we study the quantum back-reactions to the the system in the Hartree approximation.\n\nThe equations of motion for and are\n\n ¨ϕ+3H˙ϕ+(m2+g2ψ2)ϕ=0 (6) ¨ψ+3H˙ψ+(−M2+g2ϕ2+λψ2)ψ=0. (7)\n\nWith the assumption of vacuum dominated potential, the Hubble expansion rate is nearly constant during inflation, . It is more convenient to use the number of e-foldings as the clock, and the background fields equations are now written as\n\n ϕ′′+3ϕ′+(α+g2ψ2H20)ϕ=0 (8) ψ′′+3ψ′+(−β+g2ϕ2H20+λψ2H20)ψ=0, (9)\n\nwhere the dimensionless parameters and are defined as\n\n α≡m2H20=3λm2m2p2πM4≃3Ne,β≡M2H20=3λm2p2πM2, (10)\n\nand the prime denotes the differentiation with respect to the number of e-foldings.\n\nUsing the rapid waterfall condition Eq. (5) combined with the expression for the total number of e-foldings Eq. (3) one obtains\n\n β≫Ne. (11)\n\nSimilarly, one can check that .\n\nTo simplify the notation, we take the critical point as the reference point and define . We use the convention that at the start of inflation for , , at the time of phase transition and at the end of inflation . With this convention before phase transition whereas afterwards. As assumed, the field is much heavier than during inflation so it rapidly rolls down to and one can simply solve Eq. 8\n\n ϕ(n)≃ϕcexp(−r n) (12)\n\nwith\n\n r=(32−√94−α)≃α3≃1Ne. (13)\n\nEquivalently, one also has\n\n (14)", null, "Figure 1: The background classical dynamics for hybrid inflation potential Eq. (1) with M=.67×10−7, m=2.5×10−10 and λ=g2=2.5×10−11 in the units where mp=1. The upper solid blue curve shows the amplitude of curvature perturbations, the middle solid green curve shows the background ϕ field evolution and the lower solid red curve shows the behavior of background ψ field, all three curves are obtained from full numerical analysis. The dashed dark blue curve shows our analytical solution for ψ field. The vertical axis is logarithmic and the horizontal axis is the number of e-foldings.\n\nLet us now turn to the dynamics of field. As it can be confirmed from our full numerical results we are in the limit where so the equation for simplifies to\n\n ψ′′+3ψ′+β(e−2rn−1)ψ=0. (15)\n\nWe divide the solution into two regions. First, we solve this equation for the period before phase transition, . The solution of this equation during this period is\n\n (16)\n\nwhere and are the Bessel functions, and are constants of integrations and . As explained previously, we are in the vacuum dominated limit so and . Using the approximations for the Bessel functions with large arguments one finds\n\n ψ≃e−(3−r)n/2 [c′1cos(√βre−rn−12νπ−14π)+c′2sin(√βre−rn−12νπ−14π)]. (17)\n\nThis solution means that has oscillatory behavior with exponentially decaying frequency and exponentially decaying amplitude. This behavior can be seen in Fig. 1.\n\nNow, we solve the evolution for the period after phase transition till end of inflation, . Since inflation ends in few e-folds after reaches the critical point, it is more appropriate to use the small approximation in Eq. (15) and\n\n ψ′′+3ψ′−(2βrn)ψ=0, (18)\n\nwhich has the following solution\n\n ψ≃e−3n/2⎡⎢⎣C1Ai⎛⎜⎝ϵ2/3ψn+94ϵ4/3ψ⎞⎟⎠+C2Bi⎛⎜⎝ϵ2/3ψn+94ϵ4/3ψ⎞⎟⎠⎤⎥⎦, (19)\n\nin which\n\n ϵψ≡√2rβ≃√23αβ, (20)\n\nand Ai(x) and Bi(x) are Airy functions of first and second kind respectively. For , which is the case in our analysis to satisfy the water-fall condition, the second term in the argument of Airy function can be ignored. Since Ai(x) is a damping function for we just keep the term containing Bi(x). In a good approximation the solution can be read as\n\n ψ(n)≃ψ(Nc)Bi(ϵ2/3ψn) e−3n/2. (21)\n\nwhere . By using the asymptotic behavior of for large , ,\n\n Bi(z)∝e2/3z3/2√π4√z, (22)\n\nand keeping just the exponential dependence one finds that for\n\n ψ≃ψiexp(−3−r2Nc) 1ϵ1/6ψn1/4 exp(−32n+2ϵψ3n3/2), (23)\n\nwhere to get the final result Eq.(12) have been used.", null, "Figure 2: Here the contents of Fig. 1 are shown in the last few e-foldings after the phase transition at Nc≃56. Here we see that, considering only the classical fields dynamics, R grows rapidly at the end of inflation until the condition (56) is met when R′≃0 at N≃63.5. This is followed by a short period of inflation when condition (57) is met and the non-linear corrections to ψ dynamics settles it down to the global minimum ending inflation.\n\n## Iii The entropy perturbations\n\nHere we look into entropy perturbation following Gordon00 closely. In the field space of one can perform local fields rotations such that\n\n δσ=cosθδϕ+sinθδψ,δs=−sinθδϕ+cosθδψ, (24)\n\nsuch that and . In this picture, and represent, respectively, the adiabatic and the entropic perturbations. One can check that the evolution of the curvature perturbation for a mode with momentum is\n\n ˙R=H˙Hk2a2Ψ+2H˙σ˙θδs, (25)\n\nwhere\n\n ˙θ=−Vs/˙σ (26)\n\nwith . Eq. (25) indicates that in the presence of large entropy perturbations or sharp turns in field space, the curvature perturbations can change on super-horizon scales.\n\nThe equation of entropy perturbations is Gordon00\n\n ¨δs+3H˙δs+(k2a2+Vss+3˙θ2)δs=˙θ˙σk22πGa2Ψ, (27)\n\nin which .", null, "Figure 3: Dynamics of angle in the (ϕ,ψ) field space with the same parameters as in Fig. 1 : The upper solid black curve and the lower solid blue curve, respectively, show the full numerical solutions of |lnθ| and |lnθ′|. The upper dashed green curve and the lower dashed red curve, respectively, show the evolution of the corresponding quantities obtained from our analytical solutions, Eqs. (28) and (29). Inflation ends when θ∼1 indicated in the graph by the vertical line.\n\nWe are interested in solving this equation for . As we shall see later, these conditions are true until the end of inflation which can also be seen from Fig. 3. For the later references, from Eq. (12) and Eq.(23), one can easily find that\n\n θ≃tanθ=ψ′ϕ′=−1r(ψiϕi)exp(−3−3r2Nc)(ϵ5/6ψn1/4) exp(2ϵψ3n3/2−3−2r2n), (28)\n\nwhich for or (after phase transition) results in:\n\n θ′≃tanθ′=−1r(ψiϕi)exp(−3−3r2Nc)(ϵ11/6ψn3/4) exp(2ϵψ3n3/2−3−2r2n). (29)\n\nNeglecting in Eq. (27), the equation governing the evolution of entropy fluctuations is the same as the equation of background field, Eq. (15). This indicates that, although the entropy perturbations are suppressed before the phase transition, but they becomes highly tachyonic during the phase transition. The subsequent tachyonic enhancement of the entropy perturbations overcome their initial suppression prior to phase transition. This is the key effect to obtain large curvature perturbations induced from the entropy perturbations at the classical level. However, as we shall see in V.2, one should also take into account the quantum back-reactions which can uplift the tachyonic instability of the background fields and modify the evolution significantly.\n\nThe differences between and evolutions therefore are only in the initial conditions. For a given mode, we consider the evolution of the entropy fluctuation when it leaves the Hubble radius at the moment of Hubble radius crossing , so for , similar to Eq. (23), one obtains\n\n δs(n)≃δs∗ (ϕiϕc)1/2 exp(3−r2N∗)exp(−32Nc)exp(−3n2+2ϵψ3n3/2). (30)\n\nWe note that the contribution represents the tachyonic enhancement of the entropy perturbations during the waterfall phase transition whereas the factors indicates the suppression of the entropy perturbations from the time of horizon crossing till the onset of phase transition. Both of these competitive behaviors can be seen in Fig. 1.\n\nTo find the final amplitude of entropy perturbations, we need to find their amplitude at the time of Hubble radius crossing, . To do this we note that the entropy perturbations, which are basically the waterfall field perturbations, are very heavy during inflation. As explained previously, this causes their suppression before the phase transition which should be taken into account. During inflation so . Rewriting the mode equation in conformal time and defining the equation of entropy perturbation is\n\n ∂2vk∂η2+[k2−(2−m2ψ(ψ)H2)1η2]vk=0 (31)\n\nin which\n\n m2ψ(ϕ)H2≃β(ϕ2ϕ2c−1). (32)\n\nSince we are interested in momenta which exit the horizon during first few e-folds of inflation the above ratio is nearly constant and\n\n m2ψ(ϕ)H2≃β(ϕ2iϕ2c−1)≡~β∼β. (33)\n\nBy these considerations and noting that one has\n\n ∂2vk∂η2+[k2+~βη2]vk=0. (34)\n\nAs the frequency of this equation changes adiabatically, one can use the WKB approximation with the Bunch-Davis vacuum for initial times or and obtains\n\n vk≃14√4k2+4~β/η2 e±i∫√k2+~β/η2dη. (35)\n\nUsing this solution the amplitude of the entropy perturbation at the time of Hubble radius crossing, , is obtained to be\n\n δs∗≃H√2k31~β1/4∣∣∗. (36)\n\nThe extra factor represents the suppression of the entropy perturbations before the phase transition.\n\nAssuming that the scale factor at the start of inflation, e-folds before the end of inflation, is unity, , one also has . The final amplitude of entropy perturbations at the end of inflation therefore is\n\n δsf≃ 1~β1/4H√2k3(kH)(3−r)/2∗(ϕiϕc)1/2e−3/2Nc exp(−32nf+2ϵψ3n3/2f), (37)\n\nwhere denotes the number of e-foldings from the start of phase transition till end of inflation. For a vacuum dominated potential with a quick phase transition, we have .\n\nThe final amplitude of the entropy perturbation is\n\n PS=AS(kH∗)(3−r), (38)\n\nwith\n\n AS=(H22π˙ϕ)21~β1/2(ϕiϕc)e−3Nc exp(−3nf+4ϵψ3n3/2f). (39)\n\nFrom Eq. (38 ) one observes that the entropy perturbations are highly blue-tilted with . In next section we show that at the classical level these highly blue-tilted entropy perturbation induce large blue-tilted spectrum on super-horizon curvature perturbations.\n\n## Iv Power Spectrum of Curvature Perturbations\n\nWe now have all the materials to calculate the final power spectrum of curvature perturbations. For this purpose we need to know the amplitude of adiabatic curvature perturbations at horizon crossing as the initial conditions and integrate the evolutions of curvature perturbation from the time of horizon crossing till end of inflation. The final amplitude of curvature perturbation, therefore, is\n\n Rf=R0+∫nf0R′dn, (40)\n\nwhere represents the adiabatic curvature perturbations in the absence of entropy perturbations.\n\nStarting with Eq. (25), the evolution of curvature perturbations for the super-horizon modes, induced by the entropy perturbations, can be written as\n\n R′=2θ′σ′ δs. (41)\n\nAs one can see from above equation both and can source the curvature perturbations. We also note that represents the acceleration of the field , specially during the phase transition. As can be seen from the full numerical analysis, our classical background is such that during inflation and phase transition, . However, shortly after phase transition rises quickly from its value during inflation to its final value at the global minimum when inflation ends. The rapid rise of and cause inflation to end when become large which can also be seen in bf Fig. 3. Equivalently, this can be interpreted as when the classical as well as quantum back-reactions from and interactions induce large masses for and such that they roll rapidly to the global minimum, ending inflation. Therefore, in the analysis below we work in the limit where and consider the end of inflation when .\n\nTo calculate the evolution of curvature perturbation from Eq. (41) we need to estimate and . The derivative of in field space is\n\n θ′=tanθ′=ψ′′ϕ′−ψ′ϕ′′ϕ′2. (42)\n\nSince , the first term is much larger than the second term by a factor of and . To calculate we observe that its equation has the same form as the background equation and therefore\n\n δs(N)=Ωsψψ(N), (43)\n\nin which\n\n Ωsψ≡δs(N∗)ψ(N∗)=δs∗ψi (kH)3−r2∗ eiδϕ. (44)\n\nThe term represents the phase difference between the oscillations of and the entropy perturbations. This phase difference vanishes after time averaging when we find the final curvature power spectrum.\n\nCombining the above expressions for and , the final curvature perturbation is integrated to\n\n R(Ne)=R0−2∫nf0Ωsψϕ′2ψ′′ψ dn. (45)\n\nAs the function grows more rapidly than the linear exponential, from Eq. 21 one obtains\n\n ψ′′(n)≃ψ(Nc) e−3/2n Bi′′(ϵ2/3ψn). (46)\n\nUsing Eq. (110) and noting that from Eq. (12), the final curvature perturbations is obtained to be\n\n R=R0−2ψ(Nc)2 Ωsψϵ2ψr2ϕ2c∫nf0n e−(3−2r)n Bi2(ϵ2/3ψn) dn. (47)\n\nThe amplitude of quantum fluctuations of adiabatic perturbations at the moment of horizon crossing is\n\n Qσ∗=H√2k3∣∣∗, (48)\n\nwhere represents the Sasaki-Mukhanov variables for the adiabatic perturbations Gordon00 . Using the form of given in Eq. (44), the curvature perturbation calculated from Eq. (47) is\n\n R(nf)=R0⎛⎝1−C(kH)3−3r2∗I(nf)⎞⎠, (49)\n\nwhere is the initial value of adiabatic curvature perturbations at the moment of horizon crossing\n\n R0=H˙ϕH√2k3∣∣∗, (50)\n\nand\n\n C=2ϵ2ψr~β1/4 ψiϕie(−3+3r)Nc. (51)\n\nFurthermore, the integral has the following form\n\n I(nf)=∫nf0n′ e−(3−2r)n′ Bi2(ϵ2/3ψn′) dn′. (52)\n\nEq. (49) has some interesting features. If the second term in the big bracket in Eq. (49) is larger than unity, then the induced curvature perturbations from the entropy perturbations dominate over the adiabatic curvature perturbations. Furthermore, the dominant momentum dependence in the big bracket comes from the initial amplitudes of entropy perturbations and the integral is nearly constant for all momenta. This implies that the induced curvature perturbations from the entropy perturbations are highly blue-tilted.\n\nUsing the integral approximation given by Eq. ( 109) one can calculate the integral approximately and\n\n I(nf)≃√nf2ϵψ e−(3−2r)nf Bi2(ϵ2/3ψn). (53)\n\nPlugging this into Eq. (49), and using the asymptotic behavior of Airy function of second kind, the curvature perturbations at the end of inflation is calculated to be\n\n R(nf)=R0⎡⎢⎣1−(kH)3−3r2∗⎛⎜⎝ϵ2/3ψπr~β1/4⎞⎟⎠(ψiϕi)e(−3+3r)Ncexp(4/3ϵψn3/2f−(3−2r)nf)⎤⎥⎦ (54)\n\nTo get the final curvature power spectrum we need to know , the time of end of inflation. To determine this we proceed as follows. The exponential growth of the modes (background as well as quantum fluctuations) can violate our background solutions for and . The exponential growth of the background field have two important effects. First, through the interaction term , it can induce large mass for field which speeds up its rolling toward the global minimum and violate the slow-roll conditions. The second important effect is that the back-reaction from term induces large positive mass for which uplifts the tachyonic mass of field leading to the deceleration of this field. These two effects jointly terminate both inflation and the growth of the super-horizon curvature perturbations. To see the latter effect we start from Eqs. (42) and (41) where\n\n R′∼ψ′′ψϕ′2. (55)\n\nWe observe that the deceleration of the field as well as the fast-rolling of the field jointly cause the termination of the super-horizon curvature perturbations as explained above.\n\n## V Classical vs. Quantum Mechanical Back-reactions\n\nIn this section we study the back-reactions of classical (zero momentum) mode of field as well as the quantum back-reactions of inhomogeneities produced during phase transition on the dynamics of the system. We examine which of the above two mechanisms dominate sooner to terminate inflation.\n\nWe will demonstrate that the quantum mechanical back-reactions dominate completely over the classical back-reactions and hence the end-point of inflation is determined by quantum back-reactions. However, as we shall see later, the variance of the quantum fluctuations (on all scales) after phase transition has the same time dependence (-dependence) as the classical trajectory and the difference is just in a proportionality factor. In order to demonstrate why the classical back-reactions are secondary we start the analysis with the classical back-reactions.\n\n### v.1 Classical Back-reactions\n\nDue to smallness of background field before phase transition our analysis in II.2 concentrated only on the linear level. Now we add the back-reactions of the non-linear terms and on the dynamics of the system at the classical level.\n\nThe correction from the interaction term becomes important in evolution, Eq.(8), when\n\n α≃g2ψ2(n)H20. (56)\n\nOn the other hand, the non-linear self-interaction term in background evolution,  Eq. (9), becomes important when\n\n βr≃λψ2(n)H20. (57)\n\nHowever, as a consequence of the vacuum domination condition Eq.(11). This indicates that the back-reaction of the field on the inflaton dynamics becomes important sooner before its self-interaction corrections affect its own dynamics. Putting it another way, during hybrid inflation where and are the conventional slow-roll parameters. Due to back-reactions of on inflaton mass the condition is met sooner before find the chance to become order of unity due to its initial smallness during inflation. We have numerically checked that when the condition Eq. (56) is satisfied then . This followed by a very short period of inflation when condition Eq. (57) is met and inflation ends.\n\nCombining Eq.(56) and Eq.(23) one finds the end-point of inflation to be\n\n exp(43ϵψn3/2f−3nf)≃ϵ1/3ψ(αβ)(ϕiψi)2e3Nc. (58)\n\nUsing this in Eq. (54), the amplitude of curvature perturbation at the end of inflation is\n\n Rf≃R0⎡⎣1−(kH)(3−3r)/2∗⎛⎝ϵψαπrβ~β1/4⎞⎠(ϕiψi)⎤⎦. (59)\n\nConsequently, the final power spectrum of curvature perturbations, , is calculate to be\n\n (60)\n\nwhere is the power spectrum of the adiabatic curvature perturbations. As explained previously, during inflation is very heavy so rolls to its instantaneous minimum very quickly and . Below we demonstrate that such that the curvature perturbations induced from the entropy perturbations dominate completely over the adiabatic curvature perturbations. To see this note that for the field to be nearly zero such that its fluctuations do not contribute to the curvature perturbation around e-foldings before the end of inflation, that is at the start of inflation, one requires that\n\n R′R≃θ′δs∗δσ∗≪1,→ψ′′ϕ′δs∗δσ∗≪1. (61)\n\nUsing Eq.(15) and Eq.(8) one can easily find that\n\n θ′i≃βψirϕi (62)\n\nThis in turn results in\n\n ψiϕi≪r~β1/4β. (63)\n\nNow compare the second term in the big bracket in Eq. (60) to the initial curvature perturbations\n\n (64)\n\nNoting that with for super-horizon modes, and using the inequality Eq. (63) and one obtains\n\n ΔPRPR0≫(9e3N∗π2β~β)Ne. (65)\n\nNoting that and , one concludes that . This indicates that, in the limit where only the classical back-reactions are considered to determine the endpoint of inflation, the curvature perturbations induced from the entropy perturbations dominate completely over the original adiabatic curvature perturbations. We shall see in next subsection that this conclusion is not stable against quantum back-reactions.", null, "Figure 4: Here the final power spectrum of curvature perturbations, at the classical level, as a function of comoving momentum for the parameters used in Fig. 1. are shown. The momenta exited approximately in first 6 e-folds. Blue squares shows the full numerical results whereas the solid red curve shows the best fit. The best power law fit is acquired for PR=Akns−1 with ns≃4.01(3.98−4.04) at 95%CL.\n\nWith , the spectral index of curvature perturbations, , is\n\n nR−1=n0R−1+3−3r, (66)\n\nin which is the spectral index calculated from the adiabatic curvature perturbations using , , where and are the standard slow-roll parameters. Using the approximation one has\n\n nR−1=3−1Ne. (67)\n\nThis indicates that the curvature perturbations receive a large blue-tilted spectrum from the entropy perturbations and .\n\nFinally, to calculate the time of end of inflation, , from Eq. (58) we obtain\n\n nf≃[94ϵψNc+34ϵψln(ϵ1/3ψ(αβ)(ϕiψi)2)]2/3 (68)\n\nFor , the second term in the big bracket above can be ignored and one approximately has\n\n nf≃(94ϵψNc)2/3. (69)\n\nIt is also interesting to calculate the angle in phase space at the end of inflation, . Using Eqs. (28) and (58) we obtain\n\n θf≃3e−32rNc∼1. (70)\n\nThis verify our previous claim that inflation ends classically when . This can also be seen from Fig 3.\n\nAs an example consider the vacuum dominated hybrid inflation with parameters , and . From Eq. (67) the spectral index is which is in good agreement with obtained from the full numerical analysis. Also from Fig. 4 one can see that there are good agreements between our analytical results and the results obtained from the full numerical analysis at the classical level. Finally, which is in agreement with the numerical results (see Fig. 2, and so ).\n\n### v.2 Quantum Fluctuations Back-reactions\n\nOur analysis so far concentrated on the classical evolution of the system. Due to tachyonic instability during the phase transition quanta of particles inhomogeneities, , are produce Felder:2000hj ; Felder:2001kt which can back-react on the classical backgrounds as in preheating models Kofman:1994rk ; Traschen:1990sw ; Shtanov:1994ce . If the mechanism of particle creation due to tachyonic instability is very efficient, the back-reaction of the produced particles can induce large mass for inflaton field in the form of where is the expectation value of in the Hartree approximation. This violates the slow-roll conditions and rapidly rolls towards the global minimum. Furthermore, the large back-reactions of also induce a large mass for the background which can uplift its tachyonic mass. Therefore, one has to take into account the quantum back-reactions and see which of the classical or quantum back-reactions dominate first to terminate inflation.\n\nTo handle the quantum back-reactions, we use the Hartree approximation and calculate the effects of all modes which become tachyonic during the phase transition. Let us compute which modes become tachyonic after the phase transition. The equation of the fluctuations at the linear order is\n\n δψ′′k+3δψ′k+(k2a2H2−2βrn)δψk=0 (71)\n\nTherefore modes which satisfy the inequality\n\n kkc≲√2βr=ϵψ (72)\n\nbecome tachyonic soon after the phase transition. Here we defined as the critical mode which leaves the horizon at the time of waterfall phase transition.\n\nWe divide the tachyonic modes into two categories: first, large modes , corresponding to , which exit the Hubble radius sometime before the phase transition and second, the small modes , corresponding to , which do not exit the Hubble radius till time of phase transition. Using Eq. (35), the amplitude of both modes at the time of critical point is\n\n |vk(n=0)|≃14√|4k2−2/η2|, (73)\n\nand therefore for large modes one has\n\n |δψLk(n=0)|≃e−3Nc/2√2H. (74)\n\nThis is in agreement with our previous result, Eq.(37), for . On the other hand, for the small modes which remain sub-horizon till the time of phase transition one has\n\n |δψSk(n=0)|≃e−Nc√2k. (75)\n\nWith this division of the modes, and using Eq. (37), is calculated to be\n\n ⟨δψ2⟩ ≃ [∫kc0d3k(2π)312H0e−3Nc+∫ϵψkckcd3k(2π)312ke−2Nc] exp(−3n+4ϵψ3n3/2) (76) ≡ ⟨δψ2⟩L+⟨δψ2⟩S,\n\nin which we have ignored factors of order unity. The first integral, representing , comes from the large modes which leave the horizon sometimes prior to phase transition whereas the second integral, representing , indicates the contributions from the very small scales modes which are sub-horizon till phase transition.\n\nWe observe that both and have similar -dependence which becomes important in our discussion below when we compare the classical and quantum back-reactions. However,\n\n ⟨δψ2⟩S∼ϵ2ψ⟨δψ2⟩L. (77)\n\nAs the second integral in Eq. (76) is much bigger than the first one. This means that the cumulative contributions of the modes which becomes tachyonic but remained sub-horizon till the time of phase transition are more important in quantum back-reactions and\n\n ⟨δψ2⟩H20≃⟨δψ2⟩SH20∼ϵ2ψ4π2 exp(4ϵψ3n3/2−3n). (78)\n\nAs explained before, there are two competitive corrections which terminate inflation and the evolution of large scale curvature perturbations during the phase transition. The First one is the classical corrections from the interactions which induce large mass for the inflaton, violating the slow-roll condition during the phase transition. This is the effect which was studied in subsection (V.1), specifically Eq. (58). The second contribution is the quantum back-reaction corrections into inflaton mass via . We need to see which of the above two corrections dominate first. Comparing Eq. (58) and Eq. (78), one finds that these two terms have similar time-dependence (-dependence) and\n\n ⟨δψ2(n)⟩ψ2(n)≃⟨δψ2(n)⟩Sψ2(n)≃ϵ2ψg2β(ϕiψi)2e3Nc. (79)\n\nUsing Eq.(11) and Eq.(63), one finds that\n\n ⟨δψ2(n)⟩ψ2(n)≫(g2β3/2)e3Nc. (80)\n\nAs , one concludes that for any reasonable value of coupling the quantum back-reactions dominate completely over the classical back-reactions to terminate inflation. Therefore, the end-point of inflation is determined by the quantum back-reaction effects from and\n\n exp(43ϵψn3/2f−3nf)≃8π2αg2ϵ2ψ. (81)\n\nIt’s now instructive to calculate , the time when the quantum back-reactions terminate inflation. If we follow the estimation of Felder:2000hj ; Felder:2001kt performed in flat backgrounds, we obtain\n\n nf∼1ϵψln8π2αg2. (82)\n\nFor the parameters of our numerical investigations such as in Fig 1 one has . However, taking into account the background cosmological expansion, from our Eq. (81), one finds\n\n nf∼(1ϵψln8π2αϵ2ψg2)2/3, (83)\n\nwhich for the parameters of our numerical studies this gives . This clearly indicates that quantum back-reactions dominate sooner than the classical back-reaction which happens at .\n\nAnother important point which should be considered is that the quantum fluctuations, , can change the effective classical trajectory. As we demonstrated in Eq. (79) the expectation value of the quantum fluctuations dominates completely over the “zero momentum” mode and they will induce new effective trajectory in the field space. The fact that dominates over the background classical field contributions in waterfall dynamics also indicates that using the “zero momentum” as the classical trajectory is not reliable. Quantum modes with tachyonic mass become highly occupied soon after the phase transition and in some senses they become classical. This suggests that we may take as the effective classical field trajectory. Below we justify this proposal so one can introduce an effective classical trajectory defined by .\n\nTo justify our proposal, we start with the conventional method for the evolution of super-horizon curvature perturbations Wands:2000dp . In the study of curvature perturbations in hybrid inflation this method was pioneered in Lyth:2010ch and Lyth:2010zq (see also Gong:2010zf ). In this method, the change in the comoving curvature perturbation on super-horizon scales is given by\n\n ˙Rck=Hδpckρ+p, (84)\n\nwhere is the pressure perturbations on comoving slices. In our formalism we relate to the entropy perturbations or . In the analysis here it is assumed that the field is frozen at the background so classically. Our goal is to demonstrate that on super-horizon scales Eq. (84) reduces to our starting equation for the evolution of curvature perturbation, Eq. (25), obtained in Gordon00 for the two field inflationary system, with the appropriate definition of and replacing .\n\nOne can simply check that , where taking into account the quantum effects, one also has Gong:2010zf . However, as in our previous analysis, the velocity along the inflation trajectory in the field space can be well approximated by till end of inflation. Therefore\n\n ˙Rck=H˙σδpck˙σ. (85)\n\nAs the potential is vacuum dominated we can also neglect the gravitational back-reactions on . Neglecting the self interaction term, and noting that the comoving slice coincides with the surface Gong:2010zf , the contribution of -field in the pressure becomes\n\n δpc=12˙δψ2+12(M2−g2ϕ2)δψ2. (86)\n\nUsing Eq. 30 for the time evolution of mode (which is the same as mode) one finds that the contribution of the kinetic term above is approximately equal to the potential term and\n\n δpck≃(M2−g2ϕ2)(" ]

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[ "Is 72 a prime number? What are the divisors of 72?\n\n## Parity of 72\n\n72 is an even number, because it is evenly divisible by 2: 72 / 2 = 36.\n\nFind out more:\n\n## Is 72 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 72 is about 8.485.\n\nThus, the square root of 72 is not an integer, and therefore 72 is not a square number.\n\n## What is the square number of 72?\n\nThe square of a number (here 72) is the result of the product of this number (72) by itself (i.e., 72 × 72); the square of 72 is sometimes called \"raising 72 to the power 2\", or \"72 squared\".\n\nThe square of 72 is 5 184 because 72 × 72 = 722 = 5 184.\n\nAs a consequence, 72 is the square root of 5 184.\n\n## Number of digits of 72\n\n72 is a number with 2 digits.\n\n## What are the multiples of 72?\n\nThe multiples of 72 are all integers evenly divisible by 72, that is all numbers such that the remainder of the division by 72 is zero. There are infinitely many multiples of 72. The smallest multiples of 72 are:\n\n• 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 72 too, since 0 × 72 = 0\n• 72: indeed, 72 is a multiple of itself, since 72 is evenly divisible by 72 (we have 72 / 72 = 1, so the remainder of this division is indeed zero)\n• 144: indeed, 144 = 72 × 2\n• 216: indeed, 216 = 72 × 3\n• 288: indeed, 288 = 72 × 4\n• 360: indeed, 360 = 72 × 5\n• etc.\n\n## How to determine whether an integer is a prime number?\n\nTo determine the primality of a number, several algorithms can be used. The most naive technique is to test all divisors strictly smaller to the number of which we want to determine the primality (here 72). First, we can eliminate all even numbers greater than 2 (and hence 4, 6, 8…). Then, we can stop this check when we reach the square root of the number of which we want to determine the primality (here the square root is about 8.485). Historically, the sieve of Eratosthenes (dating from the Greek mathematics) implements this technique in a relatively efficient manner.\n\nMore modern techniques include the sieve of Atkin, probabilistic algorithms, and the cyclotomic AKS test.\n\n## Numbers near 72\n\n• Preceding numbers: …70, 71\n• Following numbers: 73, 74\n\n### Nearest numbers from 72\n\n• Preceding prime number: 71\n• Following prime number: 73\nFind out whether some integer is a prime number" ]

