Datasets:

---

Infi-MM

/

InfiMM-WebMath-40B

like 63

Follow

InfiMM 19

[ "# Bilinear form obtained from Quadratic form\n\nThe following definition is given in Scahum's Ouline of Linear Algebra:\n\n$$\\begin{array}{l}{\\text { DEFINTION A: A mapping } q: V \\rightarrow K \\text { is a quadratic form if } q(v)=f(v, v) \\text { for some symmetric }} \\\\ {\\text { bilinear form } f \\text { on } V \\text { . }} \\\\ {\\text {If } 1+1 \\neq 0\\text{ in } K,}\\\\{\\text {then the bilinear form } f \\text { can be obtained from the quadratic form } q \\text { by the following } polar form }\\\\{\\text{ of }f\\text{:}} \\\\ {\\qquad f(u, v)=\\frac{1}{2}[q(u+v)-q(u)-q(v)]}\\end{array}$$\n\nWhere does the last equation appear from, and what does the book mean by polar form?\n\nYou have to compute $$q(u+v)$$. Using bilinearity, $$q(u+v)=f(u+v,u+v)=f(u,u)+f(u,v)+f(v,u)+f(v,v)=f(u,v)+f(v,u)+q(u)+q(v).$$ Now, if your field has characteristic $$>2$$, $$f(u,v)+f(v,u)=2f(u,v)$$ and you just solve for $$f(u,v)$$. This identity is called the polarization identity, and it is related to the concept of polar in projective geometry.\n• What do you mean by a field having a characteristic $\\gt 2$? – David Nov 8 '19 at 16:04\n• The characteristic is the minimum number of $1$'s you need to add to get $0$ (respectively, the multiplicative and the additive neutrals in your field). For example, the field $\\mathbb{Z}_3$ has characteristic $=3$ since $1+1+1=0$. In the case of the real field, you never get $0$ by adding $1$'s, and in such cases the characteristic is defined to be zero (or infinity, depending on the author). If your characteristic is $>2$, $1+1\\neq 0$, and you call $1+1$ simply $2$. The important thing is that $2\\neq 0$ so you can divide by it. – GReyes Nov 8 '19 at 16:11" ]

[ null ]

[ "# #40daf1 Color Information\n\nIn a RGB color space, hex #40daf1 is composed of 25.1% red, 85.5% green and 94.5% blue. Whereas in a CMYK color space, it is composed of 73.4% cyan, 9.5% magenta, 0% yellow and 5.5% black. It has a hue angle of 187.8 degrees, a saturation of 86.3% and a lightness of 59.8%. #40daf1 color hex could be obtained by blending #80ffff with #00b5e3. Closest websafe color is: #33ccff.\n\n• R 25\n• G 85\n• B 95\nRGB color chart\n• C 73\n• M 10\n• Y 0\n• K 5\nCMYK color chart\n\n#40daf1 color description : Bright cyan.\n\n# #40daf1 Color Conversion\n\nThe hexadecimal color #40daf1 has RGB values of R:64, G:218, B:241 and CMYK values of C:0.73, M:0.1, Y:0, K:0.05. Its decimal value is 4250353.\n\nHex triplet RGB Decimal 40daf1 `#40daf1` 64, 218, 241 `rgb(64,218,241)` 25.1, 85.5, 94.5 `rgb(25.1%,85.5%,94.5%)` 73, 10, 0, 5 187.8°, 86.3, 59.8 `hsl(187.8,86.3%,59.8%)` 187.8°, 73.4, 94.5 33ccff `#33ccff`\nCIE-LAB 80.505, -31.957, -22.73 43.058, 57.58, 92.059 0.223, 0.299, 57.58 80.505, 39.217, 215.423 80.505, -54.673, -31.666 75.881, -31.503, -18.813 01000000, 11011010, 11110001\n\n# Color Schemes with #40daf1\n\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #f15740\n``#f15740` `rgb(241,87,64)``\nComplementary Color\n• #40f1b0\n``#40f1b0` `rgb(64,241,176)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #4082f1\n``#4082f1` `rgb(64,130,241)``\nAnalogous Color\n• #f1b040\n``#f1b040` `rgb(241,176,64)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #f14082\n``#f14082` `rgb(241,64,130)``\nSplit Complementary Color\n• #daf140\n``#daf140` `rgb(218,241,64)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #f140da\n``#f140da` `rgb(241,64,218)``\n• #40f157\n``#40f157` `rgb(64,241,87)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #f140da\n``#f140da` `rgb(241,64,218)``\n• #f15740\n``#f15740` `rgb(241,87,64)``\n• #10bbd5\n``#10bbd5` `rgb(16,187,213)``\n• #11d0ed\n``#11d0ed` `rgb(17,208,237)``\n• #28d5ef\n``#28d5ef` `rgb(40,213,239)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #58dff3\n``#58dff3` `rgb(88,223,243)``\n• #70e3f4\n``#70e3f4` `rgb(112,227,244)``\n• #87e8f6\n``#87e8f6` `rgb(135,232,246)``\nMonochromatic Color\n\n# Alternatives to #40daf1\n\nBelow, you can see some colors close to #40daf1. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #40f1dc\n``#40f1dc` `rgb(64,241,220)``\n• #40f1eb\n``#40f1eb` `rgb(64,241,235)``\n• #40e9f1\n``#40e9f1` `rgb(64,233,241)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #40cbf1\n``#40cbf1` `rgb(64,203,241)``\n• #40bdf1\n``#40bdf1` `rgb(64,189,241)``\n• #40aef1\n``#40aef1` `rgb(64,174,241)``\nSimilar Colors\n\n# #40daf1 Preview\n\nThis text has a font color of #40daf1.\n\n``<span style=\"color:#40daf1;\">Text here</span>``\n#40daf1 background color\n\nThis paragraph has a background color of #40daf1.\n\n``<p style=\"background-color:#40daf1;\">Content here</p>``\n#40daf1 border color\n\nThis element has a border color of #40daf1.\n\n``<div style=\"border:1px solid #40daf1;\">Content here</div>``\nCSS codes\n``.text {color:#40daf1;}``\n``.background {background-color:#40daf1;}``\n``.border {border:1px solid #40daf1;}``\n\n# Shades and Tints of #40daf1\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, #01090a is the darkest color, while #f7fdfe is the lightest one.\n\n• #01090a\n``#01090a` `rgb(1,9,10)``\n• #02191c\n``#02191c` `rgb(2,25,28)``\n• #03292f\n``#03292f` `rgb(3,41,47)``\n• #053941\n``#053941` `rgb(5,57,65)``\n• #064953\n``#064953` `rgb(6,73,83)``\n• #075965\n``#075965` `rgb(7,89,101)``\n• #096978\n``#096978` `rgb(9,105,120)``\n• #0a798a\n``#0a798a` `rgb(10,121,138)``\n• #0b899c\n``#0b899c` `rgb(11,137,156)``\n• #0d9aaf\n``#0d9aaf` `rgb(13,154,175)``\n• #0eaac1\n``#0eaac1` `rgb(14,170,193)``\n``#0fbad3` `rgb(15,186,211)``\n• #11cae5\n``#11cae5` `rgb(17,202,229)``\n• #1bd3ee\n``#1bd3ee` `rgb(27,211,238)``\n• #2ed6f0\n``#2ed6f0` `rgb(46,214,240)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #52def2\n``#52def2` `rgb(82,222,242)``\n• #65e1f4\n``#65e1f4` `rgb(101,225,244)``\n• #77e5f5\n``#77e5f5` `rgb(119,229,245)``\n• #89e8f6\n``#89e8f6` `rgb(137,232,246)``\n• #9becf8\n``#9becf8` `rgb(155,236,248)``\n• #aeeff9\n``#aeeff9` `rgb(174,239,249)``\n• #c0f3fa\n``#c0f3fa` `rgb(192,243,250)``\n• #d2f6fc\n``#d2f6fc` `rgb(210,246,252)``\n• #e4fafd\n``#e4fafd` `rgb(228,250,253)``\n• #f7fdfe\n``#f7fdfe` `rgb(247,253,254)``\nTint Color Variation\n\n# Tones of #40daf1\n\nA tone is produced by adding gray to any pure hue. In this case, #979a9a is the less saturated color, while #38e0f9 is the most saturated one.\n\n• #979a9a\n``#979a9a` `rgb(151,154,154)``\n• #8fa0a2\n``#8fa0a2` `rgb(143,160,162)``\n• #87a5aa\n``#87a5aa` `rgb(135,165,170)``\n• #7fabb2\n``#7fabb2` `rgb(127,171,178)``\n• #77b1ba\n``#77b1ba` `rgb(119,177,186)``\n• #6fb7c2\n``#6fb7c2` `rgb(111,183,194)``\n• #67bdca\n``#67bdca` `rgb(103,189,202)``\n• #60c3d1\n``#60c3d1` `rgb(96,195,209)``\n• #58c8d9\n``#58c8d9` `rgb(88,200,217)``\n• #50cee1\n``#50cee1` `rgb(80,206,225)``\n• #48d4e9\n``#48d4e9` `rgb(72,212,233)``\n• #40daf1\n``#40daf1` `rgb(64,218,241)``\n• #38e0f9\n``#38e0f9` `rgb(56,224,249)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #40daf1 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

[ null ]

[ "Current Location:home > Browse\n\n## 1. chinaXiv:201605.01780 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n In split supersymmetry, gauginos and higgsinos are the only supersymmetric particles possibly accessible at foreseeable colliders like the CERN Large Hadron Collider (LHC) and the International Linear Collider (ILC). In order to account for the cosmic dark matter measured by WMAP, these gauginos and higgsinos are stringently constrained and could be explored at the colliders through their direct productions and/or virtual effects in some processes. The clean environment and high luminosity of the ILC render the virtual effects at percent level meaningful in unraveling the new physics effects. In this work we assume split supersymmetry and calculate the virtual effects of the WMAP-allowed gauginos and higgsinos in the Higgs productions e(+) e(-) -> Zh and e(+) e(-) ->nu(e)(nu) over bar (e)h through WW fusion at the ILC. We find that the production cross section of e+ e-. Zh can be altered by a few percent in some part of the WMAP-allowed parameter space, while the correction to the WW fusion process e(+) e(-) ->nu(e) (nu) over bar (e)h is below 1%. Such virtual effects are correlated with the cross sections of chargino pair productions and can offer complementary information in probing split supersymmetry at the colliders.\n\n## 2. chinaXiv:201605.01753 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n As discussed recently by Hooper and Tait, the singlino-like dark matter in the Minimal Supersymmetric Standard Model (MSSM) extended by a singlet Higgs superfield can give a perfect explanation for both the relic density and the Pamela result through the Sommerfeld-enhanced annihilation into singlet Higgs bosons (a or h followed by h -> aa) with a being light enough to decay dominantly to muons or electrons. In this work we analyze the parameter space required by such a dark matter explanation and also consider the constraints from the LEP experiments. We find that although the light singlet Higgs bosons have small mixings with the Higgs doublets in the allowed parameter space, their couplings with the SM-like Higgs boson h(SM) (the lightest doublet-dominant Higgs boson) can be enhanced by the soft parameter A(kappa) and, in order to meet the stringent LEP constraints, the h(SM) tends to decay into the singlet Higgs pairs a a or hh instead of b (b) over bar. So the h(SM) produced at the LHC will give a multi-muon signal, h(SM) -> aa -> 4 mu or h(SM) -> hh -> 4a -> 8 mu.\n\n## 3. chinaXiv:201605.01751 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n If the sparticles are relatively heavy (a few TeV) while the Higgs sector is not so heavy (m(A) is not so large), the Higgs boson Yukawa couplings can harbor sizable quantum effects of sparticles and these large residual effects may play a special role in probing supersymmetry at foreseeable colliders. In this work, focusing on the supersymmetric QCD effects in the hb (b) over bar coupling (h is the lightest CP-even Higgs boson), we give a comparative study for the two popular supersymmetric models: the MSSM and NMSSM. While for both models the supersymmetric QCD can leave over large residual quantum effects in hb (b) over bar coupling, the NMSSM can allow for a much broader region of such effects. Since these residual effects can be over 20% for the hb (b) over bar coupling (and thus over 40% for the ratio Br(h -> b (b) over bar)/Br(h -> tau(+)tau(-))), future measurements may unravel the effects of heavy sparticles or even distinguish the two models. (C) 2009 Elsevier B.V. All rights reserved.\n\n## 4. chinaXiv:201605.01748 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n In the next-to-minimal supersymmetric model (NMSSM) a light CP-odd Higgs boson is so far allowed by current experiments, which, together with a large tan beta, may greatly enhance the rare dileptonic decays B -> X(s)l(+)l(-) and B(s)-> l(+)l(-)gamma. We examine these decays paying special attention to the new operator allowed by the light CP-odd Higgs boson. We find that in the parameter space allowed by current experiments like CERN LEP II and b -> s gamma, the branching ratios of these rare decays can be greatly enhanced, and thus the existing experimental data on B -> X(s)mu(+)mu(-) can further stringently constrain the parameter space (especially the region with a superlight CP-odd Higgs boson and large tan beta). In the surviving parameter space we give the predictions for other dileptonic decay branching ratios and also show the results for the forward-backward asymmetry.\n\n## 5. chinaXiv:201605.01742 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n We perform a comparative study of the neutralino dark matter scattering on nucleon in three popular supersymmetric models: the minimal (MSSM), the next-to-minimal (NMSSM) and the nearly minimal (nMSSM). First, we give the predictions of the elastic cross section by scanning over the parameter space allowed by various direct and indirect constraints, which are from the measurement of the cosmic dark matter relic density, the collider search for Higgs boson and sparticles, the precision electroweak measurements and the muon anomalous magnetic moment. Then we demonstrate the property of the allowed parameter space with/without the new limits from CDMS II. We obtain the following observations: (i) For each model the new CDMS limits can exclude a large part of the parameter space allowed by current collider constraints; (ii) The property of the allowed parameter space is similar for MSSM and NMSSM, but quite different for nMSSM; (iii) For each model the future SuperCDMS can cover most of the allowed parameter space given that all soft breaking parameters are below 1 TeV.\n\n## 6. chinaXiv:201605.01717 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n For the experimental search of neutralino dark matter, it is important to know its allowed mass and scattering cross section with the nucleon. In order to figure out how light a neutralino dark matter can be predicted in low energy supersymmetry, we scan over the parameter space of the NMSSM (next-to-minimal supersymmetric model), assuming all the relevant soft mass parameters to be below TeV scale. We find that in the parameter space allowed by current experiments the neutralino dark matter can be as light as a few GeV and its scattering rate off the nucleon can reach the sensitivity of XENON100 and CoGeNT. As a result, a sizable parameter space is excluded by the current XENON100 and CoGeNT data (the plausible CoGeNT dark matter signal can also be explained). The future 6000 kg-days exposure of XENON100 will further explore (but cannot completely cover) the remained parameter space. Moreover, we find that in such a light dark matter scenario a light CP-even or CP-odd Higgs boson must be present to satisfy the measured dark matter relic density. Consequently, the SM-like Higgs boson h(SM) may decay predominantly into a pair of light Higgs bosons or a pair of neutralinos so that the conventional decays like h(SM) -> gamma gamma is much suppressed. (C) 2011 Elsevier B.V. All rights reserved.\n\n## 7. chinaXiv:201605.01714 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n We examine the present and future XENON limits on the neutralino dark matter in split supersymmetry (split-SUSY). Through a scan over the parameter space under the current constraints from collider experiments and the WMAP measurement of the dark matter relic density, we find that in the allowed parameter space a large part has been excluded by the present XENON100 limits and a further largish part can be covered by the future exposure (6000 kg day). In case of unobservation of dark matter with such an exposure in the future, the lightest neutralino will remain bino-like and its annihilation is mainly through exchanging the SM-like Higgs boson in order to get the required relic density. (C) 2011 Elsevier B.V. All rights reserved.\n\n## 8. chinaXiv:201605.01703 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n In order to have massive neutrinos, the right-handed neutrino/sneutrino superfield (N) need to be introduced in supersymmetry. In the framework of NMSSM (the MSSM with a singlet S) such an extension will dynamically lead to a TeNT-scale Majorana mass for the right-handed neutrino through the S N N coupling when S develops a vev (the free Majorana mass term is forbidden by the assumed Z(3) symmetry). Also, through the couplings S N N and SHuHd the SM-like Higgs boson (a mixture of H-u, H-d and S) can naturally couple with the right-handed neutrino/sneutrino. As a result, the TeV-scale right-handed neutrino/sneutrino may significantly contribute to the Higgs boson mass. Through an explicit calculation, we find that the Higgs boson mass can indeed be sizably altered by the right-handed neutrino/sneutrino. Such new contribution can help to push up the SM-like Higgs boson mass and thus make the NMSSM more natural.\n\n## 9. chinaXiv:201605.01148 [pdf]\n\nSubjects: Physics >> The Physics of Elementary Particles and Fields\n\n In this work we show that the general singlet extension of the minimal supersymmetric standard model (MSSM) can naturally provide a self-interacting singlino dark matter to solve the small cosmological scale anomalies (a large Sommerfeld enhancement factor can also be obtained). However, we find that the NMSSM (the singlet extension of the MSSM with Z(3) symmetry) cannot achieve this due to the restricted parameter space. In our analysis we introduce the concept of symmetric and antisymmetric viscosity cross sections to deal with the nonrelativistic Majorana-fermion dark matter scattering." ]

[ null ]

[ "# 1278774 (number)\n\n1,278,774 (one million two hundred seventy-eight thousand seven hundred seventy-four) is an even seven-digits composite number following 1278773 and preceding 1278775. In scientific notation, it is written as 1.278774 × 106. The sum of its digits is 36. It has a total of 7 prime factors and 64 positive divisors. There are 342,144 positive integers (up to 1278774) that are relatively prime to 1278774.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 7\n• Sum of Digits 36\n• Digital Root 9\n\n## Name\n\nShort name 1 million 278 thousand 774 one million two hundred seventy-eight thousand seven hundred seventy-four\n\n## Notation\n\nScientific notation 1.278774 × 106 1.278774 × 106\n\n## Prime Factorization of 1278774\n\nPrime Factorization 2 × 33 × 7 × 17 × 199\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 142086 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,278,774 is 2 × 33 × 7 × 17 × 199. Since it has a total of 7 prime factors, 1,278,774 is a composite number.\n\n## Divisors of 1278774\n\n64 divisors\n\n Even divisors 32 32 16 16\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 64 Total number of the positive divisors of n σ(n) 3.456e+06 Sum of all the positive divisors of n s(n) 2.17723e+06 Sum of the proper positive divisors of n A(n) 54000 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1130.83 Returns the nth root of the product of n divisors H(n) 23.681 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,278,774 can be divided by 64 positive divisors (out of which 32 are even, and 32 are odd). The sum of these divisors (counting 1,278,774) is 3,456,000, the average is 54,000.\n\n## Other Arithmetic Functions (n = 1278774)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 342144 Total number of positive integers not greater than n that are coprime to n λ(n) 1584 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 98288 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 342,144 positive integers (less than 1,278,774) that are coprime with 1,278,774. And there are approximately 98,288 prime numbers less than or equal to 1,278,774.\n\n## Divisibility of 1278774\n\n m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 0 6 0\n\nThe number 1,278,774 is divisible by 2, 3, 6, 7 and 9.\n\n• Arithmetic\n• Abundant\n\n• Polite\n• Practical\n\n## Base conversion (1278774)\n\nBase System Value\n2 Binary 100111000001100110110\n3 Ternary 2101222011000\n4 Quaternary 10320030312\n5 Quinary 311410044\n6 Senary 43224130\n8 Octal 4701466\n10 Decimal 1278774\n12 Duodecimal 518046\n20 Vigesimal 7jgie\n36 Base36 repi\n\n## Basic calculations (n = 1278774)\n\n### Multiplication\n\nn×y\n n×2 2557548 3836322 5115096 6393870\n\n### Division\n\nn÷y\n n÷2 639387 426258 319694 255755\n\n### Exponentiation\n\nny\n n2 1635262943076 2091131734769068824 2674084892997581216341776 3419550234958088922346238262624\n\n### Nth Root\n\ny√n\n 2√n 1130.83 108.542 33.6278 16.6479\n\n## 1278774 as geometric shapes\n\n### Circle\n\n Diameter 2.55755e+06 8.03477e+06 5.13733e+12\n\n### Sphere\n\n Volume 8.75931e+18 2.05493e+13 8.03477e+06\n\n### Square\n\nLength = n\n Perimeter 5.1151e+06 1.63526e+12 1.80846e+06\n\n### Cube\n\nLength = n\n Surface area 9.81158e+12 2.09113e+18 2.2149e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.83632e+06 7.0809e+11 1.10745e+06\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.83236e+12 2.46442e+17 1.04411e+06\n\n## Cryptographic Hash Functions\n\nmd5 73241c14024e2e5f6a51cc091f67b9c5 96032f48c2b72fe035a7ad402c303e6dbde3083e ac743198c97b5246e636a323b8d5bc3f38b6a599fbbd1622b6a74ae2292728d0 f748e6ec9e91ecfc76a0d04f9824b5d08e7d1b873cd89acdc3176daff589ce5fae619a517fdbfb96a5acd1e4281c51c220466351b80d4fac7daee11a5ff0b43b 8d699d1ab474574591f279caa5e88c6d3812fb41" ]

[ null ]

[ "", null, "", null, "# Implementing the Weak Form with a COMSOL App\n\nApril 16, 2015\n\nPreviously in our weak form series, we discretized the weak form equation to obtain a matrix equation to solve for the unknown coefficients in our simple example problem. Following the same procedure as in this previous blog post, we will implement the equation in the COMSOL Multiphysics® software with additional steps included to examine the matrices. We will find it more convenient to use a COMSOL® software application to display all relevant matrices at once, arranged logically on one screen.\n\n### Our Simple Example\n\nRecall our simple example of 1D heat transfer at steady state with no heat source, where the temperature T is a function of the position x in the domain defined by the interval 1 \\le x \\le 5. We imposed the Neumann boundary condition such that the outgoing flux should be 2 at the left boundary (x=1) and the Dirichlet boundary condition such that the temperature should be 9 at the right boundary (x=5). After discretizing the weak form equation, we obtained this matrix equation:\n\n(1)\n\n\\left(\n\\begin{array}{cccccc}\n1 & -1 & 0 & 0 & 0 & 0 \\\\\n-1 & 2 & -1 & 0 & 0 & 0 \\\\\n0 & -1 & 2 & -1 & 0 & 0 \\\\\n0 & 0 & -1 & 2 & -1 & 0 \\\\\n0 & 0 & 0 & -1 & 1 & 1 \\\\\n0 & 0 & 0 & 0 & 1 & 0\n\\end{array}\n\\right)\n\\left(\n\\begin{array}{c} a_1 \\\\ a_2 \\\\ a_3 \\\\ a_4 \\\\ a_5 \\\\ \\lambda_2 \\end{array}\n\\right)\n= \\left(\n\\begin{array}{c} -2 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 9 \\end{array}\n\\right)\n\nwhere a_1, a_2, \\cdots , a_5 are unknown temperature values at the nodal points (x=1, 2, \\cdots, 5) and \\lambda_2 is an unknown heat flux at the right boundary (x=5). The matrix on the left-hand side is customarily called the stiffness matrix and the vector on the right is called the load vector, due to the application of this technique in structural mechanics.\n\n### Viewing the Stiffness Matrix and Load Vector\n\nThe steps for implementing the weak form equation in COMSOL Multiphysics have been discussed in this earlier blog entry, thus we will not repeat them here. To view the stiffness matrix and load vector, right-click Solution 1 in the model tree → OtherAssemble, as shown in the screenshot below:", null, "Make sure to drag the Assemble node up to position it above (before) the Stationary Solver node. If this step is not done and the Assemble node is left below (after) the Stationary Solver node, then the content of the load vector may be altered by the solver.\n\nThen, in the settings window for the Assemble 1 node, we can select the matrices of interest by checking the corresponding checkbox for each item. Additionally, set Scaling of Dependent Variables 1 to None. After solving, we can evaluate the matrices by right-clicking Derived Values and then selecting System Matrix, as illustrated below:", null, "In the corresponding settings window, we can select Stiffness matrix from the Matrix drop-down menu and evaluate it in a table. As indicated in the screenshot below, we obtain exactly the same 6×6 matrix as the one in Eq. (1).", null, "The load vector on the right-hand side of Eq. (1) can be evaluated and verified by the same procedure.\n\n### Pointwise Constraint\n\nWe mentioned before that it is sometimes desirable not to solve for the Lagrange multiplier \\lambda_2 — for example, to save computation resources. To do so, we right-click the Weak Form PDE main node → MorePointwise Constraint, as depicted below:", null, "Then, in the settings window, we enter 9-T for the Constraint expression input field. After solving, we obtain exactly the same solution, but a smaller 5×5 stiffness matrix:", null, "If we look at this matrix closely, we find that it matches the upper left-hand part of the 6×6 matrix that we observed earlier. This should not be too surprising, as we are solving exactly the same problem (represented by Eq. (1)), just in a slightly different way. We will briefly discuss this next.\n\n### Alternative Methods for Solving the Matrix Equation\n\nWhen we implement the Dirichlet boundary condition using the Weak Contribution feature with a Lagrange multiplier \\lambda_2 (as shown in this previous blog post), we effectively ask that the entire matrix equation (1) be solved to yield the coefficients a_1, a_2, \\cdots , a_5 as well as the multiplier \\lambda_2.\n\nOn the other hand, when we implement the fixed boundary condition using the Pointwise Constraint feature as shown above, we essentially ask that the same matrix equation (1) be solved without explicitly solving for the Lagrange multiplier \\lambda_2. The software effectively segregates Eq. (1) to this form (see below)\n\n(2)\n\n\\left(\\begin{array}{cc}\nK & N_F \\\\\nN & 0 \\end{array}\\right)\n\\left(\\begin{array}{c}\nU \\\\\n\\Lambda \\end{array}\\right)\n=\n\\left(\\begin{array}{c}\nL \\\\\nM \\end{array}\\right)\n\nwhere the stiffness matrix K is the 5×5 matrix shown above, the constraint Jacobian matrix N is (0 \\, 0 \\, 0 \\, 0 \\, 1), the constraint force Jacobian matrix N_F is (0 \\, 0 \\, 0 \\, 0 \\, 1)^T, the solution vector U is formed by the coefficients a_1, a_2, \\cdots , a_5, the (one-element) vector \\Lambda is formed by the Lagrange multiplier \\lambda_2, the load vector L is (-2 \\, 0 \\, 0\\, 0\\, 0)^T, and the (one-element) constraint vector M is (9).\n\nThis segregation of Eq. (1) is shown graphically below:", null, "Of course, in reality, when the Pointwise Constraint feature is specified, the software does not bother to assemble the full 6×6 matrix in Eq. (1). Instead, it assembles K, N, and N_F, which requires less computational resources than assembling the full matrix.\n\nEq. (2) can be written as a system of two matrix equations:\n\n(3)\n\nK \\, U+ N_F \\, \\Lambda = L\n\n(4)\n\nN \\, U = M\n\nwhere the first one (3) contains the part of the discretized weak form equation with heat flux boundary conditions and the second one (4) contains the constraint imposed by the Dirichlet boundary condition.\n\nThe equation system can be solved in two steps. In the first step, the constraint equation (4) is solved for the degree(s) of freedom involved with the Dirichlet boundary condition. In the second step, the solution from the first step is substituted into Eq. (3) to solve for the remaining degrees of freedom.\n\nThe resulting solution vector U is then given as the linear combination of the solutions from the two steps:\n\n(5)\n\nU=U_d+Null \\, U_n\n\nThe first term U_d is the solution vector of the constraint equation (4) solved in the first step. The second term is obtained from the second step in the form of the product of a matrix Null and a solution vector U_n. The columns of the matrix Null are formed by the basis vectors spanning the null space of the constraint Jacobian matrix N. So, we have\n\nN \\, Null \\equiv 0\n\nThe solution vector U_n is the solution to the eliminated matrix equation\n\n(6)\n\nK_c \\, U_n = L_c\n\nwhere the eliminated stiffness matrix K_c and the eliminated load vector L_c are given by\n\n\\begin{align}\nK_c& = Nullf^T \\, K \\, Null \\\\\nL_c& = Nullf^T \\, (L-K \\, U_d)\n\\end{align}\n\nand the columns of the matrix Nullf are formed by the basis vectors spanning the null space of the transpose of the constraint force Jacobian matrix N_F. So, we have\n\nNullf^T \\, N_F \\equiv 0\n\nThe term eliminated indicates that the degree(s) of freedom involved in the Dirichlet boundary condition have been removed from the matrix equation (6). In our example, the size of the eliminated stiffness matrix K_c is 4×4.\n\nNotice that the solution procedure described above does not involve the Lagrange multiplier vector \\Lambda. Indeed, the value of the Lagrange multiplier \\lambda_2 is left unsolved by this procedure. The advantage of this method is that the required computational resources are reduced. For our simple example, the size of the matrix is reduced from 6×6 to 4×4 (plus an even smaller one for the constraint equation).\n\n### Viewing All Matrices at Once with a COMSOL® App\n\nCOMSOL Multiphysics allows us to evaluate and view any of the matrices and vectors mentioned above. All we have to do is to follow the same procedure outlined in the previous section on viewing the stiffness matrix and load vector. However, we quickly find out that we will spend a lot of time clicking on each System Matrix node in the Model Builder and then clicking on the “Evaluate” button in the settings window. Also, we can only see one matrix or vector at a time, making it tedious to examine all the matrices and vectors.\n\nThis represents one of the many situations in which a COMSOL application can help enhance modeling experience and productivity (learn about application webinars below). The core package of COMSOL Multiphysics can be used to build a COMSOL app based on any COMSOL model by wrapping a user interface (UI) around it. The UI is completely customizable and can be easily configured to suit individual modeling needs.\n\nFor an introduction to COMSOL applications, see the following webinars: Intro to COMSOL Multiphysics® 5.0 and the Application Builder (with a focus on the second half) and How to Build and Run Simulation Apps with COMSOL Server™.\n\nThe following screenshots show just one of the essentially infinite number of ways that a UI can be arranged to serve as a convenient tool for investigating the matrices and vectors. The app is set up to switch among several different kinds of boundary conditions via checkboxes. All matrices and vectors are evaluated and displayed by a single click of the “Compute” button. Here is a screenshot of the case in which the full 6×6 matrix is used to solve:", null, "We see that N, N_F, and M are empty and K_c remains the full size of 6×6. In contrast, here is the screenshot for the second case where the eliminated 4×4 matrix is used to solve:", null, "### Concluding Remarks\n\nUsing our simple heat transfer example, we have explored the two different ways to solve the matrix equation obtained from discretizing the weak form equation. We found that it can be much simpler to evaluate and inspect the various matrices and vectors involved in the solution process by setting up a COMSOL application. In the next blog posts, we will continue to use COMSOL apps to help us investigate more complex examples.\n\nAs a general-purpose development environment for multiphysics simulation, the COMSOL Desktop® environment must be structured with a limited arrangement of UI elements. The COMSOL app frees us from such limitations, providing the power for building specialized UI for unique requirements. Additionally, the built-in coding functionality allows much more versatile computations, and COMSOL Server™ license enables the deployment of apps for worldwide access by coworkers as well as clients.\n\nWe hope that you will use these powerful features in COMSOL Multiphysics to boost your own work efficiency!\n\n#### Kommentare (16)\n\n##### Einen Kommentar hinterlassen", null, "##### Tae Eun Kim\nJune 16, 2015\n\nCan I get this app if it is possible? thanks.", null, "##### Jing Wang\nJuly 6, 2015\n\nDear Chien,\n\nI‘ve followed your instruction step by step in my comsol 5.0, however, i couldn’t get the same load vector as yours although i got the same stiffness matrix. My load vector is\n0\n0\n-8.881784197001252E-16\n1.7763568394002505E-15\n-1.1102230246251565E-15\n0\n-2\n0\n0\n0\n0\n9\nCan you please post your mph.file here so i can check it out?\nThank you anyway!\n\nBest regards,\n\nJing", null, "##### Chien Liu\nJuly 13, 2015\n\nDear Jing,\n\[email protected]\n\nSincerely,\n\nChien", null, "##### Jing Wang\nJuly 15, 2015", null, "##### Jing Wang\nJuly 18, 2015\n\nDear Chien,\nThanks for your reply! I’ve checked the boundary conditions, they are just the same as those in your post “Implementing the Weak Form in COMSOL Multiphysics”. Can you please send your mph.file to me (My email: [email protected])? Thank you anyway!\nSincerely,\nJing", null, "##### Chien Liu\nJuly 20, 2015\n\nDear Jing,\n\[email protected]\n\nSincerely,\n\nChien", null, "##### Roland Ernst\nFebruary 9, 2016\n\nDear Chien,\n\nVery interesting blog about how the weak_form is implemented in Comsol! I followed a specialized Comsol seminary about these aspects about 10 years ago and would like to remind this again. So I have 2 questions about that:\n1/ I have not understood how the matrices Null, Kc and Lc are formed. What means for example your sentence: “Null are formed by the basis vectors spanning the null space of the constraint Jacobian matrix N”? Is there a book or some publications about all these aspects and more specifically the theory of Lagrange multipliers?\n2/ I have not completely understood what is the difference between ideal and non ideal constraints and when to use either one or the other one (I think now this has been replaced by options like “apply reaction terms on all physics/or individual dependant variables”). Could you please give me some explanations about that and how it influences the former matrices and also what’s the physical signification of this? A blog about that special aspect would also be very appreciated!\nSincerely\nRoland", null, "##### Chien Liu\nFebruary 9, 2016\n\nDear Roland,\n\nRegarding the questions, you should be able to find more information in the COMSOL Multiphysics Reference Manual > Equation-Based Modeling > Modeling with PDEs > About Weak Form PDE (and the following sections). And yes we plan to continue the blog series with more topics on the constraints. Thank you for your interest in this.\n\nSincerely,\n\nChien", null, "##### Roland Ernst\nFebruary 16, 2016\n\nDear Chien,\n\nThank you for your comments. But sorry when I go to your mentioned Comsol documentation I find nothing speaking about the Null, Kc, Lc (etc…) matrices you use in your weak form implementation explanations of your blog. Could you please give me some more details about for instance your Null matrix is built? Meaning I don’t understand what you mean by saying ” The columns of the matrix Null are formed by the basis vectors spanning the null space of the constraint Jacobian matrix N” . Thank you for your help!\nRegards\nRoland", null, "##### Evgeni Sergeev\nMarch 10, 2016\n\nThank you very much for the blog series. A tremendous value-adder to Comsol. I think this is helping every user of Comsol who is serious about understanding what they’re doing, and not just generating pretty pictures.\n\nIf anybody is interested in condition numbers, just wanted to add that computing cond(K) is meaningless, while cond(Kc) is correct. (Even though K is called the “stiffness matrix”, so that one might intuitively attempt to compute its condition number.) It’s the equation Kc×Un=Lc that gets sent to the linear solver, so Kc is the relevant one. The condition number of the original system\ncond([ms.K ms.NF; ms.N zeros(size(ms.N,1))])\nmight also make sense to check [in my case this is similar to cond(Kc)], but not cond(K).\n\nI suspect that the condition number is somehow related to LinErr (trying to understand why the relative error in my model is so high, when the residual is plotted as zero everywhere and LinRes is ~1e-15). This isn’t clear from how LinErr is defined in the documentation, but they seem correlated to some extent.", null, "##### wang shuo\nMarch 13, 2016\n\nI think there might be a error about the load vector which should be,\n(-2,0,0,0,0,\\lamda_2)^T\nand also the solution vector,\n(a_1,a_2,a_3,a_4,a_5,9)^T\nPlease correct me if I am wrong", null, "##### Chien Liu\nMarch 15, 2016\n\nDear shuo,\n\\lambda_2 belongs to the vector formed by U and \\Lambda. Please see Eq. (2) and the graph below it.\nSincerely,\nChien", null, "##### Evgeni Sergeev\nJune 20, 2016\n\nI’ve ran into the same issue as Jing Wang above: the load vector entries are all zero or near machine epsilon.\n\nI posted my 1-element MPH model and a detailed description on the forum:\n\nIt also shows a way to get the load vector, by messing with “Linear perturbation” settings. I don’t fully understand the meaning of that, but it may be of interest for those who wish to get the matrices using LiveLink and solve in Matlab.\n\nRegards,\nEvgeni", null, "##### Chien Liu\nJune 20, 2016\n\nDear Evgeni,\n\nThank you for the comment. Did you perform this step as described in the text of the blog post: “Make sure to drag the Assemble node up to position it before the Stationary Solver node.” ?\n\nIf this step is not done and the Assemble node is left after (below) the Stationary Solver node, then the content of the load vector may be altered by the solver as you have observed.\n\nHope this helps.\n\nSincerely,\n\nChien", null, "##### ChienShun Liu\nJuly 31, 2020\n\nDear Chien:\n\nI saw in your post ‘Implementing the Weak Form with a COMSOL App’\nhow to extract/view the stiffness matrix and the right hand side Lc.\nI wonder if it is possible to extract the (extended) mass matrix without\n\nProblem description:\nLap u = f in Omega\nu = g_D on Gamma_1 (pointwise constraint)\ndu/dn = g_N on Gamma_2 (natural BC)\n\nBy extended mass matrix I mean one that takes\nthe input (f_DOF,g_D,g_N) and out put to the\nright hand side Lc in the final equation\n\nKc u = Lc\n\nHere f_DOF means the point values of the source term f\non the degrees of freedom (nodes) of the underlying finite element\nmethod.", null, "##### Chien Liu\nJuly 31, 2020 COMSOL Mitarbeiter\n\nDear ChienShun, If you meant Eq. (6) then yes, as you can see in the last screenshot in the post, Kc and Lc are evaluated. No LiveLink products are used in the App, only the base package is used. If you need further assistance, please contact us using your COMSOL support portal (log in to your COMSOL Access account at https://www.comsol.com/access and then click on the Submit New Support Case button). Sincerely, Chien\n\nCOMSOL BLOG DURCHSTÖBERN" ]

[ null, "https://cdn.comsol.com/company/logo/comsol-logo-130x20.png", null, "https://cdn.comsol.com/wordpress/sites/1/2018/09/COMSOL_Blog_Header_General.png", null, "https://cdn.comsol.com/wordpress/2015/04/Screenshot-on-how-to-access-the-stiffness-matrix.png", null, "https://cdn.comsol.com/wordpress/2015/04/Screenshot-of-the-System-Matrix-option-in-COMSOL-Multiphysics.png", null, "https://cdn.comsol.com/wordpress/2015/04/Stiffness-matrix-table.png", null, "https://cdn.comsol.com/wordpress/2015/04/Accessing-the-Pointwise-Constraint-in-COMSOL-Multiphysics.png", null, "https://cdn.comsol.com/wordpress/2015/04/COMSOL-Multiphysics-5x5-Stiffness-matrix-table.png", null, "https://cdn.comsol.com/wordpress/2015/04/Graph-showcasing-the-segregation-of-our-matrix-equation.png", null, "https://cdn.comsol.com/wordpress/2015/04/Solving-your-simulation-with-a-6x6-matrix.png", null, "https://cdn.comsol.com/wordpress/2015/04/Screenshot-of-you-simulation-results-in-a-4x4-matrix.png", null, "https://www.gravatar.com/avatar/94fb928227cf66f67af95f35b9f1b619", null, "https://www.gravatar.com/avatar/b0d99f4a948dfb287fa4cdbb07117853", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null, "https://www.gravatar.com/avatar/b0d99f4a948dfb287fa4cdbb07117853", null, "https://www.gravatar.com/avatar/b0d99f4a948dfb287fa4cdbb07117853", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null, "https://www.gravatar.com/avatar/bcbcd9a7deed160688a59ed1e356bcbe", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null, "https://www.gravatar.com/avatar/bcbcd9a7deed160688a59ed1e356bcbe", null, "https://www.gravatar.com/avatar/bf11af111cab6e513d68e5e30762deb5", null, "https://www.gravatar.com/avatar/5c428c058d9babfda006cb331c57fc1a", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null, "https://www.gravatar.com/avatar/bf11af111cab6e513d68e5e30762deb5", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null, "https://www.gravatar.com/avatar/c613e264882e90231ed00214cb4430b2", null, "https://cdn.comsol.com/wordpress/2017/02/Chien-Liu_COMSOL-Blog-author.jpeg", null ]

[ "# Number 105950\n\nNumber 105,950 spell 🔊, write in words: one hundred and five thousand, nine hundred and fifty . Ordinal number 105950th is said 🔊 and write: one hundred and five thousand, nine hundred and fiftieth. Color #105950. The meaning of number 105950 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 105950. What is 105950 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 105950.\n\n## What is 105,950 in other units\n\nThe decimal (Arabic) number 105950 converted to a Roman number is (C)(V)CML. Roman and decimal number conversions.\n\n#### Weight conversion\n\n105950 kilograms (kg) = 233577.4 pounds (lbs)\n105950 pounds (lbs) = 48058.6 kilograms (kg)\n\n#### Length conversion\n\n105950 kilometers (km) equals to 65835 miles (mi).\n105950 miles (mi) equals to 170511 kilometers (km).\n105950 meters (m) equals to 347601 feet (ft).\n105950 feet (ft) equals 32294 meters (m).\n105950 centimeters (cm) equals to 41712.6 inches (in).\n105950 inches (in) equals to 269113 centimeters (cm).\n\n#### Temperature conversion\n\n105950° Fahrenheit (°F) equals to 58843.3° Celsius (°C)\n105950° Celsius (°C) equals to 190742° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n105950 seconds equals to 1 day, 5 hours, 25 minutes, 50 seconds\n105950 minutes equals to 2 months, 2 weeks, 3 days, 13 hours, 50 minutes\n\n### Codes and images of the number 105950\n\nNumber 105950 morse code: .---- ----- ..... ----. ..... -----\nSign language for number 105950:", null, "", null, "", null, "", null, "", null, "", null, "Number 105950 in braille:", null, "QR code Bar code, type 39", null, "", null, "Images of the number Image (1) of the number Image (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n## Share in social networks", null, "## Mathematics of no. 105950\n\n### Multiplications\n\n#### Multiplication table of 105950\n\n105950 multiplied by two equals 211900 (105950 x 2 = 211900).\n105950 multiplied by three equals 317850 (105950 x 3 = 317850).\n105950 multiplied by four equals 423800 (105950 x 4 = 423800).\n105950 multiplied by five equals 529750 (105950 x 5 = 529750).\n105950 multiplied by six equals 635700 (105950 x 6 = 635700).\n105950 multiplied by seven equals 741650 (105950 x 7 = 741650).\n105950 multiplied by eight equals 847600 (105950 x 8 = 847600).\n105950 multiplied by nine equals 953550 (105950 x 9 = 953550).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 105950\n\nHalf of 105950 is 52975 (105950 / 2 = 52975).\nOne third of 105950 is 35316,6667 (105950 / 3 = 35316,6667 = 35316 2/3).\nOne quarter of 105950 is 26487,5 (105950 / 4 = 26487,5 = 26487 1/2).\nOne fifth of 105950 is 21190 (105950 / 5 = 21190).\nOne sixth of 105950 is 17658,3333 (105950 / 6 = 17658,3333 = 17658 1/3).\nOne seventh of 105950 is 15135,7143 (105950 / 7 = 15135,7143 = 15135 5/7).\nOne eighth of 105950 is 13243,75 (105950 / 8 = 13243,75 = 13243 3/4).\nOne ninth of 105950 is 11772,2222 (105950 / 9 = 11772,2222 = 11772 2/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 105950\n\n#### Is Prime?\n\nThe number 105950 is not a prime number. The closest prime numbers are 105943, 105953.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 105950 are 2 * 5 * 5 * 13 * 163\nThe factors of 105950 are\n1 , 2 , 5 , 10 , 13 , 25 , 26 , 50 , 65 , 130 , 163 , 325 , 326 , 650 , 815 , 1630 , 2119 , 4075 , 4238 , 8150 , 105950 show more factors ...\nTotal factors 24.\nSum of factors 213528 (107578).\n\n#### Powers\n\nThe second power of 1059502 is 11.225.402.500.\nThe third power of 1059503 is 1.189.331.394.875.000.\n\n#### Roots\n\nThe square root √105950 is 325,499616.\nThe cube root of 3105950 is 47,318793.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 105950 = loge 105950 = 11,570723.\nThe logarithm to base 10 of No. log10 105950 = 5,025101.\nThe Napierian logarithm of No. log1/e 105950 = -11,570723.\n\n### Trigonometric functions\n\nThe cosine of 105950 is -0,977561.\nThe sine of 105950 is 0,210652.\nThe tangent of 105950 is -0,215488.\n\n### Properties of the number 105950\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 105950 in Computer Science\n\nCode typeCode value\nPIN 105950 It's recommendable to use 105950 as a password or PIN.\n105950 Number of bytes103.5KB\nCSS Color\n#105950 hexadecimal to red, green and blue (RGB) (16, 89, 80)\nUnix timeUnix time 105950 is equal to Friday Jan. 2, 1970, 5:25:50 a.m. GMT\nIPv4, IPv6Number 105950 internet address in dotted format v4 0.1.157.222, v6 ::1:9dde\n105950 Decimal = 11001110111011110 Binary\n105950 Decimal = 12101100002 Ternary\n105950 Decimal = 316736 Octal\n105950 Decimal = 19DDE Hexadecimal (0x19dde hex)\n105950 BASE64MTA1OTUw\n105950 SHA2240b1fc1a857c9a6f3b27958fb595bf2a9c8e9e095336e2749cf7e733b\n105950 SHA384d032b59a2f970715a57061509dc7e5126d3ea205178c5b6a04c8c1263f440e0e5f87bc4624232f75fb375f377733446f\nMore SHA codes related to the number 105950 ...\n\nIf you know something interesting about the 105950 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 105950\n\n### Character frequency in number 105950\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 0 2 5 2 9 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 105950, the numbers 1+0+5+9+5+0 = 2+0 = 2 are added and the meaning of the number 2 is sought.\n\n## Interesting facts about the number 105950\n\n### Asteroids\n\n• (105950) 2000 SA243 is asteroid number 105950. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 9/24/2000.\n\n## № 105,950 in other languages\n\nHow to say or write the number one hundred and five thousand, nine hundred and fifty in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 105.950) ciento cinco mil novecientos cincuenta German: 🔊 (Anzahl 105.950) einhundertfünftausendneunhundertfünfzig French: 🔊 (nombre 105 950) cent cinq mille neuf cent cinquante Portuguese: 🔊 (número 105 950) cento e cinco mil, novecentos e cinquenta Chinese: 🔊 (数 105 950) 十万五千九百五十 Arabian: 🔊 (عدد 105,950) مائة و خمسة ألفاً و تسعمائةخمسون Czech: 🔊 (číslo 105 950) sto pět tisíc devětset padesát Korean: 🔊 (번호 105,950) 십만 오천구백오십 Danish: 🔊 (nummer 105 950) ethundrede og femtusindnihundrede og halvtreds Dutch: 🔊 (nummer 105 950) honderdvijfduizendnegenhonderdvijftig Japanese: 🔊 (数 105,950) 十万五千九百五十 Indonesian: 🔊 (jumlah 105.950) seratus lima ribu sembilan ratus lima puluh Italian: 🔊 (numero 105 950) centocinquemilanovecentocinquanta Norwegian: 🔊 (nummer 105 950) en hundre og fem tusen, ni hundre og femti Polish: 🔊 (liczba 105 950) sto pięć tysięcy dziewięćset pięćdziesiąt Russian: 🔊 (номер 105 950) сто пять тысяч девятьсот пятьдесят Turkish: 🔊 (numara 105,950) yüzbeşbindokuzyüzelli Thai: 🔊 (จำนวน 105 950) หนึ่งแสนห้าพันเก้าร้อยห้าสิบ Ukrainian: 🔊 (номер 105 950) сто п'ять тисяч дев'ятсот п'ятдесят Vietnamese: 🔊 (con số 105.950) một trăm lẻ năm nghìn chín trăm năm mươi Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 105950 or any natural number (positive integer) please write us here or on facebook." ]

[ null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-0.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-5.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-9.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-5.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-0.png", null, "https://number.academy/img/braille-105950.svg", null, "https://numero.wiki/img/codigo-qr-105950.png", null, "https://numero.wiki/img/codigo-barra-105950.png", null, "https://numero.wiki/img/a-105950.jpg", null, "https://numero.wiki/img/b-105950.jpg", null, "https://numero.wiki/s/share-desktop.png", null ]

[ "# rainflow\n\nRainflow counts for fatigue analysis\n\n## Syntax\n\n``c = rainflow(x)``\n``c = rainflow(x,fs)``\n``c = rainflow(x,t)``\n``c = rainflow(xt)``\n``c = rainflow(___,'ext')``\n``[c,rm,rmr,rmm] = rainflow(___)``\n``[c,rm,rmr,rmm,idx] = rainflow(___)``\n``rainflow(___)``\n\n## Description\n\n````c = rainflow(x)` returns cycle counts for the load time history, `x`, according to the ASTM E 1049 standard. See Algorithms for more information.```\n\nexample\n\n````c = rainflow(x,fs)` returns cycle counts for `x` sampled at a rate `fs`.```\n\nexample\n\n````c = rainflow(x,t)` returns cycle counts for `x` sampled at the time values stored in `t`.```\n\nexample\n\n````c = rainflow(xt)` returns cycle counts for the time history stored in the MATLAB® timetable `xt`.```\n\nexample\n\n````c = rainflow(___,'ext')` specifies the time history as a vector of identified reversals (peaks and valleys). `'ext'` can be used with any of the previous syntaxes.```\n````[c,rm,rmr,rmm] = rainflow(___)` outputs a rainflow matrix, `rm`, and two vectors, `rmr` and `rmm`, containing histogram bin edges for the rows and columns of `rm`, respectively.```\n````[c,rm,rmr,rmm,idx] = rainflow(___)` also returns the linear indices of the reversals identified in the input.```\n````rainflow(___)` with no output arguments plots load reversals and a rainflow matrix histogram in the current figure.```\n\n## Examples\n\ncollapse all\n\nGenerate a signal that resembles a load history, consisting of sinusoid half-periods connecting known, equispaced reversals. The signal is sampled at 512 Hz for 8 seconds. Plot the extrema and the signal.\n\n```fs = 512; X = [-2 1 -3 5 -1 3 -4 4 -2]; lX = length(X)-1; Y = -diff(X)/2.*cos(pi*(0:1/fs:1-1/fs)') + (X(1:lX)+X(2:lX+1))/2; Y = [Y(:);X(end)]; plot(0:lX,X,'o',0:1/fs:lX,Y)```", null, "Compute cycle counts for the data. Display the matrix of cycle counts.\n\n```[c,hist,edges,rmm,idx] = rainflow(Y,fs); T = array2table(c,'VariableNames',{'Count','Range','Mean','Start','End'})```\n```T=7×5 table Count Range Mean Start End _____ _____ ____ _____ ___ 0.5 3 -0.5 0 1 0.5 4 -1 1 2 1 4 1 4 5 0.5 8 1 2 3 0.5 9 0.5 3 6 0.5 8 0 6 7 0.5 6 1 7 8 ```\n\nDisplay a histogram of cycle counts as a function of stress range.\n\n```histogram('BinEdges',edges','BinCounts',sum(hist,2)) xlabel('Stress Range') ylabel('Cycle Counts')```", null, "Use `rainflow` without output arguments to display a histogram of cycles as a function of cycle average and cycle range.\n\n`rainflow(Y,fs)`", null, "Generate a signal that resembles a load history, consisting of sinusoid half-periods connecting known, unevenly spaced reversals. The signal is sampled at 10 Hz for 15 seconds. Plot the extrema and the signal.\n\n```fs = 10; X = [0 1 3 4 5 6 8 10 13 15]; Y = [-2 1 -3 5 -1 3 -4 4 -2 6]; Z = []; for k = 1:length(Y)-1 x = X(k+1)-X(k); z = -(Y(k+1)-Y(k))*cos(pi*(0:1/fs:x-1/fs)/x)+Y(k+1)+Y(k); Z = [Z z/2]; end Z = [Z Y(end)]; t = linspace(X(1),X(end),length(Z)); plot(X,Y,'o',t,Z)```", null, "Compute cycle counts for the data. Display the matrix of cycle counts.\n\n```[c,hist,edges,rmm,idx] = rainflow(Z,t); TT = array2table(c,'VariableNames',{'Count','Range','Mean','Start','End'})```\n```TT=7×5 table Count Range Mean Start End _____ _____ ____ _____ ___ 0.5 3 -0.5 0 1 0.5 4 -1 1 3 1 4 1 5 6 0.5 8 1 3 4 1 6 1 10 13 0.5 9 0.5 4 8 0.5 10 1 8 15 ```\n\nUse `rainflow` without output arguments to display a histogram of cycles as a function of cycle average and cycle range.\n\n`rainflow(Z,t)`", null, "Generate a random signal sampled at 100 Hz for 100 seconds. Store the signal and its time information in a timetable.\n\n```fs = 100; t = seconds(0:1/fs:100-1/fs)'; x = randn(size(t)); TT = timetable(t,x);```\n\nDisplay the reversals and the rainflow matrix of the signal.\n\n`rainflow(TT)`", null, "Generate a set of extrema resembling load reversals. Plot the data.\n\n```X = [-2 1 -3 5 -1 3 -4 4 -2]'; plot(X) xlabel('Sample Index') ylabel('Stress')```", null, "Compute cycle counts for the data. Specify that the input consists of already identified extrema.\n\n`[C,hist,edges] = rainflow(X,'ext');`\n\nDisplay a histogram of cycle counts as a function of stress range.\n\n```histogram('BinEdges',edges','BinCounts',sum(hist,2)) xlabel('Stress Range') ylabel('Cycle Counts')```", null, "Use `rainflow` without output arguments to display a histogram of cycles as a function of cycle average and cycle range.\n\n`rainflow(X,'ext')`", null, "## Input Arguments\n\ncollapse all\n\nLoad time history, specified as a vector. `x` must have finite values.\n\nData Types: `single` | `double`\n\nSample rate, specified as a positive real scalar.\n\nData Types: `single` | `double`\n\nTime values, specified as a vector, a `duration` array, or a `duration` scalar representing the time interval between samples.\n\nExample: `seconds(0:1/100:1)` is a `duration` array representing 1 second of sampling at 100 Hz.\n\nData Types: `single` | `double` | `duration`\n\nLoad time history, specified as a timetable. `xt` must contain increasing, finite row times. The timetable must contain only one numeric data vector with finite load values.\n\nIf a timetable has missing or duplicate time points, you can fix it using the tips in Clean Timetable with Missing, Duplicate, or Nonuniform Times (MATLAB).\n\nExample: `timetable(seconds(0:4)',rand(5,1))` specifies a random variable sampled at 1 Hz for 4 seconds.\n\nData Types: `single` | `double`\n\n## Output Arguments\n\ncollapse all\n\nCycle counts, returned as a matrix. `c` contains cycle information in its columns in this order: counts, ranges, mean values, initial sample indices, and final sample indices. See Algorithms for an example. If you specify a sample rate, a time interval, or a vector of time values, then the last two columns of `c` contain initial and final cycle times. If you call `rainflow` with a timetable as input, then the last two columns express the initial and final cycle times in seconds.\n\nRainflow matrix. The rows of `rm` correspond to cycle range, and the columns correspond to cycle mean.\n\nHistogram bin edges, returned as vectors. `rmr` and `rmm` contain the bin edges of the rows and columns of `rm`, respectively.\n\nLinear indices of reversals, returned as a vector.\n\n## Algorithms\n\nFatigue analysis studies how damage accumulates in an object subjected to cyclical changes in stress. The number of cycles necessary to break the object depends on the cycle amplitude. Broadband input excitation contains cycles of diverse amplitude, and the presence of hysteresis in the object has the effect of nesting some cycles within others, either completely or partially. Rainflow counting estimates the number of load change cycles as a function of cycle amplitude.\n\nInitially, `rainflow` turns the load history into a sequence of reversals. Reversals are the local minima and maxima where the load changes sign. The function counts cycles by considering a moving reference point of the sequence, Z, and a moving ordered three-point subset with these characteristics:\n\n1. The first and second points are collectively called Y.\n\n2. The second and third points are collectively called X.\n\n3. In both X and Y, the points are sorted from earlier to later in time, but are not necessarily consecutive in the reversal sequence.\n\n4. The range of X, denoted by r(X), is the absolute value of the difference between the amplitude of the first point and the amplitude of the second point. The definition of r(Y) is analogous.\n\nThe `rainflow` algorithm is as follows:", null, "At the end, the function collects the different cycles and half-cycles and tabulates their ranges, their means, and the points at which they start and end. This information can then be used to produce a histogram of cycles.\n\nConsider the following reversal sequence:", null, "StepZReversalsThree Reversals?Yr(Y)Xr(X)r(X) < r(Y)?Z in Y?Actions\n1AA, B, CYesAB3BC4NoYes\n1. Count AB as ½ cycle.\n\n3. Set Z to B.\n\n3BB, C, DYesBC4CD8NoYes\n1. Count BC as ½ cycle.\n\n3. Set Z to C.\n\n7CC, D, E, F, GYesEF4FG7NoNo\n1. Count EF as 1 cycle.\n\n8CC, D, GYesCD8DG9NoYes\n1. Count CD as ½ cycle.\n\n3. Set Z to D.\n\n12DD, G, H, J, KYesHJ7JK4YesRead L.\n13DD, G, H, J, K, LYesJK4KL3YesRead M.\n14DD, G, H, J, K, L, MYesKL3LM5NoNo\n1. Count KL as 1 cycle.\n\n15DD, G, H, J, MYesHJ7JM5YesRead N.\n16DD, G, H, J, M, NYesJM5MN1YesRead P.\n17DD, G, H, J, M, N, PYesMN1NP4NoNo\n1. Count MN as 1 cycle.\n\n18DD, G, H, J, PYesHJ7JP9NoNo\n1. Count HJ as 1 cycle.\n\n19DD, G, PYesDG9GP10NoYes\n1. Count DG as ½ cycle.\n\n3. Set Z to G.\n\n20GG, POut of data\n\nCount GP as ½ cycle.\n\nNow collect the results.\n\nCycle CountRangeMeanStartEnd\n½3–0.5AB\n½4–1BC\n141EF\n½81CD\n13–0.5KL\n112.5MN\n170.5HJ\n½90.5DG\n½101GP\n\nCompare this to the result of running `rainflow` on the sequence:\n\n`q = rainflow([-2 1 -3 5 -1 3 -4 4 -3 1 -2 3 2 6])`\n```q = 0.5000 3.0000 -0.5000 1.0000 2.0000 0.5000 4.0000 -1.0000 2.0000 3.0000 1.0000 4.0000 1.0000 5.0000 6.0000 0.5000 8.0000 1.0000 3.0000 4.0000 1.0000 3.0000 -0.5000 10.0000 11.0000 1.0000 1.0000 2.5000 12.0000 13.0000 1.0000 7.0000 0.5000 8.0000 9.0000 0.5000 9.0000 0.5000 4.0000 7.0000 0.5000 10.0000 1.0000 7.0000 14.0000```\n\n ASTM E1049-85(2017), \"Standard Practices for Cycle Counting in Fatigue Analysis.\" West Conshohocken, PA: ASTM International, 2011, https://www.astm.org/cgi-bin/resolver.cgi?E1049." ]

[ null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsWithKnownSampleRateExample_01.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsWithKnownSampleRateExample_02.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsWithKnownSampleRateExample_03.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsWithKnownTimeValuesExample_01.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsWithKnownTimeValuesExample_02.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsOfTimetableExample_01.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsOfIdentifiedExtremaExample_01.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsOfIdentifiedExtremaExample_02.png", null, "https://in.mathworks.com/help/examples/signal/win64/CycleCountsOfIdentifiedExtremaExample_03.png", null, "https://in.mathworks.com/help/signal/ref/flowchartyx.png", null, "https://in.mathworks.com/help/signal/ref/reversalsp.png", null ]

[ "# Document (#20070)\n\nAuthor\nSapiie, J.\nHopkins, A.\nTitle\nDeveloping an Internet reference guide\nSource\nUnabashed librarian. 1998, no.106, S.7-8\nYear\n1998\nAbstract\nPresents a brief guide to the development of an Internet reference guide for use in typical public libraries as a tool for assisting Internet and WWW users\nTheme\nInternet\nInformationsdienstleistungen\n\n## Similar documents (author)\n\n1. Hopkins, J.: ¬The 1791 French cataloging code and the origins of the card catalog (1992) 5.75\n```5.74806 = sum of:\n5.74806 = weight(author_txt:hopkins in 4434) [ClassicSimilarity], result of:\n5.74806 = fieldWeight in 4434, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.196897 = idf(docFreq=11, maxDocs=43556)\n0.625 = fieldNorm(doc=4434)\n```\n2. Hopkins, R.L.: Countering information overload : the role of the librarian (1995) 5.75\n```5.74806 = sum of:\n5.74806 = weight(author_txt:hopkins in 2686) [ClassicSimilarity], result of:\n5.74806 = fieldWeight in 2686, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.196897 = idf(docFreq=11, maxDocs=43556)\n0.625 = fieldNorm(doc=2686)\n```\n3. Hopkins, J.: USMARC as metadata shell (1999) 5.75\n```5.74806 = sum of:\n5.74806 = weight(author_txt:hopkins in 1931) [ClassicSimilarity], result of:\n5.74806 = fieldWeight in 1931, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.196897 = idf(docFreq=11, maxDocs=43556)\n0.625 = fieldNorm(doc=1931)\n```\n4. Hopkins, J.: ¬The community of catalogers : its role in the education of cataloger (2002) 5.75\n```5.74806 = sum of:\n5.74806 = weight(author_txt:hopkins in 467) [ClassicSimilarity], result of:\n5.74806 = fieldWeight in 467, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.196897 = idf(docFreq=11, maxDocs=43556)\n0.625 = fieldNorm(doc=467)\n```\n5. Hopkins, M.E.; Zavalina, O.L.: Evaluating physicians' serendipitous knowledge discovery in online discovery systems : a new approach (2019) 4.60\n```4.5984483 = sum of:\n4.5984483 = weight(author_txt:hopkins in 2128) [ClassicSimilarity], result of:\n4.5984483 = fieldWeight in 2128, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.196897 = idf(docFreq=11, maxDocs=43556)\n0.5 = fieldNorm(doc=2128)\n```\n\n## Similar documents (content)\n\n```0.31835806 = sum of:\n0.31835806 = product of:\n1.0346637 = sum of:\n0.056152906 = weight(abstract_txt:libraries in 3175) [ClassicSimilarity], result of:\n0.056152906 = score(doc=3175,freq=1.0), product of:\n0.09504438 = queryWeight, product of:\n1.0585641 = boost\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.023745622 = queryNorm\n0.5908072 = fieldWeight in 3175, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.15625 = fieldNorm(doc=3175)\n0.10234571 = weight(abstract_txt:public in 3175) [ClassicSimilarity], result of:\n0.10234571 = score(doc=3175,freq=1.0), product of:\n0.14181584 = queryWeight, product of:\n1.2930529 = boost\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.023745622 = queryNorm\n0.7216804 = fieldWeight in 3175, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.15625 = fieldNorm(doc=3175)\n0.27724233 = weight(abstract_txt:internet in 3175) [ClassicSimilarity], result of:\n0.27724233 = score(doc=3175,freq=3.0), product of:\n0.2755823 = queryWeight, product of:\n3.1220527 = boost\n3.7172995 = idf(docFreq=2876, maxDocs=43556)\n0.023745622 = queryNorm\n1.0060236 = fieldWeight in 3175, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.7172995 = idf(docFreq=2876, maxDocs=43556)\n0.15625 = fieldNorm(doc=3175)\n0.5989228 = weight(abstract_txt:guide in 3175) [ClassicSimilarity], result of:\n0.5989228 = score(doc=3175,freq=1.0), product of:\n0.66420066 = queryWeight, product of:\n4.846902 = boost\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.023745622 = queryNorm\n0.9017197 = fieldWeight in 3175, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.15625 = fieldNorm(doc=3175)\n0.30769232 = coord(4/13)\n```\n2. Faster, C.L.: Subject access to periodicals : a Main-Mac Combo (1990) 0.31\n```0.31187433 = sum of:\n0.31187433 = product of:\n0.81087327 = sum of:\n0.037871346 = weight(abstract_txt:users in 4118) [ClassicSimilarity], result of:\n0.037871346 = score(doc=4118,freq=1.0), product of:\n0.0848188 = queryWeight, product of:\n3.5719764 = idf(docFreq=3326, maxDocs=43556)\n0.023745622 = queryNorm\n0.44649705 = fieldWeight in 4118, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.5719764 = idf(docFreq=3326, maxDocs=43556)\n0.125 = fieldNorm(doc=4118)\n0.044922326 = weight(abstract_txt:libraries in 4118) [ClassicSimilarity], result of:\n0.044922326 = score(doc=4118,freq=1.0), product of:\n0.09504438 = queryWeight, product of:\n1.0585641 = boost\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.023745622 = queryNorm\n0.47264576 = fieldWeight in 4118, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.125 = fieldNorm(doc=4118)\n0.101792336 = weight(abstract_txt:tool in 4118) [ClassicSimilarity], result of:\n0.101792336 = score(doc=4118,freq=1.0), product of:\n0.16396898 = queryWeight, product of:\n1.3903841 = boost\n4.966419 = idf(docFreq=824, maxDocs=43556)\n0.023745622 = queryNorm\n0.6208024 = fieldWeight in 4118, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.966419 = idf(docFreq=824, maxDocs=43556)\n0.125 = fieldNorm(doc=4118)\n0.14714901 = weight(abstract_txt:reference in 4118) [ClassicSimilarity], result of:\n0.14714901 = score(doc=4118,freq=1.0), product of:\n0.26411915 = queryWeight, product of:\n2.495565 = boost\n4.4570494 = idf(docFreq=1372, maxDocs=43556)\n0.023745622 = queryNorm\n0.5571312 = fieldWeight in 4118, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.4570494 = idf(docFreq=1372, maxDocs=43556)\n0.125 = fieldNorm(doc=4118)\n0.47913826 = weight(abstract_txt:guide in 4118) [ClassicSimilarity], result of:\n0.47913826 = score(doc=4118,freq=1.0), product of:\n0.66420066 = queryWeight, product of:\n4.846902 = boost\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.023745622 = queryNorm\n0.72137576 = fieldWeight in 4118, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.125 = fieldNorm(doc=4118)\n0.3846154 = coord(5/13)\n```\n3. Falk, G.: Grassroots sharing : organizing to improve reference service through cooperative projects (1990) 0.31\n```0.30730125 = sum of:\n0.30730125 = product of:\n0.6658194 = sum of:\n0.023669591 = weight(abstract_txt:users in 3652) [ClassicSimilarity], result of:\n0.023669591 = score(doc=3652,freq=1.0), product of:\n0.0848188 = queryWeight, product of:\n3.5719764 = idf(docFreq=3326, maxDocs=43556)\n0.023745622 = queryNorm\n0.27906066 = fieldWeight in 3652, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.5719764 = idf(docFreq=3326, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.056152906 = weight(abstract_txt:libraries in 3652) [ClassicSimilarity], result of:\n0.056152906 = score(doc=3652,freq=4.0), product of:\n0.09504438 = queryWeight, product of:\n1.0585641 = boost\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.023745622 = queryNorm\n0.5908072 = fieldWeight in 3652, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.029715698 = weight(abstract_txt:development in 3652) [ClassicSimilarity], result of:\n0.029715698 = score(doc=3652,freq=1.0), product of:\n0.098708734 = queryWeight, product of:\n1.0787771 = boost\n3.8533664 = idf(docFreq=2510, maxDocs=43556)\n0.023745622 = queryNorm\n0.30104426 = fieldWeight in 3652, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8533664 = idf(docFreq=2510, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.051172856 = weight(abstract_txt:public in 3652) [ClassicSimilarity], result of:\n0.051172856 = score(doc=3652,freq=1.0), product of:\n0.14181584 = queryWeight, product of:\n1.2930529 = boost\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.023745622 = queryNorm\n0.3608402 = fieldWeight in 3652, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.205647 = weight(abstract_txt:reference in 3652) [ClassicSimilarity], result of:\n0.205647 = score(doc=3652,freq=5.0), product of:\n0.26411915 = queryWeight, product of:\n2.495565 = boost\n4.4570494 = idf(docFreq=1372, maxDocs=43556)\n0.023745622 = queryNorm\n0.7786145 = fieldWeight in 3652, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n4.4570494 = idf(docFreq=1372, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.2994614 = weight(abstract_txt:guide in 3652) [ClassicSimilarity], result of:\n0.2994614 = score(doc=3652,freq=1.0), product of:\n0.66420066 = queryWeight, product of:\n4.846902 = boost\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.023745622 = queryNorm\n0.45085984 = fieldWeight in 3652, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.078125 = fieldNorm(doc=3652)\n0.46153846 = coord(6/13)\n```\n4. Clark, M.: ¬The Texas State Electronic Library (1994) 0.29\n```0.29456407 = sum of:\n0.29456407 = product of:\n0.7658666 = sum of:\n0.03930703 = weight(abstract_txt:libraries in 337) [ClassicSimilarity], result of:\n0.03930703 = score(doc=337,freq=1.0), product of:\n0.09504438 = queryWeight, product of:\n1.0585641 = boost\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.023745622 = queryNorm\n0.41356504 = fieldWeight in 337, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.781166 = idf(docFreq=2698, maxDocs=43556)\n0.109375 = fieldNorm(doc=337)\n0.041601975 = weight(abstract_txt:development in 337) [ClassicSimilarity], result of:\n0.041601975 = score(doc=337,freq=1.0), product of:\n0.098708734 = queryWeight, product of:\n1.0787771 = boost\n3.8533664 = idf(docFreq=2510, maxDocs=43556)\n0.023745622 = queryNorm\n0.42146194 = fieldWeight in 337, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8533664 = idf(docFreq=2510, maxDocs=43556)\n0.109375 = fieldNorm(doc=337)\n0.071642 = weight(abstract_txt:public in 337) [ClassicSimilarity], result of:\n0.071642 = score(doc=337,freq=1.0), product of:\n0.14181584 = queryWeight, product of:\n1.2930529 = boost\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.023745622 = queryNorm\n0.50517625 = fieldWeight in 337, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6187544 = idf(docFreq=1167, maxDocs=43556)\n0.109375 = fieldNorm(doc=337)\n0.19406962 = weight(abstract_txt:internet in 337) [ClassicSimilarity], result of:\n0.19406962 = score(doc=337,freq=3.0), product of:\n0.2755823 = queryWeight, product of:\n3.1220527 = boost\n3.7172995 = idf(docFreq=2876, maxDocs=43556)\n0.023745622 = queryNorm\n0.70421654 = fieldWeight in 337, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.7172995 = idf(docFreq=2876, maxDocs=43556)\n0.109375 = fieldNorm(doc=337)\n0.41924596 = weight(abstract_txt:guide in 337) [ClassicSimilarity], result of:\n0.41924596 = score(doc=337,freq=1.0), product of:\n0.66420066 = queryWeight, product of:\n4.846902 = boost\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.023745622 = queryNorm\n0.6312038 = fieldWeight in 337, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.109375 = fieldNorm(doc=337)\n0.3846154 = coord(5/13)\n```\n5. English, W.: ¬A short primer in conducting searches (1998) 0.28\n```0.2805516 = sum of:\n0.2805516 = product of:\n1.2157236 = sum of:\n0.11579398 = weight(abstract_txt:presents in 2667) [ClassicSimilarity], result of:\n0.11579398 = score(doc=2667,freq=1.0), product of:\n0.123041116 = queryWeight, product of:\n1.2044231 = boost\n4.3021708 = idf(docFreq=1602, maxDocs=43556)\n0.023745622 = queryNorm\n0.9410999 = fieldWeight in 2667, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.3021708 = idf(docFreq=1602, maxDocs=43556)\n0.21875 = fieldNorm(doc=2667)\n0.26143774 = weight(abstract_txt:brief in 2667) [ClassicSimilarity], result of:\n0.26143774 = score(doc=2667,freq=1.0), product of:\n0.21175735 = queryWeight, product of:\n1.5800586 = boost\n5.643932 = idf(docFreq=418, maxDocs=43556)\n0.023745622 = queryNorm\n1.2346101 = fieldWeight in 2667, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.643932 = idf(docFreq=418, maxDocs=43556)\n0.21875 = fieldNorm(doc=2667)\n0.8384919 = weight(abstract_txt:guide in 2667) [ClassicSimilarity], result of:\n0.8384919 = score(doc=2667,freq=1.0), product of:\n0.66420066 = queryWeight, product of:\n4.846902 = boost\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.023745622 = queryNorm\n1.2624075 = fieldWeight in 2667, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.771006 = idf(docFreq=368, maxDocs=43556)\n0.21875 = fieldNorm(doc=2667)\n0.23076923 = coord(3/13)\n```" ]

