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http://txpp2v.com/index.php-params=game-75-Arithmetic-Sequence-.htm | [
"You are Playing: Other Games » Arithmetic Sequence\n\nRating:\n• Currently 3/5\n• 1\n• 2\n• 3\n• 4\n• 5\n« Click stars to rate.",
null,
"Description:\nThe next three numbers should form the correct sequence of arithmetic operations with the use, between the first two numbers (the third number is the result), one of the four arithmetic operations: +, -,:, or x.\nInstructions:\nHere is a sample sequence:\n1 + 0=1 +1=2 /2=1 +5=6 x2=12 /4=3 ...\nFor such sequence a player should choose numbers in order:\n1,0,1,2,5,2,4."
] | [
null,
"http://txpp2v.com/img/arithmetic-sequence.png.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9132826,"math_prob":0.9980277,"size":364,"snap":"2021-43-2021-49","text_gpt3_token_len":103,"char_repetition_ratio":0.14166667,"word_repetition_ratio":0.0,"special_character_ratio":0.32417583,"punctuation_ratio":0.23076923,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9708063,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T15:34:32Z\",\"WARC-Record-ID\":\"<urn:uuid:dfc0f7ae-bd93-4eb5-b720-151f02e98e7b>\",\"Content-Length\":\"41051\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0178ca96-ba9a-45fd-80ab-5e1933c02842>\",\"WARC-Concurrent-To\":\"<urn:uuid:54bcdc14-fd4e-42c5-be1b-d516b9b52e7d>\",\"WARC-IP-Address\":\"172.104.97.181\",\"WARC-Target-URI\":\"http://txpp2v.com/index.php-params=game-75-Arithmetic-Sequence-.htm\",\"WARC-Payload-Digest\":\"sha1:BSV5PVZ64R3Z7TMOHKAOEN2G2WGENYVB\",\"WARC-Block-Digest\":\"sha1:SEB3PKG4EWGCNWW2F74RWHYA42KD5RIZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585322.63_warc_CC-MAIN-20211020152307-20211020182307-00707.warc.gz\"}"} |
https://vegibit.com/python-zip-function/ | [
"# Python zip() Function",
null,
"The zip() function in Python is a neat little function that takes two or more sequences as inputs and allows you to iterate through those sequences at the same time. To show how zip() can be used when in your programs, we’ll take a look at several examples of zip() in this tutorial. When we say iterables we are referring to the various iterables in Python such as a list, tuple, string, etc. Let’s get started with some examples of the zip() function now.\n\n## zip() Example one\n\n``````meats = ['Steak', 'Pork', 'Duck', 'Turkey']\ntoppings = ['Butter', 'Garlic', 'Olive Oil', 'Cranberry']\nfor meat, topping in zip(meats, toppings):\nprint(f'{meat} topped with {topping}')``````\n```Steak topped with Butter\nPork topped with Garlic\nDuck topped with Olive Oil\nTurkey topped with Cranberry```\n\n## zip() Example two\n\n``````numbers = [1, 2, 3, 4]\nstr_numbers = ['One', 'Two', 'Three', 'Four']\nresult = zip(numbers, str_numbers)\nprint(result)\nprint(list(result))``````\n```\n[(1, 'One'), (2, 'Two'), (3, 'Three'), (4, 'Four')]```\n\n## zip() with Dictionary\n\n``````colors = ['Red', 'White', 'Blue']\ncars = ['Corvette', 'Bronco', 'Mustang']\n\nmy_dict = {}\nfor color, car in zip(colors, cars):\nmy_dict[color] = car\nprint(my_dict)``````\n`{'Red': 'Corvette', 'White': 'Bronco', 'Blue': 'Mustang'}`\n\n### Calling zip() with no iterable\n\n``````result = zip()\nprint(result)``````\n`<zip object at 0x0000025021AD51C0>`\n\n### Passing more than two iterables to zip()\n\n``````numerals = [1, 2, 3]\nstr_numerals = ['One', 'Two', 'Three']\nroman_numerals = ['I', 'II', 'III']\nresult = zip(numerals, str_numerals, roman_numerals)\nprint(list(result))``````\n`[(1, 'One', 'I'), (2, 'Two', 'II'), (3, 'Three', 'III')]`\n\n### Cast zip object to list of tuples\n\n``````states = ['Massachusetts', 'Colorado', 'California', 'Florida']\ncapitals = ['Boston', 'Denver', 'Sacremento', 'Tallahassee']\nzipped = zip(states, capitals)\nziplist = list(zipped)\nprint(ziplist)``````\n`[('Massachusetts', 'Boston'), ('Colorado', 'Denver'), ('California', 'Sacremento'), ('Florida', 'Tallahassee')]`\n\n### Cast zip object to a dictionary\n\n``````states = ['Massachusetts', 'Colorado', 'California', 'Florida']\ncapitals = ['Boston', 'Denver', 'Sacremento', 'Tallahassee']\nzipped = zip(states, capitals)\nzipdict = dict(zipped)\nprint(zipdict)``````\n`{'Massachusetts': 'Boston', 'Colorado': 'Denver', 'California': 'Sacremento', 'Florida': 'Tallahassee'}`\n\n### Using zip() with the Python next() function\n\n``````states = ['Massachusetts', 'Colorado', 'California', 'Florida']\ncapitals = ['Boston', 'Denver', 'Sacremento', 'Tallahassee']\nzipped = zip(states, capitals)\nwhile True:\ntry:\ntup = next(zipped)\nprint(tup, \"capital is\", tup)\nexcept StopIteration:\nbreak``````\n```Massachusetts capital is Boston\nColorado capital is Denver\nCalifornia capital is Sacremento\nFlorida capital is Tallahassee```\n\n### How to unzip a zipped object\n\n``````states = ['Massachusetts', 'Colorado', 'California', 'Florida']\ncapitals = ['Boston', 'Denver', 'Sacremento', 'Tallahassee']\nzipped = zip(states, capitals)\nprint(zipped)\n\nfirst, second = zip(*zipped)\nprint(first)\nprint(second)``````\n```<zip object at 0x000001A6ED61E1C0>\n('Massachusetts', 'Colorado', 'California', 'Florida')\n('Boston', 'Denver', 'Sacremento', 'Tallahassee')```\n\n### Different length iterables\n\nAll of the examples we have looked at so far use iterables that all have the same number of elements in them. What happens when you try to use the zip() function on iterables where one is longer than the other? The default behavior for Python is to limit the operation to the smaller-sized iterable. Here is an example where one iterable has 3 elements and the other has five.\n\n``````one = [1, 2, 3, 4, 5]\ntwo = ['a', 'b', 'c']\n\nresult = zip(one, two)\nprint(list(result))``````\n`[(1, 'a'), (2, 'b'), (3, 'c')]`\n\nIf you would want to change the behavior so that the longer of the iterables is used, you can use the zip_longest() function which is part of the itertools package.\n\n``````import itertools as it\n\none = [1, 2, 3, 4, 5]\ntwo = ['a', 'b', 'c']\n\nresult = it.zip_longest(one, two)\nprint(list(result))``````\n`[(1, 'a'), (2, 'b'), (3, 'c'), (4, None), (5, None)]`\n\nYou can see that the zip_longest() function simply inserts a value of None for the slots where an empty value exists due to the different sized iterables. If you would like to specify a different value, you can make use of the fillvalue parameter like so.\n\n``````import itertools as it\n\none = [1, 2, 3, 4, 5]\ntwo = ['a', 'b', 'c']\n\nresult = it.zip_longest(one, two, fillvalue='😊')\nprint(list(result))``````\n`[(1, 'a'), (2, 'b'), (3, 'c'), (4, '😊'), (5, '😊')]`\n\n### Python zip() Function Summary\n\nThe zip() function takes iterables and aggregates them into a tuple and returns it.\n\nThe syntax of the zip() function is:\n\nzip(*iterables)\n\nThe iterables parameter can be built-in iterables such as list, string, dict, or user-defined iterables.\n\nThe zip()function returns an iterator of tuples based on the iterable objects.\n\n• If we do not pass any parameter, zip() returns an empty iterator\n• If a single iterable is passed, zip() returns an iterator of tuples with each tuple having only one element.\n• If multiple iterables are passed, zip() returns an iterator of tuples with each tuple having elements from all the iterables.\n\nSuppose, two iterables are passed to zip(); one iterable containing three and other containing five elements. Then, the returned iterator will contain three tuples. It’s because iterator stops when the shortest iterable is exhausted."
