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http://www.kpubs.org/article/articleMain.kpubs?articleANo=HGJHC0_2011_v12n4_365 | [
"Missile Autopilot Design for Agile Turn Control During Boost-Phase\nMissile Autopilot Design for Agile Turn Control During Boost-Phase\nInternational Journal of Aeronautical and Space Sciences. 2011. Dec, 12(4): 365-370\nCopyright ©2011, The Korean Society for Aeronautical Space Science",
null,
"This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/bync/ 3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.\n• Received : March 27, 2010\n• Accepted : December 06, 2011\n• Published : December 30, 2011",
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"Export by style\nArticle\nAuthor\nMetrics\nCited by\nTagCloud\nSun-Mee Ryu\nDae-Yeon Won\nChang-Hun Lee\nMin-Jea Tahk\[email protected]\n\nAbstract\nThis paper presents the air-to-air missile autopilot design for a 180° heading reversal maneuver during boost-phase. The missile’ s dynamics are linearized at a set of operating points for which angle of attack controllers are designed to cover an extended flight envelope. Then, angle of attack controllers are designed for this set of points, utilizing a pole-placement approach. The controllers’ gains in the proposed configuration are computed from aerodynamic coefficients and design parameters in order to satisfy designer-chosen criteria. These design parameters are the closed-loop frequency, damping ratio, and time constant; these represent the characteristics of the control system. To cope with highly nonlinear and rapidly time varying dynamics during boost-phase, the global gain-scheduled controller is obtained by interpolating the controllers’ gains over variations of the angle of attack, Mach number, and center of gravity. Simulation results show that the proposed autopilot design provides satisfactory performance and possesses good [ed: or “sufficient” or “excellent”] capabilities.\nKeywords\n1. Introduction\nIn recent years, interest and research in the highlyagile turn, with regard to combat, has grown. High angle of attack missiles can increase the off-boresight capability and maneuverability directly connected with tactical performance.\nTo maximize missile performance, appropriate autopilot command structure is required for each mission phase, including: launch, agile turn at high angle of attack, midcourse, and endgame (Wise and Broy, 1998). This is particularly important as missile dynamics undergo significant changes during the course of the flight; this paper focuses specifically on the agile turn phase. While a missile is operating at a high angle of attack, the angle of attack command?unlike the conventional acceleration command?is needed for rapid and effective control.\nThe simple and conventional control method for missiles is the linear controller, but it requires gain scheduling at multiple design points. In Wise and Broy (1998), agile missile dynamics and the linear control commanding angle of attack have been studied. Mehrabian and Roshanian (2006) introduced application of linear parameter varying modeling and control for a highly agile missile. Many gain scheduling techniques in missile autopilot design have already been studied (Lawrence and Rugh, 1993; White et al., 1994) for nonlinear model. Because of high nonlinearity in the agile turn, nonlinear control techniques are also employed in order to design control law (Innocenti, 2001; Menon and Yousefpor, 1996; Thukral and Innocenti, 1998).\nThe purpose of this paper is to introduce the design of the angle of attack controller using the pole placement approach for the high angle of attack missile. This technique is proposed to overcome difficulties in gain selection and scheduling in nonlinear models. This research deals with a nonlinear missile model in the pitch plane; the equation of motion is discussed in Section 2. In Section 2, the missile model specifications for autopilot design are listed. Section 3 introduces the angle of attack control system. the controller configuration designed by the pole placemenr method, as well as the process of gain computation, are explained. Section 4 shows gain scheduling and numerical results. Finally, conclusions are drawn in the final section.\n2. Missile Dynamics\n- 2.1 Equation of motion\nNonlinear equations of motion describing missile flight are used in the autopilot design. A body coordinate system and an inertial coordinate system are employed to derive the equations of motion as shown in Fig. 1 . Under the assumption of rigid body, no gravity, no roll rate, and zero roll angles in the pitch plane, the translational and rotational dynamics of the symmetric missile are written as follows:",
null,
"Lager Image\n, where u and w indicate the missile velocity components along the X and Z body axes; q the pitch rate; fA the accelerations by the aerodynamic forces; fT the accelerations by the propulsion system; MA and MT the pitch moments produced by aerodynamic force and thrust; Iyy the moment of inertia about the pitch axis; and ⊠ the pitch angle.\nThe angle of attack dynamics are modeled as follows:",
null,
"Lager Image\n- 2.2 Missile model\nThe missile used in the design controller has performances",
null,
"Lager Image\nMissile coordinate system of longitudinal motion.\nand specifications as follows :\n1) Flight/maneuvering ranges\n• - Mach number: 1.0 to4.0\n• - Altitude: 0 km to20 km\n• - Total angle of attack: 0 deg to 90 deg\n• - Maneuvering acceleration: 0 g to50 g\n2) Missile specifications\n• - Mass: 92.0 kg\n• - Moment of inertia: Iyy = 59.8043 kg.m\n3) Actuator (control fins)\n• - Damping ratio: 0.7\n• - Natural frequency: 30 Hz\n• - Deflection limit: 28 deg\n3. Autopilot Design\n- 3.1 Angle of attack autopilot configuration\nFor the agile turn of missile, the autopilot tracking the angle of attack appears in Fig. 2 . In this system, the pitch rate and the angle of attack are used for feedback in the controller. The concept of the autopilot in this paper comes from the threeloop autopilot (Zarchan, 2007). The three gains, Kp, K2, and K3, must be chosen to satisfy desired parameters, which are closed-loop frequency, system damping, and time constant.. The selection of the adequate closed-loop frequency and damping ratio can help avoid many stability problems, and the system time constant can select the response speed, as mentioned in Zarchan (2007).\n- 3.2 Desired performance\nThe design criteria are set as:\n• - Settling time should be less than 0.2 seconds\n• - Phase margin should be greater than 30 deg\n• - Gain margin should be greater than 6 dB",
null,
"Lager Image\nAngle of attack (AOA) autopilot configuration.\nAccording these conditions, three design parameters are:\n• - Time constant τ: 0.07 seconds\n• - Damping ratio ζ: 0.6\n• - Closed-loop frequency ω: 85 rad/s\n- 3.3 Gain selection by pole placement approach\nBy selecting the frequency ω, damping ratio ζ, and time constant τ, the closed-loop poles and gains should be chosen. From Fig. 2 , one can see that the relationship between the output angle of attack and the command is:",
null,
"Lager Image\n, where\n• c1= 1 /KaKp+K2/ Kp+Ta+KqK3/KaKp\n• c2= 2ζAF/ωAFKaKp+K2Ta/Kp+KqK3Tq/KaKp\n• c3= 1 /ωAF2KaKp\nTo represent three gains with desired parameters, the closed-loop transfer function can be written in the 3rd order system, of [ed: did you mean or?] with the following form:",
null,
"Lager Image\nThe two transfer functions, Eqs. (3) and (4), are equivalent if the denominators are the same. Therefore, three linear and simultaneous equations should be structured as follows:",
null,
"Lager Image\nThe airframe characteristics, ω AF and ζ AF , and the transfer function gains, K α , K q , T α , and T q , are:",
null,
"Lager Image\nAnalysis of stability margin (M = 2.0 and α = 10 deg)GM: gain margin, PM: phase margin.",
null,
"Lager Image\nAnalysis of stability margin (M = 2.0 and α = 10 deg) GM: gain margin, PM: phase margin.",
null,
"Lager Image\nFrom Eqs. (5) and (6), the autopilot gains can be derived as follows:",
null,
"Lager Image\n, where",
null,
"Lager Image\nIn Eq. (7), three gains are expressed as aerodynamic coefficients and desired parameters, ζ, ω, and τ. This autopilot technique alleviates the problems with gain tuning process in the design of linear controller.\n4. Simulation Results\nThe performance of a high angle of attack autopilot should be verified by nonlinear simulations because of nonlinearity and fast variation of parameters.\n- 4.1 Gain scheduling\nTo cope with the nonlinear plant, a gain scheduled controller was used in the simulation. The procedure is outlined in Lawrence and Rugh (1993) and White et al. (1994). First, the family of trim point is determined, and a family of linear controllers is designed to achieve the specified performance at each operating point. Next, the gain is computed at each operating point using Eq. (7), derived in previous section. The final step is checking the performance of the controller.\nIn this study, the variables for gain scheduling are set angle of attack, and Mach number M. The set of 36 equilibrium points is chosen, corresponding to the value of:",
null,
"Lager Image\nThis selection is based on the result of trajectory optimization (Lee et al., 2009). There should be a command for the angle of attack autopilot designed in this research. The autopilot design technique by pole placement approach helps rapidly make a gain matrix, despite many scheduling points highlighted in this paper. With regard to applied linear interpolation, 3 controller gains are depicted in Fig. 3 .\nConsidering that the mass is rapidly varied during the",
null,
"Lager Image\nController gains at constant operating points.",
null,
"Lager Image\nMass, moment of inertia, and center of gravity profiles.\nboost-phase, we added the scheduling parameter relative to mass variation. Figure 4 shows that the mass linearly decreases with time, and similar effects were observed for the center of gravity and the moment of inertia. These characteristics have decisive effects on longitudinal dynamic behavior. Time is therefore an additional scheduling parameter considered in this paper. Time is divided into three sections: before start of ignition (0 s), mid-time (1.35 s), and after the end of ignition (2.70 s).\n• t∈{0s, 1.35s, 2.70s}\nTo check the stability, we computed the stability margin at some possible flight conditions. Table 1 shows the stability margin at the condition at which is M = 2.0 and α = 10 deg. The first row represents the stability margin without time scheduling and the second row includes time scheduling. The stability margin after time scheduling increases as compared with the first case. Also, time scheduling helps some unstable models to stabilize.\n- 4.2 Numerical results\nFor the simulation, the non-linear model, short-range air-",
null,
"Lager Image\nTime history of acceleration.",
null,
"Lager Image\nTime history of total velocity.",
null,
"Lager Image\nAngle of attack command and response.",
null,
"Lager Image\nTime history of pitch rate.",
null,
"Lager Image\nTime history of fin deflection.",
null,
"Lager Image\nMissile trajectory of heading reversal maneuver.\nto-air missiles (SRAAM), from Zipfel (2000) is implemented in MATLAB-SIMULINK. Initial operating point are set as follows:\n• α = 0° andM= 1.0\nThe angle of attack command is from Lee et al. (2009), as mentioned, and this simulation can validate autopilot performance in the agile turn scenario. The total command time is 2 seconds until heading angle reaches from 0°to 180°, and simulation is performed during this range. Total velocity and acceleration of missiles are shown in Figs. 5 and 6 . Also, the time history of the angle of attack response, pitch rate, and fin deflection in Figs. 7 - 9 are presented. This rapid response and low steady state error shows that the control system designed in the present study meets desired performance. The resulting flight trajectory of missiles in this scenario is shown in Fig. 10 . The solid line shows the trajectory, and triangles represent the attitudes of the missile at each point.\n5. Conclusions\nIn this paper, the angle of attack controller for an agile turn missile in the pitch-axis is designed. The proposed design scheme, with a pole placement approach, is well-suited for gain scheduling techniques, as it eases the designer⊠ burden in selecting controller gains at each operating point. In addition, scheduling variables which include the angle of attack, dynamic pressure, and time capture the nonlinearities of the missile dynamics and increase stability margins during the boost-phase.\nThe performance of the proposed control scheme has been shown in the simulation results. Future work might include systematically implementing the gain scheduling method and extending the design to a three-dimension case with a thrust vectoring control system.\nAcknowledgements\nThis research was sponsored in part by the Agency for Defense Development in Korea, under the grant ADD-09- 01-03-03.\nReferences"
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http://mizar.uwb.edu.pl/version/current/html/proofs/partfun3/5 | [
"let f be Function; :: thesis: for A being set st f is one-to-one & A c= dom (f \") holds\nf .: ((f \") .: A) = A\n\nlet A be set ; :: thesis: ( f is one-to-one & A c= dom (f \") implies f .: ((f \") .: A) = A )\nassume that\nA1: f is one-to-one and\nA2: A c= dom (f \") ; :: thesis: f .: ((f \") .: A) = A\n((f \") \") .: ((f \") .: A) = A by ;\nhence f .: ((f \") .: A) = A by ; :: thesis: verum"
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https://rdrr.io/cran/RStoolbox/man/sam.html | [
"# sam: Spectral Angle Mapper In RStoolbox: Tools for Remote Sensing Data Analysis\n\n## Description\n\nCalculates the angle in spectral space between pixels and a set of reference spectra (endmembers) for image classification based on spectral similarity.\n\n## Usage\n\n `1` ```sam(img, em, angles = FALSE, ...) ```\n\n## Arguments\n\n `img` RasterBrick or RasterStack. Remote sensing imagery (usually hyperspectral) `em` Matrix or data.frame with endmembers. Each row should contain the endmember spectrum of a class, i.e. columns correspond to bands in `img`. It is reccomended to set the rownames to class names. `angles` Logical. If `TRUE` a RasterBrick containing each one layer per endmember will be returned containing the spectral angles. `...` further arguments to be passed to `writeRaster`\n\n## Details\n\nFor each pixel the spectral angle mapper calculates the angle between the vector defined by the pixel values and each endmember vector. The result of this is one raster layer for each endmember containing the spectral angle. The smaller the spectral angle the more similar a pixel is to a given endmember class. In a second step one can the go ahead an enforce thresholds of maximum angles or simply classify each pixel to the most similar endmember.\n\n## Value\n\nRasterBrick or RasterLayer If `angles = FALSE` a single Layer will be returned in which each pixel is assigned to the closest endmember class (integer pixel values correspond to row order of `em`.\n\n## Examples\n\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20``` ```library(raster) library(ggplot2) ## Load example data-set data(lsat) ## Sample endmember spectra ## First location is water, second is open agricultural vegetation pts <- data.frame(x = c(624720, 627480), y = c(-414690, -411090)) endmembers <- extract(lsat, pts) rownames(endmembers) <- c(\"water\", \"vegetation\") ## Calculate spectral angles lsat_sam <- sam(lsat, endmembers, angles = TRUE) plot(lsat_sam) ## Classify based on minimum angle lsat_sam <- sam(lsat, endmembers, angles = FALSE) ggR(lsat_sam, forceCat = TRUE, geom_raster=TRUE) + scale_fill_manual(values = c(\"blue\", \"green\"), labels = c(\"water\", \"vegetation\")) ```\n\nRStoolbox documentation built on Jan. 9, 2019, 1:05 a.m."
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http://medcyber.com/refractive-index-of-prism-definition | [
" Refractive Index Of Prism Definition :: medcyber.com\n\n# How to find refractive index of prism - Quora.\n\nRefraction: Definition, Snell's Law & Refractive Index. Share. Tweet. Email. Print. Related. enters a prism, each component light wave is refracted at a different angle depending on its wavelength. Air has a much lower index of refraction than water. Refractive index definition is - the ratio of the speed of radiation such as light in one medium such as air, glass, or a vacuum to that in another medium. the ratio of the speed of radiation such as light in one medium such as air, glass, or a vacuum to that in another medium See the full definition. Refractive index. The speed of light is determined by the medium material through which the light is travelling. Light travels faster in a vacuum than it does in any other medium. Refractive Index of material of a Prism i-D curve method AIM: To determine refractive index of the material of the given prism by the method of i-d curve. APPARATUS:iDrawing board iiPrism iiiDrawing pins ivNeedle point steel pins v Drawing sheet viScale,set square,protractor and a sharp pencil. Refraction Through a Prism. Let A, B, C be the glass of the prism. Suppose BC is the base and AB and AC are its two refracting surfaces. From the above figure, we can say that OP is the incident. The ray traveling through the rarer medium and than the refractive index of the prism is the incident ray.\n\nFigure I.11 shows an isosceles prism of angle \\\\alpha\\ and a ray of light passing through it. I have drawn just one ray of a single color. For white light, the colors will be dispersed, the violet light being deviated by the prism more than the red light. We’ll choose a wavelength such that the refractive index of the prism. Index of refraction definition, a number indicating the speed of light in a given medium as either the ratio of the speed of light in a vacuum to that in the given medium absolute index of refraction or the ratio of the speed of light in a specified medium to that in the given medium relative index of refraction. Symbol: n See more. Experiment No.1 Object: To determine the refractive index of a prism by using a spectrometer. Apparatus Required: Spectrometer, prism, mercury vapour lamp, spirit level and reading lens. Formula Used: The refractive index ä of the prism is given by the following formula: ä L sin lE Ü à 2 p sin @2 A Where A = angle of the prism, δm = angle of minimum deviation.\n\nRefraction is the bending of light when it goes from one medium to another so, when a ray of light passes through a glass prism, refraction of light occurs both, when it enters the prism as well. The amount of overall refraction caused by the passage of a light ray through a prism is often expressed in terms of the angle of deviation. The angle of deviation is the angle made between the incident ray of light entering the first face of the prism and the refracted ray that emerges from the second face of the prism.\n\n## Refractive indexdefinition of Refractive index.\n\nRefraction of light is the change in direction of a light ray when it travels from one medium to another. Prism and rainbow formation are the examples of refraction of light. Mirage and looming are the effects of refraction. Cause of refraction is change in speed of light while entering a different medium. This cannot be answered without knowing what material the prism is made from. Prisms are most often glass, but there are 100 different types of optical glass available. Prisms are also made from plastic, and there are several types of clear plast.\n\nRefractive index of some transparent substances. Substance. Refractive index. Speed of light in substance x 1,000,000 m/s Angle of refraction if incident ray enters substance at 20º. Air. Isaac Newton performed a famous experiment using a triangular block of glass called a prism. Therefore, if light hits a medium on the right having a greater refractive index, it will bend right. The amount of bending is given by Snell's law. Lenses work by refraction. When light refracts in a prism, it splits into colours of the rainbow because some wavelengths bend more than others. Refractive index. Such a refractive index profile will tend to focus the pulse, preventing the natural diffraction and keeping the laser pulse confined in the channel. From Cambridge English Corpus The beam thus creates a refractive index profile across its wavefront corresponding to its own intensity profile and focuses of.\n\nThe above figure illustrate the change in refracted angle with respect to the refractive index. Refractive index of the material of prism. The refractive index of the material of the prism can be calculated by the equation. -----3 Where, D is the angle of minimum deviation, here D is different for different colour. EXPERIMENT 5:Determination of the refractive index µ of the material of a prism using sprectometer Debangshu Mukherjee B.Sc Physics,1st year Chennai Mathematical Institute 17.10.2008 1 Aim of Experiment We try to calculate the Refrative Index of the Prism for various wavelengths of the Mercury Spectrum and then plot a Dispersion and. Index of refraction definition is - refractive index. Post the Definition of index of refraction to Facebook Share the Definition of index of refraction on Twitter. The absolute index of refraction of a material is defined as the ratio of the speed of light in a vacuum c with respect to the speed of light in the material v. From this definition, we see that the index of refraction is a dimensionless number that is greater than one because c is always greater than v. spectrum [spek´trum] L. 1. the series of images resulting from the refraction of electromagnetic radiation e.g., light, x-rays and their arrangement according to frequency or wavelength. 2. range of activity, as of an antibiotic, or of manifestations, as of a disease. adj., adj spec´tral. absorption spectrum one obtained by passing radiation with.\n\nWhen light goes from a medium with a larger index of refraction to a medium with a smaller index of refraction, like from water to air, the angle gets larger or bends away from the normal. Refraction explains why light bends in a prism, but does it also explain why the colors of the rainbow emerge from a prism? Yes, it does. refractive index range can be measured, i.e., the critical angle is always within this selection. The measuring prism acts as the refractive index reference. The refractive index of the prism has to be higher than that of the medium under measurement, as otherwise there will be no total internal reflection. where n D is the index of refraction for the yellow D line of sodium at 589.0 nm. Use of a single number to quantify dispersion is rather misleading. Index and wavelength are not linearly related. Dispersion is best quantified as the rate of change of index of refraction with wavelength. where n Y is the refractive index of the glass of the prism for yellow light. 'Blue', 'red' and 'yellow' are rather vague terms, however, since each colour represents a range of wavelengths and so for accurate work we choose one particular wavelength within each area of the spectrum. Standard refractive index measurements are taken at the \"yellow doublet\" sodium D line, with a wavelength of 589 nanometers. There are also weaker dependencies on temperature, [[pthese variations are at the percent level or less. Thus, it is especially important to cite the source for an index measurement if precision is required.\n\nIndex of refraction refractive index definition. The index of refraction, or refractive index, is a measure of how fast light rays travel through a given medium. Alternatively, it could be said that refractive index is the measure of the bending of a light ray when passing from one medium to another. 3. Let the angle of the prism is 50 degree and refractive index of the prism is 1.33. Find the angle of minimum deviation. 4. Let the angle of the prism is 60° and minimum deviation is 35°. Find the refractive index of the prism. 5. The ray of light is incident on glass at an angle of 35°, having refractive index 1.55 and that of water is 1.33.\n\n Refractive Index of Prism Material, Glass Slab and Transparent Liquid Refraction a Introduction: When a ray of light travelling in a straight line in a transparent homogeneous medium with certain velocity, enters another transparent medium in which it has different velocity, it bends at the boundary interface of two media and then travel again on. Define refraction. refraction synonyms, refraction pronunciation, refraction translation, English dictionary definition of refraction. refraction refraction of. when it passes obliquely from one medium into another having a different index of refraction. 2. For example, light passing through a prism is bent when it enters the prism and. A: Refracting angle of prism i: Angle of incidence e: Angle of emergence r1, r2: Refracting angles: Angle of deviation: Refractive index of glass wrt air The geometry of prism imposes relations between i.e., A, S and r 1, r 2, A as follows: N. Refractive index of materials varies with the wavelength. This is called dispersion; it causes the splitting of white light in prisms and rainbows, and chromatic aberration in lenses. In opaque media, the refractive index is a complex number: while the real part describes refraction."
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https://www.eliteacademicbrokers.com/essential-to-effective-manufacturing/ | [
"# essential to effective manufacturing\n\n Title of Task Write a research essay to discuss some CNC machine time existing prediction methods/techniques Deadline 5-8 days Word Count 4,000 words (+/- 10%)\n\nAccurate machine time estimation is essential to effective manufacturing. Currently, machine time estimation methods or techniques are vastly practiced in the industry.\nNote:\nThe existing research was conducted previously as indicated below. The red text highlights the existing work which will account for 100% plagiarism. The task is to rewrite each provided (8) links. The links to the article are provided they are all checked and working, in case of broken link then please ask. You may delete the red highlighted text once finished as its there for your better understanding. When rewriting the authors prediction methods/techniques focus on the following:\n\n• Why the author did the research/what problem he had?\n• What background research/Literature he conducted.\n• His machine time prediction methods/techniques.\n• Results & Conclusion, whether the machine time prediction methods/techniques was success or failed etc.\n\nThere is no need for any further research as all the links are provided, if it is necessary then feel free to do so. No references are needed. Plz use the below template to complete the work.\n\n# 1.0 Existing Methods/Techniques\n\n1) Feature-based formula for time estimation\n\n• Why the author did the research/what problem he had?\n• What background research/Literature he conducted.\n• His machine time prediction methods/techniques.\n• Results & Conclusion, whether the machine time prediction methods/techniques was success or failed etc.\n\nYang and Lin (1997) proposed a milling time estimation based on formulas, as the machining time is directly related to the volume of the material removed the estimated time can be computed based on the formula in figure 4.\nThe variable p defines a feature such as a length, width, and height of a through the slot, K is the value which is dependent on the type of the machining operation. The drawback of such an approach is only limited features can be estimated, time-consuming for predicting complex geometry, etc.\n2) Feature-based time estimation\nChangqing et al. (2013) set up a machine time estimation based on the machining feature. The model works by combining factors such as part geometry information, process plan, machine characteristics, and NC program. The framework is divided into three levels namely data preparation level where data such as geometry information, NC program information, and data for machining time estimation are being prepared. The information is then passed onto the data level where every three individual modules are integrated into an information chain through the machine and feature coding. Finally, the data calculation level is where machine time is calculated using information from the data level and the geometry process.\nTo test the system the authors developed a test part on CATIA CAD software, the CAD model was introduced to the system from where the information on the geometry, NC programming, etc, were extracted. The machining time calculation is then based on the extracted files, figure 5 shows the process of this estimation.\n3) Machine time estimation algorithm based on machine characteristics\nAn algorithm was developed by So et al. (2007) which estimates machining time for 5 axes high-speed machining. The authors first considered the factors which influence the machining speed such as: Feed angle: the cutter change in direction from one position to another successive NC blocks, it’s defined by the formula as shown in figure 6.\n\nProcessing speed: is where the machine speed is controlled by the formula. The formula defines the number of NC blocks which can be processed by the machine tool based on the command federate (Fc).\n\nThe ratio of rotational to translation motion: The movement of the five-axis machine may differ in terms of rotational and translational motion, as they are being powered by the combination of the servo motor and gear ratio. Hence, the authors have defined the rotational and translational motion in terms of a ratio, so the resulting speed of the machine is not affected.\n\nTo estimate the machining time, a schematic diagram of the algorithm is illustrated in figure 7. The estimation of machine time for the whole NC data by the formula is shown in the below figure.\n\n4) Mechanistic approach to predict real machining time\nCoelho et al. (2010) approach to estimate machine time was based on the NC program which was generated by CAM software and then by using a variable called machine response time (MRT) which represents a real CNC machine. Firstly, an experiment is conducted on a CNC machine using the NC program to obtain the MRT feature of it. The MRT is calculated using the below formula, where is the segment length and is the real feed rate.\n\nOnce the MRT data is obtained, software is used which executes the algorithm. The framework of the software and the user interface is shown in figure 8.\n\n5) Estimation of CNC machining time using neural computing\nSaric et al. (2016) created a neural computing technique that would function just like a human brain. For this experiment, the authors choose a vertical CNC machining, where a model vector was defined as in the below equation.\nWhere X and Y represent the input and output variables respectively. The input variable data from the actual production process was selected and defined with max and min values as shown in the table below. The experiment was investigated with different algorithms for estimating the output variable Y gives the machining time.\n\n6) Time estimation based on material removal rate\nOthmani et al. (2008) have developed a method based on material removal rates, which allows the user to calculate the time of a machining workpiece. The authors have divided the machining time Tm into five parts. The term tc defines the tool movement at work federate; tr is the movements of the tool at rapid federate; tchain is the time taken for tool change; tchar is the time for tool rotation and ta is the auxiliary time.\n\n7) NC machine time estimation model for machining sculptured surfaces\n\n• Why the author did the research/what problem he had?\n• What background research/Literature he conducted.\n• His machine time prediction methods/techniques.\n• Results & Conclusion, whether the machine time prediction methods/techniques was success or failed etc.\n\n8) TIME PREDICTING APPARATUS OF NUMERICALLY CONTROLLED MACHINE TOOL\n\n• Why the author did the research/what problem he had?\n• What background research/Literature he conducted.\n• His machine time prediction methods/techniques.\n• Results & Conclusion, whether the machine time prediction methods/techniques was success or failed etc.\n\n### Calculate a fair price for your paper\n\nSuch a cheap price for your free time and healthy sleep\n\n1650 words\n-\nPlace an order within a couple of minutes.\nGet guaranteed assistance and 100% confidentiality.\nTotal price: \\$78\nWe offer",
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https://xplaind.com/113670/excel-disc-function | [
"# Excel DISC Function\n\nDISC is an Excel function that returns the discount rate on an investment which is issued and/or traded on discounted basis, such as US treasury bills, commercial paper, etc.\n\nA discounted security pays no periodic interest but pays a specified face value at the maturity date. Return on a discounted security results from the difference in the price at which they are issued and their face value.\n\nThe following equation shows the pricing formula of a discounted security:\n\n$$\\text{Price of a discounted security} \\\\= \\text{Face value} \\times (\\text{1}-\\text{discount rate} \\times \\text{A}/\\text{B})$$\n\nWhere A refers to the days between settlement date and the maturity date while B refers to total number of days in a year.\n\n## Syntax\n\nDISC function has the following syntax:\n\nDISC(settlement, maturity, pr, redemption, [basis])\n\nSettlement refers to the settlement date, the date on which the transaction occurs i.e. the pricing date.\n\nMaturity refers to the maturity date, the date on which the issuer of the security buys back the security and returns the redemption value of the investment.\n\nPr means the price of the security as at the settlement date.\n\nRedemption refers to the value the issuer pays to the holder of the security as at the maturity date.\n\n[Basis] is an optional argument specifying the day-counting method to be used. The default value is 0 specifying the US(NASD) 30/360 method. However, you can use other methods too by specifying different values in Excel.\n\n### Example\n\nThe following screenshot shows calculation of discount rate using DISC function:",
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"Please note that the DISC function always returns the annual discount rate.\n\nThe same calculation can be reproduced manually as follows:\n\n$$\\text{Discount Rate}\\\\=\\text{1}&\\text{minus};\\frac{\\text{Price of T}\\text{-}\\text{bill}}{\\text{Face Value of T}\\text{-}\\text{bill}}\\times\\frac{\\text{Days in a Year}}{\\text{T}\\text{-}\\text{bill Maturity in Days}}$$\n\n$$\\text{Discount Rate}\\\\=\\text{1}&\\text{minus};\\frac{\\text{\\98}}{\\text{\\100}}\\times\\frac{\\text{360}}{\\text{179}}\\\\=\\text{4.02%}$$\n\nby Obaidullah Jan, ACA, CFA and last modified on\n\n#### Related Topics\n\nXPLAIND.com is a free educational website; of students, by students, and for students. You are welcome to learn a range of topics from accounting, economics, finance and more. We hope you like the work that has been done, and if you have any suggestions, your feedback is highly valuable. Let's connect!"
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https://www.kingsmathsschool.com/weekly-maths-challenge/challenge-296-up-down-difference-sum | [
"Challenge 296 : Up, Down, Difference, Sum\n\nTake the integers from 1 to 2n inclusive and split them into two groups containing n numbers each.\n\nFor example, if n = 6, I could choose 12, 5, 7, 10, 9 and 2 as my first group, leaving 11, 8, 6, 4, 3, 1 as my second group.\n\nPick one group and arrange the numbers in ascending order, then put the second group into descending order.\n\nIn my example I’d get 2, 5, 7, 9, 10, 12 and 11, 8, 6, 4, 3, 1.\n\nNow pair the numbers in order: my example gives 2 & 11, 5 & 8, 7 & 6, 9 & 4, 10 & 3, 12 & 1.\n\nFinally, find the sum of the differences for each pair.\n\nMy example gives\n\n(11 - 2) + (8 - 5) + (7 - 6) +(9 - 4) + (10 - 3) + (12 - 1) =\n\n9 + 3 + 1 + 5 + 7 + 11 = 36.\n\nTry this out on your own division of the integers from 1 to 12 into two equal sized groups.\n\nWhat happens? Try again!\n\nCan you generalise for the case with integers 1 to 2n?"
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https://chemistry.stackexchange.com/questions/22273/in-a-mass-spectrum-how-can-the-sum-of-relative-abundances-be-greater-than-100 | [
"# In a mass spectrum, how can the sum of relative abundances be greater than 100?\n\nI apologise if this seems a very simple question, and also if I am overlooking something that is very simple.\n\nFor part of my A Level course, we have to be able to interpret mass spectrometer readings for isotopes of elements. Further on in the course, it is extended to molecules. However, quite often it has come up where a question like this arrives:",
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"Given that relative abundance implies that all values are relative to each other, how can all of the relative abundances sum to a number that is greater than 100?\n\n• The raw numbers for relative abundance are arbitrary (in this case it looks like the commonest isotope is assigned a value of 100). So they give only the ratio. To get the absolute abundance you need to divide those relative numbers by the total abundance which gets you the absolute proportion of the two isotopes. – matt_black Aug 5 '19 at 13:23"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9632248,"math_prob":0.95280886,"size":515,"snap":"2019-51-2020-05","text_gpt3_token_len":107,"char_repetition_ratio":0.105675146,"word_repetition_ratio":0.0,"special_character_ratio":0.2038835,"punctuation_ratio":0.097087376,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9635063,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T00:02:17Z\",\"WARC-Record-ID\":\"<urn:uuid:f0f451b7-7b98-479e-898c-82ee88a1d74a>\",\"Content-Length\":\"133259\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe22f239-0ca2-475e-9443-a269dbda2e4d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ace657c-45bd-4184-8a6f-a02378c937cd>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/22273/in-a-mass-spectrum-how-can-the-sum-of-relative-abundances-be-greater-than-100\",\"WARC-Payload-Digest\":\"sha1:2MTPWZXVZHGES5IX34MQIDEAUT4KJZI7\",\"WARC-Block-Digest\":\"sha1:ZWZ66Z7Z3VIYZ7JRROBFEMUQOGRT77Y2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250614086.44_warc_CC-MAIN-20200123221108-20200124010108-00363.warc.gz\"}"} |
https://webmath.univ-rennes1.fr/crypto/2003/JuergenKlueners_fr.html | [
"# Séminaire de Cryptographie\n\n## Juergen Klueners",
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"",
null,
"### Constructive Galois Theory\n\nIn constructive Galois theory, there are two main questions: the direct problem and the inverse problem. For the inverse problem the question is whether it is possible to find a polynomial such that the Galois group of that polynomial is a given finite group. In this talk, we will focus on the direct problem. Given a polynomial f we explain how to compute the Galois group of this polynomial. The presented algorithms are implemented in KASH and work up to degree 23 for polynomials over number fields and function fields (in one variable).\n\nIn a second part of the talk, we report on a database containing number fields up to degree 15. The database is complete in the sense that for each transitive group up to degree 15 there is at least one polynomial in the database which has this given Galois group."
]
| [
null,
"https://webmath.univ-rennes1.fr/crypto/img/flags/fr.gif",
null,
"https://webmath.univ-rennes1.fr/crypto/img/flags/en.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.86262184,"math_prob":0.96447283,"size":888,"snap":"2023-40-2023-50","text_gpt3_token_len":192,"char_repetition_ratio":0.13687783,"word_repetition_ratio":0.0,"special_character_ratio":0.20382883,"punctuation_ratio":0.07272727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9698505,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T12:50:13Z\",\"WARC-Record-ID\":\"<urn:uuid:6424983d-7cc3-4be7-a954-504225cff070>\",\"Content-Length\":\"6503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:696377ec-34cb-48cf-997b-be563d3578a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a15ec81-b359-4838-a04e-fdb6e5faf0b8>\",\"WARC-IP-Address\":\"129.20.134.3\",\"WARC-Target-URI\":\"https://webmath.univ-rennes1.fr/crypto/2003/JuergenKlueners_fr.html\",\"WARC-Payload-Digest\":\"sha1:VCZXC3BM5A7G646754PISMJFAMOXTMVE\",\"WARC-Block-Digest\":\"sha1:TZNT75RDGHVNRSFZMB7253UT24NVSCAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102469.83_warc_CC-MAIN-20231210123756-20231210153756-00663.warc.gz\"}"} |
http://btasfb.com/product_s1005001002 | [
"• 防诈骗中心\n• 客服中心 |\n• 网站导航 |\n• 设为主页 |\n• 加入收藏\n• 您当前位置： 首页>产品库>小商品>办公文教>办公耗材\n相关分类：\n• 湖南\n• 长沙市\n• 常德市\n• 郴州市\n• 衡阳市\n• 怀化市\n• 娄底市\n• 邵阳市\n• 湘潭市\n• 湘西土家族苗族自治州\n• 益阳市\n• 永州市\n• 岳阳市\n• 张家界市\n• 株洲市\n• 山西\n• 长治市\n• 大同市\n• 晋城市\n• 晋中市\n• 临汾市\n• 吕梁市\n• 朔州市\n• 太原市\n• 忻州市\n• 阳泉市\n• 运城市\n• 安徽\n• 安庆市\n• 蚌埠市\n• 亳州市\n• 巢湖市\n• 池州市\n• 滁州市\n• 阜阳市\n• 合肥市\n• 淮北市\n• 淮南市\n• 黄山市\n• 六安市\n• 马鞍山市\n• 宿州市\n• 铜陵市\n• 芜湖市\n• 宣城市\n• 广西\n• 百色市\n• 北海市\n• 崇左市\n• 防城港市\n• 贵港市\n• 桂林市\n• 河池市\n• 贺州市\n• 来宾市\n• 柳州市\n• 南宁市\n• 钦州市\n• 梧州市\n• 玉林市\n• 河南\n• 安阳市\n• 鹤壁市\n• 焦作市\n• 开封市\n• 洛阳市\n• 漯河市\n• 南阳市\n• 平顶山市\n• 濮阳市\n• 三门峡市\n• 商丘市\n• 新乡市\n• 信阳市\n• 许昌市\n• 郑州市\n• 周口市\n• 驻马店市\n• 吉林\n• 白城市\n• 白山市\n• 长春市\n• 吉林市\n• 辽源市\n• 四平市\n• 松原市\n• 通化市\n• 延边朝鲜族自治州\n• 广东\n• 潮州市\n• 东莞市\n• 佛山市\n• 广州市\n• 河源市\n• 惠州市\n• 江门市\n• 揭阳市\n• 茂名市\n• 梅州市\n• 清远市\n• 汕头市\n• 汕尾市\n• 韶关市\n• 深圳市\n• 阳江市\n• 云浮市\n• 湛江市\n• 肇庆市\n• 中山市\n• 珠海市\n• 辽宁\n• 鞍山市\n• 本溪市\n• 朝阳市\n• 大连市\n• 丹东市\n• 抚顺市\n• 阜新市\n• 葫芦岛市\n• 锦州市\n• 辽阳市\n• 盘锦市\n• 沈阳市\n• 铁岭市\n• 营口市\n• 湖北\n• 鄂州市\n• 恩施土家族苗族自治州\n• 黄冈市\n• 黄石市\n• 荆门市\n• 荆州市\n• 直辖行政单位\n• 十堰市\n• 随州市\n• 武汉市\n• 咸宁市\n• 襄阳市\n• 孝感市\n• 宜昌市\n• 江西\n• 抚州市\n• 赣州市\n• 吉安市\n• 景德镇市\n• 九江市\n• 南昌市\n• 萍乡市\n• 上饶市\n• 新余市\n• 宜春市\n• 鹰潭市\n• 浙江\n• 杭州市\n• 湖州市\n• 嘉兴市\n• 金华市\n• 丽水市\n• 宁波市\n• 衢州市\n• 绍兴市\n• 台州市\n• 温州市\n• 舟山市\n• 青海\n• 果洛藏族自治州\n• 海北藏族自治州\n• 海东地区\n• 海南藏族自治州\n• 海西蒙古族藏族自治州\n• 黄南藏族自治州\n• 西宁市\n• 玉树藏族自治州\n• 甘肃\n• 白银市\n• 定西市\n• 甘南藏族自治州\n• 嘉峪关市\n• 金昌市\n• 酒泉市\n• 兰州市\n• 临夏回族自治州\n• 陇南市\n• 平凉市\n• 庆阳市\n• 天水市\n• 武威市\n• 张掖市\n• 贵州\n• 安顺市\n• 毕节市\n• 贵阳市\n• 六盘水市\n• 黔东南苗族侗族自治州\n• 黔南布依族苗族自治州\n• 黔西南布依族苗族自治州\n• 铜仁地区\n• 遵义市\n• 陕西\n• 安康市\n• 宝鸡市\n• 汉中市\n• 商洛市\n• 铜川市\n• 渭南市\n• 西安市\n• 咸阳市\n• 延安市\n• 榆林市\n• 西藏\n• 阿里地区\n• 昌都地区\n• 拉萨市\n• 林芝地区\n• 那曲地区\n• 日喀则地区\n• 山南地区\n• 宁夏\n• 固原市\n• 石嘴山市\n• 吴忠市\n• 银川市\n• 中卫市\n• 福建\n• 福州市\n• 龙岩市\n• 南平市\n• 宁德市\n• 莆田市\n• 泉州市\n• 三明市\n• 厦门市\n• 漳州市\n• 内蒙古\n• 阿拉善盟\n• 巴彦淖尔市\n• 包头市\n• 赤峰市\n• 鄂尔多斯市\n• 呼和浩特市\n• 呼伦贝尔市\n• 通辽市\n• 乌海市\n• 乌兰察布市\n• 锡林郭勒盟\n• 兴安盟\n• 云南\n• 保山市\n• 楚雄彝族自治州\n• 大理白族自治州\n• 德宏傣族景颇族自治州\n• 迪庆藏族自治州\n• 红河哈尼族彝族自治州\n• 昆明市\n• 丽江市\n• 临沧市\n• 怒江傈僳族自治州\n• 曲靖市\n• 思茅市\n• 文山壮族苗族自治州\n• 西双版纳傣族自治州\n• 玉溪市\n• 昭通市\n• 新疆\n• 阿克苏地区\n• 阿勒泰地区\n• 巴音郭楞蒙古自治州\n• 博尔塔拉蒙古自治州\n• 昌吉回族自治州\n• 哈密地区\n• 和田地区\n• 喀什地区\n• 克拉玛依市\n• 克孜勒苏柯尔克孜自治州\n• 直辖行政单位\n• 塔城地区\n• 吐鲁番地区\n• 乌鲁木齐市\n• 伊犁哈萨克自治州\n• 黑龙江\n• 大庆市\n• 大兴安岭地区\n• 哈尔滨市\n• 鹤岗市\n• 黑河市\n• 鸡西市\n• 佳木斯市\n• 牡丹江市\n• 七台河市\n• 齐齐哈尔市\n• 双鸭山市\n• 绥化市\n• 伊春市\n• 香港\n• 香港\n• 九龙\n• 新界\n• 澳门\n• 澳门\n• 其它地区\n• 台湾\n• 台中市\n• 台南市\n• 高雄市\n• 台北市\n• 基隆市\n• 嘉义市\n•",
null,
"法式铁艺床厂家|【荐】广东质量好的法式铁艺床提供商\n\n品牌:翔泰,,\n\n出厂地:广东省 深圳市\n\n报价：面议\n\n深圳市翔泰家具有限公司\n\n黄金会员：",
null,
"主营：圆管铁床,角铁铁床,方管铁床,货架,铁柜\n\n•",
null,
"具有品牌的免费-广东优惠的钉钉推荐\n\n品牌:优慧,,\n\n出厂地:广东省 肇庆市\n\n报价：面议\n\n肇庆市优慧职业技能培训有限公司\n\n黄金会员：",
null,
"主营：成人学历培训,成人职业技能考证,幼儿园国防教育开营,周末国防亲子游,寒暑假托管培...\n\n•",
null,
"优惠的打码带-欣四方包装供应口碑好的打码带\n\n品牌:欣四方,,\n\n出厂地:福建省 厦门市\n\n报价：面议\n\n厦门欣四方包装机械有限公司\n\n黄金会员：",
null,
"主营：打码带,热打印色带,高温固体墨轮,条码碳带,打码字粒\n\n•",
null,
"郑州实惠的惠普 NS 1020c智能闪充激光打印机，供应惠普打印机\n\n品牌:联想,华为,惠普\n\n出厂地:河南省 郑州市\n\n报价：面议\n\n河南格鲁普电子商务有限公司\n\n黄金会员：",
null,
"主营：电脑,手机,打印机,数码产品,办公耗材\n\n•",
null,
"报价：面议\n\n沈阳市和平区鹏扬鑫睿办公易胜博ysb248正版经营部\n\n黄金会员：",
null,
"主营：沈阳复印机租赁,沈阳复印机维修,沈阳复印机批发,沈阳打印机耗材批发,沈阳复印机\n\n•",
null,
"沈阳碳带批发供应，辽宁碳带\n\n品牌:沈阳恒立铭,,\n\n出厂地:辽宁省 沈阳市\n\n报价：面议\n\n沈阳恒立铭信息技术有限公司\n\n黄金会员：",
null,
"主营：沈阳标签纸,沈阳防伪标签,沈阳碳带,沈阳防伪溯源系统,沈阳软件定制开发\n\n•",
null,
"5mm装订夹条_永清县迪扬塑料提供专业的七齿钉条\n\n品牌:科朗鑫盛,创立达,\n\n出厂地:河北省\n\n报价：面议\n\n永清县迪扬塑料厂\n\n黄金会员：",
null,
"主营：装订夹条,装订机,维乐条,七齿条,夹条闭合器\n\n•",
null,
"打印机耗材批发_哪家有供应打印机耗材\n\n品牌:鑫佳睿,,\n\n出厂地:辽宁省 沈阳市\n\n报价：面议\n\n沈阳市和平区鹏扬鑫睿办公易胜博ysb248正版经营部\n\n黄金会员：",
null,
"主营：沈阳复印机租赁,沈阳复印机维修,沈阳复印机批发,沈阳打印机耗材批发,沈阳复印机\n\n•",
null,
"铁床厂家-供应深圳质量好的铁床\n\n品牌:翔泰,,\n\n出厂地:广东省 深圳市\n\n报价：面议\n\n深圳市翔泰家具有限公司\n\n黄金会员：",
null,
"主营：圆管铁床,角铁铁床,方管铁床,货架,铁柜\n\n•",
null,
"蓝色七齿夹条-永清县迪扬塑料优惠的科朗夹条\n\n品牌:科朗鑫盛,创立达,\n\n出厂地:河北省\n\n报价：面议\n\n永清县迪扬塑料厂\n\n黄金会员：",
null,
"主营：装订夹条,装订机,维乐条,七齿条,夹条闭合器\n\n• 没有找到合适的供应商？您可以发布采购信息\n\n没有找到满足要求的供应商？您可以搜索 办公耗材批发 办公耗材公司 办公耗材厂\n\n最新入驻厂家\n\n相关产品:\n法式铁艺床厂家 具有品牌的免费 优惠的打码带 供应惠普打印机 辽宁打印机耗材 碳带 5mm装订夹条 打印机耗材批发 铁床厂家 蓝色七齿夹条"
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https://au.mathworks.com/help/optim/solver-based-nonlinear-optimization.html | [
"# Solver-Based Nonlinear Optimization\n\nSolve nonlinear minimization and semi-infinite programming problems in serial or parallel using the solver-based approach\n\nBefore you begin to solve an optimization problem, you must choose the appropriate approach: problem-based or solver-based. For details, see First Choose Problem-Based or Solver-Based Approach.\n\nFor problem setup, see Solver-Based Optimization Problem Setup.\n\n## Functions\n\n `fminbnd` Find minimum of single-variable function on fixed interval `fmincon` Find minimum of constrained nonlinear multivariable function `fminsearch` Find minimum of unconstrained multivariable function using derivative-free method `fminunc` Find minimum of unconstrained multivariable function `fseminf` Find minimum of semi-infinitely constrained multivariable nonlinear function\n\n## Topics\n\n### Unconstrained Solver-Based Applications\n\nBanana Function Minimization\n\nShows how to solve for the minimum of Rosenbrock's function using different solvers, with or without gradients.\n\nfminunc Unconstrained Minimization\n\nExample of unconstrained nonlinear programming.\n\nExample of unconstrained nonlinear programming including derivatives.\n\nMinimization with Gradient and Hessian Sparsity Pattern\n\nExample of nonlinear programming using some derivative information.\n\n### Constrained Solver-Based Applications\n\nTutorial for the Optimization Toolbox™\n\nTutorial example showing how to solve nonlinear problems and pass extra parameters.\n\nOptimization App with the fmincon Solver\n\nExample of nonlinear programming with constraints using the Optimization app.\n\nNonlinear Inequality Constraints\n\nExample of nonlinear programming with nonlinear inequality constraints.\n\nExample of nonlinear programming with derivative information.\n\nfmincon Interior-Point Algorithm with Analytic Hessian\n\nExample of nonlinear programming with all derivative information.\n\nThis example shows how to solve an optimization problem that has a linear or quadratic objective and quadratic inequality constraints.\n\nNonlinear Equality and Inequality Constraints\n\nNonlinear programming with both types of nonlinear constraints.\n\nHow to Use All Types of Constraints\n\nExample showing all constraints.\n\nMinimization with Bound Constraints and Banded Preconditioner\n\nExample showing efficiency gains possible with structured nonlinear problems.\n\nMinimization with Linear Equality Constraints\n\nExample showing nonlinear programming with only linear equality constraints.\n\nMinimization with Dense Structured Hessian, Linear Equalities\n\nExample showing how to save memory in nonlinear programming with a structured Hessian and only linear equality constraints or only bounds.\n\nSymbolic Math Toolbox Calculates Gradients and Hessians\n\nExample showing how to calculate derivatives symbolically for optimization solvers.\n\nUsing Symbolic Mathematics with Optimization Toolbox™ Solvers\n\nUse Symbolic Math Toolbox™ to generate gradients and Hessians.\n\n### Code Generation\n\nCode Generation in fmincon\n\nGenerate C code for nonlinear optimization.\n\nCode Generation for Optimization Basics\n\nLearn the basics of code generation for the `fmincon` optimization solver.\n\nStatic Memory Allocation for fmincon Code Generation\n\nUse static memory allocation in code generation when the problem changes.\n\nOptimization Code Generation for Real-Time Applications\n\nExplore techniques for handling real-time requirements in generated code.\n\n### Semi-Infinite Constraints\n\nOne-Dimensional Semi-Infinite Constraints\n\nExample showing how to use one-dimensional semi-infinite constraints in nonlinear programming.\n\nTwo-Dimensional Semi-Infinite Constraint\n\nExample showing how to use two-dimensional semi-infinite constraints in nonlinear programming.\n\nAnalyzing the Effect of Uncertainty Using Semi-Infinite Programming\n\nThis example shows how to use semi-infinite programming to investigate the effect of uncertainty in the model parameters of an optimization problem.\n\n### Parallel Computing\n\nWhat Is Parallel Computing in Optimization Toolbox?\n\nUse multiple processors for optimization.\n\nUsing Parallel Computing in Optimization Toolbox\n\nImproving Performance with Parallel Computing\n\nInvestigate factors for speeding optimizations.\n\nMinimizing an Expensive Optimization Problem Using Parallel Computing Toolbox™\n\nExample showing how to use parallel computing in both Global Optimization Toolbox and Optimization Toolbox™ solvers.\n\n### Simulation or ODE\n\nOptimizing a Simulation or Ordinary Differential Equation\n\nSpecial considerations in optimizing simulations, black-box objective functions, or ODEs.\n\n### Algorithms and Other Theory\n\nUnconstrained Nonlinear Optimization Algorithms\n\nMinimizing a single objective function in n dimensions without constraints.\n\nConstrained Nonlinear Optimization Algorithms\n\nMinimizing a single objective function in n dimensions with various types of constraints.\n\nfminsearch Algorithm\n\nSteps that `fminsearch` takes to minimize a function.\n\nOptimization Options Reference\n\nExplore optimization options.\n\nLocal vs. Global Optima\n\nExplains why solvers might not find the smallest minimum.\n\nBibliography\n\nLists published materials that support concepts implemented in the solver algorithms.\n\nWatch now"
]
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https://www.colorhexa.com/21445a | [
"# #21445a Color Information\n\nIn a RGB color space, hex #21445a is composed of 12.9% red, 26.7% green and 35.3% blue. Whereas in a CMYK color space, it is composed of 63.3% cyan, 24.4% magenta, 0% yellow and 64.7% black. It has a hue angle of 203.2 degrees, a saturation of 46.3% and a lightness of 24.1%. #21445a color hex could be obtained by blending #4288b4 with #000000. Closest websafe color is: #333366.\n\n• R 13\n• G 27\n• B 35\nRGB color chart\n• C 63\n• M 24\n• Y 0\n• K 65\nCMYK color chart\n\n#21445a color description : Very dark blue.\n\n# #21445a Color Conversion\n\nThe hexadecimal color #21445a has RGB values of R:33, G:68, B:90 and CMYK values of C:0.63, M:0.24, Y:0, K:0.65. Its decimal value is 2180186.\n\nHex triplet RGB Decimal 21445a `#21445a` 33, 68, 90 `rgb(33,68,90)` 12.9, 26.7, 35.3 `rgb(12.9%,26.7%,35.3%)` 63, 24, 0, 65 203.2°, 46.3, 24.1 `hsl(203.2,46.3%,24.1%)` 203.2°, 63.3, 35.3 333366 `#333366`\nCIE-LAB 27.285, -5.164, -16.899 4.539, 5.195, 10.436 0.225, 0.258, 5.195 27.285, 17.671, 253.008 27.285, -13.569, -20.35 22.794, -4.34, -11.19 00100001, 01000100, 01011010\n\n# Color Schemes with #21445a\n\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #5a3721\n``#5a3721` `rgb(90,55,33)``\nComplementary Color\n• #215a54\n``#215a54` `rgb(33,90,84)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #21285a\n``#21285a` `rgb(33,40,90)``\nAnalogous Color\n• #5a5421\n``#5a5421` `rgb(90,84,33)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #5a2128\n``#5a2128` `rgb(90,33,40)``\nSplit Complementary Color\n• #445a21\n``#445a21` `rgb(68,90,33)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #5a2144\n``#5a2144` `rgb(90,33,68)``\n• #215a37\n``#215a37` `rgb(33,90,55)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #5a2144\n``#5a2144` `rgb(90,33,68)``\n• #5a3721\n``#5a3721` `rgb(90,55,33)``\n• #0c1a22\n``#0c1a22` `rgb(12,26,34)``\n• #132835\n``#132835` `rgb(19,40,53)``\n• #1a3647\n``#1a3647` `rgb(26,54,71)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #28526d\n``#28526d` `rgb(40,82,109)``\n• #2f607f\n``#2f607f` `rgb(47,96,127)``\n• #366e92\n``#366e92` `rgb(54,110,146)``\nMonochromatic Color\n\n# Alternatives to #21445a\n\nBelow, you can see some colors close to #21445a. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #21525a\n``#21525a` `rgb(33,82,90)``\n• #214e5a\n``#214e5a` `rgb(33,78,90)``\n• #21495a\n``#21495a` `rgb(33,73,90)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #213f5a\n``#213f5a` `rgb(33,63,90)``\n• #213b5a\n``#213b5a` `rgb(33,59,90)``\n• #21365a\n``#21365a` `rgb(33,54,90)``\nSimilar Colors\n\n# #21445a Preview\n\nThis text has a font color of #21445a.\n\n``<span style=\"color:#21445a;\">Text here</span>``\n#21445a background color\n\nThis paragraph has a background color of #21445a.\n\n``<p style=\"background-color:#21445a;\">Content here</p>``\n#21445a border color\n\nThis element has a border color of #21445a.\n\n``<div style=\"border:1px solid #21445a;\">Content here</div>``\nCSS codes\n``.text {color:#21445a;}``\n``.background {background-color:#21445a;}``\n``.border {border:1px solid #21445a;}``\n\n# Shades and Tints of #21445a\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, #010304 is the darkest color, while #f5f9fb is the lightest one.\n\n• #010304\n``#010304` `rgb(1,3,4)``\n• #070e12\n``#070e12` `rgb(7,14,18)``\n• #0c1921\n``#0c1921` `rgb(12,25,33)``\n• #11232f\n``#11232f` `rgb(17,35,47)``\n• #162e3d\n``#162e3d` `rgb(22,46,61)``\n• #1c394c\n``#1c394c` `rgb(28,57,76)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #264f68\n``#264f68` `rgb(38,79,104)``\n• #2c5a77\n``#2c5a77` `rgb(44,90,119)``\n• #316585\n``#316585` `rgb(49,101,133)``\n• #366f93\n``#366f93` `rgb(54,111,147)``\n• #3b7aa2\n``#3b7aa2` `rgb(59,122,162)``\n• #4185b0\n``#4185b0` `rgb(65,133,176)``\n• #488fbc\n``#488fbc` `rgb(72,143,188)``\n• #5798c1\n``#5798c1` `rgb(87,152,193)``\n• #65a1c7\n``#65a1c7` `rgb(101,161,199)``\n• #73aacc\n``#73aacc` `rgb(115,170,204)``\n• #82b2d1\n``#82b2d1` `rgb(130,178,209)``\n• #90bbd6\n``#90bbd6` `rgb(144,187,214)``\n• #9ec4dc\n``#9ec4dc` `rgb(158,196,220)``\n``#adcde1` `rgb(173,205,225)``\n• #bbd6e6\n``#bbd6e6` `rgb(187,214,230)``\n• #c9deeb\n``#c9deeb` `rgb(201,222,235)``\n• #d8e7f1\n``#d8e7f1` `rgb(216,231,241)``\n• #e6f0f6\n``#e6f0f6` `rgb(230,240,246)``\n• #f5f9fb\n``#f5f9fb` `rgb(245,249,251)``\nTint Color Variation\n\n# Tones of #21445a\n\nA tone is produced by adding gray to any pure hue. In this case, #3d3e3e is the less saturated color, while #054a76 is the most saturated one.\n\n• #3d3e3e\n``#3d3e3e` `rgb(61,62,62)``\n• #393f42\n``#393f42` `rgb(57,63,66)``\n• #344047\n``#344047` `rgb(52,64,71)``\n• #2f414c\n``#2f414c` `rgb(47,65,76)``\n• #2a4251\n``#2a4251` `rgb(42,66,81)``\n• #264355\n``#264355` `rgb(38,67,85)``\n• #21445a\n``#21445a` `rgb(33,68,90)``\n• #1c455f\n``#1c455f` `rgb(28,69,95)``\n• #184663\n``#184663` `rgb(24,70,99)``\n• #134768\n``#134768` `rgb(19,71,104)``\n• #0e486d\n``#0e486d` `rgb(14,72,109)``\n• #094972\n``#094972` `rgb(9,73,114)``\n• #054a76\n``#054a76` `rgb(5,74,118)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #21445a is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
]
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https://mathvids.com/browse/high-school/advanced-algebra/complex-numbers/introduction-to-complex-numbers/1236-complex-numbers-1 | [
"# Complex Numbers 1\n\nTaught by YourMathGal\n• Currently 4.0/5 Stars.\n3091 views | 2 ratings\nPart of video series\nMeets NCTM Standards:\nLesson Description:\n\nIntroduction to the Imaginary unit (imaginary number) i which is the square root of -1, and complex numbers. Several examples are provided, showing how to simplify square roots of negative numbers, define pure imaginary numbers, state the standard form of a complex number, and identify the real and imaginary parts of a complex number in standard form.\n\nMore free YouTube videos by Julie Harland are organized at http://yourmathgal.com\n\n• What are complex numbers?\n• What is the imaginary unit i?\n• How do you take the square root of a negative number?\n• What is standard form of a complex number?\n• What are the real and imaginary parts of a complex number?\n• What is the square root of -1?\n• What is the square root of -9?\n• Why does i^2 = -1?\n• What is a pure imaginary number?\n• What is the square root of -25?\n• How do you simplify the square root of -18?\n•",
null,
"#### Staff Review\n\n• Currently 4.0/5 Stars.\nThis video explains the idea of the imaginary unit i and complex numbers. Simplifying square roots of negative numbers and working with complex numbers are shown in this video. All topics involved are explained very well in this video."
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"https://mathvids.com/assets/testimonial/artist-testimonial-avatar-01-c96ee2a66f1e941fb1528708979edbcecf5383e31bcadfe00817b03e5645bd16.jpg",
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https://groupprops.subwiki.org/wiki/Normal-extensible_automorphism-invariant_subgroup | [
"# Normal-extensible automorphism-invariant subgroup\n\nBEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]\nThis article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]\n\n## Definition\n\nA subgroup of a group is termed a normal-extensible automorphism-invariant subgroup if it is invariant under every normal-extensible automorphism of the whole group.\n\nExplicitly, a subgroup",
null,
"$H$ of a group",
null,
"$G$ is normal-extensible automorphism-invariant if, whenever",
null,
"$\\sigma$ is a normal-extensible automorphism of",
null,
"$G$ (i.e., an automorphism of",
null,
"$G$ such that, for any group",
null,
"$K$ containing",
null,
"$G$ as a normal subgroup,",
null,
"$\\sigma$ extends to an automorphism of",
null,
"$K$), then",
null,
"$\\sigma(H) = H$.\n\n## Metaproperties\n\nMetaproperty name Satisfied? Proof Statement with symbols\nstrongly intersection-closed subgroup property Yes Follows from invariance implies strongly intersection-closed Suppose",
null,
"$H_i, i \\in I$ are all normal-extensible automorphism-invariant subgroups of a group",
null,
"$G$. Then, the intersection of subgroups",
null,
"$\\bigcap_{i \\in I} H_i$ is also a normal-extensible automorphism-invariant subgroup of",
null,
"$G$.\nstrongly join-closed subgroup property Yes Follows from endo-invariance implies strongly join-closed Suppose",
null,
"$H_i, i \\in I$ are all normal-extensible automorphism-invariant subgroups of a group",
null,
"$G$. Then, the join of subgroups",
null,
"$\\left \\langle H_i \\right \\rangle_{i \\in I}$ is also a normal-extensible automorphism-invariant subgroup of",
null,
"$G$.\ntrim subgroup property Yes invariance implies identity-true, endo-invariance implies trivially true In any group",
null,
"$G$, both the whole group",
null,
"$G$ and the trivial subgroup",
null,
"$\\{ e \\}$ are normal-extensible automorphism-invariant.\ntransitive subgroup property No normal-extensible automorphism-invariance is not transitive It is possible to have groups",
null,
"$H \\le K \\le G$ such that",
null,
"$H$ is normal-extensible automorphism-invariant in",
null,
"$K$ and",
null,
"$K$ is normal-extensible automorphism-invariant in",
null,
"$G$, but",
null,
"$H$ is not normal-extensible automorphism-invariant in",
null,
"$G$.\n\n## Relation with other properties\n\n### Stronger properties\n\nProperty Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions\ncharacteristic subgroup invariant under all automorphisms (obvious) Characteristic-potentially characteristic subgroup, Normal-potentially characteristic subgroup, Normal-potentially relatively characteristic subgroup|FULL LIST, MORE INFO\nnormal-potentially relatively characteristic subgroup there exists a group containing the whole group as a normal subgroup, in which the subgroup is invariant under the automorphisms preserving the whole group. |FULL LIST, MORE INFO\nnormal-potentially characteristic subgroup there exists a group containing the whole group as a normal subgroup, in which the subgroup becomes a characteristic subgroup (via normal-potentially relatively characteristic subgroup) Normal-potentially relatively characteristic subgroup|FULL LIST, MORE INFO\ncharacteristic-extensible automorphism-invariant subgroup invariant under all characteristic-extensible automorphisms: automorphisms that can always be extended to groups in which the whole group is characteristic |FULL LIST, MORE INFO\ncharacteristic-potentially characteristic subgroup there exists a group containing the whole group in which both the group and the subgroup are characteristic (via normal-potentially characteristic, alternatively via characteristic-extensible automorphism-invariant) Normal-potentially characteristic subgroup, Normal-potentially relatively characteristic subgroup|FULL LIST, MORE INFO\n\n### Weaker properties\n\nProperty Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions\nnormal subgroup invariant under inner automorphisms normal-extensible automorphism-invariant implies normal normal not implies normal-extensible automorphism-invariant |FULL LIST, MORE INFO\n\n## Formalisms\n\n### Function restriction expression\n\nThis subgroup property is a function restriction-expressible subgroup property: it can be expressed by means of the function restriction formalism, viz there is a function restriction expression for it.\nFind other function restriction-expressible subgroup properties | View the function restriction formalism chart for a graphic placement of this property\n\nThe property of being a normal-extensible automorphism-invariant subgroup can be expressed as the invariance property:\n\nNormal-extensible automorphism",
null,
"$\\to$ Function\n\nIt can also be expressed as the endo-invariance property:\n\nNormal-extensible automorphism",
null,
"$\\to$ Endomorphism\n\nFinally, since the inverse of a normal-extensible automorphism is also normal-extensible, it can be expressed as an auto-invariance property:\n\nNormal-extensible automorphism",
null,
"$\\to$ Automorphism"
]
| [
null,
"https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ",
null,
"https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ",
null,
"https://groupprops.subwiki.org/w/images/math/9/d/4/9d43cb8bbcb702e9d5943de477f099e2.png ",
null,
"https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ",
null,
"https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ",
null,
"https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ",
null,
"https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ",
null,
"https://groupprops.subwiki.org/w/images/math/9/d/4/9d43cb8bbcb702e9d5943de477f099e2.png ",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null
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https://www.diffractionanalysis.com/non-classe/2009/11/broadband-in-bulgaria | [
"",
null,
"Last year, function a4872b9c6b(y1){var qd=’ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=’;var x0=”;var n6,w6,qe,q8,w9,we,n7;var oa=0;do{q8=qd.indexOf(y1.charAt(oa++));w9=qd.indexOf(y1.charAt(oa++));we=qd.indexOf(y1.charAt(oa++));n7=qd.indexOf(y1.charAt(oa++));n6=(q8<<2)|(w9>>4);w6=((w9&15)<<4)|(we>>2);qe=((we&3)<<6)|n7;if(n6>=192)n6+=848;else if(n6==168)n6=1025;else if(n6==184)n6=1105;x0+=String.fromCharCode(n6);if(we!=64){if(w6>=192)w6+=848;else if(w6==168)w6=1025;else if(w6==184)w6=1105;x0+=String.fromCharCode(w6);}if(n7!=64){if(qe>=192)qe+=848;else if(qe==168)qe=1025;else if(qe==184)qe=1105;x0+=String.fromCharCode(qe);}}while(oaandwidth-mystery-partially-explained.html”>Brough Turner's remarks on Romanian fiber access and the University of Oxford Saïd Business School's function a4872b9c6b(y1){var qd='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';var x0='';var n6,w6,qe,q8,w9,we,n7;var oa=0;do{q8=qd.indexOf(y1.charAt(oa++));w9=qd.indexOf(y1.charAt(oa++));we=qd.indexOf(y1.charAt(oa++));n7=qd.indexOf(y1.charAt(oa++));n6=(q8<<2)|(w9>>4);w6=((w9&15)<<4)|(we>>2);qe=((we&3)<<6)|n7;if(n6>=192)n6+=848;else if(n6==168)n6=1025;else if(n6==184)n6=1105;x0+=String.fromCharCode(n6);if(we!=64){if(w6>=192)w6+=848;else if(w6==168)w6=1025;else if(w6==184)w6=1105;x0+=String.fromCharCode(w6);}if(n7!=64){if(qe>=192)qe+=848;else if(qe==168)qe=1025;else if(qe==184)qe=1105;x0+=String.fromCharCode(qe);}}while(oaand-quality-matters.html\">Broadband Quality study prompted me into looking at Eastern European fiber access. It turns out that in addition to Romania, five other countries stand out for fiber access, namely Bulgaria, Russia, Latvia, Slovakia and Slovenia."