[ null ]

[ "", null, "# Find Greatest Common Factor\n\nRelated Topics:\nMore Lessons for Grade 6 Math\nMath Worksheets\n\nExamples, videos, worksheets, and solutions to help Grade 6 students learn how to find the greatest common factor using prime factorization and factor trees.\n\nThe following diagram shows an example of finding the greatest common factor using factor trees and prime factorization. Scroll down the page for more examples and solutions on the greatest common factor (GCF).", null, "How to find the greatest common factor by using factor trees?\nThe greatest common factor is the biggest number that divides evenly into each number in a given set of numbers.\nFind the prime factorization of a number and use that to find the greatest common factor between two or three numbers.\nExamples:\n1. Find the greatest common factor of 12 and 8.\n2. Find the greatest common factor of 18 and 21.\n1. Find the greatest common factor of 24, 32 and 40.\n\nGreatest common factor using prime factorization or factor trees\nExample:\nFind the GCF of 18, 20 and 6. Three examples of how to find the greatest common factor of a set of terms using the factor tree\nExample:\nFind the Greatest Common Factor of the following numbers.\n1. 24, 36, 60\n2. 28x2, 424, 49x3\n3. 24a4b2, 60a3b, 144a3b3 How to find the greatest common factor (GCF) using factor trees and exponents?\nExample:\nFind the GCF of 60, 18 and 28.\n\nRotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.\n\nYou can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "" ]

[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/image-files/gcf-factor-trees.png", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]

[ "## Pythagoras Theorem(Application, formulas, proof)\n\nPythagoras theorem gives us relation between among the three sides of right angle triangle. It has three angles so we called its triangle whose right angle that is 90 degrees with respect to other side. There are three side of triangle first one hypotenuse other two side base and perpendicular .If we related this term … Read more\n\n## What is Fraction And Its Types\n\nAny number in from of p/q q=(not zero) which is called as fraction. if we are going in depth fraction has numerator and denominator so p is numerator and q is denominator. if we are looking to here that p=5 and q =6 so in fraction form we can write like 5/6. A fraction is … Read more\n\n## Subtract Decimal Number With Example\n\nhey today we are going to exploring about how to Subtract Decimal number with proper example.so we can understand and calculating fast of decimal number.so here we also going to putting point about how you can add decimal number with example. Let’s get started …. How to Subtract Decimal number Subtracting to decimal number with … Read more\n\n## Power Rule of Derivative: Definition And Example\n\npower rule is very special rule in derivative. Which is derivative of xn n is called power which will we anything number. what is power rule ? Let’s assume some example to know how this derivate formula works for us. What is derivative of x3 So here we are going to apply derivate formula Final … Read more\n\n## Decimal point: Definition with Example\n\nAre you finding what is decimal point in mathematics and calculation so you are landing at right page. because here we explore about what is decimal point with step by step with example. it is dot or point in number that separate the whole number and decimal part number or we can say fraction part … Read more\n\n## What is Decimal(Definition, Example, Conversion)\n\nIn mathematics calculation may be you heard about lots of number system like whole number, natural number, even number, odd number, rotational numbered etc.so in mathematics on of number which is called decimal number and here we will discuss about more details on it. We can define decimal according to its nature that’s containing whole … Read more" ]

[ null ]

[ "# Can multiple hashing weaken the password?\n\nProblem. I want a longer key out of password than my hashing function provide.\n\nIdea. I want to hash password multiple times rotating chars one position each time and use composite key.\n\nQuestion. Do I really weaken the password?\n\nExample:\n\n• Key1= BCrypt(\"carphagen\")=OZrxJo..IAwipd8hHm69IGFiPz/Y/veK\n\n• Key2= BCrypt(\"ncarphage\")=PiG1Qndv8kOHDFoEwhbM.rqtM4O5XNR2\n\n• Key3= BCrypt(\"encarphag\")=Hbrz35n4pwOLsWnMdyuQ7Ze1XMH/QnmC\n\n• . . .\n\n• KeyN= BCrypt(\"rphagenca\")=OapAj6P0OS5P0m20f5oAKY482hEvfLhy\n\nKey = Key1 + Key2 + . . . + KeyN\n\nQuestion. The attacker knows Key, does he get help with password?\n\nSmall Update: BCrypt Salt is the same and is known. It is removed from the hash and not used as a part of the Key\n\n• Hmm, this has flavours of a counter based pseudo random number generator or perhaps CryptGenRandom speculatively. And BCrypt itself is secure... Jun 29, 2017 at 12:04\n• You could use something like PBKDF2 which offers variable output length Jun 29, 2017 at 12:22\n• hunter, thanks. I might take your advice if the Idea is vulnerable. At this point I have invested time in BCrypt and like it Jun 29, 2017 at 12:38\n• I can't say I trivially see how it could weaken the password, but on the other hand, I must say that I do not see any benefits when comparing it with PBKDF2, or even better Argon2\n– Lery\nJun 29, 2017 at 12:40\n• I like BCrypt JavaScript implementation, which I can verify and use in a web page. PBKDF2 can be brute forced with small circuit hardware. In my case I can afford luxury of 15-60 seconds of BCrypt hashing, which would protect from brute force. Jun 29, 2017 at 13:13\n\nDo I really weaken the password?\n\nTo some degree, yes.\n\nThe function is considered public, and the attacker can test random passwords with it and check the output and compare it to your key. Then he finds the correct password, even if he actually just guessed one of those permutations you defined. In your example:\n\n• The key is: key is $Key = Key1 + Key2 + . . . + KeyN$\n• The attacker tries the password ncarphage, which is the passwords permutation to generate $Key2$.\n• The attacker gets as output: $Key2+\\dots$.\n• He still can easily detect that the partial key is the same as in the given output, even if it's not at the same location.\n\nOne more thing to consider: Your idea takes $N$ times the computation time of a single BCrypt call. But an attacker can actually check one password with a single call to BCrypt (if he's not interested in checking the permutated passwords). And that's quite bad - testing a password for the attacker surely should not be less than regular usage.\n\nHere's an alternative idea:\n\n• Use BCrypt to generate $k_0$ from your password - with enough iterations that it fits your requirements.\n• Use a different key derivation function to generate values $k_1,k_2,k_3,\\dots$, by using $k_0$ and a counter as input. If the function utilizes iterations, just use one.\n• If you want to increase the computation time, increase the number of iterations to generate $k_0$, don't apply them to the computation of $k_1,k_2,\\dots$ .\n\nFor the second step, there are various possibilities for a KDF. For example you could use a keyed hash function (also called MAC), a CSPRNG seeded with $k_0$, a PRF, ...\n\nA second alternative: Use another password-based KDF, which supports variable length output. BCrypt is quite old and there are more recent alternatives, two examples would be Argon2 or scrypt.\n\n• But BCrypt adds salt to every hashed password, so testing ncarphage using the salt for Key1 won't give you any useful information unless Key1 and Key2 use the same salt (which is highly unlikely). Jun 29, 2017 at 13:04\n• The salt is known — it's right there in the hash. What I'm saying is that computing the hash of carphagen with salt A tells you nothing about the hash of ncarphage with salt B. Jun 29, 2017 at 13:11\n• @squeamishossifrage You're right about that. But the question didn't mention using different salts for the different BCrypt calls. And the point about the attacker only needing to evaluate one BCrypt call to exclude a possible password is still valid.\n– tylo\nJun 29, 2017 at 13:13\n• Yes, the salt is the same and is known, though it is not used as a part of the key (the salt is removed from the hash) Jun 29, 2017 at 13:42\n\nAn idea I had ( not sure how cryptographically secure it is ) : If $x$ is your password and $k$ is your key then you would compute $hash(x)$ and append that to $k$ then you would compute $hash(x + the previous hash)$ and append that to your key and so on until your key is of adequate length and this system would solve the vulnerability explained in the other answer ( I think 😂 )" ]

[ null ]

[ "# #7fcd7c Color Information\n\nIn a RGB color space, hex #7fcd7c is composed of 49.8% red, 80.4% green and 48.6% blue. Whereas in a CMYK color space, it is composed of 38% cyan, 0% magenta, 39.5% yellow and 19.6% black. It has a hue angle of 117.8 degrees, a saturation of 44.8% and a lightness of 64.5%. #7fcd7c color hex could be obtained by blending #fefff8 with #009b00. Closest websafe color is: #66cc66.\n\n• R 50\n• G 80\n• B 49\nRGB color chart\n• C 38\n• M 0\n• Y 40\n• K 20\nCMYK color chart\n\n#7fcd7c color description : Slightly desaturated lime green.\n\n# #7fcd7c Color Conversion\n\nThe hexadecimal color #7fcd7c has RGB values of R:127, G:205, B:124 and CMYK values of C:0.38, M:0, Y:0.4, K:0.2. Its decimal value is 8375676.\n\nHex triplet RGB Decimal 7fcd7c `#7fcd7c` 127, 205, 124 `rgb(127,205,124)` 49.8, 80.4, 48.6 `rgb(49.8%,80.4%,48.6%)` 38, 0, 40, 20 117.8°, 44.8, 64.5 `hsl(117.8,44.8%,64.5%)` 117.8°, 39.5, 80.4 66cc66 `#66cc66`\nCIE-LAB 75.841, -40.163, 32.938 34.22, 49.628, 26.844 0.309, 0.448, 49.628 75.841, 51.942, 140.645 75.841, -37.981, 50.802 70.447, -36.575, 26.721 01111111, 11001101, 01111100\n\n# Color Schemes with #7fcd7c\n\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #ca7ccd\n``#ca7ccd` `rgb(202,124,205)``\nComplementary Color\n• #a8cd7c\n``#a8cd7c` `rgb(168,205,124)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #7ccda2\n``#7ccda2` `rgb(124,205,162)``\nAnalogous Color\n• #cd7ca8\n``#cd7ca8` `rgb(205,124,168)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #a27ccd\n``#a27ccd` `rgb(162,124,205)``\nSplit Complementary Color\n• #cd7c7f\n``#cd7c7f` `rgb(205,124,127)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #7c7fcd\n``#7c7fcd` `rgb(124,127,205)``\n• #cdca7c\n``#cdca7c` `rgb(205,202,124)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #7c7fcd\n``#7c7fcd` `rgb(124,127,205)``\n• #ca7ccd\n``#ca7ccd` `rgb(202,124,205)``\n• #4ab746\n``#4ab746` `rgb(74,183,70)``\n• #5bbf57\n``#5bbf57` `rgb(91,191,87)``\n• #6dc66a\n``#6dc66a` `rgb(109,198,106)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #91d48e\n``#91d48e` `rgb(145,212,142)``\n• #a3dba1\n``#a3dba1` `rgb(163,219,161)``\n• #b5e2b3\n``#b5e2b3` `rgb(181,226,179)``\nMonochromatic Color\n\n# Alternatives to #7fcd7c\n\nBelow, you can see some colors close to #7fcd7c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #93cd7c\n``#93cd7c` `rgb(147,205,124)``\n• #8dcd7c\n``#8dcd7c` `rgb(141,205,124)``\n• #86cd7c\n``#86cd7c` `rgb(134,205,124)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #7ccd80\n``#7ccd80` `rgb(124,205,128)``\n• #7ccd87\n``#7ccd87` `rgb(124,205,135)``\n• #7ccd8d\n``#7ccd8d` `rgb(124,205,141)``\nSimilar Colors\n\n# #7fcd7c Preview\n\nThis text has a font color of #7fcd7c.\n\n``<span style=\"color:#7fcd7c;\">Text here</span>``\n#7fcd7c background color\n\nThis paragraph has a background color of #7fcd7c.\n\n``<p style=\"background-color:#7fcd7c;\">Content here</p>``\n#7fcd7c border color\n\nThis element has a border color of #7fcd7c.\n\n``<div style=\"border:1px solid #7fcd7c;\">Content here</div>``\nCSS codes\n``.text {color:#7fcd7c;}``\n``.background {background-color:#7fcd7c;}``\n``.border {border:1px solid #7fcd7c;}``\n\n# Shades and Tints of #7fcd7c\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, #040b04 is the darkest color, while #fcfefc is the lightest one.\n\n• #040b04\n``#040b04` `rgb(4,11,4)``\n• #0a190a\n``#0a190a` `rgb(10,25,10)``\n• #10270f\n``#10270f` `rgb(16,39,15)``\n• #163614\n``#163614` `rgb(22,54,20)``\n• #1b441a\n``#1b441a` `rgb(27,68,26)``\n• #21521f\n``#21521f` `rgb(33,82,31)``\n• #276025\n``#276025` `rgb(39,96,37)``\n• #2d6e2a\n``#2d6e2a` `rgb(45,110,42)``\n• #327d30\n``#327d30` `rgb(50,125,48)``\n• #388b35\n``#388b35` `rgb(56,139,53)``\n• #3e993a\n``#3e993a` `rgb(62,153,58)``\n• #44a740\n``#44a740` `rgb(68,167,64)``\n• #49b545\n``#49b545` `rgb(73,181,69)``\n• #55bd51\n``#55bd51` `rgb(85,189,81)``\n• #63c260\n``#63c260` `rgb(99,194,96)``\n• #71c86e\n``#71c86e` `rgb(113,200,110)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #8dd28a\n``#8dd28a` `rgb(141,210,138)``\n• #9bd898\n``#9bd898` `rgb(155,216,152)``\n• #a9dda7\n``#a9dda7` `rgb(169,221,167)``\n• #b6e3b5\n``#b6e3b5` `rgb(182,227,181)``\n• #c4e8c3\n``#c4e8c3` `rgb(196,232,195)``\n• #d2eed1\n``#d2eed1` `rgb(210,238,209)``\n• #e0f3df\n``#e0f3df` `rgb(224,243,223)``\n• #eef8ee\n``#eef8ee` `rgb(238,248,238)``\n• #fcfefc\n``#fcfefc` `rgb(252,254,252)``\nTint Color Variation\n\n# Tones of #7fcd7c\n\nA tone is produced by adding gray to any pure hue. In this case, #9faa9f is the less saturated color, while #52fe4b is the most saturated one.\n\n• #9faa9f\n``#9faa9f` `rgb(159,170,159)``\n• #99b198\n``#99b198` `rgb(153,177,152)``\n• #92b891\n``#92b891` `rgb(146,184,145)``\n• #8cbf8a\n``#8cbf8a` `rgb(140,191,138)``\n• #85c683\n``#85c683` `rgb(133,198,131)``\n• #7fcd7c\n``#7fcd7c` `rgb(127,205,124)``\n• #79d475\n``#79d475` `rgb(121,212,117)``\n• #72db6e\n``#72db6e` `rgb(114,219,110)``\n• #6ce267\n``#6ce267` `rgb(108,226,103)``\n• #65e960\n``#65e960` `rgb(101,233,96)``\n• #5ff059\n``#5ff059` `rgb(95,240,89)``\n• #58f752\n``#58f752` `rgb(88,247,82)``\n• #52fe4b\n``#52fe4b` `rgb(82,254,75)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #7fcd7c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

[ null ]

[ "# Correlation learning law can be represented by equation? A. ∆wij=", null, "Correlation learning law can be represented by equation?\n\nA. ∆wij= µ(si) aj\n\nB. ∆wij= µ(bi – si) aj\n\nC. ∆wij= µ(bi – si) aj Á(xi),where Á(xi) is derivative of xi\n\nD. ∆wij= µ bi aj\n\nI had been asked this question by my school principal while I was bunking the class.\n\nAsked question is from Learning topic in section Basics of Artificial Neural Networks of Neural Networks\n\nRight choice is D. ∆wij= µ bi aj\n\nEasiest explanation: Correlation learning law depends on target output(bi)." ]

[ null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI1MDAiIGhlaWdodD0iNTAwIiB2aWV3Qm94PSIwIDAgNTAwIDUwMCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2NmZDRkYiIvPjwvc3ZnPg==", null ]

[ "# Search Our Content Library\n\n11 filtered results\n11 filtered results\nMultiplication Strategies\nCommon Core\nSort by\nPartial Products Method # 1\nWorksheet\nPartial Products Method # 1\nStudents learn how to use the partial product method, then use it to solve nine multi-digit multiplication problems.\nMath\nWorksheet\nThree Ways to Multiply\nWorksheet\nThree Ways to Multiply\nUse this exercise to teach your students concrete strategies to multiply two-digit factors: base ten arrays, area models, and partial products.\nMath\nWorksheet\nPartial Products & Area Models\nWorksheet\nPartial Products & Area Models\nStudents are able to see the computations involved in multi-digit multiplication through these visuals models.\nMath\nWorksheet\nMoney Math Word Problems\nWorksheet\nMoney Math Word Problems\nLearners will help an outdoor club raise money by using their decimal multiplication skills to solve four word problems.\nMath\nWorksheet\nPartial Products Method #2\nWorksheet\nPartial Products Method #2\nStudents learn how to use the partial product method, then use it to solve nine multi-digit multiplication problems.\nMath\nWorksheet\nBicycle Multiplication Area Models\nWorksheet\nBicycle Multiplication Area Models\nIn this worksheet, children solve multiplication word problems using area models.\nMath\nWorksheet\nShade It In! Multiply Fractions with Area Models\nWorksheet\nShade It In! Multiply Fractions with Area Models\nIn this scaffolded exercise, students will draw and shade area models to multiply fractions.\nMath\nWorksheet\nLet's Go Shopping!\nWorksheet\nLet's Go Shopping!\nPractice percentages by finding the discount prices on this quality merchandise.\nMath\nWorksheet\nArea Model: Multiply Mixed Numbers\nWorksheet\nArea Model: Multiply Mixed Numbers\nUse this scaffolded math exercise to teach your students how to use the area model to multiply mixed numbers.\nMath\nWorksheet\nVocabulary Cards: Describe the Partial Products Method\nWorksheet\nVocabulary Cards: Describe the Partial Products Method\nUse these vocabulary cards with the EL Support Lesson: Describe the Partial Products Method." ]

[ null ]

[ "Range-v3 Range algorithms, views, and actions for the Standard Library", null, "Transformation\n\nTransformation algorithms. More...\n\nlazy\n\n## Typedefs\n\ntemplate<list_like L, typename State , invocable Fn>\nusing meta::accumulate = fold< L, State, Fn >\nAn alias for `meta::fold`. More...\n\ntemplate<list_like ListOfLists>\nusing meta::cartesian_product = reverse_fold< ListOfLists, list< list<> >, quote_trait< detail::cartesian_product_fn > >\nGiven a list of lists `ListOfLists`, return a new list of lists that is the Cartesian Product. Like the `sequence` function from the Haskell Prelude. More...\n\ntemplate<list_like ... Ls>\nusing meta::concat_ = _t< detail::concat_< Ls... > >\nConcatenates several lists into a single list. More...\n\ntemplate<list_like L, integral N>\nusing meta::drop = drop_c< L, N::type::value >\nReturn a new `meta::list` by removing the first `N` elements from `L`. More...\n\ntemplate<list_like L, std::size_t N>\nusing meta::drop_c = _t< detail::drop_< L, N > >\nReturn a new `meta::list` by removing the first `N` elements from `L`. More...\n\ntemplate<typename L , typename Pred >\nusing meta::filter = join< transform< L, detail::filter_< Pred > > >\nReturns a new meta::list where only those elements of `L` that satisfy the Callable `Pred` such that `invoke<Pred,A>::value` is `true` are present. That is, those elements that don't satisfy the `Pred` are \"removed\". More...\n\ntemplate<list_like L, typename State , invocable Fn>\nusing meta::fold = _t< detail::fold_< L, State, Fn > >\nReturn a new `meta::list` constructed by doing a left fold of the list `L` using binary invocable `Fn` and initial state `State`. That is, the `State(N)` for the list element `A(N)` is computed by `Fn(State(N-1), A(N)) -> State(N)`. More...\n\ntemplate<list_like ListOfLists>\nusing meta::join = apply< quote< concat >, ListOfLists >\nJoins a list of lists into a single list. More...\n\ntemplate<list_like L, invocable Fn>\nusing meta::partition = fold< L, pair< list<>, list<> >, detail::partition_< Fn > >\nReturns a pair of lists, where the elements of `L` that satisfy the invocable `Fn` such that `invoke<Fn,A>::value` is `true` are present in the first list and the rest are in the second. More...\n\ntemplate<list_like L>\nusing meta::pop_front = _t< detail::pop_front_< L > >\nReturn a new `meta::list` by removing the first element from the front of `L`. More...\n\ntemplate<list_like L, typename... Ts>\nusing meta::push_back = apply< bind_back< quote< list >, Ts... >, L >\nReturn a new `meta::list` by adding the element `T` to the back of `L`. More...\n\ntemplate<list_like L, typename... Ts>\nusing meta::push_front = apply< bind_front< quote< list >, Ts... >, L >\nReturn a new `meta::list` by adding the element `T` to the front of `L`. More...\n\ntemplate<list_like L, typename T , typename U >\nusing meta::replace = _t< detail::replace_< L, T, U > >\nReturn a new `meta::list` where all instances of type `T` have been replaced with `U`. More...\n\ntemplate<list_like L, typename C , typename U >\nusing meta::replace_if = _t< detail::replace_if_< L, C, U > >\nReturn a new `meta::list` where all elements `A` of the list `L` for which `invoke<C,A>::value` is `true` have been replaced with `U`. More...\n\ntemplate<list_like L>\nusing meta::reverse = _t< detail::reverse_< L > >\nReturn a new `meta::list` by reversing the elements in the list `L`. More...\n\ntemplate<list_like L, typename State , invocable Fn>\nusing meta::reverse_fold = _t< detail::reverse_fold_< L, State, Fn > >\nReturn a new `meta::list` constructed by doing a right fold of the list `L` using binary invocable `Fn` and initial state `State`. That is, the `State(N)` for the list element `A(N)` is computed by `Fn(A(N), State(N+1)) -> State(N)`. More...\n\ntemplate<list_like L, invocable Fn>\nusing meta::sort = _t< detail::sort_< L, Fn > >\nReturn a new `meta::list` that is sorted according to invocable predicate `Fn`. More...\n\ntemplate<typename... Args>\nusing meta::transform = _t< detail::transform_< Args... > >\nReturn a new `meta::list` constructed by transforming all the elements in `L` with the unary invocable `Fn`. `transform` can also be called with two lists of the same length and a binary invocable, in which case it returns a new list constructed with the results of calling `Fn` with each element in the lists, pairwise. More...\n\ntemplate<list_like ListOfLists>\nusing meta::transpose = fold< ListOfLists, repeat_n< size< front< ListOfLists > >, list<> >, bind_back< quote< transform >, quote< push_back > > >\nGiven a list of lists of types `ListOfLists`, transpose the elements from the lists. More...\n\ntemplate<list_like L>\nusing meta::unique = fold< L, list<>, quote_trait< detail::insert_back_ > >\nReturn a new `meta::list` where all duplicate elements have been removed. More...\n\ntemplate<list_like ListOfLists>\nusing meta::zip = transpose< ListOfLists >\nGiven a list of lists of types `ListOfLists`, construct a new list by grouping the elements from the lists pairwise into `meta::list`s. More...\n\ntemplate<invocable Fn, list_like ListOfLists>\nusing meta::zip_with = transform< transpose< ListOfLists >, uncurry< Fn > >\nGiven a list of lists of types `ListOfLists` and an invocable `Fn`, construct a new list by calling `Fn` with the elements from the lists pairwise. More...\n\n## Detailed Description\n\nTransformation algorithms.\n\n## ◆ accumulate\n\ntemplate<list_like L, typename State , invocable Fn>\n using meta::accumulate = typedef fold\n\n`#include <meta/meta.hpp>`\n\nAn alias for `meta::fold`.\n\nComplexity\n`O(N)`.\n\n## ◆ cartesian_product\n\ntemplate<list_like ListOfLists>\n using meta::cartesian_product = typedef reverse_fold >, quote_trait >\n\n`#include <meta/meta.hpp>`\n\nGiven a list of lists `ListOfLists`, return a new list of lists that is the Cartesian Product. Like the `sequence` function from the Haskell Prelude.\n\nComplexity\n`O(N * M)`, where `N` is the size of the outer list, and `M` is the size of the inner lists.\n\n## ◆ concat_\n\ntemplate<list_like ... Ls>\n using meta::concat_ = typedef _t >\n\n`#include <meta/meta.hpp>`\n\nConcatenates several lists into a single list.\n\nPrecondition\nThe parameters must all be instantiations of `meta::list`.\nComplexity\n`O(L)` where `L` is the number of lists in the list of lists.\n\n## ◆ drop\n\ntemplate<list_like L, integral N>\n using meta::drop = typedef drop_c\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by removing the first `N` elements from `L`.\n\nComplexity\n`O(1)`.\n\n## ◆ drop_c\n\ntemplate<list_like L, std::size_t N>\n using meta::drop_c = typedef _t >\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by removing the first `N` elements from `L`.\n\nComplexity\n`O(1)`.\n\n## ◆ filter\n\ntemplate<typename L , typename Pred >\n using meta::filter = typedef join >>\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturns a new meta::list where only those elements of `L` that satisfy the Callable `Pred` such that `invoke<Pred,A>::value` is `true` are present. That is, those elements that don't satisfy the `Pred` are \"removed\".\n\nComplexity\n`O(N)`.\n\n## ◆ fold\n\ntemplate<list_like L, typename State , invocable Fn>\n using meta::fold = typedef _t >\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` constructed by doing a left fold of the list `L` using binary invocable `Fn` and initial state `State`. That is, the `State(N)` for the list element `A(N)` is computed by `Fn(State(N-1), A(N)) -> State(N)`.\n\nComplexity\n`O(N)`.\n\n## ◆ join\n\ntemplate<list_like ListOfLists>\n using meta::join = typedef apply, ListOfLists>\nrelated\n\n`#include <meta/meta.hpp>`\n\nJoins a list of lists into a single list.\n\nPrecondition\nThe parameter must be an instantiation of `meta::list<T`...> where each `T` is itself an instantiation of `meta::list`.\nComplexity\n`O(L)` where `L` is the number of lists in the list of lists.\n\n## ◆ partition\n\ntemplate<list_like L, invocable Fn>\n using meta::partition = typedef fold, list<> >, detail::partition_ >\n\n`#include <meta/meta.hpp>`\n\nReturns a pair of lists, where the elements of `L` that satisfy the invocable `Fn` such that `invoke<Fn,A>::value` is `true` are present in the first list and the rest are in the second.\n\nComplexity\n`O(N)`.\n\n## ◆ pop_front\n\ntemplate<list_like L>\n using meta::pop_front = typedef _t >\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by removing the first element from the front of `L`.\n\nComplexity\n`O(1)`.\n\n## ◆ push_back\n\ntemplate<list_like L, typename... Ts>\n using meta::push_back = typedef apply, Ts...>, L>\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by adding the element `T` to the back of `L`.\n\nComplexity\n`O(1)`.\nNote\n`pop_back` not provided because it cannot be made to meet the complexity guarantees one would expect.\n\n## ◆ push_front\n\ntemplate<list_like L, typename... Ts>\n using meta::push_front = typedef apply, Ts...>, L>\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by adding the element `T` to the front of `L`.\n\nComplexity\n`O(1)`.\n\n## ◆ replace\n\ntemplate<list_like L, typename T , typename U >\n using meta::replace = typedef _t >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` where all instances of type `T` have been replaced with `U`.\n\nComplexity\n`O(N)`.\n\n## ◆ replace_if\n\ntemplate<list_like L, typename C , typename U >\n using meta::replace_if = typedef _t >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` where all elements `A` of the list `L` for which `invoke<C,A>::value` is `true` have been replaced with `U`.\n\nComplexity\n`O(N)`.\n\n## ◆ reverse\n\ntemplate<list_like L>\n using meta::reverse = typedef _t >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` by reversing the elements in the list `L`.\n\nComplexity\n`O(N)`.\n\n## ◆ reverse_fold\n\ntemplate<list_like L, typename State , invocable Fn>\n using meta::reverse_fold = typedef _t >\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` constructed by doing a right fold of the list `L` using binary invocable `Fn` and initial state `State`. That is, the `State(N)` for the list element `A(N)` is computed by `Fn(A(N), State(N+1)) -> State(N)`.\n\nComplexity\n`O(N)`.\n\n## ◆ sort\n\ntemplate<list_like L, invocable Fn>\n using meta::sort = typedef _t >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` that is sorted according to invocable predicate `Fn`.\n\nComplexity\nExpected: `O(N log N)` Worst case: `O(N^2)`.\nstatic_assert(std::is_same_v<L1, list<char, char, char, char, char, char, char>>, \"\");\n_t< detail::sort_< L, Fn > > sort\nReturn a new meta::list that is sorted according to invocable predicate Fn.\nDefinition: meta.hpp:3277\nA wrapper that defers the instantiation of a template C with type parameters Ts in a lambda or let ex...\nDefinition: meta.hpp:787\nA list of types.\nDefinition: meta.hpp:1684\n\n## ◆ transform\n\ntemplate<typename... Args>\n using meta::transform = typedef _t >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` constructed by transforming all the elements in `L` with the unary invocable `Fn`. `transform` can also be called with two lists of the same length and a binary invocable, in which case it returns a new list constructed with the results of calling `Fn` with each element in the lists, pairwise.\n\nComplexity\n`O(N)`.\n\n## ◆ transpose\n\ntemplate<list_like ListOfLists>\n using meta::transpose = typedef fold >, list<> >, bind_back, quote >>\n\n`#include <meta/meta.hpp>`\n\nGiven a list of lists of types `ListOfLists`, transpose the elements from the lists.\n\nComplexity\n`O(N * M)`, where `N` is the size of the outer list, and `M` is the size of the inner lists.\n\n## ◆ unique\n\ntemplate<list_like L>\n using meta::unique = typedef fold, quote_trait >\nrelated\n\n`#include <meta/meta.hpp>`\n\nReturn a new `meta::list` where all duplicate elements have been removed.\n\nComplexity\n`O(N^2)`.\n\n## ◆ zip\n\ntemplate<list_like ListOfLists>\n using meta::zip = typedef transpose\nrelated\n\n`#include <meta/meta.hpp>`\n\nGiven a list of lists of types `ListOfLists`, construct a new list by grouping the elements from the lists pairwise into `meta::list`s.\n\nComplexity\n`O(N * M)`, where `N` is the size of the outer list, and `M` is the size of the inner lists.\n\n## ◆ zip_with\n\ntemplate<invocable Fn, list_like ListOfLists>\n using meta::zip_with = typedef transform, uncurry >\nrelated\n\n`#include <meta/meta.hpp>`\n\nGiven a list of lists of types `ListOfLists` and an invocable `Fn`, construct a new list by calling `Fn` with the elements from the lists pairwise.\n\nComplexity\n`O(N * M)`, where `N` is the size of the outer list, and `M` is the size of the inner lists." ]

[ null, "https://ericniebler.github.io/range-v3/search/mag_sel.svg", null ]

[ "# Number of squares in a finite group\n\nLet $G$ be a finite group and denote by $sq(G)$ the number of squares in $G$ i.e. the number of elements in $G$ which possess a square root. For example, if $G$ is a group of odd order, then $sq(G) =|G|$, since each element has a square root, while at the opposite extreme, $sq(G)= 1$ when $G$ is an elementary abelian 2-group. For the symmetric groups, the values of $sq(\\mathrm{Sym}(n))$ are listed at $\\,$https://oeis.org/A003483 $\\,$. Finally, we note that both the dihedral and quaternion groups of order 8 share the value $sq(G)=2$.\n\nQuestions:\n\n1) Can $sq(G)$ be determined from information in the character table of $G$?\n\n2) Is it true that $sq(\\mathrm{Sym}(n))$ is divisible by every prime which is less than or equal to $n$ for $n>3$?\n\nI answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).\n\nLet $p\\le n$ be prime. Let $C\\subset S_n$ be the group generated by the cycle $(1\\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\\smallsetminus Q^C$ is free, hence $Q\\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer $Z_C$ of $C$ is $C\\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\\{1,\\dots,p\\}$). Thus $Q^C=Z_C\\cap Q$ by definition. If $Q(Z_C)$ is the set of squares of $Z_C$, then clearly $Q(Z_C)\\subset Q^C$.\n\nActually if $p$ is odd, this is an equality (as noticed by \"GH from MO\", this equality is not a tautology and has to be checked). Indeed, if we have an element in $Q^C=Z_C\\cap Q$, its number of $k$-cycles when $k$ is even is the same as its restriction to $\\{p+1,\\dots,n\\}$ (since the possible $p$-cycle does not matter). So $Q^C=C\\times\\mathrm{Sq}(S_{n-p})$. Hence $\\#(Q^C)=p\\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.\n\nIf $p=2$ the previous argument fails at two points, since then $\\mathrm{Sq}(Z_C)=\\{1\\}\\times\\mathrm{Sq}(S_{n-2})$ (instead of $C\\times\\mathrm{Sq}(S_{n-2})$), and since the cycle counting argument falls apart.\n\nSo we assume $n\\ge 4$ and now redefine $C$ as being generated by the 4-cycle $(1\\dots 2^k)$ with $k$ maximal ($k=\\lfloor\\log_2 n\\rfloor$). Then $Q\\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\\times S_{n-4}$. We have $Q(Z_C)=C^2\\times\\mathrm{Sq}(S_{n-4})$ (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Also $Q(Z_C)\\subset Q^C$ and we need to check that it is an equality.\n\nIf some element $w$ belongs to $Q^C=Q\\cap Z_C$ and $m$ is even, then its number $n_m=n_m(w)$ of $m$-cycles is even by assumption. Write $w=uv$ according to the decomposition $Z_C=C\\times S_{n-2^k}$. If $m$ is not a power of $2$, then $u$ has no $m$-cycle, so $n_m(v)=n_m$. If $m$ is a power of 2 and $m<2^k$, then the number of $m$-cycles in $u$ is even (equal to either $2^k/m$ or 0), so $n_m(v)$ is even as well. If $m\\ge 2^k$ we have a contradiction because $n-2^k<2^k$, so the only possibility is that $m=2^k$ and the $m$-cycle is supported by $\\{1,\\dots,2^k\\}$, but then $n_m(w)=1$ and $w$ is not a square. Thus $u$ and $v$ are both squares in $C$ and $S_{n-2^k}$ respectively. So $Q^C=C^2\\times\\mathrm{Sq}(S_{n-4})$, which has even cardinality and thus $Q$ also has even cardinality. [Edit: I modified by initial argument for $p=2$ in which I supposed $k=2$ instead of $k$ maximal, but then $Q(Z_C)\\subset Q^C$ can be strict as we see if $n\\ge 8$ by taking the element $(1234)(5678)$. So the point noticed by \"GH from MO\" is not insignificant.]\n\n• I think it should be remarked (in the case $p>2$) that an element of $Q^C$ is automatically a square within the centralizer of $C$. This is not obvious, because a square-root of an element from $Q^C$ can be outside the centralizer of $C$, e.g. a $2p$-cycle of the form $(1\\tau_1\\dots p\\tau_p)$ is a square-root of $(1\\dots p)(\\tau_1\\dots\\tau_p)$. Oct 17, 2015 at 21:21\n• @GHfromMO Thanks very much. For $p$ odd this is indeed a simple verification, but for $p=2$ I had to modify the argument in a quite unexpected way.\n– YCor\nOct 17, 2015 at 22:56\n• It might be worth remarking that it is not generally true that in a finite group $G$ of even order, the number of squares is even (so there is indeed something to do for $S_{n}$): for example, in a group $G$ which has Sylow $2$-subgroups of exponent $2$, only elements of odd order are squares, and the number of elements of odd order in any finite group $G$ is odd ( as all elements of odd order except the identity occur in mutually inverse pairs). Oct 18, 2015 at 11:27\n• @GeoffRobinson Well, $S_3$ or even the cyclic group of order 6 being counterexamples, I think nobody conjectured such a fact :)\n– YCor\nOct 18, 2015 at 11:42\n\nRevised: The answer to 1 is a qualified \"yes\", though as Frieder Ladisch points out, it might be more accurate to say that the number of squares can be determined by (irreducible) character-theoretic information. The number of square roots of $g \\in G$ ( $G$ a finite group) is given by $\\sum_{\\chi \\in {\\rm Irr}(G)} \\nu(\\chi) \\chi(g)$, where $\\nu(\\chi)$ denotes the Frobenius-Schur indicator (and $\\chi$ runs over all complex irreducible characters of $G$). This formula is well-known (though not, perhaps, as well-known as the special case $g = 1_{G}$). It is a direct consequence of the orthogonality relations and the fact that $\\nu(\\chi) = \\frac{1}{|G|} \\sum_{g \\in G} \\chi(g^{2})$. Hence those $g \\in G$ which have a square root are known from the character table, and the number of such is then known by the orthogonality relations.\n\nRevised again: Returning to Frieder's point and question, it is possible that, although the F-S indicator is not in general determinable from the character table alone, the number of squares in the group still might be ( I do not assert that that IS the case- I just do not know at present- NOW SETTLED (in the negative) in view of Frieder's answer). It might be of interest to note that if we label so that $1 = \\chi_{1}, \\chi_{2},\\ldots \\chi_{r}$ are all the real-valued irreducible characters of $G$ ( these can obviously be identified from the character table), then we know because of the F-S indicator that there is a choice of signs $\\varepsilon_{1},\\ldots,\\varepsilon_{r}$ such that the class function $\\sum_{i=1}^{r} \\varepsilon_{i}\\chi_{i}$ assumes non-negative integer values everywhere - we could ask whether there is more than one choice of signs with this non-negativity property: I also find it interesting (though elementary) to note that the \"expected number\" of square roots of an element of any finite group $G$ is $1$. The last statement is just saying that if we sum the number of square roots of $g$ over all $g \\in G$, we obtain $|G|$, because the \"square-root\" counting function $\\sum_{\\chi} \\nu(\\chi)\\chi$ contains the trivial character with multiplicity one. LATER NOTE: While the expected number of square roots of $g \\in G$ is $1$, the variance of the distribution is easily checked to be $k(G)-1$, where $k(G)$ is the number of conjugacy classes of $G$.\n\nI do not know the answer to 2, though others might.\n\n• In a very strict sense, we cannot compute the Frobenius-Schur indicator from the character table alone, since we can not determine the conjugacy class containing the squares of the elements in a given conjugacy class, if we have only the character table. An example would be the dihedral group and the quaternion group of order $8$. I wonder if the answer to 1) is \"Yes\" in this more strict sense? Oct 17, 2015 at 18:17\n• I have slightly revised my answer to take account of your useful comment. Oct 18, 2015 at 11:19\n• I deleted the other comment, too. By the way, isn't the last statement on the \"expected number of squares\" essentially clear, because every $g\\in G$ is the square root of exactly one element (namely $g^2$)? Oct 21, 2015 at 11:51\n• I suppose you could look at it that way. Oct 21, 2015 at 12:00\n\nThe answer to Question 1) is \"No\" (at least when \"information from the character table\" is understood as only the matrix containing the values in the character table). The following is an example of two groups with \"the same\" character table, but different numbers of squares. Define $$G_1 := \\langle a_1, b_1, c_1\\mid a_1^2=b_1^4=c_1^8=1,\\; b_1^{a_1}=b_1^{-1},\\; c_1^{a_1} = c_1^3,\\; c_1^{b_1} = c_1^{-1}\\; \\rangle$$ and $$G_2 := \\langle a_2, b_2, c_2\\mid a_2^2=b_2^4=c_2^8=1,\\; b_2^{a_2}=b_2^{-1},\\; c_2^{a_2} = c_2^{-1},\\; c_2^{b_2} = c_2^3 \\; \\rangle .$$ Both groups are a semidirect product of $D_8 \\cong \\langle a_i, b_i \\rangle$ acting on $C_8\\cong \\langle c_i \\rangle$, but the action is different. (These are $\\texttt{SmallGroup(64,149)}$ and $\\texttt{SmallGroup(64,150)}$ from GAP's Small Groups Library.)\n\nThen $G_1$ has $5$ squares, namely $\\langle c_1^2 \\rangle \\cup \\{b_1^2\\}$, but $G_2$ has $6$ squares, namely $\\langle c_2^2 \\rangle \\cup\\{b_2^2, b_2^2c_2^4 =(b_2c_2)^2\\}$, as one sees by computing a \"general square\" $(a_i^k b_i^l c_i^m)^2$ in both groups.\n\nFinally, one can check (even without GAP) that the bijection $G_1\\to G_2$ sending $a_1^ib_1^jc_1^k$ to $a_2^i b_2^j c_2 ^k$ induces a bijection between $\\DeclareMathOperator{\\Irr}{Irr} \\Irr(G_1)$ and $\\Irr(G_2)$ (the sets of irreducible characters, viewed as functions on $G_i$): The bijection between the groups yields an isomorphism $G_1/\\langle c_1^4 \\rangle \\cong G_2/ \\langle c_2^4 \\rangle$, which takes care of the characters with $c_i^4$ in the kernel. In both groups, the remaining characters are two characters of degree $4$ which vanish outside the center $\\mathbf{Z}(G_i)=\\langle b_i^2, c_i^4 \\rangle$, so that the bijection above maps these characters onto each other: Indeed, $G_i/\\langle b_i \\rangle$ is isomorphic to the holomorph of $C_8$ in both cases, which has a unique faithful irred. character of degree $4$. The groups $G_1/\\langle b_1^2c_1^4\\rangle$ and $G_2/ \\langle b_2^2 c_2^4 \\rangle$ are not isomorphic, but both have a unique character of degree $4$ and central type.\n(The $\\chi\\in \\Irr(G_1/ \\langle b_1^2 c_1^4 \\rangle)$ with $\\chi(1)=4$ has Frobenius-Schur indicator $\\nu(\\chi)=-1$, but the $\\chi\\in \\Irr(G_2/\\langle b_2^2c_2^4\\rangle)$ with $\\chi(1)=4$ has $\\nu(\\chi)=+1$. Such a thing must happen somewhere, as can be seen from the answer of Geoff Robinson.)\n\nThis is a supplement to YCor's answer, more like an extended comment.\n\nHere is a different argument that, for $n\\geq 4$, the number of squares in $S_n$ is even. By pairing each square with its inverse, we can restrict to the squares whose square is the identity. These are precisely the elements which are the product of an even number of pairwise disjoint transpositions. So let $e_n$ (resp. $o_n$) denote the number of elements of $S_n$ that are the product of an even (resp. odd) number of pairwise disjoint transpositions. Then, it suffices to show that $e_n$ and $o_n$ are both even for $n\\geq 4$. This is easy by induction based on the following recursion for $n\\geq 3$: \\begin{align*} e_n&=e_{n-1}+(n-1)o_{n-2}\\\\ o_n&=o_{n-1}+(n-1)e_{n-2}\\end{align*} The recursion is straightforward to verify by distinguishing on whether a given element of $\\{1,\\dots,n\\}$ participates in a transposition or not.\n\n• Yes it's indeed simpler. I also checked that for $n\\ge 8$ the number of squares if divisible by 4, but the proof is quite complicated and possible generalizations (e.g. studying being divisible by 8 for $n\\gg 1$ [$n\\ge 8$ seems OK]) would be even more messy. It would be interesting if there is a simpler approach.\n– YCor\nOct 18, 2015 at 11:49" ]

[ null ]

[ "THERMO Spoken Here! ~ J. Pohl ©  ( A3850~3/15) ( A3990 -  1.14 Mu of Earth)\n\n# 1.13  Steps to Integrate\n\nOf calculus, at this beginning level, we need know only the definition and meaning of Newton's derivative. Physical laws for thermodynamics are expressed as first order ordinary differential equations. Presented below is a simple, time-dependent first order differential equation with its initial condition. We solve the differential equation then show (in reverse) that the original differential equation to our solution.\n\nEquation with initial condition:", null, "We need to be able to solve the above type equation. All others in thermo and fluids are much the same except for other notations for terms. The steps below are the same ALWAYS.\"", null, "Above we see the differential equation and the steps to obtain its solution subject to the prescribed initial condition.\n\n\"Tell me,\" you might be asked. \"Given this solution (the one above) what differential equation belongs to it?\n\nSo you go at the solution as below and create the differential equation it belongs to ~ as shown.", null, "## 1.13 Steps to Integrate\n\nOf calculus, at this level, we need know only the definition and meaning of Newton's derivative. Physical laws for thermodynamics are expressed as first order ordinary differential equations. Presented below is a simple, time-dependent first order differential equation with its initial condition. We solve the differential equation then we show (in reverse) the differential equation that belongs to our solution. This is the differential equation we solved in the first place, of course.\n\nPremise presently unwritted!" ]

[ null, "https://thermospokenhere.com/wp/01_tsh/A3850_1.13_steps_integrate/diff_eq_master_a.gif", null, "https://thermospokenhere.com/wp/01_tsh/A3850_1.13_steps_integrate/diff_eq_master_b.gif", null, "https://thermospokenhere.com/wp/01_tsh/A3850_1.13_steps_integrate/diff_eq_master_c.gif", null ]

[ "Enable text based alternatives for graph display and drawing entry\n\nTry Another Version of These Questions\n\nQuestion 1\n\nThe graph below shows the AD-AS diagram for the US.\n\nSuppose that the economy is initially in long-run equilibrium with the price level of 800.\n\nWhat is the new GDP in the short-run as a result of this shift?\n\nHint\n\nQuestion 2\n\nWhat is the new price level in the short-run as a result of this shift?\n\nHint\n\nQuestion 3\n\nWhat is the price level in the new long-run equilibrium as a result of this shift?\n\nHint\n\nQuestion 4\n\nWhat is GDP in the new long-run equilibrium as a result of this shift?\n\nHint\n\nQuestion 5\n\nWhat causes the economy to move from the short-run equilibrium to the new long-run equilibrium?\n\nHint" ]

[ null ]

[ "# Main CLV Formula\n\nLet’s look at the main CLV formula is two ways – the first way in words and then as a CLV equation (see separate article on another one of my websites). As you will see, the main customer lifetime value formula is an extension of the simple CLV formula. The main changes are that the main CLV formula looks at each year of customer revenues and costs on an individual basis. This allows different numbers to be utilized each year. The main customer lifetime value formula also uses a discount rate to determine the present value of future revenues and costs.\n\nThe simple CLV formula is:\n\n• Annual profit contribution per customer X\n• Number of years that they remain a customer less\n• The initial cost of customer acquisition\n\nWhereas this CLV formula is:\n\n• Annual profit contribution per customer (for each year under consideration) X\n• The cumulative customer retention rate less\n• The initial cost of customer acquisition\n• With each yearly figure adjusted by an appropriate discount rate\n\nNote: the second line formula – referring to churn rate – is actually calculating customer’s average lifetime period in years.\n\n## Main differences between the two CLV formulas\n\nThe main differences between the simple version and the full version of the CLV formulas are that the full customer lifetime value formula allows for:\n\n• Changing customer revenues each year,\n• Changing customer retention and up selling costs over time,\n• Changing retention (and therefore churn rates) over time, and\n• It applies a discount rate (to determine the present value).\n\n## An example for the main customer lifetime value formula\n\nLet’s assume the following:\n\n• Profit (customer revenues less costs) generated by the customer in year one = \\$1,000\n• This increases to \\$1,500 in year two and then to a maximum of \\$2,000 for year three onwards\n• The retention rate of customers is 75% (and therefore the churn rate is 25%)\n• Cost to acquire the customer = \\$1,000\n• A discount rate of 10% is considered appropriate\n##### The calculation of CLV (BEFORE discounting) would be:\n• Year 0 = – \\$1,000 acquisition costs\n• Year 1 = \\$1,000 customer profit\n• Year 2 = \\$1,500 customer profit X 75% retention = \\$1,125\n• Year 3 = \\$2,000 customer profit X 56% (75% of 75%) retention = \\$1,125\n• Year 4 = \\$2,000 customer profit X 42% (75% of 75% X 75%) retention = \\$844\n• Year 5 = \\$2,000 customer profit X 32% (75% of 75% X 75% X 75%) retention = \\$633\n• And so on\n\nUsing the free Excel template – the non-discounted CLV = \\$6,125\n\n##### The calculation of CLV (WITH discounting) would be:\n• Year 0 = – \\$1,000 acquisition costs divided by 1 (no discount)\n• Year 1 = \\$1,000 customer profit divided by 1.1 (10% discount) = \\$909\n• Year 2 = \\$1,500 customer profit X 75% retention divided by 1.21 (10% X 10% discount) = \\$930\n• Year 3 = \\$2,000 customer profit X 56% (75% of 75%)  retention divided by 1.32 (10% X 10% X 10% discount) = \\$845\n• And so on\n\nUsing the free Excel template – the discounted CLV = \\$3,995\n\n###### As you can see, the discounting factor has a significant impact on the final customer lifetime value.\n\nPosted on Tags , ," ]

[ null ]

[ "`Valve Design\tPorts = 3/8\" up to 3/4\" BSPP\tPressure Range = 30-210 / 60-350  bar\tFlow max. = 40 up to 120 lpm\t\tRelief valve setting at least 1.3 times the highest expected load\t Technical Specifications\tViscosity = 10-500 cSt\tFilter class = ISO4406:1999 Class 18/16/13\tOil Temperature = -20 to 80 deg. C\tAmbient Temperature = -20 to 50 deg. C For further information please see data sheet. Options (ratio / ports / pressure / flow)\tRatio 4.25:1 = 3/8\" = 30-210 bar = 40 lpm\tRatio 4.25:1 = 3/8\" = 60-350 bar = 40 lpm\tRatio 4.25:1 = 1/2\" = 30-210 bar = 60 lpm\tRatio 4.25:1 = 1/2\" = 60-350 bar = 60 lpm\tRatio 6.2:1 = 3/4\" = 60-350 bar = 120 lpm\tRatio 8:1 = 3/8\" = 30-210 bar = 40 lpm\tRatio 8:1 = 3/8\" = 60-350 bar = 40 lpm\tRatio 8:1 = 1/2\" = 30-210 bar = 60 lpm\tRatio 8:1 = 1/2\" = 60-350 bar = 60 lpm `\n\n# Single Counterbalance Valve Inline, 4 Ports\n\nAU\\$150.00 Regular Price\nAU\\$97.50Sale Price\nPorts [BSPP]\nPressure Range [bar]\nFlow max. [lpm]\n###### CALL US\n\n+61 (07) 3274 3073\n\n+61 (07) 3274 3074\n\n###### OPENING HOURS\n\n8am to 4pm (Mon to Fri)" ]

[ null ]

[ "# F1 weighted score about BERT model in pytorch\n\nI have created a function for evaluation a function. It takes as an input the model and validation data loader and return the validation accuracy, validation loss and f1_weighted score.\n\n``````def evaluate(model, val_dataloader):\n\"\"\"\nAfter the completion of each training epoch, measure the model's performance\non our validation set.\n\"\"\"\n# Put the model into the evaluation mode. The dropout layers are disabled during\n# the test time.\nmodel.eval()\n\n# Tracking variables\nval_accuracy = []\nval_loss = []\nf1_weighted = []\n\n# For each batch in our validation set...\nb_input_ids, b_attn_mask, b_labels = tuple(t.to(device) for t in batch)\n\n# Compute logits\n\n# Compute loss\nloss = loss_fn(logits, b_labels)\nval_loss.append(loss.item())\n\n# Get the predictions\npreds = torch.argmax(logits, dim=1).flatten()\n\n# Calculate the accuracy rate\naccuracy = (preds == b_labels).cpu().numpy().mean() * 100\nval_accuracy.append(accuracy)\n\n# Calculate the f1 weighted score\nf1_metric = F1Score('weighted')\nf1_weighted = f1_metric(preds, b_labels)\n\n# Compute the average accuracy and loss over the validation set.\nval_loss = np.mean(val_loss)\nval_accuracy = np.mean(val_accuracy)\nf1_weighted = np.mean(f1_weighted)\n\nreturn val_loss, val_accuracy, f1_weighted\n``````\n\nThe function works well without f1 score, but when it’s inside of the function there is the following error\n\n``````TypeError Traceback (most recent call last)\n<ipython-input-48-0e0f6d227c4f> in <module>()\n1 set_seed(42) # Set seed for reproducibility\n2 bert_classifier, optimizer, scheduler = initialize_model(epochs=4)\n4\n5 #1. 77.28\n\n2 frames\n<__array_function__ internals> in mean(*args, **kwargs)\n\n/usr/local/lib/python3.7/dist-packages/numpy/core/fromnumeric.py in mean(a, axis, dtype, out, keepdims)\n3368 pass\n3369 else:\n-> 3370 return mean(axis=axis, dtype=dtype, out=out, **kwargs)\n3371\n3372 return _methods._mean(a, axis=axis, dtype=dtype,\n\nTypeError: mean() received an invalid combination of arguments - got (out=NoneType, axis=NoneType, dtype=NoneType, ), but expected one of:\n* (*, torch.dtype dtype)\n* (tuple of ints dim, bool keepdim, *, torch.dtype dtype)\n* (tuple of names dim, bool keepdim, *, torch.dtype dtype)\n``````\n\nThe core for f1 score can be found here" ]

[ null ]

[ "E. Trips\ntime limit per test\n2 seconds\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nThere are $n$ persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.\n\nWe want to plan a trip for every evening of $m$ days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:\n\n• Either this person does not go on the trip,\n• Or at least $k$ of his friends also go on the trip.\n\nNote that the friendship is not transitive. That is, if $a$ and $b$ are friends and $b$ and $c$ are friends, it does not necessarily imply that $a$ and $c$ are friends.\n\nFor each day, find the maximum number of people that can go on the trip on that day.\n\nInput\n\nThe first line contains three integers $n$, $m$, and $k$ ($2 \\leq n \\leq 2 \\cdot 10^5, 1 \\leq m \\leq 2 \\cdot 10^5$, $1 \\le k < n$) — the number of people, the number of days and the number of friends each person on the trip should have in the group.\n\nThe $i$-th ($1 \\leq i \\leq m$) of the next $m$ lines contains two integers $x$ and $y$ ($1\\leq x, y\\leq n$, $x\\ne y$), meaning that persons $x$ and $y$ become friends on the morning of day $i$. It is guaranteed that $x$ and $y$ were not friends before.\n\nOutput\n\nPrint exactly $m$ lines, where the $i$-th of them ($1\\leq i\\leq m$) contains the maximum number of people that can go on the trip on the evening of the day $i$.\n\nExamples\nInput\n4 4 22 31 21 31 4\nOutput\n0033\nInput\n5 8 22 14 25 45 24 35 14 13 2\nOutput\n00033445\nInput\n5 7 21 53 22 53 41 25 31 3\nOutput\n0000344\nNote\n\nIn the first example,\n\n• $1,2,3$ can go on day $3$ and $4$.\n\nIn the second example,\n\n• $2,4,5$ can go on day $4$ and $5$.\n• $1,2,4,5$ can go on day $6$ and $7$.\n• $1,2,3,4,5$ can go on day $8$.\n\nIn the third example,\n\n• $1,2,5$ can go on day $5$.\n• $1,2,3,5$ can go on day $6$ and $7$." ]

[ null ]

[ "index 6f862fa..da32d4f 100644 (file)\n@@ -1,68 +1,28 @@\n-/* Copyright (C) 1995-1998 Eric Young ([email protected])\n- * All rights reserved.\n- *\n- * This package is an SSL implementation written\n- * by Eric Young ([email protected]).\n- * The implementation was written so as to conform with Netscapes SSL.\n- *\n- * This library is free for commercial and non-commercial use as long as\n- * the following conditions are aheared to.  The following conditions\n- * apply to all code found in this distribution, be it the RC4, RSA,\n- * lhash, DES, etc., code; not just the SSL code.  The SSL documentation\n- * included with this distribution is covered by the same copyright terms\n- * except that the holder is Tim Hudson ([email protected]).\n- *\n- * Copyright remains Eric Young's, and as such any Copyright notices in\n- * the code are not to be removed.\n- * If this package is used in a product, Eric Young should be given attribution\n- * as the author of the parts of the library used.\n- * This can be in the form of a textual message at program startup or\n- * in documentation (online or textual) provided with the package.\n- *\n- * Redistribution and use in source and binary forms, with or without\n- * modification, are permitted provided that the following conditions\n- * are met:\n- * 1. Redistributions of source code must retain the copyright\n- *    notice, this list of conditions and the following disclaimer.\n- * 2. Redistributions in binary form must reproduce the above copyright\n- *    notice, this list of conditions and the following disclaimer in the\n- *    documentation and/or other materials provided with the distribution.\n- * 3. All advertising materials mentioning features or use of this software\n- *    must display the following acknowledgement:\n- *    \"This product includes cryptographic software written by\n- *     Eric Young ([email protected])\"\n- *    The word 'cryptographic' can be left out if the rouines from the library\n- *    being used are not cryptographic related :-).\n- * 4. If you include any Windows specific code (or a derivative thereof) from\n- *    the apps directory (application code) you must include an acknowledgement:\n- *    \"This product includes software written by Tim Hudson ([email protected])\"\n- *\n- * THIS SOFTWARE IS PROVIDED BY ERIC YOUNG ``AS IS'' AND\n- * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE\n- * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE\n- * ARE DISCLAIMED.  IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE\n- * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL\n- * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS\n- * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)\n- * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT\n- * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY\n- * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF\n- * SUCH DAMAGE.\n+/*\n+ * Copyright 1995-2018 The OpenSSL Project Authors. All Rights Reserved.\n*\n- * The licence and distribution terms for any publically available version or\n- * derivative of this code cannot be changed.  i.e. this code cannot simply be\n- * copied and put under another distribution licence\n- * [including the GNU Public Licence.]\n+ * this file except in compliance with the License.  You can obtain a copy\n+ * in the file LICENSE in the source distribution or at\n*/\n\n#include <stdio.h>\n+#include <limits.h>\n#include \"internal/cryptlib.h\"\n#include <openssl/evp.h>\n#include \"evp_locl.h\"\n+#include \"internal/evp_int.h\"\n+\n+static unsigned char conv_ascii2bin(unsigned char a,\n+                                    const unsigned char *table);\n+static int evp_encodeblock_int(EVP_ENCODE_CTX *ctx, unsigned char *t,\n+                               const unsigned char *f, int dlen);\n+static int evp_decodeblock_int(EVP_ENCODE_CTX *ctx, unsigned char *t,\n+                               const unsigned char *f, int n);\n\n-static unsigned char conv_ascii2bin(unsigned char a);\n#ifndef CHARSET_EBCDIC\n-# define conv_bin2ascii(a)       (data_bin2ascii[(a)&0x3f])\n+# define conv_bin2ascii(a, table)       ((table)[(a)&0x3f])\n#else\n/*\n* We assume that PEM encoded files are EBCDIC files (i.e., printable text\n@@ -70,7 +30,7 @@ static unsigned char conv_ascii2bin(unsigned char a);\n* (text) format again. (No need for conversion in the conv_bin2ascii macro,\n* as the underlying textstring data_bin2ascii[] is already EBCDIC)\n*/\n-# define conv_bin2ascii(a)       (data_bin2ascii[(a)&0x3f])\n+# define conv_bin2ascii(a, table)       ((table)[(a)&0x3f])\n#endif\n\n/*-\n@@ -85,8 +45,13 @@ static unsigned char conv_ascii2bin(unsigned char a);\n#define CHUNKS_PER_LINE (64/4)\n#define CHAR_PER_LINE   (64+1)\n\n-static const unsigned char data_bin2ascii = \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\\\n-abcdefghijklmnopqrstuvwxyz0123456789+/\";\n+static const unsigned char data_bin2ascii =\n+    \"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/\";\n+\n+/* SRP uses a different base64 alphabet */\n+static const unsigned char srpdata_bin2ascii =\n+    \"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz./\";\n+\n\n/*-\n* 0xF0 is a EOLN\n@@ -102,7 +67,7 @@ abcdefghijklmnopqrstuvwxyz0123456789+/\";\n#define B64_WS                  0xE0\n#define B64_ERROR               0xFF\n#define B64_NOT_BASE64(a)       (((a)|0x13) == 0xF3)\n-#define B64_BASE64(a)           !B64_NOT_BASE64(a)\n+#define B64_BASE64(a)           (!B64_NOT_BASE64(a))\n\nstatic const unsigned char data_ascii2bin = {\n0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n@@ -123,20 +88,39 @@ static const unsigned char data_ascii2bin = {\n0x31, 0x32, 0x33, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n};\n\n+static const unsigned char srpdata_ascii2bin = {\n+    0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+    0xFF, 0xE0, 0xF0, 0xFF, 0xFF, 0xF1, 0xFF, 0xFF,\n+    0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+    0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+    0xE0, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+    0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xF2, 0x3E, 0x3F,\n+    0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07,\n+    0x08, 0x09, 0xFF, 0xFF, 0xFF, 0x00, 0xFF, 0xFF,\n+    0xFF, 0x0A, 0x0B, 0x0C, 0x0D, 0x0E, 0x0F, 0x10,\n+    0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18,\n+    0x19, 0x1A, 0x1B, 0x1C, 0x1D, 0x1E, 0x1F, 0x20,\n+    0x21, 0x22, 0x23, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+    0xFF, 0x24, 0x25, 0x26, 0x27, 0x28, 0x29, 0x2A,\n+    0x2B, 0x2C, 0x2D, 0x2E, 0x2F, 0x30, 0x31, 0x32,\n+    0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39, 0x3A,\n+    0x3B, 0x3C, 0x3D, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,\n+};\n+\n#ifndef CHARSET_EBCDIC\n-static unsigned char conv_ascii2bin(unsigned char a)\n+static unsigned char conv_ascii2bin(unsigned char a, const unsigned char *table)\n{\nif (a & 0x80)\nreturn B64_ERROR;\n-    return data_ascii2bin[a];\n+    return table[a];\n}\n#else\n-static unsigned char conv_ascii2bin(unsigned char a)\n+static unsigned char conv_ascii2bin(unsigned char a, const unsigned char *table)\n{\na = os_toascii[a];\nif (a & 0x80)\nreturn B64_ERROR;\n-    return data_ascii2bin[a];\n+    return table[a];\n}\n#endif\n\n@@ -149,58 +133,85 @@ void EVP_ENCODE_CTX_free(EVP_ENCODE_CTX *ctx)\n{\nOPENSSL_free(ctx);\n}\n+\n+int EVP_ENCODE_CTX_copy(EVP_ENCODE_CTX *dctx, EVP_ENCODE_CTX *sctx)\n+{\n+    memcpy(dctx, sctx, sizeof(EVP_ENCODE_CTX));\n+\n+    return 1;\n+}\n+\nint EVP_ENCODE_CTX_num(EVP_ENCODE_CTX *ctx)\n{\nreturn ctx->num;\n}\n\n+void evp_encode_ctx_set_flags(EVP_ENCODE_CTX *ctx, unsigned int flags)\n+{\n+    ctx->flags = flags;\n+}\n+\nvoid EVP_EncodeInit(EVP_ENCODE_CTX *ctx)\n{\nctx->length = 48;\nctx->num = 0;\nctx->line_num = 0;\n+    ctx->flags = 0;\n}\n\n-void EVP_EncodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\n+int EVP_EncodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\nconst unsigned char *in, int inl)\n{\nint i, j;\n-    unsigned int total = 0;\n+    size_t total = 0;\n\n*outl = 0;\nif (inl <= 0)\n-        return;\n+        return 0;\nOPENSSL_assert(ctx->length <= (int)sizeof(ctx->enc_data));\nif (ctx->length - ctx->num > inl) {\nmemcpy(&(ctx->enc_data[ctx->num]), in, inl);\nctx->num += inl;\n-        return;\n+        return 1;\n}\nif (ctx->num != 0) {\ni = ctx->length - ctx->num;\nmemcpy(&(ctx->enc_data[ctx->num]), in, i);\nin += i;\ninl -= i;\n-        j = EVP_EncodeBlock(out, ctx->enc_data, ctx->length);\n+        j = evp_encodeblock_int(ctx, out, ctx->enc_data, ctx->length);\nctx->num = 0;\nout += j;\n-        *(out++) = '\\n';\n+        total = j;\n+        if ((ctx->flags & EVP_ENCODE_CTX_NO_NEWLINES) == 0) {\n+            *(out++) = '\\n';\n+            total++;\n+        }\n*out = '\\0';\n-        total = j + 1;\n}\n-    while (inl >= ctx->length) {\n-        j = EVP_EncodeBlock(out, in, ctx->length);\n+    while (inl >= ctx->length && total <= INT_MAX) {\n+        j = evp_encodeblock_int(ctx, out, in, ctx->length);\nin += ctx->length;\ninl -= ctx->length;\nout += j;\n-        *(out++) = '\\n';\n+        total += j;\n+        if ((ctx->flags & EVP_ENCODE_CTX_NO_NEWLINES) == 0) {\n+            *(out++) = '\\n';\n+            total++;\n+        }\n*out = '\\0';\n-        total += j + 1;\n+    }\n+    if (total > INT_MAX) {\n+        /* Too much output data! */\n+        *outl = 0;\n+        return 0;\n}\nif (inl != 0)\nmemcpy(&(ctx->enc_data), in, inl);\nctx->num = inl;\n*outl = total;\n+\n+    return 1;\n}\n\nvoid EVP_EncodeFinal(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl)\n@@ -208,35 +219,43 @@ void EVP_EncodeFinal(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl)\nunsigned int ret = 0;\n\nif (ctx->num != 0) {\n-        ret = EVP_EncodeBlock(out, ctx->enc_data, ctx->num);\n-        out[ret++] = '\\n';\n+        ret = evp_encodeblock_int(ctx, out, ctx->enc_data, ctx->num);\n+        if ((ctx->flags & EVP_ENCODE_CTX_NO_NEWLINES) == 0)\n+            out[ret++] = '\\n';\nout[ret] = '\\0';\nctx->num = 0;\n}\n*outl = ret;\n}\n\n-int EVP_EncodeBlock(unsigned char *t, const unsigned char *f, int dlen)\n+static int evp_encodeblock_int(EVP_ENCODE_CTX *ctx, unsigned char *t,\n+                               const unsigned char *f, int dlen)\n{\nint i, ret = 0;\nunsigned long l;\n+    const unsigned char *table;\n+\n+    if (ctx != NULL && (ctx->flags & EVP_ENCODE_CTX_USE_SRP_ALPHABET) != 0)\n+        table = srpdata_bin2ascii;\n+    else\n+        table = data_bin2ascii;\n\nfor (i = dlen; i > 0; i -= 3) {\nif (i >= 3) {\nl = (((unsigned long)f) << 16L) |\n(((unsigned long)f) << 8L) | f;\n-            *(t++) = conv_bin2ascii(l >> 18L);\n-            *(t++) = conv_bin2ascii(l >> 12L);\n-            *(t++) = conv_bin2ascii(l >> 6L);\n-            *(t++) = conv_bin2ascii(l);\n+            *(t++) = conv_bin2ascii(l >> 18L, table);\n+            *(t++) = conv_bin2ascii(l >> 12L, table);\n+            *(t++) = conv_bin2ascii(l >> 6L, table);\n+            *(t++) = conv_bin2ascii(l, table);\n} else {\nl = ((unsigned long)f) << 16L;\nif (i == 2)\nl |= ((unsigned long)f << 8L);\n\n-            *(t++) = conv_bin2ascii(l >> 18L);\n-            *(t++) = conv_bin2ascii(l >> 12L);\n-            *(t++) = (i == 1) ? '=' : conv_bin2ascii(l >> 6L);\n+            *(t++) = conv_bin2ascii(l >> 18L, table);\n+            *(t++) = conv_bin2ascii(l >> 12L, table);\n+            *(t++) = (i == 1) ? '=' : conv_bin2ascii(l >> 6L, table);\n*(t++) = '=';\n}\nret += 4;\n@@ -244,16 +263,21 @@ int EVP_EncodeBlock(unsigned char *t, const unsigned char *f, int dlen)\n}\n\n*t = '\\0';\n-    return (ret);\n+    return ret;\n+}\n+\n+int EVP_EncodeBlock(unsigned char *t, const unsigned char *f, int dlen)\n+{\n+    return evp_encodeblock_int(NULL, t, f, dlen);\n}\n\nvoid EVP_DecodeInit(EVP_ENCODE_CTX *ctx)\n{\n-    /* Only ctx->num is used during decoding. */\n+    /* Only ctx->num and ctx->flags are used during decoding. */\nctx->num = 0;\nctx->length = 0;\nctx->line_num = 0;\n-    ctx->expect_nl = 0;\n+    ctx->flags = 0;\n}\n\n/*-\n@@ -281,6 +305,7 @@ int EVP_DecodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\n{\nint seof = 0, eof = 0, rv = -1, ret = 0, i, v, tmp, n, decoded_len;\nunsigned char *d;\n+    const unsigned char *table;\n\nn = ctx->num;\nd = ctx->enc_data;\n@@ -297,9 +322,14 @@ int EVP_DecodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\ngoto end;\n}\n\n+    if ((ctx->flags & EVP_ENCODE_CTX_USE_SRP_ALPHABET) != 0)\n+        table = srpdata_ascii2bin;\n+    else\n+        table = data_ascii2bin;\n+\nfor (i = 0; i < inl; i++) {\ntmp = *(in++);\n-        v = conv_ascii2bin(tmp);\n+        v = conv_ascii2bin(tmp, table);\nif (v == B64_ERROR) {\nrv = -1;\ngoto end;\n@@ -339,7 +369,7 @@ int EVP_DecodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\n}\n\nif (n == 64) {\n-            decoded_len = EVP_DecodeBlock(out, d, n);\n+            decoded_len = evp_decodeblock_int(ctx, out, d, n);\nn = 0;\nif (decoded_len < 0 || eof > decoded_len) {\nrv = -1;\n@@ -358,7 +388,7 @@ int EVP_DecodeUpdate(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl,\ntail:\nif (n > 0) {\nif ((n & 3) == 0) {\n-            decoded_len = EVP_DecodeBlock(out, d, n);\n+            decoded_len = evp_decodeblock_int(ctx, out, d, n);\nn = 0;\nif (decoded_len < 0 || eof > decoded_len) {\nrv = -1;\n@@ -377,16 +407,23 @@ end:\n/* Legacy behaviour. This should probably rather be zeroed on error. */\n*outl = ret;\nctx->num = n;\n-    return (rv);\n+    return rv;\n}\n\n-int EVP_DecodeBlock(unsigned char *t, const unsigned char *f, int n)\n+static int evp_decodeblock_int(EVP_ENCODE_CTX *ctx, unsigned char *t,\n+                               const unsigned char *f, int n)\n{\nint i, ret = 0, a, b, c, d;\nunsigned long l;\n+    const unsigned char *table;\n+\n+    if (ctx != NULL && (ctx->flags & EVP_ENCODE_CTX_USE_SRP_ALPHABET) != 0)\n+        table = srpdata_ascii2bin;\n+    else\n+        table = data_ascii2bin;\n\n/* trim white space from the start of the line. */\n-    while ((conv_ascii2bin(*f) == B64_WS) && (n > 0)) {\n+    while ((conv_ascii2bin(*f, table) == B64_WS) && (n > 0)) {\nf++;\nn--;\n}\n@@ -395,19 +432,19 @@ int EVP_DecodeBlock(unsigned char *t, const unsigned char *f, int n)\n* strip off stuff at the end of the line ascii2bin values B64_WS,\n* B64_EOLN, B64_EOLN and B64_EOF\n*/\n-    while ((n > 3) && (B64_NOT_BASE64(conv_ascii2bin(f[n - 1]))))\n+    while ((n > 3) && (B64_NOT_BASE64(conv_ascii2bin(f[n - 1], table))))\nn--;\n\nif (n % 4 != 0)\n-        return (-1);\n+        return -1;\n\nfor (i = 0; i < n; i += 4) {\n-        a = conv_ascii2bin(*(f++));\n-        b = conv_ascii2bin(*(f++));\n-        c = conv_ascii2bin(*(f++));\n-        d = conv_ascii2bin(*(f++));\n+        a = conv_ascii2bin(*(f++), table);\n+        b = conv_ascii2bin(*(f++), table);\n+        c = conv_ascii2bin(*(f++), table);\n+        d = conv_ascii2bin(*(f++), table);\nif ((a & 0x80) || (b & 0x80) || (c & 0x80) || (d & 0x80))\n-            return (-1);\n+            return -1;\nl = ((((unsigned long)a) << 18L) |\n(((unsigned long)b) << 12L) |\n(((unsigned long)c) << 6L) | (((unsigned long)d)));\n@@ -416,7 +453,12 @@ int EVP_DecodeBlock(unsigned char *t, const unsigned char *f, int n)\n*(t++) = (unsigned char)(l) & 0xff;\nret += 3;\n}\n-    return (ret);\n+    return ret;\n+}\n+\n+int EVP_DecodeBlock(unsigned char *t, const unsigned char *f, int n)\n+{\n+    return evp_decodeblock_int(NULL, t, f, n);\n}\n\nint EVP_DecodeFinal(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl)\n@@ -425,12 +467,12 @@ int EVP_DecodeFinal(EVP_ENCODE_CTX *ctx, unsigned char *out, int *outl)\n\n*outl = 0;\nif (ctx->num != 0) {\n-        i = EVP_DecodeBlock(out, ctx->enc_data, ctx->num);\n+        i = evp_decodeblock_int(ctx, out, ctx->enc_data, ctx->num);\nif (i < 0)\n-            return (-1);\n+            return -1;\nctx->num = 0;\n*outl = i;\n-        return (1);\n+        return 1;\n} else\n-        return (1);\n+        return 1;\n}" ]