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[ "# Number 4112\n\nNumber 4,112 spell 🔊, write in words: four thousand, one hundred and twelve . Ordinal number 4112th is said 🔊 and write: four thousand, one hundred and twelfth. The meaning of number 4112 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 4112. What is 4112 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 4112.\n\n## What is 4,112 in other units\n\nThe decimal (Arabic) number 4112 converted to a Roman number is (IV)CXII. Roman and decimal number conversions.\n The number 4112 converted to a Mayan number is", null, "Decimal and Mayan number conversions.\n\n#### Weight conversion\n\n4112 kilograms (kg) = 9065.3 pounds (lbs)\n4112 pounds (lbs) = 1865.2 kilograms (kg)\n\n#### Length conversion\n\n4112 kilometers (km) equals to 2556 miles (mi).\n4112 miles (mi) equals to 6618 kilometers (km).\n4112 meters (m) equals to 13491 feet (ft).\n4112 feet (ft) equals 1254 meters (m).\n4112 centimeters (cm) equals to 1618.9 inches (in).\n4112 inches (in) equals to 10444.5 centimeters (cm).\n\n#### Temperature conversion\n\n4112° Fahrenheit (°F) equals to 2266.7° Celsius (°C)\n4112° Celsius (°C) equals to 7433.6° Fahrenheit (°F)\n\n#### Power conversion\n\n4112 Horsepower (hp) equals to 3023.96 kilowatts (kW)\n4112 kilowatts (kW) equals to 5591.52 horsepower (hp)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n4112 seconds equals to 1 hour, 8 minutes, 32 seconds\n4112 minutes equals to 2 days, 20 hours, 32 minutes\n\n### Zip codes 4112\n\n• Zip code 04112 Portland, Maine, Cumberland, USA a map\n• Zip code 4112 Maragondon, Philippines a map\n\n### Codes and images of the number 4112\n\nNumber 4112 morse code: ....- .---- .---- ..---\nSign language for number 4112:", null, "", null, "", null, "", null, "Number 4112 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 4112 infographic", null, "### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)\n\nGregorian year 4112 is Buddhist year 4655.\nBuddhist year 4112 is Gregorian year 3569 .\nGregorian year 4112 is Islamic year 3597 or 3598.\nIslamic year 4112 is Gregorian year 4611 or 4612.\nGregorian year 4112 is Persian year 3490 or 3491.\nPersian year 4112 is Gregorian 4733 or 4734.\nGregorian year 4112 is Hebrew year 7872 or 7873.\nHebrew year 4112 is Gregorian year 352.\nThe Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.\n\n## Share in social networks", null, "## Mathematics of no. 4112\n\n### Multiplications\n\n#### Multiplication table of 4112\n\n4112 multiplied by two equals 8224 (4112 x 2 = 8224).\n4112 multiplied by three equals 12336 (4112 x 3 = 12336).\n4112 multiplied by four equals 16448 (4112 x 4 = 16448).\n4112 multiplied by five equals 20560 (4112 x 5 = 20560).\n4112 multiplied by six equals 24672 (4112 x 6 = 24672).\n4112 multiplied by seven equals 28784 (4112 x 7 = 28784).\n4112 multiplied by eight equals 32896 (4112 x 8 = 32896).\n4112 multiplied by nine equals 37008 (4112 x 9 = 37008).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 4112\n\nHalf of 4112 is 2056 (4112 / 2 = 2056).\nOne third of 4112 is 1370,6667 (4112 / 3 = 1370,6667 = 1370 2/3).\nOne quarter of 4112 is 1028 (4112 / 4 = 1028).\nOne fifth of 4112 is 822,4 (4112 / 5 = 822,4 = 822 2/5).\nOne sixth of 4112 is 685,3333 (4112 / 6 = 685,3333 = 685 1/3).\nOne seventh of 4112 is 587,4286 (4112 / 7 = 587,4286 = 587 3/7).\nOne eighth of 4112 is 514 (4112 / 8 = 514).\nOne ninth of 4112 is 456,8889 (4112 / 9 = 456,8889 = 456 8/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 4112\n\n#### Is Prime?\n\nThe number 4112 is not a prime number. The closest prime numbers are 4111, 4127.\n4112th prime number in order is 39047.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 4112 are 2 * 2 * 2 * 2 * 257\nThe factors of 4112 are 1 , 2 , 4 , 8 , 16 , 257 , 514 , 1028 , 2056 , 4112\nTotal factors 10.\nSum of factors 7998 (3886).\n\n#### Powers\n\nThe second power of 41122 is 16.908.544.\nThe third power of 41123 is 69.527.932.928.\n\n#### Roots\n\nThe square root √4112 is 64,124878.\nThe cube root of 34112 is 16,020806.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 4112 = loge 4112 = 8,321665.\nThe logarithm to base 10 of No. log10 4112 = 3,614053.\nThe Napierian logarithm of No. log1/e 4112 = -8,321665.\n\n### Trigonometric functions\n\nThe cosine of 4112 is -0,941149.\nThe sine of 4112 is 0,337993.\nThe tangent of 4112 is -0,359128.\n\n### Properties of the number 4112\n\nMore math properties ...\n\n## Number 4112 in Computer Science\n\nCode typeCode value\nPIN 4112 It's recommendable to use 4112 as a password or PIN.\n4112 Number of bytes4.0KB\nUnix timeUnix time 4112 is equal to Thursday Jan. 1, 1970, 1:08:32 a.m. GMT\nIPv4, IPv6Number 4112 internet address in dotted format v4 0.0.16.16, v6 ::1010\n4112 Decimal = 1000000010000 Binary\n4112 Decimal = 12122022 Ternary\n4112 Decimal = 10020 Octal\n4112 Decimal = 1010 Hexadecimal (0x1010 hex)\n4112 BASE64NDExMg==\n4112 MD5fdeea652a89ec3e970d22a86698ac8c4\n4112 SHA2243160e6a6ccfb16e5604dfb81a4e4cee9043824d21aff82b4f1ff4781\n4112 SHA256059f5e543bd484b00e235edc5083558c59d45706bd99c14422bda8868c6c6f90\n4112 SHA384beeeea812bb077f98e64572083d35865b90b341a901c8cf0c7bca7bb87e87bfba06bdc79f512181eac835fbc55417584\nMore SHA codes related to the number 4112 ...\n\nIf you know something interesting about the 4112 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 4112\n\n### The meaning of the number 1 (one), numerology 1\n\nCharacter frequency 1: 2\n\nNumber one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.\nMore about the meaning of the number 1 (one), numerology 1 ...\n\n### The meaning of the number 4 (four), numerology 4\n\nCharacter frequency 4: 1\n\nThe number four (4) came to establish stability and to follow the process in the world. It needs to apply a clear purpose to develop internal stability. It evokes a sense of duty and discipline. Number 4 personality speaks of solid construction. It teaches us to evolve in the tangible and material world, to develop reason and logic and our capacity for effort, accomplishment and work.\nMore about the meaning of the number 4 (four), numerology 4 ...\n\n### The meaning of the number 2 (two), numerology 2\n\nCharacter frequency 2: 1\n\nThe number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, super-sensitivity towards the needs of others.\n\nThe number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.\nMore about the meaning of the number 2 (two), numerology 2 ...\n\n## Interesting facts about the number 4112\n\n### Asteroids\n\n• (4112) Hrabal is asteroid number 4112. It was discovered by M. Mahrová from Klet Observatory on 9/25/1981.\n\n### Distances between cities\n\n• There is a 4,112 miles (6,617 km) direct distance between Anshan (China) and Zaporizhzhya (Ukraine).\n• There is a 4,112 miles (6,617 km) direct distance between Aurangabad (India) and Hamburg (Germany).\n• There is a 4,112 miles (6,617 km) direct distance between Baghdad (Iraq) and Huainan (China).\n• There is a 4,112 miles (6,617 km) direct distance between Baghdad (Iraq) and Qiqihar (China).\n• More distances between cities ...\n• There is a 4,112 miles (6,617 km) direct distance between Bangalore (India) and Kraków (Poland).\n• There is a 2,556 miles (4,112 km) direct distance between Bekasi (Indonesia) and Chengdu (China).\n• There is a 2,556 miles (4,112 km) direct distance between Cairo (Egypt) and Lomé (Togo).\n• There is a 2,556 miles (4,112 km) direct distance between Ciudad Guayana (Venezuela) and Gustavo A. Madero (Mexico).\n• There is a 2,556 miles (4,112 km) direct distance between Dombivli (India) and Chelyabinsk (Russia).\n• There is a 2,556 miles (4,112 km) direct distance between Gaziantep (Turkey) and Kampala (Uganda).\n• There is a 2,556 miles (4,112 km) direct distance between Ghāziābād (India) and Hangzhou (China).\n• There is a 2,556 miles (4,112 km) direct distance between Ghāziābād (India) and Johor Bahru (Malaysia).\n• There is a 2,556 miles (4,112 km) direct distance between Guankou (China) and Salem (India).\n• There is a 2,556 miles (4,112 km) direct distance between Guankou (China) and Tiruchirappalli (India).\n• There is a 2,556 miles (4,112 km) direct distance between Gwangju (South Korea) and Patna (India).\n• There is a 2,556 miles (4,112 km) direct distance between Chelyabinsk (Russia) and Thāne (India).\n• There is a 4,112 miles (6,617 km) direct distance between Chittagong (Bangladesh) and Sofia (Bulgaria).\n• There is a 4,112 miles (6,617 km) direct distance between Kiev (Ukraine) and Shiyan (China).\n• There is a 2,556 miles (4,112 km) direct distance between Lijiang (China) and Makassar (Indonesia).\n• There is a 4,112 miles (6,617 km) direct distance between Mombasa (Kenya) and Yangon (Burma / Myanmar).\n• There is a 4,112 miles (6,617 km) direct distance between Munich (Germany) and Nagpur (India).\n• There is a 4,112 miles (6,617 km) direct distance between Naucalpan de Juárez (Mexico) and Santiago (Chile).\n• There is a 2,556 miles (4,112 km) direct distance between Tabrīz (Iran) and Vijayawāda (India).\n\n### Mathematics\n\n• 4112 is the number of necklaces possible with 17 beads, each being one of 2 colors.\n\n## Number 4,112 in other languages\n\nHow to say or write the number four thousand, one hundred and twelve in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 4.112) cuatro mil ciento doce German: 🔊 (Anzahl 4.112) viertausendeinhundertzwölf French: 🔊 (nombre 4 112) quatre mille cent douze Portuguese: 🔊 (número 4 112) quatro mil, cento e doze Chinese: 🔊 (数 4 112) 四千一百一十二 Arabian: 🔊 (عدد 4,112) أربعة آلاف و مائة و اثنا عشر Czech: 🔊 (číslo 4 112) čtyři tisíce sto dvanáct Korean: 🔊 (번호 4,112) 사천백십이 Danish: 🔊 (nummer 4 112) firetusinde og ethundrede og tolv Hebrew: (מספר 4,112) ארבע אלף מאה ושנים עשרה Dutch: 🔊 (nummer 4 112) vierduizendhonderdtwaalf Japanese: 🔊 (数 4,112) 四千百十二 Indonesian: 🔊 (jumlah 4.112) empat ribu seratus dua belas Italian: 🔊 (numero 4 112) quattromilacentododici Norwegian: 🔊 (nummer 4 112) fire tusen, en hundre og tolv Polish: 🔊 (liczba 4 112) cztery tysiące sto dwanaście Russian: 🔊 (номер 4 112) четыре тысячи сто двенадцать Turkish: 🔊 (numara 4,112) dörtbinyüzoniki Thai: 🔊 (จำนวน 4 112) สี่พันหนึ่งร้อยสิบสอง Ukrainian: 🔊 (номер 4 112) чотири тисячi сто дванадцять Vietnamese: 🔊 (con số 4.112) bốn nghìn một trăm mười hai Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 4112 or any natural number (positive integer) please write us here or on facebook." ]

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[ "# Electrodynamics\n\n## 1st Edition\n\nAuthors:\neBook ISBN: 9780323152747\nPublished Date: 1st January 1952\nPage Count: 386\nSales tax will be calculated at check-out Price includes VAT/GST\n43.99\n30.79\n30.79\n30.79\n35.19\n30.79\n30.79\n35.19\n72.95\n51.06\n51.06\n51.06\n58.36\n51.06\n51.06\n58.36\n54.95\n38.47\n38.47\n38.47\n43.96\n38.47\n38.47\n43.96\nUnavailable\nPrice includes VAT/GST\n\n## Description\n\nLectures on Theoretical Physics provides an overview of the fundamental principles of electrodynamics. It presents biographical notes on several scientists, including Michael Faraday, James Clerk Maxwell, Heinrich Hertz, and André Marie Ampère. The book is comprised of four parts encompassing 38 chapters. Part One explains Maxwell’s equation as an axiomatic basis, in the coordinates and in differential form, but in integral form. Part Two discusses the various classes of phenomena in stationary, quasi-stationary, static, and rapidly variable fields. It also distinguishes between summation and boundary-value problems in electrostatics and magnetostatics. Part Three presents the four-dimensional form of electrodynamics as the basic introduction to the theory of relativity. It also considers the fundamental link between the dynamics of the individual electron and Maxwell’s theory. Finally, Part Four deals with the electrodynamics of moving media. This book is a valuable resource to scientists, researchers, and individuals working in the field of theoretical physics.\n\nPreface\n\nTranslator's Note\n\nPart I. Fundamentals and Basic Principles of Maxwell's Electrodynamics\n\n§1. Historical Review. Action at a Distance and Action by a Field\n\nBiographical Notes\n\n§2. Introduction to the Basic Concepts of the Electromagnetic Field\n\n§3. Maxwell's Equations in Integral Form\n\n§4. The Maxwell Equations in Differential Form and the Material Constants of the Theory\n\n1. Conductivity and Ohm's Law\n\n2. Dielectric Constant\n\n3. Permeability\n\n§5. Law of Conservation of Energy and Pointing Vector\n\n§6. The Role of the Velocity of Light in Electrodynamics\n\n§7. The Coulomb Field and the Fundamental Constants of Vacuum. Rational and Conventional Units\n\nA. Electrostatics\n\nB. Magnetostatics\n\nC. Rational and Conventional Units\n\nD. Final Determination of the Fundamental Constants ε0, μ0 in the MKSQ System\n\n§8. Four, Five, or Three Fundamental Units?\n\nA. Supplementary Note on Our System of Four Units\n\nB. The Five Units MKSQP\n\nC. The Gaussian System of Only Three Units\n\nD. Supplement Regarding Other Systems of Units\n\nPart II. Derivation of the Phenomena from the Maxwell Equations\n\n§9. The Simplest Boundary-Value Problems of Electrostatics\n\nA. Charging Problems\n\nB. Induction Problems and Method of Reciprocal Radii\n\nC. Conducting Sphere in a Uniform Field\n\nD. Dielectric Sphere in a Uniform Field\n\nE. Reflection and Refraction of Lines of Force at the Boundary of a Semi-infinite Dielectric\n\n§10. Capacity and Its Connection with Field Energy\n\nA. The Plate Condenser\n\nB. Spherical Condenser\n\nC. Capacity of an Ellipsoid of Revolution and of a Straight Piece of Wire\n\nD. Energetic Definition of Capacity\n\nE. The Capacities in an Arbitrary System of Conductors\n\n§11. General Considerations on the Electric Field\n\nA. The Law of Refraction for the Lines of Force\n\nB. On the Definition of the Vectors E and D\n\nC. The Concept of Electric Polarization; the Clausius-Mossotti Formula\n\nD. Supplement to the Calculation of the Polarization\n\nE. Permanent Polarization\n\n§12. The Field of the Permanent Bar Magnet\n\n§13. General Considerations on Magnetostatics and Corresponding Boundary-Value Problems\n\nA. The Law of Refraction of the Lines of Magnetic Excitation\n\nB. Definition of the Vectors H and Β, Particularly in Solid Bodies\n\nC. The Magnetization M in Any Non-Ferromagnetic Substance\n\nD. Dia- and Paramagnetism\n\nE. Soft Iron as Analog to the Electric Conductor\n\nF. Specific Boundary-Value Problems\n\nG. The Uniform Field within an Ellipsoid of Revolution\n\nH. The So-Called Demagnetization Factor\n\n§14. Some Remarks on Ferromagnetism\n\nA. The Weiss Domains\n\nB. The Electron Spin as Elementary Magnet\n\nC. Hysteresis Loop and Reversible Magnetization\n\nD. Thermodynamics\n\n§15. Stationary Currents and Their Magnetic Field. Method of the Vector Potential\n\nA. The Law of Biot-Savart\n\nΒ. The Magnetic Energy of the Field of Two Conductors\n\nC. Neumann's Potential as Coefficient of Mutual Induction\n\nD. The Coefficient of Self induction\n\nE. Selfinductance of the Two-Wire Line\n\nF. General Theorem Regarding Energy Transmission by Stationary Currents\n\n§16. Ampere's Method of the Magnetic Double Layer\n\nA. The Magnetic Shell for Linear Conductors\n\nB. Magnetic Energy and Magnetic Flux\n\nC. Application to the Self inductance of a Two-Wire Line\n\nD. Application to the Electromagnetic Current Measurement of Wilhelm Weber\n\n§17. Detailed Treatment of the Field of a Straight Wire and of a Coil\n\n§18. Quasi-Stationary Currents\n\nA. Energetic Interpretation of the Wave Equation\n\nΒ. The Wheatstone Bridge\n\nC. Coupled Circuits\n\nD. The Telegraph Equation\n\n§19. Rapidly Variable Fields. The Electrodynamic Potentials\n\nA. The Retarded Potentials\n\nB. The Hertzian Dipole\n\nC. Specialization for Periodic Processes\n\nD. The Characteristic Vibrations of a Metallic Spherical Oscillator\n\nE. Application to the Theory of X-Rays\n\n§20. General Considerations on the Structure of Wave Fields of Cylindrical Symmetry. Details on Alternating Current Impedance and Skin Effect\n\nA. Longitudinal and Transverse Components\n\nB. The Wave Field of Semi-infinite Space and Its Skin Effect\n\nC. The Alternating Current Impedance of a Semi-infinite Space\n\nD. The Rayleigh Resistance of a Wire\n\nE. The Alternating Current Inductance\n\nF. Further Treatment of the Alternating Current Field of a Circularly Cylindrical Wire\n\n§21. The Alternating-Current Conducting Coil\n\nA. The Field of the Coil\n\nB. Resistance and Inner Inductive Reactance of the Coil\n\nC. The Multilayer Coil\n\n§22. The Problem of Waves on Wires\n\nA. The Field within and outside of the Wire\n\nB. The Boundary Condition at Infinity\n\nC. The Boundary Condition at the Surface of the Wire\n\n§23. General Solution of the Wire-Wave Problem\n\nA. Primary Wave and Electrical Secondary Waves\n\nB. Magnetic Waves\n\nC. Asymmetric Waves of the Electromagnetic Type\n\nD. Wire Waves on a Nonconductor\n\n§24. On the Theory of Wave Guides\n\n§25. The Lecher Two-Wire Line\n\nA. The Limiting Case of Infinite Conductivity\n\nB. The Exterior of the Wires\n\nC. The Interior of the Wires\n\nD. The Boundary Condition Hv = Ηφ\n\nE. The Boundary Condition for Ex and the Law of Phase Propagation\n\nF. Supplement Regarding the Remaining Boundary Conditions\n\nG. Parallel and Push-Pull Operation\n\nPart III. Theory of Relativity and Electron Theory\n\n§26. The Invariance of the Maxwell Equations in the Four-Dimensional World\n\nA. The Four-Potential\n\nB. The Six-Vectors of Field and Excitation\n\nC. The Maxwell Equations in Four-Dimensional Form\n\nD. On the Geometric Character of the Six-Vector and Its Invariants\n\nE. Relativistically Invariant Three-Vectors\n\n§27. The Group of the Lorentz Transformations and the Kinematics of the Theory of Relativity\n\nA. The General and the Special Lorentz Transformation\n\nB. The Relative Nature of Time\n\nC. The Lorentz Contraction\n\nD. The Einstein Dilatation of Time\n\nE. The Addition Theorem for the Velocity\n\nF. c as Upper Limit for All Velocities\n\nG. Light Cone; Space-Like Vectors and Time-Like Vectors; Intrinsic Time\n\nH. The Addition Theorem for Velocities of Different Directions\n\nJ. The Principles of the Constancy of the Velocity of Light and of Charge\n\n§28. Preparation for the Electron Theory\n\nA. The Transformation of the Electric Field. Preliminaries Regarding the Lorentz Force\n\nB. The Magnetic Analog to the Lorentz Force\n\nC. The Intrinsic Field of an Electron in Uniform Motion\n\nD. An Invariant Approach to the Lorentz Force; the Four-Vector of the Force Density\n\nE. The General Orthogonal Transformation of a Tensor of the Second Rank\n\n§29. Integration of the Differential Equation of the Four-Potential\n\nA. Four-Dimensional Form of the Potential Ω\n\nB. Retarded Potentials\n\nC. The Lienard-Wiechert Approximation\n\n§30. The Field of the Accelerated Electron\n\nA. Electron in Uniform Motion\n\nB. The Accelerated Electron\n\nC. The Longitudinally Accelerated Electron\n\n§31. The Maxwell Stresses and the Stress-Energy Tensor\n\n§32. Relativistic Mechanics\n\nA. The Equivalence of Energy and Mass\n\nB. Relationship between Momentum and Energy\n\nC. The Principles of D'Alembert and Hamilton\n\nD. The Lagrange Function and Lagrange Equations\n\nE. Schwarzschild's Principle of Least Action\n\n§33. Electromagnetic Theory of the Electron\n\nPart IV. Maxwell's Theory for Moving Bodies and Other Addenda\n\n§34. Minkowski's Equations for Moving Media\n\n§35. The Ponderomotive Forces and the Stress-Energy Tensor\n\n§36. The Energy Loss of the Accelerated Electron by Radiation and Its Reaction on the Motion\n\n§37. Approaches to the Generalization of Maxwell's Equations and to the Theory of the Elementary Particles\n\n§38. General Theory of Relativity; Unified Theory of Gravitation and Electrodynamics\n\nA. Gravitational and Inertial Mass\n\nB. Observable Deductions from the General Theory of Relativity\n\nC. Unified Theory of Gravitation and Electrodynamics\n\nSymbols Employed Throughout the Text and Their Dimensions\n\nAdditional Symbols in Parts III and IV\n\nNumerical Values, Results of Measurements, and Definitions\n\nProblems for Part 1\n\nI.1. The Boundary Conditions of Maxwell's Theory\n\nI.2. The Magnetic Excitation Inside and Outside of an Infinitely Long Wire\n\nI.3. The Magnetic Excitation within an Infinitely Long Solenoid\n\nI.4. The Cosine Law of Spherical Trigonometry as Special Case of a General Vector Formula\n\nProblems for Part II\n\nII.1. The Charging Potential of a Conducting Ellipsoid of Revolution\n\nII.2. The Unilaterally Infinitely Long Rubbed Glass Rod and Its Comparison with the Conducting Paraboloid of Revolution\n\nII.3. Comparison of the Dielectric and the Conducting Sphere\n\nII.4. Edge Correction for the Plate Condenser According to Kirchhoff\n\nII.5. The Capacitance of a Leyden Jar (Cylindrical Condenser)\n\nII.6. On the Definition of the Capacitance of Two Conductors with Equal and Opposite Charges\n\nII.7. Characteristic Oscillations and Characteristic Frequencies of a Completely Conducting Cavity Bounded by a Rectangular Parallelepiped\n\nII.8. Characteristic Oscillations and Characteristic Frequencies of the Interior of a Completely Conducting Circular Cylinder of Finite Length\n\nII.9. Characteristic Oscillations within a Cavity Bounded by a Metal Sphere\n\nII.10. Determination of the Propagation Constants of Wire Waves from Kelvin's Telegraph Equation and from Rayleigh's Alternating Current Resistance\n\nProblems for Parts III and IV\n\nIII.1. The Lorentz Transformation for a Relative Motion Deviating from the x-Axis\n\nIII.2. On the Addition Theorem for Two Differently Directed Velocities\n\nIII.3. The Field of an Electron in Uniform Motion\n\nIII.4. On the Relativistic Energy Theorem for the Electron\n\nIII.5. The Electron in a Uniform Electrostatic Field\n\nIII.6. The Electron in a Uniform Magnetostatic Field\n\nIII.7. The Electron in a Uniform Electric Field and a Uniform Magnetic Field which is Parallel thereto\n\nIII.8. The Electron in a Uniform Electric Field and a Uniform Magnetic Field Perpendicular thereto\n\nIII.9. The Characteristic of the Thermionic Diode According to Langmuir and Schottky\n\nIII.10. The Acceleration of the Electron in the Betatron\n\nIV.1. The Field of Unipolar Induction\n\nAuthor Index\n\nSubject Index\n\nLectures on Theoretical Physics\n\nVolume I: Mechanics. 1952. Translated by Martin O. Stern\n\nVolume II: Mechanics of Deformable Bodies. 1950. Translated by G. Kuerti\n\nVolume IV: Optics. 1953. Translation in preparation\n\nVolume V: Thermodynamics and Statistical Mechanics\n\nVolume VI: Partial Differential Equations in Physics. Translated by Ernst G. Straus\n\nNo. of pages:\n386\nLanguage:\nEnglish" ]

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[ "### Physics Class 11 - Chapter 5BRCircular MotionBRTest - 4", null, "1.    A satellite moving round the earth constitutes?", null, "a) An inertial frame of reference", null, "b) Non inertial frame of reference", null, "c) Both inertial and non inertial", null, "d) Neither inertial nor non inertial", null, "2.    When a body is whirled in a horizontal circle by means of a string the centripetal force is supplied by?", null, "a) Velocity of the body", null, "b) Centrifugal force", null, "c) Tension of the body", null, "d) Mass of the body", null, "3.    Discovery of inverse square law is by?", null, "a) Newton’s", null, "b) Planks", null, "c) Galileo", null, "d) Einstein", null, "4.    A man of weight W is standing on an elevator which is ascending with an acceleration a. the apparent weight of the man is?", null, "a) mg - ma", null, "b) mg + ma", null, "c) mg", null, "d) None of these", null, "5.    The planet nearest to earth is", null, "a) Uranus", null, "b) Sun", null, "c) Mercury", null, "d) Venus", null, "6. Angular speed of second’s hand of a watch on rad-1 is?", null, "a) pi", null, "b) pi/180", null, "c) pi/2", null, "d) pi/30", null, "7.    The shaft of motor rotates at a constant angular speed of 360rev/min. angle turned through in 1 sec in radian is?", null, "a) 6pi", null, "b) 12pi", null, "c) 3pi", null, "d) pi", null, "8.    If acer moves with a uniform sped of 2ms-1 in a circle of radius 0.4.its angular speed is?", null, "a) 7 rads-1", null, "b) 1.5 rads-1", null, "c) 4 rads1", null, "d) 5 rads-1", null, "9.    The circumference subtends an angle?", null, "a) Pi radian", null, "b) Pi/4 radian", null, "c) 2 pi radian", null, "d) 4pi radian", null, "10.    When a body moves in a circle, the angle between its linear velocity and angular velocity is always?", null, "a) 0o", null, "b) 180o", null, "c) 90o", null, "d) 45o\nThis is more feedback!\nThis is the feedback!" ]

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[ "# Page History\n\n## Key\n\n• This line was removed.\n• Formatting was changed.\n\nIn a Poisson process, the times between successive events are described by independent identical Exponential distributions. In a renewal process, like a Poisson process, the times between successive events are independent and identical, but they can take any positive distribution. The Poisson process is thus a particular case of a renewal process.\n\nThe mathematics of the distributions of number of events in a period (equivalent to the Poisson distribution for the Poisson process) and the time to wait to observe x events (equivalent to the Gamma distribution in the Poisson process) can be quite complicated, depending on the distribution of time between events. However, Monte Carlo simulation lets us bypass the mathematics to arrive at both of these distributions.\n\nExample of a renewal process model\n\n..." ]