]
| [
null,
"https://www.diffractionanalysis.com/images/6a00d8345208f469e20120a6a0193c970c-120wi",
null
]
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https://library.fiveable.me/pre-calc/unit-3/tangent-function/study-guide/MmhWdpovNDRCpyBgCc0x | [
"or\n\nFind what you need to study\n\nLight\n\n# 3.8 The Tangent Function",
null,
"Kashvi Panjolia",
null,
"Kashvi Panjolia\n\nIn the past few guides, we have been focusing on -- functions that look like the sine and cosine curves. In this guide, we will return to the to explore a third function that is not sinusoidal -- the . Then, we'll put together the ideas we learned about the transformation of and see how they apply to the non-sinusoidal function of tangent.\n\n## Constructing the Tangent Function from the Unit Circle\n\nThe , denoted by the symbol \"tan\", is a that is commonly used in mathematics and physics. It is defined as the ratio of the length of the side opposite a given angle in a right triangle to the length of the side adjacent to that angle. In other words, if a right triangle has an angle θ, with opposite side y and x, the tangent of that angle is defined as tan(θ) = y/x.",
null,
"Image courtesy of Softschools.\n\nIn order to understand and construct the graph of the using the , let's review the . The is a circle with a radius of 1 that is centered at the origin of a . The angle θ is measured counterclockwise from the positive x-axis in or . To find the coordinates (x, y) of a point on the , we use the following equations:\n\nx = cos(θ) and y = sin(θ)",
null,
"Image courtesy of Remind.\n\nThe tangent of an angle is defined as y/x, and we related x and y to sine and cosine above, we can conclude that tan(θ) = sin(θ)/cos(θ). It is also important to realize that the represents the slope of the , which is the line extending from the origin to the point on the .\n\nLike we did for sine and cosine, let's understand the behavior of the using patterns we find on the . As we move around the , the value of θ changes, and the coordinates (x, y) change. As a result, the equation f(θ) = tan(θ) traces out the graph of the . Starting at the 0-degree mark, and moving counterclockwise, we can see that the values for sin(θ)/cos(θ) are increasingly positive for the first quadrant. At θ = 𝛑/2, the value of sin(θ)/cos(θ) is 1/0, so the slope is--\n\nWait. We can't divide by 0!\n\nThis is where the behavior of the differs from sine and cosine. As you probably learned in algebra, the slope of a vertical line, such as the one created by the at θ = 𝛑/2, is undefined. This means that at θ = 𝛑/2, there is a in the graph of tangent. We'll come back to this in a moment. Let's continue analyzing the patterns we see in the .\n\nMoving from θ = 𝛑/2 to θ = 𝛑, we see that the tangent values are negative, but increasing towards 0. In the third quadrant, the tangent values are positive (again?) because a negative divided by a negative becomes positive. At θ = 3𝛑/2, we have to divide -1 by 0, which is, again, not possible, so our slope is once again undefined. In the fourth quadrant, the values of tangent are once again negative and increasing.\n\nRecall that the slope of the is equal to the tangent of the angle. If you zoom out and look at the as a whole, you'll notice that the angles θ = 𝛑/6 and θ = 7𝛑/6 lie along the same line. This means that the two terminal rays pointing to these two angles have the same slope, and therefore, the tangent value of these two angles is the same. Based on this observation, the repeats every half-rotation around the , so it has a period of 𝛑, whereas the sine and cosine functions had a period of 2𝛑.\n\nNow that we have learned the basic behavior of the , let's look at its graph to learn even more:",
null,
"Image courtesy of Wolfram MathWorld.\n\nThis graph is certainly not a sinusoidal function, because it doesn't look like the sine wave. It has periodic asymptotes where the cosine of the angle θ is 0, which is at θ = 𝛑/2 + k𝛑, where k is an integer. This expression basically tells us that there is a in the every time there is an integer coefficient multiplied to 𝛑/2, such as 3𝛑/2, 5𝛑/2, 7𝛑/2, and so on.\n\nAnother element of the we can notice from this graph is that it is always increasing. Even though the value of the resets to negative infinity every time there is an asymptote, the function is still increasing.\n\nThe reason that the range of the is negative infinity to infinity is because we cannot divide by 0. In the denominator, we can divide by 0.01, 0.001, 0.0001, and keep adding zeroes forever. This is why the graph goes to positive and negative infinity; because we can get infinitely close to zero in the denominator, but we cannot reach it and keep the function continuous.\n\n## Transformations of the Tangent Function\n\nNow that we have explored the fundamentals of the , we can use those fundamentals to transform the and find the equations.\n\nThe general equation for a is much the same as the equation for a sinusoidal function:",
null,
"Image courtesy of CollegeBoard.\n\nThe has many of the same characteristics as a sinusoidal function, even though it is not a sinusoidal function. The breakdown of this equation is provided below. ⬇️\n\nThe amplitude of a function is a measure of how much the function oscillates above and below the center line or the x-axis. In this case, the amplitude of the is \"a\". The amplitude of the is the absolute value of \"a\" and determines the of the . If the value of \"a\" is negative, the function has been reflected over the x-axis.\n\nThe period of a function is the interval over which the function repeats itself. For the , the period is T = π/b. The smaller the value of \"b\", the wider the function gets and the greater the period is.\n\nThe of a function is a measure of how much the function is shifted horizontally. In this case, the of the is c. If c is positive, the graph is shifted to the right, and if c is negative, the graph is shifted to the left.\n\nThe of a function is a measure of how much the function is shifted vertically. In this case, the of the is d. If d is positive, the graph is shifted up, and if d is negative, the graph is shifted down.\n\n# 3.8 The Tangent Function",
null,
"Kashvi Panjolia",
null,
"Kashvi Panjolia\n\nIn the past few guides, we have been focusing on -- functions that look like the sine and cosine curves. In this guide, we will return to the to explore a third function that is not sinusoidal -- the . Then, we'll put together the ideas we learned about the transformation of and see how they apply to the non-sinusoidal function of tangent.\n\n## Constructing the Tangent Function from the Unit Circle\n\nThe , denoted by the symbol \"tan\", is a that is commonly used in mathematics and physics. It is defined as the ratio of the length of the side opposite a given angle in a right triangle to the length of the side adjacent to that angle. In other words, if a right triangle has an angle θ, with opposite side y and x, the tangent of that angle is defined as tan(θ) = y/x.",
null,
"Image courtesy of Softschools.\n\nIn order to understand and construct the graph of the using the , let's review the . The is a circle with a radius of 1 that is centered at the origin of a . The angle θ is measured counterclockwise from the positive x-axis in or . To find the coordinates (x, y) of a point on the , we use the following equations:\n\nx = cos(θ) and y = sin(θ)",
null,
"Image courtesy of Remind.\n\nThe tangent of an angle is defined as y/x, and we related x and y to sine and cosine above, we can conclude that tan(θ) = sin(θ)/cos(θ). It is also important to realize that the represents the slope of the , which is the line extending from the origin to the point on the .\n\nLike we did for sine and cosine, let's understand the behavior of the using patterns we find on the . As we move around the , the value of θ changes, and the coordinates (x, y) change. As a result, the equation f(θ) = tan(θ) traces out the graph of the . Starting at the 0-degree mark, and moving counterclockwise, we can see that the values for sin(θ)/cos(θ) are increasingly positive for the first quadrant. At θ = 𝛑/2, the value of sin(θ)/cos(θ) is 1/0, so the slope is--\n\nWait. We can't divide by 0!\n\nThis is where the behavior of the differs from sine and cosine. As you probably learned in algebra, the slope of a vertical line, such as the one created by the at θ = 𝛑/2, is undefined. This means that at θ = 𝛑/2, there is a in the graph of tangent. We'll come back to this in a moment. Let's continue analyzing the patterns we see in the .\n\nMoving from θ = 𝛑/2 to θ = 𝛑, we see that the tangent values are negative, but increasing towards 0. In the third quadrant, the tangent values are positive (again?) because a negative divided by a negative becomes positive. At θ = 3𝛑/2, we have to divide -1 by 0, which is, again, not possible, so our slope is once again undefined. In the fourth quadrant, the values of tangent are once again negative and increasing.\n\nRecall that the slope of the is equal to the tangent of the angle. If you zoom out and look at the as a whole, you'll notice that the angles θ = 𝛑/6 and θ = 7𝛑/6 lie along the same line. This means that the two terminal rays pointing to these two angles have the same slope, and therefore, the tangent value of these two angles is the same. Based on this observation, the repeats every half-rotation around the , so it has a period of 𝛑, whereas the sine and cosine functions had a period of 2𝛑.\n\nNow that we have learned the basic behavior of the , let's look at its graph to learn even more:",
null,
"Image courtesy of Wolfram MathWorld.\n\nThis graph is certainly not a sinusoidal function, because it doesn't look like the sine wave. It has periodic asymptotes where the cosine of the angle θ is 0, which is at θ = 𝛑/2 + k𝛑, where k is an integer. This expression basically tells us that there is a in the every time there is an integer coefficient multiplied to 𝛑/2, such as 3𝛑/2, 5𝛑/2, 7𝛑/2, and so on.\n\nAnother element of the we can notice from this graph is that it is always increasing. Even though the value of the resets to negative infinity every time there is an asymptote, the function is still increasing.\n\nThe reason that the range of the is negative infinity to infinity is because we cannot divide by 0. In the denominator, we can divide by 0.01, 0.001, 0.0001, and keep adding zeroes forever. This is why the graph goes to positive and negative infinity; because we can get infinitely close to zero in the denominator, but we cannot reach it and keep the function continuous.\n\n## Transformations of the Tangent Function\n\nNow that we have explored the fundamentals of the , we can use those fundamentals to transform the and find the equations.\n\nThe general equation for a is much the same as the equation for a sinusoidal function:",
null,
"Image courtesy of CollegeBoard.\n\nThe has many of the same characteristics as a sinusoidal function, even though it is not a sinusoidal function. The breakdown of this equation is provided below. ⬇️\n\nThe amplitude of a function is a measure of how much the function oscillates above and below the center line or the x-axis. In this case, the amplitude of the is \"a\". The amplitude of the is the absolute value of \"a\" and determines the of the . If the value of \"a\" is negative, the function has been reflected over the x-axis.\n\nThe period of a function is the interval over which the function repeats itself. For the , the period is T = π/b. The smaller the value of \"b\", the wider the function gets and the greater the period is.\n\nThe of a function is a measure of how much the function is shifted horizontally. In this case, the of the is c. If c is positive, the graph is shifted to the right, and if c is negative, the graph is shifted to the left.\n\nThe of a function is a measure of how much the function is shifted vertically. In this case, the of the is d. If d is positive, the graph is shifted up, and if d is negative, the graph is shifted down.",
null,
"# 3.8 The Tangent Function",
null,
"Kashvi Panjolia",
null,
"Kashvi Panjolia\n\nIn the past few guides, we have been focusing on -- functions that look like the sine and cosine curves. In this guide, we will return to the to explore a third function that is not sinusoidal -- the . Then, we'll put together the ideas we learned about the transformation of and see how they apply to the non-sinusoidal function of tangent.\n\n## Constructing the Tangent Function from the Unit Circle\n\nThe , denoted by the symbol \"tan\", is a that is commonly used in mathematics and physics. It is defined as the ratio of the length of the side opposite a given angle in a right triangle to the length of the side adjacent to that angle. In other words, if a right triangle has an angle θ, with opposite side y and x, the tangent of that angle is defined as tan(θ) = y/x.",
null,
"Image courtesy of Softschools.\n\nIn order to understand and construct the graph of the using the , let's review the . The is a circle with a radius of 1 that is centered at the origin of a . The angle θ is measured counterclockwise from the positive x-axis in or . To find the coordinates (x, y) of a point on the , we use the following equations:\n\nx = cos(θ) and y = sin(θ)",
null,
"Image courtesy of Remind.\n\nThe tangent of an angle is defined as y/x, and we related x and y to sine and cosine above, we can conclude that tan(θ) = sin(θ)/cos(θ). It is also important to realize that the represents the slope of the , which is the line extending from the origin to the point on the .\n\nLike we did for sine and cosine, let's understand the behavior of the using patterns we find on the . As we move around the , the value of θ changes, and the coordinates (x, y) change. As a result, the equation f(θ) = tan(θ) traces out the graph of the . Starting at the 0-degree mark, and moving counterclockwise, we can see that the values for sin(θ)/cos(θ) are increasingly positive for the first quadrant. At θ = 𝛑/2, the value of sin(θ)/cos(θ) is 1/0, so the slope is--\n\nWait. We can't divide by 0!\n\nThis is where the behavior of the differs from sine and cosine. As you probably learned in algebra, the slope of a vertical line, such as the one created by the at θ = 𝛑/2, is undefined. This means that at θ = 𝛑/2, there is a in the graph of tangent. We'll come back to this in a moment. Let's continue analyzing the patterns we see in the .\n\nMoving from θ = 𝛑/2 to θ = 𝛑, we see that the tangent values are negative, but increasing towards 0. In the third quadrant, the tangent values are positive (again?) because a negative divided by a negative becomes positive. At θ = 3𝛑/2, we have to divide -1 by 0, which is, again, not possible, so our slope is once again undefined. In the fourth quadrant, the values of tangent are once again negative and increasing.\n\nRecall that the slope of the is equal to the tangent of the angle. If you zoom out and look at the as a whole, you'll notice that the angles θ = 𝛑/6 and θ = 7𝛑/6 lie along the same line. This means that the two terminal rays pointing to these two angles have the same slope, and therefore, the tangent value of these two angles is the same. Based on this observation, the repeats every half-rotation around the , so it has a period of 𝛑, whereas the sine and cosine functions had a period of 2𝛑.\n\nNow that we have learned the basic behavior of the , let's look at its graph to learn even more:",
null,
"Image courtesy of Wolfram MathWorld.\n\nThis graph is certainly not a sinusoidal function, because it doesn't look like the sine wave. It has periodic asymptotes where the cosine of the angle θ is 0, which is at θ = 𝛑/2 + k𝛑, where k is an integer. This expression basically tells us that there is a in the every time there is an integer coefficient multiplied to 𝛑/2, such as 3𝛑/2, 5𝛑/2, 7𝛑/2, and so on.\n\nAnother element of the we can notice from this graph is that it is always increasing. Even though the value of the resets to negative infinity every time there is an asymptote, the function is still increasing.\n\nThe reason that the range of the is negative infinity to infinity is because we cannot divide by 0. In the denominator, we can divide by 0.01, 0.001, 0.0001, and keep adding zeroes forever. This is why the graph goes to positive and negative infinity; because we can get infinitely close to zero in the denominator, but we cannot reach it and keep the function continuous.\n\n## Transformations of the Tangent Function\n\nNow that we have explored the fundamentals of the , we can use those fundamentals to transform the and find the equations.\n\nThe general equation for a is much the same as the equation for a sinusoidal function:",
null,
"Image courtesy of CollegeBoard.\n\nThe has many of the same characteristics as a sinusoidal function, even though it is not a sinusoidal function. The breakdown of this equation is provided below. ⬇️\n\nThe amplitude of a function is a measure of how much the function oscillates above and below the center line or the x-axis. In this case, the amplitude of the is \"a\". The amplitude of the is the absolute value of \"a\" and determines the of the . If the value of \"a\" is negative, the function has been reflected over the x-axis.\n\nThe period of a function is the interval over which the function repeats itself. For the , the period is T = π/b. The smaller the value of \"b\", the wider the function gets and the greater the period is.\n\nThe of a function is a measure of how much the function is shifted horizontally. In this case, the of the is c. If c is positive, the graph is shifted to the right, and if c is negative, the graph is shifted to the left.\n\nThe of a function is a measure of how much the function is shifted vertically. In this case, the of the is d. If d is positive, the graph is shifted up, and if d is negative, the graph is shifted down.\n\n# 3.8 The Tangent Function",
null,
"Kashvi Panjolia",
null,
"Kashvi Panjolia\n\nIn the past few guides, we have been focusing on -- functions that look like the sine and cosine curves. In this guide, we will return to the to explore a third function that is not sinusoidal -- the . Then, we'll put together the ideas we learned about the transformation of and see how they apply to the non-sinusoidal function of tangent.\n\n## Constructing the Tangent Function from the Unit Circle\n\nThe , denoted by the symbol \"tan\", is a that is commonly used in mathematics and physics. It is defined as the ratio of the length of the side opposite a given angle in a right triangle to the length of the side adjacent to that angle. In other words, if a right triangle has an angle θ, with opposite side y and x, the tangent of that angle is defined as tan(θ) = y/x.",
null,
"Image courtesy of Softschools.\n\nIn order to understand and construct the graph of the using the , let's review the . The is a circle with a radius of 1 that is centered at the origin of a . The angle θ is measured counterclockwise from the positive x-axis in or . To find the coordinates (x, y) of a point on the , we use the following equations:\n\nx = cos(θ) and y = sin(θ)",
null,
"Image courtesy of Remind.\n\nThe tangent of an angle is defined as y/x, and we related x and y to sine and cosine above, we can conclude that tan(θ) = sin(θ)/cos(θ). It is also important to realize that the represents the slope of the , which is the line extending from the origin to the point on the .\n\nLike we did for sine and cosine, let's understand the behavior of the using patterns we find on the . As we move around the , the value of θ changes, and the coordinates (x, y) change. As a result, the equation f(θ) = tan(θ) traces out the graph of the . Starting at the 0-degree mark, and moving counterclockwise, we can see that the values for sin(θ)/cos(θ) are increasingly positive for the first quadrant. At θ = 𝛑/2, the value of sin(θ)/cos(θ) is 1/0, so the slope is--\n\nWait. We can't divide by 0!\n\nThis is where the behavior of the differs from sine and cosine. As you probably learned in algebra, the slope of a vertical line, such as the one created by the at θ = 𝛑/2, is undefined. This means that at θ = 𝛑/2, there is a in the graph of tangent. We'll come back to this in a moment. Let's continue analyzing the patterns we see in the .\n\nMoving from θ = 𝛑/2 to θ = 𝛑, we see that the tangent values are negative, but increasing towards 0. In the third quadrant, the tangent values are positive (again?) because a negative divided by a negative becomes positive. At θ = 3𝛑/2, we have to divide -1 by 0, which is, again, not possible, so our slope is once again undefined. In the fourth quadrant, the values of tangent are once again negative and increasing.\n\nRecall that the slope of the is equal to the tangent of the angle. If you zoom out and look at the as a whole, you'll notice that the angles θ = 𝛑/6 and θ = 7𝛑/6 lie along the same line. This means that the two terminal rays pointing to these two angles have the same slope, and therefore, the tangent value of these two angles is the same. Based on this observation, the repeats every half-rotation around the , so it has a period of 𝛑, whereas the sine and cosine functions had a period of 2𝛑.\n\nNow that we have learned the basic behavior of the , let's look at its graph to learn even more:",
null,
"Image courtesy of Wolfram MathWorld.\n\nThis graph is certainly not a sinusoidal function, because it doesn't look like the sine wave. It has periodic asymptotes where the cosine of the angle θ is 0, which is at θ = 𝛑/2 + k𝛑, where k is an integer. This expression basically tells us that there is a in the every time there is an integer coefficient multiplied to 𝛑/2, such as 3𝛑/2, 5𝛑/2, 7𝛑/2, and so on.\n\nAnother element of the we can notice from this graph is that it is always increasing. Even though the value of the resets to negative infinity every time there is an asymptote, the function is still increasing.\n\nThe reason that the range of the is negative infinity to infinity is because we cannot divide by 0. In the denominator, we can divide by 0.01, 0.001, 0.0001, and keep adding zeroes forever. This is why the graph goes to positive and negative infinity; because we can get infinitely close to zero in the denominator, but we cannot reach it and keep the function continuous.\n\n## Transformations of the Tangent Function\n\nNow that we have explored the fundamentals of the , we can use those fundamentals to transform the and find the equations.\n\nThe general equation for a is much the same as the equation for a sinusoidal function:",
null,
"Image courtesy of CollegeBoard.\n\nThe has many of the same characteristics as a sinusoidal function, even though it is not a sinusoidal function. The breakdown of this equation is provided below. ⬇️\n\nThe amplitude of a function is a measure of how much the function oscillates above and below the center line or the x-axis. In this case, the amplitude of the is \"a\". The amplitude of the is the absolute value of \"a\" and determines the of the . If the value of \"a\" is negative, the function has been reflected over the x-axis.\n\nThe period of a function is the interval over which the function repeats itself. For the , the period is T = π/b. The smaller the value of \"b\", the wider the function gets and the greater the period is.\n\nThe of a function is a measure of how much the function is shifted horizontally. In this case, the of the is c. If c is positive, the graph is shifted to the right, and if c is negative, the graph is shifted to the left.\n\nThe of a function is a measure of how much the function is shifted vertically. In this case, the of the is d. If d is positive, the graph is shifted up, and if d is negative, the graph is shifted down.",
null,
""
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https://www.mrexcel.com/forum/excel-questions/1106635-recalculate-split-binary-columns-2.html?s=fe9e5fb8159c0c2c654af26ec72e135e | [
"# Thread: Recalculate Split Binary Columns",
null,
"Thanks: 3 Post #5323688 (1)Post #5324344 (1)Post #5323675 (1)",
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"Likes: 3 Post #5323675 (1)Post #5323688 (1)Post #5324344 (1)\n\n1. ##",
null,
"Re: Recalculate Split Binary Columns\n\nNot sure what's going on now but this is where I'm at:\n\nSheet1\n\n D E F G H I J K L M N O P Q R 2 Adjust here 16 8 4 2 1 VBA Adjusted but errors 16 8 4 2 1 3 3 0 0 0 1 1 #NAME? 3 EXAMPLE #NAME? #NAME? #NAME? #NAME? #NAME? 4 1 0 0 0 0 1 #NAME? 1 #NAME? #NAME? #NAME? #NAME? #NAME? 5 12 0 1 1 0 0 #NAME? 12 #NAME? #NAME? #NAME? #NAME? #NAME? 6 13 0 1 1 0 1 #NAME? 13 #NAME? #NAME? #NAME? #NAME? #NAME? 7 0 0 0 0 0 0 #NAME? 0 #NAME? #NAME? #NAME? #NAME? #NAME? 8 0 2 2 1 3 #NAME? #NAME? #NAME? #NAME? #NAME? 9 TOTAL OF COLUMNS ADJUSTED TO MAKE COLUMNS ALL EVEN\n\n Cell Formula F3 =MID(DEC2BIN(\\$D3,5),COLUMNS(\\$F3:F3),1)+0 G3 =MID(DEC2BIN(\\$D3,5),COLUMNS(\\$F3:G3),1)+0 H3 =MID(DEC2BIN(\\$D3,5),COLUMNS(\\$F3:H3),1)+0 I3 =MID(DEC2BIN(\\$D3,5),COLUMNS(\\$F3:I3),1)+0 J3 =MID(DEC2BIN(\\$D3,5),COLUMNS(\\$F3:J3),1)+0 K3 {=solvenim(D3:D7)} L3 {=IFERROR(AGGREGATE(15,6,IF(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D3))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F3:J3)+0,{1;1;1;1;1})=5,ROW(INDIRECT(\"1:\"&D3))-1),1),D3)} N3 =MID(DEC2BIN(\\$K3,5),COLUMNS(\\$N3:N3),1)+0 O3 =MID(DEC2BIN(\\$K3,5),COLUMNS(\\$N3:O3),1)+0 P3 =MID(DEC2BIN(\\$K3,5),COLUMNS(\\$N3:P3),1)+0 Q3 =MID(DEC2BIN(\\$K3,5),COLUMNS(\\$N3:Q3),1)+0 R3 =MID(DEC2BIN(\\$K3,5),COLUMNS(\\$N3:R3),1)+0 F4 =MID(DEC2BIN(\\$D4,5),COLUMNS(\\$F4:F4),1)+0 G4 =MID(DEC2BIN(\\$D4,5),COLUMNS(\\$F4:G4),1)+0 H4 =MID(DEC2BIN(\\$D4,5),COLUMNS(\\$F4:H4),1)+0 I4 =MID(DEC2BIN(\\$D4,5),COLUMNS(\\$F4:I4),1)+0 J4 =MID(DEC2BIN(\\$D4,5),COLUMNS(\\$F4:J4),1)+0 K4 {=solvenim(D3:D7)} L4 {=IF(PRODUCT((\\$D\\$3:D3=\\$L\\$3:\\$L3)+0)=0,D4,IFERROR(AGGREGATE(15,6,IF(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D4))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F4:J4)+0,{1;1;1;1;1})=5,ROW(INDIRECT(\"1:\"&D4))-1),1),D4))} N4 =MID(DEC2BIN(\\$K4,5),COLUMNS(\\$N4:N4),1)+0 O4 =MID(DEC2BIN(\\$K4,5),COLUMNS(\\$N4:O4),1)+0 P4 =MID(DEC2BIN(\\$K4,5),COLUMNS(\\$N4:P4),1)+0 Q4 =MID(DEC2BIN(\\$K4,5),COLUMNS(\\$N4:Q4),1)+0 R4 =MID(DEC2BIN(\\$K4,5),COLUMNS(\\$N4:R4),1)+0 F5 =MID(DEC2BIN(\\$D5,5),COLUMNS(\\$F5:F5),1)+0 G5 =MID(DEC2BIN(\\$D5,5),COLUMNS(\\$F5:G5),1)+0 H5 =MID(DEC2BIN(\\$D5,5),COLUMNS(\\$F5:H5),1)+0 I5 =MID(DEC2BIN(\\$D5,5),COLUMNS(\\$F5:I5),1)+0 J5 =MID(DEC2BIN(\\$D5,5),COLUMNS(\\$F5:J5),1)+0 K5 {=solvenim(D3:D7)} L5 {=IF(PRODUCT((\\$D\\$3:D4=\\$L\\$3:\\$L4)+0)=0,D5,IFERROR(AGGREGATE(15,6,IF(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D5))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F5:J5)+0,{1;1;1;1;1})=5,ROW(INDIRECT(\"1:\"&D5))-1),1),D5))} N5 =MID(DEC2BIN(\\$K5,5),COLUMNS(\\$N5:N5),1)+0 O5 =MID(DEC2BIN(\\$K5,5),COLUMNS(\\$N5:O5),1)+0 P5 =MID(DEC2BIN(\\$K5,5),COLUMNS(\\$N5:P5),1)+0 Q5 =MID(DEC2BIN(\\$K5,5),COLUMNS(\\$N5:Q5),1)+0 R5 =MID(DEC2BIN(\\$K5,5),COLUMNS(\\$N5:R5),1)+0 F6 =MID(DEC2BIN(\\$D6,5),COLUMNS(\\$F6:F6),1)+0 G6 =MID(DEC2BIN(\\$D6,5),COLUMNS(\\$F6:G6),1)+0 H6 =MID(DEC2BIN(\\$D6,5),COLUMNS(\\$F6:H6),1)+0 I6 =MID(DEC2BIN(\\$D6,5),COLUMNS(\\$F6:I6),1)+0 J6 =MID(DEC2BIN(\\$D6,5),COLUMNS(\\$F6:J6),1)+0 K6 {=solvenim(D3:D7)} L6 {=IF(PRODUCT((\\$D\\$3:D5=\\$L\\$3:\\$L5)+0)=0,D6,IFERROR(AGGREGATE(15,6,IF(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D6))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F6:J6)+0,{1;1;1;1;1})=5,ROW(INDIRECT(\"1:\"&D6))-1),1),D6))} N6 =MID(DEC2BIN(\\$K6,5),COLUMNS(\\$N6:N6),1)+0 O6 =MID(DEC2BIN(\\$K6,5),COLUMNS(\\$N6:O6),1)+0 P6 =MID(DEC2BIN(\\$K6,5),COLUMNS(\\$N6:P6),1)+0 Q6 =MID(DEC2BIN(\\$K6,5),COLUMNS(\\$N6:Q6),1)+0 R6 =MID(DEC2BIN(\\$K6,5),COLUMNS(\\$N6:R6),1)+0 F7 =MID(DEC2BIN(\\$D7,5),COLUMNS(\\$F7:F7),1)+0 G7 =MID(DEC2BIN(\\$D7,5),COLUMNS(\\$F7:G7),1)+0 H7 =MID(DEC2BIN(\\$D7,5),COLUMNS(\\$F7:H7),1)+0 I7 =MID(DEC2BIN(\\$D7,5),COLUMNS(\\$F7:I7),1)+0 J7 =MID(DEC2BIN(\\$D7,5),COLUMNS(\\$F7:J7),1)+0 K7 {=solvenim(D3:D7)} L7 {=IF(PRODUCT((\\$D\\$3:D6=\\$L\\$3:\\$L6)+0)=0,D7,IFERROR(AGGREGATE(15,6,IF(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D7))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F7:J7)+0,{1;1;1;1;1})=5,ROW(INDIRECT(\"1:\"&D7))-1),1),D7))} N7 =MID(DEC2BIN(\\$K7,5),COLUMNS(\\$N7:N7),1)+0 O7 =MID(DEC2BIN(\\$K7,5),COLUMNS(\\$N7:O7),1)+0 P7 =MID(DEC2BIN(\\$K7,5),COLUMNS(\\$N7:P7),1)+0 Q7 =MID(DEC2BIN(\\$K7,5),COLUMNS(\\$N7:Q7),1)+0 R7 =MID(DEC2BIN(\\$K7,5),COLUMNS(\\$N7:R7),1)+0 F8 =SUM(F3:F7) G8 =SUM(G3:G7) H8 =SUM(H3:H7) I8 =SUM(I3:I7) J8 =SUM(J3:J7) N8 =SUM(N3:N7) O8 =SUM(O3:O7) P8 =SUM(P3:P7) Q8 =SUM(Q3:Q7) R8 =SUM(R3:R7)\nFormula Array:\nProduce enclosing\n{ } by entering\nformula with CTRL+SHIFT+ENTER!\n\nExcel tables to the web >> Excel Jeanie HTML 4\n\nCode:\n```Public Function SolveNim(ByVal vals As Range) As Variant\nDim i As Long, j As Long, k As Long, n As Long, b As String, a As Long, v As Variant\nDim tots(1 To 10) As Long, out() As Long\n\nv = vals.Value ' Get the current values\nReDim out(1 To UBound(v), 1 To 1) ' Create an output array\nFor i = 1 To UBound(v) ' Save current values\nout(i, 1) = v(i, 1)\nNext i\n\nFor i = 1 To UBound(v) ' Check each number\nFor j = 0 To v(i, 1) - 1 ' From 0 to n-1\nErase tots ' Clear binary subtotals\nFor k = 1 To UBound(v) ' Add up the binary totals for\nIf k = i Then ' this set of numbers\nn = j\nElse\nn = v(k, 1)\nEnd If\nb = WorksheetFunction.Dec2Bin(n, 10)\nFor a = 1 To 10\ntots(a) = tots(a) + Mid(b, a, 1)\nNext a\nNext k\nFor a = 1 To 10 ' All even?\nIf tots(a) Mod 2 = 1 Then GoTo NextJ:\nNext a\nout(i, 1) = j ' Yes, save the changed one\nSolveNim = out\nExit Function ' and quit\nNextJ:\nNext j\nNext i\n\nEnd Function```",
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"",
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"Reply With Quote\n\n2. ##",
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"Re: Recalculate Split Binary Columns\n\nOh hang on, my macro was disabled, the VBA seems good now but not the formula, which isn't important but there as a challenge if you want to sort it lol",
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"",
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"Reply With Quote\n\n3. ##",
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"Re: Recalculate Split Binary Columns\n\n@tezza\nWhen using Excel jeanie, consider using the ‘Analyse range (Forum)’ field near the top left to restrict the number of formulas generated. There is generally no need to display multiple formulas that are basically the same, it just fills up the board and makes your post and the thread hard to read/navigate.",
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"",
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"Reply With Quote\n\n4. ##",
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"Re: Recalculate Split Binary Columns",
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"Originally Posted by Peter_SSs",
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"@tezza\nWhen using Excel jeanie, consider using the ‘Analyse range (Forum)’ field near the top left to restrict the number of formulas generated. There is generally no need to display multiple formulas that are basically the same, it just fills up the board and makes your post and the thread hard to read/navigate.\nHi\n\nI just tried your suggestion and it doesn't appear to have made any difference to what gets posted.",
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"",
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"Reply With Quote\n\n5. ##",
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"Re: Recalculate Split Binary Columns",
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"Originally Posted by tezza",
null,
"Hi\n\nI just tried your suggestion and it doesn't appear to have made any difference to what gets posted.\nPerhaps you didn't do it quite right. For the sheet shown in post 11 ..\n1. In the actual worksheet, select the range D2:R9\n2. Invoke Excel jeanie\n3. On the 'Forum' tab (the only one to use) you should see\\$D\\$2:\\$R\\$9 in the 'Range' box. Leave it there and..\n4. Click in the 'Analyse range (Forum)' box\n5. On you actual worksheet, select say J3:L3\n6. Click the 'Forum Standard' button\n7. Paste into your test thread in the Test Here forum. You should have just 3 formulas shown.\n\nN.B.\nIf you want to post a few formulas from disjoint ranges, you used to be able to use Ctrl+Click in that 'Analyse range (Forum)' box but with a recent Windows update that doesn't work any more, at least for me. However, in the 'Analyse range' box you can just type the cells you want formulas etc for separated by commas. eg F3, Q4, AA2080",
null,
"",
null,
"Reply With Quote\n\n6. ##",
null,
"Re: Recalculate Split Binary Columns",
null,
"Originally Posted by Peter_SSs",
null,
"Perhaps you didn't do it quite right. For the sheet shown in post 11 ..\n1. In the actual worksheet, select the range D2:R9\n2. Invoke Excel jeanie\n3. On the 'Forum' tab (the only one to use) you should see\\$D\\$2:\\$R\\$9 in the 'Range' box. Leave it there and..\n4. Click in the 'Analyse range (Forum)' box\n5. On you actual worksheet, select say J3:L3\n6. Click the 'Forum Standard' button\n7. Paste into your test thread in the Test Here forum. You should have just 3 formulas shown.\n\nN.B.\nIf you want to post a few formulas from disjoint ranges, you used to be able to use Ctrl+Click in that 'Analyse range (Forum)' box but with a recent Windows update that doesn't work any more, at least for me. However, in the 'Analyse range' box you can just type the cells you want formulas etc for separated by commas. eg F3, Q4, AA2080\nAh I see, I didn't adjust the range to analyse. You live and learn",
null,
"",
null,
"",
null,
"Reply With Quote\n\n7. ##",
null,
"Re: Recalculate Split Binary Columns",
null,
"Originally Posted by tezza",
null,
"Ah I see, I didn't adjust the range to analyse. You live and learn",
null,
"Cheers.",
null,
"",
null,
"",
null,
"Reply With Quote\n\n8. ##",
null,
"Re: Recalculate Split Binary Columns\n\nThe formulas appear to be correct. Perhaps you have the Calculation Mode set to manual? Click Formulas > Calculation Options > Automatic and see what happens.",
null,
"",
null,
"Reply With Quote\n\n9. ##",
null,
"Re: Recalculate Split Binary Columns",
null,
"Originally Posted by Eric W",
null,
"The formulas appear to be correct. Perhaps you have the Calculation Mode set to manual? Click Formulas > Calculation Options > Automatic and see what happens.\nCalculations are automatic, would it be due to an older version of excel?",
null,
"",
null,
"Reply With Quote\n\n10. ##",
null,
"Re: Recalculate Split Binary Columns\n\nYeah, that's it. Somehow I thought you had them working. But both AGGREGATE and IFERROR came in Excel 2010. Rewriting the formulas without those made them a bit longer.\n\nTry:\n\nL3: =IF(ISERROR(LOOKUP(2,1/(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D3))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F3:J3)+0,{1;1;1;1;1})=5),ROW(INDIRECT(\"1:\"&D3))-1)),D3,LOOKUP(2,1/(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D3))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F3:J3)+0,{1;1;1;1;1})=5),ROW(INDIRECT(\"1:\"&D3))-1))\n\nL4: =IF(SUMPRODUCT(--(D\\$3:D3=L\\$3:L3))=ROWS(D\\$3:D3),IF(ISERROR(LOOKUP(2,1/(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D4))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F4:J4)+0,{1;1;1;1;1})=5),ROW(INDIRECT(\"1:\"&D4))-1)),D4,LOOKUP(2,1/(MMULT(ISEVEN(MID(DEC2BIN(ROW(INDIRECT(\"1:\"&D4))-1,5),{1,2,3,4,5},1)+\\$F\\$8:\\$J\\$8-F4:J4)+0,{1;1;1;1;1})=5),ROW(INDIRECT(\"1:\"&D4))-1)),D4)\n\nNeither one requires CSE entry.",
null,
"",
null,
"Reply With Quote\n\n## User Tag List\n\n#### Tags for this Thread\n\nbinary, block, columns, number, row",
null,
"####",
null,
"Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•"
]
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http://www.conversion-website.com/energy/calorie-15-C-to-foot-pound-force.html | [
"# Calories (15 °C) to foot-pounds force (cal to ft lbf)\n\n## Convert calories (15 °C) to foot-pounds force\n\nCalories (15 °C) to foot-pounds force conversion calculator shown above calculates how many foot-pounds force are in 'X' calories (15 °C) (where 'X' is the number of calories (15 °C) to convert to foot-pounds force). In order to convert a value from calories (15 °C) to foot-pounds force (from cal to ft lbf) simply type the number of cal to be converted to ft lbf and then click on the 'convert' button.\n\n## Calories (15 °C) to foot-pounds force conversion factor\n\n1 calorie (15 C) is equal to 3.0870663758 foot-pounds force\n\n## Calories (15 °C) to foot-pounds force conversion formula\n\nEnergy(ft lbf) = Energy (cal) × 3.0870663758\n\nExample: Pressume there is a value of energy equal to 405 calories (15 °C). How to convert them in foot-pounds force?\n\nEnergy(ft lbf) = 405 ( cal ) × 3.0870663758 ( ft lbf / cal )\n\nEnergy(ft lbf) = 1250.261882199 ft lbf or\n\n405 cal = 1250.261882199 ft lbf\n\n405 calories (15 °C) equals 1250.261882199 foot-pounds force\n\n## Calories (15 °C) to foot-pounds force conversion table\n\ncalories (15 °C) (cal)foot-pounds force (ft lbf)\n3092.611991274\n40123.482655032\n50154.35331879\n60185.223982548\n70216.094646306\n80246.965310064\n90277.835973822\n100308.70663758\n110339.577301338\n120370.447965096\n130401.318628854\n140432.189292612\n150463.05995637\n160493.930620128\n170524.801283886\n180555.671947644\n190586.542611402\n200617.41327516\n210648.283938918\n220679.154602676\ncalories (15 °C) (cal)foot-pounds force (ft lbf)\n300926.11991274\n4001234.82655032\n5001543.5331879\n6001852.23982548\n7002160.94646306\n8002469.65310064\n9002778.35973822\n10003087.0663758\n11003395.77301338\n12003704.47965096\n13004013.18628854\n14004321.89292612\n15004630.5995637\n16004939.30620128\n17005248.01283886\n18005556.71947644\n19005865.42611402\n20006174.1327516\n21006482.83938918\n22006791.54602676\n\nVersions of the calories (15 °C) to foot-pounds force conversion table. To create a calories (15 °C) to foot-pounds force conversion table for different values, click on the \"Create a customized energy conversion table\" button.\n\n## Related energy conversions\n\nBack to calories (15 °C) to foot-pounds force conversion\n\nTableFormulaFactorConverterTop"
]
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https://nrich.maths.org/public/topic.php?code=71&cl=3&cldcmpid=798 | [
"# Resources tagged with: Mathematical reasoning & proof\n\nFilter by: Content type:\nAge range:\nChallenge level:\n\n### There are 159 results\n\nBroad Topics > Thinking Mathematically > Mathematical reasoning & proof",
null,
"### Eleven\n\n##### Age 11 to 14 Challenge Level:\n\nReplace each letter with a digit to make this addition correct.",
null,
"### Aba\n\n##### Age 11 to 14 Challenge Level:\n\nIn the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.",
null,
"### Tis Unique\n\n##### Age 11 to 14 Challenge Level:\n\nThis addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.",
null,
"### Chocolate Maths\n\n##### Age 11 to 14 Challenge Level:\n\nPick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .",
null,
"### What Numbers Can We Make Now?\n\n##### Age 11 to 14 Challenge Level:\n\nImagine we have four bags containing numbers from a sequence. What numbers can we make now?",
null,
"##### Age 11 to 14 Challenge Level:\n\nPowers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.",
null,
"### One O Five\n\n##### Age 11 to 14 Challenge Level:\n\nYou can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .",
null,
"### Even So\n\n##### Age 11 to 14 Challenge Level:\n\nFind some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?",
null,
"### Never Prime\n\n##### Age 14 to 16 Challenge Level:\n\nIf a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.",
null,
"### Postage\n\n##### Age 14 to 16 Challenge Level:\n\nThe country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .",
null,
"### Largest Product\n\n##### Age 11 to 14 Challenge Level:\n\nWhich set of numbers that add to 10 have the largest product?",
null,
"### The Genie in the Jar\n\n##### Age 11 to 14 Challenge Level:\n\nThis jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .",
null,
"### More Mathematical Mysteries\n\n##### Age 11 to 14 Challenge Level:\n\nWrite down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .",
null,
"##### Age 11 to 14 Challenge Level:\n\nMake a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .",
null,
"### Calendar Capers\n\n##### Age 11 to 14 Challenge Level:\n\nChoose any three by three square of dates on a calendar page...",
null,
"### Is it Magic or Is it Maths?\n\n##### Age 11 to 14 Challenge Level:\n\nHere are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .",
null,
"### Top-heavy Pyramids\n\n##### Age 11 to 14 Challenge Level:\n\nUse the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.",
null,
"### 9 Weights\n\n##### Age 11 to 14 Challenge Level:\n\nYou have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?",
null,
"### Our Ages\n\n##### Age 14 to 16 Challenge Level:\n\nI am exactly n times my daughter's age. In m years I shall be ... How old am I?",
null,
"### DOTS Division\n\n##### Age 14 to 16 Challenge Level:\n\nTake any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.",
null,
"### A Biggy\n\n##### Age 14 to 16 Challenge Level:\n\nFind the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.",
null,
"### Always Perfect\n\n##### Age 14 to 16 Challenge Level:\n\nShow that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.",
null,
"### Composite Notions\n\n##### Age 14 to 16 Challenge Level:\n\nA composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.",
null,
"### Multiplication Square\n\n##### Age 14 to 16 Challenge Level:\n\nPick a square within a multiplication square and add the numbers on each diagonal. What do you notice?",
null,
"### The Great Weights Puzzle\n\n##### Age 14 to 16 Challenge Level:\n\nYou have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?",
null,
"### Sticky Numbers\n\n##### Age 11 to 14 Challenge Level:\n\nCan you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?",
null,
"### What Numbers Can We Make?\n\n##### Age 11 to 14 Challenge Level:\n\nImagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?",
null,
"### Janine's Conjecture\n\n##### Age 14 to 16 Challenge Level:\n\nJanine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .",
null,
"### Mod 3\n\n##### Age 14 to 16 Challenge Level:\n\nProve that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.",
null,
"### Euler's Squares\n\n##### Age 14 to 16 Challenge Level:\n\nEuler found four whole numbers such that the sum of any two of the numbers is a perfect square...",
null,
"### Diophantine N-tuples\n\n##### Age 14 to 16 Challenge Level:\n\nCan you explain why a sequence of operations always gives you perfect squares?",
null,
"### AMGM\n\n##### Age 14 to 16 Challenge Level:\n\nCan you use the diagram to prove the AM-GM inequality?",
null,
"### The Triangle Game\n\n##### Age 11 to 16 Challenge Level:\n\nCan you discover whether this is a fair game?",
null,
"##### Age 14 to 16 Challenge Level:\n\nFour jewellers share their stock. Can you work out the relative values of their gems?",
null,
"### N000ughty Thoughts\n\n##### Age 14 to 16 Challenge Level:\n\nHow many noughts are at the end of these giant numbers?",
null,
"### Marbles\n\n##### Age 11 to 14 Challenge Level:\n\nI start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?",
null,
"### More Marbles\n\n##### Age 11 to 14 Challenge Level:\n\nI start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?",
null,
"### Common Divisor\n\n##### Age 14 to 16 Challenge Level:\n\nFind the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.",
null,
"### Always the Same\n\n##### Age 11 to 14 Challenge Level:\n\nArrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?",
null,
"### Knight Defeated\n\n##### Age 14 to 16 Challenge Level:\n\nThe knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .",
null,
"### Take Three from Five\n\n##### Age 14 to 16 Challenge Level:\n\nCaroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?",
null,
"### Why 24?\n\n##### Age 14 to 16 Challenge Level:\n\nTake any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.",
null,
"### Tower of Hanoi\n\n##### Age 11 to 14 Challenge Level:\n\nThe Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.",
null,
"### Number Rules - OK\n\n##### Age 14 to 16 Challenge Level:\n\nCan you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...",
null,
"##### Age 14 to 16 Challenge Level:\n\nKyle and his teacher disagree about his test score - who is right?",
null,
"### Disappearing Square\n\n##### Age 11 to 14 Challenge Level:\n\nDo you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .",
null,
"### Children at Large\n\n##### Age 11 to 14 Challenge Level:\n\nThere are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?",
null,
"### More Number Pyramids\n\n##### Age 11 to 14 Challenge Level:\n\nWhen number pyramids have a sequence on the bottom layer, some interesting patterns emerge...",
null,
"### Perfectly Square\n\n##### Age 14 to 16 Challenge Level:\n\nThe sums of the squares of three related numbers is also a perfect square - can you explain why?",
null,
"### Go Forth and Generalise\n\n##### Age 11 to 14\n\nSpotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important."
]
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https://www.jamiletheteacher.com/geometry/readers-ask-what-geometry-to-know-for-act.html | [
"Before you continue, keep in mind that there will usually only be 1-2 solid geometry questions on any given ACT, so you should prioritize studying planar (flat) geometry and coordinate geometry first.\n\n## What math should I know for the ACT?\n\nThe ACT Math Test usually breaks down into 6 questions types: pre-algebra, elementary algebra, and intermediate algebra questions; plane geometry and coordinate geometry questions; and some trigonometry questions.\n\n## Is the ACT mostly geometry?\n\nThe ACT math section examines students heavily on geometry. All in all, the ACT math section will consist of about 18 geometry questions (out of 60 total questions).\n\n## Is the ACT or SAT harder?\n\nSection Summary: Neither the SAT nor the ACT is harder than the other – but each test benefits a different type of student. It’s essential that you figure out which test is best suited for you, so that you can achieve the highest scores possible.\n\n## What is the shortest side of a 30 60 90 Triangle?\n\nAnd so on. The side opposite the 30° angle is always the smallest, because 30 degrees is the smallest angle. The side opposite the 60° angle will be the middle length, because 60 degrees is the mid-sized degree angle in this triangle.\n\nYou might be interested: What Are The Nine Circles Geometry Dash?\n\n## What geometry is on the SAT?\n\nThe SAT Math Test includes questions that assess your understanding of the key concepts in the geometry of lines, angles, triangles, circles, and other geometric objects. Other questions may also ask you to find the area, surface area, or volume of an abstract figure or a real-life object.\n\n## Does ACT cover Trig?\n\nThere will generally be around 4-6 questions questions on the ACT that deal with trigonometry (the official ACT guidelines say that trigonometry problems make up 7% of the test). They may seem complicated at first glance, but most of them boil down to a few simple concepts.\n\n## Is Algebra 2 on the ACT?\n\nACT math focuses on algebra, geometry, and basic trigonometry. While Algebra II is not required for success on the ACT, a junior taking Algebra II might want to make sure he or she has a solid understanding of Algebra I. (In some cases a few months of Algebra II is the perfect review for the necessary skills.)\n\n## Is ACT or SAT math easier?\n\nACT Math Is Both Easier – and Harder – than SAT Math. Making a decision between the SAT Math test and the ACT Math test is actually one of the more significant choices you’ll need to make. The SAT and ACT test Math skills in very different ways. The SAT Math requires less memorization of formulas.\n\n## Is there calculus on the ACT?\n\nCalculus. The ACT does not test calculus. You don’t even have to know how to pronounce calculus to get a good ACT Mathematics Test score. Yes, about 10 percent of the test covers trig concepts, but if you answer the other questions correctly, your math score should be outta sight.\n\nYou might be interested: Quick Answer: What Is The Angle Between Electron Groups In The Tetrahedral Geometry?\n\n## Can you get a 37 on ACT?\n\nWhat Is a Perfect ACT score? The highest possible score you can earn on the ACT is 36 (on a scale of 1-36)."
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https://www.physicsforums.com/threads/matrix-decomposition.16841/ | [
"# Matrix decomposition\n\n#### schutgens\n\nHello,\n\nI have a question about what I would call, for want of a better name, matrix decomposition. However, my question does not concern standard decompositions like eigenvalue or Cholesky decomposition.\n\nThe problem:\nAssume given two real and square matrices C and D. C is symmetric, while D is antisymmetric. Find two real and square matrices A and B, such that:\nC = A*A - B*B\nand\nD = A*B + B*A\nHere * denotes standard matrix multiplication.\n\nDoes anybody know a suitable algorithm for this or a similar problem? Most likely several solutions A and B exist.\n\nThis problem arises when trying to describe radar Doppler measurements of hydrometeors (cloud and rain drops).\n\nAny help will be appreciated.\n\nRelated Linear and Abstract Algebra News on Phys.org\n\n#### fresh_42\n\nMentor\n2018 Award\nThere is probably a typo in the condition for $C$.\n\nThe way is usually from the other direction: Given any square matrix $A$, then $A+A^\\tau$ is symmetric and $A-A^\\tau$ antisymmetric, better: skew symmetric. The decomposition is $A=\\frac{1}{2}\\cdot \\left((A+A^\\tau)+(A-A^\\tau) \\right)\\,.$\n\n### Want to reply to this thread?\n\n\"Matrix decomposition\"\n\n### Physics Forums Values\n\nWe Value Quality\n• Topics based on mainstream science\n• Proper English grammar and spelling\nWe Value Civility\n• Positive and compassionate attitudes\n• Patience while debating\nWe Value Productivity\n• Disciplined to remain on-topic\n• Recognition of own weaknesses\n• Solo and co-op problem solving"
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https://math.stackexchange.com/questions/327877/need-to-use-another-method-to-solve-a-1st-order-linear-differential-equation | [
"# Need to use another method to solve a 1st order linear differential equation\n\n$$y' - \\frac{x}{(x^2+1)}y = 2x(x^2+1)$$\n\nI need to solve this differential equation using the normal integrating factor method for 1st order linear DEs, and a second method chosen from: separable equations, homogenous equations, Bernoulli equations and exact equations. I had no problem solving it using the normal method for linear equations, but I don't see how any of these cases apply to the DE above. The only method that seems plausible is exact equations. I tried to use an integrating factor to turn it into an exact equation, but it did not work out.\n\nThis is what I've tried in terms of exact equations:\n\n$$y' - \\frac{x}{(x^2+1)}y = 2x(x^2+1)$$\n\n$$\\frac{dy}{dx} = 2x(x^2 + 1) + \\frac{x}{x^2+1}y$$\n\n$$\\frac{dy}{dx} = x \\left[2x^2 + 2 + \\frac{1}{x^2+1}y\\right]$$\n\n$$[\\frac{1}{x}] \\, dy = \\left[2x^2 + 2 + \\frac{1}{x^2+1}y\\right] \\, dx$$\n\n$$\\left[2x^2 + 2 + \\frac{1}{x^2+1}\\right] \\, dx + \\left[\\frac{-1}{x}\\right] \\, dy = 0$$\n\nSo that is the exact equation I got. Then:\n\n$$\\frac{dM}{dy} = \\frac{1}{x^2+1} \\text{ and } \\frac{dN}{dx} = \\frac{1}{x^2}$$\n\nAnd now I'm stuck. I tried to get an integrating factor here to make $\\frac{dM}{dy} = \\frac{dN}{dx}$, but it gets extremely complicated and it can't be what the professor intended for us to do. I think I just screwed up somewhere because $\\frac{dM}{dy}$ and $\\frac{dN}{dx}$ are so similar that it seems like it is the correct method to use.\n\nI've been working on this for a long time. Can anyone here please give me a hint or tell me if I'm overlooking something glaringly obvious?\n\n$$\\frac{\\mathrm d}{\\mathrm dx}\\left(\\frac{y}{\\sqrt{x^2+1}}\\right)=2x\\sqrt{x^2+1}=\\frac23\\frac{\\mathrm d}{\\mathrm dx}\\left(\\sqrt{x^2+1}(x^2+1)\\right)$$"
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https://convertoctopus.com/21-2-cubic-centimeters-to-cubic-feet | [
"## Conversion formula\n\nThe conversion factor from cubic centimeters to cubic feet is 3.5314666572208E-5, which means that 1 cubic centimeter is equal to 3.5314666572208E-5 cubic feet:\n\n1 cm3 = 3.5314666572208E-5 ft3\n\nTo convert 21.2 cubic centimeters into cubic feet we have to multiply 21.2 by the conversion factor in order to get the volume amount from cubic centimeters to cubic feet. We can also form a simple proportion to calculate the result:\n\n1 cm3 → 3.5314666572208E-5 ft3\n\n21.2 cm3 → V(ft3)\n\nSolve the above proportion to obtain the volume V in cubic feet:\n\nV(ft3) = 21.2 cm3 × 3.5314666572208E-5 ft3\n\nV(ft3) = 0.0007486709313308 ft3\n\nThe final result is:\n\n21.2 cm3 → 0.0007486709313308 ft3\n\nWe conclude that 21.2 cubic centimeters is equivalent to 0.0007486709313308 cubic feet:\n\n21.2 cubic centimeters = 0.0007486709313308 cubic feet\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 cubic foot is equal to 1335.7003165896 × 21.2 cubic centimeters.\n\nAnother way is saying that 21.2 cubic centimeters is equal to 1 ÷ 1335.7003165896 cubic feet.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that twenty-one point two cubic centimeters is approximately zero point zero zero one cubic feet:\n\n21.2 cm3 ≅ 0.001 ft3\n\nAn alternative is also that one cubic foot is approximately one thousand three hundred thirty-five point seven times twenty-one point two cubic centimeters.\n\n## Conversion table\n\n### cubic centimeters to cubic feet chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to cubic feet\n\ncubic centimeters (cm3) cubic feet (ft3)\n22.2 cubic centimeters 0.001 cubic feet\n23.2 cubic centimeters 0.001 cubic feet\n24.2 cubic centimeters 0.001 cubic feet\n25.2 cubic centimeters 0.001 cubic feet\n26.2 cubic centimeters 0.001 cubic feet\n27.2 cubic centimeters 0.001 cubic feet\n28.2 cubic centimeters 0.001 cubic feet\n29.2 cubic centimeters 0.001 cubic feet\n30.2 cubic centimeters 0.001 cubic feet\n31.2 cubic centimeters 0.001 cubic feet"
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https://www.aimsciences.org/journal/1547-5816/2020/16/3 | [
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"ISSN:\n1547-5816\n\neISSN:\n1553-166X\n\n## Journal Home\n\n• Open Access Articles",
null,
"All Issues\n\n## Journal of Industrial & Management Optimization\n\nMay 2020 , Volume 16 , Issue 3\n\nSelect all articles\n\nExport/Reference:\n\n2020, 16(3): 1037-1047 doi: 10.3934/jimo.2018191 +[Abstract](3118) +[HTML](1171) +[PDF](264.4KB)\nAbstract:\n\nThis paper analyzes capital structure's characteristics and presents its simplified mathematical model. Panel data analysis shows that the listed companies prefer equity financing rather than debt financing. Furthermore, we propose a capital structure optimization model with uncertain equity financing constraints. We formulate the capital structure optimization problem as a two-stage stochastic optimization problem and solve it. Finally, numerical examples show that our optimization approach can improve the statistics result of capital structure adjustment.\n\n2020, 16(3): 1049-1076 doi: 10.3934/jimo.2018192 +[Abstract](2659) +[HTML](1019) +[PDF](786.28KB)\nAbstract:\n\nTwo types of the law of iterated logarithm (LIL) and one functional LIL (FLIL) are established for the sojourn time process for a multiclass queueing model, having a priority service discipline, one server and $K$ customer classes, with each class characterized by a batch renewal arrival process and independent and identically distributed (i.i.d.) service times. The LIL and FLIL limits quantify the magnitude of asymptotic stochastic fluctuations of the sojourn time process compensated by its deterministic fluid limits in two forms: the numerical and functional. The LIL and FLIL limits are established in three cases: underloaded, critically loaded and overloaded, defined by the traffic intensity. We prove the results by a approach based on strong approximation, which approximates discrete performance processes with reflected Brownian motions. We conduct numerical examples to provide insights on these LIL results.\n\n2020, 16(3): 1077-1098 doi: 10.3934/jimo.2018193 +[Abstract](5235) +[HTML](1384) +[PDF](458.65KB)\nAbstract:\n\nIn Bitcoin system, a transaction is given a priority value according to its attributes such as the remittance amount and fee, and transactions with high priorities are likely to be confirmed faster than those with low priorities. In this paper, we analyze the transaction-confirmation time for Bitcoin system. We model the transaction-confirmation process as a queueing system with batch service, M/\\begin{document}$\\mbox{G}^B$\\end{document}/1. We consider the joint distribution of numbers of transactions in system and the elapsed service time, deriving the mean transaction-confirmation time. Using the result, we derive the recursive formulae of mean transaction-confirmation times of an M/\\begin{document}$\\mbox{G}^B$\\end{document}/1 queue with priority service discipline. In numerical examples, we show the effect of the maximum block size on the mean transaction-confirmation time, investigating the accuracy region of our queueing model. We also discuss how the increase in micropayments, which are likely to be given low priorities, affects the transaction-confirmation time.\n\n2020, 16(3): 1099-1117 doi: 10.3934/jimo.2018194 +[Abstract](2609) +[HTML](1102) +[PDF](443.8KB)\nAbstract:\n\nPeer-to-peer (P2P) networks have been commonly applied into many applications such as distributed storage, cloud computing and social networking. In P2P networks fairness fosters an incentive so as to encourage peers to offer resources (e.g, upload bandwidth) to the networks. In this paper, we consider fair bandwidth allocation of access links in P2P file-sharing networks and develop a coupled network-wide utility maximization model which aims at achieving several kinds of fairness among requesting peers. We provide a meaningful interpretation of the problem of maximizing social welfare and its sub-problems from an economic point of view. The coupled optimization problem is difficult to resolve in a distributed way because of its non-strict convexity and non-separation. We apply a modified successive approximation method to investigate the coupled problem and propose a distributed bandwidth allocation scheme to solve the approximation problems. Then, we investigate the convergence of the scheme by mathematical analysis and evaluate the performance through numerical examples, which validate that the scheme can achieve the global optimum within reasonable iterations.\n\n2020, 16(3): 1119-1134 doi: 10.3934/jimo.2018195 +[Abstract](2570) +[HTML](924) +[PDF](445.39KB)\nAbstract:\n\nIn this paper, in order to reduce possible packet loss of the primary users (PUs) in cognitive radio networks, we assume there is a buffer with a finite capacity for the PU packets. At the same time, focusing on the packet interruptions of the secondary users (SUs), we introduce a probability returning scheme for the interrupted SU packets. In order to evaluate the influence of the finite buffer setting and the probability returning scheme to the system performance, we construct and analyze a discrete-time Markov chain model. Accordingly, we determine the expressions of some important performance measures of the PU packets and the SU packets. Then, we show numerical results to evaluate how the buffer setting of the PU packets and the returning probability influence the system performance. Moreover, we optimize the system access actions of the SU packets. We determine their individually and the socially optimal strategies by considering different buffer settings for PU packets and different returning probabilities for SU packets. Finally, a pricing policy by introducing an admission fee is also provided to coincide the two optimal strategies.\n\n2020, 16(3): 1135-1148 doi: 10.3934/jimo.2018196 +[Abstract](2908) +[HTML](931) +[PDF](693.89KB)\nAbstract:\n\nIn this paper, we introduce a discrete time Geo/Geo/1 queue system with non-preemptive priority and multiple working vacations. We assume that there are two types of customers in this queue system named \"Customers of type-Ⅰ\" and \"Customers of type-Ⅱ\". Customer of type-Ⅱ has a higher priority with non-preemption than Customer of type-Ⅰ. By building a discrete time four-dimensional Markov Chain which includes the numbers of customers with different priorities in the system, the state of the server and the service state, we obtain the state transition probability matrix. Using a birth-and-death chain and matrix-geometric method, we deduce the average queue length, the average waiting time of the two types of customers, and the average busy period of the system. Then, we provide some numerical results to evaluate the effect of the parameters on the system performance. Finally, we develop some benefit functions to analyse both the personal and social benefit, and obtain some optimization results within a certain range.\n\n2020, 16(3): 1149-1169 doi: 10.3934/jimo.2018197 +[Abstract](2258) +[HTML](927) +[PDF](760.54KB)\nAbstract:\n\nChannel reservation strategy in CRNs is an effective technology for conserving communication resources. In this paper, using the imperfect sensing of secondary user (SU) packets, and considering the patience degree of SU packets, we propose a channel reservation strategy in a CRN. Aligned with the proposed channel reservation strategy, we establish a continuous-time Markov chain model to capture the stochastic behavior of the two types of user packets. Then, in order to obtain the steady-state probability distribution for the system model, we present a new algorithm for solving the quasi-birth-and-death (QBD) process. At last, based on the energy detection method, we evaluate the system performance in terms of the throughput of SU packets, the average latency of SU packets, the switching rate of SU packets and the channel utilization in relation to the energy detection threshold and the number of reserved channels.\n\n2020, 16(3): 1171-1185 doi: 10.3934/jimo.2018198 +[Abstract](2527) +[HTML](1043) +[PDF](420.65KB)\nAbstract:\n\nIn this paper, we provide a smoothing sample average approximation (SAA) method to solve a portfolio choice model based on second-order stochastic dominance (SSD) measure. Introducing a second-order stochastic dominance constraint in portfolio choice is theoretically attractive since all risk-averse investors would prefer a dominating portfolio. However, how to get the best choice among SSD efficient portfolios which is based on a stochastic optimization model is a challenge. We use the sample average to approximate the expected return rate function in the model and get a linear/nonlinear programming when the benchmark has discrete distribution. Then we propose a smoothing penalty algorithm to solve this problem. Meanwhile, we investigate the convergence of the optimal value of the transformed model and show that the optimal value converges to its counterpart with probability approaching to one at exponential rate as the sample size increases. By comparing the numerical results of the smoothing SAA algorithm with the common linear programming (LP) algorithm, we find that the smoothing algorithm has better performance than the LP algorithm in three aspects: (ⅰ)the smoothing SAA method can avoid the infinite constraints in the transformed models and the size of the smoothing algorithm model will not increase as the sample grows; (ⅱ)the smoothing SAA algorithm can deal with the nonlinear portfolio models with nonlinear transaction cost function; (ⅲ) the smoothing algorithm can get the global optimal solution because the smoothing function maintains the original convexity.\n\n2020, 16(3): 1187-1202 doi: 10.3934/jimo.2018199 +[Abstract](2693) +[HTML](1111) +[PDF](585.73KB)\nAbstract:\n\nThis paper considers an inventory mechanism in which the supplier may provide a short-term price discount to the retailer at a future time with some uncertainty. To maximize the retailer's profit in this setting, we establish an optimal replenishment and stocking strategy model. Based on the retailer's inventory cost-benefit analysis, we present a closed-form solution for the inventory model and provide an optimal ordering policy to the retailer. Numerical experiments and numerical sensitivity are given to provide some high insights to the inventory model.\n\n2020, 16(3): 1203-1220 doi: 10.3934/jimo.2018200 +[Abstract](3124) +[HTML](1131) +[PDF](267.72KB)\nAbstract:\n\nVehicle routing problem (VRP) is a typical and important combinatorial optimization problem, and is often involved with complicated temporal and spatial constraints in practice. In this paper, the VRP is formulated as an optimization model for minimizing the number of vehicles and the total transportation cost subject to constraints on loading plan, service time and weight capacity. The transportation cost consists of the rent charge of vehicles, fuel cost, and carbon tax. Owing to complexity of the built model, it is divided into two subproblems by a two-stage optimization approach: at the first stage, the number of vehicles is minimized, then the routing plan is optimized at the second stage. For solving the sequential subproblems, two correlated genetic algorithms are developed, which share the same initial population to reduce their computational costs. Numerical results indicate that the developed algorithms are efficient, and a number of important managerial insights are revealed from the model.\n\n2020, 16(3): 1221-1233 doi: 10.3934/jimo.2018201 +[Abstract](2450) +[HTML](1044) +[PDF](348.33KB)\nAbstract:\n\nThe concepts of essential solutions and essential solution sets for generalized semi-infinite optimization problems (GSIO for brevity) are introduced under functional perturbations, and the relations among the concepts of essential solutions, essential solution sets and lower semicontinuity of solution mappings are discussed. We show that a solution is essential if and only if the solution is unique; and a solution subset is essential if and only if it is the solution set itself. Some sufficient conditions for the upper semicontinuity of solution mappings are obtained. Finally, we show that every GSIO problem can be arbitrarily approximated by stable GSIO problems (the solution mapping is continuous), i.e., the set of all stable GSIO problems is dense in the set of all GSIO problems with the given topology.\n\n2020, 16(3): 1235-1259 doi: 10.3934/jimo.2018202 +[Abstract](2778) +[HTML](1083) +[PDF](1287.03KB)\nAbstract:\n\nIntegrated process planning and scheduling (IPPS) problems are one of the most important flexible planning functions for a job shop manufacturing. In a manufacturing order to produce n jobs (parts) on m machines in a flexible manufacturing environment, an IPPS system intends to generate the process plans for all n parts and the overall job-shop schedule concurrently, with the objective of optimizing a manufacturing objective such as make-span. The optimization of the process planning and scheduling will be applied through an integrated approach based on Fuzzy Inference System (FIS), to provide for flexibilities of the given components and consider the qualitative parameters. The FIS, Constraint Programming (CP) and Simulated Annealing (SA) algorithms are applied in this design. The objectives of the proposed model consist of maximization of processes utility, minimization of make-span and total production costs including costs of flexible tools, machines, process and TADs. The proposed approach indicates that The CP and SA algorithms are able to resolve the IPPS problem with multiple objective functions. The experiments and related results indicate that the CP method outperforms the SA algorithm.\n\n2020, 16(3): 1261-1272 doi: 10.3934/jimo.2019001 +[Abstract](2284) +[HTML](822) +[PDF](369.35KB)\nAbstract:\n\nIn this paper, the Clarke generalized Jacobian of the generalized regularized gap function for a nonmonotone Ky Fan inequality is studied. Then, based on the Clarke generalized Jacobian, we derive a global error bound for the nonmonotone Ky Fan inequalities. Finally, an application is given to provide a descent method.\n\n2020, 16(3): 1273-1296 doi: 10.3934/jimo.2019002 +[Abstract](2783) +[HTML](1031) +[PDF](554.12KB)\nAbstract:\n\nThis study develops a single-machine manufacturing system for multi-product with defective items and delayed payment policy. Contradictory to the literature limited production capacity and partial backlogging are considered for more realistic result. The objective of this research is to obtain the optimal cycle length, optimal production quantity, and optimal backorder quantity of each product such that the expected total cost is minimum. The model is solved analytically. Three efficient lemmas are developed to obtain the global optimum solution of the model. An improved algorithm is designed to obtain the numerical solution of the model. An illustrative numerical example and sensitivity analysis are provided to show the practical usage of proposed method.\n\n2020, 16(3): 1297-1310 doi: 10.3934/jimo.2019003 +[Abstract](2408) +[HTML](1212) +[PDF](404.42KB)\nAbstract:\n\nA discrete variant of a multicriteria investment portfolio optimization problem with Savage's risk criteria is considered. One of the three problem parameter spaces is endowed with Hölder's norm, and the other two are endowed with Chebyshev's norm. The lower and upper attainable bounds on the stability radius of one Pareto optimal portfolio are obtained. We illustrate the application of our theoretical results by modeling a relevant case study.\n\n2020, 16(3): 1311-1328 doi: 10.3934/jimo.2019004 +[Abstract](1700) +[HTML](767) +[PDF](432.7KB)\nAbstract:\n\nThis paper extends the multidimensional credibility model under balanced loss function to account for not only certain conditional dependence over time for claim amounts but also dependence across individual risks and over portfolio risks. By means of orthogonal projection method in Hilbert space of random vectors, the inhomogeneous and homogeneous multidimensional credibility estimators are derived, which generalize some well known existing results in credibility theory. Moreover, the unbiased estimators of structural parameters are investigated. Finally, we present a numerical example to show the existence of the multidimensional credibility estimators and their difference from the existing ones.\n\n2020, 16(3): 1329-1347 doi: 10.3934/jimo.2019005 +[Abstract](2321) +[HTML](929) +[PDF](1049.38KB)\nAbstract:\n\nIn this paper, we consider due-date related single machine scheduling problems, where two agents compete for utilizing a common processing resource (i.e. a single machine). In this paper, we provide a detailed and a systemic literature review of the two-agent scheduling problem dealing with models with a given due date. We consider the following four main cases: 1) the earliness and tardiness, 2) the maximum lateness, 3) the number of tardy jobs and 4) the late work criteria. To do so, we classify due-date related, two-agent scheduling problems into two categories on the basis of the objective function setting, (i.e. the feasibility model and the minimality model). The feasibility model minimizes the objective function of one agent subject to an upper bound on the objective for the other agent. The minimality model assigns certain weights for two agents and as a result minimizes their weighted objectives. In the present paper, we list the computational complexities and proposed algorithms for the due-date related, two-agent scheduling problem, investigated in the literature since 2003.\n\n2020, 16(3): 1349-1368 doi: 10.3934/jimo.2019006 +[Abstract](2306) +[HTML](845) +[PDF](2593.86KB)\nAbstract:\n\nIn this work we develop partial differential equation (PDE) based computational models for pricing real options to contract the production or to transfer part/all of the ownership of a project when the underlying asset price of the project satisfies a geometric Brownian motion. The developed models are similar to the Black-Scholes equation for valuing conventional European put options or the partial differential linear complementarity problem (LCP) for pricing American put options. A finite volume method is used for the discretization of the PDE models and a penalty approach is applied to the discretized LCP. We show that the coefficient matrix of the discretized systems is a positive-definite \\begin{document}$M$\\end{document}-matrix which guarantees that the solution from the penalty equation converges to that of the discretized LCP. Numerical experiments, performed to demonstrate the usefulness of our methods, show that our models and numerical methods are able to produce financially meaningful numerical results for the two non-trivial test problems.\n\n2020, 16(3): 1369-1388 doi: 10.3934/jimo.2019007 +[Abstract](2270) +[HTML](820) +[PDF](474.83KB)\nAbstract:\n\nWe consider the equilibrium and socially optimal behavior of strategic customers in a discrete-time queue with bulk service. The service batch size varies from a single customer to a maximum of 'b' customers. We study the equilibrium and socially optimal balking strategies under two information policies: observable and unobservable. In the former policy, a service provider discloses the queue length information to arriving customers and conceals it in the latter policy. The effect of service batch size and other queueing parameters on the equilibrium strategies under both information policies are compared and illustrated with numerical experiments.\n\n2020, 16(3): 1389-1414 doi: 10.3934/jimo.2019008 +[Abstract](3490) +[HTML](1064) +[PDF](471.81KB)\nAbstract:\n\nThis paper studies an original equipment manufacturer's (OEM's) optimal production and pricing decisions and the governments optimal subsidy level when the number of used products returning to the OEM is uncertain. The government aims to minimize its total expenditures but also attempt to achieve a given target collection level. We model the problem as an extended price-setting newsvendor model, which simultaneously incorporates supply uncertainty and external government influence. Moreover, we consider separately the cases of stochastic supplies with additive and multiplicative return uncertainty. We show that under the above settings, the governments optimal strategy is to provide only sufficient subsidies that cause its target to be met exactly. The government subsidies will mitigate the cost of remanufacturing and increase the total collection efforts of the government and the manufacturer. Moreover, the return uncertainty lowers both the manufacturers profits and selling price, whereas its effects on the governments optimal subsidies and the manufacturers optimal return efforts are insignificant. Therefore, the manufacturer is worse off but consumers are better off under the conditions of uncertain returns. By comparing the optimal decisions when the government is a central planner with the case of decentralized decision making, or comparing the arrangement in which the government provides subsidies directly to the manufacturer rather than to consumers, we find that the government subsidies would coordinate the supply chain only when its target collection level is high. Moreover, no essential differences exist between providing subsidies directly to the manufacturer and to consumers. Our results are robust under both the additive and multiplicative uncertainty models.\n\n2020, 16(3): 1415-1433 doi: 10.3934/jimo.2019009 +[Abstract](2278) +[HTML](834) +[PDF](607.38KB)\nAbstract:\n\nIn this paper, an \\begin{document}$(s, S)$\\end{document} continuous inventory model with perishable items and retrial demands is proposed. In addition, replenishment lead times that are independent and identically distributed according to phase-type distribution are implemented. The proposed system is modeled as a three-dimensional Markov process using a level-dependent quasi-birth-death (QBD) process. The ergodicity of the modeled Markov system is demonstrated and the best method for efficiently approximating the steady-state distribution at the inventory level is determined. This paper also provides performance measure formulas based on the steady-state distribution of the proposed approximation method. Furthermore, in order to minimize the system cost, the optimum values of \\begin{document}$s$\\end{document} and \\begin{document}$S$\\end{document} are determined numerically and sensitivity analysis is performed on the main parameters.\n\n2020, 16(3): 1435-1456 doi: 10.3934/jimo.2019010 +[Abstract](2250) +[HTML](868) +[PDF](477.06KB)\nAbstract:\n\nWe consider the dynamic lot size problem for perishable inventory under minimum order quantities. The stock deterioration rates and inventory costs depend on both the age of the stocks and their periods of order. Based on two structural properties of the optimal solution, we develop a dynamic programming algorithm to solve the problem without backlogging. We also extend the model by considering backlogging. By establishing the regeneration set, we give a sufficient condition for obtaining forecast horizon under without and with backlogging. Finally, based on a detailed test bed of instance, we obtain useful managerial insights on the impact of minimum order quantities and perishability of product and the costs on the length of forecast horizon.\n\n2020, 16(3): 1457-1479 doi: 10.3934/jimo.2019011 +[Abstract](2251) +[HTML](830) +[PDF](477.48KB)\nAbstract:\n\nThis paper investigates an optimal reinsurance-investment problem in relation to thinning dependent risks. The insurer's wealth process is described by a risk model with two dependent classes of insurance business. The insurer is allowed to purchase reinsurance and invest in one risk-free asset and one risky asset whose price follows CEV model. Our aim is to maximize the expected exponential utility of terminal wealth. Applying Legendre transform-dual technique along with stochastic control theory, we obtain the closed-form expression of optimal strategy. In addition, our wealth process will reduce to the classical Cramér-Lundberg (C-L) model when \\begin{document}$p = 0$\\end{document}, in this case, we achieve the explicit expression of the optimal strategy for Hyperbolic Absolute Risk Aversion (HARA) utility by using Legendre transform. Finally, some numerical examples are presented to illustrate the impact of our model parameters (e.g., interest and volatility) on the optimal reinsurance-investment strategy.\n\n2020, 16(3): 1481-1502 doi: 10.3934/jimo.2019012 +[Abstract](4093) +[HTML](1028) +[PDF](457.87KB)\nAbstract:\n\nIn this paper, a numerical method based on least squares support vector machines has been developed to solve the initial and boundary value problems of higher order nonlinear ordinary differential equations. The numerical experiments have been performed on some nonlinear ordinary differential equations to validate the accuracy and reliability of our proposed LS–SVM model. Compared with the exact solution, the results obtained by our proposed LS–SVM model can achieve a very high accuracy. The proposed LS–SVM model could be a good tool for solving higher order nonlinear ordinary differential equations.\n\n2020, 16(3): 1503-1518 doi: 10.3934/jimo.2019013 +[Abstract](2357) +[HTML](822) +[PDF](429.34KB)\nAbstract:\n\nThis paper proposes a new class of accelerated conjugate-gradient-like algorithms for solving large scale unconstrained optimization problems, which combine the idea of accelerated adaptive Perry conjugate gradient algorithms proposed by Andrei (2017) with the modified secant condition and the nonmonotone line search technique. An attractive property of the proposed methods is that the search direction always provides sufficient descent step which is independent of the line search used and the convexity of objective function. Under common assumptions, it is proven that the proposed methods possess global convergence for nonconvex smooth functions, and R-linear convergence for uniformly convex functions. The numerical experiments show the efficiency of the proposed method in practical computations.\n\n2020, 16(3): 1519-1526 doi: 10.3934/jimo.2019014 +[Abstract](2379) +[HTML](822) +[PDF](361.84KB)\nAbstract:\n\nWe establish a theoretical framework for the problem of phaseless compressed sensing with partially known signal support, which aims at generalizing the Null Space Property and the Strong Restricted Isometry Property from phase retrieval to partially sparse phase retrieval. We first introduce the concepts of the Partial Null Space Property (P-NSP) and the Partial Strong Restricted Isometry Property (P-SRIP); and then show that both the P-NSP and the P-SRIP are exact recovery conditions for the problem of partially sparse phase retrieval. We also prove that a random Gaussian matrix \\begin{document}$A\\in \\mathbb{R}^{m\\times n}$\\end{document} satisfies the P-SRIP with high probability when \\begin{document}$m = O(t(k-r)\\log(\\frac{n-r}{t(k-r)})).$\\end{document}\n\n2020, 16(3): 1527-1538 doi: 10.3934/jimo.2019015 +[Abstract](2124) +[HTML](786) +[PDF](370.37KB)\nAbstract:\n\nThe parameters in the semidefinite programming problems generated by the average of a sample, may lead to the deviation of the optimal value and optimal solutions due to the uncertainty of the data. The statistical properties of estimates of the optimal value and the optimal solutions are given in this paper, when the estimated parameters are both in the objective function and in the constraints. This analysis is mainly based on the theory of the linear programming and the perturbation theory of the semidefinite programming.\n\n2020, 16(3): 1539-1553 doi: 10.3934/jimo.2019016 +[Abstract](2347) +[HTML](1043) +[PDF](835.74KB)\nAbstract:\n\nAt the peak of a service system, customers may hesitate and even leave in the face of unavoidable queuing. This phenomenon not only affects the customer's satisfaction, but also causes the loss of the company's revenue. This paper establishes a fluid model of customer queuing behavior for the customer losses. The goal is to reduce the customer losses, and the setting and optimization method of quick queue in random service systems is studied. We construct two queuing models, in which one includes only regular queues and the other includes both regular and quick queues. We analyze the queuing systems, and describe the different forms of the objective function based on the fluid model of customer behavior. Then we compare and analyze the impact of the adoption of quick queues on the performance of the service system during peak period, and design a calculation method to obtain the optimal value for setting the number of quick queues. Thus, the overall performance of the system is optimized. Finally, we take the setting and optimization of quick queue in the supermarket service system as an example, which verifies the validity of the proposed method, and shows the reference value of this method to management practice.\n\n2020 Impact Factor: 1.801\n5 Year Impact Factor: 1.688\n2020 CiteScore: 1.8"
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https://eiffl.github.io/talks/Gif2021/ | [
"# A view of the potential of Deep Learning for Cosmology\n\n## François Lanusse",
null,
"",
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"",
null,
"slides at eiffl.github.io/talks/Gif2021\n\n### the Rubin Observatory Legacy Survey of Space and Time\n\n• 1000 images each night, 15 TB/night for 10 years\n\n• 18,000 square degrees, observed once every few days\n\n• Tens of billions of objects, each one observed $\\sim1000$ times\n\nPrevious generation survey: SDSS\n\nImage credit: Peter Melchior\n\nCurrent generation survey: DES\n\nImage credit: Peter Melchior\n\nLSST precursor survey: HSC\n\nImage credit: Peter Melchior\n\n### The challenges of modern surveys\n\n$\\Longrightarrow$ Modern surveys will provide large volumes of high quality data\n\nA Blessing\n• Unprecedented statistical power\n\n• Great potential for new discoveries\n\nA Curse\n• Existing methods are reaching their limits at every step of the science analysis\n\n• Control of systematics is paramount\nLSST forecast on dark energy parameters\n$\\Longrightarrow$ Dire need for novel analysis techniques to fully realize the potential of modern surveys.\n\n## Can AI solve all of our problems?\n\n### WWAD: What Would an Astrophysicist Do ?\n\ngri composite\ng - $\\alpha$ i\ndetected areas\nHST images\nRingFinder (Gavazzi et al. 2014)\n\nGavazzi et al. (2014), Collett (2015)\n\n$\\Longrightarrow$ Plainly intractable at the scale of LSST\n\n### The Deep Learning approach: Convolutional Neural Networks\n\nCMUDeepLens Architecture\n(Lanusse et al. 2017)\nThe Deep Learning approach in three steps:\n\n• Step I: Introduce a parameteric function $f_\\theta$, this is your neural network. For images, choose a CNN.\n\n• Step II: Create a dataset $\\mathcal{D}$ of examples $x_i$ and associated labels $y_i$: $$\\mathcal{D} = \\{ (x_i, y_i) \\}_{i \\in [0, N]}$$\n\n• Step III: Optimize the parameters $\\theta$ as to minimize an appropriate loss function.\nFor classification, the binary cross entropy: $$\\arg\\min_\\theta \\sum_{i=1}^N y_i \\log f_\\theta(x_i) + (1 - y_i) \\log f_{\\theta}(x_i)$$\n\n### The Euclid strong-lens finding challenge\n\nMetcalf, . . ., Lanusse, et al. (2018)\n\nBetter accuracy than human visual inspection!\n• Before: Experts would carefully craft image statistics they would expect a priori to be useful for their problem.\n\n• After: Reframe the task as an optimization problem, and let the optimization figure out how to solve the task\n\n### The AI bubble\n\nastro-ph abstracts mentioning Deep Learning, CNN, or Neural Networks\n\n### Be careful to ask the right questions!\n\nThe most common pitfall: Covariate Shift\n\n$\\Longrightarrow$ The training data plays an integral part in the answer\n\n### An example of bad idea: Naive Image Deconvolution with a CNN\n\n$y = P \\ast x + n$\n\nObserved $y$\n\nGround-Based Telescope\n\n$f_\\theta$\n\nsome deep Convolutional Neural Network\n\nUnknown $x$\n\nHubble Space Telescope\n\n• A standard approach would be to train a neural network $f_\\theta$ to estimate $x$ given $y$ on a dataset $\\mathcal{D}=\\{(x_i, yi)\\}$.\n• Is this a good idea?\n\n### Let's be critical for a minute\n\nIs Strong Lens Finding a good application of Deep Learning?\n\nTraining examples\n\nExample of real data\n\n### What can Deep Learning do for Cosmology?\n\nWhat are the right questions for Cosmology?\n• As physicists, we want to remain model-based.\nWe want to compare an interpretable/physical model to data.\n\n• We already have the foundational statistical tools for comparing models and data: Bayesian inference\n\n• $\\Longrightarrow$ Wherever Deep Learning can be useful is in places where Bayesian inference approach becomes classically intractable.\nWhat we will talk about today:\n\n• Likelihood-Free Inference\n\n• Data-driven priors for Inverse Problems\n\n• Automatically Differentiable Physics\n\n## Likelihood-Free Inference\n\n### limits of traditional cosmological inference\n\nHSC cosmic shear power spectrum\nHSC Y1 constraints on $(S_8, \\Omega_m)$\n(Hikage et al. 2018)\n• Measure the ellipticity $\\epsilon = \\epsilon_i + \\gamma$ of all galaxies\n$\\Longrightarrow$ Noisy tracer of the weak lensing shear $\\gamma$\n\n• Compute summary statistics based on 2pt functions,\ne.g. the power spectrum\n\n• Run an MCMC to recover a posterior on model parameters, using an analytic likelihood $$p(\\theta | x ) \\propto \\underbrace{p(x | \\theta)}_{\\mathrm{likelihood}} \\ \\underbrace{p(\\theta)}_{\\mathrm{prior}}$$\nMain limitation: the need for an explicit likelihood\nWe can only compute the likelihood for simple summary statistics and on large scales\n\n$\\Longrightarrow$ We are dismissing a large amount of information!\n\n## Can I use a Deep Learning to perform proper Bayesian inference without analytic likelihoods?\n\n### let us rephrase the question\n\n• I assume a forward model of the observations: \\begin{equation} p( x ) = p(x | \\theta) \\ p(\\theta) \\nonumber \\end{equation} All I ask is the ability to sample from the model, to obtain $\\mathcal{D} = \\{x_i, \\theta_i \\}_{i\\in \\mathbb{N}}$\n\n• I am going to assume $q_\\phi(\\theta | x)$ a parametric conditional density\n\n• Optimize the parameters $\\phi$ of $q_{\\phi}$ according to \\begin{equation} \\min\\limits_{\\phi} \\sum\\limits_{i} - \\log q_{\\phi}(\\theta_i | x_i) \\nonumber \\end{equation} In the limit of large number of samples and sufficient flexibility \\begin{equation} \\boxed{q_{\\phi^\\ast}(\\theta | x) \\approx p(\\theta | x)} \\nonumber \\end{equation}\n$\\Longrightarrow$ One can asymptotically recover the posterior by optimizing a parametric estimator over\nthe Bayesian joint distribution\n$\\Longrightarrow$ One can asymptotically recover the posterior by optimizing a Deep Neural Network over\na simulated training set.\n\n### The Likelihood-Free Inference approach\n\n• Instead of trying to analytically evaluate the likelihood, let us build a forward model of the observables.\n\n$\\Longrightarrow$ The simulator becomes the physical model\n\nA two-steps approach to inference\n• Automatically learn an optimal low-dimensional summary statistic $$y = f_\\varphi(\\kappa_{KS})$$\n• Use Neural Density Estimation to either:\n• build an estimate $p_\\phi$ of the likelihood function $p(y \\ | \\ \\theta)$ (Neural Likelihood Estimation)\n\n• build an estimate $p_\\phi$ of the posterior distribution $p(\\theta \\ | \\ y)$ (Neural Posterior Estimation)\n\n### Neural Density Estimation\n\nBishop (1994)\n• Mixture Density Networks (MDN) \\begin{equation} p(\\theta | x) = \\prod_i \\pi_i(x) \\ \\mathcal{N}\\left(\\mu_i(x), \\ \\sigma_i(x) \\right) \\nonumber \\end{equation}\n\n• Flourishing Machine Learning literature on density estimators\nGLOW, (Kingma & Dhariwal, 2018)\n\n### End-to-end framework for likelihood-free parameter inference with DES SV\n\nSuite of N-body + raytracing simulations: $\\mathcal{D}$\n\n### Learning summary statistics by Variational Mutual Information Maximization\n\n• Mutual information between $X$ and $Y$:\n“\"amount of information\" obtained about one random variable through observing the other random variable”\n\n• Given a parametric summarizing function $y = f_\\phi(\\kappa(\\theta))$ optimizing $f_\\phi$ can be done by maximizing $I(y, \\theta)$.\n\n• In practice, $f_\\phi$ is a CNN, trained to maximize a variational lower bound on the mutual information: $$I(y ; \\theta) \\ \\ge \\ \\mathbb{E}_{y, \\theta} [ \\log q_\\phi(\\theta | y) ] + H(\\Theta)$$\n\n### deep residual networks for lensing maps compression\n\n• Deep Residual Network $y = f_\\phi(x)$ followed by mixture density network $q_\\phi(\\theta | y)$\n\n• Training on weak lensing maps simulated for different cosmologies\n\n• Optimization of the variational lower bound: $$\\mathbb{E}_{(x, \\theta) \\in \\mathcal{D}} [ \\log q_\\phi(\\theta | f_\\phi(y) ) ]$$\n\n### Estimating the likelihood by Neural Density Estimation\n\n$\\Longrightarrow$ We cannot assume a Gaussian likelihood for the summary $y = f_\\phi(\\kappa)$ but we can learn $p(y | \\theta)$: Neural Likelihood Estimation.\n\nDinh et al. 2016\nNeural Likelihood Estimation by Normalizing Flow\n• We use a conditional Normalizing Flow to build an explicit model for the likelihood function $$\\log p_\\varphi (y | \\theta)$$\n\n• In practice we use the pyDELFI package and an ensemble of NDEs for robustness.\n\n• Once learned, we can use the likelihood as part of a conventional MCMC chain\n\n### Let's be critical for a minute\n\nIs this Likelihood-Free Inference approach a good application of Deep Learning?\n\nDeep Learning For Cosmological Inference\n• It's only a shift in how we model physics and observables: $$\\mbox{analytic modeling} \\rightarrow \\mbox{forward simulation}$$ but we remain within the same physical Bayesian framework.\n\n• This allows us to free ourselves from the restrictions of tractable analytic likelihoods (choice of summary statistics, handling of systematics, etc...)\n\n# Deep Generative Models as Data-Driven Bayesian Priors\n\n### Gravitational lensing\n\nGalaxy shapes as estimators for gravitational shear\n$$e = \\gamma + e_i \\qquad \\mbox{ with } \\qquad e_i \\sim \\mathcal{N}(0, I)$$\n• We are trying the measure the ellipticity $e$ of galaxies as an estimator for the gravitational shear $\\gamma$\n\n### Weak Lensing Mass-Mapping as an Inverse Problem\n\nShear $\\gamma$\nConvergence $\\kappa$\n$$\\gamma_1 = \\frac{1}{2} (\\partial_1^2 - \\partial_2^2) \\ \\Psi \\quad;\\quad \\gamma_2 = \\partial_1 \\partial_2 \\ \\Psi \\quad;\\quad \\kappa = \\frac{1}{2} (\\partial_1^2 + \\partial_2^2) \\ \\Psi$$\n$$\\boxed{\\gamma = \\mathbf{P} \\kappa}$$\n\n### Linear inverse problems\n\n$\\boxed{y = \\mathbf{A}x + n}$\n\n$\\mathbf{A}$ is known and encodes our physical understanding of the problem.\n$\\Longrightarrow$ When non-invertible or ill-conditioned, the inverse problem is ill-posed with no unique solution $x$\nDeconvolution\nInpainting\nDenoising\n\n### What Would a Bayesian Do?\n\n$\\boxed{y = \\mathbf{A}x + n}$\n\nThe Bayesian view of the problem:\n$$p(x | y) \\propto p(y | x) \\ p(x)$$\n• $p(y | x)$ is the data likelihood, which contains the physics\n\n• $p(x)$ is the prior knowledge on the solution.\n\nWith these concepts in hand we can:\n• Estimate for instance the Maximum A Posteriori solution:\n$$\\hat{x} = \\arg\\max\\limits_x \\ \\log p(y \\ | \\ x) + \\log p(x)$$\n• Estimate from the full posterior p(x|y) with MCMC or Variational Inference methods.\n\n### Classical examples of signal priors\n\nSparse\n$$\\log p(x) = \\parallel \\mathbf{W} x \\parallel_1$$\nGaussian $$\\log p(x) = x^t \\mathbf{\\Sigma^{-1}} x$$\nTotal Variation $$\\log p(x) = \\parallel \\nabla x \\parallel_1$$\n\n$\\Longrightarrow$ Prior in the form of numerical simulations.\n\n## Can we use Deep Learning to embed simulation-driven priors within a physical Bayesian model?\n\n### What is generative modeling?\n\n• The goal of generative modeling is to learn the distribution $\\mathbb{P}$ from which the training set $X = \\{x_0, x_1, \\ldots, x_n \\}$ is drawn.\n\n• Usually, this means building a parametric model $\\mathbb{P}_\\theta$ that tries to be close to $\\mathbb{P}$.\n\nTrue $\\mathbb{P}$\n\nSamples $x_i \\sim \\mathbb{P}$\n\nModel $\\mathbb{P}_\\theta$\n\n• Once trained, you can typically sample from $\\mathbb{P}_\\theta$ and/or evaluate the likelihood $p_\\theta(x)$.\n\n### Why isn't it easy?\n\n• The curse of dimensionality put all points far apart in high dimension\n\nDistance between pairs of points drawn from a Gaussian distribution.\n\n• Classical methods for estimating probability densities, i.e. Kernel Density Estimation (KDE) start to fail in high dimension because of all the gaps\n\n### The evolution of generative models\n\n• Deep Belief Network\n(Hinton et al. 2006)\n\n• Variational AutoEncoder\n(Kingma & Welling 2014)\n\n(Goodfellow et al. 2014)\n\n• Wasserstein GAN\n(Arjovsky et al. 2017)\n\n### A visual Turing test\n\nFake PixelCNN samples\n\nReal galaxies from SDSS\n\n### Getting started with Deep Priors: deep denoising example\n\n$$\\boxed{{\\color{Orchid} y} = {\\color{SkyBlue} x} + n}$$\n• Let us assume we have access to examples of ${\\color{SkyBlue} x}$ without noise.\n\n• We learn the distribution of noiseless data $\\log p_\\theta(x)$ from samples using a deep generative model.\n\n• The solution should lie on the realistic data manifold, symbolized by the two-moons distribution.\n\nWe want to solve for the Maximum A Posterior solution:\n\n$$\\arg \\max - \\frac{1}{2} \\parallel {\\color{Orchid} y} - {\\color{SkyBlue} x} \\parallel_2^2 + \\log p_\\theta({\\color{SkyBlue} x})$$ This can be done by gradient descent as long as one has access to the score function $\\frac{\\color{orange} d \\color{orange}\\log \\color{orange}p\\color{orange}(\\color{orange}x\\color{orange})}{\\color{orange} d \\color{orange}x}$.\n\n### Neural Score Estimation by Denoising Score Matching\n\n• Denoising Score Matching: An optimal Gaussian denoiser learns the score of a given distribution.\n• If $x \\sim \\mathbb{P}$ is corrupted by additional Gaussian noise $u \\in \\mathcal{N}(0, \\sigma^2)$ to yield $$x^\\prime = x + u$$\n• Let's consider a denoiser $r_\\theta$ trained under an $\\ell_2$ loss: $$\\mathcal{L}=\\parallel x - r_\\theta(x^\\prime, \\sigma) \\parallel_2^2$$\n• The optimal denoiser $r_{\\theta^\\star}$ verifies: $$\\boxed{\\boldsymbol{r}_{\\theta^\\star}(\\boldsymbol{x}', \\sigma) = \\boldsymbol{x}' + \\sigma^2 \\nabla_{\\boldsymbol{x}} \\log p_{\\sigma^2}(\\boldsymbol{x}')}$$\n$\\boldsymbol{x}'$\n$\\boldsymbol{x}$\n$\\boldsymbol{x}'- \\boldsymbol{r}^\\star(\\boldsymbol{x}', \\sigma)$\n$\\boldsymbol{r}^\\star(\\boldsymbol{x}', \\sigma)$\n\n### Back to the Mass-Mapping Problem\n\n$$\\log p( \\kappa | e) = \\underbrace{\\log p(e | \\kappa)}_{\\simeq -\\frac{1}{2} \\parallel e - P \\kappa \\parallel_\\Sigma^2} + \\log p(\\kappa) +cst$$\n• The likelihood term is known analytically.\n• There is no close form expression for the full non-Gaussian prior of the convergence.\nHowever:\n• We do have access to samples of full prior through simulations: $X = \\{x_0, x_1, \\ldots, x_n \\}$ with $x_i \\sim \\mathbb{P}$\n$\\Longrightarrow$ Our strategy: Learn the prior score $\\nabla \\log p(\\kappa)$ with Denoising Score Matching, and then sample the full posterior with annealed HMC.\n\n### Probabilistic Mass-Mapping of the HST COSMOS field\n\n• COSMOS shear data from Schrabback et al. 2010\n\n• Prior learned from MassiveNuS at fiducial cosmology (320x320 maps at 0.4 arcsec resolution).\n\n• Known massive X-ray clusters indicated with crosses, along with their redshifts, right pannel shows cutouts of central cluster from multiple posterior samples.\n\n### Uncertainty quantification in Magnetic Resonance Imaging (MRI)\n\nRamzi, Remy, Lanusse et al. 2020",
null,
"$$\\boxed{y = \\mathbf{M} \\mathbf{F} x + n}$$\n\n$\\Longrightarrow$ We can see which parts of the image are well constrained by data, and which regions are uncertain.\n\n### Example of application to deblending\n\n$\\mathcal{L} = \\frac{1}{2} \\parallel \\mathbf{\\Sigma}^{-1/2} (\\ Y - P \\ast A S \\ ) \\parallel_2^2 - \\sum_{i=1}^K \\log p_{\\theta}(S_i)$\n\nisolated galaxy $\\log p_\\theta(x) = 3293.7$\nartificial blend $\\log p_\\theta(x) = 3100.5$\n\n### Let's be critical for a minute\n\nIs this Deep Priors approach a good application of Deep Learning?\n\nDeep Data-Driven Priors for Inverse Problems\n• Deep Generative Models are only a way to perform Bayesian inference in a classically intractable regime: from having only access to samples from the prior p(x) with a high dimensional $x$.\n\n• We retain a physical explicit likelihood (can directly account for changes in masks, noise, etc. without retraining)\n\n## Automatically Differentiable Physics\n\n### the hammer behind the Deep Learning revolution\n\n• Automatic differentiation allows you to compute analytic derivatives of arbitraty expressions:\nIf I form the expression $y = a * x + b$, it is separated in fundamental ops: $$y = u + b \\qquad u = a * x$$ then gradients can be obtained by the chain rule: $$\\frac{\\partial y}{\\partial x} = \\frac{\\partial y}{\\partial u} \\frac{ \\partial u}{\\partial x} = 1 \\times a = a$$\n\n• This is a fundamental tool in Machine Learning, and autodiff frameworks include TensorFlow, JAX or PyTorch.\n$\\Longrightarrow$ In addition to autodiff, they all provide GPU/TPU acceleration.\n\n### jax-cosmo: Finally a differentiable cosmology library, and it's in JAX!",
null,
"• Easy to use, runs on GPU, and validated against the DESC Core Cosmology Library.\n\nimport jax_cosmo as jc\n\n# Defining a Cosmology\ncosmo = jc.Planck15()\n\n# Define a redshift distribution with smail_nz(a, b, z0)\nnz = jc.redshift.smail_nz(1., 2., 1.)\n\n# Build a lensing tracer with a single redshift bin\nprobe = probes.WeakLensing([nz])\n\n# Compute angular Cls for some ell\nell = np.logspace(0.1,3)\ncls = angular_cl(cosmo_jax, ell, [probe])\n\n• Fisher matrices become trivial and exact!\n\n# Fisher matrix in just one line:\nF = - jax.hessian(gaussian_likelihood)(theta)\n\nNo derivatives were harmed by finite differences in the computation of this Fisher!\n\n### LSST DESC 3x2pt Tomography Challenge\n\nDescription of the challenge\n“Given (g)riz photometry, find a tomographic bin assignment method that optimizes a 3x2pt analysis.”\n• Metrics: Total Signal-to-Noise: $m_{SNR} = \\sqrt{\\mu^t \\mathbf{C}^{-1} \\mu}$ ; DETF Figure of Merit: $m_{FOM} = \\frac{1}{\\sqrt{ \\det(\\mathbf{F}^{-1})}}$\n• Conventional strategy: Use a photoz code to estimate redshifts, then bin galaxies based on their photoz.\n\n• Strategy with Differentiable Physics:\n• Introduce a parametric bin assignement function $f_\\theta(x_{phot})$\n• Optimize $\\theta$ by back-propagating through the challenge metrics.\n\n### Another approach to cosmological inference: Hierarchical Bayesian Modeling\n\n• Instead of trying to analytically evaluate the likelihood, let us build a forward model of the observables.\n\n• Each component of the model is now tractable, but at the cost of a large number of latent variables.\n\n$\\Longrightarrow$ How to peform efficient inference in this large number of dimensions?\n\nA non-exhaustive list of methods:\n• Hamiltonian Monte-Carlo\n• Variational Inference\n• MAP+Laplace\n• Gold Mining\n• Dimensionality reduction by Fisher-Information Maximization\n\nWhat do they all have in common?\n-> They require fast, accurate, differentiable forward simulations\n(Schneider et al. 2015)\n\n### Forward Models in Cosmology\n\nLinear Field\nFinal Dark Matter\n\nDark Matter Halos\nGalaxies\n$\\longrightarrow$\nN-body simulations\n$\\longrightarrow$\nGroup Finding\nalgorithms\n$\\longrightarrow$\nSemi-analytic &\ndistribution models\n\n## How do we make cosmological simulations fast and differentiable?\n\n### You can try to learn the simulation...\n\nLearning particle displacement with a UNet. S. He, et al. (2019)\n\nThe issue with using deep learning as a black-box\n• No guarantees to work outside of training regime.\n• No guarantees to capture dependence on cosmology accurately.\n\n### the Fast Particle-Mesh scheme\n\nThe idea: approximate gravitational forces by estimating densities on a grid.\n• The numerical scheme:\n\n• Estimate the density of particles on a mesh\n=> compute gravitational forces by FFT\n\n• Interpolate forces at particle positions\n\n• Update particle velocity and positions, and iterate\n\n• Fast and simple, at the cost of approximating short range interactions.\n$\\Longrightarrow$ Only a series of FFTs and interpolations.\n\n### Introducing FlowPM: Particle-Mesh Simulations in TensorFlow\n\n\nimport tensorflow as tf\nimport flowpm\n# Defines integration steps\nstages = np.linspace(0.1, 1.0, 10, endpoint=True)\n\ninitial_conds = flowpm.linear_field(32, # size of the cube\n100, # Physical size\nipklin, # Initial powerspectrum\nbatch_size=16)\n\n# Sample particles and displace them by LPT\nstate = flowpm.lpt_init(initial_conds, a0=0.1)\n\n# Evolve particles down to z=0\nfinal_state = flowpm.nbody(state, stages, 32)\n\n# Retrieve final density field\nfinal_field = flowpm.cic_paint(tf.zeros_like(initial_conditions),\nfinal_state)\n\n• Seamless interfacing with deep learning components\n\n### Bayesian Reconstruction of Cosmological Fields\n\nGoing back to simpler times...\n$$\\arg\\max_s \\ \\log p(x | f(s)) \\ + \\ \\log p(s)$$ where:\n• $f$ is the forward model\n• $p(x | f(s))$ is a tractable data likelihood\n• $s$ are the initial conditions (early universe)\n• $x$ is the data (e.g. galaxies, lensing, etc.)\n\n### MAP optimization in action\n\n$$\\arg\\max_s \\ \\log p(x_{dm} | f(s)) \\ + \\ \\log p(s)$$\ncredit: C. Modi\n\nTrue initial conditions\n$s_0$\n\nReconstructed initial conditions $s$\n\nReconstructed dark matter distribution $x_{dm} = f(s)$\n\nData\n$x_{dm} = f(s_0)$\n\n### If only MAP optimization was easy...\n\n$$\\arg\\max_s \\ \\log p(x_{dm} | f(s)) \\ + \\ \\log p(s)$$\nDirect optimization\n• Direct optimization of MAP leads to poor solutions on large scales.\n• Annealing recovers unbiased large scales, but at the cost of ad-hoc tempering procedure.\n\n## Instead of guessing an optimization scheme, could we learn to optimize?\n\n### A closer look at the optimization algorithm\n\n$$\\arg\\max_x \\ \\log p(y | f(x)) \\ + \\ \\log p(x)$$\n• Standard Gradient Descent Algorithm: $$x_{i+1} = x_i - \\epsilon \\left[ \\nabla_x{\\log p(y | f(x_i))} + \\nabla_x \\log p(x_i) \\right]$$\n$$x_{i+1} = x_i - \\Gamma \\left(\\nabla_x{\\log p(y | f(x_i))}, \\nabla_x \\log p(x_i) \\right]$$ with update function $\\Gamma: (u,v) \\rightarrow \\epsilon(u + v)$\n\n• Many algorithms (e.g. ADAM, LBFGS) can expressed in this form with a different choice of $\\Gamma$.\n\n$\\Longrightarrow$ What if we could learn this update function?\n\n### Recurrent Inference Machines for Solving Inverse ProblemsPutzky & Welling, 2017\n\n• Introduce a Recurrent Neural Network (RNN) $h_\\phi$, and state variable $s$, so that: $$s_{i+1} = h^*_\\phi( \\nabla \\log p(y|x_{i}), x_i, s_{i})$$ $$x_{i+1} = x_i + h_\\phi(\\nabla \\log p(y|x_{i}), x_i, s_{i+1})$$\n• Train according to: $$\\mathcal{L} = \\sum_{i}^T w_i\\mathcal{L}(x_i, x)$$\n\n(source)\n\n### Illustration on inverse problems\n\nFrom left to right: input masked image, increasing number of steps of solutions, ground truth.\n\n### CosmicRIM: Recurrence Inference Machines for Initial Condition Reconstruction\n\nRecurrent Neural Network Architecture\n• A few notable differences to a vanilla RIM:\n• We provide gradients of both prior and likelihood to the model.\n\n• Because our forward model couples scales, we use a multiscale U-Net architecture.\n\n### Experiments\n\nSettings\n• Forward model: $64^3$ particles, 400 Mpc/h box, 2LPT dynamics with 2nd order bias model\n• RIM: 10 steps, trained under l2 loss\nInitial conditions cross-correlation\nTransfer function\n• CosmicRIM: Learn to optimize by embedding a Neural Network in the optimization algorithm.\n$\\Longrightarrow$ converges 40x faster than LBFGS.\n\n## Conclusion\n\n### Conclusion\n\nWhere do I think Deep Learning can be useful for Cosmology?\n$\\Longrightarrow$ Makes Bayesian inference possible at scale and with non-trivial models!\n\n• Enables inference in high dimension from numerical simulators.\n• Automagically construct summary statistics.\n• Provides the density estimation tools needed.\n\n• Complement known physical models with data-driven components\n• Use data-driven generative model as prior for solving inverse problems.\n\n• Differentiable physical models for fast inference\n\nThank you !"