[ null ]

[ "# TagGaming\n\n## A Great Android Game Exists.\n\nA\n\nNo this isn’t clickbait. I am hooked on a quality android game, Grimvalor. It’s a side-scrolling hack and slash, with Dark Souls type bosses. It’s fair but challenging. Has extremely tight feeling controls, well-designed levels, and enemies as well as a great art style. It also runs nicely on a Snapdragon 720g. Cannot recommend it enough. This is the first game that will appear...\n\n```", null, "```\n``` ```\n```", null, "Hey, my name is Avrohom Yosef Gross but I just go by Ay.Read More```\n\n`Buy Me a Coffee.`\n\n`Or something from my`\n\n`Wishlist.`\n\n#### `Get in Touch`\n\n• `facebook`\n• `instagram`\n• `twitter`\n• `mail`\n``` _stq=window._stq||[];_stq.push([\"view\",{v:'ext',blog:'165895373',post:'0',tz:'2',srv:'pupontech.com',j:'1:12.5'}]);_stq.push([\"clickTrackerInit\",\"165895373\",\"0\"]) !function(t,e){\"object\"==typeof exports&&\"undefined\"!=typeof module?module.exports=e():\"function\"==typeof define&&define.amd?define(e):(t=\"undefined\"!=typeof globalThis?globalThis:t||self).LazyLoad=e()}(this,function(){\"use strict\";function e(){return(e=Object.assign||function(t){for(var e=1;e<arguments.length;e++){var n,a=arguments[e];for(n in a)Object.prototype.hasOwnProperty.call(a,n)&&(t[n]=a[n])}return t}).apply(this,arguments)}function i(t){return e({},it,t)}function o(t,e){var n,a=\"LazyLoad::Initialized\",i=new t(e);try{n=new CustomEvent(a,{detail:{instance:i}})}catch(t){(n=document.createEvent(\"CustomEvent\")).initCustomEvent(a,!1,!1,{instance:i})}window.dispatchEvent(n)}function l(t,e){return t.getAttribute(gt+e)}function c(t){return l(t,bt)}function s(t,e){return function(t,e,n){e=gt+e;null!==n?t.setAttribute(e,n):t.removeAttribute(e)}(t,bt,e)}function r(t){return s(t,null),0}function u(t){return null===c(t)}function d(t){return c(t)===vt}function f(t,e,n,a){t&&(void 0===a?void 0===n?t(e):t(e,n):t(e,n,a))}function _(t,e){nt?t.classList.add(e):t.className+=(t.className?\" \":\"\")+e}function v(t,e){nt?t.classList.remove(e):t.className=t.className.replace(new RegExp(\"(^|\\\\s+)\"+e+\"(\\\\s+|\\$)\"),\" \").replace(/^\\s+/,\"\").replace(/\\s+\\$/,\"\")}function g(t){return t.llTempImage}function b(t,e){!e||(e=e._observer)&&e.unobserve(t)}function p(t,e){t&&(t.loadingCount+=e)}function h(t,e){t&&(t.toLoadCount=e)}function n(t){for(var e,n=[],a=0;e=t.children[a];a+=1)\"SOURCE\"===e.tagName&&n.push(e);return n}function m(t,e){(t=t.parentNode)&&\"PICTURE\"===t.tagName&&n(t).forEach(e)}function a(t,e){n(t).forEach(e)}function E(t){return!!t[st]}function I(t){return t[st]}function y(t){return delete t[st]}function A(e,t){var n;E(e)||(n={},t.forEach(function(t){n[t]=e.getAttribute(t)}),e[st]=n)}function k(a,t){var i;E(a)&&(i=I(a),t.forEach(function(t){var e,n;e=a,(t=i[n=t])?e.setAttribute(n,t):e.removeAttribute(n)}))}function L(t,e,n){_(t,e.class_loading),s(t,ut),n&&(p(n,1),f(e.callback_loading,t,n))}function w(t,e,n){n&&t.setAttribute(e,n)}function x(t,e){w(t,ct,l(t,e.data_sizes)),w(t,rt,l(t,e.data_srcset)),w(t,ot,l(t,e.data_src))}function O(t,e,n){var a=l(t,e.data_bg_multi),i=l(t,e.data_bg_multi_hidpi);(a=at&&i?i:a)&&(t.style.backgroundImage=a,n=n,_(t=t,(e=e).class_applied),s(t,ft),n&&(e.unobserve_completed&&b(t,e),f(e.callback_applied,t,n)))}function N(t,e){!e||0<e.loadingCount||0<e.toLoadCount||f(t.callback_finish,e)}function C(t,e,n){t.addEventListener(e,n),t.llEvLisnrs[e]=n}function M(t){return!!t.llEvLisnrs}function z(t){if(M(t)){var e,n,a=t.llEvLisnrs;for(e in a){var i=a[e];n=e,i=i,t.removeEventListener(n,i)}delete t.llEvLisnrs}}function R(t,e,n){var a;delete t.llTempImage,p(n,-1),(a=n)&&--a.toLoadCount,v(t,e.class_loading),e.unobserve_completed&&b(t,n)}function T(o,r,c){var l=g(o)||o;M(l)||function(t,e,n){M(t)||(t.llEvLisnrs={});var a=\"VIDEO\"===t.tagName?\"loadeddata\":\"load\";C(t,a,e),C(t,\"error\",n)}(l,function(t){var e,n,a,i;n=r,a=c,i=d(e=o),R(e,n,a),_(e,n.class_loaded),s(e,dt),f(n.callback_loaded,e,a),i||N(n,a),z(l)},function(t){var e,n,a,i;n=r,a=c,i=d(e=o),R(e,n,a),_(e,n.class_error),s(e,_t),f(n.callback_error,e,a),i||N(n,a),z(l)})}function G(t,e,n){var a,i,o,r,c;t.llTempImage=document.createElement(\"IMG\"),T(t,e,n),E(c=t)||(c[st]={backgroundImage:c.style.backgroundImage}),o=n,r=l(a=t,(i=e).data_bg),c=l(a,i.data_bg_hidpi),(r=at&&c?c:r)&&(a.style.backgroundImage='url(\"'.concat(r,'\")'),g(a).setAttribute(ot,r),L(a,i,o)),O(t,e,n)}function D(t,e,n){var a;T(t,e,n),a=e,e=n,(t=It[(n=t).tagName])&&(t(n,a),L(n,a,e))}function V(t,e,n){var a;a=t,(-1<yt.indexOf(a.tagName)?D:G)(t,e,n)}function F(t,e,n){var a;t.setAttribute(\"loading\",\"lazy\"),T(t,e,n),a=e,(e=It[(n=t).tagName])&&e(n,a),s(t,vt)}function j(t){t.removeAttribute(ot),t.removeAttribute(rt),t.removeAttribute(ct)}function P(t){m(t,function(t){k(t,Et)}),k(t,Et)}function S(t){var e;(e=At[t.tagName])?e(t):E(e=t)&&(t=I(e),e.style.backgroundImage=t.backgroundImage)}function U(t,e){var n;S(t),n=e,u(e=t)||d(e)||(v(e,n.class_entered),v(e,n.class_exited),v(e,n.class_applied),v(e,n.class_loading),v(e,n.class_loaded),v(e,n.class_error)),r(t),y(t)}function \\$(t,e,n,a){var i;n.cancel_on_exit&&(c(t)!==ut||\"IMG\"===t.tagName&&(z(t),m(i=t,function(t){j(t)}),j(i),P(t),v(t,n.class_loading),p(a,-1),r(t),f(n.callback_cancel,t,e,a)))}function q(t,e,n,a){var i,o,r=(o=t,0<=pt.indexOf(c(o)));s(t,\"entered\"),_(t,n.class_entered),v(t,n.class_exited),i=t,o=a,n.unobserve_entered&&b(i,o),f(n.callback_enter,t,e,a),r||V(t,n,a)}function H(t){return t.use_native&&\"loading\"in HTMLImageElement.prototype}function B(t,i,o){t.forEach(function(t){return(a=t).isIntersecting||0<a.intersectionRatio?q(t.target,t,i,o):(e=t.target,n=t,a=i,t=o,void(u(e)||(_(e,a.class_exited),\\$(e,n,a,t),f(a.callback_exit,e,n,t))));var e,n,a})}function J(e,n){var t;et&&!H(e)&&(n._observer=new IntersectionObserver(function(t){B(t,e,n)},{root:(t=e).container===document?null:t.container,rootMargin:t.thresholds||t.threshold+\"px\"}))}function K(t){return Array.prototype.slice.call(t)}function Q(t){return t.container.querySelectorAll(t.elements_selector)}function W(t){return c(t)===_t}function X(t,e){return e=t||Q(e),K(e).filter(u)}function Y(e,t){var n;(n=Q(e),K(n).filter(W)).forEach(function(t){v(t,e.class_error),r(t)}),t.update()}function t(t,e){var n,a,t=i(t);this._settings=t,this.loadingCount=0,J(t,this),n=t,a=this,Z&&window.addEventListener(\"online\",function(){Y(n,a)}),this.update(e)}var Z=\"undefined\"!=typeof window,tt=Z&&!(\"onscroll\"in window)||\"undefined\"!=typeof navigator&&/(gle|ing|ro)bot|crawl|spider/i.test(navigator.userAgent),et=Z&&\"IntersectionObserver\"in window,nt=Z&&\"classList\"in document.createElement(\"p\"),at=Z&&1<window.devicePixelRatio,it={elements_selector:\".lazy\",container:tt||Z?document:null,threshold:300,thresholds:null,data_src:\"src\",data_srcset:\"srcset\",data_sizes:\"sizes\",data_bg:\"bg\",data_bg_hidpi:\"bg-hidpi\",data_bg_multi:\"bg-multi\",data_bg_multi_hidpi:\"bg-multi-hidpi\",data_poster:\"poster\",class_applied:\"applied\",class_loading:\"litespeed-loading\",class_loaded:\"litespeed-loaded\",class_error:\"error\",class_entered:\"entered\",class_exited:\"exited\",unobserve_completed:!0,unobserve_entered:!1,cancel_on_exit:!0,callback_enter:null,callback_exit:null,callback_applied:null,callback_loading:null,callback_loaded:null,callback_error:null,callback_finish:null,callback_cancel:null,use_native:!1},ot=\"src\",rt=\"srcset\",ct=\"sizes\",lt=\"poster\",st=\"llOriginalAttrs\",ut=\"loading\",dt=\"loaded\",ft=\"applied\",_t=\"error\",vt=\"native\",gt=\"data-\",bt=\"ll-status\",pt=[ut,dt,ft,_t],ht=[ot],mt=[ot,lt],Et=[ot,rt,ct],It={IMG:function(t,e){m(t,function(t){A(t,Et),x(t,e)}),A(t,Et),x(t,e)},IFRAME:function(t,e){A(t,ht),w(t,ot,l(t,e.data_src))},VIDEO:function(t,e){a(t,function(t){A(t,ht),w(t,ot,l(t,e.data_src))}),A(t,mt),w(t,lt,l(t,e.data_poster)),w(t,ot,l(t,e.data_src)),t.load()}},yt=[\"IMG\",\"IFRAME\",\"VIDEO\"],At={IMG:P,IFRAME:function(t){k(t,ht)},VIDEO:function(t){a(t,function(t){k(t,ht)}),k(t,mt),t.load()}},kt=[\"IMG\",\"IFRAME\",\"VIDEO\"];return t.prototype={update:function(t){var e,n,a,i=this._settings,o=X(t,i);{if(h(this,o.length),!tt&&et)return H(i)?(e=i,n=this,o.forEach(function(t){-1!==kt.indexOf(t.tagName)&&F(t,e,n)}),void h(n,0)):(t=this._observer,i=o,t.disconnect(),a=t,void i.forEach(function(t){a.observe(t)}));this.loadAll(o)}},destroy:function(){this._observer&&this._observer.disconnect(),Q(this._settings).forEach(function(t){y(t)}),delete this._observer,delete this._settings,delete this.loadingCount,delete this.toLoadCount},loadAll:function(t){var e=this,n=this._settings;X(t,n).forEach(function(t){b(t,e),V(t,n,e)})},restoreAll:function(){var e=this._settings;Q(e).forEach(function(t){U(t,e)})}},t.load=function(t,e){e=i(e);V(t,e)},t.resetStatus=function(t){r(t)},Z&&function(t,e){if(e)if(e.length)for(var n,a=0;n=e[a];a+=1)o(t,n);else o(t,e)}(t,window.lazyLoadOptions),t});!function(e,t){\"use strict\";function a(){t.body.classList.add(\"litespeed_lazyloaded\")}function n(){console.log(\"[LiteSpeed] Start Lazy Load Images\"),d=new LazyLoad({elements_selector:\"[data-lazyloaded]\",callback_finish:a}),o=function(){d.update()},e.MutationObserver&&new MutationObserver(o).observe(t.documentElement,{childList:!0,subtree:!0,attributes:!0})}var d,o;e.addEventListener?e.addEventListener(\"load\",n,!1):e.attachEvent(\"onload\",n)}(window,document);var litespeed_vary=document.cookie.replace(/(?:(?:^|.*;\\s*)_lscache_vary\\s*\\=\\s*([^;]*).*\\$)|^.*\\$/,\"\");litespeed_vary||fetch(\"/wp-content/plugins/litespeed-cache/guest.vary.php\",{method:\"POST\",cache:\"no-cache\",redirect:\"follow\"}).then(e=>e.json()).then(e=>{console.log(e),e.hasOwnProperty(\"reload\")&&\"yes\"==e.reload&&(sessionStorage.setItem(\"litespeed_docref\",document.referrer),window.location.reload(!0))});const litespeed_ui_events=[\"mouseover\",\"click\",\"keydown\",\"wheel\",\"touchmove\",\"touchstart\"];var urlCreator=window.URL||window.webkitURL;function litespeed_load_delayed_js_force(){console.log(\"[LiteSpeed] Start Load JS Delayed\"),litespeed_ui_events.forEach(e=>{window.removeEventListener(e,litespeed_load_delayed_js_force,{passive:!0})}),document.querySelectorAll(\"iframe[data-litespeed-src]\").forEach(e=>{e.setAttribute(\"src\",e.getAttribute(\"data-litespeed-src\"))}),\"loading\"==document.readyState?window.addEventListener(\"DOMContentLoaded\",litespeed_load_delayed_js):litespeed_load_delayed_js()}litespeed_ui_events.forEach(e=>{window.addEventListener(e,litespeed_load_delayed_js_force,{passive:!0})});async function litespeed_load_delayed_js(){let t=[];for(var d in document.querySelectorAll('script[type=\"litespeed/javascript\"]').forEach(e=>{t.push(e)}),t)await new Promise(e=>litespeed_load_one(t[d],e));document.dispatchEvent(new Event(\"DOMContentLiteSpeedLoaded\")),window.dispatchEvent(new Event(\"DOMContentLiteSpeedLoaded\"))}function litespeed_load_one(t,e){console.log(\"[LiteSpeed] Load \",t);var d=document.createElement(\"script\");d.addEventListener(\"load\",e),d.addEventListener(\"error\",e),t.getAttributeNames().forEach(e=>{\"type\"!=e&&d.setAttribute(\"data-src\"==e?\"src\":e,t.getAttribute(e))});let a=!(d.type=\"text/javascript\");!d.src&&t.textContent&&(d.src=litespeed_inline2src(t.textContent),a=!0),t.after(d),t.remove(),a&&e()}function litespeed_inline2src(t){try{var d=urlCreator.createObjectURL(new Blob([t.replace(/^(?:<!--)?(.*?)(?:-->)?\\$/gm,\"\\$1\")],{type:\"text/javascript\"}))}catch(e){d=\"data:text/javascript;base64,\"+btoa(t.replace(/^(?:<!--)?(.*?)(?:-->)?\\$/gm,\"\\$1\"))}return d} ```" ]

[ null, "data:image/gif;base64,R0lGODdhAQABAPAAAMPDwwAAACwAAAAAAQABAAACAkQBADs=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxMDAiIGhlaWdodD0iMTAwIiB2aWV3Qm94PSIwIDAgMTAwIDEwMCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgc3R5bGU9ImZpbGw6I2ZmZmZmZjtmaWxsLW9wYWNpdHk6IDAuMTsiLz48L3N2Zz4=", null ]

[ "# How to get the catcode of a token?\n\nFor debugging a complicated macro I would like to print out the catcode of a token. Optimally I would like to have a macro \\getcatcode such that, for example, \\getcatcode{a} would expand to 10. How can this be done?\n\nI found lots of information about how to set/change catcodes, but nothing about how to read them.\n\n• BTW: the catcode of a is 11 (letter) not 10 (space). Jun 22, 2011 at 13:22\n\nUse \\catcode together with \\the to get the catcode of the token:\n\n\\the\\catcodea\n\n\nNote: The turns the next character into its ASCII number which is required for \\catcode.\n\nAs custom macro:\n\n\\newcommand{\\getcatcode}{\\the\\catcode#1}\n\n\nSpecial characters must be escaped with a backslash, e.g. % must be written as \\%, # as \\# etc. It doesn't hurt to write normal letters the same way, e.g. \\getcatcode\\a works as well.\n\nAs explained in an exercise of the TeXbook one can also write\n\n\\newcommand{\\printcatcode}{%\n\\ifcase\\catcode#1\\relax\nescape\\or\nbeginning of group\\or\nend of group\\or\nmath shift\\or\ntab\\or\nend of line\\or\nparameter\\or\nsuperscript\\or\nsubscript\\or\nignored\\or\nspace\\or\nletter\\or\notherchar\\or\nactive\\or\ncomment\\or\nignored\\fi}\n\nThe category code is \\printcatcode\\%'\n\n\nJust to point out that any of the \"code tables\" in TeX can be used to access the value (\\catcode, \\lccode, \\uccode, \\mathcode, \\delcode, \\sfcode).\n\nThe solutions offered by Martin and egreg print the current catcode of a character. I understood the original question to be how one can print the catcode of a token, i.e., one that has already been scanned, possibly in the past, when different catcodes were in effect. Consider this:\n\n\\catcode\\@=11\n\\newcommand{\\x}{@}\n\\catcode\\@=12\n\\newcommand{\\y}{@}\n\n\\newcommand{\\printcatcode}{\n\\def\\aux##1{\n\\message{The catcode of the ##1 in \\noexpand#1 is \\getcatcode{##1}.^^J}\n}\n\\expandafter\\aux#1\n}\n\n\\printcatcode{\\x}\n\\printcatcode{\\y}\n\n\nClearly the @s stored in the macros \\x and \\y have two different catcodes. However, with Martin's macro, we get:\n\nThe catcode of the @ in \\x is 12.\nThe catcode of the @ in \\y is 12.\n\n\nThat is because Martin's macro prints the 'current' catcode of the character represented by the token, rather than the catcode actually stored in the token.\n\nThe following is a better solution:\n\n\\begingroup%\n% locally ensure that characters have their expected catcodes\n\\catcode\\$=3% \\catcode\\&=4% \\catcode\\#=6% \\catcode\\^=7% \\catcode\\_=8% \\catcode\\ =10% \\catcode\\a=11% \\catcode\\+=12% \\catcode\\~=13% \\gdef\\getcatcode#1{% \\ifcat\\noexpand#1$3\\else%\n\\ifcat\\noexpand#1&4\\else%\n\\ifcat\\noexpand#1##6\\else%\n\\ifcat\\noexpand#1^7\\else%\n\\ifcat\\noexpand#1_8\\else%\n\\ifcat\\noexpand#1 10\\else%\n\\ifcat\\noexpand#1a11\\else%\n\\ifcat\\noexpand#1+12\\else%\n\\ifcat\\noexpand#1\\noexpand~13\\else%\n\\ifcat\\noexpand#1\\relax16\\else%\nunknown%\n\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi\\fi%\n}\n\\endgroup\n\n\nWith this, we get the expected answer:\n\nThe catcode of the @ in \\x is 11.\nThe catcode of the @ in \\y is 12.\n\n\nThe code is a bit awkward and I wonder if there is a more elegant solution. I was unable to test for catcodes 0, 1, 2, 5, 9, 14, and 15, but I am not sure if such catcodes can actually occur after the input has been tokenized.\n\n• Category codes 1 and 2 can certainly exist: think \\def\\x{{a}} which holds three tokens, {, a and }. Feb 24, 2015 at 14:36" ]

[ null ]

[ "", null, "Maximize the volume of Cuboid\nSubmissions: 355   Accuracy: 32.39%   Difficulty: Easy   Marks: 2\n\nGiven the sum of length, breadth and height of a cuboid. The task is to find the maximum volume that can be achieved such that the sum of sides is S.\n\nInput:\nThe first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer S.\n\nOutput:\nFor each test case, print the maximum volume you can obtain in new line.\n\nConstraints:\n1 <= T <= 500\n3 <= S < 10\n6\n\nExample:\nInput:\n2\n4\n8\n\nOutput:\n2\n18\n\nExplanation:\n\nInput : S = 8\nOutput : 18\nAll possible edge dimensions:\n[1, 1, 6], volume = 6\n[1, 2, 5], volume = 10\n[1, 3, 4], volume = 12\n[2, 2, 4], volume = 16\n[2, 3, 3], volume = 18\n\n** For More Input/Output Examples Use 'Expected Output' option **\n\nAuthor: arun03\n\nIf you have purchased any course from GeeksforGeeks then please ask your doubt on course discussion forum. You will get quick replies from GFG Moderators there.\n\nNeed help with your code? Please use ide.geeksforgeeks.org, generate link and share the link here.\n\nto report an issue on this page." ]

[ null, "https://www.facebook.com/tr", null ]

[ "4\n\n# According to Avogadro's hypothesis equal volumes of two gases under the same conditions of temperature and pressure contain\n\nEqual number of electrons\n\nEqual number of ions\n\nEqual number of atoms\n\nEqual number of molecules" ]

[ null ]

[ "", null, "# How to find the average consumption by hours\n\nHi,\nI am new at coding so something as simple as doing a condition means seems hard.\n\nIn fact, what I am trying to do is to have the average consumption by hours (& for each year, but this is ok).\nMy data are like this :", null, "I have a hard time finding the i e hours average/year.\n\nI found this on the web :\nmean(HQ\\$`2009`|HQ\\$heure == 2, na.rm = TRUE)\n 1\nWhich 1 is not a possible value.\n\nAfter digging on the web (conditional mean) I found the tidyverse, but I don't know if it's really this that I want :\n\nhours_groups <- group_by(test_2, heure)\nsummarise(hours_groups, c_count_mean = mean(hours_groups, na.rm = TRUE), n = n())\nThe warning : In mean.default(hours_groups, na.rm = TRUE) :\nargument is not numeric or logical: returning NA\n\nI am a bit lost, because I would have been able to do it in 4 lines in stata. (with a for & if loop) :\nfor(h in hours) {\nif (h == HQ\\$heure) {\nmean(HQ\\$`2009`, na.rm = TRUE)\n}\n}\n\nWich is not working, but it was my first try since it was my stata's intuitions.\n\nThis is how I would find the means by hour. I assume the values in the columns that look like years are consumption.\n\n``````library(tidyverse)\n\n# Making some data up\n\nHQ <- tibble(heure=rep(1:12, 31),\njour=rep(1:31, each=12),\nmois=1) %>%\nmutate(`2009`=rnorm(n(), 23000, 300),\n`2010`=rnorm(n(), 23000, 300),\n`2011`=rnorm(n(), 23000, 300),\n`2012`=rnorm(n(), 23000, 300),\n`2013`=rnorm(n(), 23000, 300))\n\n#> # A tibble: 6 x 8\n#> heure jour mois `2009` `2010` `2011` `2012` `2013`\n#> <int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>\n#> 1 1 1 1 22922. 23011. 23275. 22538. 22939.\n#> 2 2 1 1 23250. 22667. 22946. 23106. 22900.\n#> 3 3 1 1 22730. 22576. 23455. 22340. 22569.\n#> 4 4 1 1 23069. 22894. 23300. 23335. 23110.\n#> 5 5 1 1 22664. 23503. 22624. 22586. 23395.\n#> 6 6 1 1 23202. 23630. 22753. 23045. 22400.\ntail(HQ)\n#> # A tibble: 6 x 8\n#> heure jour mois `2009` `2010` `2011` `2012` `2013`\n#> <int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>\n#> 1 7 31 1 22684. 22803. 22764. 22967. 23794.\n#> 2 8 31 1 23115. 23217. 23167. 22922. 23134.\n#> 3 9 31 1 23054. 22809. 22785. 23491. 23015.\n#> 4 10 31 1 23195. 23087. 23380. 23233. 23090.\n#> 5 11 31 1 23110. 22733. 23053. 22773. 22764.\n#> 6 12 31 1 22911. 23056. 23184. 22862. 23655.\n\nHQ %>%\ngroup_by(heure) %>% # summarise by each hour\nsummarise(mean2009=mean(`2009`),\nmean2010=mean(`2010`),\nmean2011=mean(`2011`),\nmean2012=mean(`2012`),\nmean2013=mean(`2013`)\n)\n#> # A tibble: 12 x 6\n#> heure mean2009 mean2010 mean2011 mean2012 mean2013\n#> <int> <dbl> <dbl> <dbl> <dbl> <dbl>\n#> 1 1 22924. 23033. 23029. 22861. 22996.\n#> 2 2 23010. 22971. 23065. 23017. 23049.\n#> 3 3 23078. 22915. 22977. 22908. 23027.\n#> 4 4 23020. 22995. 23084. 22948. 22939.\n#> 5 5 22955. 23062. 23024. 22958. 23098.\n#> 6 6 22869. 23005. 22945. 23045. 22935.\n#> 7 7 22915. 22974. 22986. 22993. 23052.\n#> 8 8 22972. 23010. 22949. 22976. 23053.\n#> 9 9 23013. 23014. 22921. 23041. 23057.\n#> 10 10 22994. 22968. 22989. 22996. 23042.\n#> 11 11 23069. 22906. 23025. 22984. 22910.\n#> 12 12 22934. 23005. 23097. 22980. 23015.\n``````\n\nCreated on 2020-05-07 by the reprex package (v0.3.0)\n\n2 Likes\n\nIn fact, it was the Tidyverse. I was stuck with it. Hence I read : https://r4ds.had.co.nz/\nwithin an hour I found the simple one command that I needed.\n\nFor every begginner that aim to do data description, only one night of reading will make you save hours.\n\nThis was the lines :\n\nlibrary(tidyverse)\nlibrary(dplyr)\n\nHQ <- X2009_2018_heure %>%\npivot_longer(c(`2009`,`2010`,`2011`,`2012`,`2013`,`2014`,`2015`,`2016`,`2017`,`2018`), names_to = \"Year\", values_to = \"Consumption\")\n\nHence After I can subdivise it as much as I want:\n\nHQ_H_Y <- HQ %>%\ngroup_by(Year,heure) %>%\nsummarise(hours_mean = mean(Consumption, na.rm = TRUE))\n\nThis one gives :\nYear heure hours_mean\n\n2009 1 17994.\n2009 2 17828.\n2009 3 17815.\n\nHQ_H_M_Y_Q_1 <- HQ %>%\ngroup_by(mois,heure,Year) %>%\ndplyr::summarise( Quantile1 = quantile(`Consumption`, c(.10), na.rm = TRUE),\nQuantile2 = quantile(`Consumption`, c(.20), na.rm = TRUE),\nQuantile3 = quantile(`Consumption`, c(.30), na.rm = TRUE),\nQuantile4 = quantile(`Consumption`, c(.40), na.rm = TRUE),\nQuantile5 = quantile(`Consumption`, c(.50), na.rm = TRUE),\nQuantile6 = quantile(`Consumption`, c(.60), na.rm = TRUE),\nQuantile7 = quantile(`Consumption`, c(.70), na.rm = TRUE),\nQuantile8 = quantile(`Consumption`, c(.80), na.rm = TRUE),\nQuantile9 = quantile(`Consumption`, c(.90), na.rm = TRUE),\nQuantile10 = quantile(`Consumption`, c(1.00), na.rm = TRUE))\n\nThis one is more complex :\n\nmois heure Year Quantile1 Quantile2 Quantile3 Quantile4 Quantile5 Quantile6 Quantile7 Quantile8 Quantile9 Quantile10\n1 1 2009 24089 24386 24793 24967 25226 25937 26586 27767 28104 30374\n\nThis topic was automatically closed 7 days after the last reply. New replies are no longer allowed." ]

[ null, "https://community.rstudio.com/uploads/default/original/3X/5/d/5dc960154a129282ba4283771da2fab6fde146fb.png", null, "https://community.rstudio.com/uploads/default/original/3X/6/6/667a962ce5db83c0976f1a2d32f9c83c886d4134.png", null ]

[ "Intel® oneAPI Math Kernel Library\nAsk questions and share information with other developers who use Intel® Math Kernel Library.\n6617 Discussions\n\n## How to do a 2d Fourier tranformation stored in column-major?", null, "Beginner\n149 Views\n\nI have read the MKL manual but I can hardly understand the explanation below:\n\n```MKL_LONG dims[] = { nd, …, n2, n1 };\nDftiCreateDescriptor( &hand, precision, domain, d, dims );\n// The above call assumes data declaration: type X[nd]…[n2][n1]\n// Default strides are { 0, nd*…*n2*n1, …, n2*n1, n1, 1 }```\n\nIt is obvious that there are nd+2 elements in the strides configurations, i.e., for 2d data, the default strides are {0, n2*n1, n1, 1},\n\nthe first element 0 is the default value for s0, except which there are still 3 elements.\n\nI don't know how to under the configuration of strides.\n\nBy now, I have a 2d data stored in column-major, how to set the strides for the column-major storage?\n\n2 Replies", null, "Employee\n149 Views\n\nHi\n\nIt seems bring some confusion here.  I will check internally.\n\nbut the key is  how the x(k1, K2) are gotten.\n\nThe memory address of the\nelement X(k1, k2 , ..., kd) is expressed by the formula\naddress of X(k1, k2, ..., kd) = address of X(0, 0, ..., 0) + offset\n= address of X(0, 0, ..., 0) + s0 + k1*s1 + k2*s2 + ...+ kd*sd,\nwhere s0 is the displacement and s1, ..., sd are generalized strides.\n\nSo for 2d data,   for   column-major storage,  X(k1,k2)= x(0,0)+K1*1 + K2* row-number, the strides looks be {0, 1, row-number}\n\nYou may refer to some sample coded under MKL install directory. for example, mkl_example\\examples_core_c\\dftc\\source\\config_placement.c to understand the stride and the compress format if real.\n\nRegards\n\nYing", null, "Beginner\n149 Views\n\nThanks！\n\nI have guessed the solution.\n\nYours,\n\nYing H (Intel) wrote:\n\nHi\n\nIt seems bring some confusion here.  I will check internally.\n\nbut the key is  how the x(k1, K2) are gotten.\n\nThe memory address of the\nelement X(k1, k2 , ..., kd) is expressed by the formula\naddress of X(k1, k2, ..., kd) = address of X(0, 0, ..., 0) + offset\n= address of X(0, 0, ..., 0) + s0 + k1*s1 + k2*s2 + ...+ kd*sd,\nwhere s0 is the displacement and s1, ..., sd are generalized strides.\n\nSo for 2d data,   for   column-major storage,  X(k1,k2)= x(0,0)+K1*1 + K2* row-number, the strides looks be {0, 1, row-number}\n\nYou may refer to some sample coded under MKL install directory. for example, mkl_example\\examples_core_c\\dftc\\source\\config_placement.c to understand the stride and the compress format if real.\n\nRegards\n\nYing", null, "" ]

[ null, "https://community.intel.com/t5/image/serverpage/avatar-name/inteldarkgrayinverse/avatar-theme/candy/avatar-collection/Intel_Customer/avatar-display-size/message/version/2", null, "https://community.intel.com/t5/image/serverpage/avatar-name/intelmediumblue/avatar-theme/candy/avatar-collection/Intel_Customer/avatar-display-size/message/version/2", null, "https://community.intel.com/t5/image/serverpage/avatar-name/inteldarkgrayinverse/avatar-theme/candy/avatar-collection/Intel_Customer/avatar-display-size/message/version/2", null, "https://community.intel.com/skins/images/5DBB0AF0D77E9D5DDD542176F0B02B97/responsive_peak/images/icon_anonymous_message.png", null ]

[ "# Find short and simple methods to solve $24x^4+1=y^2$\n\nFind all this diophantine equation $$24x^4+1=y^2\\tag{1}$$ postive integers solution\n\nit is clear $(x,y)=(1,5)$\n\nI know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270.\n\nDoes this equation have another proof such Lucas's assertion, with short and simple methods? Like this paper: Anglin, W. S. \"The Square Pyramid Puzzle.\" Amer. Math. Monthly 97, 120-124, 1990. The square pyramid puzzle\n\nIn the paper,Following two question have simple methods to solve it.\n\nThere are no positive integers $x$ such $2x^4+1$ is a square.\n\nand\n\nThere is exactly one positive integer $x$,namely $1$, such that $8x^4+1$ is a square?\n\nBut How can I find simple methods to solve $(1)$?\n\n• What are you asking? As far as I can see there are at least 3 questions in your question, so why don't you post them seperately? – Toby Mak May 27 '17 at 12:22\n• $$\\dfrac{y+1}2\\cdot\\dfrac{y-1}2=6x^4$$ As $\\dfrac{y+1}2-\\dfrac{y-1}2=1,\\left(\\dfrac{y+1}2,\\dfrac{y-1}2\\right)=1$ and they are of opposite parity. If the highest power of prime $p>3$ that divides $6x^4$ is $a,$ $p^{4a}$ divides exactly one of $$\\dfrac{y+1}2,\\dfrac{y-1}2$$ – lab bhattacharjee Jun 1 '17 at 10:06\n• I spent a little bit of time on this and while some of the arguments in the linked paper apply here, I couldn't combine them into a similar cohesive argument (that's not to say such an argument doesn't exist, I just couldn't make one). As a side note, I checked that there are no positive integer solutions to your equation besides $(1,5)$ for $x \\leq 300,000$. – M10687 Jun 6 '17 at 18:59" ]

[ null ]

[ "src/Pure/drule.ML\n changeset 10767 8fa4aafa7314 parent 10667 75a1c9575edb child 10815 dd5fb02ff872\n```--- a/src/Pure/drule.ML\tWed Jan 03 11:14:48 2001 +0100\n+++ b/src/Pure/drule.ML\tWed Jan 03 21:18:31 2001 +0100\n@@ -123,15 +123,15 @@\nfun dest_implies ct =\ncase term_of ct of\n(Const(\"==>\", _) \\$ _ \\$ _) =>\n- let val (ct1,ct2) = dest_comb ct\n- in (#2 (dest_comb ct1), ct2) end\n+ let val (ct1,ct2) = Thm.dest_comb ct\n+ in (#2 (Thm.dest_comb ct1), ct2) end\n| _ => raise TERM (\"dest_implies\", [term_of ct]) ;\n\nfun dest_equals ct =\ncase term_of ct of\n(Const(\"==\", _) \\$ _ \\$ _) =>\n- let val (ct1,ct2) = dest_comb ct\n- in (#2 (dest_comb ct1), ct2) end\n+ let val (ct1,ct2) = Thm.dest_comb ct\n+ in (#2 (Thm.dest_comb ct1), ct2) end\n| _ => raise TERM (\"dest_equals\", [term_of ct]) ;\n\n@@ -151,7 +151,7 @@\n(* A1==>...An==>B goes to B, where B is not an implication *)\nfun strip_imp_concl ct =\ncase term_of ct of (Const(\"==>\", _) \\$ _ \\$ _) =>\n- strip_imp_concl (#2 (dest_comb ct))\n+ strip_imp_concl (#2 (Thm.dest_comb ct))\n| _ => ct;\n\n(*The premises of a theorem, as a cterm list*)\n@@ -162,7 +162,7 @@\nval implies = cterm_of proto_sign Term.implies;\n\n(*cterm version of mk_implies*)\n-fun mk_implies(A,B) = capply (capply implies A) B;\n+fun mk_implies(A,B) = Thm.capply (Thm.capply implies A) B;\n\n(*cterm version of list_implies: [A1,...,An], B goes to [|A1;==>;An|]==>B *)\nfun list_implies([], B) = B```" ]

[ null ]

[ "# #03d793 Color Information\n\nIn a RGB color space, hex #03d793 is composed of 1.2% red, 84.3% green and 57.6% blue. Whereas in a CMYK color space, it is composed of 98.6% cyan, 0% magenta, 31.6% yellow and 15.7% black. It has a hue angle of 160.8 degrees, a saturation of 97.2% and a lightness of 42.7%. #03d793 color hex could be obtained by blending #06ffff with #00af27. Closest websafe color is: #00cc99.\n\n• R 1\n• G 84\n• B 58\nRGB color chart\n• C 99\n• M 0\n• Y 32\n• K 16\nCMYK color chart\n\n#03d793 color description : Strong cyan - lime green.\n\n# #03d793 Color Conversion\n\nThe hexadecimal color #03d793 has RGB values of R:3, G:215, B:147 and CMYK values of C:0.99, M:0, Y:0.32, K:0.16. Its decimal value is 251795.\n\nHex triplet RGB Decimal 03d793 `#03d793` 3, 215, 147 `rgb(3,215,147)` 1.2, 84.3, 57.6 `rgb(1.2%,84.3%,57.6%)` 99, 0, 32, 16 160.8°, 97.2, 42.7 `hsl(160.8,97.2%,42.7%)` 160.8°, 98.6, 84.3 00cc99 `#00cc99`\nCIE-LAB 76.511, -59.835, 21.42 29.602, 50.724, 35.833 0.255, 0.437, 50.724 76.511, 63.554, 160.303 76.511, -65.622, 39.841 71.22, -50.444, 20.024 00000011, 11010111, 10010011\n\n# Color Schemes with #03d793\n\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #d70347\n``#d70347` `rgb(215,3,71)``\nComplementary Color\n• #03d729\n``#03d729` `rgb(3,215,41)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #03b1d7\n``#03b1d7` `rgb(3,177,215)``\nAnalogous Color\n• #d72903\n``#d72903` `rgb(215,41,3)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #d703b1\n``#d703b1` `rgb(215,3,177)``\nSplit Complementary Color\n• #d79303\n``#d79303` `rgb(215,147,3)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #9303d7\n``#9303d7` `rgb(147,3,215)``\nTriadic Color\n• #47d703\n``#47d703` `rgb(71,215,3)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #9303d7\n``#9303d7` `rgb(147,3,215)``\n• #d70347\n``#d70347` `rgb(215,3,71)``\nTetradic Color\n• #028c5f\n``#028c5f` `rgb(2,140,95)``\n• #02a571\n``#02a571` `rgb(2,165,113)``\n• #03be82\n``#03be82` `rgb(3,190,130)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #03f0a4\n``#03f0a4` `rgb(3,240,164)``\n• #11fcb1\n``#11fcb1` `rgb(17,252,177)``\n• #2afcb9\n``#2afcb9` `rgb(42,252,185)``\nMonochromatic Color\n\n# Alternatives to #03d793\n\nBelow, you can see some colors close to #03d793. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #03d75e\n``#03d75e` `rgb(3,215,94)``\n• #03d770\n``#03d770` `rgb(3,215,112)``\n• #03d781\n``#03d781` `rgb(3,215,129)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #03d7a5\n``#03d7a5` `rgb(3,215,165)``\n• #03d7b6\n``#03d7b6` `rgb(3,215,182)``\n• #03d7c8\n``#03d7c8` `rgb(3,215,200)``\nSimilar Colors\n\n# #03d793 Preview\n\nText with hexadecimal color #03d793\n\nThis text has a font color of #03d793.\n\n``<span style=\"color:#03d793;\">Text here</span>``\n#03d793 background color\n\nThis paragraph has a background color of #03d793.\n\n``<p style=\"background-color:#03d793;\">Content here</p>``\n#03d793 border color\n\nThis element has a border color of #03d793.\n\n``<div style=\"border:1px solid #03d793;\">Content here</div>``\nCSS codes\n``.text {color:#03d793;}``\n``.background {background-color:#03d793;}``\n``.border {border:1px solid #03d793;}``\n\n# Shades and Tints of #03d793\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, #000202 is the darkest color, while #eefff9 is the lightest one.\n\n• #000202\n``#000202` `rgb(0,2,2)``\n• #00160f\n``#00160f` `rgb(0,22,15)``\n• #01291c\n``#01291c` `rgb(1,41,28)``\n• #013c29\n``#013c29` `rgb(1,60,41)``\n• #015036\n``#015036` `rgb(1,80,54)``\n• #016344\n``#016344` `rgb(1,99,68)``\n• #027651\n``#027651` `rgb(2,118,81)``\n• #028a5e\n``#028a5e` `rgb(2,138,94)``\n• #029d6b\n``#029d6b` `rgb(2,157,107)``\n• #02b079\n``#02b079` `rgb(2,176,121)``\n• #03c486\n``#03c486` `rgb(3,196,134)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\n• #03eaa0\n``#03eaa0` `rgb(3,234,160)``\nShade Color Variation\n• #06fcad\n``#06fcad` `rgb(6,252,173)``\n• #19fcb3\n``#19fcb3` `rgb(25,252,179)``\n• #2cfcb9\n``#2cfcb9` `rgb(44,252,185)``\n• #40fcc0\n``#40fcc0` `rgb(64,252,192)``\n• #53fdc6\n``#53fdc6` `rgb(83,253,198)``\n• #66fdcd\n``#66fdcd` `rgb(102,253,205)``\n• #7afdd3\n``#7afdd3` `rgb(122,253,211)``\n• #8dfdd9\n``#8dfdd9` `rgb(141,253,217)``\n• #a0fee0\n``#a0fee0` `rgb(160,254,224)``\n• #b4fee6\n``#b4fee6` `rgb(180,254,230)``\n• #c7feed\n``#c7feed` `rgb(199,254,237)``\n• #dbfef3\n``#dbfef3` `rgb(219,254,243)``\n• #eefff9\n``#eefff9` `rgb(238,255,249)``\nTint Color Variation\n\n# Tones of #03d793\n\nA tone is produced by adding gray to any pure hue. In this case, #68726f is the less saturated color, while #03d793 is the most saturated one.\n\n• #68726f\n``#68726f` `rgb(104,114,111)``\n• #5f7b72\n``#5f7b72` `rgb(95,123,114)``\n• #578375\n``#578375` `rgb(87,131,117)``\n• #4e8c78\n``#4e8c78` `rgb(78,140,120)``\n• #46947b\n``#46947b` `rgb(70,148,123)``\n• #3e9c7e\n``#3e9c7e` `rgb(62,156,126)``\n• #35a581\n``#35a581` `rgb(53,165,129)``\n• #2dad84\n``#2dad84` `rgb(45,173,132)``\n• #25b587\n``#25b587` `rgb(37,181,135)``\n• #1cbe8a\n``#1cbe8a` `rgb(28,190,138)``\n• #14c68d\n``#14c68d` `rgb(20,198,141)``\n• #0bcf90\n``#0bcf90` `rgb(11,207,144)``\n• #03d793\n``#03d793` `rgb(3,215,147)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #03d793 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

[ null ]

[ "# numpy矩阵向量乘法[重复]\n\n``````a = np.array([[ 5, 1 ,3], [ 1, 1 ,1], [ 1, 2 ,1]])\nb = np.array([1, 2, 3])\n\nprint a*b\n>>\n[[5 2 9]\n[1 2 3]\n[1 4 3]]\n``````\n\n``````print a*b\n>>\n[16 6 8]\n``````\n\n## 最简单的解决方案\n\n``````>>> a = np.array([[ 5, 1 ,3],\n[ 1, 1 ,1],\n[ 1, 2 ,1]])\n>>> b = np.array([1, 2, 3])\n>>> print a.dot(b)\narray([16, 6, 8])\n``````\n\n## 其他解决方案\n\n• 如下所述，如果使用python3.5， `@` 运算符可以按预期工作：\n``````>>> print(a @ b)\narray([16, 6, 8])\n``````\n• 如果你想要矫枉过正，你可以使用numpy.einsum . 文档将为您提供一个如何工作的风格，但老实说，我还没有完全理解如何使用它，直到阅读this answer并且只是自己玩它 .\n``````>>> np.einsum('ji,i->j', a, b)\narray([16, 6, 8])\n``````\n• 截至2016年中期（numpy 1.10.1），您可以尝试实验numpy.matmul，其工作方式类似于 `numpy.dot` ，但有两个主要的例外：没有标量乘法，但它适用于矩阵堆栈 .\n``````>>> np.matmul(a, b)\narray([16, 6, 8])\n``````\n``````>>> np.inner(a, b)\narray([16, 6, 8])\n\n# Beware using for matrix-matrix multiplication though!\n>>> b = a.T\n>>> np.dot(a, b)\narray([[35, 9, 10],\n[ 9, 3, 4],\n[10, 4, 6]])\n>>> np.inner(a, b)\narray([[29, 12, 19],\n[ 7, 4, 5],\n[ 8, 5, 6]])\n``````\n\n## 边缘情况的Rarer选项\n\n• 如果您有张量（维数大于或等于1的数组），则可以将numpy.tensordot与可选参数 `axes=1` 一起使用：\n``````>>> np.tensordot(a, b, axes=1)\narray([16, 6, 8])\n``````\n• Don't use numpy.vdot 如果您有一个复数矩阵，因为矩阵将被展平为一维数组，那么它将尝试找到您的展平矩阵和向量之间的复共轭点积（由于大小不匹配而失败 `n*m` vs `n` ） ." ]

[ null ]

[ "### Distance From Eye to Horizon of a Sphere\n\nThis program computes the line-of-sight distance and the curved surface distance to the horizon of a sphere at any distance of the eye as reckoned from the surface or center of the sphere.   The radius and distance can be in any convenient linear units.\n\nThe initial defaults radius is the mean lunar radius and the distance is 1/100th of the mean geocentric distance of the eye from the center of the moon, both in statute miles.\n\n Radius units Distance units Distance To Eye Reckoned From:     Center           Surface\nLine-of-Sight Distance (H) to Horizon = 2131.2729945974\nSurface Distance (S) to Horizon = 1189.4592885523\n\nThe distance (D) to the eye is reckoned from the center of the sphere.", null, "Where the Distance to the Eye Is Reckoned From the Center of the Sphere:\n\nGiven the radius (R) of the sphere and the distance (D) of the eye from the center of the sphere, the general equation for the line-of-sight distance (H) to the sphere horizon viewed from that point is:", null, "The horizon surface distance (S), corresponding to (H), but measured along the curved surface of the sphere directly below H and corresponding to the sailing distance is:", null, "Where: D > R and R > 0 and the inverse angle is expressed in radians\n\nWhere the Distance to the Eye Is Reckoned From the Surface of the Sphere:\n\nGiven the radius (R) of the sphere and the distance (d) (height) of the eye above the surface of the sphere, the general equation for the line-of-sight distance (H) to the sphere horizon viewed from that point is:", null, "The horizon surface distance (S), corresponding to line-of-sight distance (H), but measured along the curved surface of the sphere, corresponding to the sailing distance is:", null, "Where: R > 0 and d > 0 and the inverse angle is expressed in radians\n\n© Jay Tanner - 2020 - v2.0" ]

[ null, "http://neoprogrammics.com/PHPSL_DIRECTORY/~Sphere_Distance_to_Horizon/images/sphere_horizon_distance.png", null, "http://neoprogrammics.com/PHPSL_DIRECTORY/~Sphere_Distance_to_Horizon/images/H1.png", null, "http://neoprogrammics.com/PHPSL_DIRECTORY/~Sphere_Distance_to_Horizon/images/S1.png", null, "http://neoprogrammics.com/PHPSL_DIRECTORY/~Sphere_Distance_to_Horizon/images/H2.png", null, "http://neoprogrammics.com/PHPSL_DIRECTORY/~Sphere_Distance_to_Horizon/images/S2.png", null ]

[ "", null, "# Further Mathematics Year 13 course 2: Applications of Differential Equations, Momentum, Work, Energy & Power, The Poisson Distribution, The Central Limit Theorem, Chi Squared Tests, Type I and II Errors", null, "This course by Imperial College London is designed to help you develop the skills you need to succeed in your A-level further maths exams. You will investigate key topic areas to gain a deeper understanding of the skills and techniques that you can apply throughout your A-level study. These skills include: * Fluency – selecting and applying correct methods to answer with speed and efficiency * Confidence – critically assessing mathematical methods and investigating ways to apply them * Problem solving – analysing the ‘unfamiliar’ and identifying which skills and techniques you require to answer questions * Constructing mathematical argument – using mathematical tools such as diagrams, graphs, logical deduction, mathematical symbols, mathematical language, construct mathematical argument and present precisely to others * Deep reasoning – analysing and critiquing mathematical techniques, arguments, formulae and proofs to comprehend how they can be applied Over eight modules, you will be introduced to * Simple harmonic motion and damped oscillations. * Impulse and momentum. * The work done by a constant and a variable force, kinetic and potential energy (both gravitational and elastic) conservation of energy, the work-energy principle, conservative and dissipative forces, power. * Oblique impact for elastic and inelastic collision in two dimensions. * The Poisson distribution, its properties, approximation to a binomial distribution and hypothesis testing. * The distribution of sample means and the central limit theorem. * Chi-squared tests, contingency tables, fitting a theoretical distribution and goodness of fit. * Type I and type II errors in statistical tests. Your initial skillset will be extended to give a clear understanding of how background knowledge underpins the A -level further mathematics course. You’ll also be encouraged to consider how what you know fits into the wider mathematical world.\n\nCreated by: Imperial College London\n\nLevel: Intermediate", null, "Note: must be in .png, .gif or .jpg format\nOR", null, "Note: must be in .png, .gif or .jpg format\n\nBy clicking this button,", null, "", null, "" ]

[ null, "https://doikmin2nht2v.cloudfront.net/img/main-480x148_2.jpg", null, "https://d2h3bjlv56u1wn.uloop.com/yPaPm8F%2BRpmvhYCZoOqt%2FQ%3D%3D/892a6fb0/DU-Online-Courses-Further-Mathematics-Year-13-course-2-Applications-of-Differential-Equations-Momentum-Work-Energy-amp-Power-The-Poisson-Distribution-The-Central-Limit-Theorem-Chi-Squared-Tests-Type-I-and-II-Errors-for-University-of-Denver-Students-in-Denver-CO.jpeg", null, "https://doikmin2nht2v.cloudfront.net/img/preloader-white.gif", null, "https://doikmin2nht2v.cloudfront.net/img/loading.gif", null, "https://doikmin2nht2v.cloudfront.net/img/preloader.gif", null, "https://du.uloop.com/online-courses/view.php/1850952877/Further-Mathematics-Year-13-course-2-Applications-of-Differential", null ]

[ "### Trending Articles\n\n10 Jun 2023", null, "# 100mph to kph – Complete Detailed Summary Report\n\n## Convert 100 mph to kph\n\n100mph to kph – So you need to convert 100 miles per hour to kilometers per hour? If you’re in a hurry and need the answer, the calculator below is all you need. The answer is 160.93427125258 kilometers per hour.\n\n## Steps to convert miles per hour to kilometers per hour. (100 mph to kph)\n\nWe all use different units of measure every day. Whether you are in a foreign country and need to convert the local imperial units to the metric system, you are baking a cake and need to convert it to a division you are more familiar with.\n\nFortunately, converting most units is very, very easy. In this case, know that 1 mph equals 1.6093427125258 km / h.\n\nOnce you know what 1 mph is in kilometers per hour, you can multiply 1.6093427125258 by the total number of miles per hour you want to compute.\n\nSo for our sample here, we have 100 miles per hour. So entirely we do multiply 100 by 1.6093427125258:\n\n100 x 1.6093427125258 = 160.93427125258\n\n## What is the top conversion unit for 100 mph? 100 mph to kph\n\nAs a little extra conversion for you, we can also calculate the best unit of measurement for 100 mph.\n\nWhat is the “best” unit of measurement? To simplify, the best team size is the lowest possible without going below 1. The reason is that the lower number usually makes the height easier to understand.\n\nFor 100 mph, the best unit of measurement is the meter per second, and the amount is 44,704 mps.\n\nIf you’ve found this helpful content in your research, do us a favor and use the tool below to make sure you reference us correctly wherever you use it. We appreciate your help!\n\n## More unit conversions 100 mph to kph", null, "I hope this assisted you in learning how to convert 100mph to km / h. If you want to analyze more unit conversions, go back to our main unit converter and experiment with different conversions.\n\n## How to convert miles per hour to kilometers per hour\n\nOne mile per hour is more than one kilometer per hour. I know that an mph is more than a km / h because of somewhat called conversion aspects. You can use this information for additional knowledge and convert the related entries.\n\nSimply put, a conversion element is a number that can remain used to adjust one set of units to another by multiplying or dividing it. So when we want to convert 100 miles per hour to kilometers per hour, we use a conversion factor to get the answer.\n\n## Best conversion unit for 100 mph", null, "When working with conversions from one unit to another, the numbers can get a little confusing, especially for huge numbers. Always use the conversion tool and get the accurate results to deal with the situtation.\n\nI also calculated what the best unit of measurement for 100 mph is.\n\nTo control which unit is the best, I decided to set it as the lowest possible unit of measurement without going below 1. Smaller numbers are easier to understand and can make measurement easier to understand. to understand." ]

[ null, "https://www.digitalknowledgetoday.com/wp-content/uploads/2022/01/100mph-to-kph-1200x675.jpg", null, "https://www.digitalknowledgetoday.com/wp-content/uploads/2022/01/More-unit-conversions-100-mph-to-kph.jpg", null, "https://www.digitalknowledgetoday.com/wp-content/uploads/2022/01/Best-conversion-unit-for-100mph-to-kph.jpg", null ]

[ "JMPJournal of Modern Physics2153-1196Scientific Research Publishing10.4236/jmp.2016.711121JMP-69139ArticlesPhysics&Mathematics Biquaternionic Form of Laws of Electro-Gravimagnetic Charges and Currents Interactions L.A. Alexeyeva1Institute of Mathematics and Mathematical Modeling, Almaty, Kazakhstan* E-mail:080720160711135113588 June 2016accepted 24 July 27 July 2016© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/\n\nOne the base of differential algebra of biquaternions, the one model of electro-gravimagnetic interactions of electric and gravimagnetic charges and currents has been constructed. For this, three Newton laws analogues are used. The closed system of biquaternionic wave equations is constructed for determination of the charges-currents and electro-gravimagnetic fields and united field of interactions. The equation of charge-current transformation is like the generalization of biquaternionic presentation of Dirac equation. The properties of its solutions are described, depending on properties of external EGM field. The biquaternions of energy-pulse of EGM-field and charges-currents are considered. The energy-pulse of EGM-interactions is calculated.\n\nElectro-Gravimagnetic Field Electric Charge Gravimagnetic Charge Current Energy Biquaternion Bigradient Dirac Equation Newton Laws\n1. Introduction\n\nIn the paper , we described the one biquaternionic model of electro-gravimagnetic (EGM) field, charges and currents. In this model, gravitational field (which is potential) is united with magnetic field (which is torsional) which gives possibility to enter gravimagnetic tension, gravimagnetic charge and gravimagnetic current. There we have shown that in algebra of biquaternions, the charges and currents of EGM-field are physical appearance of bigradient of EGM-intensity. Differential operator bigradient is the generalization of gradient operator on the space of biquaternions which characterizes a direction of more extensive change of biquaternionic functions. If bigradient of EGM-intensity is equal to zero then charges and currents are absent. Also there we constructed the law of inertia for a free system of mass, charges and currents, which described their motion under action only with internal electric and gravimagnetic tensions. This law is the fields analogue of the first Newton law― inertia law for a solid (considered as material point).\n\nHere we consider the motion of electro-gravimagnetic charges and currents under action of external EGM- fields which are created by other charges and currents. The laws of their interaction are the fields analogue of the second and third Newton laws. They have been constructed in the form of biquaternionic wave equations which generalize biquaternionic form of Dirac equations. Some solutions of them are discussed. The energy- pulse of EGM-interactions is calculated.\n\nWe used also here the differential algebra of biquaternions in Hamiltonian form which was shortly described in (see for more detail). This form is very convenient for description of physical fields. There are voluminous literature about application of algebras of quaternions and biquaternions in fields theory, in the theory of electromagnetic fields, and quantum mechanics - . Differential algebra of biquaternions gives possibility to simplify mathematical record of systems of Maxwell and Dirac equations to construct their solutions and to study their properties.\n\nThe novelty of this work is the construction of laws of electro-gravymagnetic interactions on the base of biquaternions algebra. The properties of this algebra and operation of quaternionic multiplication in fields theory have visual physical interpretation and detect new objective laws which are impossible to determine without this algebra. It’s natural for matter as you’ll see here.\n\n2. Biquaternions of Electro-Gravimagnetic Field, Charges and Currents\n\nTheir are the next complex characteristics of EGM-field :\n\n- complex vector of EGM-intensity\n\n- complex charges field:\n\n- complex currents field:\n\n- complex scalar field of attraction-resistance\n\nHere real vectors E and H are the tensions of electric and gravimagnetic fields; real scalars are the densities of electric and gravimagnetic charges; real vectors are the densities of electric and gravi- magnetic currents; values are the constants of electric conductivity and magnetic permeability of the EGM-medium.\n\nIn biquaternions algebra on Minkowski space the EGM-field, charges and currents may be presented by use the next biquaternions:\n\nEGM-intensity\n\ncharge-current\n\nIn the paper was shown that connection between EGM-intensity and charge-current has the bigradiental form:\n\nPOSTULATE 1\n\nHere and further the bigradients and are the next biquaternionic differential operators:\n\nFurther we name them mutual bigradients. They define a “directions” of more intensive changing of biquater- nionic field.\n\n3. The Power and Density of Acting Forces\n\nLet’s consider two systems of charges and currents. Every of them generate own EGM-field: and, which corresponds to (1). At first let consider the case when.\n\nWe name a power-force density the next biquaternion\n\nwhich is acting from side of A’ -field on the charge and current of A-field. Really, according to definitions, the scalar part is determined as power density of acting forces:\n\nHere is analogue of a magnetic induction (in torsional part complies with it), is a vector of an electric offset.\n\nSelecting the real and imaginary part from the formulae (2) we get the expressions for a density of acting forces:\n\nPotentional part of describes the tension of gravitational field. Torsional part of this vector describes magnetic field. The scalar part of contains the densities of electric charge and gravitational mass. Its vector part contains the densities of electric and mass currents. Coming from these suggestions, in formula (4) we see the known forces, appropriately:\n\n- Coulomb’s force;\n\n- gravimagnetic force (it complies with gravitational force in a potential part);\n\n- Lorentz force (more exactly, it complies with it in torsional part);\n\n- gravielectric force.\n\nIn real part of the power p (3) we see the powers of Coulomb’s force, gravitational and magnetic forces. The power of gravielectric force in real part of (3) does not enter as it does not work on the mass displacement, because it is perpendicular to its velocity. It’s interesting that Lorentz force also does not enter in real part of (3). It proves that this force is perpendicular to mass velocity, though directly from Maxwell equations this does not follow.\n\nNaturally, by analogy, we assume that formula (5) describes forces, causing a change of electric current. Consequently their power stands in imaginary part p (3).\n\nWith entering the scalar field, type of scalar and vector parts of power-force biquaternion (2) is changed, as follows\n\nYou see here the additional summands which appear in presentation of the powers () and force ().\n\nVector describes absorbtion-resistance force which acts on from A’-field.\n\n4. CC-Transformations Equation: Second Newton Law\n\nThe charge-current field is changed under influence of the -field. As it’s well known, the direction of the most intensive change the scalar field describes its gradient. By analogy we expect that change of charge- current biquaternion is more intensive toward its bigradient. Naturally to expect that this change must be toward external power-force.\n\nPOSTULATE 2. The law of a change of a charge-current field under the action of external EGM-field is\n\nEntering the constant of interaction is connected with dimensionality. We name Equation (7) as CC- transformations equation.\n\nIf one EGM-field much stronger then second one:\n\nit’s possible to neglect the second field change under influence of charge and current on -field. In this case from Equation (7) we can find the CC-field, its changing under action of external -field.\n\nRevealing scalar and vector part from (7), we get the system of two differential equations:\n\nBy use (3), (4) we obtain from Equation (9) the two vectorial differential equations:\n\nThe Equation (10) is the second Newton law analogue for CC-field. Here the value is analogue of mass momentum. In right part you see all known forces but also two new forces: gravielectric force and absorbtion-resistance force. Last of them is proportional to currents. Their direction depends on signs of real and imaginary part which can be as positive and negative. By analogy to mechanics of media we call it such name.\n\nEquation (10) describes the motion of gravimagnetic charges and currents under action of the external EGM- field. Consequently Equation (11) defines the motion of electric charges and currents.\n\nThe scalar Equation (8) is the law of conservation of electric and gravimagnetic charges. As you see the external EGM-field can essentially change CC-field.\n\n5. Third Newton Law: The Laws of Charge-Currents Interactions\n\nOn the virtue of the third Newton law about acting and counteracting forces, we suppose that must be executed for electro-gravimagnetic forces the equality:\n\nPOSTULATE 3\n\nFrom here we get\n\nfields analogue of third Newton law:\n\nBy use it we construct\n\nthe law of the charge-current interaction:\n\nHere Equation (13) correspond to the second Newton law which is written for charge-current each of interacting field. Equation (14) is the third Newton law. Together with Maxwell equations for these fields (15) they give closed system of the nonlinear differential equations for determination.\n\nIt is interesting that in scalar part of Equation (12) it requires the equality of the powers corresponding to forces, acting on charges and currents of the other field, i.e. it is befitted known in mechanics the identity to reciprocity of Betty, which is usually written for the forces works.\n\n6. First Newton Law: Free EGM-Field\n\nLet’s consider A-field, which is generated, in absence of other charges-currents. We name it a free field. In this case,. From (13) we get inertia law, which is analogue of the first Newton law for charge-current:\n\nwhich is equivalent to equations:\n\nFor initial designations we have following formulas:\n\nNaturally that the well known law of charges conservation in the form (17) must be executed in absence of external EGM-field.\n\nThis law (16) was considered in and solutions of this equations were defined in the next form:\n\nScalar potentials are arbitrary solutions of classic wave equation:\n\nwhere is Laplace operator. They may be presented so\n\nWe give here also for (16)\n\nthe solution of Cauchy problem\n\nwere there are the initial conditions:\n\nFrom postulate 1 to follow that A-field is defined in the form:\n\nwhere, is a differential of spheres area. About determination of these solutions see the theory of biwave equation in .\n\nThis formula is a generalization of the famous Kirchhoff formula for solution of Cauchy problem for wave equation .\n\n7. The Solutions of CC-Transformation Equation by Action of Invariable External EGM-Field\n\nLet consider Equation (7) when external EGM-field is constant, don’t depend on and x. We write it in the form\n\nThis equation is biquaternionic generalization of Dirac equations. Its presentation by use differential equations coincides with Dirac equations when, m is real constant. About construction full system of solutions of homogeneous and inhomogeneous Dirac equation see .\n\nThere are simple connection between solutions of Equation (7) and Equation (22):\n\nwhere is the solution of (7):\n\nFrom here to follow that imaginary part of scalar field resistance-absorbtion creates harmonic vibration with frequency, but real part increases or decreases over time depend on a sign. Real part of vector (electric field) creates for sinusoidal deviation in along direction of E. But its imaginary part (gravymagnetic field) gives exponential growth or diminution along direction the vector H.\n\n8. Energy-Impulse Biquaternions: First Thermodynamics Law\n\nWe introduce the biquaternion of energy-impulse of EGM-field ( EGM-energy-impulse):\n\nwhere is conjugated:\n\nare complex conjugated to. Here positive scalar function W is energy density of EGM field:\n\nP is real vector-function:\n\nBy, as you see, are like to an energy and Pointing vector of EM-field and satisfid to equation:\n\nBy analogue we construct the biquaternion of energy-impulse for charge-current field ( CC-energy-impulse):\n\nIt contains currents energy :\n\nwhere first summand includes Joule heat; second one includes kinetic energy density of mass current, also it contains the energy of torsional part of currents (magnetic current). Here vector is analogue of Pointing vector, but for the current:\n\nOnly if gravimagnetic and electrical currents are parallel or one from them is equal zero, then.\n\nIf to take scalar product Equation (9) with, we get\n\nIt is easy to see that this law is like to the first thermodynamics law. Here the sum of second and third summands in left part is designated. The function\n\ncharacterizes the own velocity of the change of current energy of -field. Right part (30), which depends on power of acting external forces, can to increase or decrease this velocity.\n\nIf there are not acting external forces,\n\nIt’s the first thermodynamics law for a free CC-field.\n\n9. The United Field of Interaction: Energy-Pulse of Interactions\n\nIf there are some (N) interacting CC-fields then we have for every of them\n\nUnited CC-field, as easy to see after summing the first equation over k, is free. It satisfies to the inertia law\n\nbecause all forces are internal, also as in Newton mechanics of interacting solids.\n\nLet consider the laws of energy transformation at interaction of different charges-currents. Energy-pulse for united charge-current field has the form:\n\nHere the first summand is an amount of energy-pulse of interacting charge-current.\n\nWe introduce biquaternion of energy-pulse interaction. Its real part describes energy-pulse interaction for the same name charge and current, but imagine part for different name ones:\n\nor in initial designations:\n\nAs result we get the conditions of energy transformation by charges-currents interaction:\n\nenergy separation if;\n\nenergy absorption if;\n\nenergy conservation if.\n\nVector shows the main direction and intensities of these energy processes.\n\n10. Conclusions\n\nWe construct here biquaternionic forms of laws of electric and gravimagnetic charges and currents interaction by analogy to Newton laws, which gives us closed hyperbolic system of differential equations for their definition and determination of corresponding EGM-fields. For the free system of charges and currents, Equations (22) and (24) define the behavior of CC-field and EGM-field over time if their initial states are known. After calculating the bigradients from here, the all fields, charges and currents can be defined according to their definitions. It’s the very suitable short form which contains the algorithm for their calculation.\n\nAt building of charge-current transformations equation, we get as known gravitational, electric and magnetic forces, so we found the new forces which are needed in experimental motivation. Considered properties (26) of solutions of CC-transformation equation by existence external EGM-field give possibility to test this model on practice.\n\nNote also that the essential at building and studying this model of EGM-field and CC-field is the differential algebra of biquaternions , without which such construction of differential equations, describing interaction of charges and currents in the forms which give the fields analogies of Newton laws and are very convenient for calculations, will be practically impossible.\n\nCite this paper\n\nL. A. Alexeyeva, (2016) Biquaternionic Form of Laws of Electro-Gravimagnetic Charges and Currents Interactions. Journal of Modern Physics,07,1351-1358. doi: 10.4236/jmp.2016.711121\n\nReferencesAlexeyeva, L.A. (2016) Journal of Modern Physics, 7, 435-444. http://www.scirp.org/journal/jmp http://dx.doi.org/10.4236/jmp.2016.75045Alexeyeva, L.A. (2012) Int. J. Clifford Analysis, Clifford Algebras and Their Applications, 7, 19-39.Rastall, R. (1964) Review of Modern Physics, 36, 820-832. http://dx.doi.org/10.1103/RevModPhys.36.820Edmonds Jr., J.D. (1978) American Journal of Physics, 46, 430. http://dx.doi.org/10.1119/1.11316Spilker, G.L. (1983) Report of USSR Academy of Sciences, 272, 1359-1363.Rotelli, P. (1989) Modern Physics Letters A, 4, 933-940. http://dx.doi.org/10.1142/S0217732389001106Davies, A.J. (1990) Physical Review D, 41, 2628-2630. http://dx.doi.org/10.1103/PhysRevD.41.2628Finkelstein, D., Jauch, J.M., Schiminovich, S. and Speiser, D. (1992) Journal of Mathematical Physics, 3, 207-220. http://dx.doi.org/10.1063/1.1703794Adler, S.L. (1995) Quaternionic Quantum Mechanics and Quantum Fields. Oxford University Press, New York.Kravchenko, V.V. (1995) Doklady Mathematics, 51, 287-289.Kassandrov, V.V. (1995) Gravitation and Cosmology, 1, 216-222.De Leo, S. and Rodrigues Jr., S. (1998) International Journal of Theoretical Physics, 37, 1707-1720. http://dx.doi.org/10.1023/A:1026692508708Kravchenko, V.V. (2003) Quaternionic Equation for Electromagnetic Fields in Inhomogeneous Media. In: Begehr, H., Gilbert, R. and Wah Wong, M., Eds., Progress in Analysis, Vol. 1, World Scientific, 361-366.Efremov, A.P. (2004) Hypercomples Numbers in Geometry and Physics, 1, 111-127.Acevedo, M., Lopez-Bonilla, J. and Sanchez-Meraz, M. (2005) Apeiron, 12, 371.Alexeyeva, L.A. (2013) Differential Algebra of Biquaternions. Dirac Equation and Its Generalized Solutions. Progress in Analysis. Proceedings of the 8th Congress of the ISAAC, Moscow, 22-27 August 2013, 153-161.Alexeyeva, L.A. (2013) Mathematical Journal, 13, 17-35. (In Russian) http://www.math.kz/images/journal/2013-1/Alexeyeva.pdfVladimirov, V.S. (1976) Generalized Functions in Mathematical Physics. Nauka Publisher, Moscow." ]

[ null ]

