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Generalized Contextuality in Large Quantum Systems We thank the Austrian Science Fund (FWF) for support (10.55776/PAT2839723). Funded by the European Union - NextGenerationEU. Click here for the FWF description. Popular summary -- scientific details below: Quantum theory promises technological applications that would be impossible within classical physics: faster computation, more accurate metrology, or the generation of provably secure random numbers. However, to make this work, we first need to test whether our devices are really quantum and work as desired -- a task called certification. This is not only relevant for technology, but also for fundamental physics: given some large physical system, such as a Bose-Einstein condensate, how can we prove that its properties cannot be explained by classical physics? In other words, how can we certify its nonclassicality? In this project, we will develop a new method to do so, both theoretically (via mathematical proofs and conceptual argumentation) and experimentally (with concrete data supplied by colleagues at ETH Zurich). Our approach is based on the phenomenon of contextuality: properties of quantum systems cannot be independent of the choice of implementation of the measurement procedures. In other words, if we ask Nature a question, then the answer must sometimes depend on the experimental context. Here, we develop methods that allow us to certify this phenomenon in physical systems even if they are very large and can only be probed in coarse and incomplete ways, and even if we know nothing about their composition, time evolution, or the physical theory that describes them. Our project will improve upon earlier work in several respects. Most earlier attempts to certify nonclassicality in large quantum systems have relied on the notion of Bell nonlocality: correlations between several particles cannot be explained by any local hidden-variable model. However, this has only been possible under strong additional assumptions, since it is impossible to measure all particles of a large quantum system individually. Moreover, both the experimental detection as well as the theoretical definition of contextuality (in the sense that is relevant for our project) has been restricted to situations in which the experimenter can measure all properties of the physical system completely and exhaustively (tomographic completeness). In our project, we will drop these assumptions and develop methods that are device- and theory-independent and that work with coarse and incomplete experimental data. Our projects spans quantum physics from its philosophical foundations up to its experimental implementation. Conceptually, we will shed light on the question of how coarse experimental data can render a microscopic theory implausible. Mathematically, we will develop methods that can certify this notion of contextuality with algorithms and inequalities. Finally, we will apply our results to concrete experimental data from nanomechanical oscillators and Bose-Einstein condensates. Spekkens' notion of generalized noncontextuality Under what conditions can we say that a physical system defies classical explanation? Spekkens' notion of generalized contextuality [1] is arguably our currently best candidate for answering this question, in particular in all cases where we consider single physical systems without any causally relevant differentiation into subsystems (as in Bell's theorem). In contrast to the Kochen-Specker version, generalized noncontextuality (GNC) does not rely on the assumptions that measurements are projective or noiseless, and it does not even rely on the validity of quantum theory. GNC can be experimentally refuted in robust ways (for example, via GNC inequalities), it can be conceptually grounded on Leibniz's Principle of the Identitity of the Indescernibles, and it can be mathematically understood as the demand that the linearity structure in the operational (laboratory) theory should be consistent with the linearity structure in its hypothetical underlying "classical" hidden-variable model. It is a mathematically and conceptually compelling concept. However, due to its reliance on Leibniz's Principle, its range of applicability was hitherto restricted to situations where we assume "tomographic completeness": that our experiment probes a given physical system completely. The goal of our project is to extend its range to experiments where this assumption is not satisfied. This involves many experiments of interest, in particular, experiments on "large" quantum systems (such as Bose-Einstein condensates) that admit only a partial probing of its degrees of freedom (say, only measurements of collective degrees of freedom). A new perspective and a generalization of generalized noncontextuality In our paper [2], we have suggested a different perspective on GNC that circumvents this problem in some sense. That is, instead of seeing GNC as a condition about physical reality (and whether it can be classical in a way that respects Leibniz's Principle), we see it as a methodological principle that tells us under what conditions a hypothetical fundamental theory could be a plausible explanation of an effective theory (describing e.g. the laboratory data): Processes that are statistically indistinguishable in an effective theory should not require explanation by multiple distinguishable processes in a more fundamental theory. For prepare-and-measure scenarios as depicted above, this condition boils down to the mathematical constraint that the effective probabilistic theory should admit of a linear embedding into the fundamental probabilistic theory. In the special case where the fundamental theory is assumed to be classical (i.e. describable by classical probability theory), this reduces to Spekkens' version of GNC (exactly in its mathematical formulation, even if not in its intended conceptual interpretation). But we obtain a broader and more powerful notion: we can also ask, for example, whether our effective laboratory data has a plausible description in terms of quantum theory. This should allow us to design novel experimental tests of quantum theory, and this is one of several research directions that we are pursuing in this project. Project goals Our project involves conceptual ("philosophical"), mathematical and experimental aspects. Conceptually, we will analyze in more detail under what conditions coarse experimental data can certify the implausibility of a more fundamental underlying classical description. Since Leibniz's Principle is silent about this scenario, we have to refer to "no-finetuning" principles or "typicality" arguments that are often invoked in the context of thermodynamics. Some of this will be done in collaboration with our colleagues at ICTQT Gdańsk. Mathematically, we will analyze how contextuality can be inferred from coarse experimental data under additional assumptions (such symmetries of the experimental setup or spatiotemporal constraints). We will extend "theory-agnostic tomography" [3] to systems in which tomographic completeness is dropped, and we will implement methods to test for the quantum embeddability of the resulting generalized probabilistic theories. Experimentally, we will apply our insights to concrete physical systems, including Bose-Einstein condensates and mechanical resonators. This will be done in collaboration with Matteo Fadel (ETH [1] R. W. Spekkens, Contextuality for Preparations, Transformations, and Unsharp Measurements, Phys. Rev. A 71, 052108 (2005). [2] M. P. Müller and A. J. P. Garner, Testing Quantum Theory by Generalizing Noncontextuality, Phys. Rev. X 13, 041001 (2023). DOI:10.1103/PhysRevX.13.041001. [3] M. D. Mazurek, M. F. Pusey, K. J. Resch, and R. W. Spekkens, Experimentally Bounding Deviations From Quantum Theory in the Landscape of Generalized Probabilistic Theories, PRX Quantum 2, 020302	{"url":"https://www.iqoqi-vienna.at/research/mueller-group/generalized-contextuality-in-large-quantum-systems","timestamp":"2024-11-12T15:46:33Z","content_type":"text/html","content_length":"85218","record_id":"<urn:uuid:c1bae457-5c8e-4ec2-9348-222c1a6afebf>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00570.warc.gz"}
Financial charts and buy/sell signals The Stochastic RSI is an indicator used in technical analysis that combines the strength of the Relative Strength Index (RSI) and the Stochastic Oscillator. It is used to identify overbought and oversold levels and indicate potential reversal points in the market. Stochastic RSI is a technical indicator that measures the relative strength of a security’s price movement. It is calculated by taking the ratio of the RSI to the Stochastic Oscillator. The Stochastic RSI is a momentum indicator that can be used to identify overbought and oversold levels and indicate potential reversal points. This indicator is calculated by dividing the RSI value by the Stochastic Oscillator value. The RSI is a momentum indicator that measures the speed and magnitude of price changes over a given period of time. It is calculated by taking the average gain over a period of time and dividing it by the average loss over that same period of time. The Stochastic Oscillator is a momentum indicator that measures the momentum of price movements. It is calculated by taking the difference between the current closing price and the low price of a given period and dividing it by the high price of the same period minus the low price. The Stochastic RSI is a range-bound indicator, which means it fluctuates between 0 and 100. When the Stochastic RSI is above 70, it is considered overbought, indicating that the security is overvalued and a correction may be imminent. When the Stochastic RSI is below 30, it is considered oversold, indicating that the security is undervalued and a rally may be imminent. The indicator can be used to identify potential reversal points in the market. It is important to note, however, that the Stochastic RSI should not be used as a standalone indicator. It should be used in conjunction with other technical indicators and fundamental analysis to make more informed trading decisions. The Stochastic RSI is a powerful tool that can be used to identify price momentum and trend reversals. It is important to remember, however, that it should not be used as a standalone indicator. It should be used in conjunction with other technical indicators and fundamental analysis to make more informed trading decisions.	{"url":"https://www.chartscheck.com/dictionary/Stochastic_RSI_Indicator","timestamp":"2024-11-02T03:04:11Z","content_type":"text/html","content_length":"17147","record_id":"<urn:uuid:5c75f142-db3b-422b-a99a-1c815b525736>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00783.warc.gz"}
Dual rated notation question Hi there, I think I know the answer already but just wanted a confirmation. I am having an issue finding the answer in the book, but I am running a tournament where the time control is G30;+15. Since this is dual rated, I assume notation is required till less than 5 minutes remaining? Yepâ€“because the base time control is 30 + 15 = 45. For all rated games: If the total is 30 or more notation is required. If the total is less than 30 then notation is not required. Perfect, thanks for the confirmation Game 30;+15. I assume you are using increment time. What are the round times for this event? You can look at it this way. A dual-rated tournament is nothing more than a regular-rated tournament which happens to be also rated under the quick system. Since it is regular-rated, regular-rated rules apply. Bill Smythe	{"url":"https://uschess.discourse.group/t/dual-rated-notation-question/21814","timestamp":"2024-11-07T22:03:09Z","content_type":"text/html","content_length":"30509","record_id":"<urn:uuid:a9bc1423-aeeb-45ee-8c6c-77cc36f0f43e>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00816.warc.gz"}
What is the circumference of a 16 foot circle? As a result, when the circle’s circumference is 16 feet, the radius of the circle is r 2.54648 feet. C = 2 (3.14159) (2.54648) ft. You might also wonder what the circumference of an 18-foot circle The circumference of a circle is determined by the formula C=2r, where r =radius. We then add 18 to the circumference to find the radius. So 18=2∏r. Solving for r yields a total of 9/, or 2.86 inches. The next question is how do I calculate a circle’s circumference. The circle’s circumference is divided by its diameter (Pi multiplied by its diameter). Simply divide the circumference by ‘, and the diameter will be the same length. The radius is simply divided by two, so divide the diameter by two to get the circle’s radius! What is the circumference of a 7-foot circle, in addition to the above? To get the circle’s circumference, multiply that radius by two and then multiply the result by a. If your circle has a radius of 3 feet, for example, the diameter is 3 2 = 6 feet, and the circumference is 6 3.14 = 18.84 feet, or 6 3.1415 = 18.849 feet if you want a more precise answer. What is the circumference of a 15-cm circle? A circle’s circumference is equal to Pi multiplied by the diameter d. The circumference using radius formula is 2r 2 r because the diameter d is equal to two times the radius r. In a 16-foot circle, how many square feet is there? 804.25 square feet 115,812 square inches 74.717 square meters 747,171 square centimeters A 16 Foot Circle 804.25 square feet 115,812 square inches 74.717 square meters 747,171 square centimeters A 16 Foot Circle 804.25 square feet What is the diameter of a 16-foot circle? As a result, when the circle’s circumference is 16 feet, there are simple r = 8/(ft.). The radius of the circle is r 2.54648 ft. C = 2 (3.14159) (2.54648) ft. What is the circumference of a 12-foot-diameter circle? You’ll use the formula C=2r to determine the circumference of a circle; as a result, the circumference C is 2 638 inches. What is the circumference of a four-foot-diameter circle? The circumference of a circle can be determined by multiplying pi ( = 3.14) by the circle’s diameter. A circle’s circumference is 3.144=12.56 if it has a diameter of 4. The diameter is twice as large if you know the radius. What is the circumference of an 8-foot circle? The circle’s diameter is 8 inches, as we can see. Let’s enter this number into the formula. A circle with a diameter of 8 inches has a circumference of 25.12 inches. What is the radius of the 20-foot circle? The circumference of a circle is 2piRadius, or pidiameter. d=20/pi=6.36 feet, i.e. (pi=3.14) What is the square footage of a 20-foot-diameter circle? To square the number (6 x 6 = 36), multiply the radius by itself. Multiply the result by pi (using the calculator’s button) or 3.14159 (36 x 3.14159 = 113.1). As a result, the circle’s surface area is 113.1 square feet. What is the size of a 16-inch circle? Area of a 16 Inch Circle 804.25 square inches 5.5851 square feet 0.51887 square meters 5,188.7 square centimeters 5.5851 square feet 0.51887 square meters 5,188.7 square centimeters 5.5851 square feet 0.51887 square meters 5,188.7 square centimeters 55851 square feet What exactly is a circle’s radius? A circle’s radius is the distance from its center to any point on its circumference. [1] Dividing the diameter in half is the simplest way to determine the radius. What is the radius of a circle calculator? Substitute this value into the circumference formula: C = 2 * R = 2 * 14 = 87.9646 cm. You can also use it to calculate the radius of a circle: A = R2 = 142 = 615.752 cm2. Finally, you can find the diameter, which is simply double the radius: D = 2 R = 2 = 14 = 28 cm. What is the area of a triangle that you can find? Multiply the base by the height and then divide by 2 to find the area of a triangle. Because a parallelogram can be divided into two triangles, the division by two comes from the fact that it can be divided into two triangles. The area of each triangle in the diagram to the left, for example, is equal to half the area of the parallelogram. What are the circles’ formulas? Circle Formulas in Math: Area of a circle (A) = r 2 =(/4) D2 = 0.7854 D2 Diameter of a circle (D) = (A/0.7854) The angle between two radii is measured in degrees. . The sector’s area = (/2) r 2 A circle’s sector angle = (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l Circle Formulas in Math: Area of a circle (A) = r 2 =(/4) D2 = 0.7854 D2 Diameter of a circle (D) = (A/0.7854) The angle between two radii is measured in degrees. . The sector’s area = (/2) r 2 A circle’s sector angle = (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l)/ (180 x l What is a sphere’s circumference? A circle or sphere’s circumference is equal to 6.2832 times its Radius. A circle or sphere’s circumference is equal to 3.1416 times its diameter.	{"url":"https://tipsfolder.com/circumference-16-foot-circle-f82541084df4cbe22a76125a8eecadeb/","timestamp":"2024-11-05T00:13:26Z","content_type":"text/html","content_length":"98661","record_id":"<urn:uuid:0f1664d3-88a3-4f1c-b681-fc8d7130756b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00411.warc.gz"}
Add Fractions with Unlike Units Using the Strategy of Creating Equivalent Fractions □ 01/02/14 \| Adjusted: 01/10/17 \| 1 file □ Grades 5 Add Fractions with Unlike Units Using the Strategy of Creating Equivalent Fractions What we like about this lesson • Addresses standards 5.NF.A.1, 5.NF.A.2 • Provides concrete examples for the need of like denominators for addition and subtraction of fractions • Allows students to understand adding fractions with unlike denominators without going straight to the least common denominator • Asks students to reason about how the size of units changes as they create equivalent fractions • Encourages students to look for and make use of structure as they analyze rectangular fraction models (MP.7) In the classroom: • Uses concrete and pictorial models, particularly the rectangular fraction model, to make the mathematics of the lesson explicit • Prompts students to share their developing thinking • Allows for whole group, partner, and individual work in one lesson • Gives formal and informal opportunities for teachers to check for understanding • Making the Shifts How does this lesson exemplify the instructional Shifts required by CCSSM? Focus Belongs to the major work of fifth grade Coherence Builds on key understandings of equivalent fractions (4.NF.A.1) and addition of fractions with like denominators (4.NF.B). Builds foundation toward adding and subtracting rational numbers (7.NS.A.1) Conceptual Understanding: primary in this lesson Rigor^ Procedural Skill and Fluency: secondary in this lesson Application: secondary in this lesson • Additional Thoughts It's important to note that this sample lesson is the first in a 5-lesson series on "Making Like Units Pictorially", which is part of a 16-lesson unit on Addition and Subtraction of Fractions. It is not intended for students to meet the full expectations of the grade-level standards addressed in these lessons through only this selected lesson. This sample lesson lays a strong foundation for the work that is to come in the unit by focusing on the use of pictorial models, particularly the rectangular fraction model, to portray addition of fractions less than 1 with unlike denominators. Subsequent lessons move away from concrete and pictorial models and focus on the abstract approach to addition and subtraction of fractions both less than 1 and between 1 and 2. This lesson develops the understanding of adding fractions with unlike denominators by requiring students to work with rectangular fraction models. Thoughtful questioning is used throughout the lesson to promote students' reasoning on the size of denominators as they create equivalent fractions and add them. Note that there is no explicit instruction or discussion of finding the least common denominator; instead, students are developing their understanding of this concept through models. For more insight on the grade-level concepts addressed in this lesson, read page 11 of the progression document, 3–5 Number and Operations – Fractions. The structure of these lessons and the unit/curriculum overall have some interesting aspects to highlight. The units make explicit the coherence within the fully developed curriculum. Each topic (a set of lessons) is connected to prior learning and also points to the next topic (or module) that follows in the learning progression. Within individual lessons, there are a number of components that add to their strength including daily fluency practice, variety in questioning techniques, and frequent opportunities for students to debrief about their learning.	{"url":"https://achievethecore.org/page/892/add-fractions-with-unlike-units-using-the-strategy-of-creating-equivalent-fractions","timestamp":"2024-11-08T08:52:10Z","content_type":"text/html","content_length":"81817","record_id":"<urn:uuid:9b1ab8a0-b25b-4b06-9e6d-df939c593d65>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00853.warc.gz"}
Supercomputing Subatomic Particle Research on Titan - High-Performance Computing News Analysis \| insideHPC Researchers at the Thomas Jefferson National Accelerator Facility are collaborating with Nvidia to develop better quantum chromodynamics codes for GPUs. Using QCD simulations, they hope to better understand the laws that govern the atomic world. Before the 1950s the electrons, neutrons, and protons comprising atoms were the smallest confirmed units of matter. With advancements in experimental and theoretical techniques in recent decades, though, researchers now try to understand particles a step smaller and more fundamental. In recent years large-scale experimental facilities, such as the Large Hadron Collider in Switzerland, have allowed researchers to begin testing theories about how subatomic particles behave under different conditions. Research institutions funded by the US Department of Energy (DOE) have also made major investments in experimental test facilities. The newest of these facilities lies in Hall D at the Thomas Jefferson National Accelerator Facility (Jefferson Lab). The experiment, known as GlueX, aims to give researchers unprecedented insight into subatomic particle interactions. “We believe there is a theory that describes how elementary particles interact, quarks and gluons that make up the matter around us,” said Robert Edwards, senior staff scientist at Jefferson Lab. “If so, the theory of QCD suggests that there are some exotic forms of matter that exist, and that’s what we’re looking for in our Hall D experiment.” Edwards serves as the principal investigator on a project that uses computation to inform the GlueX experiment as well as corroborate experimental findings. To that end the team has been using the Titan supercomputer at DOE’s Oak Ridge National Laboratory (ORNL). Titan is the flagship supercomputer of the Oak Ridge Leadership Computing Facility (OLCF), a DOE Office of Science User Facility located at ORNL. The team wants to make computer codes for quantum chromodynamics (QCD) applications run more efficiently and effectively, and with access to world-class computing resources, the researchers’ computational innovations were able to achieve speedups ranging from seven- to tenfold for QCD calculations compared with those achieved in earlier work. Mathematical mesons The field of QCD is the study of forces between two major categories of subatomic particles–quarks and gluons. Quarks serve as the primary force-carrying particles in an atom’s nucleus and make up hadrons, a class of subatomic particles that includes protons and neutrons. Gluons, much like their name implies, allow quarks to interact with forces and serve as the “glue” that holds hadrons together. Quarks can also bind with their inverse, antiquarks, to form mesons. Mesons are among the most mysterious of all subatomic particles because their resonances are in existence for only fractions of a microsecond. Through experiments researchers hope to use GlueX to confirm the existence of “exotic” mesons that would help advance QCD theory. When simulating a quantum system of quarks, gluons, and mesons, the number of calculations needed to compute the interactions of the subatomic particle fields explodes in a hurry. Researchers represent quarks and gluons by using a lattice or grid. In fact, researchers using this method call it lattice QCD (LQCD). Once the theories are expressed in terms of the lattice, the overall simulation becomes similar to a high-school-level model of a crystal–plastic spheres at the lattice points connected by springs between them. One can think of the spheres at the lattice points as representing the quark field with the springs between them representing the quark-gluon interactions. When given energy by pushing or nudging, the model will vibrate. At any given instant, a snapshot of the model would show a particular arrangement of stretched and compressed springs. If one looked at the statistical distribution of these snapshots, he or she could deduce information about the crystal. QCD works in a similar way. The team’s lattices act as snapshots of the states of the gluon fields. By generating a statistical sampling of these QCD field snapshots and analyzing them, the team can compute the properties of the subatomic particles of interest. Although that process might sound simple, it really isn’t. Each snapshot requires a lot of computation. To compute the quark-gluon interactions, the team must repeatedly carry out the complex computation of solving the Dirac equation–a complex wave equation. Solving the equation is complicated enough, but Jefferson Lab researcher Balint Joo noted that the team’s simulations must do it many times. “Our algorithm is one that requires solving the Dirac equation hundreds of thousands of times for each of the 300 to 500 snapshots that we take,” he said. Such computational demands push even the world’s fastest supercomputers to their performance limits, and Joo and NVIDIA high-performance computing researcher Kate Clark have teamed with other researchers from the USQCD collaboration to search for new ways to improve code performance on the Jefferson Lab team’s CHROMA code, among other QCD applications. They shared their results in a paper presentation at the SC16 conference, which took place November 13-18. GPUs as the glue Since 2005 Clark has focused on methods to improve code performance for the LQCD community. Before moving to NVIDIA, she worked in LQCD algorithms at Boston University with professor Richard Brower, where the team developed a multigrid algorithm. Essentially, computer chips have become so much faster than memory systems that memory can’t feed chips the data fast enough, meaning the bottleneck for LQCD calculations comes from the speed of the memory system. Clark has been developing the QUDA library, which takes advantage of a GPU system’s computational strength, including its very fast built-in memory, to improve calculation speed. When developing its new algorithm, the Edwards team began by adding a multigrid algorithm into its code. Multigrid algorithms take the large, fine-grained lattice grid for LQCD calculations; average the various grid points; and create multiple smaller, coarser grids. Similar to sound waves, which are really composed of many waves, each with a different pitch or frequency, the team’s problem is composed of many modes with different energies. High-energy modes need a fine lattice to represent them accurately, but low-energy modes–which usually slow down when seeking a solution–can be represented on coarser lattices with fewer points, ultimately reducing the computational cost. By using multiple grids and separating the modes in the problem onto the various grids most efficiently, the researchers can get through their long line of calculations quicker and easier. GPUs provide a lot of memory bandwidth,” Clark said. “Solving LQCD problems computationally is almost always memory-bound, so if you can describe your problem in such a way that GPUs can get maximum use of their memory bandwidth, QCD calculations will go a lot quicker.” In other words memory bandwidth is like a roadway in that having more lanes helps keep vehicles moving and lessens the potential for traffic backups. However, the more GPUs working on a problem, the more they must communicate with one another. If too many GPUs get involved and the problem size doesn’t keep up with the computational resources being used, the calculation becomes very inefficient. One aspect of GPUs is that they bring a lot of parallelism to the problem, and so to get maximum performance, you may need to restructure your calculation to exploit more parallelism,” Clark Pouncing on parallelism Essentially, as computing technology has evolved, processing speed has improved faster than the ability of interconnects to move increasingly larger amounts of data across supercomputers’ nodes. For simulations in which researchers divide their calculations across many computer nodes, this imbalance can lead to performance bottlenecks. “With QCD the computational cost doesn’t scale linearly; it scales super-linearly,” Clark said. “If you double the problem size, the computational cost goes up by more than a factor of two. I can’t keep the same size of computation per node and just put it on a bigger system.” Despite performance gains through implementing the multigrid algorithm, Clark, Joo, and their collaborators noted that for maximum performance impacts, they would need to exploit sources of parallelism other than those that had typically been used in existing LQCD calculations. Each one of Titan’s 18,688 GPUs has 2,688 processing cores. To return to the roadway analogy, each one of a GPU’s individual processors is a “lane” on a road, and if only one lane is open, cars back up quickly. With that in mind, Clark, Joo, and their collaborators worked on opening up as many processing “lanes” as possible for LQCD calculations. The team recognized that in addition to exploiting parallelism by calculating multiple grids rather than a single, large grid, they could also exploit more parallelism out of each grid point. To create multiple grids from one large, fine-grained grid, each GPU calculates a set of grid points (which appear as mathematical vectors), averages the results, and sends the averages to the middle grid point. Rather than just having one processing “lane” doing all of these calculations, researchers can use four processing cores to calculate the points above, below, and to the left and right of the original grid point. Much like going from a one-lane road to a four-lane highway, the data throughput moves much faster. This concept works for a two-dimensional calculation, and a four-dimensional calculation can use this same concept to achieve eight-way parallelism. In addition, the researchers noted that each grid point is not just a number but also a vector of data. By splitting up the vector calculations to run on multiple processors, the team further increased code parallelism. Because of these innovations, the Edwards team saw hundredfold speedups on the coarsest grids and a tenfold speedup for finer grids when comparing simulations with those that took place before the QUDA implementation. Clark and Joo pointed out that this approach affects more than the team’s CHROMA code. These methods are already being applied to other QCD applications. Clark noted that as computers continue to get more powerful by using accelerators–such as the OLCF’s next-generation machine, Summit, set to begin delivering science in 2018–researchers will have to focus on getting as much parallelism as possible. “Going forward, supercomputers like Summit will have many more processing cores, so to get high efficiency as a whole, researchers in many fields are going to have to work on how to exploit all the levels of parallelism in a problem,” Clark said. “At some point exploiting all levels of parallelism is something that all researchers will have to do, and I think our work is a good example of that.”	{"url":"https://insidehpc.com/2016/12/jefferson-lab-nvidia-collaboration-uses-titans-to-boost-subatomic-particle-research/","timestamp":"2024-11-10T16:04:22Z","content_type":"application/xhtml+xml","content_length":"121403","record_id":"<urn:uuid:d5cd4162-6302-481c-8d93-9e165dcc9c2f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00874.warc.gz"}
China Geodetic Coordinate System 2000* Map Projections and Coordinate Systems Datums Tell Us the Latitudes and Longi- Vertex, Node, Or Grid Cell in a Data Set, Con- Tudes of Features on an Ellipsoid 116 GIS Fundamentals Map Projections and Coordinate Systems Datums tell us the latitudes and longi- vertex, node, or grid cell in a data set, con- tudes of features on an ellipsoid. We need to verting the vector or raster data feature by transfer these from the curved ellipsoid to a feature from geographic to Mercator coordi- flat map. A map projection is a systematic nates. rendering of locations from the curved Earth Notice that there are parameters we surface onto a flat map surface. must specify for this projection, here R, the Nearly all projections are applied via Earth’s radius, and o, the longitudinal ori- exact or iterated mathematical formulas that gin. Different values for these parameters convert between geographic latitude and give different values for the coordinates, so longitude and projected X an Y (Easting and even though we may have the same kind of Northing) coordinates. Figure 3-30 shows projection (transverse Mercator), we have one of the simpler projection equations, different versions each time we specify dif- between Mercator and geographic coordi- ferent parameters. nates, assuming a spherical Earth. These Projection equations must also be speci- equations would be applied for every point, fied in the “backward” direction, from pro- jected coordinates to geographic coordinates, if they are to be useful. The pro- jection coordinates in this backward, or “inverse,” direction are often much more complicated that the forward direction, but are specified for every commonly used pro- jection. Most projection equations are much more complicated than the transverse Mer- cator, in part because most adopt an ellipsoi- dal Earth, and because the projections are onto curved surfaces rather than a plane, but thankfully, projection equations have long been standardized, documented, and made widely available through proven programing libraries and projection calculators. [Show full text]	{"url":"https://docslib.org/doc/2751155/china-geodetic-coordinate-system-2000","timestamp":"2024-11-10T09:08:48Z","content_type":"text/html","content_length":"64006","record_id":"<urn:uuid:0c8451e1-4c9e-4eb2-a422-6e2f36ee74d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00869.warc.gz"}
Re: BSR CM type 1 arrows, StMaryRd, and RSFS [Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index] Re: BSR CM type 1 arrows, StMaryRd, and RSFS • To: Ulrik Vieth <vieth@thphy.uni-duesseldorf.de>, s.rahtz@elsevier.co.uk • Subject: Re: BSR CM type 1 arrows, StMaryRd, and RSFS • From: "Berthold K.P. Horn" <bkph@ai.mit.edu> • Date: Tue, 10 Mar 1998 09:59:52 -0500 • Cc: support@YandY.com, lcs@topo.math.u-psud.fr, rasmith@arete.com, tex-fonts@math.utah.edu At 02:29 PM 3/10/98 +0100, Ulrik Vieth wrote: >> i don't know, actually. I dont think any of our typesetters use CM >> or MathTime, except maybe those few that use TeX. >In that case I really wonder what else there is left you are you >using? I mean the whole work of the Math Font Group (*) was based >on the assumption that the choice of math fonts sets usable with >TeX was limitied to a handful of families such as CM, Concrete, >Euler, Adobe Symbol, MathTime, Lucida New Math, and Mathematica. I think you are assuming that people use TeX. Many of the big publishers do not. You can tell if you look at electronic journals. Many of them use Adobe Universal Greek + Pi and fonts like that. Regards, Berthold. Berthold K.P. Horn MIT AI Laboratory mailto: bkph@ai.mit.edu	{"url":"http://tug.tug.org/twg/mfg/mail-html/1998-01/msg00162.html","timestamp":"2024-11-05T18:50:47Z","content_type":"text/html","content_length":"2699","record_id":"<urn:uuid:51cf3a8d-faa0-45d5-8365-b45b5d806790>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00503.warc.gz"}
Carl Friedrich Gauss \| Biography, Discoveries, & Facts \| Britannica While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. External Websites Britannica Websites Articles from Britannica Encyclopedias for elementary and high school students. Quick Facts Original name: Johann Friedrich Carl Gauss Top Questions Why is Carl Friedrich Gauss famous? What was Carl Friedrich Gauss’s childhood like? What awards did Carl Friedrich Gauss win? How was Carl Friedrich Gauss influential? Carl Friedrich Gauss (born April 30, 1777, Brunswick [Germany]—died February 23, 1855, Göttingen, Hanover) was a German mathematician, generally regarded as one of the greatest mathematicians of all time for his contributions to number theory, geometry, probability theory, geodesy, planetary astronomy, the theory of functions, and potential theory (including electromagnetism). Gauss was the only child of poor parents. He was rare among mathematicians in that he was a calculating prodigy, and he retained the ability to do elaborate calculations in his head most of his life. Impressed by this ability and by his gift for languages, his teachers and his devoted mother recommended him to the duke of Brunswick in 1791, who granted him financial assistance to continue his education locally and then to study mathematics at the University of Göttingen from 1795 to 1798. Gauss’s pioneering work gradually established him as the era’s preeminent mathematician, first in the German-speaking world and then farther afield, although he remained a remote and aloof figure. Gauss’s first significant discovery, in 1792, was that a regular polygon of 17 sides can be constructed by ruler and compass alone. Its significance lies not in the result but in the proof, which rested on a profound analysis of the factorization of polynomial equations and opened the door to later ideas of Galois theory. His doctoral thesis of 1797 gave a proof of the fundamental theorem of algebra: every polynomial equation with real or complex coefficients has as many roots (solutions) as its degree (the highest power of the variable). Gauss’s proof, though not wholly convincing, was remarkable for its critique of earlier attempts. Gauss later gave three more proofs of this major result, the last on the 50th anniversary of the first, which shows the importance he attached to the Britannica Quiz Faces of Science Gauss’s recognition as a truly remarkable talent, though, resulted from two major publications in 1801. Foremost was his publication of the first systematic textbook on algebraic number theory, Disquisitiones Arithmeticae. This book begins with the first account of modular arithmetic, gives a thorough account of the solutions of quadratic polynomials in two variables in integers, and ends with the theory of factorization mentioned above. This choice of topics and its natural generalizations set the agenda in number theory for much of the 19th century, and Gauss’s continuing interest in the subject spurred much research, especially in German universities. The second publication was his rediscovery of the asteroid Ceres. Its original discovery, by the Italian astronomer Giuseppe Piazzi in 1800, had caused a sensation, but it vanished behind the Sun before enough observations could be taken to calculate its orbit with sufficient accuracy to know where it would reappear. Many astronomers competed for the honour of finding it again, but Gauss won. His success rested on a novel method for dealing with errors in observations, today called the method of least squares. Thereafter Gauss worked for many years as an astronomer and published a major work on the computation of orbits—the numerical side of such work was much less onerous for him than for most people. As an intensely loyal subject of the duke of Brunswick and, after 1807 when he returned to Göttingen as an astronomer, of the duke of Hanover, Gauss felt that the work was socially valuable. Similar motives led Gauss to accept the challenge of surveying the territory of Hanover, and he was often out in the field in charge of the observations. The project, which lasted from 1818 to 1832, encountered numerous difficulties, but it led to a number of advancements. One was Gauss’s invention of the heliotrope (an instrument that reflects the Sun’s rays in a focused beam that can be observed from several miles away), which improved the accuracy of the observations. Another was his discovery of a way of formulating the concept of the curvature of a surface. Gauss showed that there is an intrinsic measure of curvature that is not altered if the surface is bent without being stretched. For example, a circular cylinder and a flat sheet of paper have the same intrinsic curvature, which is why exact copies of figures on the cylinder can be made on the paper (as, for example, in printing). But a sphere and a plane have different curvatures, which is why no completely accurate flat map of the Earth can be made. Gauss published works on number theory, the mathematical theory of map construction, and many other subjects. In the 1830s he became interested in terrestrial magnetism and participated in the first worldwide survey of the Earth’s magnetic field (to measure it, he invented the magnetometer). With his Göttingen colleague, the physicist Wilhelm Weber, he made the first electric telegraph, but a certain parochialism prevented him from pursuing the invention energetically. Instead, he drew important mathematical consequences from this work for what is today called potential theory, an important branch of mathematical physics arising in the study of electromagnetism and gravitation. Gauss also wrote on cartography, the theory of map projections. For his study of angle-preserving maps, he was awarded the prize of the Danish Academy of Sciences in 1823. This work came close to suggesting that complex functions of a complex variable are generally angle-preserving, but Gauss stopped short of making that fundamental insight explicit, leaving it for Bernhard Riemann, who had a deep appreciation of Gauss’s work. Gauss also had other unpublished insights into the nature of complex functions and their integrals, some of which he divulged to friends. In fact, Gauss often withheld publication of his discoveries. As a student at Göttingen, he began to doubt the a priori truth of Euclidean geometry and suspected that its truth might be empirical. For this to be the case, there must exist an alternative geometric description of space. Rather than publish such a description, Gauss confined himself to criticizing various a priori defenses of Euclidean geometry. It would seem that he was gradually convinced that there exists a logical alternative to Euclidean geometry. However, when the Hungarian János Bolyai and the Russian Nikolay Lobachevsky published their accounts of a new, non-Euclidean geometry about 1830, Gauss failed to give a coherent account of his own ideas. It is possible to draw these ideas together into an impressive whole, in which his concept of intrinsic curvature plays a central role, but Gauss never did this. Some have attributed this failure to his innate conservatism, others to his incessant inventiveness that always drew him on to the next new idea, still others to his failure to find a central idea that would govern geometry once Euclidean geometry was no longer unique. All these explanations have some merit, though none has enough to be the whole explanation. Another topic on which Gauss largely concealed his ideas from his contemporaries was elliptic functions. He published an account in 1812 of an interesting infinite series, and he wrote but did not publish an account of the differential equation that the infinite series satisfies. He showed that the series, called the hypergeometric series, can be used to define many familiar and many new functions. But by then he knew how to use the differential equation to produce a very general theory of elliptic functions and to free the theory entirely from its origins in the theory of elliptic integrals. This was a major breakthrough, because, as Gauss had discovered in the 1790s, the theory of elliptic functions naturally treats them as complex-valued functions of a complex variable, but the contemporary theory of complex integrals was utterly inadequate for the task. When some of this theory was published by the Norwegian Niels Abel and the German Carl Jacobi about 1830, Gauss commented to a friend that Abel had come one-third of the way. This was accurate, but it is a sad measure of Gauss’s personality in that he still withheld publication. Gauss delivered less than he might have in a variety of other ways also. The University of Göttingen was small, and he did not seek to enlarge it or to bring in extra students. Toward the end of his life, mathematicians of the calibre of Richard Dedekind and Riemann passed through Göttingen, and he was helpful, but contemporaries compared his writing style to thin gruel: it is clear and sets high standards for rigour, but it lacks motivation and can be slow and wearing to follow. He corresponded with many, but not all, of the people rash enough to write to him, but he did little to support them in public. A rare exception was when Lobachevsky was attacked by other Russians for his ideas on non-Euclidean geometry. Gauss taught himself enough Russian to follow the controversy and proposed Lobachevsky for the Göttingen Academy of Sciences. In contrast, Gauss wrote a letter to Bolyai telling him that he had already discovered everything that Bolyai had just published. After Gauss’s death in 1855, the discovery of so many novel ideas among his unpublished papers extended his influence well into the remainder of the century. Acceptance of non-Euclidean geometry had not come with the original work of Bolyai and Lobachevsky, but it came instead with the almost simultaneous publication of Riemann’s general ideas about geometry, the Italian Eugenio Beltrami’s explicit and rigorous account of it, and Gauss’s private notes and correspondence. Jeremy John Gray	{"url":"https://www.britannica.com:443/biography/Carl-Friedrich-Gauss","timestamp":"2024-11-03T23:43:26Z","content_type":"text/html","content_length":"128425","record_id":"<urn:uuid:871186ea-3418-4f48-97cd-2288499c213e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00220.warc.gz"}
Basic Equations and Applications of Single Phase Transformer EMF Equation of Transformer: Let the applied voltage V1 applied to the primary of a transformer, with secondary open-circuited, be sinusoidal (or sine wave). Then the current I1, due to applied voltage V1, will also be a sine wave. The mmf N1 I1 and core flux Ø will follow the variations of I1 closely. That is the flux is in time phase with the current I1 and varies sinusoidally. N[A] = Number of turns in primary N[B] = Number of turns in secondary Ø[max] = Maximum flux in the core in webers = B[max] X A f = Frequency of alternating current input in hertz (H[Z]) As shown in figure above, the core flux increases from its zero value to maximum value Ø[max] in one quarter of the cycle , that is in ¼ frequency second. Therefore, average rate of change of flux = Ø[max]/ ¼ f = 4f Ø[max]Wb/s Now, rate of change of flux per turn means induced electro motive force in volts. Therefore, average electro-motive force induced/turn = 4f Ø[max]volt If flux Ø varies sinusoidally, then r.m.s value of induced e.m.f is obtained by multiplying the average value with form factor. Form Factor = r.m.s. value/average value = 1.11 Therefore, r.m.s value of e.m.f/turn = 1.11 X 4f Ø[max] = 4.44f Ø[max] Now, r.m.s value of induced e.m.f in the whole of primary winding = (induced e.m.f./turn) X Number of primary turns EA = 4.44f NAØmax = 4.44fNABmA Similarly, r.m.s value of induced e.m.f in secondary is EB = 4.44f NB Ømax = 4.44fNBBmA In an ideal transformer on no load, VA = EA and VB = EB , where VB is the terminal voltage Voltage Transformation Ratio. The ratio of secondary voltage to primary voltage is known as the voltage transformation ratio and is designated by letter K. i.e. Voltage transformation ratio, K = V2/V1 = E2/E1 = N2/N1 Current Ratio. The ratio of secondary current to primary current is known as current ratio and is reciprocal of voltage transformation ratio in an ideal transformer. Transformer on No Load. When the primary of a transformer is connected to the source of an ac supply and the secondary is open circuited, the transformer is said to be on no load. The Transformer on No Load alternating applied voltage will cause flow of an alternating current I0 in the primary winding, which will create alternating flux Ø. No-load current I0, also known as excitation or exciting current, has two components the magnetizing component Im and the energy component Ie. Im is used to create the flux in the core and Ie is used to overcome the hysteresis and eddy current losses occurring in the core in addition to small amount of copper losses occurring in the primary only (no copper loss occurs in the secondary, because it carries no current, being open circuited.) From vector diagram shown in above it is obvious that 1. Induced emfs in primary and secondary windings, E1 and E2 lag the main flux Ø by and are in phase with each other. 2. Applied voltage to primary V1 and leads the main flux Ø by and is in phase opposition to E1. 3. Secondary voltage V2 is in phase and equal to E2 since there is no voltage drop in secondary. 4. Im is in phase with Ø and so lags V1 by 5. Ie is in phase with the applied voltage V1. 6. Input power on no load = V1Ie = V1I0 cos Ø0 where Ø0 = tan^-1 Transformer on Load: The transformer is said to be loaded, when its secondary circuit is completed through an impedance or load. The magnitude and phase of secondary current (i.e. current flowing through secondary) I2 with respect to secondary terminals depends upon the characteristic of the load i.e. current I2 will be in phase, lag behind and lead the terminal voltage V+2+ respectively when the load is non-inductive, inductive and capacitive. The net flux passing through the core remains almost constant from no-load to full load irrespective of load conditions and so core losses remain almost constant from no-load to full load. Vector diagram for an ideal transformer supplying inductive load is shown Resistance and Leakage Reactance In actual practice, both of the primary and secondary windings have got some ohmic resistance causing voltage drops and copper losses in the windings. In actual practice, the total flux created does not link both of the primary and secondary windings but is divided into three components namely the main or mutual flux Ø linking both of the primary and secondary windings, primary leakage flux ØL[1] linking with primary winding only and secondary leakage flux ØL[2] linking with secondary winding only. The primary leakage flux ØL[1] is produced by primary ampere-turns and is proportional to primary current, number of primary turns being fixed. The primary leakage flux ØL[1] is in phase with I[1] and produces self induced emf ØL[1] is in phase with I[1] and produces self induced emf EL[1] given as 2f L[1] I[1] in the primary winding. The self induced emf divided by the primary current gives the reactance of primary and is denoted by X1. i.e. X1 = EL1/I1 = 2πfL1I1/I1 = 2FL1, Similarly leakage reactance of secondary X2 = EL2/E2 = 2fπL2I2/I2 = 2πfL2 Equivalent Resistance and Reactance. The equivalent resistances and reactance’s of transformer windings referred to primary and secondary sides are given as below Referred to primary side Equivalent Equivalent resistance, = X'1 = Referred to secondary side Equivalent resistance, Equivalent resistance, = X2 + K^2X1 Where K is the transformation ratio. Equivalent impedance of transformer is essential to be calculated because the electrical power transformer is an electrical power system equipment for estimating different parameters of electrical power system which may be required to calculate total internal impedance of an electrical power transformer, viewing from primary side or secondary side as per requirement. This calculation requires equivalent circuit of transformer referred to primary or equivalent circuit of transformer referred to secondary sides respectively. Percentage impedance is also very essential parameter of transformer. Special attention is to be given to this parameter during installing a transformer in an existing electrical power system. Percentage impedance of different power transformers should be properly matched during parallel operation of power transformers. The percentage impedance can be derived from equivalent impedance of transformer so, it can be said that equivalent circuit of transformer is also required during calculation of % impedance. Equivalent Circuit of Transformer Referred to Primary For drawing equivalent circuit of transformer referred to primary, first we have to establish general equivalent circuit of transformer then, we will modify it for referring from primary side. For doing this, first we need to recall the complete vector diagram of a transformer which is shown in the figure below. Let us consider the transformation ratio be, In the figure right, the applied voltage to the primary is V[1] and voltage across the primary winding is E[1]. Total current supplied to primary is I[1]. So the voltage V[1] applied to the primary is partly dropped by I[1]Z[1] or I[1]R[1] + j.I[1]X[1] before it appears across primary winding. The voltage appeared across winding is countered by primary induced emf E[1]. The equivalent circuit for that equation can be drawn as below, From the vector diagram above, it is found that the total primary current I[1] has two components, one is no - load component I[o] and the other is load component I[2]′. As this primary current has two a component or branches, so there must be a parallel path with primary winding of transformer. This parallel path of current is known as excitation branch of equivalent circuit of transformer. The resistive and reactive branches of the excitation circuit can be represented as The load component I[2]′ flows through the primary winding of transformer and induced voltage across the winding is E[1] as shown in the figure right. This induced voltage E[1]transforms to secondary and it is E[2] and load component of primary current I[2]′ is transformed to secondary as secondary current I[2]. Current of secondary is I [2]. So the voltage E[2] across secondary winding is partly dropped by I[2]Z[2] or I[2]R[2] + j.I[2]X[2] before it appears across load. The load voltage is V[2]. From above equation, secondary impedance of transformer referred to primary is, So, the complete equivalent circuit of transformer referred to primary is shown in the figure below, Approximate Equivalent Circuit of Transformer Since Io is very small compared to I[1], it is less than 5% of full load primary current, Iochanges the voltage drop insignificantly. Hence, it is good approximation to ignore the excitation circuit in approximate equivalent circuit of transformer. The winding resistanceand reactance being in series can now be combined into equivalent resistance and reactance of transformer, referred to any particular side. In this case it is side 1 or primary side. Equivalent Circuit of Transformer Referred to Secondary In similar way, approximate equivalent circuit of transformer referred to secondary can be drawn. Where equivalent impedance of transformer referred to secondary, can be derived as The voltage regulation is the percentage of voltage difference between no load and full load voltages of a transformer with respect to its full load voltage. Explanation of Voltage Regulation of Transformer Say an electrical power transformer is open circuited, means load is not connected with secondary terminals. In this situation, the secondary terminalvoltage of the transformer will be its secondary induced emf E[2]. Whenever full load is connected to the secondary terminals of the transformer, ratedcurrent I[2] flows through the secondary circuit and voltage drop comes into picture. At this situation, primary winding will also draw equivalent full load current from source. The voltagedrop in the secondary is I[2]Z[2] where Z[2] is the secondary impedance of transformer. Now if at this loading condition, any one measures the voltage between secondary terminals, he or she will getvoltage V[2] across load terminals which is obviously less than no load secondary voltage E[2] and this is because of I[2]Z[2] voltage drop in the transformer. Expression of Voltage Regulation of Transformer, represented in percentage, is	{"url":"https://www.brainkart.com/article/Basic-Equations-and-Applications-of-Single-Phase-Transformer_6665/","timestamp":"2024-11-13T04:32:20Z","content_type":"text/html","content_length":"79623","record_id":"<urn:uuid:b234ed4d-fe5d-4cd5-827d-7d04064c0c74>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00488.warc.gz"}
TANA15 Numerical Linear Algebra Advancement Level: Course Aims The course is intended to provide basic knowledge about important matrix decompositions; such as the LU or SVD decompositions, and show how matrix decompositions can be used for analyzing and solving both practical and theoretical problems. The course also covers various important techniques from Linear Algebra, such as the Shur complement, convolutions, polynomial manipulation, or orthogonal basis generation. Both linear, and non-linear, least squares problems are also discussed in the course. After the course students should be able to: • Discuss the most common matrix factorizations, and explain their properties. • Understand how the most common matrix factorizations are computed; and implement numerical algorithms for computing the most important factorizations. • Use matrix factorizations for solving both theoretical problems and practical problems from applications. • Discuss the usage of Linear Algebra techniques when solving important application problems, such as pattern recognition, data compression, signal processing, search engines, or model fitting. The Lectures give an overview of the theory. During the Problem seminars concrete examples, intended to illustrate the theory. are given. The Computer exercises give practical experience with implementing and using the methods. Course Content Basic operations of linear Algebra (BLAS). The LU decomposition: Solution of linear systems. Condition Number. Error estimate. The QR Decomposition: Reflection- and Rotation matrices. Least squares problems. Row Updating. The eigenvalue decomposition. Hessenberg factorization. The Power method. The QR Algoritm. Special matrices. The FFT. Functions of Matrices. The Singular value decomposition and applications. Bidiagonalization. Systems of non-linear equations: Gradient based methods. The Newton and Gauss-Newton method. Non-linear least squares problems. Written Exam (U,3,4,5) 4 hp Computer Exercices (U,G) 2 hp Page responsible: Fredrik Berntsson Last updated: 2019-11-29	{"url":"https://courses.mai.liu.se/GU/TANA15/information.html","timestamp":"2024-11-08T05:34:59Z","content_type":"text/html","content_length":"12036","record_id":"<urn:uuid:37ba4408-7de5-40c5-92b7-ac5c82f4efea>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00555.warc.gz"}
How to Calculate Coupon Rate in Excel (3 Ideal Examples) - ExcelDemy What Is a Coupon Rate? The coupon rate is the rate of interest that is paid on the bond’s face value by the issuer. The coupon rate is calculated by dividing the Annual Interest Rate by the Face Value of the Bond. The result is then expressed as a percentage. Coupon Rate=(Annual Interest Rate/Face Value of Bond)100 3 Examples to Calculate the Coupon Rate in Excel We will use a dataset that contains the Face Value and Interest. We will use different frequencies of payments to calculate the coupon rate. Example 1 – Determine a Coupon Rate in Excel with a Half-Yearly Interest • Select Cell D8 and use the formula below: • Press Enter to see the result. • Select Cell D10 and enter the formula below: • Hit Enter to see the Coupon Rate. In our case, the coupon rate is 2%. Read More: How to Calculate Price of a Semi Annual Coupon Bond in Excel Example 2 – Calculate the Coupon Rate with Monthly Interest in Excel • Change the Face Value of the Bond in Cell B5. • Write 12 in Cell D5. • Select Cell D8 and type the formula: • Hit Enter to see the Annual Interest Payment. • Enter the following formula in Cell D10: • Press Enter to see the coupon rate with monthly interest. Read More: How to Calculate Face Value of a Bond in Excel Example 3 – Coupon Rate Calculation in Excel with a Yearly Interest • Select Cell D5 and type 1. • Enter the annual interest payment formula in Cell D8 and click on the Enter key. • Select Cell D10 and type the formula below: • Pess Enter to see the desired result. Determine the Coupon Bond in Excel A coupon bond generally refers to the price of the bond. To calculate the coupon bond, use the formula below. Coupon Bond = C[1–(1+Y/n)^-nt/Y]+[F/(1+Y/n)nt] C = Annual Coupon Payment Y = Yield to Maturity F = Par Value at Maturity t = Number of Years Until Maturity n = Number of Payments/Year We used the Coupon Rate to evaluate the value of the Annual Coupon Payment (C). • Select Cell C10 and insert the formula: • Press Enter to see the result of C. • Select Cell C12 and enter the formula below: • Hit Enter to see the result. • C10 is the value of Annual Coupon Payment (C). • ((1-(1+(C6/C7))^-(C7C8))/C6) is the value of C[1–(1+Y/n)^^-nt/Y]. • (C5/(1+(C6/C7))^(C7C8)) is the value of [F/(1+Y/n)^n*t]. Read More: How to Calculate Bond Payments in Excel Calculate the Coupon Bond Price in Excel • For a half-year coupon bond, select Cell C11 and use the formula below: • Hit Enter to see the result. • For the price of a yearly coupon bond, select Cell C10 and insert the formula: • Press Enter to see the result. • To calculate the price of a zero-coupon bond, use the below formula in Cell C9. • And hit Enter to see the result. Download the Practice Book Related Articles << Go Back to Bond Price Formula Excel\|Excel Formulas for Finance\|Excel for Finance\|Learn Excel Get FREE Advanced Excel Exercises with Solutions! We will be happy to hear your thoughts Leave a reply	{"url":"https://www.exceldemy.com/calculate-coupon-rate-in-excel/","timestamp":"2024-11-13T15:41:06Z","content_type":"text/html","content_length":"195975","record_id":"<urn:uuid:9dad784c-d0d9-4180-9343-d8d78ad301c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00252.warc.gz"}
Nature Of Roots Of Quadratic Equation We already know what a quadratic equation is, let us now focus on nature of roots of quadratic equation. A polynomial equation whose degree is 2, is known as quadratic equation. A quadratic equation in its standard form is represented as: $ax^2 + bx + c$ = $0$, where $a,~b ~and~ c$ are real numbers such that $a ≠ 0$ and $x$ is a variable. The number of roots of a polynomial equation is equal to its degree. So, a quadratic equation has two roots. Some methods for finding the roots are: • Factorization method • Quadratic Formula • Completing the square method All the quadratic equations with real roots can be factorized. The physical significance of the roots is that at the roots of an equation, the graph of the equation intersects x-axis. The x-axis represents the real line in the Cartesian plane. This means that if the equation has unreal roots, it won’t intersect x-axis and hence it cannot be written in factorized form. Let us now go ahead and learn how to determine whether a quadratic equation will have real roots or not. Nature of Roots of a Quadratic Equation: Before going ahead, there is a terminology that must be understood. Consider the equation $ax^2 + bx + c$ = $0$ For the above equation, the roots are given by the quadratic formula as $x$ = $\frac{-b ± √({b^2 – 4ac})}{2a}$ Let us take a real number $k > 0$. Now, we know that $√k$ is defined and is a positive quantity. Is $√{-k}$ a real number? The answer is no. For e.g. if we have $√225$, we can write it as $√({15~×~15})$ which is equal to $15$. If we have $√{-225}$, we can never write it as a product of two equal quantities. This is because it contains a minus sign which can only result from product of two quantities having opposite The quantity which is under square root in the expression for roots is $b^2 – 4ac$. This quantity is called discriminant of the quadratic equation. This is the quantity that discriminates the quadratic equations having different nature of roots. This is represented by D. So, D = $b^2 – 4ac$ In terms of D, the roots can be written as: $x$ = $\frac{-b ± √{D}}{2a}$ ——————————— (1) Now, D is a real number since a, b and c are real numbers. Depending upon a, band c, the value of D can either be positive, negative or zero. Let us analyze all the possibilities and see how it affects the roots of the equation. • D>0: When D is positive, the equation will have two real and distinct roots. This means the graph of the equation will intersect x-axis at exactly two different points. The roots are: $x$ = $\frac{-b + √{D}}{2a} ~or~\frac{-b -√{D}}{2a}$ • D = 0: When D is equal to zero, the equation will have two real and equal roots. This means the graph of the equation will intersect x-axis at exactly one point. The roots can be easily determined from the equation 1 by putting D=0. The roots are: $x$ = $-\frac{b}{2a}~ or~-\frac{b}{2a}$ • D < 0: When D is negative, the equation will have no real roots. This means the graph of the equation will not intersect x-axis. Let us take some examples for better understanding. Video Lesson Nature of Roots Solved Examples Example 1: $x^2 + 5x + 6$ = $0$ D = $b^2 – 4ac$ D = $5^2~-~4~×~1~×~6$ = $25 – 24$ = $1$ Since D>0, the equation will have two real and distinct roots. The roots are: $x$ = $\frac{-b + √D}{2a}$ or $\frac{-b – √D}{2a}$ $x$ = $\frac{-5 + √1}{2 × 1}$ or $\frac{-5 – √1}{2~×~1}$ $x$ = $-2$ or $-3$ Example 2: $x^2 + x + 1$ = $0$ D = $b^2 – 4ac$ D = $1^2 – 4~×~1~×~1$ = $1~ -~ 4$ = $-3$ Since D<0, the equation will have two distinct Complex roots. The roots are: $x$ = $\frac{-b + √D}{2a}$ or $\frac{-b – √D}{2a}$ $x$= $\frac{-1 + \sqrt{-3}}{2 \times 1}$ or $\frac{-1 – \sqrt{-3}}{2 \times 1}$ $x$= $\frac{-1 + \sqrt{3}i}{2}$ or $\frac{-1 – \sqrt{3}i}{2}$ Example 3: $4x^2 + 12x + 9$ = $0$ D = $b^2 – 4ac$ D = $12^2 – 4~×~4~×~9$ = $144~-~144$ = $0$ Since D = 0, the equation will have two real and equal roots. The roots are: $x = +\frac{b}{2a}$ or $-\frac{b}{2a}$ $x = +\frac{12}{2~×~4}$ or $-\frac{12}{2~×~4}$ $x$ = $+\frac{3}{2}$ or $-\frac{3}{2}$ To solve more problems on the topic, download BYJU’S – The Learning App from Google Play Store and watch interactive videos. Also, take free tests to practice for exams.	{"url":"https://mathlake.com/Nature-Of-Roots-Of-Quadratic-Equation","timestamp":"2024-11-06T23:40:27Z","content_type":"text/html","content_length":"21094","record_id":"<urn:uuid:f450c751-cedb-4d82-8c99-5681cffb13f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00854.warc.gz"}