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[ "Math 1313 Course Objectives\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 1.3 Given a linear depreciation problem, find the rate of depreciation, the expression that expresses the book value at the end of t years and the value of the asset after a given amount of years.   Example:  A company purchased a car in 2000 for \\$13,000.  The car is depreciated linearly for 5 years.  The scrap value of the car is \\$4,000.  What is the rate of depreciation?  Write the expression that expresses the book value of the car after t years of use.  What is the value of the car in 2003?   Given the production cost, selling price of a product and the fixed costs of the company, find the cost function, revenue function, profit function, and compute the profit or loss corresponding to certain production levels.   Example  A company has a fixed cost of \\$100,000 and a production cost of \\$14 for each unit produced.  The product sells for \\$20 per unit.  What is the cost function? What is the revenue function? What is the profit function? What is the break-even point? What is the profit or loss corresponding to a production level of 12,000 and 20,000 units? 2\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 1.4 Given a word problem find break-even quantity, break-even revenue and break-even point of the company.   Example  A company has a fixed cost of \\$100,000 and a production cost of \\$14 for each unit produced.  The product sells for \\$20 per unit.  What is the break-even quantity? What is the break-even revenue? What is the break-even point? 2\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 1.5 Given a set of data or a word problem, find the equation for the least-square line of the data and use this equation to predict a certain future value.   Example:", null, "The size of the average farm in a certain town has been growing steadily over the years.  The accompanying data was collected and gives the size of the average farm y (in acres) from 1945 to 1995.  (Here x = 0 corresponds to the beginning of the year 1945.)", null, "Year, x                         0          10        20        30        40", null, "Number of Acres, y     57        63        76        88        92              a.   Find the equation of the least-squares line for these data. b.   Use the result in part (a) to estimate the size of the average farm in the year 1998. 3 Chapter.Section Objective and Examples Material Covered by End of Week # 3.2, 3.3 Set up and solve a linear programming problem.   Example:  A company manufactures two products, A and B, on two machines I and II.  It has been determined that the company will realize a profit of \\$3 on each unit of product A and a profit of \\$4 on each unit of product B.  To manufacture a unit of product A requires 6 min on machine I and 5 min on machine II.  To manufacture a unit of product B requires 9 min on machine I and 4 min on machine II.  The company has 5 hours of machine time on machine I and 3 hours of machine time on machine II in each work shift.  How many units of each product should be produced in each shift to maximize the company’s profit?  Set up the linear programming problem then solve it. 4\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 5.1-5.3 Given a certain math of finance problem, recognize what kind of problem it is and solve it.  The kind of problems given will be:  simple interest, future value or present value with simple interest, effective rate, future value or present value with compound interest, future value or present value of an annuity, amortization, or sinking fund.   Example:  A company would like to have \\$50,000 in 2 years to replace machinery.  The account they wish to invest in earns 3.45% per year compounded quarterly.  How much should they deposit into this account each quarter to have the desired funds in 2 years?  a.       What kind of problem is this? b.      Solve the problem.   Example:  Karen has decided to deposit \\$300 each month into an account that earns 2.34% per year compounded monthly.  How much will she have in this account after 3 years? a.       What kind of problem is this? b.      Solve the problem. 5 Chapter.Section Objective and Examples Material Covered by End of Week # 6.1 Given sets, list the subsets and proper subsets of a set.  Find the union, intersection and/or complement of certain given sets.   Example:  Let U = {1,2,3,4,5,6,7,a,b,c,d,e}A = {1,2,3,4,5,a,b,c}, B = {1,3,5,6,a,c,d}, and C = {2,4,7,b,d,e}, and D = {1,2,a} a.   List the subsets of D. b.   Find", null, "Use Venn diagram shading to find the union, intersection and/or complement of certain given sets.   Example:  Given the following Venn diagram, shade the given set.", null, "", null, "", null, "", null, "U                           A                        B", null, "C 6\n\nChapter.Section\n\nObjective and Examples\n\nMaterial Covered by End of Week #\n\n6.2\n\nFind the number in sets by using formulas or Venn\n\nExample:  Of 30 elementary school children, 15 read a book last summer, 17 practiced math last summer and 7 read a book and practiced math last summer.\n\nHow many of the 30 children:\n\na.   did not read a book last summer?\n\nb.   read a book but did not practice math last summer?\n\nc.   did not read a book and did not practice math last summer?\n\n7\n\nChapter.Section\n\nObjective and Examples\n\nMaterial Covered by End of Week #\n\n6.3, 6.4\n\nSolve word problems by using counting technique(s)\n\nsuch as the multiplication principle, combination or\n\npermutation.\n\n## Example:  A coin is tossed 20 times, how many    outcomes are there?\n\nExample:  In how many ways can you arrange 3 different pictures from 5 available on a wall from left to right?\n\nExample:  In how many ways can you choose 3 mystery books from a collection of 15 mystery books and 5 romance books from a collection of 20 romance books?\n\n8\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 7.1, 7.2 Given a set of data or a certain experiment, list the simple events, find the probability of each of the simple events, find the probability distribution, and find the probability of an event that consists of more than one simple event.   Example:  A pair of dice is cast.  List the simple events.  Assign probabilities to each of the simple events.  Find the probability distribution of the experiment.  Find the probability that the sum of the numbers is even. 9\n\nChapter.Section\n\nObjective and Examples\n\nMaterial Covered by End of Week #\n\n7.3\n\nGiven a word problem, use formulas or Venn\n\ndiagram shading to find the probability of the union,\n\nintersection and/or complement of certain events.\n\nExample:  Of 30 elementary school children, 15 read a book last summer, 17 practiced math last summer and 7 read a book and practiced math last summer.\n\n## What is the probability that a child selected at random\n\na.   did not read a book last summer?\n\nb.   read a book but did not practice math last summer?\n\nc.   did not read a book and did not practice math last summer?\n\n9\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 7.4 Use counting techniques to find the probability of certain events.   Example:  A box contains 25 batteries of which 5 are defective.  A random sample of 4 is chosen.  What is the probability that at least 2 are defective? 10\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 7.5 Use the conditional probability formula or tree diagrams to aid in finding certain probabilities.    Example:  A group of senators is comprised of 48 Democrats and 52 Republicans.  Seventy-one percent of the Democrats served in the military, whereas 68% of the Republicans served in the military.  What is the probability that a senator chosen at random a.   is Republican? b.   Is a Democrat and did not serve in the military? c.   served in the military? d.   did not serve in the military, given that he/she is a Democrat?   Given that certain events are independent, find the probability of the intersection of those independent events.   Example:  If A and B are independent events and P(A)=0.4 and P(B)=0.6, find P(A", null, "B). 11\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 7.6 Use tree diagrams and Bayes’ formulas to find certain conditional probabilities.   Example:  A group of senators is comprised of 48 Democrats and 52 Republicans.  Seventy-one percent of the Democrats served in the military, whereas 68% of the Republicans served in the military.  What is the probability that a senator chosen at random is a Republican, given that he/she served in the military? 11\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 8.1 Given a probability distribution, find certain probabilities and draw a histogram associated with the given probability distribution.   Construct the probability distribution of a random variable.     Example:  The probability distribution of the random variable X is shown below.", null, "", null, "x                        P(X=x)   1            0.2   2            0.3   3            0.5         a.  Find P(1 < X < 3). b.  Draw the histogram corresponding to the given probability distribution.           Example:  Given the following frequency table, construct the probability distribution associated with the random variable X.", null, "", null, "x             P(X=x)   1            45   2            20   3            32 12\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 8.2 Find the expected value of a given probability distribution or of a word problem.", null, "Example:  The following probability distribution tables describes the number of cars, x, that a certain car dealer will sell in a given day along with its associated probability.", null, "x             P(X=x)    1            0.2    2            0.3    3            0.5   Find the expected number of cars the car dealer will sell in a given day.   Given a word problem, find the odds in favor, odds against or given the odds find a certain probability.     Example:  The odds in favor of an event occurring are 4 to 5.  What is the probability of the event not occurring? 12\n\nChapter.Section\n\nObjective and Examples\n\nMaterial Covered by End of Week #\n\n8.3\n\nGiven a probability distribution or a word problem,\n\nfind the variance and standard deviation.\n\n##", null, "Example:  Given", null, "x                  P(X=x)\n\n1            0.2\n\n2           0.3\n\n3            0.5\n\nFind the variance and standard deviation.\n\nUse Chebychev’s inequality to estimate a certain\n\nprobability.\n\nExample:  The expected lifetime of a certain machine is 24 mo and the standard deviation is 3 mo.  Use Chebychev’s inequality to estimate the probability that one of these machines will last between 20 and 28 mo.\n\n13\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 8.4 Given a binomial experiment, find certain probabilities, the mean, the variance, and the standard deviation.   Example:  The probability that a certain CD player will be defective is 0.04.  If a sample of 15 CD players is chosen at random, what is the probability that the sample contains a.   no defective CD players? b.   at most 3 defective CD players? c.   Find the mean, variance and standard deviation of this experiment. 13\n\n Chapter.Section Objective and Examples Material Covered by End of Week # 8.5 Given a standard normal distribution, find certain probabilities or given the probability find the value of z.    Example:  Let Z be a standard normal random variable.  Find: a.       P(Z < 1.34) b.      P(Z> -2.33) c.       P(-0.23 < Z < 1.22)   Example:  Let Z be a standard normal random variable.  Find the value of z if:   a.       P(Z > z) = 0.8749 b.      P(-z < Z < z) = 0.4908   Given a normal distribution, possibly a word problem, standardize it to find certain probabilities. Example:  Let X be a normal random variable.  The mean is 25 and the standard deviation is 4.  Find:   a.        P(X < 30) b.       P(X > 10) c.       P(15 < X < 25) 14\n\nChapter.Section\n\nObjective and Examples\n\nMaterial Covered by End of Week #\n\n8.6\n\n# Use the normal distribution to approximate a binomial distribution.\n\nExample:  Use the normal distribution to approximate the following binomial distribution.  A biased coin is tossed 100 times.  The probability of obtaining a head is 30%.  What is the probability that the coin will land heads at least 90 times?\n\n14" ]

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[ "# Ford-Fulkerson pseudo-polynomial\n\nCan somebody explain please why Ford-Fulkerson Algorithm has pseudo-polynomial complexity? I understand that the complexity in this case strongly depends on the capacities of the edges of the network, since we have $$O(|E|*f_{max})$$, but why is the algorithm still not polynomial?\n\nLets recall that input for flow problems is a graph with capacities on their edges. Let $$G(V,E)$$ where $$V$$ denotes the vertex set of the graph and $$E$$ denotes the edge set. Let $$f_{\\text{max}}$$ denote the maximum possible flow. It is an easy caluclation that input size is $$\\mathcal{O}(V+E \\log (f_{\\text{max}}))$$ bits as writing $$f_{\\text{max}}$$ takes $$\\mathcal{O}(\\log f_{\\text{max}})$$ bits. Since $$f_{\\text{max}}$$ may be arbitrarily high (it depends neither on $$V$$ or $$E$$), this running time could be arbitrarily high, as a function of $$V$$ and $$E$$." ]

[ null ]

[ "## Algebra 2 (1st Edition)\n\n$a_n=5n-3$\nHere, we have $a_n= a_1+d{n-1}$ for the Arithmetic series. The common difference between the successive terms is $d=5$ First term is: $a_1=2$ Here, we have $a_n= 2+5 \\times {n-1}$ Hence, $a_n=5n-3$" ]

[ null ]

[ "## MHD flow and heat transfer of couple stress fluid over\n\nlatory stretching sheet embedded in a porous medium in the presence of heat source/sink. The .... surface. According to him exponential series type solution.\n\nAlexandria Engineering Journal (2016) xxx, xxx–xxx\n\nH O S T E D BY\n\nAlexandria University\n\nAlexandria Engineering Journal www.elsevier.com/locate/aej www.sciencedirect.com\n\nORIGINAL ARTICLE\n\nMHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium Nasir Ali a, Sami Ullah Khan a,*, Muhammad Sajid a, Zaheer Abbas b a b\n\nDepartment of Mathematics and Statistics, International Islamic University, Islamabad 44000, Pakistan Department of Mathematics, The Islamia University of Bahawalpur, Bahawalpur 63100, Pakistan\n\nReceived 25 December 2015; revised 9 February 2016; accepted 20 February 2016\n\nKEYWORDS Boundary layer flow; Couple stress fluid; Oscillatory stretching sheet; Homotopy analysis method\n\nAbstract This article presents the MHD flow and heat transfer of a couple stress fluid over oscillatory stretching sheet embedded in a porous medium in the presence of heat source/sink. The unsteady flow problem is reduced to two coupled partial differential equations using dimensionless variables. Homotopy analysis method is employed to obtain the solution of these equations. Based on the solution of these equations an extensive analysis is performed to investigate the effects of various flow parameters on velocity and temperature fields, skin friction coefficient and Nusselt number. It is found that the presence of couple stress in viscous fluid increases the amplitude of oscillations in velocity and skin friction coefficient. It is also noticed that temperature increases by increasing heat source parameter. Moreover, the numerical values of local Nusselt number are calculated and shown in tabular form. It is found that the Nusselt number increases by increasing Prandtl number while it decreases by increasing couple stress parameter. Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).\n\n1. Introduction It is well established fact that Navier–Stokes equations are not adequate choice in describing the flows of nonNewtonian fluids. Non-Newtonian fluids frequently occur in many industrial and technological processes. If the flow of non-Newtonian fluid is considered over a stretching sheet then it become relevant to many industrial applications such a extrusion of plastic sheets, paper production, crystal * Corresponding author. Tel.: +92 51 9019756. E-mail address: [email protected] (S. Ullah Khan). Peer review under responsibility of Faculty of Engineering, Alexandria University.\n\ngrowing and glass blowing, drawing of copper wires, continuous stretching of plastic films. The pioneering work on viscous flow due to continuous moving sheet was performed by Sakiadis . His analysis was extended for stretching sheet by Crane . Crane’s problem was extended by Rajagopal et al. for viscoelastic fluid. Gupta and Gupta extended the analysis of Crane by considering suction/blowing at the sheet surface. Dandapat and Gupta discussed the heat transfer characteristics of viscoelastic fluid over a stretching sheet. A study of uniqueness of boundary value problem arising due to a stretching boundary in a viscous fluid was performed by McLeod and Rajagopal . Rollins and Vajravelu discussed heat transfer characteristics of second-order fluid over a continuos stretching surface with internal heat generation\n\nhttp://dx.doi.org/10.1016/j.aej.2016.02.018 1110-0168 Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Please cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\nN. Ali et al. oscillatory stretching surface in the presence of heat generation or absorption. Hence we feel appropriate to consider the study of such flows, as such type of flows may find a number of industrial applications. The theory of couple stress fluid was developed by Stokes as a simple generalization of classical viscous theory that sustains couple stresses and body couples. The main effects of couple stress fluid are to introduce a size dependent effect which is not present in classical viscous theories. Blood, lubricants containing small amount of additives, electro-rheological and synthetic fluids are examples of couple stress fluid . Couple stress theory has been successfully applied to many problems in biomechanics and lubrications area by Srivastava , El-Shehawey and Mekheimer , Pal et al. , Chiang et al. , Naduvinamani et al. , Jian and Chen et al. and Lu and Lin . The presentation of the paper is as follows. Flow analysis is presented in Section 2. Homotopy analysis method is employed for solution of governing equations in Section 3. Convergence of solution is discussed in Section 4. Quantitative analysis of results is presented in Section 5. Final remarks are presented at the end of the paper in Section 6. 2. Flow analysis Let us consider the unsteady two-dimensional magnetohydrodynamics flow of an incompressible fluid over an oscillatory stretching sheet. In a rectangular coordinate system ð\u0001 x; y\u0001Þ, the sheet is assumed to coincide with y\u0001 ¼ 0 and flow takes place in the semi-infinite porous space ðy\u0001 > 0Þ of constant permeability (see Fig. 1). A constant magnetic field of strength B0 is applied perpendicular to the stretching surface. It is assumed that the sheet is stretched with velocity ux ¼ b\u0001 x sin xt in which b is the stretching rate and x denotes the angular velocity. Moreover, sheet is maintained at uniform temperature Tw ð> T1 Þ, where T1 is the temperature of ambient fluid. Now problem is to determine the velocity and temperature fields inside the fluid satisfying the boundary conditions at the wall and far away from the wall. To this end, we apply boundary layer approximation on continuity and momentum equations and write @u @v þ ¼ 0; @ x\u0001 @ y\u0001\n\nð1Þ\n\nFigure 1\n\nGeometry of the problem.\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\nMHD flow and heat transfer of couple stress fluid\n\n3\n\n\u0001 \u0003 @u @u @u @2u g @4u rB2 t/ þu þv ¼m 2\u0002 0 \u0002 0 u \u0002 u; 4 @t @ x\u0001 @ y\u0001 @ y\u0001 k q @ y\u0001 q\n\nð2Þ\n\n@T @T @T k @2T Q þu þv ¼ þ ðT \u0002 T1 Þ; @t @ x\u0001 @ y\u0001 qcp @ y\u00012 qcp\n\nð3Þ\n\nwhere u and v are velocity component along x\u0001 and y\u0001-directions, respectively, m is the kinematic viscosity, q is the density, g0 is the material constant for the couple stress fluid, Q is the heat source coefficient, r is electric conductivity and B0 is the strength of constant applied magnetic field. The flow is subjected to the following boundary conditions u ¼ ux ¼ b\u0001 x sin xt; y\u0001 ¼ 0;\n\nt > 0;\n\nu ¼ 0;\n\n@u ¼ 0; @ y\u0001\n\nv ¼ 0;\n\n@2u ¼ 0; @ y\u00012\n\nT ¼ Tw\n\nat ð4Þ\n\nT ! T1\n\nat\n\ny\u0001 ! 1:\n\nð5Þ\n\nThe first two boundary conditions in (4) result from the assumption of no-slip at the surface while the third condition is consequence of the assumption of vanishing of couple stresses at the wall. Since flow is in unbounded domain that is why augmented boundary condition is imposed at y ! 1. To nondimensionalize the flow problem, we use the dimensionless variables [30,31] rffiffiffi pffiffiffiffiffi b y¼ y\u0001; s ¼ tx; u ¼ b\u0001 xfy ðy; sÞ; v ¼ \u0002 mbfðy; sÞ; ð6Þ m hðy; sÞ ¼\n\nT \u0002 T1 : Tw \u0002 T1\n\nð7Þ\n\nIn view of above transformations, the continuity equation is identically satisfied and Eqs. (2) and (3) are transformed as follows fyyy \u0002 Sfys \u0002 f2y þ ffyy \u0002 bfy \u0002 Kfyyyyy ¼ 0;\n\nð8Þ\n\n\u0005 \u0006 hyy þ Pr fhy \u0002 Shs þ kh ¼ 0;\n\nð9Þ\n\nwhere the subscript denotes differentiation with respect to y. The boundary conditions (4) and (5) take the following form fy ð0; sÞ ¼ sin s;\n\nfð0; sÞ ¼ 0;\n\nfyyy ð0; sÞ ¼ 0;\n\nhð0; sÞ ¼ 1; ð10Þ\n\nfy ð1; sÞ ¼ 0;\n\nfyy ð1; sÞ ¼ 0;\n\nhð1; sÞ ¼ 0:\n\nð11Þ\n\nIn above equations K ¼ g0 b=qt2 is couple stress parameter, S ¼ x=b is unsteady parameter, b ¼ rB20 =qb þ t/=kb is the magneto-porous parameter, Pr ¼ lcp =k is the Prandtl number and k ¼ Q=cqcp is the heat source ðk > 0Þ or sink ðk < 0Þ parameter. The parameter K ¼ g0 qb=l2 ¼ l2 bq=l is called couple stress parameter. It has been pointed out by Stokes that the effects of couple stress are quite large for large values of K ¼ l2 bq=l, where bq=l is a typical length scale associated pffiffiffiffiffiffiffiffi with the flow and l ¼ g=l is a material coefficient which is a function of molecular dimensions of the liquid. The parameter l will be greatly varied for different liquids. For instant, the length of polymer chain may be a million time the diameter of a water molecule. This is the size dependent effect associated with the couple stress fluid theory. For small value of K the\n\neffects of couple stress diminish and fluid behaves like a Newtonian fluid. The quantities of interest are skin friction coefficient and local Nusselt number which are defined as Cf ¼\n\nsw ; qu2w\n\nNux ¼\n\nx\u0001qw ; kðTw \u0002 T1 Þ\n\nwhere\n\n\u0007 \b \u0007 \b @u g0 @ 3 u \u0002 ; sw ¼ l @ y\u0001 y\u0001¼0 q @ y\u00013 y\u0001¼0\n\nð12Þ \u0007 \b @T qw ¼ \u0002k ; @ y\u0001 y\u0001¼0\n\nð13Þ\n\nare skin friction and heat transfer at the wall, respectively. In view of transformations (6) and (7), we get Re1=2 x Cf ¼ fyy ð0; sÞ \u0002 Kfyyyy ð0; sÞ;\n\nð14Þ\n\nNux ¼ \u0002hy ð0; sÞ; Re\u00021=2 x\n\nð15Þ\n\nwhere Rex ¼ uw x\u0001=m is the local Reynolds number. 3. Solution by homotopy analysis method In many physical problems, the resulted differential equations are highly nonlinear nature. The computations of analytical or numerical solutions of such equations have set a challenge for the researchers. Homotopy analysis method (HAM) is one the best analytical methods to compute the series solution of such nonlinear partial as well as ordinary differential equations. This method tackles for strongly nonlinear problems even if a given nonlinear problem does not contain any small/large parameter. This method gives great freedom to choose and adjust the convergence region and rate of approximation. The homotopy analysis method is advantageous over routine numerical methods in the sense that it is free of rounding off errors which arise due to discretization procedure. Moreover, it does not require large computer memory and time . This method was successfully applied by researchers to solve various nonlinear problems in various disciplines of science and engineering [44–49]. It has been emphasized by Liao that the series solution obtained by HAM is purely analytic and he demonstrated this fact in number of articles for instance [50–52]. Moreover, it has been proved by Turkyilmazoglu that HAM can handle any type of nonlinearity by a suitable HAM formulation. For the series solution of Eqs. (8) and (9) subject to the boundary conditions (10) and (11), we employ homotopy analysis method (HAM). According to HAM the velocity and temperature fields can be represented as fðy; sÞ ¼ a00;0 þ\n\n1 X 1 X 1 X j an;k yk sinðjsÞ expð\u0002nyÞ;\n\nð16Þ\n\nn¼0 k¼0 j¼0\n\nhðy; sÞ ¼ b00;0 þ\n\n1 X 1 X 1 X j bn;k yk sinðjsÞ expð\u0002nyÞ;\n\nð17Þ\n\nn¼0 k¼0 j¼0 j j and bn;k are the coefficients to be determined. The where an;k appropriate initial guesses for fðy; sÞ and hðyÞ are\n\nf0 ðy; sÞ ¼\n\n1 sin sð3 \u0002 3 expð\u0002yÞ \u0002 y expð\u0002yÞÞ; 2\n\nh0 ðyÞ ¼ expð\u0002yÞ:\n\nð18Þ ð19Þ\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\n4\n\nN. Ali et al.\n\nFurther we choose £f ðfÞ ¼\n\n@ f @f \u0002 ; @y3 @y\n\n£h ðfÞ ¼\n\n@ f \u0002 f; @y2\n\n3\n\n£f ½C1 þ C2 expð\u0002yÞ þ C3 expðyÞ\u0003 ¼ 0;\n\nð20Þ\n\n£h ½C4 expðyÞ þ C5 expð\u0002yÞ\u0003 ¼ 0;\n\nð21Þ\n\nwhere Ci ði ¼ 1 \u0002 5Þ are arbitrary constants. If hf and hh denotes the nonzero auxiliary parameters then zeroth order deformation problems are constructed as follows:\n\n2\n\nas auxiliary linear operators having following properties\n\nb s; pÞ \u0002 f ðy; sÞ ¼ ph Nf ½ fðy; b s; pÞ\u0003; ð1 \u0002 pÞ£f ½ fðy; 0 f\n\nð22Þ\n\nð1 \u0002 pÞ£f ½b hðy; s; pÞ \u0002 h0 ðy; sÞ b s; pÞ\u0003; ¼ phh Nh ½b hðy; s; pÞ; fðy; b b s; pÞ ¼ 0; @ fðy; s; pÞ ¼ sin s; fð0; @y b s; pÞ @ 2 fðy; ¼ 0; ¼ 0; @y2 b hð0; s; pÞ ¼ 1;\n\nð23Þ b s; pÞ @ fðy; @y ð24Þ\n\nb hð1; s; pÞ ¼ 0;\n\nwhere p 2 ½0; 1\u0003 is an embedding parameter and nonlinear operators Nf and Nh are Figure 2 The h-curve for velocity at 7th order of approximation: with K ¼ 0:1; b ¼ 0:2; S ¼ 0:1 and s ¼ 0:5p.\n\n5b 3b b s; pÞ\u0003 ¼ \u0002K @ fðy; s; pÞ þ @ fðy; s; pÞ \u0002 S Nf ½ fðy; @y5 @y3 2b b s; pÞ @ fðy; s; pÞ b @ 2 fðy; þ fðy; s; pÞ \u0004 @[email protected] @y2 !2 b s; pÞ b s; pÞ @ fðy; @ fðy; ; \u0002 \u0002b @y @y\n\nð25Þ\n\nFigure 3 The h-curve for temperature at 7th order of approximation with K ¼ 0:1; S ¼ 0:1; b ¼ 0:2; k ¼ 0:5; Pr ¼ 0:1 and s ¼ 0:5p. Table 1 Comparison of values of f00 ð0; sÞ with Refs. with K ¼ 0 and . S\n\nb\n\ns\n\nZheng et al. \n\nAbbas et al. \n\nPresent results\n\n1:0\n\n12\n\n1:5p 5:5p 9:5p\n\n11:678565 11:678706 11:678656\n\n11:678656 11:678707 11:678656\n\n11:678565 11:678706 11:678656\n\nTable 2 Comparison of values of f00 ð0; sÞ for different values of b when K ¼ 0; S ¼ 0 and s ¼ p=2. b\n\nHayat et al. \n\nTurkyilmazoglu \n\nPresent results\n\n0 0:5 1:0 1:5 2:0\n\n\u00021:000000 \u00021:224747 \u00021:414217 \u00021:581147 \u00021:732057\n\n\u00021:00000000 \u00021:22474487 \u00021:41421356 \u00021:58113883 \u00021:73205081\n\n\u00021:000000 \u00021:224747 \u00021:414217 \u00021:581147 \u00021:732057\n\nFigure 4 Time series of the velocity profile f0 at a distance y ¼ 0:25 from the sheet in the time period s 2 ½0; 10p\u0003: (a) The effects of couple stress parameter K with S ¼ 1 and b ¼ 0:1, (b) The effects of magneto-porous parameter b with S ¼ 0:5 and K ¼ 0:1.\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\nMHD flow and heat transfer of couple stress fluid\n\n5\n\nb s; pÞ\u0003 Nh ½ b hðy; s; pÞ; fðy; ¼\n\nDifferentiating zeroth order deformations equations (22) and (23) m-times with respect to p, then setting p ¼ 0 and finally dividing by m!, we get the following mth-order deformation problem\n\n2b\n\nb @ hðy; s; pÞ b s; pÞ @ hðy; s; pÞ þ Pr fðy; @y2 @y ! @b hðy; s; pÞ þ kb hðy; s; pÞ : \u0002S @s\n\nð26Þ\n\nThe zeroth-order deformation problems defined above have the following solutions corresponding to p ¼ 0 and p ¼ 1 b s; 0Þ ¼ f ðy; sÞ; fðy; 0 b hðy; s; 0Þ ¼ h0 ðy; sÞ;\n\nb s; 1Þ ¼ fðy; sÞ; fðy; b hðy; s; 1Þ ¼ hðy; sÞ:\n\nð27Þ ð28Þ\n\nb s; pÞ and b Expanding fðy; hðy; s; pÞ in a Taylor’s series with respect to p, we get b s; pÞ ¼ fb0 ðy; sÞ þ fðy;\n\n1 X\n\nfm ðy; sÞpm ;\n\nð29Þ\n\n£f ½ fbm ðy; sÞ \u0002 ,m fm\u00021 ðy; sÞ ¼ hf Rmf ðy; sÞ;\n\nð32Þ\n\n£f ½b h m ðy; sÞ \u0002 ,m hm\u00021 ðy; sÞ ¼ hh Rhm ðy; sÞ;\n\nð33Þ\n\n@fm ðy; s; 0Þ @fm ðy; s; 0Þ ¼ @y @y y¼0 y¼1 2 @f ðy; s; 0Þ ¼ m 2 ¼ 0; @y y¼1\n\nfm ð0; s; pÞ ¼ 0;\n\nhm ð0Þ ¼ 0; hm ð1Þ ¼ 0; @ 5 fm\u00021 @ 3 fm\u00021 @ 2 fm\u00021 @f \u0002 b m\u00021 þ \u0002 S @y5 @y3 @[email protected] @y \u0007 \b m\u00021 2 X @ f f @f fm\u00021\u0002k 2k \u0002 m\u0002k\u00021 k ; þ @y @y @y k¼0\n\nð30Þ\n\nm¼1\n\nfm ðy; sÞ ¼\n\nb s; pÞ 1 @ m fðy; ; @pm m!\n\nhm ðy; sÞ ¼\n\nhðy; s; pÞ 1 @m b ; @pm m!\n\nRhm ðy; sÞ ¼ ð31Þ\n\nð35Þ\n\nRmf ðy; sÞ ¼ \u0002K\n\nm¼1 1 X b hm ðy; sÞpm ; hðy; s; pÞ ¼ b h 0 ðy; sÞ þ\n\nð34Þ\n\n\u0001 \u0003 @ 2 hm\u00021 @hm\u00021 \u0002 Pr S @y2 @s ! \u0001 m \u00021 X @hm\u00021\u0002k \u0003 þ Pr fk þ khm\u00021 ; @y k¼0\n\nð36Þ\n\nð37Þ\n\nFigure 5 Transverse profiles of the velocity field f0 for four different values of couple stress parameter K in the fifth period s 2 ½8p; 10p\u0003 for which a periodic velocity field has been reached: (a) s ¼ 8:5p, (b) s ¼ 9p, (c) s ¼ 9:5p and (d) s ¼ 10p, with S ¼ 1 and b ¼ 0:1.\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\n6\n\nN. Ali et al.\n\n,m ¼\n\n0; 1;\n\nm 6 1; m > 1;\n\nð38Þ\n\n4. Convergence of HAM solution It is well established fact that the convergence of the HAM solution is largely dependent on the proper choice of the auxiliary parameters hf and hh . For a particular set of parameters the convergence region can be obtained by plotting h-curves. Figs. 2 and 3 present two such curves showing the plausible values of hf and hh for a given set of parameters. We note from these figures that for convergent solution \u00021:2 6 hf < \u00020:2 and \u00021:5 6 hh < \u00020:2. The purpose of showing such curves is just to emphasize that in principle for any physical choice of parameters of the problem, a convergent solution can be obtained. The values of f00 ð0; sÞ obtained by HAM in the present paper are also compared with the corresponding values reported in Refs. [20,21,30,31] in Tables 1 and 2. Both tables show that our results are in excellent agreement with the existing results in the literature. 5. Results and discussion The homotopy analysis method is a powerfull technique to solve nonlinear partial and ordinary differential equations with\n\ninitial and boundary conditions. The set of nonlinear partial differential equations (8) and (9) with boundary conditions (10) and (11) is solved analytically by means of homotopy analysis method and the effects of involved parameters are illustrated through graphs. In this section, we are going to present the detail analysis of flow and heat transfer of couple stress fluid over oscillating stretching surface in the presence of heat source and sink. Fig. 4(a) and (b) are plotted to show the behavior of couple stress parameter K and magneto-porous parameter b on the time series of velocity f0 at a fixed location y ¼ 0:25 from the sheet, respectively. In Fig. 4(a) the effects of couple stress parameter K are illustrated by keeping S ¼ 1 and b ¼ 0:1. From this figure we observe that an increase in couple stress parameter K results in an increase in amplitude of velocity. This is perhaps due to the presence of couple stresses. Fig. 4 (b) points out the effects of magneto-porous parameter b by keeping other parameters constant. This figure shows that the amplitude of velocity decreases by increasing magnetoporous parameter b. Fig. 5(a)–(d) demonstrate the effects of couple stress parameter K on velocity profile f0 at four different time instants s ¼ 8:5p; s ¼ 9p; s ¼ 9:5p and s ¼ 10p. Fig. 5(a) characterizes the effects of couple stress parameter K on the velocity profile f0 at s ¼ 8:5p. This figure depicts that the velocity f0 decrease as we increase K. This decrease in velocity results in decrease in\n\nFigure 6 Transverse profiles of the velocity field f0 for four different values of magneto-porous parameter b in the fifth period s 2 ½8p; 10p\u0003 for which a periodic velocity field has been reached: (a) s ¼ 8:5p, (b) s ¼ 9p, (c) s ¼ 9:5p, and (d) s ¼ 10p, with S ¼ 0:5 and K ¼ 0:1.\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\nMHD flow and heat transfer of couple stress fluid boundary layer thickness. In Fig. 5(b), the behavior of couple stress parameter K on the f0 is shown at s ¼ 9p. This figure shows an increase in magnitude of velocity at this instant. The effects of K at s ¼ 9:5p and s ¼ 10p are illustrated in Fig. 5(c) and (d), respectively. It is observed from these figures that at s ¼ 9:5p and s ¼ 10p, the velocity f0 decreases with K. It is interesting to note that the effects of couple stress are to the increase the amplitude of velocity when f0 ¼ 1 at the surface. For other instants when f0 ¼ \u00021 or 0, the magnitude of velocity decreases due to the presence of couple stresses.\n\nFigure 7 Time series of the skin-friction coefficient Re1=2 x Cf in the first five periods s 2 ½0; 10p\u0003: (a) effects of couple stress parameter K with S ¼ 0:1 and b ¼ 0:1 and (b) effects of magneto-porous parameter b with S ¼ 0:5 and K ¼ 0:1.\n\n7 The effects of magneto-porous parameter b on velocity are illustrated in Fig. 6(a)–(d) at four different times keeping other parameters fixed. In Fig. 6(a) the effects of magneto-porous parameter b are plotted at s ¼ 8:5p. From this figure it can be examined that velocity decreases with increase in magneto-porous parameter b. The effects of magneto-porous parameter on velocity at other three time instants are illustrated in Fig. 6(c) and (d). Here, again it is observed that the magnitude of velocity decreases by increasing b. The variation of skin friction coefficient with couple stress parameter K and magneto-porous parameter b in the first five periods s 2 ½0; 10p\u0003 is shown in Fig. 7(a) and (b), respectively. Fig. 7(a) shows the time series of skin friction coefficient for different values of couple stress parameter K by taking S ¼ 0:1 and b ¼ 0:1. From this figure we observe that skin friction coefficient varies periodically due to periodic motion of the surface and its amplitude increases with couple stress parameter K. This observation reflects that skin friction for a Newtonian fluid flowing over an oscillatory stretching sheet is less in comparison with its value for a couple stress fluid performing the same motion. The effects of magneto-porous parameter b on skin friction coefficient are illustrated in Fig. 7(b). This figure depicts that amplitude of skin friction coefficient increases by increasing magneto-porous parameter b. Such effects of b and M are expected because the fact that magnetic force and drag force offered by the porous medium acts as a resistance to the flow. Figs. 8–12 present the effects of Prandtl number Pr, magneto-porous parameter b, couple stress parameter K and heat source/sink parameter k on the temperature field. Fig. 8 reveals the effects of Prandtl number Pr on temperature profile for both heat source and sink cases. It is seen that temperature is a decreasing function of Pr in both cases. The fact is that the increase in Prandtl number results in low thermal conductivity, as a result conduction as well as thermal boundary layer thickness decreases and hence we observe a decrease in temperature. It is observed that the presence of heat sink enhances the decrease in temperature by increasing Prandtl number. In Fig. 9, the influence of magneto-porous parameter b on temperature field is explained graphically by keeping other parameters fixed. This figure shows that temperature increases with magneto-porous parameter b. The influence of couple stress\n\nFigure 8 The effects of Prandtl number Pr on temperature field h at s ¼ 8:5p with K ¼ 0:1; S ¼ 0:1; b ¼ 0:2, (a) in heat source case k ¼ 0:3, (b) in heat sink case k ¼ \u00020:3.\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018\n\n8\n\nFigure 9 The effects of magneto-porous parameter b on temperature field h at s ¼ 8:5p with Pr ¼ 1:5; K ¼ 0:1; S ¼ 0:1; k ¼ 0:3.\n\nN. Ali et al.\n\nFigure 12 The effects of heat sink parameter k on temperature field h at s ¼ 8:5p with Pr ¼ 0:4; b ¼ 0:5; K ¼ 0:1; S ¼ 0:1.\n\nTable 3 The numerical values of the Local Nusselt number for different values of Prandtl number Pr, couple stress parameter K, magneto-porous parameter b and heat source parameter k at time s ¼ p=2 with S ¼ 1.\n\nFigure 10 The effects of couple stress number K on temperature field h at s ¼ 8:5p with Pr ¼ 1:5; b ¼ 5; k ¼ 0:4; S ¼ 0:6.\n\nFigure 11 The effects of heat source parameter k on temperature field h at s ¼ 8:5p with Pr ¼ 0:4; b ¼ 0:5; K ¼ 0:1; S ¼ 0:1.\n\nparameter K on temperature is shown in Fig. 10, which depicts that increase in couple stress parameter results in increase in temperature. Since we are discussing the temperature field in the presence of heat source/sink so effects of this parameter are very important for both the cases. The effects of heat source/sink parameter are illustrated in Figs. 11 and 12 by keeping Pr ¼ 0:4; b ¼ 0:5; K ¼ 0:1; S ¼ 0:1. It is noted from Fig. 11 that as we increase the strength of the heat source, the temperature increases. This is due to the fact that heat source can add more heat to the stretching sheet which increases its temperature and thus the temperature of fluid rises. Moreover, the thermal boundary layer is also found to increase by increasing the strength of heat source. Fig. 12 is\n\nPr\n\nk\n\nK\n\nb\n\nRex\u00021=2 Nux\n\n0:5 1:0 1:5 2:0 0:5\n\n0:2\n\n5:0\n\n5:0\n\n0:596791 0:634885 0:671806 0:705075 0:596791 0:547204 0:496478 0:44459 0:573683 0:573021 0:572122 0:571885 0:643221 0:630873 0:619684 0:609657\n\n0:2 0:4 0:6 0:8 0:2\n\n0:2 0:4 0:8 1:0 5:0\n\n0:5 1:5 2:5 3:5\n\nplotted to observe the effects of heat sink by keeping other parameters constant. This figure shows opposite results i.e., the temperature decreases by increasing the strength of heat sink. This is because, when strength of heat sink increases, more heat is removed from the sheet and thus a decrease in thermal boundary-layer thickness and temperature is observed. This result is of key importance for the flows where heat transfer is of prime importance. The numerical values of Nusselt number are shown in Table 3. It is found that the Nusselt number increases with Prandtl number while it decreases with increase in heat source, couple stress and magneto-porous parameter. 6. Concluding remarks Flow and heat transfer of a couple stress fluid over an oscillatory stretching sheet in the presence of heat source/sink is investigated using HAM. It is observed that flow and heat transfer characteristics are greatly influenced by the presence of couple stress and heat source/sink parameter. In fact, the skin friction coefficient increases by increasing couple stress and magneto-porous parameter. It is also interesting to note that rate of heat transfer is enhanced by increasing\n\nPlease cite this article in press as: N. Ali et al., MHD flow and heat transfer of couple stress fluid over an oscillatory stretching sheet with heat source/sink in porous medium, Alexandria Eng. J. (2016), http://dx.doi.org/10.1016/j.aej.2016.02.018" ]