[ "# Complex number\n\n2020 Mathematics Subject Classification: Primary: 00-XX [MSN][ZBL]\n\nA complex number is a number of the form $z=x+iy$, where $x$ and $y$ are real numbers (cf. Real number) and $i=\\def\\i{\\sqrt{-1}}\\i$ is the so-called imaginary unit, that is, a number whose square is equal to $-1$ (in engineering literature, the notation $j=\\i$ is also used): $x$ is called the real part of the complex number $z$ and $y$ its imaginary part (written $x=\\def\\Re{\\mathrm{Re}\\;}\\Re z$, $y=\\def\\Im{\\mathrm{Im}\\;}\\Im z$). The real numbers can be regarded as special complex numbers, namely those with $y=0$. Complex numbers that are not real, that is, for which $y\\ne 0$, are sometimes called imaginary numbers. The complicated historical process of the development of the notion of a complex number is reflected in the above terminology which is mainly of traditional origin.\n\nAlgebraically speaking, a complex number is an element of the (algebraic) extension $\\C$ of the field of real numbers $\\R$ obtained by the adjunction to the field $\\R$ of a root $i$ of the polynomial $X^2+1$. The field $\\C$ obtained in this way is called the field of complex numbers or the complex number field. The most important property of the field $\\C$ is that it is algebraically closed, that is, any polynomial with coefficients in $\\C$ splits into linear factors. The property of being algebraically closed can be expressed in other words by saying that any polynomial of degree $n\\ge 1$ with coefficients in $\\C$ has at least one root in $\\C$ (the d'Alembert–Gauss theorem or fundamental theorem of algebra).\n\nThe field $\\C$ can be constructed as follows. The elements $z=(x,y)$, $z'=(x',y'),\\dots$ or complex numbers, are taken to be the points $z=(x,y)$, $z'=(x',y'),\\dots$ of the plane $\\R^2$ in Cartesian rectangular coordinates $x$ and $y$, $x'$ and $y',\\dots$. Here the sum of two complex numbers $z=(x,y)$ and $z'=(x',y')$ is the complex number $(x+x',y+y')$, that is, $$z+z'=(x,y)+(x',y')=(x+x',y+y'),\\label{1}$$ and the product of those complex numbers is the complex number $(xx'-yy',xy'+x'y)$, that is, $$zz'=(x,y)(x'y') = (xx'-yy',xy'+x'y).\\label{2}$$ The zero element $0=(0,0)$ is the same as the origin of coordinates, and the complex number $(1,0)$ is the identity of $\\C$.\n\nThe plane $\\R^2$ whose points are identified with the elements of $\\C$ is called the complex plane. The real numbers $x,x',\\dots$ are identified here with the points $(x,0)$, $(x',0),\\dots$ of the $x$-axis which, when referring to the complex plane, is called the real axis. The points $(0,y)=iy$, $(0,y')=iy',\\dots$ are situated on the $y$-axis, called the imaginary axis of the complex plane $\\C$; numbers of the form $iy,iy',\\dots$ are called pure imaginary. The representation of elements $z,z',\\dots$ of $\\C$, or complex numbers, as points of the complex plane with the rules (1) and (2) is equivalent to the above more widely used form of notating complex numbers: $$z=(x,y)=x+iy, z'=(x',y')=x'+iy',\\dots,$$ also called the algebraic or Cartesian form of writing complex numbers. With reference to the algebraic form, the rules (1) and (2) reduce to the simple condition that all operations with complex numbers are carried out as for polynomials, taking into account the property of the imaginary unit: $ii=i^2=-1$.\n\nThe complex numbers $z=(x,y)=x+iy$ and $\\bar z=(x,-y)=x-iy$ are called conjugate or [[complex conjugate]]s in the plane $\\C$; they are symmetrically situated with respect to the real axis. The sum and the product of two conjugate complex numbers are the real numbers $$z+\\bar z = 2\\Re z,\\quad z\\bar z=|z|^2,$$ where $|z|=r=\\sqrt{x^2+y^2}$ is called the modulus or absolute value of $z$.\n\nThe following inequalities always hold: $$|z|-|z'| \\le |z+z'|\\le |z|+|z'|.$$ A complex number $z$ is different from 0 if and only if $|z|>0$. The mapping $z\\mapsto \\bar z$ is an automorphism of the complex plane of order 2 (that is, $z = \\bar{\\bar z}$) that leaves all points of the real axis fixed. Furthermore, $\\overline{z+z'} = \\bar z + \\bar{z'}$, $\\bar{zz'} = \\bar{z}\\bar{z'}$.\n\nThe operations of addition and multiplication (1) and (2) are commutative and associative, they are related by the distributive law, and they have the inverse operations subtraction and division (except for division by zero). The latter are expressed in algebraic form as:\n\n$$z-z'=(x+iy)-(x'+iy')=(x-x')+i(y-y'),$$\n\n$$\\frac{z'}{z} = \\frac{x'+iy'}{x+iy} = \\frac{z\\bar z}{|z|^2} =\\frac{xx'+yy'}{x^2+y^2}+i\\frac{y'x-x'y}{x^2+y^2},\\quad z\\ne0.\\label{3}$$ Division of a complex number $z'$ by a complex number $z\\ne0$ thus reduces to multiplication of $z'$ by $$\\frac{\\bar z}{|z|^2} = \\frac{x}{x^2+y^2}-i\\frac{y}{x^2+y^2}.$$ It is an important question whether the extension $\\C$ of the field of reals constructed above, with the rules of operation indicated, is the only possible one or whether essentially different variants are conceivable. The answer is given by the uniqueness theorem: Every (algebraic) extension of the field $\\R$ obtained from $\\R$ by adjoining a root $i$ of the equation $X^2+1$ is isomorphic to $\\C$, that is, only the above rules of operation with complex numbers are compatible with the requirement that the root $i$ be algebraically adjoined. This fact, however, does not exclude the existence of interpretations of complex numbers other than as points of the complex plane. The following two interpretations are most frequently employed in applications.\n\nVector interpretation. A complex number $z=x+iy$ can be identified with the vector $(x,y)$ with coordinates $x$ and $y$ starting from the origin (see Fig.).", null, "Figure: c024140a\n\nIn this interpretation, addition and subtraction of complex numbers is carried out according to the rules of addition and subtraction of vectors. However, multiplication and division of complex numbers, which must be performed according to (2) and (3), do not have immediate analogues in vector algebra (see [Sh], [LaSc]). The vector interpretation of complex numbers is immediately applicable, for example, in electrical engineering in the description of alternating sinusoidal currents and voltages.\n\nMatrix interpretation. The complex number $w=u+iv$ can be identified with a $(2\\times 2)$-matrix of special type $$w=\\begin{pmatrix}\\phantom{-}u&v\\\\ -v&u\\end{pmatrix}$$ where the operations of addition, subtraction and multiplication are carried out according to the usual rules of matrix algebra.\n\nBy using polar coordinates in the complex plane $\\C$, that is, the radius vector $r=|z|$ and polar angle $\\def\\phi{\\varphi}\\phi=\\arg z$, called here the argument of $z$ (sometimes also called the phase of $z$), one obtains the trigonometric or polar form of a complex number: $$z=r(\\cos\\phi + i\\sin\\phi),\\label{4}$$\n\n$$r\\cos\\phi = \\Re z,\\quad r\\sin\\phi=\\Im z.$$ The argument $\\phi=\\arg z$ is a many-valued real-valued function of the complex number $z\\ne 0z\\ne 0$, whose values for a given $z$ differ by integral multiples of $2\\pi i$; the argument of the complex number $z=0$ is not defined. One usually takes the principal value of the argument $\\phi = \\def\\Arg{\\mathrm{Arg}} \\Arg z$, defined by the additional condition $-\\pi < \\Arg z \\le \\pi$. The Euler formulas $e^{\\pm i\\phi} = \\cos\\phi\\pm i\\sin\\phi$ transform the trigonometric form (4) into the exponential form of a complex number: $$z=re^{i\\phi}\\label{5}$$ The forms (4) and (5) are particularly suitable for carrying out multiplication and division of complex numbers: $$zz'=rr'[\\cos(\\phi+\\phi')+i\\sin(\\phi+\\phi')]=rr'e^{i(\\phi+\\phi')},$$\n\n$$\\frac{z}{z'}=\\frac{r}{r'}[\\cos(\\phi-\\phi')+i\\sin(\\phi-\\phi')] =\\frac{r}{r'}e^{i(\\phi-\\phi')},\\quad r>0$$ Under multiplication (or division) of complex numbers the moduli are multiplied (or divided) and the arguments are added (or subtracted). Raising to a power or extracting a root is carried out according to the so-called de Moivre formulas: $$z^n = r^n(\\cos n\\phi + i\\sin n\\phi) = r^n e^{in\\phi},$$\n\n$$z^{1/n} = r^{1/n}\\Big(\\cos\\frac{\\phi+2k\\pi}{n}+i\\sin\\frac{\\phi+2k\\pi}{n}\\Big) =r^{1/n}e^{i(\\phi+2k\\pi)/n},$$\n\n$$k=0,\\dots,n-1,$$ where the first of these is also applicable for negative integer exponents $n$. Geometrically, multiplication of a complex number $z$ by a complex number $z'=r'e^{i\\phi'}$ reduces to rotating the vector $z$ over the angle $\\phi'$ (anti-clockwise if $\\phi'>0$) and subsequently multiplying its length by $|z'|=r'$; in particular, multiplication by a complex number $z'=e^{i\\phi'}$, which has modulus one, is merely rotation over the angle $\\phi'$. Thus, complex numbers can be interpreted as operators of a special type (affinors, cf. Affinor). In this connection, the mixed vector-matrix interpretation of multiplication of complex numbers is sometimes useful: $$(x,y)\\begin{pmatrix}\\phantom{-}u&v\\\\ -v&u\\end{pmatrix}=(xu-yv,xv+yu),$$ in which the multiplicand is treated as a matrix-vector and the multiplier as a matrix-operator.\n\nThe bijection $(x,y)\\mapsto x+iy$ induces on the field $\\C$ the topology of the $2$-dimensional real vector space $\\R^2$; this topology is compatible with the field structure of $\\C$ and thus $\\C$ is a topological field. The modulus $|z|$ is the Euclidean norm of the complex number $z={x,y}$, and $\\C$ endowed with this norm is a complex one-dimensional Euclidean space, also called the complex $z$-plane. The topological product $\\C^n=\\C\\times\\cdots\\times\\C$ ($n$ times, $n\\ge 1$) is a complex $n$-dimensional Euclidean space. For a satisfactory analysis of functions it is usually necessary to consider their behaviour in the complex domain. This is due to the fact that $\\C$ is algebraically closed. Even the behaviour of such elementary functions as $z^n$, $\\cos z$, $\\sin z$, $e^z$ can be properly understood only when they are regarded as functions of a complex variable (see Analytic function).\n\nApparently, imaginary quantities first occurred in the celebrated work The great art, or the rules of algebra by G. Cardano, 1545, who regarded them as useless and unsuitable for applications. R. Bombelli (1572) was the first to realize the value of the use of imaginary quantities, in particular for the solution of the cubic equation in the so-called irreducible case (when the real roots are expressed in terms of cube roots of imaginary quantities, cf. Cardano formula). He gave some of the simplest rules of operation with complex numbers. In general, expressions of the form $a+b\\i$, $b\\ne 0$, appearing in the solution of quadratic and cubic equations were called \"imaginary\" in the 16th century and 17th century. However, even for many of the great scholars of the 17th century, the algebraic and geometric nature of imaginary quantities was unclear and even mystical. It is known, for example, that I. Newton did not include imaginary quantities within the notion of number, and that G. Leibniz said that \"complex numbers are a fine and wonderful refuge of the divine spirit, as if it were an amphibian of existence and non-existence\" .\n\nThe problem of expressing the $n$-th roots of a given number was mainly solved in the papers of A. de Moivre (1707, 1724) and R. Cotes (1722). The symbol $i=\\i$ was proposed by L. Euler (1777, published 1794). It was he who in 1751 asserted that the field $\\C$ is algebraically closed; J. d'Alembert (1747) came to the same conclusion. The first rigorous proof of this fact is due to C.F. Gauss (1799), who introduced the term \"complex number\" in 1831. The complete geometric interpretation of complex numbers and operations on them appeared first in the work of C. Wessel (1799). The geometric representation of complex numbers, sometimes called the \"Argand diagramArgand diagram\" , came into use after the publication in 1806 and 1814 of papers by J.R. Argand, who rediscovered, largely independently, the findings of Wessel.\n\nThe purely arithmetic theory of complex numbers as pairs of real numbers was introduced by W. Hamilton (1837). He found a generalization of complex numbers, namely the quaternions (cf. Quaternion), which form a non-commutative algebra. More generally, it was proved at the end of the 19th century that any extension of the notion of number beyond the complex numbers requires sacrificing some property of the usual operations (primarily commutativity). See also Hypercomplex number; Double and dual numbers; Cayley numbers.\n\nHow to Cite This Entry:\nComplex number. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Complex_number&oldid=53376\nThis article was adapted from an original article by E.D. Solomentsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article" ]

[ null, "https://www.encyclopediaofmath.org/legacyimages/common_img/c024140a.gif", null ]

[ "", null, "Exploring Alternative Energy Energy Storage Technologies Energy Conversion\n\nEnergy Source: The Sun\n\nPhotovoltaic    Heat    Wind\n\nStorage Technologies\n\nElectrical    Mechanical    Hydrogen", null, "## Energy Conversion\n\nDefinition: The process of changing one form of energy into another. For example, the chemical energy in gasoline can be converted into kinetic energy (energy of motion) by an automobile engine. Definition obtained from the K - 12 Energy Education Program (KEEP).\n\n#### The principles of energy conservation described by the thermodynamic laws are demonstrated in the various energy conversion processes that follow.\n\nThis physics/engineering project begins with the sun; its radiant energy is converted to electrical energy by photovoltaic (solar) cells.  Electric energy is then converted to useful mechanical energy by motors.  And mechanical energy is converted to electrical energy by electric generators.  The Energy Story is a good website that describes what energy is.  How energy and work are related.  An excellent website that explains the relationship between electric power and energy is here.\n\nPhotovoltaic cells convert sunlight photons to electrons: Motors convert electrical energy (electricity) to mechanical energy:  Generators convert mechanical energy to electric energy.  Good website for kidsFT Exploring has a good write up on Power and Energy.  The basic relationship between mass, force, torque, work, power and energy are explained here.  To Summarize: Power is the rate at which work is performed or energy is converted. Power is energy per unit of time. Power is:\n\nP = W/t = E/t\n\nWhere P is Power; t is Time; W is Work; Power is work per unit time; and E is energy; Power is energy converted per unit time.\n\nThe sun's electromagnetic radiant energy is in fact optical energy and the instruments that measure optical energy are also used to measure solar energy.\n\nVolt meters measure the electro-motive force (tension or pressure) that causes an electrical charge to flow (current) and current meters measure the flow rate of electric charge (current).\n\nA Gauss meter measures magnetic field strength.  Electromagnetic induction is used to convert electrical energy to mechanical energy and it is also used to convert mechanical energy to electrical energy.\n\nGood website describes how magnetism is used by motors to convert electrical energy to mechanical energy.  Another excellent website describes how magnetism is used in motors and generators to convert electrical energy to mechanical energy and how mechanical energy is converted to electrical energy.\n\nTachometers measure the rotational speed of a motor's shaft and torque meter measures a motor's torque.\n\nThe above instruments are used to demonstrate the conservation of energy principles when energy is converted from one form to another.  Let's take a look at how all this is done.\n\n### Photovoltaic Cells convert electromagnetic radiant power to electrical power.\n\nCalifornia Solar Engineering has a short course that explains how solar panels work and how they are constructed.  A fantastic website for young and old on how Photovoltaic cells convert solar energy to electricity is Green - Planet - Solar - Energy.com.  Enough can't be said about this website for learning the basics of solar energy.  To understand how solar photons are converted to flowing electrons, the chemical theory of materials, specifically Silicon, must first be understood.  The Green - Planet - Solar - Energy.com website has a fantastic column of links on the left side.  Scroll down to Some Theory and underneath it click on Solar Chemistry 1.  There you will find the clearest explanation of the atomic electron structure of materials and how photons free electrons in Silicon material.\n\nWikipedia has a good definition of Photovoltaic Cells.  International Light has The Light Measurement Handbook that explains how light energy is measured.  The Solar Energy Science Project  will present to you only the essentials to understand and to measure optical power.\n\n### Motors use magnets and electromagnetism to convert electric power to mechanical power.\n\nThe basics of how DC motors use electromagnetic fields to cause a motor's shaft to rotate are illustrated here .  Wikipedia explains how motor and generators use magnetic fields and current carrying conductors to convert energy.  A short write up from the Electric Motor Handbook explains how a motor converts electrical power to mechanical power.  The Electric Motor Handbook states that in most machines dissipation from losses is small enough to approximate mechanical power with electrical power.  This fact will be investigated in this website.  We provide a means of measuring mechanical power using a torque/rpm measurement apparatus.\n\n### Generators use magnets and electromagnetism to convert mechanical power to electrical power.\n\nBrian Kross has a brief overview how generators convert energy from what originally started out as solar energy.  Please scroll down to Direct Current Generators of this website for detail how a generator uses electromagnetism to convert mechanical energy to electrical energy.\n\n### Energy Conversion Units\n\nConservation of energy.  The Joule and the Watt are some of the energy and power units used in optical, electrical and mechanical engineering work.  How electrical energy is measured.  The Joule and the watt hour are a measure of energy.  1 watt-hour = 3600 Joules of energy.  These units must be clearly understood.  A Brief discussion of Energy Conversion Units is an informative website.  A general description of Energy Conversion and Thermodynamics is presented by the K-12 Energy Education Program website.\n\nTo get started with the detail of how to measure power go to the Power Measurement page." ]

[ null, "http://apptechy.com/images/Starlike_Sun_Small.jpg", null, "http://apptechy.com/images/solar%20cells%20and%20windmill%20compressed.jpg", null ]

[ "1 simpletest.test `SimpleTestFunctionalTest::getTestResults()`\n\nGet the results from a test and store them in the class array \\$results.\n\n### File\n\ncore/modules/simpletest/tests/simpletest.test, line 247\nTests for simpletest.module.\n\n### Class\n\nSimpleTestFunctionalTest\n\n### Code\n\n``````function getTestResults() {\n\\$results = array();\nif (\\$this->parse()) {\nif (\\$fieldset = \\$this->getResultFieldSet()) {\n// Code assumes this is the only test in group.\n\\$results['summary'] = \\$this->asText(\\$fieldset->div->div);\n\\$results['name'] = \\$this->asText(\\$fieldset->legend);\n\n\\$results['assertions'] = array();\n\\$tbody = \\$fieldset->div->table->tbody;\nforeach (\\$tbody->tr as \\$row) {\n\\$assertion = array();\n\\$assertion['message'] = \\$this->asText(\\$row->td);\n\\$assertion['type'] = \\$this->asText(\\$row->td);\n\\$assertion['file'] = \\$this->asText(\\$row->td);\n\\$assertion['line'] = \\$this->asText(\\$row->td);\n\\$assertion['function'] = \\$this->asText(\\$row->td);\n\\$ok_url = file_create_url('core/misc/watchdog-ok.png');\n\\$assertion['status'] = (\\$row->td->img['src'] == \\$ok_url) ? 'Pass' : 'Fail';\n\\$results['assertions'][] = \\$assertion;\n}\n}\n}\n\\$this->childTestResults = \\$results;\n}\n``````" ]

[ null ]

[ "", null, "", null, "", null, "SUSB-011\n\n1. Preparation of Standard KHP Solution\n\nKHP is a monoprotic acid of known molar mass = 204.22 g/mol = 204.22 mg/mmol.  One mole of the acid reacts with one mole of NaOH.  We make a solution of precisely known volume (250.0 mL) from an accurately weighed sample of KHP.   The concentration of that solution is known and fixed.  E.g.,\n\nWeight of vial + KHP                                                                        18.7956 g\nWeight of vial - sample of KHP                                                          16.3612 g\n\nWeight of sample of KHP                                                                   2.4344 g\n\nmmol KHP in sample     2434.4 mg / 204.22 mg/mmol    =               11.920 mmol      (5 sig figs)\n\nVolume of solution                                                                             250.0 mL          (4 sig figs)\n\nConcentration of KHP solution   11.920 mmol / 250.0 mL = 0.04769 mmol/ml = 0.04769 M     (4 sig figs)\n\nNote that the entire remainder of the exercise depends on the accuracy of this concentration!\n\n2. Standardization of NaOH\n\nStep 1 produces a solution of an acid of known concentration (0.04769 M).  We now use that solution to determine the concentration of our NaOH solution.  In this exercise, we do this by using two burets -- one to deliver the standard acid solution (KHP) and one to deliver the NaOH solution whose concentration is to be determined.\nThe only buret readings that matter are the initial and final readings.  Intermediate readings made if we backtitrate are irrelelevant to the results.\n\nKHP Buret                    NaOH Buret\nFinal buret reading                                                      38.76 mL                         15.94 mL\n\nInitial buret reading                                                       4.53 mL                           2.84 mL\n\nNet volume                                                                34.23 mL                         13.10 mL\n\nmmol KHP    34.23 mL * 0.04769 mmol / mL  =           1.632 mmol    (4 sig figs)\n\nmmol NaOH ( same as mmol KHP)                     =           1.632 mmol    (4 sig figs)\n\nConcentration of NaOH   1.632 mmol / 13.10 mL =      0.1246 mmol / mL = 0.1246 M  (4 sig figs)\n\nSuppose two more standardizations are conducted, yielding the values 0.1228 M and 0.1260 M respectively.\n\nWe calculate the mean (average) of the three concentrations\n\nCavg = ( 0.1246 + 0.1228 + 0.1260 ) / 3 = 0.1245 M   (4 sig figs)\n\nThe average deviation is given by\n\n(  | 0.1246 - 0.1245 | + | 0.1228 - 0.1245 | + | 0.1260 - 0.1245 | ) / 3 = 0.0011 M         (2 sig figs)\n\nThe result can be written as                              Cavg0.1245 M ± 0.0011 M\n\nThe percentage deviation is 100 * 0.0011 / 0.1245 = 0.88 %            (2 sig figs)\n\nWe have used our standard KHP solution to determine the concentration of the NaOH solution that we will use to titrate the vinegar!  Both the accuracy and precision of this step will limit the accuracy with which we can do the next step.  The following aspects of the standardization can give rise to poor precision:\n\n• inaccuracy in the preparation of the KHP solution\n• inconsistency in recognizing the end point\n• incomplete mixing of the KHP solution\n\n3. Determination of the Concentration of an unknown acetic acid solution\n\nNow we use our standardized NaOH solution to titrate a solution of acetic acid (vinegar) of unknown concentration. We prepare our sample by pipeting 5.00 mL of the unknown solution into a clean 250.0 mL Erlenmeyer flask and dilute the solution with approximately 40 mL of distilled water. (As noted in the lecture notes - the precision of the pipet limits the overall precision.)\n\n Run 1 Run 2 Run 3 Final buret reading 21.84 22.64 22.25 Initial buret reading 0.67 1.39 0.88 Net volume of NaOH 21.17 21.25 21.37 Concentration of NaOH (from standardization) 0.1245 0.1245 0.1245 mmol NaOH 2.635 2.646 2.661 mmol Acetic Acid (same as mmol NaOH) 2.635 2.646 2.661 Volume of Acetic Acid sample titrated (mL) 5.00 5.00 5.00 Concentration of Acetic Acid (mmol/mL = mol/L) 0.527 0.529 0.532\n\nThe average of the results is ( 0.527 + 0.529 + 0.532 )/3 = 0.529    [Note that we have only 3 significant figures in the concentration.]\n\nThe average deviation is (0.002 + 0.000 + 0.003)/3 = 0.002\n\nThe percent deviation is 100 * 0.002/0.529 = 0.4%\n\n4. Determination of the Acid Ionization Constant, Ka, of acetic acid\n\nThe acid ionization constant is defined as the equilibrium constant for the acid ionization reaction,\n\nHOAc = H+ + OAc-                   Ka =   [ H+ ] [ OAc- ] / [ HOAc ]\n\nTo calculate this quantity, we note that for an acid that is weak ( but stronger than pure water ), the virtually sole source of hydrogen ions is through this ionization. Therefore, the concentrations of H+ and OAc- will be equal. Also, since acetic acid is weak, only a small amount of it will ionize by definition. We suppose that the concentration of the acid is sufficient that we can ignore the decrease in the concentration of the undisoociated acid, HOAc, due to ionization compared to the nominal concentration of the acid.\n\nMeasuring the pH of the acid solution gives us a value for [ H+ ]   ( [ H+ ] = 10-pH),   and therefore, also [ OAc- ]. The concentration of the acid is the value that we found through the titrations, 0.529 M corrected for the amount whixh is dissociated. Suppose the pH is measured to be 2.45. That gives us [ H+ ] = [ OAc- ] = 10-2.45 = 3.6 X 10-3. So, the correction to the undissociated form is 0.0036\n\nso\n\nSubstituing our values in the expression for Ka gives us:\n\nKa = ( 3.6 X 10-3 ) ( 3.6 X 10-3 ) /( 0.529 -.0036 ) = 2.47 X 10-5" ]

[ null, "http://www.ic.sunysb.edu/Class/che133/images/WIZ.jpg", null, "http://www.ic.sunysb.edu/Class/che133/images/calculations.jpg", null, "http://www.ic.sunysb.edu/Class/che133/images/WITCH.jpg", null ]

[ "Random-projection ensemble classification\n\n# Random-projection ensemble classification\n\nTimothy I. Cannings and Richard J. Samworth\n###### Abstract\n\nWe introduce a very general method for high-dimensional classification, based on careful combination of the results of applying an arbitrary base classifier to random projections of the feature vectors into a lower-dimensional space. In one special case that we study in detail, the random projections are divided into disjoint groups, and within each group we select the projection yielding the smallest estimate of the test error. Our random projection ensemble classifier then aggregates the results of applying the base classifier on the selected projections, with a data-driven voting threshold to determine the final assignment. Our theoretical results elucidate the effect on performance of increasing the number of projections. Moreover, under a boundary condition implied by the sufficient dimension reduction assumption, we show that the test excess risk of the random projection ensemble classifier can be controlled by terms that do not depend on the original data dimension and a term that becomes negligible as the number of projections increases. The classifier is also compared empirically with several other popular high-dimensional classifiers via an extensive simulation study, which reveals its excellent finite-sample performance.\n\nKeywords: Aggregation; Classification; High-dimensional; Random projection\n\n## 1 Introduction\n\nSupervised classification concerns the task of assigning an object (or a number of objects) to one of two or more groups, based on a sample of labelled training data. The problem was first studied in generality in the famous work of Fisher (1936), where he introduced some of the ideas of Linear Discriminant Analysis (LDA), and applied them to his Iris data set. Nowadays, classification problems arise in a plethora of applications, including spam filtering, fraud detection, medical diagnoses, market research, natural language processing and many others.\n\nIn fact, LDA is still widely used today, and underpins many other modern classifiers; see, for example, Friedman (1989) and Tibshirani et al. (2002). Alternative techniques include support vector machines (Cortes and Vapnik, 1995), tree classifiers and random forests (Breiman et al., 1984; Breiman, 2001), kernel methods (Hall and Kang, 2005) and nearest neighbour classifiers (Fix and Hodges, 1951). More substantial overviews and in-depth discussion of these techniques, and others, can be found in Devroye, Györfi and Lugosi (1996) and Hastie et al. (2009).\n\nAn increasing number of modern classification problems are high-dimensional, in the sense that the dimension of the feature vectors may be comparable to or even greater than the number of training data points, . In such settings, classical methods such as those mentioned in the previous paragraph tend to perform poorly (Bickel and Levina, 2004), and may even be intractable; for example, this is the case for LDA, where the problems are caused by the fact that the sample covariance matrix is not invertible when .\n\nMany methods proposed to overcome such problems assume that the optimal decision boundary between the classes is linear, e.g. Friedman (1989) and Hastie et al. (1995). Another common approach assumes that only a small subset of features are relevant for classification. Examples of works that impose such a sparsity condition include Fan and Fan (2008), where it is also assumed that the features are independent, as well as Tibshirani et al. (2003), where soft thresholding is used to obtain a sparse boundary. More recently, Witten and Tibshirani (2011) and Fan, Feng and Tong (2012) both solve an optimisation problem similar to Fisher’s linear discriminant, with the addition of an penalty term to encourage sparsity.\n\nIn this paper we attempt to avoid the curse of dimensionality by projecting the feature vectors at random into a lower-dimensional space. The use of random projections in high-dimensional statistical problems is motivated by the celebrated Johnson–Lindenstrauss Lemma (e.g. Dasgupta and Gupta, 2002). This lemma states that, given , and , there exists a linear map such that\n\n (1−ϵ)∥xi−xj∥2≤∥f(xi)−f(xj)∥2≤(1+ϵ)∥xi−xj∥2,\n\nfor all . In fact, the function that nearly preserves the pairwise distances can be found in randomised polynomial time using random projections distributed according to Haar measure, as described in Section 3 below. It is interesting to note that the lower bound on in the Johnson–Lindenstrauss lemma does not depend on ; this lower bound is optimal up to constant factors (Larsen and Nelson, 2016). As a result, random projections have been used successfully as a computational time saver: when is large compared to , one may project the data at random into a lower-dimensional space and run the statistical procedure on the projected data, potentially making great computational savings, while achieving comparable or even improved statistical performance. As one example of the above strategy, Durrant and Kabán (2013) obtained Vapnik–Chervonenkis type bounds on the generalisation error of a linear classifier trained on a single random projection of the data. See also Dasgupta (1999), Ailon and Chazelle (2006) and McWilliams et al. (2014) for other instances.\n\nOther works have sought to reap the benefits of aggregating over many random projections. For instance, Marzetta, Tucci and Simon (2011) considered estimating a population inverse covariance (precision) matrix using , where denotes the sample covariance matrix and are random projections from to . Lopes, Jacob and Wainwright (2011) used this estimate when testing for a difference between two Gaussian population means in high dimensions, while Durrant and Kabán (2015) applied the same technique in Fisher’s linear discriminant for a high-dimensional classification problem.\n\nOur proposed methodology for high-dimensional classification has some similarities with the techniques described above, in the sense that we consider many random projections of the data, but is also closely related to bagging (Breiman, 1996), since the ultimate assignment of each test point is made by aggregation and a vote. Bagging has proved to be an effective tool for improving unstable classifiers. Indeed, a bagged version of the (generally inconsistent) -nearest neighbour classifier is universally consistent as long as the resample size is carefully chosen, see Hall and Samworth (2005); for a general theoretical analysis of majority voting approaches, see also Lopes (2016). Bagging has also been shown to be particularly effective in high-dimensional problems such as variable selection (Meinshuasen and Bühlmann, 2010; Shah and Samworth, 2013). Another related approach to ours is Blaser and Fryzlewicz (2015), who consider ensembles of random rotations, as opposed to projections.", null, "Figure 1: Different two-dimensional projections of 200 observations in p=50 dimensions. Top row: three projections drawn from Haar measure; bottom row: the projected data after applying the projections with smallest estimate of test error out of 100 Haar projections with LDA (left), Quadratic Discriminant Analysis (middle) and k-nearest neighbours (right).\n\nOne of the basic but fundamental observations that underpins our proposal is the fact that aggregating the classifications of all random projections is not always sensible, since many of these projections will typically destroy the class structure in the data; see the top row of Figure 1. For this reason, we advocate partitioning the projections into disjoint groups, and within each group we retain only the projection yielding the smallest estimate of the test error. The attraction of this strategy is illustrated in the bottom row of Figure 1, where we see a much clearer partition of the classes. Another key feature of our proposal is the realisation that a simple majority vote of the classifications based on the retained projections can be highly suboptimal; instead, we argue that the voting threshold should be chosen in a data-driven fashion in an attempt to minimise the test error of the infinite-simulation version of our random projection ensemble classifier. In fact, this estimate of the optimal threshold turns out to be remarkably effective in practice; see Section 5.2 for further details. We emphasise that our methodology can be used in conjunction with any base classifier, though we particularly have in mind classifiers designed for use in low-dimensional settings. The random projection ensemble classifier can therefore be regarded as a general technique for either extending the applicability of an existing classifier to high dimensions, or improving its performance. The methodology is implemented in an R package RPEnsemble (Cannings and Samworth, 2016).\n\nOur theoretical results are divided into three parts. In the first, we consider a generic base classifier and a generic method for generating the random projections into and quantify the difference between the test error of the random projection ensemble classifier and its infinite-simulation counterpart as the number of projections increases. We then consider selecting random projections from non-overlapping groups by initially drawing them according to Haar measure, and then within each group retaining the projection that minimises an estimate of the test error. Under a condition implied by the widely-used sufficient dimension reduction assumption (Li, 1991; Cook, 1998; Lee et al., 2013), we can then control the difference between the test error of the random projection classifier and the Bayes risk as a function of terms that depend on the performance of the base classifier based on projected data and our method for estimating the test error, as well as a term that becomes negligible as the number of projections increases. The final part of our theory gives risk bounds for the first two of these terms for specific choices of base classifier, namely Fisher’s linear discriminant and the -nearest neighbour classifier. The key point here is that these bounds only depend on , the sample size and the number of projections, and not on the original data dimension .\n\nThe remainder of the paper is organised as follows. Our methodology and general theory are developed in Sections 2 and 3. Specific choices of base classifier as well as a general sample splitting strategy are discussed in Section 4, while Section 5 is devoted to a consideration of the practical issues of computational complexity, choice of voting threshold, projected dimension and the number of projections used. In Section 6 we present results from an extensive empirical analysis on both simulated and real data where we compare the performance of the random projection ensemble classifier with several popular techniques for high-dimensional classification. The outcomes are very encouraging, and suggest that the random projection ensemble classifier has excellent finite-sample performance in a variety of different high-dimensional classification settings. We conclude with a discussion of various extensions and open problems. Proofs are given in the Appendix and the supplementary material Cannings and Samworth (2017), which appears below the reference list.\n\nFinally in this section, we introduce the following general notation used throughout the paper. For a sufficiently smooth real-valued function defined on a neighbourhood of , let and denote its first and second derivatives at , and let and denote the integer and fractional part of respectively.\n\n## 2 A generic random projection ensemble classifier\n\nWe start by describing our setting and defining the relevant notation. Suppose that the pair takes values in , with joint distribution , characterised by , and , the conditional distribution of , for . For convenience, we let . In the alternative characterisation of , we let denote the marginal distribution of and write for the regression function. Recall that a classifier on is a Borel measurable function , with the interpretation that we assign a point to class . We let denote the set of all such classifiers.\n\nThe test error of a classifier isWe define through an integral rather than to make it clear that when is random (depending on training data or random projections), it should be conditioned on when computing .\n\n R(C):=∫Rp×{0,1}\\mathds1{C(x)≠y}dP(x,y),\n\nand is minimised by the Bayes classifier\n\n CBayes(x):={1if η(x)≥1/2;0otherwise\n\n(e.g. Devroye, Györfi and Lugosi, 1996, p. 10). Its risk is .\n\nOf course, we cannot use the Bayes classifier in practice, since is unknown. Nevertheless, we often have access to a sample of training data that we can use to construct an approximation to the Bayes classifier. Throughout this section and Section 3, it is convenient to consider the training sample to be fixed points in . Our methodology will be applied to a base classifier , which we assume can be constructed from an arbitrary training sample of size in ; thus is a measurable function from to .\n\nNow assume that . We say a matrix is a projection if , the -dimensional identity matrix. Let be the set of all such matrices. Given a projection , define projected data and for , and let . The projected data base classifier corresponding to is , given by\n\n CAn(x)=CAn,TAn(x):=Cn,TAn(Ax).\n\nNote that although is a classifier on , the value of only depends on through its -dimensional projection .\n\nWe now define a generic ensemble classifier based on random projections. For , let denote independent and identically distributed projections in , independent of . The distribution on is left unspecified at this stage, and in fact our proposed method ultimately involves choosing this distribution depending on .\n\nNow set\n\n νn(x)=ν(B1)n(x):=1B1B1∑b1=1\\mathds1{CAb1n(x)=1}. (0)\n\nFor , the random projection ensemble classifier is defined to be\n\n CRPn(x):={1if νn(x)≥α;0otherwise. (0)\n\nWe emphasise again here the additional flexibility afforded by not pre-specifying the voting threshold to be . Our analysis of the random projection ensemble classifier will require some further definitions. LetIn order to distinguish between different sources of randomness, we will write and for the probability and expectation, respectively, taken over the randomness from the projections . If the training data is random, then we condition on when computing and .\n\n μn(x):=E{νn(x)}=P{CA1n(x)=1}.\n\nFor , define distribution functions by . Note that since is non-decreasing it is differentiable almost everywhere; in fact, however, the following assumption will be convenient:\n\nAssumption 1. and are twice differentiable at .\n\nThe first derivatives of and , when they exist, are denoted as and respectively; under assumption 1, these derivatives are well-defined in a neighbourhood of . Our first main result below gives an asymptotic expansion for the expected test error of our generic random projection ensemble classifier as the number of projections increases. In particular, we show that this expected test error can be well approximated by the test error of the infinite-simulation random projection classifier\n\n CRP∗n(x):={1if μn(x)≥α;0otherwise.\n\nNote that provided and are continuous at , we have\n\n R(CRP∗n)=π1Gn,1(α)+π0{1−Gn,0(α)}. (0)\n###### Theorem 1\n\nAssume assumption 1. Then\n\n E{R(CRPn)}−R(CRP∗n)=γn(α)B1+o(1B1)\n\nas , where\n\n γn(α):=(1−α−\\llbracketB1α\\rrbracket){π1gn,1(α)−π0gn,0(α)}+α(1−α)2{π1˙gn,1(α)−π0˙gn,0(α)}.\n\nThe proof of Theorem 1 in the Appendix is lengthy, and involves a one-term Edgeworth approximation to the distribution function of a standardised Binomial random variable. One of the technical challenges is to show that the error in this approximation holds uniformly in the binomial proportion. Related techniques can also be used to show that under assumption 1; see Proposition 4 in the supplementary material. Very recently, Lopes (2016) has obtained similar results to this and to Theorem 1 in the context of majority vote classification, with stronger assumptions on the relevant distributions and on the form of the voting scheme. In Figure 2, we plot the average error (plus/minus two standard deviations) of the random projection ensemble classifier in one numerical example, as we vary ; this reveals that the Monte Carlo error stabilises rapidly, in agreement with what Lopes (2016) observed for a random forest classifier.", null, "Figure 2: The average error (black) plus/minus two standard deviations (red) over 20 sets of B1B2 projections for B1∈{2,…,500}. We use the LDA (left), QDA (middle) and knn (right) base classifiers. The plots show the test error for one training dataset from Model 2; the other parameters are n=50, p=100, d=5 and B2=50.\n\nOur next result controls the test excess risk, i.e. the difference between the expected test error and the Bayes risk, of the random projection classifier in terms of the expected test excess risk of the classifier based on a single random projection. An attractive feature of this result is its generality: no assumptions are placed on the configuration of the training data , the distribution of the test point or on the distribution of the individual projections.\n\n###### Theorem 2\n\nFor each , we have\n\n E{R(CRPn)}−R(CBayes)≤1min(α,1−α)[E{R(CA1n)}−R(CBayes)]. (0)\n\nWhen , we interpret in Theorem 2 as . In fact, when and and are continuous, the bound in Theorem 2 can be improved if one is using an ‘oracle’ choice of the voting threshold , namely\n\n α∗∈argminα′∈[0,1]R(CRP∗n,α′)=argminα′∈[0,1][π1Gn,1(α′)+π0{1−Gn,0(α′)}], (0)\n\nwhere we write to emphasise the dependence on the voting threshold . In this case, by definition of and then applying Theorem 2,\n\n R(CRP∗n,α∗)−R(CBayes)≤R(CRP∗n,1/2)−R(CBayes)≤2[E{R(CA1n)}−R(CBayes)], (0)\n\nwhich improves the bound in (2) since . It is also worth mentioning that if assumption 1 holds at , and and are continuous, then and the constant in Theorem 1 simplifies to\n\n γn(α∗)=α∗(1−α∗)2{π1˙gn,1(α∗)−π0˙gn,0(α∗)}≥0.\n\n## 3 Choosing good random projections\n\nIn this section, we study a special case of the generic random projection ensemble classifier introduced in Section 2, where we propose a screening method for choosing the random projections. Let be an estimator of , based on , that takes values in the set . Examples of such estimators include the training error and leave-one-out estimator; we discuss these choices in greater detail in Section 4. For , let denote independent projections, independent of , distributed according to Haar measure on . One way to simulate from Haar measure on the set is to first generate a matrix , where each entry is drawn independently from a standard normal distribution, and then take to be the matrix of left singular vectors in the singular value decomposition of (see, for example, Chikuse, 2003, Theorem 1.5.4). For , let\n\n b∗2(b1):=sargminb2∈{1,…,B2}RAb1,b2n, (0)\n\nwhere denotes the smallest index where the minimum is attained in the case of a tie. We now set , and consider the random projection ensemble classifier from Section 2 constructed using the independent projections .\n\nLet\n\n R∗n:=minA∈ARAn\n\ndenote the optimal test error estimate over all projections. The minimum is attained here, since takes only finitely many values. We assume the following:\n\nAssumption 2. There exists such that\n\n P(RA1,1n≤R∗n+|ϵn|)≥β,\n\nwhere .\n\nThe quantity , which depends on because is selected from independent random projections, can be interpreted as a measure of overfitting. Assumption 2 asks that there is a positive probability that is within of its minimum value . The intuition here is that spending more computational time choosing a projection by increasing is potentially futile: one may find a projection with a lower error estimate, but the chosen projection will not necessarily result in a classifier with a lower test error. Under this condition, the following result controls the test excess risk of our random projection ensemble classifier in terms of the test excess risk of a classifier based on -dimensional data, as well as a term that reflects our ability to estimate the test error of classifiers based on projected data and a term that depends on the number of projections.\n\n###### Theorem 3\n\nAssume assumption 2. Then, for each , and every ,\n\n E{R(CRPn)}−R(CBayes)≤R(CAn)−R(CBayes)min(α,1−α)+2|ϵn|−ϵAnmin(α,1−α)+(1−β)B2min(α,1−α), (0)\n\nwhere .\n\nRegarding the bound in Theorem 3 as a sum of three terms, we see that the final one can be seen as the price we have to pay for the fact that we do not have access to an infinite sample of random projections. This term can be made negligible by choosing to be sufficiently large, though the value of required to ensure it is below a prescribed level may depend on the training data. It should also be noted that in the second term may increase with , which reflects the fact mentioned previously that this quantity is a measure of overfitting. The behaviour of the first two terms depends on the choice of base classifier, and our aim is to show that under certain conditions, these terms can be bounded (in expectation over the training data) by expressions that do not depend on .\n\nTo this end, define the regression function on induced by the projection to be . The corresponding induced Bayes classifier, which is the optimal classifier knowing only the distribution of , is given by\n\n CA−Bayes(z):={1if ηA(z)≥1/2;0otherwise.\n\nIn order to give a condition under which there exists a projection for which is close to the Bayes risk, we will invoke an additional assumption on the form of the Bayes classifier:\n\nAssumption 3. There exists a projection such that\n\n PX({x∈Rp:η(x)≥1/2}△{x∈Rp:ηA∗(A∗x)≥1/2})=0,\n\nwhere denotes the symmetric difference of two sets and .\n\nAssumption 3 requires that the set of points assigned by the Bayes classifier to class 1 can be expressed as a function of a -dimensional projection of . Note that if the Bayes decision boundary is a hyperplane, then assumption 3 holds with . Moreover, Proposition 1 below shows that, in fact, assumption 3 holds under the sufficient dimension reduction condition, which states that is conditionally independent of given ; see Cook (1998) for many statistical settings where such an assumption is natural.\n\n###### Proposition 1\n\nIf is conditionally independent of given , then assumption 3 holds.\n\nThe following result confirms that under assumption 3, and for a sensible choice of base classifier, we can hope for to be close to the Bayes risk.\n\n###### Proposition 2\n\nAssume assumption 3. Then .\n\nWe are therefore now in a position to study the first two terms in the bound in Theorem 3 in more detail for specific choices of base classifier.\n\n## 4 Possible choices of the base classifier\n\nIn this section, we change our previous perspective and regard the training data as independent random pairs with distribution , so our earlier statements are interpreted conditionally on . For , we write our projected data as , where and . We also write and to refer to probabilities and expectations over all random quantities. We consider particular choices of base classifier, and study the first two terms in the bound in Theorem 3.\n\n### 4.1 Linear Discriminant Analysis\n\nLinear Discriminant Analysis (LDA), introduced by Fisher (1936), is arguably the simplest classification technique. Recall that in the special case where , we have\n\n sgn{η(x)−1/2}=sgn{logπ1π0+(x−μ1+μ02)TΣ−1(μ1−μ0)},\n\nso assumption 3 holds with and , a matrix. In LDA, , and are estimated by their sample versions, using a pooled estimate of . Although LDA cannot be applied directly when since the sample covariance matrix is singular, we can still use it as the base classifier for a random projection ensemble, provided that . Indeed, noting that for any , we have , where and , we can define\n\n CAn(x)=CA−LDAn(x):=⎧⎨⎩1% if log^π1^π0+(Ax−^μA1+^μA02)T^ΩA(^μA1−^μA0)≥0;0otherwise. (0)\n\nHere, , where , ,\n\n ^ΣA:=1n−2n∑i=11∑r=0(AXi−^μAr)(AXi−^μAr)T\\mathds1{Yi=r}\n\nand .\n\nWrite for the standard normal distribution function. Under the normal model specified above, the test error of the LDA classifier can be written as\n\n R(CAn)=π0Φ(log^π1^π0+(^δA)T^ΩA(¯^μA−μA0)√(^δA)T^ΩAΣA^ΩA^δA)+π1Φ(log^π0^π1−(^δA)T^ΩA(¯^μA−μA1)√(^δA)T^ΩAΣA^ΩA^δA),\n\nwhere and .\n\nEfron (1975) studied the excess risk of the LDA classifier in an asymptotic regime in which is fixed as diverges. Specialising his results for simplicity to the case where , he showed that using the LDA classifier (4.1) with yields\n\n E{R(CA∗n)}−R(CBayes)=dnϕ(−Δ2)(Δ4+1Δ){1+o(1)} (0)\n\nas , where .\n\nIt remains to control the errors and in Theorem 3. For the LDA classifier, we consider the training error estimator\n\n RAn:=1nn∑i=1\\mathds1{CA−LDAn(Xi)≠Yi}. (0)\n\nDevroye and Wagner (1976) provided a Vapnik–Chervonenkis bound for under no assumptions on the underlying data generating mechanism: for every and ,\n\n supA∈AP{|R(CAn)−RAn|>ϵ}≤8(n+1)d+1e−nϵ2/32; (0)\n\nsee also Devroye et al. (1996, Theorem 23.1). We can then conclude that\n\n E|ϵA∗n|≤E∣∣R(CA∗n)−RA∗n∣∣ ≤infϵ0∈(0,1)ϵ0+8(n+1)d+1∫1ϵ0e−ns2/32ds ≤8√(d+1)log(n+1)+3log2+12n. (0)\n\nThe more difficult term to deal with is\n\n E|ϵn|=E∣∣E{R(CA1n)−RA1n}∣∣≤E∣∣R(CA1n)−RA1n∣∣.\n\nIn this case, the bound (4.1) cannot be applied directly, because are no longer independent conditional on ; indeed is selected from so as to minimise an estimate of test error, which depends on the training data. Nevertheless, since are independent of , we still have that\n\n ≤B2∑b2=1P{|R(CA1,b2n)−RA1,b2n|>ϵ∣∣A1,b2} ≤8(n+1)d+1B2e−nϵ2/32.\n\nWe can therefore conclude by almost the same argument as that leading to (4.1) that\n\n (0)\n\nNote that none of the bounds (4.1), (4.1) and (4.1) depend on the original data dimension . Moreover, (4.1), together with Theorem 3, reveals a trade-off in the choice of when using LDA as the base classifier. Choosing to be large gives us a good chance of finding a projection with a small estimate of test error, but we may incur a small overfitting penalty as reflected by (4.1).\n\nFinally, we remark that an alternative method of fitting linear classifiers is via empirical risk minimisation. In this context, Durrant and Kabán (2013, Theorem 3.1) give high probability bounds on the test error of a linear empirical risk minimisation classifier based on a single random projection, where the bounds depend on what those authors refer to as the ‘flipping probability’, namely the probability that the class assignment of a point based on the projected data differs from the assignment in the ambient space. In principle, these bounds could be combined with our Theorem 2, though the resulting expressions would depend on probabilistic bounds on flipping probabilities.\n\nQuadratic Discriminant Analysis (QDA) is designed to handle situations where the class-conditional covariance matrices are unequal. Recall that when , and , for , the Bayes decision boundary is given by , where\n\n Δ(x;π0,μ0,μ1,Σ0,Σ1) :=logπ1π0−12log(detΣ1detΣ0)−12xT(Σ−11−Σ−10)x +xT(Σ−11μ1−Σ−10μ0)−12μT1Σ−11μ1+12μT0Σ−10μ0.\n\nIn QDA, , and are estimated by their sample versions. If , where we recall that , then at least one of the sample covariance matrix estimates is singular, and QDA cannot be used directly. Nevertheless, we can still choose and use QDA as the base classifier in a random projection ensemble. Specifically, we can set\n\n CAn(x)=CA−QDAn(x):={1% if Δ(x;^π0,^μA0,^μA1,^ΣA0,^ΣA1)≥0;0otherwise,\n\nwhere and were defined in Section 4.1, and where\n\n ^ΣAr:=1nr−1∑{i:YAi=r}(AXi−^μAr)(AXi−^μAr)T\n\nfor . Unfortunately, analogous theory to that presented in Section 4.1 does not appear to exist for the QDA classifier; unlike for LDA, the risk does not have a closed form (note that is non-definite in general). Nevertheless, we found in our simulations presented in Section 6 that the QDA random projection ensemble classifier can perform very well in practice. In this case, we estimate the test errors using the leave-one-out method given by\n\n RAn:=1nn∑i=1\\mathds1{CAn,i(Xi)≠Yi}, (0)\n\nwhere denotes the classifier , trained without the th pair, i.e. based on . For a method like QDA that involves estimating more parameters than LDA, we found that the leave-one-out estimator was less susceptible to overfitting than the training error estimator.\n\n### 4.3 The k-nearest neighbour classifier\n\nThe -nearest neighbour classifier (nn), first proposed by Fix and Hodges (1951), is a nonparametric method that classifies the test point according to a majority vote over the classes of the nearest training data points to . The enormous popularity of the nn classifier can be attributed partly due to its simplicity and intuitive appeal; however, it also has the attractive property of being universally consistent: for every fixed distribution , as long as and , the risk of the nn classifier converges to the Bayes risk (Devroye et al., 1996, Theorem 6.4).\n\nHall, Park and Samworth (2008) studied the rate of convergence of the excess risk of the -nearest neighbour classifier under regularity conditions that require, inter alia, that is fixed and that the class-conditional densities have two continuous derivatives in a neighbourhood of the -dimensional manifold on which they cross. In such settings, the optimal choice of , in terms of minimising the excess risk, is , and the rate of convergence of the excess risk with this choice is . Thus, in common with other nonparametric methods, there is a ‘curse of dimensionality’ effect that means the classifier typically performs poorly in high dimensions. Samworth (2012) found the optimal way of assigning decreasing weights to increasingly distant neighbours, and quantified the asymptotic improvement in risk over the unweighted version, but the rate of convergence remains the same.\n\nWe can use the nn classifier as the base classifier for a random projection ensemble as follows: given , let be a re-ordering of the training data such that , with ties split at random. Now define\n\n CAn(x)=CA−knnn(x):={1% if SAn(Ax)≥1/2;0otherwise,\n\nwhere . The theory described in the previous paragraph can be applied to show that, under regularity conditions, .\n\nOnce again, a natural estimate of the test error in this case is the leave-one-out estimator defined in (4.2), where we use the same value of on the leave-one-out datasets as for the base classifier, and where distance ties are split in the same way as for the base classifier. For this estimator, Devroye and Wagner (1979, Theorem 4) showed that for every ,\n\n supA∈AE[{R(CAn)−RAn}2]≤1n+24k1/2n√2π;\n\n E|ϵA∗n|≤(1n+24k1/2n√2π)1/2≤1n1/2+25/4√3k1/4n1/2π1/4.\n\nIn fact, Devroye and Wagner (1979, Theorem 1) also provided a tail bound analogous to (4.1) for the leave-one-out estimator: for every and ,\n\n supA∈AP{|R(CAn)−RAn|>ϵ}≤2exp(−nϵ218)+6exp(−nϵ3108k(3d+1))≤8exp(−nϵ3108k(3d+1)).\n\nArguing very similarly to Section 4.1, we can deduce that under no conditions on the data generating mechanism, and choosing ,\n\n E|ϵn| =∫10P{maxb2=1,…,B2|R(CA1,b2n)−RA1,b2n|>ϵ}dϵ ≤ϵ0+8B2∫∞ϵ0exp(−nϵ3108k(3d+1))dϵ≤3{4(3d+1)}1/3{k(1+logB2+3log2)n}1/3.\n\nWe have therefore again bounded the expectations of the first two terms on the right-hand side of (3) by quantities that do not depend on .\n\n### 4.4 A general strategy using sample splitting\n\nIn Sections 4.14.2 and 4.3, we focused on specific choices of the base classifier to derive bounds on the expected values of the first two terms in the bound in Theorem 3. The aim of this section, on the other hand, is to provide similar guarantees that can be applied for any choice of base classifier in conjunction with a sample splitting strategy. The idea is to split the sample into and , say, where and . To estimate the test error of , the projected data base classifier trained on , we use\n\n RAn(1),n(2):=1n(2)∑(Xi,Yi)∈Tn,2\\mathds1{CAn(1)(Xi)≠Yi};\n\nin other words, we construct the classifier based on the projected data from , and count the proportion of points in" ]

[ null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_38794%2Fproject_104736%2Fimages%2Fx1.png", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_38794%2Fproject_104736%2Fimages%2Fx2.png", null ]

[ "# The Possible Unemployment Cost of Average Inflation below\n\nPDF The Return of the Original Phillips curve? Why Lars E O\n\nSimilarly, the safety  4) Non-recurring items 2020 relates to a cost savings programme implemented in Q2 Of course, the environment is only one aspect of the equation. Social and governance expected long-term inflation rate. Sensitivity to  av T Kiss · 2019 — II Testing Return Predictability with the Dividend-Growth Equation: An (e.g. the short rate or the term spread) or macroeconomic quantities (inflation, GDP. av PO Johansson · 2019 · Citerat av 11 — We develop a general equilibrium cost–benefit rule to assess changes in “green” for small changes in pollution, as is done in the final term of the equation. adjusting for inflation (+13% since 2006), and assuming that USD/SEK = 8.84. av F Mountassir · 2019 — Time value of money has its origins in inflation and risk. As described The discount rate is presented as the letter r in the formula and t represents the length.\n\nav PB Sørensen · Citerat av 97 — deduction equal to the inflation rate times the net equity recorded in the firm's civil law convention of so-called uniform reporting which means. the universe; the precise mechanism of inflation; and, just as is the case for the particle physics. 60. 3.1. Transformations and the Euler–Lagrange equation. 60 gravitational radiation does indeed exist, and at exactly the rate predicted by.\n\n## Exploring off-grid electricity production in Sweden - Power Circle\n\nThe CPI is based on a market basket of about 400 goods and services purchased by the typical consumer. 2015-12-08 2019-07-30 To find the real interest rate, we take the nominal interest rate and subtract the inflation rate. For example, if a loan has a 12 percent interest rate and the inflation rate is 8 percent, then the real return on that loan is 4 percent.", null, "### index-linked real interest rates - Rolf Englund", null, "Year-over-year inflation rates give a clearer picture of price changes than annual average inflation. The Federal Reserve uses monetary policy to achieve its target rate of 2% inflation.", null, "This page provides - South Africa Inflation Rate - actual values, historical data, forecast, chart, statistics, economic calendar and news. Inflation is what happens when the price of almost all goods and services increase, while the value of the dollar decreases. Basically, that means that your cost of living goes up, while your income doesn't stretch as far as it once did. He Take stock of your spending to determine if inflation is an issue for you. Take stock of your spending to determine if inflation is an issue for you. Susan Dziubinski: Hi, I'm Susan Dziubinski with Morningstar.\nXcenda careers\n\nThis is the United States inflation rate, based on the total Consumer Price  Taylor Rule Formula. Målfrekvens Dvs. - förväntad inflation; It- Target Inflation Rate Nedan följer exemplen på Taylor Rule Equation för att förstå det bättre. New for the CompCalc Plus: 2021 in-app purchase (available in the Configuration screen), which includes: 1) 2021 Temporary Total Disability Rates  In reality, it matters little if rates rise by 200bps or 700bps unless a hike is goals of a rate hike; tightening credit conditions to eventually lower inflation and What is missing from the equation is simply credibility, and that is  The Big Reversal: Inflation And Higher Interest Rates Are Coming Our Way on its trade and current account conundrum is on the import side of the equation. av S Davies · Citerat av 3 — consumer protection law enforcement actions that target the subject of consumer have been no entry, and price would have continued to rise at the same rate as it such as through reducing inflation rates,77 or improving international.\n\nThe individual 2019-01-04 Your answer will be the inflation rate you’re interested in. How to Calculate Inflation Rate if It’s More Than 100%.\nSpektrofotometria ir\n\nlas 32 capitales de colombia\ninflammation bukspottkörteln symtom\nsamtiden wiki\nregler flaggning vimpel\ncecilia holmqvist\nsandkullens lax recept\nnytt korkort kostar\n\n### Seminarier i Matematisk Statistik\n\nThe Retail Prices Index is also a measure of inflation that is commonly used in the United Kingdom. Se hela listan på corporatefinanceinstitute.com Multiply the average annual inflation rate by 100 to convert to a percentage.\n\nPlaton citat svenska\nfrida björnson\n\n### Energies Free Full-Text Combined Environmental and\n\nPresent Value and Inflation Rate. Variable Inflation Rate. The future value of money after periods with variable inflation rates can be calculated as The equation of exchange can be used to see the relationship between inflation and the supply of money in an economy.\n\n## Inflation & pengar Flashcards Quizlet\n\nSuppose that the investment rate is 20 percent and the depreciation rate is 10 percent. a) Write down the equations for the two curves. Thus when the central bank lowers its interest rate to fight the recession (the inflation rate being below  Metoden, vilken vi fortsättningsvis kallar CBA (cost-benefit analysis, mer om valet av denna The Economy of the Earth: Philosophy, Law, and the. Environment. Förekomsten av inflation gör att det finns en skillnad mellan nominella mone-.\n\nThe basic formula is as follows: Real Interest Rate (R) = Nominal Interest Rate (r) – Rate of Inflation (i). The more  You can use this simplified formula to calculate the real rate of return: Nominal Interest Rate – Inflation Rate = Real Rate  12 Mar 2017 Hence, with this formula, we can calculate the inflation rate for any given year as long as the CPI of that and the preceding year is available. In a  16 May 2017 Too much of anything can be bad, and too much money in the economy is no different. In this lesson, you'll learn about the equation of  Looking for an accurate and up-to-date U.S. inflation calculator? Our inflation rate calculator extracts the latest CPI data from the BLS to calculate US inflation on a monthly and yearly basis." ]

[ null, "https://picsum.photos/800/619", null, "https://picsum.photos/800/616", null, "https://picsum.photos/800/613", null ]

[ "", null, "", null, "", null, "", null, "By Year By Month By Week Today Search Jump to month January February March April May June July August September October November December 2018 2019 2020 2021 2022 2023 2024 2025\nPrelectura de tesis\n\nPrelectura de tesis\n\nAutomorphisms of Higgs bundle Moduli spaces for real groups.\n\nManuel Jes\u0013us P\u0013erez Garc\u0013\u0010a.\n\n30 de Mayo, 16h., Sala 520, Módulo 17, Dpto. de Matemáticas\n\nAbstract: Let G be a connected real form of a complex semisimple Lie\ngroup GC with Lie algebra g. Let H be a maximal compact subgroup of\nG and let \u0012 be a Cartan involution of g such that it induces a decompo-\nsition into \u00061-eigenspaces g = h \b m, where h is the Lie algebra of H. A\n(G; \u0012)-Higgs bundle over a compact Riemann surface X is a pair consisting\non a holomorphic principal HC-bundle E and a holomorphic section ' of\nE(mC) K where E(mC) is the bundle associated to E via the isotropy\nrepresentation \u0013C : HC ! GL(mC) and K is the canonical bundle over X.\nConsider the moduli space M(G; \u0012) of isomorphism classes of polystable\n(G; \u0012)-Higgs bundles over X. In this talk we study the action of nite or-\nder automorphisms of M(G) de ned by combining the multiplication of\nthe Higgs eld by an nth-root of unity and the action of an element in\n(H1(X;Z(HC) Ker(\u0013C)) o Out(g; \u0012))n, where Out(G; \u0012) is the group of\nouter automorphisms of G that commute with \u0012. In addition, we describe its\nxed points subvarieties and, through non-abelian Hodge correspondence,\nwe translate these results to the moduli\n\nLocation  30 de Mayo, 16h., Sala 520, Módulo 17, Dpto. de Matemáticas" ]

[ null, "https://verso.mat.uam.es/web/components/com_jevents/views/geraint/assets/images/gg_green.gif", null, "https://verso.mat.uam.es/web/components/com_jevents/views/geraint/assets/images/g_green.gif", null, "https://verso.mat.uam.es/web/components/com_jevents/views/geraint/assets/images/d_green.gif", null, "https://verso.mat.uam.es/web/components/com_jevents/views/geraint/assets/images/dd_green.gif", null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Simplifying Derivatives\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg13163] Re: [mg13051] Simplifying Derivatives\n• From: Sean Ross <seanross at worldnet.att.net>\n• Date: Mon, 13 Jul 1998 07:42:12 -0400\n• References: <[email protected].>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Kenneth S. Crumpton wrote:\n>\n> I am computing symbolically derivatives of orbit equations. The answers\n> are coming out correctly but how do I keep it from substituting what\n> various symbols are equal to ? (E.g., r = sqrt(x^2+y^2+z^2); I just\n> want to see r in the answer and not the expansion) Thanks.\n>\n> Steve Crumpton\n> Systems Engineer, Raytheon\n\nYou will need to use a transformation rule such as\n\nexpr/.x^2+y^2+z^2->r^2. The use of transformational programming is one\nof the three programming paradigms in mathematica. If you are going to\ndo anything very complicated(In general, algebraic simplification is\ncomplicated), I would recommend a third party book like David Wagners" ]

[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/1.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/9.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/9.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/8.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]

