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RTK GPS GNSS Waypoint Navigation & 3D-SLAM \| MYBOTSHOP.DE One of the most important skills for autonomous use-cases in outdoor robots is Global Positioning System (GPS) Waypoint Navigation. GPS Waypoint Navigation refers to the ability to give a robot a collection of GPS waypoints (i.e., a set of latitude/longitude coordinates) and have the robot traverse independently from its present location to each of the defined waypoints. Not colliding with obstacles along the way is also desirable. Important Aspects of GPS Navigation Localization and navigation are the two most important aspects of GPS waypoint navigation. Localization is the process of estimating a robot's position using the robot's sensors (such as the UGV wheel encoders, IMU data, and GPS readings). The navigation system is in charge of sending wheel velocity commands to the UGV so that it may move to the target location in this case being a GPS coordinate. MYBOTSHOP GPS Navigation Package The MYBOTSHOP GPS package utilizes some of a collection of ROS packages as well as several python libraries to enable any robot to move from one target location to another. Although a few packages are available open-source for the GPS navigation, none of them work out of the box and requires extensive tuning of parameters. The tuning of parameters is not limited to the GPS package but also involves tuning the navigation package (i.e. the movebase/movebase flex) package. Needless to say, one can build their own navigation package and bypass the tuning, however, they may lose the intricate perks of the avoidance collision and pre-configured path planners. Today we will discuss some of the intricacies of the GPS navigation package in light of the MYBOTSHOP GPS package. Starting off, we will enumerate the required and optional hardware as well as the software. Next, we will discuss some of the required configurations and finally, we will illustrate working of the GPS navigation package. Un-manned ground vehicle (UGV) \| Un-manned aerial vehicle (UAV) Global positiong system (GPS) device Recommended GPS: EMLID REACH RS2 Recommended LiDAR: OUSTER Ubuntu 18.04 (Tested with this version) ROS-melodic (Tested with this version) MYBOTSHOP GPS package MYBOTSHOP GPS Navigation Package Demo In this demo, we initially record the GPS coordinates of using a logitech controller for a very short distance. After collecting the GPS coordinates, we run the autonomous GPS navigation package. Recording the GPS Coordinates GIF. We use the Logitech controller joystick to move the Jackal and store the GPS coordinates of four different locations represented by the pole-flag icon. Execution Autonomous GPS Navigation GIF. Once the GPS coordinates have been collected, we run the GPS navigation package. The xy goal tolerance for the give run is 0.25m" MYBOTSHOP GPS Configuration The first and foremost important step before starting the configuration is ensuring that your robot control drivers are set properly and can be operated without any issue. Once IMU calibration is especially important for out door navigation as it cannot locally orient itself using typical location markers which are available for indoor navigation. The GPS coordinates only provide the latitude and longtitude coordinates for the position of the robot. The orientation is calculated using the IMU and works via magnetic calibration. It is similar to how we draw an 8 in the air with our mobilephones for calibrating the compass. It is recommended to calibrate the robot's IMU each time the robot is power-cycled to ensure it moves in the correct direction. The GPS hardware significantly matters when performing the point to point navigation. Utilizing a normal GPS has very low accuracy and causes jumps in the position of the robot which messes with its internal localization parameters. To counter this, a percision GPS like the Recommended GPS: EMLID REACH RS2 may be used, however, it still requires some configuration to enable it to provides its iconic 4 mm accuracy. To enable the EMLID REACH RS2 percision, select the EMLID REACH RS2 in the GPS & 3D Slam Documentation. Note, that if you do not enable the percision mode, the EMLID acts like a normal GPS with large inaccuracies. Also, while the robot is moving the GPS provides accuracies in cm instead of mm as it requires some time to triangulate its exact positon. One thing to keep in mind is that currently, with the loss of GPS, the robot follows its normal recovery behavior if enabled which is to keep rotating about its current position. The localization package is very important and requires custom parameters to run with the GPS. These parameters are pre-tuned with the MYBOTSHOP GPS package. If you wish to tune the parameters for your robot yourself, you may refer to the detailed documentation of the robot_localization package. Movebase Package In move-base it is critical to configure the size of your robot, the frame-id, the costmaps, and the planners. Details on these parameters are available in the GPS & 3D Slam Documentation. At times jerky motion may be experienced by the robot for which you may create a custom package which takes the command velocites and smoothes them or utilize the open-source package of yocs. MYBOTSHOP GPS SLAM Documentation A comprehensive overview is available for the MBOTSHOP GPS \|package in the GPS & 3D Slam Documentation. Please email at support@mybotshop.de in case you find any incorrect statements, facts, typos, or if you think some additional information should be added.
Electromagnetic Waves Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Physics - Electromagnetic Waves The X-ray beam can be deflected by: Subtopic: Electromagnetic Spectrum \| In an electromagnetic wave in free space the root mean square value of the electric field is {E}_{rms}=6 V/m. The peak value of the magnetic field is: ×{10}^{-8}T 2.83×{10}^{-8}T 0.70×{10}^{-8}T 4.23×{10}^{-8} T Subtopic: Properties of EM Waves \| Out of the following options which one can be used to produce a propagating electromagnetic wave? 1. A stationary charge 2. A chargeless particle 3. An accelerating charge 4. A charge moving at constant velocity Subtopic: Generation of EM Waves \| The energy of the EM wave is of the order of 15 KeV. To which part of the spectrum does it belong? \gamma The electric field associated with an electromagnetic wave in vacuum is given by E= 40cos \left(kz-6×{10}^{8}t\right), where E, z, and t are in volt/m, meter, and second respectively. The value of wave vector k is: 2 {m}^{-1} 0.5 {m}^{-1} 6 {m}^{-1} {m}^{-1} The ratio of the amplitude of the magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to (1) the speed of light in vacuum (2) reciprocal of speed of light in vacuum (3) the ratio of magnetic permeability to the electric susceptibility of vacuum \left[\mathrm{E}={\mathrm{E}}_{0}\stackrel{^}{\mathrm{k}}, \mathrm{B}=\stackrel{˙}{{\mathrm{B}}_{0}\stackrel{^}{\mathrm{I}}}\right] \left[\mathrm{E}={\mathrm{E}}_{0}\stackrel{^}{\mathrm{j}}, \mathrm{B}={\mathrm{B}}_{0}\stackrel{^}{\mathrm{j}}\right] \left[\mathrm{E}={\mathrm{E}}_{0}\stackrel{^}{\mathrm{j}}, \mathrm{B}={\mathrm{B}}_{0}\stackrel{^}{\mathrm{k}}\right] \left[\mathrm{E}={\mathrm{E}}_{0}\stackrel{^}{\mathrm{i}}, \mathrm{B}={\mathrm{B}}_{\mathrm{o}}\stackrel{^}{\mathrm{j}}\right] The decreasing order of the wavelength of infrared, microwave, ultraviolet and gamma rays is (1) gamma rays, ultraviolet, infrared, microwaves (2) microwaves, gamma rays, infrared, ultraviolet (3) infrared, microwave, ultraviolet, gamma rays (4) microwave, infrared, ultraviolet, gamma rays Which of the following statements is false for the properties of electromagnetic waves? 1. Both electric and magnetic field vectors attain the maxima and minima at the same place and the same time 2. The energy in an electromagnetic wave is divided equally between electric and magnetic vectors 3. Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of the wave 4. These waves do not require any material medium for propagation The electric field of an electromagnetic wave in free space is given by \stackrel{\to }{\mathbf{E}}=10 \mathrm{cos}\left({10}^{7}t+kx\right)\stackrel{^}{\mathbf{j}} \mathrm{V}/\mathrm{m}, where t and x are in seconds and meters respectively. It can be inferred that (1) The wavelength \mathrm{\lambda } is 188.4 m. (2) The wave number k is 0.33 rad/m. (3) The wave amplitude is 10 V/m. (4) The wave is propagating along +x direction Which one of the following pairs of statements is correct?
Constant False Alarm Rate (CFAR) Detection - MATLAB & Simulink - MathWorks France Cell Averaging CFAR Detection CFAR Detection Using Automatic Threshold Factor CFAR Detection Using Custom Threshold Factor CFAR Detection Threshold Comparison Between CFAR and Classical Neyman-Pearson Detector CFAR Detection for Range-Doppler Images This example introduces constant false alarm rate (CFAR) detection and shows how to use CFARDetector and CFARDetector2D in the Phased Array System Toolbox™ to perform cell averaging CFAR detection. One important task a radar system performs is target detection. The detection itself is fairly straightforward. It compares the signal to a threshold. Therefore, the real work on detection is coming up with an appropriate threshold. In general, the threshold is a function of both the probability of detection and the probability of false alarm. In many phased array systems, because of the cost associated with a false detection, it is desirable to have a detection threshold that not only maximizes the probability of detection but also keeps the probability of false alarm below a preset level. There is extensive literature on how to determine the detection threshold. Readers might be interested in the Signal Detection in White Gaussian Noise and Signal Detection Using Multiple Samples examples for some well known results. However, all these classical results are based on theoretical probabilities and are limited to white Gaussian noise with known variance (power). In real applications, the noise is often colored and its power is unknown. CFAR technology addresses these issues. In CFAR, when the detection is needed for a given cell, often termed as the cell under test (CUT), the noise power is estimated from neighboring cells. Then the detection threshold, T T=\alpha {P}_{n} {P}_{n} is the noise power estimate and \alpha is a scaling factor called the threshold factor. From the equation, it is clear that the threshold adapts to the data. It can be shown that with the appropriate threshold factor, \alpha , the resulting probability of false alarm can be kept at a constant, hence the name CFAR. The cell averaging CFAR detector is probably the most widely used CFAR detector. It is also used as a baseline comparison for other CFAR techniques. In a cell averaging CFAR detector, noise samples are extracted from both leading and lagging cells (called training cells) around the CUT. The noise estimate can be computed as [1] {P}_{n}=\frac{1}{N}\sum _{m=1}^{N}{x}_{m} N is the number of training cells and {x}_{m} is the sample in each training cell. If {x}_{m} happens to be the output of a square law detector, then {P}_{n} represents the estimated noise power. In general, the number of leading and lagging training cells are the same. Guard cells are placed adjacent to the CUT, both leading and lagging it. The purpose of these guard cells is to avoid signal components from leaking into the training cell, which could adversely affect the noise estimate. The following figure shows the relation among these cells for the 1-D case. With the above cell averaging CFAR detector, assuming the data passed into the detector is from a single pulse, i.e., no pulse integration involved, the threshold factor can be written as [1] \alpha =N\left({P}_{fa}^{-1/N}-1\right) {P}_{fa} is the desired false alarm rate. In the rest of this example, we show how to use Phased Array System Toolbox to perform a cell averaging CFAR detection. For simplicity and without losing any generality, we still assume that the noise is white Gaussian. This enables the comparison between the CFAR and classical detection theory. We can instantiate a CFAR detector using the following command: cfar = phased.CFARDetector('NumTrainingCells',20,'NumGuardCells',2); In this detector we use 20 training cells and 2 guard cells in total. This means that there are 10 training cells and 1 guard cell on each side of the CUT. As mentioned above, if we assume that the signal is from a square law detector with no pulse integration, the threshold can be calculated based on the number of training cells and the desired probability of false alarm. Assuming the desired false alarm rate is 0.001, we can configure the CFAR detector as follows so that this calculation can be carried out. exp_pfa = 1e-3; cfar.ThresholdFactor = 'Auto'; cfar.ProbabilityFalseAlarm = exp_pfa; The configured CFAR detector is shown below. cfar = phased.CFARDetector with properties: Method: 'CA' NumTrainingCells: 20 ThresholdFactor: 'Auto' ProbabilityFalseAlarm: 1.0000e-03 OutputFormat: 'CUT result' ThresholdOutputPort: false NoisePowerOutputPort: false We now simulate the input data. Since the focus is to show that the CFAR detector can keep the false alarm rate under a certain value, we just simulate the noise samples in those cells. Here are the settings: The data sequence is 23 samples long, and the CUT is cell 12. This leaves 10 training cells and 1 guard cell on each side of the CUT. The false alarm rate is calculated using 100 thousand Monte Carlo trials. rs = RandStream('mt19937ar','Seed',2010); npower = db2pow(-10); % Assume 10dB SNR ratio Ntrials = 1e5; CUTIdx = 12; % Noise samples after a square law detector rsamp = randn(rs,Ncells,Ntrials)+1irandn(rs,Ncells,Ntrials); x = abs(sqrt(npower/2)rsamp).^2; To perform the detection, pass the data through the detector. In this example, there is only one CUT, so the output is a logical vector containing the detection result for all the trials. If the result is true, it means that a target is present in the corresponding trial. In our example, all detections are false alarms because we are only passing in noise. The resulting false alarm rate can then be calculated based on the number of false alarms and the number of trials. x_detected = cfar(x,CUTIdx); act_pfa = sum(x_detected)/Ntrials act_pfa = 9.4000e-04 The result shows that the resulting probability of false alarm is below 0.001, just as we specified. As explained in the earlier part of this example, there are only a few cases in which the CFAR detector can automatically compute the appropriate threshold factor. For example, using the previous scenario, if we employ a 10-pulses noncoherent integration before the data goes into the detector, the automatic threshold can no longer provide the desired false alarm rate. xn = xn + abs(sqrt(npower/2)rsamp).^2; % noncoherent integration x_detected = cfar(xn,CUTIdx); act_pfa = 0 One may be puzzled why we think a resulting false alarm rate of 0 is worse than a false alarm rate of 0.001. After all, isn't a false alarm rate of 0 a great thing? The answer to this question lies in the fact that when the probability of false alarm is decreased, so is the probability of detection. In this case, because the true false alarm rate is far below the allowed value, the detection threshold is set too high. The same probability of detection can be achieved with our desired probability of false alarm at lower cost; for example, with lower transmitter power. In most cases, the threshold factor needs to be estimated based on the specific environment and system configuration. We can configure the CFAR detector to use a custom threshold factor, as shown below. release(cfar); cfar.ThresholdFactor = 'Custom'; Continuing with the pulse integration example and using empirical data, we found that we can use a custom threshold factor of 2.35 to achieve the desired false alarm rate. Using this threshold, we see that the resulting false alarm rate matches the expected value. cfar.CustomThresholdFactor = 2.35; A CFAR detection occurs when the input signal level in a cell exceeds the threshold level. The threshold level for each cell depends on the threshold factor and the noise power in that derived from training cells. To maintain a constant false alarm rate, the detection threshold will increase or decrease in proportion to the noise power in the training cells. Configure the CFAR detector to output the threshold used for each detection using the ThresholdOutputPort property. Use an automatic threshold factor and 200 training cells. cfar.ThresholdOutputPort = true; cfar.NumTrainingCells = 200; Next, create a square-law input signal with increasing noise power. Npoints = 1e4; rsamp = randn(rs,Npoints,1)+1irandn(rs,Npoints,1); ramp = linspace(1,10,Npoints)'; xRamp = abs(sqrt(npowerramp./2).rsamp).^2; Compute detections and thresholds for all cells in the signal. [x_detected,th] = cfar(xRamp,1:length(xRamp)); Next, compare the CFAR threshold to the input signal. plot(1:length(xRamp),xRamp,1:length(xRamp),th,... find(x_detected),xRamp(x_detected),'o') legend('Signal','Threshold','Detections','Location','Northwest') ylabel('Level') Here, the threshold increases with the noise power of the signal to maintain the constant false alarm rate. Detections occur where the signal level exceeds the threshold. In this section, we compare the performance of a CFAR detector with the classical detection theory using the Neyman-Pearson principle. Returning to the first example and assuming the true noise power is known, the theoretical threshold can be calculated as T_ideal = npowerdb2pow(npwgnthresh(exp_pfa)); The false alarm rate of this classical Neyman-Pearson detector can be calculated using this theoretical threshold. act_Pfa_np = sum(x(CUTIdx,:)>T_ideal)/Ntrials act_Pfa_np = 9.5000e-04 Because we know the noise power, classical detection theory also produces the desired false alarm rate. The false alarm rate achieved by the CFAR detector is similar. cfar.ThresholdOutputPort = false; cfar.NumTrainingCells = 20; Next, assume that both detectors are deployed to the field and that the noise power is 1 dB more than expected. In this case, if we use the theoretical threshold, the resulting probability of false alarm is four times more than what we desire. npower = db2pow(-9); % Assume 9dB SNR ratio act_Pfa_np = 0.0041 On the contrary, the CFAR detector's performance is not affected. act_pfa = 0.0011 Hence, the CFAR detector is robust to noise power uncertainty and better suited to field applications. Finally, use a CFAR detection in the presence of colored noise. We first apply the classical detection threshold to the data. npower = db2pow(-10); fcoeff = maxflat(10,'sym',0.2); x = abs(sqrt(npower/2)filter(fcoeff,1,rsamp)).^2; % colored noise act_Pfa_np = 0 Note that the resulting false alarm rate cannot meet the requirement. However, using the CFAR detector with a custom threshold factor, we can obtain the desired false alarm rate. cfar.CustomThresholdFactor = 12.85; In the previous sections, the noise estimate was computed from training cells leading and lagging the CUT in a single dimension. We can also perform CFAR detection on images. Cells correspond to pixels in the images, and guard cells and training cells are placed in bands around the CUT. The detection threshold is computed from cells in the rectangular training band around the CUT. In the figure above, the guard band size is [2 2] and the training band size is [4 3]. The size indices refer to the number of cells on each side of the CUT in the row and columns dimensions, respectively. The guard band size can also be defined as 2, since the size is the same along row and column dimensions. Next, create a two-dimensional CFAR detector. Use a probability of false alarm of 1e-5 and specify a guard band size of 5 cells and a training band size of 10 cells. cfar2D = phased.CFARDetector2D('GuardBandSize',5,'TrainingBandSize',10,... 'ProbabilityFalseAlarm',1e-5); Next, load and plot a range-doppler image. The image includes returns from two stationary targets and one target moving away from the radar. [resp,rngGrid,dopGrid] = helperRangeDoppler; Use CFAR to search the range-Doppler space for objects, and plot a map of the detections. Search from -10 to 10 kHz and from 1000 to 4000 m. First, define the cells under test for this region. [~,rangeIndx] = min(abs(rngGrid-[1000 4000])); [~,dopplerIndx] = min(abs(dopGrid-[-1e4 1e4])); [columnInds,rowInds] = meshgrid(dopplerIndx(1):dopplerIndx(2),... rangeIndx(1):rangeIndx(2)); CUTIdx = [rowInds(:) columnInds(:)]'; Compute a detection result for each cell under test. Each pixel in the search region is a cell in this example. Plot a map of the detection results for the range-Doppler image. detections = cfar2D(resp,CUTIdx); helperDetectionsMap(resp,rngGrid,dopGrid,rangeIndx,dopplerIndx,detections) The three objects are detected. A data cube of range-Doppler images over time can likewise be provided as the input signal to cfar2D, and detections will be calculated in a single step. In this example, we presented the basic concepts behind CFAR detectors. In particular, we explored how to use the Phased Array System Toolbox to perform cell averaging CFAR detection on signals and range-Doppler images. The comparison between the performance offered by a cell averaging CFAR detector and a detector equipped with the theoretically calculated threshold shows clearly that the CFAR detector is more suitable for real field applications. [1] Mark Richards, Fundamentals of Radar Signal Processing, McGraw Hill, 2005
(Redirected from Lindenmayer systems) Find sources: "L-system" – news · newspapers · books · scholar · JSTOR (April 2013) (Learn how and when to remove this template message) This article reads like a textbook. Please improve this article to make it neutral in tone and meet Wikipedia's quality standards. (August 2020) 3.2 Example 2: Fractal (binary) tree 3.3 Example 3: Cantor set 4.4 Bi-directional grammars As a biologist, Lindenmayer worked with yeast and filamentous fungi and studied the growth patterns of various types of bacteria, such as the cyanobacteria Anabaena catenula. Originally, the L-systems were devised to provide a formal description of the development of such simple multicellular organisms, and to illustrate the neighbourhood relationships between plant cells. Later on, this system was extended to describe higher plants and complex branching structures. L-system structure[edit] An L-system is context-free if each production rule refers only to an individual symbol and not to its neighbours. Context-free L-systems are thus specified by a context-free grammar. If a rule depends not only on a single symbol but also on its neighbours, it is termed a context-sensitive L-system. Examples of L-systems[edit] Example 1: Algae[edit] Example 1: Algae, explained[edit] n=0: A start (axiom/initiator) n=1: A B the initial single A spawned into AB by rule (A → AB), rule (B → A) couldn't be applied n=2: A B A former string AB with all rules applied, A spawned into AB again, former B turned into A / \| \| \| \ n=3: A B A A B note all A's producing a copy of themselves in the first place, then a B, which turns ... / \| \| \| \ \| \ \ n=4: A B A A B A B A ... into an A one generation later, starting to spawn/repeat/recurse then The result is the sequence of Fibonacci words. If we count the length of each string, we obtain the famous Fibonacci sequence of numbers (skipping the first 1, due to our choice of axiom): If we would like to not skip the first 1, we can use axiom B. That would place B node before the topmost node (A) of the graph above. For each string, if we count the k-th position from the left end of the string, the value is determined by whether a multiple of the golden ratio falls within the interval {\displaystyle (k-1,k)} . The ratio of A to B likewise converges to the golden mean. This sequence is a locally catenative sequence because {\displaystyle G(n)=G(n-1)G(n-2)} {\displaystyle G(n)} is the n-th generation. Example 2: Fractal (binary) tree[edit] Example 3: Cantor set[edit] Example 4: Koch curve[edit] F+F−F−F+F+F+F−F−F+F−F+F−F−F+F−F+F−F−F+F+F+F−F−F+F+ F+F−F−F+F+F+F−F−F+F−F+F−F−F+F−F+F−F−F+F+F+F−F−F+F− Example 5: Sierpinski triangle[edit] Here, F means "draw forward", G means "draw forward", + means "turn left by angle", and − means "turn right by angle". It is also possible to approximate the Sierpinski triangle using a Sierpiński arrowhead curve L-system. rules : (A → B−A−B), (B → A+B+A) Example 6: Dragon curve[edit] rules : (F → F+G), (G → F-G) Here, F and G both mean "draw forward", + means "turn left by angle", and − means "turn right by angle". Example 7: Fractal plant[edit] See also: Barnsley fern Here, F means "draw forward", − means "turn right 25°", and + means "turn left 25°". X does not correspond to any drawing action and is used to control the evolution of the curve. The square bracket "[" corresponds to saving the current values for position and angle, which are restored when the corresponding "]" is executed. A number of elaborations on this basic L-system technique have been developed which can be used in conjunction with each other. Among these are stochastic grammars, context sensitive grammars, and parametric grammars. Stochastic grammars[edit] Context sensitive grammars[edit] Parametric grammars[edit] a(x,y) : x == 0 → a(1, y+1)b(2,3) Bi-directional grammars[edit] The bi-directional model explicitly separates the symbolic rewriting system from the shape assignment. For example, the string rewriting process in the Example 2 (Fractal tree) is independent on how graphical operations are assigned to the symbols. In other words, an infinite number of draw methods are applicable to a given rewriting system. The bi-directional model consists of 1) a forward process constructs the derivation tree with production rules, and 2) a backward process realizes the tree with shapes in a stepwise manner (from leaves to the root). Each inverse-derivation step involves essential geometric-topological reasoning. With this bi-directional framework, design constraints and objectives are encoded in the grammar-shape translation. In architectural design applications, the bi-directional grammar features consistent interior connectivity and a rich spatial hierarchy.[3] Characterisation of all the deterministic context-free L-systems which are locally catenative. (A complete solution is known only in the case where there are only two variables).[4] Types of L-systems[edit] tilings (sphinx tiling, Penrose tiling) Reaction–diffusion system – Type of mathematical model that provides diffusing-chemical-reagent simulations (including Life-like) ^ Lindenmayer, Aristid (March 1968). "Mathematical models for cellular interactions in development II. Simple and branching filaments with two-sided inputs". Journal of Theoretical Biology. 18 (3): 300–315. Bibcode:1968JThBi..18..300L. doi:10.1016/0022-5193(68)90080-5. ISSN 0022-5193. PMID 5659072. ^ Grzegorz Rozenberg and Arto Salomaa. The mathematical theory of L systems (Academic Press, New York, 1980). ISBN 0-12-597140-0 ^ Hua, H., 2017, December. A Bi‐Directional Procedural Model for Architectural Design. In Computer Graphics Forum (Vol. 36, No. 8, pp. 219-231). ^ Kari, Lila; Rozenberg, Grzegorz; Salomaa, Arto (1997). "L Systems". Handbook of Formal Languages. pp. 253–328. doi:10.1007/978-3-642-59136-5_5. ISBN 978-3-642-63863-3. L-Systems: A user friendly page to generate fractals and plants from L-Systems. OpenAlea Archived 2005-10-17 at the Wayback Machine: an open-source software environment for plant modeling,[1] which contains L-Py, an open-source python implementation of the Lindenmayer systems[2] A JAVA applet with many fractal figures generated by L-systems. Archived 2016-08-06 at the Wayback Machine Rozenberg, G.; Salomaa, A. (2001) [1994], "L-systems", Encyclopedia of Mathematics, EMS Press Laurens Lapré's L-Parser Archived 2013-09-13 at the Wayback Machine Complexity of L-System[dead link] ^ Pradal, Christophe; Fournier, Christian; Valduriez, Patrick; Cohen-Boulakia, Sarah (2015). OpenAlea: scientific workflows combining data analysis and simulation (PDF). Proceedings of the 27th International Conference on Scientific and Statistical Database Management - SSDBM '15. p. 1. doi:10.1145/2791347.2791365. ISBN 9781450337090. S2CID 14246115. ^ Boudon, Frédéric; Pradal, Christophe; Cokelaer, Thomas; Prusinkiewicz, Przemyslaw; Godin, Christophe (2012). "L-Py: An L-System Simulation Framework for Modeling Plant Architecture Development Based on a Dynamic Language". Frontiers in Plant Science. 3: 76. doi:10.3389/fpls.2012.00076. PMC 3362793. PMID 22670147. Retrieved from "https://en.wikipedia.org/w/index.php?title=L-system&oldid=1081285581"
Example problem of a right circular cone — lesson. Mathematics State Board, Class 10. The diameter of the wafer cone is $10 \ cm$ and, its height is $12 \ cm$. Calculate the curved surface area of $20$ such wafer cones. Diameter of the cone $(d)$ $=$ $10 \ cm$ Radius of the cone $(r)$ $=$ \frac{d}{2}=\frac{10}{2}=5 Height of the cone $(h)$ $=$ $12 \ cm$ Let us first find the slant height of the cone. $l = \sqrt{r^2 + h^2}$ $l = \sqrt{5^2 + 12^2}$ $l = \sqrt{25 + 144}$ $l = \sqrt{169}$ $l = 13$ $cm$ Curved surface area of the cone $=$ $\pi r l$ sq. units \frac{1430}{7} $=$ $204.28$ $cm^2$ The curved surface area of a cone is $204.28 \ cm^2$. Curved surface area of $20$ cones: $=$ $20 \times 204.28$ $=$ $4085.6$ Therefore, the curved surface area of $20$ wafer cones is $4085.6 \ cm^2$.
26 CFR § 1.170A-12 - Valuation of a remainder interest in real property for contributions made after July 31, 1969. \| CFR \| US Law \| LII / Legal Information Institute 26 CFR § 1.170A-12 - Valuation of a remainder interest in real property for contributions made after July 31, 1969. (1) Section 170(f)(4) provides that, in determining the value of a remainder interest in real property for purposes of section 170, depreciation and depletion of such property shall be taken into account. Depreciation shall be computed by the straight line method and depletion shall be computed by the cost depletion method. Section 170(f)(4) and this section apply only in the case of a contribution, not made in trust, of a remainder interest in real property made after July 31, 1969, for which a deduction is otherwise allowable under section 170. (b) Valuation of a remainder interest following only one life - (1) General rule. The value of a remainder interest in real property following only one life is determined under the rules provided in § 20.2031-7 (or for certain prior periods, § 20.2031-7A) of this chapter (Estate Tax Regulations), using the interest rate and life contingencies prescribed for the date of the gift. See, however, § 1.7520-3(b) (relating to exceptions to the use of prescribed tables under certain circumstances). However, if any part of the real property is subject to exhaustion, wear and tear, or obsolescence, the special factor determined under paragraph (b)(2) of this section shall be used in valuing the remainder interest in that part. Further, if any part of the property is subject to depletion of its natural resources, such depletion is taken into account in determining the value of the remainder interest. \left(1+\frac{i}{2}\right)\sum _{t=0}^{n-1}{v}^{t+1}\left[\left(1-\frac{{l}_{x+t+1}}{{l}_{x}}\right)-\left(1-\frac{{l}_{x+t}}{{l}_{x}}\right)\right]\left(1-\frac{1}{2n}-\frac{t}{n}\right) A, who is 62, donates to Y University a remainder interest in a personal residence, consisting of a house and land, subject to a reserved life estate in A. At the time of the gift, the land has a value of $30,000 and the house has a value of $100,000 with an estimated useful life of 45 years, at the end of which period the value of the house is expected to be $20,000. The portion of the property considered to be depreciable is $80,000 (the value of the house ($100,000) less its expected value at the end of 45 years ($20,000)). The portion of the property considered to be nondepreciable is $50,000 (the value of the land at the time of the gift ($30,000) plus the expected value of the house at the end of 45 years ($20,000)). At the time of the gift, the interest rate prescribed under section 7520 is 8.4 percent. Based on an interest rate of 8.4 percent, the remainder factor for $1.00 prescribed in § 20.2031-7(d) for a person age 62 is 0.26534. The value of the nondepreciable remainder interest is $13,267.00 (0.26534 times $50,000). The value of the depreciable remainder interest is $15,053.60 (0.18817, computed under the formula described in paragraph (b)(2) of this section, times $80,000). Therefore, the value of the remainder interest is $28,320.60. In 1972, B donates to Z University a remainder interest in his personal residence, consisting of a house and land, subject to a 20 year term interest provided for his sister. At such time the house has a value of $60,000, and an expected useful life of 45 years, at the end of which time it is expected to have a value of $10,000, and the land has a value of $8,000. The value of the portion of the property considered to be depreciable is $50,000 (the value of the house ($60,000) less its expected value at the end of 45 years ($10,000)), and this is multiplied by the fraction 20/45. The product, $22,222.22, is subtracted from $68,000, the value of the entire property, and the balance, $45,777.78, is multiplied by the factor .311805 (see § 25.2512-5A(c)). The result, $14,273.74, is the value of the remainder interest in the property. (ii) The special factor is to be computed on the basis of - \left(1+\frac{i}{2}\right)\sum _{t=0}^{n-1}{v}^{\left(t+1\right)}\left[\left(1-\frac{{1}_{x+t+1}}{{1}_{x}}\right)\left(1-\frac{{1}_{y+t+1}}{{1}_{y}}\right)-\left(1-\frac{{1}_{x+t}}{{1}_{x}}\right)\left(1-\frac{{1}_{y+t}}{{1}_{y}}\right)\right]\left(1-\frac{1}{2n}-\frac{t}{n}\right)
On the Involute-Evolute of the Pseudonull Curve in Minkowski 3-Space 2013 On the Involute-Evolute of the Pseudonull Curve in Minkowski 3-Space Ufuk Ozturk, Esra Betul Koc Ozturk, Kazim Ilarslan We have generalized the involute and evolute curves of the pseudonull curves \alpha {𝔼}_{\mathrm{1}}^{\mathrm{3}} \alpha is a spacelike curve with a null principal normal. Firstly, we have shown that there is no involute of the pseudonull curves \alpha {𝔼}_{\mathrm{1}}^{\mathrm{3}} . Secondly, we have found relationships between the evolute curve β and the pseudonull curve \alpha {𝔼}_{\mathrm{1}}^{\mathrm{3}} . Finally, some examples concerning these relations are given. Ufuk Ozturk. Esra Betul Koc Ozturk. Kazim Ilarslan. "On the Involute-Evolute of the Pseudonull Curve in Minkowski 3-Space." J. Appl. Math. 2013 1 - 6, 2013. https://doi.org/10.1155/2013/651495 Ufuk Ozturk, Esra Betul Koc Ozturk, Kazim Ilarslan "On the Involute-Evolute of the Pseudonull Curve in Minkowski 3-Space," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-6, (2013)
AI Powered Drug Discovery - Dennis It’s no secret that the traditional drug discovery model is expensive, time consuming, and has recently produced compounds with marginal efficacy in clinical trials and high failure rates. Large biopharma companies are increasingly outsourcing R&D to academic labs and small/mid size startups via licensing partnerships or acquisitions. Biopharma is among the most active industry sectors in terms of mergers and acquisitions and this trend looks to grow as companies stockpile more and more cash. This is all evidence to demonstrate that large cap biopharma companies have yet to create efficient internal R&D programs and thus have a need for high efficiency screening and discovery programs. Increasingly, companies have also focused on biologics, such as antibodies, vaccines, and DNA/RNA therapies. These are typically larger structures that rely on the body’s internal biological machinery. This is in contrast to small molecule therapeutics which are used to drug specific receptors or structures by fitting in a shape pocket. Small molecule compounds are synthesized using traditional organic chemistry techniques and include medications like Sovaldi for Hepatitis, Aspirin for pain relief, and Abilify for Schizophrenia. Small molecule therapeutics make up roughly 90% of drugs available on today’s market. However, since 2014, biologics have accounted for 93% of spending growth in discovery platforms. This makes sense, since 9 out of the top 10 best selling drugs in 2019 are biologics, compared to just 5 in 2014. Biologics have advantages over small molecule therapeutics such as higher specificity and better safety profiles, lower attrition in clinical trials, and less competition due to difficulty developing biosimilars. However, with the simpler drug discovery and development process of small molecules and superior delivery and distribution characteristics, small molecule drugs are here to stay. A Perfect Use Case for AI So what is causing the growth of biologics and stagnant investment in small molecule therapeutics programs? The current drug discovery pathway relies heavily on human derived research and capabilities. ‘Rational drug design’ comprises discovering a mechanism, identifying a druggable target, and the design of a drug to hit the target, mostly in an iterative, linear process. There is an increasing sense that the low hanging fruit has been picked, that the easy mechanisms and targets are already drugged. Still, others have suggested that we are more so held back by our ability to design molecules. This drug design process is highly automated, but increasing speed and compute power has not kept pace with costs/demand in an efficient manner. Computers are adept at simulating molecule interactions, but the binding profile is only one of many important drug design considerations. Adding to the inefficiencies, old techniques will only simulate existing molecule compounds, or try new ones, but not intelligently, merely tweaking the structure in an arbitrary manner. For each of these millions of such simulations, algorithms may use molecular mechanics and predictive binding equations to estimate the affinity of the designed ligand. These equations have limited efficacy because they rely on human derived models, and are not trained on extensive previous data libraries. This is the perfect use case for artificial intelligence. Human logic has driven us to utilize a design approach for solving drug discovery. We used first principles to develop rational drug design and to contrive equations that model molecule interactions. Yet, we are still unable to understand many aspects of biophysics and molecular interactions, so changing our model to a search approach makes much more sense. With exponential growth in compute power, data, and infrastructure, we have the computational resources to power a search strategy, where we can rely on machines to tell us what is correct. Given clean, large, and relevant datasets, AI algorithms can independently decode and organize data in ways that we wouldn’t initially think of. Prediction, classification, and even imagination are all problems an AI engine can solve now without any dependencies on biological knowledge or principles that humans may be tied to. This set of advantages that AI provides us is what is powering the rapid growth of modern computational drug design. Modern Drug Design Technologies Modern computational drug design strategies are able to factor in many more factors including toxicity, formulation, and kinetics due to tremendous growth in available training data. Using machine learning, we are able to predict key properties of molecules such as absorption, distribution, metabolism, elimination, and toxicity (ADMET properties) based off a vectorized representation of a molecule. The inability to properly predict these properties are estimated to account for up to 50% of clinical trial failures, so algorithms that can predict ADMET properties in silico before expensive clinical trials are of tremendous value. The team at Genesis Therapeutics has developed PotentialNet, a deep learning algorithm for precisely this purpose. A neural net can be trained on molecules we already know to be successful, in order to screen molecules we aren’t sure about. The human factor here is to calibrate the data into the correct format, but there is no need to rely on archaic equations to predict binding as before. Another example comes from a group from MIT who used a deep learning approach to discover halicin as a novel antibiotic. Halicin is structurally divergent from traditional antibiotics, but demonstrated curative antibiotic activity in-vivo and favorable pharmacokinetics in mice. This approach has demonstrated how deep learning networks such as these can be trained on libraries of existing compounds and their properties to predict not only important drug properties like ADMET, but also their therapeutic efficacy. The speed and efficiency of deep learning approaches versus previous brute force screening methods is a major advance here. This has significant implications especially for fields like antibiotics, where the business model doesn’t allow for large R&D programs where scale allows brute force methods to work. If we’re able to outsource the expensive creative work to machines, traditional commercial barriers to innovation can be bypassed, which is especially important for antibiotics and other low cost medications. Next Generation Computational Tools Thus far, we have seen AI deployed primarily as a prediction tool. We can predict bioactivity and we can predict important drug properties, which is a crucial need for drug screening and improving the attrition rate during clinical trials. This type of work is especially valuable for large cap biopharma companies who already have hundreds of thousands of compounds and must bear the costs for bringing drugs to market. Startups doing this type of predictive work are thus frequent targets for industry collaboration. However, the next paradigm in AI drug discovery is not the efficacy prediction of existing molecules, but the creation of generative models that are able to ‘imagine’ de novo structures, pathways, and mechanisms. The first completely AI imagined molecule to be used in a human clinical trial will start enrollment in 2020. This molecule, created by another AI drug discovery startup, Exscientia, took 12 months from conception to clinical trial enrollment, 5x faster than the traditional workflow. Their system automatically extracts key performance markers from high-dimensional phenotypic readouts and uses these to generate and optimize new iterations of compounds, to rapidly evolve compounds that satisfy key performance criteria. Platforms like these are in theory not only scalable, but also computationally less expensive than the traditional screening approach. This means that new molecules can be developed in a democratized manner, for indications traditionally avoided for not providing large enough returns on investment. As a ‘full stack’ AI drug discovery company generating its own data, Exscientia is able to leverage internal data collection capabilities to develop drug candidates in therapy areas where data is limited. Generative drug discovery thus is also an economically disruptive force, enabling the targeting of therapeutic areas that haven’t previously been addressed. Evaluating Startups in the Space When evaluating a company in this space, it’s important to consider exactly what problem the AI hopes to solve, whether it be speed, attrition, modeling accuracy, molecule design, or otherwise. The best companies will address multiple of these at once. Exscientia’s full stack drug discovery platform addresses in some way, all of these. Another key consideration is data. When dealing with biological systems, it takes much longer to validate than deep learning for computer vision in driverless cars for instance because you need to carryout experiments with real organisms eventually. This impacts data collection and model feedback iteration speeds, making progress slower. Some start-ups have set up contractual partnerships with academia and research centers, enabling them to benefit from access to proprietary data and talent to help solve this problem. Many others collaborate directly with industry, but typically only once past the stage of developing several promising compounds. Talent in particular is the biggest priority in this field. Drug discovery is a very academic field, yet you need a very multidisciplinary team of medicinal chemists, ML engineers, assay developers, etc. Every biopharma is trying to build up the internal expertise and is hiring data scientists, and you also need to compete with tech companies on the other end for data and ML engineers. A magnetic core with strong academic background, already demonstrating proof of concept is important. For example, the core technology from Genesis Therapeutics was developed during the CEO’s PhD. Successful companies require highly trained, specialist data experts and experts in AI and computational biology, and the core team needs to be able to recruit such talent. Lastly, this is a hyper competitive industry, which has both pros and cons. On one end, you need a very fast paced and competitive team that can easily detach and move on from their own research, since this field moves so quickly. It’s easy to drown, either from the success of competitors, inability to secure collaborations, or simply due to the difficulty of the task at hand. It’s possible even that the center of innovation in this field no longer happens in the U.S., as China has the most drug discovery AI research centers and the largest datasets to work with. On the other hand, this field is flooding with funding. VC dollars quadrupled in three years from 2016 to 2019, and the total R&D market is estimated to grow from 700 million in 2018 to 20 billion in the next five years. Large biopharma are seeking diversification of deals and collaborations to diversify drug discovery pipeline, and much of this growth has been and will be via AI driven solutions. The incredible pace of development demonstrates the immediate promise of using AI models to both intelligently screen and design therapeutics. We can integrate toxicity, pharmacokinetics, and formulation into the training set to improve the quality of molecules for the next steps of clinical testing. Even more promising is that the next generation of medicinal chemists can work collaboratively with algorithms, each able to creatively design compounds and mutually checking each other’s work. The bottom line here is that AI introduces a new paradigm of non-linearity in drug discovery. While previously we would try to answer the question of how to make screening, optimization, and testing faster, now we are asking how we do these processes smarter. We’re solving both attrition and speed here. We can rely on machines to transition from our limitations as designers and turn biology into a search problem. We know AI can analyze data, we didn’t know it could generate creative insights in this field, yet. Personally, I think that this shift in mindset will be the transformation that propels drug discovery to break Eroom’s Law. Biology for decades has been known to lag behind other fields, commanding huge amounts of human capital in an attempt to ‘think’ our way to cures. Finally, we have a tool that can quite literally predict biology, offloading the creative burden and covering our blindspots. In the immediate future, this technology will be used to democratize our search for small molecule therapeutics in indications previously economically unfeasible. In the long term, with the confluence of quantum computing and generation of more and more patient data, theres no obvious limit to what this platform can do. Large molecule therapeutics, protein therapeutics, and eventually even custom biologic design is not out of the question. On our existing library of drugs, we can optimize to reduce toxicity and improve other ADME properties, or even to achieve improved clinical outcomes. Drug discovery is becoming supercharged. Select Startups in AI Drug Discovery Genesis Therapeutics: A seed stage company developing tools to improve screening by better predicting pharmacological properties of drug compounds. Their AI platform, PotentialNet, predicts 20+ different ADMET properties. LabGenius: The Series A stage team at LabGenius is the first biopharmaceutical company developing next generation protein therapeutics using a machine learning-driven evolution engine (EVA™️). They use advanced deep-learning neural networks to explore protein fitness landscapes and improve multiple drug properties simultaneously. Exscientia: Series B stage company working on drug target selection and de novo molecule design using machine learning. Exscientia is another frequent industry collaborator, and is the first company to develop an entirely AI generated compound for clinical trials. Atomwise: Atomwise is a Series A stage industry collaborator that uses machine learning to improve drug hit rates by up to 10,000x and results 100 times faster than ultra high throughput screening. Atomwise’s deep convolutional neural network, AtomNet, screens more than 100 million compounds each day for potency, selectivity, and polypharmacology, and guard against off-target toxicity.