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[ "Welcome to Quickmath Solvers!\nSolve an equation, inequality or a system.\n\nExample: 2x-1=y,2y+3=x\n\n# Solving Equations\n\nThe solution of equations is the central theme of algebra. In this chapter we will study some techniques for solving equations having one variable. To accomplish this we will use the skills learned while manipulating the numbers and symbols of algebra as well as the operations on whole numbers, decimals, and fractions that you learned in arithmetics.\n\n## CONDITIONAL AND EQUIVALENT EQUATIONS\n\n### OBJECTIVES\n\nUpon completing this section you should be able to:\n\n1. Classify an equation as conditional or an identity.\n2. Solve simple equations mentally.\n3. Determine if certain equations are equivalent.\n\nAn equation is a statement in symbols that two number expressions are equal.\n\nEquations can be classified in two main types:\n\n1. An identity is true for all values of the literal and arithmetical numbers in it.\n\nExample 1 5 x 4 = 20 is an identity.\n\nExample 2 2 + 3 = 5 is an identity.\n\nExample 3 2x + 3x = 5x is an identity since any value substituted for x will yield an equality.\n\n2. A conditional equation is true for only certain values of the literal numbers in it.\n\nExample 4 x + 3 = 9 is true only if the literal number x = 6.\n\nExample 5 3x - 4 = 11 is true only if x = 5.\n\nThe literal numbers in an equation are sometimes referred to as variables.\n\nFinding the values that make a conditional equation true is one of the main objectives of this text.\n\nA solution or root of an equation is the value of the variable or variables that make the equation a true statement.\n\nThe solution or root is said to satisfy the equation.\n\nSolving an equation means finding the solution or root.\n\nMany equations can be solved mentally. Ability to solve an equation mentally will depend on the ability to manipulate the numbers of arithmetic. The better you know the facts of multiplication and addition, the more adept you will be at mentally solving equations.\n\nExample 6 Solve for x: x + 3 = 7\n\nSolution\n\nTo have a true statement we need a value for x that, when added to 3, will yield 7. Our knowledge of arithmetic indicates that 4 is the needed value. Therefore the solution to the equation is x = 4.\n\n What number added to 3 equals 7?\n\nExample 7 Solve for x: x - 5 = 3\n\nSolution\n\nWhat number do we subtract 5 from to obtain 3? Again our experience with arithmetic tells us that 8 - 5 = 3. Therefore the solution is x = 8.\n\nExample 8 Solve for x: 3x = 15\n\nSolution\n\nWhat number must be multiplied by 3 to obtain 15? Our answer is x = 5.\n\nSolution\n\nWhat number do we divide 2 by to obtain 7? Our answer is 14.\n\nExample 10 Solve for x: 2x - 1 = 5\n\nSolution\n\nWe would subtract 1 from 6 to obtain 5. Thus 2x = 6. Then x = 3.\n\nRegardless of how an equation is solved, the solution should always be checked for correctness.\n\nExample 11 A student solved the equation 5x - 3 = 4x + 2 and found an answer of x = 6. Was this right or wrong?\n\nSolution\n\nDoes x = 6 satisfy the equation 5x - 3 = 4x + 2? To check we substitute 6 for x in the equation to see if we obtain a true statement.\n\nThis is not a true statement, so the answer x = 6 is wrong.\n\nAnother student solved the same equation and found x = 5.\n\nThis is a true statement, so x = 5 is correct.\n\n Many students think that when they have found the solution to an equation, the problem is finished. Not so! The final step should always be to check the solution.\n\nNot all equations can be solved mentally. We now wish to introduce an idea that is a step toward an orderly process for solving equations.\n\n Is x = 3 a solution of x - 1 = 2? Is x = 3 a solution of 2x + I = 7? What can be said about the equations x - 1 = 2 and 2x + 1 = 7?\n\nTwo equations are equivalent if they have the same solution or solutions\n\nExample 12 3x = 6 and 2x + 1 = 5 are equivalent because in both cases x = 2 is a solution.\n\nTechniques for solving equations will involve processes for changing an equation to an equivalent equation. If a complicated equation such as 2x - 4 + 3x = 7x + 2 - 4x can be changed to a simple equation x = 3, and the equation x = 3 is equivalent to the original equation, then we have solved the equation.\n\nTwo questions now become very important.\n\n1. Are two equations equivalent?\n2. How can we change an equation to another equation that is equivalent to it?\n\nThe answer to the first question is found by using the substitution principle.\n\nExample 13 Are 5x + 2 = 6x - 1 and x = 3 equivalent equations?\n\nSolution\n\nThe answer to the second question involves the techniques for solving equations that will be discussed in the next few sections.\n\n To use the substitution principle correctly we must substitute the numeral 3 for x wherever x appears in the equation.\n\n## THE DIVISION RULE\n\n### OBJECTIVES\n\nUpon completing this section you should be able to:\n\n1. Use the division rule to solve equations.\n2. Solve some basic applied problems whose solutions involve using the division rule.\n\nAs mentioned earlier, we wish to present an orderly procedure for solving equations. This procedure will involve the four basic operations, the first of which is presented in this section.\n\nIf each term of an equation is divided by the same nonzero number, the resulting equation is equivalent to the original equation.\n\nTo prepare to use the division rule for solving equations we must make note of the following process:\n\n(We usually write 1x as x with the coefficient 1 understood.)\n\nExample 1 Solve for x: 3x = 10\n\nSolution\n\nOur goal is to obtain x = some number. The division rule allows us to divide each term of 3x = 10 by the same number, and our goal of finding a value of x would indicate that we divide by 3. This would give us a coefficient of 1 for x.\n\nCheck: 3x = 10 and x = these equivalent equations?\n\nWe substitute for x in the first equation obtaining\n\nThe equations are equivalent, so the solution is correct.\n\n Always Check!\n\nExample 2 Solve for x: 5x = 20\n\nSolution\n\n Notice that the division rule does not allow us to divide by zero. Since dividing by zero is not allowed in mathematics, expressions such as are meaningless.\n\nExample 3 Solve for x: 8x = 4\n\nSolution\n\n Errors are sometimes made in very simple situations. Don't glance at this problem and arrive at x = 2! Note that the division rule allows us to divide each term of an equation by any nonzero number and the resulting equation is equivalent to the original equation. Therefore we could divide each side of the equation by 5 and obtain , which is equivalent to the original equation. Dividing by 5 does not help find the solution however. What number should we divide by to find the solution?\n\nExample 4 Solve for x: 0.5x = 6\n\nSolution\n\nExample 6 The formula for finding the circumference (C) of a circle is C = 2πr, where π represents the radius of the circle and it is approximately 3.14. Find the radius of a circle if the circumference is measured to be 40.72 cm. Give the answer correct to two decimal places.\n\nSolution\n\nTo solve a problem involving a formula we first use the substitution principle.\n\n Circumference means \"distance around.\" It is the perimeter of a circle. The radius is the distance from the center to the circle.\n\n## THE SUBTRACTION RULE\n\n### OBJECTIVES\n\nUpon completing this section you should be able to use the subtraction rule to solve equations.\n\nThe second step toward an orderly procedure for solving equations will be discussed in this section. You will use your knowledge of like terms from chapter l as well as the techniques from section THE DIVISION RULE. Notice how new ideas in algebra build on previous knowledge.\n\nIf the same quantity is subtracted from both sides of an equation, the resulting equation will be equivalent to the original equation.\n\nExample 1 Solve for x if x + 7 = 12.\n\nSolution\n\nEven though this equation can easily be solved mentally, we wish to illustrate the subtraction rule. We should think in this manner:\n\n\"I wish to solve for x so I need x by itself on one side of the equation. But I have x + 7. So if I subtract 7 from x + 7, I will have x alone on the left side.\" (Remember that a quantity subtracted from itself gives zero.) But if we subtract 7 from one side of the equation, the rule requires us to subtract 7 from the other side as well. So we proceed as follows:\n\n Note that x + 0 may be written simply as x since zero added to any quantity equals the quantity itself.\n\nExample 2 Solve for x: 5x = 4x + 3\n\nSolution\n\nHere our thinking should proceed in this manner. \"I wish to obtain all unknown quantities on one side of the equation and all numbers of arithmetic on the other so I have an equation of the form x = some number. I thus need to subtract Ax from both sides.\"\n\n Our goal is to arrive at x = some number. Remember that checking your solution is an important step in solving equations.\n\nExample 3 Solve for x: 3x + 6 = 2x + 11\n\nHere we have a more involved task. First subtract 6 from both sides.\n\nNow we must eliminate 2x on the right side by subtracting 2x from both sides.\n\nWe now look at a solution that requires the use of both the subtraction rule and the division rule.\n\n Note that instead of first subtracting 6 we could just as well first subtract 2x from both sides obtaining 3x - 2x + 6 = 2x - 2x + 11 x + 6 = 11. Then subtracting 6 from both sides we have x + 6 - 6 = 11 - 6 x = 5. Keep in mind that our goal is x = some number.\n\nExample 4 Solve for x: 3x + 2 = 17\n\nSolution\n\nWe first use the subtraction rule to subtract 2 from both sides obtaining\n\nThen we use the division rule to obtain\n\n Always check!\n\nExample 5 Solve for x: 7x + 1 = 5x + 9\n\nSolution\n\nWe first use the subtraction rule.\n\nThen the division rule gives us\n\nExample 6 The perimeter (P) of a rectangle is found by using the formula P = 2l+ 2w, where l stands for the length and w stands for the width. If the perimeter of a rectangle is 54 cm and the length is 15 cm, what is the width?\n\nSolution\n\n Perimeter is the distance around. Do you see why the formula is P = 2l + 2w?\n\n### OBJECTIVES\n\nUpon completing this section you should be able to use the addition rule to solve equations.\n\nWe now proceed to the next operation in our goal of developing an orderly procedure for solving equations. Once again, we will rely on previous knowledge.\n\nIf the same quantity is added to both sides of an equation, the resulting equation will be equivalent to the original equation.\n\nExample 1 Solve for x if x - 7 = 2.\n\nSolution\n\nAs always, in solving an equation we wish to arrive at the form of \"x = some number.\" We observe that 7 has been subtracted from x, so to obtain x alone on the left side of the equation, we add 7 to both sides.\n\n Remember to always check your solution.\n\nExample 2 Solve for x: 2x - 3 = 6\n\nSolution\n\nKeeping in mind our goal of obtaining x alone, we observe that since 3 has been subtracted from 2x, we add 3 to both sides of the equation.\n\nNow we must use the division rule.\n\n Why do we add 3 to both sides? Note that in the example just using the addition rule does not solve the problem.\n\nExample 3 Solve for x: 3x - 4 = 11\n\nSolution\n\nWe first use the addition rule.\n\nThen using the division rule, we obtain\n\n Here again, we needed to use both the addition rule and the division rule to solve the equation.\n\nExample 4 Solve for x: 5x = 14 - 2x\n\nSolution\n\nHere our goal of obtaining x alone on one side would suggest we eliminate the 2x on the right, so we add 2x to both sides of the equation.\n\nWe next apply the division rule.\n\n Here again, we needed to use both the addition rule and the division rule to solve the equation. Note that we check by always substituting the solution in the original equation.\n\nExample 5 Solve for x: 3x - 2 = 8 - 2x\n\nSolution\n\nHere our task is more involved. We must think of eliminating the number 2 from the left side of the equation and also the lx from the right side to obtain x alone on one side. We may do either of these first. If we choose to first add 2x to both sides, we obtain\n\nWe now add 2 to both sides.\n\nFinally the division rule gives\n\n Could we first add 2 to both sides? Try it!\n\n## THE MULTIPLICATION RULE\n\n### OBJECTIVES\n\nUpon completing this section you should be able to:\n\n1. Use the multiplication rule to solve equations.\n2. Solve proportions.\n3. Solve some basic applied problems using the multiplication rule.\n\nWe now come to the last of the four basic operations in developing our procedure for solving equations. We will also introduce ratio and proportion and use the multiplication rule to solve proportions.\n\nIf each term of an equation is multiplied by the same nonzero number, the resulting equation is equivalent to the original equation.\n\nIn elementary arithmetic some of the most difficult operations are those involving fractions. The multiplication rule allows us to avoid these operations when solving an equation involving fractions by finding an equivalent equation that contains only whole numbers.\n\nRemember that when we multiply a whole number by a fraction, we use the rule\n\nWe are now ready to solve an equation involving fractions.\n\n Note that in each case only the numerator of the fraction is multiplied by the whole number.\n\nExample 4\n\nSolution\n\nKeep in mind that we wish to obtain x alone on one side of the equation. We also would like to obtain an equation in whole numbers that is equivalent to the given equation. To eliminate the fraction in the equation we need to multiply by a number that is divisible by the denominator 3. We thus use the multiplication rule and multiply each term of the equation by 3.\n\nWe now have an equivalent equation that contains only whole numbers. Using the division rule, we obtain\n\n To eliminate the fraction we need to multiply by a number that is divisible by the denominator. In the example we need to multiply by a number that is divided by 3. We could have multiplied both sides by 6, 9, 12, and so on, but the equation is simpler and easier to work with if wc use the smallest multiple.\n\nExample 5\n\nSolution\n\n See if you obtain the same solution by multiplying each side of the original equation by 16. Always check in the original equation.\n\nExample 6\n\nSolution\n\nHere our task is the same but a little more complex. We have two fractions to eliminate. We must multiply each term of the equation by a number that is divisible by both 3 and 5. It is best to use the least of such numbers, which you will recall is the least common multiple. We will therefore multiply by 15.\n\n In arithmetic you may have referred to the least common multiple as the \"lowest common denominator.\"\n\nExample 7\n\nSolution\n\nThe least common multiple for 8 and 2 is 8, so we multiply each term of the equation by 8.\n\nWe now use the subtraction rule.\n\nFinally the division rule gives us\n\n Before multiplying, change any mixed numbers to improper fractions. In this example change . Remember that each term must be multiplied by 8. Note that in this example we used three rules to find the solution.\n\nSolving simple equations by multiplying both sides by the same number occurs frequently in the study of ratio and proportion.\n\nA ratio is the quotient of two numbers.\n\nThe ratio of a number x to a number y can be written as x:y or . In general, the fractional form is more meaningful and useful. Thus, we will write the ratio of 3 to 4 as .\n\nA proportion is a statement that two ratios are equal.\n\nExample 8\n\nSolution\n\nWe need to find a value of x such that the ratio of x to 15 is equal to the ratio of 2 to 5.\n\nMultiplying each side of the equation by 15, we obtain\n\n Why do we multiply both sides by 15? Check this solution in the original equation.\n\nExample 9 What number x has the same ratio to 3 as 6 has to 9?\n\nSolution\n\nTo solve for x we first write the proportion:\n\nNext we multipy each side of the equation by 9.\n\n Say to yourself, \"2 is to 5 as x is to 10.\"Check!\n\nExample 11 The ratio of the number of women to the number of men in a math class is 7 to 8. If there are 24 men in the class, how many women are in the class?\n\nSolution\n\n Check!\n\nExample 12 Two sons were to divide an inheritance in the ratio of 3 to 5. If the son who received the larger portion got \\$20,000, what was the total amount of the inheritance?\n\nSolution\n\nWe now add \\$20,000 + \\$12,000 to obtain the total amount of \\$32,000.\n\n Check! Again, be careful in setting up the proportion. In the ratio 3/5, 5 is the larger portion. Therefore, since \\$20,000 is the larger portion, it must also appear in the denominator.\n\nExample 13 If the legal requirements for room capacity require 3 cubic meters of air space per person, how many people can legally occupy a room that measures 6 meters wide, 8 meters long, and 3 meters high?\n\nSolution\n\nSo, 48 people would be the legal room capacity.\n\n This means \"1 person is to 3 cubic meters as x people are to 144 cubic meters.\" Check the solution.\n\n## COMBINING RULES FOR SOLVING EQUATIONS\n\n### OBJECTIVES\n\nUpon completing this section you should be able to:\n\n1. Use combinations of the various rules to solve more complex equations.\n2. Apply the orderly steps established in this section to systematically solve equations.\n\nMany of the exercises in previous sections have required the use of more than one rule in the solution process. In fact, it is possible that a single problem could involve all the rules\n\nThere is no mandatory process for solving equations involving more than one rule, but experience has shown that the following order gives a smoother, more mistake-free procedure.\n\nFirst Eliminate fractions, if any, by multiplying each term of the equation by the least common multiple of all denominators of fractions in the equation.\nSecond Simplify by combining like terms on each side of the equation.\nThird Add or subtract the necessary quantities to obtain the unknown quantity on one side and the numbers of arithmetic on the other side.\nFourth Divide by the coefficient of the unknown quantity.\n\n Remember, the coefficient is the number being multiplied by the letter. (That is, in the expression 5x the coefficient is 5.)\n\n Again, make sure every term is multiplied by 3.\n\nSolution\n\nMultiplying each term by 15 yields\n\nYou may want to leave your answer as an improper fraction instead of a mixed number. Either form is correct, but the improper fraction form will be more useful in checking your solution.\n\n Note that there are four terms in this equation.\n\nExample 3 The selling price (S) of a certain article was \\$30.00. If the margin (M) was one-fifth of the cost (C), find the cost of the article. Use the formula C + M = S.\n\nSolution\n\nSince the margin was one-fifth of the cost, we may write\n\n## SUMMARY\n\n### Key Words\n\n• An equation is a statement in symbols that two number expressions are equal.\n• An identity is true for all values of the literal and arithmetic numbers in it.\n• A conditional equation is true for only certain values of the literal numbers in it.\n• A solution or root of an equation is the value of the variable that makes the equation a true statement.\n• Two equations are equivalent if they have the same solution set.\n• A ratio is the quotient of two numbers.\n• A proportion is a statement that two ratios are equal.\n\n### Procedures\n\n• If each term of an equation is divided by the same nonzero number, the resulting equation is equivalent to the original equation.\n• If the same quantity is subtracted from both sides of an equation, the resulting equation is equivalent to the original equation.\n• If the same quantity is added to both sides of an equation, the resulting equation is equivalent to the original equation.\n• If each side of an equation is multiplied by the same nonzero number, the resulting equation is equivalent to the original equation.\n• To solve an equation follow these steps:\nStep 1 Eliminate fractions by multiplying each term by the least common multiple of all denominators in the equation.\nStep 2 Combine like terms on each side of the equation.\nStep 3 Add or subtract terms to obtain the unknown quantity on one side and the numbers of arithmetic on the other.\nStep 4 Divide each term by the coefficient of the unknown quantity." ]