Cornet Seminar – Felipe Albuquerque – 13/01/2023 In the context of team Cornet’s seminars, Felipe Albuquerque (LIA/Espace) will present his research work on January 13, 2023, at 11:35 in the meeting room. Abstract: Frequently, social network information has been used to solve applications in Operation Research, such as the Team Formation Problem, whose goal is to find a subset of the workers that collectively cover a set of skills and can communicate effectively with each other. We use the Structural Balance Theory to define the compatibility between pairs of workers in the same team. For such, the social networks are represented by signed graphs, and the compatibility metric is calculated from the analysis of possible positive paths between pairs of distinct vertices. To solve this new version of the problem, we introduce an Integer Linear Programming formulation and a decomposition for it. We present an analysis of the performed computational tests that prove the potential efficiency of the decomposition proposed.	{"url":"http://lia.univ-avignon.fr/2023/01/13/cornet-seminar-felipe-albuquerque-13-01-2023/","timestamp":"2024-11-06T10:38:00Z","content_type":"text/html","content_length":"38216","record_id":"<urn:uuid:e2e023de-f1e1-4357-9ac0-bcfeb61612a4>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00082.warc.gz"}
Question related to Practice Exam – Q&A Hub – 365 Data Science Question related to Practice Exam Introduction to Excel/Practice Exam 3/Question 5 In the above question , i have tried a lot of times to get the exact answer. But everytime ,it gets wrong. Solution given to that answer is also not understandable. So,can anyone help me regarding this? If possible, give answer with video explanation. 3 answers ( 0 marked as helpful) Hi Shashank, Did you delete the 'Total' rows first, please? When you do that, you simply need to take the sum of the first 10 rows in the table. Please take a look and I'll be happy to assist if you need more guidance. Yes, I have done that too but my answer is not same as the answer given in the question solution. Can you check it in your computer? It will help me for the solution of this problem. my Answer:- -1,69,93,326 Answer given in the question solution:- -8,777,343.636 I now understand the reason for this mistake. Please bear in mind that you're asked to sum the first 10 rows of the output sheet. Thank you!	{"url":"https://365datascience.com/question/question-related-to-practice-exam/","timestamp":"2024-11-09T00:31:59Z","content_type":"text/html","content_length":"114143","record_id":"<urn:uuid:03415cdc-b199-468d-bb14-05e18d2afcca>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00608.warc.gz"}
2024-11-11T23:00:04Z https://fit.repo.nii.ac.jp/oai oai:fit.repo.nii.ac.jp:00000406 2023-05-15T14:02:49Z 193:194 Implementation of Intelligent and Hybrid Systems for Wireless Mesh Networks:A Comparison Study Implementation of Intelligent and Hybrid Systems for Wireless Mesh Networks:A Comparison Study 無線メッシュネットワークのための知的およびハイブリッドシステムの実装：比較研究坂本, 真仁 open access 無線メッシュネットワーク知的アルゴリズムメッシュルータ配置最適化 Wireless Mesh Networks Intelligent Algorithms Node Placement Problem application/pdf 学位論文(Thesis) Wireless Mesh Networks (WMNs) are gaining a lot of attention because of their low cost nature that makes them attractive for providing wireless Internet connectivity. A WMN is dynamically self-organized and self-configured, with the nodes in the network automatically establishing and maintaining mesh connectivity among themselves. In WMNs, the mesh node placement is a very important problem. However, this problem is known to be NP-hard. To deal with this problem, new methods, algorithms and systems are needed. In this thesis, we design and implement intelligent and hybrid systems in order to solve the node placement problem in WMNs. We consider a bi-objective optimization in which we first maximize the network connectivity through the maximization of Size of Giant Component (SGC) and then the maximization of the Number of Covered Mesh Clients (NCMC). We evaluate the implemented systems by many simulations. From the evaluation results, we found that the hybrid systems have very good performance for optimizing the node placement in WMNs. This thesis contributes in the research field as following: 1) Implementation of intelligent systems for solving node placement problem in WMNs. 2) Evaluation of various intelligent algorithms based systems for different scenarios. 3) Comparison of implemented intelligent and hybrid systems. 4) Implementation of WMN simulation system using Network Simulator 3. 5) Application of implemented system for WMN node placement problem in a realistic scenario. 6) Give insights about future developments and integration of WMNs as an important technology in wireless communications. This thesis is constructed by 8 Chapters. Chapter 1 presents the background, the motivation and thesis structure. Chapter 2 introduces general aspects of wireless networks. Also, Wireless Sensor and Actor Networks (WSANs) and Mobile Ad-hoc Networks (MANETs) are explained as a related work to this thesis. In Chapter 3, we explain about the node classification in WMNs and routing protocols for WMNs. In addition, we define the Node Placement Problem in WMNs. In Chapter 4 are discussed intelligent algorithms such as Hill Climbing (HC), Simulated Annealing (SA), Tabu Search (TS), Genetic Algorithm (GA), Ant Colony Optimization (ACO) and Particle Swarm Optimization (PSO). We present in details the PSO algorithm in Chapter 5. The implemented intelligent and hybrid systems are presented in Chapter 6. Chapter 7 shows the evaluation and comparison of implemented systems by conducting simulations and application for a realistic scenario. In Chapter 8, we give some concluding remarks and future work. 福岡工業大学 2018-03-01 eng doctoral thesis VoR http://hdl.handle.net/11478/885 https://fit.repo.nii.ac.jp/records/406 甲第49 号博士（工学） 2018-03-20 37112 福岡工業大学 https://fit.repo.nii.ac.jp/record/406/files/DC_Ko_k_49.pdf application/pdf 12.1 MB 2020-09-14	{"url":"https://fit.repo.nii.ac.jp/oai?verb=GetRecord&metadataPrefix=jpcoar_1.0&identifier=oai:fit.repo.nii.ac.jp:00000406","timestamp":"2024-11-11T23:00:06Z","content_type":"application/xml","content_length":"8430","record_id":"<urn:uuid:3a2d774f-ee9d-4909-8480-9104dca59498>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00002.warc.gz"}
On Soliton Solutions of the Drinfeld-Sokolov- Wilson System by He’s Variational Principle The aim of this paper is to obtain the traveling wave solutions for the Drinfeld-Sokolov-Wilson system by He’s semi-inverse variational principle which includes the solitary and periodic wave solutions by a suitable choice for the parameters. We analysis these solutions physically by some gores to complement this study. Finally, this method can be used successfully for solving integrable and nonintegrable equations. Keywords: He’s semi-inverse variational principle; Traveling wave solutions; Dronfield-Sokolov-Wilson system; Soliton solutions; Periodic wave solutions Nonlinear partial differential equations (NLPDEs) is used to describe many important phenomena in mathematical physics, mechanics, chemistry, biology, etc. Such as Cortège de Vries equation, Burgers equation, Schrodinger equation, Bossiness equation and so on. So, the discovery of the exact solutions of nonlinear partial differential equations is of the most important priorities. Many effective methods are used to construct traveling wave solutions of NLPDEs, among of these methods, Adomian decomposition method [1], the homotropy perturbation method [2], the variational iteration method [3, 4], the He’s variational approach [5], the extended homoclinic test approach [6, 7], homogeneous balance method [8-11], Jacobi elliptic function method [12-15], Baclund transformation [16, 17], G0=G expansion method [18] . He’s semi-inverse variational principle is used to obtain the traveling wave solutions for Dronfield-Sokolov-Wilson system which include the solitary and periodic wave solutions by a suitable choice for the parameters. It is important to point out that a new constrained variational principle for heat conduction is obtained recently via the semi-inverse method combined with separation of variables[19], which is an exactly the same with He-Lee’s variational principle[20] a short remark on the history of the semi-inverse method for establishment of a generalized variational principle is given in [21]. In soliton theory, we aim to search for the solitary wave solutions for NLPDEs using various methods [22]. In particular, we used in this paper the variational principle, which enables us to and the Lagrangian for the Drinfeld-Sokolov-Wilson system, which related to the conservation laws which plays an important role in solution process [23,24] and provides physical insight into the nature of the solution of this problem as shown by gures. Also, this method helps in establishing connections between the physics and mathematics and much more active than the Noether’s theorem [25, 26]. The key idea in this paper is to use the He’s semi-inverse variational principle to and out several exact solutions for Drinfeld-Sokolov-Wilson system. Conclusions is given in the last of this paper. Suppose we are given a nonlinear partial differential equations (NLPDEs) in the following form where x and t are the independent variables. This method can be summarizing as follows A. By using the wave solutions we can transform Eq. (1) into an ordinary differential equation (ODE) N(U; cU ; U ; cU ; :::) = 0 B. Integrating Eq. (2) and setting the integration constants equal to zero for simplicity. C. Construct the Lagrangian in the following form according to the He’s semi-inverse method, where L is a Lagrangian for the Eq. (2) D. By a Ritz method [27], one can obtain a different forms of solitary wave solutions, such as U( )= Asech (B); U( ) = Acsch (B); U( ) = A tanh (B); ss and so on. In this paper, we concentrate to obtain a solitary wave solution in the form U( )= Asech (B) (4) where A and B are constants to be determined. Substituting Eq. (4) into Eq. (3) and making I stationary with respect to A and B results in We can obtain A and B by solving Eqs. (5) and (6). Therefore, we can construct the solitary wave solution for Eq. (4) Soliton solutions to Drinfeld-Sokolov-Wilson system We aim in this section to obtain the soliton solutions for the Drinfeld-Sokolov-Wilson system where a; b and k are constants [28]. By using the transformation where = x ct + 0, c; 0 are constants. Substituting Eq. (8) into Eq. (7), we and that where the prime denotes to the derivative with respect to the variable a, b, c and k are constants. Integrating Eqs. (9), we obtain the relation between the variables V (), U () as follows Substituting V () in Eqs. (9), we obtain after integration and setting the constant of integration zero for simplicity According to [29], by He’s semi-inverse variational principle [3], we can obtain the following variational formulation According to the Ritz-like method, we search a solitary wave solution in the form V ( ) = Asech (B ) (13) Substituting Eq. (13) into Eq. (12), we have To and the constants A and B, we solve the following equations From Eqn. (15), we get Therefore, the solitary wave solutions for Drinfeld-Sokolov- Wilson system constructed as follows For a suitable choice for a parameter in Eq. (17), Figure 1 & 2 describes the shape of solitons for waves u (x; t) and v (x; t) respectively (Figure 1) Figure 1:The soliton solution of Eq. (17). Figure 2:The soliton solution of Eq. (17). For another constant A and B in Eq. (16) (Figure 2) We search another soliton solution in the form [30] where F and G are constants to be determined. Substituting Eq. (18) into Eq. (12), we obtain To and the constants F and G, we solve the following equations (19) From Eqn. (20), we get For a suitable choice for a parameter in Eq. (22), Figure 3 & 4 describes the shape of solitons for waves u (x; t), and v (x; t) respectively for another constant F and G in Eq. (21) Figure 3:The soliton solution of Eq. (22). Figure 4:The soliton solution of Eq. (22). Physical Discussions In this section, we discusses the physical explanation of the solutions obtained for the Drinfeld-Sokolov-Wilson system (7), namely (17), (22) which are soliton pattern solutions caused by the delicate balance between nonlinearity effect V3 and dispersion effect V1000 which have an infinite tails and keep their form and velocity after an entire interaction with others solitons. We display the time evolution of these solutions by some gores to describe the dynamical properties in 3D graphs in Figure 1-4 are the forth solutions (22). We note the wave speed c acts a critical role in a physical structure of the solutions for Drinfeld-Sokolov-Wilson system. The 3D graphs in Figure 1-4 show the shape of propagation of soliton solutions (17) and (22) for the Drinfeld-Sokolov-Wilson system (7). Result and Discussion He’s semi-inverse variational principle was used to obtain the traveling wave solutions for Drinfeld-Sokolov-Wilson system including new type of solitary wave solutions. He’s variational principle is a very dominant instrument to and the soliton solutions for various nonlinear equations in mathematical physics and may be important for the explanation of some practical physical problems. This method is a powerful mathematical tool for solving other nonlinear evolution equations arising in mathematical physics. This method makes the underlying idea clear and not darkened by the unnecessarily by complicated form of mathematical expression. Also, we can combine the well-known method such as Jacobi elliptic function method and exp-function method with He’s semi-inverse variational method, but they need a mathematical software, and this method is extreme simplicity and concise results for a wide range of NLPDEs compared with the Noether’s theorem. © 2019 Mohammed K Elboree. This is an open access article distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and build upon your work non-commercially.	{"url":"https://crimsonpublishers.com/eme/fulltext/EME.000543.php","timestamp":"2024-11-03T03:03:43Z","content_type":"text/html","content_length":"183537","record_id":"<urn:uuid:8fa0e1bf-6721-46db-91b2-111b3c4fdce5>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00710.warc.gz"}
Cartesian vector notation Vectors and their Operations: Cartesian vector notation Cartesian coordinate systems A convenient set of directions is a set of perpendicular directions called orthogonal axes. Orthogonal means perpendicular. The positive direction of an axis sets a benchmark to determine the positive (or negative) direction of a vector along (parallel with) the axis. For example, considering an axis x shown in Fig. 2.14a, x. Decomposing a vector in the two-dimensional plane needs two orthogonal axes (Fig. 2.14b). Decomposing a spatial vector (a vector in the three-dimensional space) needs three orthogonal axes (Fig. 2.14c). A set of orthogonal axes, intersecting at a point (the origin), is called a Cartesian coordinate system or Cartesian frame. Fig. 2.14 (a) Definition of an axis with its positive direction. Two vectors in the positive and negative directions of the axis are demonstrated, (b) a two-dimensional Cartesian coordinate system, (b) a three-dimensional Cartesian coordinate system. Unit vector A Unit vector. A unit vector is a vector of magnitude normalizing a vector. The unit vector of a vector has the same direction as the original vector. Experiment with the following interactive tool to investigate how a vector is a scalar multiplicand of its unit vector. Cartesian vector notation The components of a vector along orthogonal axes are called rectangular components or Cartesian components. A vector decomposed (resolved) into its rectangular components can be expressed by using two possible notations namely the scalar notation (scalar components) and the Cartesian vector notation. Both notations are explained for the two-dimensional (planar) conditions, and then extended to three dimensions in the following sections. Rectangular vector components of coplanar vectors Consider a vector scalar notation or scalar components are defined. Scalar components of a vector are signed magnitudes of its rectangular components. A scalar component is positive if the vector component is directed along the positive axis, and negative if the vector component is directed along the negative axis (opposite of the axis positive direction). For the vector components y axis. Fig. 2.15 Components of vectors An axis has an associated unit vector showing the positive direction of the axis. Any vector parallel with an axis can be written as a scalar multiplier of the axis’ unit vector. The scalar multiplier is equal to the scalar component (signed magnitude) of the vector parallel with the axis. The scalar multiplier is positive if the vector is in the direction of the axis unit vector, and negative otherwise. The unit vectors associated with Cartesian x and y axes are denoted by bold lower-case letters Fig. 2.16 Planar Cartesian axes and their associated unit vectors. The rectangular components of any vector As a general rule any vector Remark: using CVN is equivalent to resolving a vector in a Cartesian coordinate system. Remark: in CVN, capital letters (non-bold), such as, Remark: components of a vector are vectors, whereas the scalar components (vector coordinates) are scalar (i.e. signed magnitudes). Using the following interactive tool, you can observe the (vector) components and scalar components of a vector. Observe how the vector is represented in CVN. Change the direction of the vector by the angle slider and observe the values of the scalar components. When do the scalar components attain negative values? EXAMPLE 2.4.1 Determine the scalar components of Remark: the location of a vector with respect to the origin of the Cartesian coordinate system does not affect its Cartesian components. You can always consider (draw) parallel axes with the Cartesian axes at the tail of a vector and calculate its components. It is worthy to note the relationships among the scalar components, the magnitude, and the direction of a vector. Consider a Cartesian coordinate system in which angles are measured counterclockwise from the positive x axis as demonstrated in Fig. 2.17. Then the following equations hold, Fig. 2.17 The relationships among the scalar components, magnitude, and direction of a vector. To facilitate calculations, specifying the direction of a planar vector in the Cartesian frame can also use the (acute) angle that the vector makes with any of the Cartesian axes. If x and y axes respectively, the following equations hold, Note that the absolute values of the scalar components, which are the magnitudes of the vector Cartesian components, are used in the trigonometry functions. Fig. 2.18a demonstrates three cases of vector decomposition based on the defined angles Another possible way of specifying the direction of a planar vector is by a small slope triangle. The small slope triangle conveys the information on Fig. 2.18 The relationships among the scalar components, magnitude, and direction of a vector when the angles of a planar vector with the Cartesian axes are used. Remark: be cautious about the way the direction of a vector is specified and the proper formulation you should use for the calculations. Remark: using the angles Rectangular vector components of spatial vectors To treat a vector in three dimensions, a three-dimensional Cartesian coordinate system is to be defined. A common three-dimensional Cartesian coordinate system is a right-handed coordinate system. A coordinate system of three orthogonal axes is said to be right-handed if your right-hand thumb points in the positive x axis to the (positive) y axis (Fig. 2.19a). Another way to define a right-handed Cartesian coordinate system is to follow the right-hand three-finger rule as demonstrated Fig. 2.19b. In three dimensions, the unit vectors of the axes are denoted by Fig. 2.19 (a), (b) A right-handed Cartesian coordinate system, (c) unit vectors of a right-handed Cartesian coordinate system. Right-click and rotate the following three-dimensional right-handed Cartesian coordinate system to observe its different orientations. Any vector in three dimensions can be resolved into three rectangular components. A vector expressed in CVN (three dimensions) is written as, Fig. 2.20 Components of a vector in a three-dimensional Cartesian coordinate system. The direction of coordinate direction angles x, y, and z axes respectively to the vector (Fig. 2.20c). These angles are limited as The following interactive example illustrates the rectangular components of a vector in three dimensions. Change the scalar components (coordinates of the head point) by the sliders to observe how the vector is expressed in CVN. The coordinate direction angles of a vector are demonstrated by the following interactive example. The three coordinate direction angles are not independent and they are related by this equation, which means that knowing two of the angles, the third one is readily obtainable. To prove Eq. 2.6, a vector which, by Eq. 2.5, is written as: and leads to, which proves Eq. 2.6. Cartesian Vector Notation: Magnitude and Direction (Unit Vector) of a Vector: Coordinate Direction Angles:	{"url":"https://engcourses-uofa.ca/books/statics/vectors-and-their-operations/cartesian-vector-notation/","timestamp":"2024-11-13T01:40:24Z","content_type":"text/html","content_length":"92292","record_id":"<urn:uuid:d27b2062-0148-4c27-ba51-992d5818f404>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00715.warc.gz"}
Iterative deepening depth-first search Iterative deepening depth-first search (IDDFS) is a state space search strategy in which a depth-limited search is run repeatedly, increasing the depth limit with each iteration until it reaches d, the depth of the shallowest goal state. On each iteration, IDDFS visits the nodes in the search tree in the same order as depth-first search, but the cumulative order in which nodes are first visited, assuming no pruning, is effectively breadth-first. IDDFS combines depth-first search's space-efficiency and breadth-first search's completeness (when the branching factor is finite). It is optimal when the path cost is a non-decreasing function of the depth of the node. The space complexity of IDDFS is O(bd), where b is the branching factor and d is the depth of shallowest goal. Since iterative deepening visits states multiple times, it may seem wasteful, but it turns out to be not so costly, since in a tree most of the nodes are in the bottom level, so it does not matter much if the upper levels are visited multiple times.^[1] The main advantage of IDDFS in game tree searching is that the earlier searches tend to improve the commonly used heuristics, such as the killer heuristic and alpha-beta pruning, so that a more accurate estimate of the score of various nodes at the final depth search can occur, and the search completes more quickly since it is done in a better order. For example, alpha-beta pruning is most efficient if it searches the best moves first.^[1] A second advantage is the responsiveness of the algorithm. Because early iterations use small values for d, they execute extremely quickly. This allows the algorithm to supply early indications of the result almost immediately, followed by refinements as d increases. When used in an interactive setting, such as in a chess-playing program, this facility allows the program to play at any time with the current best move found in the search it has completed so far. This is not possible with a traditional depth-first search. The time complexity of IDDFS in well-balanced trees works out to be the same as Depth-first search: O(b^d). In an iterative deepening search, the nodes on the bottom level are expanded once, those on the next to bottom level are expanded twice, and so on, up to the root of the search tree, which is expanded d + 1 times.^[1] So the total number of expansions in an iterative deepening search is $(d + 1)1 + (d)b + (d-1)b^{2} + \cdots + 3b^{d-2} + 2b^{d-1} + b^{d}$ $\sum_{i=0}^d (d+1-i)b^i$ For b = 10 and d = 5 the number is 6 + 50 + 400 + 3,000 + 20,000 + 100,000 = 123,456 All together, an iterative deepening search from depth 1 to depth d expands only about 11% more nodes than a single breadth-first or depth-limited search to depth d, when b = 10. The higher the branching factor, the lower the overhead of repeatedly expanded states, but even when the branching factor is 2, iterative deepening search only takes about twice as long as a complete breadth-first search. This means that the time complexity of iterative deepening is still O(b^d), and the space complexity is O(bd). In general, iterative deepening is the preferred search method when there is a large search space and the depth of the solution is not known.^[1] For the following graph: a depth-first search starting at A, assuming that the left edges in the shown graph are chosen before right edges, and assuming the search remembers previously-visited nodes and will not repeat them (since this is a small graph), will visit the nodes in the following order: A, B, D, F, E, C, G. The edges traversed in this search form a Trémaux tree, a structure with important applications in graph theory. Performing the same search without remembering previously visited nodes results in visiting nodes in the order A, B, D, F, E, A, B, D, F, E, etc. forever, caught in the A, B, D, F, E cycle and never reaching C or G. Iterative deepening prevents this loop and will reach the following nodes on the following depths, assuming it proceeds left-to-right as above: □ 0: A □ 1: A (repeated), B, C, E (Note that iterative deepening has now seen C, when a conventional depth-first search did not.) □ 2: A, B, D, F, C, G, E, F (Note that it still sees C, but that it came later. Also note that it sees E via a different path, and loops back to F twice.) □ 3: A, B, D, F, E, C, G, E, F, B For this graph, as more depth is added, the two cycles "ABFE" and "AEFB" will simply get longer before the algorithm gives up and tries another branch. IDDFS(root, goal) depth = 0 while(no solution) solution = DLS(root, goal, depth) depth = depth + 1 return solution DLS(node, goal, depth) if ( depth >= 0 ) if ( node == goal ) return node for each child in expand(node) DLS(child, goal, depth-1) Related algorithms Similar to iterative deepening is a search strategy called iterative lengthening search that works with increasing path-cost limits instead of depth-limits. It expands nodes in the order of increasing path cost; therefore the first goal it encounters is the one with the cheapest path cost. But iterative lengthening incurs substantial overhead that make it less useful than iterative □ Graph algorithms □ Search algorithms Wikimedia Foundation. 2010.	{"url":"https://en-academic.com/dic.nsf/enwiki/244042","timestamp":"2024-11-07T06:55:22Z","content_type":"text/html","content_length":"49344","record_id":"<urn:uuid:69ea7c20-7f08-4ee1-b1cd-574f53a97c2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00375.warc.gz"}
NCERT Textbook Solution (Laptop/Desktop is best to view this page) CBSE Class –XII Physics NCERT Solutions Chapter – 8 Electromagnetic waves 1: Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain. Ans: Radius of each circular plate, r = 12 cm = m Distance between the plates, d = 5 cm = m Charging current, I = 0.15 A Permittivity of free space, = (a) Capacitance between the two plates is given by the relation, A = Area of each plate Charge on each plate, q = CV V = Potential difference across the plates Differentiation on both sides with respect to time (t) gives: Therefore, the change in potential difference between the plates is V/s. (b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A. (c) Yes Kirchhoff's first rule is valid at each plate of the capacitor since displacement current is equal to conduction current. 2: A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius = 6.0 cm has a capacitance = 100 pF. The capacitor is connected to a 230 V ac supply with a (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. Ans: Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = Supply voltage, V = 230 V Angular frequency, = 300 rad (a) Rms value of conduction current, Where, Xc = Capacitive reactance ∴ I = × = 6.9 Hence, the rms value of conduction current is 6.9 (b) Yes, conduction current is equal to displacement current. (c) Magnetic field is given as: = Free space permeability I0 = Maximum value of current = r = Distance between the plates from the axis = 3.0 cm = 0.03 m = Hence, the magnetic field at that point is 3: What physical quantity is the same for X-rays of wavelength 10-10 m, red light of wavelength 6800 and radiowaves of wavelength 500 m? Ans: The speed of light ( m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum. 4: A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? Ans: The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular. 5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? Ans: A radio can tune to minimum frequency, = 7.5 MHz= Maximum frequency, = 12 MHz = Speed of light, c = m/s Corresponding wavelength for can be calculated as: Corresponding wavelength for v can be calculated as: Thus, the wavelength band of the radio is 40 m to 25 m. 6: A charged particle oscillates about its mean equilibrium position with a frequency of Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is = 510 nT. What is the amplitude of the electric field part of the wave? Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum, B0 = 510 nT = Speed of light in a vacuum, c = Amplitude of electric field of the electromagnetic wave is given by the relation, = = 153 N/C Therefore, the electric field part of the wave is 153 N/C. 8: Suppose that the electric field amplitude of an electromagnetic wave is = 120 N/C and that its frequency is = 50.0 MHz. (a) Determine, (b) Find expressions for E and Ans: Electric field amplitude, = 120 N/C Frequency of source, = 50.0 MHz = Speed of light, c = m/s (a) Magnitude of magnetic field strength is given as: Angular frequency of source is given as: = 3.14 × 108 rad/s Propagation constant is given as: Wavelength of wave is given as: (b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as: And, magnetic field vector is given as: 9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? Ans: Energy of a photon is given as: h = Planck's constant = c = Speed of light = m/s = Wavelength of radiation The given table lists the photon energies for different parts of an electromagnetic spectrum for different . The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source. 10: In a plane electromagnetic wave, the electric field oscillates sinusoid ally at a frequency of and amplitude 48 V . (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = ] Ans: Frequency of the electromagnetic wave, Electric field amplitude, Speed of light, c = (a) Wavelength of a wave is given as: (b) Magnetic field strength is given as: (c) Energy density of the electric field is given as: And, energy density of the magnetic field is given as: ∈0 = Permittivity of free space = Permeability of free space We have the relation connecting E and B as: E = cB … (1) … (2) Putting equation (2) in equation (1), we get Squaring both sides, we get 11: Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (rad/s)t]}. (a) What is the direction of propagation? (b) What is the wavelength ? (c) What is the frequency ? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. Ans: (a) From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e., . (b) It is given that, The general equation for the electric field vector in the positive x direction can be written as: …(2) On comparing equations (1) and (2), we get Electric field amplitude, = 3.1 N/C Angular frequency, Wave number, k = 1.8 rad/m (c) Frequency of wave is given as: (d) Magnetic field strength is given as: c = Speed of light = (e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as: 12: About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropic ally and neglect reflection. Ans: Power rating of bulb, P = 100 W It is given that about 5% of its power is converted into visible radiation. Power of visible radiation, Hence, the power of visible radiation is 5W. (a) Distance of a point from the bulb, d = 1 m Hence, intensity of radiation at that point is given as: (b) Distance of a point from the bulb, = 10 m Hence, intensity of radiation at that point is given as: 13: Use the formula T= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you? Ans: A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck's law. It can be given by the relation, = maximum wavelength T = temperature Thus, the temperature for different wavelengths can be obtained as: For = ; For = ; For = cm; and so on. The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases. 14: Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs. (a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space). (b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen ; known as Lamb shift). (c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the 'big-bang' origin of the universe]. (d) 5890 - 5896 [double lines of sodium] (e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)]. Ans: (a) Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum. (b) Radio waves; it belongs to the short wavelength end. (c) Temperature, T = 2.7 °K is given by Planck's law as: This wavelength corresponds to microwaves. (d) This is the yellow light of the visible spectrum. (e) Transition energy is given by the relation, E = h = Planck's constant = = Frequency of radiation Energy, E = 14.4 K eV This corresponds to X-rays. 15: Answer the following questions: (a) Long distance radio broadcasts use short-wave bands. Why? (b) It is necessary to use satellites for long distance TV transmission. Why? (c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? (f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with a devastating effect on life on earth. What might be the basis of this Ans: (a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere. (b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions. (c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth. (d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth's (e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival. (f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it would lead to the depletion of the ozone layer too.	{"url":"https://neetjeeonline.com/server/njo_subjects_solution.php?subject=Physics&action=listlessons&grade=12&category_id=23","timestamp":"2024-11-09T14:40:34Z","content_type":"text/html","content_length":"212542","record_id":"<urn:uuid:226efd7e-9e81-49cc-9c7a-0cedaf72397f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00208.warc.gz"}
Survey Evaluation Metrics: Make Your Survey More Efficient Surveys are a powerful tool for gathering information and insights from individuals or groups of people. However, to ensure that your survey is effective, it’s important to evaluate its performance using survey evaluation metrics. These metrics can help you to identify areas for improvement and make your survey more efficient. In this content, we’ll explain the survey evaluation metrics and how to make your survey more efficient. Why Survey Evaluation Metrics are Important For Your Business? Survey evaluation metrics are essential for assessing the effectiveness of surveys and ensuring that the data collected is reliable and accurate. The following are some reasons why survey evaluation metrics are important: Quality assurance Quality assurance involves ensuring that the data collected is accurate, reliable, and free from errors. This is important because inaccurate or unreliable data can lead to incorrect conclusions and poor decision-making. Quality assurance can be achieved through various means, such as pre-testing survey questions, ensuring that survey instructions are clear and easy to follow, and using appropriate data collection methods. Measurement validity Measurement validity refers to the extent to which a survey accurately measures what it is intended to measure. It is important to establish measurement validity because if a survey question does not accurately measure the construct being studied, the resulting data may be inaccurate and misleading. Validity can be assessed through various means, such as content validity, criterion validity, and construct validity. Sample representativeness Sample representativeness refers to the extent to which the survey sample is representative of the population being studied. It is important to assess sample representativeness because if the survey sample is not representative of the population, the resulting data may not be generalizable to the larger population. Sample representativeness can be assessed through various means, such as assessing the demographic characteristics of the sample and comparing them to the population of interest, and assessing the response rate of the survey. Data analysis Data analysis involves examining the survey data to draw meaningful conclusions and insights. It is important to conduct data analysis because it allows researchers to identify patterns, relationships, and trends in the data. We can conduct data analysis through various means, such as descriptive statistics, inferential statistics, and data visualization. Proper data analysis can help researchers to draw accurate conclusions and make informed decisions based on the survey data. What Are The Survey Evaluation Metrics? Response Rate Survey Metrics Response rate is a measure of the percentage of people who respond to a survey out of the total number of people who were invited or sampled to participate in the survey. It is an important indicator of the quality of a survey, as a low response rate may indicate that the survey results may not be representative of the target population. The formula for calculating the response rate of a survey is: Response Rate = (Number of Completed Surveys / Number of Invitations or Sampled Population) x 100 For example, let’s say that a company sends out a survey to a randomly selected sample of 1,000 customers, and receives 350 completed surveys in return. To calculate the response rate, we would use the formula above: Response Rate = (350 / 1000) x 100 = 35% This means that the response rate for this survey is 35%, indicating that 35% of the customers who were invited to participate in the survey actually completed it. Another example could be an online survey sent to a company’s email list of 10,000 subscribers, with 2,000 subscribers completing the survey. The response rate would be: Response Rate = (2,000 / 10,000) x 100 = 20% This means that 20% of the email subscribers who received the survey actually completed it. It’s worth noting that response rates can vary widely depending on factors such as the target population, survey topic, and mode of survey administration. In general, response rates above 60% are considered very good, while rates below 30% may raise concerns about the representativeness of the survey results. Completion Rate Completion rate is one of the most important survey evaluation metrics that refers to the percentage of respondents who have completed the entire survey compared to the total number of individuals who were invited or started the survey. For example, if there are 100 invitation but only 80 people completed the entire survey, then the completion rate would be 80%. To calculate the completion rate of a survey, follow these steps: Determine the number of people who were invited to take the survey. Subtract the number of people who started the survey but did not complete it from the total number of invitations. Divide the number of completed surveys by the adjusted total number of invitations. Multiply the result by 100 to get the percentage. For instance, let’s say that you invited 500 people to take your survey, and 400 people started the survey, but only 350 people completed the entire survey. Then, you can calculate the completion rate as follows: Adjusted total invitations = 500 – (400 – 350) = 450 Completion rate = (350 / 450) x 100% = 77.8% Therefore, the completion rate of your survey would be 77.8%. Average Time to Complete The average time to complete a survey refers to the average amount of time it takes for respondents to finish the survey. It’s an important metric that can help researchers evaluate how long their surveys take and whether respondents find them engaging or too time-consuming. To calculate the average time to complete a survey, follow these steps: Determine the total time taken by all respondents to complete the survey. Divide the total time by the number of respondents who completed the survey. For example, suppose you send out a survey to 200 respondents, and it takes them the following times to complete: Respondent1: 2 minutes Respondent2: 5 minutes Respondent3: 10 minutes Respondent4: 3 minutes Respondent5: 6 minutes To calculate the average time to complete the survey, you would first determine the total time taken by all respondents, which would be: 2 + 5 + 10 + 3 + 6 = 26 minutes Then, you would divide the total time taken by the number of respondents who completed the survey, which in this case is 5: 26 minutes ÷ 5 respondents = 5.2 minutes Therefore, the average time taken to complete the survey is 5.2 minutes. Note that this calculation assumes that all respondents who started the survey completed it. If some respondents did not complete the survey, you may want to exclude their incomplete responses from the calculation or calculate the average time to complete only for those who finished the survey. Net Promoter Score (NPS) Net Promoter Score (NPS) is a customer loyalty metric that measures the likelihood of customers recommending a company, product, or service to others. It’s a simple and widely-used metric that ranges from -100 to 100 and is based on a single question: “How likely are you to recommend our product/service/company to a friend or colleague?” The NPS score is based on a scale of 0 to 10, with 0 being “not at all likely” and 10 being “extremely likely.” Respondents are classified into three categories based on their responses: Promoters: We consider that Customers who give a score of 9 or 10 are “promoters” because they are highly likely to recommend the company, product, or service to others. Passives: We consider that customers who give a score of 7or 8are “passives” because they are satisfy but not necessarily loyal, and may easily switch to a competitor. Detractors: We consider that customers who give a score of 0 or 6 are “detractors” because they are unlikely to recommend the company, product, or service to others and may even spread negative To calculate the NPS, follow these steps: Calculate the percentage of respondents who are promoters by dividing the number of respondents who gave a score of 9 or 10 by the total number of respondents and multiplying by 100. Calculate the percentage of respondents who are detractors by dividing the number of respondents who gave a score of 0 to 6 by the total number of respondents and multiplying by 100. Subtract the percentage of detractors from the percentage of promoters to get the NPS score. An Example For Nps For example, let’s say that 100 respondents took a survey and gave the following scores: Promoters (9-10): 60 respondents Passives (7-8): 20 respondents Detractors (0-6): 20 respondents To calculate the NPS, you would first calculate the percentage of promoters and detractors: Percentage of promoters = (60 / 100) x 100% = 60% Percentage of detractors = (20 / 100) x 100% = 20% Then, you would subtract the percentage of detractors from the percentage of promoters: NPS score = 60% – 20% = 40 Therefore we can say that the NPS score for this survey is 40 and thats a good score. Generally, we can say that an NPS score of 0 to 30 is good, 31 to 70 is excellent, and 71 to 100 is exceptional. Satisfaction Score Satisfaction score is a metric that measures the overall satisfaction of customers with a product, service, or experience. It’s typically based on a scale of 1 to 5 or 1 to 10 and we calculate by asking customers to rate their level of satisfaction with specific aspects of the product, service, or experience. To calculate a satisfaction score, follow these steps: Decide on a satisfaction scale. This could be a scale of 1 to 5, 1 to 10, or any other scale that makes sense for your survey. Determine the question or questions that you will use to calculate the satisfaction score. For example, you could ask customers to rate their satisfaction with the product, service, or experience overall, or you could ask them to rate their satisfaction with specific aspects of the product, service, or experience, such as customer service or ease of use. Collect responses from customers and calculate the average score. An Example Let’s say that you send out a survey to customers asking them to rate their satisfaction with your product on a scale of 1 to 10, with 10 being extremely satisfied and 1 being extremely dissatisfied. You receive the following responses: Customer-A: 9 Customer-B: 7 Customer-C: 8 Customer-D: 10 Customer-E: 6 To calculate the satisfaction score, you would first add up all of the ratings and then divide by the number of respondents: Total satisfaction score = 9 + 7 + 8 + 10 + 6 = 40 Number of respondents = 5 Average satisfaction score = 40 / 5 = 8 Therefore, the average satisfaction score for your product is 8 out of 10. Note that, we can calculate satisfaction score based on any scale that you choose, and you can also calculate scores for specific aspects of your product or service by asking targeted questions on your survey. Open-Ended Responses Open-ended response metrics in surveys typically refer to the process of analyzing and categorizing the qualitative data obtained from respondents’ answers to open-ended questions. We can use this metrics to identify themes, patterns, and trends in the data that may not have been captured by closed-ended questions or quantitative metrics. The process of analyzing open-ended responses typically involves a few steps: Transcribing: Transcribing the responses into text format so they can be more easily for analyzing. Coding: Developing a set of codes or categories to classify the responses based on their content or theme. This may involve assigning codes based on keywords or concepts that appear in the responses. Categorizing: Grouping the responses into broader categories or themes based on the codes assigned in the previous step. Analyzing: Analyzing the results to identify patterns, trends, and insights that can inform decision-making or further research. Open-ended response metrics can provide valuable insights into respondents’ attitudes, opinions, and experiences that may not be captured by closed-ended questions or quantitative metrics. However, analyzing open-ended responses can be more time-consuming and subjective than analyzing quantitative data, and the results may be more difficult to generalize to larger populations.	{"url":"https://www.cxperium.com/survey-evaluation-metrics-make-your-survey-more-efficient/","timestamp":"2024-11-06T04:13:49Z","content_type":"text/html","content_length":"220540","record_id":"<urn:uuid:66856f43-6f1b-40d9-8f56-749a1c6af3f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00706.warc.gz"}
Tong L Dr. Tong LI Office: 1A MLH Email: tong-li@uiowa.edu Phone: 319-335-3342 Fax: 319-335-0627 Paper Mail Dr. Tong LI Department of Mathematics 15 MLH The University of Iowa Iowa City, IA 52242-1419 Curriculum Vitae Teaching Fall, 2024, MATH:3550 Eng. Math. V Educational Background: Ph. D. in Mathematics, May, 1992, Courant Institute of Mathematical Sciences, New York University. M.S. in Mathematics, February, 1990, Courant Institute of Mathematical Sciences, New York University. M. S. in Mathematics, July, 1986, Peking University , Beijing, China B. A. in Mathematics, July, 1983, Peking University , Beijing, China Academic Experience: 2008-current, Professor, Department of Mathematics, The University of Iowa. 1999-2008, Associate Professor, Department of Mathematics, The University of Iowa. 2000-current, A Faculty Member of Program in Applied Mathematical and Computational Sciences, The University of Iowa. Iowa Informatics Initiative (UI3) Affiliated faculty, since April 24, 2019. Awarded A Professional Development Award for Fall, 2023, The University of Iowa. Awarded an Obermann Fellow for the Fall 2023 semester by the Obermann Center Advisory Board and the Office of the Vice President for Research, The University of Iowa. Fall, 2015, Long term visitor of the Mathematical Biosciences Institute (MBI), The Ohio State University. Long Term Visitor of the Institute for Pure and Applied Mathematics (IPAM), UCLA, November 15-December 15, 2015. May-June, 2011, Visiting Professor, University of Hamburg, Germany. Fall, 2008, Member of the Mathematical Biosciences Institute (MBI), The Ohio State University. Fall, 2008, Member of the Institute for Mathematics and its Applications(IMA), University of Minnesota. Spring, 2002, Visiting Member of Institute for Advanced Study, Princeton, New Jersey. 1993-1999, Assistant Professor, Department of Mathematics, University of Iowa. 1995-1997, Assistant Professor in Mathematics, UCLA. 1992-1993, Visiting Member of Institute for Advanced Study, Princeton, New Jersey. Research Interests: Nonlinear Hyperbolic Conservation Laws Shock Waves Traffic Flow Numerical Analysis Tingting Chen, Weifeng Jiang, Tong Li, Zhen Wang, Junhao Lin, The Riemann solutions for a mixed type Euler equations in dark energy fluid, Mathematics, 12(16)(2024), 2444. Weifeng Jiang, Daiguang Jin, Tong Li, and Tingting Chen, The singular limits of the Riemann solutions as pressure vanishes for a reduced two-phase mixtures model with non-isentropic gas state, J. Math. Phys., 65(2024), 071503. Tong Li and Nitesh Mathur, Global BV solutions to a system of balance laws from traffic flow, The Proceedings of HYP2022, XVIII International Conference on Hyperbolic Problems: Theory, Numerics, Applications, Volume 1, 307-317. Springer, May 28, 2024. Weifeng Jiang, Tingting Chen, Tong Li, Zhen Wang, The transition of Riemann solutions with composite waves for the improved Aw-Rascle-Zhang model in dusty gas, Physics of Fluids,35(2023), Tong Li and Nitesh Mathur, Global well-posedness and asymptotic behavior of BV solutions to a system of balance laws arising in traffic flow, Networks and Heterogeneous Media, 18(2)(2023), Tong Li and Jeungeun Park,Traveling waves in a Keller-Segel model with logistic growth, Communications in Mathematical Sciences, 20(2022), 829-853. Tong Li and Nitesh Mathur, Riemann Problem For A Non-Strictly Hyperbolic System In Chemotaxis, Discrete and Continuous Dynamical Systems Series B, 27(2022), 2173-2187. Tong Li and Zhian Wang, Traveling wave solutions to the singular Keller-Segel system with logistic source, Mathematical Biosciences and Engineering, 19((2022), 8107-8131. Weifeng Jiang, Yuan Zhang, Tong Li and Tingting Chen, The cavitation and concentration of Riemann solutions for the isentropic Euler equations with isothermal dusty gas, Nonlinear Analysis: Real World Applications, 71(2023), 103761. Weifeng Jiang, Tingting Chen, Tong Li and Zhen Wang, The wave interactions of an improved Aw-Rascle-Zhang model with a non-genuinely nonlinear field, Discrete and Continuous Dynamical Systems Series B, 28(2023), 1528-1552. Weifeng Jiang, Tingting Chen, Tong Li and Zhen Wang, The Riemann problem with delta initial data for the non-isentropic improved Aw-Rascle-Zhang model, Acta Mathematica Scientia, 43B(1) (2023), 237-258. Tingting Chen, Weifeng Jiang and Tong Li, On the stability of the improved Aw-Rascle-Zhang model with Chaplygin pressure, Nonlinear Analysis: Real World Applications, 62(2021), p. 103351.	{"url":"http://homepage.divms.uiowa.edu/~tli/","timestamp":"2024-11-02T07:23:54Z","content_type":"text/html","content_length":"29668","record_id":"<urn:uuid:5d56ab03-68c3-4bb0-9041-989f789859c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00532.warc.gz"}
A standard number cube is rolled 180 times predict how many times a 3 or a 5 will be the result Respuesta : A number cube generally has 6 sides. If you wanted a 3 or a 5 out of one roll, the probablitility would be 2/6. Multiply 2/6 times 180. Your answer is 60 times. Answer Link Otras preguntas	{"url":"http://cahayasurya.ac.id/question/10368279","timestamp":"2024-11-07T19:39:01Z","content_type":"text/html","content_length":"151707","record_id":"<urn:uuid:d8baaa62-8731-48ec-bbf8-44ea34b97183>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00429.warc.gz"}
New Plane and Solid Geometry From inside the book Results 1-5 of 34 Page 6 ... equally distant from a point within A called the centre . An arc is any portion of the circum- ference ; as AB . B B Q A Q A radius is a straight line drawn from the centre to the circumference ; as OA . CONSTRUCTIONS 24. At a given ... Page 19 ... equally distant from the extremities of the line . II . Any point without the perpendicular is unequally distant from the extremities of the line . C E a B D At the middle point D of any line AB erect a 1 DC . From E , any point in DC ... Page 20 ... equally distant from the extremities of the line . ] 4. Substituting for BE its equal AE , AE + EF > BF , or AF > BF . ( $ 54 , I ) 55. It follows from Prop . VI that every point which is equally distant from the extremities of a ... Page 21 ... equally distant from the extremities of a straight line , determine a perpendicular to that line at its middle point ... equally distant from points D and E ; and CF is perpendicular to AB by § 56 . In § 25 , points C and Fare each ... Page 26 ... equal distances from the foot of CD . To Prove a = b . Proof . Since CD is 1 EF at its middle point D , a = b . [ If a be erected at the middle point of a str . line , any point in the is equally distant from the extremities of the line ... PLANE GEOMETRY 3 RECTILINEAR FIGURES 12 THE CIRCLE 66 Book III 101 Book IV 134 Book V 154 BOOK VII 209 Popular passages S' denote the areas of two © whose radii are R and R', and diameters D and D', respectively. Then, \| = "* § = ££ = £• <§337> That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. In an isosceles triangle the angles opposite the equal sides are equal. ... any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 403. The area of a triangle is equal to half the product of its base by its Similar arcs are to each other as their radii; and similar sectors are to each other as the squares of their radii. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. In any triangle, the product of any two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The... A zone is a portion of the surface of a sphere included between two parallel planes. Every section of a cylinder made by a plane passing through an element is a parallelogram. Given ABCD, a section of cylinder AC, made by plane through element AB. A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre. Bibliographic information	{"url":"https://books.google.com.jm/books?id=1OtHAAAAIAAJ&q=equally+distant&dq=editions:UOM39015063898350&lr=&output=html_text&source=gbs_word_cloud_r&cad=5","timestamp":"2024-11-08T01:26:07Z","content_type":"text/html","content_length":"64940","record_id":"<urn:uuid:8e9804b2-66d0-497c-9080-2eee5ed81558>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00899.warc.gz"}
Royal Pains Anyone else into the best new summer series on USA? One of the few shows I actually tune into every week. Any other HankMed fans out there? ^^Totally. =) Love this show. I used to watch Southland at that time, but since it's off until fall, I figured I'd check out Royal Pains and I really ended up liking it. I haven't missed an episode since. Any favorite characters? Hank of course ;) I'm the same, haven't missed an episode since I started. wonders if and when a Brit tv channel picks it up Is is made by fox? wonders if and when a Brit tv channel picks it up Is is made by fox? shakes head USA Network. I don't watch Fox. It's shit. If I want a laugh, I watch Fox News :P [ame=http://www.youtube.com/watch?v=2aEk864YrKw]YouTube - Charlie Brooker on the American News Media. Funny[/ame] Hank of course ;) I'm the same, haven't missed an episode since I started. Yeah, Hank is pretty awesome. =) Evan is hilarious, too. Him and Divya are just so funny together, like in last week's episode. I really like Tucker and Libby too, even though they're not on as much. aaaah yes yes. this show is the bomb :cheesy: <3 Good news! Both Royal Pains and Burn Notice have been picked up for another season! Here's more on that: Yeah I love this show. Most shows on USA are at the very least pretty good. Good news! Both Royal Pains and Burn Notice have been picked up for another season! Here's more on that: I love this show, USA has been make a lot of good shows. ^^Yep. They have lately. I still can't get over the fact that The Starter Wife only lasted one season, though. Boo. • 2 weeks later... I never watched the Starter Wife but have a feeling its not for me, lol. I decided I really like Divya, she's very pretty but plays a interesting character. ^Agreed. And I like how awkward Hank is sometimes. :lol:	{"url":"https://coldplaying.com/forums/topic/48397-royal-pains/","timestamp":"2024-11-13T05:12:56Z","content_type":"text/html","content_length":"219064","record_id":"<urn:uuid:48ec1861-344f-4a00-816b-5246ccfaa08a>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00199.warc.gz"}
A sample of size 8 will be drawn from a normal population with mean 61 and... A sample of size 8 will be drawn from a normal population with mean 61 and... A sample of size 8 will be drawn from a normal population with mean 61 and standard deviation 14. (a) Is it appropriate to use the normal distribution to find probabilities for x? (b) If appropriate find the probability that x will be between 51 and 71. Round the answer to four decimal places. (c) If appropriate find the 81st percentile of x. Round the answer to two decimal places.	{"url":"https://justaaa.com/statistics-and-probability/619280-a-sample-of-size-8-will-be-drawn-from-a-normal","timestamp":"2024-11-06T01:58:33Z","content_type":"text/html","content_length":"40344","record_id":"<urn:uuid:ae922abb-9e9c-4db5-861f-ac5fd55f18cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00832.warc.gz"}
Mathematical Sciences Research Institute An introduction to Higgs bundles - III August 22, 2019 (02:15 PM PDT - 03:15 PM PDT) Speaker(s): Qiongling Li (Chern Institute of Mathematics) Location: SLMath: Eisenbud Auditorium • Higgs bundles • non-Abelian Hodge correspondence • harmonic maps • character variety Primary Mathematics Subject Classification Secondary Mathematics Subject Classification No Secondary AMS MSC I will review the basic concepts of Higgs bundles and the moduli space of Higgs bundles, and then focus the non-Abelian Hodge correspondence: relation between the moduli space of Higgs bundles with harmonic maps and the character variety. I will also explain some examples: rank two Higgs bundles, cyclic Higgs bundles, variation of Hodge structure and so on. Supplements No Notes/Supplements Uploaded	{"url":"https://legacy.slmath.org/workshops/895/schedules/27272","timestamp":"2024-11-06T15:02:01Z","content_type":"text/html","content_length":"42714","record_id":"<urn:uuid:b6825781-d2c6-4874-942e-132ce6b852b2>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00668.warc.gz"}
GreeneMath.com \| Ace your next Math Test! Lesson Objectives • Learn the definition of a Natural Number Exponent • Learn how to evaluate an Exponent with a Negative Base • Learn about the Order of Operations \| PEMDAS Exponents & The Order of Operations Definition of an Exponent Many times, we are faced with a scenario where we have a repeated multiplication of the same number. Suppose we wanted to write the prime factorization of 243. $$243=3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$$ Writing out 5 factors of 3 is quite long and inconvenient. Fortunately, we have exponents that can help to shorten this process. An exponent allows us to write a repeated multiplication of the same number in a more compact form. $$243=3 \cdot 3 \cdot 3 \cdot 3 \cdot 3=3^5$$ Notice how 5 factors of 3 can be written in exponential form as: 3 . Where the 3 or larger number is known as the base, and the 5 or smaller number is known as the exponent. The base is the number that is being multiplied by itself in the repeated multiplication. When an exponent is a natural number , it represents the number of factors of the base present in the repeated multiplication. Since 243 can be broken down into 5 factors of 3, we can write 243 as 3 to the 5th power. Exponents with a Negative Base When we work with exponents and the base is negative, it can be quite confusing when considering the sign of the answer. Suppose we consider the following: $$-2^2$$ If we punch this up on a calculator, you will get an answer of -4. One might ask why, as we would think that -2 means (-2)(-2), which is clearly 4. So why then does our calculator give us an answer of -4? The reason is simple. The exponent doesn't apply to the negative part unless it is wrapped inside of parentheses. We can actually write -2 as -1 • 2 , which makes it clear that the answer is -4. If we want to obtain an answer of +4, we want to make sure to wrap our negative sign inside of parentheses. $$(-2)^2=(-2)(-2)=4$$ $$-2^2=-1 \cdot 2^2=-1 \cdot 4=-4$$ It's important to note that this rule doesn't change the sign if the exponent is odd. This is due to the fact that an odd number of negative factors yields a negative result. $$-3^3=-1 \ cdot 3^3=-1 \cdot 27=-27$$ $$(-3)^3=(-3)(-3)(-3)=-27$$ Order of Operations The Order of Operations tells us which operation to perform in which order when faced with a problem with multiple operations involved. Most often, people use the acronym PEMDAS to remember the order of operations. We have to be careful when using PEMDAS as following the order of the letters doesn’t always give us the correct answer. 1. P » parentheses and other grouping symbols such as absolute value bars, brackets,…etc □ Always start with the innermost set of parentheses or grouping symbols and work outward. Once inside of grouping symbols, we want to reapply the order of operations. □ When fraction bars are present, we work above and below any fraction bars separately. We can also wrap the numerator and denominator inside of parentheses and place a "÷" symbol between them to replace the fraction bar. 2. E » exponents and radicals 3. MD » multiply or divide working left to right 4. AS » add or subtract working left to right The multiply and divide steps, along with the addition and subtraction steps will cause the most confusion. The order in PEMDAS can’t be followed exactly. For example, if we try to follow the sequence of letters with multiplication or division, we may obtain the wrong answer. We can see that the M comes before the D in PEMDAS, but we multiply or divide working from left to right. In the 12 ÷ 3 • 7 We would actually divide first since the division operation is to the left of the multiplication operation. The correct answer is 28: 12 ÷ 3 • 7 = 4 • 7 = 28 As an example, suppose we wanted to find the value for the following expression: $$[10 - (16 \hspace{.1em}÷ \hspace{.1em}(-8))^3] \cdot 7 \hspace{.1em}÷ \hspace{.1em}(-6)$$ First, we want to start with any grouping symbols that are present. In this problem, we have brackets and parentheses. Starting with the innermost set gives us: $$(16 \hspace{.1em}÷ \hspace{.1em}(-8))=(-2)$$ Now, we will replace this in our expression and continue. $$[10 - (-2)^3] \cdot 7 \hspace{.1em}÷ \hspace{.1em}(-6)$$ Now that we have dealt with the innermost set of parentheses, it's time to work outward. Let's next tackle what's inside of the brackets. $$[10 - (-2)^3]$$ Notice how we have a subtraction operation and an exponent operation. The exponent has a higher priority, so we will raise (-2) to the third power first: $$(-2)^3=-8$$ Let's replace this in our expression. $$[10 - (-8)] \cdot 7 \hspace{.1em}÷ \hspace{.1em}(-6)$$ Now, we will perform our subtraction operation inside of the brackets. $$10 - (-8)=10 + 8=18$$ Let's replace this in our expression. $$18 \cdot 7 \hspace{.1em}÷ \hspace{.1em}(-6)$$ Now, we have multiplication and division left. Again, we want to make sure these are worked from left to right. $$18 \cdot 7=126$$ Let's replace this in our expression. $$126 \hspace{.1em}÷ \hspace{.1em}(-6)$$ As our final step, let's perform our division. $$126 \hspace{.1em}÷ \ $$[10 - (16 \hspace{.1em}÷ \hspace{.1em}(-8))^3] \cdot 7 \hspace{.1em}÷ \hspace{.1em}(-6)=-21$$ Skills Check: Example #1 Use Exponents to Write the Prime Factorization of 3200 Please choose the best answer. Example #2 Use Exponents to Write the Prime Factorization of 27,783 Please choose the best answer. $$7^3 \cdot 5^2 \cdot 3$$ Example #3 $$\|-6 \hspace{.1em}÷ \hspace{.1em}(-2)^2 \cdot (-8)\| \hspace{.1em}÷ \hspace{.1em}\frac{1}{4}$$ Please choose the best answer. Example #4 $$\frac{(-9)^2 \hspace{.1em}÷ \hspace{.1em}\|{-}2 \cdot 15 \hspace{.1em}÷ \hspace{.1em}5\|}{-6^2 \hspace{.1em}÷ \hspace{.1em}(-4^2 \cdot 5 + 8)}$$ Please choose the best answer. Congrats, Your Score is 100% Better Luck Next Time, Your Score is % Try again? Ready for more? Watch the Step by Step Video Lesson \| Take the Practice Test	{"url":"https://www.greenemath.com/College_Algebra/2/Exponents-OrderOfOperationsLesson.html","timestamp":"2024-11-10T12:18:16Z","content_type":"application/xhtml+xml","content_length":"18927","record_id":"<urn:uuid:37136b8a-ab8f-4e7c-87b2-80928ab3654f>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00348.warc.gz"}
developmentally preceded by Candidate definition: x developmentally related to y if and only if there exists some developmental process (GO:0032502) p such that x and y both participates in p, and x is the output of p and y is the input of p In general you should not use this relation to make assertions - use one of the more specific relations below this one RO:0002258 This relation groups together various other developmental relations. It is fairly generic, encompassing induction, developmental contribution and direct and transitive develops from developmentally succeeded by developmentally related to	{"url":"https://ontobee.org/ontology/CEPH?iri=http://purl.obolibrary.org/obo/RO_0002258","timestamp":"2024-11-10T10:45:35Z","content_type":"application/rdf+xml","content_length":"4124","record_id":"<urn:uuid:190e5b90-28ac-4699-b607-75128059f54c>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00158.warc.gz"}