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https://www.dsprelated.com/thread/8336/calculating-noise-floor-of-digitization-system-using-fft | [
"## Calculating noise floor of digitization system using FFT",
null,
"Started by 4 years ago9 replieslatest reply 4 years ago3445 views\n\nHi all\n\nI am trying to calculate self noise of a digitization system with differential inputs.For this purpose, i select a differential channel, set input range to +/-5 V, short both differential inputs and performed following steps:\n\n1. Collect 'x' no of samples of voltage signal\n2. Perform FFT of size 'x' on collected samples\n3. Divide each complex output of FFT by 'x'\n4. Find the absolute of each complex output of FFT\n5. Multiply the output of above step by 1.414 to get Vrms against each frequency bin\n6. Take log10 and multiply by 20 of each output in above step\n7. Add 10xlog10(x/2) in each output of step above to compensate FFT gain against noise\n\nThe result is self noise floor graph of a channel in terms of dBVrms Vs frequency.\n\nKindly guide me whether this method is accurate enough to estimate self noise of a digitizer?\n\nThanks\n\n[ - ]",
null,
"Hi Naumankalia,\n\nWhat is the difference between your scheme and just applying a full-scale sinewave and then taking the windowed fft? What is your definition of self noise?\n\nNote that in the DSP world, we typically refer to the signal as x(n), where n is the sample number. Usually the total number of samples is called N.\n\nRegarding computation of the power of a captured signal, see my post at:\n\nhttps://www.dsprelated.com/showarticle/1004.php\n\nThe post includes windowing the signal, and normalizes the window amplitude so that it does not effect the power calculation.\n\nregards,\n\nNeil\n\n[ - ]",
null,
"Thanks for reply. I am sorry i could not clearly explain the purpose of this scheme. My main purpose is to determine practically (not theoretically) Hardware noise of my digitization system (which i call self noise of system) across a frequency band. Reason to do this is to determine the minimum input signal level which will be detectable by digitization system beyond which it will merge in own noise of system.\n\n[ - ]",
null,
"Hello ,\n\nThe digitizer probably consists of ADC and DSP, according to your algorithm for noise calculation.\n\nRegarding ADC, the quantization noise within fs/2 is 6*N (ADC number of bits) + 4.77dB relative to the peak voltage, this is HW limitation.\n\nIf you do oversampling, you have to take that into account as 10*log(fs/2/BW).\n\nRegarding DSP, it is function of your FFT setting, this computation limitation.\n\nTypically, the computation is negligible relative to HW limitation, if the FFT and Algorithms parameters are set appropriately.\n\nBest regards,\n\nShahram Shafie\n\n[ - ]",
null,
"Thanks for reply. I am sorry i could not clearly explain the purpose of this scheme. My main purpose is to determine practically (not theoretically) Hardware noise of my digitization system (which i call self noise of system) across a frequency band. Reason to do this is to determine the minimum input signal level which will be detectable by digitization system beyond which it will merge in own noise of system.\n[ - ]",
null,
"Hi, Though it's not clear the term 'self noise', i'm assuming you're looking for noise / harmonics content in your signal. if that is the case, you can use the THD + N measurement. if you're not expecting harmonic distortion you can directly go with SNR measurement.\n\nthe terms are well defined and illustrated in https://www.analog.com/media/en/training-seminars/...\n\nDo remember the following while you perform FFT:\n\n1. select the right window. especially, if you are using sine tone to measure THD and the freq is not a multiple of sampling rate. using right win will help to keep the bin slim.\n\n2. if you're automatically detecting the freq index based on the FFT bin, make use of three adjacent dominant peaks to decide more 'accurate' freq index. there are several approaches to get this value more accurately. This would be needed of the sine generated by a source which is Independence on the system you use for measurement.\n\n-Chalil\n\n[ - ]",
null,
"Thanks for reply. I am sorry i could not clearly explain the purpose of this scheme. My main purpose is to determine practically (not theoretically) Hardware noise of my digitization system (which i call self noise of system) across a frequency band. Reason to do this is to determine the minimum input signal level which will be detectable by digitization system beyond which it will merge in own noise of system.\n[ - ]",
null,
"you haven't described what your input is apart from saying it is +/- 5V.\n\nIf you want to estimate noise created by your ADC system then why not just try zero input(i.e. no signal) first. Next try single tone.\n\nzero input should show what will be added to your signal over any bandwidth you choose.\n\nsingle tone will tell you about quantisation noise for full scale input swing.\n\n[ - ]",
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https://www.redblobgames.com/x/2145-spaceship-flyer/ | [
"# Spaceship flyer\n\nfrom Red Blob Games\n8 Nov 2021\n\nThe spaceship is:\n\nIn 2009 I had done a monte-carlo approach where I tried many different thruster configurations $$\\begin{bmatrix} t_{1} & t_{2} & t_{3} & t_{4} \\end{bmatrix}$$ and picked something close. But my gut feeling was that it was a linear programming problem. This time, in 2021, I tried solving it as a linear programming problem but I ran into an issue: I’m looking for something that has the correct angle in output space, and among solutions with that angle, I want to find the one that has the most movement, and secondarily the least fuel. But I’m not sure how to make that a linear programming problem! It might be that I need to convert that angle into a normal vector and then add a constraint that we can’t move off the correct plane. I’m still thinking about that.\n\nI’m visualizing the thruster capabilities like this:\n\nAnother option is to treat this as a simpler problem, a linear system of equations that’s underconstrained. However there too I’m not quite sure how to deal with the angle problem.\n\n$\\begin{bmatrix} u_{1} & u_{2} & u_{3} & u_{4} \\\\ v_{1} & v_{2} & v_{3} & v_{4} \\\\ r_{1} & r_{2} & r_{3} & r_{4} \\\\ \\end{bmatrix} \\times \\begin{bmatrix} t_{1} \\\\ t_{2} \\\\ t_{3} \\\\ t_{4} \\end{bmatrix} = \\begin{bmatrix} u \\\\ v \\\\ r \\end{bmatrix}$\n\nMaybe I can pick a normalized version of the angle as the desired output, and then I can rescale that until I hit a limit with the thrusters. I’m reading this lecture (pdf) to learn how this math works. It looks like given my physics matrix $$P$$ and my desired normalized vector $$n$$ I have to solve for $$P P^{T} w = n$$ and then the thrusters are $$x = P^{T} w$$. That sounds… simple? The thrusters are for a normalized vector though, and I can just scale up $$x$$ until one of the thrusters is 1.0. But the problem with this math is that there’s nothing ensuring the thruster values are ≥ 0. So I don’t think this that useful for my problem. Linear programming is what gives me that constraint.\n\nRich G suggested I might try gradient descent, as it’s more general purpose and probably can handle what I’m wanting simply. But what would I actually do here? For each thruster row $$u_{i} v_{i} r_{i}$$ I can see if adding ε times that thruster gets me closer to the desired angle, and I can add more of the thrusters that get me closer?