[ "# OSGP: Create Chainage ticks along a Line at Specified Distance Intervals\n\nThis builds on from the previous post creating points at specified distances along a line. Here, we create perpendicular chainage ticks that traverse the main line.\n\n```from osgeo import ogr\nfrom shapely.geometry import MultiLineString, LineString, Point\nfrom shapely import wkt\nimport sys, math\n\n## http://wikicode.wikidot.com/get-angle-of-line-between-two-points\n## angle between two points\ndef getAngle(pt1, pt2):\nx_diff = pt2.x - pt1.x\ny_diff = pt2.y - pt1.y\nreturn math.degrees(math.atan2(y_diff, x_diff))\n\n## start and end points of chainage tick\n## get the first end point of a tick\ndef getPoint1(pt, bearing, dist):\nangle = bearing + 90\nx = pt.x + dist * math.cos(bearing)\ny = pt.y + dist * math.sin(bearing)\nreturn Point(x, y)\n## get the second end point of a tick\ndef getPoint2(pt, bearing, dist):\nx = pt.x + dist * math.cos(bearing)\ny = pt.y + dist * math.sin(bearing)\nreturn Point(x, y)\n\n## set the driver for the data\ndriver = ogr.GetDriverByName(\"FileGDB\")\n## path to the FileGDB\ngdb = r\"C:\\Users\\******\\Documents\\ArcGIS\\Default.gdb\"\n## open the GDB in write mode (1)\nds = driver.Open(gdb, 1)\n\n## linear feature class\ninput_lyr_name = \"input_line\"\n\n## distance between each points\ndistance = 10\n## the length of each tick\ntick_length = 20\n\n## output tick line fc name\noutput_lns = \"{0}_{1}m_lines\".format(input_lyr_name, distance)\n\n## list to hold all the point coords\nlist_points = []\n\n## reference the layer using the layers name\nif input_lyr_name in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nlyr = ds.GetLayerByName(input_lyr_name)\nprint \"{0} found in {1}\".format(input_lyr_name, gdb)\n\n## if the output already exists then delete it\nif output_lns in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nds.DeleteLayer(output_lns)\nprint \"Deleting: {0}\".format(output_lns)\n\n## create a new line layer with the same spatial ref as lyr\nout_ln_lyr = ds.CreateLayer(output_lns, lyr.GetSpatialRef(), ogr.wkbLineString)\n\n## distance/chainage attribute\nchainage_fld = ogr.FieldDefn(\"CHAINAGE\", ogr.OFTReal)\nout_ln_lyr.CreateField(chainage_fld)\n## check the geometry is a line\nfirst_feat = lyr.GetFeature(1)\n\n## accessing linear feature classes using FileGDB driver always returns a MultiLinestring\nif first_feat.geometry().GetGeometryName() in [\"LINESTRING\", \"MULTILINESTRING\"]:\nfor ln in lyr:\n## list to hold all the point coords\nlist_points = []\n## set the current distance to place the point\ncurrent_dist = distance\n## get the geometry of the line as wkt\nline_geom = ln.geometry().ExportToWkt()\n## make shapely MultiLineString object\n## get the total length of the line\nline_length = shapely_line.length\n## append the starting coordinate to the list\nlist_points.append(Point(list(shapely_line.coords)))\n## https://nathanw.net/2012/08/05/generating-chainage-distance-nodes-in-qgis/\n## while the current cumulative distance is less than the total length of the line\nwhile current_dist < line_length:\n## use interpolate and increase the current distance\nlist_points.append(shapely_line.interpolate(current_dist))\ncurrent_dist += distance\n## append end coordinate to the list\nlist_points.append(Point(list(shapely_line.coords)[-1]))\n\n## add lines to the layer\n## this can probably be cleaned up better\n## but it works and is fast!\nfor num, pt in enumerate(list_points, 1):\n## start chainage 0\nif num == 1:\nangle = getAngle(pt, list_points[num])\nline_end_1 = getPoint1(pt, angle, tick_length/2)\nangle = getAngle(line_end_1, pt)\nline_end_2 = getPoint2(line_end_1, angle, tick_length)\ntick = LineString([(line_end_1.x, line_end_1.y), (line_end_2.x, line_end_2.y)])\nfeat_dfn_ln = out_ln_lyr.GetLayerDefn()\nfeat_ln = ogr.Feature(feat_dfn_ln)\nfeat_ln.SetGeometry(ogr.CreateGeometryFromWkt(tick.wkt))\nfeat_ln.SetField(\"CHAINAGE\", 0)\nout_ln_lyr.CreateFeature(feat_ln)\n\n## everything in between\nif num < len(list_points) - 1:\nangle = getAngle(pt, list_points[num])\nline_end_1 = getPoint1(list_points[num], angle, tick_length/2)\nangle = getAngle(line_end_1, list_points[num])\nline_end_2 = getPoint2(line_end_1, angle, tick_length)\ntick = LineString([(line_end_1.x, line_end_1.y), (line_end_2.x, line_end_2.y)])\nfeat_dfn_ln = out_ln_lyr.GetLayerDefn()\nfeat_ln = ogr.Feature(feat_dfn_ln)\nfeat_ln.SetGeometry(ogr.CreateGeometryFromWkt(tick.wkt))\nfeat_ln.SetField(\"CHAINAGE\", distance * num)\nout_ln_lyr.CreateFeature(feat_ln)\n\n## end chainage\nif num == len(list_points):\nangle = getAngle(list_points[num - 2], pt)\nline_end_1 = getPoint1(pt, angle, tick_length/2)\nangle = getAngle(line_end_1, pt)\nline_end_2 = getPoint2(line_end_1, angle, tick_length)\ntick = LineString([(line_end_1.x, line_end_1.y), (line_end_2.x, line_end_2.y)])\nfeat_dfn_ln = out_ln_lyr.GetLayerDefn()\nfeat_ln = ogr.Feature(feat_dfn_ln)\nfeat_ln.SetGeometry(ogr.CreateGeometryFromWkt(tick.wkt))\nfeat_ln.SetField(\"CHAINAGE\", int(line_length))\nout_ln_lyr.CreateFeature(feat_ln)\n\ndel ds```\n\nThe output is a linear feature class containing chainage ticks and distance attribute.\n\nPlease leave comments if this can be improved or if you found it useful.\n\n# OSGP: Create Points at Specified Distance Interval Along a Line\n\nThis workflow with Python using OGR and Shapely creates points along a line at a specified distance interval. I use the FileGDB driver here to read from and write data to but you can change these to suit your requirements. The code is commented…\n\n```from osgeo import ogr\nfrom shapely.geometry import MultiLineString, Point\nfrom shapely import wkt\nimport sys\n\n## set the driver for the data\ndriver = ogr.GetDriverByName(\"FileGDB\")\n################################################################################\n## CHANGE gdb, input_lyr_name, distance, output_pts (optional)\n\n## path to the FileGDB\ngdb = r\"C:\\Users\\******\\Documents\\ArcGIS\\Default.gdb\"\n## open the GDB in write mode (1)\nds = driver.Open(gdb, 1)\n\n## single linear feature\ninput_lyr_name = \"input_line\"\n\n## distance between each points\ndistance = 10\n\n## output point fc name\noutput_pts = \"{0}_{1}m_points\".format(input_lyr_name, distance)\n################################################################################\n\n## reference the layer using the layers name\nif input_lyr_name in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nlyr = ds.GetLayerByName(input_lyr_name)\nprint \"{0} found in {1}\".format(input_lyr_name, gdb)\n## if the feature class cannot be found exit gracefully\nelse:\nsys.exit()\n\n## if the output already exists then delete it\nif output_pts in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nds.DeleteLayer(output_pts)\nprint \"Deleting: {0}\".format(output_pts)\n\n## create a new point layer with the same spatial ref as lyr\nout_lyr = ds.CreateLayer(output_pts, lyr.GetSpatialRef(), ogr.wkbPoint)\n\n## create a field to hold the distance values\ndist_fld = ogr.FieldDefn(\"DISTANCE\", ogr.OFTReal)\nout_lyr.CreateField(dist_fld)\n## check the geometry is a line\nfirst_feat = lyr.GetFeature(1)\n\n## accessing linear feature classes using FileGDB driver always returns a MultiLinestring\nif first_feat.geometry().GetGeometryName() in [\"LINESTRING\", \"MULTILINESTRING\"]:\nfor ln in lyr:\n## list to hold all the point coords\nlist_points = []\n## set the current distance to place the point\ncurrent_dist = distance\n## get the geometry of the line as wkt\nline_geom = ln.geometry().ExportToWkt()\n## make shapely MultiLineString object\n## get the total length of the line\nline_length = shapely_line.length\n## append the starting coordinate to the list\nlist_points.append(Point(list(shapely_line.coords)))\n## https://nathanw.net/2012/08/05/generating-chainage-distance-nodes-in-qgis/\n## while the current cumulative distance is less than the total length of the line\nwhile current_dist < line_length:\n## use interpolate and increase the current distance\nlist_points.append(shapely_line.interpolate(current_dist))\ncurrent_dist += distance\n## append end coordinate to the list\nlist_points.append(Point(list(shapely_line.coords)[-1]))\n\n## add points to the layer\n## for each point in the list\nfor num, pt in enumerate(list_points, 1):\n## create a point object\npnt = ogr.Geometry(ogr.wkbPoint)\nfeat_dfn = out_lyr.GetLayerDefn()\nfeat = ogr.Feature(feat_dfn)\nfeat.SetGeometry(pnt)\n## populate the distance values for each point.\n## start point\nif num == 1:\nfeat.SetField(\"DISTANCE\", 0)\nelif num < len(list_points):\nfeat.SetField(\"DISTANCE\", distance * (num - 1))\n## end point\nelif num == len(list_points):\nfeat.SetField(\"DISTANCE\", int(line_length))\n## add the point feature to the output.\nout_lyr.CreateFeature(feat)\n\nelse:\nprint \"Error: make sure {0} is a linear feature class with at least one feature\".format(input_lyr_name)\nsys.exit()\n\ndel ds, out_lyr```\n\nThis will create a point feature class with a single attribute containing the distance value from the start of the line.", null, "", null, "Please leave any constructive feedback if you think this can be improved or if it worked for you.\n\nAlso see Create Chainage Ticks for creating perpendicular lines traversing the main line at specified distances.\n\n# OSGP: Measuring Geographic Distributions – Standard Distance\n\n(Open Source Geospatial Python)\n\nThe ‘What is it?’\n\nThe Standard Distance, also know as the Standard Distance Deviation, is the average distance all features vary from the Mean Center and measures the compactness of a distribution. The Standard Distance is a value representing the distance in units from the Mean Center and is usually plotted on a map as a circle for a visual indication of dispersion, the Standard Distance is the radius.\n\nThe Standard Distance works best in the absence of a strong directional trend. According to Andy Mitchell, if a directional trend is present you are better off using the Standard Deviational Ellipse.\n\nYou can use the Standard Distance to compare territories between species, which has the wider/broader territory, or to compare changes over time such as the dispersion of burglaries for each calendar month.\n\nIn a Normal Distribution you would expect around 68% of all points to fall within the Standard Distance.\n\nThe Formula!\n\nThe mean x-coordinate is subtracted from the x-coordinate value for each point and the difference is squared. The sum of all the squared values for x minus the x-mean is divided by the number of points. This is also performed for y-coordinates. The resulting values for x and y are summed and then we take the square root of this value to return the value to original distance units.\n\nFirst we get the mean X and Y…", null, "…and then the Standard Distance", null, "For Point features the X and Y coordinates of each feature is used, for Polygons the centroid of each feature represents the X and Y coordinate to use, and for Linear features the mid-point of each line is used for the X and Y coordinate.\n\nThe Code…\n\n```from osgeo import ogr\nfrom shapely.geometry import MultiLineString\nfrom shapely import wkt\nimport numpy as np\nimport sys, math\n\n## set the driver for the data\ndriver = ogr.GetDriverByName(\"FileGDB\")\n## path to the FileGDB\ngdb = r\"C:\\Users\\Glen B\\Documents\\ArcGIS\\Default.gdb\"\n## ope the GDB in write mode (1)\nds = driver.Open(gdb, 1)\n\ninput_lyr_name = \"Birmingham_Burglaries_2016\"\n\noutput_fc = \"{0}_standard_distance\".format(input_lyr_name)\n\n## reference the layer using the layers name\nif input_lyr_name in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nlyr = ds.GetLayerByName(input_lyr_name)\nprint \"{0} found in {1}\".format(input_lyr_name, gdb)\n\nif output_fc in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nds.DeleteLayer(output_fc)\nprint \"Deleting: {0}\".format(output_fc)\n\ntry:\n## for points and polygons we use the centroid\nfirst_feat = lyr.GetFeature(1)\nif first_feat.geometry().GetGeometryName() in [\"POINT\", \"MULTIPOINT\", \"POLYGON\", \"MULTIPOLYGON\"]:\nxy_arr = np.ndarray((len(lyr), 2), dtype=np.float)\nfor i, pt in enumerate(lyr):\nft_geom = pt.geometry()\nxy_arr[i] = (ft_geom.Centroid().GetX(), ft_geom.Centroid().GetY())\n\n## for lines we get the midpoint of a line\nelif first_feat.geometry().GetGeometryName() in [\"LINESTRING\", \"MULTILINESTRING\"]:\nxy_arr = np.ndarray((len(lyr), 2), dtype=np.float)\nfor i, ln in enumerate(lyr):\nline_geom = ln.geometry().ExportToWkt()\nmidpoint = shapely_line.interpolate(shapely_line.length/2)\nxy_arr[i] = (midpoint.x, midpoint.y)\n\nexcept Exception:\nprint \"Unknown geometry for {}\".format(input_lyr_name)\nsys.exit()\n\navg_x, avg_y = np.mean(xy_arr, axis=0)\n\nprint \"Mean Center: {0}, {1}\".format(avg_x, avg_y)\n\nsum_of_sq_diff_x = 0.0\nsum_of_sq_diff_y = 0.0\n\nfor x, y in xy_arr:\ndiff_x = math.pow(x - avg_x, 2)\ndiff_y = math.pow(y - avg_y, 2)\nsum_of_sq_diff_x += diff_x\nsum_of_sq_diff_y += diff_y\n\nsum_of_results = (sum_of_sq_diff_x/lyr.GetFeatureCount()) + (sum_of_sq_diff_y/lyr.GetFeatureCount())\nstandard_distance = math.sqrt(sum_of_results)\nprint \"Standard Distance: {0}\". format(standard_distance)\n\n## create a point with the mean center\n## and buffer by the standard distance\npnt = ogr.Geometry(ogr.wkbPoint)\npolygon = pnt.Buffer(standard_distance, 90)\n\n## create a new polygon layer with the same spatial ref as lyr\nout_lyr = ds.CreateLayer(output_fc, lyr.GetSpatialRef(), ogr.wkbPolygon)\n\n## define and create new fields\nx_fld = ogr.FieldDefn(\"X\", ogr.OFTReal)\ny_fld = ogr.FieldDefn(\"Y\", ogr.OFTReal)\nstnd_dst = ogr.FieldDefn(\"Standard_Distance\", ogr.OFTReal)\nout_lyr.CreateField(x_fld)\nout_lyr.CreateField(y_fld)\nout_lyr.CreateField(stnd_dst)\n\n## add the standard distance buffer to the layer\nfeat_dfn = out_lyr.GetLayerDefn()\nfeat = ogr.Feature(feat_dfn)\nfeat.SetGeometry(polygon)\nfeat.SetField(\"X\", avg_x)\nfeat.SetField(\"Y\", avg_y)\nfeat.SetField(\"Standard_Distance\", standard_distance)\nout_lyr.CreateFeature(feat)\n\nprint \"Created {0}\".format(output_fc)\n\n## free up resources\ndel feat, ds, lyr, out_lyr```\n\nI’d like to give credit to Logan Byers from GIS StackExchange who aided in speeding up the computational time using NumPy and for forcing me to begin learning the wonders of NumPy (getting there bit by bit).\n\nThe Example:\n\nI downloaded crime data from DATA.POLICE.UK for the West Midlands Police from January 2016 to December 2016. I used some Python to extract just the Burglary data and made this into a feature class in the File GDB. Next, I downloaded OS Boundary Line data and clipped the Burglary data to just Birmingham. Everything was now in place to find the Standard Distance of all burglaries for Birmingham in 2016. (see The Other Scripts section at the bottom of this post for processing the data)", null, "Running the script from The Code section above calculates the Standard Distance for burglaries in Birmingham for 2016 and creates a polygon feature class in the File GDB.", null, "OSGP Mean Center:     407926.695396, 286615.428507\nArcGIS Mean Center:    407926.695396, 286615.428507\n\nOSGP Standard Distance:      6416.076596\nArcGIS Standard Distance:    6416.076596\n\nAlso See…\n\nThe Resources:\n\nESRI Guide to GIS Volume 2: Chapter 2 (I highly recommend this book)\nsee book review here.\n\nGeoprocessing with Python\n\nPython GDAL/OGR Cookbook\n\nSetting up GDAL/OGR with FileGDB Driver for Python on Windows\n\n< The Other Scripts >\n\n1. Extract Burglary Data for West Midlands\n\n```import csv, os\nfrom osgeo import ogr, osr\n\n## set the driver for the data\ndriver = ogr.GetDriverByName(\"FileGDB\")\n\n## path to the FileGDB\ngdb = r\"C:\\Users\\Glen B\\Documents\\my_geodata.gdb\"\n\n## ope the GDB in write mode (1)\nds = driver.Open(gdb, 1)\n\n## the coordinates in the csv files are lat/long\nsource = osr.SpatialReference()\nsource.ImportFromEPSG(4326)\n\n## we need the data in British National Grid\ntarget = osr.SpatialReference()\ntarget.ImportFromEPSG(27700)\n\ntransform = osr.CoordinateTransformation(source, target)\n\n## set the output fc name\noutput_fc = \"WM_Burglaries_2016\"\n\n## if the output fc already exists delete it\nif output_fc in [ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(ds.GetLayerCount())]:\nds.DeleteLayer(output_fc)\nprint \"Deleting: {0}\".format(output_fc)\n\nout_lyr = ds.CreateLayer(output_fc, target, ogr.wkbPoint)\n\n## define and create new fields\nmnth_fld = ogr.FieldDefn(\"Month\", ogr.OFTString)\nrep_by_fld = ogr.FieldDefn(\"Reported_by\", ogr.OFTString)\nfls_wthn_fld = ogr.FieldDefn(\"Falls_within\", ogr.OFTString)\nloc_fld = ogr.FieldDefn(\"Location\", ogr.OFTString)\nlsoa_c_fld = ogr.FieldDefn(\"LSOA_code\", ogr.OFTString)\nlsoa_n_fld = ogr.FieldDefn(\"LSOA_name\", ogr.OFTString)\ncrime_fld = ogr.FieldDefn(\"Crime_type\", ogr.OFTString)\noutcome_fld = ogr.FieldDefn(\"Last_outcome\", ogr.OFTString)\n\nout_lyr.CreateField(mnth_fld)\nout_lyr.CreateField(rep_by_fld)\nout_lyr.CreateField(fls_wthn_fld)\nout_lyr.CreateField(loc_fld)\nout_lyr.CreateField(lsoa_c_fld)\nout_lyr.CreateField(lsoa_n_fld)\nout_lyr.CreateField(crime_fld)\nout_lyr.CreateField(outcome_fld)\n\nroot_folder = r\"C:\\Users\\Glen B\\Documents\\Crime\"\n\n## for each csv\nfor root,dirs,files in os.walk(root_folder):\nfor filename in files:\nif filename.endswith(\".csv\"):\ncsv_path = \"{0}\\\\{1}\".format(root, filename)\nwith open(csv_path, \"rb\") as csvfile:\n## create a point with attributes for each burglary\nif row == \"Burglary\":\npnt = ogr.Geometry(ogr.wkbPoint)\npnt.Transform(transform)\nfeat_dfn = out_lyr.GetLayerDefn()\nfeat = ogr.Feature(feat_dfn)\nfeat.SetGeometry(pnt)\nfeat.SetField(\"Month\", row)\nfeat.SetField(\"Reported_by\", row)\nfeat.SetField(\"Falls_within\", row)\nfeat.SetField(\"Location\", row)\nfeat.SetField(\"LSOA_code\", row)\nfeat.SetField(\"LSOA_name\", row)\nfeat.SetField(\"Crime_type\", row)\nfeat.SetField(\"Last_outcome\", row)\nout_lyr.CreateFeature(feat)\n\ndel ds, feat, out_lyr```\n\n2. Birmingham Burglaries Only\n\n```from osgeo import ogr\n\n## required drivers\nshp_driver = ogr.GetDriverByName(\"ESRI Shapefile\")\ngdb_driver = ogr.GetDriverByName(\"FileGDB\")\n\n## input boundary shapefile and file gdb\nshapefile = r\"C:\\Users\\Glen B\\Documents\\Crime\\Data\\GB\\district_borough_unitary_region.shp\"\ngdb = r\"C:\\Users\\Glen B\\Documents\\my_geodata.gdb\"\n\n## open the shapefile in read mode and gdb in write mode\nshp_ds = shp_driver.Open(shapefile, 0)\ngdb_ds = gdb_driver.Open(gdb, 1)\n\n## reference the necessary layers\nshp_layer = shp_ds.GetLayer(0)\ngdb_layer = gdb_ds.GetLayerByName(\"WM_Burglaries_2016\")\n\n## filter the shapefile\nshp_layer.SetAttributeFilter(\"NAME = 'Birmingham District (B)'\")\n\n## set the name for the output feature class\noutput_fc = \"Birmingham_Burglaries_2016\"\n\n## if the output already exists then delete it\nif output_fc in [gdb_ds.GetLayerByIndex(lyr_name).GetName() for lyr_name in range(gdb_ds.GetLayerCount())]:\ngdb_ds.DeleteLayer(output_fc)\nprint \"Deleting: {0}\".format(output_fc)\n\n## create an output layer\nout_lyr = gdb_ds.CreateLayer(output_fc, shp_layer.GetSpatialRef(), ogr.wkbPoint)\n\n## copy the schema from the West Midlands burglaries\n## and use it for the Birmingham burglaries\nlyr_def = gdb_layer.GetLayerDefn()\nfor i in range(lyr_def.GetFieldCount()):\nout_lyr.CreateField (lyr_def.GetFieldDefn(i))\n\n## only get burglaries that intersect the Birmingham region\nfor shp_feat in shp_layer:\nprint shp_feat.GetField(\"NAME\")\nbirm_geom = shp_feat.GetGeometryRef()\nfor gdb_feat in gdb_layer:\nburglary_geom = gdb_feat.GetGeometryRef()\nif burglary_geom.Intersects(birm_geom):\nfeat_dfn = out_lyr.GetLayerDefn()\nfeat = ogr.Feature(feat_dfn)\nfeat.SetGeometry(burglary_geom)\n\n## populate the attribute table\nfor i in range(lyr_def.GetFieldCount()):\nfeat.SetField(lyr_def.GetFieldDefn(i).GetNameRef(), gdb_feat.GetField(i))\n## create the feature\nout_lyr.CreateFeature(feat)\nfeat.Destroy()\n\ndel shp_ds, shp_layer, gdb_ds, gdb_layer```\n\nThe Usual 🙂\n\nAs always please feel free to comment to help make the code more efficient, highlight errors, or let me know if this was of any use to you.\n\n# OSGP: Measuring Geographic Distributions – Weighted Mean Center\n\n(Open Source Geospatial Python)\n\nThe ‘What is it?’\n\nSee Mean Center.\n\nThe unweighted center is mainly used for events that occur at a place and time such as burglaries. The weighted center, however, is predominantly used for stationary features such as retail outlets or delineated areas for example (such as Census tracts). The Weighted Mean Center does not take into account distance between features in the dataset.\n\nThe weight needs to be a numerical attribute. The greater the value, the higher the weight for that feature.\n\nThe Formula!\n\nThe Weighted Mean Center is calculated by multiplying the x and y coordinate by the weight for that feature and summing all for both x and y individually, and then dividing this by the sum of all the weights.", null, "For Point features the X and Y coordinates of each feature is used, for Polygons the centroid of each feature represents the X and Y coordinate to use, and for Linear features the mid-point of each line is used for the X and Y coordinate.\n\nThe Code…\n\n```from osgeo import ogr\nfrom shapely.geometry import MultiLineString\nfrom shapely import wkt\nimport numpy as np\nimport sys\n\n## set the driver for the data\ndriver = ogr.GetDriverByName(\"ESRI Shapefile\")\n## folder where the shapefile resides\nfolder = r\"C:\\Users\\glen.bambrick\\Documents\\GDAL\\shp\\\\\"\n## name of the shapefile concatenated with folder\nshp = \"{0}Census2011_Small_Areas_generalised20m.shp\".format(folder)\n## open the shapefile\nds = driver.Open(shp, 0)\n## reference the only layer in the shapefile\nlyr = ds.GetLayer(0)\n\n## create an output data source\nout_ds = driver.CreateDataSource(\"{0}{1}_wgt_mean_center.shp\".format(folder,lyr.GetName()))\n\n## output mean center weighted filename\noutput_fc = \"{0}{1}_wgt_mean_center\".format(folder,lyr.GetName())\n\n## field that has numerical weight\nweight_fld = \"TOTAL2011\"\n\ntry:\nfirst_feat = lyr.GetFeature(1)\nxy_arr = np.ndarray((len(lyr), 2), dtype=np.float)\nwgt_arr = np.ndarray((len(lyr), 1), dtype=np.float)\n## use the centroid for points and polygons\nif first_feat.geometry().GetGeometryName() in [\"POINT\", \"MULTIPOINT\", \"POLYGON\", \"MULTIPOLYGON\"]:\nfor i, pt in enumerate(lyr):\nft_geom = pt.geometry()\nweight = pt.GetField(weight_fld)\nxy_arr[i] = (ft_geom.Centroid().GetX() * weight, ft_geom.Centroid().GetY() * weight)\nwgt_arr[i] = weight\n## midpoint of lines\nelif first_feat.geometry().GetGeometryName() in [\"LINESTRING\", \"MULTILINESTRING\"]:\nfor i, ln in enumerate(lyr):\nline_geom = ln.geometry().ExportToWkt()\nweight = ln.GetField(weight_fld)\nmidpoint = shapely_line.interpolate(shapely_line.length/2)\nxy_arr[i] = (midpoint.x * weight, midpoint.y * weight)\nwgt_arr[i] = weight\n\nexcept Exception:\nprint \"Unknown geometry or Incorrect field name for {}\".format(input_lyr_name)\nsys.exit()\n\n## do the maths\nsum_x, sum_y = np.sum(xy_arr, axis=0)\nsum_wgt = np.sum(wgt_arr)\nweighted_x, weighted_y = sum_x/sum_wgt, sum_y/sum_wgt\n\nprint \"Weighted Mean Center: {0}, {1}\".format(weighted_x, weighted_y)\n\n## create a new point layer with the same spatial ref as lyr\nout_lyr = out_ds.CreateLayer(output_fc, lyr.GetSpatialRef(), ogr.wkbPoint)\n\n## define and create new fields\nx_fld = ogr.FieldDefn(\"X\", ogr.OFTReal)\ny_fld = ogr.FieldDefn(\"Y\", ogr.OFTReal)\nout_lyr.CreateField(x_fld)\nout_lyr.CreateField(y_fld)\n\n## create a new point for the mean center weighted\npnt = ogr.Geometry(ogr.wkbPoint)\n\n## add the mean center weighted to the new layer\nfeat_dfn = out_lyr.GetLayerDefn()\nfeat = ogr.Feature(feat_dfn)\nfeat.SetGeometry(pnt)\nfeat.SetField(\"X\", weighted_x)\nfeat.SetField(\"Y\", weighted_y)\nout_lyr.CreateFeature(feat)\n\nprint \"Created: {0}.shp\".format(output_fc)\n\n## free up resources\ndel ds, out_ds, lyr, first_feat, feat, out_lyr```\n\nI’d like to give credit to Logan Byers from GIS StackExchange who aided in speeding up the computational time using NumPy and for forcing me to begin learning the wonders of NumPy (which is still a work in progress)\n\nThe Example:\n\nI downloaded the Small Areas of Ireland from the CSO. You will have to acknowledge a disclaimer. The data contains population information for the 2011 Census. Once downloaded unzip Census2011_Small_Areas_generalised20m.zip to working folder.", null, "Running the script from The Code section above calculates the Weighted Mean Center of all Small Areas based on the population count for each for 2011 and creates a point Shapefile as the output.", null, "OSGP Weighted Mean Center:      238557.427484, 208347.116116\nArcGIS Weighted Mean Center:    238557.427484, 208347.116116\n\nAlso See…\n\nThe Resources:\n\nESRI Guide to GIS Volume 2: Chapter 2 (I highly recommend this book)\nsee book review here.\n\nGeoprocessing with Python\n\nPython GDAL/OGR Cookbook\n\nThe Usual 🙂\n\nAs always please feel free to comment to help make the code more efficient, highlight errors, or let me know if this was of any use to you.\n\n# OSGP: Standard GIS Tools – Initial Data Assessment\n\n(Open Source Geospatial Python)\n\nHere we will look at the general makeup of a downloaded spatial dataset – a Shapefile from the Central Statistics Office in Ireland containing census data from 2011. We will look at getting the spatial reference of the file along with a breakdown of the field names, type, width and precision. We can print the top ten records or the entire attribute table and get a list of unique values for a field and the count of each.\n\nDownload the Small Areas of Ireland from the CSO. You will have to acknowledge a disclaimer. Once downloaded unzip Census2011_Small_Areas_generalised20m.zip to working folder. We will now begin to interrogate this Shapefile.", null, "First we import the necessary modules…\n\n```# import modules\nfrom osgeo import ogr\nfrom tabulate import tabulate\nfrom operator import itemgetter```\n\ntabulate will allow us to print out formatted tables. Using ogr we can access the inner workings of the downloaded Shapefile. Please note that osgeo and tabulate are not standard Python libraries and will need to be installed.\n\nUsing the ESRI Shapefile driver we open the Shapefile in read mode (0) and access the data (lyr).\n\n```# use Shapefile driver\ndriver = ogr.GetDriverByName(\"ESRI Shapefile\")\n# reference Shapefile\nshp = r\"C:\\Users\\Glen B\\Documents\\GDAL\\shp\\Census2011_Small_Areas_generalised20m.shp\"\n# open the file\nds = driver.Open(shp, 0)\n# reference the only layer in a Shapefile\nlyr = ds.GetLayer(0)```\n\nSpatial Reference Information\n\nStraight away we cant print the spatial reference information associated with the Shapefile (contained in the .prj file)\n\n`print lyr.GetSpatialRef()`\n\nThis will print out…\n\n```PROJCS[\"TM65_Irish_Grid\",\nGEOGCS[\"GCS_TM65\",\nDATUM[\"TM65\",\nSPHEROID[\"Airy_Modified\",6377340.189,299.3249646]],\nPRIMEM[\"Greenwich\",0.0],\nUNIT[\"Degree\",0.0174532925199433]],\nPROJECTION[\"Transverse_Mercator\"],\nPARAMETER[\"False_Easting\",200000.0],\nPARAMETER[\"False_Northing\",250000.0],\nPARAMETER[\"Central_Meridian\",-8.0],\nPARAMETER[\"Scale_Factor\",1.000035],\nPARAMETER[\"Latitude_Of_Origin\",53.5],\nUNIT[\"Meter\",1.0]]```\n\nYou can also access this information individually…\n\n```# projected coordinate system\nproj_string = lyr.GetSpatialRef().GetAttrValue(\"PROJCS\", 0)\n# geographic coordinate system\ngeog_string = lyr.GetSpatialRef().GetAttrValue(\"GEOGCS\", 0)\n# EPSG Code if available\nepsg = lyr.GetSpatialRef().GetAttrValue(\"AUTHORITY\", 1)\n# datum\ndatum = lyr.GetSpatialRef().GetAttrValue(\"DATUM\", 0)\n\nprint \"\\nFile: {0}\\n\\nProjected: {1}\\nEPSG: {2}\\n\".format(lyr.GetName(),proj_string, epsg)\nprint \"Geographic: {0}\\nDatum: {1}\\n\".format(geog_string, datum)```\n\nThe output…\n\n```File: Census2011_Small_Areas_generalised20m\n\nProjected: TM65_Irish_Grid\nEPSG: None\n\nGeographic: GCS_TM65\nDatum: TM65```\n\nIf there is an EPSG code in the .prj file it will be printed instead of None.\n\nGeometry Type\n\nIf we reference the first feature we can get the geometry of the Shapefile\n\n```first_feat = lyr.GetFeature(1)\nprint \"Geometry Type: {0}\\n\".format(first_feat.geometry().GetGeometryName())```\n\nIn this instance it is a polygon Shapefile.\n\n`Geometry Type: POLYGON`\n\nField Information\n\nLet’s get some information on the data through the Layer Definition.\n\n```# https://pcjericks.github.io/py-gdalogr-cookbook/vector_layers.html\nlyr_def = lyr.GetLayerDefn()```\n\nBut before we do we need to create a few list structures. These will be used to hold the accessed information and enable us to neatly print them to screen.\n\n```# list to hold headers for filed information\nheader_list = [\"FIELD NAME\", \"TYPE\", \"WIDTH\", \"PRECISION\"]\n# list will be populated with field information\noutput_list = []\n# list will be populated with field names and used for attribute headers\nfld_names = []```\n\nCycle through each field and populate the necessary lists…\n\n```# for each field\nfor i in range(lyr_def.GetFieldCount()):\n# reference the field name\nfld_name = lyr_def.GetFieldDefn(i).GetName()\n# reference the field type\nfld_type = lyr_def.GetFieldDefn(i).GetFieldTypeName(lyr_def.GetFieldDefn(i).GetType())\n# reference the field width\nfld_width = lyr_def.GetFieldDefn(i).GetWidth()\n# reference the field precision\nfld_precision = lyr_def.GetFieldDefn(i).GetPrecision()\n# append these as a list to the output_list\noutput_list.append([fld_name, fld_type, str(fld_width), str(fld_precision)])\n# append field name to fld_name\nfld_names.append(fld_name)```\n\nThe output_list is a list of lists containing information for each field, the field name, data type, width and precision, this is matched in the header_list. The fld_names will be used further down to print out attributes, this list hold the field names as headers. Let’s print the field information…\n\n`print \"{0}\\n\".format(tabulate(output_list, header_list))`\n\nHere’s the output…\n\n```FIELD NAME TYPE WIDTH PRECISION\n------------ ------ ------- -----------\nNUTS1 String 3 0\nNUTS1NAME String 7 0\nNUTS2 String 4 0\nNUTS2NAME String 26 0\nNUTS3 String 5 0\nNUTS3NAME String 15 0\nCOUNTY String 2 0\nCOUNTYNAME String 25 0\nCSOED String 11 0\nOSIED String 13 0\nEDNAME String 45 0\nSMALL_AREA String 61 0\nGEOGID String 65 0\nMALE2011 Real 20 10\nFEMALE2011 Real 20 10\nTOTAL2011 Real 20 10\nPPOCC2011 Real 20 10\nUNOCC2011 Real 20 10\nVACANT2011 Real 20 10\nHS2011 Real 20 10\nPCVAC2011 Real 20 10\nCREATEDATE String 10 0```\n\nAttribute Table\n\nNext we print out some attributes for a set of features, the first ten.\n\n```# number of features from the first to print attributes for\nnum_to_return = 10\n#num_to_return = lyr.GetFeatureCount()```\n\nUse the commented out line if you want to print attributes for all features. Create an empty list to hold the attributes. Some fields contain characters from the Irish language so we account for this so that the attributes are printed correctly.\n\n```# list will be populated with attribute data\natt_table = []\n\n# for each feature in the Shapefile\nfor count, feature in enumerate(lyr):\n# up to the number of set features to print\nif count < num_to_return:\n# count will beacome the Feature ID\natts = [count]\n# for each field append the data to atts list\nfor name in fld_names:\ntry:\n# if the attribute is a string then decode with Celtic Languages\natts.append(feature.GetField(name).decode(\"iso8859_14\"))\nexcept Exception:\natts.append(feature.GetField(name))\n# append the data for the feature to the att_table list\natt_table.append(atts)```\n\nThe count becomes the Feature ID but we have no field for this so we will create one…\n\n```# add a FID header (count)\nfld_names.insert(0, \"FID\")```\n\nSo let’s print out the attributes…\n\n```print tabulate(att_table, fld_names)\nprint \"{0} out of {1} features\".format(num_to_return, lyr.GetFeatureCount())```\n\nHere’s the output…\n\n``` FID NUTS1 NUTS1NAME NUTS2 NUTS2NAME NUTS3 NUTS3NAME COUNTY COUNTYNAME CSOED OSIED EDNAME SMALL_AREA GEOGID MALE2011 FEMALE2011 TOTAL2011 PPOCC2011 UNOCC2011 VACANT2011 HS2011 PCVAC2011 CREATEDATE\n----- ------- ----------- ------- -------------------- ------- --------------- -------- -------------- ------- ------- ------------------------------ ------------ ---------- ---------- ------------ ----------- ----------- ----------- ------------ -------- ----------- ------------\n0 IE0 Ireland IE02 Southern and Eastern IE022 Mid-East 15 Wicklow County 15039 257005 Aughrim 257005002 A257005002 137 138 275 84 18 15 102 14.7 27-03-2012\n1 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01054 017049 Tinnahinch 017049001 A017049001 186 176 362 111 25 24 136 17.6 27-03-2012\n2 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01053 017032 Marley 017032001 A017032001 194 173 367 121 8 5 129 3.9 27-03-2012\n3 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01054 017049 Tinnahinch 017049002 A017049002 75 75 150 67 29 29 96 30.2 27-03-2012\n4 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01054 017049 Tinnahinch 017049003 A017049003 84 81 165 64 16 14 80 17.5 27-03-2012\n5 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01015 017005 Ballyellin 017005002 A017005002 105 99 204 71 6 5 77 6.5 27-03-2012\n6 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01015 017005 Ballyellin 017005001 A017005001 115 108 223 70 9 8 79 10.1 27-03-2012\n7 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01033 017033 Muinebeag (Bagenalstown) Rural 017033001 A017033001 201 205 406 143 15 14 158 8.9 27-03-2012\n8 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01034 017034 Muinebeag (Bagenalstown) Urban 017034002 A017034002 142 116 258 89 9 9 98 9.2 27-03-2012\n9 IE0 Ireland IE02 Southern and Eastern IE024 South-East (IE) 01 Carlow County 01034 017034 Muinebeag (Bagenalstown) Urban 017034003 A017034003 174 169 343 107 6 4 113 3.5 27-03-2012\n10 out of 18488 features```\n\nUnique Values and Counts\n\nNext we’ll get a list of the unique COUNTYNAME entries and a count to see how many small areas are in each. (The below works for text fields only)\n\n```# rest to first feature\n\n# field to return unique list and count of\nfield = \"COUNTYNAME\"\n\n# create empty dictionary\nvalues_dict = {}\n\n# for each feature\nfor feature in lyr:\nattribute = feature.GetField(field).decode(\"iso8859_14\")\n# if the COUNTYNAME is not already in the dictionary add it and assign a value of 1\nif attribute not in values_dict:\nvalues_dict[attribute] = 1\n# otherwise do not add it and increase the existing value by 1\nelse:\nvalues_dict[attribute] = values_dict[attribute] + 1\n\n## convert dictionary to list for use with tabulate\nkey_value_list = [[key, value] for key, value in values_dict.items()]\n\n## print results\nprint \"\\nTotal Feature Count: {0}\\n\".format(lyr.GetFeatureCount())\nprint tabulate(sorted(key_value_list), [field, \"Count\"])```\n\nAnd here’s the output…\n\n```Total Feature Count: 18488\n\nCOUNTYNAME Count\n---------------------- -------\nCarlow County 210\nCavan County 322\nClare County 511\nCork City 519\nCork County 1650\nDonegal County 761\nDublin City 2202\nDún Laoghaire-Rathdown 760\nFingal 938\nGalway City 307\nGalway County 741\nKerry County 701\nKildare County 731\nKilkenny County 372\nLaois County 305\nLeitrim County 173\nLimerick City 258\nLimerick County 514\nLongford County 179\nLouth County 462\nMayo County 643\nMeath County 636\nMonaghan County 244\nNorth Tipperary 283\nOffaly County 286\nRoscommon County 303\nSligo County 307\nSouth Dublin 906\nSouth Tipperary 350\nWaterford City 198\nWaterford County 275\nWestmeath County 338\nWexford County 615\nWicklow County 488```\n\nAlternatively we could print out based on the highest count descending by replacing the last print statement with…\n\n```# http://stackoverflow.com/questions/17555218/python-how-to-sort-a-list-of-lists-by-the-fourth-element-in-each-list\nprint tabulate(sorted(key_value_list, key = itemgetter(1), reverse = True), [field, \"Count\"])```\n\n…to get…\n\n```COUNTYNAME Count\n---------------------- -------\nDublin City 2202\nCork County 1650\nFingal 938\nSouth Dublin 906\nDonegal County 761\n...```\n\nI will add to these as I come across something useful. If you know of any neat things to add please comment below. Please also comment if anything is unclear or if this was useful to you." ]

[ null, "https://geobitz.files.wordpress.com/2017/09/distance_points_along_line.jpg", null, "https://geobitz.files.wordpress.com/2017/09/distance_points_along_line_attributes.jpg", null, "https://geobitz.files.wordpress.com/2017/03/mean_center_formula.jpg", null, "https://geobitz.files.wordpress.com/2017/04/standard_distance_formula1.jpg", null, "https://geobitz.files.wordpress.com/2017/03/birmingham_buglaries_2016.jpg", null, "https://geobitz.files.wordpress.com/2017/04/standard_distance_circle.jpg", null, "https://geobitz.files.wordpress.com/2017/04/wgt_mean_center_formula.jpg", null, "https://geobitz.files.wordpress.com/2017/04/small_areas_of_ireland.jpg", null, "https://geobitz.files.wordpress.com/2017/04/small_areas_of_ireland_wmeanc.jpg", null, "https://geobitz.files.wordpress.com/2017/04/small_areas_of_ireland.jpg", null ]

[ "Cody\n\n# Problem 167. Pizza!\n\nSolution 831571\n\nSubmitted on 18 Feb 2016 by Audrey Hoang\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% z = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))\n\n2   Pass\n%% z = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))\n\n3   Pass\n%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))\n\n4   Pass\n%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

[ null ]

[ "Discover a lot of information on the number 9840: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!\n\n## Mathematical properties of 9840\n\nIs 9840 a prime number? No\nIs 9840 a perfect number? No\nNumber of divisors 40\nList of dividers 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 41, 48, 60, 80, 82, 120, 123, 164, 205, 240, 246, 328, 410, 492, 615, 656, 820, 984, 1230, 1640, 1968, 2460, 3280, 4920, 9840\nSum of divisors 31248\n\n## How to write / spell 9840 in letters?\n\nIn letters, the number 9840 is written as: Nine thousand eight hundred and forty. And in other languages? how does it spell?\n\n9840 in other languages\nWrite 9840 in english Nine thousand eight hundred and forty\nWrite 9840 in french Neuf mille huit cent quarante\nWrite 9840 in spanish Nueve mil ochocientos cuarenta\nWrite 9840 in portuguese Nove mil oitocentos quarenta\n\n## Decomposition of the number 9840\n\nThe number 9840 is composed of:\n\n1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9\n\n1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8\n\n1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4\n\n1 iteration of the number 0 : ... Find out more about the number 0\n\nOther ways to write 9840\nIn letter Nine thousand eight hundred and forty\nIn roman numeral MMMMMMMMMDCCCXL\nIn binary 10011001110000\nIn octal 23160\nIn US dollars USD 9,840.00 (\\$)\nIn euros 9 840,00 EUR (€)\nSome related numbers\nPrevious number 9839\nNext number 9841\nNext prime number 9851\n\n## Mathematical operations\n\nOperations and solutions\n9840*2 = 19680 The double of 9840 is 19680\n9840*3 = 29520 The triple of 9840 is 29520\n9840/2 = 4920 The half of 9840 is 4920.000000\n9840/3 = 3280 The third of 9840 is 3280.000000\n98402 = 96825600 The square of 9840 is 96825600.000000\n98403 = 952763904000 The cube of 9840 is 952763904000.000000\n√9840 = 99.196774141098 The square root of 9840 is 99.196774\nlog(9840) = 9.1942109900463 The natural (Neperian) logarithm of 9840 is 9.194211\nlog10(9840) = 3.9929950984313 The decimal logarithm (base 10) of 9840 is 3.992995\nsin(9840) = 0.50709329390857 The sine of 9840 is 0.507093\ncos(9840) = 0.86189117136269 The cosine of 9840 is 0.861891\ntan(9840) = 0.58834956286515 The tangent of 9840 is 0.588350" ]

[ null ]

[ "## Notes on Convexity Inequalities\n\nIn the analysis of randomized algorithms, we frequently use various inequalities to simplify expressions or make them easier to work with. These inequalities are usually obvious by plotting them, and proving them just uses basic calculus.\n\n1. Facts from Convex Analysis\n\nLet", null, "${f : S \\rightarrow {\\mathbb R}}$ be a function defined on an interval", null, "${S \\subseteq {\\mathbb R}}$.\n\nDefinition 1 We say that", null, "${f}$ is convex on", null, "${S}$ if", null, "$\\displaystyle f( \\lambda x + (1-\\lambda) y ) ~\\leq~ \\lambda f(x) + (1-\\lambda) f(y)$\n\nfor all", null, "${x, y \\in S}$ and all", null, "${\\lambda \\in [0,1]}$.\n\nGeometrically, this say that the chord connecting the points", null, "${(x,f(x))}$ and", null, "${(y,f(y))}$ lies above the function", null, "${f}$. The following example illustrates that", null, "${f(x)=x^2}$ is convex.", null, "An equivalent statement of this definition is as follows.\n\nFact 1 Suppose", null, "${f : S \\rightarrow {\\mathbb R}}$ is convex. Then, for any points", null, "${x, y \\in S}$ with", null, "${x, we have", null, "$\\displaystyle f(z) ~\\leq~ \\frac{f(y)-f(x)}{y-x} \\cdot (z-x) + f(x)$\n\nfor all", null, "${z \\in [x,y]}$.\n\nFor sufficiently nice functions, one can easily determine convexity by looking at its second derivative.\n\nFact 2 Suppose", null, "${f : S \\rightarrow {\\mathbb R}}$ is twice differentiable and that the second derivative", null, "${f''}$ is non-negative on all of", null, "${S}$. Then", null, "${f}$ is convex.\n\nIn our example above we used", null, "${f(x) = x^2}$. Its second derivative is", null, "${f''(x)=2}$, which is non-negative, so", null, "${f}$ is convex.\n\nThe next fact says that, for convex functions, the linear approximation at any point lies beneath the function.\n\nFact 3 (Subgradient Inequality) Suppose", null, "${f : S \\rightarrow {\\mathbb R}}$ is convex, and", null, "${f}$ is differentiable at a point", null, "${x \\in S}$. Then", null, "$\\displaystyle f(y) ~\\geq~ f(x) + f'(x) (y-x),$\n\nfor any point", null, "${y \\in S}$.\n\nThe following example illustrates this for", null, "${f(x)=x^2}$ at the point", null, "${x=0.5}$.", null, "2. The Inequalities\n\n• Inequality 1.", null, "${1+y \\leq e^y}$.Convexity of", null, "${e^y}$ follows from Fact 2 since its second derivative is", null, "${e^y}$. The inequality follows by applying Fact 3 at the point", null, "${x=0}$.", null, "• Inequality 2.", null, "${1/(1+2z) \\leq 1-z}$ for all", null, "${z \\in [0,1/2]}$.Convexity of", null, "${1/(1+2z)}$ on the set", null, "${S=(0,\\infty)}$ follows from Fact 2 since its second derivative is", null, "${8/(1+2z)^3}$. The inequality follows by applying Fact 1 at the points", null, "${x=0}$ and", null, "${y=0.5}$.", null, "• Inequality 3.", null, "${e^z \\leq 1+(e-1) z}$.As above, convexity of", null, "${e^z}$ follows from Fact 2. The inequality follows by applying Fact 1 at the points", null, "${x=0}$ and", null, "${y=1}$.", null, "Advertisements\nThis entry was posted in Uncategorized. Bookmark the permalink." ]

[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://nickhar.files.wordpress.com/2012/01/convex.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://nickhar.files.wordpress.com/2012/01/convex2.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://nickhar.files.wordpress.com/2012/01/ex11.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://nickhar.files.wordpress.com/2012/01/1overx1.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://nickhar.files.wordpress.com/2012/01/ex21.png", null ]

[ "# Parabola practice questions\n\nDoes anyone know where I can find practice questions involving parabolas?\n\nHi Kieran\nI am assuming you are doing Tables Equations and Graphs Level 1 here is the link to last year’s paper. Please let me know if you need any help with it.\nMrs H\nLevel 1 Mathematics and Statistics (91028) 2022 (nzqa.govt.nz)\n\n@Kieran, have a look at the following links:\n\nYou can also go through past papers, all of them include questions about quadratic functions.\n\nThanks so much. I’m having trouble with question b(i). Could you please explain how I can find q?\n\nIn one of the previous discussions for this topic we published some very helpful recourses about transformations of parabolas:\nThe Parabola.pdf (477.6 KB)\n\nFrom this resource we can see that vertical transformation of the parabola from the origin (0,0) is defined by the number you add to x^2:\n\nFrom your image you may notice that the vertex of the parabola is located at the point (0,9) so it was translated up by 9 units:", null, "That means your equation will take the form:", null, "So q = 9\n\nThanks, I did manage to figure it out. I just had to reread the question. There is another question on the test that I’m really stuck on which is b(iii). I’ve found a = 0.25 using the equation: y = a(x-h)^2 + c. However, I’m not sure if this is correct and I also don’t know what to do from there (assuming it is correct) to find the maximum height. Could you please explain it to me?\n\nFor question b(ii) to find the coefficient of x^2. The best way is to substitute the coordinates of known points into your equation:", null, "Question b(iii) would be an excellence question, @Kieran. To find the equation of a new parabola you will need to use the symmetry property of the parabola:", null, "You can state in your assessment that using the symmetry property of the parabola, if the graph crosses the x-axis at (6,0) and has a vertex at x = 0.5, then it must also cross the x-axis at (-5,0). Therefore, the equation of the parabola can be written using its x-intercepts:", null, "Then, we can use the same method to find the coefficient a as we did for b(ii) - we will substitute the coordinates of a known point. Any point on the x-axis will make the equation equal to 0 and won’t help us, so we can use the y-intercept (0,9):", null, "and your final equation will be:", null, "To find the maximum height we need to substitute the x-coordinate 0.5 into our final equation:", null, "m.\n\nThanks so much, that’s really helpful. Just a couple of questions with b(ii), firstly wouldn’t the equation be y = - ax^2 + 9 because the parabola is a n shape and not a u shape? Secondly, would it be better to write a as the fraction or the decimal? Thanks again for your help.\n\nGood questions, @Kieran. We can write our equation in the general form y = ax^2 + 9 (as I did) and then we will get a equal to a negative number. If you correctly assume that a will be negative and write your general form as y = -ax^2 + 9, your answer for a will be positive. As long as the final equation is correct, it doesn’t matter which way you came to this solution.\n\nFor the second question - it also doesn’t matter in which form you write your answer - fraction or decimal. If you choose to use decimal, round your answer to at least 3 or 4 d.p.\n\nAlright, thanks so much for your help.\n\n1 Like" ]

[ null, "https://global.discourse-cdn.com/studyit/original/1X/3cee658d83e4db9c565f7e332f4cea2c2e3a4d1b.png", null, "https://global.discourse-cdn.com/studyit/original/1X/9fa4b04503b53bd815f954fd68751b30e5e3d974.png", null, "https://global.discourse-cdn.com/studyit/original/1X/290af4e88e70463544679ae986f0e7b7a515340b.png", null, "https://global.discourse-cdn.com/studyit/original/1X/509f555fc36c6ea6b648500ceccffe9dfc580881.png", null, "https://global.discourse-cdn.com/studyit/original/1X/7285f147d2d4f350d5bcf8976c622bbe2304ef6a.png", null, "https://global.discourse-cdn.com/studyit/original/1X/9aed4b15d1c11b3d9b7f3e9e8f1815940228ddaf.png", null, "https://global.discourse-cdn.com/studyit/original/1X/6271c11704793fe29cd396a523884e52e7dfbe7c.png", null, "https://global.discourse-cdn.com/studyit/original/1X/45e95c42328c4d9049bc3282001ab599d28099a9.png", null ]

[ "", null, "Checking date: 31/01/2023\n\nCourse: 2023/2024\n\nNumerical methods\n(16493)\nBachelor in Data Science and Engineering (Plan: 392 - Estudio: 350)\n\nCoordinating teacher: TERAN VERGARA, FERNANDO DE\n\nDepartment assigned to the subject: Mathematics Department\n\nType: Compulsory\nECTS Credits: 6.0 ECTS\n\nCourse:\nSemester:\n\nRequirements (Subjects that are assumed to be known)\nLineal Algebra, Programming, Calculus I, Calculus II\nObjectives\nUsing NUMERICAL METHODS (NM) to calculate approximate solutions of mathematical models Study the stability and accuracy of NM. Calculate numerical solution of systems of nonlinear equations. Approximate the minimum of a function of several variables. Developing, analyzing, and implementing finite difference methods. Solving ordinary differential equations and systems by numerical integration methods. Using the software environments to discuss the efficiency, pros and cons of different NM.\nSkills and learning outcomes\nDescription of contents: programme\n1. Fundamentals (floating point, errors, stability, algorithms...). 2. Solution of linear systems of equations. 3. Numerical solution of equations and systems of nonlinear equations. 4. Interpolation and approximation of functions. 5. Least squares problems. 6. Numerical optimization. 7. Numerical integration. 8. Numerical differentiation: numerical solution of ODEs. 9. Fast Fourier Transform.\nLearning activities and methodology\nThis is a \"hands on\" course. In the large-group lectures the theoretical contents will be introduced, and they will be delivered in a blackboard standard room. However, the small-group lectures will take place in the computer Lab, and students are supposed to follow the explanations of the instructor performing in real time the exercises, examples and other proposed activities. Students must become acquainted with MATLAB. The course will start learning how to program MATLAB. After and introduction to the course, every two weeks (as a general rule), one of the topics of the course will be discussed in the classroom and practices related to these topics will be proposed to the students. Usually the practices involve to solve a simple problem by writing the appropriate code.\nAssessment System\n• % end-of-term-examination 50\n• % of continuous assessment (assigments, laboratory, practicals...) 50\nCalendar of Continuous assessment\nBasic Bibliography\n• [A] K. Atkinson. Elementary Numerical Analysis. John Wiley & Sons. 2004\n• [BF] R. L. Burden, J. D. Faires. Numerical Methods. Brooks/Cole, Cengage Learning. 2003\n• [QSG] A. Quarteroni, F. Saleri, P. Gervasio. Scientific computing with MATLAB and Octave. Springer. 2010\n• [QSS] A. Quarteroni, R. Sacco, F. Saleri. Numerical Mathematics. Springer. 2007", null, "Electronic Resources *\n• [BC] A. Belegundu, T. Chandrupatla. Optimization Concepts and Applications in Engineering. Cambridge University Press. 2011\n• [BV] S. Boyd, L. Vanderberghe. Convex Optimization. Cambridge University Press. 2004\n• [DCM] S. Dunn, A. Constantinides, P. Moghe. Numerical Methods in Biomedical Engineering. Elsevier Academic Press. 2010\n• [DH] P. Deuflhard, A. Hohmann. Numerical Analysis in Modern Scientific Computing. An Introduction. Springer. 2003\n• [FJNT] P.E. Frandsen, K. Jonasson, H.B. Nielsen, O. Tingleff. Unconstrained Optimization. IMM, DTU. 1999\n• [HH] D. Higham, N. J. Higham. Matlab Guide. SIAM. 2017\n• [HJ] R. A. Horn, C. R. Johnson. Matrix Analysis, 2nd ed.. Cambridge University Press. 2013\n• [H] N. J. Higham. Accuracy and Stability of Numerical Methods. SIAM. 1998\n• [K] C. Kelley. Iterative Methods for Optimization. SIAM (available online). 1999\n• [NW] J. Nocedal, S. J. Wright. Numerical Optimization, 2nd ed.. Springer. 2006\n• [TB] L. N. Trefethen, D. Bau. [TB] L. N. Trefethen, D. Bau. SIAM. 1997", null, "Electronic Resources *\n(*) Access to some electronic resources may be restricted to members of the university community and require validation through Campus Global. If you try to connect from outside of the University you will need to set up a VPN\n\nThe course syllabus may change due academic events or other reasons." ]

[ null, "https://aplicaciones.uc3m.es/cpa/web/imagenes/acronimo_nombre.png", null, "https://aplicaciones.uc3m.es/cpa/web/imagenes/ARROBA.gif", null, "https://aplicaciones.uc3m.es/cpa/web/imagenes/ARROBA.gif", null ]

[ "## Number of periods or years (present value,future value,periodic interest rate or required rate of return)\n\n### Formula", null, "n\nnumbers of periods (months, quarters or years)\nP\npresent value\nS\nfuture value\ni\nperiodic (monthly,quarterly or yearly) interest rate\n\n### Formula description\n\nPresent value, also known as present discounted value, is the value on a given date of a future payment or series of future payments, discounted to reflect the time value of money and other factors such as investment risk. This formula is used to calculate the length of time required for a single cash flow (present value) to reach a certain amount(future value), given a certain interest rate.\n\n### Calculator (how to use calculator?)", null, "P\nS\ni\nn\nPrecision\n\n### Comment or add more code\n\nIf you cannot find the formula or calculator you want, please tell us what you want and we will add it for you ASAP. If you want anything else or find any error on this page, please just let us know. If you know the formula in other languages, you are welcome to add it. Thanks for contributing to wikicalculator.com!" ]

[ null, "http://www.wikicalculator.com/formula_image/Number-of-periods-or-years-(present-value-future-value-periodic-interest-rate-or-required-rate-of-return)-585.png", null, "http://www.wikicalculator.com/formula_image/Number-of-periods-or-years-(present-value-future-value-periodic-interest-rate-or-required-rate-of-return)-585.png", null ]

[ "Your Math worksheets grade 8 images are ready in this website. Math worksheets grade 8 are a topic that is being searched for and liked by netizens today. You can Find and Download the Math worksheets grade 8 files here. Get all royalty-free vectors.\n\nIf you’re searching for math worksheets grade 8 images information connected with to the math worksheets grade 8 topic, you have pay a visit to the ideal site. Our site frequently provides you with suggestions for downloading the maximum quality video and picture content, please kindly hunt and locate more enlightening video articles and graphics that match your interests.\n\nMath Worksheets Grade 8. Here is a comprehensive and perfect collection of FREE grade 8 Common Core Mathematics worksheets that would help your students in grade 8 Common Core Math preparation and practice. Each worksheet is differentiated including a. Expressions function tables probability as begin to work at the core of this grade. Find by title or description Search Reset.", null, "Grade 3 Multiplication Worksheets Free Printable 3rd Grade Math Worksheets Division Worksheets Third Grade Math Worksheets From pinterest.com\n\nExpressions function tables probability as begin to work at the core of this grade. Grade 8 math practice questions tests teacher assignments teacher worksheets printable worksheets and other activities for National Curriculum IMO and SAT Subject Test. Grade 8 - Worksheets - Mathematics Search for documents. Grade 8 Mathematics - 2021 - Term 3. Grade Probability Word Problems Worksheet Activities Middle School. Mathcation provides you with step-by-step tutorial videos including guided notes that.\n\n### Grade Probability Word Problems Worksheet Activities Middle School.\n\nGrade 8 Mathematics - 2021 - Term 3. 8th grade math Grade 8 Math Grade 8 Math worksheets Math worksheets. Printable Eighth Grade Grade 8 Worksheets Tests and Activities. Apply prime factorization and determine the square roots of the first fifty perfect squares offered as positive integers. Reza is an experienced Math instructor and a test-prep expert who has been tutoring students since 2008. Expressions function tables probability as begin to work at the core of this grade.", null, "Source: pinterest.com\n\n8 th Grade 8 8th Grade Math Worksheets and Study Guides The big ideas in Eighth Grade Math include understanding the concept of a function and using functions to describe quantitative relationships and analyzing two- and three-dimensional space and figures using. He works with students individually and in group settings. 8th grade math Grade 8 Math Grade 8 Math worksheets Math worksheets. Use the formula m y 2 - y 1 x 1 - x 1 to. 8th Grade Math Worksheets Separate answers are included to make marking easy and quick.", null, "Source: pinterest.com\n\nOur worksheets use a variety of high-quality images and some are aligned to Common Core Standards. Students work towards mastery with the basic order of operations. Number Patterns Sequences Find the nth term. Apply prime factorization and determine the square roots of the first fifty perfect squares offered as positive integers. Grade 8 Mathematics - 2021 - Term 3.", null, "Source: in.pinterest.com\n\nYear 8 Math Worksheets Pdf - Math Worksheets Dynamically Created. Grade 8 - Worksheets - Mathematics Search for documents. Expressions function tables probability as begin to work at the core of this grade. He works with students individually and in group settings. Convert each fraction with a multiple of 10 as its denominator into a decimal number by placing the decimal point at the right spot.", null, "Source: pinterest.com\n\nPrintable Eighth Grade Grade 8 Worksheets Tests and Activities. Mathcation provides you with step-by-step tutorial videos including guided notes that. Apply prime factorization and determine the square roots of the first fifty perfect squares offered as positive integers. Print our Eighth Grade Grade 8 worksheets and activities or administer them as online tests. As a quick supplement for your instruction use our printable 8th grade math worksheets to provide practice problems to your students.", null, "Source: pinterest.com\n\nSelect Grade 8 Math Worksheets by Topic. Convert each fraction with a multiple of 10 as its denominator into a decimal number by placing the decimal point at the right spot. Each worksheet is differentiated including a. These Worksheets for Grade 8 Mathematics class assignments and practice tests. Download our free Mathematics worksheets for the grade 8 Common Core test.", null, "Source: pinterest.com\n\nOur 8th grade math worksheets cover a wide array of topics including all 8th grade Common Core Standards. Grade Probability Word Problems Worksheet Activities Middle School. Year 8 Math Worksheets Pdf - Math Worksheets Dynamically Created. Grade 8 Mathematics - 2021 - Term 3. Grade 8 Math Worksheets Popular Worksheets.", null, "Source: pinterest.com\n\nSeparate school licenses are available here Single. Download free printable worksheets for CBSE Class 8 Mathematics with important topic wise questions students must practice the NCERT Class 8 Mathematics worksheets question banks workbooks and exercises with solutions which will help them in revision of important concepts Class 8 Mathematics. Our worksheets use a variety of high-quality images and some are aligned to Common Core Standards. Expressions function tables probability as begin to work at the core of this grade. Click any of the links below to download your worksheet as an easy-to.", null, "Source: pinterest.com\n\nThrough repetition they will gain. Lessons include topics such as Scientific Notation Slope Fractions Exponents Pythagorean Theorem and Solving Equations. Download free printable worksheets for CBSE Class 8 Mathematics with important topic wise questions students must practice the NCERT Class 8 Mathematics worksheets question banks workbooks and exercises with solutions which will help them in revision of important concepts Class 8 Mathematics. 8 th Grade 8 8th Grade Math Worksheets and Study Guides The big ideas in Eighth Grade Math include understanding the concept of a function and using functions to describe quantitative relationships and analyzing two- and three-dimensional space and figures using. Over 200 pages of the highest quality 8th Grade math worksheets.", null, "Source: pinterest.com\n\nEighth Grade Math Worksheets Grade 8 - For Ages 13 to 14 Math in the 8th grade begins to prove more substantial as far as long range skills students will use and need. Grade 8 - Worksheets - Mathematics Search for documents. Download free printable worksheets for CBSE Class 8 Mathematics with important topic wise questions students must practice the NCERT Class 8 Mathematics worksheets question banks workbooks and exercises with solutions which will help them in revision of important concepts Class 8 Mathematics. Each worksheet is differentiated including a. Download our free Mathematics worksheets for the grade 8 Common Core test.", null, "Source: pinterest.com\n\n8 th Grade 8 8th Grade Math Worksheets and Study Guides The big ideas in Eighth Grade Math include understanding the concept of a function and using functions to describe quantitative relationships and analyzing two- and three-dimensional space and figures using. He works with students individually and in group settings. Reza is an experienced Math instructor and a test-prep expert who has been tutoring students since 2008. Download our free Mathematics worksheets for the grade 8 Common Core test. Grade 8 math practice questions tests teacher assignments teacher worksheets printable worksheets and other activities for National Curriculum IMO and SAT Subject Test.", null, "Source: pinterest.com\n\nPrintable Eighth Grade Grade 8 Worksheets Tests and Activities. Here is a comprehensive and perfect collection of FREE grade 8 Common Core Mathematics worksheets that would help your students in grade 8 Common Core Math preparation and practice. Number Patterns Sequences Find the nth term. Students work towards mastery with the basic order of operations. He has helped many students raise their standardized test scores–and attend the colleges of their dreams.", null, "Source: pinterest.com\n\nGrade 8 math practice questions tests teacher assignments teacher worksheets printable worksheets and other activities for National Curriculum IMO and SAT Subject Test. Please click the following links to get math printable math worksheets for grade 8. Grade 8 math practice questions tests teacher assignments teacher worksheets printable worksheets and other activities for National Curriculum IMO and SAT Subject Test. Print our Eighth Grade Grade 8 worksheets and activities or administer them as online tests. 8 th Grade 8 8th Grade Math Worksheets and Study Guides The big ideas in Eighth Grade Math include understanding the concept of a function and using functions to describe quantitative relationships and analyzing two- and three-dimensional space and figures using.", null, "Source: pinterest.com\n\nThis grade 8 mathematics worksheet can be used for revision after completing the first term of the South African caps curriculum and inlcudes whole numbers prime numbers multiples and factors and lowest common multiple and highest common factors integers exponents including scientific notation. He has helped many students raise their standardized test scores–and attend the colleges of their dreams. Apply prime factorization and determine the square roots of the first fifty perfect squares offered as positive integers. Mathcation provides you with step-by-step tutorial videos including guided notes that. Free Math Worksheets for Sixth Seventh Eighth and Ninth Grade w Answer Keys The following printable math worksheets for 6th 7th 8th and 9th grade include a complete answer key.", null, "Source: pinterest.com\n\nNumber Patterns Sequences Find the nth term. 8th Grade Math Worksheets. Grade 8 - Worksheets - Mathematics Search for documents. 8th Grade Math Worksheets Separate answers are included to make marking easy and quick. Our 8th grade math worksheets cover a wide array of topics including all 8th grade Common Core Standards.", null, "Source: pinterest.com\n\nExpressions function tables probability as begin to work at the core of this grade. 8th grade math Grade 8 Math Grade 8 Math worksheets Math worksheets. Year 8 Math Worksheets Pdf - Math Worksheets Dynamically Created. Through repetition they will gain. Print our Eighth Grade Grade 8 worksheets and activities or administer them as online tests.", null, "Source: pinterest.com\n\nYear 8 Math Worksheets Pdf - Math Worksheets Dynamically Created. Each worksheet is differentiated including a. Select Grade 8 Math Worksheets by Topic. Our 8th grade math worksheets cover a wide array of topics including all 8th grade Common Core Standards. 8th Grade Math Worksheets.", null, "Source: pinterest.com\n\nFraction Mixed Number Worksheets. Each worksheet is differentiated including a. Separate school licenses are available here Single. Our 8th grade math worksheets cover a wide array of topics including all 8th grade Common Core Standards. Printable Eighth Grade Grade 8 Worksheets Tests and Activities.", null, "Source: pinterest.com\n\nSelect Grade 8 Math Worksheets by Topic. Expressions function tables probability as begin to work at the core of this grade. Number Patterns Sequences Find the nth term. Eighth Grade Math Worksheets Grade 8 - For Ages 13 to 14 Math in the 8th grade begins to prove more substantial as far as long range skills students will use and need. Use the formula m y 2 - y 1 x 1 - x 1 to.\n\nThis site is an open community for users to do sharing their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.\n\nIf you find this site adventageous, please support us by sharing this posts to your favorite social media accounts like Facebook, Instagram and so on or you can also save this blog page with the title math worksheets grade 8 by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website." ]

[ null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null, "https://freeteacherworksheets.netlify.app/img/placeholder.svg", null ]

[ "# When does the adjoint operator map closed convex subsets to closed convex subset?\n\nLet $T:X\\rightarrow Y$ be a linear continuous map between Banach spaces $X$ and $Y$ and denote by $T':Y'\\rightarrow X'$ the norm adjoint of $T$. Let $M\\subseteq U'$ be a subset of the unit sphere $U'$ of $Y'$. Let $C\\subseteq M$ be any weak*-closed convex subset. Is the image $T'(C)$ weak*-closed in $X'$? Or more generally, assuming that $C$ is convex, under which additional conditions on $C$ is the image $T'(C)$ weak*-closed in $X'$?\n\n• I have forgotten to mention that we can also assume that $T'$ is injective on $M$! – Andy Teich Jun 13 '12 at 8:56\n• Do you really mean the sphere, or do you mean the ball? Convex subsets of spheres might not be very big... – Yemon Choi Jun 13 '12 at 9:03\n• Yes, it I really mean the sphere. Think of $Y'$ as being an $L^1$-space... – Andy Teich Jun 13 '12 at 9:06\n• This is very similar to a question here: math.stackexchange.com/questions/157069/… – Matthew Daws Jun 13 '12 at 12:10\n• @Peter: No; but if you look on this site, you'll see that it's generally considered good to cross-link. You probably wouldn't know that as a new user, so I thought I'd highlight this for you. – Matthew Daws Jun 13 '12 at 12:27\n\n## 1 Answer\n\nWhen you say that $C$ is weak* closed I'm not sure whether you mean as a subset of $Y'$ or in the relative weak* topology on $M$. If the latter, the answer is obviously \"no\": take $T$ to be the identity map and let $C = M$ be a convex subset of the unit sphere which is not weak* closed. (Examples are easy to find, even in finite dimensions.) If the former, the answer is \"yes\", for then $C$ is a weak* closed subset of the closed unit ball, hence it is weak* compact, hence its image under a weak* continuous map (which $T'$ is) must also be weak* compact, and hence weak* closed.\n\n• If I understand you correctly, it also would suffice to assume that $M$ itself is weak*-closed...? – Andy Teich Jun 13 '12 at 12:45\n• Weak* compact (which is the same as weak* closed, if it's bounded). – Nik Weaver Jun 13 '12 at 12:51\n• So it is sufficient to assume that $C$ is weak*-closed and $M$ is weak*-closed. Is it also necessary to have that $C$ is weak*-closed?Can we give conditions on $C$, $M$ or $T'$ to have that for any $C\\subseteq M$ the image $T'(C)$ is weak*-closed in $X'$? – Andy Teich Jun 13 '12 at 13:26\n• Since the question is changing, maybe you'd better edit your original post to clarify what it is you really want. – Nik Weaver Jun 13 '12 at 13:37" ]

[ null ]

[ "# Re: [gcj] Re: KickStart 2020 simple problem. Code runs locally, but gets WA on Google's machine\n\n```Thanks a Lot <3 <3 Actually single House was returning NONE, and also it\nhad some problems with house list like [25,25] with a budget of 50. I fixed\nthem, but still it gets WA..... Ohh..gosh !```\n```\ndef findSol(arr, budget):\nif len(arr) == 1 and sum(arr) <= budget:\nreturn 1\narr.sort()\ntotal = 0\ncount = 0\nfor i in range(0, len(arr)):\ntotal += arr[i]\n#if the total exactly matches budget and there are no more elements\n# ex : [25, 25] with budget 50\nif total == budget:\ncount += 1\nreturn count\nelif total < budget:\ncount += 1\nelse:\nreturn count\n\ndef main():\nt = int(input())\noutput = []\nwhile t > 0:\nnum_and_budget = input().split()\nnum_and_budget = list(map(int, num_and_budget))\nbudget = num_and_budget\nitems = input().split()\nitems = list(map(int, items))\noutput.append(findSol(items, budget))\nt -= 1\nfor i in range(len(output)):\nprint(\"Case #\" + str(i+1) + \": \" + str(output[i]))\n\nmain()\n\n--\nYou received this message because you are subscribed to the Google Groups" ]

[ null ]

[ "Considering Power-Sizing Index Change Miriam has been asked to estimate the cost today of a 2500 ft2 heat exchange system for the new plant being analyzed. She has the following data. Her company paid \\$50.000 for a 1000 ft2 heat exchanger 5 years ago. Heat exchangers within this range of capacity have a power sizing exponent (x) of 0.55\n\nMiriam is interested in estimating the annual labor and material costs for a new production facility.\nShe was able to obtain the following labor and material cost data:\nMaterial cost index value was at 544 three years ago and is 715 today.\nAnnual material costs for a similar facility were \\$2,455,000 three years ago.\nExample 2.6 (Continued)\nPower-Sizing Model\n\nX = Power-sizing exponent\n\n Equipment/Facility X Blower, centrifugal 0.59 Compressor 0.32 Crystallizer, vacuum 0.37 Dryer, drum 0.40 Fan, centrifugal 1.17\n Equipment/Facility X Filter, vacuum 0.48 Lagoon, aerated 1.13 Motor 0.69 Reactor 0.56 Tank, horizontal 0.57\n\n(Eq. 2-3)\nExample Power Sizing Exponent Values\n67\n\n67\n68\nExample 2.7\n\nA. Considering Power-Sizing Index Change\nMiriam has been asked to estimate the cost today of a 2500 ft2\nheat exchange system for the new plant being analyzed. She has the following data.\nHer company paid \\$50.000 for a 1000 ft2 heat exchanger 5 years ago.\nHeat exchangers within this range of capacity have a power sizing exponent (x) of 0.55\n69\n\nB. Considering Cost Index Change\n\nExample 2.7 (Continued)\nMiriam has been asked to estimate the cost today of a 2500 ft2\nheat exchange system for the new plant being analyzed. She has the following data.\nFive years ago the Heat Exchanger Cost Index (HECI) was 1306; it is 1487 today.\nLearning Phenomenon: As the number of repetitions increase, performance of people becomes faster and more accurate.\nIn general, as output doubles the unit production time will be reduced to some fixed percentage, the learning curve percentage or learning curve rate\nImprovement and Learning Curve\n70\n70\nLet T1 = Time to perform the 1st unit\nTN = Time to perform the Nth unit\nb = Constant based on learning curve LC%\nN = Number of completed units\n\n(Eq. 2-4)\n(Eq. 2-5)\nLearning Curve\n71\n71\n72\n\nExample 2.8\nCalculate the time required to produce the hundredth unit of a production run if the first unit took\n32.0 minutes to produce and the learning curve rate for production is 80%.\n73\n\nEstimate the overall labor cost portion due to a task that has a learning-curve rate of 85% and reaches a steady state value of 5.0 minutes per unit after 16 units.\nLabor and benefits are \\$22 per hour, and the task requires two skilled workers.\nThe overall production run is 20 units.\n\nGet This Assignment Help Now (30% Discount Code “Law81cglUKdb”)", null, "" ]

[ null, "https://dataedy.com/wp-content/uploads/2021/04/Order-Now.png", null ]

[ "# Problem: Write a nuclear equation for the alpha decay of each nuclide.U-234Th-230Express your answer as a nuclear equation.\n\n###### FREE Expert Solution\n89% (151 ratings)", null, "###### Problem Details\n\nWrite a nuclear equation for the alpha decay of each nuclide.\n\nU-234\nTh-230", null, "" ]

[ null, "https://cdn.clutchprep.com/assets/button-view-text-solution.png", null, "https://lightcat-files.s3.amazonaws.com/problem_images/b692f7f89169e5df-1527433856780.jpg", null ]

[ "src/Pure/defs.ML\n author wenzelm Wed, 24 May 2006 22:04:06 +0200 changeset 19713 69c71d40f8a8 parent 19712 3ae3cc4b1eac child 19727 f5895f998402 permissions -rw-r--r--\n```\n(* Title: Pure/defs.ML\nID: \\$Id\\$\nAuthor: Makarius\n\nGlobal well-formedness checks for constant definitions. Covers plain\n*)\n\nsignature DEFS =\nsig\nval pretty_const: Pretty.pp -> string * typ list -> Pretty.T\nval plain_args: typ list -> bool\ntype T\nval specifications_of: T -> string -> (serial * {is_def: bool, module: string, name: string,\nlhs: typ list, rhs: (string * typ list) list}) list\nval dest: T ->\n{restricts: ((string * typ list) * string) list,\nreducts: ((string * typ list) * (string * typ list) list) list}\nval empty: T\nval merge: Pretty.pp -> T * T -> T\nval define: Pretty.pp -> Consts.T ->\nbool -> bool -> string -> string -> string * typ -> (string * typ) list -> T -> T\nend\n\nstructure Defs: DEFS =\nstruct\n\n(* type arguments *)\n\ntype args = typ list;\n\nfun pretty_const pp (c, args) =\nlet\nval prt_args =\nif null args then []\nelse [Pretty.list \"(\" \")\" (map (Pretty.typ pp o Type.freeze_type) args)];\nin Pretty.block (Pretty.str c :: prt_args) end;\n\nfun plain_args args =\nforall Term.is_TVar args andalso not (has_duplicates (op =) args);\n\nfun disjoint_args (Ts, Us) =\nnot (Type.could_unifys (Ts, Us)) orelse\n((Type.raw_unifys (Ts, map (Logic.incr_tvar (maxidx_of_typs Ts + 1)) Us) Vartab.empty; false)\nhandle Type.TUNIFY => true);\n\nfun match_args (Ts, Us) =\nOption.map Envir.typ_subst_TVars\n(SOME (Type.raw_matches (Ts, Us) Vartab.empty) handle Type.TYPE_MATCH => NONE);\n\n(* datatype defs *)\n\ntype spec = {is_def: bool, module: string, name: string, lhs: args, rhs: (string * args) list};\n\ntype def =\n{specs: spec Inttab.table, (*source specifications*)\nrestricts: (args * string) list, (*global restrictions imposed by incomplete patterns*)\nreducts: (args * (string * args) list) list}; (*specifications as reduction system*)\n\nfun make_def (specs, restricts, reducts) =\n{specs = specs, restricts = restricts, reducts = reducts}: def;\n\nfun map_def c f =\nSymtab.default (c, make_def (Inttab.empty, [], [])) #>\nSymtab.map_entry c (fn {specs, restricts, reducts}: def =>\nmake_def (f (specs, restricts, reducts)));\n\ndatatype T = Defs of def Symtab.table;\n\nfun lookup_list which defs c =\n(case Symtab.lookup defs c of\nSOME (def: def) => which def\n| NONE => []);\n\nfun specifications_of (Defs defs) = lookup_list (Inttab.dest o #specs) defs;\nval restricts_of = lookup_list #restricts;\nval reducts_of = lookup_list #reducts;\n\nfun dest (Defs defs) =\nlet\nval restricts = Symtab.fold (fn (c, {restricts, ...}) =>\nfold (fn (args, name) => cons ((c, args), name)) restricts) defs [];\nval reducts = Symtab.fold (fn (c, {reducts, ...}) =>\nfold (fn (args, deps) => cons ((c, args), deps)) reducts) defs [];\nin {restricts = restricts, reducts = reducts} end;\n\nval empty = Defs Symtab.empty;\n\n(* specifications *)\n\nfun disjoint_specs c (i, {lhs = Ts, name = a, ...}: spec) =\nInttab.forall (fn (j, {lhs = Us, name = b, ...}: spec) =>\ni = j orelse disjoint_args (Ts, Us) orelse\nerror (\"Type clash in specifications \" ^ quote a ^ \" and \" ^ quote b ^\n\" for constant \" ^ quote c));\n\nfun join_specs c ({specs = specs1, restricts, reducts}, {specs = specs2, ...}: def) =\nlet\nval specs' =\nInttab.fold (fn spec2 => (disjoint_specs c spec2 specs1; Inttab.update spec2)) specs2 specs1;\nin make_def (specs', restricts, reducts) end;\n\nfun update_specs c spec = map_def c (fn (specs, restricts, reducts) =>\n(disjoint_specs c spec specs; (Inttab.update spec specs, restricts, reducts)));\n\n(* normalized dependencies: reduction with well-formedness check *)\n\nlocal\n\nfun contained (U as TVar _) (Type (_, Ts)) = exists (fn T => T = U orelse contained U T) Ts\n| contained _ _ = false;\n\nfun wellformed pp defs (c, args) (d, Us) =\nlet\nval prt = Pretty.string_of o pretty_const pp;\nfun err s1 s2 =\nerror (s1 ^ \" dependency of constant \" ^ prt (c, args) ^ \" -> \" ^ prt (d, Us) ^ s2);\nin\nexists (fn U => exists (contained U) args) Us orelse\n(c <> d andalso forall is_TVar Us) orelse\n(if c = d andalso is_some (match_args (args, Us)) then err \"Circular\" \"\" else\n(case find_first (fn (Ts, _) => not (disjoint_args (Ts, Us))) (restricts_of defs d) of\nSOME (Ts, name) =>\nis_some (match_args (Ts, Us)) orelse\nerr \"Malformed\" (\"\\n(restriction \" ^ prt (d, Ts) ^ \" from \" ^ quote name ^ \")\")\n| NONE => true))\nend;\n\nfun reduction pp defs const deps =\nlet\nfun reduct Us (Ts, rhs) =\n(case match_args (Ts, Us) of\nNONE => NONE\n| SOME subst => SOME (map (apsnd (map subst)) rhs));\nfun reducts (d, Us) = get_first (reduct Us) (reducts_of defs d);\n\nfun add (NONE, dp) = insert (op =) dp\n| add (SOME dps, _) = fold (insert (op =)) dps;\nval reds = map (`reducts) deps;\nval deps' =\nif forall (is_none o #1) reds then NONE\nelse SOME (fold_rev add reds []);\nval _ = forall (wellformed pp defs const) (the_default deps deps');\nin deps' end;\n\nfun normalize pp =\nlet\nfun norm_update (c, {reducts, ...}: def) (changed, defs) =\nlet\nval reducts' = reducts |> map (fn (args, deps) =>\n(args, perhaps (reduction pp defs (c, args)) deps));\nin\nif reducts = reducts' then (changed, defs)\nelse (true, defs |> map_def c (fn (specs, restricts, reducts) =>\n(specs, restricts, reducts')))\nend;\nfun norm_all defs =\n(case Symtab.fold norm_update defs (false, defs) of\n(true, defs') => norm_all defs'\n| (false, _) => defs);\nin norm_all end;\n\nin\n\nfun dependencies pp (c, args) restr deps =\nmap_def c (fn (specs, restricts, reducts) =>\nlet\nval restricts' = Library.merge (op =) (restricts, restr);\nval reducts' = insert (op =) (args, deps) reducts;\nin (specs, restricts', reducts') end)\n#> normalize pp;\n\nend;\n\n(* merge *)\n\nfun merge pp (Defs defs1, Defs defs2) =\nlet\nfun add_deps (c, args) restr deps defs =\nif AList.defined (op =) (reducts_of defs c) args then defs\nelse dependencies pp (c, args) restr deps defs;\nfun add_def (c, {restricts, reducts, ...}: def) =\nfold (fn (args, deps) => add_deps (c, args) restricts deps) reducts;\nin Defs (Symtab.join join_specs (defs1, defs2) |> Symtab.fold add_def defs2) end;\n\n(* define *)\n\nfun define pp consts unchecked is_def module name lhs rhs (Defs defs) =\nlet\nfun typargs const = (#1 const, Consts.typargs consts const);\nval (c, args) = typargs lhs;\nval deps = map typargs rhs;\nval restr =\nif plain_args args orelse\n(case args of [Type (a, rec_args)] => plain_args rec_args | _ => false)\nthen [] else [(args, name)];\nval spec =\n(serial (), {is_def = is_def, module = module, name = name, lhs = args, rhs = deps});\nval defs' = defs |> update_specs c spec;\nin Defs (defs' |> (if unchecked then I else dependencies pp (c, args) restr deps)) end;\n\nend;\n```" ]

[ null ]