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[ "# Multidimensional CFA with RStan\n\n## 2018/11/28\n\nFor starters, here’s the Stan code. And here’s the R script: Stan code. If you are already familiar with RStan, the basic concepts you need to combine are standard multilevel models with correlated random slopes and heteroskedastic errors.\n\nI will embed R code into the demonstration. The required packages are lavaan, lme4 and RStan.\n\nI like to understand most statistical methods as regression models. This way, it’s easy to understand the claims underlying a large number of techniques. It’s an approach that works for multilevel, SEM, and IRT models. Here, I’ll be focusing on confirmatory factor analysis (CFA), so I’ll begin by developing CFA from a model that is easy to fit in any multilevel regression software:\n\n$$y_{pi} = \\theta_p + \\beta_i + \\epsilon$$\n\nwhere $$y_{pi}$$ is the response of person $$p$$ to item $$i$$, $$\\theta_p$$ is the random intercept that varies by person and is assumed to be normal with mean zero and standard deviation $$\\alpha$$ $$\\big(\\mathcal{N}(0, \\alpha^2)\\big)$$, $$\\beta_i$$ are the coefficients of the item dummies, and $$\\epsilon$$ is the residual (level 1) error assumed normal with mean zero and standard deviation $$\\sigma,\\ \\mathcal{N}(0, \\sigma^2)$$.\n\nAnyone familiar with multilevel regression can fit this basic random-intercept model. Shape your person by item data to “long form” where each row contains a person ID, an item ID and the item response. Make the item response the outcome, make each item into a dummy variable as predictors (at level 1) and place a random intercept on the person. Your random intercept standard deviation is $$\\alpha$$, residual standard deviation (or level one error) is $$\\sigma$$ and the item dummies are $$\\beta_i$$. If your software permits, remove the fixed intercept and use all item dummies. Finally, $$\\theta_p$$ represents the latent score for each individual $$p$$.\n\nTo demonstrate, I use the HolzingerSwineford dataset. There are 9 items, items 1 to 3 load on factor 1, 4 to 6 on factor 2, and 7 to 9 on factor 3.\n\nlibrary(lavaan)\n\ndat <- HolzingerSwineford1939\ndat$sex <- dat$sex - 1 # Who is 1, no idea\ndat$grade <- dat$grade - min(dat$grade, na.rm = TRUE) dat$ID <- 1:nrow(dat)\n\n# Make data long\ndat.l <- tidyr::gather(dat, item, score, x1:x9)\ndat.l$item.no <- as.integer(gsub(\"x\", \"\", dat.l$item))\n\nlibrary(lme4)\n\nlmer(score ~ 0 + factor(item.no) + (1 | ID), dat.l, REML = FALSE)\n\n# Random effects:\n# Groups Name Std.Dev.\n# ID (Intercept) 0.5758\n# Residual 0.9694\n# Number of obs: 2709, groups: ID, 301\n\n\nThe model above fit with ML not REML is identical to a unidimensional CFA where we constrain all the item loadings to be the same value and the item errors to be the same value. The random intercept standard deviation, $$\\alpha$$, is the factor loading for all items and $$\\sigma^2$$ is the item error variance for all items. To get the standardized loading which we will call $$\\lambda$$, use:\n\n$$\\lambda = \\frac{\\alpha}{\\sqrt{\\alpha^2+\\sigma^2}} = \\frac{0.5758}{\\sqrt{0.5758^2+0.9694^2}}=0.5107$$\n\nThe lavaan model below should confirm this to be true. Note that in the lavaan syntax, the factor is standardized to have variance of 1 using std.lv = TRUE. The item dummies (not shown) from lme4 are also the CFA item intercepts.\n\nparameterEstimates(sem(\n\"F1 =~ a * x1 + a * x2 + a * x3 + a * x4 + a * x5 + a * x6 + a * x7 + a * x8 + a * x9\\n\nx1 ~~ f * x1\\nx2 ~~ f * x2\\nx3 ~~ f * x3\\nx4 ~~ f * x4\\n\nx5 ~~ f * x5\\nx6 ~~ f * x6\\nx7 ~~ f * x7\\nx8 ~~ f * x8\\nx9 ~~ f * x9\",\ndat, std.lv = TRUE\n), standardized = TRUE)[c(1:2, 10:11), c(1:5, 12)]\n\n# lhs op rhs label est std.all\n# 1 F1 =~ x1 a 0.576 0.511\n# 2 F1 =~ x2 a 0.576 0.511\n# 10 x1 ~~ x1 f 0.940 0.739\n# 11 x2 ~~ x2 f 0.940 0.739\n\n\nThe standardized loading has a similar formula to the (residual) intraclass correlation coefficient (ICC) which is $$\\frac{\\alpha^2}{\\alpha^2+\\sigma^2}$$. So we have that $$\\lambda=\\sqrt{\\mathrm{ICC}}$$. This is in line with thinking of standardized loadings as correlation coefficients and ICCs as $$R^2$$.\n\nNow, everyone familiar with CFA knows we almost never constrain all item loadings and item errors to have the same value (for what it’s worth, this is what we do in the Rasch model with binary outcomes). But we will maintain the CFA as regression for as long as possible. Let us extend the model to include multiple factors. To include multiple factors, we create an indicator column in long form that uniquely identifies the factor an item belongs to. And instead of using a single random intercept, we use the factor dummies as random slopes without the random intercept. In this model, we assume:\n\n$$y_{pi} = \\mathbf{\\theta}_{pf} + \\beta_i + \\epsilon$$\n\nwhere $$\\mathbf{\\theta}_{pf}$$ instead of being the same at all times for any individual now also depends on the specific factor $$f$$. Precisely, we assume the latent scores are multivariate normal, $$\\mathbf{\\theta}_{pf}\\sim\\mathcal{N}_f(\\mathbf{0,\\Sigma})$$, where $$f$$ is the number of factors, and $$\\mathbf{\\Sigma}$$ represents the covariance matrix of the random slopes. The square root of the diagonal of this covariance matrix (i.e. standard deviations of random slopes) will again be the unstandardized factor loadings, $$\\alpha_f$$. The off-diagonal covariances when standardized (i.e. correlations) will be the interfactor correlations. We can again fit this model in lme4:\n\n# Assign item to factors\ndat.l$Fs <- ((dat.l$item.no - 1) %/% 3) + 1\n\nlmer(score ~ 0 + factor(item) + (0 + factor(Fs) | ID), dat.l, REML = FALSE)\n\n# Random effects:\n# Groups Name Std.Dev. Corr\n# ID factor(Fs)1 0.7465\n# factor(Fs)2 0.9630 0.41\n# factor(Fs)3 0.6729 0.38 0.30\n# Residual 0.7909\n\n\nThe corresponding lavaan model is:\n\nparameterEstimates(sem(\n\"F1 =~ a * x1 + a * x2 + a * x3\\nF2 =~ b * x4 + b * x5 + b * x6\\nF3 =~ c * x7 + c * x8 + c * x9\\n\nx1 ~~ f * x1\\nx2 ~~ f * x2\\nx3 ~~ f * x3\\nx4 ~~ f * x4\\nx5 ~~ f * x5\\n\nx6 ~~ f * x6\\nx7 ~~ f * x7\\nx8 ~~ f * x8\\nx9 ~~ f * x9\",\ndat, std.lv = TRUE\n), standardized = TRUE)[c(1:10, 22:24), c(1:5, 12)]\n\n# lhs op rhs label est std.all\n# 1 F1 =~ x1 a 0.746 0.686\n# 2 F1 =~ x2 a 0.746 0.686\n# 3 F1 =~ x3 a 0.746 0.686\n# 4 F2 =~ x4 b 0.963 0.773\n# 5 F2 =~ x5 b 0.963 0.773\n# 6 F2 =~ x6 b 0.963 0.773\n# 7 F3 =~ x7 c 0.673 0.648\n# 8 F3 =~ x8 c 0.673 0.648\n# 9 F3 =~ x9 c 0.673 0.648\n# 10 x1 ~~ x1 f 0.626 0.529\n# 22 F1 ~~ F2 0.407 0.407\n# 23 F1 ~~ F3 0.385 0.385\n# 24 F2 ~~ F3 0.301 0.301\n\n\nWe see that the factor loadings in CFA are the random slope standard deviations from multilevel. And the interfactor correlation matrix matches the random slope correlations from multilevel.\n\nThe final model we can fit with some, not all, multilevel regression software is:\n\n$$y_{pi} = \\mathbf{\\theta}_{pf} + \\beta_i + \\epsilon_i$$\n\nwhere we permit the residual error, $$\\epsilon_i$$, to be different depending on the item. This is a heteroskedastic regression model. One would have to switch to from lme4 to nlme or glmmTMB in R to estimate it; it should be possible in HLM. In lavaan, the model syntax would be:\n\n# Drop the error variance constraints\n\"F1 =~ a * X1 + a * X2 + a * X3\\nF2 =~ b * X4 + b * X5 + b * X6\\nF3 =~c * X7 + c * X8 + c * X9\"\n\n\nThe latest model is very close to a standard CFA model. The final change is that we need to permit the item loadings to vary by item, not by factor. Once we do this, we can no longer fit the model with multilevel regression software. The equation becomes:\n\n$$y_{pi} = \\alpha_i\\theta_{pf} + \\beta_i + \\epsilon_i$$\n\n$$\\mathbf{\\theta}_{pf}$$ is still multivariate normal, but with variance $$\\mathbf{1},\\ \\mathbf{\\theta}_{pf}\\sim\\mathcal{N}_f(\\mathbf{0,1})$$. The most interesting inclusion is that we multiply the latent scores by an item coefficient, $$\\alpha_i$$.1 Note that $$\\alpha$$ consistently represents the unstandardized loadings in this demonstration. Here, instead of being constant or varying by factor, it varies by item.\n\nBayesian software can fit a complicated model like this. We’d have to specify priors for the different components of this equation. Page 144 of the Stan manual provides some priors for the 2PL item-response theory model that we can apply here as a starting point:\n\n• loadings will be log-normal constraining them to be positive, $$\\alpha_i\\sim\\mathrm{Lognormal}(0,\\sigma_\\alpha)$$. For $$\\sigma_\\alpha$$, we can assume it’s half-Cauchy, $$\\sigma_\\alpha\\sim C^{+}(0, 2.5)$$.\n• there is usually enough information to estimate item intercepts/difficulties, we can use a uniform prior\n• item errors can be assumed to be inverse gamma, $$\\epsilon_i\\sim\\mathrm{InvGamma}(1,1)$$.\n• the latent scores are multivariate normal, $$\\mathbf{\\theta}_{pf}\\sim\\mathcal{N}_f(\\mathbf{0,1})$$\n\nThe one point requiring additional comments is the multivariate normal prior for $$\\mathbf{\\theta}_{pf}$$. For a positive-definite correlation matrix, $$\\mathbf{R}$$, we can apply the Cholesky decomposition: $$\\mathbf{R} = \\mathbf{L}\\mathbf{L}^*$$. In Stan, it is computationally more convenient to apply a prior on $$\\mathbf{L}$$ and generate the multivariate normal latent scores using $$\\mathbf{L}$$. The recommended choice of prior for $$\\mathbf{L}$$ is the LKJ-prior.2\n\nIn Stan syntax, the required data are:\n\ndata {\nreal g_alpha; // for inverse gamma\nreal g_beta; // for inverse gamma\nint<lower = 0> N; // scalar, number of person times number of items\nint<lower = 0> Ni; // scalar, number of items\nint<lower = 0> Np; // scalar, number of persons\nint<lower = 0> Nf; // scalar, number of factors\nvector[N] response; // vector, long form of item responses\n// all remaining entries are data in long form\n// with consecutive integers beginning at 1 acting as unique identifiers\nint<lower = 1, upper = Ni> items[N];\nint<lower = 1, upper = Np> persons[N];\nint<lower = 1, upper = Nf> factors[N];\n}\n\n\nThe estimated parameters are:\n\nparameters {\nvector<lower = 0>[Ni] item_vars; // item vars heteroskedastic\nvector[Ni] betas; // item intercepts, default uniform prior\nvector[Nf] thetas[Np]; // person scores for each factor\ncholesky_factor_corr[Nf] L; // Cholesky decomp of corr mat of random slopes\n}\n\n\nWe need some transformed parameters to capture the mean and variance of the item responses. This is where we provide the regression equation verbatim:\n\ntransformed parameters {\nvector[N] yhat;\nvector[N] item_sds_i;\n\nfor (i in 1:N) {\nyhat[i] = alphas[items[i]] * thetas[persons[i], factors[i]] + betas[items[i]];\nitem_sds_i[i] = sqrt(item_vars[items[i]]);\n}\n}\n\n\nAnd for the priors:\n\nmodel {\nvector[Nf] A = rep_vector(1, Nf); // Vector of random slope variances\nmatrix[Nf, Nf] A0;\n\nL ~ lkj_corr_cholesky(Nf);\nA0 = diag_pre_multiply(A, L);\nthetas ~ multi_normal_cholesky(rep_vector(0, Nf), A0);\n\nalphas ~ lognormal(0, sigma_alpha);\nsigma_alpha ~ cauchy(0, 2.5); // hyperparm\nitem_vars ~ inv_gamma(g_alpha, g_beta);\n\nresponse ~ normal(yhat, item_sds_i);\n}\n\n\nFinally, we can compute the standardized loadings and the interfactor correlation matrix, R:\n\ngenerated quantities {\nmatrix[Nf, Nf] R;\n\nR = tcrossprod(L);\nfor (i in 1:Ni) {\n}\n}\n\n\nWe could make a few modifications:\n\n• We can standardize the item responses prior to modeling to improve computational stability\n• Then apply a normal prior to the item intercepts\n\nThe syntax to run the model would then be:\n\n# First, let's fit the model in lavaan:\ncfa.lav.fit <- sem(\n\"F1 =~ x1 + x2 + x3\\nF2 =~ x4 + x5 + x6\\nF3 =~ x7 + x8 + x9\",\ndat, std.lv = TRUE, meanstructure = TRUE)\n\nlibrary(rstan)\n\noptions(mc.cores = parallel::detectCores()) # Use multiple cores\nrstan_options(auto_write = TRUE) # One time Stan program compilation\n\ncfa.mm <- stan_model(stanc_ret = stanc(file = \"bayes_script/cfa.stan\")) # Compile Stan code\n\ncfa.stan.fit <- sampling(\ncfa.mm, data = list(\nN = nrow(dat.l), Ni = length(unique(dat.l$item)), Np = length(unique(dat.l$ID)), Nf = length(unique(dat.l$Fs)), items = dat.l$item.no, factors = dat.l$Fs, persons = dat.l$ID, response = dat.l$score, g_alpha = 1, g_beta = 1), control = list(adapt_delta = .99999, max_treedepth = 15)) What are some loadings? print(cfa.stan.fit, pars = c(\"alphas\", \"loadings_std\"), probs = c(.025, .5, .975), digits_summary = 3) # mean se_mean sd 2.5% 50% 97.5% n_eff Rhat # alphas 0.889 0.003 0.078 0.733 0.890 1.041 790 1.002 # alphas 0.991 0.002 0.056 0.885 0.988 1.101 1263 1.002 # alphas 1.102 0.002 0.062 0.980 1.102 1.224 1056 1.001 # alphas 0.692 0.003 0.075 0.548 0.692 0.846 799 1.005 # loadings_std 0.751 0.002 0.052 0.643 0.752 0.848 601 1.003 # loadings_std 0.848 0.001 0.023 0.801 0.849 0.890 1275 1.003 # loadings_std 0.851 0.001 0.023 0.803 0.852 0.891 1176 1.001 # loadings_std 0.672 0.003 0.059 0.552 0.673 0.786 556 1.007 # For comparison, the lavaan loadings are: parameterEstimates(cfa.lav.fit, standardized = TRUE)[1:9, c(1:5, 11)] # lhs op rhs est se std.all # 1 F1 =~ x1 0.900 0.081 0.772 # 4 F2 =~ x4 0.990 0.057 0.852 # 5 F2 =~ x5 1.102 0.063 0.855 # 9 F3 =~ x9 0.670 0.065 0.665 For the interfactor correlations: print(cfa.stan.fit, pars = c(\"R[1, 2]\", \"R[1, 3]\", \"R[2, 3]\"), probs = c(.025, .5, .975), digits_summary = 3) # mean se_mean sd 2.5% 50% 97.5% n_eff Rhat # R[1,2] 0.435 0.001 0.065 0.303 0.437 0.557 2019 0.999 # R[1,3] 0.451 0.003 0.081 0.289 0.450 0.607 733 1.005 # R[2,3] 0.271 0.001 0.071 0.130 0.272 0.406 2599 1.000 # From lavaan: parameterEstimates(cfa.lav.fit, standardized = TRUE)[22:24, c(1:5, 11)] # lhs op rhs est se std.all # 22 F1 ~~ F2 0.459 0.064 0.459 # 23 F1 ~~ F3 0.471 0.073 0.471 # 24 F2 ~~ F3 0.283 0.069 0.283 For the error variances: print(cfa.stan.fit, pars = c(\"item_vars\"), probs = c(.025, .5, .975), digits_summary = 3) # mean se_mean sd 2.5% 50% 97.5% n_eff Rhat # item_vars 0.829 0.003 0.095 0.652 0.828 1.026 1292 1.000 # item_vars 0.383 0.001 0.049 0.292 0.381 0.481 1552 1.002 # item_vars 0.459 0.001 0.059 0.351 0.456 0.581 1577 1.001 # item_vars 0.575 0.004 0.085 0.410 0.575 0.739 532 1.008 # From lavaan: parameterEstimates(cfa.lav.fit, standardized = TRUE)[10:18, 1:5] # lhs op rhs est se # 12 x3 ~~ x3 0.844 0.091 # 13 x4 ~~ x4 0.371 0.048 # 14 x5 ~~ x5 0.446 0.058 # 18 x9 ~~ x9 0.566 0.071 For the item intercepts: print(cfa.stan.fit, pars = c(\"betas\"), probs = c(.025, .5, .975), digits_summary = 3) # mean se_mean sd 2.5% 50% 97.5% n_eff Rhat # betas 6.087 0.001 0.068 5.954 6.089 6.219 2540 1.001 # betas 2.248 0.001 0.066 2.122 2.248 2.381 1980 1.002 # betas 2.182 0.003 0.063 2.058 2.182 2.302 625 1.008 # betas 4.185 0.002 0.066 4.054 4.186 4.315 1791 1.001 # From lavaan: parameterEstimates(cfa.lav.fit, standardized = TRUE)[25:33, 1:5] # lhs op rhs est se # 26 x2 ~1 6.088 0.068 # 27 x3 ~1 2.250 0.065 # 30 x6 ~1 2.186 0.063 # 31 x7 ~1 4.186 0.063 So we are able to replicate the results from lavaan. From here, you can extend the model in interesting ways to arrive at other results. For example, if you want to do a regression of the factors, you can use the posterior of the correlation matrix and the solve() function to arrive at the coefficients for the factors in the regression. Here, I regress factor 1 on both factors 2 and 3: R <- extract(cfa.stan.fit, c(\"R[1, 2]\", \"R[1, 3]\", \"R[2, 3]\")) R <- cbind(R$R[1,2], R$R[1,3], R$R[2,3])\ncoefs <- matrix(NA, nrow(R), ncol(R) - 1)\nfor (i in 1:nrow(R)) {\nm <- matrix(c(1, R[i, 3], R[i, 3], 1), 2, 2)\ncoefs[i, ] <- solve(m, R[i, 1:2])\n}; rm(i, m)\nt(apply(coefs, 2, function (x) {\nc(estimate = mean(x), sd = sd(x), quantile(x, c(.025, .25, .5, .75, .975)))\n}))\n# estimate sd 2.5% 25% 50% 75% 97.5%\n# [1,] 0.3362981 0.07248634 0.1918812 0.2877936 0.3387682 0.3875141 0.4725508\n# [2,] 0.3605951 0.08466494 0.1996710 0.3027047 0.3594806 0.4164141 0.5308578\n\n\n1. This makes the model non-linear in its parameters showing why software for generalized linear mixed models cannot fit this model. ↩︎\n\n2. I’m guessing older software manuals may recommend the inverse-Wishart prior for $$\\mathbf{\\Sigma}$$. But see HERE as starting point for comments on LKJ-prior. As a funny side note, LKJ is an acronym for the researchers who recommended this approach, the last author is Harry, last name, Joe. And he developed an earlier form of the method. So in the paper by LKJ, you see references to Joe’s method which seems out of place for an academic paper :). ↩︎" ]

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[ "# Neutron 6: 2nd Degree (N6-30)", null, "", null, "", null, "", null, "", null, "A spherical solution for N6. Note: At one time I had a sinusoidal solution off the core structure of N6-30 which had 3 P.V.B.'s, but because it was difficult to form the core structure for this shell, I removed the sinusoidal string and put on fillers to test this structure. The results are what you see. So, there is at least one other solution available off the core structure of this shell.\n\n## Stereograms of N6-30 Shell", null, "", null, "Required Snap Points: 66 Available Snap Points: 66 Pinned Vector Bosons: 0 L1 = 29.10494525\" (Structural) L2 = 29.10494525\" (Structural) L3 = 16.42129106\" (Filler) Model Diameter: 9.26439181\" Loop Ratio: (1:2): 1 (1:3): 1.772390802 (2:3): 1.772390802 Loop Equation: 3(L1) + 3(L2) + 8(L3) = 306\" Snap Point Equation: 3(6) + 3(8) + 8(3) = 66 R.S.P.\n\n# Neutron 6: 3rd Degree, Polar Star (N6-30P)", null, "", null, "This P6 shell locates all 12 Down quarks at the two poles of the shell. It stacks into a complementary shell formation with P6-30PB and P6-40PS. From the typical structures carbon 12 forms, I believe this neutron shell typically takes the lower (inner) shell position with P6-30PB as the outer shell.\n\n## Stereograms of N6-30P Shell", null, "", null, "Required Snap Points: 66 Available Snap Points: 66 Pinned Vector Bosons: 0 L1 = 27.94145579\" (Structural) L2 = 16.84933181\" (Filler) L3 = 18.62763719\" (Polar) Model Diameter: 10.26995824\" Loop Ratios: (1:2): 1.658312395 (1:3): 1.5 (2:3): 0.904534034 Loop Equation: 6(L1) + 6(L2) + 2(L3) = 306\" Snap Point Equation: 6(6) + 6(3) + 2(6) = 66 R.S.P.\n\n# Neutron 6: 4th Degree, Polar Star (N6-40P)", null, "Similar to the solution above, but with one polar and one equatorial circlet rather than two polar circlets. The down quarks all reside at the equator and the pole opposite the one shown. (Side and opposite pole not shown because I never bothered to actually build the model.) This neutron shell will stack with P5-35P to form a Boron 11 nuclei having all 5 of the proton shells P.V.B.'s clustered at one pole of the P5-35P shell. It will also stack into complementary shells with P6-30PA to form a highly polarized nuclear isomer of carbon 12.\n\n Required Snap Points: 66 Available Snap Points: 66 Pinned Vector Bosons: 0 L1 = 26.74941211\" (Structural) L2 = 30.8875714\" (Equatorial) L3 = 16.13050242\" (Filler) L4 = 17.83294141\" (Polar) Model Diameter: 9.831819336\" Loop Ratios: (1:2): 0.8660251 (1:3): 1.658312395 (1:4): 1.5 (2:3): 1.914854887 (2:4): 1.732051415 (3:4): 0.904534034 Loop Equation: 6(L1) + 1(L2) + 6(L3) + 1(L4) = 306\" Snap Point Equation: 6(6) + 1(6) + 6(3) + 1(6) = 66 R.S.P.\n\n# Neutron 6: 4th Degree, Belt Sinusoidal (N6-42S)", null, "", null, "", null, "Here's an exotic banded solution set for N6 containing an interesting arrangement of strings. The 4 structural loops are equatorial, all intersecting on the down quarks at either pole. A sinusoidal string wraps the equator with pseudo-equatorial circlets making contact at the minima and maxima of the sine. This sinusoidal string is fairly easy to mensurate because the amplitude of the sine can be easily calculated.\n\n## Stereograms of N6-42S Shell", null, "", null, "Required Snap Points: 66 Available Snap Points: 64 Pinned Vector Bosons: 2 L1 = 34.78010941\" (Structural) L2 = 33.30825351\" (Pseudo-Equatorial) L3 = 14.15742257\" (Filler) L4 = 43.63336506\" (Sinusoidal) Model Diameter: 11.07085267\" Loop Ratios: (1:2): 1.044188925 (1:3): 2.456669585 (1:4): 0.797098949 (2:3): 2.352706035 (2:4): 0.7633666 (3:4): 0.32446323 Loop Equation: 4(L1) + 2(L2) + 4(L3) + 1(L4) = 306\" Snap Point Equation: 4(6) + 2(8) + 4(4) + 1(8) = 64 A.S.P. P.V.B.'s:(+ 2 P.V.B.'s = 66 R.S.P.)\n\nCopyright 1997 by Arnold J. Barzydlo\nGo Back" ]

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[ "", null, "# Z Score Table Printable That Are Impeccable\n\nz score table printable That are Impeccable\n\nUncover values upon the remaining of the necessarily mean within this detrimental Z rating desk. Desk entries for z signify the community down below the bell curve in direction of the remaining of z. Destructive ratings inside the z-desk correspond in direction of the values which are considerably less than the signify.", null, "desk figures printable ranking likelihood slip-up tail remaining community data pieces straight print acquiring textbook suggest style speculation compute amongst\n\nCumulative likelihood for Optimistic z-values are mentioned within just the just after desk: Name: std natural desk.xls Produced Day: 9/8/2005 8:46:41 AM …\n\ntags: printable z ranking desk figures, printable z desk chart, printable z desk data Similar For Printable Z Desk Data. Printable Large Flower Template Totally free Printable Free of charge Printables For Banners | Free of charge Printable. Printable Child Wrapping Paper", null, "ranking desk printable chart data optimistic print rankings a2 instructors ztable all-natural distribution pdf examine sheet directly handout\n\nDown load absolutely free printable worksheets, printable coloring web pages, printable papers, printable traces, printable graphs, printable envelopes, printable labels and many others for all your artwork assignments for the distinctive scenarios. Immediately simply click your mouse – then conserve the impression and employ your printer towards print it out!", null, "desk ranking all-natural data distribution detrimental ap chart printable regular quizlet constructive curve math tables random suggest axis functions ztable\n\nDetrimental Z rating desk Retain the services of the detrimental Z rating desk in this article in direction of obtain values upon the still left of the suggest as can be observed inside the graph along with. Corresponding values which are significantly less than the necessarily mean are marked with a adverse ranking inside of the z-desk and respresent the Room below the bell curve towards theContinue Examining\n\nRegular Pure DISTRIBUTION: Desk Values Characterize Space towards the Remaining of the Z rating. Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .50000 .50399 .50798 .51197 .51595 …\n\nREAD  77 ways to play tenzi printable That are Gorgeous", null, "desk ranking printable studies pdf replica chart print directly slideshare tags natural tabel future amount mouse", null, "natural desk verify distribution significance percentile tests statistical signify opportunity statistic possibilities approximation data speculation curve binomial local values making use of\n\nYou might also search for different z score table printable. A few examples include:\n\n• Z Table Chart\n• Z-Table PDF\n• Z-Score Chart\n• Full Z-Score Table\n• Positive Z-Score Table\n• Z-Table Stats\n• Negative Z-Score Table\n• Z-Score Table Calculator\n• Standard Negative Z-Score Table\n• T-score Table\n• Normal Z-Score Table\n• Z-Score P-Value Table\n\nLooking for answers about z score table printable? You’ll most likely find them here! Below are the FAQ which contain a list of questions.\n\n### How do you find Z score using Z table?\n\nHow to Find Probabilities for Z with the Z-TableGo to the row that represents the ones digit and the first digit after the decimal point (the tenths digit) of your z-value.Go to the column that represents the second digit after the decimal point (the hundredths digit) of your z-value.Intersect the row and column from Steps 1 and 2.\n\n### What is the z score table?\n\nA Z-Score Table, is a table that shows the percentage of values (or area percentage) to the left of a given z-score on a standard normal distribution.(The label for columns contains the second decimal of the z-score.) The intersection of the rows and columns gives the probability or area under the normal curve.\n\n### How do you calculate the Z score?\n\nTo find the Z score of a sample, you’ll need to find the mean, variance and standard deviation of the sample. To calculate the z-score, you will find the difference between a value in the sample and the mean, and divide it by the standard deviation.\n\n### What is Z table in normal distribution?\n\nThe z-table is short for the “Standard Normal z-table”. The Standard Normal model is used in hypothesis testing, including tests on proportions and on the difference between two means. The area under the whole of a normal distribution curve is 1, or 100 percent.\n\nREAD  The Best printable recipe book\n\n### How do you make a Z table?\n\nGo to ABAP dictionary (SE11) to create a SAP table. Enter the name of the table to be created and press enter. Enter a proper short description for the table and maintain delivery class as ‘A'(Application Table). Now press on Fields tab to maintain the fields of the table.Jan 22, 2011\n\n### What is Z score used for?\n\nStandard Score. The standard score (more commonly referred to as a z-score) is a very useful statistic because it (a) allows us to calculate the probability of a score occurring within our normal distribution and (b) enables us to compare two scores that are from different normal distributions.\n\n### What z score represents the 90th percentile?\n\nComputing PercentilesPercentileZ25th-0.67550th075th0.67590th1.2827 more rowsJul 24, 2016\n\n### What does a negative z score mean?\n\nA Z-score of 1.0 would indicate a value that is one standard deviation from the mean. Z-scores may be positive or negative, with a positive value indicating the score is above the mean and a negative score indicating it is below the mean.Jun 24, 2019\n\n### How do you read at table?\n\nHow to Read a Table. A table can be read from left to right or from top to bottom. If you read a table across the row, you read the information from left to right. In the Cats and Dogs Table, the number of black animals is 2 + 2 = 4.", null, "" ]

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[ "# CVA as a running spread - risk annuity calculation in the Monte Carlo framework\n\nI have simulated future term structures in the one-factor Hull-White model and calculated the CVA of a particular trade (let's say, now I have it in absolute value, in dollars). However, I want to represent this CVA value as a running spread. As far as I understood from the Gregory's book (2011) \"Counterparty credit risk - The new challenge for global financial markets\", to get the annual spread value I have to divide my absolute CVA value by the CDS risky annuity (and then also by the par value to get the value in %). This is something where I am stuck.\n\nThe formula of the risk annuity given in the book is (1-exp(-(r+h)(T-t)))/(r+h) where r is the constant continiously compounded interest rate and h is the hazard rate. For my case T-t = 12 years, and the yearly CDS spread is 300 bps. However, in my example r is not constant (I have the upward-sloping initial interest rate term structure).\n\nCan anyone advice me how to estimate CVA as a running spread in this case? If u also can offer a good paper as a reference, I would also be very grateful." ]