\n\ndesired_vec\ncurrent_vec\nsum_of_weights = 0\nweights = * thrusters.length\nfor t <= thrusters.length:\nweight[t] = max(0, dot(current_vec - thrusters[t], [1, 1, 1]))\nsum_of_weights += weight[t]\n// the [1, 1, 1] could be weighted higher for the parts that should be zero\nfor t <= thrusters.length:\ncurrent_vec += weight[t] / sum_of_weights * step_size * thrusters[t]\n\n\nGradient descent seems like overkill here because it’s a linear function! So I think I can take large step sizes. But I also kind of want to minimize fuel consumption. Does that happen automatically? I think it does if I choose one thruster each round, whichever gives me the most improvement for the least amount of fuel. But I think it doesn’t in general because one thruster may cause us to have to compensate with another, while a third thruster might’ve gotten there without needing compensation. Does that make sense? Hm. I think I should just try it and see how it goes.\n\nIn any case I think independent of which algorithm I use to choose the thrusters, I also would like to visualize the thruster capabilities, and I think the visualization I used in 2009 is probably the starting point. I need to intersect each pair of thruster configurations with a zero-plane and then I can draw those points:\n\nfor conf in configurations\nfor t < thrusters.length\nlet a = conf with thruster t on\nlet b = conf with thruster t off\nfor axis in ['au', 'av', 'aθ']\nif a[axis] * b[axis] < 0\n// signs are opposite, so there must be a mix of these where they add to 0\nmix = abs(a[axis] / (b[axis] - a[axis]))\nlet newconfig = a * mix + b * (1-mix)\n// verify that newconfig[axis] is 0!!\n\nEmail me , or tweet @redblobgames, or comment:"
] | [
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http://www.fact-archive.com/encyclopedia/Loop_quantum_gravity | [
"",
null,
"Search\n\n# The Online Encyclopedia and Dictionary",
null,
"",
null,
"",
null,
"## Encyclopedia",
null,
"## Dictionary",
null,
"## Quotes",
null,
"",
null,
"# Loop quantum gravity\n\nLoop quantum gravity (LQG), also known as loop gravity, quantum geometry and canonical quantum general relativity, is a proposed quantum theory of spacetime which attempts to blend together the seemingly incompatible theories of quantum mechanics and general relativity. It was developed in parallel with loop quantization, a rigorous framework for nonperturbative quantization of diffeomorphism-invariant gauge theories.\n\n Contents\n\nLeading advocates of loop quantum gravity consider it the main competitor of string theory as a theory of quantum gravity. The current position (as of 2004) is not symmetrical, in that string theorists are generally not concerned with loop quantum gravity, or they are skeptical that there is any real competition. String theory has its many critics, but has been the dominant force in quantum gravity since the mid-1980s.\n\n## Loop quantum gravity in general, and its ambitions\n\nLQG in itself was initially less ambitious than string theory, purporting only to be a quantum theory of gravity. String theory, on the other hand, appears to predict not only gravity but also various kinds of matter and energy that lie inside spacetime. Many string theorists believe that it is not possible to quantize gravity in 3+1 dimensions without creating these artifacts. But this is not proven, and it is also not proven that the matter artifacts of string theory are exactly the same as observed matter. Should LQG succeed as a quantum theory of gravity, the known matter fields would have to be incorporated into the theory a posteriori. Lee Smolin, one of the fathers of LQG, has explored the possibility that string theory and LQG are two different approximations to the same ultimate theory.\n\nThe main claimed successes of loop quantum gravity are: (1) that it is a nonperturbative quantization of 3-space geometry, with quantized area and volume operators; (2) that it includes a calculation of the entropy of black holes; and (3) that it is a viable gravity-only alternative to string theory. However, these claims are not universally accepted. While many of the core results are rigorous mathematical physics, their physical interpretations are speculative. LQG may or may not be viable as a refinement of either gravity or geometry; entropy is calculated for a kind of hole which may or may not be a black hole.\n\n## The incompatibility between quantum mechanics and general relativity\n\nMain article: quantum gravity\n\nQuantum field theory studied on curved (non-Minkowskian) backgrounds has shown that some of the core assumptions of quantum field theory cannot be carried over. In particular, the vacuum, when it exists, is shown to depend on the path of the observer through space-time (see Unruh effect).\n\nHistorically, there have been two reactions to the apparent inconsistency of quantum theories with the necessary background-independence of general relativity. The first is that the geometric interpretation of general relativity is not fundamental, but emergent. The other view is that background-independence is fundamental, and quantum mechanics needs to be generalized to settings where there is no a priori specified time.\n\nLoop quantum gravity is an effort to formulate a background-independent quantum theory. Topological quantum field theory is a background-independent quantum theory, but it lacks causally-propagating local degrees of freedom needed for 3 + 1 dimensional gravity.\n\n## History of LQG\n\nMain article: history of loop quantum gravity\n\nIn 1986 physicist Abhay Ashtekar reformulated Einstein's field equations of general relativity using what have come to be known as Ashtekar variables, a particular flavor of Einstein-Cartan theory with a complex connection. He was able to quantize gravity using gauge field theory. In the Ashtekar formulation, the fundamental objects are a rule for parallel transport (technically, a connection) and a coordinate frame (called a vierbein) at each point. Because the Ashtekar formulation was background-independent, it was possible to use Wilson loops as the basis for a nonperturbative quantization of gravity. Explicit (spatial) diffeomorphism invariance of the vacuum state plays an essential role in the regularization of the Wilson loop states.\n\nAround 1990, Carlo Rovelli and Lee Smolin obtained an explicit basis of states of quantum geometry, which turned out to be labelled by Penrose's spin networks. In this context, spin networks arose as a generalization of Wilson loops necessary to deal with mutually intersecting loops. Mathematically, spin networks are related to group representation theory and can be used to construct knot invariants such as the Jones polynomial.\n\nBeing closely related to topological quantum field theory and group representation theory, LQG is mostly established at the level of rigour of mathematical physics.\n\n## The ingredients of loop quantum gravity\n\n### Loop quantization\n\nAt the core of loop quantum gravity is a framework for nonperturbative quantization of diffeomorphism-invariant gauge theories, which one might call loop quantization. While originally developed in order to quantize vacuum general relativity in 3+1 dimensions, the formalism can accommodate arbitrary spacetime dimensionalities, fermions (Baez and Krasnov), an arbitrary gauge group (or even quantum group), and supersymmetry (Smolin), and results in a quantization of the kinematics of the corresponding diffeomorphism-invariant gauge theory. Much work remains to be done on the dynamics, the classical limit and the correspondence principle, all of which are necessary in one way or another to make contact with experiment.