Convert compensator coefficient to amplitude and phase imbalance - MATLAB iqcoef2imbal - MathWorks France Estimate I/Q Imbalance from Compensator Coefficient Convert compensator coefficient to amplitude and phase imbalance [A,P] = iqcoef2imbal(C) [A,P] = iqcoef2imbal(C) converts compensator coefficient C to its equivalent amplitude and phase imbalance. Use iqcoef2imbal to estimate the amplitude and phase imbalance for a given complex coefficient. The coefficients are an output from the step function of the IQImbalanceCompensator. Create a raised cosine transmit filter to generate a 64-QAM signal. txFilt = comm.RaisedCosineTransmitFilter; txSig = step(txFilt,dataMod); Normalize the power of the received signal Remove the I/Q imbalance using the comm.IQImbalanceCompensator System object™. Set the compensator object such that the complex coefficients are made available as an output argument. hIQComp = comm.IQImbalanceCompensator('CoefficientOutputPort',true); [compSig,coef] = step(hIQComp,rxSig); Estimate the imbalance from the last value of the compensator coefficient. [ampImbEst,phImbEst] = iqcoef2imbal(coef(end)); Compare the estimated imbalance values with the specified ones. Notice that there is good agreement. [ampImb phImb; ampImbEst phImbEst] complex-valued scalar or vector Coefficient used to compensate for an I/Q imbalance, specified as a complex-valued vector. Example: 0.4+0.6i Example: [0.1+0.2i; 0.3+0.5i] Amplitude imbalance in dB, returned as a real-valued vector with the same dimensions as C. Phase imbalance in degrees, returned as a real-valued vector with the same dimensions as C. The function iqcoef2imbal is a supporting function for the comm.IQImbalanceCompensator System object™. Given a scaling and rotation factor, G, compensator coefficient, C, and received signal, x, the compensated signal, y, has the form y=G\left[x+C\mathrm{conj}\left(x\right)\right]\text{\hspace{0.17em}}. In matrix form, this can be rewritten as Y=R\text{ }X\text{\hspace{0.17em}}, where X is a 2-by-1 vector representing the imbalanced signal [XI, XQ] and Y is a 2-by-1 vector representing the compensator output [YI, YQ]. The matrix R is expressed as R=\left[\begin{array}{cc}1+\mathrm{Re}\left\{C\right\}& \mathrm{Im}\left\{C\right\}\\ \mathrm{Im}\left\{C\right\}& 1-\mathrm{Re}\left\{C\right\}\end{array}\right] For the compensator to perfectly remove the I/Q imbalance, R = K-1 because X=K\text{\hspace{0.17em}}S , where K is a 2-by-2 matrix whose values are determined by the amplitude and phase imbalance and S is the ideal signal. Define a matrix M with the form M=\left[\begin{array}{cc}1& -\alpha \\ \alpha & 1\end{array}\right] Both M and M-1 can be thought of as scaling and rotation matrices that correspond to the factor G. Because K = R-1, the product M-1 R K M is the identity matrix, where M-1 R represents the compensator output and K M represents the I/Q imbalance. The coefficient α is chosen such that K\text{ }M=L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{Q}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{Q}\right)\end{array}\right] where L is a constant. From this form, we can obtain Igain, Qgain, θI, and θQ. For a given phase imbalance, ΦImb, the in-phase and quadrature angles can be expressed as \begin{array}{c}{\theta }_{I}=-\left(\pi /2\right)\left({\Phi }_{Imb}/180\right)\\ {\theta }_{Q}=\pi /2+\left(\pi /2\right)\left({\Phi }_{Imb}/180\right)\end{array} Hence, cos(θQ) = sin(θI) and sin(θQ) = cos(θI) so that L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{Q}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{Q}\right)\end{array}\right]=L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{I}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{I}\right)\end{array}\right] The I/Q imbalance can be expressed as \begin{array}{c}K\text{ }M=\left[\begin{array}{cc}{K}_{11}+\alpha {K}_{12}& -\alpha {K}_{11}+{K}_{12}\\ {K}_{21}+\alpha {K}_{22}& -\alpha {K}_{21}+{K}_{22}\end{array}\right]\\ =L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{I}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{I}\right)\end{array}\right]\end{array} \left({K}_{21}+\alpha {K}_{22}\right)/\left({K}_{11}+\alpha {K}_{12}\right)=\left(-\alpha {K}_{11}+{K}_{12}\right)/\left(-\alpha {K}_{21}+{K}_{22}\right)=\mathrm{sin}\left({\theta }_{I}\right)/\mathrm{cos}\left({\theta }_{I}\right) The equation can be written as a quadratic equation to solve for the variable α, that is D1α2 + D2α + D3 = 0, where \begin{array}{c}{D}_{1}=-{K}_{11}{K}_{12}+{K}_{22}{K}_{21}\\ {D}_{2}={K}_{12}^{2}+{K}_{21}^{2}-{K}_{11}^{2}-{K}_{22}^{2}\\ {D}_{3}={K}_{11}{K}_{12}-{K}_{21}{K}_{22}\end{array} When \|C\| ≤ 1, the quadratic equation has the following solution: \alpha =\frac{-{D}_{2}-\sqrt{{D}^{2}-4{D}_{1}{D}_{3}}}{2{D}_{1}} Otherwise, when \|C\| > 1, the solution has the following form: \alpha =\frac{-{D}_{2}+\sqrt{{D}^{2}-4{D}_{1}{D}_{3}}}{2{D}_{1}} Finally, the amplitude imbalance, AImb, and the phase imbalance, ΦImb, are obtained. \begin{array}{c}{K}^{\prime }=K\left[\begin{array}{cc}1& -\alpha \\ \alpha & 1\end{array}\right]\\ {A}_{Imb}=20{\mathrm{log}}_{10}\left({{K}^{\prime }}_{11}/{{K}^{\prime }}_{22}\right)\\ {\Phi }_{Imb}=-2{\mathrm{tan}}^{-1}\left({{K}^{\prime }}_{21}/{{K}^{\prime }}_{11}\right)\left(180/\pi \right)\end{array} If C is real and \|C\| ≤ 1, the phase imbalance is 0 and the amplitude imbalance is 20log10((1–C)/(1+C)) If C is real and \|C\| > 1, the phase imbalance is 180° and the amplitude imbalance is 20log10((C+1)/(C−1)). If C is imaginary, AImb = 0. iqimbal \| iqimbal2coef
EUDML \| Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains. EuDML \| Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains. Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains. Fleckinger, Jacqueline; Harrell, Evans M.II; de Thélin, François Fleckinger, Jacqueline, Harrell, Evans M.II, and de Thélin, François. "Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains.." Electronic Journal of Differential Equations (EJDE) [electronic only] 2001 (2001): Paper No. 77, 14 p., electronic only-Paper No. 77, 14 p., electronic only. <http://eudml.org/doc/121852>. @article{Fleckinger2001, author = {Fleckinger, Jacqueline, Harrell, Evans M.II, de Thélin, François}, keywords = {-Laplacian; Riccati transformation; uncertainty principle; -Laplacian}, title = {Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains.}, AU - Fleckinger, Jacqueline AU - Harrell, Evans M.II AU - de Thélin, François TI - Asymptotic behavior of solutions for some nonlinear partial differential equations on unbounded domains. KW - -Laplacian; Riccati transformation; uncertainty principle; -Laplacian p -Laplacian, Riccati transformation, uncertainty principle, p Articles by Fleckinger Articles by Harrell Articles by de Thélin
EUDML \| Resonance and strong resonance for semilinear elliptic equations in . EuDML \| Resonance and strong resonance for semilinear elliptic equations in . Resonance and strong resonance for semilinear elliptic equations in {ℝ}^{N} López Garza, Gabriel; Rumbos, Adolfo J. López Garza, Gabriel, and Rumbos, Adolfo J.. "Resonance and strong resonance for semilinear elliptic equations in .." Electronic Journal of Differential Equations (EJDE) [electronic only] 2003 (2003): Paper No. 124, 22 p., electronic only-Paper No. 124, 22 p., electronic only. <http://eudml.org/doc/123634>. @article{LópezGarza2003, author = {López Garza, Gabriel, Rumbos, Adolfo J.}, keywords = {Resonance; strong resonance; concentration-compactness}, title = {Resonance and strong resonance for semilinear elliptic equations in .}, AU - López Garza, Gabriel AU - Rumbos, Adolfo J. TI - Resonance and strong resonance for semilinear elliptic equations in . KW - Resonance; strong resonance; concentration-compactness Resonance, strong resonance, concentration-compactness Articles by López Garza Articles by Rumbos
Harmonic Subtangent Structures Adara M. Blaga, "Harmonic Subtangent Structures", Journal of Mathematics, vol. 2014, Article ID 603078, 5 pages, 2014. https://doi.org/10.1155/2014/603078 Adara M. Blaga1 1Department of Mathematics, West University of Timişoara, Boulevard V. Pârvan No. 4, 300223 Timişoara, Romania The concept of harmonic subtangent structures on almost subtangent metric manifolds is introduced and a Bochner-type formula is proved for this case. Conditions for a subtangent harmonic structure to be preserved by harmonic maps are also given. Inspired by the paper of Jianming [1], we introduce the notion of harmonic almost subtangent structure and underline the connection between harmonic subtangent structures and harmonic maps. It is well known that harmonic maps play an important role in many areas of mathematics. They often appear in nonlinear theories because of the nonlinear nature of the corresponding partial differential equations. In theoretical physics, harmonic maps are also known as sigma models. Remark also that harmonic maps between manifolds endowed with different geometrical structures have been studied in many contexts: Ianus and Pastore treated the case of contact metric manifolds [2], Bejan and Benyounes the almost para-Hermitian manifolds [3], Sahin the locally conformal Kähler manifolds [4], Ianus et al. the quaternionic Kähler manifolds [5], Jaiswal the Sasakian manifolds [6], Fetcu the complex Sasakian manifolds [7], Li the Finsler manifolds [8], and so forth. Fotiadis studied the noncompact case, describing the problem of finding a harmonic map between noncompact manifolds [9]. Let be a smooth, -dimensional real manifold for which we denote by the real algebra of smooth real functions on , by the Lie algebra of vector fields on , and by the -module of tensor fields of -type on . An element of is usually called vector 1-form or affinor. Recall the concept of almost tangent geometry. Definition 1 (see [10]). is called almost tangent structure on if it has a constant rank and The pair is called almost tangent manifold. The name is motivated by the fact that (1) implies the nilpotence exactly as the natural tangent structure of tangent bundles. Denoting it results in . If in addition, we assume that is integrable, that is, then is called tangent structure and is called tangent manifold. From [11] we deduce some aspects of tangent manifolds:(i)the distribution defines a foliation;(ii)there exist local coordinates on such that ; that is, We call canonical coordinates and the change of canonical coordinates is given by So another description can be obtained in terms of -structures. Namely, a tangent structure is a -structure with [12] and is the invariance group of matrix ; that is, if and only if . The natural almost tangent structure of is an example of tangent structure having exactly the expression (3) if are the coordinates on and are the coordinates in the fibers of . A class of examples is obtained by duality [12]: if is an (integrable) endomorphism with , then its dual , given by for , is (integrable) endomorphism with . If the condition in the Definition 1 is weakened, requiring that only squares to , we call almost subtangent structure. In this case, . 2. Harmonic Subtangent Structures Let be an almost subtangent metric manifold of dimension , that is, a -dimensional smooth manifold endowed with an almost subtangent structure which is compatible with a pseudo-Riemannian metric (i.e., , for any , and let be the Levi-Civita connection associated with . Consider the exterior differential and codifferential operators defined for any tangent bundle-valued -form by for an orthonormal frame field and the Hodge-Laplace operator on : Jianming studied in [1] some properties of harmonic complex structures and we discussed in [13] the paracosymplectic case. Definition 2. An almost subtangent structure is called harmonic if . If is compact, from the definition it follows that is harmonic if and only if and which is equivalent to , for any , and , being the Levi-Civita connection associated with the pseudo-Riemannian structure . Proposition 3. On a compact almost subtangent manifold, any harmonic almost subtangent structure is integrable (i.e., it is a subtangent structure). Proof. Let , . Then As implies , we get which shows the integrability of . Remark 4. As expected, the harmonicity of an almost subtangent structure is not always preserved under conformal transformations. Indeed, let be a harmonic subtangent structure (with respect to ) and for a smooth positive function on the -dimensional manifold , let . Then the Levi-Civita connection associated with is , for any , . The necessary and sufficient condition for to be harmonic (with respect to ) is but so the first relation is equivalent to Taking an orthonormal frame field on with , , and computing the second relation is equivalent to In conclusion, is also harmonic with respect to if and only if Now we want to see how a Bochner-type formula can be written on an almost subtangent metric manifold. We know that for any tangent bundle-valued differential form, , the following Weitzenböck formula holds [14]: where and , , for an orthonormal frame field and , , , the Riemann curvature tensor field. We will also use the notations and , , , , . Now, on the almost subtangent metric manifold , taking equal to , for any vector field , we have We can state the following theorem. Theorem 5. Let be an almost subtangent metric manifold and assume that is harmonic subtangent structure. Then a Bochner-type formula reduces to for an orthonormal frame field on with , . Proof . A similar computation like in [1] leads us to Therefore, as is harmonic if , from (17), we obtain Notice that if is only almost subtangent structure, from the proof of the theorem, we deduce that If is compact, integrating this relation with respect to the canonical measure, we obtain the following characterization of a harmonic almost subtangent structure. Corollary 6. Let be a compact almost subtangent metric manifold. Then the almost subtangent structure is harmonic if and only if Example 7. Concerning the existence of almost tangent structures of order (i.e., those with ) on the spheres, Rosendo and Gadea [15] proved that the only spheres that admit such structures are and . Moreover, they proved that the only spheres that admit almost tangent structures (of different orders) are (of order ), (of order ), and (of order or ). For these cases, let and . Computing and taking into account that , for any , from Corollary 6, we get 3. Harmonic Maps and Harmonic Subtangent Structures Let and be two almost subtangent metric manifolds - and -dimensional, respectively. Denote by and , respectively, the Levi-Civita connections associated with and , respectively. Consider a smooth map and let be the tension field of , where is an orthonormal frame field on . Proposition 8. Let be a smooth map between almost subtangent metric manifolds such that . Then for an orthonormal frame field on the -dimensional manifold . Proof. Express and replace it in the left side of the relation. Proposition 9. Let be a smooth map between almost subtangent metric manifolds such that . If for any , , then Proof. For any , , and for , Definition 10. A smooth map is said to be harmonic if its tension field vanishes. Proposition 11. Let be a smooth map between almost subtangent metric manifolds such that . If is harmonic map, then for an orthonormal frame field on the -dimensional manifold . Moreover, if for any , , then for an orthonormal frame field on the -dimensional manifold . Corollary 12. Let be a smooth map between almost subtangent metric manifolds such that and is harmonic subtangent structure. (1)If for any , , then Moreover, if is surjective submersion, then is harmonic subtangent structure, too.(2)If is harmonic map, then for an orthonormal frame field on the -dimensional manifold . The author thanks the referees for the valuable suggestions they made in order to improve the paper. She also acknowledges the support by the Research Grant PN-II-ID-PCE-2011-3-0921. W. Jianming, “Harmonic complex structures,” Chinese Annals of Mathematics A, vol. 30, no. 6, pp. 761–764, 2009. View at: Google Scholar S. Ianus and A. M. 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Amines Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers An organic compound A on reduction gives compound B which on reaction with chloroform and potassium hydroxide forms C. The compound C on catalytic reduction gives N-methylaniline. The compound A is Subtopic: Urea & Nitro Compound \| Acetamide and ethyl amine can be distinguished by reacting with 1. aq. HCl and heat 2. aq. NaOH and heat 3. acidified KMnO4 4. bromine water Subtopic: Amines - Preparation & Properties \| Identification of Primary, Secondary & Tertiary Amines \| Mechanism \| Which of the following compound gives dye test 3. Diphenylamine Subtopic: Diazonium Salts: Preparation, Properties & Uses \| CH3CH2Cl \stackrel{\mathrm{NaCN}}{\to } \stackrel{\mathrm{Ni}/{\mathrm{H}}_{2}}{\to } \stackrel{\mathrm{Acetic} \mathrm{anhydride}}{\to } In the above reaction sequence, Z is 1. CH3CH2CH2NHCOCH3 3. CH3CH2CH2CONHCH3 4. CH3CH2CH2CONHCOCH3 Subtopic: Cyanides & Isocyanides \| Predict the product, Subtopic: Amines - Preparation & Properties \| Aniline in a set of reactions yield the following products The structure of the product D would be - 1. C6H5CH2NH2 2. C6H5NHCH2CH3 3. C6H5NHOH Subtopic: Cyanides & Isocyanides \| Diazonium Salts: Preparation, Properties & Uses \| Intermediates formed during the reaction of RCONH2 with Br2 and KOH are - 1. RCONHBr and RNCO 2. RNHCOBr and RNCO 3. RNHBr and RCONHBr 4. RCONBr2 In a reaction of aniline a colored product C was obtained. The structure of C would be Product Q in the above reaction is \stackrel{{\mathrm{Ac}}_{2}\mathrm{O}}{\to } \underset{{\mathrm{CH}}_{2}\mathrm{COOH}}{\overset{{\mathrm{Br}}_{2}}{\to }} \underset{{\mathrm{H}}^{+}}{\overset{{\mathrm{H}}_{2}\mathrm{O}}{\to }} C in the above mentioned reaction is - Subtopic: Mechanism \|
What is LHC:ATLAS? Exploring The Universe With LHC Glossary Accelerating particles close to the speed of light and colliding them hoping something interesting will happen. ATLAS Structure Listen to collisions detected by ATLAS The Large Hadron Collider (LHC) is the result of a multinational collaboration with over 10,000 scientists and 100 countries.[1] By colliding charged particles, mostly protons, at speeds close to that of the speed of light, the LHC has made discoveries ranging from the confirmation of the existence and properties of the Higgs boson[2], to the likely non-existence of supersymmetrical particles [3] [4], and, recently, the discovery of two baryons composed of one bottom quark and two up quarks, and one bottom quark and two down quarks. [5] A picture of the caliometers of the ATLAS detector. ATLAS is one of the four main detectors at LHC, and measures the decay products from the collisions. When protons hit head on[6] in inelastic relativistic collisions[7], they create new particles which spread around in the detector. The atlas detector is constructed as a cylinder with multiple layers. The different layers can measure different properties of the decay products. The ATLAS detector is constructed as a cylinder through which the particle beams shoot. Upon a head-on collision,[8] the protons decay to other particles which scatter in all directions. This is possible while conserving momentum precisely because the protons collided head on. Since we cannot predict where the decay go, the detectors are constructed as approximate onion-like shells around the beam direction as illustrated below.[9][10] Schematic of ATLAS from [email protected] In the innermost shell, we have tracking devices such as the Xenon gas tubes, which can track as charged particles go through them, and the PIXEL/SCT trackers. Which is explained well in the following video: Knowing the trajectory of the particles along with strong magnets which curve the trajectory allows us, with some difficulty when dealing with multiple particles, to calculate the relativistic momentum of the charged particles. In the next shell, the particles pass through calorimeters which tries to measure the energy of the particles by absorbing them passively through high density materials such as lead, and actively through liquid argon for example. Electromagnetic and Hadronic calorimeters can stop most particles, but neutrinos and muons. The next shells detects Cherenkov and transition radiation, and at the outermost shell, we have the the muon spectrometers. [11][12] On the picture above, we see the inner shells where the trackers can see the charged particles such as the proton, electron and muon. Notice, how the magnets bend the paths of the particles. In the calorimeters, we see most particles being absorbed and decay to other particles. By observing the decay particles, and considering quantum symmetries, we can figure out the types of decay particles as has been colored in on the image. Notice further that we cannot directly detect neutrinos in the detector. Instead, we have to rely on inference from looking at the missing total energy in the system because we know there's conservation of energy even if some of it goes to creating new particles. You can listen to the audio of detected collisions from ATLAS courtesy of Ewan Hill below. The audio is generated by associating the values from different detectors of a subset of the collisions with MIDI tones and normalizing the frequencies to be within the audible range. 30 seconds is approximately equivalent to one event. [13] The Telegraph. ↩︎ Observation of a new boson at a mass of 125 GeV with the CMS experiment at the LHC ↩︎ What next for the CMSSM and the NUHM: Improved prospects for superpartner and dark matter detection ↩︎ Implications of a 125 GeV Higgs for the MSSM and Low-Scale SUSY Breaking ↩︎ LHCb experiment discovers two, perhaps three, new particles ↩︎ Most collisions only strays resulting in an elastic collision where the particles get a slight deviation in \eta : The pseudorapidity. The fraction of head on collisions versus straying collisions is well known, so the number of straying collisions are measured, so we can normalize the data to get the number of head-on collisions. ↩︎ Since we collide the protons with speeds approaching the speed of light, we have to account for the significant relativistic effects. ↩︎ It's slightly misleading to talk about single collisions as at every event, which only lasts a few i, about \mu \approx 50 collisions happen. ↩︎ This is not entirely accurate portrayal as there are places where the detector is less accurate (for example the muon detectors) depending on where in the detector the particles collide. Source: Conversations with Troels Petersen (NBI). ↩︎ ATLAS Fact Sheet ↩︎ How a detector works ↩︎ The Inner Detector ↩︎ Musician and Mega-Machine: Compositions Driven by Real-Time Particle Collision Data from the ATLAS Detector ↩︎ The hunt that could make or break the standard model.
Current Electricity, Popular Questions: ICSE Class 12-science PHYSICS, Physics Part I - Meritnation 42. Reading of ammeter {A}_{1}, {A}_{2} and {A}_{3} will be respectively:- 1. 1A, 0A, 1A 3. 1A, 0.5A, 0.5A Calculate the de-Broglie's wavelength associated with an electron of energy 200eV .What will be the change in its wavelength if the accelerating potential is increased to four times its earlier value? As shown in the diagram, network of resistors R1 and R2 extends off to infinity to the right. Find the equivalent resistance. 41. In given circuit current through AB is zero, then what will be the value of unknown resistance 'X' 1. 10\Omega 2. 5\Omega 3. 40\Omega 4. 15\Omega in alpha particle (2He4) is revolving in a circular orbit of radius 3.14A with speed of 8106m/s .then the equivalent current is ? a wire of 5.8 metre long,2 millimetre diameter carries 750 milliampere current when 22 mV potential difference is applied at its ends if drift speed of electrons is found 1.7 into 10^-5 metre per second then find current density and number of free electrons per unit volume current density=2.41)^5A/m square n=8.8*10^28 Niveditha Vudayagiri asked a question An equiconvex lens of refractive index mu1, focal length f and radius of curvature R is immersed in a liquid of refractive indec mu2. Find the focal length of the lend in terms of original focal length and the refractive index of glass of lens and that of medium. hc = 1242eV nm. how ? h ki value toh 6.636 smthng h fer yhan kya li h apne ? G.sumuka G asked a question Drift Speed of Electrons is of the order of 10-3 ms-1, but current is established in the circuit with the speed of Light. Explain Which concepts of Electrostatics are must do to learn Current Electricity? \left(1\right) \frac{12\mathrm{R}}{13} \left(2\right) \frac{\mathrm{R}}{13}\phantom{\rule{0ex}{0ex}}\left(3\right) \frac{5\mathrm{R}}{13} \left(4\right) \frac{15\mathrm{R}}{13} Abinash Pati asked a question How is i= qf?? Two material bars A and B of equal area of cross-section, are connected in series to a DC supply. A is made of usual resistance wire and B of an n-type semiconductor. (b) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected? Give the order of magnitude of number density of free electrons in a metal. Diksha B asked a question Arti Princess asked a question In example no 3.6 at page 116 of NCERT Physics, the value of R is found to b 5/6 ohm.. bt i think that it's incorrect.. it shud b 5/4 ohm... Even if v put the valur of R=5/6 ohm in the formula I=V/R then v dnt get the current I as 4 A.. But on putting R=5/4 ohm, v get the correct value of I too. So it shud b 5/4 and not 5/6.... Isnt it.....???????? Compute the voltage drop across a 2 KW electric heater element whose resistance when hot is 20 ohm. FIND THE EQUIVALENT CAPACITANCE BETWEEN POINTS A AND B . A battery of five lead acid accumulators ,each of emf 4 V and internal resistance 1 ohm connected in series is charged by 100 V dc source. Calxulate the following: 1)the series resistance to be used in the circuit to have a current of 5 A Plzz answer fast experts.. No links plzz... The network PQRS shown in the circuit diagram has the battery of 4V and 5V and negligible internal resistance. A milliammeter of 20\Omega resistance is connected between P and R. Calculate the reading in the ammeter. Pls describe Callender-Griffith's Bridge and Carey-Phoster's Bridge Mandeep Panjeta asked a question Is mobility a scalar quantity or vector? Calculate the equivalent resistance between the points M and N . Also calculate current in the arm AB . If r = 3ohm and potential difference across MN is 10 volt. Why current density is a vector quantity when electric current is a scalar quantity? Archit Anand K.p. asked a question Please explain Q26 5) In the circuit shown, cells are of equal emf E but of different internal resistance {r}_{1}=6\Omega and {r}_{2}=4\Omega . Reading of the ideal voltmeter connected across cell 1 is zero. Value of the external resistance R in ohm is equal to. \left(A\right) 2\phantom{\rule{0ex}{0ex}}\left(B\right) 2.4\phantom{\rule{0ex}{0ex}}\left(C\right) 10\phantom{\rule{0ex}{0ex}}\left(D\right) 24 a current of 5A flows in an electrical circuit.how much charge will flow through a point of the circuit in 10 mins. ans is 3000C Mohammed Mufeeth asked a question 43) Calculate the potential difference between B and D points in the given figure. An e.m.f. of 12 V is connected to the circuit. Haphi K. Shiing asked a question two identical cells whether joined in series or in parallel gives the same current when connected to an external resistance of 1ohm. find the internal resistance of each cell. 10. In the following circuit diagram the value of resistance X for the potential difference between B and D is zero -
Thermodynamics Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers A system is taken from state A to state B along two different paths 1 and 2. If the heat absorbed and work done by the system along these two paths are {Q}_{1}, {Q}_{2} and {W}_{1}, {W}_{2} {Q}_{1}={Q}_{2} {W}_{1}={W}_{2} {Q}_{1}-{W}_{1}={Q}_{2}-{W}_{2} {Q}_{1}+{W}_{1}={Q}_{2}+{W}_{2} In a given process dW = 0, dQ < 0, then for the gas: 1. Temperature - increases 2. Volume - decreases 3. Pressure - decreases 4. Pressure - increases A given mass of gas expands from state A to state B by three paths 1, 2 and 3 as shown in the figure. If {\mathrm{W}}_{1}, {\mathrm{W}}_{2} \mathrm{and} {\mathrm{W}}_{3} respectively be the work done by the gas along the three paths, then: {\mathrm{W}}_{1} > {\mathrm{W}}_{2} > {\mathrm{W}}_{3} {\mathrm{W}}_{1} < {\mathrm{W}}_{2} < {\mathrm{W}}_{3} {\mathrm{W}}_{1} = {\mathrm{W}}_{2} = {\mathrm{W}}_{3} {\mathrm{W}}_{1} < {\mathrm{W}}_{2} = {\mathrm{W}}_{3} Subtopic: Work Done by a Gas \| The ratio of the relative rise in pressure for adiabatic compression to that for isothermal compression is \gamma \frac{1}{\gamma } (3) 1- \gamma \frac{1}{1-\gamma } Subtopic: Types of Processes \| A sink, that is the system where heat is rejected, is essential for the conversion of heat into work. From which law the above inference follows? (1) zeroth Subtopic: Second Law of Thermodynamics \| For the indicator diagram given below, select the wrong statement: 1. Cycle - II is a heat engine cycle. 2. Net work is done on the gas in cycle I. 3. Work done is positive for cycle I. 4. Work done is positive for cycle II. Subtopic: Cyclic Process \| An ideal gas with adiabatic exponent y is heated at constant pressure and it absorbs Q heat. What fraction of this heat is used to perform external work? \mathrm{\gamma } \frac{1}{\mathrm{\gamma }} 1 - \frac{1}{\mathrm{\gamma }} \mathrm{\gamma } - 1 Temperature is defined by 3. Third Law of thermodynamics Subtopic: Basic Terms \| If 32 gm of {O}_{2} 27°\mathrm{C} is mixed with 64 gm of {\mathrm{O}}_{2} 327°\mathrm{C} in an adiabatic vessel, then the final temperature of the mixture will be : 200°\mathrm{C} 227°\mathrm{C} 314.5°\mathrm{C} 235.5°\mathrm{C} Subtopic: Molar Specific Heat \| {W}_{1} is the work done in compressing an ideal gas from a given initial state through a certain volume isothermally and {W}_{2} is the work done in compressing the same gas from the same initial state through the same volume adiabatically, then: {W}_{1}={W}_{2} {W}_{1}<{W}_{2} {W}_{1}>{W}_{2} {W}_{1}=2{W}_{2}
Acoustic Scene Recognition Using Late Fusion - MATLAB & Simulink - MathWorks España \begin{array}{l}\underset{}{\overset{\sim }{x}}=\lambda {x}_{i}+\left(1-\lambda \right){x}_{j}\\ \underset{}{\overset{\sim }{y}}=\lambda {y}_{i}+\left(1-\lambda \right){y}_{j}\end{array} Define the CNN architecture. This architecture is based on [1] and modified through trial and error. See List of Deep Learning Layers to learn more about deep learning layers available in MATLAB®. Define trainingOptions for the CNN. These options are based on [3] and modified through a systematic hyperparameter optimization workflow. Call trainNetwork to train the network. Call predict to predict responses from the trained network using the held-out test set. Call confusionchart to visualize the accuracy on the test set. For each 10-second audio clip, call predict to return the labels and the weights, then map it to the corresponding predicted location. Call confusionchart to visualize the accuracy on the test set. trainNetwork \| trainingOptions \| classify \| layerGraph \| batchNormalizationLayer \| convolution2dLayer
Performance Measurements of Solid-Oxide Electrolysis Cells for Hydrogen Production \| J. Electrochem. En. Conv. Stor \| ASME Digital Collection C. M. Stoots, J. S. Herring, P. A. Lessing, P. A. Lessing J. J. Hartvigsen, J. J. Hartvigsen O’Brien, J. E., Stoots, C. M., Herring, J. S., Lessing, P. A., Hartvigsen, J. J., and Elangovan, S. (February 1, 2005). "Performance Measurements of Solid-Oxide Electrolysis Cells for Hydrogen Production." ASME. J. Fuel Cell Sci. Technol. August 2005; 2(3): 156–163. https://doi.org/10.1115/1.1895946 An experimental study has been completed to assess the performance of single solid-oxide electrolysis cells operating over a temperature range of 800 to 900°C. The experiments were performed over a range of steam inlet partial pressures (2.3–12.2 kPa), carrier gas flow rates (50–200 sccm), and current densities (−0.75–0.25A∕cm2) using single electrolyte-supported button cells of scandia-stabilized zirconia. Steam consumption rates associated with electrolysis were measured directly using inlet and outlet dew-point instrumentation. Cell operating potentials and cell current were varied using a programmable power supply and monitored continuously. Values of area-specific resistance and thermal efficiency are presented as a function of current density. Cell performance is shown to be continuous from the fuel-cell mode to the electrolysis mode of operation. The effects of steam starvation and thermal cycling on cell performance parameters are discussed. solid oxide fuel cells, hydrogen economy, current density, solid electrolytes, stability, scandium, zirconium, steam Current density, Electrolysis, Hydrogen, Hydrogen production, Solid oxide fuel cells, Steam, Temperature, Fuel cells, Flow (Dynamics), Gas flow (IAEA), 1999, “ ,” IAEA-TECDOC-1085. EG&G Technical Services, Science Applications International Corp. , 6th Edition, DOE∕NETL 2002∕1179. Maskalick High Temperature Electrolysis Cell Performance Characterization Hydrogen Production by High-Temperature Electrolysis of Steam Proc. of Technical Committee Meeting, International Atomic Energy Agency, Oarai, Japan, Oct. 19–20 , Trieste, IAEA-TECDOC-761, pp. ,” JAERI-Research 97-064, Japan Atomic Energy Research Institute. Hatagishi Power Generation and Steam Electrolysis Characteristics of an Electrochemical Cell With a Zirconia- or Ceria-Based Electrolyte
Electrical energy — lesson. Science State Board, Class 10. Electricity is used in both households and industries. The amount of electricity consumed is determined by two factors. They are The amount of electricity consumed The duration of usage The product of electric power and its usage time is used to calculate the electrical energy consumed. If an electric appliance of power $200\ watt$ is utilised for $3\ hours$, then the total electric power consumption is Electrical energy consumption is quantified and stated in watt-hours, even though the SI unit is watt-second. But in reality, a larger electrical energy unit is required. The higher units of electrical energy are watt-hour ($W\ h$) or kilowatt-hour ($kW\ h$). In terms of joules, 1\phantom{\rule{0.147em}{0ex}}\mathit{watt}\phantom{\rule{0.147em}{0ex}}\mathit{hour}\phantom{\rule{0.147em}{0ex}}=3600\phantom{\rule{0.147em}{0ex}}J Kilowatt is the larger unit of power that is most commonly used in the devices. In terms of megajoules, 1\phantom{\rule{0.147em}{0ex}}\mathit{kW}\phantom{\rule{0.147em}{0ex}}h\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}3.6\phantom{\rule{0.147em}{0ex}}M\phantom{\rule{0.147em}{0ex}}J The energy consumed for domestic purposes is measured using the electric meter in the units of a kilowatt-hour. Another unit called horse power ($hp$) is also used to express electric power. It is one of the units in the foot-pound second (FPS) system or English system. Two bulbs are have the ratings as $60\ W$, $220\ V$ and $40\ W$, $220\ V$ respectively. Which one has a greater resistance? Electric power, P\phantom{\rule{0.147em}{0ex}}=\frac{{V}^{2}}{R} For the same value of $V$, $R$ is inversely proportional to $P$. Therefore, the lesser the power, the greater the resistance. Hence, the bulb with $40\ W$, $220\ V$ rating has greater resistance.
Electronics \| Special Issue : Machine Learning for Wireless Networks - Recent Advances and Future Trends Machine Learning for Wireless Networks - Recent Advances and Future... Special Issue "Machine Learning for Wireless Networks - Recent Advances and Future Trends" Dr. Shankar Kathiresan Department of Computer Applications, Alagappa University, Karaikudi 630 002, India Interests: healthcare applications; secret image sharing scheme; digital image security; cryptography; internet of things; optimization algorithms Department of Computer Science and Engineering, Maharaja Agrasen institute of Technology (GGSIPU), Delhi 110086, India Interests: software engineering; software usability; human computer interaction; algorithm computing; soft computing; neural networks; testing Special Issue in Sensors: Software-Defined Networking for Sensor Networks and Internet of Things Special Issue in Sensors: Sensor Networks and IoT for E-health Applications Special Issue in Sustainability: Advanced Application of Sustainable Transportation: Intelligent and Autonomous Traffic Monitoring, Control and Management Systems for Smart Cities Intelligent Computing and Communication Lab, Sejong University, Sejong 05006, Korea Interests: sensor localization; image sensors; MAC and routing protocols for wireless sensor networks; cognitive radio Wireless Sensor Networks; RFID system; IoT; smart city; deep learning and digital convergence Special Issue in Sensors: Industry 4.0: From Future of IoT to Industrial IoT Interests: healthcare systems; augmented reality; big data; deep learning; internet of things; data mining Special Issue in Symmetry: Applications of Internet of Things Our society is experiencing a digitization revolution, with a drastic growth of Internet users and connected devices. Next-generation wireless networks should provide ultra-reliable, low-latency communication and intelligently control the internet of things (IoT) devices in real-time scenarios. Wireless network applications like in real-time traffic data, sensor reading from driverless cars, or Netflix entertainment recommendations generate extreme volumes of data that must be collected and processed in real time. These communication requirements and core intelligence can only be achieved through the integration of machine learning techniques in the wireless infrastructure and end-user devices. In recent times, machine learning algorithms have gained significant interest in the area of wireless networking and communication. Machine learning-driven algorithms and models can enable wireless network analysis and resource management and be of advantage in handling the development in volume of communication and computation for evolving networking applications. Nevertheless, the application of machine learning techniques for heterogeneous wireless networks is still under debate. More endeavors are needed to link the gap between machine learning and wireless networking research. The objective of this Special Issue is to explore recent advancements in machine learning concepts to address practical challenges in wireless networks. This Special Issue will bring together researchers and academics to present new results in network modeling and architecture, networking applications, security and privacy, resource management, load balancing, and various challenges related to the design for future wireless networks with the help of machine learning. This Special Issue “Machine Learning for Wireless Networks – Recent Advances and Future Trends” will solicit papers on various disciplines, including but not limited to the following: machine learning algorithms for network scheduling and control; machine learning based energy-efficient networking techniques; machine learning-based network resource allocation and optimization in wireless networks; new supervised machine learning methods for wireless networks; new unsupervised machine learning methods for wireless networks; novel reinforcement learning methods for wireless networks; new optimization methods for machine learning for wireless networks; machine learning-based innovative intelligent computing architecture/algorithms for wireless networks; machine learning based big data analytic frameworks for networking data; machine learning-based intelligent routing algorithms for traffic management in wireless networks; machine learning-based resource allocation for shared/virtualized networks using machine learning; machine learning-based quality of service (QoS) management in wireless networks; nature-inspired algorithms for wireless networks; machine learning-based blockchain for wireless networks; machine learning-based node localization in wireless networks Ahmad Gendia Peak-to-average power ratio (PAPR) reduction in multiplexed signals in orthogonal frequency division multiplexing (OFDM) systems has been a long-standing critical issue. Clipping and filtering (CF) techniques offer good performance in terms of PAPR reduction at the expense of a relatively high computational cost [...] Read more. Peak-to-average power ratio (PAPR) reduction in multiplexed signals in orthogonal frequency division multiplexing (OFDM) systems has been a long-standing critical issue. Clipping and filtering (CF) techniques offer good performance in terms of PAPR reduction at the expense of a relatively high computational cost that is inherent in the repeated application of fast Fourier transform (FFT) operations. The ever-increasing demand for low-latency operation calls for the development of low-complexity novel solutions to the PAPR problem. To address this issue while providing an enhanced PAPR reduction performance, we propose a synchronous neural network (NN)-based solution to achieve PAPR reduction performance exceeding the limits of conventional CF schemes with lower computational complexity. The proposed scheme trains a neural network module using hybrid collections of samples from multiple OFDM symbols to arrive at a signal mapping with desirable characteristics. The benchmark NN-based approach provides a comparable performance to conventional CF. However, it can underfit or overfit due to its asynchronous nature which leads to increased out-of-band (OoB) radiations, and deteriorating bit error rate (BER) performance for high-order modulations. Simulations’ results demonstrate the effectiveness of the proposed scheme in terms of the achieved cubic metric (CM), BER, and OoB emissions. Full article Wireless vehicular communications are a promising technology. Most applications related to vehicular communications aim to improve road safety and have special requirements concerning latency and reliability. The traditional channel estimation techniques used in the IEEE 802.11 standard do not properly perform over vehicular [...] Read more. Wireless vehicular communications are a promising technology. Most applications related to vehicular communications aim to improve road safety and have special requirements concerning latency and reliability. The traditional channel estimation techniques used in the IEEE 802.11 standard do not properly perform over vehicular channels. This is because vehicular communications are subject to non-stationary, time-varying, frequency-selective wireless channels. Therefore, the main goal of this work is the introduction of a new channel estimation and equalization technique based on a Semi-supervised Extreme Learning Machine (SS-ELM) in order to address the harsh characteristics of the vehicular channel and improve the performance of the communication link. The performance of the proposed technique is compared with traditional estimators, as well as state-of-the-art machine-learning-based algorithms over an urban scenario setup in terms of bit error rate. The proposed SS-ELM scheme outperformed the extreme learning machine and the fully complex extreme learning machine algorithms for the evaluated scenarios. Compared to traditional techniques, the proposed SS-ELM scheme has a very similar performance. It is also observed that, although the SS-ELM scheme requires the largest operation time among the evaluated techniques, its execution time is still far away from the latency requirements specified by the standard for safety applications. Full article Srihari Kannan In this paper, Deep Neural Networks (DNN) with Bat Algorithms (BA) offer a dynamic form of traffic control in Vehicular Adhoc Networks (VANETs). The former is used to route vehicles across highly congested paths to enhance efficiency, with a lower average latency. The [...] Read more. In this paper, Deep Neural Networks (DNN) with Bat Algorithms (BA) offer a dynamic form of traffic control in Vehicular Adhoc Networks (VANETs). The former is used to route vehicles across highly congested paths to enhance efficiency, with a lower average latency. The latter is combined with the Internet of Things (IoT) and it moves across the VANETs to analyze the traffic congestion status between the network nodes. The experimental analysis tests the effectiveness of DNN-IoT-BA in various machine or deep learning algorithms in VANETs. DNN-IoT-BA is validated through various network metrics, like packet delivery ratio, latency and packet error rate. The simulation results show that the proposed method provides lower energy consumption and latency than conventional methods to support real-time traffic conditions. Full article Analysis of Interconnected Arrivals on Queueing-Inventory System with Two Multi-Server Service Channels and One Retrial Facility T. Harikrishnan Present-day queuing inventory systems (QIS) do not utilize two multi-server service channels. We proposed two multi-server service channels referred to as T1S (Type 1 n-identical multi-server) and T2S (Type 2 m-identical multi-server). It includes an optional interconnected service connection [...] Read more. T1S T2S (Type 2 m-identical multi-server). It includes an optional interconnected service connection between T1S T2S , which has a finite queue of size N. An arriving customer either uses the inventory (basic service or main service) for their demand, whom we call T1 , or simply uses the service only, whom we call T2 . Customer T1 will utilize the server T1S , while customer T2 T2S T1 can also get the second optional service after completing their main service. If there is a free server with a positive inventory, there is a chance that T1 customers may go to an infinite orbit whenever they find that either all the servers are busy or no sufficient stock. The orbital customer can request for T1S service under the classical retrial policy. Q\left(=S-s\right) items are replaced into the inventory whenever it falls into the reorder level s such that the inequality always holds n<s . We use the standard \left(s,Q\right) ordering policy to replace items into the inventory. By varying S and s, we investigate to find the optimal cost value using stationary probability vector \varphi . We used the Neuts Matrix geometric approach to derive the stability condition and steady-state analysis with R-matrix to find \varphi . Then, we perform the waiting time analysis for both T1 T2 customers using Laplace transform technique. Further, we computed the necessary system characteristics and presented sufficient numerical results. Full article Using Ultrasonic Sensors and a Knowledge-Based Neural Fuzzy Controller for Mobile Robot Navigation Control Shiou-Yun Jeng Hsueh-Yi Lin This study proposes a knowledge-based neural fuzzy controller (KNFC) for mobile robot navigation control. An effective knowledge-based cultural multi-strategy differential evolution (KCMDE) is used for adjusting the parameters of KNFC. The KNFC is applied in PIONEER 3-DX mobile robots to achieve automatic navigation [...] Read more. This study proposes a knowledge-based neural fuzzy controller (KNFC) for mobile robot navigation control. An effective knowledge-based cultural multi-strategy differential evolution (KCMDE) is used for adjusting the parameters of KNFC. The KNFC is applied in PIONEER 3-DX mobile robots to achieve automatic navigation and obstacle avoidance capabilities. A novel escape approach is proposed to enable robots to autonomously avoid special environments. The angle between the obstacle and robot is used and two thresholds are set to determine whether the robot entries into the special landmarks and to modify the robot behavior for avoiding dead ends. The experimental results show that the proposed KNFC based on the KCMDE algorithm has improved the learning ability and system performance by 15.59% and 79.01%, respectively, compared with the various differential evolution (DE) methods. Finally, the automatic navigation and obstacle avoidance capabilities of robots in unknown environments were verified for achieving the objective of mobile robot control. Full article
Ben Green (mathematician) - Wikipedia Ben Green (mathematician) Find sources: "Ben Green" mathematician – news · newspapers · books · scholar · JSTOR (January 2022) (Learn how and when to remove this template message) This article is about the mathematician. For the British World War II internee, see Ben Greene. For those of a similar name, see Benjamin Green (disambiguation). (BA, MMath, PhD) Topics in Arithmetic Combinatorics (2003) Ben Joseph Green FRS (born 27 February 1977) is a British mathematician, specialising in combinatorics and number theory. He is the Waynflete Professor of Pure Mathematics at the University of Oxford. Ben Green was born on 27 February 1977 in Bristol, England. He studied at local schools in Bristol, Bishop Road Primary School and Fairfield Grammar School, competing in the International Mathematical Olympiad in 1994 and 1995.[1] He entered Trinity College, Cambridge in 1995 and completed his BA in mathematics in 1998, winning the Senior Wrangler title. He stayed on for Part III and earned his doctorate under the supervision of Timothy Gowers, with a thesis entitled Topics in arithmetic combinatorics (2003). During his PhD he spent a year as a visiting student at Princeton University. He was a research Fellow at Trinity College, Cambridge between 2001 and 2005, before becoming a Professor of Mathematics at the University of Bristol from January 2005 to September 2006 and then the first Herchel Smith Professor of Pure Mathematics at the University of Cambridge from September 2006 to August 2013. He became the Waynflete Professor of Pure Mathematics at the University of Oxford on 1 August 2013. He was also a Research Fellow of the Clay Mathematics Institute and held various positions at institutes such as Princeton University, University of British Columbia, and Massachusetts Institute of Technology. The majority of Green's research is in the fields of analytic number theory and additive combinatorics, but he also has results in harmonic analysis and in group theory. His best known theorem, proved jointly with his frequent collaborator Terence Tao, states that there exist arbitrarily long arithmetic progressions in the prime numbers: this is now known as the Green–Tao theorem.[2] Amongst Green's early results in additive combinatorics are an improvement of a result of Jean Bourgain of the size of arithmetic progressions in sumsets,[3] as well as a proof of the Cameron–Erdős conjecture on sum-free sets of natural numbers.[4] He also proved an arithmetic regularity lemma[5] for functions defined on the first {\displaystyle N} natural numbers, somewhat analogous to the Szemerédi regularity lemma for graphs. From 2004–2010, in joint work with Terence Tao and Tamar Ziegler, he developed so-called higher order Fourier analysis. This theory relates Gowers norms with objects known as nilsequences. The theory derives its name from these nilsequences, which play an analogous role to the role that characters play in classical Fourier analysis. Green and Tao used higher order Fourier analysis to present a new method for counting the number of solutions to simultaneous equations in certain sets of integers, including in the primes.[6] This generalises the classical approach using Hardy–Littlewood circle method. Many aspects of this theory, including the quantitative aspects of the inverse theorem for the Gowers norms,[7] are still the subject of ongoing research. Green has also collaborated with Emmanuel Breuillard on topics in group theory. In particular, jointly with Terence Tao, they proved a structure theorem[8] for approximate groups, generalising the Freiman-Ruzsa theorem on sets of integers with small doubling. Green also has work, joint with Kevin Ford and Sean Eberhard, on the theory of the symmetric group, in particular on what proportion of its elements fix a set of size {\displaystyle k} Green and Tao also have a paper[10] on algebraic combinatorial geometry, resolving the Dirac-Motzkin conjecture (see Sylvester–Gallai theorem). In particular they prove that, given any collection of {\displaystyle n} points in the plane that are not all collinear, if {\displaystyle n} is large enough then there must exist at least {\displaystyle n/2} lines in the plane containing exactly two of the points. Kevin Ford, Ben Green, Sergei Konyagin, James Maynard and Terence Tao, initially in two separate research groups and then in combination, improved the lower bound for the size of the longest gap between two consecutive primes of size at most {\displaystyle X} .[11] The form of the previously best-known bound, essentially due to Rankin, had not been improved for 76 years. More recently Green has considered questions in arithmetic Ramsey theory. Together with Tom Sanders he proved that, if a sufficiently large finite field of prime order is coloured with a fixed number of colours, then the field has elements {\displaystyle x,y} {\displaystyle x,y,x{+}y,xy} all have the same colour.[12] Green has also been involved with the new developments of Croot-Lev-Pach-Ellenberg-Gijswijt on applying a polynomial method to bound the size of subsets of a finite vector space without solutions to linear equations. He adapted these methods to prove, in function fields, a strong version of Sárközy's theorem.[13] Green has been a Fellow of the Royal Society since 2010,[14] and a Fellow of the American Mathematical Society since 2012.[15] Green was chosen by the German Mathematical Society to deliver a Gauss Lectureship in 2013. He has received several awards: 2004: Clay Research Award 2005: Salem Prize 2005: Whitehead Prize[16] 2008: European Mathematical Society prize recipient 2014: Sylvester Medal, awarded by the Royal Society. ^ Ben Green's results at International Mathematical Olympiad ^ Green, Ben; Tao, Terence (2008). "The Primes Contain Arbitrarily Long Arithmetic Progressions". Annals of Mathematics. 167 (2): 481–547. arXiv:math/0404188. doi:10.4007/annals.2008.167.481. JSTOR 40345354. S2CID 1883951. ^ Green, B. (1 August 2002). "Arithmetic progressions in sumsets". Geometric & Functional Analysis GAFA. 12 (3): 584–597. doi:10.1007/s00039-002-8258-4. ISSN 1016-443X. S2CID 120755105. ^ GREEN, BEN (19 October 2004). "The Cameron–Erdos Conjecture". Bulletin of the London Mathematical Society. 36 (6): 769–778. arXiv:math/0304058. doi:10.1112/s0024609304003650. ISSN 0024-6093. S2CID 119615076. ^ Green, B. (1 April 2005). "A Szemerédi-type regularity lemma in abelian groups, with applications". Geometric & Functional Analysis GAFA. 15 (2): 340–376. arXiv:math/0310476. doi:10.1007/s00039-005-0509-8. ISSN 1016-443X. S2CID 17451915. ^ Green, Benjamin; Tao, Terence (2010). "Linear equations in primes". Annals of Mathematics. 171 (3): 1753–1850. doi:10.4007/annals.2010.171.1753. JSTOR 20752252. ^ Green, Ben; Tao, Terence; Ziegler, Tamar (2012). "An inverse theorem for the Gowers U s+1 [N]-norm". Annals of Mathematics. 176 (2): 1231–1372. doi:10.4007/annals.2012.176.2.11. JSTOR 23350588. ^ Breuillard, Emmanuel; Green, Ben; Tao, Terence (1 November 2012). "The structure of approximate groups". Publications Mathématiques de l'IHÉS. 116 (1): 115–221. arXiv:1110.5008. doi:10.1007/s10240-012-0043-9. ISSN 0073-8301. S2CID 119603959. ^ Eberhard, Sean; Ford, Kevin; Green, Ben (23 December 2015). "Permutations Fixing a k-set". International Mathematics Research Notices. 2016 (21): 6713–6731. arXiv:1507.04465. Bibcode:2015arXiv150704465E. doi:10.1093/imrn/rnv371. ISSN 1073-7928. S2CID 15188628. ^ Green, Ben; Tao, Terence (1 September 2013). "On Sets Defining Few Ordinary Lines". Discrete & Computational Geometry. 50 (2): 409–468. arXiv:1208.4714. doi:10.1007/s00454-013-9518-9. ISSN 0179-5376. S2CID 15813230. ^ Ford, Kevin; Green, Ben; Konyagin, Sergei; Maynard, James; Tao, Terence (16 December 2014). "Long gaps between primes". arXiv:1412.5029 [math.NT]. ^ Green, Ben; Sanders, Tom (1 March 2016). "Monochromatic sums and products". Discrete Analysis. 5202016 (1). arXiv:1510.08733. doi:10.19086/da.613. ISSN 2397-3129. S2CID 119140038. ^ Green, Ben (23 November 2016). "Sárközy's Theorem in Function Fields". The Quarterly Journal of Mathematics. 68 (1): 237–242. arXiv:1605.07263. doi:10.1093/qmath/haw044. ISSN 0033-5606. S2CID 119150134. ^ List of Fellows of the American Mathematical Society. Retrieved 19 January 2013. ^ "List of LMS prize winners – London Mathematical Society". Ben Green personal homepage at Oxford Ben Green faculty page at Oxford Ben Green Homepage at Trinity College, Cambridge Clay Research Award 2004 announcement Ben Green at the Mathematics Genealogy Project math.NT/0404188 – Preprint on arbitrarily long arithmetic progressions on primes Retrieved from "https://en.wikipedia.org/w/index.php?title=Ben_Green_(mathematician)&oldid=1089349808" Scientists from Bristol Waynflete Professors of Pure Mathematics
Hydroxymethylglutaryl-CoA synthase - Wikipedia 3-hydroxy-3-methylglutaryl-Coenzyme A synthase 1 (soluble) 3-hydroxy-3-methylglutaryl-Coenzyme A synthase 2 (mitochondrial) Hydroxymethylglutaryl-coenzyme A synthase N terminal staphylococcus aureus 3-hydroxy-3-methylglutaryl-coa synthase HMG_CoA_synt_N Hydroxymethylglutaryl-coenzyme A synthase C terminal HMG_CoA_synt_C In molecular biology, hydroxymethylglutaryl-CoA synthase or HMG-CoA synthase EC 2.3.3.10 is an enzyme which catalyzes the reaction in which acetyl-CoA condenses with acetoacetyl-CoA to form 3-hydroxy-3-methylglutaryl-CoA (HMG-CoA). This reaction comprises the second step in the mevalonate-dependent isoprenoid biosynthesis pathway. HMG-CoA is an intermediate in both cholesterol synthesis and ketogenesis. This reaction is overactivated in patients with diabetes mellitus type 1 if left untreated, due to prolonged insulin deficiency and the exhaustion of substrates for gluconeogenesis and the TCA cycle, notably oxaloacetate. This results in shunting of excess acetyl-CoA into the ketone synthesis pathway via HMG-CoA, leading to the development of diabetic ketoacidosis. acetyl-CoA + H2O + acetoacetyl-CoA {\displaystyle \rightleftharpoons } (S)-3-hydroxy-3-methylglutaryl-CoA + CoA The 3 substrates of this enzyme are acetyl-CoA, H2O, and acetoacetyl-CoA, whereas its two products are (S)-3-hydroxy-3-methylglutaryl-CoA and CoA. In humans, the protein is encoded by the HMGCS1 gene on chromosome 5. 5.1.1 Cytosolic 5.1.2 Mitochondrial This enzyme belongs to the family of transferases, specifically those acyltransferases that convert acyl groups into alkyl groups on transfer. The systematic name of this enzyme class is acetyl-CoA:acetoacetyl-CoA C-acetyltransferase (thioester-hydrolysing, carboxymethyl-forming). Other names in common use include (S)-3-hydroxy-3-methylglutaryl-CoA acetoacetyl-CoA-lyase, (CoA-acetylating), 3-hydroxy-3-methylglutaryl CoA synthetase, 3-hydroxy-3-methylglutaryl coenzyme A synthase, 3-hydroxy-3-methylglutaryl coenzyme A synthetase, 3-hydroxy-3-methylglutaryl-CoA synthase, 3-hydroxy-3-methylglutaryl-coenzyme A synthase, beta-hydroxy-beta-methylglutaryl-CoA synthase, HMG-CoA synthase, acetoacetyl coenzyme A transacetase, hydroxymethylglutaryl coenzyme A synthase, and hydroxymethylglutaryl coenzyme A-condensing enzyme. HMG-CoA synthase contains an important catalytic cysteine residue that acts as a nucleophile in the first step of the reaction: the acetylation of the enzyme by acetyl-CoA (its first substrate) to produce an acetyl-enzyme thioester, releasing the reduced coenzyme A. The subsequent nucleophilic attack on acetoacetyl-CoA (its second substrate) leads to the formation of HMG-CoA.[1] This enzyme participates in 3 metabolic pathways: synthesis and degradation of ketone bodies, valine, leucine and isoleucine degradation, and butanoate metabolism. HMG-CoA synthase occurs in eukaryotes, archaea, and certain bacteria.[2] Eukaryotes[edit] In vertebrates, there are two different isozymes of the enzyme (cytosolic and mitochondrial); in humans the cytosolic form has only 60.6% amino acid identity with the mitochondrial form of the enzyme. HMG-CoA is also found in other eukaryotes such as insects, plants, and fungi.[3] Cytosolic[edit] The cytosolic form is the starting point of the mevalonate pathway, which leads to cholesterol and other sterolic and isoprenoid compounds. Mitochondrial[edit] The mitochondrial form is responsible for the biosynthesis of ketone bodies. The gene for the mitochondrial form of the enzyme has three sterol regulatory elements in the 5' flanking region.[4] These elements are responsible for decreased transcription of the message responsible for enzyme synthesis when dietary cholesterol is high in animals: the same is observed for 3-hydroxy-3-methylglutaryl-CoA and the low density lipoprotein receptor. In bacteria, isoprenoid precursors are generally synthesised via an alternative, non-mevalonate pathway, however a number of Gram-positive pathogens utilise a mevalonate pathway involving HMG-CoA synthase that is parallel to that found in eukaryotes.[5][6] As of late 2007, 4 structures have been solved for this class of enzymes, with PDB accession codes 1XPK, 1XPL, 1XPM, and 2P8U. HMG-CoA+synthase at the US National Library of Medicine Medical Subject Headings (MeSH) ^ Theisen MJ, Misra I, Saadat D, Campobasso N, Miziorko HM, Harrison DH (November 2004). "3-hydroxy-3-methylglutaryl-CoA synthase intermediate complex observed in "real-time"". Proc. Natl. Acad. Sci. U.S.A. 101 (47): 16442–7. doi:10.1073/pnas.0405809101. PMC 534525. PMID 15498869. ^ Bahnson BJ (November 2004). "An atomic-resolution mechanism of 3-hydroxy-3-methylglutaryl-CoA synthase". Proc. Natl. Acad. Sci. U.S.A. 101 (47): 16399–400. Bibcode:2004PNAS..10116399B. doi:10.1073/pnas.0407418101. PMC 534547. PMID 15546978. ^ Bearfield JC, Keeling CI, Young S, Blomquist GJ, Tittiger C (April 2006). "Isolation, endocrine regulation and mRNA distribution of the 3-hydroxy-3-methylglutaryl coenzyme A synthase (HMG-S) gene from the pine engraver, Ips pini (Coleoptera: Scolytidae)". Insect Molecular Biology. 15 (2): 187–95. doi:10.1111/j.1365-2583.2006.00627.x. PMID 16640729. S2CID 46317830. ^ Goldstein J.L., Brown M.S. (1990) Regulation of the mevalonate pathway. Nature 343, 425-430 ^ Steussy CN, Robison AD, Tetrick AM, Knight JT, Rodwell VW, Stauffacher CV, Sutherlin AL (December 2006). "A structural limitation on enzyme activity: the case of HMG-CoA synthase". Biochemistry. 45 (48): 14407–14. doi:10.1021/bi061505q. PMID 17128980. ^ Steussy CN, Vartia AA, Burgner JW, Sutherlin A, Rodwell VW, Stauffacher CV (November 2005). "X-ray crystal structures of HMG-CoA synthase from Enterococcus faecalis and a complex with its second substrate/inhibitor acetoacetyl-CoA". Biochemistry. 44 (43): 14256–67. doi:10.1021/bi051487x. PMID 16245942. RUDNEY H (1957). "The biosynthesis of beta-hydroxy-beta-methylglutaric acid". J. Biol. Chem. 227 (1): 363–77. doi:10.1016/S0021-9258(18)70822-3. PMID 13449080. Retrieved from "https://en.wikipedia.org/w/index.php?title=Hydroxymethylglutaryl-CoA_synthase&oldid=1046342638"
Kinetic Theory of Gases Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Physics - Kinetic Theory of Gases An isolated system- 1. is a specified region where transfers of energy and mass take place 2. is a region of constant mass and only energy is allowed through the closed boundaries 3. is one in which mass within the system is not necessarily constant. 4. cannot transfer either energy or mass to or from the surroundings. Subtopic: Ideal Gas Equation \| The pressure and temperature of two different gases are P and T having the volume V for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be- [Pb. PMT 1997, 98; DPMT 1999; MH CET 2003] If the mean free path of atoms is doubled then the pressure of the gas will become: Subtopic: Mean Free Path \| A diatomic molecule has how many degrees of freedom- Subtopic: Law of Equipartition of Energy \| A thermally insulated piston divides a container into two compartments. Volume, temperature, and pressure in the right compartment are 2V, T, and 2P, while in the left compartment the respective values are V, T, and P. If the piston can slide freely, then in the final equilibrium position, the volume of the right-hand compartment will be- \frac{3V}{5} \frac{9V}{4} \frac{12V}{5} The graph which represents the variation of mean kinetic energy of molecules with temperature t°C Subtopic: Kinetic Energy of an Ideal Gas \| {V}_{rms},{V}_{av},{V}_{mp} are root mean square, average and most probable speeds of molecules of a gas obeying Maxwellian velocity distribution. Which of the following statements is correct? {V}_{rms}<{V}_{av}<{V}_{mp} {V}_{rms}>{V}_{av}>{V}_{mp} {V}_{rms} < {V}_{av}>{V}_{mp} {V}_{avg}>{V}_{rms}<{V}_{mp} Subtopic: Types of Velocities \| Gas is found to obey the law {P}^{2}V = constant. The initial temperature and volume are {T}_{0} {V}_{0} . If the gas expands to a volume 3{V}_{0} , its final temperature becomes: \frac{{T}_{0}}{3} \frac{{T}_{0}}{\sqrt{3}} 3{T}_{0} For a gas, the r.m.s speed at 800 K is: 1. Four times the value at 200 K 2. Half the value at 200 K 3. Twice the value at 200 K 4. Same as at 200 K The average kinetic energy of a helium atom at 30°C is: [MP PMT 2004] 1. Less than 1 eV 2. A few KeV 3. 50-60 eV 4. 13.6 eV
Solve Constrained Nonlinear Optimization, Problem-Based - MATLAB & Simulink - MathWorks United Kingdom Convert Objective Function Using fcn2optimexpr This example shows how to find the minimum of a nonlinear objective function with a nonlinear constraint by using the problem-based approach. For a video showing the solution to a similar problem, see Problem-Based Nonlinear Programming. To find the minimum value of a nonlinear objective function using the problem-based approach, first write the objective function as a file or anonymous function. The objective function for this example is f\left(x,y\right)={e}^{x}\left(4{x}^{2}+2{y}^{2}+4xy+2y-1\right). Create the optimization problem variables x and y. Create the objective function as an expression in the optimization variables. Create an optimization problem with obj as the objective function. Create a nonlinear constraint that the solution lies in a tilted ellipse, specified as \frac{xy}{2}+\left(x+2{\right)}^{2}+\frac{\left(y-2{\right)}^{2}}{2}\le 2. Create the constraint as an inequality expression in the optimization variables. Include the constraint in the problem. Create a structure representing the initial point as x = –3, y = 3. Try a different start point. Plot the ellipse, the objective function contours, and the two solutions. The solutions are on the nonlinear constraint boundary. The contour plot shows that these are the only local minima. The plot also shows that there is a stationary point near [–2,3/2], and local maxima near [–2,0] and [–1,4]. For some objective functions or software versions, you must convert nonlinear functions to optimization expressions by using fcn2optimexpr. See Supported Operations for Optimization Variables and Expressions and Convert Nonlinear Function to Optimization Expression. Pass the x and y variables in the fcn2optimexpr call to indicate which optimization variable corresponds to each objfunx input. Create an optimization problem with obj as the objective function just as before. The remainder of the solution process is identical.