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[ "# How do I find Tension of a suspended object, when given only weight and acceleration?\n\n1. Pulling up on a rope, you lift a 4.00-kg bucket of water from a well with an acceleration of 2.40m/s2. What is the Tension in the rope?\n\n2. I know the basic equations like net force = mass X acceleration. Is the tension my net force? Or is it my normal force? I basically have no clue what equation to use.\n\n3. I tried Fnet=ma, assuming Tension is the net force. With that, I got 9.6N. Is this correct?\n\n## The Attempt at a Solution\n\nRelated Introductory Physics Homework Help News on Phys.org\nPhanthomJay\nHomework Helper\nGold Member\n\n1. Pulling up on a rope, you lift a 4.00-kg bucket of water from a well with an acceleration of 2.40m/s2. What is the Tension in the rope?\n\n2. I know the basic equations like net force = mass X acceleration.\ncorrect\nIs the tension my net force? Or is it my normal force? I basically have no clue what equation to use.\nyou have the right equation, but you need to identify the net force. The tension force is just one of the forces acting on the bucket. What's the other one?\n\nThe only information I was given was the acceleration and mass. I don't know what other force there would be except maybe normal force..?\n\nPhanthomJay\nHomework Helper\nGold Member\n\nThe only information I was given was the acceleration and mass. I don't know what other force there would be except maybe normal force..?\nNormal forces act perpendicular to the objects and are generally pushing type forces. The bucket's hanging, so there is no normal force. But you are missing a very basic force which acts on all objects due to the force of the earth on the object. It's a gravitational force. What is it??\n\nWeight? So do I take the mass given (4kg) and multiply by gravity (9.81)? What do I do with this weight then? Subtract from the net force (9.6)?\n\nWaitt! I figured it out, do you ADD the weight to the net force?\n\nPhanthomJay\nHomework Helper\nGold Member\n\nWaitt! I figured it out, do you ADD the weight to the net force?\nThe tension force, T, acts up (tension forces always pull away from the object). Weight always acts down. So you have T up, and mg down. The net force must be up, since you are lifting and accelerating the bucket upward. Solve for T. (Note: For sample, if you pull right at 50N and I pull left at 20N, the net force is 30N right. Don't overthink the net force. If you pull up at T newtons and the weight pulls down at 40N, the net force is T-40, up.)\n\nLast edited:\n\nSo does Fnet=T+W? Since the net force is the total of all the forces acting on an object? Like if the T is the same amount as the W, it would cancel out and give me 0 for Fnet. But since the bucket is accelerating upwards, the T must be greater than weight. But if that is true, wouldn't W have to be negative to be able to cancel out when you have an object that doesn't move?\n\nPhanthomJay" ]

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[ "## Problems with add_local_term and H.c\n\nGeneral discussions on the ideas behind the algorithm.\nFabio_Mendez\nPosts: 3\nJoined: 01 Feb 2022, 01:50\n\n### Problems with add_local_term and H.c\n\nHi,\nI'm trying to decrease the number of terms appearing in the Hamiltonian MPO by using H. c. in add_coupling_term. However, when I use plus_hc I always get an error. I have tried this with different models, and I get the same error. I provide an example below.\nI have a ladder of SpinHalfFermions on a Chain lattice with the following hopping coupling terms:\nself.add_local_term(-adj_mat[i, j], [('Cdu', (i, 0)), ('Cu', (j, 0))])\nself.add_local_term(adj_mat[i, j], [('Cu', (i, 0)), ('Cdu', (j, 0))])\nself.add_local_term(-adj_mat[i, j], [('Cdd', (i, 0)), ('Cd', (j, 0))])\nself.add_local_term(adj_mat[i, j], [('Cd', (i, 0)), ('Cdd', (j, 0))])\nThis case works without any problem. However, when I change to:\nself.add_local_term(-adj_mat[i, j], [('Cdu', (i, 0)), ('Cu', (j, 0))],plus_hc=True)\nself.add_local_term(adj_mat[i, j], [('Cd', (i, 0)), ('Cdd', (j, 0))],plus_hc=True)\nI receive the error:\nIndexError: tuple index out of range\nAm I doing something wrong?\n\nF. P. M. Méndez-Cordoba.\nJohannes\nThe index error indicates that you didn't actually set the option explicit_plus_hc=True, otherwise it would not have occured..." ]

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[ "# Quantization of Lie bialgebras, III\n\nPavel Etingof and David Kazhdan Department of MathematicsHarvard UniversityCambridge, MA 02138, USAe-mail: etingofmath.harvard.edukazhdanmath.harvard.edu\n\n## Introduction\n\nThis article is the third part of the series of papers on quantization of Lie bialgebras which we started in 1995. However, its object of study is much less general than in the previous two parts. While in the first and second paper we deal with an arbitrary Lie bialgebra, here we study Lie bialgebras of -valued functions on a punctured rational or elliptic curve, where is a finite dimensional simple Lie algebra. Of course, the general result of the first paper, which says that any Lie bialgebra admits a quantization, applies to this particular case. However, this result is not sufficiently effective, as the construction of quantization utilizes a Lie associator, which is computationally unmanageable. The goal of this paper is to give a more effective quantization procedure for Lie bialgebras associated to punctured curves, i.e. a procedure which will not use an associator. We will describe a general quantization procedure which reduces the problem of quantization of the algebra of -valued functions on a curve with many punctures to the case of one puncture, and apply this procedure in a few special cases to obtain an explicit quantization.\n\nThe main object of study in this paper are Lie bialgebras associated to rational and elliptic curves with punctures, which can be described as follows.\n\nWe work over an algebraically closed field of characteristic zero. Let be a 1-dimensional algebraic group over (i.e. , or an elliptic curve), and be an additive formal parameter near the origin. Let be a rational -valued function on with the Laurent expansion near of the form , satisfying the classical Yang-Baxter equation (2.1) (here are dual bases of with respect to the half-Killing form). Such a function is called a classical r-matrix.\n\nConsider the vector space . Let . Define a Lie bialgebra structure on by the formulas\n\n [τ13(u),τ23(v)]=[r12(u−v),τ13(u)+τ23(v)],δ(τ(u))=[τ12(u),τ13(u)].\n\nIt is convenient to understand the classical r-matrix as a bilinear form on with values in , defined by the rule , where is the invariant form on , and is understood as an element of (that is, the function is expanded as ). We can regard as an element of , where the tensor product is understood in the completed sense. The form satisfies a set of axioms which are dual to the axioms of a pseudotriangular structure on a Lie bialgebra [Dr1]. Thus we call a copseudotriangular structure.\n\nNow we describe Lie bialgebras corresponding to punctured curves. Let be the set of poles of . For any collection of points on such that when , define the Lie bialgebra to be the direct sum , where are Lie subbialgebras in identified with , and the commutation relations between and are given by the formula\n\n [τ13i(u),τ23j(v)]=[r12(u−v+zi−zj),τ13i(u)+τ23j(v)],\n\nwhere is the image of under the identification . Here is regarded as an element of . This is possible as is regular at by the choice of .\n\nThe simplest example of this situation occurs when and (the Yang’s r-matrix). In this case it is easy to check that is the Lie algebra , with cobracket dual to the standard bracket in [Dr1], and is the Lie algebra of rational functions on the line with values in which have no poles outside and vanish at infinity, with a Lie coalgebra structure dual to the standard bracket in [Dr1].\n\nThe Lie bialgebra has two essential properties:\n\n1. Local factorization in a product of equal components.\n\n(a) factorization: As a Lie coalgebra, admits a decomposition , where are Lie subbialgebras of .\n\n(b) locality: .\n\n(c) equal components: All the Lie bialgebras are identified with the same Lie bialgebra .\n\n2. Expression of commutator via the classical r-matrix.\n\nThe commutator between the components and is given in terms of the classical r-matrix via formula (2).\n\nThus, we see that the structure of is completely determined by the pair of a Lie bialgebra and a copseudotriangular structure on it.\n\nOur purpose in this paper is to describe an “explicit” quantization of . The construction of such quantization consists of two parts.\n\nPart 1. We define the notion of a copseudotriangular structure on a Hopf algebra , by dualizing the notion of a pseudotriangular structure, introduced by Drinfeld [Dr1]. We show that any nondegenerate copseudotriangular Lie bialgebra can be quantized, i.e. that there exists a copseudotriangular Hopf algebra whose quasiclassical limit is . This is done using the methods of [EK1, EK2]\n\nPart 2. We define the notion of a factored Hopf algebra, which is a quantization of the above notion of a factored Lie bialgebra, and give a construction of a factored Hopf algebra corresponding to points starting from a copseudotriangular Hopf algebra . We show that if is a quantization of then is a quantization of .\n\nIn some cases, this quantization of can be described by explicit formulas. For example, in the case of the Yang’s r-matrix the quantization of is , where is the dual algebra to the Yangian , with opposite product, and is the pseudotriangular structure on defined by Drinfeld. In this case, is defined by (3.5),(3.6), where is the Yang’s quantum R-matrix.\n\nLet us describe briefly the structure of the paper.\n\nIn Chapter I we introduce the notion of a locally factored Lie bialgebra, which is a Lie bialgebra with properties 1(a) and 1(b), and the corresponding quantum notion of a locally factored Hopf algebra. We also introduce the notion of a weak ()-copseudotriangular Lie bialgebra, which is a Lie bialgebra with a derivation and a form having analogous properties to the form above, and the corresponding quantum notion of a weakly ()-copseudotriangular Hopf algebra. We explain how to introduce a factored Lie bialgebra structure on given a weak -copseudotriangular structure on a Lie bialgebra , and how to introduce a factored Hopf algebra structure on given a weak -copseudotriangular structure on a Hopf algebra . Then we explain how to quantize a given weak -copseudotriangular structure on a Lie bialgebra. This allows us to describe the quantization (constructed in [EK1, EK2]) of the factored algebra purely in terms of the quantization of .\n\nIn Chapter II we describe in detail the Lie bialgebras and discussed above, and consider various examples (rational, trigonometric, and elliptic). The results of this Chapter are not original.\n\nIn Chapter III we give the main results of the paper. In the first part we completely describe the quantization of in the case of the Yang’s r-matrix in terms of the Yangian and its universal R-matrix. In the second part we consider the case and , and describe quantizations of the simplest rational, trigonometric, and elliptic algebras explicitly by generators and relations containing the quantum R-matrices . This construction is analogous to the Faddeev-Reshetikhin-Takhtajan construction of quantum [FRT], and the similar construction of the Yangian of [Dr1].\n\nRemark. Algebras similar to were considered in [Ch1] and [AGS1, AGS2].\n\nIn a subsequent paper we will use the algebras to describe a deformed version of the holomorphic part of the Wess-Zumino-Witten conformal field theory (in genus zero). More specifically, we will consider representation theory of , the notion of quantum conformal blocks, quantum vertex operators, and show how the quantum Knizhnik-Zamolodchikov equations of Frenkel-Reshetikhin appear naturally in this context.\n\n## Acknowledgements\n\nThe authors would like to thank I.Cherednik for useful discussions and references, and V.Schomerus for careful reading of the manuscript. The authors are grateful to V.Drinfeld who pointed out an error in the original formulation of Proposition 3.9. The work of P.E. was supported by an NSF postdoctoral fellowship. The work of D.K. was supported by an NSF grant.\n\n## 1. Copseudotriangular Lie bialgebras and their quantization\n\nThroughout the paper, will denote an algebraically closed field of characteristic zero, and “an algebra” means “an associative -algebra with 1”.\n\n### 1.1. Factored algebras.\n\nLet be an algebra over .\n\n###### Definition\n\nA factorization of is a collection of algebras , which are subalgebras in , such that for any the multiplication map defines a bijection . Such a factorization is called local if the image of in under the multiplication map is a subalgebra of for all .\n\nIt is clear that a factorization of is local if and only if for all .\n\nWe call an algebra equipped with a (local) factorization a (locally) factored algebra.\n\nLet be algebras. We want to describe locally factored algebras with factors .\n\nSuppose we are given a locally factored algebra , with factors , , and product . Let be the restriction of to . By the definition, the map is injective for , and the images of and are the same. Therefore, for any , , we can define a linear isomorphism by the formula , where is the permutation of components.\n\nIt is easy to see that the maps satisfy the following conditions:\n\n(i)\n\n XijXikXjk=XjkXikXij,i\n\n(ii)\n\n Xij(a⊗bc)=(1⊗m)(X12ijX13ij(a⊗b⊗c)), Xij(ab⊗c)=(m⊗1)(X23ijX13ij(a⊗b⊗c)),\n\n(iii)\n\n Xij(a⊗1)=a⊗1,a∈Ai, Xij(1⊗a)=1⊗a,a∈Aj.\n\nRemark. Condition (i) is vacuous if .\n\nConversely, let be an arbitrary collection of invertible operators satisfying (i)-(iii), and be the ideal in the free product generated by the relations , , . Denote by the quotient of by this ideal.\n\nLet be the linear map defined by the rule .\n\n###### Proposition 1.1\n\nThe map is a bijection.\n\nTo prove this proposition, we define a product on the space by\n\n (a1⊗...⊗an)⋅(b1⊗...⊗bn)=(m⊗...⊗m)(X2n−1,nn−1,n...Xn+2,323...Xn+2,n2nXn+1,212...Xn+1,n1n(a1⊗...⊗an⊗b1⊗...⊗bn)),\n\nwhere the superscripts denote the components where the corresponding -operator acts.\n\n###### Lemma 1.2\n\nThe product defined by (1.4) endows with the structure of an associative algebra, in which the unit is .\n\n###### Demonstration Proof\n\nThe associativity follows directly from properties (i),(ii) of . The unit axiom follows from property (iii) of .\n\nWe denote the algebra of Lemma 1.2 by . We have a natural homomorphism , induced by the embeddings .\n\n###### Lemma 1.3\n\nis an isomorphism.\n\n###### Demonstration Proof\n\nThis homomorphism is surjective, as is generated by . It is also injective, as is obviously spanned by the elements of the form , . The Lemma is proved.\n\n###### Demonstration Prooof of Proposition 1.1\n\nIt is clear that , so is an isomorphism.\n\n###### Corollary 1.4\n\nThe assignment is a 1-1 correspondence between locally factored algebra structures on , with factors , and collections of invertible operators satisfying (i)-(iii).\n\nThe same definitions and results apply to the case when is an algebra over which is a deformation of an algebra over . In this case, the sign denotes the completed (in the h-adic topology) tensor product over .\n\n### 1.2. Factored Lie bialgebras and Hopf algebras\n\nLet be a Lie bialgebra over .\n\n###### Definition\n\nA factorization of is a decomposition , where are Lie subbialgebras. Such a factorization is called local if is a Lie subbialgebra of for , i.e. if for .\n\nLet be a Hopf algebra.\n\n###### Definition\n\nA factorization of is a collection of Hopf algebras , which are Hopf subalgebras in , such that for any the multiplication map defines a bijection . Such a factorization is called local if the image of in under the multiplication map is a Hopf subalgebra of for , i.e. if for .\n\nWe will call a Lie bialgebra (Hopf algebra) with a (local) factorization a (locally) factored Lie bialgebra (Hopf algebra).\n\nIt is clear that if a Lie bialgebra is the quasiclassical limit of a quantized universal enveloping (QUE) algebra ([Dr1]), then any factorization of (as a QUE algebra) defines a factorization of . Moreover, if the first factorization is local, so is the second one.\n\nFix a universal Lie associator . Let be the functor of quantization from the category of Lie bialgebras to the category of QUE algebras, defined in [EK2] using . Suppose we are given a factorization . Using the functoriality of , we obtain embeddings of Hopf algebras , so we can regard as Hopf subalgebras of . These Hopf subalgebras define a factorization . Moreover, it follows from the functoriality of that if the factorization of is local, so is the factorization of .\n\nThus, we have the following proposition.\n\n###### Proposition 1.5\n\nTo any (local) factorization of a Lie bialgebra one can assign its quantization, i.e. a (local) factorization of , whose quasicalssical limit is the initial factorization of .\n\nNow we give an analogue of Corollary 1.4 for Hopf algebras.\n\nLet be Hopf algebras. Then the free product has a natural Hopf algebra structure induced by the Hopf algebra structures on .\n\n###### Proposition 1.6\n\nThe assignment is a 1-1 correspondence between locally factored Hopf algebra structures on , with factors , and such collections of invertible operators satisfying (i)-(iii), that the ideal is a Hopf ideal in .\n\nClear.\n\n### 1.3. Copseudotriangular Lie bialgebras\n\nLet be a Lie bialgebra over , with commutator and cocommutator .\n\nRecall [Dr1] that a quasitriangular structure on is an element satisfying the classical Yang-Baxter equation\n\n [r12,r13]+[r12,r23]+[r13,r23]=0,\n\nsuch that\n\n δ(x)=[x⊗1+1⊗x,r],x∈b.\n\nIt is easy to see that relation (1.5) can be replaced by any one of the two relations\n\n (δ⊗1)(r)=[r13,r23],(1⊗δ)(r)=[r13,r12].\n\nIndeed, given (1.6), relation (1.5) is equivalent to either of (1.7).\n\nLet be a finite dimensional quasitriangular Lie bialgebra, and . In this case, the element defines a bilinear form , by\n\n β(x,y)=(r,x⊗y), x,y∈a.\n\nThis form has the following properties, which are equivalent to properties (1.6),(1.7) of :\n\n β([xy],z)=β(x⊗y,δ(z)),β(x,[zy])=β(δ(x),y⊗z);\n [yx]=β12(x⊗δ(y))+β13(δ(x)⊗y).\n###### Definition\n\nA bilinear form on a Lie bialgebra satisfying (1.9)-(1.10) is called a coquasitriangular structure on .\n\nThus, if is finite-dimensional, a coquasitriangular structure on is the same thing as a quasitriangular structure on .\n\nThe following generalization of this definition is essentially due to Drinfeld, [Dr1].\n\nLet be a Lie bialgebra over . Let be a derivation of as a Lie bialgebra. Define the map by\n\n αu(x)=eu∂x.\n###### Definition\n\nA -copseudotriangular structure on a Lie bialgebra is a bilinear form\n\n β:a⊗a→k((u)),\n\nwhich satisfies the following conditions:\n\n β([xy],z)=β(x⊗y,δ(z)),β(x,[zy])=β(δ(x),y⊗z).\n [y,αu(x)]=β12(x⊗δ(y))(u)+αu(β13(δ(x)⊗y))(u).\n β(αv(x),y)(u)=β(x,α−v(y))(u)=β(x,y)(u+v).\n\nIf is a subalgebra, and in addition to (1.13)-(1.15) takes values in , we say that is a -copseudotriangular -structure.\n\nConsider the linear map . It is easy to check that this map is a derivation of as a Lie bialgebra [Dr1]. We call it the canonical derivation of . If , we will call a copseudotriangular structure (without specification of ).\n\nExample. If then a -copseudotriangular -structure is the same thing as a coquasitriangular structure, which vanishes on the image of .\n\n###### Proposition 1.7\n\nLet be a -copseudotriangular structure on such that for a suitable we have for any . Then is constant (does not depend on ), and therefore defines a coquasitriangular structure on , which vanishes on the image of .\n\n###### Demonstration Proof\n\nConsider the form defined by . Since takes values in , this expression makes sense. From (1.15) we get that , so is constant. Identities (1.13)-(1.14) for imply (1.9)-(1.10) for . Thus, is a quasitriangular structure on . It is obvious that vanishes on the image of .\n\nRemark. Proposition 1.7 shows that interesting (i.e. not coquasitriangular) examples of -copseudotriangular structures can only arise when can have a pole of arbitrary order at , for which has to be infinite dimensional.\n\n### 1.4. Copseudotriangular Hopf algebras\n\nLet be a Hopf algebra over , such that is commutative, and is a deformation of . Let be the product, unit, coproduct, counit, and the antipode of .\n\nLet be a derivation of . Define a Hopf algebra homomorphism\n\n αu:=eu∂:A→A[[u]].\n###### Definition\n\nA -bilinear form is called a -copseudotriangular structure on if it satisfies the following conditions:\n\n B(xy,z)=B(x⊗y,Δ(z)),B(x,zy)=B(Δ(x),y⊗z),x,y,z∈A;\n m((αu⊗1)[B13(Δ(x)⊗Δ(y))(u)])=mop((αu⊗1)[B24(Δ(x)⊗Δ(y))(u)]);\n B(αv(x),y)(u)=B(x,α−v(y))(u)=B(x,y)(u+v);\n B(1,x)=B(x,1)=ε(x),x∈A;B(x,y)=ε(x)ε(y)+O(h),\n\nwhere means that is evaluated on the -th and -th components of the product.\n\nLet be a subalgebra of . We say that is a -copseudotriangular -structure on if takes values in .\n\nRemark. Commutativity of is essential to enable the equality in presence of condition (1.18).\n\nSince is commutative, we have . Let . It is clear that is a derivation of . We call the canonical derivation of . If , we will call a copseudotriangular structure (without specification of ).\n\n### 1.5. Weak ∂-copseudotriangular structures\n\nWe will need the following weaker version of the notion of a -copseudotriangular structure.\n\n###### Definition\n\n(i) Let be a Lie bialgebra over . A -valued bilinear form on is called a weak -copseudotriangular structure on if it satisfies equations (1.13),(1.15), and the equation\n\n β([y,αu(x)],z)(v)=β12(u)β34(v)(x⊗δ(y)⊗z)−β14(v)β23(u)((αu⊗1)(δ(x))⊗y⊗z).\n\n(ii) Let be a Hopf algebra, as in Section 1.4. A -valued bilinear form on is called a weak -copseudotriangular structure on if it satisfies equations (1.17),(1.19),(1.20) and the equation\n\n B(m((αu⊗1)[B13(Δ(x)⊗Δ(y))(u)]),z)(v)=B(mop((αu⊗1)[B24(Δ(x)⊗Δ(y))(u)]),z)(v).\n\nIf or takes values in , it is called a weak -copseudotriangular -structure.\n\nIf (respectively, ), we will call (respectively, ) a weak copseudotriangular structure (without specification of ).\n\nRemark. Equations (1.21), (1.22) are obtained by applying the functionals , to equations (1.14),(1.18). Thus, for a left-nondegenerate bilinear form (i.e. a form with trivial left kernel), the property to be a weak -copseudotriangular structure is the same as to be a -copseudotriangular structure.\n\n### 1.6. ∂-copseudotriangular structure on h-formal groups\n\nLet be a Lie coalgebra, be the ring of functions on the corresponding formal group, and be a deformation of as a topological Hopf algebra. Let be the maximal ideal in , i.e. the kernel of the projection .\n\nLet be a -copseudotriangular -structure on . We define the quasiclassical limit of as follows.\n\nLet be the h-adic completion of the direct sum (see [Dr1]). Then is a quantized universal enveloping algebra.\n\nLet be the Lie bialgebra over which is the quasiclassical limit of [Dr1]. Let be the commutator and the cocommutator in , and be the quasiclassical limit of . In particular if , then (see [Dr1], Section 8).\n\nConsider the pairing . It is clear that , so we define by .\n\nLet . It is clear that descends to a bilinear form , which vanishes on on the left and on the right. As is naturally identified with , we get a bilinear form .\n\n###### Proposition 1.8\n\nThe form satisfies equations (1.13)-(1.15).\n\n###### Demonstration Proof\n\nRelations (1.13)-(1.15) for are easily obtained from relations (1.17)-(1.19) for .\n\nThus, Proposition 1.8 states that the quasiclassical limit of is endowed with a natural -copseudotriangular -structure . We call the quasiclassical limit of , and a quantization of .\n\nSimilar definitions and statements apply to weak -copseudotriangular structures.\n\n### 1.7. Factorizations associated to ∂-copseudotriangular structures\n\nConsider the field . For , we have a subfield . Consider the embedding by the formula\n\n f(t)→f(u−v):=∑m≥0f(m)(u)(−v)m/m!.\n\nUsing this embedding, we define subfields .\n\nLet be a Lie bialgebra over with a weak -copseudotriangular structure . Let be a Lie bialgebra over the field obtained by extension of scalars from . In this section we will define a structure of a factored Lie bialgebra on the space .\n\nLet , . For any , we define a linear map by the formula\n\n μij(x,y)=β13(δ(x)⊗y)(ui−uj)⊕β12(x⊗δ(y))(ui−uj)\n\nFor , such that , , , set\n\n [a1+...+an,b1+...+bn]=n∑j=1[aj,bj]−∑i\n\nThe space has a natural Lie coalgebra structure , coming form the Lie coalgebra structures on .\n\n###### Proposition 1.9\n\nThe bracket is a Lie bracket on , and is a Lie bialgebra structure on .\n\n###### Demonstration Proof\n\nSkew symmetry of is obvious. The Jacobi identity and the cocycle condition for is verified by a direct computation.\n\nProposition 1.9 shows that any weak -copseudotriangular structure on defines a natural structure of a factored Lie bialgebra on .\n\nNow consider the quantum analogue of this construction. Let be a Hopf algebra with a weak copseudoriangular structure . Define linear operators by the formula\n\n Xl(a⊗b)=B13(Δ(a)⊗Δ(b)), Xr(a⊗b)=B24(Δ(a)⊗Δ(b)),X=XlX−1r.\n\nBecause of property (1.20) of , we have .\n\n###### Proposition 1.10\n\n(i) For any .\n\nAlso, satisfy the following equations:\n\n(ii) The quantum Yang-Baxter equation\n\n Y12(u1−u2)Y13(u1−u3)Y23(u2−u3)=Y23(u2−u3)Y13(u1−u3)Y12(u1−u2).\n\n(iii)\n\n Y(a⊗bc)=(1⊗m)(Y12Y13(a⊗b⊗c)), Y(ab⊗c)=(m⊗1)(Y23Y13(a⊗b⊗c));\n\n(iv) , , .\n\n###### Demonstration Proof\n\nThe identity is obvious from the definition. The Yang-Baxter equation follows from the properties (1.17)-(1.19) of . Identities (iii),(iv) follow directly from properties (1.17),(1.20) of .\n\nLet be the h-adic completion of . , . For , let be the operators defined by the formula . Proposition 1.10 implies that satisfy properties (i)-(iii) in Section 1.1, so they define a factored algebra .\n\n###### Proposition 1.11\n\nThe ideal is a Hopf ideal with respect to the coproduct on induced by the coproduct on .\n\n###### Demonstration Proof\n\nIt is convenient to write the relations of the ideal in the form , , . Then it is easy to verify directly that this relation is invariant under .\n\nProposition 1.11 shows that has a natural Hopf algebra structure. We denote this Hopf algebra by .\n\nNow let be an h-formal group, and the corresponding QUE algebra. Let be a -copseudotriangular structure on . Although does not extend to , we have the following proposition.\n\n###### Proposition 1.12\n\nThe operator extends to an invertible, h-adically continuous operator .\n\n###### Demonstration Proof\n\nIt is easy to see that for any and belong to the coset . On the other hand, as , we have . Therefore, . Using Proposition 1.10, we see that for any . Therefore, extends to . Similarly, we can extend to .\n\nProposition 1.12 enables us to construct the factored Hopf algebra . We denote this Hopf algebra by .\n\nRemark. It is easy to see that is an h-formal group over , and is the corresponding QUE algebra, so this notation is consistent with the previous notation.\n\nNow let us consider copseudotriangular structures and factorizations which are defined over the ring of functions on some algebraic variety.\n\nLet be a connected 1-dimensional algebraic group over with a fixed invariant differential , a finite subset of , and be the algebra of regular functions on . We can regard as a subalgebra in using the canonical formal parameter near the origin (the parameter whose differential is ).\n\nLet be the variety of all such that . Let be the ring of regular functions on . We have a natural embedding , which acts by taking the Laurent expansion of a function near the origin, consecutively in the variables .\n\nLet be a weak -copseudotriangular -structure on a Lie bialgebra . Then the Lie bialgebra over has a natural -structure. Indeed, the -submodule is a Lie bialgebra over , and . For any , define the Lie bialgebra , where is the ideal of functions vanishing at . Then is a factored Lie bialgebra over , with factors isomorphic to .\n\nSimilarly, one defines factored Hopf algebras , , , .\n\n###### Proposition 1.16\n\nIf is a quantization of a Lie bialgebra , and is a quantization of a weak -copseudotriangular structure on , then the QUE algebra is a quantization of the Lie bialgebra . If in addition , are -structures, then are quantizations of .\n\nEasy.\n\n### 1.8. Quantization of weak ∂-copseudotriangular structures\n\nIn this section we will show that any weak -copseudotriangular structure on a Lie bialgebra admits a quantization.\n\nLet be a Lie bialgebra over with a weak -copseudotriangular structure . Consider the Lie bialgebra over defined above. Define a linear map by\n\n θn(a1+...+an)(b)=β(a1,b)(u1)+...+β(an,b)(un).\n###### Proposition 1.17\n\n(i) is a homomorphism of Lie bialgebras (where is with opposite cocommutator, for any Lie bialgebra ).\n\n(ii) .\n\n(iii) , .\n\n(iv) is injective if and only if is left-nondegenerate.\n\n###### Demonstration Proof\n\nProperties (i)-(iv) follow from (1.13),(1.15),(1.21). We check only the case of property (i), which is less obvious than others. For , (i) follows from (1.13). Consider the case . It is clear that we can assume . In this case (i) reduces to the identity\n\n θ2(μ(x,y))=[θ1(y)(v),θ1(x)(u)],x,y∈a.\n\nUsing the definition (1.23) of , we rewrite (1.29) in the form\n\n θ1(u)(β13(δ(x)⊗y)(u−v))+θ1(v)(β12(x⊗δ(y))(u−v))=[θ1(v)(y),θ1(u)(x)].\n\nEvaluating both sides of (1.30) on an element , we rewrite (1.30) in the form\n\n β13(u−v)β24(u)(δ(x)⊗y⊗z)+β12(u−v)β34(v)(x⊗δ(y)⊗z)+β13(u)β24(v)(x⊗y⊗δ(z))=0.\n\nOn the other hand, and using (1.13), (1.15), we can rewrite (1.21) in the form\n\n β13(u)β24(u+v)(δ(x)⊗y⊗z)+" ]