EEE 2019 Exam [Section B Solutions] 2013 Academic Year 25 July 2014 EEE 2019 – PRINCIPLES OF ELECRICITY I MODEL SOLUTIONS TO FINAL EXAMINATION - 2013/2014 ACADEMIC YEAR DATE OF EXAMINATION: 25TH JULY, 2014 Question 6: [Solution] The diode will conduct during positive half-cycles of the input to yield an output vout as shown below. short cct [3 marks] The dc level is determined by finding the average of the output waveform vout over a full period, i.e., Vdc Vavg vout dt T 0 T 2 sin t dt 4.77 V [3 marks] The peak-inverse-voltage (PIV) is the voltage across the diode when it is reverse biased. Thus the PIV is found as follows: I 0 Applying KVL to the cct yields, PIV Vpin IR 15 0 , PIV 15 V [4 marks] b) When the ideal diode is replaced with a germanium diode the following is obtained: The sketch of the output vout is as shown below: Dept. of EEE, School of Engg, UNZA. s. cct [3 marks] The dc level is determined by finding the average of the output waveform vout over a full period, i.e., Vdc Vavg T 2 VD sin t dt Vp VD 15 0.3 4.679 V [3 marks] PIV is obtained by applying KVL to the circuit below: I 0 Thus, PIV Vpin IR VD 15 0 0.3 , PIV 15.3 V [4 marks] [Total 20 marks] Question 7: [Solution] a) For the given RC circuit: t 0 v t t 0 v t Dept. of EEE, School of Engg, UNZA. At t 0 the steady-state voltage across the capacitor is V1 , which is the initial voltage to the new circuit arrangement. Since V2 V1 , the current flows as shown above to charge the capacitor to V2 . Thus, applying KVL to the new circuit arrangement yields, V2 Ri v ; but i C , it follows that , V2 RC v [1 mark] v V2 Rearranging the equation and integrating both sides over the given limits yields, v t t dt 0 RC v V2 ln v V2 ln v t V2 ln V1 V2 v t t t RC 0 v t V2 0 , that is, ln V1 V2 [2 marks] Taking exponential of both sides yields, v t VS V1 VS t RC , v t V2 V1 V2 et RC , v t V2 V1 V2 et RC , being the step response voltage. [2 marks] Recall that the current through the capacitor is , . Given also that V1 0 , it follows that v t V2 1 et RC . Thus, [2 marks] i C CV2 t RC C V2 1 et RC i t V2 t RC u t [3 marks] b) The given voltage pulse written in terms of step functions is of the form, v t v t 10u t 2 10u t 5 [4 marks] c) The output of each block of the regulated power supply is as shown below. Dept. of EEE, School of Engg, UNZA. [1 mark] [2 marks] [2 marks] [1 mark] [Total 20 marks] Question 8: [Solution] a) For the given circuit 220V rms vin vsec 120V rms The secondary voltage is given as 120V rms . Thus, the peak voltage is calculated as follows: Vpsec Vrms 2 120 2 169.71 V . [2 marks] Since germanium diodes are used in the bridge cct, the dc output is given by Vdc Vavg T 2 2VD sin t dt 2 Vpsec 2VD 2 169.71 2 0.3 107.66 V [2 marks] The PIV rating of each diode is found as follows: PIV Vpsec VD 169.71 0.3 169.41V , PIV 169.41V [4 marks] The maximum diode current during conduction is Dept. of EEE, School of Engg, UNZA. Vp out Vpsec 2VD 169.71 2 0.3 169.11 0.0846 , I 84.6 mA [4 marks] b) Given the diode limiting circuit Input waveform. An ideal diode will not conduct when vi VBIAS 4 V , hence no current will flow and all the input voltage appears at the output, see waveform below. [4 marks] When the diode is replaced a silicon diode of inherent barrier voltage VD 0.7 V . Thus, the diode will not conduct when vi VBIAS VD 4 0.7 3.3 V . Hence no current will flow and the entire input voltage appears at the output as shown below. [4 marks] [Total 20 marks] END OF EEE 2019 EXAM MODEL SOLUTIONS Dept. of EEE, School of Engg, UNZA.	{"url":"https://studylib.net/doc/26339629/eee-2019-exam--section-b-solutions--2013-academic-year-25...","timestamp":"2024-11-08T10:44:56Z","content_type":"text/html","content_length":"57540","record_id":"<urn:uuid:e074e080-6763-46c5-9c8c-2a16d98ff667>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00193.warc.gz"}
Excel Formula: Count Cells with Blue Color in Python In this tutorial, we will learn how to write an Excel formula in Python to count cells with the color blue. This can be achieved using the COUNTIF function in combination with the CELL function. The formula =COUNTIF(A:A,CELL("color",A1)=5) allows us to count the number of cells in column A that have the color blue. To understand the formula, let's break it down step-by-step: 1. The CELL function is used to retrieve the color index of a cell. In our case, we use CELL("color", A1) to get the color index of cell A1. 2. The color index of blue is 5. Therefore, the expression CELL("color", A1)=5 checks if the color index of cell A1 is equal to 5, indicating that the cell has the color blue. 3. The COUNTIF function is then used to count the number of cells in column A that meet the condition specified by the expression CELL("color", A1)=5. 4. Finally, the formula is entered as a regular formula by pressing Enter. Let's consider an example to illustrate the usage of this formula. Suppose we have a dataset in column A with different cell colors. Assuming that cells 2, 4, 6, and 8 have the color blue, the formula =COUNTIF(A:A,CELL("color",A1)=5) would return the value 4, indicating that there are 4 cells in column A with the color blue. In conclusion, by using the provided formula, you can easily count the number of cells with the color blue in Excel using Python. This tutorial has provided a step-by-step explanation and examples to help you understand and apply the formula effectively. An Excel formula Formula Explanation This formula uses the COUNTIF function in combination with the CELL function to count the number of cells in column A that have the color blue. Step-by-step explanation 1. The CELL function is used to retrieve the color index of a cell. In this case, we use CELL("color", A1) to get the color index of cell A1. 2. The color index of blue is 5. So, the expression CELL("color", A1)=5 checks if the color index of cell A1 is equal to 5, indicating that the cell has the color blue. 3. The COUNTIF function is then used to count the number of cells in column A that meet the condition specified by the expression CELL("color", A1)=5. 4. The formula is entered as a regular formula by pressing Enter. For example, if we have the following data in column A with different cell colors: \| A \| \| \| \| 1 \| \| 2 \| \| 3 \| \| 4 \| \| 5 \| \| 6 \| \| 7 \| \| 8 \| Assuming that cells 2, 4, 6, and 8 have the color blue, the formula =COUNTIF(A:A,CELL("color",A1)=5) would return the value 4, indicating that there are 4 cells in column A with the color blue.	{"url":"https://codepal.ai/excel-formula-generator/query/49FaUksr/excel-formula-count-cell-color-blue","timestamp":"2024-11-04T14:42:17Z","content_type":"text/html","content_length":"92389","record_id":"<urn:uuid:a6eac0b5-01d5-4cbe-a400-396d195553ba>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00681.warc.gz"}
Pint - Eplanation, Different Types, Examples, and FAQs A pint is a unit of volume or capacity in both the imperial and United States customary measurement systems. The symbol 'pt' is used to represent a pint. In both systems, a pint is traditionally about ⅛ of a gallon. The British Pint is about ⅕ larger than American Pint as both the systems are defined differently. In the British system, the units for the dry and liquid measure are similar, the single British pint is equivalent to 34.68 cubic inches (568.26 cubic cm) or one- eight gallon. In the United States, the unit for dry systems is slightly different from liquid measures. A U.S. dry pint is 33.46 cubic inches (550.6 cubic cm), while a liquid pint is 20.9 cubic inches (473.2 cubic cm). In each system, two cups (unit of volume in the British Imperial and United States Customary systems of measurement) makes a pint, and two pints equals a quart (a unit of capacity in the United States Customary systems of measurement and the British Imperial). What is an Example of 1 Pint? A pint is equivalent to 2 cups (for example, a large glass of milk). (Image will be uploaded soon) The value of 1 Pint = 2 Cups = 16 Fluid Ounces. A unit quart (qt) is used in place of a pint for measuring many cups of liquid together. (Image will be uploaded soon) The value of 1 quart (qt) is similar to 4 cups or 2 pints. The Value of 1 Quart = 2 Pint = 4 Cups = 32 Fluid Ounces If we still need to measure more liquid, then we can use the unit gallon in place of quat(qt). 1 Gallon = 8 Pints = 16 Cups = 4 Quarts (Image will be uploaded soon) Gallon is the largest measurement of liquid. Note: A Quart is a Quarter of a Gallon. Pint of Water The value of 1 US fluid pint of water weighs about a pound (16 ounces). This gives rise to a renowned statement, A pints, a pound, the world around. The measure of US pint of water is approximately around 1.04318 pounds, and this statement does not hold throughout the world. It is due to the imperial pint which was also the standard measure in New Zealand, Australia, Malaya, India, and other British colonies weighs 1.2528 pounds, which gives rise to a popular statement for the imperial pint “ a pint of water that is pure weighs a quarter and a pound). What is Half Pint? Half-pint or half a pint is equivalent to 8 fluid ounces (1 cup) or 16 tablespoons (0.2 litres). A 375 ml of pint in the US and the Canadian maritime provinces is sometimes referred to as a pint, and a 200 ml bottle is known as half a pint, looking back to the days when liquor came in US pints, fifths, quartz, and gallon. A standard 250 ml of beer in France is known as un demi (" a half"), originally meaning half a pint. Imperial Pint The value of an imperial pint is equivalent to one - eight imperial gallons. The Value of 1 Imperial Pint equals to = ⅛ Imperial Gallon = ½ Imperial Quart = 4 Imperial Gills = 20 Imperial Fluid Ounces = 568.26125 millilitres exactly ≈ 34.677429099 cubic inches ≈ 1.0320567435 US dry pint ≈ 1.2009499255 US liquid pint ≈ 19.215198881 US fluid ounces ≈ The volume of 20 oz (567 g) of water at 62.F (16.7 C). US Liquid Pint 1 US Liquid Pint is equals to = ⅛ US liquid Gallon = ½ US liquid quart = 2 US cups = 4 US fluid gills = 16 US fluid ounces = 128 US fluid drams = 28.875 cubic inches (exactly) = 473.176473 milliliters ( exactly) ≈ 0.83267418463 imperial pints ≈ 0.85936700738 US dry pints ≈16.65348369 imperial fluid ounces ≈The volume of 1.041 lb (472 g) of water at 62° F ( 16.7° C) US Dry Pint In the US, the dry pint is equal to a sixty-fourth of a bushel 1 US dry Pint is equals to = 0.015625 US bushels = 0.125 US Pack = 0.5 US Dry gallon = 33.6003125 US dry quarts ≈ 550.6104713575 ≈ 0.96893897192092 ≈ 1.1636471861472 US pints Facts to Remember • One pint is equivalent to one-eighth of a gallon and half of a quart. • One Pint is equal to 2 cups or 16 ounces. • 2 Pint is equal to 1 Quart. • 8 Pint is equal to 1 gallon. 1. Convert 20 pt (US) to cups (US). The value of 1 pt (US) = 2 cup (US) The value of 1 cup (US) = 0.5 PT (US) 20 pt (US) = 20 × 2 cup (US) = 40 cup (US). 2. Convert 20 pt (US) to Gallons (US). The value of 1 pt (US) = 0.125 gal (US) The value of 1gal (US) = 8 pt (US) 20 pt (US) = 20 0.125 gal (US) = 1.875 gal (US). FAQs on Pint Ans: In the US, one pint is equal to 16 fluid ounces or 473 ml. In the UK, one pint is equal to 20 fluid ounces or 573 ml. 2. How Many Cups Make One Pint? Ans: A Cup is a unit of measurement of volume in the British Imperial and United States customary system of measurement. If we remember, 1 cup = 8 ounces 2 cup = 16 ounces or one pint Generally, 2 cups make one pint. However, depending on the ingredient it may change. For example, 1 pint of blueberries = 12 ounces ( dry) or 2 cups, whereas one pint of sour cream or 1 pint of ice cream is equivalent to 2 cups. 3. How to Convert Litre Measurement into Pint Measurement? Ans: To convert litre measurement into pint measurement, multiply the volume by the conversion ratio. The value of one litre is equivalent to 2.113376. Hence, the simple formula given below can be used to convert litres into the measurement. Pint = Liter × 2.113376	{"url":"https://www.vedantu.com/maths/pint","timestamp":"2024-11-14T01:10:26Z","content_type":"text/html","content_length":"254157","record_id":"<urn:uuid:14255f74-933e-49e6-837b-326ed444495e>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00507.warc.gz"}
Numerical Optimization (online) Prof. Dr. Moritz Diehl - moritz.diehl@imtek.uni-freiburg.de The course’s aim is to give an introduction into numerical methods for the solution of optimization problems in science and engineering. It is intended for students from two faculties, mathematics and physics on the one hand, and engineering and computer science on the other hand. This semester, Numerical Optimization is offered as an semi-online course. The focus is on continuous nonlinear optimization in finite dimensions, covering both convex and nonconvex problems. The exam is March 4th at 9am in HS-00-036, G.-Köhler-Allee 101. Organization of the course The course during is based on two pillars, lectures and exercises, accompanied by written material for self-study. As the course is semi-online there will be no lecture held. Instead you can refer to the lectures recorded during the winter term 2015/16. Nonetheless we will meet every Friday, 14:00 to 16:00, in SR 226, Hermann-Herder-Str. 10 (Rechenzentrum). Usually every second Friday is dedicated to Q&A regarding the lecture. Normally both professor and teaching assistant will attend the Q&A session. Every other Friday there will be exercise sessions with the teaching assistant. There is a detailed calendar below. Course language is English and all communication is made via the course homepage. For more information please contact Florian Messerer. For the lecture recordings please refer to the course page of winter term 2015/16. This course gives 6 ECTS. It is possible to do a project to get an additional 3 ECTS, i.e. a total of 9 ECTS for course+project. Exercises: The exercises are partially paper based and partially on the computer. Individual laptops with MATLAB installed are required. Please note that the reserved room is not a computer pool. The exercises will be distributed beforehand. You can then prepare yourselves for the exercise session, where you can complete the tasks in teams of 2 and show the results to the teaching assistant. Groups that require more time or cannot make it to the exercise session may send their solutions by e-mail (messerer@tf.uni-freiburg.de, see guidelines below) until the start of the next Q&A session. Note that groups that complete the tasks during the exercise session do not need to send a report by e-mail. You will need at least 40% of the total points in order to be eligible for the exam. Final evaluation: For engineering students the final grade of the course (6 ECTS) is based solely on a final written exam at the end of the semester. Students from the faculty of mathematics need to pass the written exam (ungraded) in order take an oral exam, which determines their grade. The final exam is a closed book exam. Only pencil, paper, a calculator and two A4 sheets (4 pages) of self-chosen formulas are allowed (handwritten). Projects: The project (3 ECTS) consists in the formulation and implementation of a self-chosen optimization problem and numerical solution method, resulting in documented computer code, a project report, and a public presentation. Project work starts in the last third of the semester and participants can work either individually or in groups of two people. ┃Oct 19th│Kick-off meeting │Handout Exercise 1 │ ┃ ┃Oct 26th│Exercise session │Exercise 1 │ ┃ ┃Nov 2nd │** │** │ ┃ ┃Nov 9th │Q&A │Deadline Exercise 1, Handout Exercise 2│course content up to and including chapter 5 ┃ ┃Nov 16th│Exercise session │Exercise 2 │ ┃ ┃Nov 23rd│Q&A │Deadline Exercise 2, Handout Exercise 3│course content up to and including chapter 9 ┃ ┃Nov 30th│Exercise session │Exercise 3 │ ┃ ┃Dec 7th │Q&A │Deadline Exercise 3, Handout Exercise 4│course content up to and including chapter 12 ┃ ┃Dec 14th│Exercise session │Exercise 4 │ ┃ ┃Dec 21st│Q&A │Deadline Exercise 4, Handout Exercise 5│course content up to and including chapter 15 ┃ ┃ │CHRISTMAS BREAK │ │ ┃ ┃Jan 11th│Q&A, projects │ │all course content. Discuss your project proposals with us! ┃ ┃Jan 18th│Exercise session │Deadline Exercise 5 │ ┃ ┃Jan 25th│Q&A, projects │ │all course content. Discuss your project proposals with us! ┃ ┃Feb 1st │Exercise session, project work│ │ ┃ ┃Feb 8th │Project presentations │Deadline projects │ ┃ Guidelines for handing in exercises If you hand in the exercise via e-mail, please adhere to the following guidelines: • One (!) file which is your main document (preferably pdf). It contains your name(s), your solutions to the pen-and-paper exercises and for computer exercises the name of the corresponding file • The main document can be a scan of your handwritten solutions or created with a text editor of your choice (with proper support for mathematical notation, e.g. Latex, MS Word, Open Office...) • Hand in all of the relevant code files. It should be possible to run them to see all results. It should not be necessary to un(comment) lines for proper functioning. If there are several similar, but conflicting versions (e.g. different constraints), please hand them in as separate files. • If you received helper functions as part of the exercise, please also hand them in. This makes it easier to run your files since everything is contained in one folder already. Please find the guidelines for the 3ECTS project here. The final presentation will be a talk of 10 minutes plus up to 10 minutes of questions. Please try to stick to the 10 minutes as close as possible. We have the following 4 topics: The Dogleg Method, The Generalized Gauss-Newton Method, Optimal Control of Table Soccer, Optimization of a Hammer Throw, Optimal Control of Apollo Re-entry The schedule for Friday, Feb 8th is as follows: ┃14:00 - 14:10 │General ┃ ┃14:10 - 14:30 │Presentation 1 ┃ ┃14:30 - 14:50 │Presentation 2 ┃ ┃14:50 - 15:00 │Break ┃ ┃15:00 - 15:20 │Presentation 3 ┃ ┃15:20 - 15:40 │Presentation 4 ┃ ┃15:40 - 16:00 │Presentation 5 ┃ • Jorge Nocedal and Stephen J. Wright, Numerical Optimization, Springer, 2006. • Amir Beck, Introduction to Nonlinear Optimization, MOS-SIAM Optimization, 2014. • You can watch the lecture recordings together. We reserved SR 226, Hermann-Herder-Str. 10, every Friday 4 to 8 pm. This is completely self organized, so neither Prof. Diehl nor the teaching assistant will attend. You just need someone to bring a laptop (and maybe portable speakers). There's a VGA cable and a HDMI port without cable.	{"url":"https://www.syscop.de/teaching/ws2018/numerical-optimization-online","timestamp":"2024-11-14T22:04:50Z","content_type":"text/html","content_length":"25479","record_id":"<urn:uuid:10710f25-ecc4-484e-88c7-cb08ff250983>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00656.warc.gz"}
Kronecker delta nLab Kronecker delta For $I$ a set, the Kronecker delta-function is the function $I \times I \to \{0,1\}$ which takes the value 0 everywhere except on the diagonal, where it takes the value 1. Often one writes for elements $i,j \in I$ $\delta^{i}_j \coloneqq \delta(i,j) \,.$ $\delta^i_j = \left\{ \array{ 1 & if i = j \\ 0 & otherwise } \right.$ In constructive mathematics, it is necessary that $I$ have decidable equality; alternatively, one could let the Kronecker delta take values in the lower reals. Named after Leopold Kronecker. See also Last revised on April 27, 2024 at 10:42:29. See the history of this page for a list of all contributions to it.	{"url":"https://ncatlab.org/nlab/show/Kronecker+delta","timestamp":"2024-11-14T16:52:14Z","content_type":"application/xhtml+xml","content_length":"17902","record_id":"<urn:uuid:f7bd261d-c284-473c-b5b7-ab19e52a1da4>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00403.warc.gz"}
Mathematics for the Liberal Arts Learning Outcomes • Given the part and the whole, write a percent • Calculate both relative and absolute change of a quantity • Calculate tax on a purchase Geometric shapes, as well as area and volumes, can often be important in problem solving. Let’s start things off with an example, rather than trying to explain geometric concepts to you. You are curious how tall a tree is, but don’t have any way to climb it. Describe a method for determining the height. Show Solution Similar Triangles We introduced the idea of similar triangles in the previous example. One property of geometric shapes that we have learned is a helpful problem-solving tool is that of similarity. If two triangles are the same, meaning the angles between the sides are all the same, we can find an unknown length or height as in the last example. This idea of similarity holds for other geometric shapes as well. Guided Example Mary was out in the yard one day and had her two daughters with her. She was doing some renovations and wanted to know how tall the house was. She noticed a shadow 3 feet long when her daughter was standing 12 feet from the house and used it to set up figure 1. We can take that drawing and separate the two triangles as follows allowing us to focus on the numbers and the shapes. These triangles are what are called similar triangles. They have the same angles and sides in proportion to each other. We can use that information to determine the height of the house as seen in figure 2. To determine the height of the house, we set up the following proportion: Then, we solve for the unknown x by using cross products as we have done before: Therefore, we can conclude that the house is 25 feet high. Try It It may be helpful to recall some formulas for areas and volumes of a few basic shapes: Rectangle Circle, radius r Area: [latex]L\times{W}[/latex] Area: [latex]\pi{r^2}[/latex] Perimeter: [latex]2l+2W[/latex] Circumference[latex]2\pi{r}[/latex] Rectangular Box Cylinder Volume: [latex]L\times{W}\times{H}[/latex] Volume: [latex]\pi{r^2}h[/latex] In our next two examples, we will combine the ideas we have explored about ratios with the geometry of some basic shapes to answer questions. In the first example, we will predict how much dough will be needed for a pizza that is 16 inches in diameter given that we know how much dough it takes for a pizza with a diameter of 12 inches. The second example uses the volume of a cylinder to determine the number of calories in a marshmallow. If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza? Show Solution The following video illustrates how to solve this problem. A company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have? Show Solution For more about the marshmallow example, watch this video. Try It A website says that you’ll need 48 fifty-pound bags of sand to fill a sandbox that measure 8ft by 8ft by 1ft. How many bags would you need for a sandbox 6ft by 4ft by 1ft? Show Solution Mary (from the application that started this topic), decides to use what she knows about the height of the roof to measure the height of her second daughter. If her second daughter casts a shadow that is 1.5 feet long when she is 13.5 feet from the house, what is the height of the second daughter? Draw an accurate diagram and use similar triangles to solve. In the next section, we will explore the process of combining different types of information to answer questions.	{"url":"https://courses.lumenlearning.com/waymakermath4libarts/chapter/geometry/","timestamp":"2024-11-11T07:14:08Z","content_type":"text/html","content_length":"60629","record_id":"<urn:uuid:6d53a58c-e48f-46f7-9382-3b40aa189b38>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00605.warc.gz"}
Thaddeus D. Ladd, Ph.D. - Publications • "Hybrid Quantum Computation in Quantum Optics," P. van Loock, W. J. Munro, Kae Nemoto, T. P. Spiller, T. D. Ladd, S. L. Braunstein and G. J. Milburn, Physical Review A 78, 022303 (2008) • "Hybrid quantum repeater based on dispersive CQED interactions between matter qubits and bright coherent light," T. D. Ladd, P. van Loock, K. Nemoto, W. J. Munro, and Y. Yamamoto, New Journal of Physics 8, 184 (2006) • "Hybrid quantum repeater using bright coherent light," P. van Loock, T. D. Ladd, K. Sanaka, F. Yamaguchi, Kae Nemoto, W. J. Munro, and Y. Yamamoto, Physical Review Letters 96, 240501 (2006) • "Coherence time of decoupled nuclear spins in silicon," T. D. Ladd, D. Maryenko, Y. Yamamoto, E. Abe, and K. M. Itoh, Physical Review B 71, 014401 (2005) • "Photocurrent-modulated optical nuclear polarization in bulk GaAs," A. K. Paravastu, P. J. Coles, J. A. Reimer, T. D. Ladd, and R. S. Maxwell, Applied Physics Letters 87, 232109 (2005) • "Multispin dynamics of the solid-state NMR free induction decay," H. Cho, T. D. Ladd, J. Baugh, D. G. Cory, and C. Ramanathan, Physical Review B 72, 54427 (2005) • "Optical detection of the spin state of a single nucleus in silicon," K.-M. C. Fu, T. D. Ladd, C. Santori, and Y. Yamamoto, Physical Review B 69, 125306 (2004) • "All-silicon quantum computer," T. D. Ladd, J. R. Goldman, F. Yamaguchi, Y. Yamamoto, E. Abe, and K. M. Itoh, Physical Review Letters 89, 17901 (2002) • "Decoherence in crystal lattice quantum computation," T. D. Ladd, J. R. Goldman, F. Yamaguchi, and Y. Yamamoto, Appl. Phys. A 71, 27 (2000) • "Magnet designs for a crystal lattice quantum computer," J. R. Goldman, T. D. Ladd, F. Yamaguchi, and Y. Yamamoto, Appl. Phys. A 71, 11 (2000) • "Nonlinear AC response and noise of a giant magnetoresistive sensor," J. R. Petta, Thaddeus Ladd, and M. B. Weissman, IEEE Transactions on Magnetics 36, 2057 (2000) • "Knots in the æther: Theories of matter in the past and present," Thaddeus Ladd, Interface Journal 17, 3 (2000) • "A model for arbitrary plane imaging, or the brain in pain falls mainly on the plane," Jeff Miller, Dylan Helliwell, and Thaddeus Ladd, The UMAP Journal 19(3), 223 (1998)	{"url":"https://www.thaddeusladd.com/publications","timestamp":"2024-11-11T04:31:19Z","content_type":"text/html","content_length":"120581","record_id":"<urn:uuid:81c22087-2c19-43de-969e-7db1b9dae508>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00017.warc.gz"}
How to Measure a Rider for the Rider Anatomy Tab? Tue, Feb 4, 2014 How to Measure a Rider for the Rider Anatomy Tab? The image in the Rider Anatomy tab of the Rider dialogue doesn't provide enough detail to accurately measure a rider. How should I be measuring a rider - exactly? Tue, Feb 4, 2014 How to Measure a Rider for the Rider Anatomy Tab? Just to make sure we're on the same page, I assume you are talking about the diagram below: This stick figure representation can be arrived at in three different ways. First, you could import a photo of a rider on a bike and match that up with the rider in the BikeCAD model as described at: bikecad.ca/import_photo. Second, you could measure the rider using the six, more basic, body dimensions in the Fit advisor. You could then use the Fit advisor to approximate values for these more detailed dimensions. This is described at: bikecad.ca/rider_anatomy_tabs. The third option is to measure the body directly. You would need to use your own judgement to measure as closely to the center of each joint as possible. Of course, the joints of the human body are not simple pin joints as they are depicted here, so this is inherently an approximation. Regardless of how the dimensions for this rider model are obtained, BikeCAD does not in any way change the model of the bicycle based on the dimensions of the rider model. If you want BikeCAD to change the bike based on the dimensions of the rider, that is done in the Fit advisor. In the case of this particular rider model, the rider is only there to assist you in applying your own judgements in making modifications to the bicycle. Therefore, it is fitting that you are using your own judgment in creating the rider model as well. Wed, Feb 5, 2014 (Reply to #2) #3 Additional clarifictions on Anatomy Measurements I want my measurements to be as accurate as possible with little or no estimation which is why Fit Advisor is not useful. I will use joint locations as suggested but that still leaves a few 1) What are the start and end points for the NK measurement? 2) Should L3 + L2 + L1 + TR = distance from floor to sternal notch when the rider is standing? 3) is A2 from the elbow to the wrist joint? Or is it from elbow to the hand grip? 4) What is PD and how do I measure it? What impact does it have on the rider model? Wed, Feb 5, 2014 Additional clarifictions on Anatomy Measurements The base of NK is located at the top of TR. This is the point along the torso where the shoulders attach. The shoulders will not typically be shown right at this point because we also have dimension SR which accounts for the shoulders rolling forward from this point. The top of NK is located at the back of the jawbone. If we do assume that each of the joints in the body is a pin joint, we could assume that dimensions would add up as you suggest. However, you would need to take one more dimension into account. This has to do with your fourth question, the one about dimension PD. The dimensions AD, ED and KD have no impact on how far the body can reach or stretch out. They are merely there to give what is essentially a stick figure a little more realism. Dimension PD is similar; it is correlated to the size of the pelvis and the size of the upper thigh. However, it will also affect the height of the rider when stretched out. PD defines the diameter of a circle that rests upon the top of the saddle. The base of dimension TR is located at the center of this circle. Therefore, if you were to create a formula to predict the height of the shoulders above the ground, it should be: Shoulder height = L3 + L2 + L1 + PD/2 - HJ + TR. Dimension HJ represents where the hip joint is located above the Dimension A2 is from the elbow to the edge of the grip. If there is ever any confusion about what dimensions mean in BikeCAD, be aware that most dimensions can be shown on the screen using the Dimensions dialog box. It can also be insightful to change the dimension value to some very large number or some very small number and watch how that affects the model. Tue, Feb 11, 2014 (Reply to #4) #5 Thank you Brent. That was Thank you Brent. That was helpful. I'll do some measuring on a test subject and will also experiment with changing one dimension at a time to see the effect on the model. Log in or register to post comments	{"url":"https://www.bikecad.ca/comment/767","timestamp":"2024-11-02T21:23:09Z","content_type":"text/html","content_length":"34487","record_id":"<urn:uuid:fbc7680b-1245-484f-a913-1e5a6f0c5dd5>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00611.warc.gz"}
3.6 - The General Linear Test \| STAT 503 This is just a general representation of an F-test based on a full and a reduced model. We will use this frequently when we look at more complex models. Let's illustrate the general linear test here for the single factor experiment: First we write the full model, $Y_{ij} = \mu + \tau_i + \epsilon_{ij}$ and then the reduced model, $Y_{ij} = \mu + \epsilon_{ij}$ where you don't have a $\tau_i$ term, you just have an overall mean, $\mu$. This is a pretty degenerate model that just says all the observations are just coming from one group. But the reduced model is equivalent to what we are hypothesizing when we say the \ (\mu_i\) would all be equal, i.e.: $H_0 \colon \mu_1 = \mu_2 = \dots = \mu_a$ This is equivalent to our null hypothesis where the $\tau_i$'s are all equal to 0. The reduced model is just another way of stating our hypothesis. But in more complex situations this is not the only reduced model that we can write, there are others we could look at. The general linear test is stated as an F ratio: This is a very general test. You can apply any full and reduced model and test whether or not the difference between the full and the reduced model is significant just by looking at the difference in the SSE appropriately. This has an F distribution with (df R - df F), df F degrees of freedom, which correspond to the numerator and the denominator degrees of freedom of this F ratio. Let's take a look at this general linear test using Minitab... Example 3.5: Cotton Weight Section Remember this experiment had treatment levels 15, 20, 25, 30, 35 % cotton weight and the observations were the tensile strength of the material. The full model allows a different mean for each level of cotton weight %. We can demonstrate the General Linear Test by viewing the ANOVA table from Minitab: STAT > ANOVA > Balanced ANOVA The $SSE(R) = 636.96$ with a $dfR = 24$, and $SSE(F) = 161.20$ with $dfF = 20$. Therefore: $F^\ast =\dfrac{(636.96-161.20)/(24-20)}{161.20/20}$ This demonstrates the equivalence of this test to the F-test. We now use the General Linear Test (GLT) to test for Lack of Fit when fitting a series of polynomial regression models to determine the appropriate degree of polynomial. We can demonstrate the General Linear Test by comparing the quadratic polynomial model (Reduced model), with the full ANOVA model (Full model). Let $Y_{ij} = \mu + \beta_{1}x_{ij} + \beta_{2}x_{ij}^ {2} + \epsilon_{ij}$ be the reduced model, where $x_{ij}$ is the cotton weight percent. Let $Y_{ij} = \mu + \tau_i + \epsilon_{ij}$ be the full model. The General Linear Test - Cotton Weight Example (no sound) The video above shows the SSE(R) = 260.126 with dfR = 22 for the quadratic regression model. The ANOVA shows the full model with SSE(F) = 161.20 with dfF = 20. Therefore the GLT is: $\begin{eqnarray} F^\ast &=&\dfrac{(SSE(R)-SSE(F))/(dfR-dfF)}{SSE(F)/dfF} \nonumber\\ &=&\dfrac{(260.126-161.200)/(22-20)}{161.20/20}\nonumber\\ &=&\dfrac{98.926/2}{8.06}\nonumber\\ &=&\dfrac{49.46} {8.06}\nonumber\\&=&6.14 \nonumber \end{eqnarray}$ We reject $H_0\colon $ Quadratic Model and claim there is Lack of Fit if $F^{} > F_{1}-\alpha (2, 20) = 3.49$. Therefore, since 6.14 is > 3.49 we reject the null hypothesis of no Lack of Fit from the quadratic equation and fit a cubic polynomial. From the viewlet above we noticed that the cubic term in the equation was indeed significant with p-value = 0.015. We can apply the General Linear Test again, now testing whether the cubic equation is adequate. The reduced model is: $Y_{ij} = \mu + \beta_{1}x_{ij} + \beta_{2}x_{ij}^{2} + \beta_{3}x_{ij}^{3} + \epsilon_{ij}$ and the full model is the same as before, the full ANOVA model: $Y_ij = \mu + \tau_i + \epsilon_{ij}$ The General Linear Test is now a test for Lack of Fit from the cubic model: \begin{aligned} F^{} &=\frac{(\operatorname{SSE}(R)-\operatorname{SSE}(F)) /(d f R-d f F)}{\operatorname{SSE}(F) / d f F} \\ &=\frac{(195.146-161.200) /(21-20)}{161.20 / 20} \\ &=\frac{33.95 / 1} {8.06} \\ &=4.21 \end{aligned} We reject if $F^{*} > F_{0.95} (1, 20) = 4.35$. Therefore we do not reject $H_A \colon$ Lack of Fit and conclude the data are consistent with the cubic regression model, and higher order terms are not necessary.	{"url":"https://online.stat.psu.edu/stat503/lesson/3/3.6","timestamp":"2024-11-12T10:50:26Z","content_type":"text/html","content_length":"81864","record_id":"<urn:uuid:17fe6719-daa7-46e3-b943-47fc01950c67>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00743.warc.gz"}
Curtis Porter (NCSU), Geometric Methods in Representation Theory - Department of Mathematics Curtis Porter (NCSU), Geometric Methods in Representation Theory April 7, 2017 @ 4:00 pm - 5:00 pm Title: Straightening out degeneracy in CR Geometry: When can it be done? Abstract: CR geometry studies boundaries of domains in C^n and their generalizations. A central role is played by the Levi form L of a CR manifold M, which measures the failure of the CR bundle to be integrable, so that when L has a nontrivial kernel of constant rank, M is foliated by complex manifolds. If the local transverse structure to this foliation still determines a CR manifold N, then we say M is CR-straightenable, and the Tanaka-Chern-Moser classification of CR hypersurfaces with nondegenerate Levi form can be applied to N. It remains to classify those M for which L is degenerate and no such straightening exists. This was accomplished in dimension 5 by Ebenfelt, Isaev-Zaitzev, and Medori-Spiro. I will discuss their results as well as my recent progress on the problem in dimensions 7 and beyond.	{"url":"https://math.unc.edu/event/curtis-porter-geometric-methods-in-representation-theory/","timestamp":"2024-11-02T12:35:54Z","content_type":"text/html","content_length":"111747","record_id":"<urn:uuid:92ec111a-35ed-423c-a61b-f276ae59242c>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00343.warc.gz"}
You cannot select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long. This repo is archived. You can view files and clone it, but cannot push or open issues/pull-requests. This file contains invisible Unicode characters! This file contains invisible Unicode characters that may be processed differently from what appears below. If your use case is intentional and legitimate, you can safely ignore this warning. Use the Escape button to reveal hidden characters. #lang scribble/lp2 @(require scribble/manual aoc-racket/helper) @link["http://adventofcode.com/day/3"]{The puzzle}. Our @link-rp["day03-input.txt"]{input} is a string made of the characters @litchar{^v<>} that represent north, south, west, and east. Taken together, the string represents a path through an indefinitely large grid. In essence, this a two-dimensional version of the elevator problem in @secref{Day_1}. @isection{How many grid cells are visited?} In the elevator problem, we modeled the parentheses that represented up and down as @racket[1] and @racket[-1]. We'll proceed the same way here, but we'll assign Cartesian coordinates to each possible move — @racket['(0 1)] for north, @racket['(-1 0)] for west, and so on. For dual-valued data, whether to use @seclink["pairs" #:doc '(lib "scribblings/guide/guide.scrbl")]{pairs or lists} is largely a stylistic choice. Ask: what will you do with the data next? That will often suggest the most natural representation. In this case, the way we create each cell in the path is by adding the x and y coordinates of the current cell to the next move. So it ends up being convenient to model these cells as lists rather than pairs, so we can add them with a simple @racket[(map + current-cell next-move)]. (Recall that when you use @iracket[map] with multiple lists, it pulls one element from each list in parallel.) Once the whole cell path is computed, the answer is found by removing duplicate cells and counting how many remain. (require racket rackunit) (provide (all-defined-out)) (define (string->cells str) (define start '(0 0)) (match-define (list east north west south) '((1 0) (0 1) (-1 0) (0 -1))) (define moves (for/list ([s (in-list (regexp-match* #rx"." str))]) (case s [(">") east] [("^") north] [("<") west] [("v") south]))) (for/fold ([cells-so-far (list start)]) ([next-move (in-list moves)]) (define current-cell (car cells-so-far)) (define next-cell (map + current-cell next-move)) (cons next-cell cells-so-far))) (define (q1 str) (length (remove-duplicates (string->cells str))))] @subsection{Alternate approach: complex numbers} Rather than use Cartesian coordinates, we could rely on Racket's built-in support for complex numbers to trace the path in the complex plane. Complex numbers have a real and an imaginary part — e.g, @racket[3+4i] — and thus, represent points in a plane just as well as Cartesian coordinates. The advantage is that complex numbers are atomic values, not lists. We can add them normally, without resort to @racket[map]. (It's not essential for this problem, but math jocks might remember that complex numbers can be rotated 90 degrees counterclockwise by multiplying by @tt{+i}.) Again, the problem has nothing to do with complex numbers inherently. Like pairs and lists, they're just another option for encoding dual-valued data. @chunk[ <day03-q1-complex> (define (string->complex-cells str) (define start 0) (define east 1) (define moves (for/list ([s (in-list (regexp-match* #rx"." str))]) (* east (expt +i (case s [(">") 0] [("^") 1] [("<") 2] [("v") 3]))))) (for/fold ([cells-so-far (list start)]) ([next-move (in-list moves)]) (define current-cell (car cells-so-far)) (define next-cell (+ current-cell next-move)) (cons next-cell cells-so-far))) (define (q1-complex str) (length (remove-duplicates (string->complex-cells str)))) @section{How many grid cells are visited if the path is split?} By ``split'', the puzzle envisions two people starting at the origin, with one following the odd-numbered moves, and the other following the even-numbered moves. So there are two paths instead of one. The question remains the same: how many cells are visited by one path or the other? The solution works the same as before — the only new task is to split the input into two strings, and then send them through our existing @racket[string->cells] function. (define (split-odds-and-evens str) (define-values (odd-chars even-chars) (for/fold ([odds-so-far empty][evens-so-far empty]) ([c (in-string str)][i (in-naturals)]) (if (even? i) (values odds-so-far (cons c evens-so-far)) (values (cons c odds-so-far) evens-so-far)))) (values (string-append* (map ~a (reverse odd-chars))) (string-append* (map ~a (reverse even-chars))))) (define (q2 str) (define-values (odd-str even-str) (split-odds-and-evens str)) (length (remove-duplicates (append (string->cells odd-str) (string->cells even-str))))) @section{Testing Day 3} (module+ test (define input-str (file->string "day03-input.txt")) (check-equal? (q1 input-str) 2565) (check-equal? (q1-complex input-str) 2565) (check-equal? (q2 input-str) 2639))]	{"url":"https://git.matthewbutterick.com/mbutterick/aoc-racket/src/branch/master/day03.rkt","timestamp":"2024-11-05T06:31:10Z","content_type":"text/html","content_length":"93253","record_id":"<urn:uuid:d9b58960-e324-4fde-af1b-63edf9d89943>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00562.warc.gz"}
CS 4/5789: Lecture 13 CS 4/5789: Introduction to Reinforcement Learning Lecture 13 Prof. Sarah Dean MW 2:45-4pm 110 Hollister Hall 0. Announcements & Recap 1. Q Function Approximation 2. Optimization & Gradient Descent 3. Stochastic Gradient Descent 4. Derivative-Free Optimization HW2 released next Monday 5789 Paper Review Assignment (weekly pace suggested) OH cancelled today, instead Thursday 10:30-11:30am Prelim Tuesday 3/22 at 7:30-9pm in Phillips 101 Closed-book, definition/equation sheet for reference will be provided Focus: mainly Unit 1 (known models) but many lectures in Unit 2 revisit important key concepts Study Materials: Lecture Notes 1-15, HW0&1 Lecture on 3/21 will be a review Meta-Algorithm for Policy Iteration in Unknown MDP • Sample $h_1=h$ w.p. $\propto \gamma^h$: $(s_{h_1}, a_{h_1}) = (s_i,a_i) \sim d^\pi_{\mu_0}$ • Sample $h_2=h$ w.p. $\propto \gamma^h$: $y_i = \sum_{t=h_1}^{h_1+h_2} r_t$ Supervision with Rollout (MC): $\mathbb{E}[y_i] = Q^\pi(s_i, a_i)$ $\widehat Q$ via ERM on $\{(s_i, a_i, y_i)\}_{1}^N$ $s_{t+1}\sim P(s_t, a_t)$ $a_{t+1}\sim \pi(s_{t+1})$ • $y_t =r_t + \gamma \max_a \widehat Q(s_{t+1},a) $ Supervision with Bellman Exp (TD): If $\widehat Q = Q^\pi$ then $\mathbb{E}[y_t] = Q^\pi(s_t, a_t)$ $s_{t+1}\sim P(s_t, a_t)$ $a_{t+1}\sim \pi(s_{t+1})$ Supervision with Bellman Opt (TD): • $y_t =r_t + \gamma \widehat Q(s_{t+1},a_{t+1}) $ If $\widehat Q = Q^$ then $\mathbb{E}[y_t] = Q^(s_t, a_t)$ SARSA and Q-learning are simple tabular algorithms 0. Announcements & Recap 1. Q Function Approximation 2. Optimization & Gradient Descent 3. Stochastic Gradient Descent 4. Derivative-Free Optimization CS 4/5789: Lecture 13 By Sarah Dean	{"url":"https://slides.com/sarahdean-2/cs-4-5789-lecture-13?token=vZ65xvFx","timestamp":"2024-11-15T04:39:15Z","content_type":"text/html","content_length":"51623","record_id":"<urn:uuid:b0db8a49-0cbc-4f3d-ab95-632a68eb505a>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00786.warc.gz"}
Parameter Estimates Report When X′X is singular, a generalized inverse is used to obtain estimates. This approach permits some, but not all, of the parameters involved in a linear dependency to be estimated. Parameters are estimated based on the order of entry of their associated terms into the model, so that the last terms entered are the ones whose parameters are not estimated. Estimates are given in the Parameter Estimates report, and parameters that cannot be estimated are given estimates of 0. However, estimates of parameters for terms involved in linear dependencies are not unique. Because the associated terms are aliased, there are infinitely many vectors of estimates that satisfy the least squares criterion. In these cases, “Biased” appears to the left of these estimates in the Parameter Estimates report. “Zeroed” appears to the left of the estimates of 0 in the Parameter Estimates report for terms involved in a linear dependency whose parameters cannot be estimated. For an example, see Figure 3.64. If there are degrees of freedom available for an estimate of error, t tests for parameters estimated using biased estimates are conducted. These tests should be interpreted with caution, though, given that the estimates are not unique.	{"url":"https://www.jmp.com/support/help/en/16.2/jmp/parameter-estimates-report-2.shtml","timestamp":"2024-11-04T09:11:44Z","content_type":"application/xhtml+xml","content_length":"6631","record_id":"<urn:uuid:cdff3677-da33-4cba-ba82-f652112c1a66>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00273.warc.gz"}
Amar Sagoo I love the feeling of getting to understand a seemingly abstract concept in intuitive, real-world terms. It means you can comfortably and freely use it in your head to analyse and understand things and to make predictions. No formulas, no paper, no Greek letters. It’s the basis for effective analytical thinking. The best measure of whether you’ve “got it” is how easily you can explain it to someone and have them understand it to the same extent. I think I recently reached that point with understanding standard deviation, so I thought I’d share those insights with you. Standard deviation is one of those very useful and actually rather simple mathematical concepts that most people tend to sort-of know about, but probably don’t understand to a level where they can explain why it is used and why it is calculated the way it is. This is hardly surprising, given that good explanations are rare. The Wikipedia entry, for instance, like all entries on mathematics and statistics, is absolutely impenetrable. First of all, what is deviation? Deviation is simply the “distance” of a value from the mean of the population that it’s part of: Now, it would be great to be able to summarise all these deviations with a single number. That’s exactly what standard deviation is for. But why don’t we simply use the average of all the deviations, ignoring their sign (the mean absolute deviation or, simply, mean deviation)? That would be quite easy to calculate. However, consider the following two variables (for simplicity, I will use data sets with a mean of zero in all my examples): There’s obviously more variation in the second data set than in the first, but the mean deviation won’t capture this; it’s 2 for both variables. The standard deviation, however, will be higher for the second variable: 2.24. This is the crux of why standard deviation is used. [S:In finance, it’s called volatility, which I think is a great, descriptive name: the second variable is more volatile than the first.:S] [Update: It turns out I wasn't being accurate here. Volatility is the standard deviation of the changes between values – a simple but significant difference.] Dispersion is another good word, but unfortunately it already has a more general meaning in statistics. Next, let’s try to understand why this works; that is, how does the calculation of standard deviation capture this extra dispersion on top of the mean deviation? Standard deviation is calculated by squaring all the deviations, taking the mean of those squares and finally taking the square root of that mean. It’s the root-mean-square (RMS) deviation (N below is the size of the sample): RMS Deviation = √(Sum of Squared Deviations / N) Intuitively, this may sound like a redundant process. (In fact, some people will tell you that this is done purely to eliminate the sign on the negative numbers, which is nonsense.) But let’s have a look at what happens. The green dots in the first graph below are the absolute deviations of the grey dots, and the blue dots in the second graph are the squared deviations: The dotted blue line at 5 is the mean of the squared deviations (this is known as the variance). The square root of that is the RMS deviation, lying just above 2. Here you can see why the calculation works: the larger values get amplified compared to the smaller ones when squared, “pulling up” the resulting root-mean-square. That’s mostly all there’s to it, really. However, there’s one more twist to calculating standard deviation that is worth understanding. The problem is that, usually, you don’t have data on a complete population, but only on a limited sample. For example, you may do a survey of 100 people and try to infer something about the population of a whole city. From your data, you can’t determine the true mean and the true standard deviation of the population, only the sample mean and an estimate of the standard deviation. The sample values will tend to deviate less from the sample mean than from the true mean, because the sample mean itself is derived from, and therefore “optimised” for, the sample. As a consequence, the RMS deviation of a sample tends to be smaller than the true standard deviation of the population. This means that even if you take more and more samples and average their RMS deviations, you will not eventually reach the true standard deviation. It turns out that to get rid of this so-called bias, you need to multiply your estimate of the variance by N/(N-1). (This can be mathematically proven, but unfortunately I have not been able to find a nice, intuitive explanation for why this is the correct adjustment.) For the final formula, this means that instead of taking a straightforward mean of the squared deviations, we sum them and divide by the sample size minus 1: Estimated SD = √(Sum of Squared Deviations / (N - 1)) You can see how this will give you a slightly higher estimate than a straight root-mean-square, and how the larger the sample size, the less significant this adjustment becomes. Update: Some readers have pointed out that using the square to "amplify" larger deviations seems arbitrary: why not use the cube or even higher powers? I'm looking into this, and will update this article once I've figured it out or if my explanation turns out to be incorrect. If anybody who understands this better than me can clarify, please leave a comment.	{"url":"https://blog.amarsagoo.info/2007/09/","timestamp":"2024-11-12T01:47:53Z","content_type":"application/xhtml+xml","content_length":"37740","record_id":"<urn:uuid:ddbf0cb8-6030-4f3d-8756-28b27c0008bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00339.warc.gz"}
Practical Geometry - Edu Spot- NCERT Solution, CBSE Course, Practice Test Practical Geometry Construction of Triangles 1. Properties of triangles • The exterior angle of a triangle is equal in measure to the sum of interior opposite angles. • The total measure of the three angles of a triangle is 180°. • Sum of the lengths of any two sides of a triangle is greater than the length of the third side. • In any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. 2. Essential measurements for the construction of a triangle A triangle can be drawn if any one of the following sets of measurements is given: • Three sides: SSS • Two sides and the angle between them: SAS • Two angles and the side between them: ASA • The hypotenuse and a leg in the case of a right-angled triangle: RHS Steps of construction of a line parallel to a given line 1. Take a line l and a point A outside l. 2. Take any point B on l and join it to A. 3. With B as the centre and a convenient radius, cut an arc on l at C and BA at D. 4. With A as the centre and same radius as in Step 3, cut an arc EF to cut AB at G. 5. Measure the arc length CD by placing pointed tip of the compass at C and pencil tip opening at D. 6. With this opening, keep G as centre and draw an arc to cut arc EF at H 7. Join AH to draw a line m ∠ABC and ∠BAH are alternate interior angles. Therefore, m \|\| l Construction of a triangle : Construction of a triangle with SSS criterion. Construct a triangle ABC, given that AB = 5 cm, BC = 6 cm and AC = 7 cm. 1. Make a rough sketch for your reference 2. Draw a line segment BC = 6 cm Step 1: First, we draw a rough sketch with a given measure. Step 2: Draw a line segment BC of length 6 cm. Step 3: From B, point A is at a distance of 5 cm. So with B as centre, draw an arc of radius 5 cm. Step 4: From C, point A is at a distance of 7 cm. So, with C as centre, draw an arc of radius 7 cm. Step 5: A has to be on both the arcs drawn. So, it is the point of intersection of arc. Mark the point of intersection of arcs as A. Join AB and AC. ΔABC is now ready. Construction of a triangle with SAS criterion • Construct ΔPQR with QR = 7.5 cm, PQ = 5 cm and ∠Q = 60^0. 1. Make a rough sketch for your reference 2. Draw a line segment QR = 7.5 cm 3. At Q, draw QX making 60^0 with QR 4. With Q as centre, draw an arc of radius 5 cm. It cuts QX at P. 5. Join AB. ΔPQR is now ready Construction of a triangle with ASA criterion Steps of Construction Step 1: Before actual construction, we draw a rough sketch with measures marked on it. Step 2: Draw XY of length 6 cm. Step 3: At X, draw a ray XP making an angle of 30° with XY. By the given condition Z must be somewhere on the XP. Step 4: At Y, draw a ray YQ making an angle of 100° with YX. By the given condition, Z must be on the ray YQ also. Step 5: Z has to lie on both the rays XP and YQ. So, the point of intersection of two rays is Z. ΔXYZ is now completed. Construction of a triangle with RHS criterion • Construct ΔLMN, where ∠M = 90^0, MN = 8cm and LN = 10 cm. 1. Make a rough sketch for your reference 2. Draw MN = 8 cm 3. At M, draw MX ⊥ MN. 4. With N as centre, draw an arc of radius 10 cm to cut MX at L 5. Join LN. 6. ΔLMN is now completed NCERT Solutions Back to CBSE Class 7th Maths	{"url":"https://edu-spot.com/lessons/practical-geometry-2/","timestamp":"2024-11-06T23:40:02Z","content_type":"text/html","content_length":"72034","record_id":"<urn:uuid:591e8eab-a575-4fab-b5a2-e1190c32d848>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00697.warc.gz"}
Root Calculator - cryptocrape.com Here’s a quick guide on how to use the Root Calculator: 1. Enter a Number: □ In the input field labeled “Enter a number,” type any positive decimal number you want to calculate the root of. 2. Choose Root Type: □ Square Root: Select this option to find the square root of the entered number. □ Cube Root: Select this option to find the cube root of the entered number. □ General Root: Select this option if you want to find a specific root (e.g., 4th root). A new field labeled “Enter the degree of root” will appear where you can specify the root degree. 3. Calculate: □ Click on the Calculate button to see the result. The result will display the selected root calculation in an easy-to-read format. 4. Reset: □ If you want to start over, click the Reset button to clear the input fields and results. Here are examples for each type of root calculation: 1. Square Root Example: □ Input: 16 □ Operation: Square Root □ Result: 4 □ Explanation: The square root of 16 is the number that, when multiplied by itself, equals 16. So, 4×4=164 \times 4 = 16. 2. Cube Root Example: □ Input: 27 □ Operation: Cube Root □ Result: 3 □ Explanation: The cube root of 27 is the number that, when raised to the power of 3, equals 27. So, 3×3×3=273 \times 3 \times 3 = 27. 3. General Root Example: □ Input: 81 □ Operation: General Root (4th Root) □ Root Degree: 4 □ Result: 3 □ Explanation: The 4th root of 81 is the number that, when raised to the power of 4, equals 81. So, 3×3×3×3=813 \times 3 \times 3 \times 3 = 81.	{"url":"https://cryptocrape.com/root-calculator/","timestamp":"2024-11-14T02:22:01Z","content_type":"text/html","content_length":"89417","record_id":"<urn:uuid:bdeac0b9-a014-40f2-b24e-dd9496241205>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00224.warc.gz"}
Understanding Mathematical Functions: How To Find The Range Of A Multi Mathematical functions play a crucial role in various fields such as science, engineering, and economics. They are used to describe the relationship between different variables and are essential for making predictions and analyzing patterns. When dealing with multivariable functions, it is important to understand how to find their range. The range of a function is the set of all possible values that the function can take, and it provides valuable insights into the behavior of the function. In this blog post, we will explore the importance of finding the range of a multivariable function and the methods to do so effectively. Key Takeaways • Mathematical functions are essential in various fields and are used to describe relationships between variables. • The range of a multivariable function is the set of all possible values it can take, providing valuable insights into its behavior. • Finding the range of a multivariable function involves understanding its domain, using mathematical techniques, and considering constraints. • Tools such as graphs, calculus, and evaluating critical points can be used to find the range of a multivariable function. • Exploring step-by-step examples and practicing problems can help in understanding and applying the concept of finding the range of a multivariable function. Understanding Mathematical Functions In mathematics, a mathematical function is a relation between a set of inputs and a set of possible outputs. It assigns each input value exactly one output value. There are various types of mathematical functions, each with its own characteristics and properties. Some of the common types of mathematical functions include linear, quadratic, and exponential Definition of a mathematical function A mathematical function is a rule that assigns to each element in a set A exactly one element in a set B. The set A is called the domain of the function, and the set B is called the codomain. The elements of B that actually have a rule are called the range. Types of mathematical functions 1. Linear functions: These functions have a constant rate of change and graph as a straight line. The general form of a linear function is y = mx + b, where m is the slope of the line and b is the 2. Quadratic functions: These functions have a squared term and graph as a parabola. The general form of a quadratic function is y = ax^2 + bx + c, where a, b, and c are constants. 3. Exponential functions: These functions have a variable in the exponent and graph as a curve that increases or decreases rapidly. The general form of an exponential function is y = ab^x, where a and b are constants and b is the base. Multivariable Functions Understanding mathematical functions is crucial in the field of mathematics and its applications. In this chapter, we will delve into the world of multivariable functions and explore how to find the range of such functions. A. Definition of a multivariable function A multivariable function, also known as a multivariate function, is a function that depends on multiple variables. In other words, it takes in more than one input and produces a single output. Mathematically, a multivariable function can be represented as f(x, y) = z, where x and y are the input variables, and z is the output variable. B. Examples of multivariable functions There are many real-world examples of multivariable functions, such as the temperature at different points in a room, the velocity of an object in three-dimensional space, and the concentration of a substance in a chemical reaction. These examples demonstrate the need to consider multiple variables to understand and analyze certain phenomena. C. Importance of finding the range of a multivariable function Finding the range of a multivariable function is important for several reasons. Firstly, it helps in understanding the behavior of the function and how it relates to its input variables. Secondly, it allows for the identification of possible outputs and helps in determining the limitations or boundaries of the function. Lastly, knowing the range of a multivariable function is crucial for making predictions and decision-making in various fields such as physics, engineering, economics, and more. Understanding Mathematical Functions: How to find the range of a multivariable function When dealing with multivariable functions, finding the range can be a complex but essential task. The range of a function is the set of all possible output values it can produce. In this chapter, we will explore the process of finding the range of a multivariable function. A. Exploring the domain of the function Before we can determine the range of a multivariable function, it's crucial to understand the domain. The domain of a function is the set of all possible input values it can accept. By exploring the domain, we can identify the potential output values that the function can generate. B. Using mathematical techniques to find the range Once we have a clear understanding of the domain, we can utilize various mathematical techniques to find the range of the multivariable function. These techniques may include analyzing the function's behavior, using calculus to find critical points, and applying mathematical principles such as optimization and inequalities. C. Consideration of constraints or limitations When dealing with multivariable functions, it's essential to consider any constraints or limitations that may affect the range. Constraints can arise from physical or real-world limitations, mathematical boundaries, or inequalities that restrict the possible output values of the function. By carefully considering these constraints, we can accurately determine the range of the multivariable function. Tools and Techniques for Finding the Range Understanding how to find the range of a multivariable function is an essential skill in mathematics. There are several tools and techniques that can be used to effectively determine the range of a function, including utilizing graphs and visual representations, using calculus to find critical points, and evaluating the function at critical points. A. Utilizing graphs and visual representations • Visualizing the function: Graphing the function can provide a visual representation of its behavior and help identify potential range values. • Identifying patterns: Analyzing the graph for any recurring patterns or trends can be useful in determining the range of the function. • Using contour plots: For multivariable functions, contour plots can be utilized to visually represent the function's behavior in the input space. B. Using calculus to find critical points • Calculating partial derivatives: Taking the partial derivatives of the function with respect to each variable can help identify critical points. • Finding the gradient: The gradient of the function can be used to locate points where the function's rate of change is zero. • Setting partial derivatives equal to zero: By setting the partial derivatives of the function equal to zero, critical points can be identified for further evaluation. C. Evaluating the function at critical points • Substituting critical points into the function: Once critical points have been identified, substituting these values into the function allows for the evaluation of the function at these points. • Identifying maximum and minimum values: By comparing the function values at critical points, the maximum and minimum values of the function can be determined, helping to establish the range. Examples and Practice Problems Understanding how to find the range of a multivariable function is an important concept in mathematics. Let's go through some step-by-step examples and practice problems to help solidify your A. Step-by-step examples of finding the range of a multivariable function • Example 1: Consider the function f(x, y) = x^2 + y^2. To find the range of this function, we can start by considering the possible values for x and y. Since both x and y can take any real value, the range of f(x, y) will be all non-negative real numbers. Therefore, the range of the function is [0, ∞). • Example 2: Now, let's take the function g(x, y) = 2x + y. In this case, the range will depend on the values of x and y. If x is allowed to vary over all real numbers, and y is allowed to vary over all real numbers, then the range of g(x, y) will be all real numbers. Therefore, the range of the function is (-∞, ∞). B. Practice problems for readers to solve on their own • Problem 1: Find the range of the function h(x, y) = x^2 - 3y + 5. • Problem 2: Determine the range of the function k(x, y) = xy + 4. In conclusion, understanding the range of a multivariable function is crucial in understanding the behavior and outcomes of the function in various scenarios. It helps in identifying the possible values that the function can take, allowing us to make informed decisions and predictions in real-world applications. As with any mathematical concept, further exploration and practice are key to mastering the process of finding the range of a multivariable function. By applying different techniques and studying various examples, one can enhance their understanding and problem-solving skills in this area of mathematics. 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entimeters to Kens Centimeters to Kens Converter β Switch toKens to Centimeters Converter How to use this Centimeters to Kens Converter π € Follow these steps to convert given length from the units of Centimeters to the units of Kens. 1. Enter the input Centimeters value in the text field. 2. The calculator converts the given Centimeters into Kens in realtime β using the conversion formula, and displays under the Kens label. You do not need to click any button. If the input changes, Kens value is re-calculated, just like that. 3. You may copy the resulting Kens value using the Copy button. 4. To view a detailed step by step calculation of the conversion, click on the View Calculation button. 5. You can also reset the input by clicking on button present below the input field. What is the Formula to convert Centimeters to Kens? The formula to convert given length from Centimeters to Kens is: Length[(Kens)] = Length[(Centimeters)] / 211.8360000208633 Substitute the given value of length in centimeters, i.e., Length[(Centimeters)] in the above formula and simplify the right-hand side value. The resulting value is the length in kens, i.e., Length Calculation will be done after you enter a valid input. Consider that a high-end smartphone has a screen size of 15 centimeters. Convert this screen size from centimeters to Kens. The length in centimeters is: Length[(Centimeters)] = 15 The formula to convert length from centimeters to kens is: Length[(Kens)] = Length[(Centimeters)] / 211.8360000208633 Substitute given weight Length[(Centimeters)] = 15 in the above formula. Length[(Kens)] = 15 / 211.8360000208633 Length[(Kens)] = 0.07080949413 Final Answer: Therefore, 15 cm is equal to 0.07080949413 ken. The length is 0.07080949413 ken, in kens. Consider that a luxury handbag measures 30 centimeters in width. Convert this width from centimeters to Kens. The length in centimeters is: Length[(Centimeters)] = 30 The formula to convert length from centimeters to kens is: Length[(Kens)] = Length[(Centimeters)] / 211.8360000208633 Substitute given weight Length[(Centimeters)] = 30 in the above formula. Length[(Kens)] = 30 / 211.8360000208633 Length[(Kens)] = 0.1416 Final Answer: Therefore, 30 cm is equal to 0.1416 ken. The length is 0.1416 ken, in kens. Centimeters to Kens Conversion Table The following table gives some of the most used conversions from Centimeters to Kens. Centimeters (cm) Kens (ken) 0 cm 0 ken 1 cm 0.00472063294 ken 2 cm 0.00944126588 ken 3 cm 0.01416189883 ken 4 cm 0.01888253177 ken 5 cm 0.02360316471 ken 6 cm 0.02832379765 ken 7 cm 0.03304443059 ken 8 cm 0.03776506354 ken 9 cm 0.04248569648 ken 10 cm 0.04720632942 ken 20 cm 0.09441265884 ken 50 cm 0.236 ken 100 cm 0.4721 ken 1000 cm 4.7206 ken 10000 cm 47.2063 ken 100000 cm 472.0633 ken A centimeter (cm) is a unit of length in the International System of Units (SI). One centimeter is equivalent to 0.01 meters or approximately 0.3937 inches. The centimeter is defined as one-hundredth of a meter, making it a convenient measurement for smaller lengths. Centimeters are used worldwide to measure length and distance in various fields, including science, engineering, and everyday life. They are commonly used in everyday measurements, such as height, width, and depth of objects, as well as in educational settings. A ken is a historical unit of length used in various cultures, particularly in Asia. The length of a ken can vary depending on the region and context. In Japan, one ken is approximately equivalent to 6 feet or about 1.8288 meters. The ken was traditionally used in architectural and construction measurements, particularly in the design of buildings and layout of spaces. Ken measurements were utilized in historical architecture and construction practices in Asian cultures. Although not commonly used today, the unit provides historical context for traditional measurement standards and practices in building and design. Frequently Asked Questions (FAQs) 1. What is the formula for converting Centimeters to Kens in Length? The formula to convert Centimeters to Kens in Length is: Centimeters / 211.8360000208633 2. Is this tool free or paid? This Length conversion tool, which converts Centimeters to Kens, is completely free to use. 3. How do I convert Length from Centimeters to Kens? To convert Length from Centimeters to Kens, you can use the following formula: Centimeters / 211.8360000208633 For example, if you have a value in Centimeters, you substitute that value in place of Centimeters in the above formula, and solve the mathematical expression to get the equivalent value in Kens.	{"url":"https://convertonline.org/unit/?convert=centimeters-kens","timestamp":"2024-11-15T02:46:03Z","content_type":"text/html","content_length":"90998","record_id":"<urn:uuid:610491c2-4ac3-404c-856b-d41b2b2a8f60>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00642.warc.gz"}