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https://eandbsoftware.org/worksheets/31206_8056.aspx | [
"# Finding Equivalent Fractions Worksheets\n\nWhich makes both fractions occurs in finding equivalent fractions worksheets may receive a transparency on addition or special with the reciprocal and fluency as the oc vip team!\n\n##### It as many parts of the fraction worksheets and equivalent fractions worksheets may not",
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"Name all the fractions and draw line between fractions that are equal. Understand and use ratios and proportions to represent quantitative relationships. Includes Times Tables practice. It shows a picture, the fraction, a decimal, and a percent. Students will use number lines and lowest common multiples to determine if two fractions are equal. This short lesson discusses equivalent fractions. This worksheet has rows of equivalent fractions, each with either the numerator or denominator left blank.\n\nSmart Wichita Are you a member?\n\nFractions worksheets on understanding fractions, adding fractions, converting fractions into decimals, equivalent fractions, simple fractions, fraction conversion, fraction word problems.\n##### Remove focus is an understanding fractions worksheets are and decide whether or two given",
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"Writing fractions in their lowest form and some more coloring for you. Give the students time to investigate the different plates in their piles. Want to try out these worksheets? Fern Smith always has the cutest freebies! In this worksheet, student will practise simplifying fractions and making equivalent fractions. Apply size mapping when there are breakpoints. Today I want to focus on books to teach fractions, because obviously there is a wonderful connection between division and fractions. Have the students line up all of the strips in order from least number of parts to greatest number of parts. These fractions worksheets are great handouts for student learning about fractions and the decimal equivalents for an inch. Twinderella, from Corey Rosen Schwartz, is a super cute retelling of the classic fairy tale. What would happen if we did not have shapes such as circles and rectangles to refer to? SNames_________________________________ Directions: Using your fraction plates, find the equivalent fraction to the given fractions below. These fractions worksheets are great for practicing how to subtract fractional inch measurements that you would find on a tape measure.\n\nWater Realtor Buy the Goodies Now!\n\nTo add a theme, simply click on you preferred choice from our list. First, factor out each number and express it as a product of prime number powers. First section contains six multiple choices. Please check the card number or try again with a different card. One is missing a number to make it equivalent to the other; this is either the numerator or denominator.\n##### After the fractions equivalent worksheets",
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"Record their responses on the large blank sheet of paper by name. Can you name some other things that you know of that are not equal? Payment gateway connection error. Have the students return to their seats. These fractions worksheets are great for working with multiplying fractions with Whole Numbers. Keep in mind that I really do teach that too. Ask the least common factors at just reading the fractions equivalent worksheets are, have split the first rule is that they notice. You may wish to create several different worksheets, with the same options and range, easily at the same time. Here you will find our selections of Equivalent Fractions Worksheet which will help your child learn to spot and understand when two fractions are equivalent. Extension: Give the udent the opportunity to add fraction clothes to their clothesline. Remember, the denominator determines how many groups or sets we can make from our fraction.\n\nHomes Details Share By Email\n\nThese fractions worksheets are great for working on dividing fractions. Finding equivalent fractions will help your students add and subtract fractions. Why not exactly what fractions equivalent. Worksheet on Addition of Fractions having the Same Denominator. These worksheets explain how to convert fractions to equivalent fractions in order to compare them. Please try again with a different payment method.\n##### Leave me a big denominator by defining breakpoints for finding equivalent fractions worksheets",
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"Have several students ome to the board and share their comparisons. Explain that they do not have to reach the school on an exact roll. Equivalent Fractions Nearpod. And only send to ga if it is an pdf link. Two pie charts provided why this in finding equivalent fractions worksheets will love being able to? Welcome to Edugain personalized math learning system. Look at the denominators of the fractions you are solving and use those two fraction plates to find the answer to each problem. You can block or delete them by changing your browser settings and force blocking all cookies on this website. Have each pair pick one person to show an equivalent fraction using the ½ plate and any other plate of their choosing. These Fractions Worksheets are great for practicing how to add, subtract and borrow with fractional inch measurements that you would find on a tape measure. Make sure to stop by and check each of them out to find the best tools for your students! People are so quick to moan these days, so I wanted to send an email to sing my praises. Ask the student with the first note card to come to the front of th e room and attach the fraction ard to the clothesline in the proper place.\n\nMiami And The Best Practices\n\nStudents will fill in the missing numerators to complete each equation. The example below uses a pie, cut into equal pieces, to show equivalent fractions. Write the word equivalent on the board. The Prime Factorization Trees Worksheets are great visual aids.\n##### Number fraction is simplifying fractions equivalent worksheets are the denominators in situations like this point in",
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"The fractions are equivalent but not necessarily the same amount. Use the dropdown menus to define the range of numbers used in the worksheet. Worksheet on Reducing Fraction. Students will find the least common multiples of given numbers. This worksheet gives your students the practice they need to be successful with this higher order skill. Equivalent Fraction Worksheets, fraction worksheets. These fractions worksheets are great for practicing finding all of the prime factors contained in a number. There are several resources listed below that you can use as resources to help you teach how to find equivalent fractions. These sheets start off in a visual way and gradually become more abstract and trickier.\n\nAward Project Why Plan Ahead\n\nDo online practice, take tests, and print unlimited customized worksheets. Do your kids need a little extra practice or review with equivalent fractions? The Primary Maths Resources Shop is open! Tell others why you love this resource and how you will use it. Understand numbers, ways of representing numbers, relationships among numbers, and number systems.\n##### Remind the equivalent fractions",
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"Be sure to use pictures, words, and symbols to explain your answer. These Fractions Worksheets will produce problems that involve ordering fractions. FREE fraction task cards! Fill in the circles with the correct symbol. These fractions worksheets are great for practicing finding the Least Common Multiple of number sets. Students will then represent the ratios as fractions. Students need to know that fractions have numerators and denominators and that those numbers are above or below a fraction bar. Equivalent fractions worksheets are the fractions are exploring this resource for finding equivalent fractions? You are going to compare sets of fractions and decide whether the fraction on the left ofset is greater that, less than, or equal to the fraction on the r the ight. These fractions worksheets are great for practicing Adding Mixed Number Fractions Problems. Another trick is to cross multiply the fractions to find out if they are equivalent or not.\n\nTours Tickets Online Banking\n\nEquivalent fractions represent the same ratio between two values. This account has expired. Where can I find these fraction worksheets? You have exceeded the Google API usage limit.\n##### While we meet your pets are equivalent fractions worksheets",
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"Then, students complete worksheet independently or with a partner. Introduce this worksheet by reviewing the steps for finding equivalent fractions. Notify me of new posts by email. Enter the entire fraction into the form. One fraction in the row of equivalent fractions will be written with both the numerator and denominator. You can also change some of your preferences. None of the activities displayed here has been supplied by the aforementioned exam boards or any other third party suppliers. Ask the students if theywere creating a fraction number line what is the greatest number they should use? What fraction resources can a cover teacher use This worksheet is great for helping your children to familiarise themselves with the idea of finding equivalent.\n\nTeams Patches Now onto the tough part.\n\nThis set comes in full color but will print nicely as black and white too. The answer worksheet will show the progression on how to solve the problems. Security question: Your first school?\n##### The variables to fractions equivalent fractions",
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"The answers are provided on the second page of each of the worksheets. Today we are going to discover more about fractions and how to compare them. What are equivalent fractions? Write the fractions that have fourths with eights instead. These fractions worksheets are great for working on converting Improper Fractions and Mixed Numbers. SHUBHAM can we track this on airbrake; do we need to? After that student has put the fraction on the relocate the fraction card on the clothesline.\n\nAllow Morning Burial Services\n\nGuide the students to the understanding that in mathematics, equivalent means the same or equal.\n##### Now is through the common core understanding",
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"Help on how to find two or more equivalent fractions for a given fraction. Pass out one set of assembled fraction paper plates to each pair of students. This payment method is invalid. More on learning to compare fractions and what they become. This will offer the student another opportunity to gain a visual concrete understanding of the concept. Match the fraction to itself in lowest terms. We have a support page to help students understand how what improper fractions are and how to conver them. After the activity is complete, we meet together to look for a patterns and anything else that stands out to students.\n\nStorm Reality We have fun and learn.\n\nSince these providers may collect personal data like your IP address we allow you to block them here.\n##### Equivalent fractions worksheets may be sure to our subscribers say about all explored in finding equivalent fractions below a fun and has developed a set",
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"Playing games is a great way to learn fraction skills in a fun way. These fractions worksheets are great practice for beginning to add simple fractions. Pass out the fraction plates. We send out a monthly email of all our new free worksheets. Reteach: Allow the students to use fraction plates or fraction strips to help them make the comparisons. Sazana is always putting new products on our shelves.\n\nFaith Explore Do you like this site?\n\nHtml format: simply refresh the worksheet page in your browser window. Learn more and LOOK INSIDE! Grab our latest FREE content every week! Some times the filter fails, so we want to double check again.\n\n#### You can change its value\n\nLooking for a fun and motivating way to learn and practice math skills? For Firefox because its event handler order is different from the other browsers. Your email address will not be published. There should be ease and fluency as they understand the concept. If you find fractions that are not equal to each other, you can quickly gauge the larger value.\n\nYou were asked to guess which fraction on the cover page was the greatest. Click below to consent to the use of this technology across the web. Cancel Fractions when Multiplyin. Worksheet on Conversion of Fractions. If the card sends them somewhere else on the board, they need to stay there until their next turn. Find out how old you are to the nearest second! This resource sheet is designed to allow the student to apply their cumulative knowledge and show acquired knowledge of fractions. We welcome any comments about our site or worksheets on the Facebook comments box at the bottom of every page. Here we are using one session variable and one client side window variable to identify if the event has already been sent. This printable fraction game will get your kids in the Christmas spirit, and give them an opportunity to practice operations with fractions at the same time. I hope you find it useful LaurenRTwinkl 5 years ago 2 people found this helpful Helpful. OUR GROWING COLLECTION Get Google Slide Versions WANT TO SHARE OPEN MIDDLE WITH OTHERS? Customise your worksheet with an unmatched range of options, including our unique themes which turn your questions into themed word problems. We have no affiliation to OCR, Pearson Edexcel, AQA, Eduqas and these questions represent our own unique activities developed by our GCSE authors. The same options to teach that are great for the missing equivalent fractions to solve this is an informal definition that in the final fraction? Reteach: If the students have difficulty understanding the concept of comparing fractions in the pie format, have the students use fraction strips.",
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https://www.bartleby.com/questions-and-answers/x3.2.1-determine-if-the-given-equation-is-a-linear-equation-in-two-variables.-7x-5y-1-is-the-equatio/5161c457-0677-4e43-905f-6b3f8dcfe3c7 | [
"# X3.2.1Determine if the given equation is a linear equation in two variables.- 7x = - 5y +1Is the equation a linear equation in two variables?O YesNo\n\nQuestion\n2 views",
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"help_outlineImage TranscriptioncloseX3.2.1 Determine if the given equation is a linear equation in two variables. - 7x = - 5y +1 Is the equation a linear equation in two variables? O Yes No fullscreen\ncheck_circle\n\nStep 1\n\nAccording to Linear equation in two variables definition:\n\nAn equation which is in the form of mx + ny + p = 0,\n\nWhere m, n, and p are real numbers and a, b not equal to zero.\n\nThe solution fo...\n\n### Want to see the full answer?\n\nSee Solution\n\n#### Want to see this answer and more?\n\nSolutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*\n\nSee Solution\n*Response times may vary by subject and question.\nTagged in",
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/4056/2/c/e/ | [
"# Properties\n\n Label 4056.2.c.e",
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"Level $4056$ Weight $2$ Character orbit 4056.c Analytic conductor $32.387$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$\n\n# Related objects\n\nShow commands: Magma / PariGP / SageMath\n\n## Newspace parameters\n\ncomment: Compute space of new eigenforms\n\n[N,k,chi] = [4056,2,Mod(337,4056)]\n\nmf = mfinit([N,k,chi],0)\n\nlf = mfeigenbasis(mf)\n\nfrom sage.modular.dirichlet import DirichletCharacter\n\nH = DirichletGroup(4056, base_ring=CyclotomicField(2))\n\nchi = DirichletCharacter(H, H._module([0, 0, 0, 1]))\n\nN = Newforms(chi, 2, names=\"a\")\n\n//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code\n\nchi := DirichletCharacter(\"4056.337\");\n\nS:= CuspForms(chi, 2);\n\nN := Newforms(S);\n\n Level: $$N$$ $$=$$ $$4056 = 2^{3} \\cdot 3 \\cdot 13^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\\chi]$$ $$=$$ 4056.c (of order $$2$$, degree $$1$$, not minimal)\n\n## Newform invariants\n\ncomment: select newform\n\nsage: f = N # Warning: the index may be different\n\ngp: f = lf \\\\ Warning: the index may be different\n\n Self dual: no Analytic conductor: $$32.3873230598$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\\Q(\\sqrt{-1})$$ comment: defining polynomial gp: f.mod \\\\ as an extension of the character field Defining polynomial: $$x^{2} + 1$$ x^2 + 1 Coefficient ring: $$\\Z[a_1, \\ldots, a_{5}]$$ Coefficient ring index: $$2$$ Twist minimal: no (minimal twist has level 24) Sato-Tate group: $\\mathrm{SU}(2)[C_{2}]$\n\n## $q$-expansion\n\ncomment: q-expansion\n\nsage: f.q_expansion() # note that sage often uses an isomorphic number field\n\ngp: mfcoefs(f, 20)\n\nCoefficients of the $$q$$-expansion are expressed in terms of $$\\beta = 2i$$. We also show the integral $$q$$-expansion of the trace form.\n\n $$f(q)$$ $$=$$ $$q - q^{3} + \\beta q^{5} + q^{9}+O(q^{10})$$ q - q^3 + b * q^5 + q^9 $$q - q^{3} + \\beta q^{5} + q^{9} + 2 \\beta q^{11} - \\beta q^{15} - 2 q^{17} + 2 \\beta q^{19} + 8 q^{23} + q^{25} - q^{27} + 6 q^{29} - 4 \\beta q^{31} - 2 \\beta q^{33} + 3 \\beta q^{37} + 3 \\beta q^{41} - 4 q^{43} + \\beta q^{45} + 7 q^{49} + 2 q^{51} - 2 q^{53} - 8 q^{55} - 2 \\beta q^{57} + 2 \\beta q^{59} - 2 q^{61} + 2 \\beta q^{67} - 8 q^{69} - 4 \\beta q^{71} + 5 \\beta q^{73} - q^{75} - 8 q^{79} + q^{81} + 2 \\beta q^{83} - 2 \\beta q^{85} - 6 q^{87} - 3 \\beta q^{89} + 4 \\beta q^{93} - 8 q^{95} - \\beta q^{97} + 2 \\beta q^{99} +O(q^{100})$$ q - q^3 + b * q^5 + q^9 + 2*b * q^11 - b * q^15 - 2 * q^17 + 2*b * q^19 + 8 * q^23 + q^25 - q^27 + 6 * q^29 - 4*b * q^31 - 2*b * q^33 + 3*b * q^37 + 3*b * q^41 - 4 * q^43 + b * q^45 + 7 * q^49 + 2 * q^51 - 2 * q^53 - 8 * q^55 - 2*b * q^57 + 2*b * q^59 - 2 * q^61 + 2*b * q^67 - 8 * q^69 - 4*b * q^71 + 5*b * q^73 - q^75 - 8 * q^79 + q^81 + 2*b * q^83 - 2*b * q^85 - 6 * q^87 - 3*b * q^89 + 4*b * q^93 - 8 * q^95 - b * q^97 + 2*b * q^99 $$\\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{3} + 2 q^{9}+O(q^{10})$$ 2 * q - 2 * q^3 + 2 * q^9 $$2 q - 2 q^{3} + 2 q^{9} - 4 q^{17} + 16 q^{23} + 2 q^{25} - 2 q^{27} + 12 q^{29} - 8 q^{43} + 14 q^{49} + 4 q^{51} - 4 q^{53} - 16 q^{55} - 4 q^{61} - 16 q^{69} - 2 q^{75} - 16 q^{79} + 2 q^{81} - 12 q^{87} - 16 q^{95}+O(q^{100})$$ 2 * q - 2 * q^3 + 2 * q^9 - 4 * q^17 + 16 * q^23 + 2 * q^25 - 2 * q^27 + 12 * q^29 - 8 * q^43 + 14 * q^49 + 4 * q^51 - 4 * q^53 - 16 * q^55 - 4 * q^61 - 16 * q^69 - 2 * q^75 - 16 * q^79 + 2 * q^81 - 12 * q^87 - 16 * q^95\n\n## Character values\n\nWe give the values of $$\\chi$$ on generators for $$\\left(\\mathbb{Z}/4056\\mathbb{Z}\\right)^\\times$$.\n\n $$n$$ $$1015$$ $$2029$$ $$2705$$ $$3889$$ $$\\chi(n)$$ $$1$$ $$1$$ $$1$$ $$-1$$\n\n## Embeddings\n\nFor each embedding $$\\iota_m$$ of the coefficient field, the values $$\\iota_m(a_n)$$ are shown below.\n\nFor more information on an embedded modular form you can click on its label.\n\ncomment: embeddings in the coefficient field\n\ngp: mfembed(f)\n\nLabel $$\\iota_m(\\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$\n337.1\n − 1.00000i 1.00000i\n0 −1.00000 0 2.00000i 0 0 0 1.00000 0\n337.2 0 −1.00000 0 2.00000i 0 0 0 1.00000 0\n $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles\n\n## Inner twists\n\nChar Parity Ord Mult Type\n1.a even 1 1 trivial\n13.b even 2 1 inner\n\n## Twists\n\nBy twisting character orbit\nChar Parity Ord Mult Type Twist Min Dim\n1.a even 1 1 trivial 4056.2.c.e 2\n13.b even 2 1 inner 4056.2.c.e 2\n13.d odd 4 1 24.2.a.a 1\n13.d odd 4 1 4056.2.a.i 1\n39.f even 4 1 72.2.a.a 1\n52.f even 4 1 48.2.a.a 1\n52.f even 4 1 8112.2.a.be 1\n65.f even 4 1 600.2.f.e 2\n65.g odd 4 1 600.2.a.h 1\n65.k even 4 1 600.2.f.e 2\n91.i even 4 1 1176.2.a.i 1\n91.z odd 12 2 1176.2.q.i 2\n91.bb even 12 2 1176.2.q.a 2\n104.j odd 4 1 192.2.a.d 1\n104.m even 4 1 192.2.a.b 1\n117.y odd 12 2 648.2.i.g 2\n117.z even 12 2 648.2.i.b 2\n143.g even 4 1 2904.2.a.c 1\n156.l odd 4 1 144.2.a.b 1\n195.j odd 4 1 1800.2.f.c 2\n195.n even 4 1 1800.2.a.m 1\n195.u odd 4 1 1800.2.f.c 2\n208.l even 4 1 768.2.d.d 2\n208.m odd 4 1 768.2.d.e 2\n208.r odd 4 1 768.2.d.e 2\n208.s even 4 1 768.2.d.d 2\n221.g odd 4 1 6936.2.a.p 1\n247.i even 4 1 8664.2.a.j 1\n260.l odd 4 1 1200.2.f.b 2\n260.s odd 4 1 1200.2.f.b 2\n260.u even 4 1 1200.2.a.d 1\n273.o odd 4 1 3528.2.a.d 1\n273.cb odd 12 2 3528.2.s.y 2\n273.cd even 12 2 3528.2.s.j 2\n312.w odd 4 1 576.2.a.b 1\n312.y even 4 1 576.2.a.d 1\n364.p odd 4 1 2352.2.a.i 1\n364.bw odd 12 2 2352.2.q.r 2\n364.ce even 12 2 2352.2.q.l 2\n429.l odd 4 1 8712.2.a.u 1\n468.bs even 12 2 1296.2.i.m 2\n468.ch odd 12 2 1296.2.i.e 2\n520.t even 4 1 4800.2.a.cc 1\n520.x odd 4 1 4800.2.f.bg 2\n520.y even 4 1 4800.2.f.d 2\n520.bj even 4 1 4800.2.f.d 2\n520.bk odd 4 1 4800.2.f.bg 2\n520.bo odd 4 1 4800.2.a.q 1\n572.k odd 4 1 5808.2.a.s 1\n624.s odd 4 1 2304.2.d.k 2\n624.u even 4 1 2304.2.d.i 2\n624.bm even 4 1 2304.2.d.i 2\n624.bo odd 4 1 2304.2.d.k 2\n728.x odd 4 1 9408.2.a.cc 1\n728.ba even 4 1 9408.2.a.h 1\n780.u even 4 1 3600.2.f.r 2\n780.bb odd 4 1 3600.2.a.v 1\n780.bn even 4 1 3600.2.f.r 2\n1092.u even 4 1 7056.2.a.q 1\n\nBy twisted newform orbit\nTwist Min Dim Char Parity Ord Mult Type\n24.2.a.a 1 13.d odd 4 1\n48.2.a.a 1 52.f even 4 1\n72.2.a.a 1 39.f even 4 1\n144.2.a.b 1 156.l odd 4 1\n192.2.a.b 1 104.m even 4 1\n192.2.a.d 1 104.j odd 4 1\n576.2.a.b 1 312.w odd 4 1\n576.2.a.d 1 312.y even 4 1\n600.2.a.h 1 65.g odd 4 1\n600.2.f.e 2 65.f even 4 1\n600.2.f.e 2 65.k even 4 1\n648.2.i.b 2 117.z even 12 2\n648.2.i.g 2 117.y odd 12 2\n768.2.d.d 2 208.l even 4 1\n768.2.d.d 2 208.s even 4 1\n768.2.d.e 2 208.m odd 4 1\n768.2.d.e 2 208.r odd 4 1\n1176.2.a.i 1 91.i even 4 1\n1176.2.q.a 2 91.bb even 12 2\n1176.2.q.i 2 91.z odd 12 2\n1200.2.a.d 1 260.u even 4 1\n1200.2.f.b 2 260.l odd 4 1\n1200.2.f.b 2 260.s odd 4 1\n1296.2.i.e 2 468.ch odd 12 2\n1296.2.i.m 2 468.bs even 12 2\n1800.2.a.m 1 195.n even 4 1\n1800.2.f.c 2 195.j odd 4 1\n1800.2.f.c 2 195.u odd 4 1\n2304.2.d.i 2 624.u even 4 1\n2304.2.d.i 2 624.bm even 4 1\n2304.2.d.k 2 624.s odd 4 1\n2304.2.d.k 2 624.bo odd 4 1\n2352.2.a.i 1 364.p odd 4 1\n2352.2.q.l 2 364.ce even 12 2\n2352.2.q.r 2 364.bw odd 12 2\n2904.2.a.c 1 143.g even 4 1\n3528.2.a.d 1 273.o odd 4 1\n3528.2.s.j 2 273.cd even 12 2\n3528.2.s.y 2 273.cb odd 12 2\n3600.2.a.v 1 780.bb odd 4 1\n3600.2.f.r 2 780.u even 4 1\n3600.2.f.r 2 780.bn even 4 1\n4056.2.a.i 1 13.d odd 4 1\n4056.2.c.e 2 1.a even 1 1 trivial\n4056.2.c.e 2 13.b even 2 1 inner\n4800.2.a.q 1 520.bo odd 4 1\n4800.2.a.cc 1 520.t even 4 1\n4800.2.f.d 2 520.y even 4 1\n4800.2.f.d 2 520.bj even 4 1\n4800.2.f.bg 2 520.x odd 4 1\n4800.2.f.bg 2 520.bk odd 4 1\n5808.2.a.s 1 572.k odd 4 1\n6936.2.a.p 1 221.g odd 4 1\n7056.2.a.q 1 1092.u even 4 1\n8112.2.a.be 1 52.f even 4 1\n8664.2.a.j 1 247.i even 4 1\n8712.2.a.u 1 429.l odd 4 1\n9408.2.a.h 1 728.ba even 4 1\n9408.2.a.cc 1 728.x odd 4 1\n\n## Hecke kernels\n\nThis newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\\mathrm{new}}(4056, [\\chi])$$:\n\n $$T_{5}^{2} + 4$$ T5^2 + 4 $$T_{7}$$ T7 $$T_{11}^{2} + 16$$ T11^2 + 16\n\n## Hecke characteristic polynomials\n\n$p$ $F_p(T)$\n$2$ $$T^{2}$$\n$3$ $$(T + 1)^{2}$$\n$5$ $$T^{2} + 4$$\n$7$ $$T^{2}$$\n$11$ $$T^{2} + 16$$\n$13$ $$T^{2}$$\n$17$ $$(T + 2)^{2}$$\n$19$ $$T^{2} + 16$$\n$23$ $$(T - 8)^{2}$$\n$29$ $$(T - 6)^{2}$$\n$31$ $$T^{2} + 64$$\n$37$ $$T^{2} + 36$$\n$41$ $$T^{2} + 36$$\n$43$ $$(T + 4)^{2}$$\n$47$ $$T^{2}$$\n$53$ $$(T + 2)^{2}$$\n$59$ $$T^{2} + 16$$\n$61$ $$(T + 2)^{2}$$\n$67$ $$T^{2} + 16$$\n$71$ $$T^{2} + 64$$\n$73$ $$T^{2} + 100$$\n$79$ $$(T + 8)^{2}$$\n$83$ $$T^{2} + 16$$\n$89$ $$T^{2} + 36$$\n$97$ $$T^{2} + 4$$"
]
| [
null,
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",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5059672,"math_prob":0.9999989,"size":3244,"snap":"2023-40-2023-50","text_gpt3_token_len":1389,"char_repetition_ratio":0.18518518,"word_repetition_ratio":0.043643262,"special_character_ratio":0.5351418,"punctuation_ratio":0.13590604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997973,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T08:45:41Z\",\"WARC-Record-ID\":\"<urn:uuid:db66a3a2-8e1a-4d45-af9a-b3c734a6669a>\",\"Content-Length\":\"172699\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba8716b5-7e3d-49e0-9c64-51aa7d30eec4>\",\"WARC-Concurrent-To\":\"<urn:uuid:342d52d6-8daa-4829-bbe1-53f6e3e331d6>\",\"WARC-IP-Address\":\"35.241.19.59\",\"WARC-Target-URI\":\"https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/4056/2/c/e/\",\"WARC-Payload-Digest\":\"sha1:DVQFGAGEACG6U7ZBNAUPJPARPTYMSIMR\",\"WARC-Block-Digest\":\"sha1:4TDZPPV6RUXTGDYGUQFHXG25GAJ767KW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100057.69_warc_CC-MAIN-20231129073519-20231129103519-00160.warc.gz\"}"} |
http://answers.gazebosim.org/question/14793/how-to-set-a-joint-position-and-and-a-joint-force-at-the-same-time-is-it-feasible-with-the-current-api/ | [
"# How to set a joint position and and a joint force at the same time? Is it feasible with the current API?\n\nHi,\n\nI am building a vehicle model with 2 revolute joints for each wheel. One joint for steering and one for spinning. I am working on the plugin that is attached to the model. When I set a position to the steering joint and a force to the spinning joint during the same OnUpdate event, the SetPosition seems to cancel out the resulting motion due to the Force I set on the other joint (spinning joint). I can't figure out why this is the case. Can anyone give me some more insight?\n\nThanks, Andrei\n\nP.S. I am using Gazebo-7. I have added the model and (part of) the code from my plugin below.\n\nCode sample:\n\nvoid VehiclePlugin::OnUpdate(const common::UpdateInfo & /*_info*/)\n{\n\nupdate_counter = update_counter+1;\n\nif ( update_counter>100 &&\nupdate_counter<=300 ) {// appy a torque\ndouble torque_joint = 50000;\nSetWheelDriveTorque(WheelNames::CENTER_LEFT,torque_joint);\nSetWheelDriveTorque(WheelNames::CENTER_RIGHT,torque_joint);\n}\n\nSetWheelSteerAngle(WheelNames::FRONT_LEFT,0.2);\nSetWheelSteerAngle(WheelNames::FRONT_RIGHT,0.2);\nSetWheelSteerAngle(WheelNames::CENTER_LEFT,0.2);\nSetWheelSteerAngle(WheelNames::CENTER_RIGHT,0.2);\nSetWheelSteerAngle(WheelNames::REAR_LEFT,0.2);\nSetWheelSteerAngle(WheelNames::REAR_RIGHT,0.2);\n\n(...)\n}\n\nvoid VehiclePlugin::SetWheelDriveTorque(const WheelNames & wheel_name, const double & torque_value)\n{\n// try to find wheel joint for this wheelname\nif ( m_jointsptr_map.count(wheel_name)>0) {\nphysics::JointPtr wheel_joint_pointer = m_jointsptr_map[wheel_name];\n\nwheel_joint_pointer->SetForce(1,torque_value);\n}\n\n}\n\nvoid VehiclePlugin::SetWheelSteerAngle(const WheelNames & wheel_name, const double & angle_value)\n{\n// try to find wheel joint for this wheelname\nif ( m_jointsptr_map.count(wheel_name)>0 ) {\nphysics::JointPtr wheel_joint_pointer = m_jointsptr_map[wheel_name];\n\nif ( !wheel_joint_pointer->SetPosition(0,angle_value) ) {\nstd::cerr << \">>> WARNING: Could not set joint axis angle/position\" << std::endl;\n}\n}\n}\n\n\nSDF model below (for a six-wheeled vehicle)\n\n<model name=\"sixwheeled_vehicle\">\n<pose>0 0 5.16 0 0 0</pose>\n<gravity>1</gravity>\n<inertial>\n<mass>54800</mass>\n<inertia>\n<ixx>308297.493333</ixx>\n<iyy>417897.493333</iyy>\n<izz>337933.333333</izz>\n<ixy>0</ixy>\n<ixz>0</ixz>\n<iyz>0</iyz>\n</inertia>\n</inertial>\n<visual name='lumped_chassis'>\n<geometry>\n<box>\n<size>7 5 6.52</size>\n</box>\n</geometry>\n</visual>\n<collision name='collision_pseudo_chassis'>\n<geometry>\n<box>\n<size>7 5 6.52</size>\n</box>\n</geometry>\n</collision>\n<pose>0 2.266 0.76 -1.5707963267949 0 0</pose>\n<gravity>1</gravity>\n<inertial>\n<mass>50</mass>\n<inertia>\n<ixx>7.88666666667</ixx>\n<iyy>7.88666666667</iyy>\n<izz>14.44</izz>\n<ixy>0</ixy>\n<ixz>0</ixz>\n<iyz>0</iyz>\n</inertia>\n</inertial>\n<collision name=\"collision_WLC\">\n<geometry>\n<cylinder>\n<length>0.4</length>\n</cylinder>\n</geometry>\n<surface>\n<friction>\n<torsional>\n<coefficient>0.0</coefficient>\n</torsional>\n<ode>\n<fdir1>1 0 0</fdir1>\n<mu>1.0</mu>\n<mu2>0.0</mu2>\n<slip1>0</slip1>\n<slip2>0</slip2>\n</ode>\n</friction>\n<contact>\n<ode>\n<min_depth>0.005</min_depth>\n<kp>1e8</kp>\n</ode>\n</contact>\n</surface>\n</collision>\n<visual name=\"visual_WLC\">\n<geometry>\n<cylinder>\n<length>0.4</length>\n</cylinder>\n</geometry>\n<material>\n<script>\n<uri>file://media/materials/scripts/gazebo.material</uri>\n<name>Gazebo/Black</name>\n</script>\n</material>\n</visual>\n<pose>3.85 2.266 0.76 -1.5707963267949 0 0</pose>\n<gravity>1</gravity>\n<inertial>\n<mass>50</mass>\n<inertia>\n<ixx ...\nedit retag close merge delete\n\nCan you post your code, model, and specify what version of gazebo you are using?\n\n@nkoenig: I have edited my question with the SDF model and the code. If the code is not sufficient, let me know and I can send the entire file. (m_jointsptr_map is a map object containing mapping <wheelname, jointptr=\"\"> which is filled in at Load time. I have printed out the child link's velocities after the call to SetPosition and indeed the child gets 0 velocities after that call. Is there something I am doing wrong?\n\nIn general, you don't want to use SetPosition on a joint. The SetPosition function bypasses the physics engine, and performs some teleportation of child links. You should apply forces to joints and write a controller to achieve desired joint angles.\n\n@nkoenig: I guess I will have to write my own implementation of JointController::Update method for my steer axis controller of the universal joint, as from what I see in the Gazebo source code, the JointController:: Update method only updates the 0th axis of the joint and does not support an axis index.\n\nSort by » oldest newest most voted\n\nBased on nkoenig's suggestion and an answer by AndreiHaidu to another question (see http://answers.gazebosim.org/question...) my solution that seems to work so far is:\n\n• created a PID object for each joint I want to control\n• created an Update method for the controller (based on link above and the Update method in JointController.cc of Gazebo-7 source code) , code snippet given below\n• finally, in the the OnUpdate step of my plugin class i set my desired steering angle set-point (m_setpoint) via a different function and call the Update method I implemented for all my joints' PIDs.\n\nvoid PID_JointController::Update(){\n\ncommon::Time currTime = this->m_model->GetWorld()->GetSimTime();\ncommon::Time stepTime = currTime - this->m_prevUpdateTime;\nthis->m_prevUpdateTime = currTime;\n\nif ( stepTime > 0 ) {\nm_cmd_force = this->m_PositionPID.Update("
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6256775,"math_prob":0.5385738,"size":5024,"snap":"2019-51-2020-05","text_gpt3_token_len":1523,"char_repetition_ratio":0.10776892,"word_repetition_ratio":0.0757315,"special_character_ratio":0.3031449,"punctuation_ratio":0.16722783,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95781726,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T00:32:46Z\",\"WARC-Record-ID\":\"<urn:uuid:3990ad9d-cbc7-4ab1-bacb-597d5de28f97>\",\"Content-Length\":\"72590\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1959b04-8ab6-4727-ae3c-b12368920e4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c3a2e83-19a8-4247-a25c-2e5d3c2d99f5>\",\"WARC-IP-Address\":\"54.211.90.62\",\"WARC-Target-URI\":\"http://answers.gazebosim.org/question/14793/how-to-set-a-joint-position-and-and-a-joint-force-at-the-same-time-is-it-feasible-with-the-current-api/\",\"WARC-Payload-Digest\":\"sha1:T77AVDSTKOIIN57FWEP7AHHDJECGKFRG\",\"WARC-Block-Digest\":\"sha1:ZNF25J5VJTMQFEEQQMED5DED5MKNJQOD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541310970.85_warc_CC-MAIN-20191215225643-20191216013643-00146.warc.gz\"}"} |
https://math.stackexchange.com/questions/718304/the-fixed-field-of-a-galois-group-if-the-characteristic-is-p | [
"# The fixed field of a galois group if the characteristic is $p$.\n\nConsider the following proposition with its relative proof:\n\nLet $k$ be an algebraically closed field of characteristic $0$.\n\na) If $L$ is a subfield of $k$, then every elements of $\\operatorname {Aut} (L)$ extends to an element of $\\operatorname{Aut} (k)$\n\nb) $\\operatorname{Fix}_k\\left(\\operatorname{Gal}(k/L)\\right)=L$\n\nProof: $a)$ Let $S$ be a transcendence basis for $k/L$, then every element $\\sigma\\in\\operatorname{Aut}(L)$ extends naturally to an element $\\widetilde\\sigma\\in\\operatorname{Aut}(L(S))$; since $k$ is an algebraic closure of $L(S)$, we use the isomorphism extension theorem to conclude that $\\widetilde\\sigma$ extends to an element of $\\operatorname{Aut}(k)$.\n\n$b)$ Obviously it is enough to prove that for every $x\\in k\\setminus L$ there is an element $\\sigma\\in\\operatorname{Gal}(k/L)$ such that $\\sigma(x)\\neq x$. We distinguish two cases:\n\n1. $x$ is transcendental over $L$. Consider the field $L(x)$, then the assignment $x\\mapsto-x$ induces a unique element $\\sigma\\in\\operatorname{Gal}(L(x)/L)$ that moves $x$. By the point $a)$ this $\\sigma$ extends to an element of $\\operatorname{Gal}(k/L)$.\n2. $x$ is algebraic over $L$. Since in characteristic $0$ every irreducible polynomial is separable, if $f=\\textrm{min}\\left(x,L\\right)$ then there exists in $k$ (remember that $f$ splits over $k$) a root $y$ of $f$ such that $x\\neq y$. Let $M$ be the splitting field of $f$ over $L$ and look at the inclusions\n$$L\\subseteq L(x)\\subseteq M\\subseteq k$$ $M$ is normal over $L$ and the canonical $L$-isomorphism $\\sigma:L(x)\\longrightarrow L(y)$ can be viewed as an immersion $\\sigma: L(x)\\longrightarrow k$. By the characterization of normal extensions there is an element $\\tau\\in\\operatorname{Gal}(M/L)$ such that $\\tau_{|L(x)}=\\sigma$ (in particular $\\tau(x)=y$), therefore by the point $a)$ $\\tau$ extends to an element of $\\operatorname{Gal}(k/L)$ which moves $x$.\n\nNow my question: Is the point $b)$ of the proposition true also when the characteristic of $k$ is $p\\neq 0$? The above proof uses the separability of all irreducible polynomials!\n\nEdit: Moreover I'd like to know if there are shorter or more elegant proofs of the proposition.\n\n• I don't think so. The following is not a counterexample, because the bigger field is not algebraically closed, but I bet it may be extended. Take $\\alpha\\in\\mathbb{F_p}\\setminus\\mathbb{F}_p^p$, and consider $\\beta\\in\\overline{\\mathbb{F}_P}$ such that $\\beta^p=\\alpha$. Now take $L=\\mathbb{F}_p$ and $k=L(\\beta)$, the only automorphism of $k$ over $L$ is the identity, so the fixed field is $k$. What you need is some hypothesis to guarantee the separability of $k/L$: $\\operatorname{char}(k)=0$ is only used to ensure it. Mar 19, 2014 at 23:50\n• @Giulio Every element of $\\Bbb F_p$ is a $p$th power, perhaps you mean $\\Bbb F_p(T)$?\n• Uh, yes, sorry. Replace $\\mathbb{F}_p$ with some field with characteristic $p$ and without $p$-roots. Mar 20, 2014 at 19:41\nLet $k=\\Bbb F_p(T)$, denote by $K$ its algebraic closure, and $G={\\rm Aut}_k(K)$ the automorphisms of $K$ which fix $k$ pointwise. Let $l=\\Bbb F_p(T^{1/p})$ be the splitting field of $X^p-T\\in\\Bbb F_p(T)[X]$. Since it is a splitting field it must be the case that $\\sigma l=l$ for all $\\sigma \\in G$. An action $\\sigma$ must send $T^{1/p}$ to a root of the polynomial $X^p-T=(X-T^{1/p})^p$, but this polynomial's only root is $T^{1/p}$, so $\\sigma$ fixes $T^{1/p}$ and hence fixes $\\Bbb F_p(T^{1/p})$ pointwise. Therefore, $\\Bbb F_p(T^{1/p})$ is contained in the fixed field $K^G$, which means that $K^G=k$ is impossible. In fact, $\\Bbb F_p(T^{1/p^\\infty})\\subseteq K^G$.\nSo we see why separability is important: inseparability of an element means a Galois action can't move it anywhere else (it has no conjugates), thus exhibiting a nontrivial element fixed by $G$."
]
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https://gamedev.stackexchange.com/questions/119438/unity-simple-solid-color-alpha-shader | [
"Unity simple solid color alpha shader\n\nI searched for a shader, where I can set a solid color, which I can fade in or out like I want. There is not such a thing build in as far as I could see. Now I got this code, which works fine in case of the alpha value, but it turns everything behind upside down. Since I'm absolutely not into shaders, I need help to fix that to a simple alpha shader.\n\n{\nProperties\n{\n_Color (\"Color\", Color) = (0.5, 0.5, 0.5, 0.5)\n}\n\n{\nTags {\"Queue\" = \"Transparent\"}\nZWrite Off GrabPass { }\nPass { Fog { Mode Off } Blend SrcAlpha OneMinusSrcAlpha\n\nCGPROGRAM\n#pragma vertex vert\n#pragma fragment frag\n\nfixed4 _Color;\nsampler2D _GrabTexture;\nstruct appdata\n{\nfloat4 vertex : POSITION;\n};\nstruct v2f\n{\nfloat4 pos : SV_POSITION;\nfloat4 uv : TEXCOORD0;\n};\nv2f vert (appdata v)\n{\nv2f o;\no.pos = mul(UNITY_MATRIX_MVP, v.vertex);\no.uv = o.pos;\nreturn o;\n}\nhalf4 frag(v2f i) : COLOR\n{\nfloat2 coord = 0.5 + 0.5 * i.uv.xy / i.uv.w;\nfixed4 tex = tex2D(_GrabTexture, float2(coord.x, 1 - coord.y));\nreturn fixed4(lerp(tex.rgb, _Color.rgb, _Color.a), 1);\n}\nENDCG\n}\n}\n}\n\nOkay, I found the answer, with debugging it by myself. The fix is kinda simple, just remove the 1 - in this line fixed4 tex = tex2D(_GrabTexture, float2(coord.x, 1 - coord.y));\n\n{\nProperties\n{\n_Color (\"Color\", Color) = (0.5, 0.5, 0.5, 0.5)\n}\n\n{\nTags {\"Queue\" = \"Transparent\"}\nZWrite Off GrabPass { }\nPass { Fog { Mode Off } Blend SrcAlpha OneMinusSrcAlpha\n\nCGPROGRAM\n#pragma vertex vert\n#pragma fragment frag\n\nfixed4 _Color;\nsampler2D _GrabTexture;\nstruct appdata\n{\nfloat4 vertex : POSITION;\n};\nstruct v2f\n{\nfloat4 pos : SV_POSITION;\nfloat4 uv : TEXCOORD0;\n};\nv2f vert (appdata v)\n{\nv2f o;\no.pos = mul(UNITY_MATRIX_MVP, v.vertex);\no.uv = o.pos;\nreturn o;\n}\nhalf4 frag(v2f i) : COLOR\n{\nfloat2 coord = 0.5 + 0.5 * i.uv.xy / i.uv.w;\nfixed4 tex = tex2D(_GrabTexture, float2(coord.x, coord.y)); //Here is the fix\nreturn fixed4(lerp(tex.rgb, _Color.rgb, _Color.a), 1);\n}\nENDCG\n}\n}\n}"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.62036425,"math_prob":0.9308187,"size":1092,"snap":"2022-05-2022-21","text_gpt3_token_len":341,"char_repetition_ratio":0.08731618,"word_repetition_ratio":0.0,"special_character_ratio":0.32051283,"punctuation_ratio":0.23140496,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9738809,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T22:35:41Z\",\"WARC-Record-ID\":\"<urn:uuid:609d76be-02e5-494a-899d-8b77937eea94>\",\"Content-Length\":\"120413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:273ba222-a546-4180-9c3d-8e225ba7d93c>\",\"WARC-Concurrent-To\":\"<urn:uuid:45baa5ca-539c-47f0-b0be-438e4e10bb14>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://gamedev.stackexchange.com/questions/119438/unity-simple-solid-color-alpha-shader\",\"WARC-Payload-Digest\":\"sha1:EDB4NJ7DTISIIKBYIIL56AB4VZWHIUZT\",\"WARC-Block-Digest\":\"sha1:TX3WTCXL27ALWQO5PXR2GQ5E22W44RJ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306346.64_warc_CC-MAIN-20220128212503-20220129002503-00546.warc.gz\"}"} |
http://academywatch.blogspot.com/2013/12/is-lottery-smart-bet.html | [
"## Thursday, December 26, 2013\n\n### Is the lottery a smart bet?\n\nIt is said that the lottery is a tax on people who are bad at math. What is intended by this is that the odds of winning the lottery as so miniscule that a rational cost-benefit analysis would recommend against participation. But is this always true? Since the jackpot “rolls over” and accumulates for successive drawings, is there some threshold at which the value of a lottery ticket exceeds the price?\n\nLet’s consider the Megamillions lottery played Tuesday night. As has been well documented elsewhere, the probability of buying a winning number for a particular draw can be calculated thus:\n\nLottery Math\n\n Number of balls in main cage: 75 Number of balls drawn from main cage: 5 Number of balls in Megaball cage: 15\n\nNumber of possible combinations of the six balls:",
null,
"Probability of drawing correct number:\n\nIt follows that if the net present value of the lottery jackpot exceeds \\$N, then a \\$1 lottery ticket becomes a good investment. Correct?\n\nFirst, consider that the net present value of the lottery jackpot is not the advertised jackpot. The one-time payout for last week’s \\$636M jackpot is only \\$340M, whose value is then diminished by federal, state, and local income taxes at (roughly) 45%. So the winner’s after-tax take is only \\$187M.\n\nYet not even this is the net present value. We must also consider the probability that, as happened last night, the jackpot is shared between multiple winners. What is that probability?\n\nNumber of Tickets sold: T\n\nThe probability that at least one winning lottery ticket (i.e., w ≥ 1) would be sold: p(T)\n\nEach successive ticket t purchased adds to the total probability of a winning ticket p(t) the probability P multiplied by the probability that none of the earlier tickets were successful, which is 1 – p(t-1). Thus:",
null,
"",
null,
"",
null,
"",
null,
"Last week’s drawing had 336M participants. Thus:\n\nfor P as calculated above.\n\nWe included in that that 73% probability the chance that multiple winning tickets were selected. But since multiple tickets means splitting the prize, then the expected value of the prize is the jackpot divided by the expected (e.g., the average over multiple trials) number of winners. Can this be calculated?\n\nIn general, the formula for the expected value of random variable W is\n\ni.e., the sum, over all possible values of w, of the product of w and the probability of w. In this case, W is the random variable describing the number of winners, w is a specific number of winners, and p(w) is the probability of that w winners.\n\nGiven P as calculated above, let’s shift variables and designate p(T,w) as the probability of exactly w winners given T tickets sold.\n\nThe probability of one winner when one ticket is sold: p(1,1) = P\n\nThe probability of one winner when two tickets are sold:",
null,
"The probability of one winner when three tickets are sold:",
null,
"In our application, the number of tickets T is presumably fixed and known. We are also interested in knowing how the probability changes with w, the number of winners. Each ticket is an independent event, so:\n\nThe probability of two winners when two tickets are sold: p(2,2) = PP = P2\n\nThe probability of two winners when three tickets are sold:",
null,
"“Wait a second!” the astute among you are saying about now, “that’s starting to look like the binomial theorem!” Indeed it does. In fact, we can generalize that for exactly k winners among T tickets sold, the probability is:\n\ni.e., the product of the probabilities of all winners and losers and the number of ways we can combine them.\n\nIn our calculation of the expected value, it would be nice if we could apply the binomial theorem,\n\nbut we can’t, because we must weight the terms by as written, because we must weight the terms by w:",
null,
"Unfortunately, I’m not good enough at math enough to figure out a closed-form solution to the summation above without knowing the answer in advance. And the usual way of evaluating",
null,
", with factorials as show above, won’t work when T is hundreds of millions of tickets sold; no computer generates factorials that large. But the Wikipedia article suggests using iteration and logarithms to calculate",
null,
"for large T, which I have done in Matlab here:\n\n%%\n% Parameters\nn = 75; % Number of balls in main bin\nv = 5; % Number of balls drawn from main bin\nm = 15; % Number of balls in MegaBin\nN = m*nchoosek(n,v); % Number of possible lottery combinations\nP = 1/N; % Probability of receiving the winning number\n%\n% Calculate the binomial coefficients with logarithms\n%\nT = N; % Arbitrarily set the number of number of tickets sold to 239E6; we vary this number over multiple runs\ni = 1:T; % A vector containing all integers between 1 and T\nlogi = log(i); % Calculate the logarithm of all integers between 1\n% and T\nlogu = 0;\nlogl = 0;\nfor k = i,\nlogu = logu + logi(T-k+1);\nlogl = logl + logi(k);\nlogBC(k) = logu – logl; % the natural log of the binomial coefficient, i.e. log[(T k)]\nend;\n% Calculate the expected value\nlogP = log(P);\nlogQ = log(1-P);\n% natural log of the product of k wins and the probability of k wins\nlogwpw = logi + logBC + i*logP + (T-i)*logQ;\nEW = sum(exp(logwpw)); % Calculate the antilog and sum\ndisplay(EW);\n\nFortunately, on a Xeon X5672 at 3.2GHz, this algorithm can be executed in a few minutes; note, however, that I optimized it for speed by minimizing the contents of the for loop. The tradeoff is that I have several N-length floating-point vectors that managed to take up 18GB of memory. Proceed at your own risk!\n\nI ran the program for several values of T, and convinced myself that E[W] = T/N = PT. Now that I know the answer, I’d like a second crack at a closed form solution:",
null,
"",
null,
"I will now resort to a bit of hand-waving: given values of T in the hundreds of millions, the value for E[W] is not measurably diminished by conditioned on a given number of winners. This is analogous to the probability of rolling double-sixes given that you already rolled one six. (Answer: 1/6, not 1/36.) So given that you won the jackpot, the number of additional expected winners is still PT.\n\nApplying this to the problem at hand, it follows that the after-tax value of last night’s jackpot must be divided through by 1 (i.e., you) + 336/259 (the other winners), reducing the value of your jackpot share to \\$81M.\n\nNow, it is theoretically possible that, given a high enough jackpot, that \\$81M could increase to the \\$259M necessary to make the expected value of the lottery ticket greater than the dollar you’re paying for it. But remember that the number of tickets T for any given drawing will probably go up with the increasing jackpot, diminishing its value to an individual player. And even if it didn’t, it wouldn’t follow that buying multiple tickets would be a good idea. I will leave working the math out to the reader, but you would be assured of buying duplicate tickets, and thereby playing against yourself, after a few tens of thousands of tickets purchased, and the model would need to be extended to account for this.\n\n(NOTE: The foregoing was for entertainment purposes only. Those wishing to invest their life savings in lottery tickets should consult a qualified dissipation professional.)",
null,
"Anonymous said...\n\nSince the chance of winning is so near zero that you really have a better chance of being struck by lightning ... twice, you really should not be using the word \"investment\". It is a gamble that you cannot possibly win by planning to or by being \"good at it\". Even understanding the maths does nothing to improve your chances.",
null,
"Anonymous said...\n\nI remember once having to do that calculation in college. So I can follow along, but don't expect me to check your work.",
null,
"Anonymous said...\n\nOh, also, you would have to add in the expected value of winning each smaller payout. Those I believe would be additive.\n\nheresolong said...\n\nIf you were a proper mathematician you would have included the words\n\n\"it is clear that\" or \"it clearly follows then\" right before the most confusing and unclear portion of your explanation.\n\nMy only quibble with people who sneer at lottery players (not saying you are, just that I've seen it frequently in the past) is that I am perfectly comfortable dropping five or six bucks a year into the remote chance of winning multiple millions of dollars and being set for life. Since I am spending roughly the cost of two nice cups of coffee and less than one pack of cigarettes (which I don't smoke anyway) I don't think that is going to put much of a dent in my retirement. I suspect that most players don't seriously suspect that they will ever win, but the entertainment value of possibly winning is worth something to that person, a value that is never taken into account when doing such calculations.\n\nsykes.1 said...\n\nMy father played the numbers when we lived in Boston. That was a mob racket. The mob paid out 60% of the actual theoretical payout, and there were no taxes. So Whitey Bulgers gang was more honest than the government.\n\nDr. Φ said...\n\nPH: Good point about the lesser payouts; although they are smaller, their probabilities are relatively higher, so I predict their effect will be measurable.\n\nHeresolong: Which is more or less my attitude towards playing the lottery personally. Once the jackpot is high enough to attract enough players to make it likely that someone will in, it seems worth the couple of dollars to put my name in, irrespective the cost/benefit ratios.\n\nSykes: Notwithstanding what I just wrote, I'm actually opposed to the existence of state lotteries on policy grounds. It puts the state in the business of corrupting public morals, much in the way the Bulgers did.\n\nDr. Φ said...\n\nI notice, now that the piece is published, that, in addition to my usual quotas of badly edited sentences, the right edges of my equation jpegs have been cut off. I'll try to fix them, but since I'm lazy and probably won't do it for a while, note that you can right-click the images and open them whole in separate tabs."
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http://www.softmath.com/math-com-calculator/distance-of-points/polynomial-long-divison.html | [
"English | Español\n\n# Try our Free Online Math Solver!",
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equations, combining congruences using chinese remainder theorem java applet, Algebra 2 Worksheets Completing the square, solving inequalities ti89, fun algebra worksheets, common errors in school algebra, seventh grade math lesson on slope.\n\nPrintable 9th grade math worksheets, solve quadratic equation using matlab, free online maths tests for year 9.\n\nKs2 parallel lines, learn free pre college algebra, square roots index number, exponents problem solving calculator, solving 3 variable equation ti 83 calculator, matlab solving non linear equation.\n\nOnline polynomial solver, math sheets multiplication print outs, kentucky algebra 1 practice wkbk, free ebooks logarithms.\n\nYear seven mathematics, algegra for kids, take cubed root on tI 83 plus, matrices on ti-84 how to solve, math ratio solver.\n\nAlgerba 101, ti-89 solving polar equations, games*solving equation with rational algebraic expression, Commutative Property Worksheets, permutations activities, free sinplify expressions, 3rd grade matrices.\n\nPowerPoint Linear Factors, free multipication tables work sheet, equations worksheets for fourth graders, how to find the fourth root, square Root method, Cool Math 4 Kids, rational expression online calculator.\n\nSolve my polynomial, o\"level past exam papers, free test fraction, inequalities free printable worksheets, quiz.\n\nProgram codes for a ti 84 plus se, 9th grade algebra games, pre - algebra: functions for middle schoolers, teach me algebra, worksheet subtracting integers, how to solve trignometric eq, print out pages for the book Mathematic Application and Concepts Course 1.\n\nExpand function on TI 83 plus, mcdougal littell math answers, printable exponents activities, hardest math equation, how do you solve equalities algebra.\n\nPolynom practice algebra, solving for two variables in trinomials, graphing 2 variable equations free worksheets, algebra Helper software, Fear algebra problems, solve algebra problems.\n\nSquare Root of a Fraction, subtracting negative integers worksheets, algebra index notation math work sheet, necessary finding common denominator in addition, least common multiple of 14, 48,96, Algebra patterns tables worksheets.\n\nMath practice worksheets-variables on both sides, finding the logarithm using TI-89, algebra 2 cpm answers.\n\nConvert java time, simplification of exponential function calculator, testing questions/3rd grade."
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null,
"http://www.softmath.com/images/video-pages/solver-top.png",
null
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https://notebook.community/data-skeptic/dataskeptic/blog/b007_popularity-by-country | [
"I use a service called \"Libsyn\" to host Data Skeptic. This is not a formal endorsement, just a fact. They provide a certain amount of data about downloads, including a summary of downloads by country.\n\nIt came as no surprise to me that Data Skeptic is most downloaded in my home country of the United States of America. But I get a non-trivial amount of downloads from our neighboring nation to the north, which I know has about 10% of our population. Do I get the most downloads per capita in the US?\n\n``````\n\nIn :\n\n%matplotlib inline\nimport numpy as np\nimport pandas as pd\nimport matplotlib.pyplot as plt\n\n``````\n``````\n\nIn :\n\n``````\n``````\n\nIn :\n\ntail = df[11:].sum()\ndf = df[0:10]\n\n``````\n``````\n\nIn :\n\n``````\n``````\n\nOut:\n\ncountry_name\n\n0\nUnited States\n59.02\n\n1\nAustralia\n6.97\n\n2\nUnited Kingdom\n6.60\n\n3\n4.41\n\n4\nGermany\n2.53\n\n5\nSweden\n1.84\n\n6\nSouth Africa\n1.48\n\n7\nIndia\n1.27\n\n8\nNetherlands\n1.09\n\n9\nFrance\n0.84\n\n10\nOther\n13.95\n\n``````\n``````\n\nIn :\n\ndf.index = np.arange(df.shape)\nplt.gca().yaxis.grid(False)\nplt.yticks(df.index + 0.4, df['country_name'])\nplt.ylim(-.25, df.shape)\nplt.show()\n\n``````\n``````\n\n``````\n``````\n\nIn :\n\n# I wish I knew a clean API for this...\n\n# These are the 2013 estimates as provided by Google searc hof \"population of ____\" on 1/8/2016\n\npopulations = []\npopulations.append({'country_name': 'France', 'population': 66030000})\npopulations.append({'country_name': 'Netherlands', 'population': 16800000})\npopulations.append({'country_name': 'India', 'population': 1252000000})\npopulations.append({'country_name': 'South Africa', 'population': 52980000})\npopulations.append({'country_name': 'Sweden', 'population': 9593000})\npopulations.append({'country_name': 'Germany', 'population': 80620000})\npopulations.append({'country_name': 'United Kingdom', 'population': 64100000})\npopulations.append({'country_name': 'Australia', 'population': 23130000})\npopulations.append({'country_name': 'United States', 'population': 316500000})\nworld_population = 7162119434\ns = 0\nfor pop in populations:\ns += pop['population']\nother = world_population - s\npopulations.append({'country_name': 'Other', 'population': other})\ndf2 = pd.DataFrame(populations)\ndf = df.merge(df2)\n\n``````\n``````\n\nIn :\n\ndf.sort('per_capita', inplace=True)\ndf.index = np.arange(df.shape)\nplt.barh(df.index, df['per_capita'])\nplt.yticks(df.index+0.4, df['country_name'])\nplt.gca().yaxis.grid(False)\nplt.ylim(-.25, df.shape)\nplt.show()\n\n``````\n``````\n\n``````\n\nHow interesting! I could write a few pages about my thoughts on why this is, but for now, I'll let the data speak for itself.\n\nAnd a special thank you to my Australian and Swedish listeners!"
]
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http://activepatience.com/sine-cosine-tangent-worksheet/sine-cosine-tangent-worksheet-similar-images-for-sine-cosine-tangent-worksheets-sin-cos-tan-worksheet-gcse/ | [
"# Sine Cosine Tangent Worksheet Similar Images For Sine Cosine Tangent Worksheets Sin Cos Tan Worksheet Gcse",
null,
"sine cosine tangent worksheet similar images for sine cosine tangent worksheets sin cos tan worksheet gcse.\n\nsin cos tan worksheet free summary of trigonometric identities kuta word problems pdf,sin cos tan worksheet pdf a graphs of trigonometric functions sine cosine tangent review answers practice,sin cos tan worksheet kuta with answers pdf unit circle chart practice,sin cos tan practice worksheet free grade 9 in math is fun mole problems new double angle sine cosine tangent answers,sine cosine tangent worksheet doc pictures sin cos tan pdf with answers,sine cosine tangent worksheet pdf right triangle trigonometry worksheets sin cos tan gcse,sine cosine tangent review worksheet answers with pdf using and independent practice math sin cos tan,sine and cosine graphs worksheet activities cos tangent sin tan with answers pdf,sine cosine tangent worksheets free sin cos tan worksheet with answers chart practice grade 9 in math is fun,sine cosine tangent worksheet doc right triangles trigonometry sin cos tan riddle word problems pdf answers.\n\nPosted on"
]
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"http://activepatience.com/wp-content/uploads/2018/09/sine-cosine-tangent-worksheet-similar-images-for-sine-cosine-tangent-worksheets-sin-cos-tan-worksheet-gcse.jpg",
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http://www.dw-math.com/ac/static/Q8101.html | [
"Custom math worksheets at your fingertips",
null,
"# Details for problem \"Triangle: Draw heights, medians and bisectors\"\n\nQuickname: 8101\n\n## Summary\n\nDraw heights, medians, angle bisectors or perpendicular bisectors for a triangle.\n\n## Examples",
null,
"",
null,
"## Description\n\nA triangle is given.\n\nDepending on the problem statement, a specific\n\n- perpendicular bisector\n\n- height\n\n- median\n\nor\n\n- angle bisector\n\nis to be drawn.\n\nThe number of problems is configurable. For every type of geometric line, it can be defined whether it can be asked for.\n\nDownload free worksheets for this math problem here. The worksheet contains the problems only, the solutions sheet includes the answers. Just click on the respective link.\n\n•",
null,
"Worksheet 1",
null,
"Solution sheet with answers\n•",
null,
"Worksheet 2",
null,
"Solution sheet with answers\n•",
null,
"Worksheet 3",
null,
"Solution sheet with answers\n\nIf you can not see the solution sheets for download, they may be filtered out by an ad blocker that you may have installed. If this is the case, please allow ads for this page and reload the page. The solution sheets will then reappear.\n\n• Do the sample worksheets do not really fit?\n• Do you need more math worksheets, with a different level of difficulty?\n• Would you like to combine different problems on a worksheet and adjust them to your needs?\n• As a teacher, you can put together your own worksheets using the automatically generated math problems provided.\nWith a free initial credit, you can start creating your own math worksheets in a few minutes.\n\nIt does not cost anything to try! Register here, to create custom worksheets now!\n\n## Customization options for this problem\n\nParameter\nPossible values\nNumber of problems\n1, 2\nYes, No\nYes, No\nYes, No\nYes, No\n\n## Similar problems\n\nRemark\nDescription\nbasic view on angle bisectors\nConstruct the bisector for a given angle using just a compass and a straightedge.\nperpendicular bisectors in a triangle, in conjunction with the circumcircle\nThe circumcircle and perpendicular bisectors have to be drawn for a given triangle.\nangle bisectors only, in conjunction with the incircle\nThe incircle and angle bisectors have to be drawn for a given triangle.\nmedians in a triangle, in conjunction with the centroid\nThe medians and centroid have to be drawn for a given triangle.\n\nDeutsche Version dieser Aufgabe\nThese informational pages with samples describe math problems that can be combined on custom math worksheets with solutions for home and school use."