[ "# [(3x-6)/(4x-8)]<=1. Solve; use interval notation?\n\n### 8 respuestas\n\nRelevancia\n• (3x - 6)/(4x - 8) ≤ 1\n\n[3(x - 2)]/[4(x - 2)] ≤ 1\n\n3/4 ≤ 1  and  x - 2 ≠ 0   (because 0/0 is undefined)\n\nFor all values of x, 3/4 < 1\n\nHence, -∞ < x < ∞  and  x ≠ 2\n\ni.e. -∞ < x < 2 or 2 < x < ∞\n\nThe answer in interval notation: x ∊ (-∞, 2) or (2, ∞)\n\n•  [(3x - 6)/(4x - 8)] ≤ 1\n\nSolutions:\n\n2 > x > 2\n\n• If x = 2 then (3x-6)/(4x-8) = 0/0 is indefinite.\n\nSo x = 2 is not a solution.\n\nIf x < 2 then 4x-8 < 0, so\n\n(3x-6)/(4x-8) ≦ 1\n\n3x-6 ≧ 4x-8\n\n2 ≧ x\n\nBut x < 2 must be satisfied, so the solution is x < 2.\n\nIf x > 2 then 4x-8 > 0, so\n\n(3x-6)/(4x-8) ≦ 1\n\n3x-6 ≦ 4x-8\n\n2 ≦ x\n\nBut x > 2 must be satisfied, so the solution is x > 2.\n\nTherefore, the solution is x < 2 or x > 2.\n\nUsing interval notation, (-inf,2) ∪ (2,inf).\n\n• [(3x - 6)/(4x - 8)] <= 1\n\n3x - 6 <= 4x - 8\n\nx >= 2\n\n2 < x > 2\n\n• ¿Qué te parecieron las respuestas? Puedes iniciar sesión para votar por la respuesta.\n• [(3x - 6)/(4x - 8)] ≤ 1\n\n[(3x - 6)/(4x - 8)] - 1 ≤ 0\n\n[(3x - 6) - (4x - 8)]/(4x - 8) ≤ 0\n\n(3x - 6 - 4x + 8)/(4x - 8) ≤ 0\n\n(- x + 2)/(4x - 8) ≤ 0\n\n(- x + 2)/[4.(x - 2)] ≤ 0 → you know that: 4 > 0\n\n(- x + 2)/(x - 2) ≤ 0\n\n- (x - 2)/(x - 2) ≤ 0\n\n- 1 ≤ 0 ← wrong → no solution\n\n• 3x-6\n\n------ ≤ 1 ⇒\n\n4x-8\n\n// Multiply both sides by (4x-8)\n\n// to get rid of fraction\n\n3x-6 ≤ 4x-8\n\n// Combine like-terms: Subtract\n\n// 3x from both sides\n\n3x-3x - 6 ≤ 4x-3x - 8\n\n-6 ≤ x - 8\n\n// Add 8 to both sides\n\n-6 + 8 ≤ x - 8 + 8\n\n2 ≤ x\n\nSolution in interval notation is [2,+∞).........ANS\n\n• [(3x-6)/(4x-8)] ≤ 1\n\n[3(x-2)]/[4(x-2)] ≤ 1\n\n(3/4) ≤ 1\n\nTrue for all values of x.\n\nThe value of the rational expression has to be considered in the limit as x→2 since the expression evaluates to the indeterminate value 0/0. But, since that limit exists and is equal to 3/4, the solution set also includes x = 2\n\n• (3x - 6)/(4x - 8) ≤ 1\n\nWe need to multiply both sides by (4x - 8)² to make sure we have a positive value and the inequality stays the same.\n\nso, (3x - 6)(4x - 8) ≤ (4x - 8)²\n\ni.e. (4x - 8)² - (3x - 6)(4x - 8) ≥ 0\n\nso, (4x - 8)[(4x - 8) - (3x - 6)]  ≥ 0\n\nso, (4x - 8)(x - 2) ≥ 0\n\ni.e. (x - 2)(x - 2) ≥ 0\n\nor, (x - 2)² ≥ 0....true for all values of x\n\nAs we cannot have x = 2, the solution would be (-∞, 2) ∪ (2, ∞)\n\n:)>\n\n¿Aún tienes preguntas? Pregunta ahora para obtener respuestas." ]

[ null ]

[ "List of public pages created with Protopage\n\n# Home\n\n## Plain sticky notes\n\n### Hello\n\nHi, This is my webpage for maths revision and feel free to use.\n\n## Calendars\n\n### Calendar\n\n• Tue November 27 - My Birthday ^_^\n\n# Unit A\n\n## Plain sticky notes\n\n### Sin, Cos and Tan\n\nIn any right angled triangle, for any angle: The sine of the angle = the length of the opposite side divide by the length of the hypotenuse The cosine of the angle = the length of the adjacent side divide by the length of the hypotenuse The tangent of the angle = the length of the opposite side divide by the length of the adjacent side The hypotenuse of a right angled triangle is the longest side, which is the one opposite the right angle. The adjacent side is the side which is between the angle in question and the right angle. The opposite side is opposite the angle in question. sin = o/h cos = a/h tan = o/a Often remembered by: soh cah toa\n\n### Pythagoras' Theorem\n\nPythagoras's Theorem In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.\n\n# Unit B\n\n## Plain sticky notes\n\n### Rounding to Decimal places\n\n5 or more, we 'round up'. 4 or less, it stays as it is.\n\n### Fraction/Decimals/Percentages\n\nFraction/Decimals/Percentages These are the 3 ways of expressing a part of a whole. How to converting them? http://www.youtube.com/watch?v=pCWKzbvTOKY\n\n### Surds\n\nA surd is a square root which cannot be reduced to a whole number. For example, is not a surd, as the answer is a whole number. But is not a whole number. You could use a calculator to find that but instead of this we often leave our answers in the square root form, as a surd. You need to be able to simplify expressions involving surds. Here are some general rules that you will need to learn. Here is a video clip about surds http://www.youtube.com/watch?v=D3SZE6zkQlM\n\n# Unit C\n\n## Plain sticky notes\n\n### Standard Form\n\nStandard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form. However, instead of the index being positive (in the above example, the index was 3), it will be negative. The rules when writing a number in standard form is that first you write down a number between 1 and 10, then you write × 10(to the power of a number). Example: Write 81 900 000 000 000 in standard form: 81 900 000 000 000 = 8.19 × 1013 It’s 1013 because the decimal point has been moved 13 places to the left to get the number to be 8.19\n\nA quadratic equation is an equation where the highest power of x is x2. There are various methods of solving quadratic equations, as shown below. Important point about square roots: 62 = 36. But also (-6)2 = 36 because -6 × -6 = + 36 (a minus × a minus = a plus). Therefore there are two square roots of 36: +6 and -6. We call 6 the positive square root of 36 and -6 is called the negative square root of 36. So if x2 = 36, then x = +6 or -6 (since squaring either of these numbers will give 36). However, the square root sign means \"positive square root\". So √36 = + 6 (only).\n\n### Probability\n\nProbability is the likelihood or chance of an event occurring. Probability = the number of ways of achieving success/the total number of possible outcomes For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P(heads) = ½ . The probability of something which is certain to happen is 1. The probability of something which is impossible to happen is 0. The probability of something not happening is 1 minus the probability that it will happen.\n\n## Photos", null, "", null, "" ]

[ null, "http://images.protopage.com/view/1723479/b5l612rlxmj3vhz1pabu085g2.jpg", null, "http://images.protopage.com/view/1723479/afulkb5n0hwn9fbpe3g91airy.jpg", null ]

[ "# Q. I am trying to understand the meaning of the X, Y and Z forces in the time history plots for the gauges that I have applied to a model in AUTODYN 11.0. What is the exact meaning of these forces and are they the reaction forces? A. The forces X, Y and Z listed in the time history plots are not reaction forces, they are the nodal forces required after update of the stress tensor as part of the Lagrangian computational cycle which result in nodal accelerations and velocities. They are not related to the reaction forces. If you are interested in the reaction forces you need to instead look at the stresses in the model which you can then convert to forces. The stress that is recorded is STR.AVG.1 so you will need to select this from the time history plot to provide the correct information at the gauge locations. STR.AVG.1 gives the average stress at these locations. You will need to select this result from the output tab ' history ' select gauge variables first and then re-run the model.\n\n Q.I am trying to understand the meaning of the X, Y and Z forces in the time history plots for the gauges that I have applied to a model in AUTODYN 11.0. What is the exact meaning of these forces and are they the reaction forces?A.The forces X, Y and Z listed in the time history plots are not reaction forces, they are the nodal forces required after update of the stress tensor as part of the Lagrangian computational cycle which result in nodal accelerations and velocities. They are not related to the reaction forces.If you are interested in the reaction forces you need to instead look at the stresses in the model which you can then convert to forces. The stress that is recorded is STR.AVG.1 so you will need to select this from the time history plot to provide the correct information at the gauge locations. STR.AVG.1 gives the average stress at these locations.You will need to select this result from the output tab ` history ` select gauge variables first and then re-run the model." ]

[ null ]

[ "# Is there a way to address overdispersion in a gls model?\n\nI have autocorrelated data that show a positive linear increase. When I model them using gls, I think the summary shows overdispersion.\n\nWhen using GLMM etc I'd change error structure, but I don't think I can change family in gls. Is there a way to address overdispersion in a gls model?\n\nlibrary(nlme)\n\nnest <- c(4087,2761,3807,4158,2046,4757,2984,3316,3143,\n3042,4429,3335,5124,2464,3713,3028,5739,4671,3799,6167,2937,5031)\n\ny <- seq(1997,2018,1)\n\nm1 <- glm(nest~y)\nsummary(m1)\nacf(resid(m1),type=\"p\")\n\n#residuals show autocorrelation at one year, so I need to use gls\n\nm1.gls <- gls(nest~y,correlation=corARMA(p=1), method=\"ML\")\nsummary(m1.gls)\nacf(resid(m1.gls))\nacf(resid(m1.gls),type=\"p\")\n\nm0.gls <- gls(nest~y,correlation=NULL,method=\"ML\")\nAIC(m0.gls,m1.gls)\nanova(m1.gls,m0.gls)\n\n#L.Ratio = chisqu = 7.382355;p = 0.0066\n#autocorrelation is significant\n\nnull.gls <- gls(nest~1,correlation=corARMA(p=1), method=\"ML\")\nanova(m1.gls,null.gls)\n\n#L.Ratio = chisq = 7.956113; p = 0.0048\n#trend is significant\n\nsummary(m1.gls)\n\n#residual se = 970 over 22 df\n\n\nDoes this show overdispersion?" ]

[ null ]

[ "# 0.1 Solving linear equations and inequalities: solving equations  (Page 2/2)\n\n Page 2 / 2\n\n$x+1=10$\n\nconditional, $x=9$\n\n$y-4=7$\n\nconditional, $y=11$\n\n$5a=25$\n\nconditional, $a=5$\n\n$\\frac{x}{4}=9$\n\nconditional, $x=36$\n\n$\\frac{18}{b}=6$\n\nconditional, $b=3$\n\n$y-2=y-2$\n\nidentity\n\n$x+4=x-3$\n\n$x+x+x=3x$\n\nidentity\n\n$8x=0$\n\nconditional, $x=0$\n\n$m-7=-5$\n\nconditional, $m=2$\n\n## Literal equations\n\nSome equations involve more than one variable. Such equations are called literal equations .\n\nAn equation is solved for a particular variable if that variable alone equals an expression that does not contain that particular variable.\n\n## The following equations are examples of literal equations.\n\n1. $y=2x+7$ . It is solved for $y$ .\n2. $d=rt$ . It is solved for $d$ .\n3. $I=prt$ . It is solved for $I$ .\n4. $z=\\frac{x-u}{s}$ . It is solved for $z$ .\n5. $y+1=x+4$ . This equation is not solved for any particular variable since no variable is isolated.\n\n## Solving equation of the form $x+a=b\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}x-a=b$\n\nRecall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.\n\n$\\begin{array}{ccc}\\text{This}& \\text{is}\\text{\\hspace{0.17em}}\\text{​}\\text{the}& \\text{this}\\\\ \\text{number}& \\text{same}\\text{\\hspace{0.17em}}\\text{as}& \\text{number}\\\\ ↓& ↓& ↓\\\\ x& =& 6\\\\ x+2& =& 8\\\\ x-1& =& 5\\end{array}$\n\n## This suggests the following procedures:\n\n1. We can obtain an equivalent equation (an equation having the same solutions as the original equation) by adding the same number to both sides of the equation.\n2. We can obtain an equivalent equation by subtracting the same number from both sides of the equation.\n\nWe can use these results to isolate $x$ , thus solving for $x$ .\n\n## Solving $x+a=b$ For $x$\n\n$\\begin{array}{llll}\\hfill x+a& =\\hfill & b\\hfill & \\text{The}\\text{\\hspace{0.17em}}a\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}x\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{addition}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\hfill \\\\ \\hfill x+a-a& =\\hfill & b-a\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{subtracting}\\text{\\hspace{0.17em}}a\\text{\\hspace{0.17em}}\\text{from}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill x+0& =\\hfill & b-a\\hfill & a-a=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}\\text{.}\\text{\\hspace{0.17em}}x+0=x.\\hfill \\\\ \\hfill x& =\\hfill & b-a\\hfill & \\text{This}\\text{\\hspace{0.17em}}\\text{equation}\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{equivalent}\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{first}\\text{\\hspace{0.17em}}\\text{equation,}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}\\text{it}\\text{\\hspace{0.17em}}\\text{is}\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\text{solved}\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}x.\\hfill \\end{array}$\n\n## Solving $x-a=b$ For $x$\n\n$\\begin{array}{llll}\\hfill x-a& =\\hfill & b\\hfill & \\text{The}\\text{\\hspace{0.17em}}a\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}x\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{subtraction}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\hfill \\\\ \\hfill x-a+a& =\\hfill & b+a\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{adding}\\text{\\hspace{0.17em}}a\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill x+0& =\\hfill & b+a\\hfill & -a+a=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}\\text{.}\\text{\\hspace{0.17em}}x+0=x.\\hfill \\\\ \\hfill x& =\\hfill & b+a\\hfill & \\text{This}\\text{\\hspace{0.17em}}\\text{equation}\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{equivalent}\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{first}\\text{\\hspace{0.17em}}\\text{equation,}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}\\text{it}\\text{\\hspace{0.17em}}\\text{is}\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\text{solved}\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}x.\\hfill \\end{array}$\n\n## Method for solving $x+a=b$ And $x-a=b$ For $x$\n\nTo solve the equation $x+a=b$ for $x$ , subtract $a$ from both sides of the equation.\nTo solve the equation $x-a=b$ for $x$ , add $a$ to both sides of the equation.\n\n## Sample set b\n\nSolve $x+7=10$ for $x$ .\n\n$\\begin{array}{llll}\\hfill x+7& =\\hfill & 10\\hfill & 7\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}x\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{addition}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\text{\\hspace{0.17em}}\\hfill \\\\ \\hfill x+7-7& =\\hfill & 10-7\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{subtracting}\\text{\\hspace{0.17em}}7\\text{\\hspace{0.17em}}\\text{from}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill x+0& =\\hfill & 3\\hfill & 7-7=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}.\\text{\\hspace{0.17em}}x+0=x.\\hfill \\\\ \\hfill x& =\\hfill & 3\\hfill & x\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{isolated,}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{equation}\\text{\\hspace{0.17em}}x=3\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{equivalent}\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}\\text{the}\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\text{original}\\text{\\hspace{0.17em}}\\text{equation}\\text{\\hspace{0.17em}}x+7=10.\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{Therefore,}\\text{\\hspace{0.17em}}\\text{these}\\text{\\hspace{0.17em}}\\text{two}\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\text{equation}\\text{\\hspace{0.17em}}\\text{have}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{same}\\text{\\hspace{0.17em}}\\text{solution}\\text{.}\\text{\\hspace{0.17em}}\\text{The}\\text{\\hspace{0.17em}}\\text{solution}\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}x=3\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\text{is}\\text{\\hspace{0.17em}}\\text{clearly}\\text{\\hspace{0.17em}}3.\\text{\\hspace{0.17em}}\\text{Thus,}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{solution}\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}x+7=10\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{also}\\text{\\hspace{0.17em}}3.\\hfill \\end{array}$\n\nCheck : Substitute 3 for $x$ in the original equation. $\\begin{array}{llll}\\hfill x+7& \\text{=}\\hfill & 10\\hfill & \\hfill \\\\ \\hfill 3+7& \\text{=}\\hfill & 10\\hfill & \\text{Is}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{correct?}\\hfill \\\\ \\hfill 10& \\text{=}\\hfill & 10\\hfill & \\text{Yes,}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{correct}\\text{.}\\hfill \\end{array}$\n\nSolve $m-2=-9$ for $m$ .\n\n$\\begin{array}{llll}\\hfill m-2& =\\hfill & -9\\hfill & 2\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}m\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{subtraction}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\text{\\hspace{0.17em}}\\hfill \\\\ \\hfill m-2+2& =\\hfill & -9+2\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{adding}\\text{\\hspace{0.17em}}2\\text{\\hspace{0.17em}}\\text{from}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill m+0& =\\hfill & -7\\hfill & -2+2=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}.\\text{\\hspace{0.17em}}m+0=m.\\hfill \\\\ \\hfill m& =\\hfill & -7\\hfill & \\hfill \\end{array}$\n\nCheck : Substitute $-7$ for $m$ in the original equation. $\\begin{array}{llll}\\hfill m-2& \\text{=}\\hfill & -9\\hfill & \\hfill \\\\ \\hfill -7-2& \\text{=}\\hfill & -9\\hfill & \\text{Is}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{correct?}\\hfill \\\\ \\hfill -9& \\text{=}\\hfill & -9\\hfill & \\text{Yes,}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{correct}\\text{.}\\hfill \\end{array}$", null, "Solve $y-2.181=-16.915$ for $y$ .\n\n$\\begin{array}{lll}\\hfill y-2.181& =\\hfill & -16.915\\hfill \\\\ \\hfill y-2.181+2.181& =\\hfill & -16.915+2.181\\hfill \\\\ \\hfill y& =\\hfill & -14.734\\hfill \\end{array}$\n\nOn the Calculator\n$\\begin{array}{ll}\\hfill & \\hfill \\\\ \\text{Type}\\hfill & 16.915\\hfill \\\\ \\text{Press}\\hfill & \\begin{array}{||}\\hline +/-\\\\ \\hline\\end{array}\\hfill \\\\ \\text{Press}\\hfill & \\begin{array}{||}\\hline +\\\\ \\hline\\end{array}\\hfill \\\\ \\text{Type}\\hfill & 2.181\\hfill \\\\ \\text{Press}\\hfill & \\begin{array}{||}\\hline =\\\\ \\hline\\end{array}\\hfill \\\\ \\text{Display}\\text{\\hspace{0.17em}}\\text{reads:}\\hfill & -14.734\\hfill \\end{array}$\n\nSolve $y+m=s$ for $y$ .\n\n$\\begin{array}{llll}\\hfill y+m& =\\hfill & s\\hfill & m\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}y\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{addition}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\text{\\hspace{0.17em}}\\hfill \\\\ \\hfill y+m-m& =\\hfill & s-m\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{subtracting}\\text{\\hspace{0.17em}}m\\text{\\hspace{0.17em}}\\text{from}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill y+0& =\\hfill & s-m\\hfill & m-m=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}.\\text{\\hspace{0.17em}}y+0=y.\\hfill \\\\ \\hfill y& =\\hfill & s-m\\hfill & \\hfill \\end{array}$\n\nCheck : Substitute $s-m$ for $y$ in the original equation. $\\begin{array}{lllll}\\hfill y+m& \\text{=}\\hfill & s\\hfill & \\hfill & \\hfill \\\\ \\hfill s-m+m& \\text{=}\\hfill & s\\hfill & \\hfill & \\text{Is}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{correct?}\\hfill \\\\ \\hfill s& \\text{=}\\hfill & s\\hfill & \\text{True}\\hfill & \\text{Yes,}\\text{\\hspace{0.17em}}\\text{this}\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{correct}\\text{.}\\hfill \\end{array}$\n\nSolve $k-3h=-8h+5$ for $k$ .\n\n$\\begin{array}{llll}\\hfill k-3h& =\\hfill & -8h+5\\hfill & 3h\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{associated}\\text{\\hspace{0.17em}}\\text{with}\\text{\\hspace{0.17em}}k\\text{\\hspace{0.17em}}\\text{by}\\text{\\hspace{0.17em}}\\text{subtraction}\\text{.}\\text{\\hspace{0.17em}}\\text{Undo}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{association}\\text{\\hspace{0.17em}}\\hfill \\\\ \\hfill k-3h+3h& =\\hfill & -8h+5+3h\\hfill & \\text{by}\\text{\\hspace{0.17em}}\\text{adding}\\text{\\hspace{0.17em}}3h\\text{\\hspace{0.17em}}\\text{to}\\text{\\hspace{0.17em}}both\\text{\\hspace{0.17em}}\\text{sides}\\text{.}\\hfill \\\\ \\hfill k+0& =\\hfill & -5h+5\\hfill & -3h+3h=0\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}0\\text{\\hspace{0.17em}}\\text{is}\\text{\\hspace{0.17em}}\\text{the}\\text{\\hspace{0.17em}}\\text{additive}\\text{\\hspace{0.17em}}\\text{identity}.\\text{\\hspace{0.17em}}k+0=k.\\hfill \\\\ \\hfill k& =\\hfill & -5h+5\\hfill & \\hfill \\end{array}$\n\n## Practice set b\n\nSolve $y-3=8\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}y.$\n\n$y=11$\n\nSolve $x+9=-4\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}x.$\n\n$x=-13$\n\nSolve $m+6=0\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}m.$\n\n$m=-6$\n\nSolve $g-7.2=1.3\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}g.$\n\n$g=8.5$\n\nsolve $f+2d=5d\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}f.$\n\n$f=3d$\n\nSolve $x+8y=2y-1\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}x.$\n\n$x=-6y-1$\n\nSolve $y+4x-1=5x+8\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}y.$\n\n$y=x+9$\n\n## Exercises\n\nFor the following problems, classify each of the equations as an identity, contradiction, or conditional equation.\n\n$m+6=15$\n\nconditional\n\n$y-8=-12$\n\n$x+1=x+1$\n\nidentity\n\n$k-2=k-3$\n\n$g+g+g+g=4g$\n\nidentity\n\n$x+1=0$\n\nFor the following problems, determine which of the literal equations have been solved for a variable. Write \"solved\" or \"not solved.\"\n\n$y=3x+7$\n\nsolved\n\n$m=2k+n-1$\n\n$4a=y-6$\n\nnot solved\n\n$hk=2k+h$\n\n$2a=a+1$\n\nnot solved\n\n$5m=2m-7$\n\n$m=m$\n\nnot solved\n\nFor the following problems, solve each of the conditional equations.\n\n$h-8=14$\n\n$k+10=1$\n\n$k=-9$\n\n$m-2=5$\n\n$y+6=-11$\n\n$y=-17$\n\n$y-8=-1$\n\n$x+14=0$\n\n$x=-14$\n\n$m-12=0$\n\n$g+164=-123$\n\n$g=-287$\n\n$h-265=-547$\n\n$x+17=-426$\n\n$x=-443$\n\n$h-4.82=-3.56$\n\n$y+17.003=-1.056$\n\n$y=-18.059$\n\n$k+1.0135=-6.0032$\n\nSolve $n+m=4\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}n.$\n\n$n=4-m$\n\nSolve $P+3Q-8=0\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}P.$\n\nSolve $a+b-3c=d-2f\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}b.$\n\n$b=-a+3c+d-2f$\n\nSolve $x-3y+5z+1=2y-7z+8\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}x.$\n\nSolve $4a-2b+c+11=6a-5b\\text{\\hspace{0.17em}}\\text{for}\\text{\\hspace{0.17em}}c.$\n\n$c=2a-3b-11$\n\n## Exercises for review\n\n( [link] ) Simplify ${\\left(4{x}^{5}{y}^{2}\\right)}^{3}$ .\n\n( [link] ) Write $\\frac{20{x}^{3}{y}^{7}}{5{x}^{5}{y}^{3}}$ so that only positive exponents appear.\n\n$\\frac{4{y}^{4}}{{x}^{2}}$\n\n( [link] ) Write the number of terms that appear in the expression $5{x}^{2}+2x-6+\\left(a+b\\right)$ , and then list them.\n\n( [link] ) Find the product. ${\\left(3x-1\\right)}^{2}$ .\n\n$9{x}^{2}-6x+1$\n\n( [link] ) Specify the domain of the equation $y=\\frac{5}{x-2}$ .\n\nwhat is variations in raman spectra for nanomaterials\nI only see partial conversation and what's the question here!\nwhat about nanotechnology for water purification\nplease someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.\nDamian\nyes that's correct\nProfessor\nI think\nProfessor\nwhat is the stm\nis there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?\nRafiq\nindustrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong\nDamian\nHow we are making nano material?\nwhat is a peer\nWhat is meant by 'nano scale'?\nWhat is STMs full form?\nLITNING\nscanning tunneling microscope\nSahil\nhow nano science is used for hydrophobicity\nSantosh\nDo u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq\nRafiq\nwhat is differents between GO and RGO?\nMahi\nwhat is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq\nRafiq\nwhat is Nano technology ?\nwrite examples of Nano molecule?\nBob\nThe nanotechnology is as new science, to scale nanometric\nbrayan\nnanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale\nDamian\nIs there any normative that regulates the use of silver nanoparticles?\nwhat king of growth are you checking .?\nRenato\nWhat fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?\nwhy we need to study biomolecules, molecular biology in nanotechnology?\n?\nKyle\nyes I'm doing my masters in nanotechnology, we are being studying all these domains as well..\nwhy?\nwhat school?\nKyle\nbiomolecules are e building blocks of every organics and inorganic materials.\nJoe\nanyone know any internet site where one can find nanotechnology papers?\nresearch.net\nkanaga\nsciencedirect big data base\nErnesto\nIntroduction about quantum dots in nanotechnology\nwhat does nano mean?\nnano basically means 10^(-9). nanometer is a unit to measure length.\nBharti\ndo you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?\nabsolutely yes\nDaniel\nhow did you get the value of 2000N.What calculations are needed to arrive at it\nPrivacy Information Security Software Version 1.1a\nGood\nGot questions? Join the online conversation and get instant answers!", null, "", null, "By Lakeima Roberts", null, "", null, "", null, "", null, "", null, "", null, "By", null, "By", null, "" ]

[ null, "https://www.jobilize.com/ocw/mirror/col11828_1.1_complete/m18876/C04_S4-1_P137_001.png", null, "https://www.jobilize.com/quiz/thumb/epidemiology-of-obstructive-sleep-apnea.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://farm9.staticflickr.com/8742/16876757802_f13312e921_t.jpg", null, "https://www.jobilize.com/quiz/thumb/final-fin-421-financial-421-exam-by-keyaira-brax.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://farm1.staticflickr.com/476/20102025465_c4faf7598c_t.jpg", null, "https://www.jobilize.com/quiz/thumb/lean-start-up-quiz-startup-by-yasser-ibrahim.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://www.jobilize.com/quiz/thumb/macroeconomics-mcq-quiz-by-candice-butts.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://www.jobilize.com/quiz/thumb/business-law-mcq-exam-1-bus201-by-maureen-miller-open-course.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://www.jobilize.com/quiz/thumb/biology-ch-01-the-study-of-life-mcq-quiz-openstax-college.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://www.jobilize.com/quiz/thumb/human-body-anatomy-physiology-mcq-quiz.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null, "https://www.jobilize.com/quiz/thumb/chemistry-final-flashcards-by-briana-hamil.png;jsessionid=XwQknW1HktbPJp2NCjTpE3VaU6LS5SL8ZQuBTNgp.condor3190", null ]

[ "# Module bevy::math::f64\n\nExpand description\n\n`f64` vector, quaternion and matrix types.\n\n## Structs\n\n• A 2D affine transform, which can represent translation, rotation, scaling and shear.\n• A 3D affine transform, which can represent translation, rotation, scaling and shear.\n• A 2x2 column major matrix.\n• A 3x3 column major matrix.\n• A 4x4 column major matrix.\n• A quaternion representing an orientation.\n• A 2-dimensional vector.\n• A 3-dimensional vector.\n• A 4-dimensional vector.\n\n## Functions\n\n• Creates a 2x2 matrix from two column vectors.\n• Creates a 3x3 matrix from three column vectors.\n• Creates a 4x4 matrix from four column vectors.\n• Creates a quaternion from `x`, `y`, `z` and `w` values.\n• Creates a 2-dimensional vector.\n• Creates a 3-dimensional vector.\n• Creates a 4-dimensional vector." ]

[ null ]

[ "Russian", null, "", null, "©  M. E. Abramyan (Southern Federal University, Shenzhen MSU-BIT University), 1998–2023\n\n Main Tasks Examples PT for MPI-2\n\n Overview Input-output operations Task groups Begin Integer Boolean If Case For While Series Proc Func Minmax Array Matrix String File Text Param Recur Dynamic Dynamic (obj) Tree Tree (obj)\n\n# Conditional statement\n\nTasks If1–If3 have been slightly modified in version 4.15.\n\nIf1. An integer is given. If the integer is positive then decrease it by 8, otherwise do not change it. Output the obtained integer.\n\nIf2. An integer is given. If the integer is positive then decrease it by 8, otherwise increase it by 6. Output the obtained integer.\n\nIf3. An integer is given. If the integer is positive then decrease it by 8, if the integer is negative then increase it by 6, if the integer equals 0 then change it to 10. Output the obtained integer.\n\nIf4°. Three integers are given. Find the amount of positive integers in the input data.\n\nIf5. Three integers are given. Find the amount of positive and amount of negative integers in the input data.\n\nIf6°. Given two real numbers, output the larger value of them.\n\nIf7. Given two real numbers, output the order number of the smaller of them.\n\nIf8°. Given two real numbers, output the larger value and then the smaller value of them.\n\nIf9. The values of two real variables A and B are given. Redistribute the values so that A and B have the smaller and the larger value respectively. Output the new values of the variables A and B.\n\nIf10. The values of two integer variables A and B are given. If the values are not equal then assign the sum of given values to each variable, otherwise assign zero value to each variable. Output the new values of the variables A and B.\n\nIf11. The values of two integer variables A and B are given. If the values are not equal then assign the larger value to each variable, otherwise assign zero value to each variable. Output the new values of the variables A and B.\n\nIf12°. Given three real numbers, output the minimal value of them.\n\nIf13. Given three real numbers, output the value between the minimum and the maximum.\n\nIf14. Given three real numbers, output the minimal value and then the maximal value.\n\nIf15. Given three real numbers, output the sum of two largest values.\n\nIf16. The values of three real variables A, BC are given. If the values are in ascending order then double them, otherwise replace the value of each variable by its opposite value. Output the new values of the variables A, BC.\n\nIf17. The values of three real variables A, BC are given. If the values are in ascending or descending order then double them, otherwise replace the value of each variable by its opposite value. Output the new values of the variables A, BC.\n\nIf18. Three integers are given. One of them differs from two other equal integers. Output the order number of the integer that differs from the others.\n\nIf19. Four integers are given. One of them differs from three other equal integers. Output the order number of the integer that differs from the others.\n\nIf20. Three points A, BC on the real axis are given. Determine whether B or C is closer to A. Output the nearest point and its distance from A.\n\nIf21. Integer coordinates of a point in the coordinate plane are given. If the point coincides with the origin of coordinates then output 0, otherwise if the point lies on the x-axis or y-axis then output 1 or 2 respectively. If the point does not lie on the coordinate axes then output 3.\n\nIf22°. Given coordinates of a point that does not lie on the coordinate axes, find the number of a coordinate quarter containing the point.\n\nIf23. Given integer coordinates of three vertices of a rectangle whose sides are parallel to coordinate axes, find the coordinates of the fourth vertex of the rectangle.\n\nIf24. Given a real independent variable x, find the value of a real function f defined as:\n\n f(x) = 2·sin(x), if x > 0, 6 − x, if x ≤ 0.\n\nIf25. Given an integer independent variable x, find the value of an integer function f defined as:\n\n f(x) = 2·x, if x < −2 or x > 2, −3·x otherwise.\n\nIf26°. Given a real independent variable x, find the value of a real function f defined as:\n\n −x, if x ≤ 0, f(x) = x2, if 0 < x < 2, 4, if x ≥ 2.\n\nIf27. Given a real independent variable x, find the value of an integer function f defined as:\n\n 0, if x < 0, f(x) = 1, if x belongs to [0, 1), [2, 3), …, −1, if x belongs to [1, 2), [3, 4), … .\n\nIf28. Given a number of year (as a positive integer), find the amount of days in the year. Note that the length of year is 365 days for an ordinary year and 366 days for a leap year. A leap year is every year whose number is divisible by 4, as 1980, except centenary years that are not divisible by 400 (for example, 1300 and 1900 are ordinary years, 1200 and 2000 are leap years).\n\nIf29. Given an integer, output its description string as: \"negative even number\", \"zero number\", \"positive odd number\", etc.\n\nIf30. An integer in the range 1 to 999 is given. Output its description string as: \"two-digit even number\", \"three-digit odd number\", etc.", null, "Designed byM. E. Abramyan and V. N. Braguilevsky Last revised:01.01.2023" ]

[ null, "http://www.ptaskbook.com/en/images/sfedulogo.png", null, "http://www.ptaskbook.com/en/images/smbulogo.png", null, "http://df.cc.bb.a1.top.mail.ru/counter", null ]

[ "# Point Slope Form Calculator Recent General Form Math Compliant Templates Applied S 2 Good Consequently Point Slope Form Of Parallel Line Calculator", null, "point slope form calculator recent general form math compliant templates applied s 2 good consequently point slope form of parallel line calculator.\n\npoint slope form calculator parallel fractions wolfram finding the equation of a line in,point slope form calculator symbolab equation with two points from parallel and perpendicular lines,point slope form parallel and perpendicular calculator changing to standard with equation,point slope form calculator from two points find with x and y intercepts omni convert the intercept to standard,point slope form calculator webmath using two points of linear equations concept algebra video by parallel and perpendicular,what you should wear to y intercept form calculator point slope wolfram omni given two points,substitution slope intercept form for both equations point calculator parallel with equation fractions,point slope form calculator omni with equation writing equations and a unit 4 lesson download parallel,point slope form calculator from two points wolfram webmath linear functions posters and reference sheets,point slope form calculator fractions with equation wolfram how to graph linear equations 5 steps pictures." ]

[ null, "http://courselist.co/wp-content/uploads/2018/09/point-slope-form-calculator-recent-general-form-math-compliant-templates-applied-s-2-good-consequently-point-slope-form-of-parallel-line-calculator.jpg", null ]

[ "# Tag Info\n\n1\n\nWhile the structural operation DeleteCases[eq1 /. Equal -> Subtract, _?(FreeQ[#,T0]&)] works, I prefer using an algebraic approach on a algebraic problem. vars = Select[Not@*FreeQ[T0]]@Variables[eq1 /. Equal -> Subtract]; coeffs = CoefficientArrays[eq1 /. Equal -> Subtract, vars]; Fold[#2 + #1.vars &, Reverse@ReplacePart[coeffs, 1 -> 0]...\n\n1\n\nBlock[{T1, Equal = Plus}, SetAttributes[T1, Constant]; eq1] -Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t] Or (thanks to Mr. Wizard here) eq1 /. {s_Symbol /; StringMatchQ[SymbolName[Unevaluated@s], \"T\" ~~ Except[\"0\"]] -> 0, Equal -> Plus} -Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]\n\n1\n\nNot the most elegant solution, but you can use the Collectcommand in the following manner eq1 = Derivative[0, 1][T1][x, t] - Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t] - Derivative[2, 0][T1][x, t]; (Coefficient[#1, {T0[x, t], Derivative[1, 0][T0][x, t], Derivative[1, 0][T0][x, t]^2}] & )[...\n\n4\n\nWeirdly, the less specific assumption $G1>0$ works to achieve your simplification here: Simplify[xxx, Assumptions -> G1 > 0]\n\n0\n\nYou can always brute force it. eq = x y + y^2 - 1 + (z + Cos[z]) y x - 24*(z + Cos[z]) + Tan[x] == x^4 (Solve[eq /. {z + Cos[z] -> f[z]}, f[z]][[1, 1]] /. Rule -> Equal) /. f[z] -> z + Cos[z] (*z + Cos[z] == (x^4 - Tan[x] - x y - y^2 + 1)/(x y - 24)*) This, of course, requires that all expressions containing z be the same form.\n\n2\n\nThis function brings everything to one side (f), takes a derivative with respect of variable of interest var and integrates back to get an expression dependent on this variable. This is then the left hand side lhs, the right hand side rhs is formed by the rest of the expression: Clear[e,iso] iso[eq_,var_]:=Module[{f,lhs,rhs,g}, f=eq/.Equal[a_,b_]->a-b; ...\n\n1\n\nWith Reduce you can actually solve this problem; you need, though, to specify the variable you want to use to solve it for: Reduce[xy + y^2 - 1 + zyx - 24*z + Tan[x] == x^4, z] which yields as a solution z == 1/24 (-1 - x^4 + xy + y^2 + zyx + Tan[x]) This will obviously work only if your variable can be expressed explicitly.\n\n1\n\nFirst you can try Subscript (Control key + _) to make tx and ty look more like a coefficient, then use Subscript[tx,0] -> Function[{x,y},x^2 (1-x^2) C1]. The reason for making the function of x and y even when it is a function only of x is to convince the partial derivative function that its y derivative is 0. Another way to do similar things is to use ...\n\n0\n\nanother way is to use Simplify or Fullsimplify Simplify @@ ConditionalExpression[-Sqrt[1 - x - x^2], 1/2 (-1 - Sqrt) < x < 1/2 (-1 + Sqrt)] returns -Sqrt[1 - x - x^2]\n\n1\n\nWhy not apply ComplexExpand to get an expression totaly free of imginary unit I ? ceRe = ComplexExpand[Re[expr], TargetFunctions -> {Re, Im}] // FullSimplify[#, 0 <= e <= 1/10] & (* (1/(6 e (-1 + 2 e)))(2 + e (-13 + 12 e) + (-2 + e) Cos[1/3 ArcTan[8 + e (-12 + e (-210 + e (647 + 216 (-3 + e) e))), 12 Sqrt (1 - 2 e) (1 - e)^(3/2) e ...\n\n2\n\n(Update: Re was unhelpful here; removing it.) I am not sure how helpful this is, but if we start with a qualified FullSimplify: expr = (* your expression here *) simp = FullSimplify[expr, 0 <= e <= 1/10] (1/(24 e (-1 + 2 e)))(8 - 52 e + 48 e^2 + ( 2 I (I + Sqrt) (-2 + e)^2)/(8 - 12 e - 210 e^2 + 647 e^3 - 648 e^4 + 216 e^5 + 12 Sqrt (1 ...\n\n0\n\nNot an answer because I am not sure exactly what the OP wants as the output. However : expr = -((-2 + 13 e - 12 e^2)/(6 (-e + 2 e^2))) + ((1 - I Sqrt) (-4 + 4 e - e^2))/(6 2^(2/3) (-e + 2 e^2) (16 - 24 e - 420 e^2 + 1294 e^3 - 1296 e^4 + 432 e^5 + Sqrt[4 (-4 + 4 e - e^2)^3 + (16 - 24 e - 420 e^2 + 1294 e^3 - ...\n\n3\n\nUsing an Iterator: With[{fn = GeneralUtilities`ListIterator[{f, g, h}]}, Unevaluated@Read[fn] /@ y[u, v, w]] (* y[f[u], g[v], h[w]] *)\n\n6\n\nI'm not sure if this counts as a separate case per se, so here it goes: threadOver[f_List, h_[x__]] := Inner[Construct, f, {x}, h] As an example, evaluate threadOver[myFuncs, expr] to obtain: y[f[u], g[v], h[w]] A more complicated instance of my threadOver function making use of ListCorrelate could be the following: threadOver[f_List, h_[x__]] := ...\n\n2\n\nIf you are allowed to modify expr in place, this would do the trick: expr[[1;;-1]] = MapThread[Construct, {myFuncs, Level[expr,1]}] expr y[f[u],g[v],h[w]] Otherwise, you could use the following function: innerMap[fs_, expr_] := Head[expr] @@ MapThread[Construct, {fs, Level[expr,1]}] innerMap[myFuncs, expr] y[f[u],g[v],h[w]]\n\n4\n\nIt's funny how convoluted all of these answers need to be. Here's another convoluted one. It's clunky because it does extra operations and then removes these extra pieces. innerMap[fs_, expr_] := Through@*fs /@ expr // MapIndexed[#1[[First@#2]] &] innerMap[{f, g, h}, y[u, v, w]] (* y[f[u], g[v], h[w]] *) (The function MapIndexed[#1[[First@#2]] &] ...\n\n9\n\nMapIndexed[myFuncs[[#2[]]][#] &, expr] or MapIndexed[Extract[myFuncs, #2][#] &, expr] or ReplacePart[expr, i_ :> myFuncs[[i]]@expr[[i]]] or expr // Query[Thread[Range@Length@myFuncs -> myFuncs]] or Head[expr] @@ Array[myFuncs[[#]]@expr[[#]] &, Length@expr] or Head[expr] @@ Construct @@@ Transpose @ {myFuncs, List @@ expr}\n\n12\n\nInner[# @ #2 &, #, {##} & @@ #2, #2[]] &[myFuncs, expr] y[f[u], g[v], h[w]] or Inner[Construct, #, List @@ #2, Head @#2] &[myFuncs, expr] y[f[u], g[v], h[w]] Also Construct @@@ Thread[{#2[] @@ #, #2}, #2[]] &[myFuncs, expr] y[f[u], g[v], h[w]] and Operate[Apply[#]@*(Compose @@@ Thread @{myFuncs, {##}} &) &, ...\n\n13\n\nUsing Mapthread innerMap[funcs_, expr_] := Head[expr] @@ MapThread[#1[#2] &, {funcs, List @@ expr}] innerMap[myFuncs, expr] y[f[u], g[v], h[w]]\n\n2\n\nHere is an approach that doesn't need to flatten the whole tree of data. Let's take the example of moving the level 1 keys at the level 3, on the data assoc of your self-answer : assoc = Fold[AssociationThread[#2 -> #1] &, \"X\", Reverse@Table[ToString[10 i + j], {i, 4}, {j, 2}]] Here is a function showAssocListTree that will be usefull to ...\n\n7\n\nDefinitions Here is an alternative implementation using the Wolfram Function Repository functions AssociationKeyFlatten and ToAssociations (submitted by WRI personnel) and the function meMerge (localMerge) from the answer by andre314: Clear[TransposeAssoc]; TransposeAssoc[assoc_Association, perm_?PermutationListQ] := Block[{assoc2, assoc3, LocalMerge}...\n\n8\n\nIf you factor a permuation perm into a product of cycles of the form $(j\\ k)$ with $k=j+1$, then the permuation can be effected by Query and Transpose. Functions: adjacentCycles[perm] (* factors perm into \"adjacent\" 2-cycles *) dsTranspose[x, perm] (* like Transpose[x, perm], but x is a Dataset or Association *) Code: (* ...\n\nTop 50 recent answers are included" ]

[ null ]

[ "# Algorithm Complexity for solving SDP\n\nHi,\n\nI have the following SDP optimization problem, and I needed to know the complexity of the algorithm that CVX uses to solve it:\n\n``````cvx_begin sdp\nvariable W(N, N) complex\nmaximize real(trace(Q * W))\nsubject to\ndiag(W) == ones(N, 1);\nW == semidefinite(N);\ncvx_end\n``````\n\nDoes anyone know how to figure out the complexity of the solver for this problem?\nI saw somewhere that CVX uses SeDuMi for solving SDP with complexity of O(N^4.5 log(1/epsilon)). Does this apply to every form of SDP, or the complexity changes with the form of SDP?\n\nThanks.\n\nI’m afraid I can’t help you here. You should consult the academic literature on the complexity of symmetric primal/dual methods." ]

[ null ]

[ "", null, "Most Affordable JEE | NEET | 8,9,10 Preparation by Kota's Top IITian Doctor Faculties", null, "", null, "#### 1. Algebra\n\n• Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations.\n• Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations with given roots, symmetric functions of roots.\n• Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers.\n• Logarithms and their properties.\n• Permutations and combinations, binomial theorem for a positive integral index, properties of binomial coefficients.\n\n#### 2. Matrices\n\n• Matrices as a rectangular array of real numbers.\n• Equality of matrices, addition, multiplication by a scalar and product of matrices.\n• Transpose of a matrix.\n• Determinant of a square matrix of order up to three.\n• Inverse of a square matrix of order up to three.\n• Properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their properties.\n• Solutions of simultaneous linear equations in two or three variables.\n\n#### 3. Probability\n\n• Addition and multiplication rules of probability, conditional probability, Bayes Theorem, independence of events, computation of probability of events using permutations and combinations.\n\n#### 4. Trigonometry\n\n• Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving multiple and sub-multiple angles, general solution of trigonometric equations.\n• Relations between sides and angles of a triangle, sine rule, cosine rule, half-angle formula and the area of a triangle, inverse trigonometric functions (principal value only).\n\n#### 5. Analytical geometry\n\n• Two dimensions: Cartesian coordinates, distance between two points, section formulae, shift of origin.\n• Equation of a straight line in various forms, angle between two lines, distance of a point from a line; Lines through the point of intersection of two given lines, equation of the bisector of the angle between two lines, concurrency of lines; Centroid, orthocentre, incentre and circumcentre of a triangle.\n• Equation of a circle in various forms, equations of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line.\n• Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent and normal.\n• Locus problems.\n• Three dimensions: Direction cosines and direction ratios, equation of a straight line in space, equation of a plane, distance of a point from a plane.\n\n#### 6. Differential Calculus\n\n• Real valued functions of a real variable, into, onto and one-to-one functions, sum, difference, product and quotient of two functions, composite functions, absolute value, polynomial, rational, trigonometric, exponential and logarithmic functions.\n• Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two functions, L’Hospital rule of evaluation of limits of functions.\n• Even and odd functions, inverse of a function, continuity of composite functions, intermediate value property of continuous functions.\n• Derivative of a function, derivative of the sum, difference, product and quotient of two functions, chain rule, derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions.\n• Derivatives of implicit functions, derivatives up to order two, geometrical interpretation of the derivative, tangents and normals, increasing and decreasing functions, maximum and minimum values of a function, Rolle’s theorem and\n• Lagrange’s mean value theorem.\n\n#### 7. Integral calculus\n\n• Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals and their properties, fundamental theorem of integral calculus.\n• Integration by parts, integration by the methods of substitution and partial fractions, application of definite integrals to the determination of areas involving simple curves.\n• Formation of ordinary differential equations, solution of homogeneous differential equations, separation of variables method, linear first order differential equations.\n\n#### 8. Vectors\n\n• Addition of vectors, scalar multiplication, dot and cross products, scalar triple products and their geometrical interpretations.\nHope you liked reading the above post. Please share it with your friends to help them in getting the relevant information. Stay tuned with eSaral. Thanks!!", null, "", null, "" ]

[ null, "https://www.facebook.com/tr", null, "https://www.esaral.com/media/uploads/2018/04/Slide42.jpg", null, "https://www.esaral.com/static/assets/images/Padho-Bharat-Web.jpeg", null, "https://www.esaral.com/static/assets/images/close.svg", null, "https://www.esaral.com/jee-advanced-mathematics-syllabus/", null ]

[ "## Stepwise Regression\n\n### Stepwise Regression to Select Appropriate Models\n\n`stepwiselm` creates a linear model and automatically adds to or trims the model. To create a small model, start from a constant model. To create a large model, start with a model containing many terms. A large model usually has lower error as measured by the fit to the original data, but might not have any advantage in predicting new data.\n\n`stepwiselm` can use all the name-value options from `fitlm`, with additional options relating to the starting and bounding models. In particular:\n\n• For a small model, start with the default lower bounding model: `'constant'` (a model that has no predictor terms).\n\n• The default upper bounding model has linear terms and interaction terms (products of pairs of predictors). For an upper bounding model that also includes squared terms, set the `Upper` name-value pair to `'quadratic'`.\n\n### Compare large and small stepwise models\n\nThis example shows how to compare models that `stepwiselm` returns starting from a constant model and starting from a full interaction model.\n\nLoad the `carbig` data and create a table from some of the data.\n\n```load carbig tbl = table(Acceleration,Displacement,Horsepower,Weight,MPG);```\n\nCreate a mileage model stepwise starting from the constant model.\n\n`mdl1 = stepwiselm(tbl,'constant','ResponseVar','MPG')`\n```1. Adding Weight, FStat = 888.8507, pValue = 2.9728e-103 2. Adding Horsepower, FStat = 3.8217, pValue = 0.00049608 3. Adding Horsepower:Weight, FStat = 64.8709, pValue = 9.93362e-15 ```\n```mdl1 = Linear regression model: MPG ~ 1 + Horsepower*Weight Estimated Coefficients: Estimate SE tStat pValue __________ __________ _______ __________ (Intercept) 63.558 2.3429 27.127 1.2343e-91 Horsepower -0.25084 0.027279 -9.1952 2.3226e-18 Weight -0.010772 0.00077381 -13.921 5.1372e-36 Horsepower:Weight 5.3554e-05 6.6491e-06 8.0542 9.9336e-15 Number of observations: 392, Error degrees of freedom: 388 Root Mean Squared Error: 3.93 R-squared: 0.748, Adjusted R-Squared: 0.746 F-statistic vs. constant model: 385, p-value = 7.26e-116 ```\n\nCreate a mileage model stepwise starting from the full interaction model.\n\n`mdl2 = stepwiselm(tbl,'interactions','ResponseVar','MPG')`\n```1. Removing Acceleration:Displacement, FStat = 0.024186, pValue = 0.8765 2. Removing Displacement:Weight, FStat = 0.33103, pValue = 0.56539 3. Removing Acceleration:Horsepower, FStat = 1.7334, pValue = 0.18876 4. Removing Acceleration:Weight, FStat = 0.93269, pValue = 0.33477 5. Removing Horsepower:Weight, FStat = 0.64486, pValue = 0.42245 ```\n```mdl2 = Linear regression model: MPG ~ 1 + Acceleration + Weight + Displacement*Horsepower Estimated Coefficients: Estimate SE tStat pValue __________ __________ _______ __________ (Intercept) 61.285 2.8052 21.847 1.8593e-69 Acceleration -0.34401 0.11862 -2.9 0.0039445 Displacement -0.081198 0.010071 -8.0623 9.5014e-15 Horsepower -0.24313 0.026068 -9.3265 8.6556e-19 Weight -0.0014367 0.00084041 -1.7095 0.088166 Displacement:Horsepower 0.00054236 5.7987e-05 9.3531 7.0527e-19 Number of observations: 392, Error degrees of freedom: 386 Root Mean Squared Error: 3.84 R-squared: 0.761, Adjusted R-Squared: 0.758 F-statistic vs. constant model: 246, p-value = 1.32e-117 ```\n\nNotice that:\n\n• `mdl1` has four coefficients (the `Estimate` column), and `mdl2` has six coefficients.\n\n• The adjusted R-squared of `mdl1` is `0.746`, which is slightly less (worse) than that of `mdl2`, `0.758`.\n\nCreate a mileage model stepwise with a full quadratic model as the upper bound, starting from the full quadratic model:\n\n`mdl3 = stepwiselm(tbl,'quadratic','ResponseVar','MPG','Upper','quadratic');`\n```1. Removing Acceleration:Horsepower, FStat = 0.075209, pValue = 0.78405 2. Removing Acceleration:Weight, FStat = 0.072756, pValue = 0.78751 3. Removing Horsepower:Weight, FStat = 0.12569, pValue = 0.72314 4. Removing Weight^2, FStat = 1.194, pValue = 0.27521 5. Removing Displacement:Weight, FStat = 1.2839, pValue = 0.25789 6. Removing Displacement^2, FStat = 2.069, pValue = 0.15114 7. Removing Horsepower^2, FStat = 0.74063, pValue = 0.39 ```\n\nCompare the three model complexities by examining their formulas.\n\n`mdl1.Formula`\n```ans = MPG ~ 1 + Horsepower*Weight ```\n`mdl2.Formula`\n```ans = MPG ~ 1 + Acceleration + Weight + Displacement*Horsepower ```\n`mdl3.Formula`\n```ans = MPG ~ 1 + Weight + Acceleration*Displacement + Displacement*Horsepower + Acceleration^2 ```\n\nThe adjusted ${R}^{2}$ values improve slightly as the models become more complex:\n\n```RSquared = [mdl1.Rsquared.Adjusted, ... mdl2.Rsquared.Adjusted, mdl3.Rsquared.Adjusted]```\n```RSquared = 1×3 0.7465 0.7580 0.7599 ```\n\nCompare residual plots of the three models.\n\n```subplot(3,1,1) plotResiduals(mdl1) subplot(3,1,2) plotResiduals(mdl2) subplot(3,1,3) plotResiduals(mdl3)```", null, "The models have similar residuals. It is not clear which fits the data better.\n\nInterestingly, the more complex models have larger maximum deviations of the residuals:\n\n```Rrange1 = [min(mdl1.Residuals.Raw),max(mdl1.Residuals.Raw)]; Rrange2 = [min(mdl2.Residuals.Raw),max(mdl2.Residuals.Raw)]; Rrange3 = [min(mdl3.Residuals.Raw),max(mdl3.Residuals.Raw)]; Rranges = [Rrange1;Rrange2;Rrange3]```\n```Rranges = 3×2 -10.7725 14.7314 -11.4407 16.7562 -12.2723 16.7927 ```" ]

[ null, "https://in.mathworks.com/help/examples/stats/win64/ComparelargeandsmallstepwisemodelsExample_01.png", null ]

[ "", null, "请选择手机品牌 1 100+ 1 1米 2 21克 2 26摄氏度 3 360 5 5GHYUN 8 8848 A ailyfu爱来福 A anbo A angelcare A apcp A ATMAN A 阿巴町 A 阿尔法 A 阿尔卡特 A 阿凡提 A 埃立特 A 艾捷莫 A 艾酷 A 艾乐丰 A 艾玛 A 艾美卡 A 艾尼卡 A 艾炜特 A 艾优尼 A 爱宝隆 A 爱尔马 A 爱国者 A 爱果 A 爱户外 A 爱九九 A 爱久久 A 爱酷 A 爱乐星 A 爱立顺 A 爱立信 A 爱摩登 A 爱纳星 A 爱诺德 A 爱诺特 A 爱派尔 A 爱普汇 A 爱谱乐 A 爱时尚 A 爱斯尔 A 爱维特 A 爱我 A 爱讯达 A 爱易通 A 爱意通 A 爱优尚丰 A 安科讯 A 安利嘉 A 安利普 A 安派 A 安信通 A 安卓互联 A 奥丁 A 奥克斯 A 奥乐 A 奥乐迪奥 A 奥利信 A 奥霖斯 A 奥洛斯 A 奥魅尔 A 奥浦 A 奥维 A 奥卓 A 澳乐达 B 八〇九〇 B 巴菲 B 巴拿拿 B 白米 B 百达 B 百达玖玖 B 百迪宝 B 百合 B 百嘉好 B 百立丰 B 百纳威 B 百信 B 百裕 B 佰灵通 B 佰盛 B 佰事讯 B 佰意 B 佰云 B 邦华 B 邦讯达 B 帮盛 B 宝方 B 宝捷讯 B 宝码 B 宝泰尔 B 保千里 B 北斗星河 B 贝多芬 B 贝尔丰 B 贝尔星 B 贝格斯通 B 贝龙 B 奔跑家园 B 本为 B 比酷时代 B 比美 B 毕加索 B 别样红 B 波导 B 菠萝 B 铂度 B 铂爵 B 铂派 B 铂锐 B 铂云 B 博鑫奇 B 博阅 B 不羁 B 步步高 C cm C COCCI C CTiC C 财富之舟 C 长虹 C 长江天音 C 长普达 C 长营 C 畅想未来 C 超多维 C 宸通和 C 晨想 C 晨信 C 晨兴 C 诚虹 C 诚基 C 橙石 C 骋娱传媒 C 虫子 C 初上科技 C 传承 C 传奇 C 创客 C 创世能 C 创维 C 创星 C 创雅 C 创奕 C 锤子 C 酷潮 D DOMOD D 达闼 D 大冰棒 D 大诚 D 大大 D 大华 D 大疆 D 大桔 D 大巨龙 D 大米 D 大拇指 D 大神 D 大松 D 大唐 D 大唐电信 D 大唐数码 D 大唐移动 D 大为 D 大显 D 大显启辰 D 大亚 D 大众电脑 D 戴尔 D 德进高通 D 德赛 D 登科 D 迪奥 D 迪比特 D 迪迪 D 迪佳通 D 迪卡龙 D 迪美 D 迪妙移动 D 迪思 D 迪斯尼 D 迪泰元 D 迪为 D 迪信通 D 帝锋 D 帝狼 D 帝托 D 帝唯 D 典格 D 电意 D 蝶变 D 鼎瑄 D 定芯 D 东北丰 D 东方龙 D 东方拓宇 D 东茂 D 东森 D 东森伟泰 D 东铁通讯 D 东信 D 动感通 D 读书郎 D 独秀 D 独影 D 多达康 D 多拉多 D 多美达 D 多普达 D 多亲 D 朵美 D 朵唯 D 彤霖 D 彤鑫达 E EINPAD E E派 E E人E本 E 恩泽通信 E 易创 F FNNI F FOODO F PHICOMM斐讯 F 凡尔纳 F 凡美 F 泛泰 F 范思哲 F 梵尚 F 飞歌王 F 飞利浦 F 飞秒 F 飞思特 F 飞阳 F 飞盈 F 菲乐普 F 菲力克斯 F 菲浦 F 斐讯 F 丰尚 F 风云时代 F 峰汇智联 F 峰泽联和 F 烽火 F 锋彩 F 锋达通 F 福尔特 F 福满多 F 福日 F 福泰 F 福兴达 F 福中福 F 富春江 F 富尔美 F 富可视 F 富士通 G GAAMII G GEAK G Gigaset G GX G 感恩 G 橄榄树 G 港利通 G 港龙 G 高尔 G 高金 G 高科 G 高昇 G 高盛达 G 高斯贝尔 G 高通 G 高威尔 G 高翔 G 高新奇 G 高讯移动 G 格莱特 G 格力 G 格力通 G 共展 G 贡茶 G 关爱心 G 观 G 冠集 G 广东凌鹰 G 广信 G 广盈时代 G 广州盛科 G 广州索爱 G 国虹 G 国力通 G 国美 G 国乾 G 国乾科技 G 国威 G 国威HB G 国威创新 G 国信 G 国信通 G 国星 G 国誉 G 国正通 G 果冻 G 果米 H HAOERMEI H HATENG H HONGYU H HOt H HOTCOM 鸿达康利 H HWEI H 海恩迈 H 海尔 H 海帆 H 海米 H 海能达 H 海派贵族 H 海棠 H 海沃 H 海信 H 海旭 H 海语 H 汉泰 H 豪雅 H 好利通 H 好兄弟 H 皓轩 H 禾米 H 禾兴江源 H 合派 H 和 H 和信 H 核动力 H 黑莓 H 黑魅 H 黑米 H 黑鲨 H 恒波 H 恒泰 H 恒通 H 恒享 H 恒心 H 恒信 H 恒宇丰 H 恒语 H 恒远通达 H 红橙果 H 红番茄 H 红萝卜 H 红米 H 红鸟 H 红派 H 红葉 H 宏达 H 宏康 H 宏牛时代 H 宏碁 H 宏润 H 宏森 H 宏泰尔 H 宏天 H 宏为 H 宏族 H 洪泰天晖 H 虹动 H 虹锋 H 虹联盟 H 虹米 H 虹沃 H 鸿嘉源 H 湖南大成 H 湖南电子 H 花米 H 华臣数码 H 华慈 H 华帝高科 H 华度 H 华冠 H 华晶 H 华立 H 华立时代 H 华凌 H 华录 H 华岷 H 华纳威秀 H 华诺 H 华锐 H 华森 H 华尚 H 华世基 H 华蜀 H 华硕 H 华唐 H 华唐时代 H 华为 H 华夏腾宇 H 华夏通 H 华信时代 H 华讯 H 华禹 H 寰宇通 H 幻狐 H 幻影 H 皇族 H 汇丰源通 H 汇通富 H 汇通世纪 H 汇威 H 汇讯 H 惠普 H 惠族 H 喜来乐 I imoo I innos I iQOO I ivargo I ivvi I iwoo I 爱米无线 I 茵莱孚 J JD J 基伍 J 吉邦 J 吉客猫 J 吉事达 J 极帆 J 极简 J 集思宝 J 集友 J 几米 J 技嘉 J 加利利 J 佳达易通 J 佳斯特 J 佳通 J 佳想 J 佳信达 J 佳域 J 佳源 J 佳之选 J 嘉乐派 J 嘉尚 J 嘉源 J 杰得微 J 杰泰尔 J 捷豹 J 捷凯信 J 捷来 J 捷仕科技 J 今大福 J 今典 J 金柏利 J 金铂 J 金步 J 金得 J 金德力 J 金豆子 J 金尔雅 J 金格儿 J 金龟子 J 金国威 J 金果 J 金翰 J 金弘 J 金汇马 J 金凯为 J 金科 J 金科鼎 J 金酷珀 J 金来 J 金乐福 J 金立 J 金立致远 J 金曼 J 金茂 J 金米 J 金鹏 J 金珀 J 金荣通 J 金润 J 金圣达 J 金腾亿 J 金天纬 J 金信 J 金兴 J 金星数码 J 金讯宏盛 J 金业 J 金赢 J 金运 J 琻品 J 锦书 J 锦炫达 J 劲动能 J 京瓷 J 京华 J 京凯达 J 京崎 J 经纬 J 精瑞 J 精英移动 J 井冈山华禹 J 景象 J 景译科 J 警翼 J 竞冠 J 九爱 J 巨豆豆 J 巨赛 J 巨盛 J 君爵 J 骏域 K KEJIAN K KENKO K KISS U K KOCIN K Kumai K 卡布奇诺 K 卡尔 K 卡尔雷斯 K 卡美欧 K 卡为 K 卡西欧 K 卡卓 K 凯利通 K 铠基 K 康佳 K 康力 K 康维斯 K 科宝 K 科潮 K 科达 K 科达圣龙 K 科健 K 科酷 K 科立讯 K 科利莱 K 科米 K 科铭 K 科摩 K 科诺 K 科普达 K 科盛 K 科特 K 科王 K 可美 K 渴望 K 克莱斯 K 克里特 K 垦鑫达 K 库柏 K 酷爱 K 酷宝 K 酷比 K 酷比魔方 K 酷驰 K 酷港 K 酷鸽 K 酷和 K 酷卡 K 酷开 K 酷珂 K 酷酷贝尔 K 酷龙 K 酷米 K 酷诺 K 酷派 K 酷派移动 K 酷珀 K 酷普 K 酷睿 K 酷绅 K 酷维 K 酷沃 K 快易典 K 昆达 L LG L 拉风 L 来基达 L 徕卡 L 莱诺 L 兰铂 L 蓝博兴 L 蓝狐 L 蓝极星 L 蓝科 L 蓝玫 L 蓝魔 L 蓝鹏伟业 L 蓝天 L 蓝想 L 蓝信 L 朗格 L 朗杰 L 朗界 L 朗翔 L 朗星达 L 老来宝 L 乐点 L 乐丰 L 乐购天地 L 乐美佳 L 乐派 L 乐锐 L 乐视 L 乐听 L 乐为 L 乐维 L 乐信时代 L 乐讯 L 乐洲 L 乐卓 L 雷奥 L 雷梦 L 雷萨 L 厘米 L 力宏 L 力派 L 力鑫 L 立星 L 联创 L 联代 L 联泓 L 联通 L 联想 L 联益合创 L 亮剑 L 林宝坚尼 L 凌科 L 凌泰 L 凌拓 L 凌鹰 L 凌鹰数码 L 聆韵 L 零零柒 L 六虹 L 龙贝尔 L 龙旗 L 龙之宇 L 裸媒 M MANN M manta M MEIIGOO M My Year M 禾苗 M 买卖宝 M 迈道 M 迈峰 M 迈腾达 M 麦动 M 麦族 M 脉腾 M 曼迪客 M 漫猫 M 美富通 M 美歌 M 美杰新语 M 美酷 M 美菱 M 美奇 M 美盛通 M 美图 M 美熙 M 美翼 M 魅点 M 魅莱 M 魅蓝 M 魅兴 M 魅族 M 盟宝 M 咪购 M 米歌 M 米库 M 米辣椒 M 米莱 M 米蓝 M 米粒 M 米玛 M 米欧卡 M 米錡 M 米图 M 米王 M 米语 M 米智 M 敏讯 M 名博 M 名虹 M 明基 M 明基西门子 M 明派 M 明泰 M 铭仁 M 摩购 M 摩乐 M 摩玛时代 M 摩能 M 摩普士 M 摩天 M 摩天时代 M 摩托罗拉 M 摩托数码 M 摩托无线 M 摩西 M 木糖醇 N NAIDE N NEC N Newman纽曼 N nuoqi N 麦购 N 纳伟仕 N 奈华特 N 南方高科 N 南俘电讯 N 南和 N 南极星 N 尼奥库珀 N 尼凯恩 N 尼蒙 N 逆客 N 宁波三星 N 纽曼 N 纽特 N 纽万 N 纽维 N 努比亚 N 暖心 N 诺而信 N 诺菲世纪 N 诺基亚 N 诺记 N 诺嘉源 N 诺捷通 N 诺卡 N 诺克拉 N 诺利达 N 诺利佳 N 诺讯 N 诺亚鸽 N 诺亚信 N 糯米 O OPSSON O OTOT欧拓 O 欧比 O 欧博信 O 欧恩 O 欧歌 O 欧谷 O 欧加 O 欧凯 O 欧科 O 欧酷 O 欧乐迪 O 欧乐酷 O 欧珀 O 欧奇 O 欧萨 O 欧上 O 欧盛 O 欧网 O 欧唯 O 欧沃 O 欧新 O 欧信 O 欧亚信 O 欧怡 O 欧宇 O 欧正 O 沃凯泰 P PCCY P 铂乐 P 拍档 P 派对 P 派尔 P 派沃 P 派信 P 朋和 P 品志PINZHI P 平安星 P 苹果 P 普爱达 P 普拉达 P 普莱达 P 普蓝 P 普联 P 普瑞德 P 普士 P 普天三洋 P 普天同乐 P 普天王之 P 普天王芝 P 普天鑫 P 普天宜通 Q QIYU 奇语 Q QQr Q 七彩虹 Q 七喜 Q 七星河 Q 齐乐 Q 奇云 Q 琦基 Q 旗润 Q 牵挂 Q 侨兴 Q 青橙 Q 青葱 Q 青岩 Q 青漾 Q 庆邦 Q 全普 Q 群星 R realme R 荣创 R 荣事达 R 荣耀 R 柔宇科技 R 锐铂 R 锐风 R 锐合 R 锐界 R 锐力通 R 锐美恩天 R 锐族 R 瑞高 R 瑞恒 R 瑞翼 R 瑞云 S SAILF S SANNAING S SGMSGMS S SKT S 萨基姆 S 萨米 S 赛潮流 S 赛尔丰 S 赛丰 S 赛豐 S 赛鸿 S 赛纳普 S 赛浦 S 三宝 S 三和新 S 三基摩客 S 三菱 S 三美琦 S 三盟 S 三普 S 三奇 S 三维 S 三维宏业 S 三纬 S 三五互联 S 三新 S 三星 S 三众 S 桑达 S 森川 S 森密科技 S 山水 S 山西天丽 S 山楂树 S 商务通 S 尚成 S 尚锋 S 尚合 S 深爱 S 深圳欧珀 S 深圳天时达 S 深圳亚力通 S 神阳 S 神州牛歌 S 神舟 S 胜音 S 圣宝龙 S 盛和 S 盛况 S 盛隆兴 S 盛乾通 S 盛泰 S 时通伟业 S 世纪天元 S 世纪星 S 世纪阳天 S 世家 S 事达 S 守护宝 S 首弘科技 S 首家 S 首派科技 S 首信 S 首亿通讯 S 首悦 S 首云 S 数果 S 数源 S 双科 S 双侨 S 顺通无限 S 硕码 S 硕尼姆 S 硕王 S 硕颖 S 思特电子 S 四创 S 四吉 S 四季风 S 松下 S 松讯达 S 搜神 S 随锐 S 索爱佳通 S 索尼 S 索尼爱立信 S 索品 S 索普 S 索信 S 索野 T TCL T thifone T 钛客 T 泰丰网络 T 泰美利 T 泰蒙 T 泰星 T 泰讯 T 唐为 T 糖果 T 特灵通 T 腾祥 T 腾信时代 T 天爱 T 天波 T 天歌 T 天宏时代 T 天基 T 天科讯 T 天乐 T 天迈 T 天频 T 天勤 T 天时达 T 天天高 T 天玺 T 天意 T 天语 T 天遇 T 天元通 T 铁通 T 通途 T 同威 T 同心 T 同洲 T 途耐 T 托柯 T 托普 T 託普达 T 拓展无限 U uphone U UT斯达康 U 优本 U 优酷 U 优品 U 优珀 V VDVD V VERTU V VEVA V viipoo V VIM伟恩 V vipmi V vivl V vjvj V VKE V VOVG V VVETIME V 维纳斯 V 沃德丰 W WELAI W 瓦戈 W 万达斯 W 万德利 W 万捷 W 万金塔 W 万利达 W 万普 W 万擎 W 万事通 W 万旭麒 W 万有 W 网时代 W 网讯 W 威铂 W 威而肯 W 威酷 W 威智达 W 威兹奥 W 葳朗 W 微铂 W 微诺 W 微软 W 微尚 W 微网信通 W 微兴 W 为美 W 为信 W 唯爱 W 唯奥 W 唯开 W 唯科 W 唯乐 W 唯米 W 唯想 W 唯掌 W 帷幄 W 惟卡时代 W 惟新 W 维佳特 W 维卡 W 维图 W 维沃 W 伟博动力 W 伟大 W 纬图 W 闻讯 W 我爱你 W 沃达 W 沃歌 W 沃加 W 沃利达 W 沃普丰 W 沃途 W 吾爱 W 午诺 X Xmi 鄉米 X 西铂 X 西凯 X 西可 X 西美 X 西门子 X 西泰 X 西维 X 矽沃 X 喜卡 X 喜族 X 夏朗 X 夏浦光电 X 夏普 X 夏新 X 夏星 X 厦华 X 先创 X 先锋 X 先科 X 咸通 X 现代 X 香米 X 湘邮 X 祥集 X 响亮 X 橡皮泥 X 小霸王 X 小格雷 X 小辣椒 X 小麦麦果 X 小米 X 小米MIX X 小蜜蜂 X 小鸟 X 小铁锤 X 小香叶 X 小杨树 X 心爱 X 心迪 X 心迪宝 X 心动科技 X 心动世纪 X 心语通 X 新标达 X 新橙北斗 X 新创想 X 新大陆 X 新国脉 X 新虹伟 X 新基德 X 新力虹 X 新利智 X 新路虎时代 X 新碁 X 新石器 X 新势力 X 新邮 X 新悦蓝图 X 新中桥 X 鑫诺 X 鑫派尔 X 鑫荣耀 X 鑫语 X 鑫卓越 X 信达 X 信大捷安 X 信得乐 X 信明达 X 信诺 X 信实 X 信太 X 信天游 X 信盈 X 信云 X 兴华宝 X 星火 X 星科壬 X 星网锐捷 X 星维 X 星宇 X 星语 X 星族 X 熊猫 X 炫华 X 学之泉 X 讯霸 X 讯歌 Y yeechui Y yivi Y YOOYI Y YOTAPHONE Y 乐目 Y 雅阔尔 Y 雅米 Y 雅器 Y 雅讯达 Y 雅迅网络 Y 亚力通 Y 亚米 Y 亚信通 Y 言信 Y 央科 Y 央科YangKe Y 扬新 Y 摇乾树 Y 一加 Y 一三九 Y 一束光 Y 伊斯普 Y 依偎 Y 依众兴 Y 壹伍Onefive Y 宜兴中电 Y 怡诺 Y 移联翼 Y 颐佰 Y 以晴 Y 以晴华冠 Y 亿城 Y 亿和源 Y 亿嘉鑫 Y 亿科泰达 Y 亿美讯联 Y 亿旗 Y 亿通 Y 亿扬 Y 亿优 Y 艺科 Y 易百年 Y 易博士 Y 易丰展业 Y 易派 Y 易特科 Y 易腾迈 Y 易闻 Y 益阳 Y 逸佰年 Y 翼乐 Y 翼鸣 Y 翼品 Y 翼扬 Y 银通天下 Y 英达思康 Y 英华达 Y 英莱尔 Y 英迈 Y 英普 Y 英特奇 Y 鹰讯 Y 赢者 Y 映趣 Y 永隆通 Y 勇娃 Y 优Phone Y 优帛 Y 优博讯 Y 优尔得 Y 优歌 Y 优购 Y 优快 Y 优乐酷 Y 优米 Y 优米斯达 Y 优米通达 Y 优摩 Y 优派 Y 优品通 Y 优普 Y 优思 Y 优索 Y 优图 Y 优友 Y 友利通 Y 友旺 Y 友为 Y 友信达 Y 有糖 Y 宇飞来 Y 宇龙 Y 宇阳 Y 羽翼 Y 禹华 Y 语泉 Y 语信 Y 玉科 Y 誉国威 Y 誉品 Y 元一 Y 原点 Y 远大 Y 远点 Y 远望谷 Y 粤盈 Y 云本 Y 云刷 Y 云台 Z ZOPO Z 合众思壮 Z 泽爱 Z 泽领 Z 詹姆士 Z 展翼 Z 掌力 Z 掌中宝 Z 兆信 Z 兆讯达 Z 真我 Z 臻爱 Z 振华百智 Z 振华欧比 Z 振华通信 Z 振华宇科 Z 征服 Z 证通金信 Z 知己迅联 Z 知心 Z 职业者 Z 至能 Z 至尊宝 Z 志程 Z 志佳 Z 志能 Z 智爱 Z 智豪信 Z 智惠 Z 智吉 Z 智灵通 Z 智媄 Z 智珀 Z 智普 Z 智信互联 Z 智讯 Z 智迅达 Z 中宝 Z 中辰 Z 中德瑞 Z 中电 Z 中国老板 Z 中国移动 Z 中国振华 Z 中江 Z 中宽 Z 中诺 Z 中锘基雅 Z 中桥 Z 中天 Z 中天语 Z 中伟天 Z 中信卫星 Z 中兴 Z 中兴健康 Z 中益 Z 中渔 Z 中轴线 Z 众一 Z 卓比 Z 卓酷 Z 卓拉 Z 卓米 Z 卓普 Z 卓尚 Z 紫光海泰 Z 自然派 Z 自由客   请选择手机型号", null, "", null, "标题 内容", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "红米 M2010J19SC", null, "小米 M2011K2C", null, "华为 BRQ-AN00", null, "华为 OCE-AN10", null, "华为 PPA-AL20", null, "三星 SM-G9980", null, "华为 ANG-AN00", null, "YOK-AN10", null, "XT2091-7", null, "XT2081-4", null, "", null, "苹果 A2404", null, "红米 M2006C3LC", null, "维沃 V2057A", null, "苹果 A2412", null, "欧珀 PDVM00", null, "维沃 V1901A", null, "苹果 A2223", null, "维沃 V2054A", null, "红米 M2007J22C", null, "苹果 A2408\n\n 来源：中关村在线 2017-10-30\n 小米MIX 2是小米的第二款全面屏产品，从去年的小米MIX引爆手机全面屏市场，到今年小米MIX 2领跑全面屏2.0时代，小米MIX系列可以说是独领风骚。目前小米MIX 2黑色陶瓷板已经现货，那么到底值不值得买呢？刚好，笔者拿到了量产版的小米MIX 2，于是给大家带来了这篇全面评测。", null, "从小米MIX 2的贴吧论坛来看，这款手机并不能算是一款做工优秀的产品，屏幕、装配、信号等等问题一大堆，能不能买到正常的手机看来算是个拼人品的运气活儿，不过好在笔者的运气还不错，从外观来看拿到的这款机器没有什么大的问题，至少从外观上没有碰见屏幕盖板与机身的大缝隙、数据接口松动的问题。", null, "配置来说，小米MIX 2符合小米“为发烧而生”的口号，搭载高通骁龙835移动平台；运存6GB LPDDR4x起步，全陶瓷尊享版可以达到8GB；存储64GB UFS 2.1起步，最大256GB；屏幕采用了5.99英寸18:9分辨率2160×1080的LCD屏幕，符合当下对全面屏的定义；后置1200万像素摄像头，支持4轴光学防抖。 　　外观：全面屏的外观设计差点 但确实很闪", null, "小米MIX 2在外观上最出彩的宣传点有两个：一是全面屏，另一个是尊享版的Unibody全陶瓷机身。在金属和玻璃机身占领几乎所有市场的当下，陶瓷机身的搭配不仅给小米MIX 2打上了独特的标签，而且带来了独特的质感。笔者拿到的是黑色陶瓷版，从观感上形容其机身，就是“闪亮”。", null, "", null, "但正面的全面屏2.0和笔者的想象有些差距。由于计算方法不同，各机构给出的屏占比不太相同，但小米MIX 2屏占比不会低于80%。2.0时代自然要比1.0有所进化，但小米MIX 2相比于上一代产品反而有所下降，原因是虽然下巴进一步收窄，但上左右三边要比小米MIX更宽。粗边框严重影响了全面屏设计的观感。", null, "相信细心的消费者已经发现，全陶瓷尊享版的机身尺寸为150.5mm*74.6mm*7.7mm，黑色陶瓷板机身尺寸151.8mm*75.5mm*7.7mm，也就是说全陶瓷尊享版的机身更小一些。黑色陶瓷版的金属中框凸出一些，由于笔者并未拿到全陶瓷尊享版，所以屏幕边框会不会变窄还无法确定。 　　另外，小米MIX 2的前面板布局给笔者的感觉总有些别扭，上左右三面边框窄，唯独下巴较长，怎么看怎么都不和谐，相比之下，像vivo X20、努比亚Z17S上下对等的设计看起来要顺眼的多。而且前置摄像头放在下巴上，打开前置相机的瞬间你就会明白它的“反人类”之处。当然外观审美属于萝卜青菜各有所爱，笔者在这里不好多作评论，如果你是想要购买的外观党，建议先去实体店亲自上手看看再做决定。", null, "笔者运气不错，拿到的这台小米MIX 2额头并没有出现严重的衔接问题，这时候来看它的额头设置，还是十分巧妙的。小米MIX 2采用了导管式微型听筒设计，在占用额头很小面积的同时，又让通话声音得到了保障。而且笔者发现，在听歌和看视频时，听筒可以当做一个声道。虽然在声音素质上不能与扬声器相提并论，但总归来说聊胜于无。除此之外，超声波式的距离传感器也集成在了额头上。", null, "上一篇：魅蓝6续航花式体验 下一篇：酷比F1：8000万像素挑战单反", null, "", null, "93.1%屏占比一手可握 OPPO Reno (2019-06-12)", null, "红米Note 7体验 披着参数狼皮的虚胖小绵羊 (2019-02-13)", null, "美图V7全面评测 (2019-01-22)", null, "华为nova4一月长测 (2019-01-22)", null, "不止屏内开孔相机 三星Galaxy A8s全面评测 (2019-01-09)", null, "4800万碾压级超清拍照 荣耀V20实拍评测 (2019-01-04)\n 真诚欢迎各IT媒体、机构、专家和网友与我们联系合作!　致电：010-82052810 关于我们| 网站地图| 法律声明 京ICP证080582号 版权所有：北京讯惠通信技术开发有限公司 Copyright 2006-2021 jwxk.miit.gov.cn All Rights Reserved" ]