EUDML \| Algebraic linking numbers of knots in 3-manifolds. EuDML \| Algebraic linking numbers of knots in 3-manifolds. Algebraic linking numbers of knots in 3-manifolds. Schneiderman, Rob. "Algebraic linking numbers of knots in 3-manifolds.." Algebraic & Geometric Topology 3 (2003): 921-968. <http://eudml.org/doc/123621>. @article{Schneiderman2003, author = {Schneiderman, Rob}, keywords = {concordance invariant; knots; linking number; 3-manifold}, title = {Algebraic linking numbers of knots in 3-manifolds.}, AU - Schneiderman, Rob TI - Algebraic linking numbers of knots in 3-manifolds. KW - concordance invariant; knots; linking number; 3-manifold concordance invariant, knots, linking number, 3-manifold 3 {S}^{3} Articles by Schneiderman
Box-Cox transformation - MATLAB boxcox - MathWorks India Transform a Data Series Contained in a Financial Times Series Object transfts Using a fints object for the tsobj argument of boxcox is not recommended. Use fts2timetable to convert a fints object to a timetable object and then use timetable2table and table2array. [transdat,lambda] = boxcox(data) [transfts,lambda] = boxcox(tsobj) transdat = boxcox(lambda,data) transdat = boxcox(lambda,tsobj) [transdat,lambda] = boxcox(data) transforms the data vector data using the Box-Cox transformation method into transdat. It also estimates the transformation parameter λ. [transfts,lambda] = boxcox(tsobj) transforms the financial time series object tsobj using the Box-Cox transformation method into transfts. It also estimates the transformation parameter λ. transdat = boxcox(lambda,data) transform the data using a certain specified λ for the Box-Cox transformation. This syntax does not find the optimum λ that maximizes the LLF. transdat = boxcox(lambda,tsobj) transform the tsobj using a certain specified λ for the Box-Cox transformation. This syntax does not find the optimum λ that maximizes the LLF. Use boxcox to transform the data series contained in a financial time series object into another set of data series with relatively normal distributions. Create a financial time series object from the supplied whirlpool.dat data file. whrl = ascii2fts('whirlpool.dat', 1, 2, []); Fill any missing values denoted with NaN's in whrl with values calculated using the linear method. f_whrl = fillts(whrl); Transform the nonnormally distributed filled data series f_whrl into a normally distributed one using Box-Cox transformation. bc_whrl = boxcox(f_whrl); Compare the result of the Close data series with a normal (Gaussian) probability distribution function and the nonnormally distributed f_whrl. The bar chart on the top represents the probability distribution function of the filled data series, f_whrl, which is the original data series whrl with the missing values interpolated using the linear method. The distribution is skewed toward the left (not normally distributed). The bar chart on the bottom is less skewed to the left. If you plot a Gaussian probability distribution function (PDF) with similar mean and standard deviation, the distribution of the transformed data is very close to normal (Gaussian). When you examine the contents of the resulting object bc_whrl, you find an identical object to the original object whrl but the contents are the transformed data series. positive column vector Data, specified as a positive column vector. numeric \| structure Lambda, specified as a scalar numeric or structure. If the input data is a vector, lambda is a scalar. If the input is a financial time series object (tsobj), lambda is a structure with fields similar to the components of the object. For example, if tsobj contains series names Open and Close, lambda has fields lambda.Open and lambda.Close. transdat — Data Box-Cox transformation Data Box-Cox transformation, returned as a vector. transfts — Financial time series Box-Cox transformation Financial time series Box-Cox transformation, returned as a vector. lambda — Lambda transformation parameter Lambda transformation parameter, returned as a numeric. boxcox transforms nonnormally distributed data to a set of data that has approximately normal distribution. The Box-Cox transformation is a family of power transformations. If λ is not = 0, then data\left(\lambda \right)=\frac{dat{a}^{\lambda }-1}{\lambda } If λ is = 0, then data\left(\lambda \right)=\mathrm{log}\left(data\right) The logarithm is the natural logarithm (log base e). The algorithm calls for finding the λ value that maximizes the Log-Likelihood Function (LLF). The search is conducted using fminsearch.
Comments on tag 0280—Kerodon Subsection 5.1.6: Equivalence of Inner Fibrations (cite) Comment #1141 by Daniel Gratzer on December 25, 2021 at 12:20 In the first sentence of the proof of Prop 5.1.6.5 F should be assumed to be an equivalence of inner fibrations, not a categorical equivalence of simplicial sets. Comment #1143 by Kerodon on December 27, 2021 at 17:57 In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0280. The letter 'O' is never used.
Uracil dehydrogenase - Wikipedia Uracil dehydrogenase (EC 1.1.99.19, uracil oxidase) is an enzyme with systematic name uracil:(acceptor) oxidoreductase.[1] This enzyme catalyses the following chemical reaction uracil + acceptor {\displaystyle \rightleftharpoons } barbiturate + reduced acceptor Also oxidizes thymine. The enzyme acts on the hydrated derivative of the substrate. ^ Hayaishi O, Kornberg A (May 1952). "Metabolism of cytosine, thymine, uracil, and barbituric acid by bacterial enzymes". The Journal of Biological Chemistry. 197 (2): 717–32. PMID 12981104. Uracil+dehydrogenase at the US National Library of Medicine Medical Subject Headings (MeSH) Retrieved from "https://en.wikipedia.org/w/index.php?title=Uracil_dehydrogenase&oldid=918620501"
IsTournament - Maple Help Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : IsTournament IsTournament(G) IsTournament returns true if the input graph is a tournament. It returns false otherwise. A tournament is a directed graph G that satisfies the following property: for every pair of vertices u and v in G exactly one of the directed edges [u,v] [v,u] is in G. \mathrm{with}⁡\left(\mathrm{GraphTheory}\right): \mathrm{T1}≔\mathrm{Digraph}⁡\left({[1,2],[2,3],[3,1]}\right) \textcolor[rgb]{0,0,1}{\mathrm{T1}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: a directed unweighted graph with 3 vertices and 3 arc\left(s\right)}} \mathrm{IsTournament}⁡\left(\mathrm{T1}\right) \textcolor[rgb]{0,0,1}{\mathrm{true}} \mathrm{T2}≔\mathrm{Digraph}⁡\left({[1,2],[1,3],[2,3],[3,1]}\right) \textcolor[rgb]{0,0,1}{\mathrm{T2}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 2: a directed unweighted graph with 3 vertices and 4 arc\left(s\right)}} \mathrm{IsTournament}⁡\left(\mathrm{T2}\right) \textcolor[rgb]{0,0,1}{\mathrm{false}} \mathrm{T3}≔\mathrm{Digraph}⁡\left({[1,2],[2,3]}\right) \textcolor[rgb]{0,0,1}{\mathrm{T3}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 3: a directed unweighted graph with 3 vertices and 2 arc\left(s\right)}} \mathrm{IsTournament}⁡\left(\mathrm{T3}\right) \textcolor[rgb]{0,0,1}{\mathrm{false}} RandomGraphs[RandomTournament]
Represent Tustin pilot model - Simulink - MathWorks í•œêµ The Tustin Pilot Model block represents the pilot model that A. Tustin describes in The Nature of the Operatorâ€™s Response in Manual Control, and its Implications for Controller Design [1]. When modeling human pilot models, use this block for the least accuracy, compared to that provided by the Crossover Pilot Model and Precision Pilot Model blocks. This block requires less input than those blocks, and provides better performance. However, the results might be less accurate. \frac{u\left(s\right)}{e\left(s\right)}=\frac{{K}_{p}\left(1+Ts\right)}{s}{e}^{â\mathrm{Ï}s}. Ï„ Transport delay time caused by the pilot neuromuscular system. [1] Tustin, A., The Nature of the Operatorâ€™s Response in Manual Control, and its Implications for Controller Design. Convention on Automatic Regulators and Servo Mechanisms. May, 1947.
BERT large model (uncased) Pretrained model on English language using a masked language modeling (MLM) objective. It was introduced in this paper and first released in this repository. This model is uncased: it does not make a difference between english and English. 1024 hidden dimension {'sequence': "[CLS] hello i'm a role model. [SEP]", {'sequence': "[CLS] hello i'm a fitness model. [SEP]", 'token_str': 'fitness'}] model = BertModel.from_pretrained("bert-large-uncased") model = TFBertModel.from_pretrained("bert-large-uncased") [{'sequence': '[CLS] the man worked as a bartender. [SEP]', 'token_str': 'bartender'}, {'sequence': '[CLS] the man worked as a mechanic. [SEP]', {'sequence': '[CLS] the man worked as a lawyer. [SEP]', {'sequence': '[CLS] the man worked as a carpenter. [SEP]', 'token_str': 'carpenter'}] [{'sequence': '[CLS] the woman worked as a waitress. [SEP]', {'sequence': '[CLS] the woman worked as a nurse. [SEP]', {'sequence': '[CLS] the woman worked as a bartender. [SEP]', {'sequence': '[CLS] the woman worked as a secretary. [SEP]', 'token_str': 'secretary'}] {\beta }_{1}=0.9\beta_\left\{1\right\} = 0.9 {\beta }_{2}=0.999\beta_\left\{2\right\} = 0.999 Datasets used to train bert-large-uncased Spaces using bert-large-uncased merve/fill-in-the-blank merve/uncertainty-calibration merve/data-leak merve/anonymization merve/measuring-fairness merve/dataset-worldviews merve/hidden-bias merve/measuring-diversity merve/private-and-fair
Complex conjugate - Wikipedia In mathematics, the complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign. That is, (if {\displaystyle a}nd {\displaystyle b} are real, then) the complex conjugate of {\displaystyle a+bi} {\displaystyle a-bi.} {\displaystyle z} {\displaystyle {\overline {z}}.} Geometric representation (Argand diagram) of {\displaystyle z} {\displaystyle {\overline {z}}} in the complex plane. The complex conjugate is found by reflecting {\displaystyle z} across the real axis. In polar form, the conjugate of {\displaystyle re^{i\varphi }} {\displaystyle re^{-i\varphi }.} This can be shown using Euler's formula. The product of a complex number and its conjugate is a real number: {\displaystyle a^{2}+b^{2}} {\displaystyle r^{2}} in polar coordinates). 3 Use as a variable The complex conjugate of a complex number {\displaystyle z} {\displaystyle {\overline {z}}} {\displaystyle z^{}.} The first notation, a vinculum, avoids confusion with the notation for the conjugate transpose of a matrix, which can be thought of as a generalization of the complex conjugate. The second is preferred in physics, where dagger (†) is used for the conjugate transpose, as well as electrical engineering and computer engineering, where bar notation can be confused for the logical negation ("NOT") Boolean algebra symbol, while the bar notation is more common in pure mathematics. If a complex number is represented as a {\displaystyle 2\times 2} matrix, the notations are identical.[clarification needed] The following properties apply for all complex numbers {\displaystyle z} {\displaystyle w,} unless stated otherwise, and can be proved by writing {\displaystyle z} {\displaystyle w} {\displaystyle a+bi.} {\displaystyle {\begin{aligned}{\overline {z+w}}&={\overline {z}}+{\overline {w}},\\{\overline {z-w}}&={\overline {z}}-{\overline {w}},\\{\overline {zw}}&={\overline {z}}\;{\overline {w}},\quad {\text{and}}\\{\overline {\left({\frac {z}{w}}\right)}}&={\frac {\overline {z}}{\overline {w}}},\quad {\text{if }}w\neq 0.\end{aligned}}} Conjugation does not change the modulus of a complex number: {\displaystyle \left\|{\overline {z}}\right\|=\|z\|.} Conjugation is an involution, that is, the conjugate of the conjugate of a complex number {\displaystyle z} {\displaystyle z.} {\displaystyle {\overline {\overline {z}}}=z.} {\displaystyle z{\overline {z}}={\left\|z\right\|}^{2}.} This allows easy computation of the multiplicative inverse of a complex number given in rectangular coordinates: {\displaystyle z^{-1}={\frac {\overline {z}}{{\left\|z\right\|}^{2}}},\quad {\text{ for all }}z\neq 0.} {\displaystyle {\overline {z^{n}}}=\left({\overline {z}}\right)^{n},\quad {\text{ for all }}n\in \mathbb {Z} } {\displaystyle \exp \left({\overline {z}}\right)={\overline {\exp(z)}}} {\displaystyle \ln \left({\overline {z}}\right)={\overline {\ln(z)}}{\text{ if }}z{\text{ is non-zero }}} {\displaystyle p} is a polynomial with real coefficients and {\displaystyle p(z)=0,} {\displaystyle p\left({\overline {z}}\right)=0} {\displaystyle \varphi } is a holomorphic function whose restriction to the real numbers is real-valued, and {\displaystyle \varphi (z)} {\displaystyle \varphi ({\overline {z}})} are defined, then {\displaystyle \varphi \left({\overline {z}}\right)={\overline {\varphi (z)}}.\,\!} {\displaystyle \sigma (z)={\overline {z}}} {\displaystyle \mathbb {C} } {\displaystyle \mathbb {C} } is a homeomorphism (where the topology on {\displaystyle \mathbb {C} } is taken to be the standard topology) and antilinear, if one considers {\displaystyle \mathbb {C} } as a complex vector space over itself. Even though it appears to be a well-behaved function, it is not holomorphic; it reverses orientation whereas holomorphic functions locally preserve orientation. It is bijective and compatible with the arithmetical operations, and hence is a field automorphism. As it keeps the real numbers fixed, it is an element of the Galois group of the field extension {\displaystyle \mathbb {C} /\mathbb {R} .} This Galois group has only two elements: {\displaystyle \sigma } and the identity on {\displaystyle \mathbb {C} .} Thus the only two field automorphisms of {\displaystyle \mathbb {C} } Use as a variableEdit Once a complex number {\displaystyle z=x+yi} {\displaystyle z=re^{i\theta }} is given, its conjugate is sufficient to reproduce the parts of the {\displaystyle z} Real part: {\displaystyle x=\operatorname {Re} (z)={\dfrac {z+{\overline {z}}}{2}}} Imaginary part: {\displaystyle y=\operatorname {Im} (z)={\dfrac {z-{\overline {z}}}{2i}}} Modulus (or absolute value): {\displaystyle r=\left\|z\right\|={\sqrt {z{\overline {z}}}}} Argument: {\displaystyle e^{i\theta }=e^{i\arg z}={\sqrt {\dfrac {z}{\overline {z}}}},} {\displaystyle \theta =\arg z={\dfrac {1}{i}}\ln {\sqrt {\frac {z}{\overline {z}}}}={\dfrac {\ln z-\ln {\overline {z}}}{2i}}} {\displaystyle {\overline {z}}} can be used to specify lines in the plane: the set {\displaystyle \left\{z:z{\overline {r}}+{\overline {z}}r=0\right\}} is a line through the origin and perpendicular to {\displaystyle {r},} since the real part of {\displaystyle z\cdot {\overline {r}}} is zero only when the cosine of the angle between {\displaystyle z} {\displaystyle {r}} is zero. Similarly, for a fixed complex unit {\displaystyle u=e^{ib},} {\displaystyle {\frac {z-z_{0}}{{\overline {z}}-{\overline {z_{0}}}}}=u^{2}} determines the line through {\displaystyle z_{0}} parallel to the line through 0 and {\displaystyle u.} These uses of the conjugate of {\displaystyle z} For matrices of complex numbers, {\textstyle {\overline {\mathbf {AB} }}=\left({\overline {\mathbf {A} }}\right)\left({\overline {\mathbf {B} }}\right),} {\textstyle {\overline {\mathbf {A} }}} represents the element-by-element conjugation of {\displaystyle \mathbf {A} .} [2] Contrast this to the property {\textstyle \left(\mathbf {AB} \right)^{}=\mathbf {B} ^{}\mathbf {A} ^{},} {\textstyle \mathbf {A} ^{}} represents the conjugate transpose of {\textstyle \mathbf {A} .} One may also define a conjugation for quaternions and split-quaternions: the conjugate of {\textstyle a+bi+cj+dk} {\textstyle a-bi-cj-dk.} {\displaystyle {\left(zw\right)}^{}=w^{}z^{}.} There is also an abstract notion of conjugation for vector spaces {\textstyle V} over the complex numbers. In this context, any antilinear map {\textstyle \varphi :V\to V} {\displaystyle \varphi ^{2}=\operatorname {id} _{V}\,,} {\displaystyle \varphi ^{2}=\varphi \circ \varphi } {\displaystyle \operatorname {id} _{V}} {\displaystyle V,} {\displaystyle \varphi (zv)={\overline {z}}\varphi (v)} {\displaystyle v\in V,z\in \mathbb {C} ,} {\displaystyle \varphi \left(v_{1}+v_{2}\right)=\varphi \left(v_{1}\right)+\varphi \left(v_{2}\right)\,} {\displaystyle v_{1}v_{2},\in V,} is called a complex conjugation, or a real structure. As the involution {\displaystyle \varphi } is antilinear, it cannot be the identity map on {\displaystyle V.} {\textstyle \varphi } {\textstyle \mathbb {R} } -linear transformation of {\textstyle V,} if one notes that every complex space {\displaystyle V} has a real form obtained by taking the same vectors as in the original space and restricting the scalars to be real. The above properties actually define a real structure on the complex vector space {\displaystyle V.} Composition algebra – Type of algebras, possibly non associative Conjugate (square roots) Hermitian function – Type of complex function Wirtinger derivatives – Concept in complex analysis ^ a b Friedberg, Stephen; Insel, Arnold; Spence, Lawrence (2018), Linear Algebra (5 ed.), ISBN 978-0134860244 , Appendix D ^ Arfken, Mathematical Methods for Physicists, 1985, pg. 201 ^ Budinich, P. and Trautman, A. The Spinorial Chessboard. Springer-Verlag, 1988, p. 29 Retrieved from "https://en.wikipedia.org/w/index.php?title=Complex_conjugate&oldid=1086813401"
The Fundamental Postulates Example: Compton Spread When we read What is LHC:ATLAS?, I mentioned that since the speed of the particles approaches the speed of light, we have to account for the relativistic effects. "However, we do have Galilean relativity which has worked perfectly fine for me" you might inject, and you'd be right. Galilean relativity serves us just well in most of our lives. In fact, for speeds far below the speed of light Einstein relativity and Galilean relativity are nearly indistinguishable from each other. [1] However, in certain circumstances such as when you're using a particle accelerator, or even something as mundane as using GPS, Galilean relativity is not enough, and you have to use Einstein's relativity. "Okay, so I need this Einstein's relativity. What is it?" you ask begrudgingly. Einstein's special relativity (refered to just as special relativity from here)[2] can be derived entirely from two fundamental postulates. All intertial reference frames are equivalent. This version of the postulate, however, follows a somewhat circular argument as an object which does not accelerate does not have a net force, however, the only way we know that it doesn't have a net force is by checking if the object accelerates. This inconsistency lead Einstein to develop his general theory of relativity which solves the problem by working for any reference system, and not just inertial systems. However, if you ignore this detail, you'll be fine. The speed of light in vacuum, c , is a universal constant in all reference frames. While today, we know this to be true as we have measured it empirically,[3][4] Einstein couldn't possibly have known this at his time, you rebut. And you'd be right. In fact, it's possible to derive that given that light is a electromagnetic wave that light must travel at c from Maxwell's equations. However, Einstein describes in his notes from 1949 himself finding it as a solution to an apparent paradox he stumbled upon when he was 16. This approach leads to the stronger claim that c is constant in all reference frames. The standard configuration is a configuration of coordinate systems often used in special relativity to ease, and somewhat normalize the notation. The standard configuration consists of an rest system S (x,y) [5], and another system S' (x',y') moving with v x -axis with respect to S .[6] Furthermore, at t=0 x=x'=0 This setup is illustrated below. Since handling the individual dimensions separately can quickly become a mess, it's an often used convention to combine the 3+1 dimensions into a four vector with the unit vectors: \textbf{e}_t,\textbf{e}_x,\textbf{e}_y,\textbf{e}_z The index of the time dimension can wary between authors. One convention is to use the zeroth dimension as the time dimension, so the unit vectors become \textbf{e}_0,\textbf{e}_1,\textbf{e}_2,\textbf{e}_3 which gives us the four vector \textbf{a}=\sum_{a=0}^3 a^\alpha \textbf{e}_\alpha. By using Einstein notation, we get \textbf{a}=a^\alpha \textbf{e}_\alpha where it's assumed we sum over repeated greek letters. We have adopted the convention where we denote four vectors with bold face such as, \textbf{a} , while we continue to denote three vectors with an arrow such as, \vec{a} For some more syntatic sugar, let \eta_{\alpha \beta} = \textbf{e}_\alpha \cdot \textbf{e}_\beta = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}. \textbf{a}\cdot \textbf{b}=\eta_{\alpha \beta} a^\alpha b^\beta. The reason we have a -1 1 in the time dimension is to avoid messing up causality which the philosophers tell me is a good thing. One neat thing about using four vectors is that the dot product between any two four vectors (or a chain of them) is invariant. So \textbf{a} \cdot \textbf{b} = \textbf{a}' \cdot \textbf{b}' . In the standard configuration, the componentwise Lorentz transformation is given by \begin{aligned} &a'^{0} = \gamma(a^{0} - \beta a^1 ), \\ &a'^{1} = \gamma(a^{1} - \beta a^0 ), \\ &a'^{2} = a^2, \\ &a'^{3} = a^3. \\ \end{aligned} While deriving every formula of special relativity is beyond the scope of this project, deriving time contraction seems worthwhile due to it's broad applicability. Let's consider a light-clock consisting of two perfect mirrors put parallel with a distance L to each other, and let a foton travelling at speed c bounce between the mirrors. Obviously, the time it takes for the foton to bounce from the bottom mirror to the top mirror and back again is going to be t = \frac{2L}{c} because time equals distance over speed. L = \frac{ct}{2} Now, let's consider the same system, but this time the mirrors move along the x v while the foton continous to bounce between the mirrors. Let the time it takes for the foton to complete one full bounce be t' , and let the distance the mirrors have moved before the foton comes back be 2s . Then we know that 2s = v\cdot t'. s=\frac{v\cdot t'}{2} We know that the foton travels along h twice for a total distance of 2h , so we find that 2h=c\cdot t. We now have all the sides in a right-angled triangled, and per pythagoras, we know that \big(\frac{vt'}{2}\big)^2 + \big(\frac{ct}{2}\big)^2 = \big(\frac{ct'}{2}\big)^2. c^2t^2 = c^2 t'^2 - v^2t'^2. By dividing by c^2 t^2 = t'^2 - \frac{v^2t'^2}{c^2}=t'^2 \big( 1 - \frac{v^2}{c^2} \big). t' t' = t \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} t' = t \gamma. Just like classcal mechanics, we still have conservation of momemtum and total energy in relativistic collisions. We consider a system consisting of a photon and an electron. In the initial state, the photon has a four momentum vector \textbf{p}_\gamma , and the electron is at rest, \textbf{p}_e = m_e c(1,\vec{0}) In the final state, the photon has the four momentum vector \bar{\textbf{p}}_\gamma , and the electron has the four momentum vector \bar{\textbf{p}}_e The setup is illustrated below. This is an example of Compton spread. The angle of spreading \theta depends on the wavelength of the incoming photon with the relation \bar{\lambda}=\lambda+\lambda_c(1-\cos(\theta)) \lambda_c = \frac{h}{m_e c} This chapter has leaned on the content from "Speciel Relativitetsteori" by Ulrik Uggerhøj whenever a source has not been cited. Note: While there'll always be a small factor of difference between Einstein and Galilean relativity, for small velocities, the difference will be within the margin of error. ↩︎ We use special and not general relativity because it's simpler, and sufficient for our purposes, and also because I'm taking a course in special, and not general relativity. ↩︎ Actually, we're so sure about c being an universal constant that we have defined the meter to be derived from c which has the interesting consequence that every time we attempt to 'measure' c , we are actually just measuring the accuracy of the equipment. ↩︎ There are several different ways of measuring c including some which can be done at home. One interesting way of measuring c is by using the aberration of light first discovered by James Bradley who found that light travels approximately 10,210 times faster than the Earth around the sun. ↩︎ Note: While this is done in 2+1 dimensions, the exact same principles apply in 3+1 dimensions; you just tag along a z axis. ↩︎ Note: The prime symbol ( ' ) has nothing to do with differentiation in special relativity. Except when it does, but in those rare cases the author usually marks it clearly to avoid ambiguity. ↩︎
Limitations of SMA Simple Moving Average FAQs What Is a Simple Moving Average (SMA)? Simple moving averages calculate the average of a range of prices by the number of periods within that range. A simple moving average (SMA) is an arithmetic moving average calculated by adding recent prices and then dividing that figure by the number of time periods in the calculation average. For example, one could add the closing price of a security for a number of time periods and then divide this total by that same number of periods. Short-term averages respond quickly to changes in the price of the underlying security, while long-term averages are slower to react. There are other types of moving averages, including the exponential moving average (EMA) and the weighted moving average (WMA). \begin{aligned} &\text{SMA}=\dfrac{A_1 + A_2 + ... + A_n}{n} \\ &\textbf{where:}\\ &A_n=\text{the price of an asset at period } n\\ &n=\text{the number of total periods}\\ \end{aligned} SMA=nA1+A2+...+Anwhere:An=the price of an asset at period nn=the number of total periods For example, this is how you would calculate the simple moving average of a security with the following closing prices over a 15-day period. Week One (5 days): 20, 22, 24, 25, 23 Week Two (5 days): 26, 28, 26, 29, 27 Week Three (5 days): 28, 30, 27, 29, 28 A 10-day moving average would average out the closing prices for the first 10 days as the first data point. The next data point would drop the earliest price, add the price on day 11, then take the average, and so on. Likewise, a 50-day moving average would accumulate enough data to average 50 consecutive days of data on a rolling basis. A simple moving average is customizable because it can be calculated for different numbers of time periods. This is done by adding the closing price of the security for a number of time periods and then dividing this total by the number of time periods, which gives the average price of the security over the time period. A simple moving average smooths out volatility and makes it easier to view the price trend of a security. If the simple moving average points up, this means that the security's price is increasing. If it is pointing down, it means that the security's price is decreasing. The longer the time frame for the moving average, the smoother the simple moving average. A shorter-term moving average is more volatile, but its reading is closer to the source data. One of the most popular simple moving averages is the 200-day SMA. However, there is a danger to following the crowd. As The Wall Street Journal explains, since thousands of traders base their strategies around the 200-day SMA, there is a chance that these predictions could become self-fulfilling and limit price growth. Moving averages are an important analytical tool used to identify current price trends and the potential for a change in an established trend. The simplest use of an SMA in technical analysis is using it to quickly determine if an asset is in an uptrend or downtrend. Another popular, albeit slightly more complex, analytical use is to compare a pair of simple moving averages with each covering different time frames. If a shorter-term simple moving average is above a longer-term average, an uptrend is expected. On the other hand, if the long-term average is above a shorter-term average then a downtrend might be the expected outcome. Two popular trading patterns that use simple moving averages include the death cross and a golden cross. A death cross occurs when the 50-day SMA crosses below the 200-day SMA. This is considered a bearish signal, indicating that further losses are in store. The golden cross occurs when a short-term SMA breaks above a long-term SMA. Reinforced by high trading volumes, this can signal further gains are in store. The major difference between an exponential moving average (EMA) and a simple moving average is the sensitivity each one shows to changes in the data used in its calculation. More specifically, the EMA gives a higher weighting to recent prices, while the SMA assigns an equal weighting to all values. The two averages are similar because they are interpreted in the same manner and are both commonly used by technical traders to smooth out price fluctuations. Since EMAs place a higher weighting on recent data than on older data, they are more reactive to the latest price changes than SMAs are, which makes the results from EMAs more timely and explains why the EMA is the preferred average among many traders. Limitations of Simple Moving Average It is unclear whether or not more emphasis should be placed on the most recent days in the time period or on more distant data. Many traders believe that new data will better reflect the current trend the security is moving with. At the same time, other traders feel that privileging certain dates over others will bias the trend. Therefore, the SMA may rely too heavily on outdated data since it treats the 10th or 200th day's impact the same as the first or second day's. Similarly, the SMA relies wholly on historical data. Many people (including economists) believe that markets are efficient—that is, that current market prices already reflect all available information. If markets are indeed efficient, using historical data should tell us nothing about the future direction of asset prices. How Are Simple Moving Averages Used in Technical Analysis? Traders use simple moving averages (SMAs) to chart the long-term trajectory of a stock or other security, while ignoring the noise of day-to-day price movements. This allows traders to compare medium- and long-term trends over a larger time horizon. For example, if the 200-day SMA of a security falls below its 50-day SMA, this is usually interpreted as a bearish death cross pattern and a signal of further declines. The opposite pattern, the golden cross, indicates potential for a market rally. How Do You Calculate a Simple Moving Average? To calculate a simple moving average, the number of prices within a time period is divided by the number of total periods. For instance, consider shares of Tesla closed at $10, $11, $12, $11, $14 over a five day period. The simple moving average of Tesla’s shares would equal $10 + $11 + $12 + $11 + $14 divided by 5, equaling $11.6. What Is the Difference Between a Simple Moving Average and an Exponential Moving Average? While a simple moving average gives equal weight to each of the values within a time period, an exponential moving average places greater weight on recent prices. Exponential moving averages are typically seen as a more timely indicator of a price trend, and because of this, many traders prefer using this over a simple moving average. Common short-term exponential moving averages include the 12-day and 26-day. The 50-day and 200-day exponential moving averages are used to indicate long-term trends. The Wall Street Journal. "Does Chart Analysis Really Work?" Accessed Feb. 1, 2022.
The s-Block Elements Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Highly soluble hydroxide in water is formed by - 1. Ni2+ What would you observe if excess of dilute NaOH solution is added and shaken with an aqueous solution of aluminium chloride? 1. A permanent white precipitate is formed immediately 2. No change at first but a white precipitate is formed on standing 3. A white precipitate is formed which later dissolves 4. A green precipitate which turns red on standing in air Subtopic: Preparations,Properties & Uses of S Block Elements \| Strongest reducing agent among the following is: When crystal of caustic soda is exposed to air, a liquid layer is deposited because: 1. crystal melts 2. crystal loses water 3. crystal absorbs moisture and CO2 4. crystal sublimes Which one of the following is formed on dissolving I2 in aqueous solution of KI ? 3. KI3 The correct statement among the following regarding CsBr3 is - 1. It is a covalent compound. 2. It contains Cs2+ and Br- ions. 3. It contains Cs+, Br- and Br2 lattice molecules. 4. It contains Cs+ and {\mathrm{Br}}_{3}^{-} Subtopic: Chemical Properties \| Compounds of Ca and Na -Preparations,Properties & Uses \| Hypo is chemically: 1. Na2S2O3.2H2O If NaOH is added to an aqueous solution of Zn2+ ions, a white precipitate appears and on adding axcess NaOH, the precepitate dissolves. In this solution zinc exists in the: 1. cationic part 2. anionic part 3. both in cationic and anionic parts 4. there is no zinc left in the solution Brine is chemically: 1. conc. solution of Na2CO3 2. conc. solution of Na2SO4 3. conc. solution of NaCl 4. conc. solution of alum Subtopic: Physical Properties \| Chemical Properties \| NO2 is obtained by heating:
Electrostatic Potential and Capacitance Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown. Let {V}_{A},{V}_{B},{V}_{C} be the potentials at points A, B and C respectively. Then {V}_{A}<{V}_{B}<{V}_{C} {V}_{A}>{V}_{B}>{V}_{C} {V}_{C}>{V}_{B}={V}_{A} {V}_{A}={V}_{B}={V}_{C} Subtopic: Electric Potential \| Four particles each having charge q are placed at the vertices of a square of side a. The value of the electric potential at the midpoint of one of the side will be \frac{1}{4\pi {ϵ}_{0}}\frac{2q}{a}\left(2+\frac{2}{\sqrt{5}}\right) \frac{1}{4\pi {ϵ}_{0}}\frac{2q}{a}\left(2-\frac{2}{\sqrt{5}}\right) \frac{1}{4\pi {ϵ}_{0}}\frac{2q}{a}\left(1+\frac{1}{\sqrt{5}}\right) If E be the electric field inside a parallel plate capacitor due to Q and -Q charges on the two plates, then electrostatic force on plate having charge -Q due to the plate having charge +Q will be (1) -QE \frac{-QE}{2} (3) QE \frac{-QE}{4} Subtopic: Combination of Capacitors \| If W be the amount of heat produced in the process of charging an uncharged capacitor then the amount of energy stored in it is \frac{W}{2} Subtopic: Energy stored in Capacitor \| A metallic sphere of capacitance {C}_{1} , charged to electric potential {V}_{1} is connected by a metal wire to another metallic sphere of capacitance {C}_{2} charged to electric potential {V}_{2} . The amount of heat produced in the connecting wire during the process is \frac{{C}_{1}{C}_{2}}{2\left({C}_{1}+{C}_{2}\right)}{\left({V}_{1}+{V}_{2}\right)}^{2} \frac{{C}_{1}{C}_{2}}{2\left({C}_{1}+{C}_{2}\right)}{\left({V}_{1}-{V}_{2}\right)}^{2} \frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}{\left({V}_{1}-{V}_{2}\right)}^{2} The electric potential at the surface of a charged solid sphere of insulator is 20V. The value of electric potential at its centre will be The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become \frac{KC}{2\left(K+1\right)} \frac{2KC}{K+1} \frac{5KC}{4K+1} \frac{4KC}{3K+1} Subtopic: Dielectrics in Capacitors \| The electric potential at a point at distance 'r' from a short dipole is proportional to {r}^{2} {\mathrm{r}}^{-1} {r}^{-2} {r}^{1} A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is \frac{3\mathrm{V}}{19\mathrm{R}} \frac{\mathrm{V}}{6\mathrm{R}} \frac{3\mathrm{V}}{32\mathrm{R}} \frac{3\mathrm{V}}{16\mathrm{R}} {C}_{1}=10\mu F and {C}_{2}=30\mu F are connected in series across a source of emf 20KV. The potential difference across {C}_{1} (1) 5 KV
For obtaining high COP, the pressure range of compressor should be \| Answers Updates & Alerts ↴ UPSC Answer Key 2019 UPPSC 2018 Answer Key Complete List of Ministers In Modi Govt. 2017 UPSC CAPF Answer Key 2017 Who is Who 2020 Types of Speciation \| Unisexual and Bisexual Flowers \| DISANET \| National Child Labour Project Scheme \| Tropical Deciduous Forest \| Hoolock Gibbon: National Parks \|&nbsp Sectors of Indian Economy \| How Legislative Council is Formed \| Insectivorous Plants \| Causes of Revolt of 1857 in Points \| Ordinance Making Power of President and Governors \| Mangroves Cover in India \| Coral Reefs in India \| Seasons and Climate in India \| Constitutional and Statutory Bodies in India \| Rock Edicts of Ashoka \| Types of Motions in Parliamentary Procedure \| Types of Majorities in Parliament \| Pressure Groups in World and India \| Order of Precedence \| Types of Rocks \| Treaties in Indian History \| Types of Rocks \| Features of Indian Economy \| Geographical Indications \| Ancient History Terminology and Meanings \| Schedules of the Constitution \| Nuclear Power Plants in India List \| Types of Clouds for SSC CGL \| Nuclear Power Plants in India List \| Types of Ocean Currents \| Structure: Layers of Earth \| APPSC Group 1 Answer Key \| SSC JE 2017 QUESTION PAPER \| One Word Substitution for SSC and Bank Exams \| MPSC Answer Key 2017 2017 \| Layers of Atmosphere \| MPSC Answer Key 2017 2017 \| BPSC Solved Question Paper 2017 \| MPPSC Solved Question Paper 2017 \| GST Bill India \| Delhi Sultanate \| Law Commission of India \| Oscar Awards Winner 2017 List \| Diseases in Crops \| Common Drugs and Uses \| Different Types of Deserts in World \| UPSC Civil Services Exam-2016 Answer Key \| UPSC CAPF Exam-2016 Answer Key \| UPPSC UPPER Subordinate Exam-2016 Answer Key \| Regulatory Bodies \| National Parks in India \| Home » Coefficient of Performance COP , Refrigeration and Air-Conditioning » For obtaining high COP, the pressure range of compressor should be For obtaining high COP, the pressure range of compressor should be Q. For obtaining high COP, the pressure range of compressor should be For obtaining high COP, the pressure range of compressor should be low. The coefficient of performance or COP (sometimes CP or CoP) of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs. The COP usually exceeds 1. {\displaystyle {\rm {COP}}={\frac {Q}{W}}} To improve the coefficient of performance, it requires that the compressor work should decrease and the refrigeration effect should increase. It means that decrease in condenser pressure and temperature so the refrigeration effect will increase and compressor input work due to this cop will increase. Thanks for reading For obtaining high COP, the pressure range of compressor should be Tags Coefficient of Performance COP, Refrigeration and Air-Conditioning
Identifying Single Cointegrating Relations - MATLAB & Simulink - MathWorks Nordic Modern approaches to cointegration testing originated with Engle and Granger [66]. Their method is simple to describe: regress the first component y1t of yt on the remaining components of yt and test the residuals for a unit root. The null hypothesis is that the series in yt are not cointegrated, so if the residual test fails to find evidence against the null of a unit root, the Engle-Granger test fails to find evidence that the estimated regression relation is cointegrating. Note that you can write the regression equation as {y}_{1t}-{b}_{1}{y}_{2t}-...-{b}_{d}{y}_{dt}-{c}_{0}=\beta \prime {y}_{t}-{c}_{0}={\epsilon }_{t} \beta =\left[\begin{array}{cc}1& -b\prime \end{array}\right]\prime is the cointegrating vector and c0 is the intercept. A complication of the Engle-Granger approach is that the residual series is estimated rather than observed, so the standard asymptotic distributions of conventional unit root statistics do not apply. Augmented Dickey-Fuller tests (adftest) and Phillips-Perron tests (pptest) cannot be used directly. For accurate testing, distributions of the test statistics must be computed specifically for the Engle-Granger test. The Engle-Granger method has several limitations. First of all, it identifies only a single cointegrating relation, among what might be many such relations. This requires one of the variables, {y}_{1t} , to be identified as "first" among the variables in {y}_{t} . This choice, which is usually arbitrary, affects both test results and model estimation. To see this, permute the three interest rates in the Canadian data and estimate the cointegrating relation for each choice of a "first" variable.
Gaps between zeros of the derivative of the Riemann $\xi $-function Gaps between zeros of the derivative of the Riemann \xi Hung Manh Bui1 1 Mathematical Institute University of Oxford Oxford, OX1 3LB England Assuming the Riemann hypothesis, we investigate the distribution of gaps between the zeros of {\xi }^{\prime }\left(s\right) . We prove that a positive proportion of gaps are less than 0.796 times the average spacing and, in the other direction, a positive proportion of gaps are greater than 1.18 times the average spacing. We also exhibit the existence of infinitely many normalized gaps smaller (larger) than 0.7203 1.5 En supposant l’hypothèse de Riemann, on examine la distribution d’écarts entre les zéros du {\xi }^{\prime }\left(s\right) . On démontre qu’une proportion positive d’écarts sont inférieurs à 0.796 fois l’écart moyen et que dans l’autre direction, une proportion positive d’écarts sont 1.18 fois supérieurs à l’écart moyen. On montre également l’existence d’un nombre infini d’écarts normalisés qui sont inférieurs (supérieurs) à 0.7203 1.5 Classification: 11M26, 11M06 Hung Manh Bui&hairsp;1 author = {Hung Manh Bui}, title = {Gaps between zeros of the derivative of the {Riemann} $\xi $-function}, TI - Gaps between zeros of the derivative of the Riemann $\xi $-function %T Gaps between zeros of the derivative of the Riemann $\xi $-function Hung Manh Bui. Gaps between zeros of the derivative of the Riemann $\xi $-function. Journal de Théorie des Nombres de Bordeaux, Volume 22 (2010) no. 2, pp. 287-305. doi : 10.5802/jtnb.716. https://jtnb.centre-mersenne.org/articles/10.5802/jtnb.716/ [1] J. Bian, The pair correlation of zeros of {\xi }^{\left(k\right)}\left(s\right) and discrete moments of \zeta \left(s\right) . PhD thesis, University of Rochester, 2008. [2] H. M. Bui, Large gaps between consecutive zeros of the Riemann zeta-function. Preprint, available on Arxiv at http://arxiv.org/abs/0903.4007 [3] H. M. Bui, M. B. Milinovich, Nathan Ng, A note on the gaps between consecutive zeros of the Riemann zeta-function. To appear in Proc. Amer. Math. Soc. Available on Arxiv at http://arxiv.org/abs/0910.2052 \| MR: 2680043 [4] J. B. Conrey, A. Ghosh, D. Goldston, S. M. Gonek, D. R. Heath-Brown, On the distribution of gaps between zeros of the zeta-function. Quart. J. Math. Oxford 36 (1985), 43–51. \| MR: 780348 \| Zbl: 0557.10028 [5] J. B. Conrey, A. Ghosh, S. M. Gonek, A note on gaps between zeros of the zeta function. Bull. London Math. Soc. 16 (1984), 421–424. \| MR: 749453 \| Zbl: 0536.10033 [6] J. B. Conrey, A. Ghosh, S. M. Gonek, Large gaps between zeros of the zeta-function. Mathematika 33 (1986), 212–238. \| MR: 882495 \| Zbl: 0615.10048 [7] T. Craven, G. Csordas, W. Smith, The zeros of derivatives of entire functions and the Pólya-Wiman conjecture. Ann. Math. 125 (1987), 405–431. \| MR: 881274 \| Zbl: 0625.30036 [8] H. Davenport, Multiplicative number theory. GTM 74, Springer-Verlag, 2000. \| MR: 1790423 \| Zbl: 1002.11001 [9] D. W. Farmer, S. M. Gonek, Pair correlation of the zeros of the derivative of the Riemann \xi -function. Preprint, available on Arxiv at http://arxiv.org/abs/0803.0425 [10] D. W. Farmer, R. Rhoades, Differentiation evens out zero spacings. Trans. Amer. Math. Soc. 357 (2005), 3789–3811. \| MR: 2146650 \| Zbl: 1069.30005 [11] A. Fujii, On the distribution of the zeros of the Riemann zeta-function in short intervals. Bull. Amer. Math. Soc. 81 (1975), 139–142. \| MR: 354575 \| Zbl: 0297.10026 [12] R. R. Hall, A new unconditional result about large spaces between zeta zeros. Mathematika 52 (2005), 101–113. \| MR: 2261847 \| Zbl: 1119.11050 [13] H. L. Montgomery, The pair correlation of zeros of the zeta function. Analytic Number Theory, Proc. Sym. Pure Math. 24 (1973), 181–193. \| MR: 337821 \| Zbl: 0268.10023 [14] H. L. Montgomery, A. M. Odlyzko, Gaps between zeros of the zeta function. Topics in Classical Number Theory, Coll. Math. Soc. Janos Bolyai 34 (1984), 1079–1106, North-Holland. \| MR: 781177 \| Zbl: 0546.10033 [15] H. L. Montgomery, R. C. Vaughan, The large sieve. Mathematika 20 (1973), 119–134. \| MR: 374060 \| Zbl: 0296.10023 [16] J. Mueller, On the difference between consecutive zeros of the Riemann zeta function. J. Number Theory 14 (1982), 327–331. \| MR: 660377 \| Zbl: 0483.10035 [17] Nathan Ng, Large gaps between the zeros of the Riemann zeta function. J. Number Theory 128 (2008), 509–556. \| MR: 2389854 \| Zbl: 1182.11038 [18] A. Selberg, Note on a paper by L. G. Sathe. J. Indian Math. Soc. 18 (1954), 83–87. \| MR: 67143 \| Zbl: 0057.28502 [19] K. Soundararajan, On the distribution of gaps between zeros of the Riemann zeta-function. Quart. J. Math. Oxford 47 (1996), 383–387. \| MR: 1412563 \| Zbl: 0858.11047 [20] E. C. Titchmarsh, The theory of the Riemann zeta-function. Revised by D. R. Heath-Brown, Clarendon Press, second edition, 1986. \| MR: 882550 \| Zbl: 0601.10026
Biomorphs \| Sayed's Blog In a previous post I recreate Dawkin's Weasel Program. In this post we will recreate a more visual demonstration of evolution - biomorphs. A biomorph is a computer generated shape, with an appearance that can be influenced by a few variables. Richard Dawkins treats these variables as genes, and produces multiple biomorphs. They look like this: He uses them to demonstrate evolution by asking members of an audience to try to evolve their own biomorphs, by selecting the one they like the most, and cloning their "genes" multiple times, with some variation to produce biomorphs that look slightly different. He was able to produce a large variety of different biomorphs. Here are some of his favourite: To draw his Biomorphs, Dawkins imitates embryology using recursion: Embryonic development is far too elaborate a process to simulate realistically on a small computer. We must represent it by some simplified analogue. We must find a simple picture-drawing rule that the computer can easily obey, and which can then be made to vary under the influence of 'genes'. What drawing rule shall we choose? Textbooks of computer science often illustrate the power of what they call 'recursive' programming with a simple tree-growing procedure. The computer starts by drawing a single vertical line. Then the line branches into two. Then each of the branches splits into two sub- branches. Then each of the sub-branches splits into sub-sub-branches, and so on. It is 'recursive' because the same rule (in this case a branching rule) is applied locally all over the growing tree. No matter how big the tree may grow, the same branching rule goes on being applied at the tips of all its twigs. Drawing the biomorphs I'll be using Javascript and HTML canvas to recreate Dawkin's biomorphs. Let's start with enough HTML to get a canvas on the screen and get a rendering context. First create a file called biomorphs.html. <title>Biomorphs</title> Now I will make a javascript function to create biomorphs. First we write the code to draw a line at 45 degrees. If we have a line with length s, and a angle of a, and a starting position of x and y, then the final position, according to trigonometry, is x+sin(a) y+cos(a) In our case, however, since the origin of the canvas coordinate system is on the top left, we will be using y-cos(a) so the angles appear as expected. function toRadians(angle) { return angle * Math.PI / 180; function drawLine(x, y, endX, endY) { function drawBranch(x, y) { var endX = x + Math.sin(toRadians(45)) * 40; var endY = y - Math.cos(toRadians(45)) * 40; drawLine(x, y, endX, endY); drawBranch(100, 100); Since the sin and cos functions in Javascript use radians, I've created a helper function that converts degrees to radians. This is because people tend to prefer degrees, and 360 is divisible by many whole numbers, allowing for the easy creation of variety of biomorphs. I've also created a function to draw a line between two points, since that does not exist in Javascript canvas. Running them in the browser should show this: Now let's remove the hard-coded values for the angle: function drawBranch(x, y, size, angle) { var endX = x + Math.sin(toRadians(angle)) * size; var endY = y - Math.cos(toRadians(angle)) * size; drawBranch(100, 100, 40, 45); After drawing the first line, two more lines should to branch from the end. Like this: Before we go about making this function recursive, let's try to implement it by hand. y = endY; endX = x + Math.sin(toRadians(angle+angle)) * size; endY = y - Math.cos(toRadians(angle+angle)) * size; endX = x + Math.sin(toRadians(angle-angle)) * size; endY = y - Math.cos(toRadians(angle-angle)) * size; Most of the code is repeated. The difference is the angle has changed. The first branch is 45 degrees to the left, and the other branch it is 45 degrees to the right. This means we only have to modify the angle when calling recursively, and add a base condition to return when the number of iterations has been met to prevent an infinite loop. function drawBranch(x, y, size, angle, angleDiff, iterations) { if (iterations == 0) return; drawBranch(endX, endY, size, angle+angleDiff, angleDiff, iterations-1); drawBranch(endX, endY, size, angle-angleDiff, angleDiff, iterations-1); drawBranch(100, 100, 40, 45, 45, 2); Now, by changing the number of iterations, we have a whole new image. drawBranch(100, 200, 40, 0, 45, 3); drawBranch(endX, endY, size0.8, angle+angleDiff, angleDiff, iterations-1); drawBranch(endX, endY, size0.8, angle-angleDiff, angleDiff, iterations-1); drawBranch(50, 200, 20, 0, 45, 3); Now replace the drawBranch calls with: drawBranch(col * 110, row * 110, 20, 0, i6, 8); One last gene - the size factor: function drawBranch(x, y, size, sizeFactor, angle, angleDiff, iterations) { drawBranch(endX, endY, sizesizeFactor, sizeFactor, angle+angleDiff, angleDiff, iterations-1); drawBranch(endX, endY, sizesizeFactor, sizeFactor, angle-angleDiff, angleDiff, iterations-1); drawBranch(col 110, row * 110, 20, 0.8, 0, i6, 8); Remember, every change to the parameters that drawBranch accepts means we have to pass it in when calling it. The evolution bit Now we have a function that, given some parameters, can draw many different shapes. But we have no way of evolving it yet. To do that, we will need a concept of genes. I'll use arrays to replicate this. First I will create a Biomorph class that takes in an array of genes. const GENE_INDICIES = { SIZE_FACTOR: 1, ANGLE_DIFF: 2, class Biomorph { constructor(genes) { this.genes = genes; this.genes[GENE_INDICIES.SIZE], this.genes[GENE_INDICIES.SIZE_FACTOR], this.genes[GENE_INDICIES.ANGLE_DIFF], this.genes[GENE_INDICIES.ITERATIONS] Now replace the loop for drawing the branches with this: var biomorph = new Biomorph([20, 0.8, 25, 8]); biomorph.draw(100, 100); And running it should show this: Now let's add the ability to "reproduce" a biomorph: reproduce() { var newGenes = []; for (var i = 0; i < this.genes.length; i++) { var geneValue = this.genes[i]; var difference = Math.random() 0.3 - 0.15; if (Math.random() * 100 > 20) geneValue = 1 + difference; newGenes.push(geneValue); return new Biomorph(newGenes); We change a value by no more than 25%. To make this work we also have to make sure the genes stay within valid ranges - otherwise they get too big: Math.min(this.genes[GENE_INDICIES.SIZE], 30), Math.min(this.genes[GENE_INDICIES.SIZE_FACTOR], 0.9), Math.min(Math.round(this.genes[GENE_INDICIES.ITERATIONS]), 12) Now we can test this by creating a biomorph and reproducing it then drawing both to the screen. var child = biomorph.reproduce(); child.draw(300, 100); Next we need to fill a grid with child biomorphs. First let's create a method to populate the grid. In this method we set add the parent biomorph to the middle cell, and add children of the parent in all other cells. function populateGrid(parent) { var grid = Array.from(Array(3), () => new Array(5)); for (var x = 0; x < grid[y].length; x++) { if (y == (grid.length+1)/2 - 1 && x == (grid[y].length+1)/2 - 1) { grid[y][x] = parent; grid[y][x] = parent.reproduce(); This takes in a parent so that when we select a biomorph, we create a new grid with the selected biomorph as the parent. grid[y][x].draw((x+1) 110, (y+1) * 110 + 25); grid = populateGrid(biomorph); Next we need a way to select a biomorph. We'll do this by adding an event handler to the canvas. To get the mouse coordinates relative to the canvas, we get the clientX and clientY from the event. var x = evt.clientX; var y = evt.clientY; And to convert from the x and y coordinates to indices in the grid, we offset the coordinates and divide by the size of the cell. var col = Math.floor((x - 55)/110); var row = Math.floor((y - 55)/110); Now that we have the indices in the grid, all we have to do is extract the selected biomorph, repopulate the grid using that biomorph as the parent, and redraw the grid. We also have to check that the row and column are valid indices within the grid: if (row >= 0 && row < grid.length && col >= 0 && col < grid[row].length) { var parent = grid[row][col]; grid = populateGrid(parent); To prevent the previous biomorphs from being drawn over, we clear the canvas: This is enough to demonstrate the general principle but there is a lot more that could be done. For example, by adding "segmentation", and playing with symmettry and adding a gradient scale, Dawkins was able to create much more complex shapes: Another way we could extend it is by saving a selected biomorph - that's something Richard Dawkins decided to do after trying to recreate a biomorph.