\n\nIn a nutshell, loop quantization is the result of applying C*-algebraic quantization to a non-canonical algebra of gauge-invariant classical observables. Non-canonical means that the basic observables quantized are not generalized coordinates and their conjugate momenta. Instead, the algebra generated by spin network observables (built from holonomies) and field strength fluxes is used.\n\nLoop quantization techniques are particularly successful in dealing with topological quantum field theories, where they give rise to state-sum/spin-foam models such as the Turaev-Viro model of 2+1 dimensional general relativity. A much studied topological quantum field theory is the so-called BF theory in 3+1 dimensions, because classical general relativity can be formulated as a BF theory with constraints, and it is hoped that a consistent quantization of gravity may arise from perturbation theory of BF spin-foam models.\n\n### Lorentz invariance\n\nFor detailed discussion see the Lorentz covariance page\n\nLQG is a quantization of a classical Lagrangian field theory which is equivalent to the usual Einstein-Cartan theory in that it leads to the same equations of motion describing general relativity with torsion. As such, it can be argued that LQG respects local Lorentz invariance. Global Lorentz invariance is broken in LQG just as in general relativity. A positive cosmological constant can be realized in LQG by replacing the Lorentz group with the corresponding quantum group.\n\n### Diffeomorphism invariance and background independence\n\nGeneral covariance (also known as diffeomorphism invariance) is the invariance of physical laws (for example, the equations of general relativity) under arbitrary coordinate transformations. This symmetry is one of the defining features of general relativity. LQG preserves this symmetry by requiring that the physical states must be invariant under the generators of diffeomorphisms. The interpretation of this condition is well understood for purely spatial diffemorphisms; however the understanding of diffeomorphisms involving time (the Hamiltonian constraint) is more subtle because it is related to dynamics and the so-called problem of time in general relativity, and a generally accepted calculational framework to account for this constraint is yet to be found.\n\nWhether or not Lorentz invariance is broken in the low-energy limit of LQG, the theory is formally background independent. The equations of LQG are not embedded in or presuppose space and time (except for its topology that cannot be changed), but rather they are expected to give rise to space and time at large distances compared to the Planck length. It has not been yet shown that LQG's description of spacetime at the Planckian scale has the right continuum limit described by general relativity with possible quantum corrections.\n\n## Open problems\n\n### The classical limit\n\nAny successful theory of quantum gravity must provide physical predictions that closely match known observation, and reproduce the results of quantum field theory and gravity. To date Einstein's theory of general relativity is the most successful theory of gravity. It has been shown that quantizing the field equations of general relativity will not necessarily recover those equations in the classical limit. It remains unclear whether LQG yields results that match general relativity in the domain of low-energy, macroscopic and astronomical realm. To date, LQG has been shown to yield results that match general relativity in 1+1 and 2+1 dimensions where the metric tensor carries no physical degrees of freedom. To date, it has not been shown that LQG reproduces classical gravity in 3+1 dimensions. Thus, it remains unclear whether LQG successfully merges quantum mechanics with general relativity.\n\n### Time\n\nAdditionally, in LQG, time is not continuous but discrete and quantized, just as space is: there is a minimum moment of time, Planck time, which is on the order of 10−43 seconds, and shorter intervals of time have no physical meaning. This carries the physical implication that relativity's prediction of time dilation due to accelerating speed or gravitational field, must be quantized, and must consist of multiples of Planck time units. (This helps resolve the time zero singularity problem: \"the big bang\".)\n\n### Particle physics\n\nWhile classical particle physics posit particles traveling through space and time that is continuous and therefore infinitely divisible, LQG predicts that space-time is quantized or granular. The two different models of space and time affects the way ultra high energy cosmic rays interacts with the background, with quantized spacetime predicting that the threshold for allowable energies for such high energy particles be raised. Such particles have been observed, however, alternative explanations have not been ruled out.\n\nLQG does not constrain the spectrum of non-gravitational forces and elementary particles. Unlike the situation in string theory, all of them must be added to LQG by hand. It has proved difficult to incorporate elementary scalar fields, Higgs mechanism, and CP-violation into the framework of LQG.\n\n### Quantum field theory\n\nQuantum field theory is background dependent. One problem LQG may be able to address in QFT is the ultraviolet catastrophe.\n\nThe term ultraviolet catastrophe is also applied to similar situations in quantum electrodynamics. There, summing over all energies results in an infinite value because the higher energy terms do not decrease quickly.\n\nIn LQG, the background is quantized, and there is apparently no physical \"room\" for the ultraviolet infinities of quantum field theory to occur. This argument, however, may be compromised if LQG does not admit a limiting smooth geometry at long distance scales. LQG is constructed as an alternative to perturbative quantum field theory on a fixed background. In its present form it does not allow a perturbative calculation of graviton scattering or other processes.\n\n### Graviton\n\nIn quantum field theories, the graviton is a hypothetical elementary particle that transmits the force of gravity in most quantum gravity systems. In order to do this gravitons have to be always-attractive (gravity never pushes), work over any distance (gravity is universal) and come in unlimited numbers (to provide high strengths near stars). In quantum theory, this defines an even-spin (spin 2 in this case) boson with a rest mass of zero.\n\nIt remains open to debate whether loop quantum gravity requires, or does not require, the graviton, or whether the graviton can be accounted for in its theoretical framework. As of today, the appearance of smooth space and gravitons in LQG has not been demonstrated, and hence questions about graviton scattering cannot be answered.