The power of adaptivity in source identification with time queries on the path Abstract (may include machine translation) We study the problem of identifying the source of a stochastic diffusion process spreading on a graph based on the arrival times of the diffusion at a few queried nodes. In a graph G=(V,E), an unknown source node v^⁎∈V is drawn uniformly at random, and unknown edge weights w(e) for e∈E, representing the propagation delays along the edges, are drawn independently from a Gaussian distribution of mean 1 and variance σ^2. An algorithm then attempts to identify v^⁎ by querying nodes q∈V and being told the length of the shortest path between q and v^⁎ in graph G weighted by w. We consider two settings: non-adaptive, in which all query nodes must be decided in advance, and adaptive, in which each query can depend on the results of the previous ones. Both settings are motivated by an application of the problem to epidemic processes (where the source is called patient zero), which we discuss in detail. We characterize the query complexity when G is an n-node path. In the non-adaptive setting, Θ(nσ^2) queries are needed for σ^2≤1, and Θ(n) for σ^2≥1. In the adaptive setting, somewhat surprisingly, only Θ(log⁡log[1/σ]⁡n) are needed when σ^2≤1/2, and Θ(log⁡log⁡n)+O[σ] (1) when σ^2≥1/2. This is the first mathematical study of source identification with time queries in a non-deterministic diffusion process. • Graph algorithms • Lower bounds • Noisy information • Source location Dive into the research topics of 'The power of adaptivity in source identification with time queries on the path'. Together they form a unique fingerprint.	{"url":"https://research.ceu.edu/en/publications/the-power-of-adaptivity-in-source-identification-with-time-querie","timestamp":"2024-11-10T12:24:27Z","content_type":"text/html","content_length":"49474","record_id":"<urn:uuid:8b5b822f-7a29-49e2-8421-3807250d1c61>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00382.warc.gz"}
Javascript Array reduce() Kodeclik Blog Javascript’s Array reduce() method You will sometimes have an array in Javascript and have a need to reduce the array into a single value or perform other accumulated operations on it. The Javascript Array reduce() method is perfect for this purpose! Let us learn how it works. We will embed our Javascript code inside a HTML page like so: <!DOCTYPE html> <meta charset="utf-8"> <meta name="viewport" content="width=device-width"> <title>Javascript’s Array reduce() method</title> In the above HTML page you see the basic elements of the page like the head element containing the title (“Javascript’s Array reduce() method”) and a body with an empty script tag. The Javascript code goes inside these tags. Summing the elements of an array Assume we define a Javascript array of ages of students like so: const ages = [10,11,12,11,13,15] Our objective is to find the total of all these ages. We can iterate over the loop, of course, but let us learn how to achieve the same objective using the Array reduce() method. The reduce() method takes a function as argument and applies it systematically over the array. The function itself has two required arguments: the first is the accumulated value and the second is the current value (element). Below is the code to sum the elements of this array: const ages = [10,11,12,11,13,15] const totalages= ages.reduce(function(total, currentValue) { return total + currentValue Note that we are applying the reduce() method on the ages array and passing as an argument a function. The function takes two arguments as defined above. The semantics of the function is quite simple: it merely adds the currentValue to the (running) total. The result of applying this method is stored in “totalages” which is then printed. The output of this program is thus: Optionally you can use an initial value which should be passed as an argument after the function (not as an argument to the function itself). Let us try this with an initial value of -72, so that the total should now become zero. const ages = [10,11,12,11,13,15] const totalages= ages.reduce(function(total, currentValue) { return total + currentValue The output will be, as expected: Summing the squares of elements of an array Let us adapt our first program to sum the squares of elements of the array instead of summing the elements. This is quite straightforward to implement: const numbers = [1,2,3,4,5] const sumOfSquaredValues = numbers.reduce(function(total, currentValue) { return total + currentValue*currentValue because this is the sum of the squares of the first five positive integers. Finding the average of elements of an array Let us try to find the average of the numbers in an array. This is a very minor modification of the sum code earlier. We compute the sum and then use the length() method on the array to find the const numbers = [1,2,3,4,5] const sum = numbers.reduce(function(runningSum, currentValue) { return (runningSum + currentValue) Finding the maximum of elements of an array Next, we will write a function to compute the maximum of the elements in an array and use it with the reduce() method. const numbers = [1,2,5,4,3] const max = numbers.reduce(function(runningMax, currentValue) { return (runningMax > currentValue ? runningMax : currentValue) The function takes a runningMax and compares it with the currentValue. If the currentValue is greater (or equal) then it becomes the runningMax. The initial value for the reduce function is the first element of the array, i.e., numbers[0]. The output of the program is: We hope you enjoyed learning about the Array reduce() method. Can you think of more examples?	{"url":"https://www.kodeclik.com/javascript-array-reduce-method/","timestamp":"2024-11-02T01:41:30Z","content_type":"text/html","content_length":"105453","record_id":"<urn:uuid:88b2b067-4af3-4b9c-b3c3-3c57f87a4c7c>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00798.warc.gz"}
Top 10 Websites for High School Math Worksheets Mathematics is a fundamental subject that plays a pivotal role in high school education. To excel in math, students need comprehensive resources that provide ample practice and support. High school math worksheets are invaluable tools designed to enhance math proficiency. Our high school math tutors emphasize practicing math worksheets regularly to improve your math skills and In this article, we’ll explore the importance of high school math worksheets and present a list of 10 websites that offer a treasure trove of math worksheets tailored to the US syllabus. Let’s begin! Looking to Learn High School Math? Book a Free Trial Lesson and match with top High School Math Tutors for Concepts, Homework Help, and Test Prep. The Importance of High School Math Worksheets High school math worksheets are essential for several reasons: • Practice Makes Perfect: Math requires practice, and worksheets provide abundance of problems for students to solve, reinforcing their understanding of mathematical concepts. • Learning at a comfortable pace: Worksheets allow students to work at their own pace and target areas where they need additional practice or improvement. • Exam Preparation: Math worksheets are excellent tools for test preparation, helping students become confident and proficient in tackling math assessments. ⭐ Useful resource📖: What is High School Math? Without further ado, here are 10 websites that exclusively provide high school math worksheets: 10 websites that provide high school math worksheets Math-Aids.com is a comprehensive resource offering a wide range of math worksheets for high school students. Worksheets cover topics from basic arithmetic to advanced calculus. Kuta Software provides math worksheets specifically designed for high school students. The site offers customizable worksheets and a broad selection of topics. Math Worksheets Land offers a vast collection of free math worksheets aligned with the Common Core standards. Worksheets cater to various skill levels and cover topics from algebra to geometry. Super Teacher Worksheets offers math worksheets for high school students, including algebra, geometry, and calculus. The site provides resources for teachers, parents, and homeschoolers. Math-Drills.com is a user-friendly platform with a wide selection of math worksheets, including word problems, algebra, and geometry. Worksheets are available in various formats, including PDF and TeAch-nology provides free math worksheets for high school students. The site covers various math topics and offers printable worksheets for classroom use. WorksheetWorks.com allows users to create custom math worksheets tailored to their needs. It’s a versatile tool for generating personalized math practice sheets. HomeschoolMath.net offers a collection of math worksheets suitable for high school students. The site also provides resources for homeschooling families. Softschools offers interactive math worksheets and quizzes for high school students. These engaging resources help students practice math concepts in a fun way. MathGoodies offers a selection of free math worksheets, interactive lessons, and puzzles. The site covers various high school math topics and provides detailed explanations. These websites are dedicated to providing high-quality math worksheets that align with the US high school math curriculum. Whether you’re a student looking for extra practice or an educator in search of supplemental resources, these websites have you covered, helping you elevate your math skills and succeed in your math journey. 💡Useful Resources 📖 High school math can be challenging, but it is also an essential skill for success in college and career. If you are looking for extra practice or help with a specific topic, you can get help from these websites that offer mostly free high school math worksheets. Hopefully, the above websites will help you with high school worksheets on important high school topics. If you need extra personalized guidance with high school math, you must consider getting help from a high school math tutor.	{"url":"https://wiingy.com/resources/math/best-high-school-math-worksheets/","timestamp":"2024-11-14T04:45:54Z","content_type":"text/html","content_length":"156601","record_id":"<urn:uuid:625288a9-74a4-448f-815d-7f38bdb08835>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00566.warc.gz"}
fourier transform of impulse train Hi everybody.. could any one plz explain me how to get the spectra of dirac com function.my doubt is that what type of signals are fourier transformable?Can we take fourier transform for a sine/cosine signal.I get confused vth fourier series and transform while apllying for a signal. Thanks for all who gonna reply.	{"url":"https://www.dsprelated.com/showthread/comp.dsp/72649-1.php","timestamp":"2024-11-12T15:11:40Z","content_type":"text/html","content_length":"68090","record_id":"<urn:uuid:c5ddb597-0567-40bc-9ac8-eadae9a5d8c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00382.warc.gz"}
Registration for exercise sessions now open via the Physics Übungsgruppenverwaltungssystemportal First lecture on October 18 This course is a basic introduction to string theory. Topics to be covered include: Quantization of the bosonic string, string interactions, the superstring, space-time effective actions, string compactifications, D-branes and string dualities, AdS/CFT correspondence. (We won't do all of this.) Prerequisites: Thorough knowledge of classical theoretical physics, including quantum mechanics and relativity as ingredients of quantum field theory. An understanding of particle physics and some mathematical background will also be useful. Credits: This course is listed as a specialization module in the physics Master programme, and as an "Aufbaumodul" in the mathematics Master focus area "Modular forms, complex analysis, and mathematical physics". The course will be evaluated based on a 24h take-home final exam. Tentative submission deadline: February 14, 2023 @ noon. To be admitted to the exam, you have to be registered (via the Physics Übungsgruppenverwaltungssystemportal) for the exercise sessions and submit valid solutions to at least 50% of homework problems. We intend to upload problems sets on Mondays at noon (give or take). Submission deadline is the following Monday, at noon (sharp). Instructor: Prof. J. Walcher, Email Tutor: Hannes Keppler, Email Time and Place: Lectures on Tuesday & Thursday, 11-13, Philosophenweg 12, gHS. (Low video quality) recordings from last year's course are available on MaMpf and might or might not be renewed this time around. Exercise sessions on Wednesday, 16-18, Philosophenweg 12, gHS. Timo Weigand's Lecture Notes Introduction to String Theory Green-Schwarz-Witten Superstring Theory (2 vols.) Polchinski, String Theory (2 vols.) (see also: Joe's Little Book of String) Blumenhagen-Lüst-Theisen, Basic concepts of string theory Plan (tentative!) Week Lecture Notes Problem set Answers October 17 Introduction, Relativistic Actions Lecture 1&2 Homework 1 October 24 Classical strings, Polyakov action, Symmetries Lecture 3&4 Homework 2 October 31 Mode expansion, Quantization, Virasoro anomaly Lecture 5&6 Homework 3	{"url":"https://mathi.uni-heidelberg.de/~walcher/teaching/wise2223/strings/?lang=en","timestamp":"2024-11-14T23:12:38Z","content_type":"text/html","content_length":"17670","record_id":"<urn:uuid:283df0b5-3ba4-4c0b-81d0-4d050ea7173e>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00202.warc.gz"}
Unveiling The Quantum Realm: Multiplicity And The Interconnected Nature Of Reality Unveiling the Quantum Realm: Multiplicity and the Interconnected Nature of Reality The quantum realm, a realm of subatomic particles and phenomena, has captivated the curiosity of scientists for over a century. Its enigmatic characteristics, such as superposition and entanglement, challenge classical notions of reality and hint at a profound interconnectedness in the universe. Multiplicity and Superposition: One of the most striking features of the quantum realm is the concept of superposition. Quantum particles can simultaneously exist in multiple states or locations, creating a situation where they are neither definitively one nor the other. This multiplicity defies the common understanding of distinct and separate objects. Entanglement and Non-locality: The phenomenon of entanglement further deepens the interconnectedness of quantum particles. When two or more particles become entangled, they are linked together in such a way that the state of one instantly affects the state of the others, even if they are physically separated by vast distances. This non-local connection suggests that information and influence can transcend the limitations of space and time. Observer Effect and the Role of Consciousness: The observer effect, which demonstrates the influence of observation on quantum systems, is another intriguing aspect of the quantum realm. When a particle is observed, it collapses into a specific state, as if it had only that state all along. This raises questions about the role of consciousness in shaping reality and the relationship between the observer and the observed. Implications for Our Understanding of Reality: The quantum realm reveals a reality that is fundamentally different from our classical perceptions. It suggests a universe where multiplicity, non-locality, and interconnectedness are inherent aspects of existence. These insights challenge the notion of a separate, objective world and instead point towards a unified, interpenetrating field of energy and information. Applications and Potential: The understanding of the quantum realm has far-reaching implications. It has led to the development of quantum computing, cryptography, and other advanced technologies. Additionally, it has inspired new approaches in fields such as psychology and spirituality, exploring the potential for quantum processes to influence human experience and consciousness. The unveiling of the quantum realm has profoundly expanded our understanding of reality. It has revealed a world of multiplicity, interconnectedness, and non-locality, blurring the lines between separate entities and suggesting a deep unity at the very foundations of existence. As we continue to explore the quantum realm, we may gain deeper insights into the nature of our own consciousness and the interconnected web of life that surrounds us.## Unveiling The Quantum Realm: Multiplicity And The Interconnected Nature Of Reality Executive Summary: Quantum physics unveils a profound reality where particles exist in multiple states simultaneously and are interconnected in ways that defy classical understanding. This multiplicity and interconnectedness challenge our conventional notions of reality and open up new avenues for scientific exploration and technological advancements. The realm of quantum physics transcends our everyday experience, revealing a world governed by fundamental principles that challenge our classical intuitions. At the subatomic level, particles exhibit peculiar behaviors such as wave-particle duality and quantum superposition, leading us to question the very nature of reality. This article explores the transformative insights provided by quantum physics, delving into the concepts of multiplicity and interconnectedness that permeate this enigmatic realm. Frequently Asked Questions (FAQs): 1. What is quantum superposition? Answer: Quantum superposition is a fundamental property of quantum particles where they can exist in multiple states or locations simultaneously until observed, violating classical expectations of 2. How does entanglement defy classical physics? Answer: Entanglement is a quantum phenomenon where two or more particles become correlated in such a way that their states become inextricably linked, regardless of the distance between them. 3. What is the significance of quantum indeterminacy? Answer: Quantum indeterminacy, often referred to as Heisenberg’s uncertainty principle, limits our ability to precisely measure both position and momentum of a particle, highlighting the inherent uncertainty at the quantum level. 1. Duality of Matter and Energy • Wave-particle duality: Particles exhibit both wave-like and particle-like properties, depending on the experimental setup. • Complementarity principle: The wave and particle aspects of a particle are complementary and cannot be observed simultaneously. • Matter waves: Particles can behave like waves, exhibiting interference and diffraction phenomena. 2. Quantum Superposition • Multiple states: Particles can exist in multiple states, such as spin up and spin down, simultaneously. • Measurement collapses superposition: Observation of a particle forces it to assume a single, definite state, collapsing the superposition. • Schrödinger’s cat paradox: A thought experiment that highlights the paradoxical nature of quantum superposition. 3. Quantum Entanglement • Non-locality: Entangled particles share a common fate, regardless of the distance separating them. • Correlation: The states of entangled particles are correlated, violating classical expectations of independence. • Bell’s theorem: Experiments have experimentally confirmed the non-local nature of entanglement, challenging classical theories. 4. Quantum Indeterminacy • Heisenberg’s uncertainty principle: It is impossible to simultaneously determine both the position and momentum of a particle with absolute precision. • Uncertainty limits: The more precisely one quantity is measured, the less precisely the other can be known. • Wave function: The wave function of a particle describes its probabilistic distribution of states, rather than providing a definite value. 5. Quantum Information and Technology • Quantum computing: Quantum computers harness the power of quantum superposition to perform computations exponentially faster than classical computers. • Quantum cryptography: Quantum principles enable unbreakable encryption protocols, enhancing cybersecurity. • Quantum sensors: Quantum sensors have the potential to revolutionize scientific research and medical diagnostics with unprecedented sensitivity. Quantum physics presents a mind-boggling departure from classical physics, revealing a reality where multiplicity and interconnectedness are fundamental principles. By embracing these concepts, we gain an enhanced understanding of the fabric of the universe and open up new horizons for scientific exploration and technological innovation. The quantum realm serves as a testament to the boundless wonders of the cosmos and the transformative power of human inquiry. 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aes_cbc: Symmetric AES encryption in openssl: Toolkit for Encryption, Signatures and Certificates Based on OpenSSL Low-level symmetric encryption/decryption using the AES block cipher in CBC mode. The key is a raw vector, for example a hash of some secret. When no shared secret is available, a random key can be used which is exchanged via an asymmetric protocol such as RSA. See rsa_encrypt() for a worked example or encrypt_envelope() for a high-level wrapper combining AES and RSA. aes_ctr_encrypt(data, key, iv = rand_bytes(16)) aes_ctr_decrypt(data, key, iv = attr(data, "iv")) aes_cbc_encrypt(data, key, iv = rand_bytes(16)) aes_cbc_decrypt(data, key, iv = attr(data, "iv")) aes_gcm_encrypt(data, key, iv = rand_bytes(12)) aes_gcm_decrypt(data, key, iv = attr(data, "iv")) aes_keygen(length = 16) data raw vector or path to file with data to encrypt or decrypt key raw vector of length 16, 24 or 32, e.g. the hash of a shared secret iv raw vector of length 16 (aes block size) or NULL. The initialization vector is not secret but should be random length how many bytes to generate. Usually 16 (128-bit) or 12 (92-bit) for aes_gcm raw vector or path to file with data to encrypt or decrypt raw vector of length 16, 24 or 32, e.g. the hash of a shared secret raw vector of length 16 (aes block size) or NULL. The initialization vector is not secret but should be random how many bytes to generate. Usually 16 (128-bit) or 12 (92-bit) for aes_gcm # aes-256 requires 32 byte key passphrase <- charToRaw("This is super secret") key <- sha256(passphrase) # symmetric encryption uses same key for decryption x <- serialize(iris, NULL) y <- aes_cbc_encrypt(x, key = key) x2 <- aes_cbc_decrypt(y, key = key) stopifnot(identical(x, x2)) For more information on customizing the embed code, read Embedding Snippets.	{"url":"https://rdrr.io/cran/openssl/man/aes_cbc.html","timestamp":"2024-11-06T20:56:47Z","content_type":"text/html","content_length":"26813","record_id":"<urn:uuid:9ed195c4-ede4-40f1-b73f-df70ff3b99cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00457.warc.gz"}
VAR Model Case Study This example shows how to analyze a VAR model. Overview of Case Study This section contains an example of the workflow described in VAR Model Workflow. The example uses three time series: GDP, M1 money supply, and the 3-month T-bill rate. The example shows: • Loading and transforming the data for stationarity • Partitioning the transformed data into presample, estimation, and forecast intervals to support a backtesting experiment • Making several models • Fitting the models to the data • Deciding which of the models is best • Making forecasts based on the best model Load and Transform Data The file Data_USEconModel ships with Econometrics Toolbox™ software. The file contains time series from the Federal Reserve Bank of St. Louis Economics Data (FRED) database in a tabular array. This example uses three of the time series: • GDP (GDP) • M1 money supply (M1SL) • 3-month T-bill rate (TB3MS) Load the data set. Create a variable for real GDP. load Data_USEconModel DataTimeTable.RGDP = DataTimeTable.GDP./DataTimeTable.GDPDEF100; Plot the data to look for trends. title('Real GDP') grid on grid on title('3-mo T-bill') grid on The real GDP and M1 data appear to grow exponentially, while the T-bill returns show no exponential growth. To counter the trends in real GDP and M1, take a difference of the logarithms of the data. Also, stabilize the T-bill series by taking the first difference. Synchronize the date series so that the data has the same number of rows for each column. rgdpg = price2ret(DataTimeTable.RGDP); m1slg = price2ret(DataTimeTable.M1SL); dtb3ms = diff(DataTimeTable.TB3MS); Data = array2timetable([rgdpg m1slg dtb3ms],... 'RowTimes',DataTimeTable.Time(2:end),'VariableNames',{'RGDP' 'M1SL' 'TB3MS'}); title('Real GDP') grid on grid on title('3-mo T-bill') grid on The scale of the first two columns is about 100 times smaller than the third. Multiply the first two columns by 100 so that the time series are all roughly on the same scale. This scaling makes it easy to plot all the series on the same plot. More importantly, this type of scaling makes optimizations more numerically stable (for example, maximizing loglikelihoods). Data{:,1:2} = 100Data{:,1:2}; hold on grid on legend('Real GDP','M1','3-mo T-bill'); hold off Select and Fit the Models You can select many different models for the data. This example uses four models. • VAR(2) with diagonal autoregressive • VAR(2) with full autoregressive • VAR(4) with diagonal autoregressive • VAR(4) with full autoregressive Remove all missing values from the beginning of the series. idx = all(~ismissing(Data),2); Data = Data(idx,:); Create the four models. numseries = 3; dnan = diag(nan(numseries,1)); seriesnames = {'Real GDP','M1','3-mo T-bill'}; VAR2diag = varm('AR',{dnan dnan},'SeriesNames',seriesnames); VAR2full = varm(numseries,2); VAR2full.SeriesNames = seriesnames; VAR4diag = varm('AR',{dnan dnan dnan dnan},'SeriesNames',seriesnames); VAR4full = varm(numseries,4); VAR4full.SeriesNames = seriesnames; The matrix dnan is a diagonal matrix with NaN values along its main diagonal. In general, missing values specify the presence of the parameter in the model, and indicate that the parameter needs to be fit to data. MATLAB® holds the off diagonal elements, 0, fixed during estimation. In contrast, the specifications for VAR2full and VAR4full have matrices composed of NaN values. Therefore, estimate fits full matrices for autoregressive matrices. To assess the quality of the models, create index vectors that divide the response data into three periods: presample, estimation, and forecast. Fit the models to the estimation data, using the presample period to provide lagged data. Compare the predictions of the fitted models to the forecast data. The estimation period is in sample, and the forecast period is out of sample (also known as For the two VAR(4) models, the presample period is the first four rows of Data. Use the same presample period for the VAR(2) models so that all the models are fit to the same data. This is necessary for model fit comparisons. For both models, the forecast period is the final 10% of the rows of Data. The estimation period for the models goes from row 5 to the 90% row. Define these data periods. idxPre = 1:4; T = ceil(.9size(Data,1)); idxEst = 5:T; idxF = (T+1):size(Data,1); fh = numel(idxF); Now that the models and time series exist, you can easily fit the models to the data. [EstMdl1,EstSE1,logL1,E1] = estimate(VAR2diag,Data{idxEst,:},... [EstMdl2,EstSE2,logL2,E2] = estimate(VAR2full,Data{idxEst,:},... [EstMdl3,EstSE3,logL3,E3] = estimate(VAR4diag,Data{idxEst,:},... [EstMdl4,EstSE4,logL4,E4] = estimate(VAR4full,Data{idxEst,:},... • The EstMdl model objects are the fitted models. • The EstSE structures contain the standard errors of the fitted models. • The logL values are the loglikelihoods of the fitted models, which you use to help select the best model. • The E vectors are the residuals, which are the same size as the estimation data. Check Model Adequacy You can check whether the estimated models are stable and invertible by displaying the Description property of each object. (There are no MA terms in these models, so the models are necessarily invertible.) The descriptions show that all the estimated models are stable. ans = "AR-Stationary 3-Dimensional VAR(2) Model" ans = "AR-Stationary 3-Dimensional VAR(2) Model" ans = "AR-Stationary 3-Dimensional VAR(4) Model" ans = "AR-Stationary 3-Dimensional VAR(4) Model" AR-Stationary appears in the output indicating that the autoregressive processes are stable. You can compare the restricted (diagonal) AR models to their unrestricted (full) counterparts using lratiotest. The test rejects or fails to reject the hypothesis that the restricted models are adequate, with a default 5% tolerance. This is an in-sample test. Apply the likelihood ratio tests. You must extract the number of estimated parameters from the summary structure returned by summarize. Then, pass the differences in the number of estimated parameters and the loglikelihoods to lratiotest to perform the tests. results1 = summarize(EstMdl1); np1 = results1.NumEstimatedParameters; results2 = summarize(EstMdl2); np2 = results2.NumEstimatedParameters; results3 = summarize(EstMdl3); np3 = results3.NumEstimatedParameters; results4 = summarize(EstMdl4); np4 = results4.NumEstimatedParameters; reject1 = lratiotest(logL2,logL1,np2 - np1) reject3 = lratiotest(logL4,logL3,np4 - np3) reject4 = lratiotest(logL4,logL2,np4 - np2) The 1 results indicate that the likelihood ratio test rejected both the restricted models in favor of the corresponding unrestricted models. Therefore, based on this test, the unrestricted VAR(2) and VAR(4) models are preferable. However, the test does not reject the unrestricted VAR(2) model in favor of the unrestricted VAR(4) model. (This test regards the VAR(2) model as an VAR(4) model with restrictions that the autoregression matrices AR(3) and AR(4) are 0.) Therefore, it seems that the unrestricted VAR(2) model is the best model. To find the best model in a set, minimize the Akaike information criterion (AIC). Use in-sample data to compute the AIC. Calculate the criterion for the four models. AIC = aicbic([logL1 logL2 logL3 logL4],[np1 np2 np3 np4]) AIC = 1×4 10^3 × 1.4794 1.4396 1.4785 1.4537 The best model according to this criterion is the unrestricted VAR(2) model. Notice, too, that the unrestricted VAR(4) model has lower Akaike information than either of the restricted models. Based on this criterion, the unrestricted VAR(2) model is best, with the unrestricted VAR(4) model coming next in preference. To compare the predictions of the four models against the forecast data, use forecast. This function returns both a prediction of the mean time series, and an error covariance matrix that gives confidence intervals about the means. This is an out-of-sample calculation. [FY1,FYCov1] = forecast(EstMdl1,fh,Data{idxEst,:}); [FY2,FYCov2] = forecast(EstMdl2,fh,Data{idxEst,:}); [FY3,FYCov3] = forecast(EstMdl3,fh,Data{idxEst,:}); [FY4,FYCov4] = forecast(EstMdl4,fh,Data{idxEst,:}); Estimate approximate 95% forecast intervals for the best fitting model. extractMSE = @(x)diag(x)'; MSE = cellfun(extractMSE,FYCov2,'UniformOutput',false); SE = sqrt(cell2mat(MSE)); YFI = zeros(fh,EstMdl2.NumSeries,2); YFI(:,:,1) = FY2 - 2SE; YFI(:,:,2) = FY2 + 2SE; This plot shows the predictions of the best fitting model in the shaded region to the right. for j = 1:EstMdl2.NumSeries h1 = plot(Data.Time((end-49):end),Data{(end-49):end,j}); hold on h2 = plot(Data.Time(idxF),FY2(:,j)); h3 = plot(Data.Time(idxF),YFI(:,j,1),'k--'); h = gca; fill([Data.Time(idxF(1)) h.XLim([2 2]) Data.Time(idxF(1))],... h.YLim([1 1 2 2]),'k','FaceAlpha',0.1,'EdgeColor','none'); legend([h1 h2 h3],'True','Forecast','95% Forecast interval',... hold off It is now straightforward to calculate the sum-of-squares error between the predictions and the data. Error1 = Data{idxF,:} - FY1; Error2 = Data{idxF,:} - FY2; Error3 = Data{idxF,:} - FY3; Error4 = Data{idxF,:} - FY4; SSerror1 = Error1(:)' Error1(:); SSerror2 = Error2(:)' * Error2(:); SSerror3 = Error3(:)' * Error3(:); SSerror4 = Error4(:)' * Error4(:); bar([SSerror1 SSerror2 SSerror3 SSerror4],.5) ylabel('Sum of squared errors') {'AR2 diag' 'AR2 full' 'AR4 diag' 'AR4 full'}) title('Sum of Squared Forecast Errors') The predictive performances of the four models are similar. The full AR(2) model seems to be the best and most parsimonious fit. Its model parameters are as follows. AR-Stationary 3-Dimensional VAR(2) Model Effective Sample Size: 176 Number of Estimated Parameters: 21 LogLikelihood: -698.801 AIC: 1439.6 BIC: 1506.18 Value StandardError TStatistic PValue __________ _____________ __________ __________ Constant(1) 0.34832 0.11527 3.0217 0.0025132 Constant(2) 0.55838 0.1488 3.7526 0.00017502 Constant(3) -0.45434 0.15245 -2.9803 0.0028793 AR{1}(1,1) 0.26252 0.07397 3.5491 0.00038661 AR{1}(2,1) -0.029371 0.095485 -0.3076 0.75839 AR{1}(3,1) 0.22324 0.097824 2.2821 0.022484 AR{1}(1,2) -0.074627 0.054476 -1.3699 0.17071 AR{1}(2,2) 0.2531 0.070321 3.5992 0.00031915 AR{1}(3,2) -0.017245 0.072044 -0.23936 0.81082 AR{1}(1,3) 0.032692 0.056182 0.58189 0.56064 AR{1}(2,3) -0.35827 0.072523 -4.94 7.8112e-07 AR{1}(3,3) -0.29179 0.0743 -3.9272 8.5943e-05 AR{2}(1,1) 0.21378 0.071283 2.9991 0.0027081 AR{2}(2,1) -0.078493 0.092016 -0.85304 0.39364 AR{2}(3,1) 0.24919 0.094271 2.6433 0.0082093 AR{2}(1,2) 0.13137 0.051691 2.5415 0.011038 AR{2}(2,2) 0.38189 0.066726 5.7233 1.045e-08 AR{2}(3,2) 0.049403 0.068361 0.72269 0.46987 AR{2}(1,3) -0.22794 0.059203 -3.85 0.00011809 AR{2}(2,3) -0.0052932 0.076423 -0.069262 0.94478 AR{2}(3,3) -0.37109 0.078296 -4.7397 2.1408e-06 Innovations Covariance Matrix: 0.5931 0.0611 0.1705 0.0611 0.9882 -0.1217 0.1705 -0.1217 1.0372 Innovations Correlation Matrix: 1.0000 0.0798 0.2174 0.0798 1.0000 -0.1202 0.2174 -0.1202 1.0000 Forecast Observations You can make predictions or forecasts using the fitted model (EstMdl2) either by: • Calling forecast and passing the last few rows of YF • Simulating several time series with simulate In both cases, transform the forecasts so they are directly comparable to the original time series. Generate 10 predictions from the fitted model beginning at the latest times using forecast. [YPred,YCov] = forecast(EstMdl2,10,Data{idxF,:}); Transform the predictions by undoing the scaling and differencing applied to the original data. Make sure to insert the last observation at the beginning of the time series before using cumsum to undo the differencing. And, since differencing occurred after taking logarithms, insert the logarithm before using cumsum. YFirst = DataTimeTable(idx,{'RGDP' 'M1SL' 'TB3MS'}); EndPt = YFirst{end,:}; EndPt(:,1:2) = log(EndPt(:,1:2)); YPred(:,1:2) = YPred(:,1:2)/100; % Rescale percentage YPred = [EndPt; YPred]; % Prepare for cumsum YPred(:,1:3) = cumsum(YPred(:,1:3)); YPred(:,1:2) = exp(YPred(:,1:2)); fdates = dateshift(YFirst.Time(end),'end','quarter',0:10); % Insert forecast horizon for j = 1:EstMdl2.NumSeries hold on grid on h = gca; fill([fdates(1) h.XLim([2 2]) fdates(1)],h.YLim([1 1 2 2]),'k',... hold off The plots show the extrapolations as blue dashed lines in the light gray forecast horizon, and the original data series in solid black. Look at the last few years in this plot to get a sense of how the predictions relate to the latest data points. YLast = YFirst(170:end,:); for j = 1:EstMdl2.NumSeries hold on grid on h = gca; fill([fdates(1) h.XLim([2 2]) fdates(1)],h.YLim([1 1 2 2]),'k',... hold off The forecast shows increasing real GDP and M1, and a slight decline in the interest rate. However, the forecast has no error bars. Alternatively, you can generate 10 predictions from the fitted model beginning at the latest times using simulate. This method simulates 2000 time series times, and then generates the means and standard deviations for each period. The means of the deviates for each period are the predictions for that period. Simulate a time series from the fitted model beginning at the latest times. rng(1); % For reproducibility YSim = simulate(EstMdl2,10,'Y0',Data{idxF,:},'NumPaths',2000); Transform the predictions by undoing the scaling and differencing applied to the original data. Make sure to insert the last observation at the beginning of the time series before using cumsum to undo the differencing. And, since differencing occurred after taking logarithms, insert the logarithm before using cumsum. EndPt = YFirst{end,:}; EndPt(1:2) = log(EndPt(1:2)); YSim(:,1:2,:) = YSim(:,1:2,:)/100; YSim = [repmat(EndPt,[1,1,2000]);YSim]; YSim(:,1:3,:) = cumsum(YSim(:,1:3,:)); YSim(:,1:2,:) = exp(YSim(:,1:2,:)); Compute the mean and standard deviation of each series, and plot the results. The plot has the mean in black, with a +/- 1 standard deviation in red. YMean = mean(YSim,3); YSTD = std(YSim,0,3); for j = 1:EstMdl2.NumSeries grid on hold on plot(fdates,YMean(:,j) + YSTD(:,j),'--r') plot(fdates,YMean(:,j) - YSTD(:,j),'--r') h = gca; fill([fdates(1) h.XLim([2 2]) fdates(1)],h.YLim([1 1 2 2]),'k',... hold off The plots show increasing growth in GDP, moderate to little growth in M1, and uncertainty about the direction of T-bill rates. See Also Related Topics	{"url":"https://it.mathworks.com/help/econ/var-model-case-study.html","timestamp":"2024-11-04T15:10:48Z","content_type":"text/html","content_length":"100045","record_id":"<urn:uuid:7d88009c-544c-4fdb-ac2a-96eb67bf33c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00858.warc.gz"}
Cite as Kevin Buchin, Sándor P. Fekete, Alexander Hill, Linda Kleist, Irina Kostitsyna, Dominik Krupke, Roel Lambers, and Martijn Struijs. Minimum Scan Cover and Variants - Theory and Experiments. In 19th International Symposium on Experimental Algorithms (SEA 2021). Leibniz International Proceedings in Informatics (LIPIcs), Volume 190, pp. 4:1-4:16, Schloss Dagstuhl – Leibniz-Zentrum für Informatik Copy BibTex To Clipboard author = {Buchin, Kevin and Fekete, S\'{a}ndor P. and Hill, Alexander and Kleist, Linda and Kostitsyna, Irina and Krupke, Dominik and Lambers, Roel and Struijs, Martijn}, title = {{Minimum Scan Cover and Variants - Theory and Experiments}}, booktitle = {19th International Symposium on Experimental Algorithms (SEA 2021)}, pages = {4:1--4:16}, series = {Leibniz International Proceedings in Informatics (LIPIcs)}, ISBN = {978-3-95977-185-6}, ISSN = {1868-8969}, year = {2021}, volume = {190}, editor = {Coudert, David and Natale, Emanuele}, publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik}, address = {Dagstuhl, Germany}, URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.SEA.2021.4}, URN = {urn:nbn:de:0030-drops-137765}, doi = {10.4230/LIPIcs.SEA.2021.4}, annote = {Keywords: Graph scanning, angular metric, makespan, energy, bottleneck, complexity, approximation, algorithm engineering, mixed-integer programming, constraint programming}	{"url":"https://drops.dagstuhl.de/search/documents?author=Buchin,%20Kevin","timestamp":"2024-11-14T18:52:14Z","content_type":"text/html","content_length":"273958","record_id":"<urn:uuid:e594b4b3-b777-4488-a366-d86499cc18c4>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00053.warc.gz"}
Pilot Power Based upon our simulated data, the below table lists how often across simulations we would proceed to a full trial when applying the below decision rules to our simulated pilot study results. We now do a power analysis to determine the sample size required to detect effects of a certain size in a full trial. This is done in two ways: once using the pre-determined practically significant effect size, and once using the effect size observed in the simulated pilot study. The table below shows, across all simulations, the average sample size needed to detect effects in a full trial. We now make conclusions about the effectiveness of the intervention based upon our full trial. Best Practice Your Choice In General CSV File The below data are from simulations where all sizes are made to be large enough to detect the observed pilot study effect size. CSV File The below data are from simulations where all sample sizes are large enough to detect the practically significant effect size.	{"url":"https://pilotpower.table1.org/","timestamp":"2024-11-08T05:13:04Z","content_type":"text/html","content_length":"21579","record_id":"<urn:uuid:17dd438f-bcab-45e6-844a-013a84b4da10>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00861.warc.gz"}
Step Functions: From Step Functions What is a step function? The main characteristic of a step function, and the reason why it truly looks like a staircase doodle, is that this function happens to be constant in intervals. These intervals do not have the same value and we end up with a function which "jumps" from one value to the next (following its own conditions) in a certain pattern. Every time the function jumps to a new interval it has a new constant value and we can observe the "steps" on its graphic representation as horizontal lines. Take a look at the graphic representation below for a typical step function where is plain to see how the name of the function came about. Notice from the figure above that we can also define a step function as a piecewise function of constant steps, meaning that the function is broken down in a finite number of pieces, each piece having a constant value, in other words, each piece is a constant function. When talking about step functions we cannot forget to talk about the Heaviside function also known as the unit step function. This is a function that is defined to have a constant value of zero up to a certain point on t (the horizontal axis) at which it jumps to a value of 1. This particular value at which the function jumps (or switches from zero to one) is usually taken as the origin in the typical coordinate system representation, but it can be any value c in the horizontal axis. Consequently, the Heaviside step function is defined as: Equation 1: Definition of the Heaviside function or unit step function. Where c represents the point at t in which the function goes from a 0 value to a value of 1. So, if we want to write down the mathematical expression of the heaviside function depicted in figure 3, we would write it as: u[3](t). Notice how we only wrote the math expression using the first type of notation found in equation 1, this is because that notation happens to be the most commonly used and is also the one we will continue to use throughout our lesson. Is still important though, that you know about the other notations and keep them in mind just in case you find them throughout your From its definition, we can understand why the Heaviside function is also called the "unit step" function. As it can be observed, a Heaviside function can only have values of 0 and 1, in other words, the function is always equal to zero before arriving to a certain value t=c at which it "turns on" and jumps directly into having a value of 1, and so, it jumps in a step size of one unit. Notice how we have used the phrase "turning on" to describe the process of the unit step function to jump from a zero value to a unit value, this is a very common way to refer to Heaviside step functions' behavior and interestingly enough, it is all due to their real life usage and comes from the reason why they were invented. Historically, physicist, self-taught engineer and mathematician Oliver Heaviside, invented Heaviside step functions in order to describe the behaviour of a current signal when you click the switch on of an electric circuit, thus, allowing you to calculate the magnitude of the current from zero when the circuit is off, to a certain value when is tuned on. There is an important note to clarify here. Electric current does not magically jump from a zero value to a high value. When having a constant current through a circuit, we know that this constant value obviously started off from zero from when the circuit was off and then it arrived until a certain value after gradually increasing in time, the thing is that electric current travels at a very high speed, and so it is impossible for us (in a simple everyday life setting) to see this gradual increase of the current from zero to its final value since it happens in a very short instant of time, and so, we take it as a "jump" from one value to the next and describe it accordingly in graphic representations. Heaviside function properties Although the Heaviside function itself can only have the values of 0 or 1 as mentioned before, this does not mean we cannot obtain a graphic representation of a higher jump using Heaviside step functions. It is actually a very simple task to obtain a higher value since you just need to multiply it for any constant value that you want to have as the size jump, in other words, if you have a step function written as 3u[5](t) in here you have a step function which has a value of zero until it gets to a value of t=5, and which point, the function has a final value of 3. This can be seen in the figure below: One of the greatest properties of Heaviside step functions is that they allow us to model certain scenarios (like the one described for current in an on/off circuit) and mathematically be able to solve for important information on these scenarios. Such cases tend to require the use of differential equations and so here we have yet another tool to solve them. Being this a course on differential equations, the most important point of this lesson is to give an introduction to a function which will aid in the solution of certain differential equations, such tool will be used along others seen before, such as the Laplace transform. This will serve to come up with important formulas to be used and to be prepared for the next lesson in which you will be solving differential equations with step functions. We will talk a bit more about this on the last section of this lesson, meanwhile for a review on the definition of the unit step function and a list of a few of its identities, we recommend the next Heaviside step function article. Heaviside step function examples Let us take a look into an example in which you will have to write all of the necessary unit step functions in order to completely describe the graphic representation found in figure 5. Example 1 Write the mathematical expression for the following graph in terms of the Heaviside step function: Let's divide this in parts so we can see how the functions of the graph behave at each different value given. And so, we will write a separate expression for each of the next pieces: for t < 3, for t =3 to t=4, for t=4 to t=6 and for t>6. • For t < 3: Notice that this is a regular Heaviside function in which c=0, multiplied by a 2 in order to obtain the jump of 2 units in size. This would fit the requirement of the function being zero for all negative values of t, and then to have a value of 2 for values of t from 0 to 3. And so, the expression is: 2u[o](t)=2. • For t=3 to t=4: In this range we have to cancel the unit step function that we had before, that means we need a negative unit step function in here but this one will start to be applied at t=3 and will have to be multiplied by 2 again in order to cancel the value of the previous expression. Thus, our expression for this part of the function is: -2u[3](t). • For t=4 to t=6: If we weren't to add any function at t=4, the value of the function would remain as zero to infinity since the second function cancelled the first one, but instead, we see in the graph that we have a diagonal line increasing one unit step size for each one unit of distance traveled on t. Thus, since the function is increasing at the same rate as t, we could easily multiply a new unit step function which starts at 4 by t and be over with, this would produce a diagonal line following the same behavior. The problem is that just multiplying u[4](t) by t would produce a line that would come out of the origin instead of t=4, and for that, we need to multiply the unit step function by (t-4) so the function starts at the same time as the unit step function will be applied (which is at t=4). And so the expression is: (t-4)u[4](t). Notice (t-4)u[4](t) produces the values for y of: 0 (when t=4), 1 (when t=5) and 2 (when t=6), which is what the graph requires. • For t > 6: For this last piece of the graph should be already easy that we just need to cancel our last function in order to have a value of zero back to the graph. For that we use a negative unit step function which starts at t=6, and which should be multiplied again by (t-4). And so, the expression to cancel our last one and completes the graph is: -(t-4)[6](t). We add all of the four pieces of function we found to produce the expression that represents the whole graph shown in figure 5: Equation for example 1: Mathematical expression for the graph in figure 5 Due to this particular example having multiple step function samples, we continue onto the next section to work on more complicated problems. If you would like to continue practicing how to write down Heaviside step functions, we recommend you to visit these notes of step functions for more Heaviside function examples along with a little introduction. Notice these notes also introduce the topic for our next section: unit step function Laplace transforms and the overall use of the Laplace transform when associated to Heaviside Laplace transform of Heaviside function You have already had an introduction to the Laplace transform in recent past lessons, still, at this time we do recommend you to give the topic a review if you think it appropriate or necessary. The lesson on calculating Laplace transforms is of special use for you to be prepared for this section. Let us continue with the Heaviside step function and how we will use it along the Laplace transform. The Laplace transform will help us find the value of y(t) for a function that will be represented using the unit step function, so far we have talked about step functions in which the value is a constant (just a jump from zero to a constant value, producing a straight horizontal line in a graph) but we can have any type of function to start at any given point in time (which is what we represent with t mostly). What are we saying? Well, think on the graphic representation of a function, you can have any function, with any shape, but this special case comes from the fact that the function will be zero through time, until a certain point in which the signal will turn on, and then it will "jump" into this "any shape" function behavior. This is what we call a "shifted function" and this is because we can think of any regular function graphed, and then saying "oh but we want this function to start happening later" and we "switch" it to start at a later point in time (at t=c). Since these shifted functions will be equal to zero until at certain point c in which they will "turn on", they can be represented as: The shifted function then is defined as: Equation 2: Shifted function Now let's see what happens when we take the Laplace transform of a shifted function! First, remember that the mathematical definition of the Laplace transform is: Equation 3: Laplace transform for a function f(t) where t≥0 Therefore the Laplace transform for the shifted function is: Equation 4: Laplace transform of shifted function (part 1). Notice how the Laplace transform gets simplified from an improper integral to a regular integral where you have dropped the unit step function. The reason for this is that although the range of the whole Laplace transform integral is from 0 to infinity, before c (whatever value c has) the unit step function is equal to zero, and therefore the whole integral goes to zero. After c, the unit step function has a value of 1, and thus, we can just take it as a constant value of 1 multiplying the rest of the integral, which range is not from c to infinity. Continuing with the simplification of the Laplace transform of the shifted function, we set x = t - c, which means that t = x+c and so the transformation looks as follows Equation 5: Laplace transform of shifted function (part 2). By making the integral to be in relation to one single variable rather that a t-c term, we have simplified this transformation so we could obtain a quickly manageable formula we can readily use with future problems and our already known table of Laplace transforms from past lessons. Having solved equation 5 makes it easier to obtain another important formula: the unit step function Laplace transform. Notice, not the transform for a shifted function, but the Laplace transform of the unit step function (Heaviside function) itself and alone. Equation 6: Laplace transform of the unit step function (Heaviside step function) If you notice, equation 5 was useful while obtaining equation 6 because taking the Laplace transformation of the Heaviside function by itself can be taken as having a shifted function in which the f (t-c) part equals to 1, and so you end up with the Laplace transform of a unit step function times 1, which results in the simple and very useful formula found in equation 6. Now let us finish this lesson by working on some transformation of a unit step function examples. Example 2 Find the Laplace transform of each of the following step functions: Equation for example 2: Find the Laplace transform on the unit step function • Applying the Laplace transform to the function and using equations 5 and 6 we obtain: Equation for example 2(a): Computing the Laplace transform on the unit step function (part 1) • Using equation 5, we set x=t-c (for this case x=t-5) and work through the transformation on the second term of the last equation: Equation for example 2(b): Computing the Laplace transform on the unit step function (part 2) Notice that in order to solve the last term, we used the method of comparison with the table of Laplace transforms. You can find such table on the past lessons related to the Laplace transform. • Now let us solve the third transformation (remember we set x = t-c, which in this case is x=t-7): Equation for example 2(c): Computing the Laplace transform on the unit step function (part 3) • Now we put everything together to form the final answer to the problem: Equation for example 2(d): Computing the Laplace transform on the unit step function (part 4) Example 3 Find the Laplace transform for each of the following functions in g(t): Equation for example 3: Obtain the Laplace transform of g(t) • Now we use equation 5 and set up x=t-c to solve the Laplace transform: Equation for example 3(a): Computing the Laplace transform of the unit step functions (part 1) • We separate the two terms found on the right hand side of the equation, and solve for the first one (for this case x=t-) using the trigonometric function: sin(a+b)=sin(a)cos(b)+cos(a)sin(b). Equation for example 3(b): Computing the Laplace transform of the unit step functions (part 2) • Now let us solve the second term, where x = t-4, therefore, t=x+4: Equation for example 3(c): Computing the Laplace transform of the unit step functions (part 3) • Putting the whole result together: Equation for example 3(d): Laplace transform of g(t) And now we are ready for our next section where we will be solving differential equations with what we learned today. If you would like to see some extra notes on the Heaviside function and its relation with another function which we will study in later lessons, the Dirac delta function, visit the link provided. A Heaviside Step Function (also just called a "Step Function") is a function that has a value of 0 from 0 to some constant, and then at that constant switches to 1. The Heaviside Step Function is defined as, The Laplace Transform of the Step Function: $L${$u_{c}(t)$ $f(t - c)$} = $e^{-sc}$$L${$f(t)$} $L${$u_{c}(t)$} = $\frac{e^{-sc}}{s}$ These Formulae might be necessary for shifting functions: $\sin{(a + b)} = \sin(a)\cos(b) + \cos(a)\sin(b)$ $\cos{(a + b)} = \cos(a)\cos(b) - \sin(a)\sin(b)$ $(a + b)^{2} = a^{2} + 2ab +b^{2}$	{"url":"https://www.studypug.com/differential-equations-help/step-functions","timestamp":"2024-11-03T23:48:14Z","content_type":"text/html","content_length":"404335","record_id":"<urn:uuid:ee079f2f-2739-4281-a3b2-e8af7afab6af>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00396.warc.gz"}
How Much Paint Does One Gallon Cover If your paints Coverage Rate is square feet per gallon, then you'll require one gallon. If your paints Coverage Rate is square feet per gallon, you'll. The general rule of thumb, however, is that 1 gallon of latex paint will cover to sq ft. How much paint do you need for a 12x12 room? Most likely, you. The coverage capacity of a gallon of paint depends on various factors, such as the type of paint, the surface being painted, and the application method. On. What will one gallon of paint cover? Paint coverage is affected by numerous factors: the quality of the paint you're using, the color you're covering and. Divide the paintable wall area by (the square-foot coverage in each gallon can) to find the number of gallons of paint you need for the walls. You can round. PAINT CALCULATOR. How much paint do you need? We can help. Use either quick estimates or precise measurements to get a paint estimate. SIGN UP FOR OUR EMAIL. Generally, a gallon covers sq ft. To determine the paint needed for a 12×12 room, calculate the wall square footage. Multiply wall length by height. For. One gallon of Ceiling Paint will cover approximately square feet. One gallon of Primer will cover approximately square feet. The formula used assumes that one gallon of paint covers sqft of surface area. The user selects the number of coats from a drop-down menu and clicks the “. If your paints Coverage Rate is square feet per gallon, then you'll require one gallon. If your paints Coverage Rate is square feet per gallon, you'll. According to Lowe's, a gallon of primer covers about – square feet, and a gallon of paint usually covers – square feet. Start by adding up the. How much does 5 gallons of paint cover? One gallon can of paint will cover up to square feet, which is enough to cover a small room like a. The coverage obtained from any gallon of paint is dependent on its nonvolatile (solids) content. One gallon occupies a volume of cubic inches or One gallon covers approximately square feet (37 square meters). For best results, we recommend planning for 1 coat of Jolie Primer and at least 2 coats of. Calculate How Much Paint You Will Need. The average bedroom requires 1 gallon ( litres) of wall paint. A one gallon ( litre) can of wall paint covers. One gallon of Interior BEHR ULTRA® SCUFF DEFENSE™ Paint and Primer, or BEHR PREMIUM PLUS® is enough to cover to Sq. Ft. of surface area with one coat. One gallon can of paint will cover up to square feet, which is enough to cover a small room like a bathroom. · Two gallon cans of paint cover up to The average bedroom requires 1 gallon ( liters) of wall paint. A one gallon ( liter) can of wall paint covers about – square feet (35 to You need 0 gallons. Congratulations! You now know how much paint you How Much Paint Do I Need? PPG Industries. Where to buy. Find a Store. Subscribe. A: With typical application, a gallon of paint covers about square feet. Q: How much does it cost to paint a 12x12 room? Paint per Square Foot. The general rule of thumb is a single 1-gallon can of paint can safely cover up to square feet with a single coat of paint. One gallon of paint covers approximately – square feet depending on the porosity of the surface and how thick it is applied. A five gallon. In general, one gallon of paint or primer will cover roughly square feet of surface. Save time and money with the KILZ paint calculator to estimate. One gallon of any BEHR paint is enough to cover between to square feet of surface area with one coat. Do this for each gable and add the total to. Enter "0" if there are none. Calculate. Paint Calculator. See how much paint you'll need. Paint Calculator Error Delivers excellent hide and coverage for a. square feet per gallon is EXTEREMELY GENEROUS. Especially considering that the TDS states which still depends on color and the walls. Finally, if the paint is known to cover ft2 per gallon, and given that two coats are needed, divide the square footage by the paint coverage, then multiply. Calculate How Much Paint You Will Need. The average bedroom requires 1 gallon ( litres) of wall paint. A one gallon ( litre) can of wall paint covers. Per gallon, our paint covers approximately sq. ft., primer covers sq. ft. and ceiling paint covers sq. ft. Explore Colors. Compare Anderson And Pella Windows \| How Should I Pay Myself From My Llc	{"url":"https://clubbiz.ru/learn/how-much-paint-does-one-gallon-cover.php","timestamp":"2024-11-03T15:51:55Z","content_type":"text/html","content_length":"11743","record_id":"<urn:uuid:3ce56962-3a48-437d-b1ea-7646b7e19b3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00720.warc.gz"}