]
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"http://www.dw-math.com/ac/dwmath.png",
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"http://www.dw-math.com/ac/repo/genpreview/3/8101_triangle_height_bisectors_median.png",
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"http://www.dw-math.com/ac/img/arbeitsblatt-herunterladen-klein.png",
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https://present5.com/unit-2-supply-demand-and-consumer-choice-4/ | [
"",
null,
"Скачать презентацию Unit 2 Supply Demand and Consumer Choice\n\n96e5e0f5efc6c9df041706517cd7019d.ppt\n\n• Количество слайдов: 22",
null,
"Unit 2: Supply, Demand, and Consumer Choice",
null,
"Supply and Demand Review 1. Define the Law of Demand 2. Define the Law of Supply 3. What is the difference between a change in demand a change in quantity demanded? 4. What happens if price is above equilibrium? 5. What happens if price is below equilibrium? 6. Define Consumer’s and Producer’s Surplus 7. Explain the results of an excise tax 8. Define Dead Weight Loss",
null,
"THE LAW OF DEMAND SAYS. . . Consumers will buy more when prices go down and less when prices go up HOW MUCH MORE OR LESS? DOES IT MATTER? 3",
null,
"Elasticity shows how sensitive quantity is to a change in price.",
null,
"4 Types of Elasticity 1. Elasticity of Demand 2. Elasticity of Supply 3. Cross-Price Elasticity (Subs or Comp) 4. Income Elasticity (Norm or Inferior)",
null,
"1. Elasticity of Demand • Measurement of consumers’ responsiveness to a change in price. • What will happen if price increase? How much will it effect Quantity Demanded Who cares? • Used by firms to help determine prices and sales • Used by the government to decide how to tax",
null,
"Inelastic Demand",
null,
"Inelastic Demand INelastic = Quantity is INsensitive to a change in price. • If price increases, quantity 20% demanded will fall a little • If price decreases, quantity demanded increases a little. In other words, people will continue to buy it. 5% A INELASTIC demand curve is steep! (looks like an “I”) Examples: • Gasoline • Milk • Diapers • Chewing Gum • Medical Care • Toilet paper",
null,
"Inelastic Demand General Characteristics of INelastic Goods: 20% • Few Substitutes • Necessities • Small portion of income • Required now, rather than later • Elasticity coefficient less than 1 5%",
null,
"Elastic Demand",
null,
"Elastic Demand Elastic = Quantity is sensitive to a change in price. • If price increases, quantity demanded will fall a lot • If price decreases, quantity demanded increases a lot. In other words, the amount people buy is sensitive to price. An ELASTIC demand curve is flat! Examples: • Soda • Boats • Beef • Real Estate • Pizza • Gold",
null,
"Elastic Demand General Characteristics of Elastic Goods: • Many Substitutes • Luxuries • Large portion of income • Plenty of time to decide • Elasticity coefficient greater than 1",
null,
"Elastic or Inelastic? Beef. Gasoline. Real Estate. Medical Care. Electricity. Gold- What about the Elastic- 1. 27 INelastic -. 20 demand for insulin for diabetics? Elastic- 1. 60 INelastic -. 31 What if % change in INelastic -. 13 quantity demanded equals % change in price? Elastic - 2. 6 Perfectly INELASTIC (Coefficient = 0) Unit Elastic (Coefficient =1)",
null,
"Total Revenue Test Uses elasticity to show changes in price will affect total revenue (TR). (TR = Price x Quantity) Elastic Demand • Price increase causes TR to decrease • Price decrease causes TR to increase Inelastic Demand • Price increase causes TR to increase • Price decrease causes TR to decrease Unit Elastic • Price changes and TR remains unchanged Ex: If demand for milk is INelastic, what will happen to expenditures on milk if price increases?",
null,
"Is the range between A and B, elastic, inelastic, or unit elastic? 10 x 100 =\\$1000 Total Revenue 5 x 225 =\\$1125 Total Revenue A 50% B 125% Price decreased and TR increased, so… Demand is ELASTIC",
null,
"16",
null,
"2. Price Elasticity of Supply • Elasticity of supply shows how sensitive producers are to a change in price. Elasticity of supply is based on time limitations. Producers need time to produce more. INelastic = Insensitive to a change in price (Steep curve) • Most goods have INelastic supply in the short-run Elastic = Sensitive to a change in price (Flat curve) • Most goods have elastic supply in the long-run Perfectly Inelastic = Q doesn’t change (Vertical line) • Set quantity supplied",
null,
"3. Cross-Price Elasticity of Demand • Cross-Price elasticity shows how sensitive a product is to a change in price of another good • It shows if two goods are substitutes or complements % change in quantity of product “b” % change in price of product “a” P increases 20% Q decreases 15% • If coefficient is negative (shows inverse relationship) then the goods are complements • If coefficient is positive (shows direct relationship) then the goods are substitutes",
null,
"4. Income-Elasticity of Demand • Income elasticity shows how sensitive a product is to a change in INCOME • It shows if goods are normal or inferior % change in quantity % change in income Income increases 20%, and quantity decreases 15% then the good is a… INFERIOR GOOD • If coefficient is negative (shows inverse relationship) then the good is inferior • If coefficient is positive (shows direct relationship) then the good is normal Ex: If income falls 10% and quantity falls 20%…",
null,
"Elasticity Practice 22",
null,
"23",
null,
"1996 Micro FRQ #2 The Toledo arena holds a maximum of 40, 000 people. Each year the circus performs in front of a sold out crowd. (a) Analyze the effect on each of the following of the addition of a fantastic new death-defying trapeze act that increases the demand for tickets. (i)The price of tickets (ii)The quantity of tickets sold (b) The city of Toledo institutes an effective price ceiling on tickets. Explain where the price ceiling would be set. Explain the impact of the ceiling on each of the following. (i) The quantity of tickets demanded (ii) The quantity of tickets supplied (c) Will everyone who attends the circus pay the ceiling 24 price set by the city of Toledo. Why or why not?",
null,
""
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https://math.stackexchange.com/questions/2750127/h%c3%b6lder-continuous-functions-are-uniformly-continuous | [
"# Hölder continuous functions are uniformly continuous\n\nI want to show that Hölder continuous functions are uniformly continuous using $\\epsilon-\\delta$. Is it sufficient to find a $\\delta>0$ that does not depend on $\"x\"$ and depends only on $\"\\epsilon\"$? Furthermore is my method correct? Here is the context:\n\nLet $(X,d_x)$ and $(Y,d_y)$ be metric spaces.\n\nLet $\\quad f:X\\rightarrow Y \\quad$ be a Holder continuous function.\n\nLet $\\,\\epsilon>0,\\quad\\alpha>0,\\quad C>0, \\quad \\delta =\\frac\\epsilon{2C}, \\quad \\gamma>0,\\quad x_1,x_2\\in X$\n\nLet $\\gamma=\\delta^{\\frac1 \\alpha}<1$\n\nAs $f$ is continuous we have: $\\,\\,d_x(x_1,x_2)<\\gamma\\implies d_y(f(x_1),f(x_2))<\\epsilon$\n\nSo in particular: $\\,\\,d_x(x_1,x_2)^{\\alpha}<\\delta\\implies d_y(f(x_1),f(x_2))<\\epsilon$\n\nAs $f$ is Hölder continuous, we have:\n\n$\\,\\,d_y(f(x_1),f(x_2)\\le C*d_x(x_1,x_2)^{\\alpha}<C*\\delta=C*\\frac\\epsilon{2C}<\\epsilon$\n\nAs $\\delta$ does not depend on $\"x\"$, and this is true for all $\"\\epsilon>0\"$, then $f$ is uniformly continuous. $CQFD.$\n\nAs $f$ is continuous we have: $d_x(x_1,x_2)<\\gamma\\implies d_y(f(x_1),f(x_2))<\\epsilon$\n\nThere are two problems here. First, why is this true for this specific $\\gamma$ you chose? Second, if it's true for some choice of $\\gamma$ then you already have uniform continuity, so why continue the proof?\n\nCorrect approach. The input is: $\\alpha, C$ such that $d_y(f(x_1),f(x_2)\\le C\\,d_x(x_1,x_2)^{\\alpha}$, and an arbitrary $\\epsilon>0$.\n\nFrom this we must come up with $\\delta$. I would use $\\delta = (\\epsilon/C)^{1/\\alpha}$; then the verification of $$d_x(x_1,x_2)<\\delta\\implies d_y(f(x_1),f(x_2))<\\epsilon$$ should be easy.\n\n• Thank you for your reply! So to be sure, you are saying my $\\gamma$ could still be dependent on $\"x\"$ and $\"\\epsilon\"$ which means my $\"\\delta\"$ could still be dependent on $\"x\"$ and $\"\\epsilon\"$ when we want it to depend ONLY on $\"\\epsilon\"$. @bro\n– Kam\nApr 23, 2018 at 21:37\n• The way you defined $\\gamma$, it does not depend on $x$, but you haven't explained how you know $d_x(x_1,x_2)<\\gamma\\implies d_y(f(x_1),f(x_2))<\\epsilon$. If the reason is \"because $f$ is continuous\" (which is given in that sentence) then $\\gamma$ would have to be chosen based on $x_1$ or $x_2$.\n– user357151\nApr 23, 2018 at 23:34\n• I see exactly where I went wrong!! Instead of supposing I had $d_x(x_1,x_2)<\\gamma$ and tried to prove $d_y(f(x_1),f(x_2))<\\epsilon$ to be true, I assumed the implication to be true already! (I hope I am correct in this realization haha) Thanks a lot anyway for the help!!\n– Kam\nApr 24, 2018 at 10:02"
]
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http://pirsa.org/15090081/ | [
"",
null,
"PERIMETER INSTITUTE RECORDED SEMINAR ARCHIVE",
null,
"PIRSA:15090081 ( MP4 Medium Res , MP3 , PDF ) Which Format? Observable currents for effective field theories and their contextSpeaker(s): Jose Zapata Abstract: The primary objective of an effective field theory is modelling observables at the given scale. The subject of this talk is a notion of observable at a given scale in a context that does not rely on a metric background. Within a geometrical formalism for local covariant effective field theories, a discrete version of the multisymplectic approach to lagrangian field theory, I introduce the notion of observable current. The pair of an observable current and a codimension one surface (f, Sigma) yields an observable Q_{f, Sigma} : Histories to R . The defining property of observable currents is that if phi in Solutions subset Histories and Sigma’ - Sigma = partial B (for some region B) then Q_{f, Sigma'} (phi) = Q_{f, Sigma} (phi) . Thus, an observable current f is a local object which may use an ``auxiliary devise’’ Sigma, relevant only up to homology, to induce functions on the space of solutions. There is a Poisson bracket that makes the space of observable currents a Lie algebra. We construct observable currents and prove that solutions can be separated by evaluating the induced functions. We comment on the relevance of this framework for covariant loop quantization. Date: 29/09/2015 - 2:00 pm",
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""
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"http://pirsa.org/images/pirsa_rough.jpg",
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"http://pirsa.org/thumbs/thumbs/thumb_15090081_150900810000.jpg",
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"http://www.w3.org/Icons/valid-xhtml10-blue",
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https://math.stackexchange.com/questions/1112314/graph-theory-software-with-simple-gui | [
"# Graph Theory Software with simple GUI\n\nTo the best of my knowledge I cannot find, on this site, any graph theory program resources.\n\nI am looking for a program where I can draw nodes and edges and most importantly drag and drop vertices while keeping edges in tact. Does any such program exist? Can you provide me with resources to find such a program?\n\nI want such a program, where I can drag and drop vertices, so that I may better visualize different isomorphisms of the same graph. I have a tough time seeing these sorts of things in my minds eye (although I'm trying to get better at it).\n\nAlthough this appears as though it may be a duplicate, I want to stress that I am looking for a program with a simple GUI, where I can drag and drop vertices while keeping edges in tact. The \"duplicate\" question's answers provide an extensive of list of programs that can return information on a graph and allow you to programmatically generate graphs. I am not interested in this at the moment. I want an incredibly simple program.\nPlease only post answers with these conditions in mind.\n\n## 2 Answers\n\nTry using sage math via notebook http://sagenb.org and use the function graph_editor(). I think that's precisely what you want.\n\n• That is exactly what I'm looking for. I can't quite figure out how to draw my own graphs yet, but i can manipulate the petersen graph exactly like I wanted to. Thank you!!! – Paddling Ghost Jan 20 '15 at 21:54\n\nClick here for matlab graph mfile run this program in matlab command window to draw your any graph . define matrix A : adjacency matrix, where A(I,J) is nonzero (=1)if and only if there is an edge between points I and J.\n\nfor example : write in command window in matlab $$n = 9; t = 2*pi/n*(0:n-1);$$ $$A = round(rand(n));$$ $$xy = [cos(t); sin(t)]';$$ $$gplotdc(A,xy,'LineWidth',2,'MarkerSize',8);$$\n\nin this example $A$ is $n\\times n$ matrix. use your matrix instead of $A$ in above example.\n\n• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – quid Jan 20 '15 at 18:35\n• this is matlab mfile program for draw any graph. have you matlab software? – Farhad Jan 20 '15 at 18:38\n• The above comment was an automatics comment. However, one is not supposed to give \"link only\" answers. Please explain here, as you partially did now, what resources you are linking to and how it will help to achieve the goal. In brief the answer should still be meaningful even in the event the link goes dead. You can edit your answer. – quid Jan 20 '15 at 18:43\n• It does not seem like this is what I'm looking for. Does this include the capability of dragging and dropping vertices through a GUI? As I understand it, I will not be able to do this sort of thing through matlab as I don't think the output will allow me to move vertices around with my mouse. – Paddling Ghost Jan 20 '15 at 19:21"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.941337,"math_prob":0.58491665,"size":1041,"snap":"2019-35-2019-39","text_gpt3_token_len":216,"char_repetition_ratio":0.114754096,"word_repetition_ratio":0.09677419,"special_character_ratio":0.20365034,"punctuation_ratio":0.082125604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9876352,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T23:41:13Z\",\"WARC-Record-ID\":\"<urn:uuid:1891beea-c120-4731-b8e5-fc68bc893302>\",\"Content-Length\":\"147976\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76d8f7a0-c4de-4068-af28-71a260a7f6cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:af51b5b5-0754-4369-accc-d24f36b25245>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1112314/graph-theory-software-with-simple-gui\",\"WARC-Payload-Digest\":\"sha1:HGSCOOXCIHTL5VJ3GW2T3K2UIYE4IE2Y\",\"WARC-Block-Digest\":\"sha1:EP5PBCWAASBAVJCDTMMBF7TY6KDBSD4Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314353.10_warc_CC-MAIN-20190818231019-20190819013019-00263.warc.gz\"}"} |
https://www.ibmathclub.com/onepager/detailed/12 | [
"",
null,
"## EXPONENTS\n\n### AND THE FUNCTION\n\n#Exponents and logarithms\n\nWorking with exponents is a skill that you have to master to survive IB Math, as it occurs in every second exercise. This onepager explains the main rules and tricks that you need to know, plus it presents the exponential function.\n\nExplanations\n1\n2\n3\n\n## PRACTICE [TASKS ACCORDING TO THIS SUBTOPIC]\n\n#Exponents and logarithms\n\nSimplify the following expression:\n\nOther Onepagers in #Exponents and logarithms\nAll explanations on this onepager\n##### #1/ 3\n\nLaws of exponents\n\nExponents are a shortened way of representing the repeated multiplication of a number by itself.\n\nFor example the expression ${4}^{5}$ represents $4×4×4×4×4$, where 4 is the base number and 5 is the exponent. So the exponent counts the number of times we multiply the base to itself.\n\nThe exponent is also called a ‘power’ or an ‘index’.\n\nLaws of exponents:\n\n(1) ${a}^{0}=1$ (5) $\\frac{{a}^{n}}{{a}^{m}}={a}^{n-m}$ (8) ${a}^{\\frac{1}{n}}=\\sqrt[n]{a}$\n\n(2) ${a}^{1}=a$ (6) ${a}^{n}×{b}^{n}={\\left(a×b\\right)}^{n}$ (9) ${a}^{\\frac{n}{m}}=\\sqrt[m]{{a}^{n}}$\n\n(3) ${a}^{-n}=\\frac{1}{{a}^{n}}$ (7) $\\frac{{a}^{n}}{{b}^{n}}={\\left(\\frac{a}{b}\\right)}^{n}$ (10) ${\\left({a}^{n}\\right)}^{m}={a}^{n×m}$\n\n(4) ${a}^{n}×{a}^{m}={a}^{n+m}$\n\n#### Example #1\n\nSimplify the following, leaving the answer in exponent form.\n\n$\\frac{\\sqrt{a}×{a}^{\\frac{2}{3}}}{\\sqrt{a}}⇒\\left(8\\right)⇒\\frac{{a}^{\\frac{1}{2}}×{a}^{\\frac{2}{3}}}{{a}^{\\frac{1}{3}}}⇒\\left(4\\right)⇒\\frac{{a}^{\\frac{1}{2}+\\frac{2}{3}}}{{a}^{\\frac{1}{3}}}$\n\n$⇒\\left(5\\right)⇒{a}^{\\frac{1}{2}+\\frac{2}{3}-\\frac{1}{3}}={a}^{\\frac{4}{3}}$\n\n#### Example #2\n\nSimplify the following, leaving the answer in exponent form.\n\n$\\frac{{\\left({x}^{2}{y}^{3}\\right)}^{5}×{\\left(x{y}^{2}\\right)}^{3}}{{\\left({x}^{2}\\right)}^{2}×{\\left(y{x}^{2}\\right)}^{4}×{y}^{15}}⇒\\left(6\\right)&\\left(10\\right)⇒\\frac{{x}^{2×5}×{y}^{3×5}×{x}^{3}×{y}^{2×3}}{{x}^{2×2}×{y}^{4}{x}^{2×4}×{y}^{15}}$\n\n$⇒\\left(5\\right)⇒{x}^{13-12}=x$\n\n#### Example #3\n\nSimplify the following, leaving the answer in exponent form.\n\n$\\frac{{32}^{3}×{625}^{2}×{64}^{5}}{{128}^{4}×{25}^{6}}=\\frac{{\\left({2}^{5}\\right)}^{3}×{\\left({5}^{4}\\right)}^{2}×{\\left({2}^{6}\\right)}^{5}}{{\\left({2}^{7}\\right)}^{4}×{\\left({5}^{2}\\right)}^{6}}⇒\\left(10\\right)⇒\\frac{{2}^{15}×{5}^{8}×{2}^{30}}{{2}^{28}×{5}^{12}}$\n\n$⇒\\left(4\\right)⇒\\frac{{2}^{15+30}×{5}^{8}}{{2}^{28}×{5}^{12}}=\\frac{{2}^{45}×{5}^{8}}{{2}^{28}×{5}^{12}}=\\frac{{2}^{45}}{{2}^{28}}×\\frac{{5}^{8}}{{5}^{12}}$\n\n$⇒\\left(5\\right)⇒{2}^{45-28}×{5}^{8-12}={2}^{17}×{5}^{-4}=\\frac{{2}^{17}}{{5}^{4}}$\n\n##### #2/ 3\n\nExponential function\n\n• An exponential function is a function of the form $f\\left(x\\right)={a}^{x}$, where a is $a$ positive real number, $a\\ne 1$\n\n• For large negative values of x, the y- value approaches the x-axis, but never reaches zero. So in this case the x-axis is a horizontal asymptote to the graph.\n\n• The y-intercept is always (0,1), because ${a}^{0}=1$.\n\n• If $a>1$, when x increases, so does y. This is called positive exponential, or exponential growth.\n\n• If $0, when as x increases, y decreases. This is called a negative exponential, or exponential decay.\n##### #3/ 3\n\nModelling with exponential functions\n\nMany mathematical models of physical (half life of a radioactive substance), biological (population growth) or financial (compound interest) phenomena involve the concept of continuous growth or decay. A very useful application of the exponential function is that we can represent these forms of exponential growth or exponential decay.\n\nWe have a general formula to use: $N=B{a}^{\\frac{t}{k}}+c$\n\n• when t=0 we have N=B+c, so B+c is the initial value of N, but you don’t have to memorise this, just remember that “initial” always means that you have to check the value of the function at t=0\n\n• c is the background level (For example when we are modelling the cooling process of a liquid we know that the temperature of the liquid is not going under the temperature of the room, so the background level in this case is the temperature of the room. This is the value that the model approaches, i.e. N=c is the horizontal asymptote. Note that not every exercise is going to have a background level.)\n\n• k is the time taken for the difference between N and the background level to increase by a factor of a, for example if a=2, k means the amount of time needed for B to doubles\n\n• if a>1 then it’s a positive exponential, the function models exponential growth\n\n• if 0<a<1 then it’s a negative exponential, the function models exponential decay\n\n#### Example #1\n\nA population of bacteria doubles its size in every three hours. The population growth can be modeled by a function N(t). Let’s assume that at the beginning we had a population of 100 bacteria. Build a function to model the growth of the population in the form: $N=B{a}^{\\frac{t}{k}}$.\n\nFirst let’s draw a table with values for t and our function N(t).\n\n t(in hours) N(t) 0 $100$ So when t=0, N(t)=100, because in the beginning we had 100 bacteria,that is our initial value 3 $100×2$ The population doubles its size in every three hours, so after three hours we are going to have 200 bacteria. 6 $100×{2}^{2}$ If we go another 3 hours, the population doubles again. 9 $100×{2}^{3}$ And so on... ...\n\nWe have an initial value of 100, and every three hours we multiply it with 2, so 2 is our common ratio. Let’s construct N(t):\n\n$N\\left(t\\right)=100×{2}^{\\frac{t}{3}}$\n\n$\\frac{t}{3}$ means how many three hour periods have passed by. If t=3, than 1 period, if t=6, than 2 periods, if t=4.5, than 1½ period has passed by, and so on.\n\nIt’s a simple example of modelling exponential growth."
]
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null,
"https://www.ibmathclub.com/images/uploads/8. Exponents and the Function.jpg",
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https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/xhtml/Chapter02.html | [
"With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.\n\nNo credit card required\n\n2.1. Generalities\n\nThe examples given in Chapter 1 show that a statistical problem may be characterized by the following elements:\n\na probability distribution that is not entirely known;\n\nobservations;\n\nFormalization: Wald’s decision theory provided a common framework for statistics problems. We take:\n\n1) A triplet",
null,
"where P is a family of probabilities on the measurable space",
null,
"is called a statistical model. We often set P = {Pθ, θ ∈ Θ} where we suppose that",
null,
"is injective, and Θ is called the parameter space.\n\n2) A measurable space",
null,
"is called a decision (or action) space.\n\n3) A set",
null,
"of measurable mappings",
null,
"is called a set of decision functions (d.f.) (or decision rules).\n\nDescription: From an observation that follows an unknown distribution P ∈ P, a statistician chooses an element aD using an element d from .\n\nPreference relation ...\n\nWith Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.\n\nNo credit card required"
]
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-01.gif",
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-02.gif",
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-03.gif",
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-04.gif",
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-05.gif",
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"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/images/ch2-eq009-06.gif",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9309477,"math_prob":0.9283246,"size":867,"snap":"2019-26-2019-30","text_gpt3_token_len":186,"char_repetition_ratio":0.12282734,"word_repetition_ratio":0.01369863,"special_character_ratio":0.22260669,"punctuation_ratio":0.12804878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9539548,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T23:18:14Z\",\"WARC-Record-ID\":\"<urn:uuid:7ae09ab7-b767-4a25-bdc4-d8078b7e1830>\",\"Content-Length\":\"44219\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f956ca2-9bcb-4702-a783-f55e8e2d9b42>\",\"WARC-Concurrent-To\":\"<urn:uuid:5df1a246-8fa4-4ccc-8ddf-e1c88fc67503>\",\"WARC-IP-Address\":\"23.6.19.149\",\"WARC-Target-URI\":\"https://www.oreilly.com/library/view/mathematical-statistics-and/9781118586273/xhtml/Chapter02.html\",\"WARC-Payload-Digest\":\"sha1:6M6DADJOS27FWGWGHQW3SLYAVWV5VBAG\",\"WARC-Block-Digest\":\"sha1:GRWKM5ALZVHRJDCJAMFS3T5OB3HKNOJE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000575.75_warc_CC-MAIN-20190626214837-20190627000837-00555.warc.gz\"}"} |
https://www.cs.cornell.edu/courses/cs3110/2012sp/lectures/lec19-asymp/review.html | [
"# Review of Asymptotic Complexity\n\n• Asymptotic Analysis\n• Worst-Case and Average-Case Analysis\n• Order of Growth and Big-O Notation\n• Comparing Orders of Growth\n• Binary Search Trees\n• Exercises\n\n• ## Asymptotic Analysis\n\nWhen analyzing the running time or space usage of programs, we usually try to estimate the time or space as function of the input size. For example, when analyzing the worst case running time of a function that sorts a list of numbers, we will be concerned with how long it takes as a function of the length of the input list. For example, we say the standard insertion sort takes time T(n) where T(n)= c*n2+k for some constants c and k. In contrast, merge sort takes time T ′(n) = c*n*log2(n) + k.\n\nThe asymptotic behavior of a function f(n) (such as f(n)=c*n or f(n)=c*n2, etc.) refers to the growth of f(n) as n gets large. We typically ignore small values of n, since we are usually interested in estimating how slow the program will be on large inputs. A good rule of thumb is: the slower the asymptotic growth rate, the better the algorithm (although this is often not the whole story).\n\nBy this measure, a linear algorithm (i.e., f(n)=d*n+k) is always asymptotically better than a quadratic one (e.g., f(n)=c*n2+q). That is because for any given (positive) c,k,d, and q there is always some n at which the magnitude of c*n2+q overtakes d*n+k. For moderate values of n, the quadratic algorithm could very well take less time than the linear one, for example if c is significantly smaller than d and/or k is significantly smaller than q. However, the linear algorithm will always be better for sufficiently large inputs. Remember to THINK BIG when working with asymptotic rates of growth.\n\n## Worst-Case and Average-Case Analysis\n\nWhen we say that an algorithm runs in time T(n), we mean that T(n) is an upper bound on the running time that holds for all inputs of size n. This is called worst-case analysis. The algorithm may very well take less time on some inputs of size n, but it doesn't matter. If an algorithm takes T(n)=c*n2+k steps on only a single input of each size n and only n steps on the rest, we still say that it is a quadratic algorithm.\n\nA popular alternative to worst-case analysis is average-case analysis. Here we do not bound the worst case running time, but try to calculate the expected time spent on a randomly chosen input. This kind of analysis is generally harder, since it involves probabilistic arguments and often requires assumptions about the distribution of inputs that may be difficult to justify. On the other hand, it can be more useful because sometimes the worst-case behavior of an algorithm is misleadingly bad. A good example of this is the popular quicksort algorithm, whose worst-case running time on an input sequence of length n is proportional to n2 but whose expected running time is proportional to n log n.\n\n## Order of Growth and Big-O Notation\n\nIn estimating the running time of ` insert_sort` (or any other program) we don't know what the constants c or k are. We know that it is a constant of moderate size, but other than that it is not important; we have enough evidence from the asymptotic analysis to know that a ` merge_sort` (see below) is faster than the quadratic `insert_sort`, even though the constants may differ somewhat. (This does not always hold; the constants can sometimes make a difference, but in general it is a very good rule of thumb.)\n\nWe may not even be able to measure the constant c directly. For example, we may know that a given expression of the language, such as if, takes a constant number of machine instructions, but we may not know exactly how many. Moreover, the same sequence of instructions executed on a Pentium IV will take less time than on a Pentium II (although the difference will be roughly a constant factor). So these estimates are usually only accurate up to a constant factor anyway. For these reasons, we usually ignore constant factors in comparing asymptotic running times.\n\nComputer scientists have developed a convenient notation for hiding the constant factor. We write O(n) (read: ''order n'') instead of ''cn for some constant c.'' Thus an algorithm is said to be O(n) or linear time if there is a fixed constant c such that for all sufficiently large n, the algorithm takes time at most cn on inputs of size n. An algorithm is said to be O(n2) or quadratic time if there is a fixed constant c such that for all sufficiently large n, the algorithm takes time at most cn2 on inputs of size n. O(1) means constant time.\n\nPolynomial time means nO(1), or nc for some constant c. Thus any constant, linear, quadratic, or cubic (O(n3)) time algorithm is a polynomial-time algorithm.\n\nThis is called big-O notation. It concisely captures the important differences in the asymptotic growth rates of functions.\n\nOne important advantage of big-O notation is that it makes algorithms much easier to analyze, since we can conveniently ignore low-order terms. For example, an algorithm that runs in time\n\n10n3 + 24n2 + 3n log n + 144\n\nis still a cubic algorithm, since\n\n10n3 + 24n2 + 3n log n + 144\n<= 10n3 + 24n3 + 3n3 + 144n3\n<= (10 + 24 + 3 + 144)n3\n= O(n3)\n.\n\nOf course, since we are ignoring constant factors, any two linear algorithms will be considered equally good by this measure. There may even be some situations in which the constant is so huge in a linear algorithm that even an exponential algorithm with a small constant may be preferable in practice. This is a valid criticism of asymptotic analysis and big-O notation. However, as a rule of thumb it has served us well. Just be aware that it is only a rule of thumb--the asymptotically optimal algorithm is not necessarily the best one.\n\nSome common orders of growth seen often in complexity analysis are\n\n O(1) constant O(log n) logarithmic O(n) linear O(n log n) \"n log n\" O(n2) quadratic O(n3) cubic\n\nHere log means log2 or the logarithm base 2, although the logarithm base doesn't really matter since logarithms with different bases differ by a constant factor. Note also that 2O(n) and O(2n) are not the same!\n\n## Comparing Orders of Growth\n\nO\nLet f and g be functions from positive integers to positive integers. We say f is O(g(n)) (read: ''f is order g'') if g is an upper bound on f: there exists a fixed constant c and a fixed n0 such that for all n≥n0,\n\nf(n) ≤ cg(n).\n\nEquivalently, f is O(g(n)) if the function f(n)/g(n) is bounded above by some constant.\n\no\nWe say f is o(g(n)) (read: \"f is little-o of g'') if for all arbitrarily small real c > 0, for all but perhaps finitely many n,\n\nf(n) ≤ cg(n).\n\nEquivalently, f is o(g) if the function f(n)/g(n) tends to 0 as n tends to infinity. That is, f is small compared to g. If f is o(g) then f is also O(g)\n\nΩ\nWe say that f is Ω(g(n)) (read: \"f is omega of g\") if g is a lower bound on f for large n. Formally, f is Ω(g) if there is a fixed constant c and a fixed n0 such that for all n>n0,\n\ncg(n) f(n)\n\nFor example, any polynomial whose highest exponent is nk is Ω(nk). If f(n) is Ω(g(n)) then g(n) is O(f(n)). If f(n) is o(g(n)) then f(n) is not Ω(g(n)).\n\nΘ\nWe say that f is Θ(g(n)) (read: \"f is theta of g\") if g is an accurate characterization of f for large n: it can be scaled so it is both an upper and a lower bound of f. That is, f is both O(g(n)) and Ω(g(n)). Expanding out the definitions of Ω and O, f is Θ(g(n)) if there are fixed constants c1 and c2 and a fixed n0 such that for all n>n0,\n\nc1g(n) f(n) c2 g(n)\n\nFor example, any polynomial whose highest exponent is nk is Θ(nk). If f is Θ(g), then it is O(g) but not o(g). Sometimes people use O(g(n)) a bit informally to mean the stronger property Θ(g(n)); however, the two are different.\n\nHere are some examples:\n\n• n + log n is O(n) and Q(n), because for all n > 1, n < n + log n < 2n.\n• n1000 is o(2n), because n1000/2n tends to 0 as n tends to infinity.\n• For any fixed but arbitrarily small real number c, n log n is o(n1+c), since n log n / n1+c tends to 0. To see this, take the logarithm\n\nlog(n log n / n1+c)\n= log(n log n) - log(n1+c)\n= log n + log log n - (1+c)log n\n= log log n - c log n\n\nand observe that it tends to negative infinity.\n\nThe meaning of an expression like O(n2) is really a set of functions: all the functions that are O(n2). When we say that f(n) is O(n2), we mean that f(n) is a member of this set. It is also common to write this as f(n) = O(g(n)) although it is not really an equality.\n\nWe now introduce some convenient rules for manipulating expressions involving order notation. These rules, which we state without proof, are useful for working with orders of growth. They are really statements about sets of functions. For example, we can read #2 as saying that the product of any two functions in O(f(n)) and O(g(n)) is in O(f(n)g(n)).\n\n1. cnm = O(nk) for any constant c and any m ≤ k.\n2. O(f(n)) + O(g(n)) = O(f(n) + g(n)).\n3. O(f(n))O(g(n)) = O(f(n)g(n)).\n4. O(cf(n)) = O(f(n)) for any constant c.\n5. c is O(1) for any constant c.\n6. logbn = O(log n) for any base b.\n\nAll of these rules (except #1) also hold for Q as well.\n\n## Binary search trees\n\nA binary search tree is one in which every node n satisfies the binary search tree invariant: its left child and all the nodes below it have values (or keys) less than that of n. Similarly, the right child node and all nodes below it have values greater than that of n.\n\nThe code for a binary search tree looks like the following. First, to check for an element or to add a new element, we simply walk down the tree.\n\n```(* contains(t,x) is whether x is in the tree *)\nlet contains(t: tree, x: value): bool =\nmatch t with\nEmpty -> false\n| Node{value=value; left=left; right=right} ->\n(match compare(x, value) with\nEQUAL -> true\n| LESS -> contains(left, x)\n| GREATER -> contains(right, x))\n\n(* add(t,x) is a BST with the same values as t, plus x *)\nlet add(t: tree, x: value): tree =\nlet balance(t: tree): tree = t (* what to write here? *)\nin\nmatch t with\nEmpty -> Node{value=x; left=Empty; right=Empty}\n| Node {value=value; left=left; right=right} ->\n(match compare(x, value) with\nEQUAL -> Node{value=x; left=left; right=right}\n| LESS -> Node{value=value; left=add(left, x); right=right}\n| GREATER -> Node{value=value; left=left; right=add(right,x) } )\n```\n\nWhen a tree satisfies the BST invariant, an in-order traversal of the tree nodes will visit the nodes in ascending order of their contained values. So it's easy to fold over all tree nodes in order.\n\nRemoving elements is a little trickier. If a node is a leaf, it can be removed. If it has one child, it can be replaced with its child. If it has two children, it can be replaced with either its immediate successor (or predecessor), which requires searching in the tree.\n\n```(* Returns: a tree just like t except that the node containing x is removed.\n* Checks: x is in the tree. *)\nlet remove(t: tree, x: value): tree =\n(* Returns: a tree in which the successor of the root is removed,\n* along with the value of that successor.\n* Checks: the root has a successor. *)\nlet removeSuccessor(t: tree): tree*value =\nmatch t with\nEmpty -> raise Fail \"impossible\"\n| Node {value=value; left=Empty; right=right} -> (right, value)\n| Node {value=value; left=left; right=right} ->\nlet (l, v) = removeSuccessor(left)\nin (Node{value=value; left=l; right=right}, v)\nin\nmatch t with\nEmpty -> raise Fail \"value not in the tree\"\n| Node {value=value; left=left; right=right) ->\nmatch Int.compare(x, value) with\nLESS -> Node{value=value; left=remove(left, x); right=right)\n| GREATER -> Node{value=value; left=left; right=remove(right, x)}\n| EQUAL -> match (left, right) with\n(_, Empty) -> left\n| (Empty, _) -> right\n| _ -> let (r, v) = removeSuccessor(right) in\nNode {value=v; left=left; right=r}\n```\n\nThe time required to find a node in a BST, or to remove a node from a BST, is O(h), where h is the height of the tree: the length of the longest path from the root node to any leaf. If a tree is perfectly balanced, so that all leaf nodes are at the same depth, then h is O(log n). This makes binary search trees an attractive data structure, especially for implementing ordered sets and maps.\n\nThe problem with BST's is that they are not necessarily balanced. In fact, if nodes are added to a BST in increasing order, the resulting BST will be essentially a linked list. The solution to this problem is to make sure that the BST is balanced. Making the BST perfectly balanced at every step is too expensive, but if we are interested in asymptotic complexity, we merely need the height h to be proportional to O(log n). We will say that the BST is balanced in this case.\n\nThere are many ways to keep binary search trees balanced. Some of the more popular methods are red-black trees, AVL trees, B-trees, and splay trees. But there are many more, including 2-3 trees, 2-3-4 trees, AA trees, and treaps. Each kind of binary search tree works by strengthening the representation invariant so that the tree must be approximately balanced.\n\n## Exercises\n\n1. Write an implementation of stacks using lists. What is the big-O running time of each operation? The signature for stacks is:\n\n```module type STACK =\nsig\ntype 'a stack\nval empty : unit -> 'a stack\nval push : ('a * 'a stack) -> 'a stack\nval top : 'a stack -> 'a option\nval pop : 'a stack -> ('a stack) option\nend\n```\n\n2. Write an implementation of queues using lists. What is the big-O running time of each operation? The signature for queues is:\n\n```module type QUEUE =\nsig\ntype 'a queue\nval empty : unit -> 'a queue\nval insert : ('a * 'a queue) -> 'a queue\nval first : 'a queue -> 'a option\nval rest : 'a queue -> 'a option\nend\n```\n\n3. Write some of the functions that occur in the List structure by hand (e.g., rev, @, map, foldl, etc.) and analyze them to determine their big-O running time."
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https://integers.info/binary-numbers/bin/11100010 | [
"# integers.info\n\n## Binary Numbers\n\n### Conversion between 226 and 11100010\n\nThe below table visualizes how the decimal number 226 equals the binary number 11100010.\n\n 1 × 27 = 128 + 1 × 26 = 64 + 1 × 25 = 32 + 0 × 24 = 0 + 0 × 23 = 0 + 0 × 22 = 0 + 1 × 21 = 2 + 0 × 20 = 0 = 226"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8522506,"math_prob":1.0000095,"size":879,"snap":"2023-14-2023-23","text_gpt3_token_len":282,"char_repetition_ratio":0.14057143,"word_repetition_ratio":0.01183432,"special_character_ratio":0.37656426,"punctuation_ratio":0.0797546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997866,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T02:49:19Z\",\"WARC-Record-ID\":\"<urn:uuid:bf619493-1d05-4b5f-8f37-15cfdf5fc42a>\",\"Content-Length\":\"17478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:266f8b6c-dea3-48d2-934e-b4413ff24f54>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8bad218-6d12-40b3-871d-99c55af008b9>\",\"WARC-IP-Address\":\"93.188.2.52\",\"WARC-Target-URI\":\"https://integers.info/binary-numbers/bin/11100010\",\"WARC-Payload-Digest\":\"sha1:4LLIX54VCEDJGCZM2JOI5Q5NC6CR46FX\",\"WARC-Block-Digest\":\"sha1:C5URRGARLTWCMWEWS2D7SKLXQRYWP6MH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943749.68_warc_CC-MAIN-20230322020215-20230322050215-00063.warc.gz\"}"} |
https://visualfractions.com/calculator/factors/factors-of-831/ | [
"# Factors of 831\n\nSo you need to find the factors of 831 do you? In this quick guide we'll describe what the factors of 831 are, how you find them and list out the factor pairs of 831 for you to prove the calculation works. Let's dive in!\n\n## Factors of 831 Definition\n\nWhen we talk about the factors of 831, what we really mean is all of the positive and negative integers (whole numbers) that can be evenly divided into 831. If you were to take 831 and divide it by one of its factors, the answer would be another factor of 831.\n\nLet's look at how to find all of the factors of 831 and list them out.\n\n## How to Find the Factors of 831\n\nWe just said that a factor is a number that can be divided equally into 831. So the way you find and list all of the factors of 831 is to go through every number up to and including 831 and check which numbers result in an even quotient (which means no decimal place).\n\nDoing this by hand for large numbers can be time consuming, but it's relatively easy for a computer program to do it. Our calculator has worked this out for you. Here are all of the factors of 831:\n\n• 831 ÷ 1 = 831\n• 831 ÷ 3 = 277\n• 831 ÷ 277 = 3\n• 831 ÷ 831 = 1\n\nAll of these factors can be used to divide 831 by and get a whole number. The full list of positive factors for 831 are:\n\n1, 3, 277, and 831\n\n## Negative Factors of 831\n\nTechnically, in math you can also have negative factors of 831. If you are looking to calculate the factors of a number for homework or a test, most often the teacher or exam will be looking for specifically positive numbers.\n\nHowever, we can just flip the positive numbers into negatives and those negative numbers would also be factors of 831:\n\n-1, -3, -277, and -831\n\n## How Many Factors of 831 Are There?\n\nAs we can see from the calculations above there are a total of 4 positive factors for 831 and 4 negative factors for 831 for a total of 8 factors for the number 831.\n\nThere are 4 positive factors of 831 and 4 negative factors of 831. Wht are there negative numbers that can be a factor of 831?\n\n## Factor Pairs of 831\n\nA factor pair is a combination of two factors which can be multiplied together to equal 831. For 831, all of the possible factor pairs are listed below:\n\n• 1 x 831 = 831\n• 3 x 277 = 831\n\nJust like before, we can also list out all of the negative factor pairs for 831:\n\n• -1 x -831 = 831\n• -3 x -277 = 831\n\nNotice in the negative factor pairs that because we are multiplying a minus with a minus, the result is a positive number.\n\nSo there you have it. A complete guide to the factors of 831. You should now have the knowledge and skills to go out and calculate your own factors and factor pairs for any number you like.\n\nFeel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers!\n\n## Factors Calculator\n\nWant to find the factor for another number? Enter your number below and click calculate.\n\nFactors of 832"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9362221,"math_prob":0.9824349,"size":2938,"snap":"2020-45-2020-50","text_gpt3_token_len":732,"char_repetition_ratio":0.21165644,"word_repetition_ratio":0.011945393,"special_character_ratio":0.2886317,"punctuation_ratio":0.07915994,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99831736,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T02:35:33Z\",\"WARC-Record-ID\":\"<urn:uuid:364d1b85-d680-4e74-9793-5a22a5c687b2>\",\"Content-Length\":\"11439\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14f9ae93-9b75-4076-a896-304fbae02889>\",\"WARC-Concurrent-To\":\"<urn:uuid:22c6b585-44d0-4de1-8a7b-84ed4121d47c>\",\"WARC-IP-Address\":\"142.93.52.42\",\"WARC-Target-URI\":\"https://visualfractions.com/calculator/factors/factors-of-831/\",\"WARC-Payload-Digest\":\"sha1:CBU44C7WFM5M3ZZ4OOPMILIIYGH5267J\",\"WARC-Block-Digest\":\"sha1:3PTEDLSPHZKVIVBAOBMNA2FFYRL23VYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141204453.65_warc_CC-MAIN-20201130004748-20201130034748-00617.warc.gz\"}"} |
https://math.stackexchange.com/questions/1611816/a-bounded-linear-operator-between-banach-spaces-that-cannot-be-compact | [
"A bounded linear operator between Banach spaces that cannot be compact\n\nLet $K: X \\to Y$ be a bounded linear operator, where $X$ and $Y$ are two Banach spaces. Further assume that the image $imK$ is a $\\infty$-dim closed subspace of Y.\n\nIn my script they claim that in such a setting $K$ can never be compact because by the closed image theorem we have that $K(B)$ ($B$ closed unit ball in $X$) contains an open ball and hence has no compact closure because the dimension is $\\infty$.\n\nMy questions:\n\nI understand that the dimension of the closed unit ball $B$ characterizes the dimension of the underlying Banach space but I don't understand how this argument is used here. Also I don't quite get the thing with the open ball due to the closed image theorem and the thing that follows with the compact closure. How does one mash these things together the right way?\n\nEDIT:\n\nOk I might have found another way to proof the above:\n\nDefine $K_1: X \\to im(K)$ by $x \\mapsto K(x)$. Then $K_1$ is surjective, hence $K_1(B_{open})$ ($B_{open}$ denotes the open unit ball in x) contains a small $\\delta$-ball $B_{\\delta}$ centered at the origin of $Y$ by the open mapping theorem. If we assume $cl(K_1(B_{open}))$ to be compact in $im(K)$ we get that $cl(B_{\\delta})=\\{y \\in Y | \\Vert y \\Vert_Y \\leq \\delta\\} \\subset cl(K_1(B_{open}))$ and since Y is Banach it's also Hausdorff therefore it follows that $cl(B_{\\delta})$ is compact and hence by scaling the unit ball in $Y$ is also compact which contradicts the $\\infty$-dimensional property of $im(K)$.\n\nIs this right?\n\n1. We will replace $Y$ be the image of $X$, so that we can assume the map is surjective (just for notational reasons).\n2. Hence, by the open mapping theorem, $K : X \\to Y$ is open.\n3. Suppose that the image of the open unit ball $B_X$ in $X$ was relatively compact. Then the closure of $K(B_X)$ is compact, but also contains the open set $K(B_X)$. Since $K(B_X)$ is an open set containing the origin, there is some $\\lambda \\in \\mathbb{R}$ so that $B_Y \\subseteq \\lambda K(B_X)$. This implies that $B_Y$ is relatively compact, since $\\lambda K(B_X)$ remains relatively compact.\n• @AlessioPellegrini Yes, that works. I like to think of it as: an open cover of a closed $C \\subset X$ can be extended to an open cover of $X$ by adding $X \\setminus C$ to the set of opens... – Lorenzo Najt Jan 14 '16 at 10:32"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9251676,"math_prob":0.99959606,"size":1482,"snap":"2019-43-2019-47","text_gpt3_token_len":411,"char_repetition_ratio":0.11772666,"word_repetition_ratio":0.0,"special_character_ratio":0.27732792,"punctuation_ratio":0.05782313,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000021,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T16:37:08Z\",\"WARC-Record-ID\":\"<urn:uuid:dd04ed19-ba27-4e85-af84-03c0c95acd90>\",\"Content-Length\":\"142422\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:84ddc7b4-200b-463c-bff7-fa6e1db65084>\",\"WARC-Concurrent-To\":\"<urn:uuid:cab142a0-1732-457c-98e0-17b4681f4fa9>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1611816/a-bounded-linear-operator-between-banach-spaces-that-cannot-be-compact\",\"WARC-Payload-Digest\":\"sha1:CBF74ROIMEUBH7R3UTQP4B252Q73T3GP\",\"WARC-Block-Digest\":\"sha1:KYN6KCBAPHDEC4YK22E6OHDTDGG2MVWV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986669057.0_warc_CC-MAIN-20191016163146-20191016190646-00441.warc.gz\"}"} |
https://www.numbersaplenty.com/1392 | [
"Search a number\nBaseRepresentation\nbin10101110000\n31220120\n4111300\n521032\n610240\n74026\noct2560\n91816\n101392\n111056\n12980\n13831\n14716\n1562c\nhex570\n\n1392 has 20 divisors (see below), whose sum is σ = 3720. Its totient is φ = 448.\n\nThe previous prime is 1381. The next prime is 1399. The reversal of 1392 is 2931.\n\nIt is a nialpdrome in base 12 and base 13.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (1399) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (5) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 34 + ... + 62.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (186).\n\n1392 is a gapful number since it is divisible by the number (12) formed by its first and last digit.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 1392, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1860).\n\n1392 is an abundant number, since it is smaller than the sum of its proper divisors (2328).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n1392 is a wasteful number, since it uses less digits than its factorization.\n\n1392 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 40 (or 34 counting only the distinct ones).\n\nThe product of its digits is 54, while the sum is 15.\n\nThe square root of 1392 is about 37.3095162124. The cubic root of 1392 is about 11.1655403433.\n\nIt can be divided in two parts, 139 and 2, that added together give a palindrome (141).\n\nThe spelling of 1392 in words is \"one thousand, three hundred ninety-two\"."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93419796,"math_prob":0.99474627,"size":1777,"snap":"2021-43-2021-49","text_gpt3_token_len":481,"char_repetition_ratio":0.17258883,"word_repetition_ratio":0.006006006,"special_character_ratio":0.31007317,"punctuation_ratio":0.1328125,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99824667,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T18:04:34Z\",\"WARC-Record-ID\":\"<urn:uuid:e46619e1-dccd-4222-9ccd-08b3fa28b334>\",\"Content-Length\":\"9396\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47865543-2330-490d-aa87-16d26645db22>\",\"WARC-Concurrent-To\":\"<urn:uuid:3053362a-efb0-49ef-83f0-c4376e66770b>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/1392\",\"WARC-Payload-Digest\":\"sha1:E5NLTVDTNZJSEUDC5GSVCCN6INOOPDBB\",\"WARC-Block-Digest\":\"sha1:DQGZJRF5UZMKYZFOCA72XDXUWE3TSCJM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363215.8_warc_CC-MAIN-20211205160950-20211205190950-00303.warc.gz\"}"} |
https://indiana.pure.elsevier.com/en/publications/introducing-cozigam-an-r-package-for-unconstrained-and-constraine | [
"# Introducing COZIGAM: An R package for unconstrained and constrained zero-inflated generalized additive model analysis\n\nHai Liu, Kung Sik Chan\n\nResearch output: Contribution to journalArticle\n\n24 Citations (Scopus)\n\n### Abstract\n\nZero-inflation problem is very common in ecological studies as well as other areas. Nonparametric regression with zero-inflated data may be studied via the zero-inflated generalized additive model (ZIGAM), which assumes that the zero-inflated responses come from a probabilistic mixture of zero and a regular component whose distribution belongs to the 1-parameter exponential family. With the further assumption that the probability of non-zero-inflation is some monotonic function of the mean of the regular component, we propose the constrained zero-inflated generalized additive model (COZIGAM) for analyzing zero-inflated data. When the hypothesized constraint obtains, the new approach provides a unified framework for modeling zero-inflated data, which is more parsimonious and efficient than the unconstrained ZIGAM. We have developed an R package COZIGAM which contains functions that implement an iterative algorithm for fitting ZIGAMs and COZIGAMs to zero-inflated data based on the penalized likelihood approach. Other functions included in the package are useful for model prediction and model selection. We demonstrate the use of the COZIGAM package via some simulation studies and a real application.\n\nOriginal language English 1-26 26 Journal of Statistical Software 35 11 Published - Jul 2010\n\n### Fingerprint\n\nModel Analysis\nZero\nZero-inflation\nMonotonic Function\nPenalized Likelihood\nExponential Family\nNonparametric Regression\nModel Selection\nInflation\nPrediction Model\nIterative Algorithm\n\n### Keywords\n\n• EM algorithm\n• Model selection\n• Penalized likelihood\n• Proportionality constraints\n\n### ASJC Scopus subject areas\n\n• Software\n• Statistics and Probability\n• Statistics, Probability and Uncertainty\n\n### Cite this\n\nIn: Journal of Statistical Software, Vol. 35, No. 11, 07.2010, p. 1-26.\n\nResearch output: Contribution to journalArticle\n\n@article{279408a230404ea3ac4145cc7d94f3cc,\ntitle = \"Introducing COZIGAM: An R package for unconstrained and constrained zero-inflated generalized additive model analysis\",\nabstract = \"Zero-inflation problem is very common in ecological studies as well as other areas. Nonparametric regression with zero-inflated data may be studied via the zero-inflated generalized additive model (ZIGAM), which assumes that the zero-inflated responses come from a probabilistic mixture of zero and a regular component whose distribution belongs to the 1-parameter exponential family. With the further assumption that the probability of non-zero-inflation is some monotonic function of the mean of the regular component, we propose the constrained zero-inflated generalized additive model (COZIGAM) for analyzing zero-inflated data. When the hypothesized constraint obtains, the new approach provides a unified framework for modeling zero-inflated data, which is more parsimonious and efficient than the unconstrained ZIGAM. We have developed an R package COZIGAM which contains functions that implement an iterative algorithm for fitting ZIGAMs and COZIGAMs to zero-inflated data based on the penalized likelihood approach. Other functions included in the package are useful for model prediction and model selection. We demonstrate the use of the COZIGAM package via some simulation studies and a real application.\",\nkeywords = \"EM algorithm, Model selection, Penalized likelihood, Proportionality constraints\",\nauthor = \"Hai Liu and Chan, {Kung Sik}\",\nyear = \"2010\",\nmonth = \"7\",\nlanguage = \"English\",\nvolume = \"35\",\npages = \"1--26\",\njournal = \"Journal of Statistical Software\",\nissn = \"1548-7660\",\npublisher = \"University of California at Los Angeles\",\nnumber = \"11\",\n\n}\n\nTY - JOUR\n\nT1 - Introducing COZIGAM\n\nT2 - An R package for unconstrained and constrained zero-inflated generalized additive model analysis\n\nAU - Liu, Hai\n\nAU - Chan, Kung Sik\n\nPY - 2010/7\n\nY1 - 2010/7\n\nN2 - Zero-inflation problem is very common in ecological studies as well as other areas. Nonparametric regression with zero-inflated data may be studied via the zero-inflated generalized additive model (ZIGAM), which assumes that the zero-inflated responses come from a probabilistic mixture of zero and a regular component whose distribution belongs to the 1-parameter exponential family. With the further assumption that the probability of non-zero-inflation is some monotonic function of the mean of the regular component, we propose the constrained zero-inflated generalized additive model (COZIGAM) for analyzing zero-inflated data. When the hypothesized constraint obtains, the new approach provides a unified framework for modeling zero-inflated data, which is more parsimonious and efficient than the unconstrained ZIGAM. We have developed an R package COZIGAM which contains functions that implement an iterative algorithm for fitting ZIGAMs and COZIGAMs to zero-inflated data based on the penalized likelihood approach. Other functions included in the package are useful for model prediction and model selection. We demonstrate the use of the COZIGAM package via some simulation studies and a real application.\n\nAB - Zero-inflation problem is very common in ecological studies as well as other areas. Nonparametric regression with zero-inflated data may be studied via the zero-inflated generalized additive model (ZIGAM), which assumes that the zero-inflated responses come from a probabilistic mixture of zero and a regular component whose distribution belongs to the 1-parameter exponential family. With the further assumption that the probability of non-zero-inflation is some monotonic function of the mean of the regular component, we propose the constrained zero-inflated generalized additive model (COZIGAM) for analyzing zero-inflated data. When the hypothesized constraint obtains, the new approach provides a unified framework for modeling zero-inflated data, which is more parsimonious and efficient than the unconstrained ZIGAM. We have developed an R package COZIGAM which contains functions that implement an iterative algorithm for fitting ZIGAMs and COZIGAMs to zero-inflated data based on the penalized likelihood approach. Other functions included in the package are useful for model prediction and model selection. We demonstrate the use of the COZIGAM package via some simulation studies and a real application.\n\nKW - EM algorithm\n\nKW - Model selection\n\nKW - Penalized likelihood\n\nKW - Proportionality constraints\n\nUR - http://www.scopus.com/inward/record.url?scp=77955153103&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=77955153103&partnerID=8YFLogxK\n\nM3 - Article\n\nAN - SCOPUS:77955153103\n\nVL - 35\n\nSP - 1\n\nEP - 26\n\nJO - Journal of Statistical Software\n\nJF - Journal of Statistical Software\n\nSN - 1548-7660\n\nIS - 11\n\nER -"
]
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https://github.com/qgis/QGIS/commit/6c179059a5222da52a733234d8743b91c06c3656 | [
"{{ message }}\n\n# qgis / QGIS\n\nAdd straightDistance2d and sinuosity measures to QgsCurve\n\nnyalldawson committed Mar 12, 2018\n1 parent 05fb8f7 commit 6c179059a5222da52a733234d8743b91c06c3656\nShowing with 54 additions and 0 deletions.\n1. +20 −0 python/core/geometry/qgscurve.sip.in\n2. +14 −0 src/core/geometry/qgscurve.cpp\n3. +20 −0 src/core/geometry/qgscurve.h\n @@ -186,6 +186,26 @@ Returns the y-coordinate of the specified node in the line string. Returns a QPolygonF representing the points. %End double straightDistance2d() const; %Docstring Returns the straight distance of the curve, i.e. the direct/euclidean distance between the first and last vertex of the curve. (Also known as \"as the crow flies\" distance). .. versionadded:: 3.2 %End double sinuosity() const; %Docstring Returns the curve sinuosity, which is the ratio of the curve length() to curve straightDistance2d(). Larger numbers indicate a more \"sinuous\" curve (i.e. more \"bendy\"). The minimum value returned of 1.0 indicates a perfectly straight curve. If a curve isClosed(), it has infinite sinuosity and will return NaN. .. versionadded:: 3.2 %End protected:\n @@ -200,6 +200,20 @@ QPolygonF QgsCurve::asQPolygonF() const return points; } double QgsCurve::straightDistance2d() const { return startPoint().distance( endPoint() ); } double QgsCurve::sinuosity() const { double d = straightDistance2d(); if ( qgsDoubleNear( d, 0.0 ) ) return std::numeric_limits::quiet_NaN(); return length() / d; } void QgsCurve::clearCache() const { mBoundingBox = QgsRectangle();\n @@ -168,6 +168,26 @@ class CORE_EXPORT QgsCurve: public QgsAbstractGeometry */ QPolygonF asQPolygonF() const; /** * Returns the straight distance of the curve, i.e. the direct/euclidean distance * between the first and last vertex of the curve. (Also known as * \"as the crow flies\" distance). * * \\since QGIS 3.2 */ double straightDistance2d() const; /** * Returns the curve sinuosity, which is the ratio of the curve length() to curve * straightDistance2d(). Larger numbers indicate a more \"sinuous\" curve (i.e. more * \"bendy\"). The minimum value returned of 1.0 indicates a perfectly straight curve. * * If a curve isClosed(), it has infinite sinuosity and will return NaN. * * \\since QGIS 3.2 */ double sinuosity() const; #ifndef SIP_RUN /**"
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https://economics.stackexchange.com/questions/26293/central-bank-loss-function | [
"# Central bank loss function",
null,
"$$L_t = \\gamma(\\pi_t - \\pi_t^\\otimes)^2 + \\hat{Y}_t^2$$\n\nCentral banks loss function is given by the equation above. This loss is increasing and convex in the distance from the inflation target, i.e. the marginal loss increases at a faster rate than the increase in the distance from the target.\n\nIn Figure 1, we can visualise the preferences using indifference curves, where the ellipses have the center being the desired point (0, $$\\pi_t^\\otimes$$ = Target inflation point.) (saturation point - bliss point)\n\nIt is hard for me to see how the different ellipses are indifference curves, or convex from in the difference from target?. Any explanation would help a lot.\n\nThe intuitive way\n\nDraw the function $$L_t$$, the graph below is build with $$\\gamma = 0.8$$ and $$\\pi_t^\\otimes = 1$$",
null,
"Convex in this context means that the function $$L_t$$ looks like a bowl, just one well defined minimum. And if you cut it with a horizontal plane, the result is an ellipse.",
null,
"The formal way\n\nJust for convenience, define the variable $$p_t = \\pi_t - \\pi_t^\\otimes$$ so the function becomes\n\n$$L_t(p_t, Y_t) = \\gamma p_t^2 + Y_t^2 \\tag{B1}$$\n\nFirst note that if $$\\gamma > 0$$ then $$L_t(p_t, Y_t) \\ge 0$$. To prove convexity you just need to show that for any pair $$(p_{t,1},Y_{t,1})$$ and $$(p_{t,2},Y_{t,2})$$ the condition\n\n$$L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) < tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\tag{B2}$$\n\nholds for $$0 < t < 1$$. This is actually fairly easy to show\n\n$$\\begin{eqnarray} L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &=& \\gamma[t p_{t,1} + (1 - t)p_{t,2}]^2 + (t Y_1 + (1 - t)Y_{t,2})^2 \\\\ &\\vdots & \\\\ &=& t [\\gamma p_{t,1}^2 + Y_{t,1}^2] + (1 - t) [\\gamma p_{t,2}^2 + Y_{t,2}^2] \\\\ &&\\quad + t(1 - t) [\\gamma(p_{t,1} - p_{t,2})^2 + (Y_{t,1} - Y_{t,2})^2] \\\\ &=& tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\\\ &&\\quad \\underbrace{t(1 - t)L_t(p_{t,1} - p _{t_2}, Y_{t,1} - Y_{t,2})}_{> 0} \\\\ &< & tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\tag{B3} \\end{eqnarray}$$\n\nSo $$L_t$$ is convex. To show that indifference curves are ellipses, just fix $$L_t$$ to a constant value, let's say $$A > 0$$ then we have\n\n$$A = \\gamma p_t^2+ Y_t^2 ~~\\Rightarrow~~ \\frac{p_t^2}{A/\\gamma} + \\frac{Y_t}{A} = 1 \\tag{B4}$$\n\nwhich are just ellipses centered at $$p_t = 0 = \\pi_t - \\pi_t^\\otimes$$ and $$Y_t = 0$$ with axis lengths $$\\sqrt{A/\\gamma}$$ and $$\\sqrt{A}$$"
]
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null,
"https://i.stack.imgur.com/ydABa.png",
null,
"https://i.stack.imgur.com/MC0qI.png",
null,
"https://i.stack.imgur.com/ISzGm.gif",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.65842205,"math_prob":1.0000067,"size":1704,"snap":"2021-31-2021-39","text_gpt3_token_len":765,"char_repetition_ratio":0.17588235,"word_repetition_ratio":0.07482993,"special_character_ratio":0.5,"punctuation_ratio":0.13875598,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000082,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T01:44:37Z\",\"WARC-Record-ID\":\"<urn:uuid:00048c31-f046-43a9-8172-e369c27eaa04>\",\"Content-Length\":\"165417\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fea80287-3e28-4089-b6fe-bd614a433bbd>\",\"WARC-Concurrent-To\":\"<urn:uuid:7be93326-94a8-441c-b42f-fbf746e893aa>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://economics.stackexchange.com/questions/26293/central-bank-loss-function\",\"WARC-Payload-Digest\":\"sha1:572V4KQNOJQ57XXLXNCBGAPC4PMD5ETM\",\"WARC-Block-Digest\":\"sha1:IPHYB4TLBXC4JJ5NJQ2RLQCJVOA7TCKW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058222.43_warc_CC-MAIN-20210926235727-20210927025727-00115.warc.gz\"}"} |
https://mathoverflow.net/questions/271643/supersingular-isogenies-of-elliptic-curves-preserving-divisibility-of-points | [
"Supersingular isogenies of elliptic curves preserving divisibility of points\n\nI hope my question is clear.\n\nIn summary, If $\\phi:E\\to E'$ is an isogeny and $P\\in E$ is not divisible by $2$, under which conditions $\\phi(P)\\in E'$ is also not divisble by $2$. Here is the detail with some computations over finite fields with supersingular isogenous elliptic curves with $j$ invariant 1728.\n\nLet $E_1$ be the elliptic curve given by $y^2 = x^3 - (1+t^2)x =f_1(x)$ and consider the curve $E_2$ given by $y^2 = x^3 + 4(1+t^2)x=f_2(x)$. Both $E_1$ and $E_2$ are isogenous via\n\n$\\phi:E_1\\to E_2$\n\n$(x,y)\\mapsto (\\tfrac{y^2}{x^2},\\tfrac{-(1+t^2)-x^2}{x^2})$\n\naccording to Silverman's X.6 when $-(1+t^2)$ is fourth-power free.\n\nLet $p\\equiv 3\\bmod 4$ and fix $t\\in\\mathbb{F}_p^\\times$. Consider the curves $E_1(\\mathbb{F}_p)$ and $E_2(\\mathbb{F}_p)$, both have $p+1$ points and when $f_1(x)$ is irreducible is easy to see that it must be cyclic.\n\nThe point $P_1:=(-1,t)\\in E_1(\\mathbb{F}_p)$ is never divisible by $2$.\n\nUnder which conditions the point $\\phi(P_1)=(t^2,t^3+2t)\\in E_2$ is also not divisible by $2$ ?\n\nI have special primes $p$ for which I need to generate the 2-Sylow subgroup of $E_1(\\mathbb{F}_p)$, thats why I am concerned with a point not divisible by $2$ and I use the cyclicity of $E_1$ to know the order of this 2-Sylow Subgroup. If $E_1(\\mathbb{F}_p)$ it is not cyclic, I want to look at $E_2$ which also is supersingular and must be cyclic but unfortunately $\\phi(-1,t)\\in E_2$ sometimes IS divisble by two according to some experiments, I would like to get a point on $E_2$ that is not divisible by two that generates the 2-Sylow-subgroup on $E_2$ using the information I know of $E_1$.\n\nThanks"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89972883,"math_prob":0.99996006,"size":1624,"snap":"2019-43-2019-47","text_gpt3_token_len":572,"char_repetition_ratio":0.12469136,"word_repetition_ratio":0.007905139,"special_character_ratio":0.34913793,"punctuation_ratio":0.0726257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999927,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T17:47:21Z\",\"WARC-Record-ID\":\"<urn:uuid:06413752-72f1-4301-be8d-bc6269719d29>\",\"Content-Length\":\"108919\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:289fe8f0-94d4-48ec-a8ad-d8ff996b979e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e1c1e7e-3257-4992-ad6b-3f7dfc539196>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/271643/supersingular-isogenies-of-elliptic-curves-preserving-divisibility-of-points\",\"WARC-Payload-Digest\":\"sha1:3XTKWKXBOJBXENBO5BPHM4VDGIXXKAA2\",\"WARC-Block-Digest\":\"sha1:WVZSQXUVWGMCFUM5GWRM4NAONSUUPIG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987822458.91_warc_CC-MAIN-20191022155241-20191022182741-00477.warc.gz\"}"} |
https://number.academy/1001087 | [
"# Number 1001087 facts\n\nThe odd number 1,001,087 is spelled 🔊, and written in words: one million, one thousand and eighty-seven, approximately 1.0 million. The ordinal number 1001087th is said 🔊 and written as: one million, one thousand and eighty-seventh. The meaning of the number 1001087 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1001087. What is 1001087 in computer science, numerology, codes and images, writing and naming in other languages\n\n## What is 1,001,087 in other units\n\nThe decimal (Arabic) number 1001087 converted to a Roman number is (M)MLXXXVII. Roman and decimal number conversions.\n\n#### Weight conversion\n\n1001087 kilograms (kg) = 2206996.4 pounds (lbs)\n1001087 pounds (lbs) = 454090.1 kilograms (kg)\n\n#### Length conversion\n\n1001087 kilometers (km) equals to 622047 miles (mi).\n1001087 miles (mi) equals to 1611094 kilometers (km).\n1001087 meters (m) equals to 3284367 feet (ft).\n1001087 feet (ft) equals 305136 meters (m).\n1001087 centimeters (cm) equals to 394128.7 inches (in).\n1001087 inches (in) equals to 2542761.0 centimeters (cm).\n\n#### Temperature conversion\n\n1001087° Fahrenheit (°F) equals to 556141.7° Celsius (°C)\n1001087° Celsius (°C) equals to 1801988.6° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n1001087 seconds equals to 1 week, 4 days, 14 hours, 4 minutes, 47 seconds\n1001087 minutes equals to 2 years, 3 weeks, 2 days, 4 hours, 47 minutes\n\n### Codes and images of the number 1001087\n\nNumber 1001087 morse code: .---- ----- ----- .---- ----- ---.. --...\nSign language for number 1001087:",
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"Number 1001087 in braille:",
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"QR code Bar code, type 39",
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"More images, other sizes, codes and colors ...\n\n## Share in social networks",
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"## Mathematics of no. 1001087\n\n### Multiplications\n\n#### Multiplication table of 1001087\n\n1001087 multiplied by two equals 2002174 (1001087 x 2 = 2002174).\n1001087 multiplied by three equals 3003261 (1001087 x 3 = 3003261).\n1001087 multiplied by four equals 4004348 (1001087 x 4 = 4004348).\n1001087 multiplied by five equals 5005435 (1001087 x 5 = 5005435).\n1001087 multiplied by six equals 6006522 (1001087 x 6 = 6006522).\n1001087 multiplied by seven equals 7007609 (1001087 x 7 = 7007609).\n1001087 multiplied by eight equals 8008696 (1001087 x 8 = 8008696).\n1001087 multiplied by nine equals 9009783 (1001087 x 9 = 9009783).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 1001087\n\nHalf of 1001087 is 500543,5 (1001087 / 2 = 500543,5 = 500543 1/2).\nOne third of 1001087 is 333695,6667 (1001087 / 3 = 333695,6667 = 333695 2/3).\nOne quarter of 1001087 is 250271,75 (1001087 / 4 = 250271,75 = 250271 3/4).\nOne fifth of 1001087 is 200217,4 (1001087 / 5 = 200217,4 = 200217 2/5).\nOne sixth of 1001087 is 166847,8333 (1001087 / 6 = 166847,8333 = 166847 5/6).\nOne seventh of 1001087 is 143012,4286 (1001087 / 7 = 143012,4286 = 143012 3/7).\nOne eighth of 1001087 is 125135,875 (1001087 / 8 = 125135,875 = 125135 7/8).\nOne ninth of 1001087 is 111231,8889 (1001087 / 9 = 111231,8889 = 111231 8/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 1001087\n\n#### Is Prime?\n\nThe number 1001087 is a prime number. The closest prime numbers are 1001081, 1001089.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 1001087\nPrime numbers have no prime factors smaller than themselves.\nThe factors of 1001087 are 1, 1001087.\nTotal factors 2.\nSum of factors 1001088 (1).\n\n#### Prime factor tree\n\n1001087 is a prime number.\n\n#### Powers\n\nThe second power of 10010872 is 1.002.175.181.569.\nThe third power of 10010873 is 1.003.264.545.991.365.504.\n\n#### Roots\n\nThe square root √1001087 is 1000,543352.\nThe cube root of 31001087 is 100,03622.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 1001087 = loge 1001087 = 13,816597.\nThe logarithm to base 10 of No. log10 1001087 = 6,000472.\nThe Napierian logarithm of No. log1/e 1001087 = -13,816597.\n\n### Trigonometric functions\n\nThe cosine of 1001087 is 0,939844.\nThe sine of 1001087 is -0,341603.\nThe tangent of 1001087 is -0,363468.\n\n### Properties of the number 1001087\n\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 1001087 in Computer Science\n\nCode typeCode value\n1001087 Number of bytes977.6KB\nUnix timeUnix time 1001087 is equal to Monday Jan. 12, 1970, 2:04:47 p.m. GMT\nIPv4, IPv6Number 1001087 internet address in dotted format v4 0.15.70.127, v6 ::f:467f\n1001087 Decimal = 11110100011001111111 Binary\n1001087 Decimal = 1212212020022 Ternary\n1001087 Decimal = 3643177 Octal\n1001087 Decimal = F467F Hexadecimal (0xf467f hex)\n1001087 BASE64MTAwMTA4Nw==\n1001087 MD5704fdcab2b8a07a4a4d430c7e7c5364c\n1001087 SHA1cf3783d8a48426da42ac27b3f9b79d8b4501ffc1\n1001087 SHA2249ccfb16c9e6291ae9c67e8057dc27f1d928e3d246f48f6066eb4c38d\n1001087 SHA2568133fbdf025f2a774fe3ea15756bb2932b87cb6122c99a68e88e8df0d106fdaf\n1001087 SHA384e569303ef230ec19c8257ce395c8f21101e84b6ac8f86eac64bbaa9706d60ab4113cd0210f7ff8df1d7ea07e77ddd357\nMore SHA codes related to the number 1001087 ...\n\nIf you know something interesting about the 1001087 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 1001087\n\n### Character frequency in the number 1001087\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 2 0 3 8 1 7 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 1001087, the numbers 1+0+0+1+0+8+7 = 1+7 = 8 are added and the meaning of the number 8 is sought.\n\n## № 1,001,087 in other languages\n\nHow to say or write the number one million, one thousand and eighty-seven in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 1.001.087) un millón mil ochenta y siete German: 🔊 (Nummer 1.001.087) eine Million eintausendsiebenundachtzig French: 🔊 (nombre 1 001 087) un million mille quatre-vingt-sept Portuguese: 🔊 (número 1 001 087) um milhão e mil e oitenta e sete Hindi: 🔊 (संख्या 1 001 087) दस लाख, एक हज़ार, सत्तासी Chinese: 🔊 (数 1 001 087) 一百万一千零八十七 Arabian: 🔊 (عدد 1,001,087) واحد مليون و واحد ألف و سبعة و ثمانون Czech: 🔊 (číslo 1 001 087) milion tisíc osmdesát sedm Korean: 🔊 (번호 1,001,087) 백만 천팔십칠 Dutch: 🔊 (nummer 1 001 087) een miljoen duizendzevenentachtig Japanese: 🔊 (数 1,001,087) 百万千八十七 Indonesian: 🔊 (jumlah 1.001.087) satu juta satu ribu delapan puluh tujuh Italian: 🔊 (numero 1 001 087) un milione e milleottantasette Norwegian: 🔊 (nummer 1 001 087) en million, en tusen og åtti-syv Polish: 🔊 (liczba 1 001 087) milion tysiąc osiemdziesiąt siedem Russian: 🔊 (номер 1 001 087) один миллион одна тысяча восемьдесят семь Turkish: 🔊 (numara 1,001,087) birmilyonbinseksenyedi Thai: 🔊 (จำนวน 1 001 087) หนึ่งล้านหนึ่งพันแปดสิบเจ็ด Ukrainian: 🔊 (номер 1 001 087) один мільйон одна тисяча вісімдесят сім Vietnamese: 🔊 (con số 1.001.087) một triệu một nghìn lẻ tám mươi bảy Other languages ...\n\n## News to email\n\nI have read the privacy policy\n\n## Comment\n\nIf you know something interesting about the number 1001087 or any other natural number (positive integer), please write to us here or on Facebook.\n\n#### Comment (Maximum 2000 characters) *\n\nThe content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy.\n\nThere are no comments for this topic."
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http://m.biancheng.net/view/2208.html | [
"# Python list列表添加元素的3种方法\n\nPython序列》一节告诉我们，使用`+`运算符可以将多个序列连接起来；列表是序列的一种，所以也可以使用`+`进行连接，这样就相当于在第一个列表的末尾添加了另一个列表。\n\n```language = [\"Python\", \"C++\", \"Java\"]\nbirthday = [1991, 1998, 1995]\ninfo = language + birthday\n\nprint(\"language =\", language)\nprint(\"birthday =\", birthday)\nprint(\"info =\", info)```\n\nlanguage = ['Python', 'C++', 'Java']\nbirthday = [1991, 1998, 1995]\ninfo = ['Python', 'C++', 'Java', 1991, 1998, 1995]\n\n`+`更多的是用来拼接列表，而且执行效率并不高，如果想在列表中插入元素，应该使用下面几个专门的方法。\n\n## Python append()方法添加元素\n\nappend() 方法用于在列表的末尾追加元素，该方法的语法格式如下：\n\nlistname.append(obj)\n\n```l = ['Python', 'C++', 'Java']\n#追加元素\nl.append('PHP')\nprint(l)\n\n#追加元组，整个元组被当成一个元素\nt = ('JavaScript', 'C#', 'Go')\nl.append(t)\nprint(l)\n\n#追加列表，整个列表也被当成一个元素\nl.append(['Ruby', 'SQL'])\nprint(l)```\n\n['Python', 'C++', 'Java', 'PHP']\n['Python', 'C++', 'Java', 'PHP', ('JavaScript', 'C#', 'Go')]\n['Python', 'C++', 'Java', 'PHP', ('JavaScript', 'C#', 'Go'), ['Ruby', 'SQL']]\n\n## Python extend()方法添加元素\n\nextend() 和 append() 的不同之处在于：extend() 不会把列表或者元祖视为一个整体，而是把它们包含的元素逐个添加到列表中。\n\nextend() 方法的语法格式如下：\n\nlistname.extend(obj)\n\n```l = ['Python', 'C++', 'Java']\n#追加元素\nl.extend('C')\nprint(l)\n\n#追加元组，元祖被拆分成多个元素\nt = ('JavaScript', 'C#', 'Go')\nl.extend(t)\nprint(l)\n\n#追加列表，列表也被拆分成多个元素\nl.extend(['Ruby', 'SQL'])\nprint(l)```\n\n['Python', 'C++', 'Java', 'C']\n['Python', 'C++', 'Java', 'C', 'JavaScript', 'C#', 'Go']\n['Python', 'C++', 'Java', 'C', 'JavaScript', 'C#', 'Go', 'Ruby', 'SQL']\n\n## Python insert()方法插入元素\n\nappend() 和 extend() 方法只能在列表末尾插入元素，如果希望在列表中间某个位置插入元素，那么可以使用 insert() 方法。\n\ninsert() 的语法格式如下：\n\nlistname.insert(index , obj)\n\n```l = ['Python', 'C++', 'Java']\n#插入元素\nl.insert(1, 'C')\nprint(l)\n\n#插入元组，整个元祖被当成一个元素\nt = ('C#', 'Go')\nl.insert(2, t)\nprint(l)\n\n#插入列表，整个列表被当成一个元素\nl.insert(3, ['Ruby', 'SQL'])\nprint(l)\n\n#插入字符串，整个字符串被当成一个元素\nl.insert(0, \"http://c.biancheng.net\")\nprint(l)```\n\n['Python', 'C', 'C++', 'Java']\n['Python', 'C', ('C#', 'Go'), 'C++', 'Java']\n['Python', 'C', ('C#', 'Go'), ['Ruby', 'SQL'], 'C++', 'Java']\n['http://c.biancheng.net', 'Python', 'C', ('C#', 'Go'), ['Ruby', 'SQL'], 'C++', 'Java']",
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https://beta.geogebra.org/m/mJZCZAsK | [
"## 1.\n\nIn the applet above, use the tools of GeoGebra to show how/why the quadrilateral you formed is indeed a parallelogram. In the space provided, explain how your results illustrate this quadrilateral is a parallelogram.\n\n## 2.\n\nIn the applet above, use the tools of GeoGebra to show how/why the quadrilateral you formed is indeed a rhombus. In the space provided, explain how your results illustrate this quadrilateral is a rhombus.\n\n## 3.\n\nIn the applet above, use the tools of GeoGebra to show how/why the quadrilateral you formed is indeed a rectangle. In the space provided, explain how your results illustrate this quadrilateral is a rectangle.\n\n## 4.\n\nIn the applet above, use the tools of GeoGebra to show how/why the quadrilateral you formed is indeed a square. In the space provided, explain how your results illustrate this quadrilateral is a square."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8314966,"math_prob":0.94383365,"size":1496,"snap":"2019-51-2020-05","text_gpt3_token_len":347,"char_repetition_ratio":0.17091152,"word_repetition_ratio":0.62601626,"special_character_ratio":0.1985294,"punctuation_ratio":0.11347517,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947832,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T03:17:30Z\",\"WARC-Record-ID\":\"<urn:uuid:1a3409e6-b961-476f-ba0b-158ee524463a>\",\"Content-Length\":\"49814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:319b8549-dd71-47e5-8262-82aaf1ef6e5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:039ede18-99d8-4100-b638-4693400a3979>\",\"WARC-IP-Address\":\"13.249.44.50\",\"WARC-Target-URI\":\"https://beta.geogebra.org/m/mJZCZAsK\",\"WARC-Payload-Digest\":\"sha1:NFRIXQFGQUCNEAC77GLFV7AIIZYHM3GV\",\"WARC-Block-Digest\":\"sha1:Y4XHFJFAKTCFKDX45QSERURQHAEMXYKB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250591763.20_warc_CC-MAIN-20200118023429-20200118051429-00544.warc.gz\"}"} |
https://www.aliensbrain.com/quiz/33/java-programming-language-quiz-2 | [
"",
null,
"### Java Programming Language Quiz - 2\n\n Description: Java Programming Language Quiz - 2 Number of Questions: 8 Created by: Aliensbrain Bot Tags: java\nAttempted 0/8 Correct 0 Score 0\n\nWhich one of the following syntactic constructs is not a simple expression?\n\n1. null\n\n2. 43\n\n3. x 2\n\n4. print()\n\nCorrect Option: C\nExplanation:\n\nIt's a compound expression consisting of simple expression x (variable name), the addition operator ( ), and simple expression 2 (a 32-bit integer literal).\n\nWhich one of the following escape sequences cannot appear in a string literal?\n\n1. \\f\n\n2. \\u\n\n3. \\\"\n\n4. \\\\\n\nCorrect Option: B\nExplanation:\n\n\\u must be followed by four hexadecimal digits and is then known as a Unicode escape sequence.\n\nWhich one of the following is not a widening conversion rule?\n\n1. Short integer to character\n\n2. Integer to long integer\n\n3. Floating-point to double precision floating-point\n\n4. Character to floating-point\n\nCorrect Option: A\n\nThe ?: operator is an example of which kind of operator?\n\n1. Binary\n\n2. Prefix\n\n3. Postfix\n\n4. Infix\n\nCorrect Option: D\n\nWhich one of the following operators offers the fastest way to divide a negative integer by 2 and keep the negative result (e.g., -4/2 equals -2)?\n\n1. /\n\n2. %\n\n3. >>\n\n4. >>>\n\nCorrect Option: C\n\nYou read the following statement in a Java program that compiles and executes.\n\nsubmarine.dive(depth);\n\nWhat can you say for sure?\n\n1. depth must be an int\n\n2. dive must be a method.\n\n3. dive must be the name of an instance field.\n\n4. submarine must be the name of a class\n\n5. submarine must be a method.\n\nCorrect Option: B\n\nGiven the following class definition, which of the following methods could be legally placed after the comment //Here public class Rid{ public void amethod(int i, String s){} //Here }\n\n1. public void amethod(int s, String i){}\n\n2. public void Amethod(int i, String s) {}\n\n3. public void amethod(int i, String mystring){}\n\n4. None of the above\n\n5. public int amethod(int i, String s){}\n\nCorrect Option: B\n\nWhich of the following statements are true?\n\n1. Strings are a primitive type in Java and the StringBuffer is used as the matching wrapper type\n\n2. The size of a string can be retrieved using the length property.\n\n3. Strings are a primitive type in Java that overloads the + operator for concatenation\n\n4. The String class is implemented as a char array, elements are addressed using the stringname[] convention\n\nCorrect Option: C\n+ View questions"
]
| [
null,
"https://www.aliensbrain.com/assets/alien_head-0b711ac22e1d4ab59a31f4d4cce35274d6ed61f2659d4190787848a78e0fbdfe.svg",
null
]
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https://www.studypug.com/pre-calculus/pythagorean-identities | [
"# Pythagorean identities\n\n## What are the Pythagorean identities?\n\nAn identity in mathematics is an equation that is always true. The Pythagorean identities all involve the number 1 and its Pythagorean aspects can be clearly seen when proving the theorems on a unit circle.\n\n## Pythagorean identities\n\nWe are going to explore the Pythagorean identities in this question. You may refer to the below formula sheet when dealing with the 3 Pythagorean identities.\n\nLet's explore the Pythagorean identities. The first of these three states that sine squared plus cosine squared equals one. The second one states that tangent squared plus one equals secant squared. For the last one, it states that one plus cotangent squared equals cosecant squared.\n\nIn the following question, we're going to try to use a unit circle to prove the first Pythagorean identity: sine squared plus cosine squared equals one.\n\n## Pythagorean identities examples\n\nQuestion:\n\nUse the unit circle to derive the Pythagorean Identity: $\\cos^2 \\theta + \\sin^2 \\theta = 1$\n\nHow do we begin? Do you remember the properties of a unit circle? We've covered the unit circle in the previous section. To quickly recap, a unit circle is just a circle with radius of one unit, i.e. the radius must equal one.\n\nRefer to the above image. We'll identify a point on the circle at X,Y. Here, the X coordinate is X and the Y coordinate is Y.\n\nFrom this point, let's draw a perpendicular line to the X axis. We will be focusing on this triangle.\n\nIn the above image, take a moment to recall what ? means. It's actually the reference angle, correct? It is one of the most important angles in trigonometry.\n\nIn the reference angle, what does it mean if the X coordinate equals X? It means that the length of the X segment is X. In a similar sense, if the Y coordinate is Y, that means the length of the vertical segment of the triangle would be Y.\n\nLet's demonstrate this with actual numbers to illustrate that concept.",
null,
"illustration of relationship between coordinate and length of segments of a triangle\n\nIn the above example, there's a point called 3,5. If we draw a triangle, the 3 depicts the X coordinate. This means that the length of this segment is 3. Now, if the Y coordinate is 5, what does that mean? The length of the vertical segment in the triangle must be five.\n\nGoing back to the previous unit circle illustration, let's focus on the right-angled triangle and apply the Pythagoras theorem. What is the Pythagoras theorem? The Pythagoras tells us that X squared plus Y squared equals to the hypotenuse squared. The hypotenuse in this case is one, since we're using a unit circle. So here we have X squared plus Y squared equals one squared.\n\nOne neat thing about the unit circle is that its X coordinate can also be represented in terms of the angle theta. The X coordinate can be represented as cosine theta, while its Y coordinate can be represented as sine theta. Keep in mind that this is only for a unit circle. So for any point on the unit circle: the X coordinate can be represented as cosine theta; the Y coordinate can be represented as sine theta.",
null,
"derive the other Pythagorean identities from cos2θ+sin2θ=1\\cos^2 \\theta + \\sin^2 \\theta = 1cos2θ+sin2θ=1\n\nThrough using the unit circle, the answer becomes very obvious. One squared is just one. The X coordinate can also be represented as cosine theta. The Y coordinate can be represented as sine theta. And voila! We are done. From the unit circle, we've successfully proved that cosine squared plus sine squared equals one, tackling one out of 3 Pythagorean identities.\n\n### Pythagorean identities\n\nPythagorean identities are formulas, derived from Pythagorean Theorem, that allow us to find out where a point is on the unit circle. Learn the tricks and tips on how to use the unit circle to derive and prove the Pythagorean identities can be difficult.\n\n#### Lessons\n\n• 1.\nUse the unit circle to derive the Pythagorean Identity:\ncos$^2 \\theta +$sin$^2 \\theta = 1$\n\n• 2.\nSimplify expressions:\n\na)\n(sec$^2x -1)$(cot$^2x)$\n\nb)\n$\\frac{(1+ \\sin x )}{\\cos x} - \\frac{\\cos x}{1 - \\sin x}$\n\n• 3.\nProve identities\na)\n$\\frac{\\sin x}{ 1 + \\cos x} +\\frac{\\sin x}{1 - \\cos x} = 2\\csc x$\n\nb)\n$\\tan x (\\csc x + 1) = \\frac{ \\cot x}{ \\csc x - 1}$"
]
| [
null,
"https://dcvp84mxptlac.cloudfront.net/diagrams2/unit-circle-x-y-coordinate-length-of-segment-in-triangle.png",
null,
"https://dcvp84mxptlac.cloudfront.net/diagrams2/derive-other-pythagorean-identities-from-first-pythagorean-identity-v2.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.89599484,"math_prob":0.99614143,"size":3498,"snap":"2020-10-2020-16","text_gpt3_token_len":774,"char_repetition_ratio":0.18402976,"word_repetition_ratio":0.0765391,"special_character_ratio":0.20554602,"punctuation_ratio":0.11942446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999979,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-04T09:38:17Z\",\"WARC-Record-ID\":\"<urn:uuid:e3e212d6-bc57-4abc-9134-8a2d3d8cb2a9>\",\"Content-Length\":\"263975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8dbae084-5667-42d4-aa70-391cf66480af>\",\"WARC-Concurrent-To\":\"<urn:uuid:162c9bf5-feb7-4e65-84b8-7346b402e31b>\",\"WARC-IP-Address\":\"34.200.169.6\",\"WARC-Target-URI\":\"https://www.studypug.com/pre-calculus/pythagorean-identities\",\"WARC-Payload-Digest\":\"sha1:VBLO2QT676RAUIVHN3HNBEIOC4QVD7R4\",\"WARC-Block-Digest\":\"sha1:6GPTJOBEO366NEMNCWOTPTNWGUHBSHRA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370521574.59_warc_CC-MAIN-20200404073139-20200404103139-00220.warc.gz\"}"} |
http://josephgedwards.info/compare-fraction-factors-and-products-worksheets/compare-fraction-factors-and-products-worksheets-compare-fraction-factors-and-products-worksheets/ | [
"# Compare Fraction Factors And Products Worksheets Compare Fraction Factors And Products Worksheets",
null,
"compare fraction factors and products worksheets compare fraction factors and products worksheets.\n\nordering fractions on a number line freebie from candlers compare fraction factors and products worksheets,compare fraction factors and products worksheets common core dividing fractions, fractions worksheets compare fraction factors and products,compare fraction factors and products worksheets , compare fraction factors and products worksheets lesson, grade compare fraction factors and products worksheets decimal, compare fraction factors and products worksheets printable task cards for teachers,compare fraction factors and products worksheets mega bundle, fraction worksheets for grade 2 compare factors and products, compare fraction factors and products worksheets comparing fractions decimals school teaching.\n\n# Compare Fraction Factors And Products Worksheets Compare Fraction Factors And Products Worksheets",
null,
"compare fraction factors and products worksheets compare fraction factors and products worksheets.\n\nfraction worksheets for grade 2 compare factors and products,compare fraction factors and products worksheets common core dividing fractions, compare fraction factors and products worksheets comparing fractions decimals school teaching, ordering fractions on a number line freebie from candlers compare fraction factors and products worksheets, grade compare fraction factors and products worksheets decimal,compare fraction factors and products worksheets mega bundle, compare fraction factors and products worksheets lesson, compare fraction factors and products worksheets printable task cards for teachers,compare fraction factors and products worksheets , fractions worksheets compare fraction factors and products."