\n\n### People\n\nSee list of loop quantum gravity researchers\n\n### Research in LQG and related areas\n\n#### Active research directions\n\n• Spin foam models\n• 2+1 and 3+1 theories\n• Barrett-Crane model\n• relation to the canonical approach\n• the Barbero-Immirzi parameter\n• canonical and spin foam geometries\n• the continuum limit\n• renormalization group flows\n• the Hamiltonian constraint\n• 2+1 and 3+1 theories\n• spin-foam and canonical approach\n• quantum cosmology\n• Semi-classical corrections to Einstein equations\n• factor ordering\n• finding solutions and physical inner product\n• Thiemann's phoenix project.\n• Semi-classical issues\n• kinematical and dynamical semi-classical states\n• quantum field theory on quantum geometry\n• quantum cosmology\n• Minkowski coherent state and Minkowski spin foam\n• Loop quantum phenomenology\n• Lorentz invariance\n• Doubly-special relativity\n• quantum cosmology\n• Kodama state and de Sitter background\n• Conceptual issues\n• observables through matter coupling\n• string theory in polymer representation\n• matter couplings on semi-classical states\n• the problem of time\n• spin foam histories\n• quantum groups in LQG\n\n### Loop quantum gravity's implications\n\n#### Space atoms\n\nIn LQG, the fabric of spacetime is a foamy network of interacting loops mathematically described by spin networks. These loops are about 10-35 meters in size, called the Planck scale. The loops knot together forming edges, surfaces, and vertices, much as do soap bubbles joined together. In other words, spacetime itself is quantized. Any attempt to divide a loop would, if successful, cause it to divide into two loops each with the original size. In LQG, spin networks represent the quantum states of the geometry of relative spacetime. Looked at another way, Einstein's theory of general relativity is (as Einstein predicted) a classical approximation of a quantized geometry.\n\n#### Kinematics\n\nKinematics in loop quantum gravity is the physics of space and time at the Planck scale. It is expressed in terms of area and volume operators, and spin foam formalism.\n\nArea and volume operators\n\nOne of the key results of loop quantum gravity is quantization of areas: according to several related derivations based on loop quantum gravity, the operator of the area A of a two-dimensional surface Σ should have discrete spectrum. Every spin network is an eigenstate of each such operator, and the area eigenvalue equals",
null,
"$A_{\\Sigma} = 8\\pi G_{\\mathrm{Newton}} \\gamma \\sum_i \\sqrt{j_i(j_i+1)}$\n\nwhere the sum goes over all intersections i of Σ with the spin network. In this formula, GNewton is the gravitational constant, γ is the Immirzi parameter and",
null,
"$j_i=0,0.5,1,1.5,\\dots$ is the spin associated with the link i of the spin network. The two-dimensional area is therefore \"concentrated\" in the intersections with the spin network.\n\nSimilar quantization applies to the volume operators but the mathematics behind these derivations is less convincing.\n\nSpin foams\n\n#### Quantum cosmology\n\nAn important principle in quantum cosmology that LQG adheres to is that there are no observers outside the universe. All observers must be a part of the universe they are observing. However, because light cones limit the information that is available to any observer, the Platonic idea of absolute truths does not exist in a LQG universe. Instead, there exists a consistency of truths in that every observer will report consistent (not necessarily the same) results if truthful.\n\nAnother important principle is the issue of the \"cosmological constant\", which is the energy density inherent in a vacuum. Cosmologists working on the assumption of zero cosmological constant predicted that gravity would slow the rate at which the universe is expanding following the big bang. However, astronomical observations of the magnitude and cosmological redshift of Type I supernovae in remote galaxies implies that the rate at which the universe is expanding is actually increasing. General relativity has a constant, Lambda, to account for this, and the observations, recently supported by independent data on the cosmic microwave background, appear to require a positive cosmological constant. In string theory, there are many vacua with broken supersymmetry which have positive cosmological constant, but generically their value of Lambda is much larger than the observed value. In LQG, there have been proposals to include a positive cosmological constant, involving a state referred to as the Kodama state after Hideo Kodama , a state described by a Chern-Simons wave function. Some physicists, for example Edward Witten, have argued by analogy with other theories that this state is unphysical. This issue is considered unresolved by other physicists.\n\nStandard quantum field theory and supersymmetric string theories make a prediction based on calculation of the vacuum energy density that differs from what has actually been observed by 120 orders of magnitude. To date, this remains an unsolved mystery that a successful quantum theory of gravity would hopefully avoid\n\n#### Black hole thermodynamics\n\nWhile experimental tests for LQG may be years in the future, one conceptual test any candidate for QG must pass is that it must derive the correct formula Hawking derived for the black hole entropy.\n\nWith the proper Immirzi parameter, LQG can calculate and reproduce the Hawking formula for all black holes. While string/M-theory does not need the Immirzi parameter, it can as yet only derive the Hawking formula for extremal black holes and near-extremal black holes -- black holes with a net electric charge, which differ from the nearly neutral black holes formed from the collapse of electrically neutral matter such as neutron stars. To date, the Immirzi parameter cannot be derived from more fundamental principles, and is an unavoidable artefact of quantization of general relativity's field equations.\n\nLQG's interpretation of black hole entropy is that the spacetime fabric that makes up the black hole horizon is quantized per Planck area, and the Bekenstein-Hawking entropy represents the degrees of freedom present in each Planck quantum. LQG does not offer an explanation why the interior of the black hole carries no volume-extensive entropy. Instead, it assumes that the interior does not contribute. The spacetime is truncated at the event horizon, and consistency requires to add Chern-Simons theory at the event horizon. A calculation within Chern-Simons theory leads to the desired result for the entropy, proportional to the horizon area.\n\nAdditionally, the spectrum of radiation of particles emanating from the event horizon of a black hole has been calculated from LQG's theoretical framework and precisely predicted. This prediction disagrees with Hawking's semiclassical calculation, but the use of a semiclassical calculation that is so far unconfirmed by experiment as a benchmark for an exact nonperturbative fully quantum calculation may be problematic. Modulo the Immirzi parameter, which is the only free parameter of LQG, it matches it on average, and additionally predicts a fine structure to it, which is experimentally testable and potentially an improvement.\n\n#### The big bang\n\nSeveral LQG physicists have shown that LQG can, at least formally, get rid of the infinities and singularities present when general relativity is applied to the big bang. While standard physics tools break down, LQG have provided internally self-consistent models of a big bounce in the time preceding the big bang.