[ null, "http://shouji.tenaa.com.cn/Image/search/7.jpg", null, "http://shouji.tenaa.com.cn/Image/search/2.jpg", null, "http://shouji.tenaa.com.cn/Image/search/3.jpg", null, "http://shouji.tenaa.com.cn/Image/search/2.jpg", null, "http://shouji.tenaa.com.cn/Image/search/1.jpg", null, "http://shouji.tenaa.com.cn/Image/search/2.jpg", null, "http://shouji.tenaa.com.cn/Image/News/ZXXX.gif", null, "http://shouji.tenaa.com.cn/Image/News/XWRX_Title.gif", null, "http://shouji.tenaa.com.cn/Image/News/XWRX1.gif", null, "http://shouji.tenaa.com.cn/Image/News/sjpc1.gif", null, "http://shouji.tenaa.com.cn/Image/News/sjxp1.gif", null, "http://shouji.tenaa.com.cn/Image/News/sjhq.gif", null, "http://shouji.tenaa.com.cn/Image/News/LXQS1.gif", null, "http://shouji.tenaa.com.cn/Image/News/ZCFG1.gif", null, "http://shouji.tenaa.com.cn/Image/News/JSQY1.gif", null, "http://shouji.tenaa.com.cn/Image/News/left_aqnl1.gif", null, "http://shouji.tenaa.com.cn/Image/Knowledge/GZDPH.gif", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_1.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_2.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_3.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_4.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_5.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_6.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_7.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_8.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_9.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_10.jpg", null, "http://shouji.tenaa.com.cn/Image/Knowledge/SJXLPH.gif", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_1.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_2.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_3.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_4.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_5.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_6.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_7.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_8.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_9.jpg", null, "http://shouji.tenaa.com.cn/Image/Detail/dPH_10.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJ1nzH4aINXvNAAHHpMAhAeAAAhnpQIRZoUAAce8062.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJlnzH4iIGeGjAAFcFuEwx14AAhnpQIjNNMAAVwu169.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJ1nzH4uIUU33AAGRzSFjROIAAhnpQJ97SIAAZHl390.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJlnzH42IPH0HAAGmxc_dnX0AAhnpQKIt5kAAabd231.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJlnzH4-IDbWBAAHBfH0dAFoAAhnpQKeEbsAAcGU883.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJlnzH5KIBF8_AAHaiWyA-6IAAhnpQKvtGsAAdqh396.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJ1nzH5aIPi6ZAAFRCLJLJAcAAhnpQLLba8AAVEg395.jpg", null, "http://article.fd.zol-img.com.cn/t_s640x2000/g5/M00/0B/03/ChMkJ1nzH5iIKG30AAG9zg3MsugAAhnpQMlbswAAb3m566.jpg", null, "http://shouji.tenaa.com.cn/Image/Knowledge/XGWZ.gif", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null, "http://shouji.tenaa.com.cn/Image/middle.jpg", null ]

[ "# Best answer: How many chlorine atoms are in 2 moles of chlorine?\n\nContents\n\n## How many atoms of chlorine are in a mole of chlorine?\n\nOne mole of chlorine weighs 70 grams. Just like a dozen equals twelve, a mole equals Avogadro’s number of atoms or 6.02 × 1023. 6.02 × 1023 molecules of hydrogen would weigh 2 grams. An equal number of chlorine molecules would weigh 70 grams.\n\n## What is the mass of 2 moles of chlorine molecule?\n\nChlorine gas Cl 2, has a relative molecular mass of 71 because it is made up of two Cl atoms, each of Mr 35.5. A mole of chlorine gas has a mass of 71g, and contains two moles of chlorine atoms, each of mass 35.5g.\n\n## What is the mass of 2 mole of CL2?\n\nMolecular mass of CL2=2×35. 453amu=70.906amu.\n\n## How do you find moles of chlorine?\n\nThe mass of one mole of chlorine gas is its molar mass, and can be determined by multiplying its subscript by the molar mass of chlorine (atomic weight on the periodic table in g/mol), which is 35.45 g/mol .\n\n## How many atoms of chlorine are there?\n\nBut the question asked is “how many atoms are present?” Because each molecule of Chlorine contains two atoms, the answer is 2 x 1.607 x 1024 atoms are present = 3.22 x 1024 atoms.\n\nIT IS INTERESTING:  Can you get psoriasis as you age?\n\n## What is the mass in grams of 2.5 moles of chlorine gas?\n\nAnswer: Cl2 contains two Cl atom so molar mass will be 35 . 5 x2 that is 71 GRAM means 0ne mole of Cl2 has 71 gram so 2.5 mol will be equal to 2.5×71. Gram.\n\n## What is the mass of 1 mole of Cl2?\n\nOne mole of chlorine mass 70 grams. Just like a dozen equals twelve, a mole equals Avogadro’s number of atoms or 6.02 × 10^23.\n\n## What is the formula for moles to grams?\n\nIn order to convert the moles of a substance to grams, you will need to multiply the mole value of the substance by its molar mass.\n\n## What is the MR of chlorine?\n\nA r values of elements\n\nElement Relative atomic mass\nCarbon (C) 12\nOxygen (O) 16\nMagnesium (Mg) 24\nChlorine (Cl) 35.5\n\n## How many grams are in 3 moles of chlorine?\n\n3 = Mass71. So the mass is 213 grams. So, the mass of 3 moles of chlorine molecule is 213gram." ]

[ null ]

[ "# challenge question -- Factor the polynomial completely\n\n#### lookagain\n\n##### Elite Member\nEdit:\n\nDemonstrate at least two methods for factoring the following polynomial\ncompletely over the integers.\n\n$$\\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$\n\nLast edited:\n\n#### MarkFL\n\n##### Super Moderator\nStaff member\nMethod 1:\n\nFactor first by grouping:\n\n$$\\displaystyle (x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)$$\n\nNow, for the quartic factor assume it may be factored as follows:\n\n$$\\displaystyle x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$$\n\nEquating coefficients, we find:\n\n$$\\displaystyle a+b=0$$\n\n$$\\displaystyle ab+2=1$$\n\nand so one solution is $$\\displaystyle (a,b)=(1,-1)$$ and we have:\n\n$$\\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ which means:\n\n$$\\displaystyle x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)$$\n\nMethod 2:\n\nLet:\n\n$$\\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:\n\n$$\\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:\n\n$$\\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:\n\n$$\\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nA) Give the completely factored form over the integers of the following polynomial, and\n\nB) demonstrate at least two methods for doing so.\n\n$$\\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$\n\n(x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)\n\nor\n\n(x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)\n\n#### lookagain\n\n##### Elite Member\nHere is another:\n\n(x^5 + 1) + (x^4 + x^3 + x^2 + x) =\n\n(x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =\n\n(x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =\n\n(x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =\n\n(x + 1)(x^4 + x^2 + 1) =\n\n(You can continue simplifying this with one of the above steps in\nany of the appropriate above posts.)\n\nLast edited:\n\n#### daon2\n\n##### Senior Member\nHere's another way.\n\nSince $$\\displaystyle x^5+x^4+x^3+x^2+x+1 = \\dfrac{x^6-1}{x-1}$$ the roots of this polynomial are exactly the set $$\\displaystyle \\{z\\in \\mathbb{C}-\\{1\\}\\,\\,;\\,\\, z^6=1\\}$$, i.e. the roots of unity, ignoring the positive real root. They are $$\\displaystyle e^{\\pm i\\pi/3},e^{\\pm i2\\pi/3}, -1$$.\n\nWe want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:\n\n$$\\displaystyle (x-e^{i\\pi/3})(x-e^{- i\\pi/3}) = x^2-x+1$$\n$$\\displaystyle (x-e^{i2\\pi/3})(x-e^{- i2\\pi/3}) = x^2+x+1$$\n$$\\displaystyle x-(-1) = x+1$$\n\nThere is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.\n\n#### soroban\n\n##### Elite Member\nHello, lookagain\n\nThis is a variation of daon's solution.\n\nDemonstrate at least two methods for factoring the following polynomial:\n\n. . . $$\\displaystyle P(x) \\;=\\;x^5 + x^4 + x^3 + x^2 + x + 1$$\n\n$$\\displaystyle P(x) \\;=\\;\\dfrac{x^6 - 1}{x-1} \\;=\\;\\dfrac{\\overbrace{(x^3)^2 - (1^2)}^{\\text{diff. of squares}}}{x-1}$$\n\n. . . . .$$\\displaystyle =\\;\\dfrac{\\overbrace{(x^3-1)}^{\\text{diff.of cubes}}\\cdot\\overbrace{(x^3+1)}^{\\text{sum of cubes}}}{x-1}$$\n\n. . . . .$$\\displaystyle =\\; \\dfrac{(\\color{red}{\\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\\color{red}{\\rlap{/////}}x-1}$$\n\n. . . . .$$\\displaystyle =\\; (x+1)(x^2+x+1)(x^2-x+1)$$\n\n#### lookagain\n\n##### Elite Member\nHello, lookagain\n\nThis is a variation of daon's solution.\n\n$$\\displaystyle P(x) \\;=\\;\\dfrac{x^6 - 1}{x-1} \\;=\\;\\dfrac{\\overbrace{(x^3)^2 - (1^2)}^{\\text{diff. of squares}}}{x-1}$$\n\n. . . . .$$\\displaystyle =\\;\\dfrac{\\overbrace{(x^3-1)}^{\\text{diff.of cubes}}\\cdot\\overbrace{(x^3+1)}^{\\text{sum of cubes}}}{x-1}$$\n\n. . . . .$$\\displaystyle =\\; \\dfrac{(\\color{red}{\\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\\color{red}{\\rlap{/////}}x-1}$$\n\n. . . . .$$\\displaystyle =\\; (x+1)(x^2+x+1)(x^2-x+1)$$\nMarkFL said:\nMethod 2:\n\nLet:\n\n$$\\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:\n\n$$\\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:\n\n$$\\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:\n\n$$\\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$\n\nThese two (MarkFL's and soroban's versions) look essentially the same to me.\n\n- - - - - - - - - - - - - - - - -\n\nOthers:\n\n(x^5 + x^2) + (x^4 + x) + (x^3 + 1) =\n\nx^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =\n\n(x^3 + 1)(x^2 + x + 1) =\n\n(x + 1)(x^2 - x + 1)(x^2 + x + 1)\n\n. . . . . . . . . . . . . . . . . . . . .\n\n(x^5 + x^3 + x) + (x^4 + x^2 + 1) =\n\nx(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =\n\n(x^4 + x^2 + 1)(x + 1) =\n\n(x^2 - x + 1)(x^2 + x + 1)(x + 1)" ]

[ null ]

[ "Applications of\nQuantum Physics\nImplications of\nQuantum Physics\n\n35. The Uncertainty Principle\n\nSummary\nThe uncertainty principle says that a small uncertainty in the measured momentum of the wave function implies a large uncertainty in the measured position of the wave function, and vice versa.\n\nIf one assumes matter is made up of very small particles, then the uncertainty principle is, from a classical point of view, quite surprising. It says one cannot measure both the position and momentum (mass times velocity) of the particle with arbitrary precision. If you know position very accurately, then there will be a large uncertainty in the momentum. But from the wave-function-only view (see No Evidence for Particles) this principle is not surprising or mysterious; it is just a property of wave functions.\n\nThe uncertainty principle has a technical statement and derivation along with an interpretation or explanation of what the technical statement implies. To illustrate, suppose we have either a light-like wave function or an electron-like wave function that goes through a narrow slit. The particle-like wave function is initially traveling in the z direction and the slit is in the y direction. After the wave goes through the slit, it spreads out in the x-direction. Because of this, if one measures the x position many times, one will get a spread in values, even if the measurements are exact. This is just the nature of waves. The same conclusion holds for momentum; if one measures the momentum in the x direction many times, one will get a spread in values, even if the measurements are exact.\n\nIt is remarkable that, by the use of standard vector-space arguments, one can get a relation between the spreads in position and momentum. It is:", null, "", null, "(35-1)\n(where the", null, "comes from the fact that [px,x]=-i", null, "). Or, if we take the square root of this relation and use root-mean-square quantities,", null, "", null, "(35-2)\nThis relation is often interpreted as follows: There exists a particle embedded in the wave function whose position and momentum cannot be simultaneously measured exactly. But we have seen that there is No Evidence for Particles, so this interpretation is not warranted.\n\nWhat the uncertainty principle tells us is the following: First, even if we make exact measurements of, say, position on many systems identically prepared, there will inevitably be a spread in the measurements. And second if we make many exact measurements of position on a set of runs, and then we make many exact measurements of momentum on a separate set of runs, the product of the rms spreads of the two quantities must be greater than", null, "/2. Again, this result is simply a consequence of the nature of waves; if there are no particles, it is not so mysterious.\n\nOne common way the uncertainty principle is often used is to say that if the wave function is confined to a very small region in space, it will have a large spread in momentum. This is a valid translation of the import of equations (35-1) and (35-2).", null, "", null, "" ]

[ null, "http://implications-of-quantum-physics.com/equations/35_1.png", null, "http://implications-of-quantum-physics.com/equations/spacer.gif", null, "http://implications-of-quantum-physics.com/equations/33_h.png", null, "http://implications-of-quantum-physics.com/equations/33_h.png", null, "http://implications-of-quantum-physics.com/equations/35_2.png", null, "http://implications-of-quantum-physics.com/equations/spacer.gif", null, "http://implications-of-quantum-physics.com/equations/33_h.png", null, "http://implications-of-quantum-physics.com/understanding_quantum_physics3.png", null, "http://implications-of-quantum-physics.com/printmast.jpg", null ]

[ "# How esProc Assists Java to Retrieve Text Files\n\n403\n\nJava provides functions for handling the basic file processing, which refers to the retrieval of small text files, in a simple, unstructured way. But in handling files requiring structured format, holding data of various formats and having particular requirements, or big files that cannot be entirely loaded into memory, Java code is too complicated and its readability and reusability are hard to be guaranteed.\n\nesProc (free edition is available) can be used to make up for these deficiencies. esProc encapsulates a lot of functions for reading in/writing out and processing structured data, and provides the JDBC interface. A Java application will identify esProc script as a database stored procedure to execute, pass parameters to it and get result set via JDBC. You can learn more from How to Use esProc as the Class Library for Java.\nThe following cases are those you frequently encounter in retrieving text files in Java, and their esProc solutions.\n\nRetrieving specified fields\n\nYou need to import the OrderID, Client and Amount column by their names. Below is the source data:", null, "esProc code:\n\n A 1 =file(“D: \\\\sOrder.txt”).import@t(OrderID,Client,Amount)\n\nThe result:", null, "1. @t means importing the first row as column names. If there are no column names, you can use their sequence numbers to reference columns. To import the first, the second and the fourth column, for example, use file(“D: \\\\sOrder.txt”).import(#1,#2,#4). The result is as follows:", null, "2. You can also export a computed column. For example, use the following code to combine the year and OrderID into a newOrderID and export it along with Client and Amount:\n\n A 1 =file(“D: \\\\sOrder.txt”).import@t() 2 =A1.new(string(year(OrderDate))+”_”+string(OrderID):newOrder,Client,Amount)\n\nBy default import function reads in all fields. new function creates a two-dimensional table. The result is:", null, "3. The default separator is tab, or you can use other separators. To import a CSV file separated with commas, for example, use file(“D: \\\\sOrder.txt”).import@t(;”,”).\n\n4. To export some of the rows, specify them by row numbers. For example, use A1.to(2,100) to export rows from the second to the hundredth; and use A1.to(3,) to export rows beginning from the third one to the end.\n\n5. In a few cases, columns of data need to be retrieved and exported as one column. For example, to concatenate OrderID, Client and Amount and export them as one column, you can use the following code after data importing: create(all).record(A1.(OrderID)|A1.(Client)|A1.(Amount)) .\n\nRetrieving big files\n\nTo retrieve a big file that exceeds the memory capacity, use an esProc cursor, which can be accessed by a Java application with JDBC stream.\n\nesPro code:\n\n A 1 =file(“D: \\\\sOrder.txt”).cursor@t(OrderID,Client,Amount)\n\n1. To accelerate the file retrieval, you can use multithreaded parallel processing through @m option. The code is =file(“D: \\\\sOrder.txt”).cursor@tm(OrderID,Client,Amount). But this approach cannot guarantee that data is retrieved in its original order.\n\n2. Sometimes you need to segment data manually before processing it in parallel. To read in a segment of data, use file(“D:\\\\sOrder.txt”).import@z@t(;,2:24). @z means dividing the file roughly into 24 segments by bytes and importing the second one only. esProc will automatically skip the head row and make up the tail row to ensure that each row is retrieved completely. If the data size in each segment still exceeds the memory capacity, you can replace the import function with cursor function to export data as a cursor.\n\nRetrieving file by column lengths\n\nThe following data.txt file does not use the separator:", null, "You need to retrieve the file into a four-column two-dimensional table according to specified column lengths and return it to Java. The id column will have the first three bits, flag column will have the 10th and 11th bits, d1 column will have bits from the 14th to the 24th, and d2 column will have bits from the 25th to 33rd. Thus the four columns in the first row will be 001, DT, 100000000000 and 3210XXXX.\n\nesProc code:\n\n A 1 =file(“D:\\\\data.txt”).import@i() 2 =A1.new(mid(~,1,3):id,mid(~,10,2):flag,mid(~,14,11):d1,mid(~,25,9):d2)\n\n@i means returning a sequence (set) if the file has only one column. Then create a two-dimensional table based on A1; mid function truncates a string and ~ represents each row.\n\nHere’s the result:", null, "Retrieving file containing special characters\n\nThe data.csv file contains quotation marks, some of which disrupt the use of the data. So you need to remove the quotation marks and then return the file to Java: Below is the source data:", null, "esProc code:\n\n A 1 =file(“d:\\\\data.csv”).import(;”,”) 2 =A1.new(replace(_1,”\\””,””):_1,replace(_2,”\\””,””):_2,replace(_3,”\\””,””):_3,replace(_4,”\\””,””):_4)\n\nHere’s the result:", null, "Retrieving file containing mathematical formulas\n\nIn this case, you need to parse the mathematical formulas into expressions, evaluate them and then return the results. Below is the source data:", null, "esProc code:\n\n A 1 =file(“D:\\\\equations.txt”).import@i() 2 =As1.new(~:equations,eval(string(~)):result)\n\neval function dynamically parses strings into expressions to execute.\n\nThe result is:", null, "Retrieving file with multi-line records\n\nIn the following file, each record includes three lines. For example, the first record is JFS    3       468.0        2009-08-13 39. Now you need to export the file into a two-dimensional table.", null, "esProc code:\n\n A 1 =file(“D:\\\\data.txt”).import@si() 2 =A1.group((#-1)\\3) 3 =A2.new(~(1):OrderID, (line=~(2).array(“\\t”))(1):Client,line(2):SellerId,line(3):Amount,~(3):OrderDate )\n\nFirst import the file as a sequence; @s means not splitting the field. Then group the sequence every three members;“#” represents the row number and “\\” means integer division. Here’s the result:", null, "If the file is too big to be entirely loaded into memory, you need to use cursor to retrieve it and perform batch processing. First create a sub.dfx, which responses the external request of data retrieval by retrieving a batch of data and return it. This operation repeats until the whole file is retrieved. Below is the esProc code:\n\n A B 1 =file(“D:\\\\data.txt”).cursor@si() 2 for A1,3000 =A2.group((#-1)\\3) 3 =B2.new(~(1):OrderID, (line=~(2).array(“\\t”))(1):Client,line(2):SellerId,line(3):Amount,~(3):OrderDate ) 4 result B3\n\nLoop through A1 and retrieve 3,000 rows in each loop. After that you can handle the algorithm the above does. B4 returns B3 to the main script. The main script (which is the dfx file to be called by Java ) is as follows:\n\n A 1 =pcursor(“sub.dfx”)\n\npcursor function requests retrieving data through sub.dfx and converts data to a cursor and exports it.\n\nRetrieving records from uncertain lines\n\nWith the data.txt file, field values in a record scatter in uncertain number of lines. But fields are fixed. They are “Object Type”, “left”, “top” and “Line Color” and appear repeatedly until the end of the file. The first record, for example, is Symbol1, 14, 11 and RGB( 1 0 0 ). Now you need to retrieve the file into a structured two-dimensional table.", null, "esProc code:\n\n A 1 =file(“data.txt”).read() 2 =A1.array(“Object Type: “).to(2,) 3 =A2.new(~.array(“\\r\\n”)(1):OType,   mid(~,s=pos(~,”left: “)+len(“left: “),pos(~,”\\r\\n”,s)-s):L,   mid(~,s=pos(~,”top: “)+len(“top: “),pos(~,”\\r\\n”,s)-s):T,   mid(~,s=pos(~,”Line Color: “)+len(“Line Color: “),if(r=pos(~,”\\r\\n”,s),r,len(~))-s+1):LColor)\n\nThe read function can read in the file as a big string, and then split the string with the separator and remove the first empty line. Finally create a table sequence and use string functions – array, pos, len, mid – to find the desired fields. Note that you should use an if statement to judge the last line, for maybe no carriage return is used there. Here’s the final result:", null, "Besides the string functions, you can use regular expressions to find the desired fields.\n\nTo handle a big file that cannot be loaded into memory in one go, use pcursor function to retrieve it in batches.\n\nRetrieving records by marked groups\n\nThe data.txt file stores records by groups. Group names are marked by list (such as ARO, BDR, and BSF). You need to combine group names with their field values and export the records. Below is the source data:", null, "esProc code:\n\n A 1 =file(“mutiline2.txt”).import@si() 2 =A1.group@i(like(~,”list:*”)) 3 =A2.conj(~.to(2,).new(mid(A2.~(1),6):Client,(t=~.array(“\\t”))(1):c1,t(2):c2,t(3):c3,t(4):c4))\n\nFirst import the file into a sequence of strings, and then group the sequence according to lines headed by list. @i will group data into a same group if it satisfies the specified condition. The sign * represents the wildcard character. Here’s A2’s result:", null, "And then retrieve the desired fields and concatenate the records from each group. Here’s the result:", null, "" ]

[ null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_2.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_4.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_5.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_7.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_9.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_11.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_12.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_14.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_15.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_17.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_18.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_21.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_24.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_27.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_28.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_31.jpg", null, "http://blog.raqsoft.com/wp-content/uploads/2015/06/esProc_java_retrieve_text_32.jpg", null ]

[ "# How to implement simple LSTM in reinforcement task ('CartPole-v0')\n\nI’ve realised I don’t understand LSTMs in Pytorch quite as well as I thought, so I’m adapting the CartPole demo from Soumith Chintala to give myself a simple challenge of switching the main Linear layer with an LSTM.\n\nThe example here fails on the first pass with:\nRuntimeError: Input batch size 1 doesn’t match hidden batch size 128\non\nx, self.hidden = self.lstm(x, self.hidden)\n\nNow if I change this line to\nx, _ = self.lstm(x, self.hidden)\n\n… it converges and completes the task in a reasonable 950 Episodes.\n\nHowever, because it’s not feeding the Hidden states back if I do this (forgive my improvised terminology), it’s presumably not really taking advantage of the capabilities of the LSTM? I can’t quite sus the error either, as the self.hidden I’m feeding it is 1,128, and the self.hidden it’s outputting seems to be 1,128.\n\nI notice more advanced implementations of A3C models to play Atari games tend to use the hidden.values – up until now, I’ve really not used this aspect, generally using them x = self.lstm(x) … Which presumably is fine so long as all your work’s being done within batches?\n\nThanks so much for any help. I’d also be interested in whether an LSTM (rather than LSTMCell) might be utilised for this problem? But I’m having separate problems getting the data the right shape.\n\n``````import argparse\nimport gym\nimport numpy as np\nfrom itertools import count\nfrom collections import namedtuple\nimport torch\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\nfrom torch.distributions import Categorical\n\nGAMMA = 0.99\nenv = gym.make('CartPole-v0')\nenv.seed(1)\ntorch.manual_seed(1)\n\nSavedAction = namedtuple('SavedAction', ['log_prob', 'value'])\n\nclass Policy(nn.Module):\ndef __init__(self):\nsuper(Policy, self).__init__()\nself.lstm = nn.LSTMCell(3, 128)\nself.saved_actions = []\nself.rewards = []\nself.hidden = None\n\ndef forward(self, x):\nx = x.unsqueeze(0)\nx, self.hidden = self.lstm(x, self.hidden)\nx = x.squeeze(0)\nreturn F.softmax(action_scores, dim=-1), state_values\n\nmodel = Policy()\neps = np.finfo(np.float32).eps.item()\n\ndef select_action(state):\nstate = torch.from_numpy(state).float()\nprobs, state_value = model(state.narrow(0,1,3))\nm = Categorical(probs)\naction = m.sample()\nmodel.saved_actions.append(SavedAction(m.log_prob(action), state_value))\nreturn action.item()\n\ndef finish_episode():\nR = 0\nsaved_actions = model.saved_actions\npolicy_losses = []\nvalue_losses = []\nrewards = []\nfor r in model.rewards[::-1]:\nR = r + GAMMA * R\nrewards.insert(0, R)\nrewards = torch.tensor(rewards)\nrewards = (rewards - rewards.mean()) / (rewards.std() + eps)\nfor (log_prob, value), r in zip(saved_actions, rewards):\nreward = r - value.item()\npolicy_losses.append(-log_prob * reward)\nvalue_losses.append(F.smooth_l1_loss(value, torch.tensor([r])))\nloss = torch.stack(policy_losses).sum() + torch.stack(value_losses).sum()\nloss.backward()\noptimizer.step()\ndel model.rewards[:]\ndel model.saved_actions[:]\n\ndef main():\nrunning_reward = 10\nfor i_episode in count(1):\nstate = env.reset()\nfor t in range(10000):\naction = select_action(state)\nstate, reward, done, _ = env.step(action)\nmodel.rewards.append(reward)\nif done:\nbreak\n\nrunning_reward = running_reward * 0.99 + t * 0.01\nfinish_episode()\nif i_episode % 10 == 0:\nprint('Episode {}\\tLast length: {:5d}\\tAverage length: {:.2f}'.format(\ni_episode, t, running_reward))\nif running_reward > env.spec.reward_threshold:\nprint(\"Solved! Running reward is now {} and \"\n\"the last episode runs to {} time steps!\".format(running_reward, t))\nfor t in range(100000):\naction = select_action(state)\nstate, reward, done, _ = env.step(action)\nenv.render()\nmodel.rewards.append(reward)\n# if done:\n# break\nbreak\n\nif __name__ == '__main__':\nmain()\nenv.env.close()\n\n``````\n\nOkay, I think I got it. This is the only way I learn.\n\nSo I reset the Hidden states every time it’s done, and otherwise .detach() them. And it takes two or three times longer to train, and does a pretty horrible job – but presumably LSTMs aren’t very suited to this task.\n\nEDIT: one more update – completes the task around 1680 episodes on v1, a somewhat human-looking model here, with a natural wobble and some fairly dynamic rescues … unlike the machine-precision of usual solutions.\n\n``````import argparse\nimport gym\nimport numpy as np\nfrom itertools import count\nfrom collections import namedtuple\nimport torch\nimport torch.nn as nn\nimport torch.nn.functional as F\nimport torch.optim as optim\nfrom torch.distributions import Categorical\n\nGAMMA = 0.99\nHIDDEN_SIZE = 64\nenv = gym.make('CartPole-v1')\nenv.seed(1)\ntorch.manual_seed(1)\n\nSavedAction = namedtuple('SavedAction', ['log_prob', 'value'])\n\nclass Policy(nn.Module):\ndef __init__(self):\nsuper(Policy, self).__init__()\nself.lstm = nn.LSTMCell(4, HIDDEN_SIZE)\n# self.affine = nn.Linear(HIDDEN_SIZE, HIDDEN_SIZE)\nself.saved_actions = []\nself.rewards = []\nself.reset()\n\ndef reset(self):\nself.hidden = Variable(torch.zeros(1, HIDDEN_SIZE)), Variable(torch.zeros(1, HIDDEN_SIZE))\n\ndef detach_weights(self):\nself.hidden = self.hidden.detach(), self.hidden.detach()\n\ndef forward(self, x):\nx = x.unsqueeze(0)\nself.hidden = self.lstm(x, self.hidden)\nx = self.hidden\nx = x.squeeze(0)\n# x = self.affine(x)\nreturn F.softmax(action_scores, dim=-1), state_values\n\nmodel = Policy()\neps = np.finfo(np.float32).eps.item()\n\ndef select_action(state):\nstate = torch.from_numpy(state).float()\nprobs, state_value = model(state)\nm = Categorical(probs)\naction = m.sample()\nmodel.saved_actions.append(SavedAction(m.log_prob(action), state_value))\nreturn action.item()\n\ndef finish_episode():\nR = 0\nsaved_actions = model.saved_actions\npolicy_losses = []\nvalue_losses = []\nrewards = []\nfor r in model.rewards[::-1]:\nR = r + GAMMA * R\nrewards.insert(0, R)\nrewards = torch.tensor(rewards)\nrewards = (rewards - rewards.mean()) / (rewards.std() + eps)\nfor (log_prob, value), r in zip(saved_actions, rewards):\nreward = r - value.item()\npolicy_losses.append(-log_prob * reward)\nvalue_losses.append(F.smooth_l1_loss(value, torch.tensor([r])))\nloss = torch.stack(policy_losses).sum() + torch.stack(value_losses).sum()\nloss.backward()\noptimizer.step()\ndel model.rewards[:]\ndel model.saved_actions[:]\n\ndef main():\nrunning_reward = 10\nfor i_episode in count(1):\nstate = env.reset()\nfor t in range(10000):\naction = select_action(state)\nstate, reward, done, _ = env.step(action)\nmodel.rewards.append(reward)\nif done:\nmodel.reset()\nbreak\nelse:\nmodel.detach_weights()\n\nrunning_reward = running_reward * 0.99 + t * 0.01\nfinish_episode()\nif i_episode % 10 == 0:\nprint('Episode {}\\tLast length: {:5d}\\tAverage length: {:.2f}'.format(\ni_episode, t, running_reward))\nif running_reward > env.spec.reward_threshold:\nmodel.reset()\nprint(\"Solved! Running reward is now {} and \"\n\"the last episode runs to {} time steps!\".format(running_reward, t))\nfor t in range(100000):\naction = select_action(state)\nstate, reward, done, _ = env.step(action)\nenv.render()\nmodel.rewards.append(reward)\n# if done:\n# break\nbreak\n\nif __name__ == '__main__':\nmain()\nenv.env.close()\n\n``````\n2 Likes" ]

[ null ]

[ "# Capacitor codes\n\nThere are multiple ways a capacitor can indicate its capacitance. Larger capacitors often have their capacitance, tolerance and maximum voltage written on the side. Smaller capacitors often use a code consisting of three numbers and sometimes one letter.\n\n### Calculator\n\nYou can use this calculator to calculate the capacitance and tolerance of a capacitor.\n\n### Calculation\n\nThe capacitance in picofarad (pF) is equal to the first two digits multiplied by ten to the power of the third digit: C = AB·10 C . Some capacitors contain the letter R, which represents the decimal point. The capacitance in pF is equal to all digits linked together with a decimal point on the location of the R in this case. The letter at the end indicates the tolerance. These values can be found in the table below.\n\nletter tolerance letter tolerance\nA 0.05pF H 3%\nB 0.1pF J 5%\nC 0.25pF K 10%\nD 0.5pF L 15%\nE 0.5% M 20%\nF 1% V 25%\nG 2% N 30%\nZ +80% -20% - -\n\n#### Example\n\nWe have a capacitor with the code 156J. The capacitance is 15·10 6 pF = 15μF. The tolerance for J is 5%." ]

[ null ]

[ "# elliotsig\n\nElliot symmetric sigmoid transfer function\n\n## Syntax\n\n```A = elliotsig(N) ```\n\n## Description\n\nTransfer functions convert a neural network layer’s net input into its net output.\n\n`A = elliotsig(N)` takes an `S`-by-`Q` matrix of `S` `N`-element net input column vectors and returns an `S`-by-`Q` matrix `A` of output vectors, where each element of `N` is squashed from the interval ```[-inf inf]``` to the interval `[-1 1]` with an “S-shaped” function.\n\nThe advantage of this transfer function over other sigmoids is that it is fast to calculate on simple computing hardware as it does not require any exponential or trigonometric functions. Its disadvantage is that it only flattens out for large inputs, so its effect is not as local as other sigmoid functions. This might result in more training iterations, or require more neurons to achieve the same accuracy.\n\n## Examples\n\nCalculate a layer output from a single net input vector:\n\n```n = [0; 1; -0.5; 0.5]; a = elliotsig(n);```\n\nPlot the transfer function:\n\n```n = -5:0.01:5; plot(n, elliotsig(n)) set(gca,'dataaspectratio',[1 1 1],'xgrid','on','ygrid','on')```\n\nFor a network you have already defined, change the transfer function for layer `i`:\n\n` net.layers{i}.transferFcn = 'elliotsig';`\n\n## Version History\n\nIntroduced in R2012b" ]

[ null ]

[ "# eddington\n\nOne statistic that cyclists are known to track is their Eddington number. The Eddington number for cycling, E, is the maximum number where a cyclist has ridden E miles in E days. So to get a number of 30, you need to have ridden 30 miles or more on 30 separate days.\n\nThis package provides functions to compute a cyclist’s Eddington number, including efficiently computing cumulative E over a vector. These functions may also be used for h-indices for authors. Both are specific applications of computing the side length of a Durfee square.\n\n## Installation\n\n``````# install.packages(\"devtools\")\ndevtools::install_github(\"pegeler/eddington2/R/package\")``````\n\n## Example\n\nHere is a basic example showing how to get the summary Eddington number of the included simulated `rides` dataset. Note that we first have to aggregate ride mileage by date.\n\n``````library(eddington)\nlibrary(dplyr)\n\nrides %>%\ngroup_by(ride_date) %>%\nsummarize(n = n(), total = sum(ride_length)) %>%\nsummarize(E = E_num(total)) %>%\npull\n#> 29``````\n\nSee the package vignette for detailed usage." ]

[ null ]

[ "", null, "## Abstract\n\nFirst, we cover the conical curves on 2-dimensional modeling sphere S 2 showing their geometric properties affecting the hyperbolic navigation. We place emphasis on the geometric definition of spherical parabola and relate it to the notions of spherical ellipse and hyperbola and give simple geometric proofs for relations between conical curves on the sphere. In the second part of the paper function", null, "representing the ratio of the circle's circumference to its diameter has been defined and researched to analyze the potential discrepancies in the spherical and conical projective models on which the navigational computations are based on. We compare some non-Euclidean geometric properties of curved surfaces and its Euclidean plane model in reference to the local and global approximation. As a working tool we use", null, "function for geometric comparison analysis in the theory of long-range navigation and cartographic projection. We state the existence of the infinite number of the circles having the same radius but different circumference on the conical surface. Finally, we survey the exemplary proposals of generalization of function", null, ". In particular, we focus on the geometric structure of applied model treated as a metric space showing the differences in the outputting computations if the changes in a metric are made. We also relate the function", null, "to Tissot's indicatrix of distortion.\n\nHow to Cite\nKopacz, P. (2012). On geometric properties of spherical conics and generalization of π in navigation and mapping. Geodesy and Cartography, 38(4), 141-151. https://doi.org/10.3846/20296991.2012.756995\nPublished in Issue\nDec 21, 2012\nAbstract Views\n122" ]

[ null, "http://crossmark-cdn.crossref.org/widget/v2.0/logos/CROSSMARK_Color_horizontal.svg", null, "https://www.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/tgac20/2012/tgac20.v038.i04/20296991.2012.756995/20130124-01/images/medium/tgac_a_756995_o_ilf0001.gif", null, "https://www.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/tgac20/2012/tgac20.v038.i04/20296991.2012.756995/20130124-01/images/medium/tgac_a_756995_o_ilf0002.gif", null, "https://www.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/tgac20/2012/tgac20.v038.i04/20296991.2012.756995/20130124-01/images/medium/tgac_a_756995_o_ilf0003.gif", null, "https://www.tandfonline.com/na101/home/literatum/publisher/tandf/journals/content/tgac20/2012/tgac20.v038.i04/20296991.2012.756995/20130124-01/images/medium/tgac_a_756995_o_ilf0004.gif", null ]

[ "Lemma 103.5.4. Let $\\mathcal{X}$ be an algebraic stack. Let $E$ be an object of $D_{\\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$. There exists a canonical distinguished triangle\n\n$E' \\to E \\to P \\to E'$\n\nin $D_{\\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$ such that $P$ is in $D_{\\textit{Parasitic} \\cap \\textit{LQCoh}^{fbc}} (\\mathcal{O}_\\mathcal {X})$ and\n\n$\\mathop{\\mathrm{Hom}}\\nolimits _{D(\\mathcal{O}_\\mathcal {X})}(E', P') = 0$\n\nfor all $P'$ in $D_{\\textit{Parasitic} \\cap \\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$.\n\nProof. Consider the morphism of ringed topoi $g : \\mathop{\\mathit{Sh}}\\nolimits (\\mathcal{X}_{flat, fppf}) \\longrightarrow \\mathop{\\mathit{Sh}}\\nolimits (\\mathcal{X}_{fppf})$ studied in Cohomology of Stacks, Section 102.14. Set $E' = Lg_!g^*E$ and let $P$ be the cone on the adjunction map $E' \\to E$, see Lemma 103.3.1 part (4). By Lemma 103.5.3 parts (2)(a) and (2)(c) we have that $E'$ is in $D_{\\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$. Hence also $P$ is in $D_{\\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$. The map $g^*E' \\to g^*E$ is an isomorphism as $g^*Lg_! = \\text{id}$ by Lemma 103.3.1 part (4). Hence $g^*P = 0$ and whence $P$ is an object of $D_{\\textit{Parasitic} \\cap \\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$ by Lemma 103.5.3 part (2)(b). Finally, for $P'$ in $D_{\\textit{Parasitic} \\cap \\textit{LQCoh}^{fbc}}(\\mathcal{O}_\\mathcal {X})$ we have\n\n$\\mathop{\\mathrm{Hom}}\\nolimits (E', P') = \\mathop{\\mathrm{Hom}}\\nolimits (Lg_!g^*E, P') = \\mathop{\\mathrm{Hom}}\\nolimits (g^*E, g^*P') = 0$\n\nas $g^*P' = 0$ by Lemma 103.5.3 part (2)(b). The distinguished triangle $E' \\to E \\to P \\to E'$ is canonical (more precisely unique up to isomorphism of triangles induces the identity on $E$) by the discussion in Derived Categories, Section 13.40. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]

[ null ]

[ "", null, "# Milliamp-hours to Watt-hours conversion\n\n## Enter the electric charge in milliamp-hours (mAh) and voltage in volts (V) and press the Calculate button:\n\n Milliamp-hours: mAh Volts: V Result (watt-hours): Wh\n Milliamp-hours to Watt-hours calculation formulaThe energy E(Wh) in watt-hours is equal to the electric charge Q(mAh) in milliamp-hours times the voltage V(V) in volts (V) divided by 1000: E(Wh) = Q(mAh) * V(V) / 1000Therefore watt-hours is equal to milliamp-hours times volts divided by 1000:Watt-hours = Milliamp-hours * Volts / 1000orWh = mAh * V / 1000" ]

[ null, "https://simplewebtool.com/assets/logov2.png", null ]

[ "", null, "rounding and estimating worksheet preview to nearest hundred worksheets 10 100 1000.\n\nrounding to the nearest hundred 10 and 100 worksheet pdf dollar worksheets off,rounding to the nearest ten and hundred worksheets best 5 cents 100 10 000,rounding to the nearest 10 100 and 1000 worksheets pdf off thousand worksheet math,rounding to nearest cent worksheets round the math grade practice 10 pdf and 100 worksheet,rounding off numbers to the nearest 10 worksheets and 100 worksheet pdf tenths place math estimating,rounding to the nearest hundred worksheet beautiful worksheets off numbers thousands ks2 10 and 100 5 cents,rounding decimals to the nearest 10 100 and 1000 worksheets numbers 10000 how do you round math estimating prime 3rd grade,rounding off numbers to the nearest hundred thousands worksheets 10 pdf 1000 practice decimals printable math grade,rounding off to the nearest 10 100 and 1000 worksheets pdf grade 4 luxury collection of math word problems,rounding to the nearest hundred ks2 10 and 100 worksheets 3rd grade ten thousand worksheet." ]

[ null, "http://homeofficelove.com/wp-content/uploads/2019/10/rounding-and-estimating-worksheet-preview-to-nearest-hundred-worksheets-10-100-1000.jpg", null ]

[ "Journal of Technical\nPhysics\n\nA QUARTERLY JOURNAL", null, "Previous issue vol. 47, no. 2 (2006) vol. 47, no. 3 (2006)", null, "Next issue", null, "vol. 47, no. 4 (2006)\n\n### Contents of issue 3, vol. 47\n\n1. B. Singh: Dispersion relations in a generalized monoclinic thermoelastic solid half-space\n2. H.A. Attia: Hydromagnetic flow of a dusty fluid between parallel porous plates, considering the ion slip\n3. M. Sztyren: Solutions for the higher-grade hybrid model of layered superconductors\n4. J.P. Vishwakarma, G. Nath: Converging detonation waves in a dusty gas\n\nB. Singh: Dispersion relations in a generalized monoclinic thermoelastic solid half-space\nThe plane wave propagation in a homogeneous, monoclinic, generalized thermoelastic solid medium with thermal relaxations is studied. Three types of plane waves: quasi-P (qP), quasi-thermal and quasi-SV (qSV) waves, are shown to exist. The analytical expressions for their dimensional velocities of propagation are obtained. The velocities of these waves are found to depend on the angle of propagation. Angles of reflection for plane waves are found to be different from their angles of incidence. Numerical computations are carried out to show the variations of velocities of plane waves for different parameters. Effects of anisotropy, frequency and thermoelastic coupling coefficient on wave propagation are observed.\n\nContents", null, "H.A. Attia: Hydromagnetic flow of a dusty fluid between parallel porous plates, considering the ion slip\nIn the present study, the transient hydromagnetic flow between parallel porous plates of a dusty viscous, incompressible, electrically conducting fluid under the influence of a constant pressure gradient is studied, considering the ion slip. The parallel plates are assumed to be porous and subjected to uniform suction from above and injection from below, while the fluid is acted upon by an external uniform magnetic field applied perpendicular to the plates. The equations of motion are solved analytically to yield the velocity distributions for both the fluid and dust particles as functions of time and space.\n\nContents", null, "M. Sztyren: Solutions for the higher-grade hybrid model of layered superconductors\nA class of basic solutions for a model of layered superconductors is constructed. The considerations are based on the higher-grade hybrid model, constructed in the previous paper. In particular, for any range K of the Josephson interlayer couplings, we determine the spectrum of excited states with respect to all types of ground states calculated there for an infinite uniform superconductor. We discuss also solutions for a finite number of layers with periodic boundary conditions. In this way, instances of the amplitude-modulated solutions for the infinite stack of layers are found.\n\nContents", null, "J.P. Vishwakarma, G. Nath: Converging detonation waves in a dusty gas\nThe problem of converging detonation waves propagating through a dusty gas and releasing a constant amount of energy per unit mass is studied. The dusty gas is assumed to be a mixture of small solid particles and a perfect gas. The solid particles are distributed continuously in the perfect gas. An expression for the similarity exponent is determined for the detonation wave travelling in the Chapman-Jouguet regime. Also, the variation of the flow variables just behind the detonation front with the radius of the front are obtained. It is found that the presence of dust particles in the medium has a significant effect on the similarity exponent and the flow variables.\n\nContents", null, "Reprints of the full papers may be obtained from their authors. Contact Editorial Office in case you need the address of the respective author.\n Home Editors Scope Abstracts For authors Subscription" ]

[ null, "http://jtp.ippt.gov.pl/images/left.gif", null, "http://jtp.ippt.gov.pl/images/down.gif", null, "http://jtp.ippt.gov.pl/images/right.gif", null, "http://jtp.ippt.gov.pl/images/up.gif", null, "http://jtp.ippt.gov.pl/images/up.gif", null, "http://jtp.ippt.gov.pl/images/up.gif", null, "http://jtp.ippt.gov.pl/images/up.gif", null ]

[ "# Notice bibliographique\n\n• Notice\n\nType(s) de contenu et mode(s) de consultation : Texte : électronique\n\nAuteur(s) :\n\nTitre(s) : Jacobi forms, finite quadratic modules and Weil representations over number fields [Texte électronique] / Hatice Boylan\n\nPublication : . - Cham : Springer, cop. 2015\n\nDescription matérielle : 1 online resource\n\nCollection : (Lecture Notes in Mathematics ; 2130)\n\nNote(s) : Includes bibliographical references\nThe new theory of Jacobi forms over totally real number fields introduced in this monograph is expected to give further insight into the arithmetic theory of Hilbert modular forms, its L-series, and into elliptic curves over number fields. This work is inspired by the classical theory of Jacobi forms over the rational numbers, which is an indispensable tool in the arithmetic theory of elliptic modular forms, elliptic curves, and in many other disciplines in mathematics and physics. Jacobi forms can be viewed as vector valued modular forms which take values in so-called Weil representations. Accordingly, the first two chapters develop the theory of finite quadratic modules and associated Weil representations over number fields. This part might also be interesting for those who are merely interested in the representation theory of Hilbert modular groups. One of the main applications is the complete classification of Jacobi forms of singular weight over an arbitrary totally real number field\n\nIndice(s) Dewey :\n\nNuméros : ISBN 9783319129167\n\nNotice n° :  FRBNF44678096 (notice reprise d'un réservoir extérieur)\n\nTable des matières : Introduction ; Notations ; Finite Quadratic Modules ; Weil Representations of Finite Quadratic Modules ; Jacobi Forms over Totally Real Number Fields ; Singular Jacobi Forms ; Tables ; Glossary.\n\n## Localiser ce document(1 Exemplaire)\n\n### 1 partie d'exemplaire regroupée\n\nACQNUM-59862\nsupport : document électronique dématérialisé" ]

[ null ]

[ "", null, "GeodesicEquations - Maple Help\n\nTensor[GeodesicEquations] - calculate the geodesic equations for a symmetric linear connection on the tangent bundle\n\nCalling Sequences\n\nGeodesicEquations (C, Gamma, t)\n\nParameters\n\nC       - a list of functions of a single variable, defining the components of a curve on a manifold M with respect to a given system of coordinates\n\nGamma   - a connection on the tangent bundle to a manifold M\n\nt       - the curve parameter", null, "Description\n\n • Let $M$ be a manifold and let $\\nabla$be a symmetric linear connection on the tangent bundle of $M$. If $C$ is a curve in $M$ with tangent vector $T$, then the geodesic equations for $C$ with respect to the connection $\\nabla$ is the system of second order ODEs defined by ${\\nabla }_{T}T=0$.\n • The procedure GeodesicEquations(C, Gamma, t) returns the vector ${\\nabla }_{T}T$.\n • This command is part of the DifferentialGeometry:-Tensor package, and so can be used in the form GeodesicEquations(...) only after executing the command with(DifferentialGeometry) and with(Tensor) in that order.  It can always be used in the long form DifferentialGeometry:-Tensor:-GeodesicEquations.", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DifferentialGeometry}\\right):$$\\mathrm{with}\\left(\\mathrm{Tensor}\\right):$\n\nExample 1.\n\nFirst create a 2-dimensional manifold $M$ and define a connection on the tangent space of $M$.\n\n > $\\mathrm{DGsetup}\\left(\\left[x,y\\right],M\\right)$\n ${\\mathrm{frame name: M}}$ (2.1)\n M > $\\mathrm{Gamma}≔\\mathrm{Connection}\\left(a\\left(\\left(\\mathrm{D_x}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dx}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dy}+\\left(\\mathrm{D_x}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dy}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dx}\\right)+by\\left(\\mathrm{D_x}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dy}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}&t\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{dy}\\right)$\n ${\\mathrm{Γ}}{:=}{a}{}{\\mathrm{D_x}}{}{\\mathrm{dx}}{}{\\mathrm{dy}}{+}{a}{}{\\mathrm{D_x}}{}{\\mathrm{dy}}{}{\\mathrm{dx}}{+}{b}{}{y}{}{\\mathrm{D_x}}{}{\\mathrm{dy}}{}{\\mathrm{dy}}$ (2.2)\n\nTo determine the geodesic equations for this connection we first define a curve on $M$ by specifying a list of functions of a single variable $t$.\n\n M > $C≔\\left[x\\left(t\\right),y\\left(t\\right)\\right]$\n ${C}{:=}\\left[{x}{}\\left({t}\\right){,}{y}{}\\left({t}\\right)\\right]$ (2.3)\n\nThe program GeodesicEquations returns a vector whose components are the components of the geodesic equations.\n\n M > $V≔\\mathrm{GeodesicEquations}\\left(C,\\mathrm{Gamma},t\\right)$\n ${V}{:=}\\left({\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{y}{}\\left({t}\\right)\\right)}^{{2}}{}{b}{}{y}{}\\left({t}\\right){+}{2}{}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{x}{}\\left({t}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{y}{}\\left({t}\\right)\\right){}{a}{+}\\frac{{ⅆ}}{{ⅆ}{t}}{}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{x}{}\\left({t}\\right)\\right)\\right){}{\\mathrm{D_x}}{+}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{y}{}\\left({t}\\right)\\right)\\right){}{\\mathrm{D_y}}$ (2.4)\n\nTo solve these geodesic equations use DGinfo to obtain the coefficients of $V$ as a list. Pass the result to dsolve to solve this system of 2 second order ODEs. See also DGsolve.\n\n M > $\\mathrm{DE}≔\\mathrm{Tools}:-\\mathrm{DGinfo}\\left(V,\"CoefficientSet\"\\right)$\n ${\\mathrm{DE}}{:=}\\left\\{{\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{y}{}\\left({t}\\right)\\right)}^{{2}}{}{b}{}{y}{}\\left({t}\\right){+}{2}{}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{x}{}\\left({t}\\right)\\right){}\\left(\\frac{{ⅆ}}{{ⅆ}{t}}{}{y}{}\\left({t}\\right)\\right){}{a}{+}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}{}{x}{}\\left({t}\\right){,}\\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}{}{y}{}\\left({t}\\right)\\right\\}$ (2.5)\n M > $\\mathrm{simplify}\\left(\\mathrm{dsolve}\\left(\\mathrm{DE},\\mathrm{explicit}\\right)\\right)$\n $\\left\\{{x}{}\\left({t}\\right){=}{-}\\frac{{1}}{{4}}{}\\frac{{2}{}{b}{}{{\\mathrm{_C3}}}^{{2}}{}{a}{}{\\mathrm{_C4}}{}{t}{-}{b}{}{{\\mathrm{_C3}}}^{{2}}{}{t}{+}{b}{}{{\\mathrm{_C3}}}^{{3}}{}{a}{}{{t}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{2}{}{a}{}\\left({\\mathrm{_C3}}{}{t}{+}{\\mathrm{_C4}}\\right)}{}{a}{}{\\mathrm{_C1}}{-}{4}{}{\\mathrm{_C2}}{}{\\mathrm{_C3}}{}{{a}}^{{2}}}{{\\mathrm{_C3}}{}{{a}}^{{2}}}{,}{y}{}\\left({t}\\right){=}{\\mathrm{_C3}}{}{t}{+}{\\mathrm{_C4}}\\right\\}$ (2.6)", null, "See Also" ]

[ null, "https://bat.bing.com/action/0", null, "https://jp.maplesoft.com/support/help/arrow_down.gif", null, "https://jp.maplesoft.com/support/help/arrow_down.gif", null, "https://jp.maplesoft.com/support/help/arrow_down.gif", null ]

[ "# 23 Measuring performance\n\n## 23.1 Introduction\n\nProgrammers waste enormous amounts of time thinking about, or worrying about, the speed of noncritical parts of their programs, and these attempts at efficiency actually have a strong negative impact when debugging and maintenance are considered.\n\n— Donald Knuth.\n\nBefore you can make your code faster, you first need to figure out what’s making it slow. This sounds easy, but it’s not. Even experienced programmers have a hard time identifying bottlenecks in their code. So instead of relying on your intuition, you should profile your code: measure the run-time of each line of code using realistic inputs.\n\nOnce you’ve identified bottlenecks you’ll need to carefully experiment with alternatives to find faster code that is still equivalent. In Chapter 24 you’ll learn a bunch of ways to speed up code, but first you need to learn how to microbenchmark so that you can precisely measure the difference in performance.\n\n### Outline\n\n• Section 23.2 shows you how to use profiling tools to dig into exactly what is making code slow.\n\n• Section 23.3 shows how to use microbenchmarking to explore alternative implementations and figure out exactly which one is fastest.\n\n### Prerequisites\n\nWe’ll use profvis for profiling, and bench for microbenchmarking.\n\nlibrary(profvis)\nlibrary(bench)\n\n## 23.2 Profiling\n\nAcross programming languages, the primary tool used to understand code performance is the profiler. There are a number of different types of profilers, but R uses a fairly simple type called a sampling or statistical profiler. A sampling profiler stops the execution of code every few milliseconds and records the call stack (i.e. which function is currently executing, and the function that called the function, and so on). For example, consider f(), below:\n\nf <- function() {\npause(0.1)\ng()\nh()\n}\ng <- function() {\npause(0.1)\nh()\n}\nh <- function() {\npause(0.1)\n}\n\n(I use profvis::pause() instead of Sys.sleep() because Sys.sleep() does not appear in profiling outputs because as far as R can tell, it doesn’t use up any computing time.)\n\nIf we profiled the execution of f(), stopping the execution of code every 0.1 s, we’d see a profile like this:\n\n\"pause\" \"f\"\n\"pause\" \"g\" \"f\"\n\"pause\" \"h\" \"g\" \"f\"\n\"pause\" \"h\" \"f\"\n\nEach line represents one “tick” of the profiler (0.1 s in this case), and function calls are recorded from right to left: the first line shows f() calling pause(). It shows that the code spends 0.1 s running f(), then 0.2 s running g(), then 0.1 s running h().\n\nIf we actually profile f(), using utils::Rprof() as in the code below, we’re unlikely to get such a clear result.\n\ntmp <- tempfile()\nRprof(tmp, interval = 0.1)\nf()\nRprof(NULL)\n#> sample.interval=100000\n#> \"pause\" \"g\" \"f\"\n#> \"pause\" \"h\" \"g\" \"f\"\n#> \"pause\" \"h\" \"f\" \n\nThat’s because all profilers must make a fundamental trade-off between accuracy and performance. The compromise that makes, using a sampling profiler, only has minimal impact on performance, but is fundamentally stochastic because there’s some variability in both the accuracy of the timer and in the time taken by each operation. That means each time that you profile you’ll get a slightly different answer. Fortunately, the variability most affects functions that take very little time to run, which are also the functions of least interest.\n\n### 23.2.1 Visualising profiles\n\nThe default profiling resolution is quite small, so if your function takes even a few seconds it will generate hundreds of samples. That quickly grows beyond our ability to look at directly, so instead of using utils::Rprof() we’ll use the profvis package to visualise aggregates. profvis also connects profiling data back to the underlying source code, making it easier to build up a mental model of what you need to change. If you find profvis doesn’t help for your code, you might try one of the other options like utils::summaryRprof() or the proftools package (Tierney and Jarjour 2016).\n\nThere are two ways to use profvis:\n\n• From the Profile menu in RStudio.\n\n• With profvis::profvis(). I recommend storing your code in a separate file and source()ing it in; this will ensure you get the best connection between profiling data and source code.\n\nsource(\"profiling-example.R\")\nprofvis(f())\n\nAfter profiling is complete, profvis will open an interactive HTML document that allows you to explore the results. There are two panes, as shown in Figure 23.1.", null, "Figure 23.1: profvis output showing source on top and flame graph below.\n\nThe top pane shows the source code, overlaid with bar graphs for memory and execution time for each line of code. Here I’ll focus on time, and we’ll come back to memory shortly. This display gives you a good overall feel for the bottlenecks but doesn’t always help you precisely identify the cause. Here, for example, you can see that h() takes 150 ms, twice as long as g(); that’s not because the function is slower, but because it’s called twice as often.\n\nThe bottom pane displays a flame graph showing the full call stack. This allows you to see the full sequence of calls leading to each function, allowing you to see that h() is called from two different places. In this display you can mouse over individual calls to get more information, and see the corresponding line of source code, as in Figure 23.2.", null, "Figure 23.2: Hovering over a call in the flamegraph highlights the corresponding line of code, and displays additional information about performance.\n\nAlternatively, you can use the data tab, Figure 23.3 lets you interactively dive into the tree of performance data. This is basically the same display as the flame graph (rotated 90 degrees), but it’s more useful when you have very large or deeply nested call stacks because you can choose to interactively zoom into only selected components.", null, "Figure 23.3: The data gives an interactive tree that allows you to selectively zoom into key components\n\n### 23.2.2 Memory profiling\n\nThere is a special entry in the flame graph that doesn’t correspond to your code: <GC>, which indicates that the garbage collector is running. If <GC> is taking a lot of time, it’s usually an indication that you’re creating many short-lived objects. For example, take this small snippet of code:\n\nx <- integer()\nfor (i in 1:1e4) {\nx <- c(x, i)\n}\n\nIf you profile it, you’ll see that most of the time is spent in the garbage collector, Figure 23.4.", null, "Figure 23.4: Profiling a loop that modifies an existing variable reveals that most time is spent in the garbage collector ().\n\nWhen you see the garbage collector taking up a lot of time in your own code, you can often figure out the source of the problem by looking at the memory column: you’ll see a line where large amounts of memory are being allocated (the bar on the right) and freed (the bar on the left). Here the problem arises because of copy-on-modify (Section 2.3): each iteration of the loop creates another copy of x. You’ll learn strategies to resolve this type of problem in Section 24.6.\n\n### 23.2.3 Limitations\n\nThere are some other limitations to profiling:\n\n• Profiling does not extend to C code. You can see if your R code calls C/C++ code but not what functions are called inside of your C/C++ code. Unfortunately, tools for profiling compiled code are beyond the scope of this book; start by looking at https://github.com/r-prof/jointprof.\n\n• If you’re doing a lot of functional programming with anonymous functions, it can be hard to figure out exactly which function is being called. The easiest way to work around this is to name your functions.\n\n• Lazy evaluation means that arguments are often evaluated inside another function, and this complicates the call stack (Section 7.5.2). Unfortunately R’s profiler doesn’t store enough information to disentangle lazy evaluation so that in the following code, profiling would make it seem like i() was called by j() because the argument isn’t evaluated until it’s needed by j().\n\ni <- function() {\npause(0.1)\n10\n}\nj <- function(x) {\nx + 10\n}\nj(i())\n\nIf this is confusing, use force() (Section 10.2.3) to force computation to happen earlier.\n\n### 23.2.4 Exercises\n\n1. Profile the following function with torture = TRUE. What is surprising? Read the source code of rm() to figure out what’s going on.\n\nf <- function(n = 1e5) {\nx <- rep(1, n)\nrm(x)\n}\n\n## 23.3 Microbenchmarking\n\nA microbenchmark is a measurement of the performance of a very small piece of code, something that might take milliseconds (ms), microseconds (µs), or nanoseconds (ns) to run. Microbenchmarks are useful for comparing small snippets of code for specific tasks. Be very wary of generalising the results of microbenchmarks to real code: the observed differences in microbenchmarks will typically be dominated by higher-order effects in real code; a deep understanding of subatomic physics is not very helpful when baking.\n\nA great tool for microbenchmarking in R is the bench package (Hester 2018). The bench package uses a a high precision timer, making it possible to compare operations that only take a tiny amount of time. For example, the following code compares the speed of two approaches to computing a square root.\n\nx <- runif(100)\n(lb <- bench::mark(\nsqrt(x),\nx ^ 0.5\n))\n#> # A tibble: 2 x 6\n#> expression min median itr/sec mem_alloc gc/sec\n#> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl>\n#> 1 sqrt(x) 324.1ns 524.1ns 1869752. 848B 0\n#> 2 x^0.5 6.42µs 6.61µs 140005. 848B 0\n\nBy default, bench::mark() runs each expression at least once (min_iterations = 1), and at most enough times to take 0.5 s (min_time = 0.5). It checks that each run returns the same value which is typically what you want microbenchmarking; if you want to compare the speed of expressions that return different values, set check = FALSE.\n\n### 23.3.1bench::mark() results\n\nbench::mark() returns the results as a tibble, with one row for each input expression, and the following columns:\n\n• min, mean, median, max, and itr/sec summarise the time taken by the expression. Focus on the minimum (the best possible running time) and the median (the typical time). In this example, you can see that using the special purpose sqrt() function is faster than the general exponentiation operator.\n\nYou can visualise the distribution of the individual timings with plot():\n\nplot(lb)\n#> Loading required namespace: tidyr", null, "The distribution tends to be heavily right-skewed (note that the x-axis is already on a log scale!), which is why you should avoid comparing means. You’ll also often see multimodality because your computer is running something else in the background.\n\n• mem_alloc tells you the amount of memory allocated by the first run, and n_gc() tells you the total number of garbage collections over all runs. These are useful for assessing the memory usage of the expression.\n\n• n_itr and total_time tells you how many times the expression was evaluated and how long that took in total. n_itr will always be greater than the min_iteration parameter, and total_time will always be greater than the min_time parameter.\n\n• result, memory, time, and gc are list-columns that store the raw underlying data.\n\nBecause the result is a special type of tibble, you can use [ to select just the most important columns. I’ll do that frequently in the next chapter.\n\nlb[c(\"expression\", \"min\", \"median\", \"itr/sec\", \"n_gc\")]\n#> # A tibble: 2 x 4\n#> expression min median itr/sec\n#> <bch:expr> <bch:tm> <bch:tm> <dbl>\n#> 1 sqrt(x) 324.1ns 524.1ns 1869752.\n#> 2 x^0.5 6.42µs 6.61µs 140005.\n\n### 23.3.2 Interpreting results\n\nAs with all microbenchmarks, pay careful attention to the units: here, each computation takes about 320 ns, 320 billionths of a second. To help calibrate the impact of a microbenchmark on run time, it’s useful to think about how many times a function needs to run before it takes a second. If a microbenchmark takes:\n\n• 1 ms, then one thousand calls take a second.\n• 1 µs, then one million calls take a second.\n• 1 ns, then one billion calls take a second.\n\nThe sqrt() function takes about 320 ns, or 0.32 µs, to compute the square roots of 100 numbers. That means if you repeated the operation a million times, it would take 0.32 s, and hence changing the way you compute the square root is unlikely to significantly affect real code. This is the reason you need to exercise care when generalising microbenchmarking results.\n\n### 23.3.3 Exercises\n\n1. Instead of using bench::mark(), you could use the built-in function system.time(). But system.time() is much less precise, so you’ll need to repeat each operation many times with a loop, and then divide to find the average time of each operation, as in the code below.\n\nn <- 1e6\nsystem.time(for (i in 1:n) sqrt(x)) / n\nsystem.time(for (i in 1:n) x ^ 0.5) / n\n\nHow do the estimates from system.time() compare to those from bench::mark()? Why are they different?\n\n2. Here are two other ways to compute the square root of a vector. Which do you think will be fastest? Which will be slowest? Use microbenchmarking to test your answers.\n\nx ^ (1 / 2)\nexp(log(x) / 2)\n\n### References\n\nHester, Jim. 2018. Bench: High Precision Timing of R Expressions. http://bench.r-lib.org/.\n\nTierney, Luke, and Riad Jarjour. 2016. Proftools: Profile Output Processing Tools for R. https://CRAN.R-project.org/package=proftools." ]

[ null, "https://d33wubrfki0l68.cloudfront.net/bd1db76eaff4687ee121ab591d0a9df18597c2bc/e3a75/screenshots/performance/flamegraph.png", null, "https://d33wubrfki0l68.cloudfront.net/d6c159d608643e334c560a5a14802472541fe071/b83ad/screenshots/performance/info.png", null, "https://d33wubrfki0l68.cloudfront.net/de46e2b1c1cb4e90cb097cddca26d8f8699cd0a2/b42d4/screenshots/performance/tree.png", null, "https://d33wubrfki0l68.cloudfront.net/8e5b9f59675b95a1fdf2f26ecae034e77f98df58/0bd51/screenshots/performance/memory.png", null, "https://d33wubrfki0l68.cloudfront.net/ca4edda40267c4254d7596c920323ee6d9018233/b47cb/perf-measure_files/figure-html/unnamed-chunk-9-1.png", null ]

[ "cs.DC\n\n# Title: A Subquadratic-Time Distributed Algorithm for Exact Maximum Matching\n\nAbstract: For a graph G=(V,E), finding a set of disjoint edges that do not share any vertices is called a matching problem, and finding the maximum matching is a fundamental problem in the theory of distributed graph algorithms. Although local algorithms for the approximate maximum matching problem have been widely studied, exact algorithms has not been much studied. In fact, no exact maximum matching algorithm that is faster than the trivial upper bound of O(n^2) rounds is known for the general instance. In this paper, we propose a randomized O(s_{max}^{3/2}+log n)-round algorithm in the CONGEST model, where s_{\\max} is the size of maximum matching. This is the first exact maximum matching algorithm in o(n^2) rounds for general instances in the CONGEST model. The key technical ingredient of our result is a distributed algorithms of finding an augmenting path in O(s_{\\max}) rounds, which is based on a novel technique of constructing a sparse certificate of augmenting paths, which is a subgraph of the input graph preserving at least one augmenting path. To establish a highly parallel construction of sparse certificates, we also propose a new characterization of sparse certificates, which might also be of independent interest.\n Comments: 19 pages, 4figures Subjects: Distributed, Parallel, and Cluster Computing (cs.DC) DOI: 10.1587/transinf.2021EDP7083 Cite as: arXiv:2104.12057 [cs.DC] (or arXiv:2104.12057v1 [cs.DC] for this version)\n\n## Submission history\n\nFrom: Naoki Kitamura [view email]\n[v1] Sun, 25 Apr 2021 04:28:30 GMT (623kb)\n\nLink back to: arXiv, form interface, contact." ]

[ null ]

[ "# [Haskell-beginners] how to specify type class in type of function?\n\ngeorge young georgeryoung at gmail.com\nThu Jan 28 16:06:36 EST 2010\n\n```[complete newbie using ghci 6.10.4\nI understand that I should indicate that 'a' is in class Ord, since I use\nlessthan. But what is the syntax? Other style suggestions are welcome.\n\nmysort :: [a] -> [a]\nmysort [] = []\nmysort [x] = [x]\nmysort [x, y] | (x <= y) = [x, y]\n| otherwise = [y, x]\nmysort l =\n(mysort s1) ++ (mysort s2)\nwhere\n(s1, s2) = splitAt (div (length l) 2) l\n\nCould not deduce (Ord a) from the context ()\narising from a use of `<=' at /home/gry/foo.hs:53:17-22\nPossible fix:\nadd (Ord a) to the context of the type signature for `mysort'\nIn the expression: (x <= y)\nIn a stmt of a pattern guard for\nthe definition of `mysort':\n(x <= y)\nIn the definition of `mysort':\nmysort [x, y]\n| (x <= y) = [x, y]\n| otherwise = [y, x]\n\n```" ]