Economic Profit and Decision Making - Course Hero Microeconomics/Profit/Economic Profit and Decision Making An individual is creating his own business. He expects this business to have a total revenue of $100,000 per year and a total explicit cost of $60,000 per year. Is entering into this business a smart move? It's tempting to conclude that the answer is yes, because the yearly accounting profit is positive (\$100\text{,}000-\$60\text{,}000 = \$40\text{,}000) . The person's business would make $40,000 per year. However, the individual already has a job that pays $50,000 per year. He would have to quit his current job to start a business. His $50,000 per-year job counts as an implicit cost of starting the business. If he currently makes $50,000 per year, he would lose money by quitting his current job and starting a business that earns him $40,000. This kind of information is important to consider when making business decisions. An economic profit analysis can be helpful in the decision-making process. Positive versus negative profit is important information when evaluating a business opportunity. If economic profit is negative, an individual might do better pursuing a different opportunity. If economic profit is positive, an individual should adhere to the choice. Of course, economic profit could also be exactly equal to zero. An economic profit of zero indicates that there is not much difference between the chosen opportunity and the next-best alternative. The potential for zero economic profit is very common. It is incorrect to think of an opportunity only being valuable if it makes a profit. An economic profit of zero can be thought of as an "acceptable" level of profit. For this reason, normal profit exists when a firm's total costs and total revenue are equal. Considering the economic profit can help an individual or business decide if the risk involved in a decision is worthwhile. There are always risks in business—if a business can make good decisions about what risks to take, the economic profit of the business will grow. Many businesses take time to increase economic profit from zero, but even at an economic profit of zero, the business is not losing money. Negative economic profit over a period of time shows that despite investments and attempts to grow the business, it is not profitable. Understanding whether economic profit is at zero, negative, or positive is important when making decisions about starting, continuing, or closing down a business activity. A business that is breaking even is considered to be functioning. A business making a profit is successful. A business making a loss (negative profit) is considered to be failing. <Calculating Profit and Total Revenue>Suggested Reading
How to Prepare a Statement of Cash Flows - Course Hero Principles of Accounting/Statement of Cash Flows/How to Prepare a Statement of Cash Flows Learn all about how to prepare a statement of cash flows in just a few minutes! Fabio Ambrosio, CPA, instructor of accounting at the Central Washington University, walks through the three steps to prepare the statement of cash flows using the indirect method: determine the change in cash balance, establish new cash flows from operating activities, and compile the cash flows from investing and financing activities. Steps Using the Indirect Method There are three steps to prepare the statement of cash flows using the indirect method. Step 1: Establish the amount of the change in cash by finding the difference between beginning and ending cash shown on comparative balance sheets. Step 2: Establish the amount of the net cash flows from operating activities by analyzing the income statement, comparative balance sheets, and selected transaction data. Start with net income, and convert it to net operating activities by adjusting items that affected reported net income but did not affect cash. Step 3: Establish the amount of net cash flows from investing and financing activities by analyzing effects on applicable accounts shown on the comparative balance sheet. Step 2 requires obtaining the net income amount from the income statement. Step 2 items require an adjustment to net income (an accrual amount) to obtain net cash flow from operating activities (a cash basis amount). Various common additions and subtractions from net income are part of this adjustment in order to derive net cash flow from operating activities. Adjustments to Add to Net Income: Adjustments to Subtract from Net Income: Amortization of discount on bonds payable Amortization of premium on bonds payable Amortization of intangibles or deferred charges Decrease in deferred income taxes payable Loss on sale of property, plant, and equipment Gain on sale of property, plant, and equipment Decrease in receivables Increase in receivables Decrease in inventory Increase in inventory Decrease in prepaid expenses Increase in prepaids Increase in payables or accrued liabilities Decrease in payables or accrued liabilities An example of Joe's Farms, Inc., illustrates the three steps for preparation of the indirect statement of cash flows, using a comparative balance sheets for two years, along with selected transaction information. Comparative Balance Sheets and Selected Transactions Preparing comparative balance sheets serves as a valuable method for measuring operating, investing, and financing activities over a period of time, such as 2020-21. Step 1: The indirect method for the statement of cash flows for Joe's Farms, Inc., shows an increase in cash of $5,000, which agrees with the difference between the 2020 and 2021 cash balance \left(\${14\rm{,}200}- \${9\rm{,}200} = \${5\rm{,}000}\right) Step 2: Start the operating activities section with net income, $11,000. Convert net income to net operating activities by adjusting items that affected reported net income but did not affect cash. This includes Depreciation Expense of $8,000 on the income statement, the increase in Accounts Receivable of $13,000 from the comparative balance sheets, the increase in Inventory of $10,000 from the comparative balance sheets, and the increase in Accounts Payables of $5,000 from comparative balance sheets. Step 3: Include the effects of investing and financing activities. For investing activities, these include the purchase of building, $5,000 from comparative balance sheets; financing activities, sale of common stock, $20,000 from comparative balance sheets and selected information; and financing activities payment of dividends, $11,000 from selected information. Three Primary Steps of Indirect Method There are three primary steps to prepare a statement of cash flows: Step 1-Identify the change in cash. Step 2-Convert net income to net operating activities with adjusting items. Step 3-Include effects of investing and financing activities on cash flow. Cash Flows from Operating, Investing, and Financing After determining the net cash flow from operating activities, the next step is to obtain cash flows from investing activities. This process is the same for both the indirect and the direct methods of preparation of the statement of cash flows. Investing activities often include free cash flow, which is the discretionary cash flow amount available for purchasing additional investments, retiring debt, or increasing company liquidity. Investing cash inflows include cash from the sale of property, plant, and equipment; cash from the sale of debt or equity securities of other entities; and cash from the collection of principal on loans to other entities. Investing cash outflows for this section include cash to purchase property, plant, and equipment; cash to purchase debt or equity securities of other entities; and cash to make loans to other entities. Financing activities are determined next and generally include cash changes in long-term liability and equity items. Financing cash inflows for this section include cash from the sale of equity securities and cash from the issuance of debt (such as notes and bonds). Financing cash outflows for this section include cash to stockholders in the form of dividends and cash to redeem long-term debt or reacquire capital stock (e.g., treasury stock). Under both the investing and financing sections, add or subtract the amount that increased or decreased cash. Simply consider the amount of cash that was spent (subtraction) or the amount of cash that was received (addition). After determining the three cash-flow activity sections of operating, investing, and financing, the amounts are summed and then added to the cash at the beginning of the period in order to derive the cash at the end of the period. This net increase or decrease in cash, derived from the three activities, should reconcile the beginning and ending cash balances as reported in the comparative balance sheets. Any investing or financing transactions that did not involve the exchange of cash would go into a fourth section, called the noncash investing and financing section. An example of this would be the purchase of a fixed asset with a note payable. <Reporting Cash Flows>Statement of Cash Flows Spreadsheet
Mechanical Properties of Fluids Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity \omega . The force exerted by liquid at the other end is \frac{{\mathrm{M\omega }}^{2}\mathrm{L}}{2} {\mathrm{ML\omega }}^{2} \frac{{\mathrm{M\omega }}^{2}\mathrm{L}}{4} \frac{M{L}^{2}{\omega }^{2}}{2} Subtopic: Pressure \| The radius of a soap bubble is increased from R to 2 R. Work done in this process (T = surface tension) is 24 {\mathrm{\pi R}}^{2}\mathrm{T} 48 {\mathrm{\pi R}}^{2}\mathrm{T} 12 {\mathrm{\pi R}}^{2}\mathrm{T} 36 {\mathrm{\pi R}}^{2}\mathrm{T} Subtopic: Surface Tension \| The flow rate from a tap of diameter 1.25 cm is 3 lit/min. The coefficient of viscosity of water is {10}^{-3} Pas. The nature of flow is : 3. Neither laminar nor turbulent Subtopic: Types of Flows \| A small spherical solid ball is dropped in a viscous liquid. Its journey in the liquid is best described in the figure by : 1. Curve A 2. Curve B 3. Curve C 4. Curve D Subtopic: Stokes' Law \| Water flows through a frictionless duct with a cross-section varying as shown in the figure. Pressure p at points along the axis is represented by Subtopic: Equation of Continuity \| The area of cross-section of the wider tube shown in the figure is 800 {\mathrm{cm}}^{2}. If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is : Equation of continuity based on : The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is \frac{{\mathrm{vR}}^{2}}{{\mathrm{n}}^{2}{\mathrm{r}}^{2}} \frac{{\mathrm{vR}}^{2}}{{\mathrm{nr}}^{2}} \frac{{\mathrm{vr}}^{2}}{{\mathrm{n}}^{2}{\mathrm{R}}^{2}} \frac{{\mathrm{v}}^{2}\mathrm{R}}{\mathrm{nr}} 60% NEET - 2015 The weight of an aeroplane flying in the air is balanced by (1) Vertical component of the thrust created by air currents striking the lower surface of the wings (2) Force due to reaction of gases ejected by the revolving propeller (3) Upthrust of the air which will be equal to the weight of the air having the same volume as the plane (4) Force due to the pressure difference between the upper and lower surfaces of the wings created by different air speeds on the surfaces Subtopic: Bernoulli's Theorem \| If there were a smaller gravitational effect, which of the following forces do you think would alter in some respect? (1) Viscous forces (2) Archimedes uplift (3) Electrostatic force Subtopic: Archimedes' Principle \|
In cost accounting, the high-low method is a way of attempting to separate out fixed and variable costs given a limited amount of data. The high-low method involves taking the highest level of activity and the lowest level of activity and comparing the total costs at each level. If the variable cost is a fixed charge per unit and fixed costs remain the same, it is possible to determine the fixed and variable costs by solving the system of equations. It is worth being cautious when using the high-low method, however, as it can yield more or less accurate results depending on the distribution of values between the highest and lowest dollar amounts or quantities. Calculating the outcome for the high-low method requires a few formula steps. First, you must calculate the variable cost component and then the fixed cost component, and then plug the results into the cost model formula. First, determine the variable cost component: \begin{aligned} &\text{Variable Cost} = \frac { \text{HAC} - \text{Lowest Activity Cost} }{ \text{HAUs} - \text{Lowest Activity Units} } \\ &\textbf{where:} \\ &\text{HAC} = \text{Highest activity cost} \\ &\text{HAUs} = \text{Highest activity units} \\ &\text{Variable cost is per unit} \\ \end{aligned} Variable Cost=HAUs−Lowest Activity UnitsHAC−Lowest Activity Costwhere:HAC=Highest activity costHAUs=Highest activity unitsVariable cost is per unit Next, use the following formula to determine the fixed cost component: \begin{aligned} &\text{Fixed Cost} = \text{HAC} - ( \text{Variable Cost} \times \text{HAUs} ) \\ \end{aligned} Fixed Cost=HAC−(Variable Cost×HAUs) Use the results of the first two formulas to calculate the high-low cost result using the following formula: \begin{aligned} &\text{High-Low Cost} = \text{Fixed Cost} + ( \text{Variable Cost} \times \text{UA} ) \\ &\textbf{where:} \\ &\text{UA} = \text{Unit activity} \\ \end{aligned} High-Low Cost=Fixed Cost+(Variable Cost×UA)where:UA=Unit activity What Does the High-Low Method Tell You? The costs associated with a product, product line, equipment, store, geographic sales region, or subsidiary, consist of both variable costs and fixed costs. To determine both cost components of the total cost, an analyst or accountant can use a technique known as the high-low method. The high-low method is used to calculate the variable and fixed cost of a product or entity with mixed costs. It takes two factors into consideration. It considers the total dollars of the mixed costs at the highest volume of activity and the total dollars of the mixed costs at the lowest volume of activity. The total amount of fixed costs is assumed to be the same at both points of activity. The change in the total costs is thus the variable cost rate times the change in the number of units of activity. The high-low method is a simple way to segregate costs with minimal information. The simplicity of the approach assumes the variable and fixed costs as constant, which doesn't replicate reality. Other cost-estimating methods, such as least-squares regression, might provide better results, although this method requires more complex calculations. Example of How to Use the High-Low Method For example, the table below depicts the activity for a cake bakery for each of the 12 months of a given year. Below is an example of the high-low method of cost accounting: Cakes Baked (units) The highest activity for the bakery occurred in October when it baked the highest number of cakes, while August had the lowest activity level with only 70 cakes baked at a cost of $3,750. The cost amounts adjacent to these activity levels will be used in the high-low method, even though these cost amounts are not necessarily the highest and lowest costs for the year. We calculate the fixed and variable costs using the following steps: 1. Calculate variable cost per unit using identified high and low activity levels \begin{aligned} &\text{Variable Cost} = \frac{ \text{TCHA} - \text{Total Cost of Low Activity} }{ \text{HAU} - \text{Lowest Activity Unit} } \\ &\text{Variable Cost} = \frac { \$5,550 - \$3,750 }{ 125 - 70 } \\ &\text{Variable Cost} = \frac { \$1,800 }{ 55 } = \$32.72 \text{ per Cake} \\ &\textbf{where:} \\ &\text{TCHA} = \text{Total cost of high activity} \\ &\text{HAU} = \text{Highest activity unit} \\ \end{aligned} Variable Cost=HAU−Lowest Activity UnitTCHA−Total Cost of Low ActivityVariable Cost=125−70$5,550−$3,750Variable Cost=55$1,800=$32.72 per Cakewhere:TCHA=Total cost of high activityHAU=Highest activity unit 2. Solve for fixed costs To calculate the total fixed costs, plug either the high or low cost and the variable cost into the total cost formula: \begin{aligned} &\text{Total Cost} = ( \text{VC} \times \text{Units Produced} ) + \text{Total Fixed Cost} \\ &\$5,550 = ( \$32.72 \times 125 ) + \text{Total Fixed Cost} \\ &\$5,550 = \$4,090 + \text{Total Fixed Cost} \\ &\text{Total Fixed Cost} = \$5,550 - \$4,090 = \$1,460 \\ &\textbf{where:} \\ &\text{VC} = \text{Variable cost per unit} \\ \end{aligned} Total Cost=(VC×Units Produced)+Total Fixed Cost$5,550=($32.72×125)+Total Fixed Cost$5,550=$4,090+Total Fixed CostTotal Fixed Cost=$5,550−$4,090=$1,460where:VC=Variable cost per unit 3. Construct total cost equation based on high-low calculations above Using all of the information above, the total cost equation is as follows: \begin{aligned} &\text{Total Cost} = \text{Total Fixed Cost} + ( \text{VC} \times \text{Units Produced} ) \\ &\text{Total Cost} = \$1,460 + ( \$32.72 \times 125 ) = \$5,550 \\ \end{aligned} Total Cost=Total Fixed Cost+(VC×Units Produced)Total Cost=$1,460+($32.72×125)=$5,550 This can be used to calculate the total cost of various units for the bakery. The Difference Between the High-Low Method and Regression Analysis The high-low method is a simple analysis that takes less calculation work. It only requires the high and low points of the data and can be worked through with a simple calculator. It also gives analysts a way to estimate future unit costs. However, the formula does not take inflation into consideration and provides a very rough estimation because it only considers the extreme high and low values, and excludes the influence of any outliers. Regression analysis helps forecast costs as well, by comparing the influence of one predictive variable upon another value or criteria. It also considers outlying values that help refine the results. However, regression analysis is only as good as the set of data points used, and the results suffer when the data set is incomplete. It's also possible to draw incorrect conclusions by assuming that just because two sets of data correlate with each other, one must cause changes in the other. Regression analysis is also best performed using a spreadsheet program or statistics program. Limitations of the High-Low Method The high-low method is relatively unreliable because it only takes two extreme activity levels into consideration. The high or low points used for the calculation may not be representative of the costs normally incurred at those volume levels due to outlier costs that are higher or lower than would normally be incurred. In this case, the high-low method will produce inaccurate results. The high-low method is generally not preferred as it can yield an incorrect understanding of the data if there are changes in variable or fixed cost rates over time or if a tiered pricing system is employed. In most real-world cases, it should be possible to obtain more information so the variable and fixed costs can be determined directly. Thus, the high-low method should only be used when it is not possible to obtain actual billing data. Harvard Business School. "What Is Regression Analysis in Business Analytics?"
Solving an MDP with Q-Learning from scratch \| Deep Reinforcement Learning for Hackers (Part 1) \| Curiousily - Hacker's Guide to Machine Learning It is time to learn about value functions, the Bellman equation, and Q-learning. You will use all that knowledge to build an MDP and train your agent using Python. Ready to get that ice cream? Here’s an example of how well-trained agents can act in their environments given the proper incentive: Why do we need the discount factor \gamma ? The total reward that your agent will receive from the current time step t to the end of the task can be defined as: R_t = r_t + r_{t + 1} + \ldots + r_n That looks ok, but let’s not forget that our environment is stochastic (the supermarket might close any time now). The discount factor allows us to value short-term reward more than long-term ones, we can use it as: R_t = R_t + \gamma r_{t+1} + \ldots + \gamma^{n - t} r_n = r_t + \gamma R_{t+1} Our agent would perform great if he chooses the action that maximizes the (discounted) future reward at every step. It would be great to know how “good” a given state s is. Something to tell us: no matter the state you’re in if you transition to state s your total reward will be x , word! If you start from s \pi . That would spare us from revisiting same states over and over again. The value function does this for us. It depends on the state we’re in s and the policy \pi your agent is following. It is given by: V^{\pi}(s) = \mathbb{E}(\sum_{t \geq 0}\gamma^t r_t) \quad \forall s \in \mathbb{S} There exists an optimal value function that has the highest value for all states. It is given by: V^(s) = \max_{\pi}V^{\pi}(s) \quad \forall s \in \mathbb{S} Yet, your agent can’t control what state he ends up in, directly. He can influence it by choosing some action a . Let’s introduce another function that accepts state and action as parameters and returns the expected total reward - the Q function (it represents the “quality” of a certain action given a state). More formally, the function Q^{\pi}(s, a) gives the expected return when starting in s , performing and following \pi Again, we can define the optimal Q-function Q^(s, a) that gives the expected total reward for your agent when starting at s and picks action a . That is, the optimal Q-function tells your agent how good of a choice is picking a when at state s There is a relationship between the two optimal functions V^* Q^* V^(s) = \max_aQ^(s, a) \quad \forall s \in \mathbb{S} That is, the maximum expected total reward when starting at s Q^(s, a) over all possible actions. Q^(s, a) we can extract the optimal policy \pi^* by choosing the action a that gives maximum reward Q^(s, a) s \pi^(s) = \text{arg}\max_{a} Q^* (s, a) \quad \forall s \in \mathbb{S} There is a nice relationship between all functions we defined so far. You now have the tools to identify states and state-action pairs as good or bad. More importantly, if you can identify V^* Q^* , you can build the best possible agent there is (for the current environment). But how do we use this in practice? Learning with Q-learning Let’s focus on a single state s a . We can express Q(s, a) recursively, in terms of the Q value of the next state s' Q(s, a) = r + \gamma \max_{a'}Q(s', a') This equation, known as the Bellman equation, tells us that the maximum future reward is the reward the agent received for entering the current state s plus the maximum future reward for the next state s' . The gist of Q-learning is that we can iteratively approximate Q^* using the Bellman equation described above. The Q-learning equation is given by: Q_{t+1}(s_t, a_t) = Q_t(s_t, a_t) + \alpha(r_{t+1} + \gamma \max_{a}Q_t(s_{t + 1}, a) - Q_t(s_t, a_t)) \alpha is the learning rate that controls how much the difference between previous and new Q value is considered. Can your agent learn anything using this? At first - no, the initial approximations will most likely be completely random/wrong. However, as the agent explore more and more of the environment, the approximated Q values will start to converge to Q^* Okay, it is time to get your ice cream. Let’s try a simple case first: Simple MDP - 4 possible states The initial state looks like this: 1ZOMBIE = "z" 2CAR = "c" 3ICE_CREAM = "i" 4EMPTY = "" 6grid = [ 7 [ICE_CREAM, EMPTY], 8 [ZOMBIE, CAR] 11for row in grid: 12 print(' '.join(row)) 2 z c We will wrap our environment state in a class that holds the current grid and car position. Having a constant-time access to the car position on each step will help us simplify our code: 1class State: 3 def __init__(self, grid, car_pos): 4 self.grid = grid 5 self.car_pos = car_pos 7 def __eq__(self, other): 8 return isinstance(other, State) and self.grid == other.grid and self.car_pos == other.car_pos 10 def __hash__(self): 11 return hash(str(self.grid) + str(self.car_pos)) 14 return f"State(grid={self.grid}, car_pos={self.car_pos})" All possible actions: 1UP = 0 2DOWN = 1 3LEFT = 2 4RIGHT = 3 6ACTIONS = [UP, DOWN, LEFT, RIGHT] and the initial state: 1start_state = State(grid=grid, car_pos=[1, 1]) Your agent needs a way to interact with the environment, that is, choose actions. Let’s define a function that takes the current state with an action and returns new state, reward and whether or not the episode has completed: 1from copy import deepcopy 3def act(state, action): 5 def new_car_pos(state, action): 6 p = deepcopy(state.car_pos) 7 if action == UP: 8 p[0] = max(0, p[0] - 1) 9 elif action == DOWN: 10 p[0] = min(len(state.grid) - 1, p[0] + 1) 11 elif action == LEFT: 12 p[1] = max(0, p[1] - 1) 13 elif action == RIGHT: 14 p[1] = min(len(state.grid[0]) - 1, p[1] + 1) 16 raise ValueError(f"Unknown action {action}") 19 p = new_car_pos(state, action) 20 grid_item = state.grid[p[0]][p[1]] 22 new_grid = deepcopy(state.grid) 24 if grid_item == ZOMBIE: 25 reward = -100 26 is_done = True 27 new_grid[p[0]][p[1]] += CAR 28 elif grid_item == ICE_CREAM: 29 reward = 1000 32 elif grid_item == EMPTY: 33 reward = -1 34 is_done = False 35 old = state.car_pos 36 new_grid[old[0]][old[1]] = EMPTY 37 new_grid[p[0]][p[1]] = CAR 38 elif grid_item == CAR: 42 raise ValueError(f"Unknown grid item {grid_item}") 44 return State(grid=new_grid, car_pos=p), reward, is_done In our case, one episode is starting from the initial state and crashing into a Zombie or eating the ice cream. Ok, it is time to implement the Q-learning algorithm and get the ice cream. We have a really small state space, only 4 states. This allows us to keep things simple and store the computed Q values in a table. Let’s start with some constants: 4random.seed(42) # for reproducibility 6N_STATES = 4 7N_EPISODES = 20 9MAX_EPISODE_STEPS = 100 11MIN_ALPHA = 0.02 13alphas = np.linspace(1.0, MIN_ALPHA, N_EPISODES) 14gamma = 1.0 15eps = 0.2 17q_table = dict() We will decay the learning rate, alpha, every episode - as your agent explores more and more of the environment, he will “believe” that there is not that much left to learn. Additionally, limits for the number of training episodes and steps are defined. Dicts in Python can be a bit clunky, so we’re using a helper function q that gives the Q value for a state-action pair or for all actions, given a state: 1def q(state, action=None): 3 if state not in q_table: 4 q_table[state] = np.zeros(len(ACTIONS)) 6 if action is None: 7 return q_table[state] 9 return q_table[state][action] Choosing an action given the current state is really simple - act with random action with some small probability or the best action seen so far (using our q_table): 1def choose_action(state): 2 if random.uniform(0, 1) < eps: 3 return random.choice(ACTIONS) 5 return np.argmax(q(state)) Why your agent uses random actions, sometimes? Remember, the environment is unknown, so it has to be explored in some way - your agent will do so using the power of randomness. Up next, training your agent using the Q-learning algorithm: 1for e in range(N_EPISODES): 3 state = start_state 4 total_reward = 0 5 alpha = alphas[e] 7 for _ in range(MAX_EPISODE_STEPS): 8 action = choose_action(state) 9 next_state, reward, done = act(state, action) 10 total_reward += reward 12 q(state)[action] = q(state, action) + \ 13 alpha (reward + gamma * np.max(q(next_state)) - q(state, action)) 14 state = next_state 15 if done: 17 print(f"Episode {e + 1}: total reward -> {total_reward}") 1Episode 1: total reward -> 999 2 Episode 2: total reward -> 998 8 Episode 8: total reward -> -100 10 Episode 10: total reward -> 999 Here, we use all of the helper functions defined above to ultimately train your agent to behave (hopefully) kinda optimal. We start with the initial state, at every episode, choose an action, receive reward and update our Q values. Note that the implementation looks similar to the formula for Q-learning, discussed above. You can clearly observe that the agent learns how to obtain a higher reward, really quickly. Our MDP is really small, though, and this might be just a fluke. Moreover, looking at some episodes, you can see that the agent hit a Zombie (twice). Did it learn something? Let’s extract the policy your agent has learned by selecting the action with maximum Q value at each step, we will do that manually, like a boss. First up, the start_state: Your agent stars here at every new episode 1sa = q(start_state) 2print(f"up={sa[UP]}, down={sa[DOWN]}, left={sa[LEFT]}, right={sa[RIGHT]}") 1up=998.99, down=225.12, left=-85.10, right=586.19 UP seems to have the highest Q value, let’s take that action: 1new_state, reward, done = act(start_state, UP) The new state looks like this: Getting closer to the ice cream 1sa = q(new_state) 1up=895.94, down=842.87, left=1000.0, right=967.10 But of course, going left will get you the ice cream! Hooray! Your agent seems to know it’s way around here. Isn’t this amazing? Your agent doesn’t know anything about the “rules of the game”, yet it manages to learn that Zombies are bad and ice cream is great! Also, it tries to reach the ice cream as quickly as possible. The reward seems to the ultimate signal that drives the learning process. We’re done here! You can now build complex agents that find optimal policies quickly. Except, maybe not. This was a very simple MDP. Next, we will find how Neural Networks fit into the Reinforcement Learning framework.
Estimate state-space model by reduction of regularized ARX model - MATLAB ssregest - MathWorks Switzerland \begin{array}{l}\stackrel{˙}{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)+Ke\left(t\right)\\ y\left(t\right)=Cx\left(t\right)+Du\left(t\right)+e\left(t\right)\end{array} \left({\lambda }_{1},\sigma ±j\omega ,{\lambda }_{2}\right) \left[\begin{array}{cccc}{\lambda }_{1}& 0& 0& 0\\ 0& \sigma & \omega & 0\\ 0& -\omega & \sigma & 0\\ 0& 0& 0& {\lambda }_{2}\end{array}\right] P\left(s\right)={s}^{n}+{\alpha }_{1}{s}^{n-1}+\dots +{\alpha }_{n-1}s+{\alpha }_{n} A=\left[\begin{array}{ccc}\begin{array}{l}0\\ 1\\ 0\\ 0\\ ⋮\\ 0\end{array}& \begin{array}{l}0\\ 0\\ 1\\ 0\\ ⋮\\ 0\end{array}& \begin{array}{ccc}\begin{array}{cc}\begin{array}{l}0\\ 0\\ 0\\ 1\\ ⋮\\ 0\end{array}& \begin{array}{l}\dots \\ \dots \\ \dots \\ \dots \\ \ddots \\ \dots \end{array}\end{array}& \begin{array}{l}0\\ 0\\ 0\\ 0\\ ⋮\\ 1\end{array}& \begin{array}{l}-{\alpha }_{n}\\ -{\alpha }_{n-1}\\ -{\alpha }_{n-2}\\ -{\alpha }_{n-3}\\ \text{ }⋮\\ -{\alpha }_{1}\end{array}\end{array}\end{array}\right]
Implicit solver for continuous-time algebraic Riccati equations - MATLAB icare - MathWorks Benelux {A}^{T}XE+{E}^{T}XA+{E}^{T}XGXE-\left({E}^{T}XB+S\right){R}^{-1}\left({B}^{T}XE+{S}^{T}\right)+Q\text{ }=\text{ }0 {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}-\mathrm{XB}{\mathit{B}}^{\mathit{T}}\mathit{X}+\mathit{C}{\mathit{C}}^{\mathit{T}}=0 A=\left[\begin{array}{ccc}1& -2& 3\\ -4& 5& 6\\ 7& 8& 9\end{array}\right]\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}B=\left[\begin{array}{c}5\\ 6\\ -7\end{array}\right]\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}C=\left[\begin{array}{ccc}7& -8& 9\end{array}\right]. \mathit{G}=-\mathit{B}{\mathit{B}}^{\mathit{T}} \mathit{Q}={\mathit{C}}^{\mathit{T}}\mathit{C} {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}–\left[\mathrm{XB},{\mathit{C}}^{\mathit{T}}\right]\left[I,0;0,-I\right]\left[{\mathit{B}}^{\mathit{T}}\mathit{X};C\right]=0 {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}-\left(\mathrm{XB}+S\right){\mathit{R}}^{-1}\left({B}^{\mathit{T}}\mathit{X}+{S}^{\mathit{T}}\right)=0 \mathit{B}=\left[\mathit{B},\mathrm{0}\right],\text{\hspace{0.17em}}\mathit{S}=\left[0,{\mathit{C}}^{\mathit{T}}\right],\mathrm{and}\text{\hspace{0.17em}}\mathit{R}=\left[\mathit{I},0;0,-\mathit{I}\right] {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}-\mathrm{XB}{\mathit{B}}^{\mathit{T}}\mathit{X}+\mathit{C}{\mathit{C}}^{\mathit{T}}=0 A=\left[\begin{array}{ccc}1& -2& 3\\ -4& 5& 6\\ 7& 8& 9\end{array}\right]\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}B=\left[\begin{array}{c}5\\ 6\\ -7\end{array}\right]\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}C=\left[\begin{array}{ccc}7& -8& 9\end{array}\right]. {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}–\left[\mathrm{XB},{\mathit{C}}^{\mathit{T}}\right]\left[I,0;0,-I\right]\left[{\mathit{B}}^{\mathit{T}}\mathit{X};C\right]=0 {\mathit{A}}^{\mathit{T}}\mathit{X}+\mathrm{XA}-\left(\mathrm{XB}+S\right){\mathit{R}}^{-1}\left({B}^{\mathit{T}}\mathit{X}+{S}^{\mathit{T}}\right)=0 \mathit{B}=\left[\mathit{B},\mathrm{0}\right],\text{\hspace{0.17em}}\mathit{S}=\left[0,{\mathit{C}}^{\mathit{T}}\right],\mathrm{and}\text{\hspace{0.17em}}\mathit{R}=\left[\mathit{I},0;0,-\mathit{I}\right] {a}_{i,\text{\hspace{0.17em}}j}={\overline{a}}_{j,\text{\hspace{0.17em}}i} K\text{ }=\text{ }{R}^{-1}\left({B}^{T}XE+{S}^{T}\right). L\text{ }=\text{ }eig\left(A+GXE-BK,E\right). \left[\begin{array}{cc}\begin{array}{c}Q\\ {S}^{T}\end{array}& \begin{array}{c}S\\ R\end{array}\end{array}\right]\text{ }\ge \text{ }0 \left[\begin{array}{cc}A-B{R}^{-1}{S}^{T}& Q-S{R}^{-1}{S}^{T}\end{array}\right] M-sN\text{ }=\text{ }\left[\begin{array}{ccc}A& G& B\\ -Q& -{A}^{T}& -S\\ {S}^{T}& {B}^{T}& R\end{array}\right]-s\left[\begin{array}{ccc}E& 0& 0\\ 0& {E}^{T}& 0\\ 0& 0& 0\end{array}\right] \begin{array}{l}X\text{ }={\text{ D}}_{x}\text{ }V{U}^{-1}{\text{ D}}_{x}\text{ }{E}^{-1}\text{,}\\ K\text{ }=\text{ }-{D}_{r}{\text{ WU}}^{-1}{\text{ D}}_{x},\end{array} \begin{array}{l}{D}_{x}\text{ }=\text{ }\mathrm{diag}\left({S}_{x}\right),\\ {D}_{r}\text{ }=\text{ }\mathrm{diag}\left({S}_{r}\right).\end{array}
Gravitation Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers A satellite is moving very close to a planet of density \mathrm{\rho } . The time period of the satellite is: \sqrt{\frac{3\mathrm{\pi }}{\mathrm{\rho G}}} {\left(\frac{3\mathrm{\pi }}{\mathrm{\rho G}}\right)}^{3/2} \sqrt{\frac{3\mathrm{\pi }}{2\mathrm{\rho G}}} {\left(\frac{3\mathrm{\pi }}{2\mathrm{\rho G}}\right)}^{3/2} Subtopic: Satellite \| A projectile is fired upwards from the surface of the earth with a velocity {\mathrm{kv}}_{\mathrm{e}} {\mathrm{v}}_{\mathrm{e}} is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals \frac{\mathrm{R}}{{\mathrm{k}}^{2}} \frac{\mathrm{R}}{1 - {\mathrm{k}}^{2}} \frac{2\mathrm{R}}{1 - {\mathrm{k}}^{2}} \frac{2\mathrm{R}}{1 + {\mathrm{k}}^{2}} Subtopic: Escape velocity \| The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is Subtopic: Gravitational Potential \| A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of orbit, then 1. T is conserved 2. V is always positive 3. E is always negative 4. the magnitude of L is conserved but its direction changes continuously In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity ( \omega ) and the distance of the planet from the sun (r). The correct relation for areal velocity is: \frac{dA}{dt} \alpha \omega r \frac{dA}{dt} \alpha {\omega }^{2}r \frac{dA}{dt} \alpha \omega {r}^{2} \frac{dA}{dt} \alpha \sqrt{\omega r} Subtopic: Kepler's Laws \| If A is the areal velocity of a planet of mass M, its angular momentum is \frac{\mathrm{M}}{\mathrm{A}} {\mathrm{A}}^{2}M {\mathrm{AM}}^{2} Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is -\frac{5\mathrm{Gm}}{\mathrm{r}} -\frac{6\mathrm{Gm}}{\mathrm{r}} -\frac{9\mathrm{Gm}}{\mathrm{r}} Subtopic: Gravitational Field \| Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as: {T}^{2} \alpha \frac{1}{{U}^{3}} T \alpha \frac{1}{{U}^{3}} {T}^{2} \alpha {U}^{3} {T}^{2} \alpha \frac{1}{{U}^{2}} A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45 ° to the horizontal, what should be its speed so that it escapes from the earth? \frac{\mathrm{v}}{\sqrt{2}} \sqrt{2}\mathrm{v} Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:
seminars - Quasimultiplicativity of a typical fiber-bunched SL_n(R) cocycle over a subshift of finite type. Quasimultiplicativity of a typical fiber-bunched SL_n(R) cocycle over a subshift of finite type. 1. title: Quasimultiplicativity of a typical fiber-bunched SL_n(R) cocycle over a subshift of finite type. Abstract: Following Bonatti-Viana/Avila-Viana, a fiber-bunched SL_n(R) cocycle A over a SFT \left(Sigma,f\right) is called typical if there is a fixed point p in which A(p) has simple eigenspectrum and a homoclinic point z of p such that the holonomy loop psi_p^z = H^{s}_{z,p}H^u_{p,z} twists the eigendirections of A(p) into general position. Such are known to be sufficient conditions for the simplicity of the Lyapunov exponents for any ergodic measure with local product structure. Under the same condition, we show that the quasimultiplicativity holds for the cocycle A; suitably interpreted, the quasimultiplicativity can be considered as the submultiplicativity on the norm of the dynamically defined product A^n(x). We will also discuss some applications coming from the quasimultiplicativity. 2. title: Unique equilibrium state for geodesic flow on surfaces with no focal points. Abstract: Due to work of Bowen, it is well-known that any system with expansivity and the specification property has a unique equilibrium state for any potential with Bowen property (bounded variation). Such pair includes a hyperbolic system with a Holder potential. Recently, Climenhaga and Thompson developed a program to relax the assumptions from Bowen's work to a non-uniformly hyperbolic setting. Using their techniques, Burns-Climenhaga-Fisher-Thompson established the existence and the uniqueness of an equilibrium state for a large class of potentials over the geodesic flow on closed rank one manifolds. We show that their results can be extended to the geodesic flow over surfaces with no focal points, and will discuss properties of the unique equilibrium state coming from the geometry.
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Relay Therapeutics S-1 - Dennis Relay Therapeutics S-1 Relay Therapeutics is a clinical stage small molecule drug development company based in Cambridge, founded in 2015 to utilize protein motion screening assays for precision medicine. The company IPOed in July 2020, raising $400 M at a valuation of roughly $1.8 B. The company is interesting to me because it is yet another tech enabled screening platform, but one that was partially invented by those at D.E. Shaw Research. It also seems to me like the closest thing to a real pharmaceutical platform company that isn’t untractable to analyze. Reading S-1 filings has been a good exercise to begin to understand the work being done at the cutting edge, and also to read the closest thing to a publicly released comprehensive narrative of a company’s prospects. The company is the brainchild of 4 superstar scientists, Dr. Mark Murcko, Dr. Matt Jacobson, Dr. Dorothee Kern, and Dr. David E. Shaw, all leaders in the field of computational drug design. The platform they developed, termed ‘Dynamo’, allows the development of drugs for protein targets based on their motion and changing state through time. The idea is that conventional drug development tools such as structure based drug design (championed by Murcko at Vertex), relies on static images of proteins, which is different from the behavior of proteins in their natural state, which is more dynamic. By understanding how proteins change shape through time, Dynamo allows motion based drug design for the synthesis of new molecules with improved specificity and selectivity. This work has been catalyzed by immense progress in computation, with Dynamo being supported by the Anton 2 supercomputer, a special built computer for molecular dynamics owned by D.E. Shaw Research. Relay’s total operating expenses grew from roughly 50 M in 2018, to 84 M in 2019, to 138 M in 2020 (with 83 M in 2020 revenue). Operating expenses for 1st 3 months 2021 were 42 M compared to 26 M in 2020. As of March 31st, Relay has 742 M in total assets, with 263 M of that being cash. Link to latest 10-Q Importantly, Relay focuses on indications with low biological risk (ie. it is clear that the target proteins influence disease pathology), and low clinical execution risk (ie. patient subgroups are easily stratified using companion molecular diagnostics.). With such a strategy, the company currently has 3 lead programs in precision oncology, RLY-1971 for SHP2 dependent solid tumors, RLY-4008 for patients with FGFR2-mediated cancers, and RLY-PI3K1047 for PI3Kα H1047X mutants. This program is an oral small molecule inhibitor of the protein tyrosine phosphatase (opposite of a kinase) Src homology-2 domain-containing protein tyrosine phosphatase-2 (SHP2) that binds the protein and stabilizes it in its inactive conformation. Relay believes that “inhibition of SHP2 could block a common path that cancer cells exploit to avoid killing by other antitumor agents, thus overcoming or delaying the onset of resistance to those therapies.” Given the range of cancers that SHP2 seems to be involved in, Relay believes that RLY-1971 could become a ‘backbone therapy’, meaning that it could be a relatively safe starting point onto which other drugs can be built on for combination. SHP2 is involved downstream of other common RTK targets pictured above, and thus represents a common node to address bypass resistance, a phenomena where cancers will evolve mutations to reduce dependency on a particular pathway targeted by a therapeutic. For example, MEK inhibitors can fail after tumors shift growth factor signaling to alternate RTKs to reduce sensitivity to such therapies. As a result, Relay has successfully demonstrated synergies between their SHP2 inhibitor and other targeted therapies targeting KRAS and ALK. The current development plan is a monotherapy Phase I trial to assess dosing considerations, and subsequent combination trials with other targeted agents, the soonest of which will be with Genentech. Genentech also announced in December 2020, a worldwide license and collaboration agreement for this drug in which Relay got $75 M up front, with $25 M in near term and $695 M in long term milestones. Although there are no approved therapies targeted SHP2, the competitive landscape is quite crowded, with other companies with clinical trials including Revolution Medicines (with Sanofi), Novartis, Navire Pharma, Erasca, and Jacobio (with AbbVie). Their next program is an oral small molecule selective inhibitor of fibroblast growth factor receptor 2, or FGFR2. FGFR2 is one of 4 members of the FGFR family, and RLY-4008 selectively inhibits FGFR2, without effects on the other family members. Targeting growth factors and their receptors is not new, and there are several other FGFR2 inhibitors approved including erdafitinib and pemigatinib. However, these and other molecules in clinical trials are not selective specifically for the FGFR2 family member, and due to dual inhibition of FGFR1, these other molecules cause dose limiting hyperphosphatemia. The molecule Relay developed using their platform was designed to have improved specificity for FGFR2, up to 200 fold over FGFR1 which casuses the hyperphosphatemia. This is important as it widens the therapeutics window to allow for use in combination treatments and more therapeutically active dose regimens. Currently, response rates max out at 37% at a once daily dosing level. It seems that Relay wants to dose twice daily in their initiated Phase I clinical trial. One other differentiator of Relay’s molecule is improved potency against common FGFR2 resistance mutations, likely because their molecule hits an allosteric site. To me, this is a no brainer since it hits a different site, but Relay’s molecule will probably have its own resistance mutations that it doesn’t work on. Relay’s clinical development for the molecule is occurring in a Phase I initiated in September 2020 with patients enriched for FGFR2 altered solid tumors, using a twice daily dosing schedule. Relay makes a point of defining the target patient population as follows: “We believe FGFR2-mediated cancers affect approximately 8,000 late-line patients annually in the United States, of which fusions represent approximately 2,700, amplifications approximately 1,600, and mutations approximately 3,800. In the future, if RLY-4008 advances to earlier lines of treatment, we believe it could potentially address approximately 20,000 patients annually in the United States across the different alterations.” RLY-PI3K1047 Finally, Relay wants to develop a franchise of PI3K mutant inhibitors. Phosphoinositide 3-kinase alpha (PI3Kα) is overactive in many cancerous malignancies, and acts downstream of tyrosine kinases like RAS to promote cell growth, proliferation, and survival. PI3Kα mutants are present in 13% of all solid tumors, but is difficult to drug. The one FDA approved PI3Kα inhibitor Alpelisib (Piqray) is used for breast cancer but comes with significant toxicity. Relay’s lead program in this franchise selectively targets the PI3Kα H1047X mutant, and two additional mutations of interest are E542X and E545X. Alpelisib on the other hand, is less selective and binds the active site of PI3Kα, which inhibits both mutants and the wild type PI3Kα. This leads to on target toxicity, including hyperglycemia in 64% of patients (36% Grade 3 or 4). Additional off target toxicities include gastrointestinal toxicity in 93% of patients (9% Grade 3) and 36% of patients with rash (10% Grade 3). This combination of adverse events resulted in 64% of patients requiring dose reductions, 25% of patients discontinuing treatment, and 87% of patients requiring insulin and other anti-diabetic medication to manage hyperglycemia. Despite 11 month progression-free survival (PFS) in the SOLAR-1 Phase 3 trial of alpelisib, the median duration of dosing in the alpelisib arm was 5.5 months, indicating the majority of patients discontinued dosing prior to disease progression. The Relay molecule results in 5-10 fold inhibition for the PI3Kα substrate in in vitro assays, which doesn’t happen for Alpelisib or a competitor molecule, GDC-0077 that is in development by Genentech. RLY-PI3K1047 will be tested in a tumor agnostic setting in 2021 as a monotherapy and in disease specific settings in combination with other targeted therapies. Relay estimates 10,000 late-line patients annually for the H1047X mutant molecule in the United States, and up to 50,000 if it advances to earlier lines of treatment. For the two additional mutations of interest, E542X and E545X, Relay estimates there are 15,000 late-line and 60,000 total patients annually in the United States who might benefit. Computational Drug Design Process To build this pipeline and to fuel future discovery efforts, Relay relies on a next generation chemical development platform termed Dynamo. Importantly, the platform doesn’t sell itself as an entirely automated solution. The core of the platform is the quality of their scientists. However, there are powerful technological advancements that allow their scientists to see things that weren’t possible to see before. But the whole process follows the typical pattern of hypothesis generation, identifying a druggable binding pocket, finding hits, and optimizing leads. 1. Target modulation hypothesis. The process starts with getting a mechanistic understanding of the dynamic behavior of the protein. To do this, Relay performs a range of structural biology techniques on full length proteins including room temperature X-ray crystallography and Cryo-EM. With these experimental datasets (knowing how the atoms are arranged), Relay deploys a computational simulation platform to demonstrate how the protein moves over long time scales (molecular dynamics; knowing how atoms interact). This requires a silly amount of compute power, which Relay accesses via Anton 2, which calculates the force between each atom and every other atom in a given system at discrete time points to model behavior over time. For benchmarking experiments, Relay has simulated systems of up to 1 million atoms at time slices of 2.5 x10-15 seconds at timescales of tens of microseconds. For a sense of comparison, a 10 microsecond simulation of a 1 million atom benchmark protein (satellite tobacco mosaic virus), which requires one day of processing on the Anton 2, would require 271 days on an Nvidia V100. Using these simulations, a motion picture of the protein can be stitched together, and Relay scientists can develop hypotheses about how the protein functions. For example, in the slide below, they describe how they developed a hypothesis for their FGFR2 inhibitor: 2. Identify allosteric binding pockets. Once a hypothesis is generated, Relay drug hunters are able to identify allosteric pockets that are potentially druggable. This step involves a combination of computationally and chemist derived hypotheses which are tested in laboratory experiments on full-length proteins. 3. Hit finding/lead generation. Soon after, their scientists use a bunch of different screening techniques including cloud based virtual screens and a variety of physiologically relevant activity-based and ligand-centric in vitro screens to find hits. 4. Lead optimization. Finally, during optimization, they leverage the same long timescale MD simulations to study binding pocket dynamics and to test analogs of lead compounds to prioritize which ones to synthesize and test experimentally. These simulations also generate a lot of data, which the team uses machine learning to help parse and develop predictive models to prioritize synthesis. This is a flywheel process which allows continual improvement of models and decreased screening burden over time. The information above is a simple and quick summary of what was presented in the S-1 filing, but there is a lot more and probably updated information in corporate presentations on their website. Below are my own takeaways and ~analysis~. The quality of your molecule matters. “The ability of a drug to specifically and deeply inhibit the protein target of interest while minimizing the inhibition of other closely related protein targets can result in a profound difference in outcome. A key example of this is seen in two drugs targeting the altered protein RET. One is a non-specific drug, cabozantinib, and the other is a purpose-built, specific drug, selpercatinib. Seplercatinib increased tumor response rate to 68% from the 28% demonstrated for cabozantinib.” “Importantly, subtle differences in protein conformational dynamics (on the order of a few angstroms) have been observed in otherwise structurally similar proteins. In addition to these small-scale changes, global motion of proteins can create on and off-states that can be dynamically regulated. Defects in the conformational dynamics of proteins have been implicated in up to 40% of all diseases.” Crystal structures do not accurately represent the dynamic nature of a protein in its biological context, which could impede drug design. Relay is still a hypothesis driven drug company. The platform helps them see, but the experience of drug hunters on their team helps them build the molecules. Relay’s success seems to me more a function of throwing a lot of compute at a problem, rather than automation and AI. It is gratifying to see that it is still the scientist at the end who is developing the SAR hypotheses. Foundation Medicine and other companion diagnostics have dramatically expanded the TAM of cancer therapeutics. Segmenting patient populations makes it easier to do trials and hopefully improves outcomes. Gathering more data about patient illnesses has been a huge win for the industry. Relay’s (and I presume other precision oncology companies’) general strategy is to start in a small precision indication and gradually expand to earlier lines or other cancer types. These drugs will be pricey and it is easier to justify at smaller treatment populations. Key differentiators can be the ability to be oral and be amenable to once daily continuous dosing to reduce toxicity. Cancer patients already suffer a great deal and often can’t take medicines because of toxicity or other reasons. Creating medicines that go down easy might be an under-appreciated way to boost outcomes. Relay has also demonstrated to me how easy it is to tear down competition in the pre-clinical setting and how important it is to not be first, but be best. Earlier this year in April, Relay acquired machine learning DNA encoded libraries startup Zebi AI. The deal cost $85 M up front, plus up to $185 M in earnouts and milestones over next 3 years. This acquisition boosts screening capacity and theoretically improves predictive models by adding more data. Current valuation is 3.3 B which is a bit rich for the current pipeline and up significantly since the IPO but speaks to investor confidence in the platform and team. This is on $500 M in total fundraising to date. This seems to indicate that people think the drugs will work. Genentech deal for SHP2 program was quite small also.