\n\n#### Theory of everything: unification of the four forces\n\nGrand unification theory refers to a theory in particle physics that unifies the strong interaction and electroweak interactions. A so-called theory of everything (TOE) is a putative theory that unifies the four fundamental forces of nature: gravity, the strong nuclear force, the weak nuclear force, and electromagnetism. Since the strong and electroweak interactions are described by quantum field theory, such a theory would require gravity also to be quantized, bringing with it the inconsistencies noted above.\n\nOne candidate for a consistent quantum gravity is string theory, which in addition to gravity contains gauge vector bosons and matter particles reminiscent of those experimentally observed. This has led to attempts (so far unsuccessful) to construct TOE's within its framework. In contrast, LQG is just a theory of one part of the Universe, namely quantum gravity.\n\nUnification in field theory or string theory is difficult or impossible to test directly, due to the extremely large energy (greater than 1016 GeV) at which unification is manifest. However, indirect tests exist, such as proton decay and the convergence of the coupling constants when extrapolated to high energy through the renormalization group. The simplest unified models (without supersymmetry) have failed such tests, but many models are still viable. Incorporating the correct strength of gravity in string unification is particularly challenging. While unified theories have greater explanatory and predictive power, it may be that nature does not favour them.\n\n#### Supersymmetry and extra dimensions\n\nSee supersymmetry for detailed discussion\n\nLQG in its current formulation predicts no additional spatial dimensions, nor anything else about particle physics. Lee Smolin, one of the originators of LQG, has proposed that loop quantum gravity incorporating either supersymmetry or extra dimensions, or both, be called loop quantum gravity II, in light of experimental evidence.\n\n#### Chaos theory and classical physics\n\nSensitivity on initial conditions, in the light of chaos theory means that two nonlinear systems with however small a difference in their initial state eventually will end up with a finite difference between their states. Loop quantum gravity suggests that the Planck scale represents the physical cut-off allowed for such sensitivity.\n\n### Differences between LQG and string/M-theory\n\nString theory and LQG are the products of different communities within theoretical physics. It is not generally agreed whether they are in any sense compatible, and their differences have sometimes been represented as different ways of doing physics. This is a sharp debate, or at times presented as such: in other words matters are currently subject to dialectic rather than experimental test.\n\nString theory emerged from the particle physics community and was originally formulated as a theory that depends on a background spacetime, flat or curved, which obeys Einstein's equations. This is now known to be just an approximation to a mysterious and not well-formulated underlying theory which may or may not be background independent.\n\nIn contrast, LQG was formulated with background independence in mind. However, it has been difficult to show that classical gravity can be recovered from the theory. Thus, LQG and string theory seem somewhat complementary.\n\nString theory easily recovers classical gravity, but so far it lacks a universal, perhaps background independent, description. LQG is a background independent theory of something, but the classical limit has yet not proven tractable. This has led some people to conjecture that LQG and string theory may both be aspects of some new theory, or that, perhaps there is some synthesis of the techniques of each that will lead to a complete theory of quantum gravity. For now, this is mostly a fond hope with little evidence.\n\n### Experimental tests of LQG in the near future\n\nObservation may affect the future theoretical development in quantum gravity in the areas of dark matter and dark energy. The year 2007 will see the launch of GLAST (space-based gamma-ray spectrometry experiments), and perhaps the completion and operation of LHC.\n\nLQG predicts that more energetic photons should travel ever so slightly faster than less energetic photons; this effect would be too small to observe within our galaxy. Giovanni Amelino-Camelia points out that photons which have traveled from distant galaxies may reveal the structure of spacetime.\n\nIf GLAST detects violations of Lorentz invariance in the form of energy-dependent photon velocity, in agreement with theoretical calculations, such observations would support LQG. However, string theory would not necessarily be disfavoured.\n\n## Objections to the theory\n\nObjections to the theory of loop quantum gravity\n\n\nAs a physical theory, loop quantum gravity has been subject to some heavy criticisms. Some objections to the ideas of loop quantum gravity are given here.\n\n### Too many assumptions\n\nOBJECTION Loop quantum gravity makes too many assumptions about the behavior of geometry at very short distances. It assumes that the metric tensor is a good variable at all distance scales, and it is the only relevant variable. It even assumes that Einstein's equations are more or less exact in the Planckian regime.\n\nThe spacetime dimensionality (four) is another assumption that is not questioned, much like the field content. Each of these assumptions is challenged in a general enough theory of quantum gravity, for example all the models that emerge from string theory.\n\nThese assumptions have neither theoretical nor experimental justification. Particular examples will be listed in a separate entry.\n\nThe most basic, underlying assumption is that the existence of a meaningful classical theory, of general relativity, implies that there must exist a \"quantization\" of this theory. This is commonly challenged. Many reasons are known why some classical theories do not have a quantum counterpart. Gauge anomalies are a prominent example. General relativity is usually taken to be another example, because its quantum version is not renormalizable.\n\nIt is known, therefore, that a classical theory is not always a good starting point for a quantum theory. Theorists of loop quantum gravity work with the assumption that \"quantization\" can be done, and continue to study it even if their picture seems inconsistent.\n\n### Commentary from the renormalization group aspect\n\nOBJECTION According to the logic of the renormalization group, the Einstein-Hilbert action is just an effective description at long distances; and it is guaranteed that it receives corrections at shorter distances. String theory even allows us to calculate these corrections in many cases.\n\nThere can be additional spatial dimensions; they have emerged in string theory and they are also naturally used in many other modern models of particle physics such as the Randall-Sundrum models. An infinite amount of new fields and variables associated with various objects (strings and branes) can appear, and indeed does appear according to string theory. Geometry underlying physics may become noncommutative, fuzzy, non-local, and so on. Loop quantum gravity ignores all these 20th and 21st century possibilities, and it insists on a 19th century image of the world which has become naive after the 20th century breakthroughs.