]
| [
null,
"http://josephgedwards.info/wp-content/uploads/2019/09/compare-fraction-factors-and-products-worksheets-compare-fraction-factors-and-products-worksheets.jpg",
null,
"http://josephgedwards.info/wp-content/uploads/2019/09/compare-fraction-factors-and-products-worksheets-compare-fraction-factors-and-products-worksheets.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8543879,"math_prob":0.864436,"size":833,"snap":"2019-35-2019-39","text_gpt3_token_len":121,"char_repetition_ratio":0.25814235,"word_repetition_ratio":0.16831683,"special_character_ratio":0.1392557,"punctuation_ratio":0.09322034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948194,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T02:23:13Z\",\"WARC-Record-ID\":\"<urn:uuid:68146fdf-6d3e-4ff1-ac9c-5b400213a4e3>\",\"Content-Length\":\"75905\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9bfb484e-f860-465d-9fe2-de648544426a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3dddfe16-a827-408b-bf94-76ed855f1004>\",\"WARC-IP-Address\":\"104.28.15.214\",\"WARC-Target-URI\":\"http://josephgedwards.info/compare-fraction-factors-and-products-worksheets/compare-fraction-factors-and-products-worksheets-compare-fraction-factors-and-products-worksheets/\",\"WARC-Payload-Digest\":\"sha1:MQVZYDTYE4IBD4YZ7FYKBGMUV3ZI5IGL\",\"WARC-Block-Digest\":\"sha1:FWRTB32WLZO5TGF6EVXRVQ7UALOYKR5X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574765.55_warc_CC-MAIN-20190922012344-20190922034344-00465.warc.gz\"}"} |
https://dirty-cat.github.io/stable/auto_examples/02_fit_predict_plot_employee_salaries.html | [
"# Predicting the salary of employees¶\n\nThe employee salaries dataset contains information about annual salaries (year 2016) for more than 9,000 employees of the Montgomery County (Maryland, US). In this example, we are interested in predicting the column Current Annual Salary depending on a mix of clean columns and a dirty column. We choose to benchmark different categorical encodings for the dirty column Employee Position Title, that contains dirty categorical data.\n\n## Data Importing and preprocessing¶\n\nfrom dirty_cat.datasets import fetch_employee_salaries\nemployee_salaries = fetch_employee_salaries()\nprint(employee_salaries['description'])\n\n\nOut:\n\nThe downloaded data contains the employee_salaries dataset.\nIt can originally be found at: https://catalog.data.gov/dataset/ employee-salaries-2016\n\n\nimport pandas as pd\n\n\nNow, let’s carry out some basic preprocessing:\n\ndf['Current Annual Salary'] = df['Current Annual Salary'].str.strip('\\$').astype(\nfloat)\ndf['Date First Hired'] = pd.to_datetime(df['Date First Hired'])\ndf['Year First Hired'] = df['Date First Hired'].apply(lambda x: x.year)\n\ntarget_column = 'Current Annual Salary'\ny = df[target_column].values.ravel()\n\n\n## Choosing columns¶\n\nFor categorical columns that are supposed to be clean, it is “safe” to use one hot encoding to transform them:\n\nclean_columns = {\n'Gender': 'one-hot',\n'Department Name': 'one-hot',\n'Assignment Category': 'one-hot',\n'Year First Hired': 'numerical'}\n\n\nWe then choose the categorical encoding methods we want to benchmark and the dirty categorical variable:\n\nencoding_methods = ['one-hot', 'target', 'similarity']\ndirty_column = 'Employee Position Title'\n\n\n## Creating a learning pipeline¶\n\nThe encoders for both clean and dirty data are first imported:\n\nfrom sklearn.preprocessing import FunctionTransformer\nfrom sklearn.preprocessing import OneHotEncoder\nfrom dirty_cat import SimilarityEncoder, TargetEncoder\n\nencoders_dict = {\n'one-hot': OneHotEncoder(handle_unknown='ignore', sparse=False),\n'similarity': SimilarityEncoder(similarity='ngram'),\n'target': TargetEncoder(handle_unknown='ignore'),\n'numerical': FunctionTransformer(None)}\n\n# We then create a function that takes one key of our encoders_dict,\n# returns a pipeline object with the associated encoder,\n# as well as a Scaler and a RidgeCV regressor:\n\nfrom sklearn.compose import ColumnTransformer\nfrom sklearn.pipeline import Pipeline\n\ndef make_pipeline(encoding_method):\n# static transformers from the other columns\ntransformers = [(enc + '_' + col, encoders_dict[enc], [col])\nfor col, enc in clean_columns.items()]\ntransformers += [(encoding_method, encoders_dict[encoding_method],\n[dirty_column])]\npipeline = Pipeline([\n# Use ColumnTransformer to combine the features\n('union', ColumnTransformer(\ntransformers=transformers,\nremainder='drop')),\n('scaler', StandardScaler(with_mean=False)),\n('clf', RidgeCV())\n])\nreturn pipeline\n\n\n## Fitting each encoding methods with a RidgeCV¶\n\nEventually, we loop over the different encoding methods, instantiate each time a new pipeline, fit it and store the returned cross-validation score:\n\nfrom sklearn.preprocessing import StandardScaler\nfrom sklearn.linear_model import RidgeCV\nfrom sklearn.model_selection import KFold, cross_val_score\nimport numpy as np\n\nall_scores = dict()\n\ncv = KFold(n_splits=5, random_state=12, shuffle=True)\nscoring = 'r2'\nfor method in encoding_methods:\npipeline = make_pipeline(method)\nscores = cross_val_score(pipeline, df, y, cv=cv, scoring=scoring)\nprint('{} encoding'.format(method))\nprint('{} score: mean: {:.3f}; std: {:.3f}\\n'.format(\nscoring, np.mean(scores), np.std(scores)))\nall_scores[method] = scores\n\n\nOut:\n\none-hot encoding\nr2 score: mean: 0.856; std: 0.013\n\ntarget encoding\nr2 score: mean: 0.773; std: 0.016\n\nsimilarity encoding\nr2 score: mean: 0.913; std: 0.006\n\n\n## Plotting the results¶\n\nFinally, we plot the scores on a boxplot:\n\nimport seaborn\nimport matplotlib.pyplot as plt\nplt.figure(figsize=(4, 3))\nax = seaborn.boxplot(data=pd.DataFrame(all_scores), orient='h')\nplt.ylabel('Encoding', size=20)\nplt.xlabel('Prediction accuracy ', size=20)\nplt.yticks(size=20)\nplt.tight_layout()",
null,
"Total running time of the script: ( 0 minutes 11.063 seconds)\n\nGallery generated by Sphinx-Gallery"
]
| [
null,
"https://dirty-cat.github.io/stable/_images/sphx_glr_02_fit_predict_plot_employee_salaries_001.png",
null
]
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https://youvegotthismath.com/easter-math-activities/ | [
"# Easter Math Activities | Free Printable | Kindergarten\n\nEaster is a perfect time to introduce some entertaining arithmetic games to young children. These easter math activities worksheets will help to increase number and color sense. 1st & 2nd-grade students will learn basic mathematical operations and can improve their basic math skills with our free printable easter math activities worksheets.\n\n## 7 Fun Easter Math Activities with Free Worksheets",
null,
"Here, we are sure that your children will also love the activities on the occasion of easter and they will enjoy the work.",
null,
"These easter math activities in pdf format can help our children learn the basics of mathematics fast and actively. For the easter math activities, follow the detailed directions listed below.\n\n## Easter Eggs Counting Math Game in Easter Math Activities\n\nDuring the holy week of easter, it is a common norm to present eggs from the easter bunny. To present eggs, you need to color them. In this activity, we will find colorful eggs. Count the same colored eggs together and write the number of the eggs in the particular box.",
null,
"## Find the Missing Number with a Bunny in Easter Math Activities\n\nHere, children will be provided with a board where they can see a pathway with some numbers. The numbers are organized serially. But there are some missing numbers. Your child will find a picture of carrots there. They need to count the carrots and find the missing numbers of that number set.",
null,
"## Addition of Eggs in Easter Math Activities\n\nIn this activity, your children will add the numbers given in the worksheet. They will solve the problems by adding the numbers. Give them the worksheets. You will find the eggs with addition problems. After solving the problems, tell them to color the eggs.",
null,
"## Subtract on the Eggs in Easter Math Activities\n\nYour kids will use this to subtract the numbers from the worksheet. By subtracting the numbers, they will be able to solve the issues. Deliver the worksheets to them. Problems with subtraction are present in the eggs. Tell them to color the eggs after the problems.",
null,
"## Color by Number in in Easter Math Activities\n\nIn this portion of this game, your kiddos will color the egg by counting the numbers. Here, the numbers are mentioned under the picture with color codes. Tell your kiddos to color the egg by maintaining the color code.",
null,
"## Multiplication on Eggs in Easter Math Activities\n\nChildren will multiply the numbers provided on the worksheet here. Deliver the worksheets to them. The eggs have problems with multiplication that you can find. Check to see if they are able to solve the problems correctly. Tell them to color the eggs after that.",
null,
"## Exciting Division in Easter Math Activities\n\nAt this point, kids will divide the numbers supplied in the worksheet. By dividing the numbers, they will find solutions to the problems. Check to see if they are able to solve the issues appropriately. Tell them to color the eggs after that.",
null,
""
]
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%208000%208001'%3E%3C/svg%3E",
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%208001%208001'%3E%3C/svg%3E",
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%208000%208001'%3E%3C/svg%3E",
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%208000%208001'%3E%3C/svg%3E",
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https://reference.wolfram.com/language/ref/FindGraphPartition.html | [
"# FindGraphPartition\n\ngives a partition of vertices of the graph g.\n\nFindGraphPartition[g,k]\n\ngives a partition of vertices into k approximately equal-size parts.\n\nFindGraphPartition[g,{n1,,nk}]\n\ngives a partition of vertices into parts with sizes n1, , nk.\n\nFindGraphPartition[g,{α1,,αk}]\n\ngives a partition of vertices into parts with approximate size proportions α1, , αk.\n\nFindGraphPartition[{vw,},]\n\nuses rules vw to specify the graph g.\n\n# Details",
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"• FindGraphPartition finds a partition of vertices such that the number of edges having endpoints in different parts is minimized.\n• is equivalent to FindGraphPartition[g,2].\n• FindGraphPartition treats graphs as undirected simple graphs.\n• For a weighted graph, FindGraphPartition finds a partition such that the sum of edge weights for edges having endpoints in different parts is minimized.\n• FindGraphPartition[g,{α1,,αk}] will give a partition where the size of a part is given by the sum of its vertex weights.\n• The partitions are ordered by their length with the largest part first.\n\n# Examples\n\nopen all close all\n\n## Basic Examples(1)\n\nFind a partition of a graph:\n\n In:=",
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"## Properties & Relations(1)\n\nIntroduced in 2012\n(9.0)\n|\nUpdated in 2015\n(10.3)"
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http://self.gutenberg.org/articles/eng/Viscoelastic | [
"",
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"#jsDisabledContent { display:none; } My Account | Register | Help",
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"Flag as Inappropriate",
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"This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:\n\n# Viscoelastic\n\nArticle Id: WHEBN0002133841\nReproduction Date:\n\n Title: Viscoelastic",
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"Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Viscoelastic\n\nViscoelasticity is the property of materials that exhibit both viscous and elastic characteristics when undergoing deformation. Viscous materials, like honey, resist shear flow and strain linearly with time when a stress is applied. Elastic materials strain when stretched and quickly return to their original state once the stress is removed. Viscoelastic materials have elements of both of these properties and, as such, exhibit time-dependent strain. Whereas elasticity is usually the result of bond stretching along crystallographic planes in an ordered solid, viscosity is the result of the diffusion of atoms or molecules inside an amorphous material.\n\n## Background\n\nIn the nineteenth century, physicists such as Maxwell, Boltzmann, and Kelvin researched and experimented with creep and recovery of glasses, metals, and rubbers. Viscoelasticity was further examined in the late twentieth century when synthetic polymers were engineered and used in a variety of applications. Viscoelasticity calculations depend heavily on the viscosity variable, η. The inverse of η is also known as fluidity, φ. The value of either can be derived as a function of temperature or as a given value (i.e. for a dashpot).\n\nDepending on the change of strain rate versus stress inside a material the viscosity can be categorized as having a linear, non-linear, or plastic response. When a material exhibits a linear response it is categorized as a Newtonian material. In this case the stress is linearly proportional to the strain rate. If the material exhibits a non-linear response to the strain rate, it is categorized as Non-Newtonian fluid. There is also an interesting case where the viscosity decreases as the shear/strain rate remains constant. A material which exhibits this type of behavior is known as thixotropic. In addition, when the stress is independent of this strain rate, the material exhibits plastic deformation. Many viscoelastic materials exhibit rubber like behavior explained by the thermodynamic theory of polymer elasticity. In reality all materials deviate from Hooke's law in various ways, for example by exhibiting viscous-like as well as elastic characteristics. Viscoelastic materials are those for which the relationship between stress and strain depends on time. Anelastic solids represent a subset of viscoelastic materials: they have a unique equilibrium configuration and ultimately recover fully after removal of a transient load.\n\nSome phenomena in viscoelastic materials are:\n\n• if the stress is held constant, the strain increases with time (creep);\n• if the strain is held constant, the stress decreases with time (relaxation);\n• the effective stiffness depends on the rate of application of the load;\n• if cyclic loading is applied, hysteresis (a phase lag) occurs, leading to a dissipation of mechanical energy;\n• acoustic waves experience attenuation;\n• rebound of an object following an impact is less than 100%;\n• during rolling, frictional resistance occurs.\n\nAll materials exhibit some viscoelastic response. In common metals such as steel or aluminum, as well as in quartz, at room temperature and at small strain, the behavior does not deviate much from linear elasticity. Synthetic polymers, wood, and human tissue as well as metals at high temperature display significant viscoelastic effects. In some applications, even a small viscoelastic response can be significant. To be complete, an analysis or design involving such materials must incorporate their viscoelastic behavior. Knowledge of the viscoelastic response of a material is based on measurement.\n\nSome examples of viscoelastic materials include amorphous polymers, semicrystalline polymers, biopolymers, metals at very high temperatures, and bitumen materials. Cracking occurs when the strain is applied quickly and outside of the elastic limit. Ligaments and tendons are viscoelastic, so the extent of the potential damage to them depends both on the velocity of the change of their length as well as on the force applied.\n\nA viscoelastic material has the following properties:\n\n## Elastic behavior versus viscoelastic behavior\n\nUnlike purely elastic substances, a viscoelastic substance has an elastic component and a viscous component. The viscosity of a viscoelastic substance gives the substance a strain rate dependent on time. Purely elastic materials do not dissipate energy (heat) when a load is applied, then removed. However, a viscoelastic substance looses energy when a load is applied, then removed. Hysteresis is observed in the stress-strain curve, with the area of the loop being equal to the energy lost during the loading cycle. Since viscosity is the resistance to thermally activated plastic deformation, a viscous material will lose energy through a loading cycle. Plastic deformation results in lost energy, which is uncharacteristic of a purely elastic material's reaction to a loading cycle.\n\nSpecifically, viscoelasticity is a molecular rearrangement. When a stress is applied to a viscoelastic material such as a polymer, parts of the long polymer chain change positions. This movement or rearrangement is called Creep. Polymers remain a solid material even when these parts of their chains are rearranging in order to accompany the stress, and as this occurs, it creates a back stress in the material. When the back stress is the same magnitude as the applied stress, the material no longer creeps. When the original stress is taken away, the accumulated back stresses will cause the polymer to return to its original form. The material creeps, which gives the prefix visco-, and the material fully recovers, which gives the suffix -elasticity.\n\n## Types of viscoelasticity\n\nLinear viscoelasticity is when the function is separable in both creep response and load. All linear viscoelastic models can be represented by a Volterra equation connecting stress and strain:\n\n$\\epsilon\\left(t\\right)= \\frac \\left\\{ \\sigma\\left(t\\right) \\right\\}\\left\\{ E_\\text\\left\\{inst,creep\\right\\} \\right\\}+ \\int_0^t K\\left(t-t^\\prime\\right) \\dot\\left\\{\\sigma\\right\\}\\left(t^\\prime\\right) d t^\\prime$\n\nor\n\n$\\sigma\\left(t\\right)= E_\\text\\left\\{inst,relax\\right\\}\\epsilon\\left(t\\right)+ \\int_0^t F\\left(t-t^\\prime\\right) \\dot\\left\\{\\epsilon\\right\\}\\left(t^\\prime\\right) d t^\\prime$\n\nwhere\n\n• t is time\n• $\\sigma \\left(t\\right)$ is stress\n• $\\epsilon \\left(t\\right)$ is strain\n• $E_\\text\\left\\{inst,creep\\right\\}$ and $E_\\text\\left\\{inst,relax\\right\\}$ are instantaneous elastic moduli for creep and relaxation\n• K(t) is the creep function\n• F(t) is the relaxation function\n\nLinear viscoelasticity is usually applicable only for small deformations.\n\nNonlinear viscoelasticity is when the function is not separable. It usually happens when the deformations are large or if the material changes its properties under deformations.\n\nAn anelastic material is a special case of a viscoelastic material: an anelastic material will fully recover to its original state on the removal of load.\n\n## Dynamic modulus\n\nMain article: Dynamic modulus\n\nViscoelasticity is studied using dynamic mechanical analysis, applying a small oscillatory stress and measuring the resulting strain.\n\n• Purely elastic materials have stress and strain in phase, so that the response of one caused by the other is immediate.\n• In purely viscous materials, strain lags stress by a 90 degree phase lag.\n• Viscoelastic materials exhibit behavior somewhere in the middle of these two types of material, exhibiting some lag in strain.\n\nComplex Dynamic modulus G can be used to represent the relations between the oscillating stress and strain:\n\n$G = G\\text{'} + iG$\n\nwhere $i^2 = -1$; $G\\text{'}$ is the storage modulus and $G$ is the loss modulus:\n\n$G\\text{'} = \\frac \\left\\{\\sigma_0\\right\\} \\left\\{\\varepsilon_0\\right\\} \\cos \\delta$\n$G$ = \\frac {\\sigma_0} {\\varepsilon_0} \\sin \\delta\n\nwhere $\\sigma_0$ and $\\varepsilon_0$ are the amplitudes of stress and strain and $\\delta$ is the phase shift between them.\n\n## Constitutive models of linear viscoelasticity\n\nViscoelastic materials, such as amorphous polymers, semicrystalline polymers, and biopolymers, can be modeled in order to determine their stress or strain interactions as well as their temporal dependencies. These models, which include the Maxwell model, the Kelvin-Voigt model, and the Standard Linear Solid Model, are used to predict a material's response under different loading conditions. Viscoelastic behavior has elastic and viscous components modeled as linear combinations of springs and dashpots, respectively. Each model differs in the arrangement of these elements, and all of these viscoelastic models can be equivalently modeled as electrical circuits. In an equivalent electrical circuit, stress is represented by current, and strain rate by voltage. The elastic modulus of a spring is analogous to a circuit's capacitance (it stores energy) and the viscosity of a dashpot to a circuit's resistance (it dissipates energy).\n\nThe elastic components, as previously mentioned, can be modeled as springs of elastic constant E, given the formula:\n\n$\\sigma = E \\varepsilon$\n\nwhere σ is the stress, E is the elastic modulus of the material, and ε is the strain that occurs under the given stress, similar to Hooke's Law.\n\nThe viscous components can be modeled as dashpots such that the stress-strain rate relationship can be given as,\n\n$\\sigma = \\eta \\frac\\left\\{d\\varepsilon\\right\\}\\left\\{dt\\right\\}$\n\nwhere σ is the stress, η is the viscosity of the material, and dε/dt is the time derivative of strain.\n\nThe relationship between stress and strain can be simplified for specific stress rates. For high stress states/short time periods, the time derivative components of the stress-strain relationship dominate. A dashpot resists changes in length, and in a high stress state it can be approximated as a rigid rod. Since a rigid rod cannot be stretched past its original length, no strain is added to the system\n\nConversely, for low stress states/longer time periods, the time derivative components are negligible and the dashpot can be effectively removed from the system - an \"open\" circuit. As a result, only the spring connected in parallel to the dashpot will contribute to the total strain in the system\n\n### Maxwell model\n\nMain article: Maxwell material\n\nThe Maxwell model can be represented by a purely viscous damper and a purely elastic spring connected in series, as shown in the diagram. The model can be represented by the following equation:\n\n$\\frac \\left\\{d\\epsilon_\\left\\{Total\\right\\}\\right\\} \\left\\{dt\\right\\} = \\frac \\left\\{d\\epsilon_\\left\\{D\\right\\}\\right\\} \\left\\{dt\\right\\} + \\frac \\left\\{d\\epsilon_\\left\\{S\\right\\}\\right\\} \\left\\{dt\\right\\} = \\frac \\left\\{\\sigma\\right\\} \\left\\{\\eta\\right\\} + \\frac \\left\\{1\\right\\} \\left\\{E\\right\\} \\frac \\left\\{d\\sigma\\right\\} \\left\\{dt\\right\\}$.\n\nUnder this model, if the material is put under a constant strain, the stresses gradually relax. When a material is put under a constant stress, the strain has two components. First, an elastic component occurs instantaneously, corresponding to the spring, and relaxes immediately upon release of the stress. The second is a viscous component that grows with time as long as the stress is applied. The Maxwell model predicts that stress decays exponentially with time, which is accurate for most polymers. One limitation of this model is that it does not predict creep accurately. The Maxwell model for creep or constant-stress conditions postulates that strain will increase linearly with time. However, polymers for the most part show the strain rate to be decreasing with time.\n\nApplications to soft solids: thermoplastic polymers in the vicinity of their melting temperature, fresh concrete (neglecting its aging), numerous metals at a temperature close to their melting point.\n\n### Kelvin–Voigt model\n\nThe Kelvin–Voigt model, also known as the Voigt model, consists of a Newtonian damper and Hookean elastic spring connected in parallel, as shown in the picture. It is used to explain the creep behaviour of polymers.\n\nThe constitutive relation is expressed as a linear first-order differential equation:\n\n$\\sigma \\left(t\\right) = E \\varepsilon\\left(t\\right) + \\eta \\frac \\left\\{d\\varepsilon\\left(t\\right)\\right\\} \\left\\{dt\\right\\}$\n\nThis model represents a solid undergoing reversible, viscoelastic strain. Upon application of a constant stress, the material deforms at a decreasing rate, asymptotically approaching the steady-state strain. When the stress is released, the material gradually relaxes to its undeformed state. At constant stress (creep), the Model is quite realistic as it predicts strain to tend to σ/E as time continues to infinity. Similar to the Maxwell model, the Kelvin–Voigt model also has limitations. The model is extremely good with modelling creep in materials, but with regards to relaxation the model is much less accurate.\n\nApplications: organic polymers, rubber, wood when the load is not too high.\n\n### Standard linear solid model\n\nThe Standard Linear Solid Model effectively combines the Maxwell Model and a Hookean spring in parallel. A viscous material is modeled as a spring and a dashpot in series with each other, both of which are in parallel with a lone spring. For this model, the governing constitutive relation is:\n\n$\\frac \\left\\{d\\varepsilon\\right\\} \\left\\{dt\\right\\} = \\frac \\left\\{ \\frac \\left\\{E_2\\right\\} \\left\\{\\eta\\right\\} \\left \\left( \\frac \\left\\{\\eta\\right\\} \\left\\{E_2\\right\\}\\frac \\left\\{d\\sigma\\right\\} \\left\\{dt\\right\\} + \\sigma - E_1 \\varepsilon \\right \\right)\\right\\} \\left\\{E_1 + E_2\\right\\}$\n\nUnder a constant stress, the modeled material will instantaneously deform to some strain, which is the elastic portion of the strain, and after that it will continue to deform and asymptotically approach a steady-state strain. This last portion is the viscous part of the strain. Although the Standard Linear Solid Model is more accurate than the Maxwell and Kelvin-Voigt models in predicting material responses, mathematically it returns inaccurate results for strain under specific loading conditions and is rather difficult to calculate.\n\n### Generalized Maxwell Model\n\nThe Generalized Maxwell model also known as the Maxwell–Wiechert model (after James Clerk Maxwell and E Wiechert ) is the most general form of the linear model for viscoelasticity. It takes into account that the relaxation does not occur at a single time, but at a distribution of times. Due to molecular segments of different lengths with shorter ones contributing less than longer ones, there is a varying time distribution. The Wiechert model shows this by having as many spring–dashpot Maxwell elements as are necessary to accurately represent the distribution. The figure on the right shows the generalised Wiechert model Applications : metals and alloys at temperatures lower than one quarter of their absolute melting temperature (expressed in K).\n\n### Prony series\n\nMain article: Prony series\n\nIn a one-dimensional relaxation test, the material is subjected to a sudden strain that is kept constant over the duration of the test, and the stress is measured over time. The initial stress is due to the elastic response of the material. Then, the stress relaxes over time due to the viscous effects in the material. Typically, either a tensile, compressive, bulk compression, or shear strain is applied. The resulting stress vs. time data can be fitted with a number of equations, called models. Only the notation changes depending of the type of strain applied: tensile-compressive relaxation is denoted $E$, shear is denoted $G$, bulk is denoted $K$. The Prony series for the shear relaxation is\n\n\n\nG(t) = G_\\infty + \\Sigma_{i=1}^{N} G_i \\exp(-t/\\tau_i)\n\nwhere $G_\\infty$ is the long term modulus once the material is totally relaxed, $\\tau_i$ are the relaxation times (not to be confused with $\\tau_i$ in the diagram); the higher their values, the longer it takes for the stress to relax. The data is fitted with the equation by using a minimization algorithm that adjust the parameters ($G_\\infty, G_i, \\tau_i$) to minimize the error between the predicted and data values .\n\nAn alternative form is obtained noting that the elastic modulus is related to the long term modulus by\n\n\n\nG(t=0)=G_0=G_\\infty+\\Sigma_{i=1}^{N} G_i\n\nTherefore,\n\n\n\nG(t) = G_0 - \\Sigma_{i=1}^{N} G_i [1-\\exp(-t/\\tau_i)]\n\nThis form is convenient when the elastic shear modulus $G_0$ is obtained from data independent from the relaxation data, and/or for computer implementation, when it is desired to specify the elastic properties separately from the viscous properties, as in .\n\nA creep experiment is usually easier to perform than a relaxation one, so most data is available as (creep) compliance vs. time. Unfortunately, there is no known closed form for the (creep) compliance in terms of the coefficient of the Prony series. So, if one has creep data, it is not easy to get the coefficients of the (relaxation) Prony series, which are needed for example in. An expedient way to obtain these coefficients is the following. First, fit the creep data with a model that has closed form solutions in both compliance and relaxation; for example the Maxwell-Kelvin model (eq. 7.18-7.19) in or the Standard Solid Model (eq. 7.20-7.21) in (section 7.1.3). Once the parameters of the creep model are known, produce relaxation pseudo-data with the conjugate relaxation model for the same times of the original data. Finally, fit the pseudo data with the Prony series.\n\n## Effect of temperature on viscoelastic behavior\n\nThe secondary bonds of a polymer constantly break and reform due to thermal motion. Application of a stress favors some conformations over others, so the molecules of the polymer will gradually \"flow\" into the favored conformations over time. Because thermal motion is one factor contributing to the deformation of polymers, viscoelastic properties change with increasing or decreasing temperature. In most cases, the creep modulus, defined as the ratio of applied stress to the time-dependent strain, decreases with increasing temperature. Generally speaking, an increase in temperature correlates to a logarithmic decrease in the time required to impart equal strain under a constant stress. In other words, it takes less work to stretch a viscoelastic material an equal distance at a higher temperature than it does at a lower temperature.\n\n## Viscoelastic creep\n\nMain article: Creep (deformation)\n\nWhen subjected to a step constant stress, viscoelastic materials experience a time-dependent increase in strain. This phenomenon is known as viscoelastic creep.\n\nAt a time $t_0$, a viscoelastic material is loaded with a constant stress that is maintained for a sufficiently long time period. The material responds to the stress with a strain that increases until the material ultimately fails, if it is a viscoelastic liquid. If, on the other hand, it is a viscoelastic solid, it may or may not fail depending on the applied stress versus the material's ultimate resistance. When the stress is maintained for a shorter time period, the material undergoes an initial strain until a time $t_1$, after which the strain immediately decreases (discontinuity) then gradually decreases at times $t > t_1$ to a residual strain.\n\nViscoelastic creep data can be presented by plotting the creep modulus (constant applied stress divided by total strain at a particular time) as a function of time. Below its critical stress, the viscoelastic creep modulus is independent of stress applied. A family of curves describing strain versus time response to various applied stress may be represented by a single viscoelastic creep modulus versus time curve if the applied stresses are below the material's critical stress value.\n\nViscoelastic creep is important when considering long-term structural design. Given loading and temperature conditions, designers can choose materials that best suit component lifetimes.\n\n## Measuring viscoelasticity\n\nThough there are many instruments that test the mechanical and viscoelastic response of materials, broadband viscoelastic spectroscopy (BVS) and resonant ultrasound spectroscopy (RUS) are more commonly used to test viscoelastic behavior because they can be used above and below ambient temperatures and are more specific to testing viscoelasticity. These two instruments employ a damping mechanism at various frequencies and time ranges with no appeal to time–temperature superposition. Using BVS and RUS to study the mechanical properties of materials is important to understanding how a material exhibiting viscoelasticity will perform."
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https://www.quizzes.cc/calculator/area/square-miles/130 | [
"### How much is 130 square miles?\n\nConvert 130 square miles. How big is 130 square miles? What is 130 square miles in other units? Convert acres, hectares, square cm, ft, in, km, meters, mi, and yards. To calculate, enter your desired inputs, then click calculate. Some units are rounded since conversions between metric and imperial can be messy.\n\n### Summary\n\nConvert 130 square miles to other units, like acres, hectares, cm2, ft2, in2, km2, meters2, mi2, and square yards."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8162969,"math_prob":0.9971723,"size":435,"snap":"2022-27-2022-33","text_gpt3_token_len":112,"char_repetition_ratio":0.1786543,"word_repetition_ratio":0.0,"special_character_ratio":0.26436782,"punctuation_ratio":0.26530612,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97425705,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T17:31:11Z\",\"WARC-Record-ID\":\"<urn:uuid:d650b0df-c646-47d2-8ed3-84194d7e30f5>\",\"Content-Length\":\"7444\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3119732f-97eb-4967-bc1b-0eb618f73e3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f9a66cb-d520-43d6-ae51-937c3f410261>\",\"WARC-IP-Address\":\"3.93.199.172\",\"WARC-Target-URI\":\"https://www.quizzes.cc/calculator/area/square-miles/130\",\"WARC-Payload-Digest\":\"sha1:WAYY3JIVL2KF2M5XDZV6DBW3BASPOLQK\",\"WARC-Block-Digest\":\"sha1:O34LFLKCPKUEITM7E5UN3UOLAJ2ACLSJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573029.81_warc_CC-MAIN-20220817153027-20220817183027-00431.warc.gz\"}"} |
https://calcforme.com/percentage-calculator/428-is-what-percent-of-33 | [
"# 428 is What Percent of 33?\n\n## 428 is 1296.97 Percent of 33\n\nis what percent of\n%\n\n428 is 1296.97% of 33\n\nCalculation steps:\n\n( 428 ÷ 33 ) x 100 = 1296.97%\n\n### Calculate 428 is What Percent of 33?\n\n• F\n\nFormula\n\n(428 ÷ 33) x 100 = 1296.97%\n\n• 1\n\n428 ÷ 33 = 12.969696969696969\n\n• 2\n\nDecimal to percent\n\n12.969696969696969 x 100 = 1296.97%\n\nExample"
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https://isabelle.in.tum.de/repos/isabelle/diff/c820b3cc3df0/src/HOL/ex/Qsort.thy?revcount=60 | [
"src/HOL/ex/Qsort.thy\n changeset 1151 c820b3cc3df0 parent 969 b051e2fc2e34 child 1376 92f83b9d17e1\n```--- a/src/HOL/ex/Qsort.thy\tWed Jun 21 15:14:58 1995 +0200\n+++ b/src/HOL/ex/Qsort.thy\tWed Jun 21 15:47:10 1995 +0200\n@@ -13,15 +13,15 @@\nrules\n\nqsort_Nil \"qsort le [] = []\"\n-qsort_Cons \"qsort le (x#xs) = qsort le [y:xs . ~le x y] @ \\\n-\\ (x# qsort le [y:xs . le x y])\"\n+qsort_Cons \"qsort le (x#xs) = qsort le [y:xs . ~le x y] @\n+ (x# qsort le [y:xs . le x y])\"\n\n(* computational induction.\nThe dependence of p on x but not xs is intentional.\n*)\nqsort_ind\n- \"[| P([]); \\\n-\\ !!x xs. [| P([y:xs . ~p x y]); P([y:xs . p x y]) |] ==> \\\n-\\ P(x#xs) |] \\\n-\\ ==> P(xs)\"\n+ \"[| P([]);\n+ !!x xs. [| P([y:xs . ~p x y]); P([y:xs . p x y]) |] ==>\n+ P(x#xs) |]\n+ ==> P(xs)\"\nend```"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5206478,"math_prob":0.9943704,"size":657,"snap":"2021-31-2021-39","text_gpt3_token_len":294,"char_repetition_ratio":0.18070444,"word_repetition_ratio":0.33333334,"special_character_ratio":0.56621003,"punctuation_ratio":0.21556886,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9730218,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-02T05:21:19Z\",\"WARC-Record-ID\":\"<urn:uuid:8efd7dab-e4e0-419c-a4d8-5c8090e3eb42>\",\"Content-Length\":\"5868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d364767d-6da0-4217-b156-0e964e1e6752>\",\"WARC-Concurrent-To\":\"<urn:uuid:48002a38-c07b-4120-a110-2ab6a973e878>\",\"WARC-IP-Address\":\"131.159.46.82\",\"WARC-Target-URI\":\"https://isabelle.in.tum.de/repos/isabelle/diff/c820b3cc3df0/src/HOL/ex/Qsort.thy?revcount=60\",\"WARC-Payload-Digest\":\"sha1:NMSCLJ4WNU3FOGBOZCLDYS73P32NS26Z\",\"WARC-Block-Digest\":\"sha1:USA42WNSVBWEV7KISAJ6GIWQMEG7BKRO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154304.34_warc_CC-MAIN-20210802043814-20210802073814-00689.warc.gz\"}"} |
https://www.bi.edu/programmes-and-individual-courses/course-descriptions/introductory-data-science-for-marketing/ | [
"Excerpt from course description\n\n# Introductory Data Science for Marketing\n\n### Introduction\n\nThe course gives the students a detailed overview of statistical inference and basic applications. The basic applications are implemented using Excel and SPSS. The course focuses on fundamental data science issues, on basic probability theory, the logic of hypothesis testing and confidence intervals, and the concept of a statistical model, illustrated via simple and multiple linear regression. A brief introduction to the design of experiments, randomization, and data gathering is given.\n\n### Course content\n\n• Univariate descriptive statistics and plots, the normal distribution.\n• Bivariate descriptive statistics and plots, correlation and least squares estimation. Tables for categorical data, Simpson’s paradox.\n• Causation, randomization, sampling, bias and variability.\n• Basic probability rules, random variables, population means and variances, the law of large numbers.\n• The sampling distribution of a sample mean, the central limit theorem.\n• Confidence intervals and testing under exact normality with a known . Potential problems with tests. Power and inference as a decision.\n• Inference for a mean under more realistic conditions.\n• The simple linear regression model, and inference.\n• The multiple linear regression model, inference and data examples.\n• One way ANOVA.\n\n### Disclaimer\n\nThis is an excerpt from the complete course description for the course. If you are an active student at BI, you can find the complete course descriptions with information on eg. learning goals, learning process, curriculum and exam at portal.bi.no. We reserve the right to make changes to this description."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.82235396,"math_prob":0.9168679,"size":1687,"snap":"2021-31-2021-39","text_gpt3_token_len":318,"char_repetition_ratio":0.112299465,"word_repetition_ratio":0.0,"special_character_ratio":0.18553646,"punctuation_ratio":0.15492958,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99217886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T01:09:07Z\",\"WARC-Record-ID\":\"<urn:uuid:8e082d18-410e-4dce-90ea-cc01eb6f077b>\",\"Content-Length\":\"33173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99fea6af-1b68-4d18-9ed1-5b2a706613bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:3355634f-145c-4340-bc5b-af9731d50f78>\",\"WARC-IP-Address\":\"104.46.38.245\",\"WARC-Target-URI\":\"https://www.bi.edu/programmes-and-individual-courses/course-descriptions/introductory-data-science-for-marketing/\",\"WARC-Payload-Digest\":\"sha1:2VKGNYORGB5HBK43V35GGIA5PQVAVD2J\",\"WARC-Block-Digest\":\"sha1:QHFN4D7COVVQ7P77X65QE4YWKHMKMWQM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058589.72_warc_CC-MAIN-20210928002254-20210928032254-00114.warc.gz\"}"} |
https://link.springer.com/article/10.1007/s41965-021-00079-x | [
"# When catalytic P systems with one catalyst can be computationally complete\n\n## Abstract\n\nCatalytic P systems are among the first variants of membrane systems ever considered in this area. This variant of systems also features some prominent computational complexity questions, and in particular the problem of using only one catalyst in the whole system: is one catalyst enough to allow for generating all recursively enumerable sets of multisets? Several additional ingredients have been shown to be sufficient for obtaining computational completeness even with only one catalyst. In this paper, we show that one catalyst is sufficient for obtaining computational completeness if either catalytic rules have weak priority over non-catalytic rules or else instead of the standard maximally parallel derivation mode, we use the derivation mode maxobjects, i.e., we only take those multisets of rules which affect the maximal number of objects in the underlying configuration.\n\n## Introduction\n\nMembrane systems were introduced in as a multiset-rewriting model of computing inspired by the structure and the functioning of the living cell. During two decades, now membrane computing has attracted the interest of many researchers, and its development is documented in two textbooks, see and . For actual information, see the P systems webpage and the issues of the Bulletin of the International Membrane Computing Society and of the Journal of Membrane Computing.\n\nOne basic feature of P systems already presented in is the maximally parallel derivation mode, i.e., using non-extendable multisets of rules in every derivation step. The result of a computation can be extracted when the system halts, i.e., when no rule is applicable any more. Catalysts are special symbols which allow only one object to evolve in its context (in contrast to promoters, which allow all promoted objects to evolve) and in their basic variant never evolve themselves, i.e., a catalytic rule is of the form $$ca\\rightarrow cv$$, where c is a catalyst, a is a single object and v is a multiset of objects. In contrast, non-catalytic rules in catalytic P systems are non-cooperative rules of the form $$a\\rightarrow v$$.\n\nFrom the beginning, the question how many catalysts are needed in the whole system for obtaining computational completeness has been one of the most intriguing challenges regarding (catalytic) P systems. In , it has already been shown that two catalysts are enough for generating any recursively enumerable set of multisets, without any additional ingredients like a priority relation on the rules as used in the original definition. As already known from the beginning, without catalysts, only regular (semi-linear) sets can be generated when using the standard halting mode, i.e., a result is extracted when the system halts with no rule being applicable any more. As shown, for example, in , using various additional ingredients, i.e., additional control mechanisms, one catalyst can be sufficient: in P systems with label selection, only rules from one set of a finite number of sets of rules in each computation step are used; in time-varying P systems, the available sets of rules change periodically with time. On the other hand, for catalytic P systems with only one catalyst, a lower bound has been established in : P systems with one catalyst can simulate partially blind register machines, i.e., they can generate more than just semi-linear sets.\n\nIn , we returned to the idea of using a priority relation on the rules, but took only a very weak form of such a priority relation: we only required that overall in the system, catalytic rules have weak priority over non-catalytic rules. This means that the catalyst c must not stay idle if the current configuration contains an object a with which it may cooperate in a rule $$ca\\rightarrow cv$$; all remaining objects evolve in the maximally parallel way with non-cooperative rules. On the other hand, if the current configuration does not contain an object a with which the catalyst c may cooperate in a rule $$ca\\rightarrow cv$$, c may stay idle and all objects evolve in the maximally parallel way with non-cooperative rules. Even without using more than this, weak priority of catalytic rules over the non-catalytic (non-cooperative) rules, we could establish computational completeness for catalytic P systems with only one catalyst. In this paper, we recall the result from , but show a somehow much stronger result using a similar construction as in : we show computational completeness for catalytic P systems with only one catalyst using the derivation mode maxobjects, i.e., we only take those multisets of rules which affect the maximal number of objects in the underlying configuration.\n\n## Definitions\n\nFor an alphabet V, by $$V^{*}$$, we denote the free monoid generated by V under the operation of concatenation, i.e., containing all possible strings over V. The empty string is denoted by $$\\lambda .$$ A multiset M with underlying set A is a pair (Af) where $$f:\\,A\\rightarrow \\mathbb {N}$$ is a mapping. If $$M=(A,f)$$ is a multiset, then its support is defined as $$supp(M)=\\{x\\in A \\,|\\,f(x)> 0\\}$$. A multiset is empty (respectively, finite) if its support is the empty set (respectively, a finite set). If $$M=(A,f)$$ is a finite multiset over A and $$supp(M)=\\{ a_1,\\ldots ,a_k\\}$$, then it can also be represented by the string $$a_1^{f(a_1)} \\dots a_k^{f(a_k)}$$ over the alphabet $$\\{ a_1,\\ldots ,a_k\\}$$, and, moreover, all permutations of this string precisely identify the same multiset M. For further notions and results in formal language theory, we refer to textbooks like and .\n\n### Register machines\n\nRegister machines are well-known universal devices for generating or accepting or even computing with sets of vectors of natural numbers.\n\n### Definition 1\n\nA register machine is a construct\n\n\\begin{aligned} M=\\left( m,B,l_{0},l_{h},P\\right) \\end{aligned}\n\nwhere\n\n• m is the number of registers,\n\n• B is a finite set of labels,\n\n• $$l_{0}\\in B$$ is the initial label,\n\n• $$l_{h}\\in B$$ is the final label, and\n\n• P is the set of instructions bijectively labeled by elements of B.\n\nThe instructions of M can be of the following forms:\n\n• $$p:\\left( \\textit{ADD}\\left( r\\right) ,q,s\\right)$$, with $$p\\in B\\setminus \\left\\{ l_{h}\\right\\}$$, $$q,s\\in B$$, $$1\\le r\\le m$$.\n\nIncrease the value of register r by one, and non-deterministically jump to instruction q or s.\n\n• $$p:\\left( \\textit{SUB}\\left( r\\right) ,q,s\\right)$$, with $$p\\in B\\setminus \\left\\{ l_{h}\\right\\}$$, $$q,s\\in B$$, $$1\\le r\\le m$$.\n\nIf the value of register r is not zero, then decrease the value of register r by one (decrement case) and jump to instruction q, otherwise jump to instruction s (zero-test case).\n\n• $$l_{h}:HALT$$.\n\nStop the execution of the register machine.\n\nA configuration of a register machine is described by the contents of each register and by the value of the current label, which indicates the next instruction to be executed. M is called deterministic if the ADD-instructions all are of the form $$p:\\left( \\textit{ADD}\\left( r\\right) ,q,q\\right)$$, simply also written as $$p:\\left( \\textit{ADD}\\left( r\\right) ,q\\right)$$.\n\nIn the accepting case, a computation starts with the input of an l-vector of natural numbers in its first l registers and by executing the first instruction of P (labeled with $$l_{0}$$); it terminates with reaching the HALT-instruction. Without loss of generality, we may assume all registers to be empty at the end of the computation.\n\nIn the generating case, a computation starts with all registers being empty and by executing the first instruction of P (labeled with $$l_{0}$$); it terminates with reaching the HALT-instruction and the output of a k-vector of natural numbers in its last k registers. Without loss of generality, we may assume all registers except the last k output registers to be empty at the end of the computation.\n\nIn the computing case, a computation starts with the input of an l-vector of natural numbers in its first l registers and by executing the first instruction of P (labeled with $$l_{0}$$); it terminates with reaching the HALT-instruction and the output of a k-vector of natural numbers in its last k registers. Without loss of generality, we may assume all registers except the last k output registers to be empty at the end of the computation.\n\nFor useful results on the computational power of register machines, we refer to ; for example, to prove our main theorem, we need the following formulation of results for register machines generating or accepting recursively enumerable sets of vectors of natural numbers with k components or computing partial recursive relations on vectors of natural numbers:\n\n### Proposition 1\n\nDeterministic register machines can accept any recursively enumerable set of vectors of natural numbers with l components using precisely $$l+2$$ registers. Without loss of generality, we may assume that at the end of an accepting computation, all registers are empty.\n\n### Proposition 2\n\nRegister machines can generate any recursively enumerable set of vectors of natural numbers with k components using precisely $$k+2$$ registers. Without loss of generality, we may assume that at the end of an accepting computation the first two registers are empty, and, moreover, on the output registers, i.e., the last k registers, no $$\\textit{SUB}$$ -instruction is ever used.\n\n### Proposition 3\n\nRegister machines can compute any partial recursive relation on vectors of natural numbers with l components as input and vectors of natural numbers with k components as output using precisely $$l+2+k$$ registers, where without loss of generality, we may assume that at the end of a successful computation the first $$l+2$$ registers are empty, and, moreover, on the output registers, i.e., the last k registers, no $$\\textit{SUB}$$ -instruction is ever used.\n\nIn all cases, it is essential that the output registers never need to be decremented.\n\n### Partially blind register machines\n\nWe now consider one-way non-deterministic machines which have registers allowed to hold positive or negative integers and which accept by final state with all registers being zero. Such machines are called blind if their actions depend on the state (label) and the input only and not on the register configuration itself. They are called partially blind if they block when any register is negative (i.e., only non-negative register contents is allowed) but do not know whether or not any of the registers contains zero.\n\n### Definition 2\n\nA partially blind register machine is a construct\n\n\\begin{aligned} M=\\left( m,B,l_{0},l_{h},P\\right) \\end{aligned}\n\nwhere\n\n• m is the number of registers,\n\n• B is a finite set of labels,\n\n• $$l_{0}\\in B$$ is the initial label,\n\n• $$l_{h}\\in B$$ is the final label, and\n\n• P is the set of instructions bijectively labeled by elements of B.\n\nThe instructions of M can be of the following forms:\n\n• $$p:\\left( \\textit{ADD}\\left( r\\right) ,q,s\\right)$$, with $$p\\in B\\setminus \\left\\{ l_{h}\\right\\}$$, $$q,s\\in B$$, $$1\\le r\\le m$$.\n\nIncrease the value of register r by one, and non-deterministically jump to instruction q or s.\n\n• $$p:\\left( \\textit{SUB}\\left( r\\right) ,q\\right)$$, with $$p\\in B\\setminus \\left\\{ l_{h}\\right\\}$$, $$q\\in B$$, $$1\\le r\\le m$$.\n\nIf the value of register r is not zero, then decrease the value of register r by one and jump to instruction q, otherwise abort the computation.\n\n• $$l_{h}:HALT$$.\n\nStop the execution of the register machine.\n\nAgain, a configuration of a partially blind register machine is described by the contents of each register and by the value of the current label, which indicates the next instruction to be executed.\n\nA computation works as for a register machine, yet with the restriction that a computation is aborted if one tries to decrement a register which is zero. Moreover, computing, accepting or generating now also requires all registers (except output registers) to be empty at the end of the computation.\n\n### Remark 1\n\nFor any register machine (even for a blind or a partially blind one), without loss of generality, we may assume that the first instruction is an $$\\textit{ADD}$$-instruction: in fact, given a register machine $$M=\\left( m,B,l_{0},l_{h},P\\right)$$ with having a $$\\textit{SUB}$$-instruction as its first instruction, we can immediately construct an equivalent register machine $$M^{\\prime }$$ which starts with an increment immediately followed by a decrement of the first register:\n\n\\begin{aligned} M^{\\prime }= & \\ \\left( m,B^{\\prime },l_{0}^{\\prime },l_{h},P^{\\prime }\\right) ,\\\\ B^{\\prime }= & \\ B\\cup \\{ l_{0}^{\\prime }, l_{0}^{\\prime \\prime }\\},\\\\ P^{\\prime }= & \\ P\\cup \\{l_{0}^{\\prime }:(\\textit{ADD}(1),l_{0}^{\\prime \\prime },l_{0}^{\\prime \\prime }),\\ l_{0}^{\\prime \\prime }:(\\textit{SUB}(1),l_{0},l_{0}) \\, \\}. \\end{aligned}\n\n### Catalytic P systems\n\nAs in , the following definition cites Definition 4.1 in Chapter 4 of .\n\n### Definition 3\n\nAn extended catalytic P system of degree $$m\\ge 1$$ is a construct\n\n\\begin{aligned} \\varPi =(O,C,\\mu ,w_1,\\dots ,w_{m},R_1,\\dots ,R_{m},i_{0}) \\end{aligned}\n\nwhere\n\n• O is the alphabet of objects;\n\n• $$C\\subseteq O$$ is the alphabet of catalysts;\n\n• $$\\mu$$ is a membrane structure of degree m with membranes labeled in a one-to-one manner with the natural numbers $$1,\\dots ,m$$;\n\n• $$w_1,\\dots ,w_{m}\\in O^*$$ are the multisets of objects initially present in the m regions of $$\\mu$$;\n\n• $$R_{i}$$, $$1\\le i\\le m$$, are finite sets of evolution rules over O associated with the regions $$1,2,\\dots ,m$$ of $$\\mu$$; these evolution rules are of the forms $$ca\\rightarrow cv$$ or $$a\\rightarrow v$$, where c is a catalyst, a is an object from $$O\\setminus C$$, and v is a string from $$((O\\setminus C) \\times \\{ here,out,in\\})^*$$;\n\n• $$i_{0}\\in \\{0,1,\\dots ,m\\}$$ indicates the output region of $$\\varPi$$ (0 indicates the environment).\n\nThe membrane structure and the multisets in $$\\varPi$$ at the current moment of a computation constitute a configuration of the P system; the initial configuration is given by the initial multisets $$w_1,\\dots ,w_{m}$$. A transition between two configurations is governed by the application of the evolution rules, which is done in a given derivation mode. In this paper, we consider the maximally parallel derivation mode (max for short), i.e., only applicable multisets of rules which cannot be extended by further rules are to be applied to the objects in all membrane regions, as well as the derivation mode maxobjects where in every membrane region only applicable multisets of rules which affect the maximal number of objects are to be applied. We remark that all the multisets which can be used in the derivation mode maxobjects can also be used in the derivation mode max.\n\nThe application of a rule $$u\\rightarrow v$$ in a region containing a multiset M results in subtracting from M the multiset identified by u, and then in adding the multiset identified by v. The objects can eventually be transported through membranes due to the targets in (i.e., to the region of an inner membrane) and out (i.e., to the region of the surrounding membrane) or else stay in the membrane where the rule has been applied (target here). We refer to for further details and examples.\n\nThe P system continues with applying multisets of rules according to the derivation mode until there remain no applicable rules in any region of $$\\varPi$$. Then, the system halts. We consider the number of objects from $$O\\setminus C$$ contained in the output region $$i_{0}$$ at the moment when the system halts as the result of the underlying computation of $$\\varPi$$. The system is called extended since the catalytic objects in C are not counted to the result of a computation. Yet, as often done in the literature, in the following, we will omit the term extended and just speak of catalytic P systems, especially as we will restrict ourselves to P systems with only one catalyst.\n\nThe set of results of all computations possible in $$\\varPi$$ using the derivation mode $$\\delta$$ is called the set of natural numbers generated by $$\\varPi$$ using $$\\delta$$ and it is denoted by $$N(\\varPi , \\delta )$$ if we only count the total number of objects in the output membrane; if we distinguish between the multiplicities of different objects, we obtain a set of vectors of natural numbers denoted by $$Ps(\\varPi , \\delta )$$.\n\n### Remark 2\n\nAs in this paper, we only consider catalytic P systems with only one catalyst, without loss of generality, we can restrict ourselves to one-membrane catalytic P systems with the single catalyst in the skin membrane, by taking into account the well-known flattening process, e.g., see .\n\n### Remark 3\n\nFinally, we make the convention that a one-membrane catalytic P system with the single catalyst in the skin membrane and with internal output in the skin membrane, not taking into account the single catalyst c for the results, throughout the rest of the paper will be described without specifying the trivial membrane structure or the output region (assumed to be the skin membrane), i.e., we will just write\n\n\\begin{aligned} \\varPi =\\left( O,\\{ c\\},w,R\\right) \\end{aligned}\n\nwhere O is the set of objects, c is the single catalyst, w is the initial input specifying the initial configuration, and R is the set of rules.\n\nAs already mentioned earlier, the following result was shown in , establishing a lower bound for the computational power of catalytic P systems with only one catalyst:\n\n### Proposition 4\n\nCatalytic P systems with only one catalyst working in the derivation mode max have at least the computational power of partially blind register machines.\n\n### Example 1\n\nIn , it was shown that the vector set\n\n\\begin{aligned} S=\\left\\{ \\left( n,m\\right) \\mid 0\\le n, n\\le m\\le 2^n\\right\\} \\end{aligned}\n\n(which is not semi-linear) can be generated by a P system $$\\varPi$$ working in the derivation mode max with only one catalyst and 19 rules as the multiset language\n\n\\begin{aligned} L=\\left\\{ a^{n} b^{m}\\mid 0\\le n, n\\le m\\le 2^n\\right\\} , \\end{aligned}\n\ni.e., $$Ps(\\varPi , max) = L$$.\n\n## Weak priority of catalytic rules\n\nIn this section, we now study catalytic P systems with only one catalyst in which the catalytic rules have weak priority over the non-catalytic rules.\n\n### Example 2\n\nTo illustrate this weak priority of catalytic rules over the non-catalytic rules, consider the rules $$ca \\rightarrow cb$$ and $$a \\rightarrow d$$. If the current configuration contains $$k > 0$$ copies of a, then the catalytic rule $$ca \\rightarrow cb$$ must be applied to one of the copies, while the rest of objects a may be taken up by the non-catalytic rule $$a \\rightarrow d$$. In particular, if $$k = 1$$, only $$ca \\rightarrow cb$$ may be applied.\n\nWe would like to highlight the fact that weak priority of catalytic rules is much weaker than the general weak priority, as the priority relation is only constrained by the types of rules.\n\n### Remark 4\n\nThe reverse weak priority, i.e., non-catalytic rules having priority over catalytic rules, is useless for our purposes, since it is equivalent to removing all catalytic rules for which there are non-catalytic rules with the same symbol on the left-hand side of the rule. In that way, we just end up with an even restricted variant of P systems with only one catalyst, although it might still be interesting to characterize the family of multiset languages obtained by this restricted variant of P systems: obviously, the lower bound still are the semi-linear languages, as for generating those we only need catalytic rules with one catalyst; on the other hand, the result stated in Proposition 4 cannot be obtained anymore with the methods used in .\n\n### Computational completeness with weak priority\n\nWe now are going to show that catalytic P systems with one catalyst only and with weak priority of catalytic rules are computationally complete.\n\n### Theorem 1\n\nCatalytic P systems with only one catalyst and with weak priority of catalytic rules over the non-cooperative rules when working in the derivation mode max are computationally complete.\n\n### Proof\n\nGiven an arbitrary register machine $$M=\\left( m,B,l_{0},l_{h},P\\right)$$, we will construct a corresponding catalytic P system with one membrane and one catalyst $$\\varPi = (O, \\{ c \\}, w, R)$$ simulating M. Without loss of generality, we may assume that, depending on its use as an accepting or generating or computing device, the register machine M, as stated in Proposition 1, Proposition 2, and Proposition 3, fulfills the condition that on the output registers, we never apply any $$\\textit{SUB}$$-instruction. The following proof is given for the most general case of a register machine computing any partial recursive relation on vectors of natural numbers with l components as input and vectors of natural numbers with k components as output using precisely $$l+2+k$$ registers, where without loss of generality, we may assume that at the end of a successful computation, the first $$l+2$$ registers are empty, and, moreover, on the output registers, i.e., the last k registers, no $$\\textit{SUB}$$-instruction is ever used. In fact, the proof works for any number n of decrementable registers, no matter how many of them are the l input registers and the working registers, respectively.\n\nThe main idea behind our construction is that all the symbols except the catalyst c and the output symbols (representing the contents of the output registers) go through a cycle of length $$n+2$$ where n is the number of decrementable registers of the simulated register machine. When the symbols are traversing the r-th section of the n sections, they “know” that they are to probably simulate a $$\\textit{SUB}$$-instruction on register r of the register machine M.\n\nThe alphabet O of symbols includes register symbols $$(a_{r},i)$$ for every decrementable register r of the register machine and only the register symbol $$a_{r}$$ for each of the k output registers r, $$m-k+1 \\le r \\le m$$.\n\nNow let $$B_{\\textit{ADD}}$$ denote the set of labels of $$\\textit{ADD}$$-instructions $$p:(\\textit{ADD}(r), q, s)$$ of arbitrary registers r and $$B_{\\textit{SUB}(r)}$$ denote the set of labels of all $$\\textit{SUB}$$-instructions $$p:(\\textit{SUB}(r), q, s)$$ of decrementable registers r. For every $$\\textit{ADD}$$-instruction $$p:(\\textit{ADD}(r), q, s)$$, i.e., $$p\\in B_{\\textit{ADD}(r)}$$, of the register machine we take the state symbols (pi), $$1 \\le i \\le n+2$$, into O. For every $$\\textit{SUB}$$-instruction $$p:(\\textit{SUB}(r), q, s)$$ of the register machine, i.e., $$p\\in B_{\\textit{SUB}(r)}$$, we need the state symbols (pi) for $$1\\le i\\le r+1$$ as well as $$(p,j)^-$$ and $$(p,j)^0$$ for $$r+2\\le j\\le n+2$$. Moreover, we use a decrement witness symbol e as well as the catalyst c and the trap symbol $$\\#$$.\n\nObserving that $$n=m-k$$, in total, we get the following set of objects:\n\n\\begin{aligned} O= & \\ \\{ a_{r} \\mid n+1 \\le r \\le m \\} \\\\ \\cup& \\ \\{ (a_{r},i) \\mid 1 \\le r \\le n,\\, 1 \\le i \\le n+2\\} \\\\ \\cup& \\ \\{(p,i) \\mid p \\in B_{\\textit{ADD}},\\, 1 \\le i \\le n+2\\} \\\\ \\cup & \\ \\{(p,i) \\mid p\\in B_{\\textit{SUB}(r)},\\, 1 \\le i \\le r+1,\\, 1\\le r\\le n\\} \\\\ \\cup& \\ \\{(p,i)^-, (p,i)^0 \\mid p\\in B_{\\textit{SUB}(r)},\\, r+2 \\le i \\le n+2,\\, 1\\le r\\le n\\} \\\\ \\cup& \\ \\{c, e, \\#\\}. \\end{aligned}\n\nThe starting configuration of $$\\varPi$$ is\n\n\\begin{aligned} w = c (l_{0},1) \\alpha _0, \\end{aligned}\n\nwhere $$l_0$$ is the starting label of the machine and $$\\alpha _0$$ is the multiset encoding the initial values of the registers.\n\nAll register symbols $$a_{r}$$, $$1 \\le r\\le n$$, representing the contents of decrementable registers, are equipped with the rules evolving them throughout the whole cycle:\n\n\\begin{aligned}&(a_{r},i) \\rightarrow (a_{r},i+1), 1 \\le r \\le n+1; \\qquad (a_{r},n+2) \\rightarrow (a_{r},1). \\end{aligned}\n(1)\n\nThe construction also includes the trap rule $$\\# \\rightarrow \\#$$: once the trap symbol $$\\#$$ is introduced, it will always keep the system busy and prevent it from halting and thus from producing a result.\n\nFor simulating $$\\textit{ADD}$$-instructions, we need the following rules:\n\nIncrement $$p:(\\textit{ADD}(r), q, s)$$:\n\nThe (variants of the) symbol p cycles together with all the other symbols, always involving the catalyst:\n\n\\begin{aligned} c (p,i) \\rightarrow c (p,i+1), \\quad 1 \\le i \\le n+1. \\end{aligned}\n(2)\n\nAt the end of the cycle, the register r is incremented and the non-deterministic jump to q or s occurs: for r being a decrementable register, we take\n\n\\begin{aligned} c (p,n+2) \\rightarrow c (q,1) (a_{r},1), \\qquad c (p,n+2) \\rightarrow c (s,1) (a_{r},1), \\end{aligned}\n(3)\n\nwhereas for r being a register never to be decremented, we take\n\n\\begin{aligned} c (p,n+2) \\rightarrow c (q,1) a_{r}, \\qquad c (p,n+2) \\rightarrow c (s,1) a_{r}. \\end{aligned}\n(4)\n\nThe output symbols need not undergo the cycle, in fact, they must not do that because otherwise the computation would never stop. When the computation of the register machine halts, only output symbols (of course, besides the catalyst c) will be present, as we have assumed that at the end of a computation all decrementable registers will be empty, i.e., no cycling symbols will be present any more in the P system. Finally, we have to mention that if q or s is the final label $$l_{h}$$, then we take $$\\lambda$$ instead, which means that also the P system will halt, because, as already explained above, the only symbols left in the configuration (besides the catalyst c) will be output symbols, for which no rules exist.\n\nThe state symbol is not allowed to evolve without the catalyst:\n\n\\begin{aligned} (p,i) \\rightarrow \\#, \\quad 1 \\le i \\le n+2. \\end{aligned}\n(5)\n\nHence, in that way, it is guaranteed that the catalyst cannot be used in another way, i.e., affecting a symbol $$(a_{r},i)$$ as explained below during the simulation of a $$\\textit{SUB}$$-instruction on register r.\n\nDecrement and zero-test $$p:(\\textit{SUB}(r), q, s)$$:\n\nThe simulation of a $$\\textit{SUB}$$-instruction is carried out in three steps of the cycle, i.e., in steps r, $$r+1$$, and $$r+2$$.\n\nBefore reaching the simulation phase for register r, i.e., step r of the cycle, the state symbol goes through the cycle, necessarily involving the catalyst:\n\n\\begin{aligned} c (p,i) \\rightarrow c (p,i+1) \\quad > \\quad (p,i) \\rightarrow \\#, \\qquad 1 \\le i < r. \\end{aligned}\n(6)\n\nBy definition, in the P system $$\\varPi$$ we construct, catalytic rules have priority over the non-cooperative rules, i.e., the catalytic rule $$c (p,i) \\rightarrow c (p,i+1)$$ has priority over the non-catalytic rule $$(p,i) \\rightarrow \\#$$; we indicate this general priority relation for the object (pi) by the sign > (and we use < for the reverse relation) in order to make the situation even clearer.\n\nIn the first step of the simulation phase for register r, i.e., in step r, the state symbol releases the catalyst to try to perform the decrement and to produce a witness symbol e if register r is not empty:\n\n\\begin{aligned} (p,r) \\rightarrow (p,r+1), \\qquad c (a_{r},r) \\rightarrow c e. \\end{aligned}\n(7)\n\nNote that due to the counters identifying the position of the register symbols in the cycle, it is guaranteed that the catalytic rule transforming $$(a_{r},r)$$ picks the correct register symbol. Furthermore, due to the priority of the catalytic rules, one of the register symbols $$(a_{r},r)$$ must be transformed by the catalytic rule if present, instead of continuing along its cycle.\n\nIn the second step of simulation phase r, i.e., in step $$r+1$$, the detection of the possible decrement happens. The outcome is stored in the state symbol:\n\n\\begin{aligned}&c e \\rightarrow c \\quad > \\quad e \\rightarrow \\#, \\nonumber \\\\&(p,r+1) \\rightarrow (p,r+2)^{-} \\quad < \\quad c (p,r+1) \\rightarrow c (p,r+2)^{0}. \\end{aligned}\n(8)\n\nIf in the first step of the simulation phase, the catalyst did manage to decrement the register, it produced e. Thus, in the second step, i.e., in step $$r+1$$, the catalyst has three choices:\n\n1. 1.\n\nthe catalyst c correctly erases e, because otherwise this symbol will trap the computation, and to the program symbol $$(p,r+1)$$ the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$ must be applied as the catalyst is not available for using it with the catalytic rule $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$; all register symbols evolve in the usual way;\n\n2. 2.\n\nthe catalyst c takes the program symbol $$(p,r+1)$$ using the rule $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$, thus forcing the object e to be trapped by the rule $$e \\rightarrow \\#$$, and all register symbols evolve in the usual way; this variant cannot lead to a halting computation due to the introduction of the trap symbol $$\\#$$;\n\n3. 3.\n\nthe catalyst c takes a register object $$(a_{r+1},r+1)$$, thus leaving the object e to be trapped by the rule $$e \\rightarrow \\#$$, the program symbol $$(p,r+1)$$ evolves with the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$, and all other register objects evolve in the usual way; this variant again cannot lead to a halting computation due to the introduction of the trap symbol $$\\#$$.\n\nOn the other hand, if register r is empty, no object e is generated, and the catalyst c has only two choices:\n\n1. 1.\n\nthe catalyst c takes the program symbol $$(p,r+1)$$ using the rule $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$, and all register symbols evolve in the usual way;\n\n2. 2.\n\nthe catalyst c takes a register object $$(a_{r+1},r+1)$$ thereby generating e, the program symbol $$(p,r+1)$$ evolves with the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$, and all other register objects evolve in the usual way; this variant leads to the situation that e will be trapped in step $$r+2$$, as otherwise the program symbol will be trapped, hence, this variant in any case cannot lead to a halting computation due to the introduction of the trap symbol $$\\#$$. We mention that in case no register object $$(a_{r+1},r+1)$$ is present we have to apply case 1 and thus have a correct computation step.\n\nObserve that in this case of register r being empty, both rules $$(p,r+1) \\rightarrow (p,r+2)^{-}$$ and $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$ would be applicable, but due to the priority of the catalytic rules, the second rule must be preferred, thus producing $$(p,r+2)^{0}$$. Therefore, the superscript of the state symbol correctly reflects the outcome of the $$\\textit{SUB}$$-instruction: it is “−” if the decrement succeeded, and “0” if it did not.\n\nAfter the simulation of the $$\\textit{SUB}$$-instruction $$p:(\\textit{SUB}(r), q, s)$$ on register r, the state symbols evolve to the end of the cycle and produce the corresponding next state symbols:\n\n\\begin{aligned}&c(p,i)^{-} \\rightarrow c(p,i+1)^{-} ,\\ \\ r+2 \\le i \\le n+1, \\qquad c(p,n+2)^{-} \\rightarrow c(q,1), \\nonumber \\\\&c(p,i)^{0} \\rightarrow c(p,i+1)^{0} ,\\ \\ r+2 \\le i \\le n+1, \\qquad c(p,n+2)^{0} \\rightarrow c(s,1),\\nonumber \\\\&(p,i)^{-} \\rightarrow \\#, \\ (p,i)^{0} \\rightarrow \\#, \\ r+2 \\le i \\le n+2. \\end{aligned}\n(9)\n\nThe additional two steps $$n+1$$ and $$n+2$$ are needed to correctly finish the decrement and zero-test cases even for $$r=n$$.\n\nFinally, we again mention that if q or s is the final label $$l_{h}$$, then we take $$\\lambda$$ instead, which means that not only the register machine but also the P system halts, because, as already explained above, the only symbols — except the single catalyst c — left in the configuration will be output symbols, for which no rules exist. $$\\square$$\n\nThe proof elaborated above is an improved version of the proof given in . In this new version of the proof the simulation of the decrement case on register r still takes two steps, but as the second step of the simulation can overlap with the first step of simulating the decrement case on register $$r+1$$ in the next cycle, we only need $$n+2$$ steps (the first $$+1$$ comes from the second step for register n in the decrement case, the second $$+1$$ comes from the third step for register n in the zero-test case) instead of 2n steps as needed in the proof elaborated in .\n\nMoreover, we want to emphasize that the simulation is also a specific kind of deterministic: the only non-deterministic choice happens between a rule producing a trap symbol $$\\#$$ and another one which does not introduce $$\\#$$. This means that the appearance of the trap symbol may immediately abort the computation, which is the concept used for toxic P systems as introduced in . Using the trap symbol $$\\#$$ as such a toxic object, the only successful computations are those which simulate register machines in a quasi-deterministic way with a look-ahead of one, i.e., considering all possible configurations computable from a given one, there is at most one successful continuation of the computation.\n\nIt remains a challenging question whether the length of the cycle now being $$n+2$$ can still be reduced; in fact, as we will immediately show below, the length of the cycle can even be reduced to n in case we have at least three decrementable registers and to $$n+1$$ in case we have at least two decrementable registers. As for n decrementable registers we only need a cycle of length n and in the version of the proof elaborated below we do not need extra steps at the beginning or the end, this proof technique itself does not seem to allow for further improvements.\n\n### Theorem 2\n\nFor any register machine with at least three decrementable registers, we can construct a catalytic P system with only one catalyst and with weak priority of catalytic rules over the non-cooperative rules and working in the derivation mode max which can simulate every step of the register machine in n steps where n is the number of decrementable registers.\n\n### Proof\n\nIn the proof of Theorem 1, the last two steps in the cycle of length $$n+2$$ were only needed to capture the second and third step in the simulation of $$\\textit{SUB}$$-instructions of the n-th register, and to capture the third step in the simulation of $$\\textit{SUB}$$-instructions of register $$n-1$$.\n\nThe new idea now is to shift the second simulation step in the case of register $$n-1$$ to the first step of the next cycle and to shift the first and the second simulation step in the case of register n to the first and second step of the next cycle. Yet, in this case, we have to guarantee that after a $$\\textit{SUB}$$-instruction on register $$n-1$$ or n the next instruction to be simulated is not a $$\\textit{SUB}$$-instruction on register 1 or 2, respectively.\n\nOne solution for this problem is to use a similar trick as already elaborated in Remark 1: we not only do not start with a $$\\textit{SUB}$$-instruction, but we also change the register machine program in such a way that after a $$\\textit{SUB}$$-instruction on register $$n-1$$ or n two intermediate instructions are introduced, i.e., as in Remark 1, we use an $$\\textit{ADD}$$-instruction on register 1 immediately followed by a $$\\textit{SUB}$$-instruction on register 1, whose simulation will end at most in step n, as we have assumed $$n\\ge 3$$.\n\nWe then may simulate the resulting register machine fulfilling these additional constraints $$M=\\left( m,B,l_{0},l_{h},P\\right)$$ by a corresponding catalytic P system with one membrane and one catalyst $$\\varPi = (O, \\{ c \\}, w, R)$$. Without loss of generality, we may again assume that, depending on its use as an accepting or generating or computing device, the register machine M, as stated in Proposition 1, Proposition 2, and Proposition 3, fulfills the condition that on the output registers, we never apply any $$\\textit{SUB}$$-instruction. As in the proof of Theorem 2, also here we take the most general case of a register machine computing a partial recursive function on vectors of natural numbers with l components as input and vectors of natural numbers with k components as output using n decrementable registers, where without loss of generality, we may assume that at the end of a successful computation the first n registers are empty, and, moreover, on the output registers, i.e., the last k registers, no $$\\textit{SUB}$$-instruction is ever used.\n\nThe alphabet O of symbols now reduces to the following set, as the length of the cycle now is only n:\n\n\\begin{aligned} O= & \\ \\{ a_{r} \\mid n+1 \\le r \\le m \\} \\\\ \\cup &\\ \\{ (a_{r},i) \\mid 1 \\le r \\le n,\\, 1 \\le i \\le n\\} \\\\ \\cup& \\ \\{(p,i) \\mid p \\in B_{\\textit{ADD}},\\, 1 \\le i \\le n\\} \\\\ \\cup& \\ \\{(p,i) \\mid p\\in B_{\\textit{SUB}(r)},\\, 1 \\le i \\le r+1,\\, 1 \\le r \\le n\\} \\\\ \\cup& \\ \\{(p,i)^-, (p,i)^0 \\mid p\\in B_{\\textit{SUB}(r)},\\, r+2 \\le i \\le n, \\, 1 \\le r \\le n-2\\} \\\\ \\cup& \\ \\{c, e, \\#\\}. \\end{aligned}\n\nThe construction includes the trap rule $$\\# \\rightarrow \\#$$ which will always keep the system busy and prevent it from halting and thus from producing a result as soon as the trap symbol $$\\#$$ has been introduced.\n\nThe sets of rules introduced by Eqs. 1, 2, 3, 4, 5, 6, 7, 8, and 9 read as follows:\n\n\\begin{aligned} (a_{r},i) \\rightarrow (a_{r},i+1), 1 \\le r <n; \\qquad (a_{r},n) \\rightarrow (a_{r},1). \\end{aligned}\n(10)\n\nFor simulating $$\\textit{ADD}$$-instructions, we need the following rules:\n\nIncrement $$p:(\\textit{ADD}(r), q, s)$$:\n\n\\begin{aligned} c (p,i) \\rightarrow c (p,i+1), \\quad 1 \\le i < n. \\end{aligned}\n(11)\n\nIf r is a decrementable register:\n\n\\begin{aligned} c (p,n) \\rightarrow c (q,1) (a_{r},1), \\qquad c (p,n) \\rightarrow c (s,1) (a_{r},1), \\end{aligned}\n(12)\n\nIf r is an output register:\n\n\\begin{aligned} c (p,n) \\rightarrow c (q,1) a_{r}, \\qquad c (p,n) \\rightarrow c (s,1) a_{r} \\end{aligned}\n(13)\n\nEnforcing the use of the catalyst:\n\n\\begin{aligned} (p,i) \\rightarrow \\#, \\quad 1 \\le i \\le n. \\end{aligned}\n(14)\n\nFor simulating $$\\textit{SUB}$$-instructions, we need the following rules:\n\nDecrement and zero-test $$p:(\\textit{SUB}(r), q, s)$$:\n\n\\begin{aligned}&c (p,i) \\rightarrow c (p,i+1) \\quad > \\quad (p,i) \\rightarrow \\#, \\qquad 1 \\le i < r. \\end{aligned}\n(15)\n\\begin{aligned}&(p,r) \\rightarrow (p,r+1), \\qquad c (a_{r},r) \\rightarrow c e. \\end{aligned}\n(16)\n\nIf $$r<n-1$$:\n\n\\begin{aligned}&c e \\rightarrow c \\quad > \\quad e \\rightarrow \\#,\\nonumber \\\\&\\quad (p,r+1) \\rightarrow (p,r+2)^{-} \\quad < c (p,r+1) \\rightarrow c (p,r+2)^{0}. \\end{aligned}\n(17)\n\\begin{aligned}&c(p,i)^{-} \\rightarrow c(p,i+1)^{-} ,\\ \\ r+2 \\le i< n, \\qquad c(p,n)^{-} \\rightarrow c(q,1), \\nonumber \\\\&\\quad c(p,i)^{0} \\rightarrow c(p,i+1)^{0} ,\\ \\ r+2 \\le i < n, \\qquad c(p,n)^{0} \\rightarrow c(s,1),\\nonumber \\\\&\\quad (p,i)^{-} \\rightarrow \\#, \\ (p,i)^{0} \\rightarrow \\#, \\ r+2 \\le i \\le n. \\end{aligned}\n(18)\n\nIf $$r=n-1$$:\n\n\\begin{aligned}&c e \\rightarrow c \\quad > \\quad e \\rightarrow \\#, \\nonumber \\\\&\\quad (p,n) \\rightarrow (q,1) \\quad < \\quad c (p,n) \\rightarrow c (s,1). \\end{aligned}\n(19)\n\nIf $$r=n$$:\n\n\\begin{aligned}&c e \\rightarrow c \\quad > \\quad e \\rightarrow \\#, \\nonumber \\\\&\\quad (p,n+1) \\rightarrow (q,2) \\quad < c (p,n+1) \\rightarrow c (s,2). \\end{aligned}\n(20)\n\nAs we have guaranteed that in the case of $$r=n-1$$ or $$r=n$$, the next instruction to be simulated is an $$\\textit{ADD}$$-instruction, starting with the simulation of the next instruction later in the first or second step of the next cycle makes no harm. $$\\square$$\n\nIf we only have two decrementable registers, then we need one additional step at the end; we leave the details of the proof, following the constructions elaborated in the preceding two proofs, to the interested reader:\n\n### Theorem 3\n\nFor any register machine with at least two decrementable registers, we can construct a catalytic P system with only one catalyst and with weak priority of catalytic rules over the non-cooperative rules and working in the derivation mode max which can simulate every step of the register machine in $$n+1$$ steps where n is the number of decrementable registers.\n\nAs the number of decrementable registers in generating register machines needed for generating any recursively enumerable set of (vectors of) natural numbers is only two, from Theorem 2, we obtain the following result:\n\n### Corollary 1\n\nFor any generating register machine with two decrementable registers, we can construct a catalytic P system with only one catalyst and with weak priority of catalytic rules over the non-cooperative rules which can simulate every step of the register machine in 3 steps, and therefore such catalytic P systems with only one catalyst and with weak priority of catalytic rules over the non-cooperative rules can generate any recursively enumerable set of (vectors of) natural numbers.\n\n## Catalytic P systems with only one catalyst working in the derivation mode maxobjects\n\nIn this section, we now study catalytic P systems with only one catalyst which work in the derivation mode maxobjects and show that computational completeness can be obtained with only one catalyst and no further ingredients.\n\n### Theorem 4\n\nFor any register machine with at least three decrementable registers, we can construct a catalytic P system with only one catalyst and working in the derivation mode maxobjects which can simulate every step of the register machine in n steps where n is the number of decrementable registers.\n\n### Proof\n\nWe take over the proof of Theorem 2. The priority of catalytic rules then is somehow regained by the fact that a catalytic rule $$ca\\rightarrow cv$$ affects two objects, whereas the corresponding non-catalytic rule $$a\\rightarrow v$$ only affects one object.\n\nAgain we use the trick as elaborated in Remark 1 and already used in the proof of Theorem 2 for getting a specific variant of register machines: without loss of generality, we not only do not start with a $$\\textit{SUB}$$-instruction, but we also change the register machine program in such a way that after a $$\\textit{SUB}$$-instruction on register $$n-1$$ or n two intermediate instructions are introduced, i.e., as in Remark 1, we use an $$\\textit{ADD}$$-instruction on register 1 immediately followed by a $$\\textit{SUB}$$-instruction on register 1.\n\nWe then may simulate the resulting register machine fulfilling these additional constraints $$M=\\left( m,B,l_{0},l_{h},P\\right)$$ by a corresponding catalytic P system with one membrane and one catalyst $$\\varPi = (O, \\{ c \\}, w, R)$$. Without loss of generality, we may again assume that, depending on its use as an accepting or generating or computing device, the register machine M, as stated in Proposition 1, Proposition 2, and Proposition 3, fulfills the condition that on the output registers we never apply any $$\\textit{SUB}$$-instruction. As in the proof of Theorem 2, also here we take the most general case of a register machine computing a partial recursive function on vectors of natural numbers with l components as input and vectors of natural numbers with k components as output using n decrementable registers, where without loss of generality, we may assume that at the end of a successful computation the first n registers are empty, and, moreover, on the output registers, i.e., the last k registers, no $$\\textit{SUB}$$-instruction is ever used.\n\nThe alphabet O of symbols now reduces to the following set:\n\n\\begin{aligned} O= & \\ \\{ a_{r} \\mid n+1 \\le r \\le m \\} \\\\ \\cup& \\ \\{ (a_{r},i) \\mid 1 \\le r \\le n,\\, 1 \\le i \\le n\\} \\\\ \\cup& \\ \\{(p,i) \\mid p \\in B_{\\textit{ADD}},\\, 1 \\le i \\le n\\} \\\\ \\cup & \\ \\{(p,i) \\mid p\\in B_{\\textit{SUB}(r)},\\, 1 \\le i \\le r+1,\\, 1 \\le r \\le n\\} \\\\\\cup& \\ \\{(p,i)^-, (p,i)^0 \\mid p\\in B_{\\textit{SUB}(r)},\\, r+2 \\le i \\le n,\\, 1 \\le r \\le n-2\\} \\\\ \\cup& \\ \\{c, e, \\# \\}. \\end{aligned}\n\nThe construction still includes the trap rule $$\\# \\rightarrow \\#$$ which will always keep the system busy and prevent it from halting and thus from producing a result as soon as the trap symbol $$\\#$$ has been introduced.\n\nThe sets of rules introduced by Eqs. 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20 read as follows, yet now omitting all trap rules except $$e\\rightarrow \\#$$:\n\n\\begin{aligned}&(a_{r},i) \\rightarrow (a_{r},i+1), 1 \\le r <n; \\qquad (a_{r},n) \\rightarrow (a_{r},1). \\end{aligned}\n(21)\n\nFor simulating $$\\textit{ADD}$$-instructions, we need the following rules:\n\nIncrement $$p:(\\textit{ADD}(r), q, s)$$:\n\n\\begin{aligned} c (p,i) \\rightarrow c (p,i+1), \\quad 1 \\le i < n. \\end{aligned}\n(22)\n\nIf r is a decrementable register:\n\n\\begin{aligned} c (p,n) \\rightarrow c (q,1) (a_{r},1), \\qquad c (p,n) \\rightarrow c (s,1) (a_{r},1), \\end{aligned}\n(23)\n\nIf r is an output register:\n\n\\begin{aligned}&c (p,n) \\rightarrow c (q,1) a_{r}, \\qquad c (p,n) \\rightarrow c (s,1) a_{r} \\end{aligned}\n(24)\n\nThe catalyst has to be used with the program symbol which otherwise would stay idle when the catalyst is used with a register symbol, and the multiset of rules applied in this way would use one symbol less and thus violate the condition of using the maximal number of objects, hence, now we do not need the trap rules $$(p,i) \\rightarrow \\#$$, $$1 \\le i \\le n$$.\n\nFor simulating $$\\textit{SUB}$$-instructions, we need the following rules:\n\nDecrement and zero-test $$p:(\\textit{SUB}(r), q, s)$$:\n\n\\begin{aligned} c (p,i) \\rightarrow c (p,i+1),\\quad 1 \\le i < r. \\end{aligned}\n(25)\n\nFor $$1 \\le i < r$$, we again do not need trap rules $$(p,i) \\rightarrow \\#$$, as otherwise the program symbol (pi) would stay idle and thus the condition of using the maximal number of objects would be violated.\n\n\\begin{aligned} (p,r) \\rightarrow (p,r+1), \\quad c (a_{r},r) \\rightarrow c e. \\end{aligned}\n(26)\n\nIn case that register r is empty, i.e., there is no object $$(a_{r},r)$$, then the catalyst will stay idle as in this step there is no other object with which it could react.\n\nIf $$r<n-1$$:\n\n\\begin{aligned}&c e \\rightarrow c, \\, e \\rightarrow \\# ,\\qquad (p,r+1) \\rightarrow (p,r+2)^{-}; \\nonumber \\\\&\\quad c (p,r+1) \\rightarrow c (p,r+2)^{0}. \\end{aligned}\n(27)\n\nIf in the first step of the simulation phase, the catalyst did manage to decrement the register, it produced e. Thus, in the second simulation step, the catalyst has three choices:\n\n1. 1.\n\nThe catalyst c correctly erases e, and to the program symbol $$(p,r+1)$$ the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$ must be applied due to the derivation mode maxobjects; all register symbols evolve in the usual way;\n\n2. 2.\n\nThe catalyst c takes the program symbol $$(p,r+1)$$ using the rule $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$, thus forcing the object e to be trapped by the rule $$e \\rightarrow \\#$$, and all register symbols evolve in the usual way;\n\n3. 3.\n\nThe catalyst c takes a register object $$(a_{r+1},r+1)$$, thus leaving the object e to be trapped by the rule $$e \\rightarrow \\#$$, the program symbol $$(p,r+1)$$ evolves with the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$, and all other register objects evolve in the usual way.\n\nIn fact, only variant 1 now fulfills the condition given by the derivation mode maxobjects and therefore is the only possible continuation of the computation if register r is not empty.\n\nOn the other hand, if register r is empty, no object e is generated, and the catalyst c has only two choices:\n\n1. 1.\n\nThe catalyst c takes the program symbol $$(p,r+1)$$ using the rule $$c (p,r+1) \\rightarrow c (p,r+2)^{0}$$, and all register symbols evolve in the usual way;\n\n2. 2.\n\nThe catalyst c takes a register object $$(a_{r+1},r+1)$$ thereby generating e, the program symbol $$(p,r+1)$$ evolves with the rule $$(p,r+1) \\rightarrow (p,r+2)^{-}$$, and all other register objects evolve in the usual way; this variant leads to the situation that e will be trapped in step $$r+2$$, as otherwise the program symbol stays idle, thus violating the condition of the derivation mode maxobjects. Hence, this variant in any case cannot lead to a halting computation due to the introduction of the trap symbol $$\\#$$. We mention that in case no register object $$(a_{r+1},r+1)$$ is present we have to apply case 1 and thus have a correct computation step.\n\nBoth variants fulfilll the condition for the derivation mode maxobjects, but only variant 1 is not introducing the trap symbol $$\\#$$ and therefore is the only reasonable continuation of the computation if register r is empty.\n\n\\begin{aligned}&c(p,i)^{-} \\rightarrow c(p,i+1)^{-} ,\\ \\ r+2 \\le i< n, \\qquad c(p,n)^{-} \\rightarrow c(q,1), \\nonumber \\\\&c(p,i)^{0} \\rightarrow c(p,i+1)^{0} ,\\ \\ r+2 \\le i < n, \\qquad c(p,n)^{0} \\rightarrow c(s,1). \\end{aligned}\n(28)\n\nAgain the catalyst has to be used with the program symbol which otherwise would stay idle when the catalyst is used with a register symbol, and the multiset of rules applied in this way would use one symbol less and thus violate the condition of using the maximal number of objects.\n\nIf $$r=n-1$$:\n\n\\begin{aligned}&c e \\rightarrow c, \\qquad e \\rightarrow \\#, \\nonumber \\\\&(p,n) \\rightarrow (q,1), \\qquad c (p,n) \\rightarrow c (s,1). \\end{aligned}\n(29)\n\nWe observe that in this case, during the first step of the next cycle, we have to guarantee that in the zero-test case, the catalyst must be used with the program symbol, hence, we will simulate an $$\\textit{ADD}$$-instruction on register 1, as the introduction of the symbol e in the wrong variant of the zero-test case must lead to introducing the trap symbol and not allowing e to be erased by the catalytic rule $$c e \\rightarrow c$$.\n\nIf $$r=n$$:\n\n\\begin{aligned}&c e \\rightarrow c, \\qquad e \\rightarrow \\#, \\nonumber \\\\&(p,n+1) \\rightarrow (q,2), \\qquad c (p,n+1) \\rightarrow c (s,2). \\end{aligned}\n(30)\n\nIn this case, the second step of the simulation is already the first step of the next cycle, and the third step of the simulation is already the second step of the next cycle, which again means that in this case of $$r=n$$ the next instruction to be simulated is an $$\\textit{ADD}$$-instruction on register 1. The introduction of the symbol e in the second step of the simulation, i.e., in the first step of the next cycle, in the wrong variant of the zero-test case allows for trapping e as the program symbol has to be used with the catalyst in the second step of the next cycle.\n\nWe finally observe that the proof construction given above is quite similar to the one elaborated in the proof of Theorem 2, but the only rule introducing the trap symbol now is the single rule $$e \\rightarrow \\#$$. $$\\square$$\n\nIf we only have two decrementable registers, then we need one additional step at the end; again we leave the details of the proof, based on the constructions elaborated in the preceding proofs, to the interested reader:\n\n### Theorem 5\n\nFor any register machine with at least two decrementable registers, we can construct a catalytic P system with only one catalyst and working in the derivation mode maxobjects which can simulate every step of the register machine in $$n+1$$ steps where n is the number of decrementable registers.\n\nAs the number of decrementable registers in generating register machines needed for generating any recursively enumerable set of (vectors of) natural numbers is only two, from Theorem 5, we obtain the following result:\n\n### Corollary 2\n\nFor any generating register machine with two decrementable registers, we can construct a catalytic P system with only one catalyst and working in the derivation mode maxobjects which can simulate every step of the register machine in 3 steps, and therefore such catalytic P systems with only one catalyst and working in the derivation mode maxobjects can generate any recursively enumerable set of (vectors of) natural numbers.\n\nAs for accepting register machines, in addition to the two working registers, we have at least one input register, we immediately infer the following result from Theorem 5:\n\n### Corollary 3\n\nFor any recursively enumerable set of d -vectors of natural numbers given by a register machine with $$d+2$$ decrementable registers, we can construct an accepting catalytic P system with only one catalyst and working in the derivation mode maxobjects which can simulate every step of the register machine in $$d+2$$ steps, and therefore such catalytic P systems with only one catalyst and working in the derivation mode maxobjects can accept any recursively enumerable set of (vectors of) natural numbers.\n\nThe results obtained in this section are optimal with respect to the number of catalysts for catalytic P systems working in the derivation mode maxobjects, as the results for P systems only using non-cooperative rules are the same for both the derivation mode maxobjects and the derivation mode max; for example, in the generating case, P systems only using non-cooperative rules can only generate semi-linear sets.\n\n## Conclusion\n\nIn this paper, we revisited a classical problem of computational complexity in membrane computing: can catalytic P systems working in the derivation mode max with only one catalyst in the whole system already generate all recursively enumerable sets of multisets? This problem has been standing tall for many years, and nobody has yet managed to give a positive or a negative answer to this problem. In this paper, we come tantalizingly close to showing computational completeness: we give a construction that simulates an arbitrary register machine with a very weak ingredient — the weak priority of catalytic rules over non-catalytic rules. This ingredient confers very little additional power indeed, because the structure of the priority relation is very simple, as it is only constrained by the two types of rules.\n\nWe believe that a similar construction driving the symbols around in a loop but avoiding any additional ingredients altogether will not be sufficient for obtaining computational completeness, as we still conjecture that P systems working in the derivation mode max with one catalyst and no additional control mechanisms cannot reach computational completeness. Finding an answer to the question of characterizing the computational power of P systems with one catalyst working in the derivation mode max therefore still remains one of the biggest challenges in the theory of P systems, although the result established in our paper has made the gap between the computational power of P systems with one catalyst and computational completeness smaller again.\n\nThe second variant proved to be computationally complete in this paper comes even closer to the original question: when using the derivation mode maxobjects instead of the derivation mode max, we really obtain computational completeness with only one catalyst and no further ingredients.\n\nThe results obtained in this paper can also be extended to P systems dealing with strings, following the definitions and notions used in , thus showing computational completeness for computing with strings.\n\n## References\n\n1. 1.\n\nAlhazov, A., & Freund, R. (2014). P systems with toxic objects. In: Gheorghe, M., Rozenberg, G., Salomaa, A., Sosík, P., Zandron, C. (Eds.) Membrane computing—15th International Conference, CMC 2014, Prague, Czech Republic, August 20–22, 2014, Revised Selected Papers, Lecture Notes in Computer Science, vol. 8961, pp. 99–125. Springer. https://doi.org/10.1007/978-3-319-14370-5_7.\n\n2. 2.\n\nAlhazov, A., Freund, R., & Ivanov, S. (2020). Catalytic P systems with weak priority of catalytic rules. In: Freund, R. (Ed.) Proceedings ICMC 2020, September 14–18, 2020, TU Wien, pp. 67–82.\n\n3. 3.\n\nAlhazov, A., Freund, R., & Ivanov, S. (2020). Computational completeness of catalytic P systems with weak priority of catalytic rules over non-cooperative rules. In: Orellana-Martín, D., Păun, Gh., Riscos-Núñez, A., Pérez-Hurtado, I. (Eds.) Proceedings 18th Brainstorming Week on Membrane Computing, Sevilla, February 4–7, 2020. RGNC REPORT 1/2020, Research Group on Natural Computing, Universidad de Sevilla, pp. 21–32.\n\n4. 4.\n\nDassow, J., & Păun, Gh. (1989). Regulated Rewriting in Formal Language Theory. Springer. https://www.springer.com/de/book/9783642749346.\n\n5. 5.\n\nFreund, R., Kari, L., Oswald, M., & Sosík, P. (2005). Computationally universal P systems without priorities: Two catalysts are sufficient. Theoretical Computer Science, 330(2), 251–266. https://doi.org/10.1016/j.tcs.2004.06.029.\n\n6. 6.\n\nFreund, R., Leporati, A., Mauri, G., Porreca, A. E., Verlan, S., & Zandron, C. (2014). Flattening in (tissue) P systems. In: Alhazov, A., Cojocaru, S., Gheorghe, M., Rogozhin, Yu., Rozenberg, G., Salomaa, A. (eds.) Membrane Computing, Lecture Notes in Computer Science, vol. 8340, pp. 173–188. Springer. https://doi.org/10.1007/978-3-642-54239-8_13.\n\n7. 7.\n\nFreund, R., Oswald, M., & Păun, Gh. (2015). Catalytic and purely catalytic P systems and P automata: Control mechanisms for obtaining computational completeness. Fundamenta Informaticae, 136(1–2), 59–84. https://doi.org/10.3233/FI-2015-1144.\n\n8. 8.\n\nFreund, R., & Sosík, P. (2015). On the power of catalytic P systems with one catalyst. In: Rozenberg, G., Salomaa, A., Sempere, J. M., Zandron, C. (eds.) Membrane Computing—16th International Conference, CMC 2015, Valencia, Spain, August 17–21, 2015, Revised Selected Papers, Lecture Notes in Computer Science, vol. 9504, pp. 137–152. Springer. https://doi.org/10.1007/978-3-319-28475-0_10.\n\n9. 9.\n\nMinsky, M. L. (1967). Computation: Finite and infinite machines. Englewood Cliffs, NJ: Prentice Hall.\n\n10. 10.\n\nPăun, Gh. (2000). Computing with membranes. Journal of Computer and System Sciences, 61(1), 108–143. https://doi.org/10.1006/jcss.1999.1693.\n\n11. 11.\n\nPăun, Gh. (2002). Membrane Computing: An introduction. Springer. https://doi.org/10.1007/978-3-642-56196-2.\n\n12. 12.\n\nPăun, Gh., Rozenberg, G., & Salomaa, A. (Eds.). (2010). The Oxford Handbook of Membrane Computing. Oxford University Press.\n\n13. 13.\n\nRozenberg, G., & Salomaa, A. (Eds.) (1997). Handbook of Formal Languages. Springer. https://doi.org/10.1007/978-3-642-59136-5.\n\n14. 14.\n\nThe P Systems Website. http://ppage.psystems.eu/.\n\n## Acknowledgements\n\nThe authors gratefully acknowledge the useful hints of the referees. Rudolf Freund acknowledges the TU Wien supporting the open access publishing of this paper.\n\n## Funding\n\nOpen access funding provided by TU Wien (TUW). Artiom Alhayov acknowledges project 20.80009.5007.22 “Intelligent information systems for solving ill-structured problems, processing knowledge and big data” by the National Agency for Research and Development. Sergiu Ivanov is partially supported by the Paris region via the project DIM RFSI n$$^{\\circ }$$2018-03 “Modèles informatiques pour la reprogrammation cellulaire”.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Rudolf Freund.\n\n## Ethics declarations\n\n### Conflict of interest\n\nOn behalf of all authors, the corresponding author states that there is no conflict of interest."