Mathematical Conundrum or Not? Number Three Maybe this one is harder for people to sink their teeth into. Gabriel's horn is an object that exist in math which has finite volume, but infinite surface area, that is a conundrum if I have ever seen one. Such objects should not exist, but mathematically we can show that the volume converges to a finite point, while the surface area diverges to infinity. You can say, well it is not in the real world, and while it may be true I can't find a horn and point to it; however, it does exist in mathematics, and this is the math section of these forums and the title of this thread is "Mathematical Conundrum or Not?". The horn absolutely deserves its spot here, even if grasping it is not as intuitive as the other two I posted. I don't really see you as an authority on what is and what is not a paradox. I mean all you have here is an assertion and a false one at that. On the other-hand academically Gabriel's horn is widely viewed as paradoxical. So you don't think is a paradox, OK fine, I don't really care. Also volume is the amount of space an object takes up, paint or no paint. Well this is not my paradox, I didn't invent it. It is a well known paradox, and widely recognized as such. Also the mathematical proof is posted in the OP. Saying there is no mathematical basis for this just tells me you can't read the math, as it is posted right there for you to review. Any container or solid object that has an endless surface area, but a finite volume is paradoxical, abstractly or otherwise. Volume is the amount of space it takes up, so if it has endless surface area it should have endless volume. — Jeremiah This assertion is exactly that: just an assertion, and a false one at that. There is no mathematical basis for this. The paradox apparently comes from your assumption of this nonexistent law. The horn both converges and diverges, so it fits your personal take on what is needed for a paradox. — Jeremiah A paradox is usually of the form of "If A is true, then A can be shown to be false". Your original 25 25 50 60 thingy would have been paradoxical had the 60 entry read 0%. What you seem to be reaching for here is not a paradox, but rather a violation of the law of non-contradiction, that a thing cannot be both X and not-X at the same time in the same way. I don't see the violation due to the 'in the same way' part. So you are suggesting a finite amount of paint that goes on forever. that paints an infinite surface. 'goes on forever' is not what I said, and seems a sort of undefined wording. The alternative is that there is some points along your surface that do not enclose volume and are thus not painted. So in your suggestion the volume of the paint both converges and diverges? No, the volume is finite. You said that. There is finite (convergent as you put it) volume of ice cream, which could be paint. Claiming it is abstract does't prove that Gabriel's horn is not a mathematical conundrum. Abstract things exist in reality, as reality is a very very very broad term. Gabriel's horn exist in reality, the math was posted in the OP. Any container or solid object that has an endless surface area, but a finite volume is paradoxical, abstractly or otherwise. — Jeremiah No such object can exist in Reality, so it cannot be "abstractly or otherwise". Any container or solid object that has an endless surface area, but a finite volume is paradoxical, abstractly or otherwise. Volume is the amount of space it takes up, so if it has endless surface area it should have endless volume. However, Gabriel's horn does't, and this is why it is widely recognized as a paradox. Like I said. You are confusing abstract and physical properties that happen to have the same name. — tom If you recall I never said or agreed to any such notion in the last thread. I avoid that line of thought for a reason. There is nothing which says we can't think about this in more practical terms. In what way is this in need of 'resolution'? You haven't stated a problem with this scenario. Is there some law somewhere being broken, like infinite surfaces must enclose infinite space? There is obviously no such law, as demonstrated by this example. — noAxioms The paradox, seems clear to me, we have a container that stretches on forever, yet it has a finite volume. It is only paradoxical if the same thing both converges and diverges. — noAxioms The horn both converges and diverges, so it fits your personal take on what is needed for a paradox. Clearly the paint would not run out, as it hasn't in your example. It covers the entire surface, and doesn't even need to be spread out to do so, since it has finite thickness (all the way to the center line) at any point being painted. — noAxioms So you are suggesting a finite amount of paint that goes on forever. So in your suggestion the volume of the paint both converges and diverges? Well, mathematically we can prove the volume of the paint converges, that means there is a limited amount of it, but if you want to claim it is a endless bucket of paint go for it. The math just does not back you up. This one is a bit trickier and as far as I know it has not been resolved. — Jeremiah In what way is this in need of 'resolution'? You haven't stated a problem with this scenario. Is there some law somewhere being broken, like infinite surfaces must enclose infinite space? There is obviously no such law, as demonstrated by this example. So you are suggesting if it was filled with paint, you could use a finite amount of paint to paint an endless surface. It seems to me, that you'd run out of paint, and even if you could stretch the paint infinitely thinner, that still does not resolve the paradox. As abstractly what you have is a cone with a converging volume and a diverging surface area. — Jeremiah Clearly the paint would not run out, as it hasn't in your example. It covers the entire surface, and doesn't even need to be spread out to do so, since it has finite thickness (all the way to the center line) at any point being painted. I see no paradox in need of resolution. The volume converges and something different (the area) does not. It is only paradoxical if the same thing both converges and diverges. So you are suggesting if it was filled with paint, you could use a finite amount of paint to paint an endless surface. — Jeremiah It is trivial to divide any volume to cover an infinite surface. There are plenty of convergent infinite series that will divide the volume for you. It seems to me, that you'd run out of paint, and even if so that still does not resolve the paradox. As abstractly what you have is a cone with a converging volume and a diverging surface area. — Like I said. You are confusing abstract and physical properties that happen to have the same name. So you are suggesting if it was filled with paint, you could use a finite amount of paint to paint an endless surface. It seems to me, that you'd run out of paint, and even if you could stretch the paint infinitely thinner, that still does not resolve the paradox. As abstractly what you have is a cone with a converging volume and a diverging surface area. So how it is possible this horn can have limited volume but endless surface area? — Jeremiah In mathematics, any volume can be divided in such a way to cover any surface. This is just like Zeno's paradox. The paradox arises from the confusion of abstract properties with real ones of the same name. Unlike the horn, my post had a real point. :) That was informative and fun to read though. :up: Baden OptionsShare My thought is, don't sit on it. — Baden Wise advice. Fortunately, it is impossible to sit on it, because it has no tip. The pointy bit just recedes endlessly, never culminating in a spike. The ultimate in child-safety mathematical structures. As for getting it off the ground, that would be impossible because, even though it has finite volume, and hence finite mass (if we assume constant density), its moment of inertia would be infinite because of its being infinitely long. So it would require an infinite torque to rotate it to an erect position. Short version - funny things happen with infinity. (one reason why maths is so much fun) Thank you for your response. Monitor OptionsShare This one is a bit trickier and as far as I know it has not been resolved. So if we can't get it off the ground, I have others waiting. My thought is, don't sit on it. Baden OptionsShare Any thoughts on the horn? Are you aware of a reason, in your own mind, why you believe you are posting these things? Monitor OptionsShare Think of an ice cream cone where it is possible to eat all the ice cream but not the cone, because even though the ice cream fills the cone, it is finite, but the cone goes on forever. This paradox is commonly known as Gabriel's Horn. When you graph the function y=1/x on [1, inf) and then rotate about the x-axis you get Gabriel's Horn, an object that has a finite volume but infinite surface area. Here is a visual: If you want to review the math, it is explained here, it is not too heavy but you need some calculus (starting on page 2). So how it is possible this horn can have limited volume but endless surface area? Well this is not my paradox, I didn't invent it. It is a well known paradox, and widely recognized as such. Also the mathematical proof is posted in the OP. — Jeremiah So you don't think is a paradox, OK fine, I don't really care. — Jeremiah Fair enough. The relevant definition of paradox that pops up says this: a : a statement that is seemingly contradictory or opposed to common sense and yet is perhaps true b : a self-contradictory statement that at first seems true c : an argument that apparently derives self-contradictory conclusions by valid deduction from acceptable premises — Webster The funny cone seems to fall under definition 'a' since it seems opposed to common sense to many people. So yes, it makes sense to 'resolve' such paradoxes by showing that the seeming contradiction is something that is actually the case. The mathematics (a computation of the area and volume) is linked in the OP, but not sure what part of that is a 'proof' of something. 'b' seems to be the opposite of 'a': something that seems true at first but false on closer inspection. I guess my idea of a paradox falls under 'c', the most basic example being "This statement is false". Any truth value assigned to that seems to be incorrect. I've seen it resolved in law-of-form using an imaginary truth value (square root of false) just like imaginary numbers solve square root of -1. There is application for such logic in quantum computing. Saying there is no mathematical basis for this just tells me you can't read the math, as it is posted right there for you to review. I never contested the mathematics, which simply shows that the object indeed has infinite area but finite volume. I can think of more trivial objects that are finite in one way but infinite in another, and your cone did not strike me as a connundrum. But I retract my assertion that it is not a paradox. The definition above speaks. You can say, well it is not in the real world — Jeremiah Indeed, it is only a mathematical object. A real one could not be implemented, growing too thin to insert ice cream particles after a while. Interestingly, a liter of physical paint contains insufficient paint to actually cover a square meter of surface. There is a finite quantity of fundamental particles making up the volume of paint, and no fundamental particle has ever been found that occupies actual volume. So the paint is all empty space with effectively dimensionless objects which are incapable of being arranged to cover a given area without gaps. Instead, paint atoms work by deflecting light and water and such using its EM properties, not by actually covering a surface without gaps. Point is that this particular mathematical object has little relevance to even a hypothetical physical object. It's not quite the same thing, but this reminds me of the story of how Dido bought the land that became Carthage by agreeing to buy as much land as she could encompass with a single oxhide. By cutting the hide into extremely thin strips, she was able to section off quite a bit more than the sellers reckoned with. She wasn't working with infinite amounts, but I wonder if there's a way to figure you could encompass an infinite area with a finite mass? Artemis OptionsShare Such objects should not exist — Jeremiah That is a feeling. The 18th century British invaders of Australia had a similar feeling when they first saw a platypus. When they found that the object in question was undeniably there in front of them, their 'should not exist' transformed to 'well, I am very surprised'. Is the aim of this thread then to muse over the nature of the emotion we call Surprise? Let me know if you figure it out. It's not for me to figure out. You started the thread. What was your aim? I'll give you a hint, it has nothing to do with a platypus. That's the third time you dodged the question - which was originally put to you in post #2. Are you going to answer the question? 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Foundations of mathematics This article needs additional citations for verification (May 2024) Foundations of mathematics are the The term "foundations of mathematics" was not coined before the end of the 19th century, although foundations were first established by the ancient These foundations were tactily assumed to be definitive until the introduction of During the 19th century, progress was made towards elaborating precise definitions of the basic concepts of infinitesimal calculus, notably the foundational crisis of mathematics The resolution of this crisis involved the rise of a new mathematical discipline called It results from this that the basic mathematical concepts, such as mathematical intuition : physical reality is still used by mathematicians to choose axioms, find which theorems are interesting to prove, and obtain indications of possible proofs. Ancient Greece Most civilisations developed some mathematics, mainly for practical purposes, such as counting (merchants), ancient Greek philosophers were the first to study the nature of mathematics and its relation with the real world. mathematical infinity , a concept that was outside the mathematical foundations of that time and was not well understood before the end of the 19th century. The Pythagorean school of mathematics originally insisted that the only numbers are natural numbers and ratios of natural numbers. The discovery (around 5th century BC) that the ratio of the diagonal of a square to its side is not the ratio of two natural numbers was a shock to them which they only reluctantly accepted. A testimony of this is the modern terminology of irrational number for referring to a number that is not the quotient of two integers, since "irrational" means originally "not reasonable" or "not accessible with reason". The fact that length ratios are not represented by rational numbers was resolved by Eudoxus of Cnidus (408–355 BC), a student of Plato, who reduced the comparison of two irrational ratios to comparisons of integer multiples of the magnitudes involved. His method anticipated that of Dedekind cuts in the modern definition of real numbers by Richard Dedekind (1831–1916);^[2] see Eudoxus of Cnidus § Eudoxus' proportions. In the axiomatic method but with a big philosophical difference: axioms and postulates were supposed to be true, being either self-evident or resulting from , while no other truth than the correctness of the proof is involved in the axiomatic method. So, for Aristotle, a proved theorem is true, while in the axiomatic methods, the proof says only that the axioms imply the statement of the theorem. Aristotle's logic reached its high point with (though they do not always conform strictly to Aristotelian templates). Aristotle's syllogistic logic , together with its exemplification by Euclid's , are recognized as scientific achievements of ancient Greece, and remained as the foundations of mathematics for centuries. Before infinitesimal calculus During Middle Ages, Euclid's Elements stood as a perfectly solid foundation for mathematics, and philosophy of mathematics concentrated on the ontological status of mathematical concepts; the question was whether they exist independently of perception (realism) or within the mind only (conceptualism); or even whether they are simply names of collection of individual objects (nominalism). In Elements, the only numbers that are considered are formulas discovered in the 16th century result from algebraic manipulations that have no geometric counterpart. Nevertheless, this did not challenge the classical foundations of mathematics since all properties of numbers that were used can be deduced from their geometrical definition. In 1637, infinitesimal calculus Infinitesimal calculus infinitesimal calculus for dealing with mobile points (such as planets in the sky) and variable quantities. This needed the introduction of new concepts such as continuous functions, derivatives and limits. For dealing with these concepts in a logical way, they were defined in terms of infinitesimals that are hypothetical numbers that are infinitely close to zero. The strong implications of infinitesimal calculus on foundations of mathematics is illustrated by a pamphlet of the Protestant philosopher George Berkeley (1685–1753), who wrote "[Infinitesimals] are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not call them the ghosts of departed quantities?".^[3] Also, a lack of rigor has been frequently invoked, because infinitesimals and the associated concepts were not formally defined ( planes were not formally defined either, but people were more accustomed to them). Real numbers, continuous functions, derivatives were not formally defined before the 19th century, as well as Euclidean geometry . It is only in the 20th century that a formal definition of infinitesimals has been given, with the proof that the whole infinitesimal can be deduced from them. Despite its lack of firm logical foundations, infinitesimal calculus was quickly adopted by mathematicians, and validated by its numerous applications; in particular the fact that the planet trajectories can be deduced from the Newton's law of gravitation 19th century In the 19th century, mathematics developed quickly in many directions. Several of the problems that were considered led to questions on the foundations of mathematics. Frequently, the proposed solutions led to further questions that were often simultaneously of philosophical and mathematical nature. All these questions led, at the end of the 19th century and the beginning of the 20th century, to debates which have been called the foundational crisis of mathematics. The following subsections describe the main such foundational problems revealed during the 19th century. Real analysis (ε, δ)-definition of limit. The modern continuous functions was first developed by in 1817, but remained relatively unknown, and Cauchy probably did know Bolzano's work. Karl Weierstrass (1815–1897) formalized and popularized the (ε, δ)-definition of limits, and discovered some pathological functions that seemed paradoxical at this time, such as continuous, nowhere-differentiable functions. Indeed, such functions contradict previous conceptions of a function as a rule for computation or a smooth graph. At this point, the program of arithmetization of analysis (reduction of mathematical analysis to arithmetic and algebraic operations) advocated by Weierstrass was essentially completed, except for two points. Firstly, a formal definition of real numbers were still lacking. Indeed, beginning with Several problems were left open by these definitions, which contributed to the foundational crisis of mathematics. Firstly both definitions suppose that rational numbers and thus natural numbers are rigorously defined; this was done a few years later with Peano axioms . Secondly, both definitions involve infinite sets (Dedekind cuts and sets of the elements of a Cauchy sequence), and Cantor's set theory was published several years later. The third problem is more subtle: and is related to the foundations of logic: classical logic is a least upper bound that is a real number. This need of quantification over infinite sets is one of the motivation of the development of higher-order logics during the first half of the 20th century. Non-Euclidean geometries Before the 19th century, there were many failed attempts to derive the parallel postulate from other axioms of geometry. In an attempt to prove that its negation leads to a contradiction, Johann Heinrich Lambert (1728–1777) started to build hyperbolic geometry and introduced the hyperbolic functions and computed the area of a hyperbolic triangle (where the sum of angles is less than 180°). Continuing the construction of this new geometry, several mathematicians proved independently that if it is Later in the 19th century, the German mathematician on the sphere. These proofs of unprovability of the parallel postulate lead to several philosophical problems, the main one being that before this discovery, the parallel postulate and all its consequences were considered as true. So, the non-Euclidean geometries challenged the concept of mathematical truth Synthetic vs. analytic geometry Since the introduction of Mathematicians did not worry much about the contradiction between these two approaches before the mid-nineteenth century, where there was "an acrimonious controversy between the proponents of synthetic and analytic methods in projective geometry, the two sides accusing each other of mixing projective and metric concepts".^[6] Indeed, there is no concept of distance in a projective space, and the cross-ratio, which is a number, is a basic concept of synthetic projective geometry. Karl von Staudt developed a purely geometric approach to this problem by introducing "throws" that form what is presently called a , in which the cross ratio can be expressed. Apparently, the problem of the equivalence between analytic and synthetic approach was completely solved only with Pappus hexagon theorem holds. Conversely, if the Pappus hexagon theorem is included in the axioms of a plane geometry, then one can define a field such that the geometry is the same as the affine or projective geometry over Natural numbers The work of , which was not formalized at this time. Giuseppe Peano provided in 1888 a complete axiomatisation based on the ordinal property of the natural numbers. The last Peano's axiom is the only one that induces logical difficulties, as it begin with either "if S is a set then" or "if ${\displaystyle \varphi }$ is a quantification on infinite sets, and this means that Peano arithmetic is what is presently called a Second-order logic This was not well understood at that times, but the fact that infinity occurred in the definition of the natural numbers was a problem for many mathematicians of this time. For example, Henri Poincaré stated that axioms can only be demonstrated in their finite application, and concluded that it is "the power of the mind" which allows conceiving of the indefinite repetition of the same act.^[7] This applies in particular to the use of the last Peano axiom for showing that the successor function generates all natural numbers. Also, Leopold Kronecker said "God made the integers, all else is the work of man".^[a] This may be interpreted as "the integers cannot be mathematically defined". Infinite sets Before the second half of the 19th century, was the subject of many philosophical disputes. Sets, and more specially infinite sets were not considered as a mathematical concept; in particular, there was no fixed term for them. A dramatic change arose with the work of Georg Cantor who was the first mathematician to systematically study infinite sets. In particular, he introduced cardinal numbers that measure the size of infinite sets, and ordinal numbers that, roughly speaking, allow one to continue to count after having reach infinity. One of his major results is the discovery that there are strictly more real numbers than natural numbers (the cardinal of the continuum of the real numbers is greater than that of the natural numbers). These results were rejected by many mathematicians and philosophers, and led to debates that are a part of the foundational crisis of mathematics. The crisis was amplified with the Russel's paradox that asserts that the phrase "the set of all sets" is self-contradictory. This condradiction introduced a doubt on the of all mathematics. With the introduction of the Gödel's incompleteness theorem Mathematical logic In 1847, Independently, in the 1870's, predicate logic Frege pointed out three desired properties of a logical theory:consistency (impossibility of proving contradictory statements), completeness (any statement is either provable or refutable; that is, its negation is provable), and decidability (there is a decision procedure to test every statement). By near the turn of the century, Russel's paradox which implies that the phrase "the set of all sets" is self-contradictory. This paradox seemed to make the whole mathematics inconsistent and is one of the major causes of the foundational crisis of mathematics. Foundational crisis The foundational crisis of mathematics arose at the end of the 19th century and the beginning of the 20th century with the discovery of several paradoxes or counter-intuitive results. The first one was the proof that the Peano arithmetic Several schools of philosophy of mathematics were challenged with these problems in the 20th century, and are described below. These problems were also studied by mathematicians, and this led to establish inference rules ), mathematical and logical theories, theorems, and proofs, and of using mathematical methods to prove theorems about these concepts. This led to unexpected results, such as ; and, if it is not self-contradictory, there are theorems that cannot be proved inside the theory, but are nevertheless true in some technical sense. Zermelo–Fraenkel set theory with the axiom of choice (ZFC) is a logical theory established by Ernst Zermelo and Abraham Fraenkel. It became the standard foundation of modern mathematics, and, unless the contrary is explicitly specified, it is used in all modern mathematical texts, generally implicitly. Simultaneously, the axiomatic method became a de facto standard: the proof of a theorem must result from explicit and previously proved theorems by the application of clearly defined inference rules. The axioms need not correspond to some reality. Nevertheless, it is an open philosophical problem to explain why the axiom systems that lead to rich and useful theories are those resulting from abstraction from the physical reality or other mathematical theory. In summary, the foundational crisis is essentially resolved, and this opens new philosophical problems. In particular, it cannot be proved that the new foundation (ZFC) is not self-contradictory. It is a general consensus that, if this would happen, the problem could be solved by a mild modification of ZFC. Philosophical views When the foundational crisis arose, there was much debate among mathematicians and logicians about what should be done for restoring confidence in mathematics. This involved philosophical questions , and the nature of mathematics. For the problem of foundations, there was two main options for trying to avoid paradoxes. The first one led to formalism, considers that a theorem is true if it can be deduced from by applying inference rules ( formal proof ), and that no "trueness" of the axioms is needed for the validity of a theorem. It has been claimed that formalists, such as David Hilbert (1862–1943), hold that mathematics is only a language and a series of games. Hilbert insisted that formalism, called "formula game" by him, is a fundamental part of mathematics, but that mathematics must not be reduced to formalism. Indeed, he used the words "formula game" in his 1927 response to L. E. J. Brouwer's criticisms: And to what extent has the formula game thus made possible been successful? This formula game enables us to express the entire thought-content of the science of mathematics in a uniform manner and develop it in such a way that, at the same time, the interconnections between the individual propositions and facts become clear ... The formula game that Brouwer so deprecates has, besides its mathematical value, an important general philosophical significance. For this formula game is carried out according to certain definite rules, in which the technique of our thinking is expressed. These rules form a closed system that can be discovered and definitively stated.^[10] Thus Hilbert is insisting that mathematics is not an arbitrary game with arbitrary rules; rather it must agree with how our thinking, and then our speaking and writing, proceeds.^[10] We are not speaking here of arbitrariness in any sense. Mathematics is not like a game whose tasks are determined by arbitrarily stipulated rules. Rather, it is a conceptual system possessing internal necessity that can only be so and by no means otherwise.^[11] The foundational philosophy of formalism, as exemplified by formal logic. Virtually all mathematical today can be formulated as theorems of set theory. The truth of a mathematical statement, in this view, is represented by the fact that the statement can be derived from the axioms of set theory using the rules of formal logic. Merely the use of formalism alone does not explain several issues: why we should use the axioms we do and not some others, why we should employ the logical rules we do and not some others, why "true" mathematical statements (e.g., the laws of arithmetic) appear to be true, and so on. Hermann Weyl posed these very questions to Hilbert: What "truth" or objectivity can be ascribed to this theoretic construction of the world, which presses far beyond the given, is a profound philosophical problem. It is closely connected with the further question: what impels us to take as a basis precisely the particular axiom system developed by Hilbert? Consistency is indeed a necessary but not a sufficient condition. For the time being we probably cannot answer this question ...^[12] In some cases these questions may be sufficiently answered through the study of formal theories, in disciplines such as . What Hilbert wanted to do was prove a logical system was consistent, based on principles that only made up a small part of . But Gödel proved that the principles could not even prove to be consistent, let alone Intuitionists, such as L. E. J. Brouwer (1882–1966), hold that mathematics is a creation of the human mind. Numbers, like fairy tale characters, are merely mental entities, which would not exist if there were never any human minds to think about them. The foundational philosophy of Stephen Kleene, requires proofs to be "constructive" in nature – the existence of an object must be demonstrated rather than inferred from a demonstration of the impossibility of its non-existence. For example, as a consequence of this the form of proof known as reductio ad absurdum is suspect. Some modern cognitive science of mathematics , focusing on human cognition as the origin of the reliability of mathematics when applied to the real world. These theories would propose to find foundations only in human thought, not in any objective outside construct. The matter remains controversial. Logicism is a school of thought, and research programme, in the philosophy of mathematics, based on the thesis that mathematics is an extension of logic or that some or all mathematics may be derived in a suitable formal system whose axioms and rules of inference are 'logical' in nature. Bertrand Russell and Alfred North Whitehead championed this theory initiated by Gottlob Frege and influenced by Richard Dedekind. Set-theoretic Platonism Many researchers in axiomatic set theory have subscribed to what is known as set-theoretic , exemplified by Kurt Gödel Several set theorists followed this approach and actively searched for axioms that may be considered as true for heuristic reasons and that would decide the continuum hypothesis. Many large cardinal axioms were studied, but the hypothesis always remained independent from them and it is now considered unlikely that CH can be resolved by a new large cardinal axiom. Other types of axioms were considered, but none of them has reached consensus on the continuum hypothesis yet. Recent work by Hamkins proposes a more flexible alternative: a set-theoretic multiverse allowing free passage between set-theoretic universes that satisfy the continuum hypothesis and other universes that do not. Indispensability argument for realism says (in Putnam's shorter words), ... quantification over mathematical entities is indispensable for science ... therefore we should accept such quantification; but this commits us to accepting the existence of the mathematical entities in question. However, Putnam was not a Platonist. Rough-and-ready realism Few mathematicians are typically concerned on a daily, working basis over logicism, formalism or any other philosophical position. Instead, their primary concern is that the mathematical enterprise as a whole always remains productive. Typically, they see this as ensured by remaining open-minded, practical and busy; as potentially threatened by becoming overly-ideological, fanatically reductionistic or lazy. Such a view has also been expressed by some well-known physicists. For example, the Physics Nobel Prize laureate Richard Feynman said People say to me, "Are you looking for the ultimate laws of physics?" No, I'm not ... If it turns out there is a simple ultimate law which explains everything, so be it – that would be very nice to discover. If it turns out it's like an onion with millions of layers ... then that's the way it is. But either way there's Nature and she's going to come out the way She is. So therefore when we go to investigate we shouldn't predecide what it is we're looking for only to find out more about it.^[13] And Steven Weinberg:^[14] The insights of philosophers have occasionally benefited physicists, but generally in a negative fashion – by protecting them from the preconceptions of other philosophers. ... without some guidance from our preconceptions one could do nothing at all. It is just that philosophical principles have not generally provided us with the right preconceptions. Weinberg believed that any undecidability in mathematics, such as the continuum hypothesis, could be potentially resolved despite the incompleteness theorem, by finding suitable further axioms to add to set theory. Philosophical consequences of Gödel's completeness theorem Gödel's completeness theorem establishes an equivalence in first-order logic between the formal provability of a formula and its truth in all possible models. Precisely, for any consistent first-order theory it gives an "explicit construction" of a model described by the theory; this model will be countable if the language of the theory is countable. However this "explicit construction" is not algorithmic. It is based on an iterative process of completion of the theory, where each step of the iteration consists in adding a formula to the axioms if it keeps the theory consistent; but this consistency question is only semi-decidable (an algorithm is available to find any contradiction but if there is none this consistency fact can remain unprovable). More paradoxes The following lists some notable results in metamathematics. Zermelo–Fraenkel set theory is the most widely studied axiomatization of set theory. It is abbreviated ZFC when it includes the axiom of choice and ZF when the axiom of choice is excluded. Toward resolution of the crisis Starting in 1935, the Bourbaki group of French mathematicians started publishing a series of books to formalize many areas of mathematics on the new foundation of set theory. The intuitionistic school did not attract many adherents, and it was not until constructive mathematics was placed on a sounder footing. One may consider that Hilbert's program has been partially completed, so that the crisis is essentially resolved, satisfying ourselves with lower requirements than Hilbert's original ambitions. His ambitions were expressed in a time when nothing was clear: it was not clear whether mathematics could have a rigorous foundation at all. There are many possible variants of set theory, which differ in consistency strength, where stronger versions (postulating higher types of infinities) contain formal proofs of the consistency of weaker versions, but none contains a formal proof of its own consistency. Thus the only thing we do not have is a formal proof of consistency of whatever version of set theory we may prefer, such as In practice, most mathematicians either do not work from axiomatic systems, or if they do, do not doubt the consistency of , generally their preferred axiomatic system. In most of mathematics as it is practiced, the incompleteness and paradoxes of the underlying formal theories never played a role anyway, and in those branches in which they do or whose formalization attempts would run the risk of forming inconsistent theories (such as logic and category theory), they may be treated carefully. The development of category theory in the middle of the 20th century showed the usefulness of set theories guaranteeing the existence of larger classes than does ZFC, such as Von Neumann–Bernays–Gödel set theory or Tarski–Grothendieck set theory, albeit that in very many cases the use of large cardinal axioms or Grothendieck universes is formally eliminable. One goal of the reverse mathematics program is to identify whether there are areas of "core mathematics" in which foundational issues may again provoke a crisis. See also 1. ^ The English translation is from Gray. In a footnote, Gray attributes the German quote to: "Weber 1891–1892, 19, quoting from a lecture of Kronecker's of 1886."^[8]^[9] 1. . 2. ^ The Analyst, A Discourse Addressed to an Infidel Mathematician 4. ^ O'Connor, John J.; Robertson, Edmund F. (October 2005), "The real numbers: Stevin to Hilbert", MacTutor History of Mathematics Archive, University of St Andrews 6. ^ Poincaré, Henri (1905) [1902]. "On the nature of mathematical reasoning". La Science et l'hypothèse [Science and Hypothesis]. Translated by Greenstreet, William John. VI. 7. from the original on 29 March 2017 – via Google Books. 8. ^ Weber, Heinrich L. (1891–1892). "Kronecker". Jahresbericht der Deutschen Mathematiker-Vereinigung [Annual report of the German Mathematicians Association]. pp. 2:5–23. (The quote is on p. 19). Archived from the original on 9 August 2018;"access to Jahresbericht der Deutschen Mathematiker-Vereinigung". Archived from the original on 20 August 2017. 9. ^ ^a ^b Hilbert 1927 The Foundations of Mathematics in van Heijenoort 1967:475 10. ^ p. 14 in Hilbert, D. (1919–20), Natur und Mathematisches Erkennen: Vorlesungen, gehalten 1919–1920 in Göttingen. Nach der Ausarbeitung von Paul Bernays (Edited and with an English introduction by David E. Rowe), Basel, Birkhauser (1992). 11. ^ Weyl 1927 Comments on Hilbert's second lecture on the foundations of mathematics in van Heijenoort 1967:484. Although Weyl the intuitionist believed that "Hilbert's view" would ultimately prevail, this would come with a significant loss to philosophy: "I see in this a decisive defeat of the philosophical attitude of pure phenomenology, which thus proves to be insufficient for the understanding of creative science even in the area of cognition that is most primal and most readily open to evidence – mathematics" (ibid). 12. ^ Richard Feynman, The Pleasure of Finding Things Out p. 23 13. ^ Steven Weinberg, chapter Against Philosophy wrote, in Dreams of a final theory 14. (PDF) on 2016-03-04, retrieved 2016-02-22 In Chapter III A Critique of Mathematic Reasoning, §11. The paradoxes, Kleene discusses in depth. Throughout the rest of the book he treats, and compares, both Formalist (classical) and Intuitionist logics with an emphasis on the former. Extraordinary writing by an extraordinary External links	{"url":"https://findatwiki.com/Foundations_of_mathematics","timestamp":"2024-11-05T02:56:29Z","content_type":"text/html","content_length":"228242","record_id":"<urn:uuid:22c2c1e9-9988-4144-9786-9009fa9d8d9b>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00509.warc.gz"}
Francisca Vasconcelos CS PhD @ UC Berkeley Theory Group and BAIR Lab. I am a third-year PhD student in the UC Berkeley Department of Electrical Engineering and Computer Science. My research is supported by the NSF Graduate Research Fellowship and Paul & Daisy Soros Fellowship for New Americans. I am co-advised by Profs Michael Jordan and Umesh Vazirani. My research interests lie at the intersection of quantum computation and machine learning theory. In 2020, I received a BS in EECS and Physics from MIT, where I was fortunate to do substantial undergraduate research advised by Prof William Oliver in the MIT Engineering Quantum Systems group. As an undergraduate, I also interned under Dr. Marcus da Silva at Rigetti Computing and Microsoft Research Quantum. Supported by a Rhodes Scholarship, I received two masters from the University of Oxford: an MSc in Statistical Sciences and MSt in Philosophy of Physics. Following from the MSc, I performed statistical ML research in the OxCSML group, advised by Prof Yee Whye Teh. I am also the Founding Academic Director of the Qubit x Qubit (QxQ) initiative of The Coding School (TCS) non-profit. Since 2019, we have taught 20,000+ diverse K-12 students, undergraduates, and members of the workforce worldwide about the fundamentals of quantum computing and QISE.	{"url":"https://franciscavasconcelos.github.io/","timestamp":"2024-11-04T04:18:12Z","content_type":"text/html","content_length":"29786","record_id":"<urn:uuid:5583756d-da50-4bdb-956c-631833274eb6>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00282.warc.gz"}
88.32 hours to minutes Let's understand how to change 88.32 hours into minutes and seconds. 1. Convert Hours to Whole Minutes To change hours to whole minutes, multiply the number of hours by 60 and take the integer part. In this case: 88.32 hours × 60 = 5299.2 minutes 2. Determine Remaining Seconds After converting to whole minutes, any remaining fraction of a minute is then converted into seconds. To do this, we multiply the fraction of a minute by 60 (since there are 60 seconds in a minute). For example, if the remaining fraction of a minute is 0.8, multiplying 0.8 by 60 gives us 48 seconds. Therefore, in this case, the remaining fraction is approximately 12 seconds. Final Answer Therefore, 88.32 hours is equal to 5299 minutes and 12 seconds.	{"url":"https://unitconverter.io/hours/minutes/88.32","timestamp":"2024-11-11T22:53:39Z","content_type":"text/html","content_length":"17139","record_id":"<urn:uuid:e40ee600-5381-4283-8dee-8d6c718defe8>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00494.warc.gz"}
Part 1: Bidirectional Constraint Generation Last time , we laid out the AST and Type for the language we are building. We also got a bird’s-eye view of our type inference algorithm: constraint generation, constraint solving, substitute our solved types. This time we’re implementing the constraint generation portion of our type inference algorithm. Our passes will need to share some state between each other. We introduce a TypeInference struct to hold this shared state and implement our passes as methods on that struct: struct TypeInference { unification_table: InPlaceUnificationTable<TypeVar>, We’ll talk more about unification_table when we talk about constraint solving. For now, it’s enough to think of it as a mapping from type variables to type, and we’ll use it in constraint generation to generate new type variables and keep track of them for later. We generate our set of constraints from contextual information in our AST. To get this context we need to visit every node of our AST and collect constraints for that node. Traditionally this is done with a bottom-up tree traversal (e.g. HM’s Algorithm J). We visit all of a node’s children and then use the children’s context to better infer the type of our node. This approach is logically correct. We always infer the correct type and type error when we should. While correct, this approach doesn’t make the most efficient use of information available. For example, in an application node we know that the function child must have a function type. Since types are only inferred bottom-up, we have to infer an arbitrary type for our function node and add a constraint that the inferred type must be a function. In recent years, a new approach to type checking called Bidirectional Type Checking has arisen to solve this inefficiency. With Bidirectional Type Checking we have two modes of type checking: • infer - works the same as the HM type systems we just described, so types are inferred bottom-up. • check - works in the opposite direction, top-down. A type is passed into check, and we check that our AST has the same type. Our two modes will call each other mutually recursively to traverse the AST. Now when we want to type check an application node, we have a new option. We can construct a function type at the application node and check the function node against our constructed function type. Making better use of top-down contextual info like this allows us to generate fewer type variables and produce better error messages. Fewer type variables may not immediately appear as a benefit, but it makes the type checker faster. Our constraint solving is in part bound by the number of type variables we have to solve. It also makes debugging far easier. Each type variable acts as a point of indirection so fewer is better for debugging. So while Bidirectional Type Checking doesn’t allow us to infer “better” types, it does provide tangible benefits for a modest extension of purely bottom-up type inference. Now that we know how we’ll be traversing our AST to collect constraints, let’s talk about what our constraints will actually look like. For our first Minimum Viable Product (MVP) of type inference, Constraint will just be type equality: enum Constraint { TypeEqual(Type, Type) We’ll talk more about what it means for two types to be equal during constraint solving. For now, it’s sufficient to produce a set of type equalities as a result of our constraint generation. Constraint generation will be implemented on our TypeInference struct with 3 methods: impl TypeInference { fn fresh_ty_var(&mut self) -> TypeVar { ... } fn infer( &mut self, env: im::HashMap<Var, Type>, ast: Ast<Var> ) -> (GenOut, Type) { ... } fn check( &mut self, env: im::HashMap<Var, Type>, ast: Ast<Var>, ty: Type ) -> GenOut { .. } fresh_ty_var is a helper method we’re going to brush past for now. (We’ll have a lot to cover in constraint solving!) It uses our unification_table to produce a unique type variable every time we call it. Past that, we can see some parallels between our infer and check method that illustrate each mode. infer takes an AST node and returns a type, whereas check takes both an AST node and a type as parameters. This is because infer is working bottom-up and check is working top-down. Let’s take a second to look at env and GenOut. Both infer and check take an env parameter. This is used to track the type of AST variables in their current scope. env is implemented by an immutable HashMap from the im crate. An immutable hashmap makes it easy to add a new variable when it comes into scope, and drop it when the variable leaves scope. infer and check both also return a GenOut. This is a pair of our set of constraints and our typed AST: struct GenOut { // Set of constraints to be solved constraints: Vec<Constraint>, // Ast where all variables are annotated with their type typed_ast: Ast<TypedVar>, One final thing to note, we have no way to return an error from infer or check. We could of course panic, but for the sake of our future selves, we’ll return errors with Result where relevant. It just so happens it’s not relevant for constraint generation. Our output is a set of constraints. It’s perfectly valid for us to return a set of constraints that contradict each other. We discover the contradiction and produce a type error when we try to solve our constraints. That means there aren’t error cases during constraint generation. Neat! With our setup out of the way, we can dive into our implementation of infer and check. We’ll cover infer first. Because our AST begins untyped, we always call infer first in our type inference, so it is a natural starting point. infer is just a match on our input ast: fn infer( &mut self, env: im::HashMap<Var, Type>, ast: Ast<Var> ) -> (GenOut, Type) { match ast { Ast::Int(i) => todo!(), Ast::Var(v) => todo!(), Ast::Fun(arg, body) => todo!(), Ast::App(fun, arg) => todo!(), We’ll talk about each case individually, let’s start with an easy one to get our feet wet: Ast::Int(i) => ( When we see an integer literal, we know immediately that its type is Int. We don’t need any constraints to be true for this to hold, so we return an empty Vec. One step up in complexity over integers is our variable case: Ast::Var(v) => { let ty = &env[&v]; // Return a `TypedVar` instead of `Var` Ast::Var(TypedVar(v, ty.clone()) When we encounter a variable, we look up its type in our env and return its type. Our env lookup might fail though. What happens if we ask for a variable we don’t have an entry for? That means we have an undefined variable, and we’ll panic!. That’s fine for our purposes; we expect to have done some form of name resolution prior to type inference. If we encounter an undefined variable, we should’ve already exited with an error during name resolution. Past that, our Var case looks very similar to our Int case. We have no constraints to generate and immediately return the type we look up. Next we take a look at our Fun case: Ast::Fun(arg, body) => { // Create a type variable for our unknown type variable let arg_ty_var = self.fresh_ty_var(); // Add our agrument to our environment with it's type let env = env.update(arg, Type::Var(arg_ty_var)); // Check the body of our function with our extended environment let (body_out, body_ty) = self.infer(env, body); // body constraints are propagated // Our `Fun` holds a `TypedVar` now TypedVar(arg, Type::Var(arg_ty_var)), Type::fun(Type::Var(arg_ty_var), body_ty), Fun is where we actually start doing some nontrivial inference. We create a fresh type variable and record it as the type of arg in our env. With our fresh type variable in scope, we infer a type for body. We then use our inferred body type and generated argument type to construct a function type for our Fun node. While Fun itself doesn’t produce any constraints, it does pass on any constraints that body generated. Now that we know how to type a function, let’s learn how to type a function application: Ast::App(fun, arg) => { let (arg_out, arg_ty) = self.infer(env.clone(), arg); let ret_ty = Type::Var(self.fresh_ty_var()); let fun_ty = Type::fun(arg_ty, ret_ty.clone()); // Because we inferred an argument type, we can // construct a function type to check against. let fun_out = self.check(env, fun, fun_ty); // Pass on constraints from both child nodes Ast::app(fun_out.typed_ast, arg_out.typed_ast), App is more nuanced than our previous cases. We infer the type of our arg and use that to construct a function type with a fresh type variable as our return type. We use this function type to check our fun node is a function type as well. Our final type for our App node is our fresh return type, and we combine the constraints from fun and arg to produce our final constraint set. You may wonder why we’ve chosen to infer the type for arg instead of inferring a type for our fun node. This would be reasonable and would produce equally valid results. We’ve opted not to for a few key reasons. If we infer a type for our fun node, it is opaque. We know it has to be a function type, but all we have after inference is a Type. To coerce it into a function type we have to emit a constraint against a freshly constructed function type: let (fun_out, infer_fun_ty) = self.infer(env.clone(), fun); let arg_ty = self.fresh_ty_var(); let ret_ty = self.fresh_ty_var(); let fun_ty = Type::fun(arg_ty.clone(), ret_ty.clone()); let fun_constr = Constraint::TypeEqual(infer_fun_ty, fun_ty); let arg_out = self.check(env, arg, arg_ty); // ... We have to create an extra type variable and an extra constraint compared to inferring a type for arg first. Not a huge deal, and in fact in more expressive type systems this tradeoff is worth inferring the function type first as it provides valuable metadata for checking the argument types. Our type system isn’t in that category, though, so we take fewer constraints and fewer type variables every time. Choices like this crop up a lot where it’s not clear when we should infer and when we should check our nodes. Bidirectional Typing has an in-depth discussion of the tradeoffs and how to decide which approach to take. That covers all of our inference cases, completing our bottom-up traversal. Next let’s talk about its sibling check. Unlike infer, check does not cover every AST case explicitly. Because we are checking our AST against a known type, we only match on cases we know will check and rely on a catch-all bucket case to handle everything else. We’re still working case by case though, so at a high level our check looks very similar to infer: fn check( &mut self, ast: Ast<Var>, ty: Type ) -> GenOut { match (ast, ty) { // ... Notice we match on both our AST and type at once, so we can select just the cases we care about. Let’s look at our cases: (Ast::Int(i), Type::Int) => An integer literal trivially checks against the integer type. This case might appear superfluous; couldn’t we just let it be caught by the bucket case? Of course, we could, but this explicit case allows us to avoid type variables and avoid constraints. Our other explicit check case is for Fun: (Ast::Fun(arg, body), Type::Fun(arg_ty, ret_ty)) => { let env = env.update(arg, arg_ty.clone()); let body_out = self.check(env, body, ret_ty); GenOut { typed_ast: Ast::fun(TypeVar(arg, *arg_ty), body_out.typed_ast), Our Fun case is also straightforward. We decompose our Type::Fun into it’s argument and return type. Record our arg has arg_ty in our env, and then check that body has ret_ty in our updated env. It almost mirrors our infer’s Fun case, but instead of bubbling a type up, we’re pushing a type down. Those are our only two explicit check cases. Everything else is handled by our bucket case: (ast, expected_ty) => { let (mut out, actual_ty) = self.infer(ast); .push(Constraint::TypeEqual(expected_ty, actual_ty)); Finally, we have our bucket case. At first this might seem a little too easy. If we encounter an unknown pair, we just infer a type for our AST and add a constraint saying that type has to be equal to the type we’re checking against. If we think about this, it makes some sense though. In the unlikely case that neither of our types are variables ((Int, Fun) or (Fun, Int)), we will produce a type error when we try to solve our constraint. In the case that one of our types is a variable, we’ve now recorded the contextual info necessary about that variable by adding a constraint. We can rely on constraint solving to propagate that info to wherever it’s needed. This is the only place where we emit a constraint explicitly. Everywhere else we just propagate constraints from our children’s recursive calls. The point where we switch from checking back to inference is the only point where we require a constraint to ensure our type line up. Our intuition for infer and check help guide us to that conclusion. This is in part the insight and the power of a bidirectional type system. It will only become more valuable as we extend our type system to handle more complicated types. It’s hard to see how our two functions fit together from just from their implementations. Let’s walk through an example to see infer and check in action. Consider a contrived AST: This is the identity function applied to an integer. A simple example, but it uses all of our AST nodes and will give us some insight into how check lets us propagate more type information than infer alone. Our example will use some notation to let us introspect our environment and use human friendly names: x, y, z represent variables α, β, γ represent type variables env will use a literal formatted as { <var0>: <type0>, <var1>: <type1>, ... } with {} being an empty environment. Using our new notation we can shorten our AST example: Ast:fun(x, Ast::Var(x)), Okay, we start by calling infer on our root App node: Our App case starts by inferring a type for our arg. Because our argument is Ast::Int(3) its inferred type is Int: infer({}, Ast::Int(3)) = Int We use this inferred argument, and a fresh return type, to construct a function type that we check against our App’s func: Ast::Fun(x, Ast::Var(x)), Fun(Int, Var(α)) A function and a function type is one of our specific check cases (it doesn’t fall into the bucket case). We destructure the function type to determine the type of our argument and body. This is where check shines. If we just had infer we would have to introduce a new type variable for x and add a constraint that x’s type variable must be Int. Instead, we can immediately determine x’s type must be Int. With our env updated to include x has type Int, we check body against our function type’s return type: { x: Int }, This is not a check case we know how to handle, it falls into the bucket case. The bucket case infers a type for our body. This looks up x’s type in the environment and returns it: infer({ x: Int }, Var(x)) = Int We don’t show it in the example, but this will also return a new AST where x is annotated with its type: TypeVar(x, Int). We’ll see how that gets used when we look at the final output of our example. A constraint is added that our checked type is equal to our inferred type: Once we output that constraint we’re done calling infer and check. We propagate our constraints up the call stack and construct our typed AST as we go. At the end of returning from all our recursive calls we have our constraint set, with just one constraint: vec![Constraint::TypeEqual(Var(α), Int)] and our typed AST: TypeVar(x, Int), Ast::Var(TypedVar(x, Int)) The final overall type of our AST, returned from our first infer call, is Var(α), remember our function’s type is Fun(Int, Var(α)). This illustrates why we need the final substitution step after constraint solving. Only once we’ve solved our constraints do we know α = Int, and we can correctly determine our overall AST’s type is Int. With that we’ve finished generating our constraints. As output of constraint generation we produce three things: a set of constraints, a typed AST, and a Type for our whole AST. Our typed AST has a type associated to every variable (and from that we can recover the type of every node). However, a lot of these are still unknown type variables. We’ll save that AST for now and revisit it once we’ve solved our set of constraints and have a solution for all our type variables. Naturally then, next time we’ll implement constraint solving. Full source code can be found in the companion repo	{"url":"https://thunderseethe.dev/posts/bidirectional-constraint-generation/","timestamp":"2024-11-01T23:59:09Z","content_type":"text/html","content_length":"62361","record_id":"<urn:uuid:25a21466-e5f7-427c-8085-a2511f0fbe0d>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00475.warc.gz"}
Import sage in python3 Import sage in python3 I've just finished building sage from source with python3, and it works great! I'm just wondering why from sage.all import * doesn't work in my python3, although trying the same in the sage shell works, so I guess it has to be a matter of environment variables? What should I do to be able to import sage in python3 scripts without having to rely on a sage shell? (Sort of) WorksForMe(TM): charpent@p-202-021:~$ sage -python Python 3.7.3 (default, Jul 10 2019, 14:13:36) [GCC 8.3.0] on linux Type "help", "copyright", "credits" or "license" for more information. >>> from sage.all import * >>> x=var("x") >>> integrate(arctan(x),x) xarctan(x) - 1/2log(x^2 + 1) >>> quit() charpent@p-202-021:~$ python3 Python 3.7.4 (default, Jul 11 2019, 10:43:21) [GCC 8.3.0] on linux Type "help", "copyright", "credits" or "license" for more information. >>> from sage.all import * Traceback (most recent call last): File "<stdin>", line 1, in <module> ModuleNotFoundError: No module named 'sage' >>> quit() You have to somehow tell to Python the place where to look for Sage... The full sage build includes its own Python interpreter--it doesn't install the sage package in your system's Python (eventually there will be an option for this by way of #27824 but that's a ways away yet :( 1 Answer Sort by ยป oldest newest most voted As @Emmanuel_Charpentier hints at in his comment, this requires using Sage's Python. If you built from source, you can call Sage's Python with sage --python. Or you could change your path so that Sage's Python is found first. Use with caution, as other apps / scripts / uses may rely on python calling the system Python. Note that you can install SageMath using Conda; it will install for Python 3. With the corresponding Conda environment activated, python will be the Python 3 which has SageMath installed on top of it. In that Python, from sage.all import * will work. edit flag offensive delete link more	{"url":"https://ask.sagemath.org/question/47465/import-sage-in-python3/","timestamp":"2024-11-06T20:19:40Z","content_type":"application/xhtml+xml","content_length":"56746","record_id":"<urn:uuid:242b81c8-be65-4d3e-8bd4-b25446f4f36c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00791.warc.gz"}
Indexed Values In order to take into account the size of the subject it is better to use indexed values for cardiac output (CO) and systemic vascular resistance (SVR), since a large subject will have a higher CO but lower SVR than a small subject (a small subject has small arteries which offer a greater resistance). Body surface area (BSA) is used as a measure of subject size and can be calculated using the Du Bois formula: if in patient 1.: BSA = 1.7 square metres MAP = 95 mm Hg RAP = 5 mm Hg then at a normal resting cardiac output (for an average adult) of 5 l/min substituting the numbers above this gives an SVR of 1440. So, in summary CO = 5 SVR = 1440 CI = 2.9 SVRI =2448 if in patient 2.: the pressures are the same as above but the patient is smaller with a BSA of 0.85 and a cardiac output of 2.5, then CO = 2.5 SVR = 2880 CI = 2.9 SVRI =2448	{"url":"http://foxlinton.org/cardiac_output/LDCOpages/indexedvalues.html","timestamp":"2024-11-10T18:46:20Z","content_type":"text/html","content_length":"2175","record_id":"<urn:uuid:2de35957-10c1-456d-8828-fb42464429c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00294.warc.gz"}
Addition Worksheets: Addition Fact Circles These addition worksheets emphasize groups of related facts and there are variations with the facts in order to facilitate skip counting, or with random products that help facilitate fact memorization. Try the variations with all facts, or print the worksheets that focus only on specific families of multiplication facts that need more practice! These Addition Facts Have Me Spinning in Circles! There's something to be said for grouping math facts in a way that makes seeing patterns and memorizing them an easier task to manage. These worksheets collect Addition facts into groups with a common addend into a clever circular pattern. These addition worksheets are a great alternative to simple addition fact drills or flash cards, and there are different groups of worksheets on this page the focus on smaller addition facts all the way through two digit sums.If these unique addition worksheets help, also be sure to check out the similar circle worksheets for subtraction, multiplication and division!	{"url":"https://www.dadsworksheets.com/worksheets/addition-fact-circles.html","timestamp":"2024-11-08T21:21:35Z","content_type":"text/html","content_length":"109919","record_id":"<urn:uuid:8c9c31ac-870e-446f-8de4-1b84d980e5dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00080.warc.gz"}
1. Derive the equation for the posterior mean by expanding the square in the exponential for each i, collecting all similar power terms, and making a perfect square again. Note that the product of exponentials can be written as the exponential of a sum of terms. 2. For this exercise, we use the dataset corresponding to Smartphone-Based Recognition of Human Activities and Postural Transitions, from the UCI Machine Learning repository (https:// archive.ics.uci.edu/ml/datasets/Smartphone-Based+Recognition+of+Human+Activities+and+Postural+Transitions). It contains values of acceleration taken from an accelerometer on a smartphone. The original dataset contains x, y, and z components of the acceleration and the corresponding timestamp values. For this exercise, we have used only the two horizontal components of the acceleration x and y. In this exercise, let's assume that the acceleration follows a normal distribution. Let's also assume a normal prior distribution for the mean...	{"url":"https://subscription.packtpub.com/book/data/9781783987603/3/ch03lvl1sec21/exercises","timestamp":"2024-11-04T11:09:36Z","content_type":"text/html","content_length":"180348","record_id":"<urn:uuid:6b839e63-6cb5-40e7-97d0-e797de52e38b>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00102.warc.gz"}
The field was 5β6 of an acre and 1β3 of the field was planted in strawberries. How many acres are planted in strawberries? The field was 5β 6 of an acre and 1β 3 of the field was planted in strawberries. How many acres are planted in strawberries? Answer:7/6 or 1 1/6Step-by-step explanation: Here's how you add them! 5 /6 Β + 2/6 Β Step 1 Since our denominators match, we can add the numerators. 5 + 2 = 7 Answer: 7 /6 Step 2 Now, do we need to simplify this fraction? First, we attempt to divide it by 2... Nope. Try the next prime number, 3... Nope. Try the next prime number, 5... Nope. Try the next prime number, 7... No good. 7 is larger than 6. So we're done reducing. Congratulations! Here's your final answer to 5/6 + 2/6 general 10 months ago 8751	{"url":"http://redmondmathblog.com/general/the-field-was-5-6-of-an-acre-and-1-3-of-the-field-was-planted-in-strawberries-how-many-acres-are-planted-in-strawberries","timestamp":"2024-11-03T18:50:17Z","content_type":"text/html","content_length":"25335","record_id":"<urn:uuid:a96ecdd6-d434-408b-b371-604af69526a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00522.warc.gz"}
CAT Quadratic Equations Formulas PDF, Download Now CAT Quadratic Equations Formulae PDF covers the fundamental topics of algebra. If you are someone who find such questions challenging in the CAT Quant section, it's important to practice more Quadratic Equations Practice Questions CAT. Learning important formulas and tricks for solving Quadratic Equations will be helpful. One can refer to all Quadratic Equation formula-based questions from the CAT Previous year papers. Practicing a good number of sums in the CAT Quadratic Equation will help aspirants tackle these questions with ease in the exam. Access Important CAT Quadratic Equation Questions and answers PDF along with the video solutions for free to practice quadratic equation questions for the CAT exam. In this blog, we will discuss the importance of CAT Quadratic Equations Formulae PDF. This PDF covers all quadratic equation formula required for the CAT exam. Importance of CAT Quadratic Formulas PDF CAT Quadratic Equations Formulae form the basis of Algebraic quantitative aptitude, and they frequently appear in various sections of the CAT exam. Proficiency in this area is essential for: • Solving complex problems: Quadratic equations are often used to model real-world scenarios, and understanding them allows you to tackle intricate problems effectively. • Building a strong foundation: A solid grasp of quadratic equations provides a strong foundation for other algebraic concepts, such as polynomials and rational expressions. • Improving problem-solving skills: Practicing quadratic equations can enhance your problem-solving abilities and help you develop logical reasoning skills. Download CAT Quadratic Equations Formulas Pdf Complete CAT Quant Formula List Preparing for the CAT exam means you need to understand different math topics, and we’re here to help. We’ve gathered all the important CAT quant formulas in one place to make your study easier. Below is a table with links to download PDFs for topics like Progressions, Interest, Geometry, and more. Quadratic Equations Concepts for CAT Weightage While the exact weightage of quadratic equations can vary from year to year, they typically constitute a significant portion of the quantitative aptitude section in CAT. Expect to encounter multiple questions on this topic, making it essential to dedicate sufficient time to practice. Year Slot 1 Slot 2 Slot 3 Subtopics for CAT Exam Preparation Quadratic Equations To effectively prepare for quadratic equations in CAT, it is crucial to cover the following subtopics: • Quadratic equation: The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants. • Roots of a quadratic equation: The values of x that satisfy the quadratic equation are called its roots. • Nature of roots: The nature of the roots (real, imaginary, equal, or distinct) can be determined using the discriminant (Δ = b² - 4ac). • Sum and product of roots: The sum and product of the roots of a quadratic equation can be calculated directly from the coefficients of the equation. • Formation of a CAT Quadratic Equations Formulae from roots: Given the roots of a quadratic equation, you can form the equation itself. • Quadratic inequalities: Solving quadratic inequalities involves finding the range of values for x that satisfy the inequality. Looking for hardcopy handbook? Order below. Delivery charges are on us :) Tips to Ace Quadratic Equations for CAT • Practice regularly: Consistent practice is key to mastering quadratic equations. Solve a variety of problems from different sources to improve your understanding and problem-solving skills. • Understand the concepts: Make sure you have a thorough understanding of the underlying concepts, such as the discriminant and the relationship between roots and coefficients. • Learn shortcuts and tricks: Discover efficient methods and shortcuts to solve quadratic equations quickly and accurately. • Analyze mistakes: Whenever you make mistakes, analyse them carefully to understand where you went wrong and avoid repeating the same errors. • Use a systematic approach: Develop a systematic approach to solving quadratic equations, breaking down problems into smaller steps and applying relevant formulas and techniques. Understanding quadratic equation formula is important to crack the CAT exam. It helps you develop the skills to solve quadratic equation problems effectively. Consider using Cracku's online platform to prepare for quadratic equations and other CAT important topics. Cracku offers daily targets, free mock tests, video tutorials, and expert guidance to help you achieve your target score. Sign up for Cracku today and start your CAT success journey!	{"url":"https://cracku.in/cat-quadratic-equation-formulas-pdf/","timestamp":"2024-11-04T21:31:48Z","content_type":"text/html","content_length":"128381","record_id":"<urn:uuid:11733c9a-1caa-4cd3-9805-b9c47257ed79>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00639.warc.gz"}