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[ "# Constructing a Seismic Tripartite Graph with TikZ & pgfplots\n\nI am attempting to use TiKZ and pgfplots to construct a seismic tripartite graph. These are used by structural engineers to determine seismic response for a structure. The example I am attempting to replicate is pasted below. This is related to my post is Engineering.SE, for those interested.", null, "My working example is pasted below.\n\n\\documentclass[letter,landscape]{article}\n\\usepackage[bindingoffset=0.2in,%\nleft=0.5in,right=0.5in,top=0.5in,bottom=0.5in,%\nfootskip=.25in]{geometry}\n\\usepackage{pgfplots}\n\\pgfplotsset{compat=1.5}\n\n\\begin{document}\n\n\\begin{tikzpicture}\n\n% Primary Axes\n\\begin{loglogaxis}[\n%\nwidth=9in, height=7in,\ntitle=Tripartite Paper,\n% Frequency Axis\nxlabel={Frequency (Hz)},\nxmin=0.1, xmax=1000,\ndomain=1:1000,\nlog ticks with fixed point,\nx tick label style={/pgf/number format/1000 sep=\\,},\n% Pseudovelocity Axis\nylabel={Pseudo Response Velocity (in/sec)},\nymin=0.01, ymax=10,\ndomain=1:100,\nlog ticks with fixed point,\ny tick label style={/pgf/number format/1000 sep=\\,},\ngrid=minor\n]\n\n\\end{loglogaxis}\n\n% Secondary Axes\n\\begin{loglogaxis}[\n% Pseudoacceleration Axis\nxlabel={Acceleration (g)},\nxlabel style={rotate=45,anchor=north},\nxmin=0.0001, xmax=100,\ndomain=0.0001:100,\nrotate=45,\nlog ticks with fixed point,\nx tick label style={rotate=-45, anchor=west, /pgf/number format/1000 sep=\\,},\n% Displacment Axis\nylabel={Displacement (in)},\nylabel style={rotate=-135,anchor=south},\nymin=0.000001, ymax=10,\ndomain=0.000001:10,\nlog ticks with fixed point,\ny tick label style={rotate=45, anchor=east, /pgf/number format/1000 sep=\\,},\ngrid=minor\n]\n\n\\end{loglogaxis}\n\n\\end{tikzpicture}\n\n\\end{document}\n\n\nThis generates the following graph, which has some obvious issues.", null, "The things that I can't figure out how to fix are:\n\n1. Remove the border from the secondary axes\n2. Have the secondary axes extend beyond the extents of the primary axes (ideally ad infinitum with an anchor point at a specific place)\n3. Keep all the major axis gridlines square and the same size\n4. Put the secondary axis labels above the major gridlines (see example)\n5. Move the secondary axis tick labels in the middle of the graph\n• can you explain how are built the tilting axes and scale of these axes – rpapa May 10 '16 at 20:32\n• @rpapa, not quite sure what you mean. The secondary axes are at a 45-deg skew to the primary axes, as in the first image. Ideally, I'd like to anchor the secondary axes coordinate (0.01,0.001) to the primary axes coordinate (10,0.1). For scale, the secondary axes are scaled such that the spacing of the primary axes between 10^n times cos(45deg) equals the spacing of the secondary axes between 10^n. If that makes sense...it's hard to put in writing. – grfrazee May 10 '16 at 20:42\n• In other words, the diagonal of the square made by the major gridlines of the secondary axes is the same length as the distance between major gridlines on the primary axes. – grfrazee May 10 '16 at 21:24\n• Your best bet is just to draw all the diagonals and label them with node[mid]. – John Kormylo May 11 '16 at 2:57\n\n## 1 Answer\n\nhere is my proposal, I am not satisfied but it should allow others to complete.\n\nI do not know what to write about tilting axes\n\n\\documentclass{article}\n\n\\usepackage{tikz}\n\\usetikzlibrary{calc,intersections}\n\n\\begin{document}\n\n\\begin{tikzpicture}[scale=2.5,transform shape]\n\\def\\nbdecade{4}\n\\pgfmathsetmacro{\\fin}{\\nbdecade-1}\n\\pgfmathsetmacro{\\decalx}{\\nbdecade/2}\n\n\\begin{scope}\n\n\\def\\minx{-1}\n\\def\\maxx{3}\n\\def\\miny{-2}\n\\def\\maxy{2}\n\\foreach \\yy in {\\minx,...,\\maxx}{\n\\foreach \\xx in{1,2,4,6,8}{\n\\draw[red, name path global/.expanded = X\\xx10\\yy] ({log10(\\xx*10^(\\yy)}, {log10(10^(\\miny)})node[below=0.5em,scale=1/4,rotate=90]{X\\xx10\\yy:$\\xx \\cdot 10^{\\yy}$}coordinate(X\\xx-\\yy) -- ({log10(\\xx*10^(\\yy)},{log10(10^(\\maxy+1)});\n}\n}\n\\foreach \\yy in {\\miny,...,\\maxy}{\n\\foreach \\xx in{1,2,4,6,8}{\n\\draw[blue,name path global/.expanded=Y\\xx10\\yy] ({log10(10^(\\minx)}, {log10(\\xx*10^(\\yy)})node[left,scale=1/4]{$Y\\xx10\\yy$:$\\xx \\cdot 10^{\\yy}$}coordinate(Y\\xx-\\yy) -- (\\nbdecade,{log10(\\xx*10^(\\yy)});\n}\n}\n\n\\clip (\\minx,\\miny) rectangle ({\\maxx+1},{\\maxy+1});\n\\foreach \\yy in {\\miny,...,\\maxy}{\n\\foreach \\xx in{1,2,4,6,8}{\n\\path[name intersections={of=X1101 and Y\\xx10\\yy, by= P}];\n%\\path[name intersections={of=X1101 and Y110-1, by= P}];\n\\draw [thin,green] (P) --+ (45:4)--+(-135:4)node[sloped,pos=0.4,black,scale=1/5]{$\\xx10^{\\yy}$};\n\\draw [thin,purple] (P) --+ (-45:4)--+(135:4)node[sloped,pos=0.6,black,scale=1/5]{$\\xx10^{\\yy}$};\n}\n}\n\n\\end{scope}\n\\end{tikzpicture}\n\\end{document}", null, "" ]

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[ "# croston\n\n0th\n\nPercentile\n\n##### Forecasts for intermittent demand using Croston's method\n\nReturns forecasts and other information for Croston's forecasts applied to y.\n\nKeywords\nts\n##### Usage\ncroston(y, h = 10, alpha = 0.1, x = y)\n##### Arguments\ny\n\na numeric vector or time series of class ts\n\nh\n\nNumber of periods for forecasting.\n\nalpha\n\nValue of alpha. Default value is 0.1.\n\nx\n\nDeprecated. Included for backwards compatibility.\n\n##### Details\n\nBased on Croston's (1972) method for intermittent demand forecasting, also described in Shenstone and Hyndman (2005). Croston's method involves using simple exponential smoothing (SES) on the non-zero elements of the time series and a separate application of SES to the times between non-zero elements of the time series. The smoothing parameters of the two applications of SES are assumed to be equal and are denoted by alpha.\n\nNote that prediction intervals are not computed as Croston's method has no underlying stochastic model.\n\n##### Value\n\nAn object of class \"forecast\" is a list containing at least the following elements:\n\nmodel\n\nA list containing information about the fitted model. The first element gives the model used for non-zero demands. The second element gives the model used for times between non-zero demands. Both elements are of class forecast.\n\nmethod\n\nThe name of the forecasting method as a character string\n\nmean\n\nPoint forecasts as a time series\n\nx\n\nThe original time series (either object itself or the time series used to create the model stored as object).\n\nresiduals\n\nResiduals from the fitted model. That is y minus fitted values.\n\nfitted\n\nFitted values (one-step forecasts)\n\nThe function summary is used to obtain and print a summary of the results, while the function plot produces a plot of the forecasts.\n\nThe generic accessor functions fitted.values and residuals extract useful features of the value returned by croston and associated functions.\n\n##### References\n\nCroston, J. (1972) \"Forecasting and stock control for intermittent demands\", Operational Research Quarterly, 23(3), 289-303.\n\nShenstone, L., and Hyndman, R.J. (2005) \"Stochastic models underlying Croston's method for intermittent demand forecasting\". Journal of Forecasting, 24, 389-402.\n\nses.\n\n• croston\n##### Examples\n# NOT RUN {\ny <- rpois(20,lambda=.3)\nfcast <- croston(y)\nplot(fcast)\n\n# }\n\nDocumentation reproduced from package forecast, version 8.13, License: GPL-3\n\n### Community examples\n\nLooks like there are no examples yet." ]

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[ "# Why does my recursive function return None?\n\nPosted on\n\nSolving problem is about exposing yourself to as many situations as possible like Why does my recursive function return None? and practice these strategies over and over. With time, it becomes second nature and a natural way you approach any problems in general. Big or small, always start with a plan, use other strategies mentioned here till you are confident and ready to code the solution.\nIn this post, my aim is to share an overview the topic about Why does my recursive function return None?, which can be followed any time. Take easy to follow this discuss.\n\nWhy does my recursive function return None?\n\nI have this function that calls itself:\n\n``````def get_input():\nmy_var = input('Enter \"a\" or \"b\": ')\nif my_var != \"a\" and my_var != \"b\":\nprint('You didn't type \"a\" or \"b\". Try again.')\nget_input()\nelse:\nreturn my_var\nprint('got input:', get_input())\n``````\n\nNow, if I input just “a” or “b”, everything works fine:\n\n``````Type \"a\" or \"b\": a\ngot input: a\n``````\n\nBut, if I type something else and then “a” or “b”, I get this:\n\n``````Type \"a\" or \"b\": purple\nYou didn't type \"a\" or \"b\". Try again.\nType \"a\" or \"b\": a\ngot input: None\n``````\n\nI don’t know why `get_input()` is returning `None` since it should only return `my_var`. Where is this `None` coming from and how do I fix my function?\n\nIt is returning `None` because when you recursively call it:\n\n``````if my_var != \"a\" and my_var != \"b\":\nprint('You didn't type \"a\" or \"b\". Try again.')\nget_input()\n``````\n\n..you don’t return the value.\n\nSo while the recursion does happen, the return value gets discarded, and then you fall off the end of the function. Falling off the end of the function means that python implicitly returns `None`, just like this:\n\n``````>>> def f(x):\n... pass\n>>> print(f(20))\nNone\n``````\n\nSo, instead of just calling `get_input()` in your `if` statement, you need to `return` it:\n\n``````if my_var != \"a\" and my_var != \"b\":\nprint('You didn't type \"a\" or \"b\". Try again.')\nreturn get_input()\n``````\n\nTo return a value other than None, you need to use a return statement.\n\nIn your case, the if block only executes a return when executing one branch. Either move the return outside of the if/else block, or have returns in both options.\n\n``````def get_input():\nmy_var = input('Enter \"a\" or \"b\": ')\nif my_var != \"a\" and my_var != \"b\":\nprint('You didn't type \"a\" or \"b\". Try again.')\nreturn get_input()\nelse:\nreturn my_var\nprint('got input:', get_input())\n``````\n\ni think this code more clearly\n\n``````def get_input():\nmy_var = str(input('Enter \"a\" or \"b\": '))\nif my_var == \"a\" or my_var == \"b\":\nprint('got input:', my_var)\nreturn my_var\nelse:\nprint('You didn't type \"a\" or \"b\". Try again.')\nreturn get_input()\nget_input()\n``````\nThe answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 ." ]

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[ "# Question #efac1\n\nOct 8, 2017\n\nMalia\n\n#### Explanation:\n\nIt would be easier to compare the two amounts if the fractions have the same denominator. To do this, we:\n\n- Convert the mixed fractions to improper fractions\n\nMalia:\n$1 \\frac{1}{3} = \\frac{4}{3}$\n\nJohn:\n$1 \\frac{2}{5} = \\frac{7}{5}$\n\n- Find the LCD (Lowest Common Denominator) and convert the fractions accordingly\n\nLCD $= 3 \\cdot 5 = 15$\n\nMalia:\n$\\frac{4}{3} = \\frac{5 \\cdot 4}{15} = \\frac{20}{15}$\n\nJohn:\n$\\frac{7}{5} = \\frac{3 \\cdot 7}{15} = \\frac{21}{15}$\n\nSince, we know they have the same amount of seeds initially, we can just compare the amounts left to see who ate more. The less amount left, the more that person ate .\n\n$\\frac{20}{15} < \\frac{21}{15}$\n\n$\\therefore$Malia ate more sunflower seeds." ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: coefficients but not from polynomial (variables does\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg98008] Re: [mg97968] coefficients but not from polynomial (variables does\n• From: Bob Hanlon <hanlonr at cox.net>\n• Date: Fri, 27 Mar 2009 05:39:24 -0500 (EST)\n\n```Cases[a = Sin[x*Pi] + Sin[y], _Symbol?(! NumericQ[#] &), Infinity]\n\n{x,y}\n\nBob Hanlon\n\n---- magdamoczydlowska at gmail.com wrote:\n\n=============\nDear All,\n\nProbably my question is very simple. Suppose that we have expression\na= Sin[x]+Sin[y]. I would like receive variables from which this\nexpression is dependent, so in this case {x,y}. The Variables does not\nwork of course in this because it is not a polynomial and it gives me\n{Sin[x],Sin[y]}. Do anyone know how to solve my problem.\n\nI would be very grateful for quick response.\n\nMagdalena\n\n```\n\n• Prev by Date: Solve problem\n• Next by Date: finding all definitions matching a pattern\n• Previous by thread: Re: Solve problem\n• Next by thread: finding all definitions matching a pattern" ]

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[ "", null, "# Exact and Approximate Calculations of Acid-Base Equilibria\n\nInitializing live version", null, "Requires a Wolfram Notebook System\n\nInteract on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.\n\nIn this Demonstration, we show exact and approximate calculations for acid-base equilibria.\n\n[more]\n\nThe equation for acid dissociation (and similarly for base)", null, "is given by", null, ", where", null, "is the acid (base) concentration and", null, "is the acid (base) dissociation constant. This equation can be approximated in the case", null, "by", null, ". The first graphic shows the results for concentration and the dissociation calculated both exactly and approximately. The appropriate number of significant digits is used.\n\nIn the case of a strong acid/base concentration, the water dissociation is determined by", null, ", where", null, "is the acid (base) concentration and", null, "is the autoionization constant of water (at", null, ",", null, "). When", null, ", this can be approximated by", null, ". Use the checkbox to switch from acid to base.\n\n[less]\n\nContributed by: A. Ratti, D. Meliga, L. Lavagnino and S. Z. Lavagnino (June 2021)\nOpen content licensed under CC BY-NC-SA\n\n## Snapshots", null, "", null, "", null, "## Details\n\nSnapshot 1: Exact and approximate equations give the same result since the initial concentration is high and the extent of dissociation is very low (weak acid).\n\nSnapshot 2: Exact and approximate equations are very different since we have a very high acid dissociation constant. In this case, the acid is completely dissociated (strong acid).\n\nSnapshot 3: Incorrect use of the approximate equation gives the pH of a base although the solution is a slightly acidic.\n\nReference\n\n D. C. Harris, Quantitative Chemical Analysis, 8th ed., New York: W. H. Freeman and Company, 2010.\n\n## Permanent Citation\n\nA. Ratti, D. Meliga, L. Lavagnino and S. Z. Lavagnino\n\n Feedback (field required) Email (field required) Name Occupation Organization Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Send" ]

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[ "# Member Expressions\n\nThis section describes how to create and use member expressions in InterSystems MDX.\n\n## Details\n\nIn InterSystems MDX, a member expression can have any of the following forms:\n• An member literal, which is an explicit reference to a single member by its name:\n```[dimension_name].[hierarchy_name].[level_name].[member_name]\n```\nWhere:\n• [dimension_name] is an MDX identifier that names a dimension.\n• [hierarchy_name] is an MDX identifier that names a hierarchy within that dimension. You can omit the hierarchy name. If you do, the query uses the first level with the given name, as defined in this dimension.\nYou can also omit the hierarchy name when you refer to a calculated member, because a calculated member is not in a hierarchy.\n• [level_name] is an MDX identifier that names a level within that hierarchy. You can omit the level name. If you do, the query uses the first member with the given name, as defined within this dimension.\nYou can also omit the level name when you refer to a calculated member, because a calculated member is not in a level.\n• [member_name] is an MDX identifier that names a member of that level.\nFor example:\n```[gend].[h1].[gender].[female]\n```\nFor a measure, a member literal has the following variant form, because any measure is a member of a dimension called Measures. This dimension does not have a hierarchy or a level.\n```[MEASURES].[measure_name]\n```\nFor example:\n```[MEASURES].[%COUNT]\n```\nNote that although a measure is a member, an expression like the preceding returns a scalar value and cannot be used in all the same ways as other member expressions.\n• An explicit reference to a hierarchy as follows:\n```[dimension_name].[hierarchy_name]\n```\nThis expression returns the All member of the dimension to which the hierarchy belongs.\nFor example:\n```aged.h1\n```\nThe preceding expression returns the All member of the AgeD dimension; in the sample Patients cube, the name of this member is All Patients.\n• An expression that uses an MDX function to return a single member. For example:\n```[gend].[h1].[gender].female.NEXTMEMBER\n```\nMany MDX functions return members, including LAG, NEXTMEMBER, PARENT, and others.\nNote that %TERMLIST can return a member.\n• An expression that uses the internal key of a member to return that member, via the following syntax:\n```[dimension_name].[hierarchy_name].[level_name].&[member_key]\n```\nWhere member_key is the value of the KEY property of the member. For example:\n```birthqd.h1.quarter.&\n```\nNote that member_key is a case-sensitive literal value rather than an expression. That is, you cannot replace it with a string-valued expression.\nFor information on how KEY values are created, see the reference “Intrinsic Properties.” Also note that you can use the PROPERTIES function to find the value of the KEY property or any other property of a member.\n• An expression that uses an InterSystems MDX extension to refer to a member in another cube, via the following syntax:\n```[relationship_name].member_expression\n```\nWhere [relationship_name] is an MDX identifier that names a relationship in the cube used by the query and member_expression is a member expression suitable for that cube.\n• A reference to a pivot variable that contains a member expression. To refer to a pivot variable, use the following syntax:\n```\\$VARIABLE.variablename\n```\nWhere variablename is the logical variable name. Do not enclose this expression with square brackets. This syntax is not case-sensitive; nor is the pivot variable name.\nFor information on defining pivot variables, see “Defining and Using Pivot Variables” in Using the Analyzer.\nFor example, the following member expressions are all equivalent:\n```[gend].[h1].[gender].Female\n[gend].female\ngend.H1.gender.female\ngend.h1.FEMALE\ngend.female\n```\n\n## Calculated Members\n\nMembers can be defined within the cube definition, as part of the definition of a level. You can also create calculated members, which are typically based on other members. You can define calculated members in two ways:\n• Within the WITH clause of a query. The member is available within the rest of the query, but is not available in other queries.\n• Within the CREATE MEMBER statement. The member is available within the rest of the session (for example, within the rest of the session in the MDX shell).\n\n## Uses\n\nYou can use member expressions in the following ways:\n• As a member argument to many MDX functions.\n• As the argument to the WHERE clause.\n• As an element of a set." ]

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[ "The Sycamore gate is a two-qubit gate equivalent to FSimGate(π/2, π/6).\n\nThe unitary of this gate is\n\n``````[[1, 0, 0, 0],\n[0, 0, -1j, 0],\n[0, -1j, 0, 0],\n[0, 0, 0, exp(- 1j * π/6)]]\n``````\n\nThis gate can be performed on the Google's Sycamore chip and is close to the gates that were used to demonstrate beyond classical resuts used in this paper: https://www.nature.com/articles/s41586-019-1666-5\n\n`theta` Swap angle on the `|01⟩` `|10⟩` subspace, in radians. Determined by the strength and duration of the XX+YY interaction. Note: uses opposite sign convention to the iSWAP gate. Maximum strength (full iswap) is at pi/2.\n`phi` Controlled phase angle, in radians. Determines how much the `|11⟩` state is phased. Note: uses opposite sign convention to the CZPowGate. Maximum strength (full cz) is at pi.\n\n`phi`\n\n`theta`\n\n## Methods\n\n### `controlled`\n\nReturns a controlled version of this gate. If no arguments are specified, defaults to a single qubit control.\n\nArgs\n`num_controls` Total number of control qubits.\n`control_values` Which control computational basis state to apply the sub gate. A sequence of length `num_controls` where each entry is an integer (or set of integers) corresponding to the computational basis state (or set of possible values) where that control is enabled. When all controls are enabled, the sub gate is applied. If unspecified, control values default to 1.\n`control_qid_shape` The qid shape of the controls. A tuple of the expected dimension of each control qid. Defaults to `(2,) * num_controls`. Specify this argument when using qudits.\n\nReturns\nA `cirq.Gate` representing `self` controlled by the given control values and qubits. This is a `cirq.ControlledGate` in the base implementation, but subclasses may return a different gate type.\n\n### `num_qubits`\n\nThe number of qubits this gate acts on.\n\n### `on`\n\nReturns an application of this gate to the given qubits.\n\nArgs\n`*qubits` The collection of qubits to potentially apply the gate to.\n\nReturns: a `cirq.Operation` which is this gate applied to the given qubits.\n\n### `on_each`\n\nReturns a list of operations applying the gate to all targets.\n\nArgs\n`*targets` The qubits to apply this gate to. For single-qubit gates this can be provided as varargs or a combination of nested iterables. For multi-qubit gates this must be provided as an `Iterable[Sequence[Qid]]`, where each sequence has `num_qubits` qubits.\n\nReturns\nOperations applying this gate to the target qubits.\n\nRaises\n`ValueError` If targets are not instances of Qid or Iterable[Qid]. If the gate qubit number is incompatible.\n`TypeError` If a single target is supplied and it is not iterable.\n\n### `qubit_index_to_equivalence_group_key`\n\nReturns a key that differs between non-interchangeable qubits.\n\n### `validate_args`\n\nChecks if this gate can be applied to the given qubits.\n\nBy default checks that:\n\n• inputs are of type `Qid`\n• len(qubits) == num_qubits()\n• qubit_i.dimension == qid_shape[i] for all qubits\n\nChild classes can override. The child implementation should call `super().validate_args(qubits)` then do custom checks.\n\nArgs\n`qubits` The sequence of qubits to potentially apply the gate to.\n\nRaises\n`ValueError` The gate can't be applied to the qubits.\n\n### `with_probability`\n\nCreates a probabalistic channel with this gate.\n\nArgs\n`probability` floating point value between 0 and 1, giving the probability this gate is applied.\n\nReturns\n`cirq.RandomGateChannel` that applies `self` with probability `probability` and the identity with probability `1-p`.\n\n### `wrap_in_linear_combination`\n\nReturns a LinearCombinationOfGates with this gate.\n\nArgs\n`coefficient` number coefficient to use in the resulting `cirq.LinearCombinationOfGates` object.\n\nReturns\n`cirq.LinearCombinationOfGates` containing self with a coefficient of `coefficient`.\n\n### `__call__`\n\nCall self as a function.\n\n### `__truediv__`\n\n[{ \"type\": \"thumb-down\", \"id\": \"missingTheInformationINeed\", \"label\":\"Missing the information I need\" },{ \"type\": \"thumb-down\", \"id\": \"tooComplicatedTooManySteps\", \"label\":\"Too complicated / too many steps\" },{ \"type\": \"thumb-down\", \"id\": \"outOfDate\", \"label\":\"Out of date\" },{ \"type\": \"thumb-down\", \"id\": \"samplesCodeIssue\", \"label\":\"Samples / code issue\" },{ \"type\": \"thumb-down\", \"id\": \"otherDown\", \"label\":\"Other\" }]\n[{ \"type\": \"thumb-up\", \"id\": \"easyToUnderstand\", \"label\":\"Easy to understand\" },{ \"type\": \"thumb-up\", \"id\": \"solvedMyProblem\", \"label\":\"Solved my problem\" },{ \"type\": \"thumb-up\", \"id\": \"otherUp\", \"label\":\"Other\" }]" ]

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[ "# Subtraction Regrouping Worksheets Subtraction With Regrouping Word Problems Worksheets 2nd Grade", null, "subtraction regrouping worksheets subtraction with regrouping word problems worksheets 2nd grade.\n\nsubtraction worksheets grade 2 no regrouping 3 digit with 4th math mathematics addition 1st,no regrouping adding 4 digit numbers carrying math addition and subtraction with worksheets 3rd grade pdf without for 1,subtraction without regrouping worksheets pdf addition and with wonderfully subtracting fractions 3 digit,subtraction with regrouping worksheets year 2 grade fresh pdf 3 digit numbers,subtraction regrouping worksheets grade 1 5 digit with pdf 2,addition and subtraction with regrouping worksheets 4th grade year 2,3 digit subtraction with regrouping worksheets 1st grade free library download and print on without for 1 pdf 4th,3 digit subtraction with regrouping worksheets 3rd grade pdf 4 1st worksheet without,3 digit subtraction with regrouping worksheets 1st grade free printable 2 no addition,3 digit subtraction worksheets some regrouping subtracting fractions with pdf 4 3rd grade 5.\n\nPosted on" ]

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[ "Overflow Modes - 2021.2 English\n\nVitis High-Level Synthesis User Guide (UG1399)\n\nDocument ID\nUG1399\nft:locale\nEnglish (United States)\nRelease Date\n2021-12-15\nVersion\n2021.2 English\n Saturation AP_SAT Saturation to zero AP_SAT_ZERO Symmetrical saturation AP_SAT_SYM Wrap-around AP_WRAP Sign magnitude wrap-around AP_WRAP_SM\n\nAP_SAT\n\nSaturate the value.\n\n• To the maximum value in case of overflow.\n• To the negative maximum value in case of negative overflow.\nap_fixed<4, 4, AP_RND, AP_SAT> UAPFixed4 = 19.0; // Yields: 7.0\nap_fixed<4, 4, AP_RND, AP_SAT> UAPFixed4 = -19.0; // Yields: -8.0\nap_ufixed<4, 4, AP_RND, AP_SAT> UAPFixed4 = 19.0; // Yields: 15.0\nap_ufixed<4, 4, AP_RND, AP_SAT> UAPFixed4 = -19.0; // Yields: 0.0\n\nAP_SAT_ZERO\n\nForce the value to zero in case of overflow, or negative overflow.\n\nap_fixed<4, 4, AP_RND, AP_SAT_ZERO> UAPFixed4 = 19.0; // Yields: 0.0\nap_fixed<4, 4, AP_RND, AP_SAT_ZERO> UAPFixed4 = -19.0; // Yields: 0.0\nap_ufixed<4, 4, AP_RND, AP_SAT_ZERO> UAPFixed4 = 19.0; // Yields: 0.0\nap_ufixed<4, 4, AP_RND, AP_SAT_ZERO> UAPFixed4 = -19.0; // Yields: 0.0\n\nAP_SAT_SYM\n\nSaturate the value:\n\n• To the maximum value in case of overflow.\n• To the minimum value in case of negative overflow.\n• Negative maximum for signed ap_fixed types\n• Zero for unsigned ap_ufixed types\nap_fixed<4, 4, AP_RND, AP_SAT_SYM> UAPFixed4 = 19.0; // Yields: 7.0\nap_fixed<4, 4, AP_RND, AP_SAT_SYM> UAPFixed4 = -19.0; // Yields: -7.0\nap_ufixed<4, 4, AP_RND, AP_SAT_SYM> UAPFixed4 = 19.0; // Yields: 15.0\nap_ufixed<4, 4, AP_RND, AP_SAT_SYM> UAPFixed4 = -19.0; // Yields: 0.0\n\nAP_WRAP\n\nWrap the value around in case of overflow.\n\nap_fixed<4, 4, AP_RND, AP_WRAP> UAPFixed4 = 31.0; // Yields: -1.0\nap_fixed<4, 4, AP_RND, AP_WRAP> UAPFixed4 = -19.0; // Yields: -3.0\nap_ufixed<4, 4, AP_RND, AP_WRAP> UAPFixed4 = 19.0; // Yields: 3.0\nap_ufixed<4, 4, AP_RND, AP_WRAP> UAPFixed4 = -19.0; // Yields: 13.0\n\nIf the value of N is set to zero (the default overflow mode):\n\n• All MSB bits outside the range are deleted.\n• For unsigned numbers. After the maximum it wraps around to zero.\n• For signed numbers. After the maximum, it wraps to the minimum values.\n\nIf N>0:\n\n• When N > 0, N MSB bits are saturated or set to 1.\n• The sign bit is retained, so positive numbers remain positive and negative numbers remain negative.\n• The bits that are not saturated are copied starting from the LSB side.\n\nAP_WRAP_SM\n\nThe value should be sign-magnitude wrapped around.\n\nap_fixed<4, 4, AP_RND, AP_WRAP_SM> UAPFixed4 = 19.0; // Yields: -4.0\nap_fixed<4, 4, AP_RND, AP_WRAP_SM> UAPFixed4 = -19.0; // Yields: 2.0\n\nIf the value of N is set to zero (the default overflow mode):\n\n• This mode uses sign magnitude wrapping.\n• Sign bit set to the value of the least significant deleted bit.\n• If the most significant remaining bit is different from the original MSB, all the remaining bits are inverted.\n• If MSBs are same, the other bits are copied over.\n1. Delete redundant MSBs.\n2. The new sign bit is the least significant bit of the deleted bits. 0 in this case.\n3. Compare the new sign bit with the sign of the new value.\n• If different, invert all the numbers. They are different in this case.\n\nIf N>0:\n\n• Uses sign magnitude saturation\n• N MSBs are saturated to 1.\n• Behaves similar to a case in which N = 0, except that positive numbers stay positive and negative numbers stay negative." ]