\n\n### As a predictive theory\n\nOBJECTION Loop quantum gravity is not a predictive theory. It does not offer any possibility to predict new particles, forces and phenomena at shorter distances: all these objects must be added to the theory by hand. Loop quantum gravity therefore also makes it impossible to explain any relations between the known physical objects and laws.\n\nLoop quantum gravity is not a unifying theory. This is not just an aesthetic imperfection: it is impossible to find a regime in real physics of this Universe in which non-gravitational forces can be completely neglected, except for classical physics of neutral stars and galaxies that also ignores quantum mechanics. For example, the electromagnetic and strong force are rather strong even at the Planck scale, and the character of the black hole evaporation would change dramatically had the Nature omitted the other forces and particles. Also, the loop quantum gravity advocates often claim that the framework of loop quantum gravity regularizes all possible UV divergences of gravity as well as other fields coupled to it. That would be a real catastrophe because any quantum field theory - including all non-renormalizable theories with any fields and any interactions - could be coupled to loop quantum gravity and the results of the calculations could be equal to anything in the world. The predictive power would be exactly equal to zero, much like in the case of a generic non-renormalizable theory. There is absolutely no uniqueness found in the realistic models based on loop quantum gravity. The only universal predictions - such as the Lorentz symmetry breaking discussed below - seem to be more or less ruled out on experimental grounds.\n\n### Self-consistency\n\nOBJECTION Unlike string theory, loop quantum gravity has not offered any non-trivial self-consistency checks of its statements and it has had no impact on the world of mathematics. It seems that the people are constructing it, instead of discovering it. There are no nice surprises in loop quantum gravity - the amount of consistency in the results never exceeds the amount of assumptions and input. For example, no answer has ever been calculated in two different ways so that the results would match. Whenever a really interesting question is asked - even if it is apparently a universal question, for example: \"Can topology of space change?\" - one can propose two versions of loop quantum gravity which lead to different answers.\n\nThere are many reasons to think that loop quantum gravity is internally inconsistent, or at least that it is inconsistent with the desired long-distance limit (which should be smooth space). Too many physical wisdoms seem to be violated. Unfortunately the loop quantum gravity advocates usually choose to ignore the problems. For example, the spin foam (path-integral) version of loop quantum gravity is believed to break unitarity. The usual reaction of the loop quantum gravity practitioners is the statement that unitarity follows from time-translation symmetry, and because this symmetry is broken (by a generic background) in GR, we do not have to require unitarity anymore. But this is a serious misunderstanding of the meaning and origin of unitarity. Unitarity is the requirement that the total probability of all alternatives (the squared length of a vector in the Hilbert space) must be conserved (well, it must always be 100%), and this requirement - or an equally-strong generalization of it - must hold under any circumstances, in any physically meaningful theory, including the case of the curved, time-dependent spacetime. Incidentally, the time-translation symmetry is related, via Noether's theorem, to a time-independent, conserved Hamiltonian, which is a completely different thing than unitarity.\n\nA similar type of \"anything goes\" approach seems to be applied to other no-go theorems in physics.\n\n### Gap to high-energy physics\n\nOBJECTION Loop quantum gravity is isolated from particle physics. While extra fields must be added by hand, even this ad hoc procedure seems to be impossible in some cases. Scalar fields can't really work well within loop quantum gravity, and therefore this theory potentially contradicts the observed electroweak symmetry breaking, the violation of the CP symmetry, and other well-known and tested properties of particle physics.\n\nLoop quantum gravity also may deny the importance of many methods and tools of particle physics - e.g. the perturbative techniques; the S-matrix, and so on. Loop quantum gravity therefore potentially disagrees with 99% of physics as we know it. Unfortunately, the isolation from particle physics follows from the basic opinions of loop quantum gravity practitioners and it seems very hard to imagine that a deeper theory can be created if the successful older theories, insights, and methods (and exciting newer ones) in the same or closely related fields are ignored.\n\n### Smooth space as limiting case\n\nOBJECTION Loop quantum gravity does not guarantee that smooth space as we know it will emerge as the correct approximation of the theory at long distances; there are in fact many reasons to be almost certain that the smooth space cannot emerge, and these problems of loop quantum gravity are analogous to other attempts to discretize gravity (e.g. putting gravity on lattice).\n\nWhile string theory confirms general relativity or its extensions at long distances - where GR is tested - and modifies it at the shorter ones, loop quantum gravity does just the opposite. It claims that GR is formally exact at the Planck scale, but implies nothing about the correct behavior at long distances. It is reasonable to assume that the usual ultraviolet problems in quantum gravity are simply transmuted into infrared problems, except that the UV problems seem to be present in loop quantum gravity, too.\n\n### Clash with special relativity\n\nOBJECTION Loop quantum gravity violates the rules of special relativity that must be valid for all local physical observations. Spin networks represent a new reincarnation of the 19th century idea of the luminiferous aether - environment whose entropy density is probably Planckian and that picks a privileged reference frame. In other words, the very concept of a minimal distance (or area) is not compatible with the Lorentz contractions. The Lorentz invariance was the only real reason why Einstein had to find a new theory of gravity - Newton's gravitational laws were not compatible with his special relativity.\n\nDespite claims about the background independence, loop quantum gravity does not respect even the special 1905 rules of Einstein; it is a non-relativistic theory. It conceptually belongs to the pre-1905 era and even if we imagine that loop quantum gravity has a realistic long-distance limit, loop quantum gravity has even fewer symmetries and nice properties than Newton's gravitational laws (which have an extra Galilean symmetry, and can also be written in a \"background independent\" way - and moreover, they allow us to calculate most of the observed gravitational effects well, unlike loop quantum gravity). It is a well-known fact that general relativity is called \"general\" because it has the same form for all observers including those undergoing a general accelerated motion - it is symmetric under all coordinate transformations - while \"special\" relativity is only symmetric under a subset of special (Lorentz and Poincaré) transformations that interchange inertial observers. The symmetry under any coordinate transformation is only broken spontaneously in general relativity, by the vacuum expectation value of the metric tensor, not explicitly (by the physical laws), and the local physics of all backgrounds is invariant under the Lorentz transformations.