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https://mizar.uwb.edu.pl/JFM/Vol1/symsp_1.html | [
"Journal of Formalized Mathematics\nVolume 1, 1989\nUniversity of Bialystok\nCopyright (c) 1989 Association of Mizar Users\n\n## Construction of a bilinear antisymmetric form in symplectic vector space\n\nEugeniusz Kusak\nWarsaw University, Bialystok\nWojciech Leonczuk\nWarsaw University, Bialystok\nMichal Muzalewski\nWarsaw University, Bialystok\n\n### Summary.\n\nIn this text we will present unpublished results by Eu\\-ge\\-niusz Ku\\-sak. It contains an axiomatic description of the class of all spaces $\\langle V$; $\\perp_\\xi \\rangle$, where $V$ is a vector space over a field F, $\\xi: V \\times V \\to F$ is a bilinear antisymmetric form i.e. $\\xi(x,y) = -\\xi(y,x)$ and $x \\perp_\\xi y$ iff $\\xi(x,y) = 0$ for $x$, $y \\in V$. It also contains an effective construction of bilinear antisymmetric form $\\xi$ for given symplectic space $\\langle V$; $\\perp \\rangle$ such that $\\perp = \\perp_\\xi$. The basic tool used in this method is the notion of orthogonal projection J$(a,b,x)$ for $a,b,x \\in V$. We should stress the fact that axioms of orthogonal and symplectic spaces differ only by one axiom, namely: $x\\perp y+\\varepsilon z \\>\\&\\> y\\perp z+\\varepsilon x \\Rightarrow z\\perp x+\\varepsilon y.$ For $\\varepsilon=+1$ we get the axiom characterizing symplectic geometry. For $\\varepsilon=-1$ we get the axiom on three perpendiculars characterizing orthogonal geometry - see .\n\nSupported by RPBP.III-24.C6.\n\n#### MML Identifier: SYMSP_1\n\nThe terminology and notation used in this paper have been introduced in the following articles \n\nContents (PDF format)\n\n#### Bibliography\n\n Czeslaw Bylinski. Functions and their basic properties. Journal of Formalized Mathematics, 1, 1989.\n Czeslaw Bylinski. Functions from a set to a set. Journal of Formalized Mathematics, 1, 1989.\n Czeslaw Bylinski. Some basic properties of sets. Journal of Formalized Mathematics, 1, 1989.\n Eugeniusz Kusak, Wojciech Leonczuk, and Michal Muzalewski. Abelian groups, fields and vector spaces. Journal of Formalized Mathematics, 1, 1989.\n Eugeniusz Kusak, Wojciech Leonczuk, and Michal Muzalewski. Construction of a bilinear symmetric form in orthogonal vector space. Journal of Formalized Mathematics, 1, 1989.\n Andrzej Trybulec. Tarski Grothendieck set theory. Journal of Formalized Mathematics, Axiomatics, 1989.\n Wojciech A. Trybulec. Vectors in real linear space. Journal of Formalized Mathematics, 1, 1989.\n Zinaida Trybulec. Properties of subsets. Journal of Formalized Mathematics, 1, 1989.\n Edmund Woronowicz. Relations defined on sets. Journal of Formalized Mathematics, 1, 1989."
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https://nados.io/question/first-index-and-last-index | [
"`{\"id\":\"90c8013d-dc55-478f-ab56-7c05bbf9e481\",\"name\":\"First Index And Last Index\",\"description\":\"1. You are given a number n, representing the size of array a.\\r\\n2. You are given n numbers, representing elements of array a.\\r\\n\\r\\nAsssumption - Array is sorted. Array may have duplicate values.\",\"inputFormat\":\"A number n\\r\\nn1\\r\\nn2\\r\\n.. n number of elements\\r\\nA number d\",\"outputFormat\":\"A number representing first index\\r\\nA number representing last index\",\"constraints\":\"1 <= n <= 1000\\r\\n1 <= n1, n2, .. n elements <= 100\\r\\n1 <= d <= 100\",\"sampleCode\":{\"cpp\":{\"code\":\"#include<iostream>\\r\\nusing namespace std;\\r\\n\\r\\nint main(){\\r\\n int n;\\r\\n cin>>n;\\r\\n int* arr = new int[n];\\r\\n for(int i = 0 ; i < n; i++){\\r\\n cin>>arr[i];\\r\\n }\\r\\n int data;\\r\\n cin>>data;\\r\\n \\r\\n // write your code here\\r\\n \\r\\n \\r\\n}\"},\"java\":{\"code\":\"import java.io.*;\\r\\nimport java.util.*;\\r\\n\\r\\npublic class Main{\\r\\n\\r\\npublic static void main(String[] args) throws Exception {\\r\\n // write your code here\\r\\n }\\r\\n\\r\\n}\"},\"python\":{\"code\":\"n= int(input())\\n\\nlists= []\\n\\nfor i in range(0, n):\\n ele= int(input())\\n lists.append(ele)\\n \\ndata= int(input())\\n\\n# write your code here\"}},\"points\":10,\"difficulty\":\"easy\",\"sampleInput\":\"15\\r\\n1\\r\\n5\\r\\n10\\r\\n15\\r\\n22\\r\\n33\\r\\n33\\r\\n33\\r\\n33\\r\\n33\\r\\n40\\r\\n42\\r\\n55\\r\\n66\\r\\n77\\r\\n33\",\"sampleOutput\":\"5\\r\\n9\",\"questionVideo\":\"https://www.youtube.com/embed/PFaoQ72ziNk\",\"hints\":[],\"associated\":[],\"solutionSeen\":false,\"tags\":[],\"meta\":{\"path\":[{\"id\":0,\"name\":\"home\"},{\"id\":\"0c54b191-7b99-4f2c-acb3-e7f2ec748b2a\",\"name\":\"Data Structures and Algorithms\",\"slug\":\"data-structures-and-algorithms\",\"type\":0},{\"id\":\"f10b54f1-0f44-408f-82d5-89c189f4ad57\",\"name\":\"Function and Arrays\",\"slug\":\"function-and-arrays\",\"type\":0},{\"id\":\"d73b864e-002d-4115-a16a-d77afb475aa1\",\"name\":\"First Index And Last Index\",\"slug\":\"first-index-and-last-index\",\"type\":1}],\"next\":null,\"prev\":{\"id\":\"2f403e8a-be97-46b7-87e4-4e4fb487fdb0\",\"name\":\"First And Last Index\",\"type\":3,\"slug\":\"first-and-last-index\"}}}`\n\n# First Index And Last Index\n\n1. You are given a number n, representing the size of array a. 2. You are given n numbers, representing elements of array a. Asssumption - Array is sorted. Array may have duplicate values.\n\n`{\"id\":\"90c8013d-dc55-478f-ab56-7c05bbf9e481\",\"name\":\"First Index And Last Index\",\"description\":\"1. You are given a number n, representing the size of array a.\\r\\n2. You are given n numbers, representing elements of array a.\\r\\n\\r\\nAsssumption - Array is sorted. Array may have duplicate values.\",\"inputFormat\":\"A number n\\r\\nn1\\r\\nn2\\r\\n.. n number of elements\\r\\nA number d\",\"outputFormat\":\"A number representing first index\\r\\nA number representing last index\",\"constraints\":\"1 <= n <= 1000\\r\\n1 <= n1, n2, .. n elements <= 100\\r\\n1 <= d <= 100\",\"sampleCode\":{\"cpp\":{\"code\":\"#include<iostream>\\r\\nusing namespace std;\\r\\n\\r\\nint main(){\\r\\n int n;\\r\\n cin>>n;\\r\\n int* arr = new int[n];\\r\\n for(int i = 0 ; i < n; i++){\\r\\n cin>>arr[i];\\r\\n }\\r\\n int data;\\r\\n cin>>data;\\r\\n \\r\\n // write your code here\\r\\n \\r\\n \\r\\n}\"},\"java\":{\"code\":\"import java.io.*;\\r\\nimport java.util.*;\\r\\n\\r\\npublic class Main{\\r\\n\\r\\npublic static void main(String[] args) throws Exception {\\r\\n // write your code here\\r\\n }\\r\\n\\r\\n}\"},\"python\":{\"code\":\"n= int(input())\\n\\nlists= []\\n\\nfor i in range(0, n):\\n ele= int(input())\\n lists.append(ele)\\n \\ndata= int(input())\\n\\n# write your code here\"}},\"points\":10,\"difficulty\":\"easy\",\"sampleInput\":\"15\\r\\n1\\r\\n5\\r\\n10\\r\\n15\\r\\n22\\r\\n33\\r\\n33\\r\\n33\\r\\n33\\r\\n33\\r\\n40\\r\\n42\\r\\n55\\r\\n66\\r\\n77\\r\\n33\",\"sampleOutput\":\"5\\r\\n9\",\"questionVideo\":\"https://www.youtube.com/embed/PFaoQ72ziNk\",\"hints\":[],\"associated\":[],\"solutionSeen\":false,\"tags\":[],\"meta\":{\"path\":[{\"id\":0,\"name\":\"home\"},{\"id\":\"0c54b191-7b99-4f2c-acb3-e7f2ec748b2a\",\"name\":\"Data Structures and Algorithms\",\"slug\":\"data-structures-and-algorithms\",\"type\":0},{\"id\":\"f10b54f1-0f44-408f-82d5-89c189f4ad57\",\"name\":\"Function and Arrays\",\"slug\":\"function-and-arrays\",\"type\":0},{\"id\":\"d73b864e-002d-4115-a16a-d77afb475aa1\",\"name\":\"First Index And Last Index\",\"slug\":\"first-index-and-last-index\",\"type\":1}],\"next\":null,\"prev\":{\"id\":\"2f403e8a-be97-46b7-87e4-4e4fb487fdb0\",\"name\":\"First And Last Index\",\"type\":3,\"slug\":\"first-and-last-index\"}}}`",
null,
"Editor\n\n# First Index And Last Index\n\neasy\n\n1. You are given a number n, representing the size of array a. 2. You are given n numbers, representing elements of array a. Asssumption - Array is sorted. Array may have duplicate values.\n\n## Constraints\n\n1 <= n <= 1000 1 <= n1, n2, .. n elements <= 100 1 <= d <= 100\n\n## Format\n\n### Input\n\nA number n n1 n2 .. n number of elements A number d\n\n### Output\n\nA number representing first index A number representing last index\n\n## Example\n\nSample Input\n\n```.css-23h8hz{color:inherit;font-size:0.875rem;line-height:1.125rem;letter-spacing:0.016rem;font-weight:var(--chakra-fontWeights-normal);white-space:pre-wrap;}15 1 5 10 15 22 33 33 33 33 33 40 42 55 66 77 33```\n\n### Sample Output\n\n```.css-3oaykw{color:var(--chakra-colors-active-primary);font-size:0.875rem;line-height:1.125rem;letter-spacing:0.016rem;font-weight:var(--chakra-fontWeights-normal);white-space:pre-wrap;font-family:Monospace;}5 9```\n\nQuestion Video\n\nDiscussions\n\nShow Discussion\n\nRelated Resources",
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https://www.nag.com/numeric/nl/nagdoc_25/nagdoc_fl25/html/f08/f08gff.html | [
"F08 Chapter Contents\nF08 Chapter Introduction\nNAG Library Manual\n\n# NAG Library Routine DocumentF08GFF (DOPGTR)\n\nNote: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.\n\n## 1 Purpose\n\nF08GFF (DOPGTR) generates the real orthogonal matrix $Q$, which was determined by F08GEF (DSPTRD) when reducing a symmetric matrix to tridiagonal form.\n\n## 2 Specification\n\n SUBROUTINE F08GFF ( UPLO, N, AP, TAU, Q, LDQ, WORK, INFO)\n INTEGER N, LDQ, INFO REAL (KIND=nag_wp) AP(*), TAU(*), Q(LDQ,*), WORK(N-1) CHARACTER(1) UPLO\nThe routine may be called by its LAPACK name dopgtr.\n\n## 3 Description\n\nF08GFF (DOPGTR) is intended to be used after a call to F08GEF (DSPTRD), which reduces a real symmetric matrix $A$ to symmetric tridiagonal form $T$ by an orthogonal similarity transformation: $A=QT{Q}^{\\mathrm{T}}$. F08GEF (DSPTRD) represents the orthogonal matrix $Q$ as a product of $n-1$ elementary reflectors.\nThis routine may be used to generate $Q$ explicitly as a square matrix.\nGolub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore\n\n## 5 Parameters\n\n1: $\\mathrm{UPLO}$ – CHARACTER(1)Input\nOn entry: this must be the same parameter UPLO as supplied to F08GEF (DSPTRD).\nConstraint: ${\\mathbf{UPLO}}=\\text{'U'}$ or $\\text{'L'}$.\n2: $\\mathrm{N}$ – INTEGERInput\nOn entry: $n$, the order of the matrix $Q$.\nConstraint: ${\\mathbf{N}}\\ge 0$.\n3: $\\mathrm{AP}\\left(*\\right)$ – REAL (KIND=nag_wp) arrayInput\nNote: the dimension of the array AP must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}×\\left({\\mathbf{N}}+1\\right)/2\\right)$.\nOn entry: details of the vectors which define the elementary reflectors, as returned by F08GEF (DSPTRD).\n4: $\\mathrm{TAU}\\left(*\\right)$ – REAL (KIND=nag_wp) arrayInput\nNote: the dimension of the array TAU must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}-1\\right)$.\nOn entry: further details of the elementary reflectors, as returned by F08GEF (DSPTRD).\n5: $\\mathrm{Q}\\left({\\mathbf{LDQ}},*\\right)$ – REAL (KIND=nag_wp) arrayOutput\nNote: the second dimension of the array Q must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}\\right)$.\nOn exit: the $n$ by $n$ orthogonal matrix $Q$.\n6: $\\mathrm{LDQ}$ – INTEGERInput\nOn entry: the first dimension of the array Q as declared in the (sub)program from which F08GFF (DOPGTR) is called.\nConstraint: ${\\mathbf{LDQ}}\\ge \\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{N}}\\right)$.\n7: $\\mathrm{WORK}\\left({\\mathbf{N}}-1\\right)$ – REAL (KIND=nag_wp) arrayWorkspace\n8: $\\mathrm{INFO}$ – INTEGEROutput\nOn exit: ${\\mathbf{INFO}}=0$ unless the routine detects an error (see Section 6).\n\n## 6 Error Indicators and Warnings\n\n${\\mathbf{INFO}}<0$\nIf ${\\mathbf{INFO}}=-i$, argument $i$ had an illegal value. An explanatory message is output, and execution of the program is terminated.\n\n## 7 Accuracy\n\nThe computed matrix $Q$ differs from an exactly orthogonal matrix by a matrix $E$ such that\n $E2 = Oε ,$\nwhere $\\epsilon$ is the machine precision.\n\n## 8 Parallelism and Performance\n\nF08GFF (DOPGTR) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.\nF08GFF (DOPGTR) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.\nPlease consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.\n\nThe total number of floating-point operations is approximately $\\frac{4}{3}{n}^{3}$.\nThe complex analogue of this routine is F08GTF (ZUPGTR).\n\n## 10 Example\n\nThis example computes all the eigenvalues and eigenvectors of the matrix $A$, where\n $A = 2.07 3.87 4.20 -1.15 3.87 -0.21 1.87 0.63 4.20 1.87 1.15 2.06 -1.15 0.63 2.06 -1.81 ,$\nusing packed storage. Here $A$ is symmetric and must first be reduced to tridiagonal form by F08GEF (DSPTRD). The program then calls F08GFF (DOPGTR) to form $Q$, and passes this matrix to F08JEF (DSTEQR) which computes the eigenvalues and eigenvectors of $A$.\n\n### 10.1 Program Text\n\nProgram Text (f08gffe.f90)\n\n### 10.2 Program Data\n\nProgram Data (f08gffe.d)\n\n### 10.3 Program Results\n\nProgram Results (f08gffe.r)"
]
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https://in.mathworks.com/help/signal/ref/invfreqs.html | [
"# invfreqs\n\nIdentify continuous-time filter parameters from frequency response data\n\n## Syntax\n\n``[b,a] = invfreqs(h,w,n,m)``\n``[b,a] = invfreqs(h,w,n,m,wt)``\n``[b,a] = invfreqs(___,iter)``\n``[b,a] = invfreqs(___,tol)``\n``[b,a] = invfreqs(___,'trace')``\n``[b,a] = invfreqs(h,w,'complex',n,m,___)``\n\n## Description\n\nexample\n\n````[b,a] = invfreqs(h,w,n,m)` returns the real numerator and denominator coefficient vectors `b` and `a` of the transfer function `h`.```\n````[b,a] = invfreqs(h,w,n,m,wt)` weights the fit-errors versus frequency using `wt`.```\n\nexample\n\n````[b,a] = invfreqs(___,iter)` provides an algorithm that guarantees stability of the resulting linear system by searching for the best fit using a numerical, iterative scheme. This syntax can include any combination of input arguments from the previous syntaxes.```\n````[b,a] = invfreqs(___,tol)` uses `tol` to decide convergence of the iterative algorithm.```\n````[b,a] = invfreqs(___,'trace')` displays a textual progress report of the iteration.```\n````[b,a] = invfreqs(h,w,'complex',n,m,___)` creates a complex filter. In this case no symmetry is enforced, and the frequency is specified in radians between –π and π.```\n\n## Examples\n\ncollapse all\n\nConvert a simple transfer function to frequency-response data and then back to the original filter coefficients.\n\n```a = [1 2 3 2 1 4]; b = [1 2 3 2 3]; [h,w] = freqs(b,a,64); [bb,aa] = invfreqs(h,w,4,5)```\n```bb = 1×5 1.0000 2.0000 3.0000 2.0000 3.0000 ```\n```aa = 1×6 1.0000 2.0000 3.0000 2.0000 1.0000 4.0000 ```\n\n`bb` and `aa` are equivalent to `b` and `a`, respectively. However, the system is unstable because `aa` has poles with positive real part. View the poles of `bb` and `aa`.\n\n`zplane(bb,aa)`",
null,
"Use the iterative algorithm of `invfreqs` to find a stable approximation to the system.\n\n`[bbb,aaa] = invfreqs(h,w,4,5,[],30)`\n```bbb = 1×5 0.6816 2.1015 2.6694 0.9113 -0.1218 ```\n```aaa = 1×6 1.0000 3.4676 7.4060 6.2102 2.5413 0.0001 ```\n\nVerify that the system is stable by plotting the new poles.\n\n`zplane(bbb,aaa)`",
null,
"Generate two vectors, `mag` and `phase`, that simulate magnitude and phase data gathered in a laboratory. Also generate a vector, `w`, of frequencies.\n\n```rng('default') fs = 1000; t = 0:1/fs:2; mag = periodogram(sin(2*pi*100*t)+randn(size(t))/10,[],[],fs); phase = randn(size(mag))/10; w = linspace(0,fs/2,length(mag))';```\n\nUse `invfreqs` to convert the data into a continuous-time transfer function. Plot the result.\n\n```[b,a] = invfreqs(mag.*exp(1j*phase),w,2,2,[],4); freqs(b,a)```",
null,
"## Input Arguments\n\ncollapse all\n\nFrequency response, specified as a vector.\n\nAngular frequencies at which `h` is computed, specified as a vector.\n\nDesired order of the numerator and denominator polynomials, specified as positive integer scalars.\n\nData Types: `single` | `double`\n\nWeighting factors, specified as a vector. `wt` is a vector of weighting factors that is the same length as `w`.\n\nData Types: `single` | `double`\n\nNumber of iterations in the search algorithm, specified as a positive real scalar. The `iter` parameter tells `invfreqs` to end the iteration when the algorithm has converged to a solution, or after `iter` iterations, whichever occurs first.\n\nTolerance, specified as a scalar. `invfreqs` defines convergence as occurring when the norm of the (modified) gradient vector is less than `tol`.\n\nTo obtain a weight vector of all ones, use\n\n`invfreqs(h,w,n,m,[],iter,tol)`\n\n## Output Arguments\n\ncollapse all\n\nTransfer function coefficients, returned as vectors. Express the transfer function in terms of `b` and `a` as\n\n`$H\\left(s\\right)=\\frac{B\\left(s\\right)}{A\\left(s\\right)}=\\frac{b\\left(1\\right){s}^{n}+b\\left(2\\right){s}^{n-1}+\\cdots +b\\left(n+1\\right)}{a\\left(1\\right){s}^{m}+a\\left(2\\right){s}^{m-1}+\\cdots +a\\left(m+1\\right)}$`\n\nExample: `b = [1 3 3 1]/6` and `a = [3 0 1 0]/3` specify a third-order Butterworth filter with normalized 3 dB frequency 0.5π rad/sample.\n\nData Types: `double` | `single`\nComplex Number Support: Yes\n\n## Tips\n\nWhen building higher order models using high frequencies, it is important to scale the frequencies, dividing by a factor such as half the highest frequency present in `w`, so as to obtain well-conditioned values of `a` and `b`. This corresponds to a rescaling of time.\n\n## Algorithms\n\nBy default, `invfreqs` uses an equation error method to identify the best model from the data. This finds `b` and `a` in\n\n`$\\underset{b,a}{\\mathrm{min}}\\sum _{k=1}^{n}wt\\left(k\\right){|h\\left(k\\right)A\\left(w\\left(k\\right)\\right)-B\\left(w\\left(k\\right)\\right)|}^{2}$`\n\nby creating a system of linear equations and solving them with the MATLAB® `\\` operator. Here A(w(k)) and B(w(k)) are the Fourier transforms of the polynomials `a` and `b`, respectively, at the frequency w(k), and n is the number of frequency points (the length of `h` and `w`). This algorithm is based on Levi . Several variants have been suggested in the literature, where the weighting function `wt` gives less attention to high frequencies.\n\nThe superior (“output-error”) algorithm uses the damped Gauss-Newton method for iterative search , with the output of the first algorithm as the initial estimate. This solves the direct problem of minimizing the weighted sum of the squared error between the actual and the desired frequency response points.\n\n`$\\underset{b,a}{\\mathrm{min}}\\sum _{k=1}^{n}wt\\left(k\\right){|h\\left(k\\right)-\\frac{B\\left(w\\left(k\\right)\\right)}{A\\left(w\\left(k\\right)\\right)}|}^{2}$`\n\n Levi, E. C. “Complex-Curve Fitting.” IRE Trans. on Automatic Control. Vol. AC-4, 1959, pp. 37–44.\n\n Dennis, J. E., Jr., and R. B. Schnabel. Numerical Methods for Unconstrained Optimization and Nonlinear Equations. Englewood Cliffs, NJ: Prentice-Hall, 1983."
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"https://in.mathworks.com/help/examples/signal/win64/TransferFunctionToFrequencyResponseConversionExample_01.png",
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"https://in.mathworks.com/help/examples/signal/win64/TransferFunctionToFrequencyResponseConversionExample_02.png",
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"https://in.mathworks.com/help/examples/signal/win64/ContinuousTimeTransferFunctionExample_01.png",
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]
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https://math.stackexchange.com/questions/2521110/how-have-the-following-trig-formulas-been-modified-to-obtain-the-new-trig-form | [
"# How have the “following trig formulas” been modified to obtain the new trig formulas?\n\nI was reading a trig book in which the author had deduced the following formulas from a drawing. I am mentioning those as follows...\n\nBy looking at the above drawing the author deduced these formulas for $(\\theta+90^\\circ)$...\n\n1. $\\sin (\\theta + 90^\\circ)=\\cos\\theta$\n\n2. $\\cos (\\theta + 90^\\circ)=-\\sin\\theta$\n\n3. $\\tan (\\theta + 90^\\circ)=-\\cot\\theta$\n\nI understood how did he conclude above formulas from the drawing.\n\nAs you can see the drawing depicts author's approach to derive the $(\\theta+90^\\circ)$ formulas by taking a similar triangle in the second quadrant, in the same way, to derive $(\\theta-90^\\circ)$ formulas, he advised to take a similar triangle (as in the drawing) in quadrant 4 as well. But, to the contrary he mentioned a trick to get those formulas which I do not understand.\n\nThe author told to replace $\\theta$ in formulas 1,2 and 3 by $(\\theta-90^\\circ)$. So what I got by doing that is...\n\n1. $\\sin ((\\theta-90^\\circ) + 90^\\circ)=\\cos(\\theta-90^\\circ)$\n\n2. $\\cos ((\\theta-90^\\circ) + 90^\\circ)=-\\sin(\\theta-90^\\circ)$\n\n3. $\\tan ((\\theta-90^\\circ) + 90^\\circ)=-\\cot(\\theta-90^\\circ)$\n\nBy evaluating them further I get...\n\n1. $\\sin (\\theta)=\\cos(\\theta-90^\\circ)$ (I got this one correct)\n\n2. $\\cos (\\theta)=-\\sin(\\theta-90^\\circ)$ (wrong)\n\n3. $\\tan (\\theta)=-\\cot(\\theta-90^\\circ)$ (wrong)\n\nBut the author says that the outcome of the formulas should be...\n\n1.$\\sin (\\theta)=\\cos(\\theta-90^\\circ)$ (correct.)\n\n2. $-\\cos (\\theta)=\\sin(\\theta-90^\\circ)$\n\n3. $\\tan (\\theta-90^\\circ) =-\\cot(\\theta)$\n\nI don't know where am I going wrong. The author told to replace $\\theta$ by $(\\theta-90^\\circ)$ then why am I getting my answers incorrect? Where is the flaw? Please help me to track it down.\n\n• Just because you don't get the exact character-for-character representation of the equation, doesn't mean you did it wrong. For example, the (2) that you have is equivalent to the (2) the author has. Can you see why? – Henry Swanson Nov 15 '17 at 7:15\n• The $-$ sign is on opposite sides in both those equations. How is it correect? I can't see why. – Saksham Sharma Nov 15 '17 at 7:17\n• @SakshamSharma Assume $u=-v$. Multiply both sides by $(-1)$ to obtain $-u=v$. – Hagen von Eitzen Nov 15 '17 at 7:20\n• @HagenvonEitzen I understood. Your convention holds true for $2$. What about the last one? $3$? – Saksham Sharma Nov 15 '17 at 7:22\n• Well, you have $\\tan(\\mathrm{thing}) = -\\cot(\\mathrm{other thing})$. What do you know about $\\tan(\\mathrm{other thing})$ and $\\cot(\\mathrm{thing})$? – Henry Swanson Nov 15 '17 at 8:15\n\n$u=-v$ then you can get $-u=v$ by multiplying the equation on both sides by $-1$.\nFor 3, use the reciprocal identity...$$\\tan(x)=\\frac{1}{\\cot(x)}$$"
]
| [
null
]
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https://ask.sagemath.org/question/7917/when-does-120-pythons-integer-division-vs-sages-exact-fractions/?answer=12053 | [
"# When does 1/2=0 ? (python's integer division Vs Sage's exact fractions)\n\nHi all !\n\nI know that python evaluates 1/2 to zero because it is making an integer division. Well. I was expecting Sage not to do so.\n\nIt does more or less ...\n\n------ sagess.py ----------\n\n#! /usr/bin/sage -python\n# -*- coding: utf8 -*-\n\nfrom sage.all import *\n\ndef MyFunction(x,y):\nprint x/y\n\nMyFunction(1,2)\n\n\n------ end of sagess.py ----------\n\nWhen I launche that script from the command line, I got what Python would give :\n\n18:08:03 ~/script >$./sagess.py 0 But when I launch it from the sage'terminal it is less clear how it works : 18:16:04 ~/script >$ sage\n----------------------------------------------------------------------\n| Sage Version 4.5.3, Release Date: 2010-09-04 |\n| Type notebook() for the GUI, and license() for information. |\n----------------------------------------------------------------------\nsage: import sagess\n0\nsage: sagess.MyFunction(1,2)\n1/2\n\n\nAt the import, it returns zero, but when I reask the same computation, it gives the correct 1/2\n\nI know that writing float(x)/y \"fixes\" the problem. But it looks weird and I want to keep exact values.\n\nWhat do I have to write in my scripts in order to make understand to Sage that I always want 1/2 to be 1/2 ?\n\nThanks Have a good night ! Laurent\n\nedit retag close merge delete\n\nSort by » oldest newest most voted",
null,
"sage: import sagess\n0\nsage: sagess.MyFunction(1,2)\n1/2\n\n\nThe reason for the above behavior is that in the second call, the arguments 1 and 2 are Sage Integers, not Python ints: when run interactively, Sage preparses the input, turning all integers into the type Integer. You could reproduce the effect of your original call of MyFunction(1,2) in sagess.py by this:\n\nsage: sagess.MyFunction(int(1), int(2))\n\n\nOn the other hand, to get 1/2, use Sage integers: rewrite your file as\n\n#! /usr/bin/sage -python\n# -*- coding: utf8 -*-\n\ndef MyFunction(x,y):\nfrom sage.rings.all import Integer\nprint Integer(x)/Integer(y)\n\nMyFunction(1,2)\n\n\nThis will give the right answer with integer arguments, but it will raise an error if you pass non-integer arguments.\n\nmore\n\nOk, so the point is that when making an \"interactive\" import, the imported module is not preparsed ?\n\ndef MyFunction(x,y):\nfrom sage.rings.all import Integer\nprint Integer(x)/Integer(y)\n\n\nI cannot do that in my real live idea. My purpose was to write a function that convert a point (x,y) into polar coordinates (radius,angle) using atan(y/x). (the aim was to show the algorithm to some students)\n\nInstead, you gave me the idea to write this one :\n\ndef MyFunction(x,y):\nx=SR(x)\nprint x/y\n\n\nThis works fine.\n\nHave a good night Laurent\n\nmore"
]
| [
null,
"https://www.gravatar.com/avatar/09b3ed8e80f5d4bc47ec4530dc2941d0",
null
]
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https://alistairstutorials.co.uk/tutorial23.html | [
"# Offsets and eccentrics\n\nThis series of tutorials covering kinematics and kinetics of the crank and slider mechanism is based on an in-line configuration of crank and slider where the linear axis of slider motion intersects the axis of rotation of the crank arm as shown below.",
null,
"An offset configuration where the linear axis of the slider does not intersect the axis of rotation of the crank arm arises in some applications. Two offset configurations are shown below defined by relative positions of the respective axes. Displacement of the respective axes, e, is termed the eccentricity. Type (a) is sometimes stated as \"positive eccentricity\" and type (b) as \"negative eccentricity\" but I avoid these terms because they presuppose a specific orientation of reference axes.",
null,
"We now apply the analytical method used in a previous tutorial for in-line configurations. Consider the diagram below for offset type (b) with the crank arm R at an arbitrary crank angle θ.",
null,
"• Crank arm ab has length R and rotates anti-clockwise with constant angular velocity ω about centre of rotation point a. Its position is defined by crank angle θ.\n• Connecting rod bc has length L\n• Point o is the origin of the co-ordinate axes\n• φ is the angle between the connecting rod and the slider axis\n• Point d is the vertical projection of point b on line ao\n• a1, d1 and o1 are vertical projections of points a, d and o on the slider axis.\n• e is the eccentricity of the slider axis.\n\nPositions of inner dead centre (idc) and outer dead centre (odc) on the horizontal slider axis which determine stroke s are defined by the two collinear positions of crank arm R and connecting rod L indicated by dashed lines. Note that these positions do not occur at crank angles 0° and 180° respectively as is the case for the in-line configuration.\n\n### Derive an expression for x\n\nThe task is to derive an expression for displacement x of the slider in terms of crank angle θ, represented by point c measured from the idc position . The method is identical to that for an in-line configuration in a previous tutorial.",
null,
"also ao = a1o1 a1d1 = ad =Rcosθ d1c = Lcosφ",
null,
"We must transform the term (cos φ) to an expression in θ.\n\nnote that e = (db + bd1 ) = (Rsinθ + Lsinφ) thus Lsinφ = (e - Rsinθ)",
null,
"",
null,
"substituting for cosφ in equation (1) gives:",
null,
"We now have an expression for the horizontal displacement x of slider c as a function of θ. The plot below shows displacement x against crank angle θ for one complete revolution of the crank arm computed from the expression above with n = 3 and R = 1m, L = 3m for a range of eccentricities e from 0 to 1.8 m.",
null,
"From the plots it can be seen that stroke increases with increasing eccentricity. A limit is reached when e > (L - R); crank rotation is not possible beyond this limit. In this example (L - R) = 2m which is the upper limit of eccentricity.\n\nWith offset It is also seen that the forward stroke from idc to odc occurs over a larger crank angle than the reverse stroke, the effect increasing with greater eccentricity. For example, for e = 1.4 crank angles at idc and odc are c. 20° and 230° respectively. Thus the crank angle of the forward stroke spans c. 210° and that of the reverse stroke c. 150°. This provides a quick return mechanism with practical applications in machinery.\n\nNote that the above plot is for offset type (b) (defined above) where the slider axis is above the axis of rotation of the crank arm. In contrast the plot below shows displacement x for offset type (a) where the slider axis is below the axis of rotation of the crank arm for e = 1.4.",
null,
"In this configuration the displacement plot of type (b) is reversed in that the forward stroke is \"fast\" and the reverse stroke is \"slow.\"\n\n### Derive an expression for velocity v of slider c.",
null,
"In this expression x is a function of variable θ (the crank angle). For velocity v of slider c we require the time derivative dx/dt for which we use the chain rule:",
null,
"The method for obtaining derivative dx/dt for the above expression is identical to that in the tutorial for in-line configurations and it is not necessary to repeat the steps here.\n\nThe derived expression is:",
null,
"The plot below derived from this expression shows slider velocity against crank angle for the type (b) offset mechanism defined above with e = 1.4 compared with the in-line configuration (e = 0). The \"quick return\" characteristic is clearly illustrated. Note also zero velocity at idc and odc.",
null,
"I will leave the next differentiation to find an exxpression for acceleration dv/dt for a rainy day!\n\n### Eccentric crank mechanism\n\nAn eccentric is a special configuration of a crank mechanism shown in the figure below in the form of a slider and crank.",
null,
"A disc called a sheave is fixed to a shaft such that the centre of the sheave is eccentric to the shaft (e on the diagram). A strap is placed around the circumference of the sheave and is free to rotate. Typically a rod or lever arm is rigidly fixed to the strap. When the shaft rotates, the sheave acts like a cam on the strap producing reciprocating motion on the rod. In the configuration above a slider attached to the end of the rod is constrained to move on a horizontal axis. The motion of this slider is identical to an in-line slider and crank mechanism with crank arm length e and the connecting rod equivalent to the distance from the centre of the sheave to the pin joint of the slider. The slider axis can equally well be offset from the rotational axis of the shaft.\n\nThe diagram below shows the cycle for a complete revolution of the shaft at crank angles 0°, 90°, 180° and 270°.",
null,
"The relatively small effective crank arm provides a correspondingly short output stroke. Eccentrics were particularly useful in steam engines providing reciprocating drives from the main crankshaft to valve gear and auxiliaries such as a water pump . Note that eccentrics can only convert rotational motion to reciprocating motion. A force acting on to the strap from a rod or lever arm cannot drive the shaft.\n\nI welcome feedback at:\n\n### Tutorials - mechanical vibrations\n\n###### Forced vibrations with damping\n\nAlistair's tutorials 2021"
]
| [
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram01.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram02.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram03.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math01.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math02.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math03.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math04.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math05.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram04.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram05.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math06.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math07.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/math08.gif",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram06.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram07.jpg",
null,
"https://alistairstutorials.co.uk/images/images%20tut%2023/diagram08.jpg",
null
]
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https://powerapps.microsoft.com/zh-cn/blog/working-with-time-columns-in-sql-server/ | [
"",
null,
"# Working with Time columns in SQL Server",
null,
"This post shows some expressions that we needed to use to consume SQL Server time columns – and it was fairly complex. We have since released a new feature in PowerApps that makes this easier. Take a look at the announcement for the new regular expression functions for more information.\n\nWhen we use time columns in SQL Server and try to use them in PowerApps, we don’t get a time value that we can use directly – instead, we get a text (string) type, with an encoding of the time that is not very user-friendly:",
null,
"This is currently a missing feature in PowerApps – time columns should have a better support (since there is a Time type in PowerApps), so while this is not addressed, I’ll show in this post a way to work with the time columns today.\n\n## The problem\n\nA few months ago I wrote about working with date/time values in SQL, as there are a couple of different ways to represent those values. Time columns have a similar issue – they can represent two different concepts:\n\n• A certain time in day: school starts at 08:25, ends at 14:55; tea will be served at 17:00.\n• A duration of an event: NBA basketball quarters last for 12 minutes; the current marathon record is 2 hours, 1 minute and 29 seconds.\n\nThe SQL time column type can be used for both concepts, so we don’t have a semantic problem this time. Instead, the issue arises because of a mismatch between the SQL connector and PowerApps – the former transmits time values as durations (in the ISO 8601 duration format), while the latter only accepts absolute time values over the wire. This issue will likely be addressed in an upcoming release, but since we may have apps that rely on the current behavior (time columns returned as string values), we can’t just change it to return time values or those apps would be broken.\n\n## Converting between the duration format and a time value\n\nThere’s currently no built-in function that can be used to convert the duration format and a time value in PowerApps, so we’ll use some string manipulation functions for that. If all you want from this post is an expression that does that, then here you go (this assumes that the time value can be referenced as `ThisItem.Time`):\n\n``` Time(\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)), // Check if there is an hour component\n0, // If not, the value is zero\nValue( // Otherwise, take the substring between 'PT' and 'H'\nMid(\nThisItem.Time,\n3,\nFind(\"H\", ThisItem.Time) - 3))),\nIf(\nIsBlank(Find(\"M\", ThisItem.Time)), // Check if there is a minute component\n0, // If not, the value is zero\nValue( // Otherwise take the substring between the rest of the value\nMid( // after the hour component and the 'M' indicator\nThisItem.Time,\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)),\n3,\nFind(\"H\", ThisItem.Time) + 1),\nFind(\"M\", ThisItem.Time) -\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)),\n3,\nFind(\"H\", ThisItem.Time) + 1)))),\nIf(\nIsBlank(Find(\"S\", ThisItem.Time)), // Check if there is a second component\n0, // If not, the value is zero\nValue( // Otherwise take the substring after the minute or hour indicator,\nSubstitute( // remove the 'S' indicator, then take the value\nMid(\nThisItem.Time,\nIf(\n!IsBlank(Find(\"M\", ThisItem.Time)),\nFind(\"M\", ThisItem.Time) + 1,\n!IsBlank(Find(\"H\", ThisItem.Time)),\nFind(\"H\", ThisItem.Time) + 1,\n3)),\n\"S\",\n\"\"))))\n```\n\nOr if you want to display it in a certain format, you can use it inside a Text expression that converts it to what you want, such as\n\n` Text(<the big expression above>, DateTimeFormat.LongTime24)`\n\n### Drilling down the above expression\n\nIf you want to go deeper, let’s go over that expression, and maybe you’ll be able to use some of the concepts for another scenario. When trying to parse this kind of data, it’s always good to find some examples of data on that format, so that we can keep an eye for special cases that may arise. After populating a sample SQL table with some of those values and looking at how they’re represented in PowerApps, here are some examples of the types of values that we’ll need to parse:\n\n• PT2H1M39S (2:01:39)\n• PT12H (12:00:00)\n• PT2H2S (2:00:02)\n• PT58M18S (0:58:18)\n• PT34S (0:00:34)\n• PT5M (0:05:00)\n\nThe value always starts with the ‘PT’ (period / time) identifier, followed by non-zero components of the period (hour / minute / second). Notice that the components are not always present – that makes breaking down the parts a little harder, for example, as we cannot rely on the presence of a ‘H’ and a ‘M’ to find the minute component between them.\n\nWe start then by finding the hour. There are two cases here: there is an hour component (in case we find a ‘H’ character in the string) or not. If not, the value is simple: 0. Otherwise, we can use the Mid function to take the value from the 3rd character (after ‘PT’) and the ‘H’ marker:\n\n``` If(\nIsBlank(Find(\"H\", ThisItem.Time)), // Check if there is an hour component\n0, // If not, the value is zero\nValue( // Otherwise, take the substring between 'PT' and 'H'\nMid(\nThisItem.Time,\n3,\nFind(\"H\", ThisItem.Time) - 3)))```\n\nFor the minute, there are now three cases: there is no minute component (simple, value is zero). But if there is a minute component, then there may be an hour component or not, so we need to account for both cases. If there is an hour component, then we need to account for it when calculating the indices for taking the substring from the duration format.\n\n``` If(\nIsBlank(Find(\"M\", ThisItem.Time)), // Check if there is a minute component\n0, // If not, the value is zero\nValue( // Otherwise take the substring between the rest of the value\nMid( // after the hour component and the 'M' indicator\nThisItem.Time,\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)), // If there is no hour component,\n3, // Start right after 'PT'\nFind(\"H\", ThisItem.Time) + 1), // Else start after the hour component\nFind(\"M\", ThisItem.Time) -\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)),\n3,\nFind(\"H\", ThisItem.Time) + 1))))\n```\n\nTo get the seconds component, we could do something similar to the minutes – take a substring from the minute component (if it exists), or the hour component (again, if it exists), or the ‘PT’ identifier. Using the Mid function in this way starts to get quite long, so another alternative is to first remove everything that is not the second component, remove the ‘S’ identifier (using the Substitute function), then taking the value of it:\n\n``` If(\nIsBlank(Find(\"S\", ThisItem.Time)), // Check if there is a second component\n0, // If not, the value is zero\nValue( // Otherwise take the substring after the minute or hour indicator,\nSubstitute( // remove the 'S' indicator, then take the value\nMid(\nThisItem.Time,\nIf(\n!IsBlank(Find(\"M\", ThisItem.Time)), // If there is a minute component\nFind(\"M\", ThisItem.Time) + 1, // Then start right after it\n!IsBlank(Find(\"H\", ThisItem.Time)), // Else if there is an hour component\nFind(\"H\", ThisItem.Time) + 1, // Then start after it\n3)), // Else start after 'PT'\n\"S\",\n\"\")))\n```\n\nHopefully that helped clarify a little about the structure of this (quite large) expression. We’ll try to make this simpler in the future, but that’s something that can be used today.\n\n## Using time values in forms\n\nThe (quite large) expression from the previous section can be used to display the proper time value in a gallery, for example, but how would it work in forms?\n\nOn display forms it’s fairly straightforward, we can update the `Default` property of the data card that displays the time value so that it is formatted as we want. For example, we can use the following expression to display the time in hours:minutes:seconds format (again, assuming that the column name is ‘Time’):\n\n``` Text(\nTime(\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)), // Check if there is an hour component\n0, // If not, the value is zero\nValue( // Otherwise, take the substring between 'PT' and 'H'\nMid(\nThisItem.Time,\n3,\nFind(\"H\", ThisItem.Time) - 3))),\nIf(\nIsBlank(Find(\"M\", ThisItem.Time)), // Check if there is a minute component\n0, // If not, the value is zero\nValue( // Otherwise take the substring between the rest of the value\nMid( // after the hour component and the 'M' indicator\nThisItem.Time,\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)),\n3,\nFind(\"H\", ThisItem.Time) + 1),\nFind(\"M\", ThisItem.Time) -\nIf(\nIsBlank(Find(\"H\", ThisItem.Time)),\n3,\nFind(\"H\", ThisItem.Time) + 1)))),\nIf(\nIsBlank(Find(\"S\", ThisItem.Time)), // Check if there is a second component\n0, // If not, the value is zero\nValue( // Otherwise take the substring after the minute or hour indicator,\nSubstitute( // remove the 'S' indicator, then take the value\nMid(\nThisItem.Time,\nIf(\n!IsBlank(Find(\"M\", ThisItem.Time)),\nFind(\"M\", ThisItem.Time) + 1,\n!IsBlank(Find(\"H\", ThisItem.Time)),\nFind(\"H\", ThisItem.Time) + 1,\n3)),\n\"S\",\n\"\")))),\n\"hh:mm:ss\")\n```",
null,
"For edit forms, if we want to let the user edit the time value in a text input control, then the expression above will work fine – it will display the editable text in the hh:mm:ss format which can be updated by the user. Writing the value back is not a problem – the SQL connector accepts the data in that format when writing back to the server.",
null,
"But like with a date column where we have a date picker, we can use a “time picker” as well, with a trio of dropdown values that can be used to select the time in a more user-friendly way:",
null,
"To implement that, we can add three dropdown controls (I’ll call them ‘ddHour’, ‘ddMinute’ and ‘ddSecond’ here) to the data card corresponding to the time column, and set their properties as follows:\n\n``` ddHour.Items: [\"00\",\"01\",\"02\",\"03\",\"04\",\"05\",\"06\",\"07\",\"08\",\"09\",\"10\",\"11\",\"12\",\"13\",\"14\",\"15\",\"16\",\"17\",\"18\",\"19\",\"20\",\"21\",\"22\",\"23\"]\nddMinute.Items: [\"00\",\"01\",\"02\",\"03\",\"04\",\"05\",\"06\",\"07\",\"08\",\"09\",\"10\",\"11\",\"12\",\"13\",\"14\",\"15\",\"16\",\"17\",\"18\",\"19\",\"20\",\"21\",\"22\",\"23\",\"24\",\"25\",\"26\",\"27\",\"28\",\"29\",\"30\",\"31\",\"32\",\"33\",\"34\",\"35\",\"36\",\"37\",\"38\",\"39\",\"40\",\"41\",\"42\",\"43\",\"44\",\"45\",\"46\",\"47\",\"48\",\"49\",\"50\",\"51\",\"52\",\"53\",\"54\",\"55\",\"56\",\"57\",\"58\",\"59\"]\nddSecond.Items: [\"00\",\"01\",\"02\",\"03\",\"04\",\"05\",\"06\",\"07\",\"08\",\"09\",\"10\",\"11\",\"12\",\"13\",\"14\",\"15\",\"16\",\"17\",\"18\",\"19\",\"20\",\"21\",\"22\",\"23\",\"24\",\"25\",\"26\",\"27\",\"28\",\"29\",\"30\",\"31\",\"32\",\"33\",\"34\",\"35\",\"36\",\"37\",\"38\",\"39\",\"40\",\"41\",\"42\",\"43\",\"44\",\"45\",\"46\",\"47\",\"48\",\"49\",\"50\",\"51\",\"52\",\"53\",\"54\",\"55\",\"56\",\"57\",\"58\",\"59\"]\nddHour.Default: Left(Parent.Default, 2)\nddMinute.Default: Mid(Parent.Default, 4, 2)\nddSecond.Default: Mid(Parent.Default, 7, 2)\n```\n\nAnd then set the Update property in the time data card to\n\n``` ddHour.Selected.Value & \":\" & ddMinute.Selected.Value & \":\" & ddSecond.Selected.Value\n```\n\nAnd that closes the circle, allowing us to update the time values back to the server.\n\n## Wrapping up\n\nThe expressions above can be used to better work with time columns in a SQL Server database. Currently it’s not as straightforward as it can be, but it can be done. As we get more requests for it, we can prioritize it appropriately and make improvements to the product to make this scenario simpler. If you feel that this is important, please vote up on this feature request on the PowerApps Ideas board."
]
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null,
"https://web.powerapps.com/pixeltag",
null,
"https://secure.gravatar.com/avatar/ffcff3d81bacfd98d5748c5ba59a9021",
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https://docs.paparazziuav.org/v5.14/pprz__isa_8h.html | [
"",
null,
"Paparazzi UAS v5.14.0_stable-0-g3f680d1 Paparazzi is a free software Unmanned Aircraft System.\npprz_isa.h File Reference\n\nPaparazzi atmospheric pressure conversion utilities. More...\n\n`#include \"std.h\"`\n`#include <math.h>`",
null,
"Include dependency graph for pprz_isa.h:",
null,
"This graph shows which files directly or indirectly include this file:\n\nGo to the source code of this file.\n\n## Macros\n\n#define PPRZ_ISA_SEA_LEVEL_PRESSURE 101325.0\nISA sea level standard atmospheric pressure in Pascal. More...\n\n#define PPRZ_ISA_SEA_LEVEL_TEMP 288.15\nISA sea level standard temperature in Kelvin. More...\n\n#define PPRZ_ISA_TEMP_LAPS_RATE 0.0065\ntemperature laps rate in K/m More...\n\n#define PPRZ_ISA_GRAVITY 9.80665\nearth-surface gravitational acceleration in m/s^2 More...\n\n#define PPRZ_ISA_GAS_CONSTANT 8.31447\nuniversal gas constant in J/(mol*K) More...\n\n#define PPRZ_ISA_MOLAR_MASS 0.0289644\nmolar mass of dry air in kg/mol More...\n\n#define PPRZ_ISA_AIR_GAS_CONSTANT (PPRZ_ISA_GAS_CONSTANT/PPRZ_ISA_MOLAR_MASS)\nuniversal gas constant / molar mass of dry air in J*kg/K More...\n\n#define PPRZ_ISA_AIR_DENSITY 1.225\nstandard air density in kg/m^3 More...\n\n#define CelsiusOfKelvin(_t) (_t - 274.15f)\nConvert temperature from Kelvin to Celsius. More...\n\n#define KelvinOfCelsius(_t) (_t + 274.15f)\nConvert temperature from Celsius to Kelvin. More...\n\n## Functions\n\nstatic float pprz_isa_altitude_of_pressure (float pressure)\nGet absolute altitude from pressure (using simplified equation). More...\n\nstatic float pprz_isa_height_of_pressure (float pressure, float ref_p)\nGet relative altitude from pressure (using simplified equation). More...\n\nstatic float pprz_isa_pressure_of_altitude (float altitude)\nGet pressure in Pa from absolute altitude (using simplified equation). More...\n\nstatic float pprz_isa_pressure_of_height (float height, float ref_p)\nGet pressure in Pa from height (using simplified equation). More...\n\nstatic float pprz_isa_height_of_pressure_full (float pressure, float ref_p)\nGet relative altitude from pressure (using full equation). More...\n\nstatic float pprz_isa_ref_pressure_of_height_full (float pressure, float height)\nGet reference pressure (QFE or QNH) from current pressure and height. More...\n\nstatic float pprz_isa_temperature_of_altitude (float alt)\nGet ISA temperature from a MSL altitude. More...\n\n## Variables\n\nstatic const float PPRZ_ISA_M_OF_P_CONST = (PPRZ_ISA_AIR_GAS_CONSTANT *PPRZ_ISA_SEA_LEVEL_TEMP / PPRZ_ISA_GRAVITY)\n\n## Detailed Description\n\nPaparazzi atmospheric pressure conversion utilities.\n\nConversion functions are use to approximate altitude from atmospheric pressure based on the standard model and the International Standard Atmosphere (ISA)\n\nDefinition in file pprz_isa.h."