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[ "# 指向数组的指针pointer to array\n\n``````array{'a','b','c',...}\npointer = array[6 through 10];\n``````\n\n``````*pointer == array;\n``````\n\n``````*pointer == array;\n``````\n\n``````5 == sizeof(*pointer) \\ sizeof(type);\n``````\n\n## 10 个回复10\n\n### ===============>>#1 票数：12 已采纳\n\n``````int array = {'a','b','c','d','e','f','g','h','i','j','k','l'};\nint (*pointer)[10 - 6 + 1] = (int (*)[10 - 6 + 1])&array; /* = array[6 through 10] */\n\nprintf(\"(*pointer) = %c\\n\", (*pointer));\nprintf(\"(*pointer) = %c\\n\", (*pointer));\nprintf(\"sizeof *pointer / sizeof **pointer = %lu\\n\", (unsigned long)(sizeof *pointer / sizeof **pointer));\n``````\n\n``````unsigned char buffer;\nunsigned char *ptr;\n\n/* ... */\nptr = buffer + 500;\nread(fd, ptr, 1024); /* Try and read up to 1024 bytes at position buffer */\n``````\n\n### ===============>>#2 票数：6\n\n``````int a = {1,2,3,4};\nint *b = &a;\ncout << b << \"-\" << b << endl;\n``````\n\n### ===============>>#3 票数：4\n\n``````#include <iostream>\nusing namespace std;\n\nint main()\n{\nint a = { 0, 4, 5, 7, 4, 3, 1, 6, 2, 9 };\n\nint *p1 = &a;\nint *p2 = &a;\n\nfor(int *p = p1; p != p2; ++p)\n{\ncout << *p << endl;\n}\n\nreturn 0;\n}\n``````\n\n### ===============>>#4 票数：4\n\n``````char pointer;\n``````\n\n``````char *pointer = array + 6;\n``````\n\n``````char a;\nchar b;\nb = a;\n``````\n\n``````unsigned char bigbuf[1UL << 24]; /* or whatever you want */\nunsigned char *ptr = bigbuf;\nsize_t chunk = BUFSIZ;\nFILE *fp = fopen(\"foo.txt\", \"rb\"); /* no error checking */\nsize_t total = 0;\nif (total + chunk > sizeof bigbuf) {\n/* oops, not enough space */\n}\n}\nif (ferror(fp)) {\n} else {\n/* bigbuf now has data, total has the number of bytes */\n}\n``````\n\n``````T a;\n``````\n\n``````pointer == a\npointer == a\n...\npointer == a\n``````\n\n（或者如果你喜欢冒险，在第一个声明的LHS上用一个更大的正数代替1）。\n\n``````pointer == array;\n``````\n\n``````*pointer == array;\n``````\n\n### ===============>>#6 票数：3\n\n``````char *p = array + 6;\n``````\n\n``````array[i]\n``````\n\n``````*(array + i)\n``````\n\n### ===============>>#7 票数：2\n\n`pointer = &array` `pointer`现在将引用与`array`相同的值。 直接提取对数组切片的引用并为其保持`sizeof`语义是不可能的。 您可以使用以下内容模拟它：\n\n``````template <class T>\nclass Slice {\npublic:\nSlice(T* elements, int begin, int end) : m_elements(elements + begin), m_size(end - begin) {}\nT& operator[] (int index) {return m_elements[index];}\nint size() const {return m_size;}\nprivate:\nT* m_elements;\nint m_size;\n};\n\nint values;\nSlice<int> slice(values, 5, 10);\nvalues = 3;\nassert(slice == values);\nassert(slice.size() == 5);\n``````\n\n### ===============>>#8 票数：2\n\n``````int* ptr = new int;\nint* ptrSub = &ptr;\n``````\n\n### ===============>>#9 票数：1\n\n``````int array1;\nint array2;\n\n// reference 3 elements from array1, starting at element 5 in array1\narray2 = array1;\n``````\n\n3回复\n\n1回复\n\n3回复\n\n2回复\n\n2回复\n\n12回复\n\n3回复\n\n1回复\n\n2回复\n\n1回复" ]

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[ "#### Thank you for registering.\n\nOne of our academic counsellors will contact you within 1 working day.\n\nClick to Chat\n\n1800-5470-145\n\n+91 7353221155\n\nCART 0\n\n• 0\nMY CART (5)\n\nUse Coupon: CART20 and get 20% off on all online Study Material\n\nITEM\nDETAILS\nMRP\nDISCOUNT\nFINAL PRICE\nTotal Price: Rs.\n\nThere are no items in this cart.\nContinue Shopping\n\n# How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?\n\n12 years ago\n\nTest of divisibility for 3\n\nThe sum of the digits of any number that is divisible by '3' is divisible by 3.\n\nFor instance, take the number 54372.\nSum of its digits is 5 + 4 + 3 + 7 + 2 = 21.\nAs 21 is divisible by '3', 54372 is also divisible by 3.\n\nThere are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.\n\nThe sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits are divisible by '3'.\n\nCombining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers.\n\nCase 1\nIf we do not use '0', then the remaining 5 digits can be arranged in 5! ways = 120 numbers.\n\nCase 2\nIf we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.\n\n. The first digit from the left can be any of the 4 digits 1, 2, 4 or 5.\n\nThen the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.\n\nSo, there will be 4*4! numbers = 4*24 = 96 numbers.\n\nCombining Case 1 and Case 2, there are a total of 120 + 96 = 216 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.", null, "", null, "", null, "", null, "", null, "12 years ago\n\nFOR A NUMBER TO BE DIVISIBLE BY 3 THE SUM OF ITS DIGITS SHOULD BE DIVISIBLE BY 3 . SO, AMONG THE 6 DIGITS , 5 DIGITS\n\nWHOSE SUM IS DIVISIBLE BY 3 MUST BE SELECTED.\n\n0+1+2+3+4+5 = 15 ( a multiple of 3)\n\nSO THE FIVE DIGITS CAN BE OBTAINED BY ELIMINATING A 3 MULTIPLE .\n\n15 - 3q = 3(5 - q) . SO IT IS DIVISIBLE BY 3.\n\nAMONG 0,1,2,3,4,5 ;  0 AND 3 ARE THE MULTIPLES OF 3 .\n\nSO, THE 5 DIGITS MUST BE  (1,2,3,4,5) OR (0,1,2,4,5)\n\nFIRST LET US TAKE (1,2,3,4,5) . ACCORDING TO PERMUTATION LAW , IT CAN BE ARRANGED IN 5p5 WAYS\n\n5P5  = 5 x 4 x 3 x 2 x 1 = 120\n\n(0,1,2,4,5) . IF ZERO IS THE FIRST DIGIT IT BECOMES A 4 DIGIT NUMBER .\n\nNUMBER OF  NUMBERS HAVING 0 AS FIRST DIGIT ARE\n\n4 x 3 x 2 x 1 = 24\n\nSO, NUMBER OF 5 DIGIT NUMBER FORMED FROM (0,1,2,4,5)  ARE 120 - 24 = 96.\n\nTOTAL  NUMBER OF 5 DIGIT NUMBER FORMED FROM  (0, 1,2,3,4,5)   ARE 120 + 96 = 216." ]

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[ "### docs for parametric vi\n\nparent 3f1212d2\n ... ... @@ -37,22 +37,41 @@ class MeanFieldVI: \"\"\"Collect the operators required for Gaussian meanfield variational inference. Gaussian meanfield variational inference approximates some target distribution with a Gaussian distribution with a diagonal covariance matrix. The parameters of the approximation, in this case the mean and standard deviation, are obtained by minimizing a stochastic estimate of the Kullback-Leibler divergence between the target and the approximation. In order to obtain gradients w.r.t the parameters, the reparametrization trick is employed, which separates the stochastic part of the approximation from a deterministic function, the generator. Samples from the approximation are drawn by processing samples from a standard Gaussian through this generator. Parameters ---------- position : FIXME hamiltonian : FIXME n_samples : FIXME mirror_samples : FIXME initial_sig : FIXME comm : FIXME position : Field The initial estimate of the approximate mean parameter. hamiltonian : Energy Hamiltonian of the approximated probability distribution. n_samples : int Number of samples used to stochastically estimate the KL. mirror_samples : bool Whether the negative of the drawn samples are also used, as they are equally legitimate samples. If true, the number of used samples doubles. Mirroring samples stabilizes the KL estimate as extreme sample variation is counterbalanced. Since it improves stability in many cases, it is recommended to set `mirror_samples` to `True`. initial_sig : positive Field or positive float The initial estimate of the standard deviation. comm : MPI communicator or None If not None, samples will be distributed as evenly as possible across this communicator. If `mirror_samples` is set, then a sample and its mirror image will always reside on the same task. nanisinf : FIXME If true, nan energies which can happen due to overflows in the forward model are interpreted as inf. Thereby, the code does not crash on these occasions but rather the minimizer is told that the position it has tried is not sensible. \"\"\" def __init__(self, position, hamiltonian, n_samples, mirror_samples, initial_sig=1, comm=None, nanisinf=False): ... ... @@ -98,24 +117,47 @@ class MeanFieldVI: class FullCovarianceVI: \"\"\"Collect the operators required for Gaussian full-covariance variational inference. Gaussian meanfield variational inference approximates some target distribution with a Gaussian distribution with a diagonal covariance matrix. The parameters of the approximation, in this case the mean and a lower triangular matrix corresponding to a Cholesky decomposition of the covariance, are obtained by minimizing a stochastic estimate of the Kullback-Leibler divergence between the target and the approximation. In order to obtain gradients w.r.t the parameters, the reparametrization trick is employed, which separates the stochastic part of the approximation from a deterministic function, the generator. Samples from the approximation are drawn by processing samples from a standard Gaussian through this generator. Note that the size of the covariance scales quadratically with the number of model parameters. Parameters ---------- position : FIXME hamiltonian : FIXME n_samples : FIXME mirror_samples : FIXME initial_sig : FIXME comm : FIXME position : Field The initial estimate of the approximate mean parameter. hamiltonian : Energy Hamiltonian of the approximated probability distribution. n_samples : int Number of samples used to stochastically estimate the KL. mirror_samples : bool Whether the negative of the drawn samples are also used, as they are equally legitimate samples. If true, the number of used samples doubles. Mirroring samples stabilizes the KL estimate as extreme sample variation is counterbalanced. Since it improves stability in many cases, it is recommended to set `mirror_samples` to `True`. initial_sig : positive float The initial estimate for the standard deviation. Initially no correlation between the parameters is assumed. comm : MPI communicator or None If not None, samples will be distributed as evenly as possible across this communicator. If `mirror_samples` is set, then a sample and its mirror image will always reside on the same task. nanisinf : FIXME If true, nan energies which can happen due to overflows in the forward model are interpreted as inf. Thereby, the code does not crash on these occasions but rather the minimizer is told that the position it has tried is not sensible. \"\"\" def __init__(self, position, hamiltonian, n_samples, mirror_samples, initial_sig=1, comm=None, nanisinf=False): ... ...\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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[ "# String To Binary Convertor Online | Online tool to Convert String to Binary\n\n## Convert String To Binary Online.\n\nThe String to Binary Converter is a tool that is Very helpful and used by many Developers to Give Commands to the Computer. We Use the help of ASCII Table to Convert the Text into Binary. Therefore this Tool is Really Necessary and Useful for all.\n\n## How to Convert Text to Binary\n\nConvert text to binary ASCII code:\n\n1. Get character\n2. Get a decimal code of character from the ASCII table.\n3. Convert decimal to binary byte\n4. Continue with the next character.\n\n#### Example\n\nConvert “Plant trees” text to binary ASCII code:\n\nSolution:\n\nUse the ASCII table to get ASCII code from character.\n\n“P” => 80 = 26+24 = 010100002\n\n“l” => 108 = 26+25+23+22 = 011011002\n\n“a” => 97 = 26+25+20 = 011000012\n\nFor all the text characters you should get the binary bytes:\n\n“01010000 01101100 01100001 01101110 01110100 00100000 01110100 01110010 01100101 01100101 01110011”", null, "## How to Convert String to Binary?\n\n1. Get character\n2. Get ASCII code of character from ASCII table\n3. Convert decimal to binary byte\n4. Continue with the next character\n\n## How to use Text to Binary converter?\n\n1. Paste text in the input text box.\n2. Select a character encoding type.\n3. Select the output delimiter string.\n4. Press the Convert button.\n\n## How to convert ‘A’ character to binary?\n\nUse ASCII table: ‘A’ = 6510 = 64+1 = 26+20 = 010000012\n\n## How to convert ‘0’ character to binary?\n\nUse ASCII table: ‘0’ = 4810 = 32+16 = 25+24 = 001100002\n\n## How To Use Online Tool To Convert String to Binary and Save and Share\n\n1. Write your String in the Box given below “Enter the text to encode to binary”\n2. Click on Convert.\n3. Get the Desired Converted String to Binary in “The encoded string” Box." ]

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[ "## Phasor\n\nThe representation, beloved of engineers and physicists, of a Complex Number in terms of a Complex exponential", null, "(1)\n\nwhere i (called j by engineers) is the Imaginary Number and the Modulus and Argument (also called Phase) are", null, "", null, "", null, "(2)", null, "", null, "", null, "(3)\n\nHere,", null, "is the counterclockwise Angle from the Positive Real axis. In the degenerate case when", null, ",", null, "(4)\n\nIt is trivially true that", null, "(5)\n\nNow consider a Scalar Function", null, ". Then", null, "", null, "", null, "", null, "", null, "(6)\n\nLook at the time averages of each term,", null, "(7)", null, "(8)", null, "(9)\n\nTherefore,", null, "(10)\n\nConsider now two scalar functions", null, "", null, "", null, "(11)", null, "", null, "", null, "(12)\n\nThen", null, "", null, "", null, "", null, "", null, "(13)", null, "", null, "", null, "", null, "", null, "", null, "", null, "(14)\n\nIn general,", null, "(15)\n\nSee also Affix, Argument (Complex Number), Complex Multiplication, Complex Number, Modulus (Complex Number), Phase" ]

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[ "Topics: Coronavirus | AI | Data Science | Deep Learning | Machine Learning | Python | R | Statistics\n\nKDnuggets Home » News » 2018 » Jan » Opinions, Interviews » Machine Learning Model Metrics ( 18:n04 )\n\n# Machine Learning Model Metrics\n\nIn this article we explore how to calculate machine learning model metrics, using the example of fraud detection. We'll see lots of different ways that we can try to understand just how good our learned model is.\n\nMLlib has some nice built-in functionality to calculate the ROC curve for our binary classifier (Listing 5).\n\nListing 5. Training Performance Summary\n\n```val trainingSummary = model.summary (1)\n\nval binarySummary = trainingSummary.asInstanceOf[BinaryLogisticRegressionSummary] (2)\n\nval roc = binarySummary.roc (3)\n\nroc.show() (4)\n\n```\n\n1. Producing the summary of the model’s performance\n2. Casting that summary to the appropriate type, BinaryLogisticRegressionSummary\n3. The ROC curve for the model\n4. Printing the ROC curve for inspection\n\nThe model summary is relatively new functionality within MLlib, so it's not available for all classes of models. There are also limitations to its implementation, such as the one that requires you to use asInstanceOf to cast the summary to the correct type. Make no mistake, using asInstanceOf like this is bad Scala style; it represents a subversion of the type system. But MLlib is still being rapidly developed, so this cast operation is just a sign of an incomplete implementation within MLlib. Development on MLlib is very active, but machine learning is an enormous domain for any one library to support. New functionality is being added at a rapid pace and the overarching abstractions are being dramatically improved. Look for rough edges like this class cast to disappear in future versions of Spark.\n\nOf course, we're building massively scalable machine learning systems that operate largely autonomously. So, who has time to look at a graph and make a decision about what constitutes a good enough model? Well, one of the uses of an ROC curve is to get a single number about the performance of a model, the area under the ROC curve. The higher this number, the better the model's performance. You can even make strong assertions about a model's utility using this calculation. Remember that a random model would be expected to perform according to the line x = y on the ROC curve. The area under that line would be 0.5, so any model with an area under the curve of less than 0.5 can safely be discarded as being worse than a random model.\n\nFigures 3, 4, and 5 show the differences in the area under the curve of a good, random, and worse than random model.", null, "", null, "", null, "Listing 6 shows the implementation of validating for performance better than random.\n\nListing 6. Validating Training Performance\n\n```def betterThanRandom(model: LogisticRegressionModel) = { (1)\nval trainingSummary = model.summary (2)\n\nval binarySummary = trainingSummary.asInstanceOf[BinaryLogisticRegressionSummary] (3)\n\nval auc = binarySummary.areaUnderROC (4)\n\nauc > 0.5 (5)\n}\n\nbetterThanRandom(model) (6)\n```\n\n1. Defining a function to validate that a model is better than random\n2. The training summary\n3. Class casting\n4. The area under the ROC curve\n5. Testing if the area under the curve is greater than a random model\n6. An example call to validate a model\n\nThis validation can serve as a very useful safety feature in a machine learning system, preventing you from publishing a model that could be deeply detrimental. In the Kangaroo Kapital example, since fraud is so much rarer than normal transactions, a model that failed this test would very likely be falsely accusing a lot of angry animals of fraud for normal use of their credit cards.\n\nThis technique can be extended beyond basic sanity checks like this. If you record the historical performance of your published models, you can compare the performance of your newly trained models to them. Then a logical validation would be to not publish a model with meaningfully different performance than the current published model. We'll discuss some more techniques for model validation a bit later.\n\nWe're not done asking questions about our model yet. There are other model metrics that we can consider. The metrics that we have seen thus far try to capture an aspect of a model's performance. In particular, it's not hard to imagine a model that does a bit better on precision but not on recall or vice versa. An F measure (or sometimes F1 score) is a statistic that tries to combine the concerns of precision and recall in the same metric. Specifically, it is the harmonic mean of the precision and the recall. Listing 7 shows two ways of formulating the F measure.\n\nListing 7. F Measure\n\n```F measure = 2 * (precision * recall) / (precision + recall)\n\nF measure = (2 * true positives) / (2 * true positives + false positives + false\nnegatives)\n\n```\n\nUsing the F measure as a model metric may not always be appropriate. It trades off precision versus recall evenly, which may not correspond to the modeling and business objectives of the situation. But it does have the advantage of being a single number which can be used to implement automated decision making.\n\nFor example, one use of the F measure is to set the threshold that a logistic regression model uses for binary classification. Internally, a logistic regression model is actually producing probabilities. To turn them into predicted class labels, we'll need to set a threshold to divide positive (fraud) predictions from negative (not fraud) predictions. Figure 6 shows some example prediction values from a logistic regression model and how they could be divided into positive and negative predictions using different threshold values.", null, "While the F measure is not the only way of setting a threshold, it's a useful one, so let's see how to do it. Listing 10 shows how to set a threshold using the F measure of the model on the training set.\n\nListing 8. Setting a Threshold Using the F Measure\n\n```val fMeasure = binarySummary.fMeasureByThreshold (1)\n\nval bestThreshold = fMeasure.where(\\$\"F-Measure\" === maxFMeasure) (3)\n\nmodel.setThreshold(bestThreshold) (4)\n```\n\n1. Retrieving the F measure for every possible threshold\n2. Finding the maximum F measure\n3. Finding the threshold corresponding to the maximum F measure\n4. Setting that threshold on the model\n\nNow the learned model will use the threshold selected on the basis of F measure to distinguish between positive and negative predictions.\n\nSummary\n\nIn this article we've explored how to calculate machine learning model metrics, using the example of fraud detection. We've seen lots of different ways that we can try to understand just how good our learned model is.", null, "Get KDnuggets, a leading newsletter on AI, Data Science, and Machine Learning" ]

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[ "# Tutorial – Mass <----> mole conversions    Std 3e\n\nYou have a great web site in my web helps to give you this same information in a slightly different way.  Take a look at it now by clicking on        Mole <---> grams conversion tutorial .  Here I will go through what we have done in DO NOW’s.\n\nI am going to add something to the diagram you find at that website.  So the complete diagram will give you everything you need for solving the problems below.  Remember it takes avocadoes to make guacamole from atoms or molecules.\n\n Avogadro                                                                 gaw  or atoms or molecules   <---------------->      MOLES     <--------------->     grams                                        6.022 x 1023          (of first   | atom or           gmw                                                                                     |  molecule)                                                                                     |                                                                                     |                         (mole ratio from)    ** balanced  |   equation **                                                                                     |                                                                                     |                                                                                     |                                                Avogadro                        |                          gaw  or        atoms or molecules  <---------------->      MOLES     <--------------->     grams                                              6.022 x 1023            (of 2nd atom or               gmw                                                                               molecule)\n\nWhy the MOLES two times?   If you start with moles of one kind of molecule or atom and need to change to moles of a different molecule or atom, you use the mole ratio from the chemical equation to make the change.\n\n## PROBLEM 1\n\nIf you are give 8 grams of H2 ,  how many grams of O2 will you need to burn up all 8 grams of H2 for the following balanced chemical equation:\n\n2 H2  +  O2  ----->  2 H2O\n\nSolution:\n\nFirst look at the balanced equation and the mole ratios:\n\n2 H2  +  O2  ----->  2 H2O\n\nmole ratios               2     :   1       :          2\n\nThis tells you how many moles of  O2  will be needed for each  mole of  H2  you have.  But, we were given grams of H2 so we have to change from grams to moles before we can use the mole ratio.  So we use the gmw (gram molecular weight) of H2 to do that.  The gmw of H2 can be retrieved from the periodic table\n\n2 H’s  x  1g  =  2 g for the gmw of H2  because there are two atoms in the molecule.\n\n 8 g H2 1 mole H2 1 2 g H2\n\nWith this step, we went from grams to MOLES on our diagram.  But we need the moles of O2 not H2 .  This is the step that requires the extra MOLES I put into the diagram above.  Now it’s time to cancel the units we can and also change moles of H2  to  moles of O2  .  For that we need to use the mole ratio for the balanced equation.\n\n 8 g H2 1 mole H2 1 mol O2 1 2 g H2 2 mol H2\n\nNow that we are in moles of O2  we need to take the last step and convert to grams of O2 .  For that we use the gmw of O2  which is     2 O x 16g = 32g   which is also equal to 1 mole.\n\n 8 g H2 1 mole H2 1 mol O2 32 g O2 Answer 64 g O2 1 2 g H2 2 mol H2 1 mole O2\n\nPROBLEM 2\n\nIf you are given 96.1 g  C3H8 .  You are asked to find out how many grams of  O2  will be needed to burn up all of the 96.1 g of C3H8  , given the following unbalanced chemical equation:\n\nC3H8   +   O2   ----->   CO2   +   H2O\n\nSolution:\n\nYour first task is to balance the equation.  If you do not balance the equation, you will not know the mole ratios of the reactants and products and will not be able to convert from moles of  C3H8  to moles moles of  O2  ,   which is the 3rd step in the solution of this problem.\n\nC3H8   +   5 O2   ----->   3 CO2   +   4 H2O\n\nmole ratio          1       :      5          :           3        :       4\n\nWe will need the gmw (gram molecular weight) of both    C3H8   and    O2\n\n3 C x 12g = 36 g                                  2 O x 16g = 32 g/mole of O2\n\n8 H x  1 g =   8 g\n\n--------\n\n44 g/mole C3H8\n\nThe last piece to our puzzle is the mole ratio  of  C3H8  :   O2   given to us by the balanced formula:\n\n1 mole C3H8  =  5 mole O2\n\n 96.1 g C3H8 1 mol C3H8 5 mol O2 32 g O2 1 44 g C3H8 1 mol C3H8 1 mol O2\n\nNow do the unit cancellation, multiply all the numbers left on the top, multiply all the numbers together from the 2nd row and divide the 2nd row into the top row number\n\n 96.1 g C3H8 1 mol C3H8 5 mol O2 32 g O2 Answer 348 g O2 1 44 g C3H8 1 mol C3H8 1 mol O2" ]

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[ "# dsp.UpperTriangularSolver\n\nSolve upper-triangular matrix equation\n\n## Description\n\nThe `UpperTriangularSolver` object solves UX = B for X when U is a square, upper-triangular matrix with the same number of rows as B.\n\nTo solve UX = B:\n\n1. Create the `dsp.UpperTriangularSolver` object and set its properties.\n\n2. Call the object with arguments, as if it were a function.\n\n## Creation\n\n### Syntax\n\n``uptriang = dsp.UpperTriangularSolver``\n``uptriang = dsp.UpperTriangularSolver(Name,Value)``\n\n### Description\n\nexample\n\n````uptriang = dsp.UpperTriangularSolver` returns a linear system solver, `uptriang`, used to solve UX = B where U is an upper (or unit-upper) triangular matrix.```\n````uptriang = dsp.UpperTriangularSolver(Name,Value)` returns a linear system solver, `uptriang`, with each specified property set to the specified value. Enclose each property name in single quotes. Unspecified properties have default values.```\n\n## Properties\n\nexpand all\n\nUnless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the `release` function unlocks them.\n\nIf a property is tunable, you can change its value at any time.\n\nWhen you set this property to `true`, the linear system solver replaces the elements on the diagonal of the input, U, with ones. This property is useful when matrix U is the result of another operation, such as an LDL decomposition, that uses the diagonal elements to represent the D matrix.\n\nWhen you set this property to `true`, the linear system solver optimizes computation speed if the input U is complex, but its diagonal elements are real. Set this property to either `true` or `false`.\n\n#### Dependencies\n\nThis property applies only when you set the `OverwriteDiagonal` property to `false`.\n\n### Fixed-Point Properties\n\nSpecify the rounding method as `Ceiling`, `Convergent`, `Floor`, `Nearest`, `Round`, `Simplest`, or `Zero`.\n\nSpecify the overflow action as `Wrap` or `Saturate`.\n\nSpecify the product data type as `Full precision`, ```Same as input```, or `Custom`.\n\nSpecify the product fixed-point type as a scaled `numerictype` object with a `Signedness` of `Auto`.\n\n#### Dependencies\n\nThis property applies only when you set the `ProductDataType` property to `Custom`.\n\nSpecify the accumulator data type as `Full precision`, `Same as first input`, `Same as product`, or `Custom`.\n\nSpecify the accumulator fixed-point type as a scaled `numerictype` object with a `Signedness` of `Auto`.\n\n#### Dependencies\n\nThis property applies only when you set the `AccumulatorDataType` property to `Custom`.\n\nSpecify the output data type as `Same as first input` or `Custom`.\n\nSpecify the output fixed-point type as a scaled `numerictype` object with a `Signedness` of `Auto`.\n\n#### Dependencies\n\nThis property applies only when you set the OutputDataType property to `Custom`.\n\n## Usage\n\n### Syntax\n\n``X = uptriang(U,B)``\n\n### Description\n\nexample\n\n````X = uptriang(U,B)` computes the solution, `X`, of the matrix equation UX = B, where `U` is a square, upper-triangular matrix with the same number of rows as the matrix `B`.```\n\n### Input Arguments\n\nexpand all\n\nUpper-triangular square matrix of size M-by-M.\n\nIf the matrix is of fixed-point data type, it must be signed fixed point.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `fi`\n\nInput B in the equation UX = B, where `B` is an M-by-N matrix.\n\nIf the matrix is of fixed-point data type, it must be signed fixed point.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `fi`\n\n### Output Arguments\n\nexpand all\n\nSolution of the UX = B equation, returned as an M-by-N output matrix. The object uses only the elements in the upper triangle of input U and ignores the lower elements. When you set `OverwriteDiagonal` to `true`, the object replaces the elements on the diagonal of the input, U, with ones.\n\nIf the matrix is of fixed-point data type, it must be signed fixed point.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `fi`\n\n## Object Functions\n\nTo use an object function, specify the System object™ as the first input argument. For example, to release system resources of a System object named `obj`, use this syntax:\n\n`release(obj)`\n\nexpand all\n\n `step` Run System object algorithm `release` Release resources and allow changes to System object property values and input characteristics `reset` Reset internal states of System object\n\n## Examples\n\ncollapse all\n\nNote: If you are using R2016a or an earlier release, replace each call to the object with the equivalent step syntax. For example, `obj(x)` becomes `step(obj,x)`.\n\n```uptriang = dsp.UpperTriangularSolver; u = triu(rand(4, 4)); b = rand(4, 1);```\n\nCheck that result is the solution to the linear equations.\n\n`x1 = u\\b`\n```x1 = 4×1 -179.1887 265.6759 -29.3098 6.7624 ```\n`x = uptriang(u, b)`\n```x = 4×1 -179.1887 265.6759 -29.3098 6.7624 ```\n\n## Algorithms\n\nThis object implements the algorithm, inputs, and outputs described on the Backward Substitution block reference page. The object properties correspond to the block parameters." ]

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