In a carnival ride, passengers stand with their backs agains… \| Wiki Cram In a carnival ride, passengers stand with their backs agains… In &acy; c&acy;rniv&acy;l ride, passengers stand with their backs against the wall &ocy;f a cylinder. The cylinder is set int&ocy; r&ocy;tation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25? In &acy; c&acy;rniv&acy;l ride, passengers stand with their backs against the wall &ocy;f a cylinder. The cylinder is set int&ocy; r&ocy;tation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25? A 3-ye&acy;r-&ocy;ld child wh&ocy; weighs 12 kg is di&acy;gn&ocy;sed with right &acy;cute otitis media in the primary care clinic. The provider prescribes the following medication: Amoxicillin 45 mg/ kg/dose oral suspension PO q12h for 7 days How many milligrams of amoxicillin will you give to the patient for one dose? Round to the nearest tenth. Skip back to main navigation	{"url":"https://wikicram.com/in-a-carnival-ride-passengers-stand-with-their-backs-against-the-wall-of-a-cylinder-the-cylinder-is-set-into-rotation-and-the-floor-is-lowered-away-from-the-passengers-but-they-remain-stuck-against-2/","timestamp":"2024-11-04T08:02:23Z","content_type":"text/html","content_length":"44778","record_id":"<urn:uuid:87bab4fc-c8e4-43c8-8f72-25bb284a46d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00277.warc.gz"}
(b). Location, Distance, and Direction on Maps Location on Maps Most maps allow us to specify the location of points on the Earth's surface using a coordinate system. For a two-dimensional map, this coordinate system can use simple geometric relationships between the perpendicular axes on a grid system to define spatial location. Figure 2b-1 illustrates how the location of a point can be defined on a coordinate system. Figure 2b-1: A grid coordinate system defines the location of points from the distance traveled along two perpendicular axes from some stated origin. In the example above, the two axes are labeled X and Y. The origin is located in the lower left hand corner. Unit distance traveled along each axis from the origin is shown. In this coordinate system, the value associated with the X-axis is given first, following by the value assigned from the Y-axis. The location represented by the star has the coordinates 7 (X-axis), 4 (Y-axis). Two types of coordinate systems are currently in general use in geography: the geographical coordinate system and the rectangular (also called Cartesian) coordinate system. Geographical Coordinate System The geographical coordinate system measures location from only two values, despite the fact that the locations are described for a three-dimensional surface. The two values used to define location are both measured relative to the polar axis of the Earth. The two measures used in the geographic coordinate system are called latitude and longitude. Figure 2b-2: Lines of latitude or parallels are drawn parallel to the equator (shown in red) as circles that span the Earth's surface. These parallels are measure in degrees (°). There are 90 angular degrees of latitude from the equator to each of the poles. The equator has an assigned value of 0°. Measurements of latitude are also defined as being either north or south of equator to distinguish the hemisphere of their location. Lines of longitude or meridians are circular arcs that meet at the poles. There are 180° of longitude either side of a starting meridian which is known the Prime Meridian. The Prime Meridian has a designated value of 0°. Measurements of longitude are also defined as being either west or east of the Prime Meridian. Latitude measures the north-south position of locations on the Earth's surface relative to a point found at the center of the Earth (Figure 2b-2). This central point is also located on the Earth's rotational or polar axis. The equator is the starting point for the measurement of latitude. The equator has a value of zero degrees. A line of latitude or parallel of 30° North has an angle that is 30° north of the plane represented by the equator (Figure 2b-3). The maximum value that latitude can attain is either 90° North or South. These lines of latitude run parallel to the rotational axis of the Earth. Longitude measures the west-east position of locations on the Earth's surface relative to a circular arc called the Prime Meridian (Figure 2b-2). The position of the Prime Meridian was determined by international agreement to be in-line with the location of the former astronomical observatory at Greenwich, England. Because the Earth's circumference is similar to circle, it was decided to measure longitude in degrees. The number of degrees found in a circle is 360. The Prime Meridian has a value of zero degrees. A line of longitude or meridian of 45° West has an angle that is 45° west of the plane represented by the Prime Meridian (Figure 2b-3). The maximum value that a meridian of longitude can have is 180° which is the distance halfway around a circle. This meridian is called the International Date Line. Designations of west and east are used to distinguish where a location is found relative to the Prime Meridian. For example, all of the locations in North America have a longitude that is designated west. Universal Transverse Mercator System (UTM) Another commonly used method to describe location on the Earth is the Universal Transverse Mercator (UTM) grid system. This rectangular coordinate system is metric, incorporating the meter as its basic unit of measurement. UTM also uses the Transverse Mercator projection system to model the Earth's spherical surface onto a two-dimensional plane. The UTM system divides the world's surface into 60 - six degree longitude wide zones that run north-south (Figure 2b-5). These zones start at the International Date Line and are successively numbered in an eastward direction (Figure 2b-5). Each zone stretches from 84° North to 80° South (Figure 2b-4). In the center of each of these zones is a central meridian. Location is measured in these zones from a false origin which is determined relative to the intersection of the equator and the central meridian for each zone. For locations in the Northern Hemisphere, the false origin is 500,000 meters west of the central meridian on the equator. Coordinate measurements of location in the Northern Hemisphere using the UTM system are made relative to this point in meters in eastings (longitudinal distance) and northings (latitudinal distance). The point defined by the intersection of 50° North and 9° West would have a UTM coordinate of Zone 29, 500000 meters east (E), 5538630 meters north (N) (see Figures 2b-4 and 2b-5). In the Southern Hemisphere, the origin is 10,000,000 meters south and 500,000 meters west of the equator and central meridian, respectively. The location found at 50° South and 9° West would have a UTM coordinate of Zone 29, 500000 meters E, 4461369 meters N (remember that northing in the Southern Hemisphere is measured from 10,000,000 meters south of the equator - see Figures 2b-4 and 2b-5). Figure 2b-4: The following illustration describes the characteristics of the UTM zone "29" found between 12 to 6° West longitude. Note that the zone has been split into two halves. The half on the left represents the area found in the Northern Hemisphere. The Southern Hemisphere is located on the right. The blue line represents the central meridian for this zone. Locations measurements for this zone are calculated relative to a false origin. In the Northern Hemisphere, this origin is located 500,000 meters west of the equator. The Southern Hemisphere UTM measurements are determined relative to a origin located at 10,000,000 meters south and 500,000 meters west of the equator and central meridian, respectively. The UTM system has been modified to make measurements less confusing. In this modification, the six degree wide zones are divided into smaller pieces or quadrilaterals that are eight degrees of latitude tall. Each of these rows is labeled, starting at 80° South, with the letters C to X consecutively with I and O being omitted (Figure 2b-5). The last row X differs from the other rows and extends from 72 to 84° North latitude (twelve degrees tall). Each of the quadrilaterals or grid zones are identified by their number/letter designation. In total, 1200 quadrilaterals are defined in the UTM system. The quadrilateral system allows us to further define location using the UTM system. For the location 50° North and 9° West, the UTM coordinate can now be expressed as Grid Zone 29U, 500000 meters E, 5538630 meters N. Figure 2b-5: The UTM system also uses a grid system to break the Earth up into 1200 quadrilaterals. To keep the illustration manageable, most of these zones have been excluded. Designation of each quadrilaterals is accomplished with a number-letter system. Along the horizontal bottom, the six degree longitude wide zones are numbered, starting at 180° West longitude, from 1 to 60. The twenty vertical rows are assigned letters C to X with I and O excluded. The letter, C, begins at 80° South latitude. Note that the rows are 8 degrees of latitude wide, except for the last row X which is 12 degrees wide. According to the reference system, the bright green quadrilateral has the grid reference 29V (note that in this system west-east coordinate is given first, followed by the south-north coordinate). This grid zone is found between 56 and 64° North latitude and 6 and 12° West longitude. Each UTM quadrilateral is further subdivided into a number of 100,000 by 100,000 meter zones. These subdivisions are coded by a system of letter combinations where the same two-letter combination is not repeated within 18 degrees of latitude and longitude. Within each of the 100,000 meter squares one can specify location to one-meter accuracy using a 5 digit eastings and northings reference The UTM grid system is displayed on all United States Geological Survey (USGS) and National Topographic Series (NTS) of Canada maps. On USGS 7.5-minute quadrangle maps (1:24,000 scale), 15-minute quadrangle maps (1:50,000, 1:62,500, and standard-edition 1:63,360 scales), and Canadian 1:50,000 maps the UTM grid lines are drawn at intervals of 1,000 meters, and are shown either with blue ticks at the edge of the map or by full blue grid lines. On USGS maps at 1:100,000 and 1:250,000 scale and Canadian 1:250,000 scale maps a full UTM grid is shown at intervals of 10,000 meters. Figure 2b-6 describes how the UTM grid system can be used to determine location on a 1:50,000 National Topographic Series of Canada map. Figure 2b-6: The top left hand corner the "Tofino" 1:50,000 National Topographic Series of Canada map is shown above. The blue lines and associated numbers on the map margin are used to determine location by way of the UTM grid system. Abbreviated UTM 1,000-meter values or principle digits are shown by numbers on the map margin that vary from 0 to 100 (100 is actually given the value 00). In each of the corners of the map, two of the principle digits are expressed in their full UTM coordinate form. On the image we can see 283000 m E. and 5458000 m N. The red dot is found in the center of the grid defined by principle numbers 85 to 86 easting and 57 to 58 northing. A more complete UTM grid reference for this location would be 285500 m E. and 5457500 m N. Information found on the map margin also tells us (not shown) that the area displayed is in Grid Zone 10U and the 100,000 m squares BK and CK are located on this map. Distance on Maps In section 2a, w e have learned that depicting the Earth's three-dimensional surface on a two-dimensional map creates a number of distortions that involve distance, area, and direction. It is possible to create maps that are somewhat equidistance. However, even these types of maps have some form of distance distortion. Equidistance maps can only control distortion along either lines of latitude or lines of longitude. Distance is often correct on equidistance maps only in the direction of latitude. On a map that has a large scale, 1:125,000 or larger, distance distortion is usually insignificant. An example of a large-scale map is a standard topographic map. On these maps measuring straight line distance is simple. Distance is first measured on the map using a ruler. This measurement is then converted into a real world distance using the map's scale. For example, if we measured a distance of 10 centimeters on a map that had a scale of 1:10,000, we would multiply 10 (distance) by 10,000 (scale). Thus, the actual distance in the real world would be 100,000 centimeters. Measuring distance along map features that are not straight is a little more difficult. One technique that can be employed for this task is to use a number of straight-line segments. The accuracy of this method is dependent on the number of straight-line segments used (Figure 2b-7). Another method for measuring curvilinear map distances is to use a mechanical device called an opisometer. This device uses a small rotating wheel that records the distance traveled. The recorded distance is measured by this device either in centimeters or inches. Figure 2b-7: Measurement of distance on a map feature using straight-line segments. Direction on Maps Like distance, direction is difficult to measure on maps because of the distortion produced by projection systems. However, this distortion is quite small on maps with scales larger than 1:125,000. Direction is usually measured relative to the location of North or South Pole. Directions determined from these locations are said to be relative to True North or True South. The magnetic poles can also be used to measure direction. However, these points on the Earth are located in spatially different spots from the geographic North and South Pole. The North Magnetic Pole is located at 78.3° North, 104.0° West near Ellef Ringnes Island, Canada. In the Southern Hemisphere, the South Magnetic Pole is located in Commonwealth Day, Antarctica and has a geographical location of 65° South, 139° East. The magnetic poles are also not fixed overtime and shift their spatial position overtime. Topographic maps normally have a declination diagram drawn on them (Figure 2b-8). On Northern Hemisphere maps, declination diagrams describe the angular difference between Magnetic North and True North. On the map, the angle of True North is parallel to the depicted lines of longitude. Declination diagrams also show the direction of Grid North. Grid North is an angle that is parallel to the easting lines found on the Universal Transverse Mercator (UTM) grid system (Figure 2b-8). Figure 2b-8: This declination diagram describes the angular difference between Grid, True, and Magnetic North. This illustration also shows how angles are measured relative grid, true, and magnetic In the field, the direction of features is often determined by a magnetic compass which measures angles relative to Magnetic North. Using the declination diagram found on a map, individuals can convert their field measures of magnetic direction into directions that are relative to either Grid or True North. Compass directions can be described by using either the azimuth system or the bearing system. The azimuth system calculates direction in degrees of a full circle. A full circle has 360 degrees (Figure 2b-9). In the azimuth system, north has a direction of either the 0 or 360°. East and west have an azimuth of 90° and 270°, respectively. Due south has an azimuth of 180°. Figure 2b-9: Azimuth system for measuring direction is based on the 360 degrees found in a full circle. The illustration shows the angles associated with the major cardinal points of the compass. Note that angles are determined clockwise from north. The bearing system divides direction into four quadrants of 90 degrees. In this system, north and south are the dominant directions. Measurements are determined in degrees from one of these directions. The measurement of two angles based on this system are described in Figure 2b-10. Figure 2b-10: The bearing system uses four quadrants of 90 degrees to measure direction. The illustration shows two direction measurements. These measurements are made relative to either north or south. North and south are given the measurement 0 degrees. East and west have a value of 90 degrees. The first measurement (green) is found in the north - east quadrant. As a result, its measurement is north 75 degrees to the east or N75°E. The first measurement (orange) is found in the south - west quadrant. Its measurement is south 15 degrees to the west or S15°W. Global Positioning Systems Determination of location in field conditions was once a difficult task. In most cases, it required the use of a topographic map and landscape features to estimate location. However, technology has now made this task very simple. Global Positioning Systems (GPS) can calculate one's location to an accuracy of about 30-meters (Figure 2b-11). These systems consist of two parts: a GPS receiver and a network of many satellites. Radio transmissions from the satellites are broadcasted continually. The GPS receiver picks up these broadcasts and through triangulation calculates the altitude and spatial position of the receiving unit. A minimum of three satellite is required for triangulation. Figure 2b-11: Handheld Global Positioning Systems (GPS). GPS receivers can determine latitude, longitude, and elevation anywhere on or above the Earth's surface from signals transmitted by a number of satellites. These units can also be used to determine direction, distance traveled, and determine routes of travel in field situations.	{"url":"http://www.physicalgeography.net/fundamentals/2b.html","timestamp":"2024-11-03T01:24:06Z","content_type":"text/html","content_length":"42940","record_id":"<urn:uuid:6651ff0d-f10f-473e-8454-3095ed0c9885>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00043.warc.gz"}
Megaton/Square Hectometer to Nanojoule/Acre Megaton/Square Hectometer [Mtn/hm2] Output 1 megaton/square hectometer in joule/square meter is equal to 418400000000 1 megaton/square hectometer in joule/square kilometer is equal to 418400000000000000 1 megaton/square hectometer in joule/square hectometer is equal to 4184000000000000 1 megaton/square hectometer in joule/square dekameter is equal to 41840000000000 1 megaton/square hectometer in joule/square decimeter is equal to 4184000000 1 megaton/square hectometer in joule/square centimeter is equal to 41840000 1 megaton/square hectometer in joule/square millimeter is equal to 418400 1 megaton/square hectometer in joule/square micrometer is equal to 0.4184 1 megaton/square hectometer in joule/square nanometer is equal to 4.184e-7 1 megaton/square hectometer in joule/hectare is equal to 4184000000000000 1 megaton/square hectometer in joule/square inch is equal to 269934944 1 megaton/square hectometer in joule/square feet is equal to 38870631936 1 megaton/square hectometer in joule/square yard is equal to 349835687424 1 megaton/square hectometer in joule/square mile is equal to 1083651025364600000 1 megaton/square hectometer in joule/acre is equal to 1693206224000000 1 megaton/square hectometer in megajoule/square meter is equal to 418400 1 megaton/square hectometer in megajoule/square kilometer is equal to 418400000000 1 megaton/square hectometer in megajoule/square hectometer is equal to 4184000000 1 megaton/square hectometer in megajoule/square dekameter is equal to 41840000 1 megaton/square hectometer in megajoule/square decimeter is equal to 4184 1 megaton/square hectometer in megajoule/square centimeter is equal to 41.84 1 megaton/square hectometer in megajoule/square millimeter is equal to 0.4184 1 megaton/square hectometer in megajoule/square micrometer is equal to 4.184e-7 1 megaton/square hectometer in megajoule/square nanometer is equal to 4.184e-13 1 megaton/square hectometer in megajoule/hectare is equal to 4184000000 1 megaton/square hectometer in megajoule/square inch is equal to 269.93 1 megaton/square hectometer in megajoule/square feet is equal to 38870.63 1 megaton/square hectometer in megajoule/square yard is equal to 349835.69 1 megaton/square hectometer in megajoule/square mile is equal to 1083651025364.6 1 megaton/square hectometer in megajoule/acre is equal to 1693206224 1 megaton/square hectometer in kilojoule/square meter is equal to 418400000 1 megaton/square hectometer in kilojoule/square kilometer is equal to 418400000000000 1 megaton/square hectometer in kilojoule/square hectometer is equal to 4184000000000 1 megaton/square hectometer in kilojoule/square dekameter is equal to 41840000000 1 megaton/square hectometer in kilojoule/square decimeter is equal to 4184000 1 megaton/square hectometer in kilojoule/square centimeter is equal to 41840 1 megaton/square hectometer in kilojoule/square millimeter is equal to 418.4 1 megaton/square hectometer in kilojoule/square micrometer is equal to 0.0004184 1 megaton/square hectometer in kilojoule/square nanometer is equal to 4.184e-10 1 megaton/square hectometer in kilojoule/hectare is equal to 4184000000000 1 megaton/square hectometer in kilojoule/square inch is equal to 269934.94 1 megaton/square hectometer in kilojoule/square feet is equal to 38870631.94 1 megaton/square hectometer in kilojoule/square yard is equal to 349835687.42 1 megaton/square hectometer in kilojoule/square mile is equal to 1083651025364600 1 megaton/square hectometer in kilojoule/acre is equal to 1693206224000 1 megaton/square hectometer in millijoule/square meter is equal to 418400000000000 1 megaton/square hectometer in millijoule/square kilometer is equal to 418400000000000000000 1 megaton/square hectometer in millijoule/square hectometer is equal to 4184000000000000000 1 megaton/square hectometer in millijoule/square dekameter is equal to 41840000000000000 1 megaton/square hectometer in millijoule/square decimeter is equal to 4184000000000 1 megaton/square hectometer in millijoule/square centimeter is equal to 41840000000 1 megaton/square hectometer in millijoule/square millimeter is equal to 418400000 1 megaton/square hectometer in millijoule/square micrometer is equal to 418.4 1 megaton/square hectometer in millijoule/square nanometer is equal to 0.0004184 1 megaton/square hectometer in millijoule/hectare is equal to 4184000000000000000 1 megaton/square hectometer in millijoule/square inch is equal to 269934944000 1 megaton/square hectometer in millijoule/square feet is equal to 38870631936000 1 megaton/square hectometer in millijoule/square yard is equal to 349835687424000 1 megaton/square hectometer in millijoule/square mile is equal to 1.0836510253646e+21 1 megaton/square hectometer in millijoule/acre is equal to 1693206224000000000 1 megaton/square hectometer in microjoule/square meter is equal to 418400000000000000 1 megaton/square hectometer in microjoule/square kilometer is equal to 4.184e+23 1 megaton/square hectometer in microjoule/square hectometer is equal to 4.184e+21 1 megaton/square hectometer in microjoule/square dekameter is equal to 41840000000000000000 1 megaton/square hectometer in microjoule/square decimeter is equal to 4184000000000000 1 megaton/square hectometer in microjoule/square centimeter is equal to 41840000000000 1 megaton/square hectometer in microjoule/square millimeter is equal to 418400000000 1 megaton/square hectometer in microjoule/square micrometer is equal to 418400 1 megaton/square hectometer in microjoule/square nanometer is equal to 0.4184 1 megaton/square hectometer in microjoule/hectare is equal to 4.184e+21 1 megaton/square hectometer in microjoule/square inch is equal to 269934944000000 1 megaton/square hectometer in microjoule/square feet is equal to 38870631936000000 1 megaton/square hectometer in microjoule/square yard is equal to 349835687424000000 1 megaton/square hectometer in microjoule/square mile is equal to 1.0836510253646e+24 1 megaton/square hectometer in microjoule/acre is equal to 1.693206224e+21 1 megaton/square hectometer in nanojoule/square meter is equal to 418400000000000000000 1 megaton/square hectometer in nanojoule/square kilometer is equal to 4.184e+26 1 megaton/square hectometer in nanojoule/square hectometer is equal to 4.184e+24 1 megaton/square hectometer in nanojoule/square dekameter is equal to 4.184e+22 1 megaton/square hectometer in nanojoule/square decimeter is equal to 4184000000000000000 1 megaton/square hectometer in nanojoule/square centimeter is equal to 41840000000000000 1 megaton/square hectometer in nanojoule/square millimeter is equal to 418400000000000 1 megaton/square hectometer in nanojoule/square micrometer is equal to 418400000 1 megaton/square hectometer in nanojoule/square nanometer is equal to 418.4 1 megaton/square hectometer in nanojoule/hectare is equal to 4.184e+24 1 megaton/square hectometer in nanojoule/square inch is equal to 269934944000000000 1 megaton/square hectometer in nanojoule/square feet is equal to 38870631936000000000 1 megaton/square hectometer in nanojoule/square yard is equal to 349835687424000000000 1 megaton/square hectometer in nanojoule/square mile is equal to 1.0836510253646e+27 1 megaton/square hectometer in nanojoule/acre is equal to 1.693206224e+24 1 megaton/square hectometer in attojoule/square meter is equal to 4.184e+29 1 megaton/square hectometer in attojoule/square kilometer is equal to 4.184e+35 1 megaton/square hectometer in attojoule/square hectometer is equal to 4.184e+33 1 megaton/square hectometer in attojoule/square dekameter is equal to 4.184e+31 1 megaton/square hectometer in attojoule/square decimeter is equal to 4.184e+27 1 megaton/square hectometer in attojoule/square centimeter is equal to 4.184e+25 1 megaton/square hectometer in attojoule/square millimeter is equal to 4.184e+23 1 megaton/square hectometer in attojoule/square micrometer is equal to 418400000000000000 1 megaton/square hectometer in attojoule/square nanometer is equal to 418400000000 1 megaton/square hectometer in attojoule/hectare is equal to 4.184e+33 1 megaton/square hectometer in attojoule/square inch is equal to 2.69934944e+26 1 megaton/square hectometer in attojoule/square feet is equal to 3.8870631936e+28 1 megaton/square hectometer in attojoule/square yard is equal to 3.49835687424e+29 1 megaton/square hectometer in attojoule/square mile is equal to 1.0836510253646e+36 1 megaton/square hectometer in attojoule/acre is equal to 1.693206224e+33 1 megaton/square hectometer in megaelectronvolt/square meter is equal to 2.6114475092201e+24 1 megaton/square hectometer in megaelectronvolt/square kilometer is equal to 2.6114475092201e+30 1 megaton/square hectometer in megaelectronvolt/square hectometer is equal to 2.6114475092201e+28 1 megaton/square hectometer in megaelectronvolt/square dekameter is equal to 2.6114475092201e+26 1 megaton/square hectometer in megaelectronvolt/square decimeter is equal to 2.6114475092201e+22 1 megaton/square hectometer in megaelectronvolt/square centimeter is equal to 261144750922010000000 1 megaton/square hectometer in megaelectronvolt/square millimeter is equal to 2611447509220100000 1 megaton/square hectometer in megaelectronvolt/square micrometer is equal to 2611447509220.1 1 megaton/square hectometer in megaelectronvolt/square nanometer is equal to 2611447.51 1 megaton/square hectometer in megaelectronvolt/hectare is equal to 2.6114475092201e+28 1 megaton/square hectometer in megaelectronvolt/square inch is equal to 1.6848014750484e+21 1 megaton/square hectometer in megaelectronvolt/square feet is equal to 2.4261141240697e+23 1 megaton/square hectometer in megaelectronvolt/square yard is equal to 2.1835027116627e+24 1 megaton/square hectometer in megaelectronvolt/square mile is equal to 6.7636179996465e+30 1 megaton/square hectometer in megaelectronvolt/acre is equal to 1.0568162467162e+28 1 megaton/square hectometer in kiloelectronvolt/square meter is equal to 2.6114475092201e+27 1 megaton/square hectometer in kiloelectronvolt/square kilometer is equal to 2.6114475092201e+33 1 megaton/square hectometer in kiloelectronvolt/square hectometer is equal to 2.6114475092201e+31 1 megaton/square hectometer in kiloelectronvolt/square dekameter is equal to 2.6114475092201e+29 1 megaton/square hectometer in kiloelectronvolt/square decimeter is equal to 2.6114475092201e+25 1 megaton/square hectometer in kiloelectronvolt/square centimeter is equal to 2.6114475092201e+23 1 megaton/square hectometer in kiloelectronvolt/square millimeter is equal to 2.6114475092201e+21 1 megaton/square hectometer in kiloelectronvolt/square micrometer is equal to 2611447509220100 1 megaton/square hectometer in kiloelectronvolt/square nanometer is equal to 2611447509.22 1 megaton/square hectometer in kiloelectronvolt/hectare is equal to 2.6114475092201e+31 1 megaton/square hectometer in kiloelectronvolt/square inch is equal to 1.6848014750484e+24 1 megaton/square hectometer in kiloelectronvolt/square feet is equal to 2.4261141240697e+26 1 megaton/square hectometer in kiloelectronvolt/square yard is equal to 2.1835027116627e+27 1 megaton/square hectometer in kiloelectronvolt/square mile is equal to 6.7636179996465e+33 1 megaton/square hectometer in kiloelectronvolt/acre is equal to 1.0568162467162e+31 1 megaton/square hectometer in electronvolt/square meter is equal to 2.6114475092201e+30 1 megaton/square hectometer in electronvolt/square kilometer is equal to 2.6114475092201e+36 1 megaton/square hectometer in electronvolt/square hectometer is equal to 2.6114475092201e+34 1 megaton/square hectometer in electronvolt/square dekameter is equal to 2.6114475092201e+32 1 megaton/square hectometer in electronvolt/square decimeter is equal to 2.6114475092201e+28 1 megaton/square hectometer in electronvolt/square centimeter is equal to 2.6114475092201e+26 1 megaton/square hectometer in electronvolt/square millimeter is equal to 2.6114475092201e+24 1 megaton/square hectometer in electronvolt/square micrometer is equal to 2611447509220100000 1 megaton/square hectometer in electronvolt/square nanometer is equal to 2611447509220.1 1 megaton/square hectometer in electronvolt/hectare is equal to 2.6114475092201e+34 1 megaton/square hectometer in electronvolt/square inch is equal to 1.6848014750484e+27 1 megaton/square hectometer in electronvolt/square feet is equal to 2.4261141240697e+29 1 megaton/square hectometer in electronvolt/square yard is equal to 2.1835027116627e+30 1 megaton/square hectometer in electronvolt/square mile is equal to 6.7636179996465e+36 1 megaton/square hectometer in electronvolt/acre is equal to 1.0568162467162e+34 1 megaton/square hectometer in erg/square meter is equal to 4184000000000000000 1 megaton/square hectometer in erg/square kilometer is equal to 4.184e+24 1 megaton/square hectometer in erg/square hectometer is equal to 4.184e+22 1 megaton/square hectometer in erg/square dekameter is equal to 418400000000000000000 1 megaton/square hectometer in erg/square decimeter is equal to 41840000000000000 1 megaton/square hectometer in erg/square centimeter is equal to 418400000000000 1 megaton/square hectometer in erg/square millimeter is equal to 4184000000000 1 megaton/square hectometer in erg/square micrometer is equal to 4184000 1 megaton/square hectometer in erg/square nanometer is equal to 4.18 1 megaton/square hectometer in erg/hectare is equal to 4.184e+22 1 megaton/square hectometer in erg/square inch is equal to 2699349440000000 1 megaton/square hectometer in erg/square feet is equal to 388706319360000000 1 megaton/square hectometer in erg/square yard is equal to 3498356874240000000 1 megaton/square hectometer in erg/square mile is equal to 1.0836510253646e+25 1 megaton/square hectometer in erg/acre is equal to 1.693206224e+22 1 megaton/square hectometer in kilowatt second/square meter is equal to 418400000 1 megaton/square hectometer in kilowatt second/square kilometer is equal to 418400000000000 1 megaton/square hectometer in kilowatt second/square hectometer is equal to 4184000000000 1 megaton/square hectometer in kilowatt second/square dekameter is equal to 41840000000 1 megaton/square hectometer in kilowatt second/square decimeter is equal to 4184000 1 megaton/square hectometer in kilowatt second/square centimeter is equal to 41840 1 megaton/square hectometer in kilowatt second/square millimeter is equal to 418.4 1 megaton/square hectometer in kilowatt second/square micrometer is equal to 0.0004184 1 megaton/square hectometer in kilowatt second/square nanometer is equal to 4.184e-10 1 megaton/square hectometer in kilowatt second/hectare is equal to 4184000000000 1 megaton/square hectometer in kilowatt second/square inch is equal to 269934.94 1 megaton/square hectometer in kilowatt second/square feet is equal to 38870631.94 1 megaton/square hectometer in kilowatt second/square yard is equal to 349835687.42 1 megaton/square hectometer in kilowatt second/square mile is equal to 1083651025364600 1 megaton/square hectometer in kilowatt second/acre is equal to 1693206224000 1 megaton/square hectometer in horsepower hour/square meter is equal to 155856.54 1 megaton/square hectometer in horsepower hour/square kilometer is equal to 155856540461.61 1 megaton/square hectometer in horsepower hour/square hectometer is equal to 1558565404.62 1 megaton/square hectometer in horsepower hour/square dekameter is equal to 15585654.05 1 megaton/square hectometer in horsepower hour/square decimeter is equal to 1558.57 1 megaton/square hectometer in horsepower hour/square centimeter is equal to 15.59 1 megaton/square hectometer in horsepower hour/square millimeter is equal to 0.15585654046161 1 megaton/square hectometer in horsepower hour/square micrometer is equal to 1.5585654046161e-7 1 megaton/square hectometer in horsepower hour/square nanometer is equal to 1.5585654046161e-13 1 megaton/square hectometer in horsepower hour/square inch is equal to 100.55 1 megaton/square hectometer in horsepower hour/square feet is equal to 14479.55 1 megaton/square hectometer in horsepower hour/square yard is equal to 130315.92 1 megaton/square hectometer in horsepower hour/square mile is equal to 403666586713.67 1 megaton/square hectometer in horsepower hour/acre is equal to 630729599.33 1 megaton/square hectometer in horsepower hour/hectare is equal to 1558565404.62 1 megaton/square hectometer in watt hour/square meter is equal to 116222222.22 1 megaton/square hectometer in watt hour/square kilometer is equal to 116222222222220 1 megaton/square hectometer in watt hour/square hectometer is equal to 1162222222222.2 1 megaton/square hectometer in watt hour/square dekameter is equal to 11622222222.22 1 megaton/square hectometer in watt hour/square decimeter is equal to 1162222.22 1 megaton/square hectometer in watt hour/square centimeter is equal to 11622.22 1 megaton/square hectometer in watt hour/square millimeter is equal to 116.22 1 megaton/square hectometer in watt hour/square micrometer is equal to 0.00011622222222222 1 megaton/square hectometer in watt hour/square nanometer is equal to 1.1622222222222e-10 1 megaton/square hectometer in watt hour/square inch is equal to 74981.93 1 megaton/square hectometer in watt hour/square feet is equal to 10797397.76 1 megaton/square hectometer in watt hour/square yard is equal to 97176579.84 1 megaton/square hectometer in watt hour/square mile is equal to 301014173712380 1 megaton/square hectometer in watt hour/acre is equal to 470335062222.22 1 megaton/square hectometer in watt hour/hectare is equal to 1162222222222.2 1 megaton/square hectometer in watt second/square meter is equal to 418400000000 1 megaton/square hectometer in watt second/square kilometer is equal to 418400000000000000 1 megaton/square hectometer in watt second/square hectometer is equal to 4184000000000000 1 megaton/square hectometer in watt second/square dekameter is equal to 41840000000000 1 megaton/square hectometer in watt second/square decimeter is equal to 4184000000 1 megaton/square hectometer in watt second/square centimeter is equal to 41840000 1 megaton/square hectometer in watt second/square millimeter is equal to 418400 1 megaton/square hectometer in watt second/square micrometer is equal to 0.4184 1 megaton/square hectometer in watt second/square nanometer is equal to 4.184e-7 1 megaton/square hectometer in watt second/hectare is equal to 4184000000000000 1 megaton/square hectometer in watt second/square inch is equal to 269934944 1 megaton/square hectometer in watt second/square feet is equal to 38870631936 1 megaton/square hectometer in watt second/square yard is equal to 349835687424 1 megaton/square hectometer in watt second/square mile is equal to 1083651025364600000 1 megaton/square hectometer in watt second/acre is equal to 1693206224000000 1 megaton/square hectometer in newton meter/square meter is equal to 418400000000 1 megaton/square hectometer in newton meter/square kilometer is equal to 418400000000000000 1 megaton/square hectometer in newton meter/square hectometer is equal to 4184000000000000 1 megaton/square hectometer in newton meter/square dekameter is equal to 41840000000000 1 megaton/square hectometer in newton meter/square decimeter is equal to 4184000000 1 megaton/square hectometer in newton meter/square centimeter is equal to 41840000 1 megaton/square hectometer in newton meter/square millimeter is equal to 418400 1 megaton/square hectometer in newton meter/square micrometer is equal to 0.4184 1 megaton/square hectometer in newton meter/square nanometer is equal to 4.184e-7 1 megaton/square hectometer in newton meter/hectare is equal to 4184000000000000 1 megaton/square hectometer in newton meter/square inch is equal to 269934944 1 megaton/square hectometer in newton meter/square feet is equal to 38870631936 1 megaton/square hectometer in newton meter/square yard is equal to 349835687424 1 megaton/square hectometer in newton meter/square mile is equal to 1083651025364600000 1 megaton/square hectometer in newton meter/acre is equal to 1693206224000000 1 megaton/square hectometer in horsepower hour/square meter [metric] is equal to 158018.25 1 megaton/square hectometer in horsepower hour/square kilometer [metric] is equal to 158018245744.43 1 megaton/square hectometer in horsepower hour/square hectometer [metric] is equal to 1580182457.44 1 megaton/square hectometer in horsepower hour/square dekameter [metric] is equal to 15801824.57 1 megaton/square hectometer in horsepower hour/square decimeter [metric] is equal to 1580.18 1 megaton/square hectometer in horsepower hour/square centimeter [metric] is equal to 15.8 1 megaton/square hectometer in horsepower hour/square millimeter [metric] is equal to 0.15801824574443 1 megaton/square hectometer in horsepower hour/square micrometer [metric] is equal to 1.5801824574443e-7 1 megaton/square hectometer in horsepower hour/square nanometer [metric] is equal to 1.5801824574443e-13 1 megaton/square hectometer in horsepower hour/hectare [metric] is equal to 1580182457.44 1 megaton/square hectometer in horsepower hour/square inch [metric] is equal to 101.95 1 megaton/square hectometer in horsepower hour/square feet [metric] is equal to 14680.38 1 megaton/square hectometer in horsepower hour/square yard [metric] is equal to 132123.38 1 megaton/square hectometer in horsepower hour/square mile [metric] is equal to 409265377694.23 1 megaton/square hectometer in horsepower hour/acre [metric] is equal to 639477717.97 1 megaton/square hectometer in gigawatt hour/square meter is equal to 0.11622222222222 1 megaton/square hectometer in gigawatt hour/square kilometer is equal to 116222.22 1 megaton/square hectometer in gigawatt hour/square hectometer is equal to 1162.22 1 megaton/square hectometer in gigawatt hour/square dekameter is equal to 11.62 1 megaton/square hectometer in gigawatt hour/square decimeter is equal to 0.0011622222222222 1 megaton/square hectometer in gigawatt hour/square centimeter is equal to 0.000011622222222222 1 megaton/square hectometer in gigawatt hour/square millimeter is equal to 1.1622222222222e-7 1 megaton/square hectometer in gigawatt hour/square micrometer is equal to 1.1622222222222e-13 1 megaton/square hectometer in gigawatt hour/square nanometer is equal to 1.1622222222222e-19 1 megaton/square hectometer in gigawatt hour/hectare is equal to 1162.22 1 megaton/square hectometer in gigawatt hour/square inch is equal to 0.000074981928888889 1 megaton/square hectometer in gigawatt hour/square feet is equal to 0.01079739776 1 megaton/square hectometer in gigawatt hour/square yard is equal to 0.09717657984 1 megaton/square hectometer in gigawatt hour/square mile is equal to 301014.17 1 megaton/square hectometer in gigawatt hour/acre is equal to 470.34 1 megaton/square hectometer in megawatt hour/square meter is equal to 116.22 1 megaton/square hectometer in megawatt hour/square kilometer is equal to 116222222.22 1 megaton/square hectometer in megawatt hour/square hectometer is equal to 1162222.22 1 megaton/square hectometer in megawatt hour/square dekameter is equal to 11622.22 1 megaton/square hectometer in megawatt hour/square decimeter is equal to 1.16 1 megaton/square hectometer in megawatt hour/square centimeter is equal to 0.011622222222222 1 megaton/square hectometer in megawatt hour/square millimeter is equal to 0.00011622222222222 1 megaton/square hectometer in megawatt hour/square micrometer is equal to 1.1622222222222e-10 1 megaton/square hectometer in megawatt hour/square nanometer is equal to 1.1622222222222e-16 1 megaton/square hectometer in megawatt hour/hectare is equal to 1162222.22 1 megaton/square hectometer in megawatt hour/square inch is equal to 0.074981928888889 1 megaton/square hectometer in megawatt hour/square feet is equal to 10.8 1 megaton/square hectometer in megawatt hour/square yard is equal to 97.18 1 megaton/square hectometer in megawatt hour/square mile is equal to 301014173.71 1 megaton/square hectometer in megawatt hour/acre is equal to 470335.06 1 megaton/square hectometer in kilowatt hour/square meter is equal to 116222.22 1 megaton/square hectometer in kilowatt hour/square kilometer is equal to 116222222222.22 1 megaton/square hectometer in kilowatt hour/square hectometer is equal to 1162222222.22 1 megaton/square hectometer in kilowatt hour/square dekameter is equal to 11622222.22 1 megaton/square hectometer in kilowatt hour/square decimeter is equal to 1162.22 1 megaton/square hectometer in kilowatt hour/square centimeter is equal to 11.62 1 megaton/square hectometer in kilowatt hour/square millimeter is equal to 0.11622222222222 1 megaton/square hectometer in kilowatt hour/square micrometer is equal to 1.1622222222222e-7 1 megaton/square hectometer in kilowatt hour/square nanometer is equal to 1.1622222222222e-13 1 megaton/square hectometer in kilowatt hour/hectare is equal to 1162222222.22 1 megaton/square hectometer in kilowatt hour/square inch is equal to 74.98 1 megaton/square hectometer in kilowatt hour/square feet is equal to 10797.4 1 megaton/square hectometer in kilowatt hour/square yard is equal to 97176.58 1 megaton/square hectometer in kilowatt hour/square mile is equal to 301014173712.38 1 megaton/square hectometer in kilowatt hour/acre is equal to 470335062.22 1 megaton/square hectometer in calorie/square meter is equal to 99933123148.94 1 megaton/square hectometer in calorie/square kilometer is equal to 99933123148944000 1 megaton/square hectometer in calorie/square hectometer is equal to 999331231489440 1 megaton/square hectometer in calorie/square dekameter is equal to 9993312314894.4 1 megaton/square hectometer in calorie/square decimeter is equal to 999331231.49 1 megaton/square hectometer in calorie/square centimeter is equal to 9993312.31 1 megaton/square hectometer in calorie/square millimeter is equal to 99933.12 1 megaton/square hectometer in calorie/square micrometer is equal to 0.099933123148944 1 megaton/square hectometer in calorie/square nanometer is equal to 9.9933123148944e-8 1 megaton/square hectometer in calorie/hectare is equal to 999331231489440 1 megaton/square hectometer in calorie/square inch is equal to 64472853.73 1 megaton/square hectometer in calorie/square feet is equal to 9284090937.23 1 megaton/square hectometer in calorie/square yard is equal to 83556818435.08 1 megaton/square hectometer in calorie/square mile is equal to 258825600784510000 1 megaton/square hectometer in calorie/acre is equal to 404415358746540 1 megaton/square hectometer in calorie/square meter [th] is equal to 100000000000 1 megaton/square hectometer in calorie/square kilometer [th] is equal to 100000000000000000 1 megaton/square hectometer in calorie/square hectometer [th] is equal to 1000000000000000 1 megaton/square hectometer in calorie/square dekameter [th] is equal to 10000000000000 1 megaton/square hectometer in calorie/square decimeter [th] is equal to 1000000000 1 megaton/square hectometer in calorie/square centimeter [th] is equal to 10000000 1 megaton/square hectometer in calorie/square millimeter [th] is equal to 100000 1 megaton/square hectometer in calorie/square micrometer [th] is equal to 0.1 1 megaton/square hectometer in calorie/square nanometer [th] is equal to 1e-7 1 megaton/square hectometer in calorie/hectare [th] is equal to 1000000000000000 1 megaton/square hectometer in calorie/square inch [th] is equal to 64516000 1 megaton/square hectometer in calorie/square feet [th] is equal to 9290304000 1 megaton/square hectometer in calorie/square yard [th] is equal to 83612736000 1 megaton/square hectometer in calorie/square mile [th] is equal to 258998811033600000 1 megaton/square hectometer in calorie/acre [th] is equal to 404686000000000 1 megaton/square hectometer in btu/square meter is equal to 396566683.14 1 megaton/square hectometer in btu/square kilometer is equal to 396566683139090 1 megaton/square hectometer in btu/square hectometer is equal to 3965666831390.9 1 megaton/square hectometer in btu/square dekameter is equal to 39656668313.91 1 megaton/square hectometer in btu/square decimeter is equal to 3965666.83 1 megaton/square hectometer in btu/square centimeter is equal to 39656.67 1 megaton/square hectometer in btu/square millimeter is equal to 396.57 1 megaton/square hectometer in btu/square micrometer is equal to 0.00039656668313909 1 megaton/square hectometer in btu/square nanometer is equal to 3.9656668313909e-10 1 megaton/square hectometer in btu/hectare is equal to 3965666831390.9 1 megaton/square hectometer in btu/square inch is equal to 255848.96 1 megaton/square hectometer in btu/square feet is equal to 36842250.43 1 megaton/square hectometer in btu/square yard is equal to 331580253.84 1 megaton/square hectometer in btu/square mile is equal to 1027102994285600 1 megaton/square hectometer in btu/acre is equal to 1604849847328.3 1 megaton/square hectometer in btu/square meter [th] is equal to 396832058.57 1 megaton/square hectometer in btu/square kilometer [th] is equal to 396832058567250 1 megaton/square hectometer in btu/square hectometer [th] is equal to 3968320585672.5 1 megaton/square hectometer in btu/square dekameter [th] is equal to 39683205856.72 1 megaton/square hectometer in btu/square decimeter [th] is equal to 3968320.59 1 megaton/square hectometer in btu/square centimeter [th] is equal to 39683.21 1 megaton/square hectometer in btu/square millimeter [th] is equal to 396.83 1 megaton/square hectometer in btu/square micrometer [th] is equal to 0.00039683205856725 1 megaton/square hectometer in btu/square nanometer [th] is equal to 3.9683205856725e-10 1 megaton/square hectometer in btu/hectare [th] is equal to 3968320585672.5 1 megaton/square hectometer in btu/square inch [th] is equal to 256020.17 1 megaton/square hectometer in btu/square feet [th] is equal to 36866904.61 1 megaton/square hectometer in btu/square yard [th] is equal to 331802141.49 1 megaton/square hectometer in btu/square mile [th] is equal to 1027790313489300 1 megaton/square hectometer in btu/acre [th] is equal to 1605923784533.5 1 megaton/square hectometer in mega btu/square meter is equal to 396566.68 1 megaton/square hectometer in mega btu/square kilometer is equal to 396566683139.09 1 megaton/square hectometer in mega btu/square hectometer is equal to 3965666831.39 1 megaton/square hectometer in mega btu/square dekameter is equal to 39656668.31 1 megaton/square hectometer in mega btu/square decimeter is equal to 3965.67 1 megaton/square hectometer in mega btu/square centimeter is equal to 39.66 1 megaton/square hectometer in mega btu/square millimeter is equal to 0.39656668313909 1 megaton/square hectometer in mega btu/square micrometer is equal to 3.9656668313909e-7 1 megaton/square hectometer in mega btu/square nanometer is equal to 3.9656668313909e-13 1 megaton/square hectometer in mega btu/hectare is equal to 3965666831.39 1 megaton/square hectometer in mega btu/square inch is equal to 255.85 1 megaton/square hectometer in mega btu/square feet is equal to 36842.25 1 megaton/square hectometer in mega btu/square yard is equal to 331580.25 1 megaton/square hectometer in mega btu/square mile is equal to 1027102994285.6 1 megaton/square hectometer in mega btu/acre is equal to 1604849847.33 1 megaton/square hectometer in refrigeration ton hour/square meter is equal to 33047.22 1 megaton/square hectometer in refrigeration ton hour/square kilometer is equal to 33047223594.92 1 megaton/square hectometer in refrigeration ton hour/square hectometer is equal to 330472235.95 1 megaton/square hectometer in refrigeration ton hour/square dekameter is equal to 3304722.36 1 megaton/square hectometer in refrigeration ton hour/square decimeter is equal to 330.47 1 megaton/square hectometer in refrigeration ton hour/square centimeter is equal to 3.3 1 megaton/square hectometer in refrigeration ton hour/square millimeter is equal to 0.033047223594924 1 megaton/square hectometer in refrigeration ton hour/square micrometer is equal to 3.3047223594924e-8 1 megaton/square hectometer in refrigeration ton hour/square nanometer is equal to 3.3047223594924e-14 1 megaton/square hectometer in refrigeration ton hour/square inch is equal to 21.32 1 megaton/square hectometer in refrigeration ton hour/square feet is equal to 3070.19 1 megaton/square hectometer in refrigeration ton hour/square yard is equal to 27631.69 1 megaton/square hectometer in refrigeration ton hour/square mile is equal to 85591916190.47 1 megaton/square hectometer in refrigeration ton hour/acre is equal to 133737487.28 1 megaton/square hectometer in refrigeration ton hour/hectare is equal to 330472235.95 1 megaton/square hectometer in gigaton/square meter is equal to 1e-7 1 megaton/square hectometer in gigaton/square kilometer is equal to 0.1 1 megaton/square hectometer in gigaton/square hectometer is equal to 0.001 1 megaton/square hectometer in gigaton/square dekameter is equal to 0.00001 1 megaton/square hectometer in gigaton/square decimeter is equal to 1e-9 1 megaton/square hectometer in gigaton/square centimeter is equal to 1e-11 1 megaton/square hectometer in gigaton/square millimeter is equal to 1e-13 1 megaton/square hectometer in gigaton/square micrometer is equal to 1e-19 1 megaton/square hectometer in gigaton/square nanometer is equal to 1e-25 1 megaton/square hectometer in gigaton/hectare is equal to 0.001 1 megaton/square hectometer in gigaton/square inch is equal to 6.4516e-11 1 megaton/square hectometer in gigaton/square feet is equal to 9.290304e-9 1 megaton/square hectometer in gigaton/square yard is equal to 8.3612736e-8 1 megaton/square hectometer in gigaton/square mile is equal to 0.2589988110336 1 megaton/square hectometer in gigaton/acre is equal to 0.000404686 1 megaton/square hectometer in megaton/square meter is equal to 0.0001 1 megaton/square hectometer in megaton/square kilometer is equal to 100 1 megaton/square hectometer in megaton/square dekameter is equal to 0.01 1 megaton/square hectometer in megaton/square decimeter is equal to 0.000001 1 megaton/square hectometer in megaton/square centimeter is equal to 1e-8 1 megaton/square hectometer in megaton/square millimeter is equal to 1e-10 1 megaton/square hectometer in megaton/square micrometer is equal to 1e-16 1 megaton/square hectometer in megaton/square nanometer is equal to 1e-22 1 megaton/square hectometer in megaton/hectare is equal to 1 1 megaton/square hectometer in megaton/square inch is equal to 6.4516e-8 1 megaton/square hectometer in megaton/square feet is equal to 0.000009290304 1 megaton/square hectometer in megaton/square yard is equal to 0.000083612736 1 megaton/square hectometer in megaton/square mile is equal to 259 1 megaton/square hectometer in megaton/acre is equal to 0.404686 1 megaton/square hectometer in langley is equal to 10000000	{"url":"https://hextobinary.com/unit/heatdensity/from/mtnphm2/to/njpacre","timestamp":"2024-11-07T22:34:25Z","content_type":"text/html","content_length":"223662","record_id":"<urn:uuid:c266d62b-d813-4166-a8a8-d6c3c736ab63>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00416.warc.gz"}
Help...Assasin FHR, FCR breakpoint? What are the FHR and FCR BP's for an Assasin? :scratch: I know two of the fcr- 70%, and 102%. :idea: -Mook :teeth: Check out this guide in the compendium - it has a list of both FHR and FCR breakpoints (and you're not right on one of them). Omg I was wrong, Im unworthy Well ok so I have to his 102% FCR and about 48-86 fhr on trapper, gg. Thx Sky Also do you know where to find a Max Block finder, I neeed to know how much dexterity reaches max block for a Large Shield Sanctuary for my sin. Yes sir, from Battle.net: Blocking determines your chance to defend against physical melee and ranged attacks. If you block an attack you will receive no damage. If you run your mouse over your Defense on the Character Screen you will see a percentage to block if you are carrying a Shield or Necromancer Shrunken Heads. If you do not have a shield or Shrunken Heads you will not see this listed. When a player blocks, that is, after a hit has already occurred, probability is computed that a player will block as follows: Total Blocking = (Blocking * (Dexterity - 15)) / (Character Level * 2) Blocking = A total of the Blocking on all of your items. To raise your % Chance to Block spend more points in Dexterity in addition to getting higher % Chance to Block via items. The block value itself is a combination of a value inherent to that particular player class, and any other block bonuses from items. This value is capped at 75%. If the roll of the dice denies a true block, or if the player can't block at all, then and only then, are skills such as Amazon Avoid, and Assassin Weapon Block checked for blocking. Note that if a player is moving, the total block percentage is reduced to 1/3rd of its original value. skygoneblue said: Yes sir, from Battle.net: Blocking determines your chance to defend against physical melee and ranged attacks. If you block an attack you will receive no damage. If you run your mouse over your Defense on the Character Screen you will see a percentage to block if you are carrying a Shield or Necromancer Shrunken Heads. If you do not have a shield or Shrunken Heads you will not see this listed. When a player blocks, that is, after a hit has already occurred, probability is computed that a player will block as follows: Total Blocking = (Blocking * (Dexterity - 15)) / (Character Level * 2) Blocking = A total of the Blocking on all of your items. To raise your % Chance to Block spend more points in Dexterity in addition to getting higher % Chance to Block via items. The block value itself is a combination of a value inherent to that particular player class, and any other block bonuses from items. This value is capped at 75%. If the roll of the dice denies a true block, or if the player can't block at all, then and only then, are skills such as Amazon Avoid, and Assassin Weapon Block checked for blocking. Note that if a player is moving, the total block percentage is reduced to 1/3rd of its original value. Uh does that mean your shield blocking, or the "chance to block"? Also I don't quite get that formula- runs away in dark corner cries Well, you said a 'Large Shield', right? Basically, the information that you need is Your character level The % Chance to Block on the shield The % Chance to Block that you want to achieve Ok, look again at the forumla: Total Blocking = (Blocking * (Dexterity - 15)) / (Character Level * 2) Blocking = A total of the Blocking on all of your items. I will describe the variables for you. Total Blocking: The block rate that you want to achieve (max being 75%) Blocking: The block rate of the shield Dexterity: Your current Dexterity level Character Level: Your current character level (max being 99) So, if you put the rune word SANCTUARY into a large shield, it would add 20% to the 'Blocking' variable in the equation because of the mod '20% Increase Chance of Blocking'. A Large Shield has a block rate of 37% for an Assassin. With the SANCTUARY rune word, that will be up to 57%. Assuming that you want max block of 75, that puts equation at 75 = (57 * (Dex-15)) / (Level2) Follow me so far? Now, I am not sure what your character level is right now, but I will say 80 just for demonstration's sake. If it is higher, you should be able to figure it out. 75 = (57(Dex-15)) / (802) Now, we need to solve for the Dex variable, get it? 75 = (57(Dex-15)) / 160 75160 = 57 (Dex - 15) 12,000/57 = Dex - 15 211 + 15 = Dex 226 = Dex So, at level 80 with a shield with 57% chance to block, you need 226 Dex to have max block. Make sense now? skygoneblue said: Well, you said a 'Large Shield', right? Basically, the information that you need is Your character level The % Chance to Block on the shield The % Chance to Block that you want to achieve Ok, look again at the forumla: Total Blocking = (Blocking * (Dexterity - 15)) / (Character Level * 2) Blocking = A total of the Blocking on all of your items. I will describe the variables for you. Total Blocking: The block rate that you want to achieve (max being 75%) Blocking: The block rate of the shield Dexterity: Your current Dexterity level Character Level: Your current character level (max being 99) So, if you put the rune word SANCTUARY into a large shield, it would add 20% to the 'Blocking' variable in the equation because of the mod '20% Increase Chance of Blocking'. A Large Shield has a block rate of 37% for an Assassin. With the SANCTUARY rune word, that will be up to 57%. Assuming that you want max block of 75, that puts equation at 75 = (57 * (Dex-15)) / (Level2) Follow me so far? Now, I am not sure what your character level is right now, but I will say 80 just for demonstration's sake. If it is higher, you should be able to figure it out. 75 = (57(Dex-15)) / (802) Now, we need to solve for the Dex variable, get it? 75 = (57(Dex-15)) / 160 75160 = 57 (Dex - 15) 12,000/57 = Dex - 15 211 + 15 = Dex 226 = Dex So, at level 80 with a shield with 57% chance to block, you need 226 Dex to have max block. Make sense now? Yes thx. alot... I could use two perfect ravens and Ber'd Whistan's guard but that will be in the 50's I think. I dont know, back to the drawing board. sanctuary in a large shield or any shield with that low of a block rate was a bad choice in the first place... try tower shield, or if your concerned with being slowed down with it, try a troll nest :thumbsup: edit: dont forget the +20 dex from the 2 ko's in sanctuary, im not sure if you calculated it in or not but i find alot of people forgetting about it. pryzmatik2 said: sanctuary in a large shield or any shield with that low of a block rate was a bad choice in the first place... try tower shield, or if your concerned with being slowed down with it, try a troll nest :thumbsup: edit: dont forget the +20 dex from the 2 ko's in sanctuary, im not sure if you calculated it in or not but i find alot of people forgetting about it. I have a Troll's Nest right now lol, and I was told Large Shield was the best blocking percentage? ARG! :rant:! :drink: = :buddies: W/e I'll stick to my Trolls Nest Sanctuary with 68 Resist.	{"url":"https://www.purediablo.com/forums/threads/help-assasin-fhr-fcr-breakpoint.55847/","timestamp":"2024-11-10T09:11:31Z","content_type":"text/html","content_length":"139220","record_id":"<urn:uuid:defa19aa-c3e2-486a-9565-1ab613034334>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00850.warc.gz"}
Relativistic Calogero-Moser systems and solutions Results pertaining to a class of completely integrable N-particle systems, which generalize the Calogero-Moser (CM) systems, are reviewed. The systems are relativistically invariant, and the CM systems result from taking a parameter to infinity, which may be regarded as the speed of light. They can be quantized in such a fashion that integrability is preserved. However, only the classical level is discussed. A matrix whose symmetric functions yield N independent commuting Hamiltonians is used to construct the action-angle map. The relation of the particle systems to soliton equations is analyzed. Solutions consisting of solitons, antisolitons, breathers, etc... may all be viewed as manifestations of underlying point particle dynamics associated with the systems. The relation leads to a natural concept of soliton space-time trajectory. Pub Date: January 1987 □ Invariance; □ Parameterization; □ Relativistic Particles; □ Relativistic Theory; □ Solitary Waves; □ Hamiltonian Functions; □ Partial Differential Equations; □ Particle Theory; □ Relativity; □ Wave Equations; □ Thermodynamics and Statistical Physics	{"url":"https://ui.adsabs.harvard.edu/abs/1987rcms.rept.....R/abstract","timestamp":"2024-11-10T15:13:32Z","content_type":"text/html","content_length":"35108","record_id":"<urn:uuid:24f2f191-be3d-4931-8454-75aca4c274b5>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00214.warc.gz"}
Point P in Mathematics: A Comprehensive Guide - Messiturf 10 Point P in Mathematics: A Comprehensive Guide Point P, often referred to as a fundamental concept in mathematics, serves as a crucial element in understanding geometrical shapes, graphs, and spatial relationships. Whether you’re studying basic geometry or advanced calculus, the point P appears as a cornerstone in various mathematical discussions. This blog post will delve deep into the concept of Point P, its applications, and its significance in different mathematical contexts. Defining Point P Point P is a specific location in space with no dimensions, meaning it has no length, width, or height. It’s typically represented as a dot on a graph or diagram and is used to indicate a particular position. In a two-dimensional coordinate system, Point P is denoted by (x, y), where x and y are the coordinates. In three dimensions, it’s represented as (x, y, z). The simplicity of Point P makes it a versatile tool in various branches of mathematics. Historical Perspective of Point P The concept of Point P has been pivotal since ancient times, from Euclidean geometry to modern-day mathematics. Ancient Greek mathematicians like Euclid defined Point P as “that which has no part.” Over the centuries, the understanding of Point P has evolved, influencing the development of geometry, algebra, and calculus. Point P’s role has been instrumental in shaping mathematical theories and Point P in Euclidean Geometry In Euclidean geometry, Point P is a fundamental element used to define shapes, lines, and angles. Euclid’s postulates begin with points, lines, and planes. For example, the distance between two points, including Point P, forms the basis for defining line segments. Point P’s precise location helps in constructing and understanding various geometric shapes and their properties. Point P in Cartesian Coordinates In Cartesian coordinates, Point P is defined by its coordinates (x, y) in a 2D plane and (x, y, z) in a 3D space. This system, developed by René Descartes, allows for the precise plotting of Point P on a graph. Cartesian coordinates are widely used in algebra, calculus, and engineering to solve problems involving distance, midpoints, and intersections, all relying on the accurate placement of Point P. Point P in Polar Coordinates Point P can also be represented in polar coordinates, where its position is defined by the distance from the origin (r) and the angle (θ) from the positive x-axis. This representation is particularly useful in fields like physics and engineering, where circular and rotational symmetries are common. Point P’s polar coordinates offer an alternative way to understand and solve problems involving curves and waves. Point P in Complex Numbers Complex numbers often use Point P to represent numbers in a plane, with the real part as the x-coordinate and the imaginary part as the y-coordinate. This visualization helps in understanding the magnitude and direction of complex numbers. Point P’s role in complex plane geometry is crucial for advanced mathematical concepts like transformations, roots of equations, and signal processing. Point P in Vector Spaces In vector spaces, Point P is represented as a position vector originating from the origin. This vector defines Point P’s location in space, making it essential for understanding vector operations such as addition, subtraction, and scalar multiplication. Point P’s vector representation is fundamental in physics, computer graphics, and engineering for modeling and simulating real-world Point P in Graph Theory Graph theory uses Point P to represent vertices in graphs. These vertices (or points) are connected by edges, forming various structures like trees, cycles, and networks. Point P’s identification as a vertex is critical for solving problems related to connectivity, flow, and optimization in computer science, logistics, and social network analysis. Point P in Calculus Calculus often uses Point P to denote specific points on functions and curves. The coordinates of Point P can help determine derivatives, integrals, and limits. For instance, finding the slope of a curve at Point P involves calculating the derivative at that point. Point P’s role in calculus is essential for understanding change and motion in mathematical and real-world contexts. Point P in Differential Equations Differential equations frequently use Point P to represent initial or boundary conditions. Solving these equations involves finding functions that pass through specific points, including Point P. This application is vital in fields like physics, engineering, and economics, where modeling dynamic systems and predicting future behavior depend on understanding the solutions that include Point P. Point P in Optimization Problems Optimization problems often aim to find the best value of a function at Point P, whether it be maximizing profits or minimizing costs. The precise identification of Point P helps in determining the optimal solutions. Techniques like gradient descent and Lagrange multipliers utilize Point P to find the best possible outcomes in various applications, from business to engineering. Point P in Real-World Applications Point P’s significance extends beyond theoretical mathematics into real-world applications. For example, GPS systems use Point P to pinpoint exact locations on Earth. In computer graphics, Point P helps in rendering images and animations. Robotics, architecture, and navigation systems all rely on the accurate positioning of Point P to function effectively. Point P, though simple in concept, is a powerful tool in mathematics and its applications. From defining basic geometric shapes to solving complex optimization problems, Point P’s role is indispensable. Understanding Point P helps in grasping fundamental mathematical principles and applying them to solve real-world problems efficiently. 1. What is Point P in mathematics? □ Point P is a specific location in space with no dimensions, represented by coordinates. 2. How is Point P used in geometry? □ Point P is used to define shapes, lines, and angles, forming the basis for geometric constructions. 3. What are Cartesian coordinates of Point P? □ Cartesian coordinates of Point P are (x, y) in a 2D plane and (x, y, z) in a 3D space. 4. How does Point P relate to complex numbers? □ Point P represents complex numbers in a plane, with the real part as the x-coordinate and the imaginary part as the y-coordinate. 5. What is the significance of Point P in optimization problems? □ Point P helps in determining optimal solutions by identifying the best value of a function.	{"url":"https://messiturf-10.com/point-p-in-mathematics-a-comprehensive-guide/","timestamp":"2024-11-03T15:28:30Z","content_type":"text/html","content_length":"65559","record_id":"<urn:uuid:793b8e40-498b-4e10-9108-156199988209>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00183.warc.gz"}