Archimedes and logarithmic spirals \| Sayed's Blog Archimedes and logarithmic spirals In his book "Climbing Mount Improbable", Richard Dawkins describes a museum containing every animal that has ever existed and every animal that could ever conceivably exist. Every animal is located next to an animal that most closely resembles it. Every dimension in the museum corresponds to a dimension of variation within the animals - e.g sharpness of teeth is West to East and North to South is horn length. Since there are more than 3 ways that animals can differ, this museum would have to have multiple dimensions. One way to visualise 4 dimensions is to imagine multiple 3 dimensional cubes, with animals occupying the same position relative to the cube are identical in the first 3 qualities but differ in the fourth (e.g coat hairiness). More dimensions can be visualised by making "families" of such cubes. To illustrate his analogy more clearly, Dawkins draws attention to a subset of animals that can mostly be expressed with 3 variables - shells. Many kinds of shells in nature can be modelled as logarithmic spirals. In this post I will demonstrate how to draw different kinds of spirals using code. In a future post I will demonstrate how to model shells using spirals. This is one of the most simple spirals to draw. Let's start by creating 'archimedes.html'. <title>Archimedes Spiral</title> Then define a center to begin from. As Dawkins describes, shells start small and grow at the margins. To model this with a spiral, we will draw starting from a center, and "spiral" outwards by an increasing distance. for (let angle = 0; angle < 5 * 360; angle++) { let x = centerX + Math.sin(Math.toRadians(angle)) * distance; let y = centerY + Math.cos(Math.toRadians(angle)) * distance; distance += 0.1; This code is easier to write with degrees, but Math.sin and Math.cos expect radians, so we need to convert the angles. Math.toRadians = function (number) { return Math.PI * number / 180; This results in a somewhat blurry shell. I fix this by drawing a line between each point. left x = centerX + Math.sin(Math.toRadians(angle)) * distance; left y = centerY + Math.cos(Math.toRadians(angle)) * distance; This results in a clearer spiral. Unlike an archimedes spiral, a logarithmic spiral does not grow by a constant amount every iteration. Instead they "open out" at a constant rate. For example, the gap between turns might double at every coil. The expansion rate of the spiral is actually one of the variables/dimenions used to model the different kinds of shells - Dawkins calls this "flare". Logarithmic spirals are also known as equiangular spirals. Often different types of shell are better modelled with logarithmic spirals. Let's begin by copying some of the setup code from the archimedes spiral, in equiangular.html. <title>Logarithmic Spiral</title> Like before, we will draw a point for every degree. for (let angle = 0; angle < 12 * 360; angle++) { // increase distance here How much do we increase distance? For every full turn, we want to double the distance. We can use logarithms to determine how much to grow. We need to multiply the distance by some number 360 times (a full turn) to cause it to double. To put it mathematically, if y is the distance and x is the multiplier then: yx^{360} = 2y We can cancel out the distance to obtain: x^{360} = 2 To solve this, we can take log base 2 of two. log(x^{360}) = 1 Which, is identical to 360log(x) = 1 We can rearrange this to get log(x) = 1/360 log(x) 1/360 2^{1/360} = x Substituting that in our code, we get: distance = 2(1/360); Like before, let's reduce blurring by using lines instead. distance = (2*(1/360)); This doesn't just work for doubling, we can substitue any growth factor we like. Real shells have tubes, not lines. To mimic this, we will draw circles instead of lines. ctx.drawCircle = function(x, y, radius) { ctx.arc(x, y, radius, 0, Math.PI2, false); ctx.drawCircle(x, y, 15); The tubes usually grow wider, we can model this by making it proportional to the distance from the centre. ctx.drawCircle(x, y, distance/6); Some Cool Patterns In a future tutorial I will demonstrate how to go from here to creating all sorts of different shell shapes, but here I'd like to demonstrate different kinds of patterns that can be obtained by playing around with a few variables. ctx.drawCircle(x, y, distance%15); ctx.drawCircle(x, y, Math.abs(distance%15)); distance = -(2(1/360)); ctx.drawCircle(x, y, Math.abs(distance/6)); distance = (1.5*(1/360)); distance = -(1.5**(1/360));
Scalability Challenge - How to remove duplicates in a large data set (~100M) ? Almost every mobile app has push notification feature. We need to design a way which guards sending multiple push notifications to the same user for the same campaign. Push notification - Push notifications are sent to the users device based on push notification token generated by their mobile device. Size of token - 32B to 4KB Assume total Users - Additional Details - It’s non-performant for us to index push tokens or use them as the primary user key. If user uninstalls the app and subsequently again installs the same app, and then creates a new user profile in the same device. Normally, the mobile platform will generate a new push notif token BUT that's not guaranteed always. For a smaller number of cases, we can end up having multiple user profiles with the same push notif token. Now, to avoid sending multiple notification to same device, we need to filter out those smaller number of duplicate push notif tokens. Memory required to filter 100 MN tokens = 100M x 256 = ~25 GB Bloom Filters are data structures used to efficiently answer queries when we do not have enough "search key" space to handle all possible queries. How Bloom Filter Works? Allocate a bit array of size m Choose k independent hash functions h(x) whose range is [ 0 .. m-1 ] For query q , apply hashes and check if all the corresponding bits are ‘on’ If all the corresponding bits are 'on' means this is duplicate value. NOTE : Bits might be turned ‘on’ by hash collisions leading to false positives What is the error rate and Memory requirement in bloom filter ? With m bits, k hash functions, n input strings, we need to the false positive probability (hash collisions). Probability of setting a bit using a hash function in m bits - \frac{1}{m} Now, Probability of NOT setting a bit - (1 - \frac{1}{m}) As we pass it to k hash functions & we have n input strings so the probability of a bit is not being set after passing it to n input strings and k hash functions is - {p} = (1 - \frac{1}{m})^{kn} Now, the probability of getting error/ hash collision (by mistake setting a bit) for k hash functions is {p} = {(1 - (1 - \frac{1}{m})^{kn})^k} \approx (1 - e^{(-\frac{kn}{m})})^k for minimising error rate - \frac{dp}{dk} → {0} {k} = \frac{m}{n} \times {ln \hspace{0.2cm} (2)} Size of bit array- {m} = - \frac{n \times (ln \hspace{0.2cm} p)}{(ln \hspace{0.2cm} 2)^2} Memory required for 100 Million push tokens with 0.001 error probability - {m} = - \frac{100000000 \times (ln \hspace{0.2cm} 0.001)}{(ln \hspace{0.2cm} 2)^2} \approx {171 MB} This is massive improvement from 25 GB to 171 MB (reducing memory requirements by ~98%) Facebook uses bloom filters for typeahead search, to fetch friends and friends of friends to a user typed query. MakeMyTrip uses bloom filters for personalised discount offers based on user data and behaviour instead of calculating category ( eg - loyalCustomer, newBusCustomer, newFlightCustomer) for their ~70M unique customers Hash functions for Bloom filter should be independent and uniformly distributed. Cryptographic hashes like MD5 or SHA-1 are not good choices for performance reasons. Some of the suitable fast hashes are MurmurHash, FNV hashes andJenkin’s Hashes which are faster (~10x faster), distributed and independent. Source - Suresh Kondamudi's Discussion (CleverTap )
Multiclass classification - Wikipedia Not to be confused with multi-label classification. While many classification algorithms (notably multinomial logistic regression) naturally permit the use of more than two classes, some are by nature binary algorithms; these can, however, be turned into multinomial classifiers by a variety of strategies. Multiclass classification should not be confused with multi-label classification, where multiple labels are to be predicted for each instance. 1.1 Transformation to binary 1.1.1 One-vs.-rest 1.1.2 One-vs.-one 1.2 Extension from binary 1.2.1.1 Extreme learning machines 1.2.2 k-nearest neighbours 1.3 Hierarchical classification The existing multi-class classification techniques can be categorized into (i) transformation to binary (ii) extension from binary and (iii) hierarchical classification.[1] Transformation to binary[edit] This section discusses strategies for reducing the problem of multiclass classification to multiple binary classification problems. It can be categorized into one vs rest and one vs one. The techniques developed based on reducing the multi-class problem into multiple binary problems can also be called problem transformation techniques. One-vs.-rest[edit] One-vs.-rest[2]: 182, 338 (OvR or one-vs.-all, OvA or one-against-all, OAA) strategy involves training a single classifier per class, with the samples of that class as positive samples and all other samples as negatives. This strategy requires the base classifiers to produce a real-valued confidence score for its decision, rather than just a class label; discrete class labels alone can lead to ambiguities, where multiple classes are predicted for a single sample.[2]: 182 [note 1] In pseudocode, the training algorithm for an OvR learner constructed from a binary classification learner L is as follows: L, a learner (training algorithm for binary classifiers) labels y where yi ∈ {1, … K} is the label for the sample Xi a list of classifiers fk for k ∈ {1, …, K} For each k in {1, …, K} Construct a new label vector z where zi = yi if yi = k and zi = 0 otherwise Apply L to X, z to obtain fk Making decisions means applying all classifiers to an unseen sample x and predicting the label k for which the corresponding classifier reports the highest confidence score: {\displaystyle {\hat {y}}={\underset {k\in \{1\ldots K\}}{\arg \!\max }}\;f_{k}(x)} Although this strategy is popular, it is a heuristic that suffers from several problems. Firstly, the scale of the confidence values may differ between the binary classifiers. Second, even if the class distribution is balanced in the training set, the binary classification learners see unbalanced distributions because typically the set of negatives they see is much larger than the set of positives.[2]: 338 One-vs.-one[edit] In the one-vs.-one (OvO) reduction, one trains K (K − 1) / 2 binary classifiers for a K-way multiclass problem; each receives the samples of a pair of classes from the original training set, and must learn to distinguish these two classes. At prediction time, a voting scheme is applied: all K (K − 1) / 2 classifiers are applied to an unseen sample and the class that got the highest number of "+1" predictions gets predicted by the combined classifier.[2]: 339 Like OvR, OvO suffers from ambiguities in that some regions of its input space may receive the same number of votes.[2]: 183 Extension from binary[edit] This section discusses strategies of extending the existing binary classifiers to solve multi-class classification problems. Several algorithms have been developed based on neural networks, decision trees, k-nearest neighbors, naive Bayes, support vector machines and extreme learning machines to address multi-class classification problems. These types of techniques can also be called algorithm adaptation techniques. Multiclass perceptrons provide a natural extension to the multi-class problem. Instead of just having one neuron in the output layer, with binary output, one could have N binary neurons leading to multi-class classification. In practice, the last layer of a neural network is usually a softmax function layer, which is the algebraic simplification of N logistic classifiers, normalized per class by the sum of the N-1 other logistic classifiers. Extreme learning machines[edit] Extreme learning machines (ELM) is a special case of single hidden layer feed-forward neural networks (SLFNs) wherein the input weights and the hidden node biases can be chosen at random. Many variants and developments are made to the ELM for multiclass classification. k-nearest neighbours[edit] k-nearest neighbors kNN is considered among the oldest non-parametric classification algorithms. To classify an unknown example, the distance from that example to every other training example is measured. The k smallest distances are identified, and the most represented class by these k nearest neighbours is considered the output class label. Naive Bayes[edit] Naive Bayes is a successful classifier based upon the principle of maximum a posteriori (MAP). This approach is naturally extensible to the case of having more than two classes, and was shown to perform well in spite of the underlying simplifying assumption of conditional independence. Decision tree learning is a powerful classification technique. The tree tries to infer a split of the training data based on the values of the available features to produce a good generalization. The algorithm can naturally handle binary or multiclass classification problems. The leaf nodes can refer to any of the K classes concerned. Support vector machines are based upon the idea of maximizing the margin i.e. maximizing the minimum distance from the separating hyperplane to the nearest example. The basic SVM supports only binary classification, but extensions have been proposed to handle the multiclass classification case as well. In these extensions, additional parameters and constraints are added to the optimization problem to handle the separation of the different classes. Hierarchical classification[edit] Hierarchical classification tackles the multi-class classification problem by dividing the output space i.e. into a tree. Each parent node is divided into multiple child nodes and the process is continued until each child node represents only one class. Several methods have been proposed based on hierarchical classification. Based on learning paradigms, the existing multi-class classification techniques can be classified into batch learning and online learning. Batch learning algorithms require all the data samples to be available beforehand. It trains the model using the entire training data and then predicts the test sample using the found relationship. The online learning algorithms, on the other hand, incrementally build their models in sequential iterations. In iteration t, an online algorithm receives a sample, xt and predicts its label ŷt using the current model; the algorithm then receives yt, the true label of xt and updates its model based on the sample-label pair: (xt, yt). Recently, a new learning paradigm called progressive learning technique has been developed.[3] The progressive learning technique is capable of not only learning from new samples but also capable of learning new classes of data and yet retain the knowledge learnt thus far.[4] ^ In multi-label classification, OvR is known as binary relevance and the prediction of multiple classes is considered a feature, not a problem. ^ Mohamed, Aly (2005). "Survey on multiclass classification methods" (PDF). Technical Report, Caltech. ^ a b c d e Bishop, Christopher M. (2006). Pattern Recognition and Machine Learning. Springer. ^ Venkatesan, Rajasekar; Meng Joo, Er (2016). "A novel progressive learning technique for multi-class classification". Neurocomputing. 207: 310–321. arXiv:1609.00085. doi:10.1016/j.neucom.2016.05.006. ^ Venkatesan, Rajasekar. "Progressive Learning Technique". Retrieved from "https://en.wikipedia.org/w/index.php?title=Multiclass_classification&oldid=1023970866"
EUDML \| Some results on dominant operators. EuDML \| Some results on dominant operators. Some results on dominant operators. Yang, Youngoh. "Some results on dominant operators.." International Journal of Mathematics and Mathematical Sciences 21.2 (1998): 217-220. <http://eudml.org/doc/48119>. author = {Yang, Youngoh}, keywords = {Fredholm operator; -power class ; Weyl spectrum; dominant operator; spectral mapping theorem for analytic functions; -power class }, title = {Some results on dominant operators.}, AU - Yang, Youngoh TI - Some results on dominant operators. KW - Fredholm operator; -power class ; Weyl spectrum; dominant operator; spectral mapping theorem for analytic functions; -power class Fredholm operator, M -power class \left(N\right) , Weyl spectrum, dominant operator, spectral mapping theorem for analytic functions, M \left(N\right)
EUDML \| -bounded semigroups and implicit evolution equations. EuDML \| -bounded semigroups and implicit evolution equations. B -bounded semigroups and implicit evolution equations. Banasiak, J.. "-bounded semigroups and implicit evolution equations.." Abstract and Applied Analysis 5.1 (2000): 13-32. <http://eudml.org/doc/49505>. @article{Banasiak2000, author = {Banasiak, J.}, keywords = {evolution equation; -bounded semigroups; abstract Cauchy problem; Hille-Yoshida theory; -bounded semigroups}, title = {-bounded semigroups and implicit evolution equations.}, AU - Banasiak, J. TI - -bounded semigroups and implicit evolution equations. KW - evolution equation; -bounded semigroups; abstract Cauchy problem; Hille-Yoshida theory; -bounded semigroups evolution equation, B -bounded semigroups, abstract Cauchy problem, Hille-Yoshida theory, B -bounded semigroups Articles by Banasiak
Dark Matter, Dark Energy, and the Accelerating Universe \| Astronomy 801: Planets, Stars, Galaxies, and the Universe So far, we have discussed the origin of the Universe and the age of the Universe, but not its ultimate fate. This has been a question that has been pursued for many years, and a number of theorists were considering possible ideas for the fate of the Universe concurrently with the development of the Big Bang model. If you compare college astronomy textbooks today to those published about 20 years ago or more, though, you will find that this part of the discussion of cosmology has changed substantially since about 1998. The reason is because newly discovered evidence for dark energy complicates the matter. Dark matter plays a role in determining the fate of the Universe, too. We have already encountered dark matter during our discussion of the rotation curve of the Milky Way, but I will go into some more detail here. Recall that in the Milky Way, we find that the outer parts of the Galaxy are rotating much faster than expected if all the matter in the Galaxy is visible matter. Based on the rotation curve of the Milky Way, it appears that the Galaxy contains more dark matter than luminous matter. Beyond individual galaxies, though, there is also evidence for dark matter in clusters of galaxies. Just like the rotation curves of Galaxies, you can also study the velocities of galaxies inside massive galaxy clusters. The escape speed from an object (in this case a cluster) depends on the mass of that object. In many clusters, the velocities of the galaxies in that cluster suggest that the cluster could not remain bound if all it contains is normal matter. There must be additional dark matter in the cluster, or else many of the galaxies would escape. Additional evidence for dark matter in galaxy clusters comes from images like the one below. When we observe some clusters, we see another effect predicted by Einstein, called strong gravitational lensing. Since Einstein predicted that massive objects can warp spacetime, he showed that the light from a background object will be bent if it passes by a massive object, like a galaxy cluster. Here is an image of a cluster lensing a background galaxy, just as predicted. Figure 10.16: Hubble image of a cluster lensing a background galaxy The arcs that you see in between the yellow galaxies are distorted images of the background galaxy. The details of the lensing effect depend on the mass of the lens (that is, the more massive the lens, the more distorted the background galaxy), and show that this particular cluster contains more mass than it appears based solely on the luminous galaxies. Like the case for individual galaxies, it appears that the amount of dark matter in clusters like this is significantly larger than the amount of luminous mass. The evidence for dark energy comes from different sources. The first piece of evidence comes from the Hubble diagram as calibrated by Type Ia supernovae. These objects are so luminous that they allow us to measure their distances accurately out to redshifts of z>1 . When the first Hubble diagrams were constructed using distances obtained from Type Ia supernovae, though, astronomers found a significant deviation from expectations. The supernovae were systematically fainter than expected at large distances. Below is a Hubble Diagram using Type Ia supernovae studied by two different teams. Figure 10.17: Plot of Hubble Diagram from Type Ia supernovae Credit: High-z supernova search team You can see in both the top panel and the bottom panel that the points with redshifts greater than about 0.5 seem to deviate from the straight line. This suggested that our (at the time) accepted models for the expansion of the Universe were incorrect. We know that there is some matter in the universe (we live on a giant ball of matter called the Earth, after all), but our measurements of the luminous and dark matter in the Universe have shown that there is not enough matter to close the universe, or even to make it flat (that is {\rho }_{\text{ave}}<{\rho }_{\text{crit}} ). In the mid-1990s, the data suggested that the universe is open and that the total amount of luminous matter plus dark matter in the universe was only about 30% of the critical amount necessary for a flat universe. Given that there is some matter in the Universe, though, we expected that for objects at large distances, their distances would deviate from Hubble's Law. The reason is that the combined gravitational pull of all of the objects on each other would oppose the expansion of the Universe, causing it to decelerate. Because of deceleration, at the largest distances, objects should appear closer to us than predicted by their redshift. So for many years, the question that many astronomers were pursuing using different research techniques was "How much is the universe decelerating?". However, for supernovae, the exact opposite was found. These objects appear to be farther away from us than predicted by their redshifts and Hubble's Law. The only way for this to happen is if the expansion of the universe is accelerating, not decelerating. In order for the universe to accelerate, there must be some force pushing all of the galaxies away from each other, and this force must be strong enough to counteract the deceleration by gravity. Today, we do not know what is the exact cause of this force, just that it exists. Since we call the matter that we cannot observe directly "dark matter," we call this new mysterious force (or equivalently, the energy provided by this force) dark energy. If we add in the contribution of dark energy to the density of the universe, it appears that the combination of normal matter, dark matter, and dark energy is enough to make the universe flat ( \Omega =1 , but here \Omega ={\Omega }_{\text{matter}}\quad +\text{ }{\Omega }_{\text{darkenergy}} ). Remember, our measurements showed that matter makes up only about 30% of the critical density, so dark energy makes up the other 70%! It appears that the 30% matter is about 4% normal matter (people, planets, stars, galaxies) and about 26% is dark matter. So this means that at this time, astronomers can only directly observe about 4% of the universe, and the other 96% is divided up among this peculiar dark matter and dark energy, which we have still not identified. You may think that this is a bold claim based solely upon the distances to a handful of Type Ia supernovae, but the fluctuations in the CMB seen by WMAP also predict that normal matter only makes up approximately 4% of the Universe. Thus, the results from WMAP appear to confirm the results from Type Ia supernovae. Below is an image from the WMAP team for their predictions for the contents of the Universe. Figure 10.18: Pie charts showing the contents of the Universe Click Here for text alternative for Figure 10.18. Contents of the Universe Today Atoms 4.6% Contents of the Universe 13.7 billion years ago At this point, let us reconsider the question of what will happen to the Universe over time. Right now, it is difficult to say, because we do not understand dark energy very well. However, it appears that, given the accelerating expansion of the Universe, the Universe will grow larger and larger and colder and colder. All of the luminous objects in the Universe will eventually die out, and the Universe will eventually end in a "Big Freeze," where it will be too cold to support any life. ‹ The Cosmic Microwave Background up Additional Resources ›
Application Of Derivatives, Popular Questions: CBSE Class 12-science MATH, Math Part I - Meritnation Aadarsh Vyas asked a question Determine the interval, where f(x)=sin x - cos x, 0<x<2pie is strictly increasing or decreasing? Solve:sin px cos y = cos px sin y + p. Shaurya & 1 other asked a question SHOW THAT THE SEMI-VERTICAL ANGLE OF RIGHT CIRCULAR CONE OF GIVEN SURFACE AREA AND MAX VOLUME IS SIN INVERSE(1/3). Shubham Namdev asked a question differentiation of log e with base x? Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin-1 (1/3). Sir please solve this as soon as possible.. Mayank Rahi asked a question Show that in the parabola y^2 = 4ax, the radius of curvature at any point P is twice the part of the normal intercepted between the curve and the directrix. show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle Dravit Gupta asked a question Let f(x)=x-sinx and g(x)=x-tanx where x belongs to (0,pi/2). Then for these values of x :- a)f(x)g(x)>0 b)f(x)g(x)<0 c)f(x)/g(x)>0 The answer given is B part. The concept involved is of increasing and decreasing functions. Please explain how to do it properly. Shreyash S Sarnayak asked a question If straight line x cos(alpha) + y sin(alpha) = p touches the curve x2/a2 + y2/b2 = 1 , then prove that a2 cos2(alpha) + b2 sin2(alpha) = p2. State and prove leibnitzs theorem to find the nth derivative of a product of two function of x Bhagya Ratna asked a question Prove that the least perimeter of an isosceles triangle in which circle or radius r can be inscribed is 6root3 r Sweta Kumari asked a question The cost C of manufacturing a certain article is given by the formula C=5+48/x+3x2, where x is the number of articles manufactured. Find the max value of C. If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle b/w them is pie/3. Foo asked a question Q. Let S be the non empty set containing all 'a' for which f(x)=(4a-7)/3x3+(a-3)x2+x+5 is monotonic for all x is a element of R. Find S. The correct answer is 'a' is a element of [2,8] but my answer is a<2 and a>8 I first find f'(x) which comes out to be =(4a-7)x2+2(a-3)x+1 Then I take the discriminant>0 which gives me the answer a<2 and a>8 but if i take the discriminant<0 then i get the correct answer. Can you please explain why we should take the discriminant<0 or if i am making another mistake please tell. An open box with a square base is to be made out of a given quantity of card board of area c2 square units. Show that the maximum volume of the box is c3/ 6√3 cubic units. Kashfur asked a question Q.18. A sheet of paper is to contain 18 c{m}^{2} of printed matter. The margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Find the dimensions of the sheet which require the least amount of paper. Jorawar Singh asked a question water is dripping out from a conical funnel of semi vertical angle 45 at uniform rate of 2 cm^2/s (is its sure areea) through a tiny hole at vertical of the bottom wht is the rate of decrese of slant height when the slant height of water is 4cm Find the equation of the normal to the curve y=1+sinx/cosx at x= pi/4. al.palak... asked a question Water is leaking from a conical funnel at the rate of 5 cm3/sec.if the radius of the base of the funnel is 10cm and altitude is 20cm,Find the rate at which water level is dropping when it is 5cm from top..?? Gauri Khanna asked a question differentiate 6y with respect to x 11a) Find maximum area of an isosceles triangle inscribed in the ellipse x2 / a2 + Y2 / b2 = 1 with its vertex at one end of the major axis. Shivi Singla asked a question the tangent to thecurve 3xy^2-2x^2y=1 at(1,1) meets the curve again at which point? Ganesh Prasadh asked a question Show that the semi-vertical angle of the cone fo masimum volume and of given slant height is tan inverse root 2 Show that the volume of greatest culinder that can be inscribed in a cone of height h and semi vertical angle alpha is 4/27 pi h cube tan square alpha find maximum and minimum value of f(x)=x2-3x-4/x-8 A man of height 2 metres walks at a uniform speed of 5 km/hr away from a lamp post which is 6 metres high.Please Find the rate at which the length of his shadow increases. Find the radius of curvature of the curve y = e^x at the point where it crosses the y -axis. Find the angle between the parabolas y2 = 4ax & x2= 4by at their point of intersection other than the origin ? A tank with open surface and square base is to contain 500 cubic feet of water. Find the least cost of lining it with tin at the rate of Rs.60/sq.feet. Varun Singhal asked a question A large window has the shape of a rectangle surmounted by an equilateral triangle.if the perimeter of the window is 12m, find the dimensions of the rectangle that will produce the largest area of the window? Radhika G asked a question A solid is formed by a cylinder of radius r and height h together with two hemisphere of radius r at each end. if the volume of the solid is constant but radius r is increasing at the rate of 1/2pie meter/ minute how fast mast h (height )be changing when r and h are 10 metres Kanishk Gupta asked a question 28. the slope of tangent to the curve y= cos-1(cos x) at x = -π/4 is 4.non-existent Correct ans is-1. Plz exp. A rectangle is inscribed in a semi circle of radius 'r' with one of its sides on the diameter of the semi circle. Find the dimensions of the rectangle so that its area is maximum. Also find this area. Prove that curve(x/a)n+(y/b)n=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n belongs to N at the point (a,b).. show that the function f(x) = x3+x2+x+1 has neither a maximum value nor a minimum value A helicopter is flying along the curve y= x2+2.A soldier is placed at the point (3,2). Find the nearest distance between the soldier and the helicopter.(2010Sp) Find the equations of the normals to the curve 3x2-y2=8 parallel to the line x+3y=4. R K Mishra asked a question Prove that the radius of the base of right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half that of the cone Sachin Soam asked a question How to find the negation of p=> q ? (p implies q) An isosceles triangle of vertical angle 2a is inscribed in a circle of radius r.Show that the area of triangle is maximum when a=(pi)/6. if x= cos t + log tan t/2, y=sin t,then find the value of d^2y/dx^2 and d^2y/dx^2 at t= pi/4. Gyana Ranjan Das asked a question A given quanitity of metal is to be cast into a half cylinder with a rectangular base and semi circular ends. Finfd ratio of length of the cylinder to the diameter of its semi-circular ends. Show that a right circular cylinder which is open at the top, and has a given surface area will have the greatest volume if its height is equal to the radius of the base. Namrata Patel asked a question Anil Kulkarni asked a question prove that all the pts on the curve y2 = 2a(x+asinx/a) at which tangent is parallel to the axis of x, lie on parabola. show that the line x/p+ y/q =1, touches the curve y=e^(-x/p) at the point where it crosses the y axis show that the function f(x)= sin (2x+ pie/4) is decreasing in 3pie/8 <x<5 pie/8 Chhavi asked a question the cost of fuel for running a train is proportional to square of speed generated in km/h.it cost rs 48/h when the train is moving at speed of 16km/h.what is its economical speed if the fixed charges are rs 300/hour over and above the running cost! A point on the hypotenuse of a right angled triangle is at distances 'a' and 'b' from the sides . Show that the length of the hypotenuse is at least {a raised to 2/3 + b raised to 2/3}the whole raised to 3/2 Nupur Goel asked a question Akshaya Mahesh asked a question the curve y= (x2+ax+ b) / x-10 has a turning point at (4, 1) fine the values of a and b and show that y is maximum at this point Manpreet Singh asked a question Find the volume of the largest right circular cylinder that can inscribed in a sphere of radius r cm. Subham Verma asked a question what does y'(0) mean for a function ? Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles Answer of question no. 33(I) An inverted cone has a depth of 40 cm and a base of radius 5cm .water is poured into it at a rate of 3/2 cubic centimetres per minute.find the rate at which the level of water in the cone is rising when the depth is 4 cm .. Pls answer this. I really need help . An open tank with a square base and vertical sides is to be constructed from a metalsheet so as to hold a given quantity of water. Show that the total surface area is least when depth of the tank is half its width.(2010c) Please fast . Explain properly . Exms ???? Simarjeet Kaur asked a question Show that area of the traingle formed by the tangent and the normal at the point(a,a) on the curve y2(2a-x)=x3 and the line x=2a is 5a2/4 sq. units. Abhijit Gope asked a question at what points of the ellipse 16x^2 + 9y^2 = 400 does the ordinate decrease at the same rate at which the abcissa increases ?? Parvathi Venunathan asked a question An open tank is to be constructed with a square based and vertical sides so as to contain 500 cube metres of water. What should be the dimension of the tank, if the area of metal sheet used in its construction is to be minimum Shreshth Chawla asked a question prove : semi verticle angle of right circular cone of given volume and least curved surface area is cot^-1 (root2) . find the maxima and minima of function x3+y3-12x-3y+20 Darshan Patel & 1 other asked a question find the equation of tangent and normal to the curve x=1-cos theta, y= theta-sin theta at theta=pi/4. Amogh asked a question Q. Find the dimensions of the rectangle of maximum area that can be inscribed in the portion of the parabola {y}^{2}=4 px intercepted by the line x = a. find the point on the parabola y=x^2 + 7x+2 which is closest to the straight line y=3x-3. Prateek Bose asked a question A stick of length a cm rests against a vertical wall and the horizontal floor. If the foot of the stick slides with a constant velocity of b cm/s, then the magnitude of the velocity of the middle point of the stick when it is equally inclined with the floor and the wall, is Show that the area of the triangle formed by the tangent and the normal at tha point(a,a) on tha curve y2(2a-x)=x3 and the line x=2a,is 5a2/4. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point. Yuvraj S asked a question find the coordinate of a point on the parabola y=x2 +7x+2 which is closest to the straight line y=3x-3? Vijay Kumar Sharma & 1 other asked a question a kite is moving horizontally at a height of 151.5 meters.if the speed of kite is 10 m/s,how fast is the string being let out, when the kite is 250 m away from the boy who is flying the kite? the height of the boy is1.5m. Ayush Nakhale asked a question Aakanksha Mohgaonkar asked a question a spherical balloon is being inflated by pumping in 16cm^3/s of gas.at the instant when balloon contains 36picm^3 of gas.how fast is its radius increasing? jusha asked a question show that the condition that the curves ax2+by2=1 and a'x2+b'y2=1 should intersect orthogonally (at900) such that 1/a-1/b=1/a'-1/b' The radius of a sphere shrinks from 10 to 9.8 cm.find the approximate decrease in its volume Ann Thomas asked a question A wire of length 36cm is cut into two pieces,one of the pieces is turned in d form of a square and d other in d form of equilatral tringle.find the length of each piece so that d sum of areas of d two be minimum.? reply fast. 32^1/4=? Diyanko Bhowmik asked a question Please provide me with the solution of question ASAP Find the value with the help of logarithm table: × 0.058 [Ans. 340.6] Winnie Wachege & 1 other asked a question show that the right circular cone of least curved surface area and given volume has an altitude equal to root 2 times the radius of the base. i just dont understand..plz help me in simple words.. if in a question it is given some f(x)=..x3 +.....any expression n then asked find the loacal maximum and local minimum values of func f ?. then wht to do? why do we differential..?f'(x) , and then f''(x)?... Find the ponits on the curve y=x3 at which the slope of the tangent is equal to y-cocrdinate of the point. Mihir asked a question FIND THE dy/dx, x=at2 , y= 2at Amanpreet Singh asked a question Divide the number 4 into two positive number such that the sum of the square of one and the cube of other is minimum Anamika asked a question If the length of three sides of a trapezium ,other than the base are equal to 10 cm each, then find the area of trapezium when it is maximum? If the function f(x) = 2x^3 -9mx^2 + 12m^2x + 1 where m > 0 attains its maximum and minimum at p and q respectively such tat p^2 = q, then find m. Show that the line x/a+y/b= 1 touches the curve y=be-x/a.At a point where it crosses the y axis. Gokul Dinesh asked a question show that the equation of the tangent to the curve x2/a2+y2/b2=1 at (xo,yo) is xxo/a2+yyo/b2=1 Zulfiqar Ahmed asked a question If x=cost(3 - 2 cos^2t) and y= sint(3 - 2 sin^2t) find dy/dx at t = pi / 4 If tangent is parallel to x axis then why dy/dx =0?
In the past couple of decades, we have discovered many new exoplanets in no small part due to the Kepler space telescope. Today, there are over 4000 known exoplanets. Determining the internal compositions of the exoplanets is of great scientific importance, however, exoplanets are too far away to send instruments to them, and many are also too far away to do accurate spectroscopy. Building on the work of Seager et. al. 2007 and 2018, we show that it's possible to constrain the interior composition of exoplanets and classify the composition type from just the mass M R In order to constrain the interior composition, we must make some assumptions about the planets. Our model has 4 central assumptions: The planet's mass distribution is spherically symmetric. The planet is in hydrostatic equilibrium. The planet has a polytropic equation of state. The planet is differentiated. Spherical symmetry ensures that the density \rho only depends on the radius. So the mass of a infinitely thin sphere shell at radius r \frac{dm}{dr} = 4 \pi r^3 \rho(r). We can get the total mass of the planet by integrating M = \int_0^R 4 \pi r^3 \rho(r) dr . An object is in hydrostatic equilibrium, when the outward force from its pressure gradient F_p , is equal but opposite in magnitude to the gravitational force F_g F_p=-F_g Gravity is based upon the mass of the planet. The pressure gradient is based on the pressure between the layers of mass in the planet, which results in a force, due to the pressure difference between the layers. We can the above equation to: \frac{d P(r)}{d r}=-\frac{Gm(r)\rho(r)}{r^2}. P(r) is the pressure at radius r G \rho(r) m(r) are the density and mass of the planet respectively. In general, we have an equation of state P(r)=f(\rho(r),T(r)) f is a function that depends on the material, and T is the temperature. We neglect the temperature dependency, and we will also neglect phase changes since they have a little effect on the mass-radius relationship. The relationship between pressure ( p ) and density ( \rho ) for a compressible fluid can as such described by the Polytropic equation of state p(\rho)=K\rho^\frac{n+1}{n} K,n n being the polytropic index. Taking these assumptions together, we get the Lane-Emden equation. Since we are modelling the planets using materials such as iron, silicate and water, which are incompressible below certain pressures, the polytropic equation of state is a poor model for the relationship between density and pressure. We can extend the equation of state to approximately incorporate the model of the solids at low pressures by adding a constant \rho_0 \rho (P) = \rho_0 + KP^{\frac{n+1}{n}} \rho_0 is the incompressible density. We will call this the modified polytropic equation of state. To model the planets we use the computer model developed by Zeng and Seager and translated to Python by Troels HaugbÃ¸lle. The model assumes a 3-material differentiated planet where the planet is modelled as 3 shells that grow from the center with the most dense material closest to the center and becomes progressively less dense away from the center. We denote the fractional compositions of the materials with \alpha,\beta,\gamma \alpha is the fraction of the innermost material and \gamma the fraction of the outermost material. We have that 1=\alpha+\beta+\gamma . We have used materials that are found in planets and moons in our solar system: iron, silicate, water and hydrogen. For small earth-like planets we use a iron-silicate-water model, and silicate-water-hydrogen model for larger planets where the iron-silicate-water model is insufficient for reaching the large radii. While we are inside the shell of a given material, the equation of state at that radius is modelled by the modified polytropic equation detailed for the given material. The equation of state gives the relationship between \rho P such that we can solve the central system of coupled differential equations derived from spherical symmetry and hydrostatic equilibrium. There are 5 parameters in this model ( \alpha, \beta, \gamma, M, R ), but since \gamma=1-\alpha-\beta , we only have 4 free parameters. From any 3 of the them we can determine the whole system. However, since we only have measured the mass and radius of the planet and do not have more information from the gravitational moment or spectroscopy of the atmosphere, we cannot uniquely determine the system. Instead, we get a degeneracy. We get an infinite number of pairs of possible ( \alpha,\beta ). The computer model solves this numerically for a given \alpha\in[0,1] by integrating the central equation starting at r=R and stopping at r=0 using a 4th-order Runge-Kutta solver, and finds a \beta M(0)<\epsilon for some small tolerance \epsilon=10^{-5} . It finds \beta using bisection starting with the lower bound of \beta_l=0 and upper bound \beta_u=1-\alpha . Once we have a valid pair of ( \alpha,\beta ), we can find the associated \gamma=1-\alpha-\beta We have performed a broad analysis using the entire NASA exoplanet archive (4367 planets) and then reducing the dataset to a more manageable size. In the first reduction, we required that the planets had measured radii and masses which are used for our analysis as well as measured star luminosity and distance to star. This reduced the dataset to 433 planets. After having identified a number of CHZ ranges we used the combined most relaxed range (LB: 0.725AU, UB: 1.24AU) which reduces the dataset further to 10 planets summarised below. Of these 10 planets there is one planet, Kepler-22 b, which we are able to model using the iron-silicate-water model while 4 others were classified using the silicate-water-hydrogen model. (M_e) (R_e) Kepler-1514 b 1700 12 Kepler-1654 b 160 9.1 Kepler-22 b 36 2.3 Kepler-289 c 130 12 Kepler-47 c 8900 4.6 Kepler-62 e 36 1.6 Kepler-62 f 35 1.4 bet Pic b 6400 17 We can plot the modelable planets in a ternary diagram as seen below. We note that Kepler-289 c doesn't show up. This is because given the resolution of the numerical solver, we only get one valid pair of (\alpha,\beta) . The planet has almost no silicate, and about 1% water while 99% of the mass is hydrogen. We see that the Kepler-22 b planet stands out in particular. Kepler-22 b is an interesting planet. It is the only rocky planet in the habitable zone of its host star that we know of. Moreover, we find that it's possible that up to about 20% of its mass is water, but we don't know if the water on the surface is fluid, or a liquid, or perhaps a gas. We can estimate the surface temperature through the equilibrium temperature using Stefan-Boltzmann's law of black-body radiation. L = 4\pi R_s^2 \sigma T_e^4 R_s is the radius of the star, and T_e is the effective temperature. We can then figure out how much luminance hits the planet by multiplying by the fraction of the scaled-up sphere the earth-disc fills up. L_p = L \frac{\pi R_p^2}{4\pi D^2} D is the distance from the star to the planet, and R_p is the radius of the planet. By assuming both the planet and the host star are black-body radiators, we have that the incoming flux L_p must be equal to the outcoming flux, so we can find the equilibrium temperature T_{eq} of the planet by solving L_p = 4\pi R_s^2 \sigma T_{eq}^4. However, the planet is not a perfect black-body radiator, some of the light is reflected before it is absorbed. So assuming an Earth-like albedo, the equilibrium temperature, or the surface temperature is 262K which would mean the water would be a gas. However, that is assuming no atmosphere and greenhouse effect. If we model an Earth-like atmosphere and greenhouse effect - which is a lower-bound since the Kepler-22 b is much heavier than Earth which would give it a denser atmosphere - we get an equilibrium temperature of 287K which would give us liquid water. Which is consistent with other analyses. While other analyses have been able to rule out an Earth-like composition for Kepler-22 b, we are not able to do so, as we would need the uncertainty in both the mass and the radius, but we do not have a lower-bound for the mass. This is common for planets which have been discovered using the transit-method. However, if we plot the approximately 2000 planets which have measured mass and radius, and classify then using our gas- and rocky models, we find that there is a lot of give before Kepler-22 b enters the "uncertain" region where it an be classified using both models. We have seen that it is possible to constrain the interior composition of exoplanets from just the mass and the radius. Moreover, we are able to classify if the planets are rocky planets or have a gas layer. We looked at Kepler-22 b a rocky planet in the habitable zone and found that it is possible that there is liquid water on the planet which makes it interesting for people who are looking for extraterrestrial life. We will investigate data collected by Sara Eisenhardt of the magnetisation cu...
Matrices, Popular Questions: ICSE Class 12-commerce MATH, Math Part I - Meritnation A school wants to award its student for the values of honesty,regularity and hard work with a total cash award of RS. 6000.Three times the award money for hardwork added to that given for honesty amounts to RS 11000.The award money given for honesty and hard work together is double the one given for regularity.Represent the above situation algebraically and find the award money for each value,using matrix method.apart from these values ,namely, honesty, regularity and hardwork, suggest one more value which the school must include for awards? Tarunima Majumdar asked a question Use matrix multiplication to divide rs. 30,000 in two parts such that the total annual interest at 9% on the first part and 11% on the second part amounts rs. 3060. What is cube root of unity i.e. omega??? Mary Catherine asked a question if A=[3 0 1 find A-1 pls answer "this " question answer immdediatly!!! show tha a skew symmetric matrix of odd order has determinant =0 Shweta Parida asked a question Q79 please! 79. 300 persons are participating in a meeting in India, out of which 120 are foreigners, and the rest are Indians. Out of the Indians there are 110 men who are not judges ; 160 are men or judges, and 35 are women judges. There are no foreign judges. How many Indian women attended the meeting ? (c) 3.55 (d) 40 write a square matrix of order 2 which is both symmetric and skew symmetric? using matrices solve system of equations x+3y+4z=8, 2x+y+2z=5, 5x+y+z=7 Urwashi Keshari & 1 other asked a question find the number of all possible matrices of order 2*3 with each entry 0 or 1 Christina Roy asked a question If matrix A=[0 0 find A20,A25 2 0] A total amount of Rs7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and 8 1/2% respectively. The total annual interest from these three accounts is 550 RUPEES. Equal amounts have been deposited in the 5% and 8% savings accounts. Find the amount deposited in each of the three accounts using matrix method. Using matrices, find k so that the equation: 3x-2y+2z=1, 2x+y+3z=-1, x-3y+kz=0 may have a unique solution. Rounak Agrawal asked a question A trust caring for handicapped children gets Rs.30,000 every month from its donors. The trust spends half of the funds received for medical and educational care of the children and for that it charges2% of the spent amount from them , and deposits the balance amount in privatebank to get the money multiplied so that in future the trust goes on functioning regularly. What percent of interest should the trust get from the bank to get a total of Rs.1,800 every month? Use matrix method, to find the rate of interest. Do you think people should donateto such trusts? Shreya Mahajan asked a question Two schools A and B decided to award prices to their students for 3 values, honesty(X),punctuality(Y), andobedience(Z). School A decided to award a total of Rs.11000 for the three values to 5,4,and 3 students while school B decided to award Rs.10,700 for the 3 values to be 4,3,5 2students . If all the 3 prices together amount to be Rs. 2700, then: 1. Represent the abuve situation by a matric waequation and form linear equations using matrix multiplication. 2. Is it possible to solve the system of equations so obtained using matrices? 3. Which value do you nprefer to be rewarded and why? Kelvin George & 1 other asked a question the number of possible matrices of order 3x3 with each entry 0 or 1 is: (A)27 (B)18 (C)81 (D)512 andhow? Kala Venkatachalam asked a question For keeping fit X people believe in morning walk, Y people believe in yoga Z ppl join gym. Total no. of people are 70. Further 20% 30% 40% ppl are suffering from any disease who believe in morning walk,yoga gym. Total no of such ppl is 21. If morning walk costs Rs.0, yoga Rs500 per month gym Rs400 total expenditure is Rs23000. Formulate a matrix problem calculate the no of each type of ppl. Sai Vignesh Follow Me @sigvins... asked a question Its 6 mark question plz slove this question as quick as posible plz show me ML AGGARWAL' S math book solutions of chapter 3, page no. 156 exercise 3.5 [2 -1] [-1 -8 -10] [0 1] x X= [3 4 0] [-2 4] [10 20 10] a unit X =[a b c] Zameel Udeen asked a question 1) Using elementary transformations, find the inverse of the matrix if A= [ 3 -3 4 0 -1 1 ] , then A-1= Subhom Nath asked a question prove that the diagonal elements of a scew symmetric matrix are all zero If A is a square matrix of order 3 and A' denotes transpose of matrix A, A'A=I, and det A=1, then det(A-I) must be equal to Mohammed Affan asked a question f(x) = x2-5x+ 6 .. find f(A) if A= matrix 2 0 1 if for a matrix A , A to the power 5 = I , THEN A inverse = A TRUST INVESTED MONEY IN TWO TYPES OF BONDS. THE FIRST BOND PAY 10% INTEREST AND SECOND BOND PAYS 12% INTEREST. THE TRUST RECEIVES 2800 AS INTEREST. HOWEVER, IF TRUST HAD INTERCHANGED MONEY IN BONDS THEY WOULD HAVE GOT 100 LESS AS INTEREST. USE MATRIX METHOD TO FIND THE AMOUNT INVESTED BY THE TRUST. Luv Ur Life Wt Laugh Chaithra asked a question There are two families A and B. There are 2 men 3 women and 1 child in the family A and 1 man 1 woman and 2 children are there in family B. The recommended daily allowance of calories is men 2400, women 1900, and children 1800. Represent the above data in matrix form... Here how can we put into matrix form and please post answer with explaination... Please experts...:) Praveen Kumar asked a question Piz give me solution of rs agrawal matrices chapter 1E math 12th class 19. A company is to employ 60 labourers from either of the party X or Y, comprising in age groups as under : Rate of labour applicable to categories I, II and III are Rs. 1200, Rs. 1000 and Rs. 600 respectively. Using matrix multiplication, find which party is economically preferable. find the inverse using elementary transformation If f(x) = x2 - 4x + 1 , find f(A) when A = [2 3 ] first row and [ 1 2] second row , here x2 is x to the power of 2 or x square. Vaishnavi K asked a question find the matrix A satisfying the matrix equations [2 1] A [-3 2] = [ 1 0] Mariya Martin asked a question for a square matrix A of order 3 if \|2 adj.A\| =128 then find \|A\| Varun Dhar asked a question Two schools P and Q want to award their selsected students on the values of Discipline, Politeness and Punctuality. The school P wants to award Rs. x each, Rs.y each and Rs.z each for the three respective values to its 3,2 and 1 students with a total amount money of Rs.1000/- School Q wants to spent Rs.1500/- to award its 4, 1 and 3 students on the respective values. If the total amount of awards for one prize on each value is Rs.600/-, using matrices, find the award money for each value. Srinidhi Seshadri asked a question A is a 5xp matrix, B is a 2xq matrix, AB works out to be a 5?4 matrix. Write the value of p and q Sanchit Arora asked a question If A is a square matrix such that A^2=I,then find the simplified value of (A-I)^3+(A+I)^3-7A. Rajneesh Singh asked a question for what value of k do the equation 2x-3y+2z=a, 5x+4y-2z=-3, x-13y+kz=9, not have a unique solution prove that the product of the matrices [ cos2θ cosθsinθ ] [cosθsinθ sin2θ ] [ cos2α cosαsinα ] [cosαsin α sin2α ] is zero when and differ by an odd multiples of pi/ 2 Q.5. If A be a square matrix of the order 5 and B = Adj (A) then find Adj (5A). Pushpa Subramanian asked a question If li,mi,ni where i=1,m=2,n=3 denote the direction cosines of 3 mutually perpendicular vectors in space ,prove that AA^T=I wher A is a matrix and A^T is its transpose and I is a indentity matrix if [cos2pi/7 -sin2pi/7]^k [sin2pi/7 cos2pi/7]^k = [1 0] [1 0] find the least positive integral value ok k Vishweshvar Singh asked a question if A is a square matrix of order 3 and \|2A\|= k\|A\|, then find the value of k. Arshia Singh asked a question If A and B are skew symmetric matrices of same order and AB is symmetric then show that AB=BA Aahan Gupta asked a question if A is a square matrix such that, A2=A, then write the value of (1+A)2-3A. 26. If A = \left[\begin{array}{cc}{a}_{11}& {a}_{12}\\ {a}_{21}& {a}_{22}\end{array}\right] \mathrm{such} \mathrm{that} \|A\| =-15, \mathrm{find} {a}_{11} {C}_{21}+{a}_{12}{C}_{22} {C}_{ij} \mathrm{is} \mathrm{cofactors} \mathrm{of} {a}_{ij} \mathrm{in} A=\left[{a}_{ij}\right] 29. Given a square matrix A of order 3 × 3 such that \|A\| = 12 find the value of \|A adj A\|. On her birthday seema decided to donate some money to children of an orphanage home. If there ware 8 children less, every one would have got Rs.10 more However, if there were 16 children more, every one would have got Rs.10 less. Using matrix method, find the number of children and the amount distributed by seema. A company produces three products every day. Their production on a certain day is 45 tons . It is found that the production of third product exceeds the production of first product by 8 tonswhile the total production of first and third product is twice the production of second product . Determine the production level of each product using matrix method. if B, C ARE n ROWED SQUARE MATRICES AND IF A=B+C , BC=CB ,C2=0 ,THEN SHOW THAT FOR EVERY n is the element ofN, An+1 =Bn(B+(n+1)C). Prove that the product of matrix [cos20 cos0sin0 (below dis)cos0sin0 sin20] & [cos2alpha cosalphasinalpha (below dis)cosalphasinalpha sin20] is a null matrix where 0 & alpha diffre by an odd multiple of pie / 2. Ananya Veer asked a question using elementary row operations find the inverse of the matrix A = cosX sinX -sinX cosX An = cos nX sin nX -sin nX cos nX , n belongs to all natural no. if matrix cos2pi/7 -sin2pi/7 sin2pi/7 cos 2pi/7 the whole power k = matrix 1 0 0 1 , then write the value of x+y+xy Two schools decided to award prizes to their teachers for two qualities knowledge and guidance. School A decided to award a total of Rs. 3200 for the values to 4 and 3 teachers respectively while school B decided to award a total of Rs. 1600 for the values to 1 and 2 teachers respectively. Represent the above situation by a system of linear equations and solve using matrices. Which quality you prefer to be rewarded most and why? Find matrices A and B, if 2A - B = and 2B + A = . Vaibhavi Dwivedi asked a question why when finding the inverse of a matrix do we do A=IA for rows A=AI for column? let A=[2 3 AND F(X)= X2 -4X+7. SHOW THAT F(A) =O. USE THIS RESULT TO FIND A5. If A, B, C are three non zero square matrices of same order, find the condition on A such that AB = AC implies B = C. Datta asked a question a line can be drawn which divides the following figure into two separate pare. These two parts is could then fit together to make a square, which two numbers would you connect to make this line 1. Using matrix method, solve the following system of equations : 2/x + 3/y + 10/z = 4, 4/x - 6/y + 5/z = 1, 6/x + 9/y - 20/z = 2 2. For what value of x, the matrix [5-x x+1 2 4 ] is singular? Shanuji Foujdar asked a question If A=[ 0 -tan alfa/2 tan alfa/2 0] and I is the identity matrix of order 2,show that I+A= (I-A) [ cos alfa -sin alfa sin alfa cos alfa] please explain this question step by step?????? Mamta Punjabi asked a question i m nt able to get anytng of ths concept plzz hepl :( if AB r 2 matrices AB=B BA=A then A^2+B^2= a2+b2/c c c a b2+c2/a a b b c2 +a2 = 4abc find the value of x for which A is an identity matrix, A= cosx -sinx sinx cosx ( in sq. brackets) If A is a square matrix such that A2 = A. Show that :- (I + A)3 = 7A + I A store in a mall has three dozen shirts with ‘SAVE ENVIRONMENT’ PRINTED,TWO DOZEN SHIRTS ‘SAVE TIGER’ printed and 5 dozen shirts with ‘GROW PLANTS’ printed. The cost of each shirt is Rs. 595, Rs. 610 and Rs. 795 respectively....Sir in this question that u just answered 15 mins ago, I am not getting the answer as 70,560. Can u elaborate the calculations or just cross check once please. I am getting 83,760. Please help Quick! Aparna Lonely Girl... asked a question find the value of X and Y..if 2X+3Y=[ 2 3 , 4 0 ] (2x2 matrix) and 3X+2Y=[ 2 -2 , -1 5 ] (2x2 matrix). Shagun asked a question Solve the following system of linear equations using matrices: \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4\phantom{\rule{0ex}{0ex}}\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1\phantom{\rule{0ex}{0ex}}\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2 Dhruv Agarwal asked a question a-symmetric b-skew-symmetric c-diagonal d-scalar if A= 1 3 2 4 and B = -1 2 3 5 verify that (AB)' = B'A' Meghna Kumar asked a question Plz tell me the steps for finding the inverse of a matrix by elementary row operation Srijith Sreedharan asked a question Give an example of two non-zero 2x2 matrix A and B. such that AB=0 Find the equation of the plane parallel to plane 2x − 3y + 6z − 5 = 0 at a distance of 4 units from it Ishika Choudhary asked a question there are two families M and N . there are 2 men , 2 women, and 4 childern in family N . and 4 men , 6 women and 2 children in family M . there commended daily allowance for calories is child : 1800, women : 1900 and man : 2400 and for protiens is man :55gm, woman : 45gm and child :33gm. using matrices algebra, calculate the total requirement of protiens and calories for each of the famlies ? Moon Daw asked a question if A is any square matrix then prove that AAT is symmetric? Sara Yasmin asked a question prove that \| x y z \| \| x^2 y^2 z^2 \| \|x^3 y^3 z^3 \| = (x-y) (y-z) (z-x) (xyz) Neethu John asked a question Ten students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers,the second group has honest and law abiding students and the third group consists vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 13, while the combined strength of first and second group is 4 times that of the third group. Find the number of students in each group. Mudra asked a question Q2 plz no links Benjamin M asked a question Two schools A and B want to award their selected teachers on the values of honesty, hard work and regularity. The school A wants to awardRsx each,Rsy each andRsz each for the three respective values to 3, 2 and 1 teachers with a total award money ofRs1.28lakhs. School B wants to spendRs1.54lakhsto award its 4, 1 and 3 teachers on the respective values(by giving the same award money for the three values as before). If the total amount of award for one prize on each value isRs57000, using matrices, find the award money for each value? Slovenia solve plz! G. Praveen Kumar asked a question Raagini Mahendran asked a question Construct a 2x2 matrix A[aij] whose elements are given by aij = { i - j , if i = j i + j , if i find relation be the For what value of k, the matrix( 2-k ) 3 is not invertible. Experts, plz explain me when f(x) is 1 and f(y) is 1 then f(x+y) is 1, why not 1+1=2 in question 13 of exercise 3.2 ncert. Three schools x, y and z organized a fete (mela) for collecting funds for flood victims in which they sold hand-held fans, mats and toys made from recycled material , the sale price of each being ₹25, ₹100, ₹50 respectively. The following table shows the number of articles of each type sold: Articles school. x. Y Z Hand-held fans. 30. 40. 35 Mats. 12. 15. 20 Toys. 70. 55. 75 using matrices , find the funds collected by each school by selling the above articles and the total funds collected . Also write any one value generated by the above situations . Choco Chip asked a question i find difficulty in solving problems in elementary transformation method even though i know the rules .. please help me ! Ht Bhatia asked a question A. A is a diagonal matrix B. A is a zero matrix C. A is a square matrix using elementary row transformation find the inverse of the matrix Please explain step by step...,,, if A= [ 0 -tan(alpha/2) tan(alpha/2) 0 ] & I is the identity matrix of order 2 then show that I + A= (I - A) [cos alpha -sin alpha sin alpha cos alpha ]