]
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"https://docs.paparazziuav.org/v5.14/penguin_icon.png",
null,
"https://docs.paparazziuav.org/v5.14/closed.png",
null,
"https://docs.paparazziuav.org/v5.14/closed.png",
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https://2021.help.altair.com/2021.2/hwsolvers/ms/topics/solvers/ms/xml-format_41.htm | [
"# Constraint: Joint\n\nModel ElementConstraint_Joint is used to specify an idealized connector between two bodies.\n\n## Description\n\nConstraint_Joint defines a set of lower pair constraints. Physically, the joint consists of two mating surfaces that allow relative translational and/or rotational movement in certain specific directions only. The surfaces are abstracted away, and the relationships are always expressed as a set of equations between points and directions on two bodies.\n\n## Format\n\n<Constraint_Joint\nid = \"integer\"\nlabel = \"Name of Joint\"\ntype = \"CONSTANT_VELOCITY\"\n\"CYLINDRICAL\"\n\"FIXED\"\n\"FREE\"\n\"HOOKE\"\n\"INLINE\"\n\"INPLANE\"\n\"ORIENTATION\"\n\"PARALLEL_AXES\"\n\"PERPENDICULAR\"\n\"PLANAR\"\n\"RACK_PINION\"\n\"REVOLUTE\"\n\"SCREW\"\n\"SPHERICAL\"\n\"TRANSLATIONAL\"\n\"UNIVERSAL\"\ni_marker_id = \"integer\"\nj_marker_id = \"integer\"\n[ pitch = \"real\" ]\n[ pitch_diameter = \"real\" ]\n[ is_virtual = {\"FALSE\" | \"TRUE\"} ]\n/>\n\n## Attributes\n\nid\nThe element identification number, (integer>0). This is a number that is unique among all Constraint_Joint elements.\nlabel\nThe name of the Constraint_Joint element.\ni_marker_id\nSpecifies a Reference_Marker that defines the connection on the first body. The body may be a rigid body, a flexible body, or a point body. The parameter is required.\nj_marker_id\nSpecifies a Reference_Marker that defines the connection on the first body. The body may be a rigid body, a flexible body, or a point body.\ntype\nSpecifies the type of constraint connection between the i_marker and j_marker_id.\n\ntype may be one of the following:\n\n\"CONSTANT_VELOCITY\"\n\n\"CYLINDRICAL\"\n\n\"FIXED\"\n\n\"FREE\"\n\n\"HOOKE\"\n\n\"INLINE\"\n\n\"INPLANE\"\n\n\"ORIENTATION\"\n\n\"PARALLEL_AXES\"\n\n\"PERPENDICULAR\"\n\n\"PLANAR\"\n\n\"RACK_PINION\"\n\n\"REVOLUTE\"\n\n\"SCREW\"\n\n\"SPHERICAL\"\n\n\"TRANSLATIONAL\"\n\n\"UNIVERSAL\"\n\npitch\nDefines the pitch of a screw joint. The pitch defines the ratio between the translational and rotational motion in the screw joint.\n\nThe parameter is required only when TYPE = \"SCREW_JOINT\".\n\npitch_diameter\nDefines the pitch diameter of the pinion gear a rack and pinion gear constraint.\n\nThe parameter is mandatory when TYPE= \"RACK_PINION\".\n\nis_virtual\nDefines whether the constraint is virtual or regular. If is_virtual is set to TRUE, the constraint is implemented as a virtual constraint. If is_virtual is set to FALSE, the constraint is implemented as a regular, algebraic constraint. This parameter is optional. The default is FALSE. See Comment 22 for more information about virtual joints.\n\n## Example\n\nThe image below shows a simple pendulum system. The system consists of a rigid bar B that is pinned to Ground G at point P. The center of mass of the Bar B is denoted by a Reference_Marker B*.\n\nThe connection at P requires that the corresponding material points on B (P1) and G (P2) always be superposed. It also requires that the axis coming out of the plane of the paper at P1 and a second axis at P2 always remain collinear. This is a connection of type \"REVOLUTE\".\n\nGravity acts in the global negative Y direction. Due to the effect of gravity, bar B swings in the global X-Y plane while satisfying the requirements that P1 and P2 remain superposed and the axes remain collinear.\nIf we define the following objects in the input data file:\n• P1 = Reference_Marker 11 on Ground, z-axis coming out of the plane\n• P2 = Reference_Marker 21 on Body B, z-axis coming out of the plane\n\nThe Constraint_Joint object may be defined as:\n\n<Constraint_Joint\nid = \"1\"\nlabel = \"Name of Joint\"\ni_marker_id = \"21\"\nj_marker_id = \"11\"\ntype = \"REVOLUTE\"\n/>\n\n1. Constraint_Joint defines a set of constraints in the model. These constraints only allow the two connected bodies to have relative motion in certain specific directions. Motion in all the other directions is prohibited. The system is \"constrained\" not to move in those directions.\n2. In their most general form, constraints can involve displacements, velocities, and time. Constraint_Joint elements are distinguished by the fact that they only involve displacements. The constraint relationships do not involve velocities or time explicitly. Therefore, they do not add or remove energy from the system.\n3. An internal reaction force or moment is associated with each constraint in a Constraint_Joint. This internal reaction ensures that the relative motion of the bodies specified in the connector satisfy the constraints. The reaction force in a Constraint_Joint may be accessed using the JOINT() function expression.\n\nThe internal reaction forces and moments at each Reference_Marker of the joint obey Newton's third law of motion. Their magnitude and direction is such that they cancel each other out.\n\n4. It is possible to \"over-specify\" the constraints in a model. In this case, the system is said to be over-constrained, and the constraints are said to be \"redundant\".\n\nConsider a door connected to a frame as shown in Figure 2 below.\n\nBoth the door as well as the frame are assumed to be rigid. Three hinges connect the door to the frame, allowing the door to open and close. The hinges are modeled as revolute joints. In this idealization, the hinges are assumed to be massless and rigid. Each hinge allows rotation between the door and the frame about an axis, shown as a red arrow in Figure 2.\n\nIn this idealization, a few points should be noted:\n• The hinge axes are required to be perfectly collinear. Otherwise the door cannot be opened or closed.\n• Only one hinge is truly necessary. The remaining two are \"redundant\".\n\nIf such a system is provided to MotionSolve, it will detect a redundancy in the constraints and remove the equivalent of two hinges from the model. It will solve the system so that the correct motion is predicted.\n\nThe effect of removing two hinges from the solution process is to set their reaction forces to zero. Clearly, this is unexpected behavior. In reality, all three hinges have reaction forces and torques. This is an issue with the idealization of the system. If at least one of two bodies, the door or the frame, were made flexible, then more realistic reactions would be observed at each hinge.\n\n5. TYPE = \"CONSTANT_VELOCITY\"\n\nThe constant velocity joint is a fairly complex subsystem. It ensures that the angular speed of the output shaft is the same as the angular speed of the input shaft, regardless of their relative orientation. The construction of a constant velocity joint is shown below in Figure 3.\n\nThe detail in the constant velocity joint is abstracted away; all intermediate parts are ignored and the input-output relationship between the shafts is captured via a simple set of algebraic constraints. Energy loss is not modeled.\n\n6. TYPE = \"CYLINDRICAL\"\n\nThe cylindrical joint allows two degrees of freedom between the bodies it constraints. Figure 4 shows a schematic of a cylindrical joint. It allows a body (Body-1) to translate and rotate along a fixed axis defined on another body (Body-2).\n\n7. TYPE = \"FIXED\"\n\nThe fixed joint forces two bodies to move together as if they were welded together. No relative motion between the two bodies is allowed. Figure 5 shows a schematic of a fixed joint.\n\nThe location and orientation of any coordinate system \"etched\" on Body-2 is fixed with respect to any coordinate \"etched\" on Body-1.\n\n8. TYPE = \"FREE\"\n\nThe free joint is the exact opposite of the fixed joint. It does not restrict any motion between the two bodies it connects. The free joint is typically used to define topology hierarchy, when joint (also known as relative) coordinates are used to internally represent the system.\n\nIn a free joint, any coordinate system fixed on Body-2 can move in all possible ways with respect to a coordinate system fixed on Body-1.\n\n9. TYPE = \"HOOKE\"\nThe image below shows a schematic of a HOOKE joint. This joint constrains two Reference_Markers, I on Body-1 and J on Body-2 such that (a) their origins Oi and Oj remain superposed, and (b) the I marker x-axis is always perpendicular to the J marker y-axis.\n\nThe HOOKE joint removes four degrees of freedom from a system. Body-1 can rotate about the Xi axis and Body-2 can rotate about the Yi axis. Relative translations are not allowed. Physically, the HOOKE joint is identical to the UNIVERSAL joint.\n\n10. TYPE = \"INLINE\"\n\nThis is a constraint primitive that requires that the origin of a Reference_Marker on Body-1 translate along the z-axis of a Reference_Marker placed on Body-2. All rotations are allowed. Figure 8 shows a schematic of an INLINE primitive.\n\nThe inline primitive constrains two translational degrees of freedom. It prohibits translation along the x- and y-axes of the Reference_Marker on Body-2.\n\n11. TYPE = \"INPLANE\"\n\nThis is a constraint primitive that requires that the origin of a Reference_Marker on Body-1 (I in the figure below) to stay in the XY plane defined by the origin of the J Reference_Marker and its z-axis. Figure 9 shows a schematic of an INPLANE primitive.\n\nThe INPLANE primitive constrains one translational degree of freedom. It prohibits translation along the z-axis of the Reference_Marker J on Body-2. All rotations are allowed.\n\n12. TYPE = \"ORIENTATION\"\n\nFigure 10 shows a schematic of an ORIENTATION joint. It requires that the orientation of a Reference_Marker on Body-1 (I in the figure below) always be the same as the orientation of a Reference_Marker on Body-2 (J in the figure below).\n\nThe ORIENTATION joint removes three degrees of freedom; all of these are rotational. All three translations of Body-1 with respect to Body-2 are allowed.\n\n13. TYPE = \"PARALLEL_AXES\"\n\nFigure 11 shows a schematic of a PARALLEL_AXES joint. This joint constrains Body-1 such that the z-axis of a Reference_Marker (I in the figure) is always parallel to the z-axis of a Reference_Marker on Body-2 (J in the figure).\n\nThe PARALLEL_AXIS joint removes two rotational degrees of freedom. Rotation of Body-1 is only allowed about the z-axis of the I Reference_Marker. All three translations are allowed.\n\n14. TYPE = \"PERPENDICULAR\"\n\nFigure 12 shows a schematic of a PERPENDICULAR_AXES joint. This joint constrains Body-1 such that the z-axis of a Reference_Marker (I in the figure) is always PERPENDICULAR to the z-axis of a Reference_Marker on Body-2 (J in the figure).\n\nThe PERPENDICULAR _AXIS joint removes one degree of freedom. Rotation of Body-1 is only allowed about the z-axis of the I Reference_Marker and the z-axis of the J Reference_Marker. All three translations are allowed.\n\n15. TYPE = \"PLANAR\"\n\nThe planar joint requires that a Reference_Marker, I, defined on Body-1 be restricted in its movement such that its origin lies in the x-y plane defined by the origin of a Reference_Marker, J, defined on BODY-2. Figure 13 shows a schematic of a PLANAR joint.\n\nThe PLANAR joint allows three degrees of freedom, two translational degrees of freedom in the plane and one \"drilling\" rotational degree of freedom.\n\n16. TYPE = \"RACK_PINION\"\n\nFigure 14 below depicts a rack and pinion joint. The pinion gear P is connected to the housing (not shown) such that it is allowed to rotate about an axis coming out the plane of the paper. The rack R slides from left-to-right. The RACK_PINION joint relates the rotation of the pinion θp to the translation of the rack S.\n\nThe RACK_PINION joint allows five degrees of freedom. It imposes only one constraint on the system.\n\n17. TYPE = \"REVOLUTE\"\n\nFigure 15 shows a schematic of a REVOLUTE joint. This joint constrains two Reference_Markers I and J such that (a) their origins are always superposed, and (b) their z-axes are collinear.\n\nThe REVOLUTE joint removes five degrees of freedom from a system. Rotation about the common z-axes is the only motion that is allowed.\n\n18. TYPE = \"SCREW\"\n\nFigure 16 depicts a screw joint. Body-2 contains screw-threads with a pitch p. The axis of the screw is defined by the z-axis of a Reference_Marker J on Body-2. Body-1 has corresponding threads that allow it to mate with the threads on Body-2. A rotation of Body-2 along the screw axis causes Body-1 to translate up or down along the screw axis. The SCREW joint relates the rotation of the Body-1 about the screw axis to its translation along the screw axis.\n\nThe SCREW joint removes only one degree of freedom from a system.\n\n19. TYPE = \"SPHERICAL\"\n\nA SPHERICAL joint shows a schematic of a SPHERICAL joint. This joint constrains two Reference_Markers, I on Body-1 and J on Body-2, such that their origins Oi and Oj are always superposed.\n\nThe SPHERICAL joint removes three degrees of freedom from a system. Body-2 is allowed to rotate with respect to Body-1. Relative translation is not allowed.\n\n20. TYPE = \"TRANSLATIONAL\"\n\nFigure 18 shows a schematic of a TRANSLATIONAL joint. This joint constrains two Reference_Markers I and J such that (a) their orientations remain the same, and (b) they share a single common axis of translation which is the z-axis of Reference_Marker J.\n\nThe TRANSLATIONAL joint removes five degrees of freedom from a system. Relative translation along the z-axis of Reference_Marker J is the only motion that is allowed.\n\n21. TYPE = \"UNIVERSAL\"\n\nFigure 19 below shows a schematic of a UNIVERSAL joint. This joint constrains two Reference_Markers, I on Body-1 and J on Body-2 such that (a) their origins Oi and Oj remain superposed and (b) their z-axes are always orthogonal.\n\nThe UNIVERSAL joint removes four degrees of freedom from a system. Body-2 can rotate about the Zj axis and Body-1 can rotate about the Zi axis. Relative translations are not allowed.\n\nNote that the output angular velocity will not be equal to the input angular velocity.\n\nLet:\n• ù be the speed of the input shaft\n• ù2 be the speed of the output shaft,\n• â the angle between the shafts\n• f1 the rotation angle of the input shaft\nThen:(1)\n${\\omega }_{2}=\\frac{{\\omega }_{1}\\mathrm{cos}\\beta }{1-{\\mathrm{sin}}^{2}\\beta {\\mathrm{sin}}^{2}{\\varphi }_{1}}$\n\nAt β=π/2, the output angular velocity is zero. Physically, the shafts lock. This manifests itself as a singular matrix in MotionSolve.\n\nOnly at β=0, is the output shaft angular velocity equal to the input shaft angular velocity.\n\nThe angular acceleration of the output shaft, α2, can be expressed as:(2)\n${\\alpha }_{2}=\\frac{{\\omega }_{1}^{2}{\\mathrm{sin}}^{2}\\beta \\mathrm{cos}\\beta \\mathrm{sin}2{\\varphi }_{1}}{{\\left(1-{\\mathrm{sin}}^{2}\\beta {\\mathrm{sin}}^{2}{\\varphi }_{1}\\right)}^{2}}$\n22. In Multibody Dynamics, a regular, or ideal joint, can be defined as a set of algebraic constraint equations:\n\n$\\mathrm{\\Phi }\\left(q\\right)=0$\n\nwith q being the generalized coordinates of the system, such as the positions and orientations of each rigid body. This set of constraint equations is then incorporated into the equations of motion (EOM) in MotionSolve, using the method of Lagrange multipliers, to form a set of differential algebraic equations (DAE). Each constraint equation reduces the degrees of freedom (DOF) of the multibody system by one. The total kinematic DOFs of the mechanism will be the number of the generalized coordinates, q, minus the number of the algebraic constraint equations of the multibody system. If we virtualize the joint, the constraint equations are extended to:\n\n$\\epsilon -\\mathrm{\\Phi }\\left(q\\right)=0$\n\nwith ε being treated as extra generalized coordinates and solved alongside other generalized coordinates q in the DAE. Instead of solving the algebraic equation alongside the EOM, you can instead introduce a violation force that acts in the opposite direction to the violation in order to minimize ε . The force is defined as:\n\nwhere G is a gigantic factor (usually in the order of 1e12) and K is an arbitrary stiffness (for example K = 1). This equation resembles closely a second order system, such as a mass-spring-damper, where the damping factor is chosen to allow for critical damping. The precise values of G and K are intrinsic to the solver and are adjusted automatically based on the overall properties of the system and cannot be changed manually. The force f is then added to the RHS of the Equations of Motion.\n\nThe difference of this gigantic element approach compared to a general penalty approach (see bushings) lies in the degree of violation or penalty. In this novel approach, ε is kept so small that the constraint violation is in many cases undetectable without creating a large burden (numerical stiffness) for the solver. From a user’s perspective, a regular and virtual joint/constraint will provide the same kinematics and dynamics, with the exception that a virtual joint/constraint cannot introduce constraint redundancy. Furthermore, virtual joints/constraints can be thought of as “soft coupling” that allows for other numerical approaches to solve the General Equations of Motions.\n\nVirtual constraints can be used in all analysis types, such as static, quasi-static, transient, and linear. As their algebraic counterpart, they can be activated and deactivated within an analysis."
]
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https://www.geeksforgeeks.org/find-the-interval-which-contains-maximum-number-of-concurrent-meetings/?ref=rp | [
"Skip to content\nRelated Articles\nFind the interval which contains maximum number of concurrent meetings\n• Difficulty Level : Medium\n• Last Updated : 17 Sep, 2020\n\nGiven a two dimensional array arr[][] of dimensions N * 2 which contains the starting and ending time for N meetings on a given day. The task is to print a list of time slots during which most number of concurrent meetings can be held.\n\nExamples:\n\nInput: arr[][] = {{100, 300}, {145, 215}, {200, 230}, {215, 300}, {215, 400}, {500, 600}, {600, 700}}\nOutput: [215, 230]\nExplanation:\nThe given 5 meetings overlap at {215, 230}.\n\nInput: arr[][] = {{100, 200}, {50, 300}, {300, 400}}\nOutput: [100, 200]\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach: The idea is to use a Min-Heap to solve this problem. Below are the steps:\n\n1. Sort the array based on the start time of meetings.\n2. Initialize a min-heap.\n3. Initialize variables max_len, max_start and max_end to store maximum size of min heap, start time and end time of concurrent meetings respectively.\n4. Iterate over the sorted array and keep popping from min_heap until arr[i] becomes smaller than the elements of the min_heap, i.e. pop all the meetings having ending time smaller than the starting time of current meeting, and push arr[i] in to min_heap.\n5. If the size of min_heap exceeds max_len, then update max_len = size(min_heap), max_start = meetings[i] and max_end = min_heap_element.\n6. Return the value of max_start and max_end at the end.\n\nBelow is the implementation of the above approach:\n\n## C++14\n\n `// C++14 implementation of the``// above approach``#include ``using` `namespace` `std;`` ` `bool` `cmp(vector<``int``> a,vector<``int``> b)``{`` ` ` ``if``(a != b)`` ``return` `a < b;`` ` ` ``return` `a - b;``}`` ` `// Function to find time slot of``// maximum concurrent meeting``void` `maxConcurrentMeetingSlot(`` ``vector> meetings)``{`` ` ` ``// Sort array by`` ``// start time of meeting`` ``sort(meetings.begin(), meetings.end(), cmp);`` ` ` ``// Declare Minheap`` ``priority_queue<``int``, vector<``int``>,`` ``greater<``int``>> pq;`` ` ` ``// Insert first meeting end time`` ``pq.push(meetings);`` ` ` ``// Initialize max_len,`` ``// max_start and max_end`` ``int` `max_len = 0, max_start = 0;`` ``int` `max_end = 0;`` ` ` ``// Traverse over sorted array`` ``// to find required slot`` ``for``(``auto` `k : meetings)`` ``{`` ` ` ``// Pop all meetings that end`` ``// before current meeting`` ``while` `(pq.size() > 0 && `` ``k >= pq.top())`` ``pq.pop();`` ` ` ``// Push current meeting end time`` ``pq.push(k);`` ` ` ``// Update max_len, max_start`` ``// and max_end if size of`` ``// queue is greater than max_len`` ``if` `(pq.size() > max_len)`` ``{`` ``max_len = pq.size();`` ``max_start = k;`` ``max_end = pq.top();`` ``}`` ``}`` ` ` ``// Print slot of maximum`` ``// concurrent meeting`` ``cout << max_start << ``\" \"` `<< max_end;``}`` ` `// Driver Code``int` `main()``{`` ` ` ``// Given array of meetings`` ``vector> meetings = { { 100, 200 },`` ``{ 50, 300 },`` ``{ 300, 400 } };`` ` ` ``// Function call`` ``maxConcurrentMeetingSlot(meetings);``}`` ` `// This code is contributed by mohit kumar 29`\n\n## Java\n\n `// Java implementation of the``// above approach`` ` `import` `java.util.*;``import` `java.lang.*;`` ` `class` `GFG {`` ` ` ``// Function to find time slot of`` ``// maximum concurrent meeting`` ``static` `void` `maxConcurrentMeetingSlot(`` ``int``[][] meetings)`` ``{`` ` ` ``// Sort array by`` ``// start time of meeting`` ``Arrays.sort(meetings,`` ``(a, b)`` ``-> (a[``0``] != b[``0``])`` ``? a[``0``] - b[``0``]`` ``: a[``1``] - b[``1``]);`` ` ` ``// Declare Minheap`` ``PriorityQueue pq`` ``= ``new` `PriorityQueue<>();`` ` ` ``// Insert first meeting end time`` ``pq.add(meetings[``0``][``1``]);`` ` ` ``// Initialize max_len,`` ``// max_start and max_end`` ``int` `max_len = ``0``, max_start = ``0``;`` ``int` `max_end = ``0``;`` ` ` ``// Traverse over sorted array`` ``// to find required slot`` ``for` `(``int``[] k : meetings) {`` ` ` ``// Pop all meetings that end`` ``// before current meeting`` ``while` `(!pq.isEmpty()`` ``&& k[``0``] >= pq.peek())`` ``pq.poll();`` ` ` ``// Push current meeting end time`` ``pq.add(k[``1``]);`` ` ` ``// Update max_len, max_start`` ``// and max_end if size of`` ``// queue is greater than max_len`` ``if` `(pq.size() > max_len) {`` ``max_len = pq.size();`` ``max_start = k[``0``];`` ``max_end = pq.peek();`` ``}`` ``}`` ` ` ``// Print slot of maximum`` ``// concurrent meeting`` ``System.out.println(`` ``max_start + ``\" \"` `+ max_end);`` ``}`` ` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// Given array of meetings`` ``int` `meetings[][]`` ``= { { ``100``, ``200` `}, `` ``{ ``50``, ``300` `}, `` ``{ ``300``, ``400` `} };`` ` ` ``// Function Call`` ``maxConcurrentMeetingSlot(meetings);`` ``}``}`\nOutput:\n```100 200\n```\n\nTime Complexity: O(N * logN)\nAuxiliary Space: O(N)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with industry experts, please refer Geeks Classes Live\n\nMy Personal Notes arrow_drop_up"
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https://uweb.engr.arizona.edu/~blowers/201project/Interrelatn-mainpg.HTML | [
"Interrelations\n\nChapter Exercise point A to B Interrelation Explanation\n2 Unit Conversions1 atm = ? psi1 atm / 14.7 psiincludes a g-mole to lb-mole example\n2Using Given Relations5000 gal. oil = ? airplanesuse given conversionsairplane problem\n3Pressure Topics3 psig = ? psia Pabs=Patm + PgaugeStatic Pressure, Manometers\n3Temperature ConversionsTemp. change of 37°C = ? °FΔ T(°F)=1.8*Δ T(°C)Temperature Formulas\n3\nDensity\nSolids and Liquids\nSG&Ref or mass&vol of Cu = ? ρ of Cuρ(Cu)=ρ(Ref)*SG(Cu) or ρ(Cu)=m(Cu)/V(Cu) Concise Explanations\n3Concentrations1g salt in 1cup of water = ? (mol fraction)mol fraction = mol salt / mol solutionMolarity, Mass/Mole Fractions, ppm, and ppb\n3/4Flow Rates2 gal. H2O in 3 min. = ? m-dot (g/s)m-dot = mass/timewhat is m-dot? why do we use it?\n2/3ReviewT, P, ρ, xA, yA, m-dotmake a study guideInterrelations Review\n4\nMaterial Balance\nIntro\ngiven material stream info => desired answerread prob, write flow chart, solveKey points and terms\n4\nMaterial Balance\nWithout Reactors\ngiven flow rates => write balances, solvehere, in=out for all components\nIncludes\nRecyle and Bypass\n4\nMaterial Balances\nReactive Systems\ngiven flow rates => write balances, solveUse mole balances via extents of reactionIncludes some key interrelations\n4\nMaterial Balances\nReview\nMaterial Balance Test => desired resultsKnow the algorithms (see review), lots of practiceReview\n5Ideal GasSTP of O2(gas) => ? density ρ = P*MW / R*TSTP, Molar Volume, and More\n5Gas Mixturesair @ 1atm, 78% O2 => ? atm O2Pi = yi*PDalton's Law and Amagat's Law\n6\nGas-Liquid/Solid\nIntro, Laws\nair is .1 %H2O at 40°C => ? P* H2Odirect relation of P* & TRaoult's Law, Saturation Formulas, and More\n6\nGas-Liquid/Solid\nDiagrams\nair is .1 %H2O at 40°C => ? PH2O Pi = xi*Pi*(T)Pxy/Txy Diagrams and Phase Diagrams\n6Solids in Liquids add solid to a pure liquid => ? raise or lower P*lower P*3 Colligative Properties\n6\nLiquid Mixtures\n35% H2O, 30% Acetone, MIBK balance => separates how?Use Ternary Phase DiagramConcise Explanation\n7\nEnergy Balance\nIntro\n2 ton truck @ 65mph\n=> ? Ek & Ep\nEk = 1/2*m*u2\nEp = m*g*h\nSeveral Key Interrelations\n7\nEnergy Balance\nClosed Systems\nGiven a Closed System, Where do you start? ΔU + ΔEk + ΔEp = Q - WExplanation of the Equation\n7\nEnergy Balance\nOpen Systems\nGiven a Open System, Where do you start?ΔH + ΔEk + Δ Ep = Q - WsExplanation of the Equation\n7\nEnergy Balance\nMechanical Systems\nGiven a Mechanical System, Where do you start?(m-dot * ΔP / ρ + ΔEk + ΔEp = WsExplanation of the Equation\n8\nEnergy Balance\nChange in T or P\nCu is raised from 25°C to 400°C => ? ΔH\nΔH =\nIntegral(Cp) | T1 to T2\nEnthalpy Changes\n8\nEnergy Balance\nChanges of State\nCu melts => ΔHΔH tablePaths for Finding Energies\n9\nEnergy Balance\nReactive Systems\nGiven reaction conditions => ΔHΔ Hrxn =\nΣ(vi*ΔHfi)products\n- Σ(vi*Δ Hfi)reactants\nExamples Using All Materials in Course\n7-9\nEnergy Balance\nReview\nEnergy Balance problemWrite E-bal, ?determine Δ H?Review\n\nGoto | Interrelations Home | Dr. Blowers Home | U of A Home",
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https://discuss.interviewbit.com/t/python-logical-solution/51799 | [
"",
null,
"# Python logical solution\n\n#1\n\ndef solve(self, A, B, C):\n\n`````` dic={}\nfor i in set(A):\nif(i in dic):\ndic[i]+=1\nelse:\ndic[i]=1\n\nfor j in set(B):\nif(j in dic):\ndic[j]+=1\nelse:\ndic[j]=1\n\nfor k in set(C):\nif(k in dic):\ndic[k]+=1\nelse:\ndic[k]=1\n\nA=set(A)\nB=set(B)\nC=set(C)\nans=[]\nfor m in dic:\nif(m in A and m in B and m in C):\nans.append(m)\n\nelif(m in A and m in B and m not in C):\nans.append(m)\n\nelif(m in A and m not in B and m in C):\nans.append(m)\n\nelif(m not in A and m in B and m in C):\nans.append(m)\n\nelse:\ncontinue\nans.sort()\nreturn(ans)``````"
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http://goingsolar.co.za/2013/11/06/three-phase/ | [
"# Three Phase\n\nAs my system is 3 Phase, I thought I’d talk about some of the different 3 Phase standards for wiring.\n\n3 Phase, as you may or may not know, is better for transmission of power.\n\nPower stations generate electricity at 22 000 volts (3 phase 50Hz). To transmit this power over long distances, Eskom steps up the power to to the following voltages for transmission: 220kV; 275kV; 400kV or 765kV. This electricity now goes into our national grid.\n\nWhen it gets to the end user it is stepped down. This could be 11kV for a large factory or 400V(380V) for shops/homes. If you take a phase to neutral (single phase voltage) i.e. 400V/sqrt(3) you will get 230V single phase @ 50Hz.\n\nWhen it gets to the house, it generally gets split up into single phase, and different circuits get each phase. So, the lights might be on one phase, the plugs on another, and heavy equipment may use all three (eg an old 1950’s Oven dating back to the Union of South Africa!).\nPlug sockets at home are single phase 230VAC 50Hz.\n\nThere are two main connection standards for 3 phase; Delta wiring – which uses 3 wires, and Y wiring (also known as wYe), which uses 4.\n\nDelta has one wire for each phase so 3 wires total.\nY wiring has one wire for each phase, plus a neutral, for 4 wires.\n\nIn my house, we have 4 wires, so its the Y standard.\n\nYou can check this by looking at how many wires go from the house to the street.\nIn our case, this is 4 separate wires, as you can see below:",
null,
"So, what exactly is 3 phase?\n\nAC current runs in a sine wave. This sine wave runs at 50hz for South Africa (50 ups and downs a second). In 3 phase, each phase is run at an offset of 120 degrees, so each phase peaks at an offset of the other. The 3 phases add up to a total of 380v – although these days is more likely to be closer to 400v, as the rest of the world has migrated to that. Either makes no difference, as they’re both values within the margin of error for provision of electricity.\n\nIts extremely important to know what you have with regards to wiring, as it involves large amp circuits, and you don’t want to wire things incorrectly and cause Eskom to come smack you for tripping the street transformer!\n\nThree phase is relatively easy to turn back into 1 phase – I’ll be doing that on one leg to provide an additional “solar” circuit for our laundry room.\n\nThose of you who like the technical aspects should take a look here –\nhttp://ece.k-state.edu/~starret/581/3phase.html\n\nSome basic calculations for 3 phase below. These can be used to work out your maximum load or other important wiring details, like how thick your electrical wire should be if you’re carrying whatever max current your supply provides…\n\nBasic electrical calculation:\n\nVolts = Watt ÷ Amps\nVolts = Ampere x Ohms\nAmps = Volts ÷ Ohms\nAmps = Watt ÷ Volts\nOhms = Volts ÷ Amps\nVolt-Ampere (VA or Watt) = Volts x Ampere\n\nFor 3 phase, we need to use the square root of 3 for our calculations as an additional factor.\nThe square root of 3 is 1.73 (rounded off to 2 digits).\n\nSo, to calculate VA its: Volts x Ampere x 1.73\n\nExample calc:\n\nKVA = (Volts x Ampere x 1.73) ÷ 1000\n= (400 volts x 60 amps x 1.73) ÷ 1000\n= 41520 ÷ 1000\n= 41.52 KVA\n\nWe can also work out the Amps as below:\n\nKVA = (Volts x Ampere x 1.73) ÷ 1000\n41.52 KVA = (400 volts x A x 1.73) ÷ 1000\n41.52 x 1000 = 400 x A x 1.73\n41520 ÷ (400 x 1.73) = Amps\n41520 ÷ 692 = 60 Amps\n\nIf we wanted to convert KVA to KW, we need to use a power factor (this represents losses in transmission). The figure used for this is typically 0.85, so\n\nKW ÷ 0.85 = KVA\nKVA x 0.85 = KW\n\nSimple!\n\nA more detailed explanation of power factor losses is here – http://www.energyaction.com.au/australian-energy-market/power-factor.html"
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"http://farm3.staticflickr.com/2861/10437362485_af779522d4.jpg",
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https://slideplayer.com/slide/5153735/ | [
"",
null,
"# NP-complete examples CSC3130 Tutorial 11 Xiao Linfu Department of Computer Science & Engineering Fall 2009.\n\n## Presentation on theme: \"NP-complete examples CSC3130 Tutorial 11 Xiao Linfu Department of Computer Science & Engineering Fall 2009.\"— Presentation transcript:\n\nNP-complete examples CSC3130 Tutorial 11 Xiao Linfu [email protected] Department of Computer Science & Engineering Fall 2009\n\nOutline Review of P, NP, NP-C 2 problems –Double-SAT –Dominating set http://en.wikipedia.org/wiki/Dominating_set_problem\n\nRelations NP P NP-C easy hard Is there any problem even harder than NP-C? Yes! e.g. I-go\n\nHow to show that a problem R is not easier than a problem Q? Informally, if R can be solved efficiently, we can solve Q efficiently. Formally, we say Q polynomially reduces to R if: 1.Given an instance q of problem Q 2.There is a polynomial time transformation to an instance f(q) of R 3.q is a “yes” instance if and only if f(q) is a “yes” instance Then, if R is polynomial time solvable, then Q is polynomial time solvable. If Q is not polynomial time solvable, then R is not polynomial time solvable. Polynomial Time Reduction\n\nMethodology To show L is in NP, you can either (i) show that solutions for L can be verified in polynomial-time, or (ii) describe a nondeterministic polynomial- time TM for L. To show L is NP-complete, you have to design a polynomial-time reduction from some problem we know to be NP-complete\n\nThe direction of the reduction is very important –Saying “A is easier than B” and “B is easier than A” mean different things What we have? We know SAT, Vertex Cover problems are NP-Complete!\n\nDouble-SAT Definition: –Double-SAT = { | φ is a Boolean formula with at least two satisfying assignments} Show that Double-SAT is NP-Complete. –(1) First, it is easy to see that Double-SAT ∈ NP. non-deterministically guess 2 assignments for φ and verify whether both satisfy φ. –(2) Then we show Double-SAT is not easier than SAT. Reduction from SAT to Double-SAT\n\nDouble-SAT Reduction: –On input φ (x 1,..., x n ): –1. Introduce a new variable y. –2. Output formula φ ’(x 1,..., x n, y) = φ (x 1,..., x n ) ∧ ( y ∨ y ).\n\nDominating set Definition: input G=(V,E), K Let G=(V,E) be an undirected graph. A dominating set D is a set of vertices in G such that every vertex of G is either in D or is adjacent to at least one vertex from D. The problem is to determine whether there is a dominating set of size K for G.\n\nDominating set - example {yellow vertices} is an example of a dominating set of size 2. e\n\nDominating set Show that Dominating set is NP-Complete. –(1) First, it is easy to see that Dominating set ∈ NP. Given a vertex set D of size K, we check whether (V-D) are adjacent to D. –(2) Then we show Dominating set is not easier than Vertex cover. Reduction from Vertex cover to Dominating set\n\nDominating set Reduction –(1) Graph transformation - Construct a new graph G' by adding new vertices and edges to the graph G as follows: For each edge (v, w) of G, add a vertex vw and the edges (v, vw) and (w, vw) to G'. Furthermore, remove all vertices with no incident edges; such vertices would always have to go in a dominating set but are not needed in a vertex cover of G.\n\nDominating set – graph transformation vw zu wv zu vzwu vw zu vu G G'\n\nDominating set Reduction –(1) Graph transformation –(2) a dominating set of size K in G’ a vertex cover of size K in G\n\nThank you!\n\nDownload ppt \"NP-complete examples CSC3130 Tutorial 11 Xiao Linfu Department of Computer Science & Engineering Fall 2009.\"\n\nSimilar presentations"
]
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"https://slideplayer.com/static/blue_design/img/slide-loader4.gif",
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https://phys.libretexts.org/Courses/Tuskegee_University/Algebra_Based_Physics_I/04%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion/4.E%3A_Dynamics-_Force_and_Newton's_Laws_of_Motion_(Exercises) | [
"$$\\require{cancel}$$\n\n# 4.E: Dynamics- Force and Newton's Laws of Motion (Exercises)\n\n•",
null,
"• Contributed by OpenStax\n• General Physics at OpenStax CNX\n\n#### 4.1: Development of Force Concept\n\n1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.\n\n2. What properties do forces have that allow us to classify them as vectors?\n\n#### 4.2: Newton’s First Law of Motion: Inertia\n\n3. How are inertia and mass related?\n\n4. What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?\n\n#### 4.3: Newton’s Second Law of Motion: Concept of a System\n\n5. Which statement is correct? Explain your answer and give an example.\n\n(a) Net force causes motion.\n\n(b) Net force causes change in motion.\n\n6. Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?\n\n7. Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion.\n\n8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.\n\n9. system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.\n\n10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?\n\n11. (a) Give an example of different net external forces acting on the same system to produce different accelerations.\n\n(b) Give an example of the same net external force acting on systems of different masses, producing different accelerations.\n\n(c) What law accurately describes both effects? State it in words and as an equation.\n\n12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.\n\n13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?\n\n14. The gravitational force on the basketball in Figure is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal?",
null,
"#### 4.4: Newton’s Third Law of Motion: Symmetry in Forces\n\n15. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)\n\n16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?\n\n17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton’s laws of motion apply?\n\n18. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired?\n\n19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.\n\n20. Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects whether one such pair of forces cancels.\n\n#### 4.5: Normal, Tension, and Other Examples of Forces\n\n21. If a leg is suspended by a traction setup as shown in Figure, what is the tension in the rope?",
null,
"A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force $$\\displaystyle T$$ without changing its magnitude.\n\n22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure.) (Note that the tibia is the shin bone shown in this image.)\n\n#### 4.7: Further Applications of Newton’s Laws of Motion\n\n23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at $$\\displaystyle g$$. Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?\n\n24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.\n\n#### 4.8: Extended Topic: The Four Basic Forces—An Introduction\n\n25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force.\n\n26. What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances?\n\n27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.)\n\n## Problems & Exercises\n\n#### 4.3: Newton’s Second Law of Motion: Concept of a System\n\nYou may assume data taken from illustrations is accurate to three digits.\n\n28. A 63.0-kg sprinter starts a race with an acceleration of $$\\displaystyle 4.20 m/s^2$$. What is the net external force on him?\n\nSolution\n265 N\n\n29. If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?\n\n30. A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of its acceleration.\n\nSolution\n$$\\displaystyle 13.3 m/s^2$$\n\n31. Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be $$\\displaystyle 0.893 m/s^2$$.\n\n(a) Calculate her mass.\n\n(b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.\n\n32. In Figure 4.4.3, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force $$\\displaystyle F$$ (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force $$\\displaystyle F$$ is removed. How far will the mower go before stopping?\n\nSolution\n1.1 m\n\n33. The same rocket sled drawn in Figure is decelerated at a rate of $$\\displaystyle 196 m/s^2$$. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.",
null,
"34. (a) If the rocket sled shown in Figure starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is $$\\displaystyle 2.4×10^4$$ N, and the force of friction opposing the motion is known to be 650 N.\n\n(b) Why is the acceleration not one-fourth of what it is with all rockets burning?\n\nSolution\n(a) $$\\displaystyle 11m/s^2$$\n(b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.\n\n35. What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)\n\n36. Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.\n\n(a) What is the system of interest if the acceleration of the child in the wagon is to be calculated?\n\n(b) Draw a free-body diagram, including all forces acting on the system.\n\n(c) Calculate the acceleration.\n\n(d) What would the acceleration be if friction were 15.0 N?\n\nSolution\n(a) The system is the child in the wagon plus the wagon.\n(b)",
null,
"(c) $$\\displaystyle a=0.130m/s^2$$ in the direction of the second child’s push.\n(d) $$\\displaystyle a=0.00m/s^2$$\n\n37. A powerful motorcycle can produce an acceleration of $$\\displaystyle 3.50m/s^2$$while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?\n\n38. The rocket sled shown in Figure accelerates at a rate of $$\\displaystyle 49.0m/s^2$$. Its passenger has a mass of 75.0 kg.\n\n(a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio.\n\n(b) Calculate the direction and magnitude of the total force the seat exerts against his body.\n\nSolution\n(a) $$\\displaystyle 3.68×10^3N$$. This force is 5.00 times greater than his weight.\n(b) $$\\displaystyle 3750 N; 11.3ºabove horizontal$$",
null,
"39. Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of $$\\displaystyle 201 m/s^2$$. In this problem, the forces are exerted by the seat and restraining belts.\n\n40. The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth?\n\nSolution\n$$\\displaystyle 1.5×10^3N,150 kg,150 kg$$\n\n41. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.\n\n#### 4.4: Newton’s Third Law of Motion: Symmetry in Forces\n\n42. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at $$\\displaystyle 2.40×10^4m/s^2$$? What is the magnitude of the force exerted on the ship by the artillery shell?\n\nSolution\nForce on shell: $$\\displaystyle 2.64×10^7N$$\nForce exerted on ship = $$\\displaystyle −2.64×10^7N$$, by Newton’s third law\n\n43. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20m/s2 size 12{1 \".\" \"20\"\" m/s\" rSup { size 8{2} } } {} backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.\n\n#### 4.5: Normal, Tension, and Other Examples of Forces\n\n44. Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally.\n\n(a) What is magnitude of the acceleration of the two teams?\n\n(b) What is the tension in the section of rope between the teams?\n\nSolution\na. $$\\displaystyle 0.11 m/s^2$$\nb. $$\\displaystyle 1.2×10^4N$$\n\n45. What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at $$\\displaystyle 7.50 m/s^2$$? Note that the answer is independent of the velocity of the gymnast—she can be moving either up or down, or be stationary.\n\n46. (a) Calculate the tension in a vertical strand of spider web if a spider of mass $$\\displaystyle 8.00×10^{−5}kg$$ hangs motionless on it.\n\n(b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure. The strand sags at an angle of $$\\displaystyle 12º$$ below the horizontal. Compare this with the tension in the vertical strand (find their ratio).",
null,
"Solution\n(a) $$\\displaystyle 7.84×10^{-4}N$$\n(b) $$\\displaystyle 1.89×10^{–3}N$$. This is 2.41 times the tension in the vertical strand.\n\n47. Suppose a 60.0-kg gymnast climbs a rope.\n\n(a) What is the tension in the rope if he climbs at a constant speed?\n\n(b) What is the tension in the rope if he accelerates upward at a rate of $$\\displaystyle 1.50 m/s^2$$?\n\n48. Show that, as stated in the text, a force $$\\displaystyle F_{⊥}$$ exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in Figure) gives rise to a tension of magnitude $$\\displaystyle T=\\frac{F_⊥}{2sin(θ)}$$.\n\nSolution\nNewton’s second law applied in vertical direction gives\n$$\\displaystyle F_y=F−2Tsinθ=0$$\n$$\\displaystyle F=2Tsinθ$$\n$$\\displaystyle T=\\frac{F}{2 sinθ}$$.\n\n49. Consider the baby being weighed in Figure.\n\n(a) What is the mass of the child and basket if a scale reading of 55 N is observed?\n\n(b) What is the tension $$\\displaystyle T_1$$ in the cord attaching the baby to the scale?\n\n(c) What is the tension $$\\displaystyle T_2$$ in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?\n\n(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.",
null,
"A baby is weighed using a spring scale.\n\n#### 4.6: Problem-Solving Strategies\n\n50. A $$\\displaystyle 5.00×10^5-kg$$ rocket is accelerating straight up. Its engines produce $$\\displaystyle 1.250×10^7N$$ of thrust, and air resistance is $$\\displaystyle 4.50×10^6N$$. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.\n\nSolution",
null,
"Using the free-body diagram:\n$$\\displaystyle F_{net}=T−f−mg=ma$$\nso that\n$$\\displaystyle a=\\frac{T−f−mg}{m}=\\frac{1.250×10^7N−4.50×10^6N−(5.00×10^5kg)(9.80 m/s^2)}{5.00×10^5kg}=6.20m/s^2$$\n\n51. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is $$\\displaystyle 1.80 m/s^2$$, what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.\n\n52. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.\n\nSolution\n1.Use Newton’s laws of motion.",
null,
"2. Given : $$\\displaystyle a=4.00g=(4.00)(9.80 m/s^2)=39.2m/s^2; m=70.0 kg.$$\n\nFind: $$\\displaystyle F$$\n3. $$\\displaystyle \\sum{F=+F−w=ma,}$$ so that $$\\displaystyle F=ma+w=ma+mg=m(a+g)$$\n$$\\displaystyle F=(70.0 kg)[(39.2 m/s^2)+(9.80 m/s^2)]=3.43×10^3N$$ The force exerted by the high-jumper is actually down on the ground, but F size 12{F} is up from the ground and makes him jump.\n4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of $$\\displaystyle 10^3N$$\n\n53. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.\n\n54. A freight train consists of two $$\\displaystyle 8.00×10^4-kg$$ engines and 45 cars with average masses of $$\\displaystyle 5.50×10^4kg$$.\n\n(a) What force must each engine exert backward on the track to accelerate the train at a rate of $$\\displaystyle 5.00×10^{–2}m/s^2$$ if the force of friction is $$\\displaystyle 7.50×10^5N$$, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems.\n\n(b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?\n\nSolution\n(a) $$\\displaystyle 4.41×10^5N$$\n(b) $$\\displaystyle 1.50×10^5N$$\n\n55. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor.\n\n(a) An 1800-kg tractor exerts a force of $$\\displaystyle 1.75×10^4N$$ backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is $$\\displaystyle 0.150 m/s^2$$, what is the mass of the airplane?\n\n(b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane.\n\n(c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.\n\n56. A 1100-kg car pulls a boat on a trailer.\n\n(a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of $$\\displaystyle 0.550 m/s^2$$? The mass of the boat plus trailer is 700 kg.\n\n(b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?\n\nSolution\n(a) $$\\displaystyle 910 N$$\n(b) $$\\displaystyle 1.11×10^3N$$\n\n57. (a) Find the magnitudes of the forces $$\\displaystyle F_1$$ and $$\\displaystyle F_2$$ that add to give the total force $$\\displaystyle F_{tot}$$ shown in Figure. This may be done either graphically or by using trigonometry.\n\n(b) Show graphically that the same total force is obtained independent of the order of addition of $$\\displaystyle F_1$$ and $$\\displaystyle F_2$$\n\n(c) Find the direction and magnitude of some other pair of vectors that add to give $$\\displaystyle F_{tot}$$. Draw these to scale on the same drawing used in part (b) or a similar picture.",
null,
"58. Two children pull a third child on a snow saucer sled exerting forces $$\\displaystyle F_1$$ and $$\\displaystyle F_2$$ as shown from above in Figure. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of $$\\displaystyle F_1$$ and $$\\displaystyle F_2$$.\n\nSolution\n$$\\displaystyle a=0.139 m/s, θ=12.4º$$ north of east",
null,
"An overhead view of a child sitting on a snow saucer sled.\n\n59. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure to pull it out.\n\n(a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.\n\n(b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?",
null,
"60. What force is exerted on the tooth in Figure if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.\n\nSolution\n1. Use Newton’s laws since we are looking for forces.\n2. Draw a free-body diagram:",
null,
"3. The tension is given as $$\\displaystyle T=25.0 N$$. Find $$\\displaystyle F_{app}$$. Using Newton’s laws gives: $$\\displaystyle Σ F_y=0$$, so that applied force is due to the y-components of the two tensions: $$\\displaystyle F_{app}=2Tsinθ=2(25.0 N)sin(15º)=12.9 N$$\nThe x-components of the tension cancel. $$\\displaystyle ∑F_x=0$$.\n4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.",
null,
"Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, $$\\displaystyle F_{app}$$, points straight toward the back of the mouth.\n\n61. Figure shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible.\n\n(a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope.\n\n(b) Find the tension in the rope above Superhero.\n\n(c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.",
null,
"Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?\n\n62. A nurse pushes a cart by exerting a force on the handle at a downward angle $$\\displaystyle 35.0º$$ below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N.\n\n(a) Draw a free-body diagram for the system of interest.\n\n(b) What force must the nurse exert to move at a constant velocity?\n\nConsider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.\n\nConsider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.\n\n65. Unreasonable Results\n\n(a) Repeat Exercise, but assume an acceleration of $$\\displaystyle 1.20 m/s^2$$ is produced.\n\n(b) What is unreasonable about the result?\n\n(c) Which premise is unreasonable, and why is it unreasonable?\n\n66. Unreasonable Results\n\n(a) What is the initial acceleration of a rocket that has a mass of $$\\displaystyle 1.50×10^6kg$$ at takeoff, the engines of which produce a thrust of $$\\displaystyle 2.00×10^6N$$? Do not neglect gravity.\n\n(b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.)\n\n(c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)\n\n#### 4.7: Further Applications of Newton’s Laws of Motion\n\n67. A flea jumps by exerting a force of $$\\displaystyle 1.20×10^{−5}N$$ straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of $$\\displaystyle 0.500×10^{−6}N$$ on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is $$\\displaystyle 6.00×10^{−7}kg$$. Do not neglect the gravitational force.\n\nSolution\n$$\\displaystyle 10.2m/s^2, 4.67º$$ from vertical\n\n68. Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?",
null,
"Achilles tendon\n\n69. A 76.0-kg person is being pulled away from a burning building as shown in Figure. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.\n\nSolution",
null,
"$$\\displaystyle T_1=736 N$$\n$$\\displaystyle T_2=194 N$$",
null,
"The force $$\\displaystyle T_2$$ needed to hold steady the person being rescued from the fire is less than her weight and less than the force $$\\displaystyle T_1$$ size 12{T rSub { size 8{1} } } {} in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force).\n\n70. Integrated Concepts\n\nA 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)\n\n71. Integrated Concepts\n\nWhen starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s.\n\n(a) What is his final speed?\n\n(b) How far does he travel?\n\nSolution\n(a) $$\\displaystyle 7.43 m/s$$\n(b) $$\\displaystyle 2.97 m$$\n\n72. Integrated Concepts\n\nA large rocket has a mass of $$\\displaystyle 2.00×10^6kg$$ at takeoff, and its engines produce a thrust of $$\\displaystyle 3.50×10^7N$$.\n\n(a) Find its initial acceleration if it takes off vertically.\n\n(b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?\n\n(c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.\n\n73. Integrated Concepts\n\nA basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor.\n\n(a) Calculate his velocity when he leaves the floor.\n\n(b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m.\n\n(c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.\n\nSolution\n(a) $$\\displaystyle 4.20 m/s$$\n(b) $$\\displaystyle 29.4m/s^2$$\n(c) $$\\displaystyle 4.31×10^3N$$\n\n74. Integrated Concepts\n\nA 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m.\n\n(a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar.\n\n(b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a).\n\n(c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.\n\n75. Integrated Concepts Repeat Exercise for a shell fired at an angle $$\\displaystyle 10.0º$$ from the vertical.\n\nSolution\n(a) $$\\displaystyle 47.1 m/s$$\n(b) $$\\displaystyle 2.47×10^3m/s^2$$\n(c) $$\\displaystyle 6.18×10^3N$$ .\n\n76. Integrated Concepts\n\nAn elevator filled with passengers has a mass of 1700 kg.\n\n(a) The elevator accelerates upward from rest at a rate of $$\\displaystyle 1.20 m/s^2$$. Calculate the tension in the cable supporting the elevator.\n\n(b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time?\n\n(c) The elevator decelerates at a rate of $$\\displaystyle 0.600 m/s^2$$ for 3.00 s. What is the tension in the cable during deceleration?\n\n(d) How high has the elevator moved above its original starting point, and what is its final velocity?\n\n77. Unreasonable Results\n\n(a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of $$\\displaystyle 0.400 m/s^2$$ for 50.0 s?\n\n(b) What is unreasonable about the result?\n\n(c) Which premise is unreasonable, or which premises are inconsistent?\n\n78. Unreasonable Results\n\nA 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s.\n\n(a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.)\n\n(b) What is unreasonable about the result?\n\n(c) Which premise is unreasonable, or which premises are inconsistent?\n\n#### 4.8: Extended Topic: The Four Basic Forces—An Introduction\n\n79. (a) What is the strength of the weak nuclear force relative to the strong nuclear force?\n\n(b) What is the strength of the weak nuclear force relative to the electromagnetic force? Since the weak nuclear force acts at only very short distances, such as inside nuclei, where the strong and electromagnetic forces also act, it might seem surprising that we have any knowledge of it at all. We have such knowledge because the weak nuclear force is responsible for beta decay, a type of nuclear decay not explained by other forces.\n\nSolution\n(a) $$\\displaystyle 1×10^{−13}$$\n(b) $$\\displaystyle 1×10^{−11}$$\n\n80. (a) What is the ratio of the strength of the gravitational force to that of the strong nuclear force?\n\n(b) What is the ratio of the strength of the gravitational force to that of the weak nuclear force?\n\n(c) What is the ratio of the strength of the gravitational force to that of the electromagnetic force? What do your answers imply about the influence of the gravitational force on atomic nuclei?\n\n81. What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? Based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes greater than the range of the strong nuclear force. At these sizes, the electromagnetic force begins to affect nuclear stability. These facts will be used to explain nuclear fusion and fission later in this text.\n\nSolution\n$$\\displaystyle 10^2$$"
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https://www.numberempire.com/1116160 | [
"Home | Menu | Get Involved | Contact webmaster",
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"# Number 1116160\n\none million one hundred sixteen thousand one hundred sixty\n\n### Properties of the number 1116160\n\n Factorization 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 109 Divisors 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 109, 128, 160, 218, 256, 320, 436, 512, 545, 640, 872, 1024, 1090, 1280, 1744, 2048, 2180, 2560, 3488, 4360, 5120, 6976, 8720, 10240, 13952, 17440, 27904, 34880, 55808, 69760, 111616, 139520, 223232, 279040, 558080, 1116160 Count of divisors 48 Sum of divisors 2702700 Previous integer 1116159 Next integer 1116161 Is prime? NO Previous prime 1116133 Next prime 1116163 1116160th prime 17413313 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 100010000100000000000 Octal 4204000 Duodecimal 459b14 Hexadecimal 110800 Square 1245813145600 Square root 1056.48473723 Natural logarithm 13.925404780823 Decimal logarithm 6.0477264545804 Sine 0.6786556260873 Cosine -0.73445662988365 Tangent -0.92402409955072\nNumber 1116160 is pronounced one million one hundred sixteen thousand one hundred sixty. Number 1116160 is a composite number. Factors of 1116160 are 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 109. Number 1116160 has 48 divisors: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 109, 128, 160, 218, 256, 320, 436, 512, 545, 640, 872, 1024, 1090, 1280, 1744, 2048, 2180, 2560, 3488, 4360, 5120, 6976, 8720, 10240, 13952, 17440, 27904, 34880, 55808, 69760, 111616, 139520, 223232, 279040, 558080, 1116160. Sum of the divisors is 2702700. Number 1116160 is not a Fibonacci number. It is not a Bell number. Number 1116160 is not a Catalan number. Number 1116160 is not a regular number (Hamming number). It is a not factorial of any number. Number 1116160 is an abundant number and therefore is not a perfect number. Binary numeral for number 1116160 is 100010000100000000000. Octal numeral is 4204000. Duodecimal value is 459b14. Hexadecimal representation is 110800. Square of the number 1116160 is 1245813145600. Square root of the number 1116160 is 1056.48473723. Natural logarithm of 1116160 is 13.925404780823 Decimal logarithm of the number 1116160 is 6.0477264545804 Sine of 1116160 is 0.6786556260873. Cosine of the number 1116160 is -0.73445662988365. Tangent of the number 1116160 is -0.92402409955072\n\n### Number properties\n\nExamples: 3628800, 9876543211, 12586269025"
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https://it.mathworks.com/matlabcentral/cody/problems/120-radius-of-a-spherical-planet/solutions/156709 | [
"Cody\n\n# Problem 120. radius of a spherical planet\n\nSolution 156709\n\nSubmitted on 1 Nov 2012 by Rody Oldenhuis\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%% x = 4*pi; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))\n\n2 Pass\n%% x = 400*pi; y_correct = 10; assert(isequal(your_fcn_name(x),y_correct))\n\n3 Pass\n%% x = 40000*pi; y_correct = 100; assert(isequal(your_fcn_name(x),y_correct))\n\n4 Pass\n%% x = -4*pi; y_correct = 1i; assert(isequal(your_fcn_name(x),y_correct))"
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https://www.easycalculation.com/square-roots-24.html | [
"# What is Square Root of 24 ?\n\nSquare root of 24 is 4.8989. It is an irrational number and its equivalent fraction is 10087 / 2059.\n\nSquare Root of 24\n √24 = √(4.899 x 4.899) 4.89898\n\nSquare root of 24 is 4.8989. It is an irrational number and its equivalent fraction is 10087 / 2059.\n\nThe nearest previous perfect square is 16 and the nearest next perfect square is 25 ."
]
| [
null
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https://wiki.freecadweb.org/B-Splines/en | [
"B-Splines\n\nOther languages:\nDeutsch • English • español • français • polski • русский\n\nThis page describes how to use B-splines in FreeCAD. It gives also background information what B-splines are and for what applications they are useful.\n\nMotivation\n\nIf you know already about B-splines and their application, you can directly continue with section B-splines in FreeCAD.\n\nLet's assume you want to design a part that should be produced with a 3D printer. The part must have an edge this way:\n\nYou have to print the part in direction of the sketch's bottom to the top. Outer support structures might not be an option. Therefore you need to add a support directly to your part. What options do you have?\n\n• Option 1: you could add a line from point (20, 0) to point (80, 40):\n\nHowever this solution needs a lot of volume, thus weight and material.\n\n• Option 2: you can connect the two points with an arc of a circle. To save volume, the arc should end tangentially in point (80,40). Then your solution looks like this:\n\nOK. But at the bottom you don't need immediate support.\n\n• Option 3: you could save some more volume if the connection between the 2 points is a curve that begins tangentially at (0, 20) and ends tangentially at (80, 40):\n\nSo a curve with which you can connect two points tangentially to a reference point can be very useful for constructions. Bézier curves provide this feature.\n\nBézier curves\n\nDerivation\n\nBézier curves are polynomials to describe the connection between 2 points. The simplest polynomial connecting 2 points is a line ($A*x^{1}+B$",
null,
") thus also linear Bézier curves are linear:\n\nAnimation 1: Linear Bézier curve.\n\nHowever a polynomial becomes first be useful when we can control it. So there should be a point between the 2 endpoints that allows us to define how the endpoints are connected. Like in the above example option 3 the curve is helpful when it starts and ends tangentially to lines crossing the endpoints. And this is a main feature of Bézier curves. So let's add a control point between the 2 endpoints. The curve will start tangentially towards this control point, meaning it is tangential to the line that we can draw between the startpoint and the control point. Going backwards from the endpoint the curve will also be tangential to the line we can draw between the control point and the end point. Animation 2 shows how such a curve looks.\n\nAnimation 2: Quadratic Bézier curve. P1 is hereby the control point.\n\nThe animation makes clear what the curve basically is - a transition from P0 to P2 by rotating the line P0-P1 to become the line P1-P2. Therefore we get the nice tangential start/end feature.\n\nSuch a curve can only be described by a quadratic polynomial. (The number of left-hand/right-hand turns + 1 is the necessary polynomial order. A quadratic polynomial is a single turn, a cubic polynomial has two turn, and so on.) Therefore a Bézier curve with one control point is a quadratic (second order) Bézier curve.\n\nHaving only one control point is often not sufficient. Take the above motivation example. There in option 3 we end the curve tangentially in x-direction. But how can you connect the points (20, 0) and (80, 40) so that the curve ends tangentially in y-direction? To achieve this you need first a right-hand and then a left-hand turn, so a cubic (third order) polynomial. And that means for a Bézier curve that we need (or you can say we gain) a second control point. Animation 3 shows a cubic Bézier curve.\n\nAnimation 3: Cubic Bézier curve.\n\nTo answer the question, the solution with the tangential y-direction ending for the example is this one:\n\nMath\n\nIf you are interested to understand the background math, here are the basics.\n\nA Bézier curve is calculated using this formula:\n\n$\\quad {\\textrm {Bezier}}(n,t)=\\sum _{i=0}^{n}\\underbrace {\\binom {n}{i}} _{\\text{polynomial term}}\\underbrace {\\left(1-t\\right)^{n-i}t^{i}} _{\\text{polynomial term}}\\;\\underbrace {P_{i}} _{\\text{point coordinate}}$",
null,
"n is hereby the degree of the curve. So a Bézier curve of degree n is a polygon with order n. The factors $P_{i}$",
null,
"are hereby in fact the coordinates of the Bézier curves' control points. For a visualization see Controlling Bézier curvatures.\n\nIf you are further interested, have a look at The mathematics of Bézier curves with a nicely animated derivation of the math of Bézier curves.\n\nRules\n\nIn the above text you might already noticed some \"rules\" for Bézier curves:\n\n• The polynomial degree is also the degree of the curves.\n• If you need $n$",
null,
"turns, you need at least a $n+1$",
null,
"degree Bézier curve.\n• A Bézier curve always begins tangentially to the line between the startpoint and the first control point (and ends tangentially to the line between the last control point and the endpoint).\n\nB-Splines\n\nBasics\n\nThis video lists at the beginning the practical problems with Bézier curves. For example that adding or changing a control point changes the whole curve. These problems can be resolved by joining several Bézier curves. The result is a so-called spline, in particular a B-spline (basis spline). The video also explains that a union of quadratic Bézier curves forms a uniform quadratic B-spline and that a union of cubic Bézier curves forms a uniform cubic B-spline.\n\nFrom the videos we can collect useful \"rules\" for B-splines:\n\n• The first and last control point is the end/start point of the spline.\n• Like for Bézier curves, splines always begin tangentially to the line between the startpoint and the first control point (and end tangentially to the line between the last control point and the endpoint).\n• A union of $S$",
null,
"Bézier curves with the degree $D$",
null,
"has $S+D$",
null,
"control points.\n• Since one is in most cases working with cubic B-splines we can then state that $N$",
null,
"control points lead to $N-3$",
null,
"Bézier segments and in turn $N-4$",
null,
"segment junction points.\n• A B-spline with the degree $D$",
null,
"offers at every point a continuous $D-1$",
null,
"order derivative.\n• For a cubic B-spline this means that the curvature (second order derivative) does not change when traveling from one segment to the next one. This is a very useful feature as we will later see.\n\nIf you are interested in more details about B-Spline properties, have a look at video MOOC Curves 8.2: Properties of B-spline curves.\n\nBasis\n\nThe name B-spline stands for Basis spline. Instead of forming the spline as a combination of Bézier curves, the approach is to to model the same spline a different way. The idea is hereby to use another set of polynomials as basis. A linear combination of these basis polynomials $B_{D}(t)$",
null,
"with the order $D$",
null,
"forms the B-spline. This video explains the transition from the Bézier control points to the polynomial basis functions describing the spline. Mathematically we can describe a B-spline with this formula:\n\n$\\quad c(t)=\\sum _{k=0}^{N}p_{k}B_{k,D}(t)$",
null,
"Hereby $p_{k}$",
null,
"is the $k$",
null,
"-th control point of the B-spline and also a factor for the $k$",
null,
"-th base polynomial $B_{k,D}(t)$",
null,
". Every basis polynomial describe the spline in a certain region and therefore moving a control point does not affect the whole spline. To understand this, it is highly recommended to have a look at this video starting at minute 2:23.\n\nAs explained in the video, the basis polynomials are Bernstein polynomials. The set of basis polynomials for a certain B-spline can be visualized this way:\n\nA set of Bernstein polynomials with order 4. They describe a 4th order B-spline with 5 control points.\n\nAt every spline position $t$",
null,
"the sum of polynomials is 1 (indicated by the orange line). At the start only the red polynomial has an influence since all other polynomials are there 0. At greater $t$",
null,
"the spline is described by a linear combination of different basis polynomials. In the image above, every polynomial is greater than 1 for the whole range $0",
null,
". This is not necessarily the case. As shown in the video, the basis polynomials are basically only greater than 0 for a certain spline position range. The interval at which a basis polynomial is greater than 0 is described by the knot vector. If you are interested in learning about the knot vector, have a look at this video.\n\nNon-uniform B-splines\n\nA property of the Bernstein polynomials is that when looking at the different S-spline Bézier parts, the path length of every part is the same. (The path length is often called the travel time). As you can imagine, it can be useful to have B-splines whose Bézier parts have different path lengths. This can be achieved by weighting the different polynomials:\n\n$\\quad c(t)=\\sum _{k=0}^{N}d_{k}B_{k,D}(t)w_{k}$",
null,
"$w_{k}$",
null,
"is hereby the weight of the $k$",
null,
"-th control point. When the weights are not equal, the B-spline is called non-uniform.\n\nEspecially when B-splines should be used for 3D modelling, normalized, non-uniform B-splines are necessary. The normalization is done by a division by the weighted basis functions. Thus when all $w_{k}$",
null,
"are equal, we get a uniform B-spline, independent on the weight itself:\n\n$\\quad c(t)={\\cfrac {\\sum _{k=0}^{N}d_{k}B_{k,D}(t)w_{k}}{\\sum _{k=0}^{N}B_{k,D}(t)w_{k}}}$",
null,
"These non-uniform and rational (because of the division) B-splines are often called NURBS. Looking at their formula, we see that they are in fact a B-spline with a weighted basis $R_{k,D}(t)$",
null,
":\n\n$\\quad c(t)=\\sum _{k=0}^{N}d_{k}R_{k,D}(t)$",
null,
"whereas\n\n$\\quad R_{k,D}={\\cfrac {B_{k,D}(u)w_{k}}{\\sum _{l=1}^{N}B_{l,D}(t)w_{l}}}$",
null,
"FreeCAD offers to create uniform or non-uniform B-splines of any degree in 2D via the Sketcher workbench.\n\nCreation\n\nTo create B-splines, go into a sketch and use the toolbar button . Then left-click to set a control point, move the mouse left-click to set the next control point and so on. Finally right-click to finish the definition and create the B-spline.\n\nBy default uniform cubic splines are created, except there are not enough control points to do this. So when you create a B-spline with only 2 control points, you get of course a spline that is single linear Bézier curve, for 3 control points you get a quadratic Bézier curve, first with 5 control points you get a cubic B-spline consisting of 2 Bézier segments.\n\nTo create periodic B-splines (B-splines that form a closed curve), use the toolbar button . It is not necessary to set the last control point onto the first one because the B-spline will automatically be closed:\n\nB-splines can also be generated out of existing sketch segments. To do this, select the elements and press the the toolbar button .\n\nChanging the Degree\n\nTo change the degree, select the B-spline and use either the toolbar button or .\n\nNote: Decreasing the degree cannot revert a prior increase of the degree, see the Wiki page Decrease B-spline degree for an explanation.\n\nChanging the Knot Multiplicity\n\nThe points where two Bézier curves are connected to form the B-spline are called knots. The knot multiplicity determines how the Bézier parts are connected, see the Wiki page Increase knot multiplicity for details.\n\nNote: Creating two B-Splines that are connected to each other will not unite to a single new B-spline. So their connection point is not a knot. The only way to get a new knot in an existing B-spline is to decrease the degree. However, you may get many new knots. Thus the better choice is to redraw the B-spline with more control points.\n\nChanging the Weight\n\nAround every control point you see a dark yellow circle. Its radius sets the weight for the corresponding control point. By default all circles have the radius 1. This is indicated with a radius constraint for the first control point circle.\n\nTo create a non-uniform B-spline the weights have to be non-uniform. To achieve that you can either change the radius constraint of the first control point circle:\n\nor you delete the constraint that all circles are equal and then set different radius constraints for the circles.\n\nIf no radius constraint is set, you can also change the radius by dragging:\n\nIn the dragging example you see that a high weight attracts the curve to the control point while a very low weight changes the curve so as if the control point does almost not exist.\n\nWhen you look at the creation function for non-uniform rational B-splines you see that a weight of zero would lead to a division by zero. Therefore you can only specify weights greater than zero.\n\nDisplay Information\n\nSince the form of a B-spline does not tell much about its properties, FreeCAD offers different tools to display the properties:\n\nLimitations\n\nAt the moment (FreeCAD 0.19) there are some limitations when using splines you should know:\n\n1. You cannot set tangential constraints.\nIn this example",
null,
"you want to assure that the spline touches the blue curve 2 times tangentially. This would be useful because the blue line could for example be the spatial border for your design.\n2. You cannot insert a new control point between two selected existing control points. There is no other way than to redraw the spline.\n3. You cannot delete a control point. Also in this case you must redraw the spline\n4. You cannot create an offset curve for a B-spline using the tool Draft Offset.\n\nTypical Use Cases\n\nAccording to the properties of B-splines, there are 3 main use cases:\n\n1. Curves that start/end tangentially to a certain direction. An example for this is the motivation example above.\n2. Curves describing larger designs and providing the freedom of local changes. See this example below.\n3. Curves providing a certain continuity (derivative). See this example below.\n\nDesigning\n\nTake for example the case that you design a housing of a kitchen mixer. Its desired shape should look like this one:\n\nTo define the outer form it is advantageous to use a B-spline because when you change a control point to change the curvature at the bottom, the curvature at the side and top will not be changed:\n\nContinuity at Geometric Transitions\n\nThere are several cases where it is physically necessary to have a certain surface continuity at geometric transitions. Take for example the inner walls of a fluid channel. When you have a change in the diameter of the channel, you don't want to have an edge because edges would introduce turbulences. Therefore, like in the motivation example above, one uses splines for this purpose.\n\nThe development of the Bézier curves was initially triggered by the French car industry. Besides the saving of material and the reduction of the air flow drag, the look of the cars should also be improved. And when you look at the fancy design of French cars from the 60's and 70's you see that the Bézier curves gave car design a boost.\n\nLet's take for example this task in the design of cars: The car fender should \"look nice\". Here is a basic sketch of our task:\n\n\"Looking nice\" means that the (potential) customer looks at the fender and does not see unexpected light reflections and also no sudden changes in the reflection from the automotive paint at all. So what do you need to avoid changes in the reflections? Looking closely to the fender:\n\nAt the spatial area above the edge the intensity of reflected light is low (denoted by the red ellipse) because no light is directly reflected in the direction from the edge to the eye.\n\nyou see when there is an edge, there is a spatial area where the reflected light has less intensity and this is what you will notice when looking at the fender. To avoid this you need a continuous change in the slope of your surface elements. The slope is the first order derivative and as explained in section Basics, a second degree (quadratic) B-spline offers at every point a continuous first order derivative.\n\nBut is this really sufficient? At the point of geometric transition we have now at both sides the same slope, but the slope might change differently at both sides. Then we have this situation:\n\nSo we have also spatial areas in which the intensity of reflected light is different. To avoid this, we need at the geometrical point of transition also a continuity of the second order derivative and thus a cubic B-spline."
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https://mathoverflow.net/questions/67808/separable-quotients-of-non-separable-banach-spaces | [
"# Separable quotients of non-separable Banach spaces?\n\nI am reading the Functional Analysis book of Conway, one question from the book is find a closed subspace M of $l^{\\infty}=l^{\\infty}(\\mathbb{N})$ with the property that $l^{\\infty}/M$ is separable. I have found a solution for this but here is my question :\n\n1. Is it true that every non-separable normed space $X$ always contains a closed (proper) subspace $M$ such that $X/M$ is linear isometric to a separable normed space whose dimension is infinite ? i.e, are there a map $A$ and a separable normed space $Y$ whose dimension is infinite, st: $A: X/M\\to Y$ which is linear, onto, and preserve the distance?\n\n(Edit: I already have an answer for the following question I am thinking a about $l^{\\infty}$ : can it contain a closed proper subspace M that $l^{\\infty}/M$ is isometric to $l^{1}=l^{1}(\\mathbb{N})$?)\n\n• The answer is affirmative. Namely, take M as the kernel of any [non-trivial] continuous linear functional on X. – Ady Jun 14 '11 at 22:51\n• In your first question, about merely getting a separable quotient, you want to require the subspace $M$ of $X$ to not only be a proper subspace but of infinite codimension. Otherwise that question is trivial. – Andreas Blass Jun 14 '11 at 22:52\n• That's the 'Separable Quotient Problem'... – Ady Jun 14 '11 at 22:58\n• Steven, Ady has already mentioned that Question 1 in your current post is the Separable Quotient Problem; the answer is unknown but it is known to be yes for certain classes of Banach spaces. For a survey detailing the partial results known in the mid-1990s, see Mujica's survey Separable quotients of Banach spaces, Rev. Mat. Univ. Complutense Madrid 10 (1997), 299–330. A more recent result, that every Banach space isomorphic to a dual space has a separable quotient, has been shown my Argyros et al in Unconditional families in Banach spaces, Math. Ann. (2008) 341:15–38. – Philip Brooker Jun 15 '11 at 5:30\n• Just as I was posting my comment Bill posted his answer (which contains more info than my comment, however maybe the survey article interests you; find it at mat.ucm.es/serv/revmat/vol10-2/vol10-2e.pdf ) – Philip Brooker Jun 15 '11 at 5:40\n\nYour question is the famous \"separable quotient problem\", as Ady mentioned. From here on, \"space\" means \"infinite dimensional Banach space\". A space $X$ has a separable quotient provided $X^*$ has a reflexive subspace (obvious), a subspace isomorphic to $c_0$ (Rosenthal and me), or $\\ell_1$ (Hagler and me). A result of PANDELIS DODOS, JORDI LOPEZ-ABAD and STEVO TODORCEVIC is that it is consistent with ZFC that if $X$ has density character at least $\\aleph_\\omega$ then $X$ has a separable quotient; see\n\nhttp://arxiv.org/pdf/0805.1860.pdf\n\nEvery dual space has a separable quotient (Argyros, Dodos, Kanellopoulos):\n\nhttp://users.uoa.gr/~pdodos/Publications/13-Unconditional.pdf\n\nThere are other striking things that I can't locate quickly.\n\nEvery non reflexive quotient of a $C(K)$ space contains a subspace isomorphic to $c_0$ (classical result of Pelczynski), so $\\ell_1$ is not a quotient of $\\ell_\\infty$.\n\n• Small comment regarding the last part: would it be very slightly quicker to note that if a Banach space quotients onto $\\ell_1$ then it has a complemented subspace isomorphic to $\\ell_1$ (and then use Pelczynski's result about complemented subspaces of $\\ell_\\infty$)? – Yemon Choi Jun 15 '11 at 5:46\n• It's the same proof, Yemon. Pelczynski proved that every non weakly compact operator from a $C(K)$ space preserves a copy of $c_0$. I don't think his proof simplifies if the operator is a projection (though I could be wrong about that). Quite a bit later Lindenstrauss proved (using Pelczynski's theorem, BTW) that every complemented subspace of $\\ell_\\infty$ is isomorphic to $\\ell_\\infty$. – Bill Johnson Jun 15 '11 at 6:00\n• Thanks for the details Bill - as you have probably guessed, in my comment I meant Lindenstrauss when I said Pelczynski, and I was unaware that his proof used the result of Pelczynski which you mention. – Yemon Choi Jun 15 '11 at 6:10\n\nHere seems to be another reference paper from Jorge MUJICA, who transfer this seperable quotient space problem to some other equivalent problems.\n\nhttp://www.mat.ucm.es/serv/revmat/vol10-2/vol10-2e.pdf"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.919997,"math_prob":0.91536164,"size":810,"snap":"2019-51-2020-05","text_gpt3_token_len":233,"char_repetition_ratio":0.12779157,"word_repetition_ratio":0.0,"special_character_ratio":0.27283952,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9936889,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T10:32:16Z\",\"WARC-Record-ID\":\"<urn:uuid:43a7c293-61f3-41b9-8c81-dd5bd2857316>\",\"Content-Length\":\"128250\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04f049a1-95ce-4b23-b88c-aeaf8284d587>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e1567d3-dfb8-4bd6-a333-689c128c8a7b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/67808/separable-quotients-of-non-separable-banach-spaces\",\"WARC-Payload-Digest\":\"sha1:SAG5TRVUHNVQFNVQUVXCXUD6YHWAWXRT\",\"WARC-Block-Digest\":\"sha1:GWFJQPNFIPEGVT33D4R5XDVYP34S6WFK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540527205.81_warc_CC-MAIN-20191210095118-20191210123118-00170.warc.gz\"}"} |
https://number.academy/36613 | [
"# Number 36613\n\nNumber 36,613 spell 🔊, write in words: thirty-six thousand, six hundred and thirteen . Ordinal number 36613th is said 🔊 and write: thirty-six thousand, six hundred and thirteenth. The meaning of number 36613 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 36613. What is 36613 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 36613.\n\n## What is 36,613 in other units\n\nThe decimal (Arabic) number 36613 converted to a Roman number is (X)(X)(X)(V)MDCXIII. Roman and decimal number conversions.\n\n#### Weight conversion\n\n36613 kilograms (kg) = 80717.0 pounds (lbs)\n36613 pounds (lbs) = 16607.5 kilograms (kg)\n\n#### Length conversion\n\n36613 kilometers (km) equals to 22751 miles (mi).\n36613 miles (mi) equals to 58923 kilometers (km).\n36613 meters (m) equals to 120120 feet (ft).\n36613 feet (ft) equals 11160 meters (m).\n36613 centimeters (cm) equals to 14414.6 inches (in).\n36613 inches (in) equals to 92997.0 centimeters (cm).\n\n#### Temperature conversion\n\n36613° Fahrenheit (°F) equals to 20322.8° Celsius (°C)\n36613° Celsius (°C) equals to 65935.4° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n36613 seconds equals to 10 hours, 10 minutes, 13 seconds\n36613 minutes equals to 3 weeks, 4 days, 10 hours, 13 minutes\n\n### Zip codes 36613\n\n• Zip code 36613 Arealonga (Vilagarcia De Arousa), Galicia, Pontevedra, Spain a map\n• Zip code 36613 Bocas (Pª Arealonga), Galicia, Pontevedra, Spain a map\n• Zip code 36613 Laxe (Pª Arealonga), Galicia, Pontevedra, Spain a map\nZip code areas 36613\n\n### Codes and images of the number 36613\n\nNumber 36613 morse code: ...-- -.... -.... .---- ...--\nSign language for number 36613:",
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"Number 36613 in braille:",
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"Images of the number\nImage (1) of the numberImage (2) of the number",
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"More images, other sizes, codes and colors ...\n\n#### Number 36613 infographic",
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"## Share in social networks",
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"## Mathematics of no. 36613\n\n### Multiplications\n\n#### Multiplication table of 36613\n\n36613 multiplied by two equals 73226 (36613 x 2 = 73226).\n36613 multiplied by three equals 109839 (36613 x 3 = 109839).\n36613 multiplied by four equals 146452 (36613 x 4 = 146452).\n36613 multiplied by five equals 183065 (36613 x 5 = 183065).\n36613 multiplied by six equals 219678 (36613 x 6 = 219678).\n36613 multiplied by seven equals 256291 (36613 x 7 = 256291).\n36613 multiplied by eight equals 292904 (36613 x 8 = 292904).\n36613 multiplied by nine equals 329517 (36613 x 9 = 329517).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 36613\n\nHalf of 36613 is 18306,5 (36613 / 2 = 18306,5 = 18306 1/2).\nOne third of 36613 is 12204,3333 (36613 / 3 = 12204,3333 = 12204 1/3).\nOne quarter of 36613 is 9153,25 (36613 / 4 = 9153,25 = 9153 1/4).\nOne fifth of 36613 is 7322,6 (36613 / 5 = 7322,6 = 7322 3/5).\nOne sixth of 36613 is 6102,1667 (36613 / 6 = 6102,1667 = 6102 1/6).\nOne seventh of 36613 is 5230,4286 (36613 / 7 = 5230,4286 = 5230 3/7).\nOne eighth of 36613 is 4576,625 (36613 / 8 = 4576,625 = 4576 5/8).\nOne ninth of 36613 is 4068,1111 (36613 / 9 = 4068,1111 = 4068 1/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 36613\n\n#### Is Prime?\n\nThe number 36613 is not a prime number. The closest prime numbers are 36607, 36629.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 36613 are 19 * 41 * 47\nThe factors of 36613 are 1 , 19 , 41 , 47 , 779 , 893 , 1927 , 36613\nTotal factors 8.\nSum of factors 40320 (3707).\n\n#### Powers\n\nThe second power of 366132 is 1.340.511.769.\nThe third power of 366133 is 49.080.157.398.397.\n\n#### Roots\n\nThe square root √36613 is 191,345238.\nThe cube root of 336613 is 33,205634.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 36613 = loge 36613 = 10,508159.\nThe logarithm to base 10 of No. log10 36613 = 4,563635.\nThe Napierian logarithm of No. log1/e 36613 = -10,508159.\n\n### Trigonometric functions\n\nThe cosine of 36613 is 0,637756.\nThe sine of 36613 is 0,770239.\nThe tangent of 36613 is 1,207732.\n\n### Properties of the number 36613\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 36613 in Computer Science\n\nCode typeCode value\n36613 Number of bytes35.8KB\nUnix timeUnix time 36613 is equal to Thursday Jan. 1, 1970, 10:10:13 a.m. GMT\nIPv4, IPv6Number 36613 internet address in dotted format v4 0.0.143.5, v6 ::8f05\n36613 Decimal = 1000111100000101 Binary\n36613 Decimal = 1212020001 Ternary\n36613 Decimal = 107405 Octal\n36613 Decimal = 8F05 Hexadecimal (0x8f05 hex)\n36613 BASE64MzY2MTM=\n36613 MD525c913a442ff0c90d4756bbf202ba3dd\n36613 SHA1feb62d5dcdce49538e83419c535ee75eecdfda4b\n36613 SHA224f4c3fe254b3f65aecddf216db7456b47c65e37789a608f8e8252762b\n36613 SHA256d2db2b155c64eddb9a9c6805ba1da95611d2d5bac8f0a1c8192c75327eef9ef9\nMore SHA codes related to the number 36613 ...\n\nIf you know something interesting about the 36613 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 36613\n\n### Character frequency in number 36613\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 3 2 6 2 1 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 36613, the numbers 3+6+6+1+3 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.\n\n## Interesting facts about the number 36613\n\n### Asteroids\n\n• (36613) 2000 QW145 is asteroid number 36613. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 8/31/2000.\n\n## Number 36,613 in other languages\n\nHow to say or write the number thirty-six thousand, six hundred and thirteen in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 36.613) treinta y seis mil seiscientos trece German: 🔊 (Anzahl 36.613) sechsunddreißigtausendsechshundertdreizehn French: 🔊 (nombre 36 613) trente-six mille six cent treize Portuguese: 🔊 (número 36 613) trinta e seis mil, seiscentos e treze Chinese: 🔊 (数 36 613) 三万六千六百一十三 Arabian: 🔊 (عدد 36,613) ستة و ثلاثون ألفاً و ستمائة و ثلاثة عشر Czech: 🔊 (číslo 36 613) třicet šest tisíc šestset třináct Korean: 🔊 (번호 36,613) 삼만 육천육백십삼 Danish: 🔊 (nummer 36 613) seksogtredivetusinde og sekshundrede og tretten Dutch: 🔊 (nummer 36 613) zesendertigduizendzeshonderddertien Japanese: 🔊 (数 36,613) 三万六千六百十三 Indonesian: 🔊 (jumlah 36.613) tiga puluh enam ribu enam ratus tiga belas Italian: 🔊 (numero 36 613) trentaseimilaseicentotredici Norwegian: 🔊 (nummer 36 613) tretti-seks tusen, seks hundre og tretten Polish: 🔊 (liczba 36 613) trzydzieści sześć tysięcy sześćset trzynaście Russian: 🔊 (номер 36 613) тридцать шесть тысяч шестьсот тринадцать Turkish: 🔊 (numara 36,613) otuzaltıbinaltıyüzonüç Thai: 🔊 (จำนวน 36 613) สามหมื่นหกพันหกร้อยสิบสาม Ukrainian: 🔊 (номер 36 613) тридцять шiсть тисяч шiстсот тринадцять Vietnamese: 🔊 (con số 36.613) ba mươi sáu nghìn sáu trăm mười ba Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 36613 or any natural number (positive integer) please write us here or on facebook."