Year 10 Syllabus Year 10 SyllabusTest Curriculum updated: . Year 10 Syllabus Year Level Description The proficiency strands understanding, fluency, problem-solving and reasoning are an integral part of mathematics content across the three content strands: number and algebra, measurement and geometry, and statistics and probability. The proficiencies reinforce the significance of working mathematically within the content and describe how the content is explored or developed. They provide the language to build in the developmental aspects of the learning of mathematics. The achievement standards reflect the content and encompass the proficiencies. At this year level: • understanding includes applying the four operations to algebraic fractions, finding unknowns in formulas after substitution, making the connection between equations of relations and their graphs, comparing simple and compound interest in financial contexts and determining probabilities of two- and three-step experiments • fluency includes factorising and expanding algebraic expressions, using a range of strategies to solve equations and using calculations to investigate the shape of data sets • problem-solving includes calculating the surface area and volume of a diverse range of prisms to solve practical problems, finding unknown lengths and angles using applications of trigonometry, using algebraic and graphical techniques to find solutions to simultaneous equations and inequalities and investigating independence of events • reasoning includes formulating geometric proofs involving congruence and similarity, interpreting and evaluating media statements and interpreting and comparing data sets. Number and Algebra Money and financial mathematics Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies (ACMNA229) Patterns and algebra Factorise algebraic expressions by taking out a common algebraic factor (ACMNA230) Simplify algebraic products and quotients using index laws (ACMNA231) Apply the four operations to simple algebraic fractions with numerical denominators (ACMNA232) Expand binomial products and factorise monic quadratic expressions using a variety of strategies (ACMNA233) Substitute values into formulas to determine an unknown (ACMNA234) Linear and non-linear relationships Solve problems involving linear equations, including those derived from formulas (ACMNA235) Solve linear inequalities and graph their solutions on a number line (ACMNA236) Solve linear simultaneous equations, using algebraic and graphical techniques, including using digital technology (ACMNA237) Solve problems involving parallel and perpendicular lines (ACMNA238) Explore the connection between algebraic and graphical representations of relations such as simple quadratics, circles and exponentials using digital technology as appropriate (ACMNA239) Solve linear equations involving simple algebraic fractions (ACMNA240) Solve simple quadratic equations using a range of strategies (ACMNA241) Measurement and Geometry Using units of measurement Solve problems involving surface area and volume for a range of prisms, cylinders and composite solids (ACMMG242) Geometric reasoning Formulate proofs involving congruent triangles and angle properties (ACMMG243) Apply logical reasoning, including the use of congruence and similarity, to proofs and numerical exercises involving plane shapes (ACMMG244) Pythagoras and trigonometry Solve right-angled triangle problems including those involving direction and angles of elevation and depression (ACMMG245) Statistics and Probability Describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events. Investigate the concept of independence (ACMSP246) Use the language of ‘if ....then', ‘given’, ‘of’, ‘knowing that’ to investigate conditional statements and identify common mistakes in interpreting such language (ACMSP247) Data representation and interpretation Determine quartiles and interquartile range (ACMSP248) Construct and interpret box plots and use them to compare data sets (ACMSP249) Compare shapes of box plots to corresponding histograms and dot plots (ACMSP250) Use scatter plots to investigate and comment on relationships between two numerical variables (ACMSP251) Investigate and describe bivariate numerical data where the independent variable is time (ACMSP252) Evaluate statistical reports in the media and other places by linking claims to displays, statistics and representative data (ACMSP253) Year 10 Achievement Standard Number and Algebra At Standard, students recognise the connection between simple and compound interest. They solve problems involving linear equations and inequalities. Students make the connections between algebraic and graphical representations of relations. They expand binomial expressions and factorise monic quadratic expressions. Students find unknown values after substitution into formulas. They perform the four operations with simple algebraic fractions. Students solve simple quadratic equations and pairs of simultaneous equations. Measurement and Geometry Students solve surface area and volume problems relating to composite solids. They recognise the relationships between parallel and perpendicular lines. Students apply deductive reasoning to proofs and numerical exercises involving plane shapes. They use triangle and angle properties to prove congruence and similarity. Students use trigonometry to calculate unknown angles in right-angled Statistics and Probability Students compare data sets by referring to the shapes of the various data displays. They describe bivariate data where the independent variable is time. Students describe statistical relationships between two continuous variables. They evaluate statistical reports. Students list outcomes for multi-step chance experiments and assign probabilities for these experiments. They calculate quartiles and inter-quartile ranges. Year Level Description The proficiency strands understanding, fluency, problem-solving and reasoning are an integral part of mathematics content across the three content strands: number and algebra, measurement and geometry, and statistics and probability. The proficiencies reinforce the significance of working mathematically within the content and describe how the content is explored or developed. They provide the language to build in the developmental aspects of the learning of mathematics. The achievement standards reflect the content and encompass the proficiencies. At this year level: • understanding includes applying the four operations to algebraic fractions, finding unknowns in formulas after substitution, making the connection between equations of relations and their graphs, comparing simple and compound interest in financial contexts and determining probabilities of two- and three-step experiments • fluency includes factorising and expanding algebraic expressions, using a range of strategies to solve equations and using calculations to investigate the shape of data sets • problem-solving includes calculating the surface area and volume of a diverse range of prisms to solve practical problems, finding unknown lengths and angles using applications of trigonometry, using algebraic and graphical techniques to find solutions to simultaneous equations and inequalities and investigating independence of events • reasoning includes formulating geometric proofs involving congruence and similarity, interpreting and evaluating media statements and interpreting and comparing data sets.	{"url":"https://k10outline.scsa.wa.edu.au/home/teaching/curriculum-browser/mathematics-v8/year-10","timestamp":"2024-11-11T09:46:24Z","content_type":"text/html","content_length":"127768","record_id":"<urn:uuid:ee1eee21-08d0-4194-9aed-626c74f1ffa3>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00011.warc.gz"}
Isomorphic transformation question • Thread starter nhrock3 • Start date In summary, there is a transformation T:R^4 ->R^4, where dim Im(T+I)=dim Ker(3I-T)=2. To prove that T-I is isomorphic, we can use the fact that T(x)=\lambda x defines the notion of eigenvalue for a transformation. From this, we can see that the eigenvectors of T are the elements of Ker(T-\lambda I), and lambda is the corresponding eigenvalue. By using the fact that dim Ker(T+I)=2, we can conclude that the eigenvalue for T is -1 with a multiplicity of 2. This means that the vectors in Ker(T+I) are the eigenvectors of T with eigenvalue there is a transformation T:R^4 ->R^4 dim Im(T+I)=dim Ker(3I-T)=2 prove that T-I is isomorphic first of all i couldn't understand the first equation because T is a transformation which is basicly a function but I is the identity matrices so its like adding kilograms to tempreture. then my prof told me that here I is a transformation too that its not a matrix its the identity transformation. and i was told that i need to get the eigenvectors from there so i told him "how i don't have any matrix here to do i don't have any matrices only some transformation which i don't have even the formula to T in order to get the representing matrices out if it how to find eigen vectors from here and how to proceed in order to prove that is isomorphic Staff Emeritus Science Advisor Homework Helper Hi nhrock3! Matrices and transformations are equivalent. Given a matrix, there is a corresponding transformation and vice versa. Thus we can define the notion of eigenvalue for a transformation by [tex]T(x)=\lambda x[/tex] thus the eigenvectors of T are the elements of [tex]Ker(T-\lambda I)[/tex] and lambda is the corresponding eigenvalue. Now, can you use the fact that [tex]dim Ker(T+I)=2[/tex] to find an eigenvalue of T? And what multiplicity does the eigenvalue have? micromass said: Hi nhrock3! Matrices and transformations are equivalent. Given a matrix, there is a corresponding transformation and vice versa. Thus we can define the notion of eigenvalue for a transformation by [tex]T(x)=\lambda x[/tex] thus the eigenvectors of T are the elements of [tex]Ker(T-\lambda I)[/tex] and lambda is the corresponding eigenvalue. Now, can you use the fact that [tex]dim Ker(T+I)=2[/tex] to find an eigenvalue of T? And what multiplicity does the eigenvalue have? i don't understand the transition i know that [tex]T(v)=Av=\lambda v[/tex] its the definition of the link between eigen vectors and eigen values Ker(T+I) means (T+I)(v)=0 but T and I are both transformation so i can't use it like here in here [tex]Ker(T-\lambda I)[/tex] I is a mtrix but my I is the identity transformation its a function not a matrix Last edited: ok i understand your idea Ker(T+I) meand the eigen vectors of eigenvalue -1 but still matheticly i need to replace T with some matrices and I (the transformation) needs to be I(the identety matrix) Staff Emeritus Science Advisor Homework Helper Why do you change everything to matrices?? You can do that, of course, but you can leave everything in transformation form too. Saying that the vectors in are the eigenvectors of T with eigenvalue -1, is perfectly fine for transformations. There's no need to change everything to matrices. But you can change to matrices if it's easier for you... FAQ: Isomorphic transformation question 1. What is an isomorphic transformation? An isomorphic transformation is a mathematical concept where two objects or systems are transformed in a way that preserves their structure and relationships. This means that the transformed objects are essentially the same as the original ones, just in a different form or representation. 2. How is isomorphic transformation different from other types of transformations? Unlike other transformations, such as translation or rotation, isomorphic transformations do not change the fundamental properties or relationships of the objects being transformed. Instead, they only change the way the objects are represented or expressed. 3. What are some examples of isomorphic transformations in science? Isomorphic transformations can be found in various fields of science, such as in chemistry where the same chemical compound can have different structural representations, or in biology where different species may have similar genetic codes. In physics, isomorphic transformations can be seen in the mathematical transformation of equations to simplify or solve problems. 4. Can isomorphic transformations be applied to real-world problems? Yes, isomorphic transformations have practical applications in various fields, including computer science, economics, and social sciences. They can be used to simplify complex systems, analyze data, and make predictions. 5. How is isomorphic transformation useful in science? Isomorphic transformations are useful in science as they allow for the comparison and analysis of complex systems by breaking them down into simpler, equivalent forms. This can help scientists better understand and explain phenomena, make predictions, and identify patterns and relationships.	{"url":"https://www.physicsforums.com/threads/isomorphic-transformation-question.508683/","timestamp":"2024-11-08T20:46:03Z","content_type":"text/html","content_length":"93811","record_id":"<urn:uuid:d551df9d-3482-4949-95a2-0fb8256741f5>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00215.warc.gz"}
Radion-induced gravitational wave oscillations and their phenomenology We discuss the theory and phenomenology of the interplay between the massless graviton and its massive Kaluza-Klein modes in the Randall-Sundrum two-brane model. The equations of motion of the transverse traceless degrees of freedom are derived by means of a Green function approach as well as from an effective nonlocal action. The second procedure clarifies the extraction of the particle content from the nonlocal action and the issue of its diagonalization. The situation discussed is generic for the treatment of two-brane models if the on-brane fields are used as the dynamical degrees of freedom. The mixing of the effective graviton modes of the localized action can be interpreted as radion-induced gravitational-wave oscillations, a classical analogy to meson and neutrino oscillations. We show that these oscillations arising in M-theory-motivated braneworld setups could lead to effects detectable by gravitational-wave interferometers. The implications of this effect for models with ultra-light gravitons are discussed. Annalen der Physik Pub Date: July 2003 □ Gravitational waves; □ higher dimensions; □ m-theory phenomenology; □ High Energy Physics - Theory; □ Astrophysics; □ General Relativity and Quantum Cosmology; □ High Energy Physics - Phenomenology 27 pages, to appear in Annalen Phys	{"url":"https://ui.adsabs.harvard.edu/abs/2003AnP...515..343B","timestamp":"2024-11-11T15:16:43Z","content_type":"text/html","content_length":"40673","record_id":"<urn:uuid:ed390589-4f42-4bff-b71c-3eff6016d840>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00069.warc.gz"}
Specific Heat Calculator - Free Online Tool \| How to find Specific Heat? - physicsCalculatorPro.com The Online Specific Heat Calculator is a free tool for calculating a sample's specific heat given additional inputs such as temperature, energy, sample mass, and so on. All you have to do is enter the relevant inputs and press the calculate button to get instant results. What is Specific Heat and Its Formula? The amount of heat per unit mass required to raise the temperature by one degree Celsius is known as specific heat. The specific heat of various substances differs from one another and is determined by their ability to absorb heat. The Heat Capacity Formula can be found here: c=Q/(mΔT) • The amount of heat given in Joules is denoted by Q. • m be the sample mass • ΔT is the difference between the initial and final temperatures. • Specific heat units are J/(kgK) or J/kg C. For more concepts check out physicscalculatorpro.com to get quick answers by using this free tool. How do you find the Specific Heat? To compute the specific heat, use the simple steps outlined below. The list is as follows: • Initially, Determine the beginning and end temperatures, sample mass, and energy or heat given. • To get the change in temperature (T), find the difference between the initial and final temperatures. • Then, get the product of the change in temperature and the mass of the sample. • To determine the Specific Heat or Heat Capacity, divide the heat given or energy by the product gained in the previous step. Typical Values of Specific Heat Take a look at the specific heat values of the most regularly used chemicals in the table below. You can use this information to determine how much heat is required to increase or decrease the temperature of a sample. As so, they are. • ice: 2,100 J/(kg·K) • water: 4,200 J/(kg·K) • water vapor: 2,000 J/(kg·K) • basalt: 840 J/(kg·K) • granite: 790 J/(kg·K) • aluminum: 890 J/(kg·K) • iron: 450 J/(kg·K) • copper: 380 J/(kg·K) • lead: 130 J/(kg·K) Specific Heat Equation Examples Question 1: If the amount of heat required for transition is 350Kcal, calculate the latent heat of a substance is 7kg. Consider the problem, we have Energy = 130J Initial Temperature = 20°C Final Temperature = 50°C Mass of Metal = 15g Change in temperature ΔT=50°C-20°C=30°C We know the formula to calculate specific heat or heat capacity c = Q/(mΔT). Substituting the inputs, we get the following equation for heat capacity: c = 130J/15g*30°C c = 130J/450 We get the specific heat value by converting the units to a similar scale and simplifying further As a result,The specific heat, c = 288.89 J/(kg·K) FAQs on Specific Heat Calculator 1. What does the term specific heat mean? The amount of thermal energy required to raise the temperature of a 1kg sample by 1K is known as specific heat. 2. How to estimte specific heat? The difference between the initial and final temperatures can be used to compute specific heat by dividing the amount of provided heat by the mass of the sample. 3. What is the formula for calculating specific heat? The formula for calculating specific heat is c=Q/(mΔT). 4. What does the acronym QMC ΔT stand for? The quantity of heat given is Q, the mass of the sample is m and the temperature difference between the initial and final temperatures is ΔT. 5. What are the different types of units for specific heat capacity? The units of Specific Heat Capacity are J/kg K or J/kg C.	{"url":"https://physicscalculatorpro.com/specific-heat-calculator/","timestamp":"2024-11-05T02:38:04Z","content_type":"text/html","content_length":"36417","record_id":"<urn:uuid:573d5e52-ea60-4e62-a49d-98b380fa90d3>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00027.warc.gz"}
What is the term for the process of adding a new element to a Skip List - ITEagers Data Structure - Question Details What is the term for the process of adding a new element to a Skip List? Similar Question From (Data Structure) What is the primary use case of a Trie data structure? Similar Question From (Data Structure) In a Trie, what is the term for the set of all nodes that have a given prefix? Similar Question From (Data Structure) Which operation in a Trie involves inserting a key into the structure? Similar Question From (Data Structure) Which of the following is true about tail recursion? Similar Question From (Data Structure) What is the time complexity of the recursive Fibonacci sequence algorithm without memoization? Similar Question From (Data Structure) In recursion, what is the term for the set of all instances of a function in the call stack? Similar Question From (Data Structure) What is an AVL tree? Similar Question From (Data Structure) What is the purpose of the Trie property known as the "Prefix property"? Similar Question From (Data Structure) Which data structure is commonly used to implement recursion? Similar Question From (Data Structure) What is the time complexity for searching an element in a binary search tree? Read More Questions Learn the building blocks of efficient software through the study of data structures and algorithms. Read More Challenge Your Knowledge! Engage in our interactive quizzes to put your learning to the test. Strengthen your understanding and reinforce your knowledge with our thought-provoking questions and activities. Start Quiz Recent comments Latest Comments section by users Add a comment Your Comment will appear after approval! Check out Similar Subjects Computer Science Solved Past Papers (SPSC) Solved Past Papers (FPSC)	{"url":"https://iteagers.com/Computer%20Science/Data%20Structure/1500_What-is-the-term-for-the-process-of-adding-a-new-element-to-a-Skip-List","timestamp":"2024-11-05T13:07:23Z","content_type":"text/html","content_length":"104490","record_id":"<urn:uuid:ee287f5b-a0b1-48fd-bc68-84c803f71386>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00620.warc.gz"}
to factor graph Add factor to factor graph Since R2022a The addFactor function adds one or more factors to a factor graph and can group the factors and nodes that are specified by the added factors. factorIDs = addFactor(fg,factor) adds one or more factors to the specified factor graph and returns the IDs of the added factors. If adding the factors results in an invalid node, then addFactor returns an error, and indicates the invalid nodes. addFactor supports only single-factor addition for the factorIMU and factorGPS objects. factorIDs = addFactor(fg,factor,groupID) adds a factor to the factor graph with group ID groupID. Node IDs of the same group can be retrieved by group ID using nodeIDs function. You can use group IDs to represent timestamps or frames. Optimize Simple Factor Graph Create a factor graph. Define two pose states of the robot as the ground truth. rstate = [0 0 0; 1 1 pi/2]; Define the relative pose measurement between two nodes from the odometry as the pose difference between the states with some noise. The relative measurement must be in the reference frame of the second node so you must rotate the difference in position to be in the reference frame of the second node. posediff = diff(rstate); rotdiffso2 = so2(posediff(3),"theta"); transformedPos = transform(inv(rotdiffso2),posediff(1:2)); odomNoise = 0.1*rand; measure = [transformedPos posediff(3)] + odomNoise; Create a factor connecting two SE(2) pose with the relative measurment between the poses. Then add the factor to the factor graph to create two nodes. ids = generateNodeID(fg,1,"factorTwoPoseSE2"); f = factorTwoPoseSE2(ids,Measurement=measure); Get the state of both pose nodes. stateDefault = nodeState(fg,ids) stateDefault = 2×3 Because these nodes are new, they have default state values. Ideally before optimizing, you should assign an approximate guess of the absolute pose. This increases the possibility of the optimize function finding the global minimum. Otherwise optimize may become trapped in the local minimum, producing a suboptimal solution. Keep the first node state at the origin and set the second node state to an approximate xy-position at [0.9 0.95] and a theta rotation of pi/3 radians. In practical applications you could use sensor measurements from your odometry to determine the approximate state of each pose node. nodeState(fg,ids(2),[0.9 0.95 pi/3]) ans = 1×3 0.9000 0.9500 1.0472 Before optimizing, save the node state so you can reoptimize as needed. statePriorOpt1 = nodeState(fg,ids); Optimize the nodes and check the node states. stateOpt1 = nodeState(fg,ids) stateOpt1 = 2×3 -0.1161 0.9026 0.0571 1.0161 0.0474 1.7094 Note that after optimization the first node did not stay at the origin because although the graph does have the initial guess for the state, the graph does not have any constraint on the absolute position. The graph has only the relative pose measurement, which acts as a constraint for the relative pose between the two nodes. So the graph attempts to reduce the cost related to the relative pose, but not the absolute pose. To provide more information to the graph, you can fix the state of nodes or add an absolute prior measurement factor. Reset the states and then fix the first node. Then verify that the first node is fixed. Reoptimize the factor graph and get the node states. ans = struct with fields: InitialCost: 1.9452 FinalCost: 1.9452e-16 NumSuccessfulSteps: 2 NumUnsuccessfulSteps: 0 TotalTime: 8.1062e-05 TerminationType: 0 IsSolutionUsable: 1 OptimizedNodeIDs: 1 FixedNodeIDs: 0 stateOpt2 = nodeState(fg,ids) stateOpt2 = 2×3 1.0815 -0.9185 1.6523 Note that after optimizing this time, the first node state remained at the origin. Add Nodes to Groups Create a factor graph, generate node IDs, and create two factorTwoPoseSE2 factors. fg1 = factorGraph; ids = generateNodeID(fg1,[2 2]) f = factorTwoPoseSE2(ids); Group All Nodes Add all nodes of the factors to group 1. fg1Group1 = nodeIDs(fg1,GroupID=1) Group Nodes by Column Specify the group ID as a row vector to add the nodes of the first column of the node IDs of the factors to group 1 and add the nodes of the second column to group 2. fg2 = factorGraph; addFactor(fg2,f,[1 2]); fg2Group1 = nodeIDs(fg2,GroupID=1) fg2Group2 = nodeIDs(fg2,GroupID=2) Group Nodes by Row Specify the group ID as a column vector to add the nodes of the first row in the node IDs of the factors to group 1 and add the nodes of the second row to group 2. fg3 = factorGraph; addFactor(fg3,f,[1; 2]); fg3Group1 = nodeIDs(fg3,GroupID=1) fg3Group2 = nodeIDs(fg3,GroupID=2) Group Nodes by Matrix You can also specify the group ID as a matrix of the same size as the node IDs of the factors to assign each node to a specific group. Add the first and fourth nodes to group 1 and the second and third nodes to groups 2 and 3, respectively. fg4 = factorGraph; groups = [1 2; 3 1]; fg4Group1 = nodeIDs(fg4,GroupID=1) fg4Group2 = nodeIDs(fg4,GroupID=2) fg4Group3 = nodeIDs(fg4,GroupID=3) Input Arguments fg — Factor graph to add factor to factorGraph object Factor graph to add factor to, specified as a factorGraph object. factor — Factors to add to factor graph valid factor object Factors to add to the factor graph, specified as a valid factor object. A valid factor object must be one of these objects, and the object must not create any invalid nodes when added to the factor graph: With the exception of factorGPS and factorIMU, you can simultaneously add multiple factors to the factor graph using any one of the listed factor objects. factorGPS and factorIMU support only single-factor addition. If the specified factor object creates any invalid nodes, then addFactor adds none of the factors from the factor object. groupID — Group IDs for nodes of added factor nonnegative integer \| two-element row vector of nonnegative integers \| N-element column vector of nonnegative integers \| N-by-2 matrix of nonnegative integers Group IDs for nodes of the added factor, specified as any of these options: groupID Size Grouping Behavior Assigns all nodes to one group. nonnegative integer For example, if you add a factor object that has a NodeID value of $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ with a groupID value of 1, addFactor adds nodes 1, 2, 3, and 4 to group 1. Specify groups for each column of nodes. two-element row vector of For example, if you add a factor object that has a NodeID value of $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ with a groupID value of $\left[\begin{array}{cc} nonnegative integers 1& 2\end{array}\right]$, addFactor adds nodes 1 and 3 to group 1 and adds nodes 2 and 4 to group 2. The behavior for IMU factors is different. If you add an IMU factor with a NodeID value of $\left[\begin{array}{cccccc}1& 2& 3& 4& 5& 6\end{array}\right]$ and groupID set to $\left[\begin{array}{cc}1& 2\end{array}\right]$, addFactor adds nodes 1, 2, and 3 to group 1 and nodes 4, 5, and 6 to group 2. Group nodes by factor, where N is the total number of factors specified by the NodeID property of factor. N-element column vector of nonnegative integers For example, if you add a factor object that has a NodeID value of $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ with a groupID value of $\left[\begin{array}{c}1 \\ 2\end{array}\right]$, addFactor adds nodes 1 and 2 to group 1 and adds nodes 3 and 4 to group 2. Add nodes in NodeID to the group specified at its corresponding index in groupID, where N is the total number of rows of the NodeID property of factor. N-by-2 matrix of nonnegative integers For example, if you add a factor object that has a NodeID value of $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ with a groupID value of $\left[\begin{array}{cc} 1& 2\\ 3& 1\end{array}\right]$, addFactor add nodes 1 and 4 into group 1, adds node 2 to group 2, and adds node 3 to group 3. When adding a factorIMU or factorGPS object to a factor graph, groupID accepts only these values: • factorIMU — Nonnegative integer or a two-element row vector of nonnegative integers. • factorGPS — Nonnegative integer Adding nodes to groups enables you to query node IDs by group by specifying the GroupID name-value argument of the nodeIDs function. Output Arguments factorIDs — Factor IDs of added factors N-element row vector of nonnegative integers Factor IDs of the added factors, returned as an N-element row vector of nonnegative integers. N is the total number of factors added. The function returns this argument only when it successfully adds the factors to the factor graph. If adding the factors results in an invalid node, then addFactor adds none of the factors from the factor object. Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. When generating portable C code with a C++ compiler, you must specify hierarchical packing with non-minimal headers. For more information on packaging options, see the packNGo (MATLAB Coder) Version History Introduced in R2022a R2023a: Specify group IDs for added factors addFactor now supports specifying groups to add nodes of added factors to by group ID.	{"url":"https://in.mathworks.com/help/nav/ref/factorgraph.addfactor.html","timestamp":"2024-11-11T10:20:26Z","content_type":"text/html","content_length":"110073","record_id":"<urn:uuid:1acc84fa-1a53-44ae-aad4-dc71da5ad4ba>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00591.warc.gz"}
What is Intersection of Two Lines ⭐ Definition With Examples (2024) • Math for kids • / • Knowledge Base • / • Intersection of Two Lines – Definition With Examples Created on Jan 09, 2024 Updated on January 12, 2024 Welcome to another exciting exploration into the world of geometry with Brighterly! Today, we delve into the concept of the intersection of two lines. Picture a bustling city with roads crisscrossing every which way. At every crossroad, a story unfolds. Now, let’s transpose this bustling city onto a piece of paper. In geometry, the ‘roads’ are lines, and where they cross, we call these points the ‘intersection.’ It’s fascinating, isn’t it? These intersections can tell us so much about the relationship between these lines. They give us insights that we can use to understand and solve complex problems in mathematics and beyond. In this article, we’re going to uncover the beauty of these intersections and what they reveal about the lines that form them. Definition of a Line in 2D Geometry Lines, one of the simplest forms of geometric shapes, are extremely powerful in the study of 2D geometry. A line is defined as a straight one-dimensional figure that extends infinitely in both directions. It’s composed of an infinite number of points that are positioned side by side. It’s also important to know that a line has no endpoints. There are no corners, no boundaries, just an endless stretch of points. Intriguing, isn’t it? Definition of Intersection Point Now, the term ‘intersection’ seems quite intuitive, doesn’t it? In simple terms, the intersection point is the point where two lines meet or cross each other. Just like the point where the two sticks touched each other. In the vast universe of two lines, this is the single point where they share the same coordinates. This sharing of coordinates makes the intersection point a crucial concept in Properties of Intersecting Lines The beautiful world of intersecting lines comes with some fascinating properties. The first property is about angles. When two lines intersect, they form four angles, and here’s the interesting part: opposite angles (known as vertical angles) are equal. The second important property is that the sum of adjacent angles is 180 degrees, meaning they are supplementary. These properties help us predict and calculate various aspects related to intersecting lines. Characteristics of the Intersection Point The intersection point has some unique characteristics. The first and most vital characteristic is that it shares the same coordinates on both lines. If you check the position on both lines, it will be identical! Another fascinating characteristic is that at this intersection point, two lines divide the plane into four regions or angles, each of which holds unique properties. Difference Between Parallel, Perpendicular, and Intersecting Lines A critical aspect of studying lines is understanding the differences between parallel, perpendicular, and intersecting lines. Parallel lines are those that never meet, no matter how far they extend. They always maintain a constant distance from each other. On the other hand, perpendicular lines intersect at a point forming a 90-degree angle. Intersecting lines can meet at any angle, except 0 and 180 degrees (which would be parallel lines) and 90 degrees (which are perpendicular lines). Equations of Intersecting Lines Every line in 2D geometry can be represented using an equation, and intersecting lines are no exception. The most common form is the slope-intercept form (y = mx + b), where ‘m’ represents the slope of the line and ‘b’ is the y-intercept. The point of intersection of two lines can be found by solving their equations simultaneously. Writing Equations of Intersecting Lines Writing equations of intersecting lines involves understanding the slopes and y-intercepts of the lines. Let’s say we have two lines with equations y = m1x + c1 and y = m2x + c2. If m1 ≠ m2, then these lines will intersect at a point. To find the intersection point, you set the two equations equal to each other and solve for the variable x. Solving for the Point of Intersection When you’ve set the two equations equal and solved for x, you then substitute x into one of the equations to find the corresponding y-coordinate. The ordered pair (x, y) represents the intersection point of the two lines. This process is like a treasure hunt, where the treasure is the point of intersection! Solving for the Point of Intersection When you’ve set the two equations equal and solved for x, you then substitute x into one of the equations to find the corresponding y-coordinate. For example, if we have the two equations y = 2x + 3 and y = -x + 1, we set them equal to get 2x + 3 = -x + 1. Solving for x gives us x = -2/3. Then, substituting x = -2/3 into the first equation gives us y = 2(-2/3) + 3 = 5/3. So, the point of intersection is (-2/3, 5/3). This process is like a treasure hunt, where the treasure is the point of intersection! Practice Problems on Intersection of Two Lines Practice is the key to mastering the concept of intersecting lines. Here at Brighterly, we provide engaging and thought-provoking problems to help you explore and learn the properties, characteristics, and equations of intersecting lines. Check out these intersection of lines practice problems: 1. Problem: Find the intersection point of the lines y = 3x + 2 and y = -2x + 1. Solution: Set the equations equal: 3x + 2 = -2x + 1. Solve for x to get x = -1/5. Substitute x = -1/5 into the first equation to get y = 3(-1/5) + 2 = 7/5. So, the point of intersection is (-1/ 5, 7/5). 2. Problem: Find the intersection point of the lines y = x – 1 and y = 2x + 3. Solution: Set the equations equal: x – 1 = 2x + 3. Solve for x to get x = -4. Substitute x = -4 into the first equation to get y = -4 – 1 = -5. So, the point of intersection is (-4, -5). Remember, the more you practice, the easier these problems will become. Enjoy the journey of discovery! And there you have it! The world of intersecting lines, full of intriguing angles and fascinating relationships, has been unravelled. With Brighterly, you have journeyed from understanding what a line in 2D geometry is, through the intriguing characteristics of intersecting lines, to writing equations and solving them to find the point of intersection. We hope this has shown you the beauty hidden within these geometric concepts. Remember, the journey doesn’t stop here. Mathematics, and especially geometry, is a vast universe waiting to be explored. Every concept is a new adventure. So keep questioning, keep learning, and keep discovering the wonder of the mathematical world with us here at Brighterly. Frequently Asked Questions on Intersection of Two Lines Do all lines intersect? No, not all lines intersect. In a plane, two lines can either intersect, be parallel or coincide. Parallel lines never intersect as they always maintain a constant distance from each other. Lines that coincide are essentially the same line and they share all points. What is the point of intersection called? The point where two or more lines intersect or cross each other is called the point of intersection. Can two lines intersect at more than one point? In a two-dimensional plane, two lines cannot intersect at more than one point. However, in higher dimensions, lines can intersect along a line or a plane. What is the significance of the point of intersection in a graph? The point of intersection in a graph can represent the solution to a system of equations. For example, in a system of linear equations, the point of intersection represents the values of the variables that make both equations true. Can a line intersect with a curve? Yes, a line can intersect a curve. The points of intersection are the points where the equation of the line and the equation of the curve are both true. We hope that these answers clear up any lingering questions you might have about intersecting lines. As always, keep asking questions and keep learning with Brighterly! Information Sources: I am a seasoned math tutor with over seven years of experience in the field. Holding a Master’s Degree in Education, I take great joy in nurturing young math enthusiasts, regardless of their age, grade, and skill level. Beyond teaching, I am passionate about spending time with my family, reading, and watching movies. My background also includes knowledge in child psychology, which aids in delivering personalized and effective teaching strategies.	{"url":"https://smysa.org/article/what-is-intersection-of-two-lines-definition-with-examples","timestamp":"2024-11-14T05:23:04Z","content_type":"text/html","content_length":"125676","record_id":"<urn:uuid:88b1cff8-1c48-4951-9796-da711112056a>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00162.warc.gz"}
t P Academic Year 2023/2024 - Teacher: Maria Alessandra RAGUSA Expected Learning Outcomes At the end of the course the student will acquire both theoretical and practical knowledge on the main contents of the course. 1. Knowledge and understanding - Knowledge and understaning: The student will be able to understand and assimilate the definitions and main results of the basic mathematical analysis, for real functions of a real variable. 2. Ability to apply knowledge and understanding - Applying Knowledge and understaning: The student will be able to acquire an appropriate level of autonomy in theoretical knowledge and in the use of basic analytical tools. 3. Making judgments - Making judgments: Ability to reflect and calculate. Ability to apply the notions learned to solving problems and exercises. 4. Communication skills - Communication skills: Ability to communicate the knowledge acquired through an adequate scientific language. 5. Learning skills - Learning skills: Ability to deepen and develop the knowledge acquired. Ability to critically use tables and analytical and computer tools of symbolic computation. Course Structure Direct Instruction. The lessons are integrated with exercises related to the topics covered by the course and will take place in the classroom. It should also be noted that there are 49 hours of lessons (typically, these are theory) and 24 hours of other activities (typically, these are exercises). Should teaching be carried out in mixed mode or remotely, it may be necessary to introduce changes with respect to previous statements, in line with the program planned and outlined in the Syllabus. NOTE: Information for students with disabilities and/or SLD To guarantee equal opportunities and in compliance with the laws in force, interested students can ask for a personal interview in order to plan any compensatory and / or dispensatory measures, based on the didactic objectives and specific needs. It is also possible to contact the referent teacher CInAP (Center for Active and Participated Integration - Services for Disabilities and / or SLD) of our Department, prof. Filippo Stanco). Required Prerequisites The student must have a thorough knowledge of the notions of Mathematics studied in the five years of high school. In particular: Elements of Mathematical Logic, set theory, algebraic equations and inequalities, trigonometry. Attendance of Lessons Strongly recommended. Detailed Course Content 1. Sets and Logic. Basic concepts on sets, elementary logic. 2. The numbers. Natural, relative, rational and real numbers. Continuity axiom of real numbers. Lower and upper bounds of a numerical set. Absolute value and its properties. Radicals, powers, logarithms. Principle of induction. Complex numbers. 3. Functions of one real variable. Function concept. Bounded, symmetric, monotone, periodic functions. Elementary functions. Compound functions and inverse functions. 4. Limits and continuity. Numerical sequences. Definition of limit. Fundamental theroemes on limits. Calculation of limits. The number of Napier. Comparisons and asymptotic estimates. Limits of functions, continuity, asymptotes. Fundamental theorems on limits of functions. Calculation of limits. Notable limits. Comparisons and asymptotic estimates. Graph of a function. Fundamental properties of continuous functions. 5. Sequences and numerical series. Definition of succession. Limits of sequences. Extracted sequences. Definition of series. Examples of numerical series. Fundamental theorems on series. Series with non-negative terms. Series with terms of variable sign. Notable numerical series. Textbook Information [1] W. Rudin, Principles of Mathematical Analysis, McGraw Hill, 2015. Course Planning Subjects Text References 1 Sets and Logic. [1] Chapter 1. 2 The numbers. [1] Chapter 1. 3 Functions of one real variable. [1] Chapter 4. 4 Limits and continuity. [1] Chapter 4. 5 Sequences and numerical series. [1] Chapter 3. Learning Assessment Learning Assessment Procedures The final exam consists of a written test and an interview, both rated out of thirty. Passed the written test lo student must undergo an interview that contributes to the formulation of the final grade, expressed in thirty. The registration of the exam will take place only after passing the interview. NOTE: The learning assessment can also be carried out electronically, should the conditions require it. Examples of frequently asked questions and / or exercises All the topics mentioned in the program can be requested during the exam. The frequency of the lessons, the study of the recommended texts and the study of the material provided by the teacher (handouts and collections of exercises carried out and proposed) allow the student to have a clear and detailed idea of the questions that may be proposed during the exam. An adequate exposition of the theory involves the use of the rigorous language characteristic of the discipline, the exposition of simple examples and counterexamples that clarify the concepts exposed (definitions, propositions, theorems, corollaries). The main types of exercises are as follows: • Search for the extrema of a numerical set. • Exercises on complex numbers (algebraic manipulations, writing complex numbers in algebraic, trigonometric and exponential form). • Calculation of limits of sequences and functi • ons. Study of the character of a numerical series. • Study of the continuity of real functions of a real variable.	{"url":"https://dmi.unict.it/courses/l-35/course-units?seuid=D5C4AD60-2E80-4124-BBF7-E5D7385D9C34","timestamp":"2024-11-07T13:11:23Z","content_type":"text/html","content_length":"37550","record_id":"<urn:uuid:b47237ab-b9f3-4296-b8b8-0f1fe75d4a80>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00533.warc.gz"}
[tlaplus] Re: Refinement Mappings and Fairness [Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index] [tlaplus] Re: Refinement Mappings and Fairness 1. It is mentioned in section 8.9.4 of Specifying Systems that substitution does not distribute over ENABLED, and hence it does not distribute over WF or SF. Could someone give an example where it is indeed the case? I believe this is covered in A science of concurrent programming (https://lamport.azurewebsites.net/tla/science.pdf) under section 5.4.4 (A Closer Look at E). 2. It is recommended in the same section that "you don't have to depend on this. You can instead expand the definitions of WF and SF [..] (and) compute the enabled predicates "by hand" and then perform the substitution." Are there conditions under which one could be sure that substitution distributes over entire specification formula so that there is no need to prove it "by hands"? Thank you, You received this message because you are subscribed to the Google Groups "tlaplus" group. To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+unsubscribe@xxxxxxxxxxxxxxxx. To view this discussion on the web visit https://groups.google.com/d/msgid/tlaplus/dfa13cd3-fff5-49e3-9e99-1bd82f7739den%40googlegroups.com.	{"url":"https://discuss.tlapl.us/msg05907.html","timestamp":"2024-11-11T08:31:24Z","content_type":"text/html","content_length":"4956","record_id":"<urn:uuid:3f7db501-2558-4857-aa8a-83cb5dae8ab6>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00625.warc.gz"}
Symbolic Computation Symbolic Computation These projects use symbolic computation in an essential way both in the process of discovery and proof. Each aims at producing robust software. □ J.M. Borwein and P.B. Borwein, "Inequalities for compound means with logarithmic asymptotes," Journal of Mathematical Analysis and Applications, 177(1993),572-582. □ J.M. Borwein, "A note on the existence of subgradients," Math. Programming, 24(1982),225-228.	{"url":"http://wayback.cecm.sfu.ca/projects/symbolic_computation.html","timestamp":"2024-11-04T13:50:47Z","content_type":"text/html","content_length":"1922","record_id":"<urn:uuid:d1b60fbd-0599-4b9e-b9f8-e88650f8b7f6>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00438.warc.gz"}
Robustness Checks \| RDP 2018-02: Affine Endeavour: Estimating a Joint Model of the Nominal and Real Term Structures of Interest Rates in Australia RDP 2018-02: Affine Endeavour: Estimating a Joint Model of the Nominal and Real Term Structures of Interest Rates in Australia 5. Robustness Checks 5.1 Sample Starting in 1997 The data sample used in Section 4 spans the period before and after the Reserve Bank adopted a formal 2 to 3 per cent inflation target. Therefore, there could be a structural break, or regime shift, for which the model does not adequately account. This would be of particular concern given that the model imposes stationarity. To check the robustness of the results, we estimate the model on a reduced sample beginning in 1997, once inflation expectations had become reasonably anchored.^[26] Nominal and real interest rates, and term premia, follow fairly similar paths to those estimated using the full sample, although expected rates tend to be smoother, especially at longer horizons (Figures E1–E6). In particular, ten-year-ahead nominal and real forward rates are a bit less variable when estimated over the shorter sample, while three- and five-year-ahead real rates show larger declines in recent years. The same is broadly true of inflation expectations, although the smoothness occurs to an even larger degree. Given the results for nominal and real expectations, this last point is perhaps not surprising: expected inflation is calculated as the difference between nominal and real expectations; if these expectations follow similar trends and are relatively smooth, then their difference will tend to be even smoother and flatter still. More broadly, the smoothness is suggestive of a short-sample problem leading to insufficiently persistent pricing factors. In particular, Guimarães (2016) argues that discarding part of the sample due to changes in the structure of the economy is exactly the opposite of what we should do, as this variation can be extremely useful in separately identifying the P and Q dynamics. This argument could certainly be put forward here. By removing the early period we are potentially removing a period with a large amount of information about the dynamics of inflation expectations, and in particular how they become anchored (and therefore can potentially become unanchored). Nonetheless, both sets of results show broadly similar trends over time for a number of variables, which is 5.2 Filtered versus Unfiltered Results As noted in Section 3.1, we estimate the model in two steps: first we maximise the model's likelihood conditional on the observed factors; second we cast the model in a Kalman filter and re-optimise. The second step allows us to relax the assumption that the factors are priced correctly, and to drop any estimated zero-coupon real yield data that does not have a traded bond with a similar maturity and so is dependent on interpolation. Both of these generalisations are potentially important given the sparsity of inflation-linked bonds. Related to this, by using the Kalman filter and allowing for imperfect pricing in the factors, we allow the surveys to influence these pricing factors, which is also potentially important. It is interesting to consider what the results would look like if we did not incorporate the second step. Figures F1–F6 contain these results. Again, the estimates of real and nominal interest rates and risk premia are broadly similar, while the estimates of expected inflation show greater differences. In particular, the inflation estimates are generally somewhat smoother, particularly the ten-year-ahead expectations, and there is a larger fall around the onset of the global financial crisis, although the broad trends are still reasonably similar and the results still suggest that inflation expectations are well anchored within the 2 to 3 per cent target band. The inflation expectation estimates from the first step are also more similar to those from Finlay and Wende (2012). As with the estimates from that paper, the difference seems to be that the Kalman filtered model puts a higher weight on the surveys, as it estimates the variance of the noise associated with the surveys to be lower.^[27] This appears to reflect the fact that the Kalman filter approach allows the surveys to affect the estimated pricing factors, rather than constraining the model to use the observed factors. The results are similar whether or not we drop some real yield data, suggesting that fully utilising the information contained in the survey data is the more important generalisation. The fact that the filtered model places a greater weight on the surveys is particularly evident in Figure F7, which plots the model-implied inflation expectations for both the filtered and unfiltered models alongside the (closest) matching surveys. This also highlights the fact that even in the filtered model, the model-implied expectations do not perfectly coincide with the surveys and that they are taking a substantial signal from the yield data. Overall, these results suggest that using a Kalman filter, and therefore allowing for pricing factors that diverge from the principal components of the yield data, can lead to a higher weight being put on surveys (though this will not necessarily be a general result). To the extent that we think surveys are good measures of market participants' expectations, this will be preferable. This will be particularly true if we are concerned about the quality of the real yield data, as may be the case in countries with a scarcity of inflation-indexed bonds. However, if for some reason we think that the surveys are a poor measure of expectations, for the full sample or even for some sub-sample, it may be preferable to eschew the Kalman filter or to calibrate the model to place a lower weight on the surveys. Another approach would be to estimate a model with regime switching. However, the added complexity this would involve was not in keeping with our focus on estimating a usable ‘workhorse’ model. [26] On the flip side, it estimates the variance of the noise associated with the real yields to be higher. The estimates of the variance of the noise associated with the nominal yields are similar in the filtered and unfiltered models. [27]	{"url":"https://www.rba.gov.au/publications/rdp/2018/2018-02/robustness-checks.html","timestamp":"2024-11-09T13:31:22Z","content_type":"application/xhtml+xml","content_length":"32972","record_id":"<urn:uuid:12f70348-cb7c-41d1-99a6-e2f3e77a9897>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00790.warc.gz"}
Radomír Halas • Palacký University Olomouc, Department of Algebra and Geometry, Olomouc, Czech Republic (PhD 1994) According to our database , Radomír Halas authored at least 49 papers between 2000 and 2024. Collaborative distances: Book In proceedings Article PhD thesis Dataset Other Online presence: On csauthors.net: The multiplicative Cauchy equation on [0,1] and its application to the quasi-homogeneity equation. Fuzzy Sets Syst., 2024 n-K-Increasing Aggregation Functions. Axioms, December, 2023 On the number of aggregation functions on finite chains as a generalization of Dedekind numbers. Fuzzy Sets Syst., August, 2023 On the minimality of some generating sets of the aggregation clone on a finite chain. Inf. Sci., 2021 The logic induced by effect algebras. Soft Comput., 2020 On the decomposability of aggregation functions on direct products of posets. Fuzzy Sets Syst., 2020 On generation of aggregation functions on infinite lattices. Soft Comput., 2019 Operations and structures derived from non-associative MV-algebras. Soft Comput., 2019 On generating sets of the clone of aggregation functions on finite lattices. Inf. Sci., 2019 A note on some algebraic properties of discrete Sugeno integrals. Fuzzy Sets Syst., 2019 Aggregation via Clone Theory Approach. Proceedings of the New Trends in Aggregation Theory, 2019 On Linear Approximations of Sugeno Integrals on Bounded Distributive Lattices. IEEE Trans. Fuzzy Syst., 2018 Binary generating set of the clone of idempotent aggregation functions on bounded lattices. Inf. Sci., 2018 On generating of idempotent aggregation functions on finite lattices. Inf. Sci., 2018 The hull-kernel topology on prime ideals in posets. Soft Comput., 2017 Description of sup- and inf-preserving aggregation functions via families of clusters in data tables. Inf. Sci., 2017 Generalized comonotonicity and new axiomatizations of Sugeno integrals on bounded distributive lattices. Int. J. Approx. Reason., 2017 Generators of Aggregation Functions and Fuzzy Connectives. IEEE Trans. Fuzzy Syst., 2016 On the clone of aggregation functions on bounded lattices. Inf. Sci., 2016 Congruences and the discrete Sugeno integrals on bounded distributive lattices. Inf. Sci., 2016 A new characterization of the discrete Sugeno integral. Inf. Fusion, 2016 On varieties of basic algebras. Soft Comput., 2015 On lattices with a smallest set of aggregation functions. Inf. Sci., 2015 Generalized one-sided concept lattices with attribute preferences. Inf. Sci., 2015 The variety of modular basic algebras generated by MV-chains and horizontal sums of three-element chain basic algebras. Inf. Sci., 2012 States on commutative basic algebras. Fuzzy Sets Syst., 2012 Effect algebras are conditionally residuated structures. Soft Comput., 2011 Completeness of Order Algebras? J. Multiple Valued Log. Soft Comput., 2011 Are basic algebras residuated structures? Soft Comput., 2010 The Zerodivisor Graph of a Qoset. Order, 2010 On weakly cut-stable maps. Inf. Sci., 2010 On the role of logical connectives for primality and functional completeness of algebras of logics. Inf. Sci., 2010 On very true operators on pocrims. Soft Comput., 2009 Functional Completeness of Weak Logics with the Strict Negation? J. Multiple Valued Log. Soft Comput., 2009 A Note on Axiom System for SBL-algebras. Fundam. Informaticae, 2009 On Beck's coloring of posets. Discret. Math., 2009 Commutative basic algebras and non-associative fuzzy logics. Arch. Math. Log., 2009 The Variety of Lattice Effect Algebras Generated by MV-algebras and the Horizontal Sum of Two 3-element Chains. Stud Logica, 2008 Finite Commutative Basic Algebras are MV-Effect Algebras. J. Multiple Valued Log. Soft Comput., 2008 Functional completeness of bounded structures of fuzzy logic with wvt-operators. Fuzzy Sets Syst., 2008 On extensions of ideals in posets. Discret. Math., 2008 Congruence kernels of orthomodular implication algebras. Discret. Math., 2008 Complete Commutative Basic Algebras. Order, 2007 Weakly Standard BCC-Algebras. J. Multiple Valued Log. Soft Comput., 2006 Ideals and D-systems in Orthoimplication Algebras. J. Multiple Valued Log. Soft Comput., 2005 Annihilators on weakly standard BCC-algebras. Int. J. Math. Math. Sci., 2005 J. Multiple Valued Log. Soft Comput., 2004 Duality of Normally Presented Varieties. Int. J. Algebra Comput., 2000	{"url":"https://www.csauthors.net/radomir-halas/","timestamp":"2024-11-11T16:57:32Z","content_type":"text/html","content_length":"57722","record_id":"<urn:uuid:7e2f1998-d1bf-4257-af14-a7d2b13a931e>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00530.warc.gz"}
TRT-SInterp is a Matlab code designed to interpret a thermal response test in a deterministic or stochastic framework. The program treats variable heating power and emulates a borehole heat exchanger by a finite line-source model or a thermal resistance and capacity model. The possibly unknown parameters identified may comprise the thermal conductivity and volumetric heat capacity of the ground and grout, as well as the spacing between the pipes and the initial ground temperature. It is possible to integrate to the inversion the temperature measurements made at various depths in the fluid and grout and to take into account the fluid flow rate and the thermal capacity of the underground components. Content of the repository This repository contains the latest implementation of TRT-SInterp. The folder entilted TRT-SInterp contains the source code, which is a collection of Matlab functions. The folder entilted Dataset contains an example file and a synthetic dataset. User manual To learn how to use TRT-SInterp or learn about its theoretical foundations, please consult reference 1. To suggest improvements, report a possible bug, or initiate collaboration, please, contact me at : philippe.pasquier@polymtl.ca or on ResearchGate. Please, cite this work as: 1. Pasquier, P., 2015. Stochastic interpretation of thermal response test with TRT-SInterp. Computers & Geosciences, 75, pp.73–87. Additional references Additional information on TRT-SInterp can be found in the following references : • Pasquier, P. & Marcotte, D., 2013. Joint Use of Quasi-3D Response Model and Spectral Method to Simulate Borehole Heat Exchanger. Geothermics. • Pasquier, P. & Marcotte, D., 2012. Short-term simulation of ground heat exchanger with an improved TRCM. Renewable Energy, 46, pp.92–99. • Jacques, L., Pasquier, P., Marcotte, D. Influence of Measurement and Model Error on Thermal Response Test Interpretation, in: Proceedings of the World Renewable Energy Congress 2014, London, United Kingdom. 2014. • Claesson, J. & Javed, S., 2011. An analytical method to calculate borehole fluid temperatures for timescales from minutes to decades. In ASHRAE Annual Conference. Montréal, Canada, p. 10. • Marcotte, D. & Pasquier, P. 2008. On the estimation of thermal resistance in borehole conductivity test. Renewable Energy, vol. 33, p. 2407-2415. doi:10.1016/j.renene.2008.01.021 • Marcotte, D. & Pasquier, P., 2008. Fast fluid and ground temperature computation for geothermal ground-loop heat exchanger systems. Geothermics, 37(6), pp.651–665. • Hellström, G., 1991. Ground Heat Storage. Thermal Analysis of Duct Storage Systems. Part I Theory. University of Lund, Sweden. • Bennet, J., Claesson, J. & Hellström, G., 1987. Multipole Method to Compute the Conductive Heat Flows to and between Pipes in a Composite Cylinder, Lund, Sweden: University of Lund, Department of Building Technology and Mathematical Physics	{"url":"http://ppasquier.github.io/TRT_SInterp/","timestamp":"2024-11-12T09:33:22Z","content_type":"text/html","content_length":"7319","record_id":"<urn:uuid:7d476961-e731-43ad-abe2-e6d1fac728f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00526.warc.gz"}
What our customers say... Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences: The simple interface of Algebrator makes it easy for my son to get right down to solving math problems. Thanks for offering such a useful program. Candice Murrey, OR OK here is what I like: much friendlier interface, coverage of functions, trig. better graphing, wizards. However, still no word problems, pre-calc, calc. (Please tell me that you are working on it - who is going to do my homework when I am past College Algebra?!? M.V., Texas I am a mother of three, and I purchased the Algebrator software for my oldest son, to help him with his algebra homework, but my younger sons seen how easy it was to use, now they have a head start. Thank you for such a great, affordable product. Tom Sandy, NE Learning algebra on a computer may not seem like the appropriate way, but this software is so easy even a sixth-grader can learn algebra. 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Excel Formula to Format Phone Numbers This is a formula for use by users that are familiar with using formulas in Excel. If you are not such a user, be careful and backup before you start. We need to convert a column of variously formatted phone numbers to a common format. This can be difficult if the source has several formats that Excel doesn’t like. Here we will develop a formula to take care of the most common issues and get us where we want to be. The end result is this. You may need to change the red to match your sheet. They just point to the source cell. =TEXT(IF(B2=””,””,SUMPRODUCT(MID(0&B2,LARGE(INDEX(ISNUMBER(–MID(B2,ROW($1:$25),1))ROW($1:$25),0),ROW($1:$25))+1,1)10^ROW($1:$25)/10)),”[<=9999999999]1+(000) 000-0000;#+(###) ###-####”) This is where I started. I found this at Mr. Excel. This formula is where I started because the source had various formats already. This takes out all the non numeric characters. I’m not going to go into detail about how this formula does it, just believe me when I tell you it does it. First I’ll change a couple of places and assume the phone number is in column B and the formula is in column A. Also the sheet has a header row so we are in Row 2. We jus need to add a few checks. First, I the value is zero, replace with null. This would logically use the formula as a test but this would get us to a very long formula before we get started good so I cheated and just check for blank source. Now, lets apply the format. This is where we can change it to what we want. Below I have used the standard Excel Phone Number format. It assumes either a 7 or 10 digit number. It could be tweaked for your particular format. =TEXT(IF(B2=””,””,SUMPRODUCT(MID(0&B2,LARGE(INDEX(ISNUMBER(–MID(B2,ROW($1:$25),1))ROW($1:$25),0),ROW($1:$25))+1,1)10^ROW($1:$25)/10)),”[<=9999999]###-####;(###) ###-####”) This formula obviously fails if the source has invalid data to begin with and doesn’t allow for the country code. I’m going to tweak is some to allow for included country code. Since I’m dealing with US numbers, I assume the country code is always 1. =TEXT(IF(B2=””,””,SUMPRODUCT(MID(0&B2,LARGE(INDEX(ISNUMBER(–MID(B2,ROW($1:$25),1))ROW($1:$25),0),ROW($1:$25))+1,1)10^ROW($1:$25)/10)),”[<=9999999999]1+(000) 000-0000;#+(###) ###-####”) Under many circumstances, you would copy the formatted cells and past special (values) to the source and delete the formulas I’m going to leave you for now with this as my final solution. I look forward to any questions about added functionality.	{"url":"https://transoftlabs.com/blog/excel-formula-format-phone-numbers","timestamp":"2024-11-10T22:00:53Z","content_type":"text/html","content_length":"57520","record_id":"<urn:uuid:5aa220eb-6128-40c7-b525-5a87357cf85d>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00026.warc.gz"}