Mutual inductor model with nominal inductance optional tolerances for each winding, operating limits and faults - MATLAB - MathWorks æ—¥æœ¬ L1 tolerance application L1 tolerance distribution L1 number of standard deviations for quoted tolerance Series resistance, [R_primary R_secondary] Parallel conductance, [G_primary G_secondary] Faulted winding Location of fault node (% of total turns from - terminal) Short-circuit turns Open-circuit at fault node Conductance of faulted ground path Mutual inductor model with nominal inductance optional tolerances for each winding, operating limits and faults Simscape / Electrical / Passive / Transformers The Mutual Inductor block lets you model a mutual inductor (two-winding transformer) with nominal inductance tolerances for each winding. The model includes the following effects: You can turn these modeling options on and off independently of each other. In the unfaulted state, the following equations describe the Mutual Inductor block behavior: {v}_{1}={L}_{1}\frac{d{i}_{L1}}{dt}+\text{â}M\frac{d{i}_{L2}}{dt}+{i}_{L1}{R}_{1} {v}_{2}={L}_{2}\frac{d{i}_{L2}}{dt}+\text{â}M\frac{d{i}_{L1}}{dt}+{i}_{L2}{R}_{2} M=k\sqrt{{L}_{1}{L}_{2}} v1 and v2 are voltages across the primary and secondary winding, respectively. L1 and L2 are inductances of the primary and secondary winding. R1 and R2 are series resistances of the primary and secondary winding. M is mutual inductance. k is coefficient of coupling. To reverse one of the winding directions, use a negative value. A parallel conductance is placed across the + and – terminals of the primary and secondary windings, so that iL1 = i1 – G1v1, where G1 is the parallel conductance of the primary winding, and i1 is the terminal current into the primary. Similar definitions and equation apply to iL2. You can apply tolerances separately for each winding. Datasheets typically provide a tolerance percentage for a given inductor type. Therefore, this value is the same for both windings. The table shows how the block applies tolerances to the nominal inductance value and calculates inductance based on the selected tolerance application option for the winding, L1 tolerance application or L2 tolerance application. Uniform distribution: L Â· (1 – tol + 2Â· tolÂ· rand) Gaussian distribution: L Â· (1 + tol Â· randn / nSigma) L Â· (1 + tol ) L Â· (1 – tol ) L is nominal inductance for the primary or secondary winding, Inductance L1 or Inductance L2 parameter value. tol is fractional tolerance, Tolerance (%) /100. Inductors are typically rated with a particular saturation current, and possibly with a maximum allowable power dissipation. You can specify operating limits in terms of these values, to generate warnings or errors if the inductor is driven outside its specification. Instantaneous changes in inductor parameters are unphysical. Therefore, when the Mutual Inductor block enters the faulted state, short-circuit and open-circuit voltages transition to their faulted values over a period of time based on this formula: CurrentValue = FaultedValue – (FaultedValue – UnfaultedValue) Â· sech(âˆ†t / Ï„) âˆ†t is time since the onset of the fault condition. Ï„ is user-defined time constant associated with the fault transition. For short-circuit faults, the conductance of the short-circuit path also changes according to the sech(âˆ†t / Ï„) function from a small value (representing an open-circuit path) to a large value. The Mutual Inductor block lets you select whether the faults occur in the primary or secondary winding. The block models the faulted winding as a faulted inductor. The unfaulted winding is coupled to the faulted winding. As a result, the actual equations involve a total of three coupled windings: two for the faulted winding and one for the unfaulted winding. The coupling between the primary and secondary windings is defined by the Coefficient of coupling parameter. After voltage exceeds the maximum permissible value a certain number of times When current exceeds the maximum permissible value for longer than a specific time interval You can enable or disable these trigger mechanisms separately, or use them together if more than one trigger mechanism is required in a simulation. When more than one mechanism is enabled, the first mechanism to trigger the fault transition takes precedence. In other words, a component fails no more than once per simulation. Faultable inductors often require that you use the fixed-step local solver, rather than the variable-step solver. In particular, if you model transitions to a faulted state that include short circuits, MathWorks recommends that you use the fixed-step local solver. For more information, see Making Optimal Solver Choices for Physical Simulation. The Primary current and Secondary current variables let you specify a high-priority target for the initial inductor current in the respective winding at the start of simulation. 1+ — Positive terminal of the primary winding Electrical conserving port associated with the primary winding positive terminal. 1- — Negative terminal of the primary winding Electrical conserving port associated with the primary winding negative terminal. 2+ — Positive terminal of the secondary winding Electrical conserving port associated with the secondary winding positive terminal. 2- — Negative terminal of the secondary winding Electrical conserving port associated with the secondary winding negative terminal. Inductance L1 — Nominal inductance value in the primary winding 10 H (default) The nominal inductance value in the primary winding. Inductance value must be greater than zero. Inductance L2 — Nominal inductance value in the secondary winding 0.1 H (default) The nominal inductance value in the secondary winding. Inductance value must be greater than zero. Coefficient of coupling — Mutual inductance coupling between windings The coupling between the primary and secondary windings. This coefficient defines the mutual inductance. To reverse one of the winding directions, use a negative value. Tolerance (%) — Inductor tolerance, in percent The inductor tolerance as defined on the manufacturer datasheet. Datasheets typically provide a tolerance percentage for a given inductor type. Therefore, this value is the same for both windings. L1 tolerance application — Select how to apply tolerance to primary winding Select how to apply tolerance during simulation to the primary winding: None — use nominal value — The block does not apply tolerance, it uses the nominal inductance value. Random tolerance — The block applies random offset to the inductance value, within the tolerance value limit. You can choose Uniform or Gaussian distribution for calculating the random number by using the Tolerance distribution parameter. Apply maximum tolerance value — The inductance is increased by the specified tolerance percent value. Apply minimum tolerance value — The inductance is decreased by the specified tolerance percent value. L1 tolerance distribution — Select the distribution type for primary winding Enabled when the L1 tolerance application parameter is set to Random tolerance. L1 number of standard deviations for quoted tolerance — Used for calculating the Gaussian random number for primary winding Enabled when the L1 tolerance distribution parameter is set to Gaussian. L2 tolerance application — Select how to apply tolerance to secondary winding Select how to apply tolerance during simulation to the secondary winding: L2 tolerance distribution — Select the distribution type for secondary winding L2 number of standard deviations for quoted tolerance — Used for calculating the Gaussian random number for secondary winding Series resistance, [R_primary R_secondary] — Equivalent series resistance of the primary and secondary winding [0.001, 0.001] Ohm (default) Equivalent series resistance of the primary and secondary winding, specified as a two-element vector. The first number corresponds to the primary winding, the second number to the secondary winding. For a faulted winding, the block allocates the resistance to each segment in proportion to the number of turns in that segment. Parallel conductance, [G_primary G_secondary] — Parallel leakage path associated with the primary and secondary winding [1e-9,1e-9] 1/Ohm (default) Parallel leakage path associated with the primary and secondary winding, specified as a two-element vector. The first number corresponds to the primary winding, the second number to the secondary winding. The parallel conductances are placed directly across the + and – terminals of the primary and secondary winding, respectively. Select Yes to enable reporting when the operational limits are exceeded. The associated parameters in the Operating Limits section become visible to let you select the reporting method and specify the operating limits in terms of power and current. Saturation current — Inductor saturation current Inductor saturation current, as defined in the manufacturer datasheets. If the net current into the primary and secondary windings exceeds this value, the core material enters saturation, and the block reports an operating limits violation. That is, the block compares the limit against \|i1 + i2\|, where currents are defined as being positive when they are into the + nodes. Power rating — Maximum power dissipation in the inductor Maximum instantaneous (total) power dissipation in the resistance and conductance elements associated with the mutual inductor. If the total power (including both primary and secondary winding) exceeds this number, the block reports an operating limits violation. Enable faults — Select On to enable faults modeling Select On to enable faults modeling. The associated parameters in the Faults section become visible to let you select the reporting method and specify the trigger mechanism (temporal or behavioral). You can enable these trigger mechanisms separately or use them together. Enabled when the Enable faults parameter is set to On. Faulted winding — Select winding to use for fault modeling Primary (default) \| Secondary Select whether the faults occur in the primary or secondary winding. Location of fault node (% of total turns from - terminal) — Percentage of turns in the subinductor that is in contact with the – port of the faulted winding In practice, faults are enabled by segmenting the faulted winding into two coupled subinductors, connected in a series. The inductance is proportional to the square of the number of turns in the respective segment, and the series resistance of each subinductor is proportional to the number of turns in each segment. The parallel conductance spans both segments. This parameter indicates the percentage of turns that are assigned to the subinductor that is in contact with the – port of the faulted winding. The remaining turns are assigned to the other subinductor. The default value is 50, which means that the overall inductance of the faulted winding is divided into two equal, coupled subinductors. Short-circuit turns — Select whether fault results in one of the segments being short-circuited No (default) \| To negative terminal \| To positive terminal Select whether the fault results in one of the subinductor segments being short-circuited: No — The fault does not produce a short circuit. To negative terminal — The fault short-circuits the subinductor that is in contact with the – port of the block. To positive terminal — The fault short-circuits the subinductor that is in contact with the + port of the block. Open-circuit at fault node — Select whether to apply an open-circuit fault between the segments Select whether to apply an open-circuit fault between the two subinductor segments. The default is No. Even with an open-circuit fault, the characteristics of the subinductors are still related because they are magnetically coupled even in the faulted state. Ground fault — Select whether fault results in one of the segments being short-circuited No (default) \| Negative terminal side of fault node \| Positive terminal side of fault node Select whether, in case of fault, there is a path for current to flow towards the ground node: No — The fault does not result in a connection to ground. Negative terminal side of fault node — The side that is in contact with the – port of the block is connected to ground. Positive terminal side of fault node — The side that is in contact with the + port of the block is connected to ground. If the Open-circuit at fault node parameter is set to Yes, you need to specify which side (negative or positive) is connected to ground. If there is no open circuit, the two options behave similarly. Physically, this corresponds to a breakdown in the insulation between the windings and the grounded core or chassis. Conductance of faulted ground path — Mutual coupling between the two subinductors If there is a ground fault, this parameter represents conductance of the current path to ground. For example, if the path to ground is through the core material, then specify a small conductance value depending on the core material being used. For highly conductive core material or for chassis-shorts, specify a higher conductance value. Enabled when the Ground fault parameter is set to Negative terminal side of fault node or Positive terminal side of fault node. Fault transition time constant — Time constant for the transition to faulted state Maximum permissible voltage — Voltage threshold to fault transition Define the voltage threshold to a fault transition. If the voltage value exceeds this threshold a certain number of times, specified by the Number of events to fail when exceeding voltage parameter value, then the block starts entering the fault state. Number of events to fail when exceeding voltage — Maximum number of times the voltage exceeds the threshold Because the physical mechanism underlying voltage-based failures depends on one or more partial discharge events occurring, this parameter allows you to set the number of voltage overshoots that the inductor can withstand before the fault transition begins. Note that the block does not check the time spent in the overvoltage condition, only the number of transitions. Define the current threshold to a fault transition. If the current value exceeds this threshold for longer than the Time to fail when exceeding current parameter value, then the block starts entering the fault state. Time to fail when exceeding current — Maximum length of time the current exceeds the threshold Fault \| Inductor \| Three-Winding Mutual Inductor \| Variable Inductor
Chemical Process Control - Wikibooks, open books for an open world 1 What is Process Control? 1.1.1 Direct Acting Control 1.1.2 Reverse Acting Control 1.3 Advanced Control 2.2 Process Reaction Curve 3 Using Mathematical Models What is Process Control?Edit The manipulation of an object (actuation device) to maintain a parameter within an acceptable deviation from an ideally required condition. At it's core, process control is the transfer of variability from on variable to another. There are two basic process control philosophies, feedback and feedforward control. Feedback ControlEdit In feedback control, the controlled variable is measured and compared with a set-point. The deviation between the controlled variable and the set-point is the error signal. The error signal is then used to reduce the deviation of controlled variable from set-point. Direct Acting ControlEdit If the controlled variable increases as the manipulated variable increases, then direct acting control is used. Reverse Acting ControlEdit Feedforward ControlEdit Advanced ControlEdit The conservation laws on mass, energy and momentum are fundamental bases for the development of models of chemical processes. The general form of the law for a variable , when applied to a control volume (CV) is {\displaystyle {\frac {d(X~IN~to~CV)}{dt}}-{\frac {d(X~OUT~of~CV)}{dt}}+{\frac {d(GENERATION~OF~X~within~CV)}{dt}}-{\frac {d(DISAPPEARANCE~of~X~within~CV)}{dt}}={\frac {d(ACCUMULATION~of~X~within~CV)}{dt}}} When applied to mass this becomes the Law of Conservation of Mass. Assuming no nuclear reactions take place, then the rate of generation or disappearance of mass is zero. Hence, we have {\displaystyle {\frac {d(Mass~IN)}{dt}}-{\frac {d(Mass~OUT)}{dt}}={\frac {d(ACCUMULATION~of~Mass)}{dt}}} In symbols we may say {\displaystyle {\dot {m}}_{in}-{\dot {m}}_{out}={\frac {dM}{dt}}} where M stands for the total mass within the CV Process Reaction CurveEdit Using Mathematical ModelsEdit The mechanical device that cause the activation or movement of a final control element. A physical device whose activation or movement causes a change in a dynamic process. In process control, the most common final control elements are control valves. IMC-PID Tuning A method for PID tuning that selects tuning parameters to approximate an IMC-derived controller. A semi-graphical programming language used to represent control algorithms. The language is expressed using symbols for logic devices. The arrangement of the device symbols and their connections has the appearance of a ladder. An integral transformation from time domain to Laplace domain. Given a function of time {\displaystyle f(t)} , the Laplace transform is given by the following {\displaystyle F(s)=\int _{0}^{\infty }\!f(t)\;e^{-st}dt} {\displaystyle F(s)} to represent the Laplace transform of {\displaystyle f(t)} is a common convention; however, in dynamics and control it is common to use {\displaystyle f(t)} {\displaystyle f(s)} to represent a time-domain function and its Laplace transform, respectively. Programmable Logic Controller, a microprocessor-based electronic device for implementing control algorithms. Retrieved from "https://en.wikibooks.org/w/index.php?title=Chemical_Process_Control&oldid=3659114"
Butterfly curve (transcendental) - Wikipedia The butterfly curve. The butterfly curve is a transcendental plane curve discovered by Temple H. Fay of University of Southern Mississippi in 1989.[1] An animated construction gives an idea of the complexity of the curve (Click for enlarged version). The curve is given by the following parametric equations:[2] {\displaystyle x=\sin t\!\left(e^{\cos t}-2\cos 4t-\sin ^{5}\!{\Big (}{t \over 12}{\Big )}\right)} {\displaystyle y=\cos t\!\left(e^{\cos t}-2\cos 4t-\sin ^{5}\!{\Big (}{t \over 12}{\Big )}\right)} {\displaystyle 0\leq t\leq 12\pi } or by the following polar equation: {\displaystyle r=e^{\sin \theta }-2\cos 4\theta +\sin ^{5}\left({\frac {2\theta -\pi }{24}}\right)} The sin term has been added for purely aesthetic reasons, to make the butterfly appear fuller and more pleasing to the eye.[1] In 2006, two mathematicians using Mathematica analyzed the function, and found variants where leaves, flowers or other insects became apparent.[3] Butterfly curve (algebraic) Oscar’s Butterfly Polar Equation r = (cos 5θ)2 + sin 3θ + 0.3 for 0 ≤ θ ≤ 6π (A polar equation discovered by Oscar Ramirez, a UCLA student, in the fall of 1991.) ^ a b Fay, Temple H. (May 1989). "The Butterfly Curve". Amer. Math. Monthly. 96 (5): 442–443. doi:10.2307/2325155. JSTOR 2325155. ^ Weisstein, Eric W. "Butterfly Curve". MathWorld. ^ "On the analysis and construction of the butterfly curve using Mathematica". International Journal of Mathematical Education in Science and Technology. 39 (5): 670–678. June 2008. doi:10.1080/00207390801923240. Butterfly Curve plotted in WolframAlpha Retrieved from "https://en.wikipedia.org/w/index.php?title=Butterfly_curve_(transcendental)&oldid=1087122273"
The confirmation of the existence of the Higgs boson is probably one of the most important findings made by the LHC. Not only did it complete the Standard model, but also gave insight into the early cosmological universe by having a better understanding of how the processes could function. So not only does the LHC play an important role in furthering particle physics, but its findings are of high significance for astrophysicists who seek to understand the early universe some of whose properties are best emulated in a particle accelerator. Finding the Higgs boson is not easy. Not only does it decay about an order of magnitude faster than other elementary particles making it impossible to detect directly in ATLAS, or other detectors, while it does decay into a lot of things, all of those are very common making isolating the Higgs boson very difficult. One common decay pattern is for the Higgs boson to decay to two photons ( H\rightarrow \gamma_1 \gamma_2 ) with the 4 momntum vectors \textbf{p}^\alpha_{\gamma_1}= \frac{E_1}{c} \cdot (1,\hat{n}_1) \textbf{p}^\alpha_{\gamma_2}= \frac{E_2}{c} \cdot (2,\hat{n}_2) . Furthermore, we know the angle between the two photon beams is given by \theta \cos(\theta)=\hat{n}_1\cdot\hat{n}_2 \textbf{p}^2=-m_0^2c^2 , and photons don't have a invariant mass, we know that \textbf{p}_1^2 = \textbf{p}_2^2 = 0 We therefore find that (\textbf{p}^\alpha_{\gamma_1}+\textbf{p}^\alpha_{\gamma_2})^2=2\textbf{p}^\alpha_{\gamma_1}\cdot\textbf{p}^\alpha_{\gamma_2} (\textbf{p}^\alpha_{\gamma_1}+\textbf{p}^\alpha_{\gamma_2})^2=-2 \frac{E_1E_2}{c^2} (1-cos(\theta)). \textbf{p}^2=-m_0^2c^2 , we can find the invariant mass of the parent particle (the Higgs boson) m_H = \frac{1}{c^2} \sqrt{2E_1E_2(1-\cos(\theta))}. E_1,E_2,\theta are all measureable by the detector, we can therefore get a value of the mass of the Higgs boson. Exactly this was done on data from 2011 and 2012 where it was found that the mass of the Higgs boson was around m_H=126.8 \text{GeV} While the above algebra may seem easy, it's not quite as simple to isolate H\rightarrow \gamma \gamma events. In order to find such events a likelihood function must be developed based on statistical knowledge as well as the physical understanding of the problem. It took most of the first run of the LHC to tune to likelihood function, due to the many tunable parameters, and it's still limiting how many collisions are possible in each batch due to the added noise of more collisions. These are some of the reasons why experiments in using machine learning algoirthms such as neural networks to let computer automatically find an optimal likelihood function have begun. Below has a simple neural network been implemented to classify Higgs decay based on the Higgs ML dataset from 2014. You can use the two features to compete against the neural network in detecting Higgs decays. (The feature values have been normalised, and the y -axis is a kernel density estimation of the value. The features have been crafted and chosen from the detector data as those with the highest predictive power) Look at the graphs to determine if there's a Higgs decay. Higgs No Higgs `) $('.dot').removeClass("correct") $('.dot').removeClass("incorrect") $('.restart').css("display", "none") $('.controls').css("display", "flex") } function game_loop(){ if (currentExample == 0){ cleanup(); init(); } // progress the game currentExample += 1 // check if game has ended if (currentExample == 6){ return end_sequence(); } render_example(); } function restart(){ cleanup(); game_loop(); } function linspace(a,b,n) { if(typeof n === "undefined") n = Math.max(Math.round(b-a)+1,1); if(n<2) { return n===1?[a]:[]; } var i,ret = Array(n); n--; for(i=n;i>=0;i--) { ret[i] = Math.round(((ib+(n-i)a)/n)* 100)/100; } return ret; } if (!window.location.href.includes("edit")) { game_loop() } You can learn more about a similar network to the one above here, or you can learn more about neural networks in general here. Time and space are the same thing, and it's all relative?
Fit linear regression model - MATLAB fitlm - MathWorks Fit Linear Regression Using Data in Table Fit Linear Regression Using Specified Model Formula Fit Linear Regression Using Terms Matrix Specify Response and Predictor Variables for Linear Model mdl = fitlm(X,y) returns a linear regression model of the responses y, fit to the data matrix X. mdl = fitlm(___,modelspec) defines the model specification using any of the input argument combinations in the previous syntaxes. mdl = fitlm(___,Name,Value) specifies additional options using one or more name-value pair arguments. For example, you can specify which variables are categorical, perform robust regression, or use observation weights. \mathit{y}={\beta }_{0}+{\beta }_{1}{\mathit{X}}_{1}+{\beta }_{2}{\mathit{X}}_{2}+{\beta }_{3}{\mathit{X}}_{3}+ϵ Fit a linear regression model for miles per gallon (MPG). Specify the model formula by using Wilkinson notation. The model 'MPG~Weight+Acceleration' in this example is equivalent to set the model specification as 'linear'. For example, If you use a character vector for model specification and you do not specify the response variable, then fitlm accepts the last variable in tbl as the response variable and the other variables as the predictor variables. Fit a linear regression model using a model formula specified by Wilkinson notation. Fit a linear regression model for miles per gallon (MPG) with weight and acceleration as the predictor variables. The p-value of 0.18493 indicates that Acceleration does not have a significant impact on MPG. Remove Acceleration from the model, and try improving the model by adding the predictor variable Model_Year. First define Model_Year as a categorical variable. Specifying modelspec using Wilkinson notation enables you to update the model without having to change the design matrix. fitlm uses only the variables that are specified in the formula. It also creates the necessary two dummy indicator variables for the categorical variable Model_Year. Fit a linear regression model using a terms matrix. Terms Matrix for Table Input If the model variables are in a table, then a column of 0s in a terms matrix represents the position of the response variable. Represent the linear model 'BloodPressure ~ 1 + Sex + Age + Smoker' using a terms matrix. The response variable is in the second column of the table, so the second column of the terms matrix must be a column of 0s for the response variable. Fit a linear model. Terms Matrix for Matrix Input If the predictor and response variables are in a matrix and column vector, then you must include 0 for the response variable at the end of each row in a terms matrix. Load the carsmall data set and define the matrix of predictors. Specify the model 'MPG ~ Acceleration + Weight + Acceleration:Weight + Weight^2' using a terms matrix. This model includes the main effect and two-way interaction terms for the variables Acceleration and Weight, and a second-order term for the variable Weight. Only the intercept and x2 term, which corresponds to the Weight variable, are significant at the 5% significance level. \mathrm{MPG}={\beta }_{0}+{\beta }_{1}{Ι}_{\mathrm{Year}=76}+{\beta }_{2}{Ι}_{\mathrm{Year}=82}+ϵ {Ι}_{\mathrm{Year}=76} {Ι}_{\mathrm{Year}=82} {Ι}_{\mathrm{Year}=76} {Ι}_{\mathrm{Year}=82} \mathit{y}={\beta }_{0}{Ι}_{{\mathit{x}}_{1}=70}+\left({\beta }_{0}+{\beta }_{1}\right){Ι}_{{\mathit{x}}_{1}=76}+\left({{\beta }_{0}+\beta }_{2}\right){Ι}_{{\mathit{x}}_{2}=82}+ϵ {Ι}_{\mathrm{Year}=70} {Ι}_{\mathrm{Year}=82} Fit a linear regression model to sample data. Specify the response and predictor variables, and include only pairwise interaction terms in the model. Fit a linear model with interaction terms to the data. Specify weight as the response variable, and sex, age, and smoking status as the predictor variables. Also, specify that sex and smoking status are categorical variables. The weight of the patients do not seem to differ significantly according to age, or the status of smoking, or interaction of these factors with patient sex at the 5% significance level. By default, fitlm takes the last variable as the response variable and the others as the predictor variables. Example: 'y ~ x1 + x2^2 + x1:x2' Example: 'Intercept',false,'PredictorVars',[1,3],'ResponseVar',5,'RobustOpts','logistic' specifies a robust regression model with no constant term, where the algorithm uses the logistic weighting function with the default tuning constant, first and third variables are the predictor variables, and fifth variable is the response variable. If data is in a table or dataset array tbl, then, by default, fitlm treats all categorical values, logical values, character arrays, string arrays, and cell arrays of character vectors as categorical variables. RobustOpts — Indicator of robust fitting type 'off' (default) \| 'on' \| character vector \| string scalar \| structure Indicator of the robust fitting type to use, specified as the comma-separated pair consisting of 'RobustOpts' and one of these values. 'off' — No robust fitting. fitlm uses ordinary least squares. 'on' — Robust fitting using the 'bisquare' weight function with the default tuning constant. Character vector or string scalar — Name of a robust fitting weight function from the following table. fitlm uses the corresponding default tuning constant specified in the table. Structure with the two fields RobustWgtFun and Tune. The RobustWgtFun field contains the name of a robust fitting weight function from the following table or a function handle of a custom weight function. The Tune field contains a tuning constant. If you do not set the Tune field, fitlm uses the corresponding default tuning constant. The default tuning constants of built-in weight functions give coefficient estimates that are approximately 95% as statistically efficient as the ordinary least-squares estimates, provided the response has a normal distribution with no outliers. Decreasing the tuning constant increases the downweight assigned to large residuals; increasing the tuning constant decreases the downweight assigned to large residuals. where resid is the vector of residuals from the previous iteration, tune is the tuning constant, h is the vector of leverage values from a least-squares fit, and s is an estimate of the standard deviation of the error term given by For robust fitting, fitlm uses M-estimation to formulate estimating equations and solves them using the method of Iteratively Reweighted Least Squares (IRLS). Example: 'RobustOpts','andrews' If the value of the 'RobustOpts' name-value pair is not [] or 'ols', the model is not a least-squares fit, but uses the robust fitting function. To access the model properties of the LinearModel object mdl, you can use dot notation. For example, mdl.Residuals returns a table of the raw, Pearson, Studentized, and standardized residual values for the model. The main fitting algorithm is QR decomposition. For robust fitting, fitlm uses M-estimation to formulate estimating equations and solves them using the method of Iteratively Reweighted Least Squares (IRLS). fitlm treats a categorical predictor as follows: fitlm treats the group of L – 1 indicator variables as a single variable. If you want to treat the indicator variables as distinct predictor variables, create indicator variables manually by using dummyvar. Then use the indicator variables, except the one corresponding to the reference level of the categorical variable, when you fit a model. For the categorical predictor X, if you specify all columns of dummyvar(X) and an intercept term as predictors, then the design matrix becomes rank deficient. fitlm considers NaN, '' (empty character vector), "" (empty string), <missing>, and <undefined> values in tbl, X, and Y to be missing values. fitlm does not use observations with missing values in the fit. The ObservationInfo property of a fitted model indicates whether or not fitlm uses each observation in the fit. If any input argument to fitlm is a tall array, then all of the other inputs must be tall arrays as well. This includes nonempty variables supplied with the 'Weights' and 'Exclude' name-value pairs. The 'RobustOpts' name-value pair is not supported with tall arrays. For tall data, fitlm returns a CompactLinearModel object that contains most of the same properties as a LinearModel object. The main difference is that the compact object is sensitive to memory requirements. The compact object does not include properties that include the data, or that include an array of the same size as the data. The compact object does not contain these LinearModel properties: You can compute the residuals directly from the compact object returned by LM = fitlm(X,Y) using If the CompactLinearModel object is missing lower order terms that include categorical factors: The plotEffects and plotInteraction methods are not supported. The anova method with the 'components' option is not supported.
Trends in the Periodic Table - Course Hero General Chemistry/The Periodic Table/Trends in the Periodic Table The periodic table arranges elements in a meaningful way, which allows for trends in atomic radius, nuclear charge, electronegativity, electron affinity, and ionization energy to be easily determined. The periodic table arranges elements in increasing atomic number. The columns and rows are formed based on the electron configuration of the element. When elements are arranged as in the periodic table, some trends can be determined. These trends are based on the number of subatomic particles found in each atom and how electrons are arranged in shells. Atomic radius decreases from left to right across the periodic table and increases from top to bottom down the table. Nuclear charge increases from left to right across the periodic table and from top to bottom down the periodic table. The reason for the development of the periodic table was to arrange the elements in a way that had meaning. Thus, specific trends are apparent in the table. The major trends associated with the periodic table are atomic radius, nuclear charge, electronegativity, electron affinity, and ionization energy. Atomic radius is half the distance between the nuclei of two identical atoms. The edge of the atom is difficult to define because the position of any electron at a given time cannot be precisely known. Atomic radius is therefore described in some different ways: Covalent radius is the half the distance between the nuclei of two identical atoms in a covalent bond. Van der Waals radius is half the distance between two identical nuclei that are not bonded but are as close as possible. Atomic radius decreases from left to right across the periodic table and increases from top to bottom down the table. Atomic radius decreases from left to right because, within a period, the number of electrons increases within a shell while the number of protons increases in the nucleus. More protons and electrons mean the electromagnetic attraction between them also increases, thereby pulling the electron shell closer to the nucleus and decreasing the atomic radius. Within a group, however, electrons are added to higher valence shells. A higher valence shell is farther away from the nucleus than existing shells. These electrons at the higher valence shell are farther away from the nucleus, increasing atomic radius. Atomic radius, which is half the distance between the nuclei of like atoms, generally decreases from left to right across the periodic table. Atomic radius generally increases from top to bottom down the periodic table. Nuclear charge (Z) is the total charge of all protons within the nucleus. It is therefore equal to atomic number. Nuclear charge increases from left to right across the periodic table and from top to bottom down the periodic table. Effective nuclear charge (Zeff) is the net positive charge of an atom when electron shielding is considered. Electron shielding is the decrease in attraction from the nucleus experienced by valence electrons due to the presence of inner electrons between the nucleus and the valence electrons. Effective nuclear charge is calculated by subtracting the number of nonvalence electrons (S) from the nuclear charge (Z). {\rm{Z}}_{\rm{eff}}=\rm{Z}-\rm{S} For example, neon has 10 protons and 10 electrons. Two of these electrons are nonvalence electrons. {\rm {Z}}_{\rm{eff}}=10-2=8 for neon. Like nuclear charge, effective nuclear charge increases from left to right across the periodic table and from top to bottom down the periodic table. Nuclear charge is the total charge of all protons in the nucleus. It increases from left to right across the periodic table and from top to bottom down the periodic table. Electronegativity increases from left to right across the periodic table, with a few exceptions, and decreases from top to bottom down the periodic table. Electron affinity increases from left to right across the periodic table and decreases from top to bottom down the periodic table. Electronegativity is the tendency of an atom to attract electrons toward itself when forming bonds. It is a qualitative rather than quantitative attribute; that is, it cannot be measured numerically. In the 1930s American chemist Linus Pauling devised a scale that ranks elements according to electronegativity, which is still in use today. Most elements form bonds with other elements to satisfy the octet rule: they attempt to fill eight electrons in their valence shell. Elements on the left side of the periodic table have fewer than four valence electrons, so they tend to lose electrons when forming bonds, while those on the right side of the periodic table have more than four valence electrons, so they tend to gain electrons when forming bonds. Thus, electronegativity increases from left to right across the periodic table, with a few exceptions. The noble gases have full valence shells, so they do not commonly form bonds with other elements at all and thus have very low electronegativity. The transition metals form vertical groups 3 to 12. Between groups 3 and 11, electronegativity of transition metals generally increases. Group 12 transition metals have lower electronegativity compared to group 11. Elements of group 13, the first group of p-block elements, have electronegativities very similar to those of group 12 transition metals. Lanthanides and actinides are a general exception to electronegativity. They have much more complicated chemistry and do not show any trends. Lanthanides and actinides are considered to have no electronegativity. Electronegativity decreases from top to bottom down the periodic table. This happens because the atoms of elements lower in a group have a larger atomic radius and thus an increased distance between the valence electrons and the nucleus. As a result, atoms with a larger atomic radius exert a weaker force on nearby atoms. They are therefore less likely to form bonds with other elements compared to atoms with a smaller atomic radius. Electronegativity, the tendency for an atom to attract electrons toward itself when forming bonds, increases from left to right across the periodic table. Electronegativity also decreases from top to bottom down the periodic table. Nobles gases have full valence shells and tend not to form bonds. Noble gases are an exception to general trend in electronegativity. Electron affinity is the change in energy that occurs when an atom of a neutral gas gains an electron. It is a quantitative attribute, and thus can be measured numerically. Electron affinity indicates the ability of an atom to accept an electron. It increases from left to right across the periodic table and decreases from top to bottom down the periodic table. These trends are due to atomic radius: a smaller atomic radius is associated with a higher electron affinity, because less energy is required to add electrons to a valence shell closer to the nucleus compared to one that is farther away. However, electron affinities are difficult to measure accurately, so the trends are a generalization. Electron affinity is the ability of an atom to accept an electron. In general it increases from left to right across the periodic table and decreases from top to bottom down the periodic table. However, it is difficult to measure electron affinity accurately, so these trends only apply in a general sense. Ionization energy increases from left to right across the periodic table and decreases from top to bottom down the periodic table. Ionization energy is the amount of energy required to remove an electron from a gaseous atom or ion. It can be thought of as the opposite of electron affinity. It is a quantitative attribute, expressed either in joules (J) or electron volts (eV). Each electron in an atom has its own ionization energy; however, the energy needed to remove the first electron is usually referenced when discussing any element. The first ionization energy is the energy required to remove an electron from each atom in one mole of atoms in a gaseous state, resulting in one mole of ions with a 1+ charge. The second ionization energy is the energy required to remove a second electron from each ion in one mole of 1+ ions in a gaseous state, resulting in a mole of ions with a 2+ charge. Note that second ionization energy is not the total energy required to remove two electrons; it is the difference between the energy required to remove the first and second electrons. There are additional ionization energies for each electron in the atom. Ionization energy depends on the charge of the nucleus, the atomic radius, and the electron configuration of the atom. Elements on the left side of the periodic table have their valence shells less than half full, so they readily give up an electron. In contrast, elements on the right side of the periodic table have their valence shells more than half full, so they do not give up electrons easily. This explains why metals are more likely to form cations and nonmetals are more likely to form anions. In addition, elements with valence shells farther away from the nucleus, that is, those lower on the periodic table, give up those valence electrons more easily than elements with valence shells closer to the nucleus. This is due to not only the distance between the nucleus and these electrons, but also the effect of electron shielding of electrons in nonvalence shells. However, after the first electron is given up, the remaining electrons are held more firmly because the charge on the nucleus does not change. This is also true for each additional electron given up. Thus, ionization energy increases from left to right across the periodic table and decreases from top to bottom down the periodic table. Ionization energy is the amount of energy required to remove an electron from an atom. It increases from left to right across the periodic table and decreases from top to bottom down the periodic table. When all of the trends on the periodic table are viewed together, relationships among the properties of elements become clear. Electron affinity, effective nuclear charge, ionization energy, and electronegativity generally increase from left to right across the periodic table because of the way in which electrons fill the valence shell across a period. Effective nuclear charge and atomic radius generally increase down a group because of the increase in distance between valence electrons and the nucleus. The periodic table can be used to show trends in atomic radius, effective nuclear charge, electronegativity, ionization energy, and electron affinity. These trends can be explained by the number of protons and electrons in the elements and how the electrons are arranged in electron shells. <Development of the Periodic Table>Suggested Reading
Home : Support : Online Help : Programming : Data Types : Type Checking : TypeTools : GetType retrieve definition of user-defined type GetType(typename) The definition for the type typename is returned. This function can only be used to retrieve definitions of types that have been registered using TypeTools[AddType]. In particular, it cannot be used to retrieve built-in types. \mathrm{TypeTools}[\mathrm{AddType}]⁡\left(\mathrm{tff},'{\mathrm{identical}⁡\left(\mathrm{FAIL}\right),\mathrm{identical}⁡\left(\mathrm{false}\right),\mathrm{identical}⁡\left(\mathrm{true}\right)}'\right) \mathrm{TypeTools}[\mathrm{GetType}]⁡\left(\mathrm{tff}\right) {\textcolor[rgb]{0,0,1}{\mathrm{identical}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{\mathrm{FAIL}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{identical}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{\mathrm{false}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{identical}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{\mathrm{true}}\right)} \mathrm{TypeTools}[\mathrm{AddType}]⁡\left(\mathrm{integer7},t↦\mathrm{evalb}⁡\left(t::'\mathrm{integer}'\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{irem}⁡\left(t,7\right)=0\right)\right) a≔\mathrm{TypeTools}[\mathrm{GetType}]⁡\left(\mathrm{integer7}\right) \textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{TypeTools/integer7}} \mathrm{print}⁡\left(a\right) \textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{↦}\textcolor[rgb]{0,0,1}{\mathrm{evalb}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{::}\textcolor[rgb]{0,0,1}{\mathrm{Typesetting}}\textcolor[rgb]{0,0,1}{:-}\textcolor[rgb]{0,0,1}{\mathrm{_Hold}}\textcolor[rgb]{0,0,1}{⁡}\left(\left[\textcolor[rgb]{0,0,1}{'}\textcolor[rgb]{0,0,1}{\mathrm{integer}}\textcolor[rgb]{0,0,1}{'}\right]\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{and}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathrm{irem}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\right)
Electrochemistry Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Metals have conductivity of the order of (ohm-1 m-1):- Subtopic: Conductance & Conductivity \| An electrochemical cell is shown below Pt, H2(1 atm)\| HCI (0.1 M)CH3COOH (0.1 M)\| H2(1 atm), Pt The EMF of the cell will not be zero, because 1. EMF depends on molarities of acids used 2. pH of 0.1 M HCl and 0.1 M CH3COOH is not same 3. the temperature is constant 4. acids used in two compartments are different Subtopic: Electrode & Electrode Potential \| Saturated solution of KNO3 is used to make 'salt-bridge' because: 1. velocity of K+ is greater than that of {\mathrm{NO}}_{3}^{-} 2. velocity of {\mathrm{NO}}_{3}^{-} is greater than that of K+ 3. Velocities of both K+ and {\mathrm{NO}}_{3}^{-} are nearly the same 4. KNO3 is highly soluble in water Subtopic: Batteries & Salt Bridge \| A current is passed through two voltameters connected in series. The first voltmeter connected in series. The first voltmeter contains XSO4(aq) while the second voltmeter contains Y2SO4(aq). The relative atomic masses of X and Y are in the ratio of 2:1. The ration of the mass of X liberated to the mass of Y liberated is: Subtopic: Faraday’s Law of Electrolysis \| The mass of silver(eq. mass = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is: A silver cup is plated with silver by passing 965 coulomb of electricity. The amount of Ag deposited is: Which is the correct representation for Nernst equation ? {E}_{\mathrm{RP}} = {E}_{\mathrm{RP}}^{\circ }\mathit{ }+ \frac{0.059}{\mathrm{n}}\mathrm{log}\frac{\left[\mathrm{oxidant}\right]}{\left[\mathrm{reductant}\right]} {E}_{\mathrm{OP}} = {E}_{\mathrm{OP}}^{\circ }\mathit{ }\mathit{-} \frac{0.059}{\mathrm{n}}\mathrm{log}\frac{\left[\mathrm{oxidant}\right]}{\left[\mathrm{reductant}\right]} {E}_{\mathrm{OP}} = {E}_{\mathrm{OP}}^{\circ }\mathit{ }+ \frac{0.059}{\mathrm{n}}\mathrm{log}\frac{\left[\mathrm{reductant}\right]}{\left[\mathrm{oxidant}\right]} Subtopic: Nernst Equation \| When a copper wire is immersed in a solution of AgNO3, the colour of solution becomes blue because copper: 1. Forms a soluble complex with AgNO3 2. Is oxidised to Cu2+ 3. Is reduced to Cu2- 4. Splits up into atoic form and dissolves. Subtopic: Electrochemical Series \| The specific conductance of a 0.1N KCl solution at 23 °\mathrm{C} is 0.012 {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1} . The resistance of cell containing the solution at the same temperature was found to be 55 \mathrm{\Omega } . The cell constant will be 1. 0.142cm-1 2. 0.66 cm-1 3. 0.918 cm-1 4. 1.12cm-1 Given below are the half-cell reactions, Mn2+ + 2e- Mn will be:
Capital Model - Nexus Mutual The capital model determines the minimum amount of funds the mutual needs to hold. Nexus Mutual uses actuarial mathematics derived from the Solvency II methodology as developed by the European Insurance and Occupational Pensions Authority ("EIOPA") to set its Minimum Capital Requirement ("MCR"). There are two main components making up the MCR calculation: The Best Estimate Liability (or “BEL”), representing the expected loss on each individual cover. A Buffer, representing the funds the pool requires to survive a 'black swan' event. Similarly to a traditional insurance entity, Nexus Mutual will hold (and invest) a Capital Pool of assets in excess of the MCR in order to back its covers (see Assets and Investment). The ratio between the Capital Pool and the MCR is known as the coverage ratio and abbreviated to MCR%. Note that the MCR is subject to a lower limit below which it cannot fall. This level was set at 7000 ETH but is now dynamic given by MCR_floor. The BEL will initially be equal to the total Risk Cost across all active covers on the mutual's books. In the future, as we begin to gather experience on the outcomes of covers, we will be able to create a more accurate BEL which also allows for the remaining duration on each individual cover. Currently, the BEL for each cover represents the entire Risk Cost regardless of remaining duration - a prudent assumption resulting in higher reserves required per cover. The Buffer represents the funds that Nexus Mutual will hold, in addition to the BEL, in order to protect itself against unforeseen adverse events. The intention (with no obligation) is to follow the Solvency II framework and calibrate this amount of funds to a level where the mutual can survive a 1-in-200-year adverse event. In practice, due to the uniqueness of the Protocol Cover and Custody Cover products, some deviations from the SII texts are initially necessary. Smart Contract Cover Module The Smart Contract Cover Module is based on the exposures Nexus Mutual has to the covers it has written. The inputs to the model are: Total Cover Amounts CA(i) for each individual protocol and custodian Correlations Corr(i,j) between each pair of contracts. Scaling Factor SC which is calibrated to make the capital result as a whole more comparable to a full Solvency II calculation. The correlations between each pair of two contracts are established by parsing the respective verified smart contract code, removing comments and spacing, and establishing the proportion of identical text. These inputs are fed into the following formula to produce a Capital Requirement CR across all covered contracts within Protocol Cover and Custody Cover: CR_{SCC} = SC \times \sqrt{\sum_{i,j} Corr(i, j) \times CA(i) \times CA(j)} The currency module allows for possible fluctuations in the value of other currencies (initially only DAI) relative to the value of the base currency (ETH). 50% stresses in both directions are applied to the value of the other currencies in order to establish the impact on both the assets and the buffer requirement of the mutual. Final MCR as per Capital Model The Currency Module scenario with the lowest resulting MCR% coverage (across both the BEL and the Buffer) is picked out. This MCR% coverage is then applied to the Capital Pool in order to inform the choice of Gearing Factor.
MapleTA[Export] - Maple Help Home : Support : Online Help : Connectivity : Maple T.A. : MapleTA[Export] export a Maple T.A. course module Export( questions, filename ) The Export command accepts an individual question, or list of questions and exports them as a course module file suitable for import into Maple T.A. Questions must be in a Record format similar to that returned by the MapleTA:-Import command. r≔\mathrm{MapleTA}:-\mathrm{Import}⁡\left("coursemodule.zip"\right) \mathrm{MapleTA}:-\mathrm{Export}⁡\left([r],"newcoursemodule.zip"\right) The MapleTA[Export] command was introduced in Maple 18.
Fuzzy Trees - MATLAB & Simulink - MathWorks í•œêµ FI{S}_{i}^{n} {x}_{ij}{}^{n} {y}_{ik}{}^{n} . In the figure, n = 3, j = 1 or 2, and k = 1. If each input has m membership functions (MFs), each FIS has a complete set of m2 rules. Hence, the total number of rules is nm2 = 3 Ã— 32 = 27. In the FIS of this figure, the total number of rules is nm4 = 1 Ã— 34 = 81. Hence, the total number of rules in an incremental fuzzy tree is linear with the number of input pairs. In an incremental fuzzy tree, each input value usually contributes to the inference process to a certain extent, without being significantly correlated with the other inputs. For example, a fuzzy system forecasts the possibility of buying an automobile using four inputs: color, number of doors, horse power, and autopilot. The inputs are four distinct automobile features, which can independently influence a buyerâ€™s decision. Hence, the inputs can be ranked using the existing data to construct a fuzzy tree, as shown in the following figure. FI{S}_{{i}_{n}}^{n} {x}_{{i}_{n}j} {y}_{{i}_{n}k} . In the figure, j = 1,2 and k = 1. In other words, each FIS has two inputs and one output. If each input has m MFs, then each FIS has a complete set of m2 rules. Hence, the total number of rules for the three fuzzy systems is 3 m2 = 3 Ã— 32 = 27, which is the same as an incremental FIS for a similar configuration. The fistree object does not provide the summing node Î£. Therefore, you must add a custom aggregation method to evaluate a parallel fuzzy tree. For an example, see the "Create and Evaluate Parallel FIS Tree" example on the fistree reference page.
Board Paper Solutions for ICSE Class 12-science PHYSICS Board Paper 2017 MeritNation Set 1 Answer all questions in Part A and ten questions from Part B, choosing four questions from (Material to be supplied: Log tables including Trigonometric functions). Useful Constants and Relations: 1. Charge of a proton (e) = 1.6 × 10–19 C 2. Planck's constant (h) = 6.6 × 10–34 Js 3. Mass of an electron (m) = 9.1 × 10–31 kg 4. Permittivity of vacuum (ε0) = 8.85 × 10–12 Fm–1 5. = 9 × 109 mF–1 6. Permeability of vacuum (µ0) = 4π × 10–1 Hm–1 7. = 1 × 10–7 Hm–1 8. Speed of light in vacuum (c) = 3 × 108 ms–1 9. Unified atomic mass unit (u) = 931 MeV 10. Electron volt (leV) = 1.6 × 10–19 J (a) Choose the correct alternative (a), (b), (c) or (d) for each of the questions given below: (i) The electrostatic–potential energy of two point charges, 1 μC each, placed 1 meter apart in air is:. (a) 9 × 103J (b) 9 × 109J (c) 9 × 10–3J (d) 9 ×10–3eV (ii) A wire of resistance 'R' is cut into 'n' equal parts. These parts are then connected in parallel with each other. The equivalent resistance of the combination is: (iii) Magnetic susceptibility of platinum is 0· 0001. Its relative permeability is–: (a) 1 ·0000 (b) 0·9999 (c) 1·0001 (iv) When a light wave travels from air to glass: (a) its wavelength decreases. (b) its wavelength increases. (c) there is no change in wavelength. (d) 2·5 (b) Answer all questions given below briefly and to the point: (i) Maximum torque acting on an electric dipole of moment 3 × 10–29 Cm in a uniform electric field E is 6 × 10–25 Nm. Find E. (ii) What is meant by drift speed of free electrons? (iii) On which conservation principle is Kirchoff's Second Law of electrical networks based? (iv) Calculate magnetic flux density of the magnetic field at the centre of a circular coil of 50 turns, having radius of 0.5 m and carrying a current of 5 A. (v) An a.c. generator generates an emf 'ε' where ε = 314 Sin(50πt) volt. Calculate the frequency of the emf ε. (vi) With what type of source of light are cylindrical wave fronts associated?· (vii) How is fringe width of an interference pattern in Young's double slit experiment affected if the two slits are brought closer to each other? (viii) In a regular prism, . what is the relation between angle of incidence and angle of emergence when it is in the minimum deviation position? (x) How can the spherical aberration produced by a lens be minimized? (xi) Calculate the momentum of a photon of energy 6 × 10–19 J. (xii) According to Bohr, 'Angular momentum of an orbiting electron is quantised'. What is meant by this statement? (xiii) Why nuclear fusion reaction is also called thermo-nuclear reaction? (xiv) What is the minimum energy which a gamma-ray photon must possess in order to produce electron-positron pair? · (xv) Show the variation of voltage with.time, for a digital signal. VIEW SOLUTION (a) Show that electric potential at a point P, at a distance 'r' from a fixed point charge Q, is given by: \mathrm{V}=\left(\frac{1}{4\pi {\epsilon }_{0}}\right)\frac{Q}{r} (b) Intensity of electric field at a perpendicular distance of 0.5 m from an infinitely long line charge having linear charge density (λ) is 3.6 × 103 Vm–1· Find the value of λ. VIEW SOLUTION (a) Three capacitors C1 = 3μF, C2 = 6μF and C3 = l0μF are connected to a 50 V battery as shown in the Figure 1 below: (ii) The charge on C1. (b) Two resistors R1 = 60 Ω and R2 = 90 Ω are connected in parallel. If electric power consumed by the resistor R1 is 15 W, calculate the power consumed by the resistor R2. VIEW SOLUTION (a) Figure 2 below shows two resistors R1 and R2 connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of 300 Ω is used to measure potential difference across R1 . Find the reading of the voltmeter. (b) A moving coil galvanometer has a coil of resistance 59Ω. It shows a full scale deflection for a current of 50 mA. How will you convert it to an ammeter having a range of 0 to 3A? VIEW SOLUTION (a) In a meter bridge circuit, resistance in the left hand gap is 2Ω and an unknown resistance X is in the right hand gap as shown in Figure 3 below. The null point is found to be 40 cm from the left end of the wire. What resistance should be connected to X so that the new null point is 50 cm from the left end of the wire? (b) The horizontal component of earth's magnetic field at a place is \frac{1}{\sqrt{3}} times the vertical component. Determine the angle of dip at that place. VIEW SOLUTION (a) Using Ampere's circuital law, obtain an expression for the magnetic flux density 'B' at a point 'X' at a perpendicular distance 'r' from a long current carrying conductor. (Statement of the law is not required). (b) PQ is a long straight conductor carrying a current of 3A as shown in Figure 4 below. An electron moves with a velocity of 2 × 107 ms–1 parallel to it. Find the force acting on the electron. (a) (i) AB and CD are two parallel conductors kept 1 m apart and connected by a resistance R of 6 Ω, as shown in Figure 5 below. They are placed in a magnetic field B = 3 × 10–2 T which is perpendicular to the plane of the conductors and directed into the paper. A wire MN is placed over AB and CD and then made to slide with a velocity 2 ms–1. (Neglect the resistance of AB, CD, and MN.) (ii) In an ideal transformer, an output of 66 kV is required when an input voltage of 220 V is available. If the primary has 300 turns, how many turns should the secondary have? (b) 1n a series LCR circuit, obtain an expression for the resonant frequency. VIEW SOLUTION (a) (i) State any one property which is common to all electromagnetic waves. (ii) Arrange the following electromagnetic waves in increasing order of their frequencies (i.e. begin with the lowest frequency): Visible light, γ rays, X rays, micro waves, radio waves, infrared radiations and ultraviolet radiations. (b) (i) What is meant by diffraction of light? (ii) In Fraunhofer diffraction, what kind of source of light is used and where is it situated? VIEW SOLUTION (a) In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe' is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate: (i) Distance between the two slits. (ii) Fringe width, i.e. fringe separation. (b) (i) State Brewster's law. (ii) Find Brewster's angle for a transparent liquid having refractive index 1·5. VIEW SOLUTION (a) Find critical angle for glass and water pair, given refractive index of glass is l·62 and that of water is 1·33. (b) Starting with an expression for refraction at a single spherical surface, obtain Lens Maker's Formula. VIEW SOLUTION (a) A compound microscope consists of two convex lenses of focal length 2 cm and 5 cm· When an object is kept at a distance of 2·l cm from the objective, a virtual and magnified image is formed 25 cm from the eye piece. Calculate the magnifying power of the microscope. (b) (i) What is meant by resolving power of a telescope? (ii) State any one method of increasing the resolving power of an astronomical telescope. VIEW SOLUTION (a) (i) Plot a labelled graph of \|VS\| where VS is stopping potential versus frequency f of the incident radiation. (ii) State how will you use this graph to determine the value of Planck's constant. (b) (i) Find the de Broglie wavelength of electrons moving with a speed of 7 × 106 ms–1. (ii) Describe in brief what is observed when moving electrons are allowed to fall on a thin graphite film and the emergent beam falls on a fluorescent screen. VIEW SOLUTION (a) Draw energy level diagram for hydrogen atom, showing first four energy levels corresponding to n = l, 2, 3 and 4. Show transitions responsible for: (i) Absorption spectrum of Lyman series. (b) (i) Find maximum frequency of X-rays produced by an X-ray tube operating at a tube potential of 66 kV. (ii) State any one difference between characteristic X-rays and continuous X-rays. VIEW SOLUTION (a) Obtain a relation between half life of a radioactive substance and decay constant (λ). (b) Calculate mass defect and binding energy per nucleon of {}_{10}{}^{20}\mathrm{Ne} \mathrm{Mass} \mathrm{of} {}_{10}{}^{20}\mathrm{Ne}=19.992397\mathit{ }u\phantom{\rule{0ex}{0ex}}\mathrm{Mass} \mathrm{of} {}_{1}{}^{1}\mathrm{H}=1.007825\mathit{ }u\phantom{\rule{0ex}{0ex}}\mathrm{Mass} \mathrm{of} {}_{0}{}^{1}n=1.008665 u (a) With reference to a semi-conductor diode, what is meant by: (b) Draw a diagram to show how NAND gates can be combined to obtain an OR gate. (Truth table is not. required) VIEW SOLUTION
Beta-ureidopropionase - Wikipedia In enzymology, a beta-ureidopropionase (EC 3.5.1.6) is an enzyme that catalyzes the chemical reaction N-carbamoyl-beta-alanine + H2O {\displaystyle \rightleftharpoons } beta-alanine + CO2 + NH3 Thus, the two substrates of this enzyme are N-carbamoyl-beta-alanine and H2O, whereas its 3 products are beta-alanine, CO2, and NH3. This enzyme belongs to the family of hydrolases, those acting on carbon-nitrogen bonds other than peptide bonds, specifically in linear amides. The systematic name of this enzyme class is N-carbamoyl-beta-alanine amidohydrolase. This enzyme participates in 3 metabolic pathways: pyrimidine metabolism, beta-alanine metabolism, and pantothenate and coenzyme A biosynthesis. As of late 2007, 6 structures have been solved for this class of enzymes, with PDB accession codes 1R3N, 1R43, 2V8D, 2V8G, 2V8H, and 2V8V. CAMPBELL LL (1960). "Reductive degradation of pyrimidines. 5. Enzymatic conversion of N-carbamyl-beta-alanine to beta-alanine, carbon dioxide, and ammonia". J. Biol. Chem. 235: 2375–8. PMID 13849303. CARAVACA J, GRISOLIA S (1958). "Enzymatic decarbamylation of carbamyl beta-alanine and carbamyl beta-aminoisobutyric acid". J. Biol. Chem. 231 (1): 357–65. PMID 13538975. Traut TW, Loechel S (1984). "Pyrimidine catabolism: individual characterization of the three sequential enzymes with a new assay". Biochemistry. 23 (11): 2533–9. doi:10.1021/bi00306a033. PMID 6433973. Retrieved from "https://en.wikipedia.org/w/index.php?title=Beta-ureidopropionase&oldid=917345490"
Gradient Descent: How Machines Learn Introduction to the Basics of Neural Networks How to intuitively understand neural networks Computational Complexity Of Neural Networks Okay so you have an objective function which you want to optimize. But how do you find the extremum of the function? Meet gradient descent. Gradient descent is an iterative algorithm that slowly finds a local minimum of a function. The algorithm is a first-order optimization method since it uses the first-derivative - the gradient of the function. It works by finding the gradient of the parameters at each step and updating them a little bit in a direction anti-parallel to the gradient (and parallel if maximizing the function - called gradient ascent). f(x;\theta) be the objective function paramterized by \theta which we want to minimize. \theta_n = \theta_{n-1} - \alpha \frac{d}{d\theta} f(\theta_{n-1}) \alpha is a parameter controlling how fast we update the parameters usually a low number of the order of 10^{-3} The algorithm assumes that the function f is continuous which is not a unreasonable requirement, and it will find the global minimum if the function is convex. However, in the real world, many functions are not neat and convex, so the algorithm will only find a local minimum which can cause problems if you need the global minimum (which is a much harder problem). Luckily, for properly designed deep neural networks, a local minimum is often good enough, and there are some arguments that the global minimum will severely overfit the data since models are very expressive and data is noisy. To better understand why this algorithm works, it can be helpful to visualize what happens at every step. Function: Polynomial Sinus Initial guess: (The initial x x_0 \alpha Using the interactive visualization above, we can see that gradient descent works similarly to "placing a ball on the curve" and let it roll down the curve to the lowest point. Moreover, we see that it is sensitive to where we place the "ball". For example, if we place the ball at 2 with the polynomial function, then it won't reach the global minimum. We also see that the ball moves more quickly when the descent is steeper, and slows down as it reaches its destination and the curve is flatter. This is expected as the amount we update the parameters is proportional to the gradient. By Kasper Fredenslund Deriving the mathematics behind backpropagation.