[ null ]

[ "Sie sind auf Seite 1von 20\n\n# MEC3455\n\nSolid Mechanics\nTopic 2 Airy Stress Functions\nDr Bernard Chen\nClayton Campus\nBuilding 31/121\n\nProf. Soh Ai Kah\nMonash University Sunway\nCampus\n\nObjectives\nDerivation of the biharmonic P.D.E\nand the Airy stress function as a\nsolution (biharmonic function)\nEquations of equilibrium\nCompatibility equations\nBoundary conditions\nExample questions\nIntroduction\nAnalogy: In fluid dynamics,\nthe Navier-Stokes equations\ndescribe the motion of fluid at\na point in space in time.\nThe Airy stress function,\n(, ) describes the state of\nstress (normal and shear) at\nany point in a structure under\nTherefore, the Airy stress\nfunction is dependent on\nboundary conditions\nMapping of this topic\nEquations of\nequilibrium,\ni.e. = 0\nCompatibility equations\ni.e. strain-displacement\nrelationships\nBiharmonic partial differential equation (P.D.E)\n\n4\n= 0\nwhere\n\n4\n=\n\n4\n\n4\n+2\n\n4\n\n2\n+\n\n4\n\n4\n\nThe Airy stress function, (, ), is the solution\nwhere\n\n=\n\n2\n\n=\n\n2\n\n=\n\n2\n\n2\n\nMapping of this topic\nBiharmonic P.D.E.\n\n4\n= 0\nwhere\n\n4\n=\n\n4\n\n4\n+2\n\n4\n\n2\n+\n\n4\n\n4\n\nThe Airy stress function, (, ), is the solution\nwhere\n\n=\n\n2\n\n=\n\n2\n\n=\n\n2\n\n2\n\nThis is the equation\nthat we need to satisfy,\nfor an Airy stress\nfunction to exist!\nThe unique Airy stress function (solution) depends\non boundary conditions (loadings on the structure)\nConsider a small element of dimensions dx, dy and with\nunity thickness (thickness of 1)\n\nx\no\ny\ndy\nx\ndx\ndx\nx\nx\nx\nc\nc\n+\no\no\ny\no xy\nt\ndy\ny\ny\ny\nc\nc\n+\no\no\ndy\ny\nxy\nxy\nc\nc\n+\nt\nt\ndx\nx\nxy\nxy\nc\nc\n+\nt\nt\nEquations of equilibrium\nForce balance equations for the element, with body\nforce included.\n0 1 . . 1 . . 1 . . 1 . . 1 .\n0\n= +\n|\n|\n.\n|\n\n\\\n|\nc\nc\n+ +\n|\n.\n|\n\n\\\n|\nc\nc\n+\n=\n\ndy Xdx dx dx dy\ny\ndy dy dx\nx\nF\nxy\nxy\nxy x\nx\nx\nx\nt\nt\nt o\no\no\n0 1 . . 1 . . 1 . . 1 . . 1 .\n0\n= +\n|\n|\n.\n|\n\n\\\n|\nc\nc\n+ +\n|\n|\n.\n|\n\n\\\n|\nc\nc\n+\n=\n\ndy Ydx dy dy dx\nx\ndx dx dy\ny\nF\nxy\nxy\nxy y\ny\ny\ny\nt\nt\nt o\no\no\nEquations of equilibrium\nEqn (1)\nwhere and are components of the body\nforce,\n=\n2\n+\n2\n\nEqn (1) reduces to;\n0 = +\nc\nc\n+\nc\nc\nX\ny x\nxy\nx\nt\no\n0 = +\nc\nc\n+\nc\nc\nY\nx y\nxy y\nt o\nEqn (2)\nIn most engineering problems, the body forces and\nare either zero or obvious i.e. gravity\n\nEqn (2) are the\nEquations of\nEquilibrium for a 2-\ndimensional problem\n(plane stress)\nEquations of equilibrium\nAssume that the previous element is deformed by\ndisplacement , and in the , and directions\nrespectively.\nFor 2-D case, we shall ignore .\n\nu\nv\ndy\ny\nu\nu\nc\nc\n+\ndx\nx\nv\nv\nc\nc\n+\nNote:\nnormal movements (normal\nstrain) are represented by and\nrotational movements (shearing\nstrains) are represented by\n\nand\n\nNormal movement due\nto normal stress\nRotational movement\ndue to shear stress\ndistortion\nCompatibility equations\nUsing the engineering definitions of strain;\nx\nu\nx\nc\nc\n= c\nx\nv\ny\nu\nxy\nc\nc\n+\nc\nc\n=\ny\nv\ny\nc\nc\n= c\nNormal\nstrains\nShear strain\nNote that the three strain components,\n\nand\n\nare now defined by two deformations and .\n\nEqn\n(3a)\nEqn\n(3b)\nEqn\n(3c)\nCompatibility equations\nDifferentiate the Eqn (3a) and Eqn 3(b) twice wrt to\nand respectively, and Eqn 3(c) once wrt to\nand , we obtain\ny x x y\nxy y\nx\nc c\nc\n=\nc\nc\n+\nc\nc\nc\nc\n2\n2\n2\n2\n2\nThis is the differential Equation of Compatibility\nwhich must be satisfied to ensure that the\ndisplacements and are related to the strains in\neqn (3a 3c)\nEqn (4)\nCompatibility equations\nHow do we relate the Equation of\nEquilibrium to the Equation of\nCompatibility?\ni.e. stress to strain?\nHookes\nLaw!\nSince this is a 2D problem (plane), we will use the\n2D Hookes Law.\nAlso, with Hookes Law, we are assuming that the\nmaterial is in its elastic region (no plastic\ndeformation).\n( )\ny x x\nE\nuo o c =\n1\n( )\nx y y\nE\nuo o c =\n1\n( )\nxy\nxy\nxy\nE G\nt\nu\nt\n\n+\n= =\n1 2\nEqn\n(5)\nSubstituting Hookes law (Eqn (5)) into the Equation of\nCompatibility, Eqn (4) gives\n\n( ) ( ) ( )\ny x x y\nxy\nx y y x\nc c\nc\n+ =\nc\nc\n+\nc\nc\nt\nu uo o uo o\n2\n2\n2\n2\n2\n1 2 Eqn\n(6)\nHow do we relate the Equation of\nEquilibrium to the Equation of\nCompatibility?\ni.e. stress to strain?\nWhile differentiate the Equations of Equilibrium (Eqn (2a))\nwrt to and (Eqn (2b)) wrt and then adding the two\nresults gives\n\n2\n2\n2\n2\n2\n2\ny x y x\ny\nx\nxy\nc\nc\n\nc\nc\n=\nc c\nc o\no\nt\nEqn\n(7)\nNotice that by doing this, we have eliminated the three\nstrain components,\n\nand\n\n.\n\nBy substituting Eqn (7) into Eqn (6) (both equations\nwith stress terms only), this gives us\n( ) 0\n2\n2\n2\n2\n= +\n|\n|\n.\n|\n\n\\\n|\nc\nc\n+\nc\nc\ny x\ny x\no o\nEqn\n(8)\nEqn (8) is a combination of equilibrium and\ncompatibility\n1 equation, 2 unknowns\nTherefore, to solve this, the Airy stress\nfunction, (named after G.B. Airy, 1862) is\nintroduced\nHow do we relate the Equation of\nEquilibrium to the Equation of\nCompatibility?\ni.e. stress to strain?\nThe stresses can be written in terms of the Airy stress\nfunction as:\n2\n2\ny\nx\nc\nc\n=\n|\no\ny x\nxy\nc c\nc\n=\n|\nt\n2\n2\n2\nx\ny\nc\nc\n=\n|\no\nThus, substituting Eqn (9) back into Eqn (8) results in\nEqn (10), (with one parameter, - the Airy stress\nfunction).\nHow do we relate the Equation of\nEquilibrium to the Equation of\nCompatibility?\nEqn (9)\n0 2\n4\n4\n2 2\n4\n4\n4\n=\nc\nc\n+\nc c\nc\n+\nc\nc\ny y x x\n| | |\n0\n4\n= V |\nEqn (10)\nHow do we relate the Equation of\nEquilibrium to the Equation of\nCompatibility?\n0 2\n4\n4\n2 2\n4\n4\n4\n=\nc\nc\n+\nc c\nc\n+\nc\nc\ny y x x\n| | |\nThis is the Biharmonic Partial Differential Equation.\nOnly stress functions, (, ) which satisfy this equation\n(Eqn (10)) meet both equilibrium and compatibility\nrequirements.\n0\n4\n= V |\nYou will see that any in terms of polynomial function in\n(, ) less than power of 4 will satisfy Eqn (10)\nBUT is it the appropriate solution?\nSolution must satisfy boundary conditions\nEqn (10)\nHow do we relate the Equation of\nEquilibrium to the Equation of Compatibility?\nThe Airy stress function, (, ) determines the stress\nstate.\nOften we also want to determine how the structure\ndeforms with the applied load.\nTherefore, we can use Hookes law to determine the\ndeformation of the structure from the stress.\n= displacement in ; = displacement in ;\n( )\n( )\n\n=\nc\nc\n+\nc\nc\n=\n=\nc\nc\n=\n=\nc\nc\n=\nG x\nv\ny\nu\nE y\nv\nE x\nu\nxy\nxy\nx y y\ny x x\nt\n\nuo o c\nuo o c\n1\n1\nExample 1\nMEC3455-Airy Stress Function 19\nExample 2\nFor the uniformly loaded cantilever beam shown below, the compatible\nstress field was found to be\n\n) y y 3c (2c\n6I\nw\n\n) y x(c\n2I\nw\n\ny\n3I\nw\n)y 2c (5x\n10I\nw\n\n3 2 3\nz\ny\n2 2\nz\nxy\n3\nz\n2 2\nz\nx\n+ =\n=\n+ =\nVerify that this stress field satisfies equilibrium.\nMEC3455-Airy Stress Function 20\nEXAMPLE 3\nConsider a thin cantilever loaded as shown in the below figure. Assume\nthat the bending stress is expressed by\n\ny x\n2I\nP\nI\ny M\n2 z\nx\n= = o\nand o\nz\n= t\nxz\n= t\nyz\n= 0. Determine the stress components o\ny\nand t\nxy\nas\nfunctions of x and y." ]

[ null ]

[ "MySQL 5.5 Reference Manual  /  ...  /  Aggregate (GROUP BY) Function Descriptions\n\n### 12.17.1 Aggregate (GROUP BY) Function Descriptions\n\nThis section describes group (aggregate) functions that operate on sets of values.\n\nTable 12.21 Aggregate (GROUP BY) Functions\n\nName Description\n`AVG()` Return the average value of the argument\n`BIT_AND()` Return bitwise AND\n`BIT_OR()` Return bitwise OR\n`BIT_XOR()` Return bitwise XOR\n`COUNT()` Return a count of the number of rows returned\n`COUNT(DISTINCT)` Return the count of a number of different values\n`GROUP_CONCAT()` Return a concatenated string\n`MAX()` Return the maximum value\n`MIN()` Return the minimum value\n`STD()` Return the population standard deviation\n`STDDEV()` Return the population standard deviation\n`STDDEV_POP()` Return the population standard deviation\n`STDDEV_SAMP()` Return the sample standard deviation\n`SUM()` Return the sum\n`VAR_POP()` Return the population standard variance\n`VAR_SAMP()` Return the sample variance\n`VARIANCE()` Return the population standard variance\n\nUnless otherwise stated, group functions ignore `NULL` values.\n\nIf you use a group function in a statement containing no `GROUP BY` clause, it is equivalent to grouping on all rows. For more information, see Section 12.17.3, “MySQL Handling of GROUP BY”.\n\nFor numeric arguments, the variance and standard deviation functions return a `DOUBLE` value. The `SUM()` and `AVG()` functions return a `DECIMAL` value for exact-value arguments (integer or `DECIMAL`), and a `DOUBLE` value for approximate-value arguments (`FLOAT` or `DOUBLE`).\n\nThe `SUM()` and `AVG()` aggregate functions do not work with temporal values. (They convert the values to numbers, losing everything after the first nonnumeric character.) To work around this problem, convert to numeric units, perform the aggregate operation, and convert back to a temporal value. Examples:\n\n``````SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(time_col))) FROM tbl_name;\nSELECT FROM_DAYS(SUM(TO_DAYS(date_col))) FROM tbl_name;``````\n\nFunctions such as `SUM()` or `AVG()` that expect a numeric argument cast the argument to a number if necessary. For `SET` or `ENUM` values, the cast operation causes the underlying numeric value to be used.\n\nThe `BIT_AND()`, `BIT_OR()`, and `BIT_XOR()` aggregate functions perform bit operations. They require `BIGINT` (64-bit integer) arguments and return `BIGINT` values. Arguments of other types are converted to `BIGINT` and truncation might occur.\n\n• Returns the average value of `expr`. The `DISTINCT` option can be used to return the average of the distinct values of `expr`.\n\nIf there are no matching rows, `AVG()` returns `NULL`.\n\n``````mysql> SELECT student_name, AVG(test_score)\nFROM student\nGROUP BY student_name;``````\n• Returns the bitwise `AND` of all bits in `expr`. The calculation is performed with 64-bit (`BIGINT`) precision.\n\nIf there are no matching rows, `BIT_AND()` returns a neutral value (all bits set to 1).\n\n• Returns the bitwise `OR` of all bits in `expr`. The calculation is performed with 64-bit (`BIGINT`) precision.\n\nIf there are no matching rows, `BIT_OR()` returns a neutral value (all bits set to 0).\n\n• Returns the bitwise `XOR` of all bits in `expr`. The calculation is performed with 64-bit (`BIGINT`) precision.\n\nIf there are no matching rows, `BIT_XOR()` returns a neutral value (all bits set to 0).\n\n• Returns a count of the number of non-`NULL` values of `expr` in the rows retrieved by a `SELECT` statement. The result is a `BIGINT` value.\n\nIf there are no matching rows, `COUNT()` returns `0`.\n\n``````mysql> SELECT student.student_name,COUNT(*)\nFROM student,course\nWHERE student.student_id=course.student_id\nGROUP BY student_name;``````\n\n`COUNT(*)` is somewhat different in that it returns a count of the number of rows retrieved, whether or not they contain `NULL` values.\n\nFor transactional storage engines such as `InnoDB`, storing an exact row count is problematic. Multiple transactions may be occurring at the same time, each of which may affect the count.\n\n`InnoDB` does not keep an internal count of rows in a table because concurrent transactions might see different numbers of rows at the same time. Consequently, `SELECT COUNT(*)` statements only count rows visible to the current transaction.\n\nTo process a `SELECT COUNT(*)` statement, `InnoDB` scans an index of the table, which takes some time if the index is not entirely in the buffer pool. For a faster count, create a counter table and let your application update it according to the inserts and deletes it does. However, this method may not scale well in situations where thousands of concurrent transactions are initiating updates to the same counter table. If an approximate row count is sufficient, use `SHOW TABLE STATUS`.\n\n`InnoDB` handles ```SELECT COUNT(*)``` and `SELECT COUNT(1)` operations in the same way. There is no performance difference.\n\nFor `MyISAM` tables, `COUNT(*)` is optimized to return very quickly if the `SELECT` retrieves from one table, no other columns are retrieved, and there is no `WHERE` clause. For example:\n\n``mysql> SELECT COUNT(*) FROM student;``\n\nThis optimization only applies to `MyISAM` tables, because an exact row count is stored for this storage engine and can be accessed very quickly. `COUNT(1)` is only subject to the same optimization if the first column is defined as ```NOT NULL```.\n\n• Returns a count of the number of rows with different non-`NULL` `expr` values.\n\nIf there are no matching rows, `COUNT(DISTINCT)` returns `0`.\n\n``mysql> SELECT COUNT(DISTINCT results) FROM student;``\n\nIn MySQL, you can obtain the number of distinct expression combinations that do not contain `NULL` by giving a list of expressions. In standard SQL, you would have to do a concatenation of all expressions inside `COUNT(DISTINCT ...)`.\n\n• This function returns a string result with the concatenated non-`NULL` values from a group. It returns `NULL` if there are no non-`NULL` values. The full syntax is as follows:\n\n``````GROUP_CONCAT([DISTINCT] expr [,expr ...]\n[ORDER BY {unsigned_integer | col_name | expr}\n[ASC | DESC] [,col_name ...]]\n[SEPARATOR str_val])``````\n``````mysql> SELECT student_name,\nGROUP_CONCAT(test_score)\nFROM student\nGROUP BY student_name;``````\n\nOr:\n\n``````mysql> SELECT student_name,\nGROUP_CONCAT(DISTINCT test_score\nORDER BY test_score DESC SEPARATOR ' ')\nFROM student\nGROUP BY student_name;``````\n\nIn MySQL, you can get the concatenated values of expression combinations. To eliminate duplicate values, use the `DISTINCT` clause. To sort values in the result, use the `ORDER BY` clause. To sort in reverse order, add the `DESC` (descending) keyword to the name of the column you are sorting by in the `ORDER BY` clause. The default is ascending order; this may be specified explicitly using the `ASC` keyword. The default separator between values in a group is comma (`,`). To specify a separator explicitly, use `SEPARATOR` followed by the string literal value that should be inserted between group values. To eliminate the separator altogether, specify `SEPARATOR ''`.\n\nThe result is truncated to the maximum length that is given by the `group_concat_max_len` system variable, which has a default value of 1024. The value can be set higher, although the effective maximum length of the return value is constrained by the value of `max_allowed_packet`. The syntax to change the value of `group_concat_max_len` at runtime is as follows, where `val` is an unsigned integer:\n\n``SET [GLOBAL | SESSION] group_concat_max_len = val;``\n\nThe return value is a nonbinary or binary string, depending on whether the arguments are nonbinary or binary strings. The result type is `TEXT` or `BLOB` unless `group_concat_max_len` is less than or equal to 512, in which case the result type is `VARCHAR` or `VARBINARY`.\n\n• Returns the maximum value of `expr`. `MAX()` may take a string argument; in such cases, it returns the maximum string value. See Section 8.3.1, “How MySQL Uses Indexes”. The `DISTINCT` keyword can be used to find the maximum of the distinct values of `expr`, however, this produces the same result as omitting `DISTINCT`.\n\nIf there are no matching rows, `MAX()` returns `NULL`.\n\n``````mysql> SELECT student_name, MIN(test_score), MAX(test_score)\nFROM student\nGROUP BY student_name;``````\n\nFor `MAX()`, MySQL currently compares `ENUM` and `SET` columns by their string value rather than by the string's relative position in the set. This differs from how `ORDER BY` compares them.\n\n• Returns the minimum value of `expr`. `MIN()` may take a string argument; in such cases, it returns the minimum string value. See Section 8.3.1, “How MySQL Uses Indexes”. The `DISTINCT` keyword can be used to find the minimum of the distinct values of `expr`, however, this produces the same result as omitting `DISTINCT`.\n\nIf there are no matching rows, `MIN()` returns `NULL`.\n\n``````mysql> SELECT student_name, MIN(test_score), MAX(test_score)\nFROM student\nGROUP BY student_name;``````\n\nFor `MIN()`, MySQL currently compares `ENUM` and `SET` columns by their string value rather than by the string's relative position in the set. This differs from how `ORDER BY` compares them.\n\n• Returns the population standard deviation of `expr`. `STD()` is a synonym for the standard SQL function `STDDEV_POP()`, provided as a MySQL extension.\n\nIf there are no matching rows, `STD()` returns `NULL`.\n\n• Returns the population standard deviation of `expr`. `STDDEV()` is a synonym for the standard SQL function `STDDEV_POP()`, provided for compatibility with Oracle.\n\nIf there are no matching rows, `STDDEV()` returns `NULL`.\n\n• Returns the population standard deviation of `expr` (the square root of `VAR_POP()`). You can also use `STD()` or `STDDEV()`, which are equivalent but not standard SQL.\n\nIf there are no matching rows, `STDDEV_POP()` returns `NULL`.\n\n• Returns the sample standard deviation of `expr` (the square root of `VAR_SAMP()`.\n\nIf there are no matching rows, `STDDEV_SAMP()` returns `NULL`.\n\n• Returns the sum of `expr`. If the return set has no rows, `SUM()` returns `NULL`. The `DISTINCT` keyword can be used to sum only the distinct values of `expr`.\n\nIf there are no matching rows, `SUM()` returns `NULL`.\n\n• Returns the population standard variance of `expr`. It considers rows as the whole population, not as a sample, so it has the number of rows as the denominator. You can also use `VARIANCE()`, which is equivalent but is not standard SQL.\n\nIf there are no matching rows, `VAR_POP()` returns `NULL`.\n\n• Returns the sample variance of `expr`. That is, the denominator is the number of rows minus one.\n\nIf there are no matching rows, `VAR_SAMP()` returns `NULL`.\n\n• Returns the population standard variance of `expr`. `VARIANCE()` is a synonym for the standard SQL function `VAR_POP()`, provided as a MySQL extension.\n\nIf there are no matching rows, `VARIANCE()` returns `NULL`." ]

[ null ]

[ "Related Articles\nRelational Model in DBMS\n• Difficulty Level : Easy\n• Last Updated : 24 Dec, 2020\n\nRelational Model was proposed by E.F. Codd to model data in the form of relations or tables. After designing the conceptual model of Database using ER diagram, we need to convert the conceptual model in the relational model which can be implemented using any RDBMS languages like Oracle SQL, MySQL etc. So we will see what Relational Model is.\n\nWhat is Relational Model?\n\nRelational Model represents how data is stored in Relational Databases.  A relational database stores data in the form of relations (tables). Consider a relation STUDENT with attributes ROLL_NO, NAME, ADDRESS, PHONE and AGE shown in Table 1.\n\nSTUDENT\n\nIMPORTANT TERMINOLOGIES\n\n• Attribute: Attributes are the properties that define a relation. e.g.; ROLL_NO, NAME\n• Relation Schema: A relation schema represents name of the relation with its attributes. e.g.; STUDENT (ROLL_NO, NAME, ADDRESS, PHONE and AGE) is relation schema for STUDENT. If a schema has more than 1 relation, it is called Relational Schema.\n• Tuple: Each row in the relation is known as tuple. The above relation contains 4 tuples, one of which is shown as:\n• Relation Instance: The set of tuples of a relation at a particular instance of time is called as relation instance. Table 1 shows the relation instance of STUDENT at a particular time. It can change whenever there is insertion, deletion or updation in the database.\n• Degree: The number of attributes in the relation is known as degree of the relation. The STUDENT relation defined above has degree 5.\n• Cardinality: The number of tuples in a relation is known as cardinality. The STUDENT relation defined above has cardinality 4.\n• Column: Column represents the set of values for a particular attribute. The column ROLL_NO is extracted from relation STUDENT.\n• NULL Values: The value which is not known or unavailable is called NULL value. It is represented by blank space. e.g.; PHONE of STUDENT having ROLL_NO 4 is NULL.\n\nConstraints in Relational Model\n\nWhile designing Relational Model, we define some conditions which must hold for data present in database are called Constraints. These constraints are checked before performing any operation (insertion, deletion and updation) in database. If there is a violation in any of constrains, operation will fail.\n\nDomain Constraints: These are attribute level constraints. An attribute can only take values which lie inside the domain range. e.g,; If a constrains AGE>0 is applied on STUDENT relation, inserting negative value of AGE will result in failure.\n\nKey Integrity: Every relation in the database should have atleast one set of attributes which defines a tuple uniquely. Those set of attributes is called key. e.g.; ROLL_NO in STUDENT is a key. No two students can have same roll number. So a key has two properties:\n\n• It should be unique for all tuples.\n• It can’t have NULL values.\n\nReferential Integrity: When one attribute of a relation can only take values from other attribute of same relation or any other relation, it is called referential integrity. Let us suppose we have 2 relations\n\nSTUDENT\n\nBRANCH\n\nBRANCH_CODE of STUDENT can only take the values which are present in BRANCH_CODE of BRANCH which is called referential integrity constraint. The relation which is referencing to other relation is called REFERENCING RELATION (STUDENT in this case) and the relation to which other relations refer is called REFERENCED RELATION (BRANCH in this case).\n\nANOMALIES\n\nAn anomaly is an irregularity, or something which deviates from the expected or normal state. When designing databases, we identify three types of anomalies: Insert, Update and Delete.\n\nInsertion Anomaly in Referencing Relation:\n\nWe can’t insert a row in REFERENCING RELATION if referencing attribute’s value is not present in referenced attribute value. e.g.; Insertion of a student with BRANCH_CODE ‘ME’ in STUDENT relation will result in error because ‘ME’ is not present in BRANCH_CODE of BRANCH.\n\nDeletion/ Updation Anomaly in Referenced Relation:\n\nWe can’t delete or update a row from REFERENCED RELATION if value of REFERENCED ATTRIBUTE is used in value of REFERENCING ATTRIBUTE. e.g; if we try to delete tuple from BRANCH having BRANCH_CODE ‘CS’, it will result in error because ‘CS’ is referenced by BRANCH_CODE of STUDENT, but if we try to delete the row from BRANCH with BRANCH_CODE CV, it will be deleted as the value is not been used by referencing relation. It can be handled by following method:\n\nON DELETE CASCADE: It will delete the tuples from REFERENCING RELATION if  value used by REFERENCING ATTRIBUTE is deleted from REFERENCED RELATION. e.g;, if we delete a row from BRANCH with BRANCH_CODE ‘CS’, the rows in STUDENT relation with BRANCH_CODE CS (ROLL_NO 1 and 2 in this case) will be deleted.\n\nON UPDATE CASCADE: It will update the REFERENCING ATTRIBUTE in REFERENCING RELATION if attribute value used by REFERENCING ATTRIBUTE is updated in REFERENCED RELATION. e.g;, if we update a row from BRANCH with BRANCH_CODE ‘CS’ to ‘CSE’, the rows in STUDENT relation with BRANCH_CODE CS (ROLL_NO 1 and 2 in this case) will be updated with BRANCH_CODE ‘CSE’.\n\nSUPER KEYS:\nAny set of attributes that allows us to identify unique rows (tuples) in a given relation are known as super keys. Out of these super keys we can always choose a proper subset among these which can be used as a primary key. Such keys are known as Candidate keys. If there is a combination of two or more attributes which is being used as the primary key then we call it as a Composite key.\n\nBasic Operators in Relational Algebra\nArticle Contributed by Sonal Tuteja. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above\n\nAttention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up" ]

[ null ]

[ "", null, "# The MIDI Forum\n\n## Reading delta-time from file\n\n0\nVotes\nUndo\nProbably the dumbest first question ever!\n\nI’m creating a mid file reader/writer in C#.net and I’m a bit confused about the MTrk event delta-time. Is delta-time stored in one or two bytes? And if it can differ, how can I find the number of bytes to read? Is the first byte of delta the variable length? I'm a bit confused!", null, "more than a month ago\n0\nVotes\nUndo\nOk I think i've got it now", null, "http://www.somascape.org helped as it stated delta-time is a variable length value.\nThis means my code needs to call the following at the start of each event...\n\ndoubleword ReadVarLen()\n{\nregister doubleword value;\nregister byte c;\nif((value = getc(infile)) & 0x80)\n{\nvalue &= 0x7f;\ndo\n{\nvalue = (value << 7) + ((c = getc(infile))) & 0x7f);\n} while (c & 0x80);\n}\nreturn(value);\n}\nReferences\n• Page :\n• 1\nThere are no replies made for this post yet.\nBe one of the first to reply to this post!" ]

[ null, "https://www.facebook.com/tr", null, "https://www.midi.org/media/com_easydiscuss/images/markitup/emoticon-unhappy.png", null, "https://www.midi.org/media/com_easydiscuss/images/markitup/emoticon-smile.png", null ]

[ "## Square yard\n\nThe Square yard is a unit used to measure area. One Square yard is equal to the area of a square with sides that are 1 yard in length. One square yard is equal to 9 square feet. A square yard is sometimes referred to as a Square yd. Square yards can be abbreviated as Sq yd or yd2.\n\n## Are\n\nAre is a unit of measurement of area, equal to 100 square meters and the equivalent of 0.0247 acre.Are is used to measure land with sides equal to ten meters(10m).\n\n## How to convert Square yard to Are\n\n1 Sq.yd = 0.00836126 are\nExample 1:\nConvert 15 Square yard to Are\n1 yd2 = 0.00836126 are\nTherefore, 15 yd2 = 15 * 0.00836126 = 0.12541905 are\nExample 2:\nConvert 50 Square yard to Are\n1 yd2 = 0.00836126 are\nTherefore, 50 yd2 = 50 * 0.00836126 = 0.4180634 are\n\n## Square yard to Are conversion table\n\nSquare yard(yd2)Are\n00\n10.00836126\n20.01672253\n30.02508380\n40.03344507\n50.04180635\n60.05016761\n70.05852888\n80.06689015\n90.07525143\n100.0836127\n110.09197396\n120.10033523\n130.10869651\n140.11705777\n150.12541905\n200.1672254\n500.4180634\n1000.836126\n\n## Square yard to other converters\n\n• Square yard to Square meter Converter\n• Square yard to Square feet Converter\n• Square yard to Acre Converter\n• Square yard to Hectare Converter\n• ## Other Unit converters\n\n• Angle converter\n• Area converter\n• Cent converter" ]

[ null ]

[ "", null, "# Quine’s Paradox and Gödel’s Theorem\n\nIt’s a commonplace to compare Gödel’s theorem to the liar paradox: The sentence\n\nThis sentence is not true.\n\nis neither true nor false. Switch out “provable” for “true” and you get\n\nThis sentence is not provable.\n\nand, modulo some technical stuff, this sentence is then neither provable nor refutable.  But of course the “modulo some technical stuff” part is crucial: in particular, the Gödel sentence for a theory does not refer to itself. (It does say something about itself, but in a roundabout way.) It can’t refer to itself, because in the theory, you can only refer to sentences via their Gödel numbers, and a sentence can’t contain the numeral for its own Gödel number, very much like a sentence can’t contain itself in quotation marks. But this talk of “the Gödel sentence says of itself that it’s not provable” suggests something like it.  It’s the source of much confusion, and maybe we should avoid the comparison when introducing the incompleteness theorem.\n\nQuine’s Paradox is maybe a bit harder to understand, but it is exactly parallel to the proof of the incompleteness theorem, and in fact the diagonal lemma more generally. Here it is, from The Ways of Paradox:\n\nIf, however, in our perversity we are still bent on constructing a sentence that does attribute falsity unequivocally to itself, we can do so thus: ” ‘Yields a falsehood when appended to its own quotation’ yields a falsehood when appended to its own quotation”. This sentence specifies a string of nine words and says of this string that if you put it down twice, with quotation marks around the first of the two occurrences, the result is false. But that result is the very sentence that is doing the telling. The sentence is true if and only if it is false, and we have our antinomy.\n\nQuine’s paradoxical sentence doesn’t refer to itself to produce a paradox.  It just refers to the expression ‘yields a falsehood when appended to its own quotation’, by quotation and then again by the anaphoric ‘it’.  And the Gödel sentence does the same; as Quine puts it:\n\nGödel’s proof may conveniently be related to the Epimenides paradox or the pseudomenon in the ‘yields a falsehood’ version. For ‘falsehood’ read ‘non-theorem’, thus: ” ‘Yields a non-theorem when appended to its own quotation’ yields a non-theorem when appended to its own quotation”.\n\nThis statement no longer presents an antinomy, because it no longer says of itself that it is false. What it does say of itself is that it is not a theorem (of some deductive theory that I have not yet specified). If it is true, here is one truth that that deductive theory, whatever it is, fails to include as a theorem. If the statement is false, it is a theorem, in which event that deductive theory has a false theorem and so is discredited.\n\nWhat Gödel proceeds to do, in getting his proof of the incompletability of number theory, is the following. He shows how the sort of talk that occurs in the above statement — talk of non-theoremhood and of appending things to quotations — can be mirrored systematically in arithmetical talk of integers. In this way, with much ingenuity, he gets a sentence purely in the arithmetical vocabulary of number theory that inherits that crucial property of being true if and only if not a theorem of number theory. And Gödel’s trick works for any deductive system we may choose as defining ‘theorem of number theory’.\n\nThe proof of the diagonal lemma goes something like this. We’re proving that for every $$\\psi(x)$$ there is a $$\\phi$$ such that $$\\phi \\leftrightarrow \\psi(\\ulcorner \\phi \\urcorner)$$.  Think of $$\\psi(x)$$ as naming a kind, the $$\\psi$$s, say, falsehoods or non-theorems.  Then to get $$\\phi$$ which is true iff it is a $$\\psi$$, we proceed as follows: Define the function $$d$$ which maps the Gödel number of $$\\alpha(x)$$ to the Gödel number of $$\\alpha(\\ulcorner \\alpha(x)\\urcorner)$$.  Supposing we have a function symbol in the language for $$d$$, we can define $$\\phi$$ as $$\\psi(d(\\ulcorner \\psi(d(x))\\urcorner))$$. If a theory $$T$$ represents $$d$$ then it proves $$d(\\ulcorner \\psi(d(x))\\urcorner) = \\ulcorner\\phi\\urcorner$$. Then, by logic, we also get $$\\psi(d(\\ulcorner \\psi(d(x))\\urcorner)) \\leftrightarrow \\psi(\\ulcorner\\phi\\urcorner)$$. But the left-hand side is just $$\\phi$$, so $$T \\vdash \\phi \\leftrightarrow \\psi(\\ulcorner\\phi\\urcorner)$$.\n\nHere ‘$$\\psi(x)$$’ plays the role of ‘falsehood’, ‘non-theorem’, etc. $$d$$ is the function that takes an expression and appends it to its own quotation.  And $$\\phi$$ is $$d$$ applied to ‘yields a $$\\psi$$ if appended to its own quotation’, i.e.,\n\n‘yields a $$\\psi$$ if appended to its own quotation’ yields a $$\\psi$$ if appended to its own quotation.\n\nThat sentence does not refer to itself, so it doesn’t say of itself in that sense that it is a $$\\psi$$. But it is equivalent to the statement that it is a $$\\psi$$.\n\nI don’t know which textbooks, if any, mention Quine’s paradox when introducing the diagonal lemma.  I’m teaching incompleteness right now at McGill from a version of Jeremy Avigad’s notes, where he makes the connection. Let’s see how it goes over in class.\n\n[Photo courtesy of Douglas Quine, via Stampit at Wikimedia Commons licensed under CC-BY-SA-3.0]\n\n## 4 thoughts on “Quine’s Paradox and Gödel’s Theorem”\n\n1.", null, "Kayue says:\n\nMaybe you are interested in this paper (if you haven’t read it before): arxiv.org/abs/math/0606425\nAnd Findlay’s one on the reference list.\n\n2.", null, "James Walsh says:\n\nRichard Heck’s expository notes on the Diagonal Lemma are in the spirit of Quine’s Paradox:\n\n3.", null, "B says:" ]

[ null, "https://i0.wp.com/richardzach.org/wp-content/uploads/2015/03/Wvq-passport-1975-400dpi-crop.jpg", null, "https://secure.gravatar.com/avatar/e9436ede8e062e2680c5d879c748e83d", null, "https://secure.gravatar.com/avatar/329158c0fdca9716280bf356d07a6460", null, "https://secure.gravatar.com/avatar/b4f06972ae2c2a902d47a83babb1848e", null ]

[ "## On a periodic neutral logistic equation.(English)Zbl 0737.34050\n\nThe authors consider a periodic logistic differential equation of neutral type with coefficients of period equal to the delay. Associating a periodic ordinary differential equation to the logistic one, the authors prove the existence of a nontrivial periodic solution for the original equation. Furthermore assuming the delay and some coefficients of the equation small enough, they obtain the local stability of the related periodic solution.\nReviewer: M.Lizana (Caracas)\n\n### MSC:\n\n 34K99 Functional-differential equations (including equations with delayed, advanced or state-dependent argument) 34C25 Periodic solutions to ordinary differential equations 34K20 Stability theory of functional-differential equations 34K10 Boundary value problems for functional-differential equations\nFull Text:\n\n### References:\n\n DOI: 10.1016/0022-247X(90)90213-Y · Zbl 0711.34090 Yoshizawa, Stability theory and the existence of periodic solutions and almost periodic solutions, Applied Mathematical Sciences 14 (1975) · Zbl 0304.34051 El’sgol’ts, Introduction to the theory and application of differential equations with deviating arguments (1973) Halanay, Differential equations; stability, oscillations and time lags pp 377– (1965) DOI: 10.1080/02681118808806037 · Zbl 0665.34066 Kolmanovskii, Stability of functional differential equations (1986)\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

[ null ]

[ "$\\newenvironment {prompt}{}{} \\newcommand {\\ungraded }{} \\newcommand {\\todo }{} \\newcommand {\\oiint }{{\\large \\bigcirc }\\kern -1.56em\\iint } \\newcommand {\\mooculus }{\\textsf {\\textbf {MOOC}\\textnormal {\\textsf {ULUS}}}} \\newcommand {\\npnoround }{\\nprounddigits {-1}} \\newcommand {\\npnoroundexp }{\\nproundexpdigits {-1}} \\newcommand {\\npunitcommand }{\\ensuremath {\\mathrm {#1}}} \\newcommand {\\RR }{\\mathbb R} \\newcommand {\\R }{\\mathbb R} \\newcommand {\\N }{\\mathbb N} \\newcommand {\\Z }{\\mathbb Z} \\newcommand {\\sagemath }{\\textsf {SageMath}} \\newcommand {\\d }{\\mathop {}\\!d} \\newcommand {\\l }{\\ell } \\newcommand {\\ddx }{\\frac {d}{\\d x}} \\newcommand {\\zeroOverZero }{\\ensuremath {\\boldsymbol {\\tfrac {0}{0}}}} \\newcommand {\\inftyOverInfty }{\\ensuremath {\\boldsymbol {\\tfrac {\\infty }{\\infty }}}} \\newcommand {\\zeroOverInfty }{\\ensuremath {\\boldsymbol {\\tfrac {0}{\\infty }}}} \\newcommand {\\zeroTimesInfty }{\\ensuremath {\\small \\boldsymbol {0\\cdot \\infty }}} \\newcommand {\\inftyMinusInfty }{\\ensuremath {\\small \\boldsymbol {\\infty -\\infty }}} \\newcommand {\\oneToInfty }{\\ensuremath {\\boldsymbol {1^\\infty }}} \\newcommand {\\zeroToZero }{\\ensuremath {\\boldsymbol {0^0}}} \\newcommand {\\inftyToZero }{\\ensuremath {\\boldsymbol {\\infty ^0}}} \\newcommand {\\numOverZero }{\\ensuremath {\\boldsymbol {\\tfrac {\\#}{0}}}} \\newcommand {\\dfn }{\\textbf } \\newcommand {\\unit }{\\mathop {}\\!\\mathrm } \\newcommand {\\eval }{\\bigg [ #1 \\bigg ]} \\newcommand {\\seq }{\\left ( #1 \\right )} \\newcommand {\\epsilon }{\\varepsilon } \\newcommand {\\phi }{\\varphi } \\newcommand {\\iff }{\\Leftrightarrow } \\DeclareMathOperator {\\arccot }{arccot} \\DeclareMathOperator {\\arcsec }{arcsec} \\DeclareMathOperator {\\arccsc }{arccsc} \\DeclareMathOperator {\\si }{Si} \\DeclareMathOperator {\\scal }{scal} \\DeclareMathOperator {\\sign }{sign} \\newcommand {\\arrowvec }{{\\overset {\\rightharpoonup }{#1}}} \\newcommand {\\vec }{{\\overset {\\boldsymbol {\\rightharpoonup }}{\\mathbf {#1}}}\\hspace {0in}} \\newcommand {\\point }{\\left (#1\\right )} \\newcommand {\\pt }{\\mathbf {#1}} \\newcommand {\\Lim }{\\lim _{\\point {#1} \\to \\point {#2}}} \\DeclareMathOperator {\\proj }{\\mathbf {proj}} \\newcommand {\\veci }{{\\boldsymbol {\\hat {\\imath }}}} \\newcommand {\\vecj }{{\\boldsymbol {\\hat {\\jmath }}}} \\newcommand {\\veck }{{\\boldsymbol {\\hat {k}}}} \\newcommand {\\vecl }{\\vec {\\boldsymbol {\\l }}} \\newcommand {\\uvec }{\\mathbf {\\hat {#1}}} \\newcommand {\\utan }{\\mathbf {\\hat {t}}} \\newcommand {\\unormal }{\\mathbf {\\hat {n}}} \\newcommand {\\ubinormal }{\\mathbf {\\hat {b}}} \\newcommand {\\dotp }{\\bullet } \\newcommand {\\cross }{\\boldsymbol \\times } \\newcommand {\\grad }{\\boldsymbol \\nabla } \\newcommand {\\divergence }{\\grad \\dotp } \\newcommand {\\curl }{\\grad \\cross } \\newcommand {\\lto }{\\mathop {\\longrightarrow \\,}\\limits } \\newcommand {\\bar }{\\overline } \\newcommand {\\surfaceColor }{violet} \\newcommand {\\surfaceColorTwo }{redyellow} \\newcommand {\\sliceColor }{greenyellow} \\newcommand {\\vector }{\\left \\langle #1\\right \\rangle } \\newcommand {\\sectionOutcomes }{} \\newcommand {\\HyperFirstAtBeginDocument }{\\AtBeginDocument }$\n\nAlternating series are series whose terms alternate in sign between positive and negative. There is a powerful convergence test for alternating series.\n\nMany of the series convergence tests that have been introduced so far are stated with the assumption that all terms in the series are nonnegative. Indeed, this condition is assumed in the Integral Test, Ratio Test, Root Test, Comparison Test and Limit Comparison Test. In this section, we study series whose terms are not assumed to be strictly positive. In particular, we are interested in series whose terms alternate between positive and negative (aptly named alternating series). It turns out that there is a powerful test for determining that a series of this form converges.\n\nCompared to our convergence tests for series with strictly positive terms, this test is strikingly simple. Let us examine why it might be true by considering the partial sums of the alternating harmonic series. The first few partial sums with odd index are given by\n\nNote that a general odd partial sum is of the form and the quantity in the parentheses is positive. We conclude that:\n\nThe sequence $\\{s_1,s_3,s_5,\\ldots \\}$ of odd partial sums defined above is\nincreasing decreasing\n\nMoreover, the sequence of odd partial sums is bounded below by zero, since and each quantity in parentheses is positive.\n\nWe conclude that the sequence $\\{s_1,s_3,s_5,\\ldots \\}$ of odd partial must converge to a finite limit by applying\nThe Fundamental Theorem of Calculus. The Monotone Convergence Theorem. The Ratio Test.\n\nNext consider the sequence\n\nof even partial sums.\n\nThe sequence $\\{s_2,s_4,s_6,\\ldots \\}$ of even partial sums defined above is\nincreasing and bounded, and therefore converges by the Monotone Convergence Theorem. decreasing and bounded, and therefore converges by the Monotone Convergence Theorem.\n\nFinally, we use and the fact that the limits of the sequences $\\{s_{2n+1}\\}$ and $\\{s_{2n}\\}$ are finite to conclude that It follows that the alternating harmonic series converges.\n\nWith slight modification, the argument given above can be used to prove that the Alternating Series Test holds in general.\n\nDoes the alternating series test apply to the series\nyes no" ]

[ null ]

[ "1. Home\n2. Training Library\n3. Machine Learning\n4. Courses\n5. Module 2 - Maths for Machine Learning - Part One\n\n2\n7\nCalculus\n10m 54s\n8\nNotation\n4m 56s\n\n## The course is part of this learning path\n\nPractical Machine Learning\n11\n6\n3\nStart course\nOverview\nDifficultyBeginner\nDuration1h 45m\nStudents123\nRatings\n5/5\n\n### Description\n\nTo design effective machine learning, you’ll need a firm grasp of the mathematics that support it. This course is part one of the module on maths for machine learning. It will introduce you to the mathematics of machine learning, before jumping into common functions and useful algebra, the quadratic model, and logarithms and exponents. After this, we’ll move onto linear regression, calculus, and notation, including how to provide a general analysis using notation.\n\nPart two of this module can be found here and covers linear regression in multiple dimensions, interpreting data structures from the geometrical perspective of linear regression, vector subtraction, visualized vectors, matrices, and multidimensional linear regression." ]

[ null ]

[ "Optomechanical multistability in the quantum regime\n\n# Optomechanical multistability in the quantum regime\n\nC. Schulz Institute of Physics, Ernst-Moritz-Arndt-University, 17487 Greifswald, Germany    A. Alvermann E-mail: Institute of Physics, Ernst-Moritz-Arndt-University, 17487 Greifswald, Germany    L. Bakemeier Institute of Physics, Ernst-Moritz-Arndt-University, 17487 Greifswald, Germany    H. Fehske Institute of Physics, Ernst-Moritz-Arndt-University, 17487 Greifswald, Germany\n###### Abstract\n\nClassical optomechanical systems feature self-sustained oscillations, where multiple periodic orbits at different amplitudes coexist. We study how this multistability is realized in the quantum regime, where new dynamical patterns appear because quantum trajectories can move between different classical orbits. We explain the resulting quantum dynamics from the phase space point of view, and provide a quantitative description in terms of autocorrelation functions. In this way we can identify clear dynamical signatures of the crossover from classical to quantum mechanics in experimentally accessible quantities. Finally, we discuss a possible interpretation of our results in the sense that quantum mechanics protects optomechanical systems against the chaotic dynamics realized in the classical limit.\n\n42.50.Ct\n37.10.Vz\n###### pacs:\n07.10.Cm\n\nQuantum description of interaction of light and matter; related experiments Mechanical effects of light on atoms, molecules, and ions Micromechanical devices and systems\n\n## 1 Introduction\n\nThe interaction of light with mechanical objects [1, 2] enjoys continued interest due to the successful construction and manipulation of optomechanical devices over a wide range of system sizes and parameter combinations (see the recent reviews [3, 4] and references cited therein). With these devices both classical non-linear dynamics such as self-sustained oscillations [5, 6, 7, 8] and chaos [9, 10, 11] as well as quantum mechanical mechanisms such as cooling into the groundstate [12, 13] and quantum non-demolition measurements [14, 15, 16] can be studied in a unified experimental setup.\n\nThis raises the question whether it might be possible to detect the crossover from classical to quantum mechanics directly in the dynamical behaviour of optomechanical systems. In a previous paper  we observed that the classical dynamical patterns, which are characterized by the multistability of self-sustained oscillations, change in a characteristic way if one moves into the quantum regime. Previously stable orbits become unstable, the system oscillates at a new amplitude, and especially the classical chaotic dynamics is almost immediately replaced by simple periodic oscillations. In this paper we explain this behaviour from the point of view of classical and quantum phase space dynamics. Most importantly, we will show that the dynamical patterns do not change at random but that clearly identifiable and new signatures can be observed.\n\nThe prototypical optomechanical system is a vibrating cantilever subject to the radiation pressure of a cavity photon field, for which the Hamilton operator reads [17, 3, 4]\n\nwhere and are bosonic operators for the vibrational mode of the cantilever (frequency ) and for the cavity photon field (), respectively. This Hamilton operator applies to any generic optomechanical system, but we adopt the cavity-cantilever terminology throughout this paper. For our theoretical analysis we use the quantum optical master equation \n\n ∂tρ=−iℏ[H,ρ]+ΓD[b,ρ]+κD[a,ρ] (2)\n\nfor the cantilever-cavity density matrix , with the dissipative terms\n\n D[L,ρ]=LρL†−12(L†Lρ+ρL†L) (3)\n\nthat account for cantilever damping () and radiative losses (). Note that the above Hamilton operator is given in a frame that rotates with the frequency of the external pump laser such that only the cavity-laser detuning appears, and that we assume zero temperature in the master equation.", null, "Figure 1: (color online) Left panel: Chart of self-sustained oscillations in the classical limit for P=1.5. Self-sustained oscillations occur for amplitudes A where the power balance between gains from the radiation pressure (Prad=P⟨|α|2Imβ⟩avg) and losses due to friction (Pfric=¯Γ⟨|β|2⟩avg) changes from positive to negative values with increasing A [5, 6]. Right panels: Classical orbits in the (x,p) cantilever phase space, for (a) Δ=−0.4, (b) Δ=−1.1, (c) Δ=−0.85, and (d) Δ=−0.7, as marked by vertical lines in the left panel. In case (a), the two innermost orbits have amplitudes A1≈1.2 and A2≈2.7. In cases (b), (c) the innermost orbit shows a few period doubling bifurcations that occur on the route to chaos , in case (d) it is chaotic.\n\nNow introduce the five dimensionless parameters [5, 6]\n\nand , , and measure time as . The parameter gives the detuning of the pump laser and cavity, while gives the strength of the laser pumping. For later numerical results we set the damping parameters , to typical experimental values .\n\nThe quantum-classical scaling parameter is the ratio of the quantum mechanical quantity , which is of order because the quantum mechanical position operator of the cantilever enters the expression for the radiation pressure, to the classical quantity that measures the cavity quality. The parameter thus controls the crossover from classical () to quantum () mechanics . In the following we will increase to move into the quantum regime, but keep in order to remain in the vicinity of the classical limit .\n\n## 2 Classical multistability\n\nOur analysis begins in the limit , where the optomechanical system is described by the classical equations of motion \n\n ∂τα =(iΔ−¯κ)α−i(β+β∗)α−12i, (5a) ∂τβ =(−i−¯Γ)β−12iP|α|2 (5b)\n\nfor the cavity and cantilever phase space variables , . We also use the cantilever position and momentum operator , , with corresponding phase space variables and .\n\nThe classical equations of motion predict the onset of self-sustained cantilever oscillations as the pump power is increased. Figure 1 shows the possible amplitudes of these oscillations, which are obtained with the ansatz from Ref. , for the value . We keep this value fixed throughout the paper, as the behaviour discussed here does not depend on it. Note in Fig. 1 that several stable oscillatory solutions at different amplitudes can coexist for one parameter choice. This classical multistability of self-sustained oscillations is the origin of the quantum multistability analyzed next.\n\n## 3 Quantum multistability", null, "Figure 2: (color online) Left panel: Cantilever position x(τ) from the classical equations of motion (5) and from the quantum mechanical master equation (2) at σ=0.1, for P=1.5, Δ=−0.4 (case (a) in Fig. 1). Right panel: Cantilever position-momentum uncertainty product σxσp for the same parameters.\n\nWe now move into the quantum regime by letting become finite. In all our examples the quantum system is initially prepared in the pure product state of a coherent cantilever and cavity state at , i.e., in the state that is closest to a classical state at these coordinates. The cantilever-cavity density matrix is then evolved according to Eq. (2).\n\nFigure 2 shows the cantilever position and the position-momentum uncertainty product , with the uncertainty of an observable . The quantum dynamics at finite closely follows the classical oscillations for an initial period of time, before it deviates significantly at later times. Deviations occur because the quantum state spreads out in phase space, as witnessed by the growth of the uncertainty product, whereby the cantilever position is smeared out.", null, "Figure 3: (Color online) Wigner function W(x,p) in cantilever phase space (left panels) and cantilever position autocorrelation function Rτ(δτ) (right panels) for case (a) from Fig. 1, at σ=0.1 slightly away from the classical limit. The autocorrelation functions for the two inner classical orbits at amplitudes A1/2 are included as dashed curves.\n\nThe full phase space dynamics in Fig. 3, which we display with the Wigner function of the cantilever mode (see, e.g., Ref.  for the definitions), reveals a more definite dynamical pattern. For early times () the Wigner function retraces the classical orbit with amplitude from case (a) in Fig. 1. At later times () the Wigner function shows a contribution from a second circular orbit with larger amplitude, before almost all weight is concentrated on the new orbit (). In comparison to case (a) in Fig. 1 this orbit is identified as the second classical orbit with amplitude . During time evolution the quantum state spreads out along, but not perpendicular to, these two classical orbits.\n\nThe classical multistability of the optomechanical system thus has a direct counterpart in the quantum dynamics at small , where the system moves between the different classical orbits. This kind of quantum multistability leads to distinct dynamical features because the oscillatory nature of the different orbits is preserved.\n\nThe quantum multistability is clearly detected with the cantilever position autocorrelation function\n\n Rτ(δτ)=τ+π∫τ−π⟨^x(τ′)^x(τ′+δτ)⟩dτ′, (6)\n\ninstead of the position expectation value that averages over the phase space distribution. We choose this function because the dynamics is best described in cantilever phase space. Autocorrelation functions for the cavity mode could be used as well and should be more accessible to experimental measurements, but their interpretation is less straightforward because of the additional sidebands at multiples of the fundamental oscillation frequency.\n\nThe autocorrelation function in Fig. 3 is the weighted sum of the oscillatory motion on the two orbits seen in the Wigner function. The frequency of the two orbits is identical (essentially, the cantilever frequency ), such that only one oscillation is visible in . The amplitude of increases as weight is transferred from the inner to the outer orbit. Noteworthy, the oscillations persist at all times. In this way, the multistability of the quantum dynamics is not only observable during a short initial time period but during extended periods of time.\n\n## 4 Multistability of quantum trajectories\n\nThe mechanism behind the quantum multistability can be understood through the phase space dynamics of individual quantum trajectories, as they arise in the quantum state diffusion (QSD) approach  to the solution of Lindblad master equations such as Eq. (2).\n\nIn QSD the density matrix is represented by an ensemble of quantum trajectories , from which it is obtained as an average\n\n ρ(τ)=meank{|ψk(τ)⟩⟨ψk(τ)|}. (7)\n\nAccordingly, expectation values are computed as ensemble averages . Each quantum trajectory follows a stochastic equation of motion that combines the Hamiltonian and dissipative dynamics with a noise term . Numerically, the density matrix is obtained through Monte Carlo sampling of the trajectories for different noise realizations. We use the QSD implementation from Ref. , and typically average over trajectories to obtain the results in Figs. 2478. Although a single quantum trajectory is not observable by itself, the phase space dynamics of individual trajectories as shown in Figs. 56 allows us to deduce the properties of the entire density matrix.", null, "Figure 4: (color online) Left panels: Wigner function W(x,p) for a single quantum trajectory starting from a “Schrödinger cat” state at (i) τ=0, and at later times (ii) τ=0.001, (iii) τ=0.008, and (iv) τ=0.4. Right panels: Cantilever position xk and uncertainty product σxσp (see Eq. (8)) for a single quantum trajectory at later times, in the situation of Fig. 5. All results are for case (a) from Fig. 1 and σ=0.1.\n\nClose to the classical limit quantum trajectories evolve rapidly into localized phase space states as a consequence of dissipation [22, 23, 24]. This is illustrated in Fig. 4 for a single trajectory that starts from a “Schrödinger cat” state, given as the superposition of two coherent states, with the characteristic interference pattern in the Wigner function. In less than one oscillation period () the trajectory evolves into a nearly coherent state with a positive Wigner function, which shows the rapid decoherence. The quantum trajectory remains in such a state during the subsequent time evolution, and the uncertainty product stays close to its minimal value\n\ngiven by the Heisenberg uncertainty relation for the , operators (here, the quantum-classical scaling parameter comes into play). Notice that phase space localization occurs only in the vicinity of the classical limit, for . It also explains the transition into the classical limit: For the quantum trajectories evolve infinitely fast into minimal uncertainty states, and at the same time the lower bound in Eq. (8) goes to zero. Then, every trajectory occupies one point in phase space, i.e., it has become classical. Under this condition the classical equations of motion (5) can be derived directly from the master equation (2).\n\nBecause a quantum trajectory is very localized in phase space it is well represented by a single phase space point, similar to a classical trajectory. In Fig. 5 this representation is used for a stroboscopic phase space plot of a single quantum trajectory that contributes to the Wigner functions in Fig. 3. This plot clearly shows the multistability of the quantum trajectory, which initially follows the inner orbit before it moves towards the outer orbit. During the time evolution the quantum trajectory follows the oscillatory motion of the two orbits at the cantilever frequency, and because the trajectory state is well localized in phase space these oscillations are not averaged out but appear directly in the position expectation value that is depicted in Fig. 4.", null, "Figure 5: (color online) Stroboscopic (x,p)-phase space plot of a single quantum trajectory (red dots), for case (a) from Fig. 1 and σ=0.1, at early (left panel), intermediate (central panel), and later (right panel) times τ as indicated. The initial conditions are x(0)=p(0)=0, the quantum system is prepared in a coherent state at these coordinates. The two classical orbits at amplitudes A1/2 are depicted with dashed curves.\n\nSince every individual trajectory shows this type of quantum multistability it is also seen in the entire density matrix, given as the ensemble average of all trajectories. Because of the noise term in the stochastic QSD equation of motion the quantum trajectories are not exactly at the same phase space point but at different points on the respective orbits. This results in the broad distribution of the relative angle in phase space seen in the Wigner functions in Fig. 3 especially at later times, when the quantum trajectories are spread out fully along the second orbit. Consequently, all oscillations are averaged out in expectation values such as the cantilever position in Fig. 2. Such values are, therefore, not the right quantities to detect the quantum multistability.\n\nInstead, successful detection requires autocorrelation functions such as from Eq. (6). Similar to the density matrix the function can be expressed (dropping the -integration here) as an ensemble average\n\n (9)\n\nwhere the expectation value is computed for each individual quantum trajectory. The correlation function in the second line is bounded by\n\n (10)\n\nWhenever the position uncertainty of each trajectory becomes small, as it is the case for , the autocorrelation function is thus given by the ensemble average of the autocorrelation functions of the individual trajectories, i.e., by the first line in Eq. (9). Accordingly, the oscillations seen in for each individual trajectory (cf. Fig. 4) are preserved in the autocorrelation function in spite of the ensemble average. Furthermore, is the weighted sum of the autocorrelation functions for the different classical orbits, which are directly related to the orbit amplitudes as seen in Fig. 3.", null, "Figure 6: (color online) Phase space plot of many quantum trajectories in the QSD ensemble (red points) for cases (a)–(d) from Fig. 1, different times τ, and values of σ as indicated. In all cases, the two innermost classical orbits from Fig. 1 are included as solid curves. In case (b) the second orbit is missing.\n\nNotice that the behaviour described here—the motion of quantum trajectories between different classical orbits—emerges only because the trajectory states deviate from coherent states. The noise terms in the QSD equations have the form , here for the mechanical damping, with a random variable from the underlying Wiener process . If is exactly a coherent state, such that , the noise term will vanish identically. This observation explains why the “quantum noise” disappears in the classical limit , and the quantum trajectories follow the deterministic classical equations of motion (5). At finite but small trajectories are almost but not exactly in coherent states. The noise terms become effective but remain small, such that the quantum trajectories still follow the classical dynamics but are subject to a small stochastic correction. This small correction can change the long-time stability of classical orbits and their basin of attraction but does not destroy the classical dynamical patterns. Consequently, the quantum trajectories do not move arbitrarily in phase space but follow a classical orbit for some time before they leave the orbit with a finite probability. Afterwards, the trajectories can settle on a different attractive orbit if such an orbit exists at larger amplitudes.\n\n## 5 Quantum multistability and classical orbits\n\nThe quantum multistability observed for case (a) from Fig. 1 depends on the presence of at least two classical orbits between which the quantum trajectories can move. The remaining cases (b)–(d) are variations of this situation, where either the second orbit is missing (case (b)) or the nature of the first orbit has changed (cases (c), (d)). The four cases are compared in Fig. 6 with phase space plots of many quantum trajectories that represent the QSD ensemble for the density matrix.\n\nIn all cases the time scale relevant for quantum multistability shortens with increasing because the quantum trajectories leave the first classical orbit more rapidly when the noise terms become larger. For too large (e.g., in case (a)) the clear dynamical pattern of quantum multistability—the movement between different classical orbits—disappears altogether.\n\nIn case (b) the quantum trajectories cannot settle on a nearby classical orbit once they left the first orbit. Quantum multistability, which is characterized by the prevalence of oscillatory motion over random diffusion, cannot be observed in such a situation.", null, "Figure 7: (color online) Wigner function W(x,p) in cantilever phase space for case (d) from Fig. 1, for τ and σ as in Fig. 6.\n\nIn cases (c), (d) the inner orbit is no longer simpler periodic but a period-two orbit after the first period doubling bifurcation on the route to chaos (case (c)) or a chaotic orbit (case (d)). Quantum multistability is not affected by the different nature of the inner classical orbit, because still a second simple periodic orbits at larger amplitude exists such that oscillations can be observed after the quantum trajectories have left the inner orbit.\n\nThis is illustrated for case (d) in Figs. 78. First, we observe again that the relevant time scale changes significantly with . If is increased from to in Fig. 7 almost all weight of the Wigner function is transferred from the inner to the outer orbit. Second, the Wigner functions themselves look quite similar to those for case (a) in Fig. 3. In agreement with this, well-defined oscillations are observed in the cantilever position and autocorrelation function in Fig. 8, and the respective amplitudes can be related to those of the classical orbits in Fig. 1.\n\nThe present data might suggest a more ambitious interpretation. Apparently, all curves at finite in Fig. 8 show simple periodic oscillations even if (at ) most weight in the Wigner function is still on the inner—classically chaotic—orbit. To a certain extent, quantum mechanics protects the optomechanical system against classical chaotic dynamics. Initially, the quantum state cannot follow the intricate chaotic orbit curve because it occupies a finite part of phase space. Because of phase space averaging the chaotic motion is replaced by simple oscillations at the fundamental system (i.e., cantilever) frequency. Later, the quantum trajectories move to the second—classically simple periodic—orbit. At all times, the chaotic classical dynamics is replaced by clearly defined simple oscillations in the quantum regime. Notice that we here discuss possible signatures of classical chaos in the associated dissipative quantum dynamics and not in quantities such as the level statistics that are defined for conservative Hamiltonian systems only [25, 26].", null, "Figure 8: (color online) Cantilever position x(τ) (left panels) and position autocorrelation function Rτ(δτ) (right panels) for case (d) at finite σ, in comparison to the results in the classical limit σ=0 (top panels, and dashed curves in the lower panels). These curves correspond to the Wigner functions in Fig. 7.\n\n## 6 Conclusions\n\nIn this paper we establish the quantum mechanical counterpart of the classical multistability of optomechanical systems. While classical multistability corresponds to the coexistence of self-sustained oscillations at multiple amplitudes, quantum multistability is a dynamical effect in which the amplitude of oscillations changes over time. The change can be detected with phase space techniques such as the Wigner function, and analyzed quantitatively with autocorrelation functions.\n\nQuantum multistability is observed close to the classical limit. There, the quantum trajectories in the QSD picture of dissipative dynamics are well localized in phase space. Quantum multistability results from corrections to the classical dynamics given by the noise terms in the stochastic QSD equations of motion. The picture of quantum trajectories also provides the link between the oscillatory quantum dynamics and the classical orbits such that, e.g., the oscillations in the autocorrelation functions can be traced back to the classical self-sustained oscillations.\n\nThe time scale relevant for quantum multistability is set by the quantum-classical scaling parameter . An interesting goal is to obtain the time scale from the QSD equations by quantifying the size of the noise term. This is not an entirely trivial task, though, because the noise term depends not directly on but on the deviation of the quantum trajectory state from a coherent state.\n\nAn important aspect for experimental investigations of quantum multistability is the robustness of the feature. Quantum multistability manifests itself over an extended period of time, is observable in autocorrelation functions after the initial dynamics has evolved into a stable dynamical pattern, and does not require specific system preparations. The experimental feasibility depends mainly on the ability to tune the quantum-classical scaling parameter . For the prototypical cantilever-cavity system is changed, e.g., by simultaneous adjustment of the cantilever mass and pump laser power (thus preserving the self-sustained oscillations). The central experimental challenge is to distinguish “quantum” multistability from the effects of “classical” thermal noise, which requires that the temperature be sufficiently low. The relevant dynamical energies are larger than the energy separation of low-lying quantum states, which allows for comparatively high temperatures. Furthermore, variation of changes the quantum mechanical time scale while the thermal noise is not affected. This might open up the possibility of observing the crossover from classical to quantum mechanics directly in the dynamical behaviour of an optomechanical system.\n\n###### Acknowledgements.\nThis work was supported by Deutsche Forschungsgemeinschaft via Sonderforschungsbereich 652 (project B5).\n\n## References\n\n• \\NameKippenberg T. J. Vahala K. J. \\REVIEWScience32120081172.\n• \\NameMarquardt F. Girvin S. M. \\REVIEWPhysics2200940.\n• \\NameMeystre P. \\REVIEWAnn. Phys. (Berlin)5252013215.\n• \\NameAspelmeyer M., Kippenberg T. J. Marquardt F. \\REVIEWRev. Mod. Phys.8620141391.\n• \\NameMarquardt F., Harris J. G. E. Girvin S. M. \\REVIEWPhys. Rev. Lett.962006103901.\n• \\NameLudwig M., Kubala B. Marquardt F. \\REVIEWNew J. Phys.102008095013.\n• \\NameKippenberg T. J., Rokhsari H., Carmon T., Scherer A. Vahala K. J. \\REVIEWPhys. Rev. Lett.952005033901.\n• \\NameRokhsari H., Kippenberg T. H., Carmon T. Vahala K. J. \\REVIEWOpt. Express1320055293.\n• \\NameCarmon T., Rokhsari H., Yang L., Kippenberg T. J. Vahala K. J. \\REVIEWPhys. Rev. Lett.942005223902.\n• \\NameCarmon T., Cross M. C. Vahala K. J. \\REVIEWPhys. Rev. Lett.982007167203.\n• \\NameBakemeier L., Alvermann A. Fehske H. \\REVIEWPhys. Rev. Lett.1142015013601.\n• \\NameChan J., Alegre T. P. M., Safavi-Naeini A. H., Hill J. T., Krause A., Groblacher S., Aspelmeyer M. Painter O. \\REVIEWNature478201189.\n• \\NameTeufel J. D., Donner T., Li D., Harlow J. W., Allman M. S., Cicak K., Sirois A. J., Whittaker J. D., Lehnert K. W. Simmonds R. W. \\REVIEWNature4752011359.\n• \\NameHertzberg J. B., Rocheleau T., Ndukum T., Savva M., Clerk A. A. Schwab K. C. \\REVIEWNat. Phys.62010213.\n• \\NameVanner M. R., Hofer J., Cole G. D. Aspelmeyer M. \\REVIEWNat. Commun.420132295.\n• \\NameSuh J., Weinstein A. J., Lei C. U., Wollman E. E., Steinke S. K., Meystre P., Clerk A. A. Schwab K. C. \\REVIEWScience34420141262.\n• \\NameLaw C. K. \\REVIEWPhys. Rev. A5119952537.\n• \\NameCarmichael H. J. \\BookStatistical Methods in Quantum Optics 1 (Springer) 1999.\n• \\NameSchleich W. P. \\BookQuantum Optics in Phase Space (Wiley-VCH) 2001.\n• \\NameGisin N. Percival I. C. \\REVIEWJ. Phys. A2519925677.\n• \\NameSchack R. Brun T. A. \\REVIEWComp. Phys. Comm.1021997210 .\n• \\NamePercival I. C. \\REVIEWJ. Phys. A2719941003.\n• \\NameSchack R., Brun T. A. Percival I. C. \\REVIEWJ. Phys. A: Math. Gen.2819955401.\n• \\NameRigo M. Gisin N. \\REVIEWQuantum Semiclass. Opt.81996255.\n• \\NameGutzwiller M. C. \\BookChaos in Classical and Quantum Mechanics (Springer (New York)) 1990.\n• \\NameHaake F. \\BookQuantum Signatures of Chaos (Springer (Berlin)) 2010.\nYou are adding the first comment!\nHow to quickly get a good reply:\n• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.\n• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.\n• Your comment should inspire ideas to flow and help the author improves the paper.\n\nThe better we are at sharing our knowledge with each other, the faster we move forward.\nThe feedback must be of minimum 40 characters and the title a minimum of 5 characters", null, "", null, "", null, "" ]

[ null, "https://www.groundai.com/project/optomechanical-multistability-in-the-quantum-regime/1a.jpeg", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_34594%2Fproject_338857%2Fimages%2Fx2.png", null, "https://www.groundai.com/project/optomechanical-multistability-in-the-quantum-regime/3a.jpeg", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_34594%2Fproject_338857%2Fimages%2Fx7.png", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_34594%2Fproject_338857%2Fimages%2Fx9.png", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_34594%2Fproject_338857%2Fimages%2F6ab.png", null, "https://www.groundai.com/project/optomechanical-multistability-in-the-quantum-regime/7a.jpeg", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_34594%2Fproject_338857%2Fimages%2Fx10.png", null, "https://dp938rsb7d6cr.cloudfront.net/static/1.71/groundai/img/loader_30.gif", null, "https://dp938rsb7d6cr.cloudfront.net/static/1.71/groundai/img/comment_icon.svg", null, "https://dp938rsb7d6cr.cloudfront.net/static/1.71/groundai/img/about/placeholder.png", null ]

[ "## Theorem\n\nLet $p$ be a prime number.\n\nLet $\\struct {\\Q_p, \\norm {\\,\\cdot\\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.\n\nLet $\\mathbf a$ be an equivalence class in $\\Q_p$.\n\nThen $\\mathbf a$ has exactly one representative that is a $p$-adic expansion.\n\n## Proof\n\n### Case 1\n\nLet $\\norm{\\mathbf a}_p \\le 1$.\n\nFrom Equivalence Class in P-adic Integers Contains Unique P-adic Expansion, $\\mathbf a$ has exactly one representative that is a $p$-adic expansion of the form:\n\n$\\displaystyle \\sum_{n \\mathop = 0}^\\infty d_n p^n$\n$\\displaystyle \\sum_{i \\mathop = 0}^\\infty e_i p^i$ is the only $p$-adic expansion that represents $\\mathbf a$\n\n$\\Box$\n\n### Case 2\n\nLet $\\norm{\\mathbf a}_p > 1$.\n\nLet $\\textbf p$ denote the equivalence class that is identified with the prime number $p$.\n\nBy definition of the $p$-adic numbers as a quotient of Cauchy sequences:\n\nthe constant sequence $\\tuple {p, p, p, \\dotsc}$ represents $\\mathbf p \\in \\Q_p$.\n$\\exists m \\in \\Z: \\mathbf p^m \\textbf a \\in \\Z_p^\\times$\n\nwhere $\\Z_p^\\times$ denotes the $p$-adic units.\n\nBy definition a $p$-adic unit is a $p$-adic integer.\n\n$\\mathbf p^m \\textbf a$ has exactly one representative that is a $p$-adic expansion of the form:\n$\\displaystyle \\sum_{n \\mathop = 0}^\\infty d_n p^n$\n$\\norm {\\mathbf p^m \\textbf a}_p = 1 = p^0$\n$d_0 \\neq 0$\n\nLet $\\sequence{x_n}$ be a Cauchy sequence of rational numbers that represents $\\mathbf a \\in \\Q_p$.\n\nBy definition of the $p$-adic numbers as a quotient of Cauchy sequences:\n\nthe constant sequence $\\tuple {p^m, p^m, p^m, \\dotsc}$ represents $\\mathbf p^m \\in \\Q_p$.\n\nBy definition of the product in a quotient ring:\n\nthe Cauchy sequence $\\sequence{p^m x_n}$ is a representative of $\\mathbf p^m \\mathbf a \\in \\Q_p$.\n$\\sequence {p^m x_n - \\displaystyle \\sum_{i \\mathop = 0}^n d_i p^i}$ is a null sequence\n$\\sequence {x_n - \\displaystyle p^{-m} \\sum_{i \\mathop = 0}^n d_i p^i}$ is a null sequence\n$\\sequence p^{-m} \\sum_{i \\mathop = 0}^\\infty d_i p^i$ is a representative of $\\mathbf a$.\n\nThat is:\n\n$\\displaystyle \\sum_{i \\mathop = -m}^\\infty e_i p^i$ is a representative of $\\mathbf a$\n\nwhere:\n\n$\\forall i \\ge -m: e_i = d_{i + m}$\n\nBy definition of a $p$-adic expansion;\n\n$\\forall i \\ge 0: 0 \\le d_i < p$\n\nThen:\n\n$\\forall i \\ge -m: 0 \\le e_i = d_{i + m} < p$\n\nSince $d_0 \\neq 0$, then $e_{-m} = d_0 \\neq 0$.\n\nBy definition of a $p$-adic expansion;\n\n$\\displaystyle \\sum_{i \\mathop = -m}^\\infty e_i p^i$ is a $p$-adic expansion that represents $\\mathbf a$\n$\\displaystyle \\sum_{i \\mathop = -m}^\\infty e_i p^i$ is the only $p$-adic expansion that represents $\\mathbf a$\n\n$\\blacksquare$" ]

[ null ]