Coloring of Graphs with no Induced Six-Vertex path Date of Submission Institute Name (Publisher) Indian Statistical Institute Document Type Doctoral Thesis Degree Name Doctor of Philosophy Subject Name Computer Science Computer Science Unit (CSU-Chennai) Karthick, T. (CSU-Chennai) Abstract (Summary of the Work) Graph coloring is one among the oldest and broadly studied topics in graph theory. A coloring of a graph G is an assignment of colors to the vertices of G such that no two adjacent vertices receive the same color, and the chromatic number of G (denoted by χ(G)) is the minimum number of colors needed to color G. The clique number of G (denoted by ω(G)) is the maximum number of mutually adjacent vertices in G. In this thesis, we focus on some problems on bounding the chromatic number in terms of clique number for certain special classes of graphs with no long induced paths, namely the class of Pt-free graphs, for t ≥ 5. A hereditary class of graphs G is said to be χ-bounded if there exists a function f : N → N with f(1) = 1 and f(x) ≥ x, for all x ∈ N (called a χ-binding function for G) such that χ(G) ≤ f(ω(G)), for each G ∈ G. The smallest χ-binding function f∗ for G is defined as f∗(x) := max{χ(G) : G ∈ G and ω(G) = x}. The class G is called polynomially χ-bounded if it admits a polynomial χ-binding function. An intriguing open question is whether the class of Pt-free graphs is polynomially χ-bounded or not. This problem is open even for t = 5 and seems to be difficult. So researchers are interested in finding (smallest) polynomial χ-binding functions for some subclasses of Pt-free graphs. Here, we explore the structure of some classes of P6-free graphs and obtain (smallest/linear) χ-binding functions for such classes of graphs. Our results generalize/improve several previously known results available in the literature. Chapter 1 consists of a brief introduction on χ-bounded graphs and a short survey on known χ-bounded P6-free graphs. We also provide motivations, algorithmic issues, and relations of χ-boundedness to other well-known/related conjectures in graph theory. In Chapter 2, we study the class of (P2 + P3, P2 + P3)-free graphs, and show that the function f : N → N defined by f(1) = 1, f(2) = 4, and f(x) = max x + 3, 3x 2 − 1 , for x ≥ 3, is the smallest χ-binding function for the class of (P2 + P3, P2 + P3)-free graphs. In Chapter 3, we are interested in the structure of (P5, 4-wheel)-free graphs, and in coloring of such graphs. Indeed, we first prove that if G is a connected (P5, 4-wheel)-free graph, then either G admits a clique cut-set, or G is a perfect graph, or G is a quasi-line graph, or G has three disjoint stable sets whose union meets each maximum clique of G at least twice and the other maximal cliques of G at least once. Using this result, we prove that every (P5, 4-wheel)-free graph G satisfies χ (G) ≤ 3 2ω(G). We also provide infinitely many (P5, 4-wheel)-free graphs H with χ(H) ≥ 10 7 ω(H). It is known that every (P5,K4)-free graph G satisfies χ(G) ≤ 5, and that the bound is tight. Both the class of (P5, flag)-free graphs and the class of (P5, K5 − e)-free graphs generalize the class of (P5,K4)-free graphs. In Chapter 4, we explore the structure and coloring of (P5, K5 − e)-free graphs. In particular, we prove that if G is a connected (P5,K5 − e)-free graph with ω(G) ≥ 7, then either G is the complement of a bipartite graph or G has a clique cut-set. From this result, we show that if G is a (P5,K5 − e)-free graph with ω(G) ≥ 4, then χ(G) ≤ max{7, ω(G)}. Moreover, the bound is tight when ω(G) /∈ {4, 5, 6}. In Chapter 5, we investigate the coloring of (P5, flag)-free graphs. We prove that every (P5, flag,K5)- free graph G that contains a K4 satisfies χ(G) ≤ 8, every (P5, flag,K6)-free graph G satisfies χ(G) ≤ 8, and that every (P5, flag,K7)-free graph G satisfies χ(G) ≤ 9. Moreover, we prove that every (P5, flag)- free graph G with ω(G) ≥ 4 satisfies χ(G) ≤ max{8, 2ω(G) − 3}, and that the bound is tight for ω(G) ∈ {4, 5, 6} ProQuest Collection ID: https://www.proquest.com/pqdtlocal1010185/dissertations/fromDatabasesLayer?accountid=27563 DSpace Identifier Recommended Citation Char, Arnab Dr., "Coloring of Graphs with no Induced Six-Vertex path" (2024). Doctoral Theses. 479.	{"url":"https://digitalcommons.isical.ac.in/doctoral-theses/479/","timestamp":"2024-11-01T23:41:54Z","content_type":"text/html","content_length":"42350","record_id":"<urn:uuid:5323129f-e523-41ce-9eda-3bc423cdea1a>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00868.warc.gz"}
Let P=⎣⎡323−10−5−2α0⎦⎤, where α∈R, Suppose Q=[qij] is a m... \| Filo Let , where , Suppose is a matrix such that , where and is the identity matrix of order 3 . If and , then Not the question you're searching for? + Ask your question Since .det , so is invertible and cofactor of th element of are Was this solution helpful? Found 5 tutors discussing this question Discuss this question LIVE for FREE 13 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice more questions from Matrices View more Practice questions on similar concepts asked by Filo students View more Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text Let , where , Suppose is a matrix such that , where and is the identity matrix of order 3 . If and , then Updated On May 19, 2023 Topic Matrices Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 2 Upvotes 300 Avg. Video Duration 11 min	{"url":"https://askfilo.com/math-question-answers/let-p-left-begin-array-ccc-3-1-2-2-0-alpha-3-5-0-end-array","timestamp":"2024-11-12T13:54:06Z","content_type":"text/html","content_length":"535705","record_id":"<urn:uuid:3281a504-86c0-4fcc-9f89-7e35649aefee>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00488.warc.gz"}
Torque Optimization in AC Motors in context of ac motor torque 01 Sep 2024 Torque Optimization in AC Motors: A Review Abstract: This article reviews the concept of torque optimization in AC motors, highlighting the importance of maximizing torque while minimizing energy consumption. The article discusses the fundamental principles governing torque production in AC motors and presents various techniques for optimizing torque. Introduction: AC motors are widely used in industrial applications due to their high efficiency, reliability, and flexibility. However, the torque produced by an AC motor is a critical parameter that affects its performance and efficiency. Torque optimization is essential to ensure optimal operation of AC motors while minimizing energy consumption. Fundamental Principles: The torque produced by an AC motor can be calculated using the following formula: T = (P * N) / (2 * π * f) where T is the torque, P is the power, N is the number of poles, and f is the frequency. The torque produced by an AC motor is influenced by several factors, including the motor’s design parameters, operating conditions, and load characteristics. The most common method for optimizing torque in AC motors is to adjust the motor’s voltage and current. Voltage Optimization: The torque produced by an AC motor can be increased by increasing the voltage applied to the motor. However, this approach has limitations due to the motor’s thermal constraints and the risk of overheating. The optimal voltage for maximum torque production can be calculated using the following formula: V_opt = (T * R) / (2 * π * f) where V_opt is the optimal voltage, T is the desired torque, R is the motor’s resistance, and f is the frequency. Current Optimization: Another approach to optimize torque in AC motors is to adjust the current flowing through the motor. The optimal current for maximum torque production can be calculated using the following formula: I_opt = (T * N) / (2 * π * f) where I_opt is the optimal current, T is the desired torque, N is the number of poles, and f is the frequency. Load Characteristics: The load characteristics of an AC motor also play a crucial role in optimizing torque. The load can be classified into two categories: resistive and inductive loads. Resistive loads are characterized by a constant impedance, while inductive loads exhibit a varying impedance with frequency. Techniques for Torque Optimization: Several techniques have been proposed to optimize torque in AC motors, including: 1. Fuzzy Logic Control: This technique uses fuzzy logic rules to adjust the motor’s voltage and current based on the load characteristics. 2. Model Predictive Control: This technique uses a mathematical model of the motor to predict its behavior and adjust the voltage and current accordingly. 3. Sliding Mode Control: This technique uses a sliding mode controller to adjust the motor’s voltage and current based on the load characteristics. Conclusion: Torque optimization in AC motors is a critical aspect of ensuring optimal operation while minimizing energy consumption. The article has reviewed the fundamental principles governing torque production in AC motors and presented various techniques for optimizing torque. Further research is needed to develop more advanced control strategies that can effectively optimize torque in AC motors under varying load conditions. • [1] Bose, B. K. (2002). Power electronics and motor drives: Advances and trends. Academic Press. • [2] Leonhard, W. (1996). Control of electrical drives: Theory and practice. Springer. • [3] Mohan, N. (1989). Power electronics: Converters, power systems, and design. McGraw-Hill. Note: The formulas provided are in ASCII format as requested. However, please note that the article does not provide numerical examples or specific values for the variables involved. Related articles for ‘ac motor torque ‘ : • Reading: Torque Optimization in AC Motors in context of ac motor torque Calculators for ‘ac motor torque ‘	{"url":"https://blog.truegeometry.com/tutorials/education/f4e1d12495ea2df987f545c3ca9a80b0/JSON_TO_ARTCL_Torque_Optimization_in_AC_Motors_in_context_of_ac_motor_torque_.html","timestamp":"2024-11-13T22:07:00Z","content_type":"text/html","content_length":"17637","record_id":"<urn:uuid:f3db202c-98f0-47d8-9471-ed0f1d887498>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00525.warc.gz"}
Hyperdimensional Physics - 2 Part II Hubble's New "Runaway Planet" - A Unique Opportunity for Testing the Exploding Planet Hypothesis and Hyperdimensional Physics - Lt. Col Thomas E. Bearden, retired army officer and physicist, has been perhaps the most vocal recent proponent for restoring integrity to the scientific and historical record regarding James Clerk Maxwell -- by widely promulgating his original equations; in a series of meticulously documented papers on the subject, going back at least 20 years, Bearden has carried on a relentless one-man research effort regarding what Maxwell really claimed. His painstaking, literally thousands of man-hours of original source documentation has led directly to the following, startling conclusion: Maxwell’s original theory is, in fact, the true, so-called "Holy Grail" of physics ... the first successful unified field theory in the history of Science ... a fact apparently completely unknown to the current proponents of "Kaluza-Klein," "Supergravity," and "Superstring" ideas .... Just how successful, Bearden documents below: " ... In discarding the scalar component of the quaternion, Heaviside and Gibbs unwittingly discarded the unified EM/G [electromagnetic/ gravitational] portion of Maxwell’s theory that arises when the translation/directional components of two interacting quaternions reduce to zero, but the scalar resultant remains and infolds a deterministic, dynamic structure that is a function of oppositive directional/translational components. In the infolding of EM energy inside a scalar potential, a structured scalar potential results, almost precisely as later shown by Whittaker but unnoticed by the scientific community. The simple vector equations produced by Heaviside and Gibbs captured only that subset of Maxwell’s theory where EM and gravitation are mutually exclusive. In that subset, electromagnetic circuits and equipment will not ever, and cannot ever, produce gravitational or inertial effects in materials and equipment. "Brutally, not a single one of those Heaviside/Gibbs equations ever appeared in a paper or book by James Clerk Maxwell, even though the severely restricted Heaviside/Gibbs interpretation is universally and erroneously taught in all Western universities as Maxwell’s theory. "As a result of this artificial restriction of Maxwell’s theory, Einstein also inadvertently restricted his theory of general relativity, forever preventing the unification of electromagnetics and relativity. He also essentially prevented the present restricted general relativity from ever becoming an experimental, engineerable science on the laboratory bench, since a hidden internalized electromagnetics causing a deterministically structured local spacetime curvature was excluded. "Quantum mechanics used only the Heaviside/Gibbs externalized electromagnetics and completely missed Maxwell’s internalized and ordered electromagnetics enfolded inside a structured scalar potential. Accordingly, QM [quantum mechanics] maintained its Gibbs statistics of quantum change, which is nonchaotic a priori. Quantum physicists by and large excluded Bohm’s hidden variable theory, which conceivably could have offered the potential of engineering quantum change -- engineering physical reality itself. "Each of these major scientific disciplines missed and excluded a subset of their disciplinary area, because they did not have the scalar component of the quaternion to incorporate. Further, they completely missed the significance of the Whittaker approach, which already shows how to apply and engineer the very subsets they had excluded. "What now exists in these areas are three separate, inconsistent disciplines. Each of them unwittingly excluded a vital part of its discipline, which was the unified field part. Ironically, then, present physicists continue to exert great effort to find the missing key to unification of the three disciplines, but find it hopeless, because these special subsets are already contradictory to one another, as is quite well-known to foundations physicists. "Obviously, if one wishes to unify physics, one must add back the unintentionally excluded, unifying subsets to each discipline. Interestingly, all three needed subsets turn out to be one and the same ..." -- T.E. Bearden, "Possible Whittaker Unification of Electromagnetics, General Relativity, and Quantum Mechanics," (Association of Distinguished American Scientists 2311 Big Cove Road, Huntsville, Alabama, 35801) Given Bearden’s analysis -- what did we actually lose ... when science "inadvertently lost Maxwell ..?" If two key physics papers often cited by Bearden (which appeared decades after the death of Maxwell), are accurate ... we lost nothing less than... the "electrogravitic" control of gravity itself!! The critically-important research cited by Bearden was originally published by "Sir Edmund Whittaker" (the same cited earlier in this paper), beginning in 1903. □ the first was titled "On the partial differential equations of mathematical physics" (Mathematische Annalen, Vol. 57, 1903, p.333-335) □ the second, "On an Expression of the Electromagnetic Field due to Electrons by means of two Scalar Potential Functions" (Proceedings of the London Mathematical Society, Vol.1, 1904, p. Whittaker, a leading world-class physicist himself, single-handedly rediscovered the "missing" scalar components of Maxwell’s original quaternions, extending their (at the time) unseen implications for finally uniting "gravity" with the more obvious electrical and magnetic components known as "light." In the first paper, as Bearden described, Whittaker theoretically explored the existence of a "hidden" set of electromagnetic waves traveling in two simultaneous directions in the scalar potential of the vacuum -- demonstrating how to use them to curve the local and/or distant "spacetime" with electromagnetic radiation, in a manner directly analogous to Einstein’s later "mass-curves-space" equations. This key Whittaker paper thus lays the direct mathematical foundation for an electrogravitic theory/technology of gravity control. In the second paper, Whittaker demonstrated how two "Maxwellian scalar potentials of the vacuum" -- gravitationally curving spacetime -- could be turned back into a detectable "ordinary" electromagnetic field by two interfering "scalar EM waves"... even at a distance. Whittaker accomplished this by demonstrating mathematically that, "the field of force due to a gravitating body can be analyzed, by a spectrum analysis’ as it were, into an infinite number of constituent fields; and although the whole field of force does not vary with time, yet each of the constituent fields is an ondulatory character, consisting of a simple-disturbance propagated with uniform velocity ... [and] the waves will be longitudinal (top) ... These results assimilate the propagation of gravity to that of light ... [and] would require that gravity be propagated with a finite velocity, which however need not be the same as that of light [emphasis added], and may be enormously greater ..." (Op. Cit., "On the partial differential equations of mathematical physics") Remarkably, four years before Whittaker’s theoretical analysis of these potentials (pun intended ...), on the evening of July 3-4, 1899, Nikola Tesla (right) -- the literal inventor of modern civilization (via the now worldwide technology of "alternating current") -- experimentally anticipated "Whittaker’s interfering scalar waves" by finding them in nature; from massive experimental radio transmitters he had built on a mountain top in Colorado, he was broadcasting and receiving (by his own assertion) "longitudinal stresses" (as opposed to conventional EM "transverse waves") through the vacuum. This he was accomplishing with his own, hand-engineered equipment (produced according to Maxwell’s original, quaternion equations), when he detected an interference "return" from a passing line of thunderstorms. Tesla termed the phenomenon a "standing columnar wave," and tracked it electromagnetically for hours as the cold front moved across the West (Nikola Tesla, Colorado Springs Notes 1899-1900, Nolit, Beograd, Yugoslavia, 1978 pp. 61-62). [Many have since speculated that Tesla’s many other astonishing (to the period) technological accomplishments, many of which apparently "were lost" with his death in 1942, were based on this true understanding of Maxwell’s original, "hyperdimensional" electromagnetic ideas ...] Tesla’s experimental earlier detection notwithstanding, what Whittaker theoretically demonstrated years after Tesla was that future electrical engineers could also take Maxwell’s original 4-space, quaternion description of electromagnetic waves (the real "Maxwell’s Equations"), add his own (Whittaker’s) specific gravitational potential analysis (stemming from simply returning Maxwell’s scalar quaternions in Heaviside’s version of "Maxwell’s Equations"...), and produce a workable "unified field theory" (if not technology!) of gravity control ... unless by now, in some government "black project," they already have... And what we’ve deliberately been "leaked" over the last seven years, in repeated video images of "exotic vehicles" performing impossible, non-Newtonian maneuvers on official NASA TV shuttle coverage ... is simply the result! (click image right) Theory is one thing (Maxwell’s or Whittaker’s), but experimental results are supposedly the ultimate Arbiter of Scientific Truth. Which makes it all the more curious that Tesla’s four-year observational anticipation of Whittaker’s startling analysis of Maxwell -- the experimental confirmation of an electromagnetic "standing columnar (longitudinal) wave" in thunderstorms -- has been resolutely ignored by both physicists and electrical engineers for the past 100 years; as have the stunning NASA TV confirmations of "something" (above) maneuvering freely in Earth orbit. With that as prologue, a new generation of physicists, also educated in the grand assumption that "Heaviside’s Equations" are actually "Maxwell’s," were abruptly brought up short in 1959 with another remarkable, equally elegant experiment -- which finally demonstrated in the laboratory the stark reality of Maxwell’s "pesky scalar potentials" ... those same "mystical" potentials that Heaviside so effectively banished for all time from current (university-taught) EM theory. In that year two physicists, Yakir Aharonov and David Bohm, conducted a seminal "electrodynamics" laboratory experiment ("Significance of Electromagnetic Potentials in Quantum Theory," The Physical Review, Vol. 115, No. 3, pp. 485-491; August, 1959). Aharonov and Bohm, almost 100 years after Maxwell first predicted their existence, succeeded in actually measuring the "hidden potential" of free space, lurking in Maxwell’s original scalar quaternion equations. To do so, they had to cool the experiment to a mere 9 degrees above Absolute Zero, thus creating a total shielding around a superconducting magnetic ring [for a slightly different version of this same experiment (click image left); the oscillation of electrical resistance in the ring (bottom graph) is due to the changing electron "wave functions" -- triggered by the "hidden Maxwell scalar potential" created by the shielded magnet -- see text, below]. Once having successfully accomplished this non-trivial laboratory set up, they promptly observed an "impossible" phenomenon: Totally screened, by all measurements, from the magnetic influence of the ring itself, a test beam of electrons fired by Aharonov and Bohm at the superconducting "donut," nonetheless, changed their electronic state ("wave functions") as they passed through the observably "field-free" region of the hole -- indicating they were sensing "something," even though it could NOT be the ring’s magnetic field. Confirmed now by decades of other physicists’ experiments as a true phenomenon (and not merely improper shielding of the magnet), this "Aharonov-Bohm Effect" provides compelling proof of a deeper "spatial strain" -- a "scalar potential" -- underlying the existence of a so-called magnetic "force-field" itself. (Later experiments revealed a similar effect with shielded electrostatic fields ...) All of which provides compelling proof of "something else," underlying all reality, capable of transmitting energy and information across space and time ... even in the complete absence of an electromagnetically detectable 3-D spatial "field".... Maxwell’s quaternion ... hyperdimensional "potential." So, what does all this have to do with NASA’s announcement of a "new planet?" If a "potential" without a field can exist in space -- as Maxwell’s quaternion analysis first asserted, and Aharonov-Bohm "only" a century later ultimately found -- then, as defined by Maxwell in his comparisons of the aether with certain properties of laboratory "solids," such a potential is equivalent to an unseen, vorticular (rotating) "stress" in space. Or, in Maxwell’s own words (first written in 1873 ...): "There are physical quantities of another kind [in the aether] which are related to directions in space, but which are not vectors. Stresses and strains in solid bodies are examples, and so are some of the properties of bodies considered in the theory of elasticity and in the theory of double [rotated] refraction. Quantities of this class require for their definition nine [part of the "27-line"...] numerical specifications. They are expressed in the language of quaternions by linear and vector functions of a vector ..." -- J.C. Maxwell "A Treatise on Electricity and Magnetism" (Vol.1, 3rd Edition, New York, 1954) And stresses, when they are relieved, must release energy into their surroundings ... There is now much fevered discussion among physicists, (~100 years post-Maxwell) of the Quantum Electrodynamics Zero Point Energy (ZPE) of space -- or, "the energy of the vacuum"; to many familiar with the original works of Maxwell, Kelvin, et. al., this sounds an awful lot like the once-familiar "aether" ... merely updated and now passing under "an assumed name." Thus, creating -- then relieving -- a "stress" in Maxwell’s vorticular aether is precisely equivalent to tapping the "energy of the vacuum" -- which, according to current "quantum mechanics’ models," possesses a staggering amount of such energy per cubic inch of space. Even inefficiently releasing a tiny percentage of this "strain energy" into our three dimensions -- or, into a body existing in three-dimensional space -- could make it appear as if the energy was coming from nowhere ... "something from nothing." In other words, to an entire generation of students and astrophysicists woefully ignorant of Maxwell’s real equations, such energy would appear "Perpetual motion!" Given the prodigious amount of "vacuum energy" calculated by modern physicists (trillions of atomic bomb equivalents per cubic centimeter ...), even a relatively minor but sudden release of such vast vacuum (aether) stress potential inside a planet ... could literally destroy it... Finally answering the crucial astrophysical objection to the "exploded planet model" that Van Flandern has been encountering ... "But Tom -- just how do you blow up’ an entire world?!" The answer is now obvious: via hyperdimensional "vacuum stress energy" ... ala Whittaker and Maxwell. As we shall show, it is this "new" source of energy -- in a far more "controlled" context -- that seems also to be responsible now for not only the "anomalous infrared excesses" observed in the so-called "giant outer planets" of this solar system... it is this same source of energy (in the Hyperdimensional Physics Model) that, according to our analysis, must now be primarily responsible for the radiated energies of stars ... including the Sun itself. Since, in three dimensions, all energy eventually "degrades" to random motions -- via Kelvin and Gibb’s 19^th Century Laws of Thermodynamics (it’s called "increasing entropy") -- "stress energy" of the aether (vacuum) released inside a material object, even if it initially appears in a coherent form -- driving, for instance, the anomalous (1400 mile-per-hour!), planet-girdling winds of distant Neptune’s "jet streams" -- will eventually degrade to simple, random heat ... ultimately radiated away as "excess infrared emissions" into space. It’s the initial, astrophysical conditions under which such "Maxwellian space potentials" can be released inside a planet (or a star ...), that have been the central focus of our efforts for ten years... to create a predictive, mathematical "hyperdimensional model" of such physics. The entire question comes down to: "What set of known spatial conditions will slowly, predictably, release the potential strains of 4-space into 3-space’ ... inside a massive world ... so that when this energy inevitably degrades to heat, its radiative signature identifies the original hyperdimensional’ source?" Fortunately, we are surrounded by almost half a dozen examples close at hand: the giant, "anomalously radiating" planets of this solar system (and some major moons). Over the past decade, as we have attempted to understand their anomalous IR radiation, one thing has become clear -- to a first order, the "infrared excesses" of the giant planets all seem to correlate very nicely with one parameter each has in common -- regardless of their individual masses, elemental compositions, or distance from the Sun: their total system "angular momentum." The mass of a body and the rate at which it spins, in classical physics, determines an object’s "angular momentum." In our Hyperdimensional Model, its a bit more complicated -- because objects apparently separated by distance in this (3-space) dimension are in fact connected in a "higher" (4-space) dimension; so, in the HD model, one also adds in the orbital momentum of an object’s gravitationally-tethered satellites -- moons in the case of planets; planets, in the case of the Sun, or companion stars in the case of other stars. When one graphs the total angular momentum of a set of objects -- such as the radiating outer planets of this solar system (plus Earth and Sun) -- against the total amount of internal energy each object radiates to space, the results are striking (click image right): The more total system angular momentum a planet (or any celestial body) possesses (as defined above -- object plus satellites), the greater its intrinsic "brightness," i.e. the more "anomalous energy" it apparently is capable of "generating." And, as can be seen from this key diagram (image right), this striking linear dependence now seems to hold across a range of luminosity and momentum totaling almost three orders of magnitude ... almost 1000/1! Especially noteworthy, the Earth (not "a collapsing gas giant," by any stretch of the imagination) also seems to fit precisely this empirical energy relationship: when the angular momentum of the Moon is added to the "spin momentum" of its parent planet, the resulting correlation with measurements derived from internal "heat budget" studies of the Earth are perfectly fitted to this solar-system-wide empirical relationship -- even though the Earth’s internal energy is supposedly derived from "radioactive sources." And, as can be seen from the accompanying historical comparison (click image left), this striking solar system linear relationship is actually more tightly constrained (even at this early stage) than the original Hubble "redshift data" supporting the Big Bang! This discovery contains major implications, not only for past geophysics and terrestrial evolution ... but for future geological and climatological events -- "Earth changes," as some have termed them. These may be driven, not by rising solar interactions or by-products of terrestrial civilization (accumulating "greenhouse gases" from burning fossil fuels), but by this same "hyperdimensional physics." If so, then learning a lot more about the mechanisms of this physics -- and quickly! -- is a critical step toward intervening and eventually controlling our future well-being, if not our destiny, on (and off!) this planet ... For the "Hyperdimensional Physics" model, this simple but powerful relationship now seems to be the equivalent of Relativity’s E=MC^2 : a celestial object’s total internal luminosity seems dependent upon only one physical parameter: L=mr^2 = total system angular momentum (object, plus all satellites) There is a well-known "rule of thumb" in science, perhaps best expressed by a late Noble Laureate, physicist Richard Feynman: "You can recognize truth by its beauty and simplicity. When you get it right, it is obvious that it is right -- at least if you have any experience -- because usually what happens is that more comes out than goes in ... The inexperienced, the crackpots, and people like that, make guesses that are simple, but you can immediately see that they are wrong, so that does not count. Others, the inexperienced students, make guesses that are very complicated, and it sort of looks as if it is all right, but I know it is not true because the truth always turns out to be simpler that you thought ..." This startling relationship -- our discovery of the simple dependence of an object’s internal luminosity on its total system angular momentum -- has that "feel" about it; it is simple ... it is elegant ... in fact... it could even be true. But, as can be seen from examining the luminosity/angular momentum diagram again, there also appears to be one glaring exception to this otherwise strikingly linear relationship: The Sun itself. Independent research, involving over 30 years of attempted confirmation of the Sun’s basic energy source -- in the form of solar/terrestrial observations of tiny atomic particles called "neutrinos," supposedly coming from the center of the Sun (above image) -- have left laboratory physicists and astrophysicists with a major astronomical enigma: The Sun is not emitting anything like the number of neutrinos required by the "Standard Solar Model" for its observed energy emission; if its energy is due to "thermo-nuclear reactions" (as the Standard Model demands), then the observed "neutrino deficit" is upwards of 60%: even more remarkable, certain kinds of primary neutrinos (calculated as required to explain the bulk of the solar interior’s fusion reactions, based on laboratory measurements) turn out to be simply missing altogether! So -- what really fuels the Sun? The answer to the Sun’s apparent violation of the Standard Solar Model -- ironically, is contained in its striking "violation" of our key angular momentum/luminosity diagram (click image right): In the Hyperdimensional Model, the Sun’s primary energy source -- like the planets’ -- must be driven by its total angular momentum -- its own "spin momentum," plus the total angular momentum of the planetary masses orbiting around it. Any standard astronomical text reveals that, though the Sun contains more than 98% of the mass of the solar system, it contains less than 2% of its total angular momentum. The rest is in the planets. Thus, in adding up their total contribution to the Sun’s angular momentum budget -- if the HD model is correct -- we should see the Sun following the same line on the graph that the planets, from Earth to Neptune, do. It doesn’t. The obvious answer to this dilemma is that the HD model is simply wrong. The less obvious is that we’re missing something ... Like ... additional planets (above)! By adding another big planet (or a couple of smaller ones) beyond Pluto (several hundred times the Earth’s distance from the Sun -- below), we can move the Sun’s total angular momentum to the right on the graph (click above right image), until it almost intersects the line (allowing for a percentage, about 30%, of internal energy expected from genuine thermonuclear reactions ...). This creates the specific "HD prediction" that "the current textbook tally of the Sun’s angular momentum is deficient because ..." We haven’t discovered all the remaining members of the solar system yet! As a dividend, this promptly presents us with our first key test of the Hyperdimensional Model: 1) Find those planets! The second test of the Hyperdimensional Model is that, unlike other efforts to explain anomalous planetary energy emissions via continued "planetary collapse," or "stored primordial heat," the hyperdimensional approach specifically predicts one radical, definitive observational difference from all other existing explanations 2) HD energy generation in both planets and stars should be -- must be -- variable. This is simply implicit in the mechanism which generates the hyperdimensional energy in the first place: ever changing hyperspatial geometry. If the ultimate source of planetary (or stellar) energy is this "vorticular (rotating) spatial stress between dimensions" (ala Maxwell), then the constantly changing pattern (both gravitationally and dimensionally) of interacting satellites in orbit around a major planet/star must modulate that stress pattern as a constantly changing, geometrically twisted "aether" (ala Whittaker’s amplifications of Maxwell). In our Hyperdimensional Model, it is this "constantly changing hyperspatial geometry" that is capable (via resonant rotations with the masses in question -- either as spin, or circular orbital motions) of extracting energy from this underlying "rotating, vorticular aether" ... and then releasing it inside material objects. Initially, this "excess energy" can appear in many different forms -- high-speed winds, unusual electrical activity, even enhanced nuclear reactions -- but, ultimately, it must all degrade to simple "excess heat." Because of the basic physical requirement for resonance in effectively coupling a planet (or a star’s) "rotating 3-D mass to the underlying 4-D aether rotation," this excess energy generation must also, inevitably, vary with time -- as the changing orbital geometry of the "satellites" interacts with the spinning primary (and the underlying, "vorticular aether"...) in and For these reasons, as stated earlier, time-variability of this continuing energy exchange must be a central hallmark of this entire "HD process." [Incidentally, understanding this basic "hyperdimensional transfer mechanism," in terms of Maxwell’s original quaternions (that describe "a rotating, vorticular, four-dimensional sponge-like aether"), immediately lends itself to creating a "Hyperdimensional Technology" based on this same mechanism. The fundamental "violations" of current physics exhibited by so-called "free energy" machines -- from the explicitly-rotating "N-machine" (click above left image) to the initially frustrating time-variable aspects of "electro-chemical cold fusion"-- are now elegantly explained by appropriate application of Maxwell’s original ideas. Even more extraordinary: the recent startling demonstration, broadcast nationwide on ABC’s "Good Morning America" last year (click below image), of a "physically impossible" major reduction -- in a few minutes! -- of long-lived radioactive Uranium isotopes. Normally, such processes require billions of years to accomplish. This too is now elegantly explained by the Hyperdimensional Model-- As -- an "induced hyperspatial stress," created by the machine ... the same stress that initially (in the Model) induces "unstable isotopes" in the first place. By technologically enhancing such vacuum stress within these nuclei, via a retuning of Maxwell’s "scalar potentials," the normal radioactive breakdown process is accelerated -- literally billions of times ... The implications for an entire "rapid, radioactive nuclear waste reduction technology" -- accomplishing in hours what would normally require aeons -- is merely one immediate, desperately needed world-wide application of such "Hyperdimensional Technologies."] In our own planetary system, all the "giant" planets possess a retinue of at least a dozen satellites: one or two major ones (approximating the size of the planet Mercury) ... with several others ranging down below the diameter and mass of our own Moon ... in addition to a host of smaller objects; because of the "lever effect" in the angular momentum calculations, even a small satellite orbiting far away (or at a steep angle to the planet’s plane of rotation) can exert a disproportional effect on the "total angular momentum" equation -- just look at Pluto and the Sun. Even now, Jupiter’s four major satellites (which have collective masses approximately 1/10,000th of Jupiter itself), during the course of their complex orbital interactions, are historically known to cause time-altered behavior in a variety of well-known Jovian phenomena..., i ncluding -- "anomalous" latitude and longitude motions of the Great Red Spot itself. As we presented at the U.N. in 1992, the Great Red Spot -- a mysterious vortex located for over 300 years at that "infamous" 19.5 degrees S. Latitude (click right image), via the circumscribed tetrahedral geometry of the equally infamous "27 line problem" -- is the classic "hyperdimensional signature" of HD physics operating within Jupiter. The existence of decades of recorded "anomalous motions" of this Spot, neatly synchronized with the highly predictable motions of Jupiter’s own moons, are clearly NOT the result of conventional "gravitational" or "tidal" interactions -- in view of the relatively insignificant masses of the moons compared to Jupiter itself; but, following Maxwell and Whittaker, the hyperdimensional effects of these same moons -- via the long "lever" of angular momentum on the constantly changing, vorticular scalar stress potentials inside Jupiter -- that is a very different story ... So, Hyperdimensional Test number three: 3) Look for small, short-term amplitude-variations in the infrared emission levels of all the giant planets ... synchronized (as are the still-mysterious motions of the GRS on Jupiter) with the orbital motions and conjunctions of their moons. All NASA models for the "anomalous energy emissions" of these planets have assumed a steady output; the "snapshot" values derived from the mere few hours of Voyager fly-bys in the 1980’s are now firmly listed in astronomy texts as new "planetary constants"; the reason: the emissions are viewed by NASA as either "primordial heat," stored across the aeons; energy release from internal long-term radioactive processes; or literal, slight settling of portions of the entire planet, still releasing gravitational potential energy ... all processes that will not change perceptibly even in thousands of years! Confirmed short-term variations in the current planetary IR (infra red) outputs, of "a few hours" (or even a few days) duration -- and synchronized with the orbital periods of the planets’ satellites themselves -- would thus be stunning evidence that all the "mainstream" explanations are in trouble ... and that the Hyperdimensional Model deserves much closer scrutiny ... In this same vein: unlike all "conventional NASA explanations," in a phenomenon akin to "hyperdimensional astrology," the HD model also specifically predicts significantly larger, long-term variability in these major planetary IR outputs ... of several years duration. These (like the shorter variations triggered by the changing geometry between the satellites) should be caused by the constantly changing hyperdimensional (spatial stress) interactions between the major planets themselves ... as they continually change their geometry relative to one another, each orbiting the Sun with a different relative velocity. These changing interactive stresses in the "boundary between hyperspace and real’ space" (in the Hyperdimensional Model) now also seem to be the answer to the mysterious "storms" that, from time to time, have suddenly appeared in the atmospheres of several of the outer planets. The virtual "disappearance," in the late 80’s, of Jupiter’s Great Red Spot is one remarkable example; Saturn’s abrupt production of a major planetary "event," photographed by the Hubble Space Telescope in 1994 (above image) as a brilliant cloud erupting at 19.5 degrees N. (where else?!), is yet another. Since the prevailing NASA view is that these planets’ "excess" IR output must be constant over time, no one has bothered to look for any further correlations -- between a rising or falling internal energy emission ... and the (now, historically well-documented) semi-periodic eruptions of such "storms." They should. Go Back	{"url":"https://www.bibliotecapleyades.net/ciencia/ciencia_hyperphysics2.htm","timestamp":"2024-11-03T08:57:51Z","content_type":"text/html","content_length":"56027","record_id":"<urn:uuid:c740151d-a18d-4efc-9c6a-64247558d71b>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00042.warc.gz"}
Practice Ordinal Numbers Worksheet - OrdinalNumbers.com Practice Ordinal Numbers Worksheet – By using ordinal numbers, you can count any number of sets. They also can be used to generalize ordinal numbers. 1st The ordinal numbers are among the basic ideas in math. It is a number that indicates the place of an object within a list. An ordinal number is usually … Read more Practice Ordinal Numbers Practice Ordinal Numbers – A limitless number of sets can easily be enumerated with ordinal numbers as a tool. They can also be used to generalize ordinal quantities. 1st The basic concept of mathematics is the ordinal. It is a number that indicates where an object is in a list. The ordinal number is identified … Read more	{"url":"https://www.ordinalnumbers.com/tag/practice-ordinal-numbers-worksheet/","timestamp":"2024-11-04T19:58:49Z","content_type":"text/html","content_length":"52433","record_id":"<urn:uuid:91df7325-f5c3-438e-830b-b6e05d2fa221>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00889.warc.gz"}
EXPM1(3P) POSIX Programmer's Manual EXPM1(3P) This manual page is part of the POSIX Programmer's Manual. The Linux implementation of this interface may differ (consult the corresponding Linux manual page for details of Linux behavior), or the interface may not be implemented on Linux. expm1, expm1f, expm1l — compute exponential functions #include <math.h> double expm1(double x); float expm1f(float x); long double expm1l(long double x); The functionality described on this reference page is aligned with the ISO C standard. Any conflict between the requirements described here and the ISO C standard is unintentional. This volume of POSIX.1‐2017 defers to the ISO C standard. These functions shall compute ex-1.0. An application wishing to check for error situations should set errno to zero and call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if errno is non-zero or fetestexcept (FE_INVALID \| FE_DIVBYZERO \| FE_OVERFLOW \| FE_UNDERFLOW) is non-zero, an error has occurred. Upon successful completion, these functions return ex-1.0. If the correct value would cause overflow, a range error shall occur and expm1(), expm1f(), and expm1l() shall return the value of the macro HUGE_VAL, HUGE_VALF, and HUGE_VALL, respectively. If x is NaN, a NaN shall be returned. If x is ±0, ±0 shall be returned. If x is -Inf, -1 shall be returned. If x is +Inf, x shall be returned. If x is subnormal, a range error may occur and x should be returned. If x is not returned, expm1(), expm1f(), and expm1l() shall return an implementation-defined value no greater in magnitude than DBL_MIN, FLT_MIN, and LDBL_MIN, respectively. These functions shall fail if: The result overflows. If the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [ERANGE]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the overflow floating-point exception shall be raised. These functions may fail if: The value of x is subnormal. If the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [ERANGE]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the underflow floating-point exception shall be raised. The following sections are informative. The value of expm1(x) may be more accurate than exp(x)-1.0 for small values of x. The expm1() and log1p() functions are useful for financial calculations of ((1+x)n-1)/x, namely: when x is very small (for example, when calculating small daily interest rates). These functions also simplify writing accurate inverse hyperbolic functions. On error, the expressions (math_errhandling & MATH_ERRNO) and (math_errhandling & MATH_ERREXCEPT) are independent of each other, but at least one of them must be non-zero. SEE ALSO¶ exp(), feclearexcept(), fetestexcept(), ilogb(), log1p() The Base Definitions volume of POSIX.1‐2017, Section 4.20, Treatment of Error Conditions for Mathematical Functions, <math.h> Portions of this text are reprinted and reproduced in electronic form from IEEE Std 1003.1-2017, Standard for Information Technology -- Portable Operating System Interface (POSIX), The Open Group Base Specifications Issue 7, 2018 Edition, Copyright (C) 2018 by the Institute of Electrical and Electronics Engineers, Inc and The Open Group. In the event of any discrepancy between this version and the original IEEE and The Open Group Standard, the original IEEE and The Open Group Standard is the referee document. The original Standard can be obtained online at http://www.opengroup.org/unix /online.html . Any typographical or formatting errors that appear in this page are most likely to have been introduced during the conversion of the source files to man page format. To report such errors, see https: //www.kernel.org/doc/man-pages/reporting_bugs.html .	{"url":"https://manpages.opensuse.org/Tumbleweed/man-pages-posix/expm1.3p.en.html","timestamp":"2024-11-06T11:12:52Z","content_type":"text/html","content_length":"24092","record_id":"<urn:uuid:167d2621-c9e4-4a64-9e07-f2ae6065731b>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00876.warc.gz"}
Mesh representation of extended object Since R2020b The extendedObjectMesh represents the 3-D geometry of an object. The 3-D geometry is represented by faces and vertices. Use these object meshes to specify the geometry of a Platform for simulating lidar sensor data using monostaticLidarSensor. mesh = extendedObjectMesh('cuboid') returns an extendedObjectMesh object, that defines a cuboid with unit dimensions. The origin of the cuboid is located at its geometric center. mesh = extendedObjectMesh('cylinder') returns a hollow cylinder mesh with unit dimensions. The cylinder mesh has 20 equally spaced vertices around its circumference. The origin of the cylinder is located at its geometric center. The height is aligned with the z-axis. mesh = extendedObjectMesh('cylinder',n) returns a cylinder mesh with n equally spaced vertices around its circumference. mesh = extendedObjectMesh('sphere') returns a sphere mesh with unit dimensions. The sphere mesh has 119 vertices and 180 faces. The origin of the sphere is located at its center. mesh = extendedObjectMesh('sphere',n) additionally allows you to specify the resolution, n, of the spherical mesh. The sphere mesh has (n + 1)^2 - 2 vertices and 2n(n - 1) faces. mesh = extendedObjectMesh(vertices,faces) returns a mesh from faces and vertices. vertices and faces set the Vertices and Faces properties respectively. Vertices — Vertices of defined object N-by-3 matrix of real scalar Vertices of the defined object, specified as an N-by-3 matrix of real scalars. N is the number of vertices. The first, second, and third element of each row represents the x-, y-, and z-position of each vertex, respectively. Faces — Faces of defined object M-by-3 matrix of positive integer Faces of the defined object, specified as a M-by-3 array of positive integers. M is the number of faces. The three elements in each row are the vertex IDs of the three vertices forming the triangle face. The ID of the vertex is its corresponding row number specified in the Vertices property. Object Functions Use the object functions to develop new meshes. translate Translate mesh along coordinate axes rotate Rotate mesh about coordinate axes scale Scale mesh in each dimension applyTransform Apply forward transformation to mesh vertices join Join two object meshes scaleToFit Auto-scale object mesh to match specified cuboid dimensions show Display the mesh as a patch on the current axes Create and Translate Cuboid Mesh Create an extendedObjectMesh object and translate the object. Construct a cuboid mesh. mesh = extendedObjectMesh('cuboid'); Translate the mesh by 5 units along the negative y axis. mesh = translate(mesh,[0 -5 0]); Visualize the mesh. ax = show(mesh); ax.YLim = [-6 0]; Create and Visualize Cylinder Mesh Create an extendedObjectMesh object and visualize the object. Construct a cylinder mesh. mesh = extendedObjectMesh('cylinder'); Visualize the mesh. Create and Auto-Scale Sphere Mesh Create an extendedObjectMesh object and auto-scale the object to the required dimensions. Construct a sphere mesh of unit dimensions. sph = extendedObjectMesh('sphere'); Auto-scale the mesh to the dimensions in dims. dims = struct('Length',5,'Width',10,'Height',3,'OriginOffset',[0 0 -3]); sph = scaleToFit(sph,dims); Visualize the mesh. Version History Introduced in R2020b	{"url":"https://se.mathworks.com/help/fusion/ref/extendedobjectmesh.html","timestamp":"2024-11-10T02:58:42Z","content_type":"text/html","content_length":"94024","record_id":"<urn:uuid:3191ffff-1827-470c-83cf-9d00ea657429>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00277.warc.gz"}
Column Major Order in Data Structure with Example - Quescol Column Major Order in Data Structure with Example Column Major Order is a way to represent the multidimensional array in sequential memory. It has similar functionality as row-major order, but the way of process is different. In Column Major Order, elements of a multidimensional array are arranged sequentially column-wise which means filling all the index of the first column and then move to the next column. Let’s see an example Suppose we have some elements {1,2,3,4,5,6,7,8} which we want to insert in an array by following column-major order. So If we insert these elements in column-major order, our 2-D array will look like Here first fill index[0][0] and then index[1][0] which is just opposite to the row-major where we first fill all row nodes then move to the next row. And then for second-column index[0][1] and then index[1][1], index[0][2], index[1][2] and so on. Why Use Column Major Order? This order is crucial in programming and data structure because it affects how quickly and efficiently a computer can access and process data. In column major order, data that appears in the same column are stored close together in the computer’s memory. This closeness can speed up tasks that need to access this column data frequently. Examples in Real Life Imagine a classroom seating chart where students are listed by columns, starting from the front to the back, and then moving to the next column. If a teacher wants to call on students column by column for a quiz, this order makes it easy to see who’s next. Column Major Order Formula The Location of element A[i, j] can be obtained by evaluating expression: LOC (A [i, j]) = base_address + w * [m * j + i] base_address = address of the first element in the array. w(Size of element) = Word size means a number of bytes occupied by each element of an Array. m = Number of rows in the array. i = is the row index. j = is the column index. Note: Array Index starts from 0. Here we have taken a 2-D array A [2, 4] which has 2 rows and 4 columns. Problem to solve on column major Suppose we want to calculate the address of element A [1, 2] in column-major order and the matrix is 24. It can be calculated as follow: Now to calculate the base address of any index using column-major order we can use the process given below. It can be calculated as follow: base_address = 2000, W= 2, M=2, i=1, j=2 LOC (A [i, j]) = base_address + W [M j + i ] LOC (A[1, 2]) = 2000 + 2 [22 + 1] = 2000 + 2 [4 + 1] = 2000 + 2 * 5 = 2000 + 10 = 2010 Benefits in Math and Science Column major order is particularly beneficial in fields like mathematics and scientific computing. Many mathematical operations, like matrix multiplication or solving systems of linear equations, can be optimized if the data is stored in this way. This is because accessing column data repeatedly is faster due to the way computer memory works. Comparison with Row Major Order It’s helpful to compare column major order with row major order, where data is filled row by row. Row major order is like reading a book from left to right, then top to bottom. Most programming languages, like C and Python, use row major order for arrays. The choice between row and column major order depends on how the data will be accessed and used. For some applications, column major order is more efficient, especially in matrix computations. Choosing the Right Order The decision to use column major order over row major order should be based on the specific needs of your project. If you’re working with matrices and find yourself accessing columns more often than rows, column major order could be the way to go. This choice can lead to faster, more efficient code, particularly in applications that require heavy mathematical computation. Column major order is a powerful concept in data structure and programming, especially useful in mathematical and scientific computing. By storing data in a way that matches how it’s accessed, programs can run faster and more efficiently. Whether you’re a student, programmer, or scientist, understanding and utilizing column major order can significantly impact your work’s effectiveness and	{"url":"https://quescol.com/data-structure/column-major-order","timestamp":"2024-11-05T16:51:12Z","content_type":"text/html","content_length":"84596","record_id":"<urn:uuid:5bdd5120-1538-4c6b-ab4a-a9b6fd134f24>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00116.warc.gz"}
Numerical integration over a Green's function • Mathematica • Thread starter member 428835 • Start date In summary, the conversation discusses the issue of numerical integration over a Green's function with odd functions. One approach suggested is using a Monte Carlo Integration method, which involves generating random points and testing if they fall within the region of integration. This method is simple to implement and can handle higher dimensional integrals, but may miss localized features if the region of integration is small relative to the rectangular envelope. Hi PF! I'm numerically integrating over a Green's function along with a few very odd functions. What I have looks like this NIntegrate[-(1/((-1.` + x)^2 (1.` + x)^2 (1.` + y)^2)) 3.9787262092516675`^14 (3.9999999999999907` + x (-14.99999999999903` + x (20.00000000000097` - 9.999999999999515` x + 1.` x^3))) (-1.` + y)^2 (4.` + y) BesselJ[4, 304.6888201459785` Sqrt[1 - x^2]] BesselJ[4, 304.6888201459785` Sqrt[1 - y^2]] Cosh[ 310.00637327206255` - 304.6888201459785` x] Cosh[ 310.00637327206255` - 304.6888201459785` y] /. {x -> xx, y -> yy}, {xx, yy} \[Element] Cos[Cos[\[Alpha]] Cos[\[Pi]/180]] < xx < yy < 1, {xx, yy}]] but Mathematica throws a "numerical integration converging too slowly" error. How would you treat this? I can send you the full notebook if you're interested, though that would have to be private. Thanks so much! EDIT: For completeness the following technique worked in Mathematica and agrees with MCI method outlined in python below, though Mathematica is much much faster: redefine ##G## as a piecewise function, since after all it is (my technique above was to split the integration of ##G## over two separate domaines, one where ##x>y## and one where ##y>x##). Then run NIntegrate over the full square domain. However, under Method specify LocalAdaptive. After much research I decided to use this technique, and no more errors or holdups, all is well. Last edited by a moderator: If I were you I would code my own numerical integration. You could for example attempt Monte Carlo Integration. This involves generating random points uniformly distributed over a square region (with area A) containing your domain of integration. For each point, test if it's in the region of integration and then evaluate your integrand function, then add that value to your cumulative sum (and increment your counter N for every randomly generated case, not just the ones in your domain of integration!). jambaugh said: If I were you I would code my own numerical integration. You could for example attempt Monte Carlo Integration. This involves generating random points uniformly distributed over a square region (with area A) containing your domain of integration. For each point, test if it's in the region of integration and then evaluate your integrand function, then add that value to your cumulative sum (and increment your counter N for every randomly generated case, not just the ones in your domain of integration!). Can you explain the advantage to this approach over what NIntegrate currently does? Also, notice the region with which I am integrating: it's essentially $$\int_{\cos(\cos \pi/180)\sin(\pi/180)}^1\ int_{\cos(\cos \pi/180)\sin(\pi/180)}^y f(x,y)\, dxdy$$ which I don't know how to perform numerically since a variable is in the integration limit. Also, the integrand when plotted over the integration region ##[\cos(\cos \pi/180)\sin(\pi/180),1]\times [\cos(\cos \pi/180)\sin(\pi/180),1]## looks very well behaved as you can see: I was thinking about how to implement so, I spent about 30 minutes and wrote a quick example python script. Code included below. Of course, I used a much simpler integration problem but I think you can adapt the code easily enough. I later thought of an adaptation that would calculate the standard deviation of the estimator (sumArea/N). I might fiddle around with it and pretty it up and will upload it here if I do. (Basically, your sumArea/N = E is your estimator. You can likewise estimate its standard deviation and construct a confidence interval of whatever confidence level and precision by letting it run until that is reached. I'm sure there are plenty of elegant packages out there which would handle it neatly.) I can't speak to the advantages over NIntegrate because I simply don't know what it does. I can't speak to why Wolfram Alpha gave you the message it did. As to the advantages or disadvantages of MCI, one advantage is that it is simple to implement, and it can reach lower precision results rather quickly. It also will handle higher dimensional integrals without much extra coding or resources. (I believe it is quite efficient memory-wise.) As with all mean type estimators, its standard error will decrease in proportion ##\frac{1}{\sqrt{N}}# #. I am not an expert and defer to those who are or what you find in searches on the topic. It also, due to the use of random sampling points, will avoid certain biases which might occur due to the regularity of the traditional meshes of direct numerical integration. Contrawise, if there are very localized features it could easily miss them until it's taken a very dense sample. One point, the smaller your region of integration is relative to the rectangular envelope, the less efficient since you're region test is rejecting so many cases. Another point is that you can adapt it to use non-uniform sampling (offset with bias in the weights of the samples inversely proportional to their probability density.) This will allow you to, for example, do a MCI over an unbounded region or increase the density near pathological features. But I don't know how easy the error analysis would be. Sorry if I'm rambling. Your question got me thinking about this all day. I think I might offer a project like this to students in my department for an undergraduate capstone. Python Code Below: (It went quickly even with this slow interpreter. A c language implementation would fly.) [Edit:] Note that the # symbol begins a comment on that line in Python. from math import from random import * # Monte Carlo method, integrating z>0 hemisphere of radius 2 over the unit circle in the x,y plane. def f(pt): # The integrand function, here z = sqrt( 4-x^2 - y^2) [python uses for powers like fortran not ^ like c and most other lang.] x= pt[0] y= pt[1] return (4.0 - x2-y2)0.5 def testRegion(pt): return (pt[0]2 + pt[1]2 < 1.0) def genpoint(): # generating coordinates uniformly in the square -1<x<1, -1<y<1 (area = 4). x = 2.0random()-1.0 y = 2.0random()-1.0 return (x,y) #Intitialize Sum = 0.0 Area = 4.0 N = 0 #Main Loop. while(True): for i in range(10000): # Number of iterations before printing current result. N+=1 pt = genpoint() if testRegion(pt): Sum += f(pt) print(SumArea/N) Python code works much better posted like this: from math import from random import * # Monte Carlo method, integrating z>0 hemisphere of radius 2 over the unit circle in the x,y plane. def f(pt): # The integrand function, here z = sqrt( 4-x^2 - y^2) [python uses for powers like fortran not ^ like c and most other lang.] x= pt[0] y= pt[1] return (4.0 - x2-y2)0.5 def testRegion(pt): return (pt[0]2 + pt[1]2 < 1.0) def genpoint(): # generating coordinates uniformly in the square -1<x<1, -1<y<1 (area = 4). x = 2.0random()-1.0 y = 2.0random()-1.0 return (x,y) Sum = 0.0 Area = 4.0 N = 0 #Main Loop. for i in range(10000): # Number of iterations before printing current result. pt = genpoint() if testRegion(pt): Sum += f(pt) But in any case if you want to try MC then you can do this with NIntegrate simply by supplying Method -> "MonteCarlo" as an argument. You could also try some more sophisticated sampling-based algorithms Don't be surprised if this does not solve the convergence problem though, depending on what tolerance you are looking for. What are all those constants like 3.9999999999999907 for? This looks like it should be 4 but includes an error term due to some previous step using a numerical method. Edit: and where has ## \cos \left(\cos \left(\frac \pi {180}\right)\right) ## come from? Whilst it is possible to calculate this number it cannot have any physical or even mathematical significance. Wow, thank you both for the replies! Never used MCI before, but I'm checking it out. Very cool idea, and nice clear python script you wrote. pbuk: I copy-pasted Mathematica output without really reading it. I didn't want to mess anything up by rounding so I thought this was easier. Those numbers stem from previous numerical calculations. And the cosine terms you ask about are a typo on my part: the integration limits in post 3 should read from ##[\sin\alpha,1]\times[\sin\alpha,1]## Also, regarding Monte Carlo integration, is it preferred to compute 100 integrals each with 100 samples, or 1 integral with 10,000 samples? Seems like it should be the same, right? pbuk said: Python code works much better posted like this: [...] Thanks! I should have spent the time to look this up. The key-word highlighting makes it so much better. The error message mentioned by the OP makes me wonder if it actually was trying to use MCI. It would explain the "converging too slowly" type message since the algorithm, if MC would only detect the rate of convergence empirically given this method. The question then is can you, using [] , change the loop limits, i.e. tell it to keep going further in trying to detect convergence? (As you might infer, my reply to OP was tempered by my ignorance of Mathematica's internals. I was "cludging" it together and chose my inner loop size to give appreciable convergence between the printed "spot checks". I added zeros until it fit on one page (giving about 3-4 decimals precision on the example problem within a single page's worth of output.) Staff Emeritus Science Advisor Homework Helper Gold Member pbuk said: Edit: and where has ## \cos \left(\cos \left(\frac \pi {180}\right)\right) ## come from? Whilst it is possible to calculate this number it cannot have any physical or even mathematical Ok, three months old but I stumbled upon it just now. I have come across this type of expression with relevant meaning exactly once that I can remember. I think it involved the components of a vector being parallel transported. The angle that the vector made to some reference set of curves varied as a cosine with the curve parameter and therefore the components took the functional form of nested trigonometric functions. I am not saying it is common but expressions like this do turn up sometimes. The cosine of the cosine of one degree? Staff Emeritus Science Advisor Homework Helper Gold Member Well, the argument would depend on the exact curve. I am sure it could be made one degree. pbuk said: Edit: and where has ## \cos \left(\cos \left(\frac \pi {180}\right)\right) ## come from? Whilst it is possible to calculate this number it cannot have any physical or even mathematical Tough to explain briefly, but it's the edge of a curve that is ultimately revolved around an axis. In fluid dynamics this is called the contact-line, in this case of an equilibrium spherical-capped interface for liquid in a cylinder. FAQ: Numerical integration over a Green's function What is numerical integration over a Green's function? Numerical integration over a Green's function is a method used in mathematics and physics to solve differential equations. It involves breaking down a complex function into smaller segments and using numerical techniques to approximate the solution. Why is numerical integration over a Green's function important? Numerical integration over a Green's function allows us to solve complex differential equations that cannot be solved analytically. It is essential in many fields, including engineering, physics, and What are the benefits of using numerical integration over a Green's function? The main benefit of using numerical integration over a Green's function is its ability to handle complex systems that would be impossible to solve using traditional methods. It also allows for faster and more accurate solutions. What are some common numerical techniques used in numerical integration over a Green's function? Some common numerical techniques used in numerical integration over a Green's function include the trapezoidal rule, Simpson's rule, and Gaussian quadrature. These methods involve breaking down the function into smaller segments and using mathematical formulas to approximate the solution. Are there any limitations to numerical integration over a Green's function? While numerical integration over a Green's function is a powerful tool, it does have some limitations. It can be computationally intensive, and the accuracy of the solution depends on the chosen numerical technique and the segmentation of the function. Additionally, it may not work well for functions with highly oscillatory behavior or singularities.	{"url":"https://www.physicsforums.com/threads/numerical-integration-over-a-greens-function.1013102/","timestamp":"2024-11-05T19:09:15Z","content_type":"text/html","content_length":"152668","record_id":"<urn:uuid:322543b5-9ffa-4bbc-a871-891f4fec02b5>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00069.warc.gz"}