Surface Chemistry Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Compound that acts as inhibitor for knocking in combustion of petrol is : 1. (C2H5)4Pb Subtopic: Catalyst \| Which does not show Tyndall effect? Subtopic: Colloidal Solution \| Among the following surfactant molecules, the surfactant that forms micelles in aqueous solution at the lower CMC at ambient condition is: {\mathrm{CH}}_{3}{\left({\mathrm{CH}}_{2}\right)}_{15}{\mathrm{N}}^{+}{\left({\mathrm{CH}}_{3}\right)}_{3}{\mathrm{Br}}^{-} {\mathrm{CH}}_{3}{\left({\mathrm{CH}}_{2}\right)}_{11}{\mathrm{OSO}}_{3}^{-}{\mathrm{Na}}^{+} {\mathrm{CH}}_{3}{\left({\mathrm{CH}}_{2}\right)}_{6}{\mathrm{COO}}_{}^{-}{\mathrm{Na}}^{+} {\mathrm{CH}}_{3}{\left({\mathrm{CH}}_{2}\right)}_{11}{\mathrm{N}}^{+}{\left({\mathrm{CH}}_{3}\right)}_{3}{\mathrm{Br}}^{-} Subtopic: Emulsions \| Butter is a colloid form in which: 1. fat is dispersed in solid casein 2. fat globules are dispersed in water 3. water is dispersed in fat 4. suspension of casein is in water The cotterells precipitator is used to : 1. neutralize charge on carbon particles in air in smoke 2. coagulate carbon atoms of smoke 3. bring in cataphoresis in carbon particles Subtopic: Adsorption and Absorption \| Colloidal Solution \| Foam is a colloidal solution of: 1. gaseous particles dispersed in gas 2. gaseous particles dispersed in a liquid 3. solid particles dispersed in a liquid 4. solid particles dispersed in gas A liquid aerosol is a colloidal system of: 1. a liquid dispersed in a solid 2. a liquid dispersed in a gas 3. a gas dispersed in a liquid 4. a solid dispersed in a gas Negative catalyst or inhibitor is one: 1. which retards the rate of reaction 2. takes the reaction in forward direction 3. promotes the side reaction Alum helps in purifying water by: 1. Forming Si complex with clay particles. 2. Sulphate part which combines with the dirt and removes it. 3. Aluminium which coagulates the mud particles. 4. Making mud water soluble. Which of the following does not form anionic micelle? 1. C12H25COONa 2. C12H25SO4Na 4. C12H25(NH3)3Cl
12 th q: 12. \mathrm{If} \mathrm{\alpha },\mathrm{\beta } \mathrm{are} \mathrm{the} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation} {\mathrm{x}}^{2}-\mathrm{x}-1=0 \mathrm{then} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} \frac{1+\mathrm{\alpha }}{1-\mathrm{\alpha }}+\frac{1+\mathrm{\beta }}{1-\mathrm{\beta }}=?\phantom{\rule{0ex}{0ex}}\left(\mathrm{A}\right) 3 \left(\mathrm{B}\right) -4\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right) 7 \left(\mathrm{D}\right) 9 8) the number of real roots of equation x(x+2) (x2-1)-1=0 are \mathrm{\alpha } \mathrm{and} \mathrm{\beta } are the roots of quadratic equation (x – 2) (x – 3) + (x – 3) + (x + 1) (x +1) (x – 2) = 0. then the value of \frac{1}{\left(\mathrm{\alpha }+1\right)\left(\mathrm{\beta }+1\right)}+\frac{1}{\left(\mathrm{\alpha }-2\right)\left(\mathrm{\beta }-2\right)}+\frac{1}{\left(\mathrm{\alpha }–3\right)\left(\mathrm{\beta }–3\right)} 4 and 5 th 2. The value(s) of a for which one of the roots of x2 + (2a + 1) x + (a2 + 2) = 0 is twice the other root is \left(\mathrm{A}\right) 4 \left(\mathrm{B}\right) -4\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right) 0 \left(\mathrm{D}\right) -2 13 th q Q1. Solve for x : 4.22x +1– 9.2x + 1 = 0. Q2. Solve for x : \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=2\frac{1}{6} 10. Find the range of values of 'a' if the roots of the equation x2+a2=8x+6a are real. Q.7. Find the condition that one root of the equation a{x}^{2}+bx+c=0 may be double of the other. 30. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). Find correct roots of the equation.
MuonCalib Tutorial - Atlas Wiki MuonCalib Tutorial 1 Muon calibration tutorial 4 A deeper look into the jobOptions Muon calibration tutorial Welcome to the Muon Calibration Tutorial Page! After doing this tutorial, you will be able to run Athena, make your ntuples containing segment information. This ntuple can then be used as input to do analysis with help of the skeleton analysis package provided. This analysis package CalibSegmentAnalysis relies heavily on the Calibration framework but is Athena independent. Much information on validation, calibration and alignment can be found at the MuonSpectrometer homepage. General information about the need of calibration in the MuonSpectrometer. A clear Wiki page on the Calibration Framework by Domizia and Niels can be found on the MuonCalibrationFramework Wiki. At this site, the Calibration EDM is explained. Maybe needed to explain the concept of segments and the calibration EDM, is to be written (work in progress). A starting point of Calibration Segments and its scope in the Reconstruction EDM can be found in a presentation given by Zdenko during the ATLAS Muonweek september-2005 called Calibration Segment Ntuple In order to get started one should know that this field is prone to many changes over time. Backward compatibility is surely broken since the ntuple format changed a lot in the period of november 2005. The ntuple format is defined by the MuonCalib package (MuonCalibNtuple), and the analysis package needs a certain ntuple format to be able to fill the MuonCalib EDM classes offline. In the following sections, recipes will be provided for different kinds of ntuple formats. Choose your ntuple-format and wich version to run in Follow the recipes below to make your ntuple, customize jobOption file where needed Build the appropriate version of the CalibSegmentAnalysis package (depending on ntuple format) Run your analysis, customize the analyse routine where needed Different versions of Ntuple Content Currently, there are two different formats of the Calibration Segment Ntuple, described in the following tables: Calibration Segment Ntuple Content A. Information on patterns, segments en Mdt hits. All segments appear twice on the ntuple since a refit-flag is performed during ntuple-writing. This ntuple should be obsolete since release 11.X.0. Calibration Segment Ntuple Content B. Information on patterns, segments and hits of all technologies (Mdt's containing extra calibration information, the other technologies are not yet fully implemented). No extra refit is performed, so no double segments. In addition, truth information and event information (run and event number) is available. Calibration Segment Ntuple Content C. Is non-existent at the moment. It is sure that one will be defined in the future, the difference with Ntuple Content B will be a better description of the other technology hits, and timing information on event-level. Which recipe to use to make the Ntuple The different formats of Ntuples can be produced with different recipes. The following recipes sets up athena and guides you to running the Muon Calibration package. With this running you can proceed to the next section below. Recipe for Ntuple A in 10.5.0 Works out-of-the-box though outdated Recipe for Ntuple B in 11.0.0 This recipe requires MuonCalib to be checked out of CVS and to be recompiled. This is a time-costly procedure, but the best available until a newer release is present. RECOMMENDED Recipe for Ntuple B in 11.X.0 This release has the new MuonCalib compiled so this should run out-of-the-box. A deeper look into the jobOptions For version 10.5.0: a good example of a jobOptions file that produces segment ntuples is provided in Domizia's public: cp /afs/cern.ch/user/d/domizia/public/MuonCalib_files/myTopOptions.py . For version 11.0.0: a good example of a jobOptions file that produces segment ntuples is provided in Zdenko's public: cp /afs/cern.ch/user/z/zvankest/public/MuonCalib_files/myTopOptions.py . you can customize this jobOption file to your own needs... Change the number of events in the ntuple Change the tracking software (Moore <-> MuonBoy, and Jochem's cosmic pattern finder package in the future) Change the muon-sample (note! The file should be present in your PoolFileCatalog): in your run directory, add the file to your PoolFileCatalog by calling pool_insertFileToCatalog <physical path of your favorite POOL file> in your myTopOptions.py, replace the following line PoolRDOInput = [ "rfio:/castor//.pool.root" ] PoolRDOInput = [ "<POOl file path>" ] From the ntuple produced, one can proceed Athena-independent. A little word about the Calibration Event Data Model (EDM) The MuonCalibration-framework works with a Calibration EDM, which defines the objects used in the Calibration. The structure is more or less like this: each event contains a certain number of Patterns. Concider these as linked segments, as potential tracks, containing information such as the Chi Squared of the pattern, the track parameters (z0, r0, {\displaystyle \vartheta } {\displaystyle \varphi } , q/p). These patterns are described in the class MuonCalib::MuonGlobalPattern. Patterns are built from a number of segments. a segment is described by MuonCalib::MuonCalibSegment, which is basically a line segment which matches the hits stored in the segment. So segments describe the premature track at chamber level. A line is given by a vector and a direction. This information can be found on the segment as well, in two co-ordinate systems: the global ATLAS co-ordinates and the local chamber co-ordinates. the hits on the segment are implemented as MuonCalib::XxxCalibHitBase 's with Xxx the technology which recorded the hits (i.e. Mdt, Tgc, Csc or Rpc). Each hitBase has their own dedicated content relevant for doing calibration; thus RPCs do not have drifttimes and MDTs do not provide timemeasurements. They do have common members such as (local en global) position and MuonCalib::MuonFixedId 's. Since the Calibration EDM classes are Athena-independent, the identifiers defined for the hit classes are decoupled from the ATLAS data-base (which is Athena-dependent). In order to do an analysis on segment level from the ntuple, the content from the ntuple must be casted into the Calibration EDM. This can be done with help of a skeleton analysis package. The CalibNtupleAnalysis package Since there exists two ntuple formats, different versions of the CalibNtupleAnalysis package exist. Recipes to run the skeleton analysis are given here: CalibNtupleAnalysis-00-00-01 Capable of processing ntuple format A CalibNtupleAnalysis-00-00-02 Capable of processing ntuple format B After a succesfull Run, you will be provided with the following files: SegmentAnalysis.root Containing the Histograms and SegmentDisplays generated by the SegmentAnalysis framework. Note that one event may generate over 100 SegmentsDisplays. Customizing your favorite analysis A skeleton needs some 'meat'... Some example routines will be provided: plotATLAS(MuonCalibSegment) Plots the hits on the segment in global coordinates. In the example an extra feature is shown; one can select MDTchambers by stationtype with a simple call on the MuonFixedId. dumpSegment(MuonCalibSegment) Dumps MdtCalibSegment information. Invaluable for debugging. refitSegment(MuonCalibSegment) Refits the MdtCalibSegment with a DCSLFitter provided by the Calibration framework. showSegment(MuonCalibSegment) Calls a simple eventdisplay in which the hits are drawn as drifcircles and the segment as line. The displays can be written to a PSfile. circleResidual(MuonCalibSegment) Given the trackparameters and the hits on the segment, this routine calculates the expected residuals by extrapolating the track in the local y-z plane in which the driftradii of the hits are circles. stripResidual(MuonCalibSegment*) Given the trackparameters and the hits on the segment, this routine local x-y plane in which the RPC strips are the precision coordinates. Retrieved from "https://wiki.nikhef.nl/atlas/index.php?title=MuonCalib_Tutorial&oldid=4742"
Build a simple Neural Network with TensorFlow.js \| Deep Learning for JavaScript Hackers (Part III) \| Curiousily - Hacker's Guide to Machine Learning 14.07.2019 — Neural Networks, Deep Learning, TensorFlow, Machine Learning, JavaScript — 7 min read TL;DR Build a simple Neural Network model in TensorFlow.js to make a laptop buying decision. Learn why Neural Networks need activation functions and how should you initialize their weights. It is in the middle night, and you’re dreaming some rather alarming dreams with a smile on your face. Suddenly, your phone starts ringing, rather internationally. You pick up, half-asleep, and listen to something bizarre. A friend of yours is calling, from the other side of our planet, asking for help in picking a laptop. After all, it is Black Friday! You’re a bit dazzled by the fact that this is the first time you hear from your friend in 5 years. Still, you’re a good person and agree to help out. Maybe it is time to put your TensorFlow.js skills into practice? How about you build a model to help out your friend so you can get back to sleep? You heard that Neural Networks are pretty hot right now. It is 3 in the morning, there isn’t much need for persuasion in your mind. You’ll use a Neural Network for this one! What is a Neural Network? In a classical cliff-hanger fashion, we’ll start far away from answering this question. Neural Networks were around for a while (since 1950s)? Why did they become popular just recently (last 5-10 years)? First introduced by Warren McCulloch and Walter Pitts in A logical calculus of the ideas immanent in nervous activity Neural Networks were really popular until the mid-1980s when Support Vector Machines and other methods overtook the community. The Universal approximation theorem states that a Neural Networks can approximate any function (under some mild assumptions), even with a single hidden layer (more on that later). One of the first proves was done by George Cybenko in 1989 for sigmoid activation functions (will have a look at those in a bit). More recently, more and more advances in the field of Deep Learning made Neural Networks a hot topic again. Why? We’ll discuss that a bit later. First, let’s start with the basics! The original model, intended to model how the human brain processed visual data and learned to recognize objects, was susggested by Frank Rosenblatt in the 1950s. The Perceptron takes one or more binary inputs x_1, x_2, \ldots, x_n and produces a binary output: To compute the output you have to: have weights w_1, w_2, \ldots, w_n expressing the importance of the respective input the binary output (0 or 1) is determined by whether the weighted sum \sum_j w_j x_j is greater or lower than some threshold \text{output} = \begin{cases} 0 & \text{if } \sum_j w_j x_j \lt \text{threshold} \\ 1 & \text{otherwise} \end{cases} Let’s have a look at an example. Imagine you need to decide whether or not you need a new laptop. The most important features are its color and size (that’s what she said). So, you have two inputs: is it small (gotcha)? You can represent these factors with binary variables x_{pink} x_{small} and assign weights/importance w_{pink} w_{small} to each one. Depending on the importance you assign to each factor, you can get different models. We can simplify the Perceptron even further. We can rewrite \sum_j w_j x_j as a dot product of two vectors w \cdot x . Next, we’ll introduce the Perceptron’s bias, b = -{\text{threshold}} . Using it, we can rewrite the model as: \text{output} = \begin{cases} 0 & \text{if } w \cdot x + b \lt 0 \\ 1 & \text{otherwise} \end{cases} The bias is a measure of how easy it is for a perceptron to output 1 (to fire). Large positive bias makes outputting 1 easy, while a large negative bias makes it difficult. Let’s build the Perceptron model using TensorFlow.js: 1const perceptron = ({ x, w, bias }) => { 2 const product = tf.dot(x, w).dataSync()[0] 3 return product + bias < 0 ? 0 : 1 An offer for a laptop comes around. It is not pink, but it is small x = \begin{bmatrix}0\\1\end{bmatrix} . You’re biased towards not buying a laptop because you’re broke. You can encode that with a negative bias. You’re one of the brainier users, and you put more emphasis on size, rather than color w = \begin{bmatrix}0.5\\0.9\end{bmatrix} 1perceptron({ 2 x: [0, 1], 3 w: [0.5, 0.9], 4 bias: -0.5, Yes, you have to buy that laptop! To make learning from data possible, we want the weights of our model to change only by a small amount when presented with an example. That is, each example should cause a small change in the output. That way, one can continuously adjust the weights while presenting new data and not worrying that a single example will wipe out everything the model has learned so far. The Perceptron is not an ideal for that purpose since small changes in the inputs are propagated linearly to the output. We can overcome this using a sigmoid neuron. The sigmoid neuron has inputs x_1, x_2, \ldots, x_n that can be values between 0 and 1. The output is given by \sigma(w \cdot x + b) \sigma is the sigmoid function, defined by: \sigma(z) = \frac{1}{1+e^{-z}} Let’s have a look at it using TensorFlow.js and Plotly: 1const xs = [...Array(20).keys()].map(x => x - 10) 2const ys = tf.sigmoid(xs).dataSync() 4renderActivationFunction(xs, ys, "Sigmoid", "sigmoid-cont") Using the weights and inputs we get: \sigma = \frac{1}{1+e^{-(\sum_j w_j x_j-b)}}. Let’s dive deeper into the sigmoid neuron and understand the similarities with the Perceptron: z is a large positive number. Then e^{-z} \approx 0 \sigma(z) \approx 1 z is a large negative number. Then e^{-z} \rightarrow \infty \sigma(z) \approx 0 z is somewhat modest, we observe a significant difference compared to the Perceptron. Let’s build the sigmoid neuron model using TensorFlow.js: 1const sigmoidPerceptron = ({ x, w, bias }) => { 3 return tf.sigmoid(product + bias).dataSync()[0] Another offer for a laptop comes around. This time you can specify the degree of how close the color is to pink and how small it is. The color is somewhat pink, and the size is just about right x = \begin{bmatrix}0.6\\0.9\end{bmatrix} . The rest stays the same: 1sigmoidPerceptron({ 2 x: [0.6, 0.9], Yes, you still want to buy this laptop, but this model also outputs the confidence of its decision. Cool, right? Architecting Neural Networks A natural way to extend the models presented above is to group them in some way. One way to do that is to create layers of neurons. Here’s a simple Neural Network that can be used to make the decision of buying a laptop: Neural Networks are a collection of neurons, connected in an acyclic graph. Outputs of some neurons are used as inputs to other neurons. They are organized into layers. Our example is composed of fully-connected layers (all neurons between two adjacent layers are connected), and it is a 2 layer Neural Network (we do not count the input layer). Neural Networks can make complex decisions thanks to combination of simple decisions made by the neurons that construct them. Of course, the output layer contains the answer(s) you’re looking for. Let’s have a look at some of the ingredients that make training Neural Networks possible: The Perceptron model is just a linear transformation. Stacking multiple such neurons on each other results in a vector product and a bias addition. Unfortunately, there are a lot of functions that can’t be estimated by a linear transformation. The activation function makes it possible for the model to approximate non-linear functions (predict more complex phenomena). The good thing is, you’ve already met one activation function - the sigmoid: One major disadvantage of the Sigmoid function is the is that it becomes really flat outside the [-3, +3] range. This leads to weights getting close to 0 - no learning is happening. ReLU, introduced in the context of Neural Networks in Rectified Linear Units Improve Restricted Boltzmann Machines, have a linear output at values greater than 0 and 0 otherwise. 2const ys = tf.relu(xs).dataSync() 4renderActivationFunction(xs, ys, "ReLU", "relu-cont") One disadvantage of ReLU is that negative values “die out” and stay at 0 - no learning. Leaky ReLU, introduced in Rectifier Nonlinearities Improve Neural Network Acoustic Models, solves the dead values introduced by ReLu: 2const ys = tf.leakyRelu(xs).dataSync() 4renderActivationFunction(xs, ys, "Leaky ReLU", "leaky-relu-cont") Note that negative values get scaled instead of zeroed out. Scaling is adjustable by a parameter in tf.leakyRelu(). The process of teaching a Neural Network to make “reasonable” predictions involves adjusting the weights of the neurons multiple times. Those weights need to have initial values. How should you choose those? The initialization process must take into account the algorithm we’re using to train our model. More often than not, that algorithm is Stochastic gradient descent (SGD). Its job is to do a search over possible parameters/weights and choose those that minimize the errors our model makes. Moreover, the algorithm heavily relies on randomness and a good starting point (given by the weights). Same constant initialization Imagine that we initialize the weights using the same constant (yes, including 0). Every neuron in the network will compute the same output, which results in the same weight/parameter update. We just defeated the purpose of having multiple neurons. Too small/large value initialization Let’s initialize the weights with a set of small values. Passing those values to the activation functions will decrease them exponentially, leaving every weight equally unimportant. On the other hand, initializing with large values will lead to an exponential increase, making the weights equally unimportant again. Random small number initialization We can use a Normal distribution with a mean 0 and standard deviation 1 to initialize the weights with small random numbers. Every neuron will compute different output, which leads to different parameter updates. Of course, multiple other ways exist. Check the TensorFlow.js Initializers Should you buy the laptop? Now that you know some Neural Network kung-fu, we can use TensorFlow.js to build a simple model and decide whether you should buy a given laptop. Let’s say that for your friend, size is much more important than the degree of pinkness! You sit down and devise the following dataset: 1const X = tf.tensor2d([ 2 // pink, small 3 [0.1, 0.1], 8 [0.75, 0.4], 10 [0.6, 0.9], 11 [0.6, 0.75], 14// 0 - no buy, 1 - buy 15const y = tf.tensor([0, 0, 1, 1, 0, 0, 1, 1, 1].map(y => oneHot(y, 2))) Well done! You did well on incorporating your friend preferences. Recall the Neural Network we’re going to build: Let’s translate it into a TensorFlow.js model: 5 inputShape: [2], We have a 2-layer network with an input layer containing 2 neurons, a hidden layer with 3 neurons and an output layer containing 2 neurons. Note that we use ReLu activation function in the hidden layer and softmax for the output layer. We have 2 neurons in the output layer since we want to obtain how certain our Neural Network is in its buy/no-buy decision. 2 optimizer: tf.train.adam(0.1), We’re using binary crossentropy to measure the quality of the current weights/parameters of our model by measuring how “good” the predictions are. Our training algorithm, Stochastic gradient descent, is trying to find weights that minimize the loss function. For our example, we’re going to use the Adam optimizer. Now that our model is defined, we can use our training dataset to teach it about our friend preferences: 1await model.fit(X, y, { 3 epochs: 20, 6 console.log("Epoch " + epoch) 7 console.log("Loss: " + logs.loss + " accuracy: " + logs.acc) We’re shuffling the data before training and log the progress after each epoch is complete: 1Epoch 1 2Loss: 0.703386664390564 accuracy: 0.5 4Loss: 0.6708164215087891 accuracy: 0.5555555820465088 10Epoch 19 11Loss: 0.08228953927755356 accuracy: 1 After 20 epochs or so seems like the model has learned the preferences of your friend. You save the model and send it over to your friend. After connecting to your friend computer, you find somewhat appropriate laptop and encode the information into the model: 1const predProb = model.predict(tf.tensor2d([[0.1, 0.6]])).dataSync() After waiting a few long milliseconds, you receive an answer: The model agrees with you. It “thinks” that your friend should buy the laptop but it is not that certain about it. You did good! Your friend seems happy with the results, and you’re thinking of making millions with your model by selling it as a browser extension. Either way, you learned a lot about: Why activation functions are needed and which one to use How to initialize the weights of your Neural Network models Build a simple Neural Network to solve a (somewhat) real problem Laying back on the comfy pillow, you start thinking. Could I’ve used Deep Learning for this? Reducing Loss: Gradient Descent Gradient descent and stochastic gradient descent from scratch Types of weight intializations What if do not use any activation function in the neural network?
Hydrogen Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers In the preparation of hydrogenated oil the chemical reaction involved is called as- Subtopic: Preparation & Properties \| The most abundant isotope of hydrogen is- 4. Para-hydrogen Hydrogen is evolved by the action of cold dilute HNO3 on: 2. Mg or Mn Subtopic: Hydrogen- Types & Isotopes \| Preparation & Properties \| A molten ionic hydride on electrolysis gives: 1. H+ ion moving towards the cathode 2. H+ ion moving towards the anode 3. H2 is liberated at anode 4. H2 is liberated at cathode Subtopic: Type of Hydride \| In solid hydrogen, the intermolecular bonding is: 2. van der Waals' The lightest gas is : Subtopic: Hydrogen- Types & Isotopes \| 10 volumes of H2O2 has a strength of approximately: Subtopic: H2O2 (Hydrogen Peroxide) \| Which is not true in case of H2O2? 1. It is more stable in basic solution 2. It acts as strong oxidising agent in acid and basic solutions 3. It is decomposed by MnO2 4. It behaves as reducing agent towards KMnO4 Decmposition of H2O2 is retarded by: Density of water is maximum at: ° ° ° Subtopic: Water \|
Determine if matrix is upper triangular - MATLAB istriu - MathWorks Italia Test Upper Triangular Matrix Test Matrix of Zeros tf = istriu(A) tf = istriu(A) returns logical 1 (true) if A is an upper triangular matrix; otherwise, it returns logical 0 (false). Test A to see if it is upper triangular. istriu(A) The result is logical 1 (true) because all elements below the main diagonal are zero. Test Z to see if it is upper triangular. istriu(Z) The result is logical 1 (true) because an upper triangular matrix can have any number of zeros on the main diagonal. Input array, specified as a numeric array. istriu returns logical 0 (false) if A has more than two dimensions. A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero. A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right) is upper triangular. A diagonal matrix is both upper and lower triangular. Use the triu function to produce upper triangular matrices for which istriu returns logical 1 (true). The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istriu(A) == isbanded(A,0,size(A,2)). isdiag \| istril \| diag \| triu \| tril \| isbanded \| bandwidth
Solutions Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Chemistry - Solutions Which gas when passed through dilute blood will impart a cherry red colour to the solution? Subtopic: Introduction/ Colligative properties \| The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The percent dissociation of N2O4 at the final temperature is: Subtopic: Van’t Hoff Factor \| At 298 K, 500cm3 H2O dissolved 15.30 cm3 CH4(STP) under a partial pressure of methane of one atm. If Henry's law holds, what pressure is required to cause 0.001 mole methane to dissolve in 300cm3 water ? Subtopic: Concentration Terms & Henry's Law \| The molal depression constant for water=1.85 deg/molal and for benzene is 5.12 deg/molal. If the ratio of the latent heats of fusion of benzene to water is 3:8, calculate the freezing point of benzene. °\mathrm{C} °\mathrm{C} °\mathrm{C} °\mathrm{C} Subtopic: Elevation of Boiling Point \| Which of the following colligative properties is associated with the concentration term 'molarity'? 1. Lowering of vap.pressure Subtopic: Osmosis & Osmotic Pressure \| The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute in the solution is- Subtopic: Relative Lowering of Vapour Pressure \| Which one of the folowing pairs of solution can we expect to be isotonic at the same temperature- 1. 0.1 (M) urea and 0.1 (M) NaCl 2. 0.1 (M) urea and 0.2 (M) MgCl2 3. 0.1 (M) NaCl and 0.1 (M) Na2SO4 4. 0.1 (M) Ca(NO3)2 and 0.1 (M) Na2SO4 The latent heat of vapourisation of water is 540 cal g-1 at 100 °\mathrm{C} . Kb for water is 1. 0.56 K.mole-1 Which of the following aqueous solution has osmotic pressure nearest to that an equimolar solution of K4[Fe(CN)6]? The molar volume of liquid benzene (density= 0.877 g ml-1) increases by a factor of 2750 as it vapourises at 20 °\mathrm{C} . At 27 °\mathrm{C} when a non-volatile solute (that does not dissociate) is dissolved in 54.6cm3 of benzene, vapour pressure of this solution, is found to be 98.88 mm Hg. calculate the freezing point of the solution. Given: Enthalpy of vapourisation of benzene(l)=394.57 Jg-1. Enthalpy of fusion of benzene (l) = 10.06 kJ mol-1 Molal depression constant for benzene=5.0 K kg mol-1. Subtopic: Depression of Freezing Point \|
Tools to scan the mSUGRA phasespace - Atlas Wiki Tools to scan the mSUGRA phasespace This document shows how to use two simple macros, susygen.py and susymap.py, to scan any part of the 5D mSUGRA phasespace. These macros can only scan for information calculated by IsaSugra (mostly only the SUSY mass spectrum), as no events are generated. To automatically generate AtlFast SUSY events look at: * Generating AtlFast SUSY Events 1 IsaSugra 2 Scanning the mSUGRA phasespace IsaSugra IsaSugra is a subprogram of IsaJet that calculates the mass-spectrum and decay-channels for a number of SUSY models, including mSUGRA. As input parameters, for mSUGRA, it requires the mSUGRA coordinate and the top mass. The output can take different forms of which two are important here. The first is a text file including all SUSY and Higgs masses and decay channels. This file can be used directly in Jimmy to produce events. The second is a more easily read text file also including all masses and decay channels, but in addition also containing some SM values that may be affected by SUSY, such as {\displaystyle Br(B->s\gamma )} {\displaystyle Br(B->\mu \mu )} {\displaystyle \Delta a_{\mu }} {\displaystyle \Omega _{CDM}h^{2}} which is the cold dark matter relic density this SUSY point would produce. If the susy point under consideration is theoretically not allowed (for example because at this point no electroweak symmetry breaking is possible), IsaSugra will give an error, indicating the reason why the susy point was rejected. At Nikhef the most resent IsaSugra release (7.75) is installed at /project/atlas/users/nicoleru/Store/isajet/isasugra.x When studying a susypoint with \|A0\|>0, keep in mind that some theoretical papers use a deffinition of A0 that differs by a minus sign from the one used by programs such as IsaSugra. Scanning the mSUGRA phasespace The macro susygen.py runs IsaSugra for a certain susypoint and checks if it is theoretically valid. If it is valid it retrieves certain values from the output files. The information gathered by susygen.py is printed to a text file. At present the following values are written (in this order on one line): M0, M12, A0, tan(beta), sgn(mu), theory, lightest neutralino mass, lightest Higgs-boson mass, the LSP, BR(b->s+gamma), omegah^2, Br(Bs->mu+mu) and delta a(mu). The values are given in GeV (when applicable). The theory value is either 0 (ok) or 1 (not valid) and if 1 no further information is included. The LSP is indicated using the particle numbering of Herwig (lightest neutralino = 450, stau = 429). susymap.py is a simple program that defines a grid (with evenly separated point in any variable) in the 5D mSUGRA phasespace and starts the text file in which all information will be printed (it also adds a line indicating which variables will be printed). susymap.py will call susygen.py for each point on the grid. The result of this program is a file indicating all relevant information about a large number of susypoints. As an example of the use of these macros I have run them for a 2D grid, with tan(beta)=50, A0=0 and sgn(mu)=+1 fixed. Both m0 and m12 are varied in 25 steps from 100 to 2500 (so basically 625 point). On stoomboot (submitted in 1 job) this took approximately 3 hours. The output file is located at /project/atlas/users/nicoleru/Store/map_A0_0_tan_50_si_1.txt. The first few lines of that file are the following: m0 m12 A0 tan sign EWSB neutalino Higgsmass LSP BR(b->s+gamma) omegah^2 Br(Bs->mu+mu) dela a(mu) 100 100 0 50 1 1 100 700 0 50 1 0 292.791500 116.384600 429 2.747800e-04 1.000000e+04 1.181600e-08 2.434000e-09 100 1000 0 50 1 0 424.931700 117.990400 429 3.092800e-04 1.000000e+04 7.275100e-09 1.209400e-09 Figure 1: The m0-m12 plane for tan(beta)=50, sgn(mu)=+1 and A0=0. The black area indicated the theoretically excluded region. In the shaded area the stau is the LSP. The colored lines show the excuded regions. Using a root file to put all this information in a plot leads to figure 1. (/project/atlas/users/nicoleru/Store/rootfilefigure1.C) Of course this program can also be used to plot a A0-tan(beta) plain, or to see how certain variables change close to the point you are studying. As an example of other possible use of this program, I have extended the grid used in the first example to include different values of A0 (and changed tan(beta) to 10), namely A0={-3,-2,-1,0,1,2,3}*m12. Due to WMAP information, most of the m0-m12 plane is excluded. Only two very thin lines remain, one that borders the LSP=stau area, and one line at very low m12. Due to this, we can basically reduce the 5D mSUGRA phasespace to 4D, by relating m0 to m12. Using another root file, it is possible to plot the (lightest) Higgs-mass, or Br(B->s+gamma) or any other variable, as function of m12, for fixed A0, tan(beta) and sgn(mu). (/project/atlas/users/nicoleru/Store/rootfilefigure2.C) Figure 2: Values on the WMAP allowed line (as function of m12) for tan(beta)=10, sgn(mu)=1 and different values of A0. Both macros (susygen.py and susymap.py) should be easy to use. Just copy them to your own directory. They are both located at: /project/atlas/users/nicoleru/Store/. As long as you are only interested in the information currently optained by susygen.py, you only need to change susymap.py to form your own 5D susy grid. If you need additional information, the mass of one of the susy particles for example, it will take slightly more work, as you need to know exactly where that information is printed in the IsaSugra output files. If you have any problems getting this additional information, just ask me (Nicole). Making the plots is more difficult as I don't have one root file that, when you feed it a susymap file, simply gives you a plot similar to figure 1. I will work on that. In the mean time, the root file used to make figure 1 should easily be convertible to other grids. Retrieved from "https://wiki.nikhef.nl/atlas/index.php?title=Tools_to_scan_the_mSUGRA_phasespace&oldid=4801"
Home : Support : Online Help : Programming : Data Types : Conversion : record convert/record convert an expression to a record convert( e, 'record' ) convert( e, 'record'['packed'], 'deep' ) The calling sequence convert( e, 'record' ) converts the expression e to a record. The input expression e must be a module without local variables, a list of equations whose left-hand sides are symbols, or a table all of whose indices are symbols. No other expression may be converted to a record. The calling sequence convert( e, 'record', 'deep' ) converts the table e to a record and any entries of e that are tables are also converted to records recursively. The deep option works only for tables. Note: If the input expression e is a record, then it is returned unchanged. If the input expression e is a module, then it must not have non-exported local variables to be converted to a record. The resulting record has the names of the module exports as slot names, and their assigned values (if any) as the corresponding slot values. A list of equations of the form [{\mathrm{name}}_{1}={\mathrm{value}}_{1},{\mathrm{name}}_{2}={\mathrm{value}}_{2},...,{\mathrm{name}}_{k}={\mathrm{value}}_{k}] may be converted to a record. Each left-hand side {\mathrm{name}}_{i} must be a symbol. In this case, the resulting record is precisely the one that would be obtained by calling the Record constructor with the given equations as arguments. If the input expression e is a table, then every one among its indices must be a symbol. The resulting record has the table indices as slot names, and the corresponding tabular values as the record slot values. In any calling sequence, you can optionally use 'record'['packed'] in place of 'record' to specify that the output will be a packed Record. r≔\mathrm{Record}⁡\left(a=1,b=2\right) \textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) \mathrm{type}⁡\left(r,'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{true}} \mathrm{convert}⁡\left(r,'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) m≔\mathbf{module}\left(\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{_export}⁡\left(a,b\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}a≔1;\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}b≔2\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end module}: \mathrm{convert}⁡\left(m,'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) m≔\mathbf{module}\left(\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{_export}⁡\left(a,b\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{_local}⁡\left(u\right);\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}a≔1;\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}b≔2;\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}u≔3\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end module}: Attempting to convert a module with a local variable to a record results in an error. \mathrm{convert}⁡\left(m,'\mathrm{record}'\right) Error, (in `convert/record`) cannot convert a module with locals to a record \mathrm{convert}⁡\left([a=1,b=2],'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) \mathrm{convert}⁡\left(\mathrm{table}⁡\left([a=1,b=2]\right),'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) \mathrm{convert}⁡\left(\mathrm{table}⁡\left([a=1,b=2]\right),'\mathrm{record}'['\mathrm{packed}']\right) {\textcolor[rgb]{0,0,1}{\mathrm{Record}}}_{\textcolor[rgb]{0,0,1}{\mathrm{packed}}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right) \mathrm{convert}⁡\left(\mathrm{table}⁡\left([a=1,b=\mathrm{table}⁡\left([d=3,e=\mathrm{table}⁡\left([f=5]\right)]\right)]\right),'\mathrm{record}'\right) \textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{⁡}\left([\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{e}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{⁡}\left([\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{5}]\right)]\right)\right) The convert/record command was updated in Maple 2020. The deep option was introduced in Maple 2020.