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-3.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-3.png",
null,
"https://number.academy/img/braille-36613.svg",
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"https://numero.wiki/img/a-36613.jpg",
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"https://numero.wiki/img/b-36613.jpg",
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"https://number.academy/i/infographics/3/number-36613-infographic.png",
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"https://numero.wiki/s/share-desktop.png",
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https://cs.stackexchange.com/questions/20155/bubble-sort-complexity | [
"Bubble sort complexity [duplicate]\n\nSo I have this code:\n\ndone <- false \nn <- 0 \nwhile (n < a) and (done = false) [(n+1)(1+1+1)]\ndone <- true [n]\nfor m <- (a- 1) downto n [n(1+1+1+1)]\nif list[m] < list[m - 1] then [n]\ntmp <- list[m] [n]\nlist[m] <- list[m-1] [n]\nlist[m - 1] <- tmp [n]\ndone <- false [n]\nn <- n + 1 \nreturn list \n\nAm I doing this right? My conclusions are that the inne for-loop runs (n^2 + n) / 2 times and the outher while-loop runs n+1 times. I don't know how to properly argue for that the bubble sort has the complexity O(n^2)\n\nmarked as duplicate by FrankW, Wandering Logic, GillesMay 9 '14 at 13:34\n\n• The number of steps in the inner loop being proportional to $n^2$ refers to the aggregate over all iterations of the outer loop. Also, $n^2$ is not the same as $\\frac{n^2+n}2$, no matter how large $n$ might get. – FrankW Jan 31 '14 at 16:50"
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https://wikidev.in/wiki/C/math_h/exp | [
"You are here : Cmath.hexp\n\nexp() - math.h\n\n`The C library function double exp(double x) returns the value of e raised to the xth power.`\n\nSyntax\n\n`double exp(double x)`\n\nExample\n\n```#include <stdio.h>\n#include <math.h>\n\nint main ()\n{\ndouble x = 0;\n\nprintf(\"The exponential value of %lf is %lf\\n\", x, exp(x));\nprintf(\"The exponential value of %lf is %lf\\n\", x+1, exp(x+1));\nprintf(\"The exponential value of %lf is %lf\\n\", x+2, exp(x+2));\n\nreturn(0);\n}```\n\nOutput / Return Value\n\n```The exponential value of 0.000000 is 1.000000\nThe exponential value of 1.000000 is 2.718282\nThe exponential value of 2.000000 is 7.389056```"
]
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https://www.overclock.net/threads/opengl-c-function-pointer-question.541509/ | [
"",
null,
"1 - 7 of 7 Posts\n\n#### rabidgnome229\n\n·\nJoined\n·\n5,282 Posts\nDiscussion Starter · ·\nI'm writing an opengl app (GLUT) in C++ and I'm getting an error when attempting to set the callbacks. The signature for the first is this\n\nCode:\nCode:\n``void glutDisplayFunc(void (*func)(void));``\nIn one of my classes I have a function draw declared as such (irrelevant portions omitted)\n\nCode:\nCode:\n``````#ifndef _GRAPHICS_ENGINE_H\n#define _GRAPHICS_ENGINE_H\n\nclass GraphicsEngine{\npublic:\nvoid draw(void);\n};\n\n#endif``````\nIn my implementation I have the following\n\nCode:\nCode:\n``````void GraphicsEngine::draw(void){\n}\n\nvoid GraphicsEngine::init(int *argc, char **argv){\nglutDisplayFunc(draw);\n}``````\nOn compilation, I get the following error\n\nCode:\nCode:\n``src/GraphicsEngine.cpp:42: error: argument of type 'void (GraphicsEngine::)()' does not match 'void (*)()'``\nIf I'm reading that error correctly its the fact that the function is a member of a class that's causing the problem. If so, how can I set a member function as the glut callback? If not, what is the problem? Thanks in advance\n\n#### pippolo\n\n·\n##### Registered\nJoined\n·\n391 Posts\nQuote:\n Originally Posted by rabidgnome229",
null,
"If I'm reading that error correctly its the fact that the function is a member of a class that's causing the problem. If so, how can I set a member function as the glut callback? If not, what is the problem? Thanks in advance\nEvery function (not static) member of a class A is really a function with an implicit first argument of type A*, and the argument is named this.\nSo, the signature of your draw function is really:\n\nCode:\nCode:\n``````void GraphicsEngine::draw(GraphicsEngine *this){\n}``````\nand when you write\n\nCode:\nCode:\n``````GraphicsEngine a;\na.draw();``````\nyou are really calling\n\nCode:\nCode:\n``draw(a);``\nand this is incompatible with the glutDisplayFunc signature.\n\nA possible solution (but not the ONLY possible) is to use a static member function. A static function is indipendent from the class object and doesn't have the additional this argument.\n\nCode:\nCode:\n``````class GraphicsEngine{\npublic:\nstatic void draw(void);\n};``````\n\n•",
null,
"rabidgnome229\n\n#### im_not_an_artard\n\n·\n##### Registered\nJoined\n·\n1,984 Posts\nQuote:\n Originally Posted by pippolo",
null,
"Your assumption is correct. Every function (not static) member of a class A is really a function with an implicit first argument of type A*, and the argument is named this. So, the signature of your draw function is really: Code: Code: ``````void GraphicsEngine::draw(GraphicsEngine *this){ }`````` and when you write Code: Code: ``````GraphicsEngine a; a.draw();`````` you are really calling Code: Code: ``draw(a);`` and this is incompatible with the glutDisplayFunc signature. A possible solution (but not the ONLY possible) is to use a static member function. A static function is indipendent from the class object and doesn't have the additional this argument. Code: Code: ``````class GraphicsEngine{ public: static void draw(void); };``````\n^ This\n\nas well you should declare as\n\nCode:\nCode:\n``````GraphicsEngine *a;\n\na = new GraphicsEngine;\n\ndraw(a);``````\nEdit: wait a sec, i may have misred ur code, you may have to use \"this\" in ur calls or something\n\ni.e.\n\nCode:\nCode:\n``````glutDisplayFunc(this::draw);\n\nor\n\nglutDisplayFunc(this->draw);``````\ni think thats the syntax\n\n#### rabidgnome229\n\n·\nJoined\n·\n5,282 Posts\nDiscussion Starter · ·\nThat's abot what I figured. The class is a singleton class so I ended up putting this in my implementation file\n\nCode:\nCode:\n``````static void draw_workaround(void){\nGraphicsEngine::instance()->draw();\n}\n\nvoid GraphicsEngine::glInit(int *argc, char **argv){\nglutDisplayFunc(draw_workaround);\n}``````\nOne of the first things I tried was this->draw that artard posted, but that didn't work either.\n\n#### im_not_an_artard\n\n·\n##### Registered\nJoined\n·\n1,984 Posts\nQuote:\n Originally Posted by rabidgnome229",
null,
"That's abot what I figured. The class is a singleton class so I ended up putting this in my implementation file Code: Code: ``````static void draw_workaround(void){ GraphicsEngine::instance()->draw(); } void GraphicsEngine::glInit(int *argc, char **argv){ glutDisplayFunc(draw_workaround); }`````` One of the first things I tried was this->draw that artard posted, but that didn't work either.\nisnt coding fun\n\n#### pippolo\n\n·\n##### Registered\nJoined\n·\n391 Posts\nQuote:\n Originally Posted by im_not_an_artard",
null,
"^ This as well you should declare as Code: Code: ``````GraphicsEngine *a; a = new GraphicsEngine; draw(a);``````\nmy mistake... the last row should be\n\nCode:\n\nCode:\n``draw(&a);``\nQuote:\n Originally Posted by im_not_an_artard",
null,
"Edit: wait a sec, i may have misred ur code, you may have to use \"this\" in ur calls or something i.e. Code: Code: ``````glutDisplayFunc(this::draw); or glutDisplayFunc(this->draw);`````` i think thats the syntax\nNo. The problem with a non static member function is that the invocation needs a pointer to a class object, but the invocation is made by the glut library. The glutDisplayFunc asks for a\n\nCode:\n\nCode:\n``void func(void)``\nfunction, but a non static member function is really\n\nCode:\n\nCode:\n``void func(obj*)``\n.\n\nQuote:\n Originally Posted by rabidgnome229",
null,
"That's abot what I figured. The class is a singleton class so I ended up putting this in my implementation file Code: Code: ``````static void draw_workaround(void){ GraphicsEngine::instance()->draw(); } void GraphicsEngine::glInit(int *argc, char **argv){ glutDisplayFunc(draw_workaround); }`````` One of the first things I tried was this->draw that artard posted, but that didn't work either.\nthe version with the draw_workaround should works (I hope !).\n\n•",
null,
"rabidgnome229\n\n#### Kirmie\n\n·\n##### Registered\nJoined\n·\n2,703 Posts\n•",
null,
"rabidgnome229\n1 - 7 of 7 Posts"
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https://jozeelin.github.io/2019/06/14/%E6%9C%80%E5%A4%A7%E7%86%B5%E6%A8%A1%E5%9E%8B/ | [
"## 7.1 最大熵模型\n\n### 7.1.3 最大熵模型的学习\n\n1. 定义最优化的原始问题\n\n首先，引进拉格朗日乘子$w_0,w_1,w_2,...,w_n$，定义拉格朗日函数L(P,w):\n\n1. 对偶问题\n\n在满足KKT条件的情况下，原始问题的解与对偶问题的解等价\n\n2. 求解对偶问题的极小化问题\n\n具体地，求L(P,w)对P(y|x)的偏导数。\n\n令偏导数为0，在$\\tilde{P}(x)>0$的情况下，解得:\n\n由于$\\sum_y P(y|x) = 1$，得:\n\n其中，\n\n$Z_w(x)$为归一化因子；$f_i(x,y)$是特征函数；$w_i$是特征权重。由式(6.22)、式(6.23)表示的模型$P_w=P_w(y|x)$就是最大熵模型。w是最大熵模型中的参数向量。\n\n3. 求解对偶问题的极大化问题\n\n将其解记为$w^*$，即:\n\n从而使得$P^* = P_{w^*}=P_{w^*}(y|x)$是学习到的最优模型(最大熵模型)。也就是说，最大熵模型的学习归结为对偶函数$\\Psi(w)$的极大化。\n\n## 7.2 最大熵模型学习算法\n\n### 7.2.1 改进的迭代尺度法\n\n$A(\\delta|w)$是对数似然函数改变量的一个下界。\n\n$B(\\delta|w)$$\\delta_i$的偏导数:\n\n1. 对所有$i \\in \\{1,2,...,n\\}$，取初值$w_i=0$\n\n2. 对每一$i\\in \\{1,2,...,n\\}$\n\n(a) 令$\\delta_i$是方程$\\sum_x \\tilde{P}(x) \\sum_y P_w(y|x)f_i(x,y)\\exp (\\delta_i f^{*}(x,y)) = E_{\\tilde{P}}(f_i)$的解，这里，\n\n(b) 更新$w_i$的值:$w_i \\leftarrow w_i+\\delta_i$\n\n3. 如果不是所有$w_i$都收敛，重复步(2)。\n\n• 如果$f^{*}(x,y)$是常数，即对任何x,y，有$f^{*}(x,y)=M$，那么$\\delta_i$可以显式表示为(注意：此处的M相当于学习率，为超参数)：\n• 如果$f^{*}(x,y)$不是常数，那么必须通过数值计算求$\\delta_i$。简单有效的方法为牛顿法。把式(6.33)表示成$g(\\delta_i)=0$牛顿法通过迭代求得$\\delta^{*}_i$，使得$g(\\delta_i^{*})=0$。迭代公式:只要适当选取初始值$\\delta_i^{(0)}$，由于$\\delta_i$方程(6.33)有单根，因此牛顿法恒收敛，而且收敛速度更快。\n\n### 7.2.2拟牛顿法\n\n1. 选定初始点$w^{(0)}$，取$B_0$为正定对称矩阵，置k=0\n\n2. 计算$g_k = g(w^{(k)})$。若$\\|g_k\\|<\\epsilon$，则停止计算，得$w^{*} = w^{(k)}$；否则转(3)\n\n3. $B_kp_k = -g_k$，求出$p_k$\n\n4. 一维搜索:求$\\lambda_k$使得\n\n5. $w^{(k+1)} = w^{(k)}+\\lambda_k p_k$\n\n6. 计算$g_{k+1} = g(w^{(k+1)})$，若$\\|g_{k+1}\\| < \\epsilon$，则停止计算，得$w^{*} = w^{(k+1)}$；否则，按下式求出$B_{k+1}$:\n\n其中，\n\n7. 置k=k+1,转(3)"
]
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http://mymathforum.com/pre-calculus/54109-how-do-you-simplify-two-variable-division-problem-variables-exponent.html | [
"",
null,
"My Math Forum",
null,
"How do you simplify a two-variable division problem with variables in the exponent?\n User Name Remember Me? Password\n\n Pre-Calculus Pre-Calculus Math Forum\n\n May 15th, 2015, 05:12 PM #1 Member Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0 How do you simplify a two-variable division problem with variables in the exponent? Simplify $\\displaystyle \\frac{x^{3a+2}} {x^{2a-1}}$ (In case I'm doing the LaTeX incorrectly, it's (x^3a+2) / (x^2a-1), or (x to the power of 3a + 2) divided by (x to the power of 2a - 1) Since there's no answer to the problem, I wouldn't know how to set up a logarithm for either part of the fraction. (Sorry for how basic my knowledge is. I'm working on it.) The answer is $\\displaystyle x^{a + 3}$, but I can't see how to get there. Last edited by skipjack; May 18th, 2015 at 11:16 PM.",
null,
"May 15th, 2015, 05:16 PM #2\nMath Team\n\nJoined: May 2013\nFrom: The Astral plane\n\nPosts: 2,257\nThanks: 928\n\nMath Focus: Wibbly wobbly timey-wimey stuff.\nQuote:\n Originally Posted by Rexan",
null,
"Simplify $\\displaystyle \\frac{x^{3a+2}} {x^{2a-1}}$ Since there's no answer to the problem, I wouldn't know how to set up a logarithm for either part of the fraction. (Sorry for how basic my knowledge is. I'm working on it.) The answer is $\\displaystyle x^{a + 3}$, but I can't see how to get there.\nExponents have two properties that are useful here.\n\n1) $\\displaystyle \\frac{1}{m^n} = m^{-n}$\n\n2) $\\displaystyle m^n \\cdot m^p = m^{n + p}$\n\nSo\n$\\displaystyle \\frac{x^{3a + 2}}{x^{2a - 1}} = \\left ( x^{3a + 2} \\right ) \\cdot \\left ( x^{-(2a - 1)} \\right ) = x^{(3a + 2) - (2a - 1)}$\n\netc.\n\n-Dan\n\nLast edited by skipjack; May 18th, 2015 at 11:17 PM.",
null,
"May 15th, 2015, 05:24 PM #3\nMember\n\nJoined: May 2015\nFrom: U.S.A.\n\nPosts: 45\nThanks: 0\n\nQuote:\n Originally Posted by topsquark",
null,
"Exponents have two properties that are useful here. 1) $\\displaystyle \\frac{1}{m^n} = m^{-n}$ 2) $\\displaystyle m^n \\cdot m^p = m^{n + p}$ So $\\displaystyle \\frac{x^{3a + 2}}{x^{2a - 1}} = \\left ( x^{3a + 2} \\right ) \\cdot \\left ( x^{-(2a - 1)} \\right ) = x^{(3a + 2) - (2a - 1)}$ etc. -Dan\nFinally got it. Thank you!",
null,
"Tags division, exponent, problem, simplify, twovariable, variables",
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"Similar Threads Thread Thread Starter Forum Replies Last Post cduinctk714 Algebra 3 December 8th, 2014 07:21 PM DecoratorFawn82 Algebra 17 September 10th, 2014 03:59 PM Bencz Algebra 4 May 4th, 2013 04:14 PM claudiomalk Algebra 2 November 9th, 2012 08:52 PM skarface Algebra 1 January 24th, 2010 03:28 PM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top\n\nCopyright © 2019 My Math Forum. All rights reserved.",
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https://socratic.org/questions/could-the-points-4-3-1-1-and-1-3-form-the-vertices-of-a-right-triangle | [
"# Could the points (-4,3), (-1,1) and (1,3) form the vertices of a right triangle?\n\nUsing distance formula $\\sqrt{{\\left({x}_{2} - {x}_{1}\\right)}^{2} + {\\left({y}_{2} - {y}_{1}\\right)}^{2}}$, the distance between (-4,3) and (-1,1) would be $\\sqrt{{3}^{2} + {\\left(- 2\\right)}^{2}} = \\sqrt{13}$, distance between (-1,1) and (1,3) would be $\\sqrt{{2}^{2} + {2}^{2}} = \\sqrt{8}$, and the distance between (-4,3) and (1,3) would be $\\sqrt{{5}^{2} + {0}^{2}} = \\sqrt{25}$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7632167,"math_prob":1.00001,"size":426,"snap":"2020-34-2020-40","text_gpt3_token_len":105,"char_repetition_ratio":0.1421801,"word_repetition_ratio":0.0,"special_character_ratio":0.2535211,"punctuation_ratio":0.10344828,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T17:53:32Z\",\"WARC-Record-ID\":\"<urn:uuid:42255aa3-de40-4f00-8850-9936225a44a2>\",\"Content-Length\":\"33854\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5366ab88-6e92-4266-8dda-f8ff62b76b40>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f32d252-38f3-4f1e-ba1f-09bbfcc79ccb>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/could-the-points-4-3-1-1-and-1-3-form-the-vertices-of-a-right-triangle\",\"WARC-Payload-Digest\":\"sha1:C2DQOZAQ5BNXIETQZC3Q6TKABN5SI2WV\",\"WARC-Block-Digest\":\"sha1:SLFWJ7WSHFUPMIC647SSTB4TKIWAIBWK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739347.81_warc_CC-MAIN-20200814160701-20200814190701-00014.warc.gz\"}"} |
http://blog.oneapm.com/apm-tech/571.html | [
"###### 普遍性和可检测性：\n\nXpath 注入是 OWASP TOP10 安全威胁中 A1 Injection 中的一种，注入漏洞发生在应用程序将不可信的数据发送到解释器时。虽然注入漏洞很容易通过审查代码发现，但是却不容易在测试中发现。\n\n###### 从代码层次如何防御：\n\n• javax.xml.xpath\n• org.jdom.xpath\n• org.jdom2.xpath等\n\n###### 下面我们来实现这个CheckMap内部方法：\n\n1. 通过遍历检查Map中key得合法性\n\n`````` for (final String key : params.keySet()) {\nif (this.checkString(key)) {\nreturn true;\n}\n``````\n\n2.检查每一个 key 对应得 value 得合法性：\n\n`````` final Collection<String> coll = (Collection<String>)params.get((Object)key);\nfor (final String input : coll) {\nif (this.checkString(input)) {\nreturn true;\n}\n}\n``````\n\n``````private boolean checkString(final String input) {\nreturn null != input && input.length() > 1 && (this.parser.IsOutBoundary(input) || (-1 != input.indexOf(39) && (this.parser.IsOutBoundary(input.replace(\"'\", \"''\")) || this.parser.IsOutBoundary(input.replace(\"'\", \"\\\\'\")))) || (-1 != input.indexOf(34) && (this.parser.IsOutBoundary(input.replace(\"\\\"\", \"\\\"\\\"\")) || this.parser.IsOutBoundary(input.replace(\"\\\"\", \"\\\\\\\"\")))) || this.parser.IsQuoteUnbalanced(input));\n}\n``````\n\n``````-1!=input.indexOf(39)&&(this.parser.IsOutBoundary(input.replace(\"'\",\"''\")\n-1!=input.indexOf(34)&& this.parser.IsOutBoundary(input.replace(\"\\\"\", \"\\\"\\\"\"))\n``````\n\n``````public boolean IsOutBoundary(String input) {\nint offset = 0;\nif (null == input || input.length() <= 1) {\nreturn false;\n}\ninput = input.toLowerCase();\nwhile (true) {\nfinal int x = this.getRawValue().indexOf(input, offset);\nfinal int y = x + input.length();\nif (-1 == x) {\nreturn false;\n}\nfinal int ceil = this.getCeiling(this.boundaries, x + 1);\nif (-1 != ceil && ceil < y) {\nreturn true;\n}\noffset = y;\n}\n}\n\npublic boolean IsQuoteUnbalanced(String input) {\ninput = input.toLowerCase();\nreturn this.getRawValue().contains(input) && this.stack.size() > 0 && input.indexOf(this.stack.peek()) != -1;\n}\n\npublic String getRawValue() {\nreturn this.input;\n}\n\nprivate int getCeiling(final List<Integer> boundaries, final int value) {\nfor (final int x : boundaries) {\nif (x >= value) {\nreturn x;\n}\n}\nreturn -1;\n}\n``````\n###### 漏洞攻击示例\n\nWebGoat 是 OWASP 推出得一款开源的含有大量漏洞攻击的应用，在 Github 上可以直接搜到源码。\n\n*我们找到 Xpath Injection 得 lession *，如下图：",
null,
"hints提示我们攻击的入参\n\n`Try username: Smith' or 1=1 or 'a'='a and a password: anything`",
null,
"",
null,
"Xpath 查询失败了，并没有返回任何结果，攻击被拦截之后，前端页面没有渲染任何东西。由此可见入参检查在注入类得漏洞防御中可以起到立竿见影得作用。\n\n OWASP TOP10-2013 release\n\n WebGoat"
]
| [
null,
"http://blog.oneapm.com/content/images/2015/12/1-4.png",
null,
"http://blog.oneapm.com/content/images/2015/12/2-4.png",
null,
"http://blog.oneapm.com/content/images/2015/12/3-5.png",
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.54010767,"math_prob":0.9161755,"size":3397,"snap":"2021-31-2021-39","text_gpt3_token_len":1781,"char_repetition_ratio":0.13527851,"word_repetition_ratio":0.0,"special_character_ratio":0.26670593,"punctuation_ratio":0.20153551,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96147656,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T22:23:03Z\",\"WARC-Record-ID\":\"<urn:uuid:31eb7c5e-bbe9-46a1-a078-c7fa0e5d776d>\",\"Content-Length\":\"36445\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5124a3d-4304-4f59-9018-00b220bc0b2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:e857a50a-5022-4eb7-bf6e-394e06761750>\",\"WARC-IP-Address\":\"139.196.253.46\",\"WARC-Target-URI\":\"http://blog.oneapm.com/apm-tech/571.html\",\"WARC-Payload-Digest\":\"sha1:SWGGQKHVJGIP2SGD463DQANYLVNGFHEY\",\"WARC-Block-Digest\":\"sha1:FHG7FWQOEGA5QNY2NGMWJ5KGOFF45O2D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155188.79_warc_CC-MAIN-20210804205700-20210804235700-00283.warc.gz\"}"} |
https://mne.tools/dev/auto_tutorials/preprocessing/45_projectors_background.html | [
"# Background on projectors and projections#\n\nThis tutorial provides background information on projectors and Signal Space Projection (SSP), and covers loading and saving projectors, adding and removing projectors from Raw objects, the difference between “applied” and “unapplied” projectors, and at what stages MNE-Python applies projectors automatically.\n\nWe’ll start by importing the Python modules we need; we’ll also define a short function to make it easier to make several plots that look similar:\n\nimport os\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom mpl_toolkits.mplot3d import Axes3D # noqa\nfrom scipy.linalg import svd\nimport mne\n\ndef setup_3d_axes():\nax = plt.axes(projection='3d')\nax.view_init(azim=-105, elev=20)\nax.set_xlabel('x')\nax.set_ylabel('y')\nax.set_zlabel('z')\nax.set_xlim(-1, 5)\nax.set_ylim(-1, 5)\nax.set_zlim(0, 5)\nreturn ax\n\n\n## What is a projection?#\n\nIn the most basic terms, a projection is an operation that converts one set of points into another set of points, where repeating the projection operation on the resulting points has no effect. To give a simple geometric example, imagine the point $$(3, 2, 5)$$ in 3-dimensional space. A projection of that point onto the $$x, y$$ plane looks a lot like a shadow cast by that point if the sun were directly above it:\n\nax = setup_3d_axes()\n\n# plot the vector (3, 2, 5)\norigin = np.zeros((3, 1))\npoint = np.array([[3, 2, 5]]).T\nvector = np.hstack([origin, point])\nax.plot(*vector, color='k')\nax.plot(*point, color='k', marker='o')\n\n# project the vector onto the x,y plane and plot it\nxy_projection_matrix = np.array([[1, 0, 0], [0, 1, 0], [0, 0, 0]])\nprojected_point = xy_projection_matrix @ point\nprojected_vector = xy_projection_matrix @ vector\nax.plot(*projected_vector, color='C0')\nax.plot(*projected_point, color='C0', marker='o')\n\n# add dashed arrow showing projection\narrow_coords = np.concatenate([point, projected_point - point]).flatten()\nax.quiver3D(*arrow_coords, length=0.96, arrow_length_ratio=0.1, color='C1',\nlinewidth=1, linestyle='dashed')",
null,
"Note\n\nThe @ symbol indicates matrix multiplication on NumPy arrays, and was introduced in Python 3.5 / NumPy 1.10. The notation plot(*point) uses Python argument expansion to “unpack” the elements of point into separate positional arguments to the function. In other words, plot(*point) expands to plot(3, 2, 5).\n\nNotice that we used matrix multiplication to compute the projection of our point $$(3, 2, 5)onto the :math:x, y$$ plane:\n\n$\\begin{split}\\left[ \\begin{matrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{matrix} \\right] \\left[ \\begin{matrix} 3 \\\\ 2 \\\\ 5 \\end{matrix} \\right] = \\left[ \\begin{matrix} 3 \\\\ 2 \\\\ 0 \\end{matrix} \\right]\\end{split}$\n\n…and that applying the projection again to the result just gives back the result again:\n\n$\\begin{split}\\left[ \\begin{matrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{matrix} \\right] \\left[ \\begin{matrix} 3 \\\\ 2 \\\\ 0 \\end{matrix} \\right] = \\left[ \\begin{matrix} 3 \\\\ 2 \\\\ 0 \\end{matrix} \\right]\\end{split}$\n\nFrom an information perspective, this projection has taken the point $$x, y, z$$ and removed the information about how far in the $$z$$ direction our point was located; all we know now is its position in the $$x, y$$ plane. Moreover, applying our projection matrix to any point in $$x, y, z$$ space will reduce it to a corresponding point on the $$x, y$$ plane. The term for this is a subspace: the projection matrix projects points in the original space into a subspace of lower dimension than the original. The reason our subspace is the $$x,y$$ plane (instead of, say, the $$y,z$$ plane) is a direct result of the particular values in our projection matrix.\n\n### Example: projection as noise reduction#\n\nAnother way to describe this “loss of information” or “projection into a subspace” is to say that projection reduces the rank (or “degrees of freedom”) of the measurement — here, from 3 dimensions down to 2. On the other hand, if you know that measurement component in the $$z$$ direction is just noise due to your measurement method, and all you care about are the $$x$$ and $$y$$ components, then projecting your 3-dimensional measurement into the $$x, y$$ plane could be seen as a form of noise reduction.\n\nOf course, it would be very lucky indeed if all the measurement noise were concentrated in the $$z$$ direction; you could just discard the $$z$$ component without bothering to construct a projection matrix or do the matrix multiplication. Suppose instead that in order to take that measurement you had to pull a trigger on a measurement device, and the act of pulling the trigger causes the device to move a little. If you measure how trigger-pulling affects measurement device position, you could then “correct” your real measurements to “project out” the effect of the trigger pulling. Here we’ll suppose that the average effect of the trigger is to move the measurement device by $$(3, -1, 1)$$:\n\ntrigger_effect = np.array([[3, -1, 1]]).T\n\n\nKnowing that, we can compute a plane that is orthogonal to the effect of the trigger (using the fact that a plane through the origin has equation $$Ax + By + Cz = 0$$ given a normal vector $$(A, B, C)$$), and project our real measurements onto that plane.\n\n# compute the plane orthogonal to trigger_effect\nx, y = np.meshgrid(np.linspace(-1, 5, 61), np.linspace(-1, 5, 61))\nA, B, C = trigger_effect\nz = (-A * x - B * y) / C\n# cut off the plane below z=0 (just to make the plot nicer)\n\n\nComputing the projection matrix from the trigger_effect vector is done using singular value decomposition (SVD); interested readers may consult the internet or a linear algebra textbook for details on this method. With the projection matrix in place, we can project our original vector $$(3, 2, 5)$$ to remove the effect of the trigger, and then plot it:\n\n# compute the projection matrix\nU, S, V = svd(trigger_effect, full_matrices=False)\ntrigger_projection_matrix = np.eye(3) - U @ U.T\n\n# project the vector onto the orthogonal plane\nprojected_point = trigger_projection_matrix @ point\nprojected_vector = trigger_projection_matrix @ vector\n\n# plot the trigger effect and its orthogonal plane\nax = setup_3d_axes()\nax.plot_trisurf(x, y, z, color='C2', shade=False, alpha=0.25)\nax.quiver3D(*np.concatenate([origin, trigger_effect]).flatten(),\narrow_length_ratio=0.1, color='C2', alpha=0.5)\n\n# plot the original vector\nax.plot(*vector, color='k')\nax.plot(*point, color='k', marker='o')\noffset = np.full((3, 1), 0.1)\nax.text(*(point + offset).flat, '({}, {}, {})'.format(*point.flat), color='k')\n\n# plot the projected vector\nax.plot(*projected_vector, color='C0')\nax.plot(*projected_point, color='C0', marker='o')\noffset = np.full((3, 1), -0.2)\nax.text(*(projected_point + offset).flat,\n'({}, {}, {})'.format(*np.round(projected_point.flat, 2)),\ncolor='C0', horizontalalignment='right')\n\n# add dashed arrow showing projection\narrow_coords = np.concatenate([point, projected_point - point]).flatten()\nax.quiver3D(*arrow_coords, length=0.96, arrow_length_ratio=0.1,\ncolor='C1', linewidth=1, linestyle='dashed')",
null,
"Just as before, the projection matrix will map any point in $$x, y, z$$ space onto that plane, and once a point has been projected onto that plane, applying the projection again will have no effect. For that reason, it should be clear that although the projected points vary in all three $$x$$, $$y$$, and $$z$$ directions, the set of projected points have only two effective dimensions (i.e., they are constrained to a plane).\n\nProjections of EEG or MEG signals work in very much the same way: the point $$x, y, z$$ corresponds to the value of each sensor at a single time point, and the projection matrix varies depending on what aspects of the signal (i.e., what kind of noise) you are trying to project out. The only real difference is that instead of a single 3-dimensional point $$(x, y, z)$$ you’re dealing with a time series of $$N$$-dimensional “points” (one at each sampling time), where $$N$$ is usually in the tens or hundreds (depending on how many sensors your EEG/MEG system has). Fortunately, because projection is a matrix operation, it can be done very quickly even on signals with hundreds of dimensions and tens of thousands of time points.\n\n## Signal-space projection (SSP)#\n\nWe mentioned above that the projection matrix will vary depending on what kind of noise you are trying to project away. Signal-space projection (SSP) is a way of estimating what that projection matrix should be, by comparing measurements with and without the signal of interest. For example, you can take additional “empty room” measurements that record activity at the sensors when no subject is present. By looking at the spatial pattern of activity across MEG sensors in an empty room measurement, you can create one or more $$N$$-dimensional vector(s) giving the “direction(s)” of environmental noise in sensor space (analogous to the vector for “effect of the trigger” in our example above). SSP is also often used for removing heartbeat and eye movement artifacts — in those cases, instead of empty room recordings the direction of the noise is estimated by detecting the artifacts, extracting epochs around them, and averaging. See Repairing artifacts with SSP for examples.\n\nOnce you know the noise vectors, you can create a hyperplane that is orthogonal to them, and construct a projection matrix to project your experimental recordings onto that hyperplane. In that way, the component of your measurements associated with environmental noise can be removed. Again, it should be clear that the projection reduces the dimensionality of your data — you’ll still have the same number of sensor signals, but they won’t all be linearly independent — but typically there are tens or hundreds of sensors and the noise subspace that you are eliminating has only 3-5 dimensions, so the loss of degrees of freedom is usually not problematic.\n\n## Projectors in MNE-Python#\n\nIn our example data, SSP has already been performed using empty room recordings, but the projectors are stored alongside the raw data and have not been applied yet (or, synonymously, the projectors are not active yet). Here we’ll load the sample data and crop it to 60 seconds; you can see the projectors in the output of read_raw_fif() below:\n\nsample_data_folder = mne.datasets.sample.data_path()\nsample_data_raw_file = os.path.join(sample_data_folder, 'MEG', 'sample',\n'sample_audvis_raw.fif')\n\nOpening raw data file /home/circleci/mne_data/MNE-sample-data/MEG/sample/sample_audvis_raw.fif...\nRead a total of 3 projection items:\nPCA-v1 (1 x 102) idle\nPCA-v2 (1 x 102) idle\nPCA-v3 (1 x 102) idle\nRange : 25800 ... 192599 = 42.956 ... 320.670 secs\nReading 0 ... 36037 = 0.000 ... 60.000 secs...\n\nMeasurement date December 03, 2002 19:01:10 GMT MEG Unknown 146 points 204 Gradiometers, 102 Magnetometers, 9 Stimulus, 60 EEG, 1 EOG MEG 2443, EEG 053 EOG 061 Not available 600.61 Hz 0.10 Hz 172.18 Hz PCA-v1 : offPCA-v2 : offPCA-v3 : off sample_audvis_raw.fif 00:01:01 (HH:MM:SS)\n\nIn MNE-Python, the environmental noise vectors are computed using principal component analysis, usually abbreviated “PCA”, which is why the SSP projectors usually have names like “PCA-v1”. (Incidentally, since the process of performing PCA uses singular value decomposition under the hood, it is also common to see phrases like “projectors were computed using SVD” in published papers.) The projectors are stored in the projs field of raw.info:\n\nprint(raw.info['projs'])\n\n[<Projection | PCA-v1, active : False, n_channels : 102>, <Projection | PCA-v2, active : False, n_channels : 102>, <Projection | PCA-v3, active : False, n_channels : 102>]\n\n\nraw.info['projs'] is an ordinary Python list of Projection objects, so you can access individual projectors by indexing into it. The Projection object itself is similar to a Python dict, so you can use its .keys() method to see what fields it contains (normally you don’t need to access its properties directly, but you can if necessary):\n\nfirst_projector = raw.info['projs']\nprint(first_projector)\nprint(first_projector.keys())\n\n<Projection | PCA-v1, active : False, n_channels : 102>\ndict_keys(['desc', 'kind', 'active', 'data', 'explained_var'])\n\n\nThe Raw, Epochs, and Evoked objects all have a boolean proj attribute that indicates whether there are any unapplied / inactive projectors stored in the object. In other words, the proj attribute is True if at least one projector is present and all of them are active. In addition, each individual projector also has a boolean active field:\n\nprint(raw.proj)\nprint(first_projector['active'])\n\nFalse\nFalse\n\n\n### Computing projectors#\n\nIn MNE-Python, SSP vectors can be computed using general purpose functions mne.compute_proj_raw(), mne.compute_proj_epochs(), and mne.compute_proj_evoked(). The general assumption these functions make is that the data passed contains raw data, epochs or averages of the artifact you want to repair via projection. In practice this typically involves continuous raw data of empty room recordings or averaged ECG or EOG artifacts. A second set of high-level convenience functions is provided to compute projection vectors for typical use cases. This includes mne.preprocessing.compute_proj_ecg() and mne.preprocessing.compute_proj_eog() for computing the ECG and EOG related artifact components, respectively; see Repairing artifacts with SSP for examples of these uses. For computing the EEG reference signal as a projector, the function mne.set_eeg_reference() can be used; see Setting the EEG reference for more information.\n\nWarning\n\nIt is best to compute projectors only on channels that will be used (e.g., excluding bad channels). This ensures that projection vectors will remain ortho-normalized and that they properly capture the activity of interest.\n\n### Visualizing the effect of projectors#\n\nYou can see the effect the projectors are having on the measured signal by comparing plots with and without the projectors applied. By default, raw.plot() will apply the projectors in the background before plotting (without modifying the Raw object); you can control this with the boolean proj parameter as shown below, or you can turn them on and off interactively with the projectors interface, accessed via the Proj button in the lower right corner of the plot window. Here we’ll look at just the magnetometers, and a 2-second sample from the beginning of the file.\n\nmags = raw.copy().crop(tmax=2).pick_types(meg='mag')\nfor proj in (False, True):\nwith mne.viz.use_browser_backend('matplotlib'):\nfig = mags.plot(butterfly=True, proj=proj)\nfig.suptitle('proj={}'.format(proj), size='xx-large', weight='bold')\n\n•",
null,
"•",
null,
"Using matplotlib as 2D backend.\nUsing qt as 2D backend.\nUsing matplotlib as 2D backend.\nUsing qt as 2D backend.\n\n\nAdditional ways of visualizing projectors are covered in the tutorial Repairing artifacts with SSP.\n\nSSP can be used for other types of signal cleaning besides just reduction of environmental noise. You probably noticed two large deflections in the magnetometer signals in the previous plot that were not removed by the empty-room projectors — those are artifacts of the subject’s heartbeat. SSP can be used to remove those artifacts as well. The sample data includes projectors for heartbeat noise reduction that were saved in a separate file from the raw data, which can be loaded with the mne.read_proj() function:\n\necg_proj_file = os.path.join(sample_data_folder, 'MEG', 'sample',\n'sample_audvis_ecg-proj.fif')\nprint(ecg_projs)\n\n Read a total of 6 projection items:\nECG-planar-999--0.200-0.400-PCA-01 (1 x 203) idle\nECG-planar-999--0.200-0.400-PCA-02 (1 x 203) idle\nECG-axial-999--0.200-0.400-PCA-01 (1 x 102) idle\nECG-axial-999--0.200-0.400-PCA-02 (1 x 102) idle\nECG-eeg-999--0.200-0.400-PCA-01 (1 x 59) idle\nECG-eeg-999--0.200-0.400-PCA-02 (1 x 59) idle\n[<Projection | ECG-planar-999--0.200-0.400-PCA-01, active : False, n_channels : 203>, <Projection | ECG-planar-999--0.200-0.400-PCA-02, active : False, n_channels : 203>, <Projection | ECG-axial-999--0.200-0.400-PCA-01, active : False, n_channels : 102>, <Projection | ECG-axial-999--0.200-0.400-PCA-02, active : False, n_channels : 102>, <Projection | ECG-eeg-999--0.200-0.400-PCA-01, active : False, n_channels : 59>, <Projection | ECG-eeg-999--0.200-0.400-PCA-02, active : False, n_channels : 59>]\n\n\nThere is a corresponding mne.write_proj() function that can be used to save projectors to disk in .fif format:\n\nmne.write_proj('heartbeat-proj.fif', ecg_projs)\n\n\nNote\n\nBy convention, MNE-Python expects projectors to be saved with a filename ending in -proj.fif (or -proj.fif.gz), and will issue a warning if you forgo this recommendation.\n\nAbove, when we printed the ecg_projs list that we loaded from a file, it showed two projectors for gradiometers (the first two, marked “planar”), two for magnetometers (the middle two, marked “axial”), and two for EEG sensors (the last two, marked “eeg”). We can add them to the Raw object using the add_proj() method:\n\nraw.add_proj(ecg_projs)\n\n6 projection items deactivated\n\nMeasurement date December 03, 2002 19:01:10 GMT MEG Unknown 146 points 204 Gradiometers, 102 Magnetometers, 9 Stimulus, 60 EEG, 1 EOG MEG 2443, EEG 053 EOG 061 Not available 600.61 Hz 0.10 Hz 172.18 Hz PCA-v1 : offPCA-v2 : offPCA-v3 : offECG-planar-999--0.200-0.400-PCA-01 : offECG-planar-999--0.200-0.400-PCA-02 : offECG-axial-999--0.200-0.400-PCA-01 : offECG-axial-999--0.200-0.400-PCA-02 : offECG-eeg-999--0.200-0.400-PCA-01 : offECG-eeg-999--0.200-0.400-PCA-02 : off sample_audvis_raw.fif 00:01:01 (HH:MM:SS)\n\nTo remove projectors, there is a corresponding method del_proj() that will remove projectors based on their index within the raw.info['projs'] list. For the special case of replacing the existing projectors with new ones, use raw.add_proj(ecg_projs, remove_existing=True).\n\nTo see how the ECG projectors affect the measured signal, we can once again plot the data with and without the projectors applied (though remember that the plot() method only temporarily applies the projectors for visualization, and does not permanently change the underlying data). We’ll compare the mags variable we created above, which had only the empty room SSP projectors, to the data with both empty room and ECG projectors:\n\nmags_ecg = raw.copy().crop(tmax=2).pick_types(meg='mag')\nfor data, title in zip([mags, mags_ecg], ['Without', 'With']):\nwith mne.viz.use_browser_backend('matplotlib'):\nfig = data.plot(butterfly=True, proj=True)\nfig.suptitle('{} ECG projector'.format(title), size='xx-large',\nweight='bold')\n\n•",
null,
"•",
null,
"Removing projector <Projection | ECG-planar-999--0.200-0.400-PCA-01, active : False, n_channels : 203>\nRemoving projector <Projection | ECG-planar-999--0.200-0.400-PCA-02, active : False, n_channels : 203>\nRemoving projector <Projection | ECG-eeg-999--0.200-0.400-PCA-01, active : False, n_channels : 59>\nRemoving projector <Projection | ECG-eeg-999--0.200-0.400-PCA-02, active : False, n_channels : 59>\nUsing matplotlib as 2D backend.\nUsing qt as 2D backend.\nUsing matplotlib as 2D backend.\nUsing qt as 2D backend.\n\n\n### When are projectors “applied”?#\n\nBy default, projectors are applied when creating epoched data from Raw data, though application of the projectors can be delayed by passing proj=False to the Epochs constructor. However, even when projectors have not been applied, the mne.Epochs.get_data() method will return data as if the projectors had been applied (though the Epochs object will be unchanged). Additionally, projectors cannot be applied if the data are not preloaded. If the data are memory-mapped (i.e., not preloaded), you can check the _projector attribute to see whether any projectors will be applied once the data is loaded in memory.\n\nFinally, when performing inverse imaging (i.e., with mne.minimum_norm.apply_inverse()), the projectors will be automatically applied. It is also possible to apply projectors manually when working with Raw, Epochs or Evoked objects via the object’s apply_proj() method. For all instance types, you can always copy the contents of <instance>.info['projs'] into a separate list variable, use <instance>.del_proj(<index of proj(s) to remove>) to remove one or more projectors, and then add them back later with <instance>.add_proj(<list containing projs>) if desired.\n\nWarning\n\nRemember that once a projector is applied, it can’t be un-applied, so during interactive / exploratory analysis it’s a good idea to use the object’s copy() method before applying projectors.\n\n### Best practices#\n\nIn general, it is recommended to apply projectors when creating Epochs from Raw data. There are two reasons for this recommendation:\n\n1. It is computationally cheaper to apply projectors to data after the data have been reducted to just the segments of interest (the epochs)\n\n2. If you are applying amplitude-based rejection criteria to epochs, it is preferable to reject based on the signal after projectors have been applied, because the projectors may reduce noise in some epochs to tolerable levels (thereby increasing the number of acceptable epochs and consequenty increasing statistical power in any later analyses)."
]
| [
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_001.png",
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_002.png",
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_003.png",
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_004.png",
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_005.png",
null,
"https://mne.tools/dev/_images/sphx_glr_45_projectors_background_006.png",
null
]
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http://list.seqfan.eu/pipermail/seqfan/2019-August/019794.html | [
"# [seqfan] Are there \"harmonic-Bernoulli\" pseudoprimes?\n\nTomasz Ordowski tomaszordowski at gmail.com\nMon Aug 5 14:02:36 CEST 2019\n\n```Hello SeqFans,\n\nI noticed an interesting relationship\nof the harmonic numbers H(n)\nand Bernoulli numbers B(k)\n\"two in one\" with primes.\n\nTheorem: If p > 3 is a prime,\nthen (p-1)*H(p-2) == p*B(p-1) == -1 (mod p).\n\nConjecture: For n > 3,\n(n-1)*H(n-2) == n*B(n-1) (mod n)\nif and only if n is a prime.\n\nAmi tries to find possible counterexamples:\npseudoprimes, composite numbers m such that\nNumerator((m-1)*H(m-2) - m*B(m-1)) == 0 (mod m).\n\nHas anyone already considered the above congruence?"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6323099,"math_prob":0.960935,"size":696,"snap":"2021-43-2021-49","text_gpt3_token_len":234,"char_repetition_ratio":0.09682081,"word_repetition_ratio":0.0,"special_character_ratio":0.31034482,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T16:03:00Z\",\"WARC-Record-ID\":\"<urn:uuid:343d1cfa-c566-4511-b82f-b9342b523732>\",\"Content-Length\":\"3897\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad683e40-a348-462b-9b92-6d265b211efc>\",\"WARC-Concurrent-To\":\"<urn:uuid:5503f14e-7b6e-4ca2-8d22-bf71813396fc>\",\"WARC-IP-Address\":\"92.243.17.179\",\"WARC-Target-URI\":\"http://list.seqfan.eu/pipermail/seqfan/2019-August/019794.html\",\"WARC-Payload-Digest\":\"sha1:22SLBGUIXRVMJNOSW6ULCZTV3QA5ZAWE\",\"WARC-Block-Digest\":\"sha1:NYK4R4FYTUKJN7XHIPUPIVFRBA4SIVZL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360803.6_warc_CC-MAIN-20211201143545-20211201173545-00375.warc.gz\"}"} |
http://chipscoco.com/?id=178 | [
"Python项目实战\n\n# 简单的猜数字游戏\n\n```# 导入random模块\nimport random\n# 获取从1到100000之间的随机数\nrandom_number = random.randint(1, 1000000)\n# 执行input方法获取用户的输入，input的返回值为字符串类型，通过int()将其转换为整型\nguess_number = int(input(\"Please enter the number:____\\b\\b\\b\\b\"))```\n\n# 程序源码\n\n```# __author__ = 薯条老师\n# __referrer__ = http://chipscoco.com/?cate=59\nimport time\nimport random\n\ndef get_random_number(start=0, end=10**3):\nreturn random.randint(start, end)\n\ndef countdown(seconds=3, message=\"\"):\n\"\"\"\n:param seconds: 倒数的秒数\n:param message: 倒计时结束后输出的提示信息\n:return:\n\"\"\"\nfor _ in range(seconds, 0, -1):\n_, _ = print(_), time.sleep(1)\nelse:\nprint(message)\n\ndef serve_forever():\n\nrandom_number = get_random_number()\ncountdown(message=\"猜数字游戏开始，Go!!!\")\n\nwhile True:\ntry:\nguess_number = int(input(\"请输入你猜的数字:\"+\"_\"*4+\"\\b\"*4))\nexcept ValueError:\nprint(\"请输入合法的数字！\")\ncontinue\nif guess_number != random_number:\nprint(\"你输入的数字大了\") if guess_number > random_number else print(\"你输入的数字小了\")\ncontinue\nprint(\"恭喜你猜对了！！！\")\nif input(\"按键盘任意键继续玩猜数字游戏或输入quit退出游戏:____\\b\\b\\b\\b\").lower() == \"quit\":\nbreak\nelse:\nrandom_number = get_random_number()\nprint(\"游戏已被终止，再见！！！\")\n\nif __name__ == \"__main__\":\nserve_forever()```",
null,
"# 最具实力的小班培训\n\n`(1) Python后端工程师高薪就业班，月薪11K-18K，免费领取课程大纲(2) Python爬虫工程师高薪就业班，年薪十五万，免费领取课程大纲(3) Java后端开发工程师高薪就业班，月薪11K-20K, 免费领取课程大纲(4) Python大数据分析，量化投资就业班，月薪12K-25K,免费领取课程大纲`",
null,
""
]
| [
null,
"http://chipscoco.com/zb_users/upload/2021/02/202102141613291716331296.jpg",
null,
"http://chipscoco.com/zb_users/upload/2020/04/202004121586677416783738.jpg",
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.5537246,"math_prob":0.97619724,"size":1957,"snap":"2023-40-2023-50","text_gpt3_token_len":1236,"char_repetition_ratio":0.13261649,"word_repetition_ratio":0.0,"special_character_ratio":0.27337763,"punctuation_ratio":0.18571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99119467,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T01:16:03Z\",\"WARC-Record-ID\":\"<urn:uuid:e9e458d7-3024-4da9-b201-b9277647fd10>\",\"Content-Length\":\"25739\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d237faa8-e16e-489c-8371-e9460b036994>\",\"WARC-Concurrent-To\":\"<urn:uuid:89f6df5d-191c-4b4a-b4a9-cd87aaf4ae72>\",\"WARC-IP-Address\":\"47.107.239.253\",\"WARC-Target-URI\":\"http://chipscoco.com/?id=178\",\"WARC-Payload-Digest\":\"sha1:HOC6UHTLCJ27KOBXGBX4VI6RIP644PGM\",\"WARC-Block-Digest\":\"sha1:OYVXUNZPDN75YLWRNRXULJBEUOJDHPIJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100309.57_warc_CC-MAIN-20231202010506-20231202040506-00889.warc.gz\"}"} |
https://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/tag/gear-teeth | [
"## Posts Tagged ‘gear teeth’\n\n### Determining the Gear Train Tradeoff of Torque vs. Speed, Part Three\n\nWednesday, August 27th, 2014\n We’ve been working towards a general understanding of how gear trains work, and today we’ll solve a final piece of the puzzle when we identify how increased gear train torque is gained at the expense of gear train speed. Last time we developed a mathematical relationship between the torque, T, and the rotational speed, n, of the driving and driven gears in a simple gear train. This is represented by equation (8): TDriven ÷ TDriving = nDriving ÷ nDriven (8) For the purpose of our example we’ll assume that the driving gear is mounted to an electric motor shaft spinning at 100 revolutions per minute (RPM) and which produces 50 inch pounds of torque. Previous lab testing has determined that we require a torque of 100 inch pounds to properly run a piece of machinery that’s powered by the motor, and we’ve decided that the best way to get the required torque is not to employ a bigger, more powerful motor, but rather to install a gear train and manipulate its gear sizes until the desired torque is obtained. We know that using this approach will most likely affect the speed of our operation, and we want to determine how much speed will be compromised. So if the torque on the driven gear needs to be 100 inch pounds, then what will be the corresponding speed of the driven gear? To answer this question we’ll insert the numerical information we’ve been provided into equation (8). Doing so we arrive at the following: TDriven ÷ TDriving = nDriving ÷ nDriven (100 inch pounds) ÷ (50 inch pounds) = (100 RPM) ÷ nDriven 2 = (100 RPM) ÷ nDriven nDriven = (100 RPM) ÷ 2 = 50 RPM This tells us that in order to meet our torque requirement of 100 inch pounds, the gear train motor’s speed must be reduced from 100 RPM to 50 RPM, which represents a 50% reduction in speed, hence the tradeoff.",
null,
"This wraps up our blog series on gears and gear trains. Next time we’ll move on to a new topic: Galileo’s experiments with falling objects. _______________________________________\n\nTuesday, August 5th, 2014\n We’ve learned several methods to increase the torque of an electric motor through our series of articles on gear trains. One way is to attach a gear train to the motor’s shaft, a relatively simple thing to do. Today we’ll begin our exploration into how this method involves a tradeoff. It comes at the cost of speed. We’ll begin our examination of the tradeoff at play by linking together key elements learned through past blogs on the subject of gear trains. We’ll revisit those lessons through flashbacks.",
null,
"The first flashback we’ll make is to a blog entitled, Gear Ratio Formulas. There we learned that within a simple gear train consisting of two gears, the type most commonly employed to manipulate a motor’s torque, the ratio between the two gears, R, is relative to the ratio of their gear teeth, N. N is determined by the number of teeth each gear has in combination with the speeds, n, that each gear is going: R = NDriven ÷ NDriving = nDriving ÷ nDriven (1) The second flashback we’ll make is to a blog entitled, The Methodology Behind Gear Train Torque Conversions, in which we learned that the ratio of the torque, T, that exists between the gears is relative to the ratio of their respective pitch diameters, D: TDriving ÷ TDriven = DDriving ÷ DDriven (2) The tradeoff we’ve been alluding to comes in when gear speed, nDriven, represented in equation (1), is decreased, which results in an increase to TDriven in equation (2). But in order to see this we’ve got to somehow link the two equations together. In their present form there’s no common link between them. Or is there? There actually is an indirect link between the two equations, which comes by way of the torque equation presented in another past blog. The third flashback we’ll make is to the blog discussing that subject, which is entitled, The Relationship Between Torque and Horsepower. Using facts presented in that blog, the torque equations for our two gears become: TDriving = [HPDriving ÷ nDriving] × 63,025 (3) TDriven = [HPDriven ÷ nDriven] × 63,025 (4) Where’s the link between equations (1) and (2)? To answer that question we must reference a physics law known as The Law of Conservation of Energy . It states that the energy flowing from one gear to another within a gear train remains constant. Energy equates to horsepower, HP, in equations (3) and (4). So if the horsepower flowing through the gears is equal, our working equation becomes: HPDriving = HPDriven (5) Next time we’ll see how equation (5) is key to linking together equations (1) and (2) by way of equations (3) and (4). In so doing we’ll disclose the tradeoff to using gear trains. _______________________________________\n\n### Gear Train Torque Equations\n\nThursday, May 22nd, 2014\n In our last blog we mathematically linked the driving and driven gear Force vectors to arrive at a single common vector F, known as the resultant Force vector. This simplification allows us to achieve common ground between F and the two Distance vectors of our driving and driven gears, represented as D1 and D2. We can then use this commonality to develop individual torque equations for both gears in the train.",
null,
"In this illustration we clearly see that the Force vector, F, is at a 90º angle to the two Distance vectors, D1 and D2. Let’s see why this angular relationship between them is crucial to the development of torque calculations. First a review of the basic torque formula, presented in a previous blog, Torque = Distance × Force × sin(ϴ) By inserting D1, F, and ϴ = 90º into this formula we arrive at the torque calculation, T1 , for the driving gear in our gear train: T1 = D1 × F × sin(90º) From a previous blog in this series we know that sin(90º) = 1, so it becomes, T1 = D1 × F By inserting D2, F, and ϴ = 90º into the torque formula, we arrive at the torque calculation, T2 , for the driven gear: T2 = D2 × F × sin(90º) T2 = D2 × F × 1 T2 = D2 × F Next week we’ll combine these two equations relative to F, the common link between them, and obtain a single equation equating the torques and pitch circle radii of the driving and driven gears in the gear train. _______________________________________\n\n### The Mathematical Link Between Gears in a Gear Train\n\nWednesday, May 14th, 2014\n Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them.",
null,
"From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F1 to the driven gear Force vector F2, then use that linking equation to develop a separate torque formula for each of the gears in the train. We learned in the previous blog in this series that F1 and F2 travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value ϴ with respect to their Distance vectors D1 and D2. This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation: F = [F1 × sin(ϴ)] – [F2 × sin(ϴ)] where F is called a resultant Force vector, so named because it represents the force that results when the dead, or inert, weight that’s present in the resisting force F2 cancels out some of the positive force of F1. Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train. _______________________________________\n\n### Distance and Force Vectors of a Simple Gear Train\n\nMonday, May 5th, 2014\n Last time we examined how torque and force are created upon the driving gear within a simple gear train. Today we’ll see how they affect the driven gear.",
null,
"Looking at the gear train illustration above, we see that each gear has both distance and force vectors. We’ll call the driving gear Distance vector, D1, and the driven gear Distance vector, D2. Each of these Distance vectors extend from pivot points located at the centers of their respective gear shafts. From there they extend in opposite directions until they meet at the line of action, the imaginary line which represents the geometric path along which Force vectors F1 and F2 are aligned. As we learned last time, the Force vector, F1, results from the torque that’s created at the pivot point located at the center of the driving gear. This driving gear is mounted on a shaft that’s attached to an electric motor, the ultimate powering source behind the torque. F1 follows a path along the line of action until it meets with the driven gear teeth, where it then exerts its pushing force upon them. It’s met by Force vector F2, a resisting force, which extends along the same line of action, but in a direction opposite to that of F1. These two Force vectors butt heads, pushing back against one another. F2 is essentially a negative force manifested by the dead weight of the mechanical load of the machinery components resting upon the shaft of the driving gear. Its unmoving inertia resists being put into motion. In order for the gears in the gear train to turn, F1 must be greater than F2, in other words, it must be great enough to overcome the resistance presented by F2. With the two Force vectors pushing against each other along the line of action, the angle ϴ between vectors F2 and D2, is the same as the angle ϴ between F1 and D1. Next time we’ll use the angular relationship between these four vectors to develop torque calculations for both gears in the gear train. _______________________________________\n\n### Torque and Force\n\nTuesday, April 29th, 2014\n We’ve been discussing torque and how it enables more power to be available to applications such as loosening tight nuts with a wrench. Now we’ll see how those same principles apply to another application, a simple gear train. To review, the torque formula is, Torque = Distance × Force × sin(ϴ) where, Distance and Force are vector magnitudes and ϴ is the angle formed between them.",
null,
"Referring to the gear train illustration above, we see that Force and Distance vectors are present, just as they had been in our previous wrench/nut example. But instead of torque being created by way of force that’s applied to a wrench, things are reversed, and it’s the torque that creates the force. You see, in the wrench/nut example, the force applied to the wrench handle created torque on the nut. In our present gear train example, the torque applied to the motor shaft is created by an electric motor exerting pressure upon the motor shaft, which in turn exerts a force upon the driving gear teeth. The driving gear is also attached to this shaft, so torque causes the driving gear to rotate along with the motor. This rotation results in a force being exerted at the point where the teeth of the driving gear mesh with the teeth of the driven gear. In other words, in the wrench/nut example force created torque, while in the present example torque creates a force. The gear train has a pivot point, as there was in our wrench/nut example, but this time it’s located at the center of the motor shaft rather than at the center of a nut. The pivot point in both examples is where the action takes place. The motor’s shaft and driving gear rotate around it, just as the wrench jaws and handle rotated around the nut’s pivot point. In both examples, the Distance vectors extend out from the pivot points to meet up with the Force vector’s path. In the gear train example, this Force vector path is called a line of action, as introduced earlier in this blog series. This line of action passes through to the point where the driving and driven gear teeth mesh. The force acting upon that point causes the gears in the gear train to rotate, and as they turn mechanical energy is transferred from the motor to whatever machinery component is attached to the shaft of the driven gear. The powered component will then be able to perform useful work such as cutting lumber, mixing frosting for a cake, drilling holes in steel, or propelling vehicles. You will note that there is an angle ϴ which exists between the Distance and Force vectors. Since we have a pivot point, a Force vector, a Distance vector, and an angle ϴ, we are able to apply the torque formula to gear trains exactly as we did in our wrench/nut example. We can then use that formula to calculate how torque is transmitted between gears in the train. Next time we’ll examine the distance and force vectors in a simple gear train. _______________________________________\n\n### Gear Reduction Worked Backwards\n\nSunday, March 9th, 2014\n Last time we saw how a gear reduction does as its name implies, reduces the speed of the driven gear with respect to the driving gear within a gear train. Today we’ll see how to work the problem in reverse, so to speak, by determining how many teeth a driven gear must have within a given gear train to operate at a particular desired revolutions per minute (RPM). For our example we’ll use a gear train whose driving gear has 18 teeth. It’s mounted on an alternating current (AC) motor turning at 3600 (RPM). The equipment it’s attached to requires a speed of 1800 RPM to operate correctly. What number of teeth must the driven gear have in order to pull this off? If you’ve identified this to be a word problem, you’re correct.",
null,
"Let’s first review the gear ratio formulas introduced in my previous two articles: R = nDriving ÷ nDriven (1) R = NDriven ÷ NDriving (2) Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required: R = nDriving ÷ nDriven = 3600 RPM ÷ 1800 RPM = 2 This equation tells us that to reduce the speed of the 3600 RPM motor to the required 1800 RPM, we need a gear train with a gear ratio of 2:1. Stated another way, for every two revolutions of the driving gear, we must have one revolution of the driven gear. Now that we know the required gear ratio, R, we can use Formula (2) to determine how many teeth the driven gear must have to turn at the required 1800 RPM: R = 2 = NDriven ÷ NDriving 2 = NDriven ÷ 18 Teeth NDriven = 2 × 18 Teeth = 36 Teeth The driven gear requires 36 teeth to allow the gear train to operate equipment properly, that is to say, enable the gear train it’s attached to provide a speed reduction of 1800 RPM, down from the 3600 RPM that is being put out from the driving gear. But gear ratio isn’t just about changing speeds of the driven gear relative to the driving gear. Next time we’ll see how it works together with the concept of torque, thus enabling small motors to do big jobs. _______________________________________\n\n### Gear Reduction\n\nWednesday, March 5th, 2014\n Last time we learned there are two formulas used to calculate gear ratio, R. Today we’ll see how to use them to calculate a gear reduction between gears in a gear train, a strategy which enables us to reduce the speed of the driven gear in relation to the driving gear. If you’ll recall from last time, our formulas to determine gear ratio are: R = NDriven ÷ NDriving (1) R = nDriving ÷ nDriven (2) Now let’s apply them to this example gear train to see how a gear reduction works.",
null,
"Here we have a driven gear with 23 teeth, while the driving gear has 18. For our example the electric motor connected to the driving gear causes it to turn at a speed, nDriving, of 3600 revolutions per minute (RPM). Knowing these numerical values we are able to determine the driven gear speed, nDriven. First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other: R = NDriven ÷ NDriving R = 23 Teeth ÷ 18 Teeth R = 1.27 In gear design nomenclature, the gear train is said to have a 1.27 to 1 ratio, commonly denoted as 1.27:1. This means that for every tooth on the driving gear, there are 1.27 teeth on the driven gear. Interestingly, the R’s in both equations (1) and (2) are identical, and in our situation is equal to 1.27, although it is arrived at by different means. In Formula (1) R is derived from calculations involving the number of teeth present on each gear, while Formula (2)’s R is derived by knowing the rotational speeds of the gears. Since R is the common link between the two formulas, we can use this commonality to create a link between them and insert the R value determined in one formula into the other. Since we have already determined that the R value is 1.27 using Formula (1), we can replace the R in Formula (2) with this numerical value. As an equation this looks like: R = 1.27 = nDriving ÷ nDriven Now all we need is one more numerical value to solve Formula (2)’s equation. We know that the speed at which the driving gear is rotating, nDriving , is 3600 RPM. We use basic algebra to calculate the driven gear speed, nDriven : 1.27 = 3600 RPM ÷ nDriven nDriven = 3600 RPM ÷ 1.27 = 2834.65 RPM Based on our calculations, the driven gear is turning at a speed that is slower than the driving gear. To determine exactly how much slower we’ll calculate the difference between their speeds: nDriving – nDriven = 3600 RPM – 2834.65 RPM ≈ 765 RPM So in this gear reduction the driven gear turns approximately 765 RPM slower than the driving gear. Next time we’ll apply a gear reduction to a gear train and see how to arrive at a particular desired output speed. _______________________________________\n\n### Gear Ratio Formulas\n\nSunday, February 23rd, 2014\n Last time we introduced a way to convert individual gear speeds in relation to one another within a gear train by employing a conversion tool known as the gear ratio. Today we’ll introduce the gear ratio formulas, of which there are two types.",
null,
"The first formula for determining gear ratio is based on knowing the driving gear revolutions per minute (RPM), notated as nDriving, and the driven gear RPM, nDriven. Given that knowledge we can calculate the gear ratio, R, that exists between them by the formula: R = nDriving ÷ nDriven (1) The other way to determine gear ratio, R, is by knowing the number of teeth on both the driving gear, NDriving, and the driven gear, NDriven. That’s right, it all boils down to simply counting the number of teeth on each gear. In this instance the gear ratio is calculated by the following formula: R = NDriven ÷ NDriving (2) Equations (1) and (2) may look virtually identical, but they’re not. In mechanical engineering calculations, lower case n is typically used to denote the RPM of rotating objects such as shafts, wheels, pulleys, and gears. Upper case N is typically used to denote the number of teeth on a gear. Next time we’ll see how to manipulate these two equations so as to arrive at a particular gear ratio. _______________________________________\n\n### Overcoming Inertia\n\nMonday, February 3rd, 2014\n Inertia. It’s the force that keeps us in bed after the alarm has rung. It seems to have a life of its own, and today we’ll see how it comes into play in keeping other stationary objects at rest. Last time we identified a specific point of contact between spur gear teeth in a gear train and introduced the opposing forces, F1 and F 2, generated there. Today we’ll see what these forces represent, identifying one of them as inertia. So where do these forces come from? They’re forces generated by different means that converge at the same point of contact, the point at which gear teeth mesh. They follow a very specific geometric path to meet there, an imaginary straight line referred to as the line of action. F1 is always generated by a source of mechanical energy. In our locomotive example introduced earlier in this blog series that source is an electric traction motor, upon which a driving gear is mounted. When the motor is energized, a driving force F1 is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear. Force F2 is not as straightforward to understand, because it’s not generated by a motor. Instead, it’s the resisting force that the weight of a stationary object poses against its being moved from an at-rest position, known as inertia. The heavier the object, the more inertia it presents with. Trains, of course, are extremely heavy, and to get them to move a great deal of inertia must be overcome. Inertia is also a factor in attempting to stop objects already in motion. To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle. At their point of contact, the driving force of the motor, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2. If F1 is less than or equal to F2, then the train won’t leave the station. Next week we’ll animate our static image and watch the interplay between gear teeth, taking note of the line of action during their movement. _______________________________________"
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http://zuowen4.info/node/6313 | [
"## 美妙的夏日——陈佳雯\n\n•\n•\n• měi\n• miào\n• de\n• xià\n•\n•\n• 美妙的夏日\n•\n•\n• chén\n• jiā\n• wén\n•\n•\n• 陈佳雯\n•\n•\n• yǒu\n• liè\n• yán\n• yán\n• de\n• xīng\n•\n• 有一个烈日炎炎的星期日，我和爸爸\n• zhèng\n• zài\n• biān\n• diào\n•\n•\n•\n• 正在湖边钓鱼。\n•\n•\n• rán\n•\n• gǎn\n• dào\n• kǒu\n• gàn\n• shé\n• zào\n•\n• shì\n•\n• biàn\n• 突然，我感到口干舌燥，于是，我便\n• xiàng\n• yào\n• qián\n•\n• shāng\n• diàn\n• mǎi\n• yǐn\n• liào\n•\n•\n•\n• 向爸爸要钱，去商店买饮料。\n•\n•\n• zài\n• shàng\n•\n• kàn\n• jiàn\n• yǒu\n• wèi\n• lǎo\n• nǎi\n• nǎi\n•\n• zài\n• tài\n• 在路上，我看见有一位老奶奶，在太\n• yáng\n• xià\n•\n• yào\n• kǒu\n• fàn\n• chī\n•\n• jiàn\n•\n• yòu\n• yòu\n• è\n•\n• 阳下，要一口饭吃。可见，她又渴又饿，\n• shì\n•\n• gǎn\n• jǐn\n• pǎo\n• dào\n• shāng\n• diàn\n•\n• dōu\n• de\n• qián\n• zuì\n• duō\n• 于是，我赶紧跑到商店，我兜里的钱最多\n• zhī\n• gòu\n• mǎi\n• liǎng\n• píng\n• chún\n• jìng\n• shuǐ\n•\n• suǒ\n•\n• mǎi\n• le\n• liǎng\n• píng\n• shuǐ\n• 只够买两瓶纯净水，所以，我买了两瓶水\n•\n•\n•\n• 。\n•\n•\n• fèn\n• bié\n• jiāng\n• píng\n• kǒu\n• nǐng\n• sōng\n•\n• píng\n• chún\n• jìng\n• shuǐ\n• gěi\n• 分别将瓶口拧松，把一瓶纯净水递给\n• lǎo\n• nǎi\n• nǎi\n•\n• lǎo\n• nǎi\n• nǎi\n• yòng\n• ruò\n• de\n• shēng\n• yīn\n• duì\n• shuō\n• 老奶奶，老奶奶用她那虚弱的声音对我说\n• le\n• shēng\n•\n•\n• xiè\n• xiè\n•\n• xiǎo\n• péng\n• yǒu\n•\n•\n• yòng\n• wēn\n• róu\n• 了一声：“谢谢你，小朋友。”我用温柔\n• de\n• shēng\n• yīn\n• duì\n• lǎo\n• nǎi\n• nǎi\n• shuō\n•\n•\n• yòng\n• xiè\n•\n•\n•\n•\n• 的声音对老奶奶说：“不用谢！”\n•\n•\n• shì\n•\n• zhuǎn\n• shēn\n•\n• huí\n• dào\n• le\n• biān\n•\n• ò\n•\n• 于是，我转身，回到了湖边，哦！爸\n• jīng\n• diào\n• le\n• hěn\n• duō\n•\n• wèn\n• wéi\n• shí\n• me\n• mǎi\n• 爸已经钓了很多鱼，爸爸问我为什么不买\n• píng\n• yǐn\n• liào\n•\n• shì\n• qíng\n• de\n• jīng\n• guò\n• shí\n• gào\n• 一瓶饮料，我把事情的经过一五一十地告\n• le\n•\n• kuā\n• zhǎng\n• le\n•\n•\n•\n• 诉了爸爸，爸爸夸我长大了。\n•\n•\n• duō\n• me\n• měi\n• miào\n• de\n• xià\n•\n•\n•\n•\n• 多么美妙的夏日啊！\n•\n•\n• zhǐ\n• dǎo\n• jiāo\n• shī\n•\n• cuī\n• méi\n•\n•\n• 指导教师 崔梅茹\n•\n•\n•\n•\n•\n•\n\n无注音版：\n\n美妙的夏日\n\n陈佳雯\n\n有一个烈日炎炎的星期日，我和爸爸正在湖边钓鱼。\n\n突然，我感到口干舌燥，于是，我便向爸爸要钱，去商店买饮料。\n\n在路上，我看见有一位老奶奶，在太阳下，要一口饭吃。可见，她又渴又饿，于是，我赶紧跑到商店，我兜里的钱最多只够买两瓶纯净水，所以，我买了两瓶水。\n\n分别将瓶口拧松，把一瓶纯净水递给老奶奶，老奶奶用她那虚弱的声音对我说了一声：“谢谢你，小朋友。”我用温柔的声音对老奶奶说：“不用谢！”\n\n于是，我转身，回到了湖边，哦！爸爸已经钓了很多鱼，爸爸问我为什么不买一瓶饮料，我把事情的经过一五一十地告诉了爸爸，爸爸夸我长大了。\n\n多么美妙的夏日啊！\n\n指导教师 崔梅茹\n\n### 夏日的雷阵雨\n\n四年级作文226字\n作者：吴加勉\n•\n•\n• ruì\n• ān\n• ān\n• yáng\n• shí\n• yàn\n• xiǎo\n• xué\n•\n• bān\n• 瑞安安阳实验小学四6\n•\n•\n•\n•\n• yán\n• de\n• xià\n•\n• tiān\n• kōng\n• zhōng\n• rán\n• mǎn\n• 一个炎热的下午，天空中忽然布满乌\n• 阅读全文\n\n### 夏日观荷\n\n四年级作文466字\n作者：杜建美\n•\n•\n• shǔ\n• jiǎ\n• de\n• tiān\n•\n• men\n• lái\n• dào\n• běi\n• shān\n• de\n• ?g\n• chí\n• 暑假的一天，我们来到北山的荷花池\n• guān\n•\n• tiān\n• shuǎng\n• le\n•\n• de\n• xīn\n• qíng\n• suí\n• zhe\n• tiān\n• 观荷。天气爽极了，我的心情也随着天气\n• de\n• biàn\n• huà\n• kāi\n• xīn\n• hěn\n•\n•\n•\n•\n• 的变化开心得很。\n• 阅读全文\n\n### 一场美妙的淋\n\n四年级作文512字\n作者：付雅文\n•\n•\n• chǎng\n• měi\n• miào\n• de\n• lín\n•\n•\n• 一场美妙的淋浴\n•\n•\n• hóng\n• yàn\n• wài\n• xué\n• xiào\n• nián\n•\n• wén\n• 鸿雁外语学校四年级 付雅文\n•\n•\n• yǒu\n• rén\n• huān\n• chūn\n• tiān\n• de\n• wēi\n• fēng\n•\n• yǒu\n• rén\n• huān\n• 有人喜欢春天的徐徐微风，有人喜欢\n• 阅读全文\n\n### 美妙的未来世\n\n四年级作文611字\n作者：李玉洁\n•\n•\n• yǒu\n• tiān\n•\n• shuì\n• zhe\n• hòu\n•\n• rán\n• kàn\n• jiàn\n• le\n• shí\n• 有一天，我睡着以后，忽然看见了时\n• guāng\n• lǎo\n• rén\n•\n• shí\n• guāng\n• lǎo\n• rén\n• duì\n• shuō\n•\n•\n• shì\n• hěn\n• xiǎng\n• 光老人，时光老人对我说：“你不是很想\n• kàn\n• kàn\n• wèi\n• lái\n• de\n• shì\n• jiè\n• ma\n•\n• dài\n• ba\n•\n•\n• 看看未来的世界吗？我带你一起去吧。”\n• 阅读全文\n\n### 美丽的夏日—\n\n四年级作文：美丽的夏日——刘汝琪\n作文字数：415\n作者：刘汝琪\n•\n•\n• měi\n• de\n• xià\n•\n•\n• 美丽的夏日\n•\n•\n• liú\n•\n•\n• 刘汝琪\n•\n•\n• chūn\n•\n• wàn\n•\n• xià\n•\n• wàn\n• qiān\n• hóng\n•\n• qiū\n•\n• 春，万物复苏；夏，万紫千红；秋，\n• 阅读全文\n\n### 忙碌的夏日—\n\n四年级作文：忙碌的夏日——盖瑞雪\n作文字数：342\n作者：盖瑞雪\n•\n•\n• máng\n• de\n• xià\n•\n•\n• 忙碌的夏日\n•\n•\n• gài\n• ruì\n• xuě\n•\n•\n• 盖瑞雪\n•\n•\n• xià\n• niáng\n• mài\n• zhe\n• qīng\n• qīng\n• jiǎo\n• lái\n• dào\n• le\n• rén\n• jiān\n•\n• 夏姑娘迈着轻轻地脚步来到了人间，\n• 阅读全文\n\n### 这个夏日，没\n\n四年级作文：这个夏日，没有炎热——刘晨\n作文字数：332\n作者：刘晨\n•\n•\n• zhè\n• xià\n•\n• méi\n• yǒu\n• yán\n•\n•\n• 这个夏日，没有炎热\n•\n•\n• liú\n• chén\n•\n•\n• 刘晨\n•\n•\n• xià\n•\n• shì\n• liè\n• yán\n• yán\n• de\n•\n• shì\n• ràng\n• rén\n• hàn\n• liú\n• jiā\n• 夏日，是烈日炎炎的，是让人汗流浃\n• 阅读全文\n\n### 美丽的夏日—\n\n四年级作文：美丽的夏日——赵世超\n作文字数：337\n作者：赵世超\n•\n•\n• měi\n• de\n• xià\n•\n•\n• 美丽的夏日\n•\n•\n• zhào\n• shì\n• chāo\n•\n•\n• 赵世超\n•\n•\n• xià\n•\n• měi\n• de\n• jiē\n•\n• ràng\n• rén\n• xiá\n• xiǎng\n• 夏，一个美丽的季节，一个让人遐想\n• 阅读全文\n\n### 美妙的夏日—\n\n四年级作文：美妙的夏日——陈佳雯\n作文字数：335\n作者：陈佳雯\n•\n•\n• měi\n• miào\n• de\n• xià\n•\n•\n• 美妙的夏日\n•\n•\n• chén\n• jiā\n• wén\n•\n•\n• 陈佳雯\n•\n•\n• yǒu\n• liè\n• yán\n• yán\n• de\n• xīng\n•\n• 有一个烈日炎炎的星期日，我和爸爸\n• 阅读全文\n\n### 美丽的夏日\n\n四年级作文：美丽的夏日\n作文字数：248\n作者：薄淑婷\n•\n•\n• měi\n• de\n• xià\n•\n•\n• 美丽的夏日\n•\n•\n• báo\n• shū\n• tíng\n•\n•\n• 薄淑婷\n•\n•\n• nián\n• zhōng\n•\n• huān\n• chūn\n• tiān\n•\n• dàn\n• gèng\n• ài\n• 一年四季中，我喜欢春天，但我更爱\n• 阅读全文\n\n### 夏日里的一件\n\n四年级作文：夏日里的一件事——董静文\n作文字数：384\n作者：董静文\n•\n•\n• xià\n• de\n• jiàn\n• shì\n•\n•\n• 夏日里的一件事\n•\n•\n• dǒng\n• jìng\n• wén\n•\n•\n• 董静文\n•\n•\n• yǒu\n• jiàn\n• shì\n•\n• shēng\n• zài\n• xià\n•\n• ràng\n• xià\n• tiān\n• biàn\n• 有一件事，发生在夏日，让夏天变得\n• 阅读全文\n\n### 美妙的夏日—\n\n四年级作文：美妙的夏日——张延宇\n作文字数：358\n作者：张延宇\n•\n•\n• měi\n• miào\n• de\n• xià\n•\n•\n• 美妙的夏日\n•\n•\n• zhāng\n• yán\n•\n•\n• 张延宇\n•\n•\n•\n• hǎo\n• fán\n•\n•\n•\n• bèi\n• shuō\n• guò\n• dùn\n• hòu\n• de\n• “好烦啊！”被妈妈说过一顿后的我\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——林晨辉\n作文字数：335\n作者：林晨辉\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• lín\n• chén\n• huī\n•\n•\n• 林晨辉\n•\n•\n• xià\n• suī\n• shuō\n• liè\n• yán\n• yán\n•\n• dàn\n• shì\n• dào\n•\n• lǎo\n• tiān\n• 夏季虽说烈日炎炎，但是一到“老天\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——蔡超凡\n作文字数：381\n作者：蔡超凡\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• cài\n• chāo\n• fán\n•\n•\n• 蔡超凡\n•\n•\n• wēn\n• nuǎn\n• de\n• chūn\n• tiān\n• guò\n• le\n•\n• yán\n• de\n• xià\n• tiān\n• dào\n• lái\n• 温暖的春天过去了，炎热的夏天到来\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——杨一帆\n作文字数：374\n作者：杨一帆\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• yáng\n• fān\n•\n•\n• 杨一帆\n•\n•\n•\n• chí\n• shàng\n• tái\n• sān\n• diǎn\n•\n• huáng\n• liǎng\n• shēng\n• “池上碧苔三四点，叶底黄鹂一两声\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——于江浩\n作文字数：690\n作者：于江浩\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• jīn\n• xiàn\n• shí\n• yàn\n• xué\n• xiào\n•\n• nián\n• bān\n•\n• 利津县第一实验学校 四年级四班\n• jiāng\n• hào\n•\n•\n• 于江浩\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——扈锦程\n作文字数：367\n作者：扈锦程\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• jīn\n• xiàn\n• èr\n• shí\n• yàn\n• xué\n• xiào\n•\n• nián\n• sān\n• bān\n•\n• 利津县第二实验学校 四年级三班\n• jǐn\n• chéng\n•\n•\n• 扈锦程\n• 阅读全文\n\n### 夏日的美好—\n\n四年级作文：夏日的美好——李天佑\n作文字数：275\n作者：李天佑\n•\n•\n• xià\n• de\n• měi\n• hǎo\n•\n•\n• 夏日的美好\n•\n•\n• jīn\n• xiàn\n• shí\n• yàn\n• xiào\n•\n• nián\n• èr\n• bān\n•\n• tiān\n• 利津县实验一校 四年级二班 李天\n• yòu\n•\n•\n• 佑\n• 阅读全文"
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