Near optimal bounds in Freiman's theorem 15 May 2011 Near optimal bounds in Freiman's theorem Tomasz Schoen1 1Faculty of Mathematics and Computer Science, Adam Mickiewicz University Duke Math. J. 158(1): 1-12 (15 May 2011). DOI: 10.1215/00127094-1276283 We prove that if for a finite set A of integers we have \|A+A\|\le K\|A\| A is contained in a generalized arithmetic progression of dimension at most {K}^{1+C\left(\mathrm{log}K{\right)}^{-1/2}} and of size at most \mathrm{exp}\left({K}^{1+C\left(\mathrm{log}K{\right)}^{-1/2}}\right)\|A\| C . We also discuss a number of applications of this result. Tomasz Schoen. "Near optimal bounds in Freiman's theorem." Duke Math. J. 158 (1) 1 - 12, 15 May 2011. https://doi.org/10.1215/00127094-1276283 Tomasz Schoen "Near optimal bounds in Freiman's theorem," Duke Mathematical Journal, Duke Math. J. 158(1), 1-12, (15 May 2011)
Electrostatic Potential And Capacitance, Revision Notes: ICSE Class 12-science PHYSICS, Physics Part I - Meritnation Work done by an external force in bringing a charge q from a point R to a point P in electric field of a certain charge configuration is {U}_{P}-{U}_{R} , which is the difference in potential energy of charge q between the final and initial points. Potential energy at a point is the work done by an external force in moving a charge from infinity to that point. Electrostatic potential at any point in a region of electrostatic field is the minimum work done in carrying a unit positive charge (without acceleration) from infinity to that point. Electric potential due to a point charge of magnitude q at a distance r from the charge is given as V=\frac{q}{4\pi {\epsilon }_{0}r} Potential difference between two points P and R can…
LMIs in Control/Matrix and LMI Properties and Tools/D-Stability Max Percent Overshoot Poles - Wikibooks, open books for an open world LMIs in Control/Matrix and LMI Properties and Tools/D-Stability Max Percent Overshoot Poles LMI for Max Percent Overshoot Poles The following LMI allows for the verification that poles of a system will within a maximum percent overshoot constraint. This can also be used to place poles for max percent overshoot when the system matrix includes a controller, such as in the form A+BK. 4 The LMI: LMI for Max Percent Overshoot Poles {\displaystyle {\begin{aligned}{\dot {x}}(t)&=Ax\end{aligned}}} {\displaystyle A\in \mathbb {R} ^{n\times n}} The data required is the matrix A and the max percent overshoot {\displaystyle M_{p}} {\displaystyle z-z^{}+{{\pi } \over ln({M_{p}})}\|z+z^{}\|{\leq }0} , where z is a complex pole of A. The goal of the optimization is to find a valid P > 0 such that the following LMI is satisfied. The LMI: LMI for Max Percent Overshoot PolesEdit The LMI problem is to find a matrix P satisfying: {\displaystyle {\begin{aligned}{\begin{bmatrix}\pi (AP+(AP)^{T})&lnM_{p}(AP-(AP)^{T})\\lnM_{p}(AP-(AP)^{T})^{T}&\pi (AP+(AP)^{T})\end{bmatrix}}<0\\\end{aligned}}} If the LMI is found to be feasible, then the pole locations of A, represented as z, will meet the max percent overshoot specification of {\displaystyle z-z^{}+{{\pi } \over ln({M_{p}})}\|z+z^{}\|{\leq }0} Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/Matrix_and_LMI_Properties_and_Tools/D-Stability_Max_Percent_Overshoot_Poles&oldid=4011185"
LMIs in Control/Matrix and LMI Properties and Tools/Frobenius Norm - Wikibooks, open books for an open world LMIs in Control/Matrix and LMI Properties and Tools/Frobenius Norm Frobenius NormEdit {\displaystyle A\in \mathbb {R} ^{n\times m}} {\displaystyle \gamma \in \mathbb {R} } The Frobenius norm of {\displaystyle A} {\displaystyle \|\|A\|\|} {\displaystyle {\sqrt {tr(A^{T}A)}}={\sqrt {tr(AA^{T})}}} The Frobenius norm is less than or equal to {\displaystyle \gamma } if and only if any of the following equivalent conditions are satisfied. 1.There exists {\displaystyle S\in \mathbb {R} ^{n}} {\displaystyle {\begin{bmatrix}Z&A^{T}\\&1\\\end{bmatrix}}\geq 0,} {\displaystyle {\begin{aligned}\qquad tr(Z)\leq \gamma ^{\text{2}}.\qquad \\\end{aligned}}} {\displaystyle S\in \mathbb {R} ^{m}} {\displaystyle {\begin{bmatrix}Z&A\\&1\\\end{bmatrix}}\geq 0,} {\displaystyle {\begin{aligned}\qquad tr(Z)\leq \gamma ^{\text{2}}.\qquad \\\end{aligned}}} Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/Matrix_and_LMI_Properties_and_Tools/Frobenius_Norm&oldid=4010580"
Systems of Particles and Rotational Motion Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Physics - Systems of Particles and Rotational Motion A small mass attached to a string rotates on a frictionless table top as shown. If the tension on the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will 1. Increase by a factor of 4 2. Decrease by a factor of 2 Subtopic: Angular Momentum \| Three-point masses 'm' each, are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about axis COD is- 2m{a}^{2} \frac{2}{3}m{a}^{2} \frac{5}{4}m{a}^{2} \frac{7}{4}m{a}^{2} Subtopic: Moment of Inertia \| A particle is moving in a circular orbit with constant speed. Select wrong alternate 1. Its linear momentum is conserved 2. Its angular momentum is conserved 3. It is moving with variable velocity 4. It is moving with variable acceleration Subtopic: Linear Momentum \| ABC is a right-angled triangular plate of uniform thickness. The sides are such that AB > BC as shown in figure. $I_1,I_2$ and $I_3$ are moments of inertia about AB, BC and AC respectively. Then, which of the following relation is correct? 1. $I_1 = I_2 = I_3$ 2. $I_2 > I_1 > I_3$ One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively IA and IB such that {\mathrm{I}}_{\mathrm{A}}={\mathrm{I}}_{\mathrm{B}} {\mathrm{I}}_{\mathrm{A}}>{\mathrm{I}}_{\mathrm{B}} {\mathrm{I}}_{\mathrm{A}}<{\mathrm{I}}_{\mathrm{B}} \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{d}_{A}}{{d}_{B}} A couple produces: [NTSE 1995; CBSE PMT 1997; DCE 2004] 1. Purely linear motion 2. Purely rotational motion 3. Linear and rotational motion Subtopic: Rotational Motion: Kinematics \| A particle of mass 1 kg is kept at (1m, 1m, 1m). The moment of inertia of this particle about z-axis would be 1 \mathrm{kg}-{\mathrm{m}}^{2} 2 \mathrm{kg}-{\mathrm{m}}^{2} 3 \mathrm{kg}-{\mathrm{m}}^{2} One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is [IIT-JEE (Screening) 2001] \frac{1}{2}M{R}^{2} \frac{1}{4}M{R}^{2} \frac{1}{8}M{R}^{2} \sqrt{2}M{R}^{2} A wheel is rotating at the rate of 33 rev/min. If it comes to stop in 20 s. Then, the angular retardation will be \mathrm{\pi }\frac{\mathrm{rad}}{{\mathrm{s}}^{2}} 11\mathrm{\pi } \mathrm{rad}/{\mathrm{s}}^{2} \frac{\mathrm{\pi }}{200} rad/{s}^{2} \frac{11\mathrm{\pi }}{200} rad/{s}^{2} A solid sphere is rotating about a diameter at an angular velocity \omega . If it cools so that its radius reduces to \frac{1}{n} of its original value, its angular velocity becomes [MP PMT 2006] \frac{\omega }{n} \frac{\omega }{{n}^{2}} n\omega {n}^{2}\omega
Tungsten Carbide as a Diffusion Barrier on Silicon Nitride Active- Metal-Brazed Substrates for Silicon Carbide Power Devices \| J. Electron. Packag. \| ASME Digital Collection H. A. Mustain, H. A. Mustain Department of Electrical Engineering, 3217 Bell Engineering Center, , Fayetteville, AR 72701 William D. Brown, e-mail: siang@uark.edu Mustain, H. A., Brown, W. D., and Ang, S. S. (July 14, 2009). "Tungsten Carbide as a Diffusion Barrier on Silicon Nitride Active- Metal-Brazed Substrates for Silicon Carbide Power Devices." ASME. J. Electron. Packag. September 2009; 131(3): 034502. https://doi.org/10.1115/1.3153582 Recently, silicon nitride (Si3N4) has been receiving renewed attention because of its potential use as a substrate material for packaging of silicon carbide (SiC) power devices for high temperature applications. It is an attractive material for this application because it has moderate thermal conductivity and a low coefficient of thermal expansion, which is close to that of SiC. Materials that show promise for use as a diffusion barrier on Si3N4 substrate for bonding SiC devices to a Si3N4 substrate are refractory metals such as titanium (Ti), molybdenum (Mo), tungsten (W), and their alloys. Tungsten carbide (WC) shows promise as a diffusion barrier for bonding these devices to copper metallization on Si3N4 substrates. This paper presents the results of an investigation of a metallization stack (Si3N4/Cu/WC/Ti/Pt/Ti/Au) used to bond SiC dice to Si3N4 substrates. The dice were bonded using transient liquid phase bonding. Samples were characterized using X-ray diffraction for phase identification and Auger electron spectroscopy for depth profiling of the elemental composition of the metallization stack in the as-deposited state, and immediately following annealing. The metallization remained stable following subjection to a temperature of 400°C for 100 h in air. diffusion barriers, electron spectroscopy, electronics packaging, metallisation, silicon compounds, thermal conductivity, thermal expansion, tungsten compounds, X-ray diffraction, tungsten carbide (WC), high temperature stability, diffusion barrier, silicon nitride substrate Copper, Diffusion (Physics), Silicon, Silicon nitride ceramics, Tungsten, Annealing, High temperature, Metals, Temperature, Electron spectroscopy, Thermal conductivity, X-ray diffraction, Thermal expansion, Titanium Packaging Materials and Approaches for High Temperature SiC Power Devices Schulz-Harder Direct Copper Bonded Substrates for Semiconductor Power Devices ,” Curamik Electronics, www.curamik.comwww.curamik.com Interactions Between Au and Cu Across a Ni Barrier Layer Crystallographic and Morphological Characterization of Reactively Sputtered Ta, TaN, and TaNO Thin Films Comparative Study of Tantalum and Tantalum Nitride as a Diffusion Barrier for Cu Metallization Diffusion Barrier Properties of TaC Between Si and Cu Stable Ohmic Contact to GaAs With TiN Diffusion Barrier Tantalum as a Diffusion Barrier Between Copper and Silicon: Failure Mechanism and Effect of Nitrogen Additions The Handbook of Binary Phase Diagrams Genium Publishing Corp Transitional Metal Carbides and Nitrides The Refractory Carbides Characterization of Tungsten Carbide as Diffusion Barrier for Cu Metallization On the Thermal Conductivity of Dispersed Ceramics Structural Comparison of CIGSS Thin Film Absorber Layer Fabricated on SS and Ti Substrates
Inverse hyperbolic cosine - MATLAB acosh - MathWorks France Inverse Hyperbolic Cosine of Vector Plot the Inverse Hyperbolic Cosine Function Y = acosh(X) Y = acosh(X) returns the inverse hyperbolic cosine of the elements of X. The function accepts both real and complex inputs. All angles are in radians. Find the inverse hyperbolic cosine of the elements of vector X. The acosh function acts on X element-wise. Plot the inverse hyperbolic cosine function over the interval 1\le x\le 5 plot(x,acosh(x)) ylabel('acosh(x)') X — Hyperbolic cosine of angle Hyperbolic cosine of angle, specified as a scalar, vector, matrix, or multidimensional array. The acosh operation is element-wise when X is nonscalar. x x>1 , the inverse hyperbolic cosine satisfies {\mathrm{cosh}}^{-1}\left(x\right)=\mathrm{log}\left(x+\sqrt{{x}^{2}-1}\right). z=x+iy , as well as real values in the domain -\text{\hspace{0.17em}}\infty <z\le \text{\hspace{0.17em}}\text{\hspace{0.17em}}1 , the call acosh(z) returns complex results. acos \| cosh \| asinh \| atanh
3-Commands and Packages - Maple Help Home : Support : Online Help : Getting Started : Tutorials : 3-Commands and Packages Part 3: Commands and Packages In Part 3: Commands and Packages, you will learn more about Maple top-level commands and about how to use packages. You will also learn to use the help system. Maple has over 4000 commands that provide comprehensive, in-depth coverage of a vast range of mathematical and programming topics. In the tutorials 1-Talking to Maple and 2-Putting Your Ideas Together, you have already seen a number of Maple commands, including sin, taylor, int, exp, dsolve, solve, fsolve, rhs, and eval, and accessed many more behind-the-scenes using context-sensitive options. The Context Panel and interactive assistants all use user-accessible Maple commands to perform their tasks. Some Maple commands are top-level commands while others are organized into packages. Frequently, general purpose commands are available at the top level, and can be accessed at any time. In addition to the command listed above, top level commands include trigonometric and special functions and commands for expression manipulation such as factor, expand, and simplify. To view the extensive list of top-level functions in Maple, see Index of Functions. Tip: Most Maple commands are written in the Maple language, but a small collection are built into the compiled Maple kernel. Some of the commands you have already seen are built-in commands, such as taylor, rhs, and eval. Other useful built-in commands are shown in the following table. Useful and Efficient Commands evalf - evaluate using floating-point arithmetic \mathrm{ln}\left(2\right) \textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{2}\right) \mathrm{evalf}\left(\mathrm{ln}\left(2\right)\right) \textcolor[rgb]{0,0,1}{0.6931471806} evalb- evaluate as a Boolean expression \mathrm{evalb}\left(-11>0\right) \textcolor[rgb]{0,0,1}{\mathrm{false}} a≔2: b≔2: \mathrm{evalb}\left(a=b\right) \textcolor[rgb]{0,0,1}{\mathrm{true}} sort - sort a list of values or a polynomial \mathrm{sort}\left(\left[2, 1, 3, 1\right]\right) [\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}] \mathrm{sort}\left(1+2 {x}^{4}+3 x+{x}^{3}\right) \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{⁢}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{⁢}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1} seq - create a sequence \mathrm{seq}\left(1..4\right) \textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4} \mathrm{seq}\left({i}^{3},i=1..4\right) \textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{27}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{64} map - apply a procedure to each operand of an expression zip - zip together two data sets by applying a binary function to the components of the two data sets The function iquo returns the quotient of two integers. \mathrm{map}\left(\mathrm{ln},\left[1,2,3,4\right]\right) [\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{⁢}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{⁡}\left(\textcolor[rgb]{0,0,1}{2}\right)] \mathrm{zip}\left(\mathrm{iquo},\left[207,241,345, 1235\right],\left[17,21,30,44\right]\right) [\textcolor[rgb]{0,0,1}{12}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{28}] select, remove, and selectremove - selection or removal from an expression Those elements which satisfy the Boolean-valued command are returned. Here, we use the Boolean-valued command issqr, which tests if an integer is a perfect square. \mathrm{select}\left(\mathrm{issqr},\left\{42, 53, 64\right\}\right) {\textcolor[rgb]{0,0,1}{64}} indets - find the indeterminates of an expression {ⅇ}^{y} are considered indeterminates. Use the type `name` to return only variable names. \mathrm{indets}\left(x\cdot y+z-x\cdot {ⅇ}^{y}\right) {\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{y}}} \mathrm{indets}\left(x\cdot y+z-x\cdot {ⅇ}^{y},\mathrm{name}\right) {\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}} The tutorial 6-Data Structures includes more examples using some of these commands. For a list of commands implemented in the kernel, see index/builtin. Maple also contains packages, which are collections of commands. Some top Maple packages are listed in the table. combinatorial functions, including commands for calculating permutations and combinations of lists and partitions of integers commands for manipulating Matrices and Vectors and performing Linear Algebra For a full list of Maple packages, see Index of Packages. There are two ways to use the commands in a package: by using the long form or short form of their calling sequences. Long form: The commands in a package can always be accessed using the long form of the calling sequence. This form is PackageName:-CommandName. Short form: The short form of the calling sequence for all commands in a package can be used during the current Maple session after with(PackageName) has been entered. The short form is simply CommandName. Use the Minimize command from the Optimization package to minimize 4\cdot {x}^{2}-\mathrm{ln}\left(x\right) , given the initial point x=.5 First, we will use the long form by calling Optimization:-Minimize. The minimum is given, followed by the x -value for which this minimum is attained. \mathrm{expr}≔4\cdot {x}^{2}-\mathrm{ln}\left(x\right): \mathrm{Optimization}:-\mathrm{Minimize}\left(\mathrm{expr},\mathrm{initialpoint}=\left\{x=.5\right\}\right) [\textcolor[rgb]{0,0,1}{1.53972077083991810}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.353553390618446}]] Now, enter with(Optimization). A list of all the commands in the package is returned. (To suppress the display of this list, use a colon (:) after this command.) Now, all these commands can be used by just entering the command name. This is the short form of the calling sequence. Redo the problem, using the short form. Tip: Packages can also be loaded from the Tools menu. \mathrm{with}\left(\mathrm{Optimization}\right) [\textcolor[rgb]{0,0,1}{\mathrm{ImportMPS}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Interactive}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{LPSolve}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{LSSolve}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Maximize}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Minimize}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{NLPSolve}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{QPSolve}}] \mathrm{Minimize}\left(\mathrm{expr},\mathrm{initialpoint}=\left\{x=.5\right\}\right) [\textcolor[rgb]{0,0,1}{1.53972077083991810}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.353553390618446}]] For more information on these two methods of accessing package commands, see Using Packages. Maple has an extensive help system, including help pages, online manuals, examples, and an integrated dictionary of mathematical and engineering terms. Select the Help>Maple Help menu to display the help browser. (You can also open Maple Help by pressing [F1].) Enter the topic name (such as "integral") into the search box. Click Search. The int help page opens. You can copy and paste the examples section to your document. From the Edit menu, select Copy Examples. In your document, choose Edit>Paste. (These options are also available from the right-click menu.) By default, examples are displayed in 2-D math. To view the examples in 1-D math, click the button to toggle the display. Search and browse full help system, including help pages, dictionary, and manuals. From the help page, you can see the calling sequences, read the description, and view examples. If you copy the examples to your document, you can then modify and execute the examples. Within a document, there are two easy ways to get help on a topic: To get help on a particular topic, use the ? notation. For example, type ?solve [Enter] in math mode. For information on a topic name that already appears in your document, place the cursor on the word and press [F2]. Example: Place the cursor on the word Optimization and press [F2]. Instant access to the help page when you know the topic name or command name. ?\mathrm{solve} The Minimize command is found in the Optimization package. Help > Quick Reference (or [Ctrl][F2]) Basic overview of important topics. ?examples,index Example worksheets illustrating different mathematical and programming commands. These worksheets will open in a new tab in your Maple window. Tools>Task>Browse for task templates. Fill-in-the-blank templates organized by concept. Access Maple's manuals through the help system. Select the Help>Maple Help menu to display the help browser. Next, in the Table of Contents tab, expand the Manuals directory to access the User Manual or Programming Guide. PDFs from the Maplesoft website provide other format options for Maple's manuals. Visit the Maplesoft Documentation Center, http://www.maplesoft.com/documentation_center. Conceptual overviews as well as more in-depth explanation. index/function, index/package, UsingPackages
Experimental Simulation of a Film Cooled Turbine Blade Leading Edge Including Thermal Barrier Coating Effects \| J. Turbomach. \| ASME Digital Collection Jonathan Maikell, David Bogard, Justin Piggush, Justin Piggush , United Technologies, CT 06108 Maikell, J., Bogard, D., Piggush, J., and Kohli, A. (September 21, 2010). "Experimental Simulation of a Film Cooled Turbine Blade Leading Edge Including Thermal Barrier Coating Effects." ASME. J. Turbomach. January 2011; 133(1): 011014. https://doi.org/10.1115/1.4000537 For this study, a simulated film cooled turbine blade leading edge, constructed of a special high conductivity material, was used to determine the normalized “metal temperature” representative of actual engine conditions. The Biot number for the model was matched to that for operational engine conditions, ensuring that the normalized wall temperature, i.e., the overall effectiveness, was matched to that for the engine. Measurements of overall effectiveness were made for models with and without thermal barrier coating (TBC) at various operating conditions. This was the first study to experimentally simulate TBC and the effects on overall effectiveness. Two models were used: one with a single row of holes along the stagnation line, and the second with three rows of holes straddling the stagnation line. Film cooling was operated using a density ratio of 1.5 and for range of blowing ratios from M=0.5 M=3.0 ⁠. Both models were tested using a range of angles of attack from 0.0 deg to ±5.0 deg. As expected, the TBC coated models had significantly higher external surface temperatures, but lower metal temperatures. These experimental results provide a unique database for evaluating numerical simulations of the effects of TBC on leading edge film cooling performance. blades, cooling, gas turbines, thermal barrier coatings Coolants, Cooling, Engines, Film cooling, Flow (Dynamics), Metals, Simulation, Temperature, Thermal barrier coatings, Thermal conductivity, Turbine blades, Blades, Airfoils, Wall temperature, Density A Design Perspective on Thermal Barrier Coatings Use of the Adiabatic Wall Temperature in Film Cooling to Predict Wall Heat Flux and Temperature Adiabatic and Overall Effectiveness for a Film Cooled Blade Internal and Film Cooling of a Flat Plate With Conjugate Heat Transfer Proceedings of the 2007 ASME Turbo Expo Experimental Study of the Effect of Oscillating Stagnation Line on Overall Cooling Performance on a Leading Edge With One Row of Cooling Holes ,” MS thesis, University of Texas at Austin, Austin, TX. Experimental Study of the Effect of TBC and Angle of Attack on Overall Cooling Performance on a Leading Edge With One and Three Rows of Cooling Holes Effects of Surface Deposition, Hole Blockage, and Thermal Barrier Coating Spallation on Vane Endwall Film Cooling
Zero price elasticity of demand means \| Answers Home » CAPF 2018 , Economics , Price Elasticity of Demand , Supply & Demand , UPSC , Zero Price Elasticity » Zero price elasticity of demand means Zero price elasticity of demand means Q. Zero price elasticity of demand means Whatever the change in price, there is absolutely no change in demand for a small change in price, there is a small change in demand for a small change in price, there is a large change in demand for a large change in price, there is a small change in demand Answer: whatever the change in price, there is absolutely no change in demand. In economics, the price elasticity of demand (PED or Ed) is a measure to show the responsiveness (or elasticity) of the quantity demanded for a good or service to a change in its price, ceteris paribus. It is also known as the percentage change in quantity demanded for a good or service for a percentage change in the price of the same good or service. In short, the price elasticity of demand is given by the following formula: {\displaystyle Price\ elasticity\ of\ demand={\frac {Percentage\ change\ in\ quantity\ demanded}{Percentage\ change\ in\ price}}} As the price and the quantity demanded of a good are inversely related (i.e. an increase in price will always cause a decrease in quantity demanded and vice versa), the sign of PED is always negative. Hence, economists tend to just ignore the sign and compare the absolute values instead. The price elasticity of demand (PED) measures the change in demand for a good in response to a change in price. Price elasticity equals to zero, it means that, when price increases, supplies quantities do not increase or decrease. They simply don't react to price. The demand curve will be vertical. If elasticity is greater than 1, indicating a high responsiveness to changes in price. Computed elasticities that are less than 1 indicate low responsiveness to price changes and are described as inelastic demand. If the co-efficient of price elasticity of demand = zero, demand is perfectly inelastic i.e. demand does not vaty with a change in price. Thanks for reading Zero price elasticity of demand means Tags CAPF 2018, Economics, Price Elasticity of Demand, Supply & Demand, UPSC, Zero Price Elasticity
How to Account for a Capital Lease: 8 Steps (with Pictures) 1 Understanding Capital Leases 2 Accounting for Capital Leases A lease agreement refers to the act of one company lending an asset to another company, in exchange for periodic rent payments (like renting an apartment, for example). Capital leases are one form of lease, where the lease is basically structured as a purchase and financing agreement. Capital leases are commonly employed when businesses loan large pieces of equipment or other capital-intensive assets to each other. To account for a capital lease, familiarize yourself with the terms of the arrangement and make the appropriate journal entries. Keep in mind that new rules issued by the Financial Accounting Standards Board (FASB) went into effect in 2018 for public companies and in 2019 for all other organizations.[1] X Research source Understanding Capital Leases {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/2\/27\/Account-for-a-Capital-Lease-Step-1.jpg\/v4-460px-Account-for-a-Capital-Lease-Step-1.jpg","bigUrl":"\/images\/thumb\/2\/27\/Account-for-a-Capital-Lease-Step-1.jpg\/aid1594073-v4-728px-Account-for-a-Capital-Lease-Step-1.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"} Learn about operating leases. In order to understand a capital lease, you must first understand an operating lease, as these are the two main kinds of leases. An operating lease is a traditional lease whereby lessor (or owner of a property) transfers the right to use the property to a borrower (or lessee) for a particular period, after which it is returned. With an operating lease, the borrower assumes no risk of ownership.[2] X Research source An operating lease involves no ownership of the asset, and therefore, the asset does not appear on the company's balance sheet in any way. The only important accounting for an operating lease is the rent, or lease payment, which appears on the income statement as an expense. Operating leases are typically short compared to the life of the asset. For example, if a piece of machinery is being leased, and the life of the machine is 25 years, an operating lease may be for five years. Contrast an operating lease with a capital lease. A capital lease is the other type of lease, and unlike an operating lease, a capital lease requires the lessee to bear some of the risks and benefits of owning the asset, even though it never actually owns the asset. A capital lease occurs when the lessee records the asset on the balance sheet as if it owns the asset. The lessee would then make lease payments to the lessor, and these payments consist of interest and principal repayments, just like a loan. There are several pro's to capital leases. Just like if the business actually owned the asset, they can choose to deduct the interest component of the lease payment each year for taxes, and can also claim depreciation each year on the asset. That is to say, as the asset decreases in value each year, the business can benefit from this, whereas this would not be possible with an operating lease.[3] X Research source There are cons as well. For example, since the asset is listed on the balance sheet, this would make the company's return on assets lower. This is because since return on assets is income as a percentage of total assets, if assets increase, the return falls (assuming income stays the same). Consider the criteria for a capital lease. Under a capital lease, the lessee is essentially buying the asset from the lessor, with the lease payments functioning as a financing arrangement. If the lease meets one of these four criteria, it must be accounted for as a capital lease: The asset's ownership will be transferred to the lessee upon the agreement's maturation. The lessee is given the option of purchasing the asset at a price below the market value upon the agreement's maturation. The term of the lease agreement is greater than 75 percent of the asset's useful life. The present value of all the future rent payments is equal to or greater than 90% of the assets market value Evaluate the terms of the lease. Before making any journal entries, make sure you understand the lease agreement's terms. For example, consider a lease agreement whereby Company A leases a building to Company B for 10 years. Company B will pay a rental payment of $12,000 at the beginning of each year. The building's useful life is 12 years; therefore, this is a capital lease because the lease term is greater than 75 percent of the asset's life. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/b\/b0\/Account-for-a-Capital-Lease-Step-5.jpg\/v4-460px-Account-for-a-Capital-Lease-Step-5.jpg","bigUrl":"\/images\/thumb\/b\/b0\/Account-for-a-Capital-Lease-Step-5.jpg\/aid1594073-v4-728px-Account-for-a-Capital-Lease-Step-5.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"} Review the basic accounting process for recognizing a capital lease. Before learning the journal entries it is important to understand the basic accounting process. From an accounting perspective, when you enter into a capital lease, you are basically purchasing the asset, and then financing it using a loan. Therefore the accounting would be very similar to if you simply bought and financed an asset. [4] X Research source This means you would first need to add the asset to the balance sheet as a fixed asset, and also add the value of the asset to the balance sheet as a capital lease liability (since you do not own the asset). Over the term of the lease, regular payments consisting of interest and principal would be made. The interest portion of the payment would be recorded as an interest expense on the income statement, and the principal would reduce the balance of the capital lease liability. For example, assume you were leasing an asset worth $10,000. This means $10,000 would be listed as an asset on the balance sheet, and $10,000 would be listed as a capital lease liability. If you had a $1,000 yearly payment, and $100 was interest, $900 would go towards reducing the capital lease liability account. Over time, this means the capital lease liability account would eventually reach zero.[5] X Research source . Finally, you would need account for depreciation. Since assets depreciate over their useful life, you would need to account for the declining value of the asset each year. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/4\/4c\/Account-for-a-Capital-Lease-Step-6.jpg\/v4-460px-Account-for-a-Capital-Lease-Step-6.jpg","bigUrl":"\/images\/thumb\/4\/4c\/Account-for-a-Capital-Lease-Step-6.jpg\/aid1594073-v4-728px-Account-for-a-Capital-Lease-Step-6.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"} Start by recording the journal entries to recognize the start of the lease. The journal entries will reflect the fact that the lease is essentially a sale. For example, assume Company A leases a building to Company B for 10 years, with an annual rent payment of $12,000.[6] X Research source Assume the value of the building is $120,000. Note that the value of the asset is supposed to be equal to the present value of all future rent payments. In this example, we are assuming the value of the building is equal to the sum of all future rent payments ($12,000 times 10 years). In reality, this would be less, since those future rent payments must be discounted to account for the fact that money received in the future is worth less than money received now. To begin, open whatever accounting software you are using, debit the "Building" asset account for $120,000, and credit the Lease Payable liability account for $120,000. If these accounts aren't available in your accounting program, you must create them. This transaction recognizes the building and the lease on the balance sheet. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/a\/af\/Account-for-a-Capital-Lease-Step-7.jpg\/v4-460px-Account-for-a-Capital-Lease-Step-7.jpg","bigUrl":"\/images\/thumb\/a\/af\/Account-for-a-Capital-Lease-Step-7.jpg\/aid1594073-v4-728px-Account-for-a-Capital-Lease-Step-7.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"} Record the journal entry to recognize each rental payment. Now that the lease is recognized on the balance sheet, you must account for the rental payments. On January 1 each year, you must make a payment to recognize your lease payments to the lessor. To do this, you would debit Lease Payable for $12,000 and credit Cash for $12,000. Doing this reduces the value of the Lease payable liability account, to reflect the fact that the principle on the "loan" is being paid down. This transaction also reduces cash as you are paying the lessee. Note that this assumes there is no interest involved. If part of that $12,000 annual payment is interest, you must debit that portion to the Interest Expense account. For example, suppose that the $12,000 payment included 10 percent interest. This means that $1,090 of that payment was interest expense. Therefore, you would record a debit of $10,910 to the capital lease liability account, a debit of $1,090 to the interest expense account and a credit of $12,000 to the accounts payable account.[7] X Research source These journal entries will continue to deplete the balance of the Lease Payable account until they reach 0 at the agreement's end. You may also make monthly payments. Account for it in the same way, but repeat the process twelve times--once for each month. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/7\/7d\/Account-for-a-Capital-Lease-Step-8.jpg\/v4-460px-Account-for-a-Capital-Lease-Step-8.jpg","bigUrl":"\/images\/thumb\/7\/7d\/Account-for-a-Capital-Lease-Step-8.jpg\/aid1594073-v4-728px-Account-for-a-Capital-Lease-Step-8.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"} Record any necessary depreciation expenses. Because a capital lease is treated like a purchase agreement, the lessee will need to record depreciation on the asset in question. In the example above, you would need to depreciate the $120,000 balance in the Building account over its life. The required journal entries would vary depending on the company's depreciation schedule.[8] X Research source Depreciation involves taking the value of the asset ($120,000), and reducing its value over the course of its life (10 years). For example, using straight-line depreciation, the asset would depreciate by $12,000 per year ( {\displaystyle \$120,000/10=\$12,000} To account for this, you must charge that amount to the income statement as an expense each year. The basic procedure is to debit the depreciation expense account by $12,000, and then credit the accumulated depreciation account for $12,000. What if an organization already has a capital lease of a copier, which they keep, but add another copier and redo the contract? Generally, the old copier becomes the property of the company and the new lease is a separate contract. How do we treat the trade-in value of trucks with zero book value in our accounting records, in a subsequent capital lease? Generally, in an exchange, the basis of the new asset is the basis of the old asset plus any cash paid to get the new asset. If the basis of the old asset is zero, then the value of the new asset is the amount of the lease. How do you account for sales tax, document fees, and prorated rent on capital lease payments? Sales taxes are administrative costs, which would be deducted currently. In a capital lease, there are no rent payments to allocate. Depreciation may be allocated if the asset is employed in various activities. The example above deals with a "direct financing" capital lease, by far the most common arrangement. If the rental payments are greater than the asset's original cost, the lease becomes a "sales type" agreement, in which a profit is recognized as each payment is made. The examples above will also work when expressed in other currencies. ↑ http://www.fasb.org/jsp/FASB/Document_C/DocumentPage?cid=1176167901010&acceptedDisclaimer=true ↑ http://pages.stern.nyu.edu/~adamodar/New_Home_Page/AccPrimer/lease.htm ↑ http://www.diffen.com/difference/Capital_Lease_vs_Operating_Lease#Advantages_of_a_capital_lease ↑ http://www.double-entry-bookkeeping.com/business-loans/capital-lease-accounting/ ↑ http://www.accountingtools.com/questions-and-answers/what-is-the-accounting-for-a-capital-lease.html ↑ http://www.accountingcoach.com/depreciation/explanation A capital lease is an agreement in which you purchase an asset from another company in regular payment instalments and the ownership rights transfer to you at the end of the term. When you take out your capital lease, first debit the Building asset account for the total cost and credit the Lease Payable liability account for the same figure. Then, on the first day of each rent period, debit your Lease Payable and credit your Cash account for rent. For example, if you were financing an asset worth 120,000 dollars for 10 years, you’d debit 12,000 and credit 12,000 each year. Don’t forget to account for depreciation. For straight-line depreciation, charge 10 percent of the asset’s total value to your income statement as an expense each year. For more tips from our Financial co-author, including how to differentiate between an operating lease and a capital lease, read on! "The step-by-step detail was helpful. I just wish the article showed how the present value is obtained." "The review was simple to follow and gave me the exact information I required." Ecner James "I now understand accounting for leases."
Simple Climate Models \| METEO 469: From Meteorology to Mitigation: Understanding Global Warming We will start out our discussion of climate models with the simplest possible conceptual models for modeling Earth's climate. These models include different variants on the so-called Energy Balance Model. An Energy Balance Model or 'EBM' does not attempt to resolve the dynamics of the climate system, i.e., large-scale wind and atmospheric circulation systems, ocean currents, convective motions in the atmosphere and ocean, or any number of other basic features of the climate system. Instead, it simply focuses on the energetics and thermodynamics of the climate system. We will start out discussion of EBMs with the so-called Zero Dimensional EBM—the simplest model that can be invoked to explain, for example, the average surface temperature of the Earth. In this very simple model, the Earth is treated as a mathematical point in space—that is to say, there is no explicit accounting for latitude, longitude, or altitude, hence we refer to such a model as 'zero dimensional'. In the zero-dimensional EBM, we solve only for the balance between incoming and outgoing sources of energy and radiation at the surface. We will then build up a little bit more complexity, taking into account the effect of the Earth's atmosphere—in particular, the impact of the atmospheric greenhouse effect—through use of the so-called "gray body" variant of the EBM. Zero Dimensional EBM The zero dimensional ('0d') EBM simply models the balance between incoming and outgoing radiation at the Earth's surface. As you'll recall from your review of radiation balance in the previous section, this balance is in reality quite complicated, and we have to make a number of simplifying assumptions if we are to obtain a simple conceptual model that encapsulates the key features. For those who are looking for more technical background material, see this "Zero-dimensional Energy Balance Model" online primer (NYU Math Department). We will treat the topic at a slightly less technical level than this, but we still have to do a bit of math and physics to be able to understand the underlying assumptions and appreciate this very important tool that is used in climate studies. We will assume that the amount of short wave radiation absorbed by the Earth is simply \left(1-\alpha \right)S/4 , where S is the Solar Constant (roughly 1370 W /m2 but potentially variable over time) and α is the average reflectivity of Earth's surface looking down from space, i.e., the 'planetary albedo', accounting for reflection by clouds and the atmosphere as well as reflective surface of Earth including ice (value of roughly 0.32 but also somewhat variable over time). We will assume that the outgoing longwave radiation is given simply by treating the Earth as a 'black body' (this is a body that absorbs all radiation incident upon it). The Stefan-Boltzman law for black body radiation holds that an object emits radiation in proportion to the 4th power of its temperature, i.e., the flux of heat from the surface is given by {F}_{bb}=\epsilon \cdot \sigma \cdot {T}_{S}{}^{4} where σ is known as the Stefan-Boltzmann constant, and has the value \sigma =5.67\text{}x\text{}{10}^{-8}\left(W{m}^{-2}{K}^{-4}\right) ; ε is the emissivity of the object (unitless fraction) — a measure of how 'good' a black body the object is over the range of wavelengths in which it is emitting radiation; and Ts (K) is the surface temperature. For the relatively cold Earth, the radiation is primarily emitted in the infrared regime of the electromagnetic spectrum, and the emissivity is very close to one. We will approximate the surface temperature, TS, as representing the average 'skin temperature' of an Earth covered with 70% ocean (furthermore, we will treat the ocean as a mixed layer of average 70m depth—this ignores the impacts of heat exchange with the deep ocean, but is not a bad first approximation). We can then approximate the thermodynamic effect of the mixed layer ocean in terms of an effective heat capacity of the Earth's (land+ocean) surface, C=2.08\text{}x\text{}{10}^{8}J{K}^{-1}{m}^{-2} . The condition of energy balance can then be described in terms of the thermodynamics, which states that any change in the internal energy per unit area per unit time (< \Delta F=Cd{T}_{s}/dt ) must balance the rate of net heating, which is the difference between the incoming shortwave and outgoing longwave radiation. Mathematically, that gives: C\frac{d{T}_{s}}{dt}=\frac{\left(1-\alpha \right)S}{4}-\epsilon \cdot \sigma \cdot {T}_{s}{}^{4} Let's suppose that the incoming radiation (the first term on the right hand side) were larger than the outgoing radiation (the second term on the right hand side). Then the entire right-hand side would be positive, which means that the left-hand side, the rate of change of Ts over time, must also be positive. In other words, Ts must be increasing. This, in turn, means that the outgoing radiation must increase, which will eventually bring the two terms on the right hand side into balance. At this point, there is no longer any change of Ts with time, i.e., we achieve an equilibrium. In equilibrium, the time derivative term is, by definition, zero, and we thus must have equality between the outgoing and incoming radiation, i.e., between the two terms on the right-hand side of equation 1. This yields the purely algebraic expression \epsilon \cdot \sigma \cdot {T}_{s}{}^{4}=\frac{S\left(1-\alpha \right)}{4} The factor of 1/4 comes from the fact (see Figure 4.1, below) that the Earth is emitting radiation over the entirety of its surface area (4πR2 where R is the radius of the earth), but at any given time only receiving incoming (solar) radiation over its cross-sectional area, πR2. It turns out that since the Earth's surface temperature varies over a relatively small range (less than 30° K) about its mean long-term temperature (in the range of O° C, or 273° K), i.e., it varies only by at most 10% or so, it is valid to approximate the 4th degree term in equation (1) by a linear relationship, i.e., \epsilon \cdot \sigma \cdot {T}_{s}{}^{4}=A+B{T}_{S} A and B, thus defined, have the approximate values: A=315\text{}W{m}^{-2} B=4.6\text{}W{m}^{-2}{K}^{-1} Such an approximation is often used in atmospheric science and other areas of physics when appropriate, and is called linearization. Using this approximation, we can readily solve for TS as {T}_{S}=\left[\frac{S\left(1-\alpha \right)}{4}-A\right]/B Figure 4.1: Simple Planetary Energy Balance (assume Te = TS for our present purposes). Credit: Reprinted with permission from: A Climate Modeling Primer, A. Henderson-Sellers and K. McGuffie, Wiley, pg. 58, (1987). 0d EBM screen shot, Click on the screen shot to pull the tool up in another window You might find it rather disappointing that, after all the work we did above to develop a realistic Energy Balance Model for Earth's climate, we were way off. Our EBM indicates that, given appropriate parameter values (i.e., S=1370\text{}W/{m}^{2},\text{}\alpha =0.32 ), the Earth should be a frozen planet with TS = 255° K, rather than the far more hospitable TS= 288° K we actually observe. Our model gave a result that was a whopping 33° C (roughly 60° F) too cold! What do you think we forgot? If you said "greenhouse gases" then you were right! So, how do we include the effect of the atmospheric greenhouse effect in a simple way? That is the topic of our next section. ‹ Energy and Radiation Balance up Simple Climate Models, cont'd › Energy and Radiation Balance Simple Climate Models, cont'd
Polymers Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers The monomer of the polymer {\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{C}=\mathrm{C}{\left({\mathrm{CH}}_{3}\right)}_{2} {\mathrm{CH}}_{3}\mathrm{CH}=\mathrm{CH}·{\mathrm{CH}}_{3} {\mathrm{CH}}_{3}\mathrm{CH}={\mathrm{CH}}_{2} Subtopic: Polymers: Natural & Synthetic, Biodegradable & Non Biodegradable \| Which can absorb over 90% of its own mass of water and does not stick to wound ? 2. Gun cotton 3. Thiokol Thermoplastics are: 1. linear polymers 2. soften or melt on heating 3. molten polymer can be moulded in desired shape -\left[\mathrm{NH}{\left({\mathrm{CH}}_{2}\right)}_{6}\mathrm{NHCO}{\left({\mathrm{CH}}_{2}\right)}_{4}\mathrm{CO}\right]{-}_{n} is a - 1. Copolymer. 2. Addition polymer. 3. Thermo-setting polymer. 4. Homopolymer. Subtopic: Classification - Methods of Polymerization & Copolymerization \| Polymers: Natural & Synthetic, Biodegradable & Non Biodegradable \| Application of Polymers \| Which of the following natural products is not a polymer ? Synthetic human hair wigs are made from a copolymer of vinyl chloride and acrylonitrile and is called: 4. dynel Subtopic: Classification - Methods of Polymerization & Copolymerization \| Polymers: Natural & Synthetic, Biodegradable & Non Biodegradable \| Nylon-6,6 is a polyamide obtained by the reaction of \mathrm{COOH}\left({\mathrm{CH}}_{2}{\right)}_{4}\mathrm{COOH} + {\mathrm{H}}_{2}{\mathrm{NC}}_{6}{\mathrm{H}}_{4}{\mathrm{NH}}_{2} - \left(p\right) \mathrm{COOH}\left({\mathrm{CH}}_{2}{\right)}_{4}\mathrm{COOH} + {\mathrm{NH}}_{2}\left({\mathrm{CH}}_{2}{\right)}_{6}{\mathrm{NH}}_{2} \mathrm{COOH}\left({\mathrm{CH}}_{2}{\right)}_{4}\mathrm{COOH} + {\mathrm{NH}}_{2}\left({\mathrm{CH}}_{2}{\right)}_{4}{\mathrm{NH}}_{2} {\mathrm{COOHC}}_{6}{\mathrm{H}}_{4}\mathrm{COOH} - \left(p\right) + {\mathrm{NH}}_{2}\left({\mathrm{CH}}_{2}{\right)}_{6}{\mathrm{NH}}_{2} Subtopic: Classification - Methods of Polymerization & Copolymerization \| Which one is protein fibre ? 1. urea and formaldehyde 3. phenol and formaldehyde 4. tetramethylene glycol Terylene is a condensation polymer of ethylene glycol and 2. Phthalic acid 4. Terephthalic acid
Students are getting worse at math, and the educational system is the problem. There is a global mathematics teaching problem. The US has regressed in recent PISA tests in mathematics, and is well below the OECD average which itself has been on a steady decline since it it started recording in 2003. Young people are worse at mathematics now than 20 years ago, and are on a trend towards becoming worse still. Students are taught mathematics, but they don't learn mathematics: "60 percent of U.S. students who enter community colleges are not qualified to take a college mathematics course, even though they have graduated high school" James Stigler UCLA Note: While this article focuses on the US, many of the same observations apply to other countries as well, and the principles are universal. Students don't like mathematics. Mathematics frequently shows up as the "most hated subject" by students. So the teaching is ineffective, and fails to engage the students leading them to hate the entire field often for life. How did it come to this? How most schools teach mathematics Most schools use an archaic outdated teaching philosophy. There are at least 3 major roadblocks. Lack of big picture The curriculum leans heavily on memorization especially in the US which leads to poor adaptation to new scenarios. For example, lacking number sense may lead a student to compute 19\times6 by rote instead of recognizing the easier variant 20\times6-6=114 Teachers have long lists of requirements and skills which they need to teach in a short amount of time, so they do not have time to properly relate the concepts to other concepts showing why they are important, and motivating the students. This leads to questions such as "why is this useful?" and "why are we learning this?". The incentive structures in school are often aligned such that students will avoid making a mistake at all costs. Otherwise, it can have effects on their grades and their future prospects. This is not very productive for learning, and can lead to students failing to reach out for help. Mistakes are a natural and necessary part of learning. These are all solvable by a proactive teacher, but there is an even more fundamental problem. Lack of personalized education Classrooms work at the abstraction level of the whole class. This is due to the expense associated with considering each student individually. Teachers are resource constrained, so they have to address many students, leading too little to distinguish different students in the class. There are efforts to individualize education [1] [2], but it is limited, and can rarely extend much beyond the current classroom goals. You can give harder algebra problems, but you are still limited with how far you can go before you lose class cohesion, or become overburdened by designing personalized curriculums. Moreover, classes are typically categorized based on age not academic level which leads a large variance in student ability making the need for individualized instruction even greater. And those classes are usually required to follow a set curriculum, so even if a student is two years (or ahead) they are forced to solve problems and tackle concepts that offer no instructional value. Either because they don't understand the prerequisites or have already mastered the concepts. Mathematics is uniquely difficult to teach since every concept depends on previous concepts, so if you fail to learn one concept, it will have compounding effects as the class moves on without you. If you fall behind, it's very difficult to catch up. This extends to homework as well which is usually given collectively to the whole class ignoring any individual differences. But there is also another limitation of homework. Usually you only get feedback later when you no longer remember what your thought process limiting how much value you get from it. While there are many facets you can improve, the fundamental problem with schools is that the teacher student ratio is too low, so you have to abstract the individual students away and handle the class as more or less one unit. However, with big data analysis, we can personalize the learning experience in a way that's simply not available outside personal tutoring which is cost prohibitive for many. But what are some of the properties that is required to create a good personalized learning experience? Proper contextualization The student should know why they're learning the concept, and how the concept relates to other concepts to properly contextualize the learning experience. Principle of proximal learning Perhaps the most important principle is that a personalized learning experience should meet the student at their level. This is formalized by the theory of the zone of proximal learning. The learning material should tailored specifically for the student, and assist the student to reach beyond their current abilities. Embracing mistakes & Immediate feedback Mistakes are a natural part of learning, and should not be punished. Instead, we can use mistakes as learning opportunities with immediate targeted feedback. By clustering common mistake types, we can leverage analysis techniques to find common errors, and help the students improve. Bringing all these principles together is a large endeavour, but I have been working on a project doing just that. Njoror works by encoding the relationships between the concepts in a graph using that as a base for contextualization, and for the student-competency model. It analyses the answers from all students in order to recommend learning material that matches each student's level. Njoror also performs cluster analysis on the answers in order to identify common mistakes, and provides instant relevant feedback for the student to help them progress. Njror is still in pre-early access, but you can sign up for updates about the project, and an early invite. We build a blockchain pretty much.
Logarithmic resistor ladder - Wikipedia A logarithmic resistor ladder is an electronic circuit composed of a series of resistors and switches, designed to create an attenuation from an input to an output signal, where the logarithm of the attenuation ratio is proportional to a digital code word that represents the state of the switches. The logarithmic behavior of the circuit is its main differentiator in comparison with digital-to-analog converters in general, and traditional R-2R Ladder networks specifically. Logarithmic attenuation is desired in situations where a large dynamic range needs to be handled. The circuit described in this article is applied in audio devices, since human perception of sound level is properly expressed on a logarithmic scale. 1 Logarithmic input/output behavior 2.1 Constant input resistance 2.2 Constant output resistance 3 Circuit variations Logarithmic input/output behavior[edit] As in digital-to-analog converters, a binary word is applied to the ladder network, whose N bits are treated as representing an integer value according to the relation: {\displaystyle \mathrm {CodeValue} =\sum _{i=1}^{N}s_{i}\cdot 2^{i-1}} {\displaystyle s_{i}} represents a value 0 or 1 depending on the state of the ith switch. For a conventional DAC or R-2R network, the output signal value (its voltage) would be: {\displaystyle V_{out}=a\cdot (\mathrm {CodeValue} +b)\cdot V_{in}} {\displaystyle a}nd {\displaystyle b} are design constants and where {\displaystyle V_{in}} typically is a constant reference voltage. (DA-converters that are designed to handle a variable input voltage are termed multiplying DAC.[1]) In contrast, the logarithmic ladder network discussed in this article creates a behavior as: {\displaystyle \log(V_{out}/V_{in})=a\cdot (\mathrm {CodeValue} +b)} {\displaystyle V_{in}} is a variable input signal. Circuit implementation[edit] This example circuit is composed of 4 stages, numbered 1 to 4, and an additional leading Rsource and trailing Rload. Each stage i has a designed input-to-output voltage attenuation ratioi as: {\displaystyle Ratio_{i}={\text{if}}\;sw_{i}\;{\text{then}}\;\alpha ^{2^{i-1}}\;{\text{else}}\;1} For logarithmic scaled attenuators, it is common practice to express their attenuation in decibels: {\displaystyle dB(Ratio_{i})=20\log _{10}\alpha ^{2^{i-1}}=2^{i-1}\cdot 20\cdot \log _{10}\alpha } {\displaystyle i=1..N} {\displaystyle sw_{i}=1} This reveals a basic property: {\displaystyle dB(Ratio_{i+1})=2\cdot dB(Ratio_{i})} To show that this {\displaystyle Ratio_{i}} satisfies the overall intention: {\displaystyle \log(V_{out}/V_{in})=\log(\prod _{i=1}^{N}Ratio_{i})=\sum _{i=1}^{N}\log(Ratio_{i})=a\cdot (CodeValue+b)} {\displaystyle b=0} {\displaystyle a=\log(\alpha )} The different stages 1 .. N should function independently of each other, as to obtain 2N different states with a composable behavior. To achieve an attenuation of each stage that is independent of its surrounding stages, either one of two design choices is to be implemented: constant input resistance or constant output resistance. Constant input resistance[edit] The input resistance of any stage shall be independent of its on/off switch position, and must be equal to Rload. {\displaystyle {\begin{cases}R_{i,parr}=(R_{i,b}\cdot R_{load})/(R_{i,b}+R_{load})\\R_{i,a}+R_{i,parr}=R_{load}\\R_{i,parr}/(R_{i,a}+R_{i,parr})=Ratio_{i}\end{cases}}} With these equations, all resistor values of the circuit diagram follow easily after choosing values for N, {\displaystyle \alpha } and Rload. (The value of Rsource does not influence the logarithmic behavior) Constant output resistance[edit] The output resistance of any stage shall be independent of its on/off switch position, and must be equal to Rsource. {\displaystyle {\begin{cases}R_{i,ser}=R_{i,a}+R_{source}\\R_{i,ser}\cdot R_{i,b}/(R_{i,ser}+R_{i,b})=R_{source}\\R_{i,b}/(R_{i,ser}+R_{i,b})=Ratio_{i}\end{cases}}} Again, all resistor values of the circuit diagram follow easily after choosing values for N, {\displaystyle \alpha } and Rsource. (The value of Rload does not influence the logarithmic behavior) Circuit variations[edit] The circuit as depicted above, can also be applied in reverse direction. That correspondingly reverses the role of constant-input and constant-output resistance equations. Since the stages do not influence each other's attenuation, the stage order can be chosen arbitrarily. Such reordering can have a significant effect on the input resistance of the constant output resistance attenuator and vice versa. R-2R ladder networks used for Digital-to-Analog conversion are rather old. A historic description is in a patent[2] filed in 1955. Multiplying DA-converters with logarithmic behavior were not known for a long time after that. An initial approach was to map the logarithmic code to a much longer code word, which could be applied to the classical (linear) R-2R based DA-converter. Lengthening the codeword is needed in that approach to achieve sufficient dynamic range. This approach was implemented in a device from Analog Devices Inc.,[3] protected through a 1981 patent filing.[4] ^ "Multiplying DACs, flexible building blocks" (PDF). Analog Devices inc. 2010. Retrieved 29 March 2012. ^ US patent 3108266, Gordon, B. M., "Signal Conversion Apparatus", issued 22 October 1963 ^ "LOGDAC CMOS Logarithmic D/A Converter AD7118" (PDF). Analog Devices Inc. Archived from the original (PDF) on 25 August 2015. Retrieved 25 August 2015. ^ US patent 4521764, Burton, David P., "Signal-controllable attenuator employing a digital-to-analog converter", issued 4 June 1985 Online calculator to configure logarithmic ladder networks Retrieved from "https://en.wikipedia.org/w/index.php?title=Logarithmic_resistor_ladder&oldid=1087948655"
Wideband direction of arrival estimation - MATLAB - MathWorks Deutschland phased.GCCEstimator SensorPairSource SensorPair DelayOutputPort CorrelationOutputPort GCC Estimate of Direction of Arrival at Microphone Array GCC-PHAT Cross-Correlation Algorithm Wideband direction of arrival estimation The phased.GCCEstimator System object™ creates a direction of arrival estimator for wideband signals. This System object estimates the direction of arrival or time of arrival among sensor array elements using the generalized cross-correlation with phase transform algorithm (GCC-PHAT). The algorithm assumes that all signals propagate from a single source lying in the array far field so the direction of arrival is the same for all sensors. The System object first estimates the correlations between all specified sensor pairs using GCC-PHAT and then finds the largest peak in each correlation. The peak identifies the delay between the signals arriving at each sensor pair. Finally, a least-squares estimate is used to derive the direction of arrival from all estimated delays. To compute the direction of arrival for pairs of element in the array: Define and set up a GCC-PHAT estimator System object, phased.GCCEstimator, using the Construction procedure. Call step to compute the direction of arrival of a signal using the properties of the phased.GCCEstimator System object. sGCC = phased.GCCEstimator creates a GCC direction of arrival estimator System object, sGCC. This object estimates the direction of arrival or time of arrival between sensor array elements using the GCC-PHAT algorithm. sGCC = phased.GCCEstimator(Name,Value) returns a GCC direction of arrival estimator object, sGCC, with the specified property Name set to the specified Value. Name must appear inside single quotes (''). You can specify several name-value pair arguments in any order as Name1,Value1,...,NameN,ValueN. phased.ULA System object (default) \| Phased Array System Toolbox™ sensor array Sensor array, specified as a Phased Array System Toolbox System object. The array can also consist of subarrays. If you do not specify this property, the default sensor array is a phased.ULA System object with default array property values. SensorPairSource — Source of sensor pairs Source of sensor pairs, specified as either 'Auto' or 'Property'. 'Auto' — choose this property value to compute correlations between the first sensor and all other sensors. The first sensor of the array is the reference channel. 'Property' — choose this property value when you want to explicitly specify the sensor pairs to be used for computing correlations. Set the sensor pair indices using the SensorPair property. You can view the array indices using the viewArray method of any array System object. SensorPair — Sensor pairs [2;1] (default) \| 2-by-N positive integer valued matrix Sensor pairs used to compute correlations, specified as a 2-by-N positive integer-valued matrix. Each column of the matrix specifies a pair of sensors between which the correlation is computed. The second row specifies the reference sensors. Each entry in the matrix must be less than the number of array sensors or subarrays. To use the SensorPair property, you must also set the SensorPairSource value to 'Property'. DelayOutputPort — Option to enable delay output Option to enable output of time delay values, specified as a Boolean. Set this property to true to output the delay values as an output argument of the step method. The delays correspond to the arrival angles of a signal between sensor pairs. Set this property to false to disable the output of delays. CorrelationOutputPort — Option to enable correlation output Option to enable output of correlation values, specified as a Boolean. Set this property to true to output the correlations and lags between sensor pairs as output arguments of the step method. Set this property to false to disable output of correlations. reset Reset states of phased.GCCEstimator System object step Estimate direction of arrival using generalized cross-correlation Estimate the direction of arrival of a signal using the GCC-PHAT algorithm. The receiving array is a 5-by-5-element URA microphone array with elements spaced 0.25 meters apart. The arriving signal is a sequence of wideband chirps. The signal arrives from 17° azimuth and 0° elevation. Assume the speed of sound in air is 340 m/s. Load the chirp signal. load chirp; Create the 5-by-5 microphone URA. mic = phased.OmnidirectionalMicrophoneElement; array = phased.URA([N,N],[d,d],'Element',mic); Simulate the incoming signal using the WidebandCollector System object™. 'SampleRate',Fs,'ModulatedInput',false); signal = collector(y,arrivalAng); Estimate the direction of arrival. estimator = phased.GCCEstimator('SensorArray',array,... 'PropagationSpeed',c,'SampleRate',Fs); ang = estimator(signal) You can use generalized cross-correlation to estimate the time difference of arrival of a signal at two different sensors. A model of a signal emitted by a source and received at two sensors is given by: \begin{array}{l}{r}_{1}\left(t\right)=s\left(t\right)+{n}_{1}\left(t\right)\\ {r}_{2}\left(t\right)=s\left(t-D\right)+{n}_{2}\left(t\right)\end{array} where D is the time difference of arrival (TDOA), or time lag, of the signal at one sensor with respect to the arrival time at a second sensor. You can estimate the time delay by finding the time lag that maximizes the cross-correlation between the two signals. From the TDOA, you can estimate the broadside arrival angle of the plane wave with respect to the line connecting the two sensors. For two sensors separated by distance L, the broadside arrival angle, Broadside Angles, is related to the time lag by \mathrm{sin}\beta =\frac{c\tau }{L} where c is the propagation speed in the medium. A common method of estimating time delay is to compute the cross-correlation between signals received at two sensors. To identify the time delay, locate the peak in the cross-correlation. When the signal-to-noise ratio (SNR) is large, the correlation peak, τ, corresponds to the actual time delay D. \begin{array}{l}R\left(\tau \right)=E\left\{{r}_{1}\left(t\right){r}_{2}\left(t+\tau \right)\right\}\\ \stackrel{^}{D}\text{ }=\text{ }\underset{\tau }{\mathrm{arg}\mathrm{max}}R\left(\tau \right)\end{array} When the correlation function is more sharply peaked, performance improves. You can sharpen a cross correlation peak using a weighting function that whitens the input signals. This technique is called generalized cross-correlation (GCC). One particular weighting function normalizes the signal spectral density by the spectrum magnitude, leading to the generalized cross-correlation phase transform method (GCC-PHAT). \begin{array}{l}S\left(f\right)={\int }_{-\infty }^{\infty }R\left(\tau \right){e}^{-i2\pi f\tau }d\tau \\ \stackrel{˜}{R}\left(\tau \right)={\int }_{-\infty }^{\infty }\frac{S\left(f\right)}{\|S\left(f\right)\|}{e}^{+i2\pi f\tau }df\\ \stackrel{˜}{D}\text{ }=\text{ }\underset{\tau }{\mathrm{arg}\mathrm{max}}\text{ }\stackrel{˜}{R}\left(\tau \right)\end{array} If you use just one sensor pair, you can only estimate the broadside angle of arrival. However, if you use multiple pairs of non-collinear sensors, for example, in a URA, you can estimate the arrival azimuth and elevation angles of the plane wave using least-square estimation. For N sensors, you can write the delay time τkj of a signal arriving at the kth sensor with respect to the jth sensor by \begin{array}{l}c{\tau }_{kj}=-\left({\stackrel{\to }{x}}_{k}-{\stackrel{\to }{x}}_{j}\right)\cdot \stackrel{\to }{u}\\ \stackrel{\to }{u}=\mathrm{cos}\alpha \mathrm{sin}\theta \stackrel{^}{i}+\mathrm{sin}\alpha \mathrm{sin}\theta \stackrel{^}{j}+\mathrm{cos}\theta \stackrel{^}{k}\end{array} where u is the unit propagation vector of the plane wave. The angles α and θ are the azimuth and elevation angles of the propagation vector. All angles and vectors are defined with respect to the local axes. You can solve the first equation using least-squares to yield the three components of the unit propagation vector. Then, you can solve the second equation for the azimuth and elevation angles. [1] Knapp, C. H. and G.C. Carter, “The Generalized Correlation Method for Estimation of Time Delay.” IEEE Transactions on Acoustics, Speech and Signal Processing. Vol. ASSP-24, No. 4, Aug 1976. [2] G. C. Carter, “Coherence and Time Delay Estimation.” Proceedings of the IEEE. Vol. 75, No. 2, Feb 1987. phased.BeamscanEstimator \| phased.RootMUSICEstimator \| gccphat
Existence of Mild Solutions for the Elastic Systems with Structural Damping in Banach Spaces Hongxia Fan, Yongxiang Li, Pengyu Chen, "Existence of Mild Solutions for the Elastic Systems with Structural Damping in Banach Spaces", Abstract and Applied Analysis, vol. 2013, Article ID 746893, 6 pages, 2013. https://doi.org/10.1155/2013/746893 Hongxia Fan,1,2 Yongxiang Li,2 and Pengyu Chen2 1Department of Mathematics, Lanzhou Jiaotong University, Lanzhou 730070, China This paper deals with the existence and uniqueness of mild solutions for a second order evolution equation initial value problem in a Banach space, which can model an elastic system with structural damping. The discussion is based on the operator semigroups theory and fixed point theorem. In addition, an example is presented to illustrate our theoretical results. Our aim in this paper is to study the existence and uniqueness of mild solutions for the semilinear elastic system with structural damping in a Banach space , where means , is a constant; is a closed linear operator and generates a -semigroup on ; , , . In 1982, Chen and Russell [1] investigated the following linear elastic system described by the second order equation in a Hilbert space with inner , where (the elastic operator) and (the damping operator) are positive definite selfadjoint operators in . They reduced (2) to the first order equation in Let , with the naturally induced inner products. Then, (2) is equivalent to the first order equation in where Chen and Russell [1] conjectured that is the infinitesimal generator of an analytic semigroup on if and either of the following two inequalities holds for some : In the same paper they obtained some results in this direction. The complete proofs of the two conjectures were given by Huang [2, 3]. Then, other sufficient conditions for or its closure to generate an analytic or differentiable semigroup on were discussed in [4–10], by choosing to be an operator comparable with for , based on an explicit matrix representation of the resolvent operator of or . However, so far as we know, among the previous works, little is concerned with an elastic system with structural damping in a Banach space. Motivated by previous works, in this paper, we investigate the existence and uniqueness of mild solutions for the elastic system (1) in a frame of Banach spaces. To this end, we firstly introduce the concept of mild solutions for system (1), which is based on the discussion about associated linear system. Secondly, we prove the existence and uniqueness of mild solutions for the semilinear elastic system (1) in a Banach space . The paper is organized as follows. In Section 2, we discuss the associated linear elastic system and give its definition of mild solutions. In Section 3, we study the existence and uniqueness of mild solutions for the semilinear elastic system (1). An example to illustrate our theoretical results is given in Section 4. 2. Preliminaries on Linear Elastic Systems Let be a Banach space, we consider the linear elastic system with structural damping where means , is a constant; is a closed linear operator, and generates a -semigroup on ; , , . For the second order evolution equation it has the following decomposition That is, It follows from (9) and (11) that By (12), we have Let which means So we reduce the linear elastic system (8) to the following two abstract Cauchy problems in Banach space : It is clear that (16) and (17) are linear inhomogeneous initial value problems for and , respectively. Since is the infinitesimal generator of -semigroup . Furthermore, for any , (13) yield , . Thus, by operator semigroups theory [11], and are infinitesimal generators of -semigroups, which implies initial value problems (16) and (17) are well-posed. Throughout this paper, we assume that and generate -semigroups and on , respectively. Note that , and is the infinitesimal generator of -semigroup . It follows that It is well known [12, Chapter 4], when , the linear initial value problem (16) has a mild solution given by Similarly, if , then the mild solution of the linear initial value problem (17) expressed by Substituting (19) into (20), we get From the argument above, we obtain the following corollary. Corollary 1. If , then the initial value problem (8) has at most one solution. If it has a solution, this solution is given by (21). For every , the right-hand side of (21) is a continuous function on . It is natural to consider it as a generalized solution of (8) even it is not differentiable and dose not strictly satisfy the equation. We therefore define the following. Definition 2. Let is the infinitesimal generator of -semigroup . Then a continuous solution of the integral equation is said to be a mild solution of the initial value problem (8). Where , were defined in (18) and was specified in (15). Let be the Banach space of all continuous functions with norm , . Let be the Banach space of all linear and bounded operators on . Note that and are -semigroups on . Thus, there exist and such that In what follows, we firstly give the definition of a mild solution for the initial value problem (1) below. Secondly, we consider the existence and uniqueness of mild solutions for (1). To this end, we make the following assumptions: be continuous and there exists , such that be continuous and there exists a positive function such that The -semigroup is compact for . Theorem 4. Assume that holds, is the infinitesimal generator of -semigroup . Then for every , and , the initial value problem (1) has a unique mild solution . Proof. Define the operator by It is obvious that the mild solution of the initial value problem (1) is equivalent to the fixed point of . For any , (23), (27), and yield Using (27), (28), and induction on it follows easily that Hence Since Thus, for large enough and by well known extension of the contraction mapping principle, has a unique fixed point . This fixed point is the desired solution of the integral equation (24). Theorem 5. Suppose that assumptions and hold. Then for every , and , the initial value problem (1) has at least one mild solution . Proof. Define the operator as (27) and choose such that Let . We proceed in two main steps. Step 1. We show that . For that, let . Then for , we have which according to and (23) gives In view of the choice of , we obtain Step 2. We prove that is completely continuous. Note that is a continuous mapping from to . Thus, is continuous. Next, we show that is compact. To this end, we use the Ascoli-Arzela's theorem. For that, we first prove that is relatively compact in , for . Obviously, is compact. Let . For each and , we define the operator by Then the sets are relatively compact in since by and (18), the semigroup is compact for on . Moreover, using (23) and we have Therefore, the set is relatively compact in for all and since it is compact at we have the relatively compactness in for all . Now, let us prove that is equicontinuous. For , we have where In fact, and tend to 0 independently of when . Note that the function is continuous for . Thus, is uniformly continuous on and thus . From (23) and , we have Let . By the compactness of and (18), we can easily conclude that is compact and therefore is continuous in the uniform operator topology for . Then, is also continuous in the uniform operator topology on . Thus as . Meanwhile, is bounded on . Hence, using Lebesgue dominated convergence theorem we deduce that . Moreover, from (23) we have Hence, . In short, we have show, that is relatively compact for , is a family of equicontinuous functions. It follows from Ascoli-Arzela's theorem that is compact. By Schauder fixed point theorem has a fixed point , which obviously is a mild solution to (1). In order to illustrate our main results, we consider the following initial-boundary value problem, which is a model for elastic system with structural damping where , are all constants, is continuous. Let , we define the linear operator in by It is well known from [13] that is the infinitesimal generator of a -semigroup on . Let , , then the initial-boundary value problem (42) can be reformulated as the following abstract second order evolution equation initial value problem in : In order to solve the initial-boundary value problem (42), we also need the following assumptions:, . The partial derivative is continuous. Theorem 6. If the assumptions and are satisfied, then for any , the initial-boundary value problem (42) has a unique mild solution . Proof. From the assumptions and , it is easily seen that the conditions in Theorem 4 are satisfied. Hence, by Theorem 4, for any , the problem (44) has a unique mild solution , which means is a mild solution for initial-boundary value problem (42). The authors are grateful to the anonymous referee for his/her valuable comments and suggestions, which improve the presentation of the original paper. Research was supported by NNSFs of China (11261053, 11061031). G. Chen and D. L. Russell, “A mathematical model for linear elastic systems with structural damping,” Quarterly of Applied Mathematics, vol. 39, no. 4, pp. 433–454, 1982. View at: Google Scholar \| MathSciNet F. L. Huang, “On the holomorphic property of the semigroup associated with linear elastic systems with structural damping,” Acta Mathematica Scientia, vol. 5, no. 3, pp. 271–277, 1985. 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