\n\nLoop quantum gravity proponents often and explicitly state that they think that general relativity does not have to respect the Lorentz symmetry in any way - which displays a misunderstanding of the symmetry structure of special and general relativity (the symmetries in general relativity extend those in special relativity), as well as of the overwhelming experimental support for the postulates of special relativity. Loop quantum gravity also depends on the background in a lot of other ways - for example, the Hamiltonian version of loop quantum gravity requires us to choose a pre-determined spacetime topology which cannot change.\n\nOne can imagine that the Lorentz invariance is restored by fine-tuning of an infinite number of parameters, but nothing is known about the question whether it is possible, how such a fine-tuning should be done, and what it would mean. Also, it has been speculated that special relativity in loop quantum gravity may be superseded by the so-called doubly special relativity, but doubly special relativity is even more problematic than loop quantum gravity itself. For example, its new Lorentz transformations are non-local (two observers will not agree whether the lion is caught inside the cage) and their action on an object depends on whether the object is described as elementary or composite.\n\n### Global justification of variables\n\nOBJECTION The discrete area spectrum is not a consequence, but a questionable assumption of loop quantum gravity. The redefinition of the variables - the formulae to express the metric in terms of the Ashtekar variables (a gauge field) - is legitimate locally on the configuration space, but it is not justified globally because it imposes new periodicities and quantization laws that do not follow from the metric itself. The area quantization does not represent physics of quantum gravity but rather specific properties of this not-quite-legitimate field redefinition. One can construct infinitely many similar field redefinitions (siblings of loop quantum gravity) that would lead to other quantization rules for other quantities. It is probably not consistent to require any of these new quantization rules - for instance, one can see that these choices inevitably break the Lorentz invariance which is clearly a bad thing.\n\n### Testability of the discrete area spectrum\n\nOBJECTION The discrete area spectrum is not testable, not even in principle. Loop quantum gravity does not provide us with any \"sticks\" that could measure distances and areas with a sub-Planckian precision, and therefore a prediction about the exact sub-Planckian pattern of the spectrum is not verifiable. One would have to convert this spectrum into a statement about the scattering amplitudes.\n\n### The S-matrix\n\nOBJECTION Loop quantum gravity provides us with no tools to calculate the S-matrix, scattering cross sections, or any other truly physical observable. It is not surprising; if loop quantum gravity cannot predict the existence of space itself, it is even more difficult to decide whether it predicts the existence of gravitons and their interactions. The S-matrix is believed to be essentially the only gauge-invariant observable in quantum gravity, and any meaningful theory of quantum gravity should allow us to calculate it, at least in principle.\n\n### Ultraviolet divergences\n\nOBJECTION Loop quantum gravity does not really solve any UV problems. Quantized eigenvalues of geometry are not enough, and one can see UV singular and ambiguous terms in the volume operators and most other operators, especially the Hamiltonian constraint. Because the Hamiltonian defines all of dynamics, which contains most of the information about a physical theory, it is a serious object. The whole dynamics of loop quantum gravity is therefore at least as singular as it is in the usual perturbative treatment based on semiclassical physics.\n\nWe simply do have enough evidence that a pure theory of gravity, without any new degrees of freedom or new physics at the Planck scale, cannot be consistent at the quantum level, and loop quantum gravity advocates need to believe that the mathematical calculations leading to the infinite and inconsistent results (for example, the two-loop non-renormalizable terms in the effective action) must be incorrect, but they cannot say what is technically incorrect about them and how exactly is loop quantum gravity supposed to fix them. Moreover, the loop quantum gravity proponents seem to believe that the naive notion of \"atoms of space\" is the only way to fix the UV problems. String theory, which allows us to make real quantitative computations, proves that it is not the case and there are more natural ways to \"smear out\" the UV problems. In fact, a legitimate viewpoint implies that the discrete, sharp character of the metric tensor and other fields at very short distances makes the UV behavior worse, not better.\n\nMoreover, as explained above, the \"universal solution of the UV problems by discreteness of space\" implies at least as serious loss of predictive power as in a generic non-renormalizable theory. Even if loop quantum gravity solved all the UV problems, it would mean that infinitely many coupling constants are undetermined - a situation analogous to a non-renormalizable theory.\n\n### Black hole entropy\n\nOBJECTION Despite various claims, loop quantum gravity is not able to calculate the black hole entropy, unlike string theory. The fact that the entropy is proportional to the area does not follow from loop quantum gravity. It is rather an assumption of the calculation. The calculation assumes that the black hole interior can be neglected and the entropy comes from a new kind of dynamics attached to the surface area - there is no justification of this assumption. Not surprisingly, one is led to an area/entropy proportionality law. The only non-trivial check could be the coefficient, but it comes out incorrectly (see the Immirzi discrepancy).\n\nThe Immirzi discrepancy was believed to be proportional to the logarithm of two or three, and a speculative explanation in terms of quasinormal modes was proposed. However it only worked for one type of the black hole - a clear example of a numerical coincidence - and moreover it was realized in July 2004 that the original calculation of the Immirzi parameter was incorrect, and the correct value (described by Meissner) is not proportional to the logarithm of an integer. The value of the Immirzi parameter - even according to the optimists - remains unexplained. Another description of the situation goes as follows: Because the Immirzi parameter represents the renormalization of Newton's constant and there is no renormalization in a finite theory - and loop quantum gravity claims to be one - the Immirzi parameter should be equal to one which leads to a wrong value of the black hole entropy.\n\n### Nonseparable Hilbert space\n\nOBJECTION While all useful quantum theories in physics are based on a separable Hilbert space; i.e. a Hilbert space with a countable basis, loop quantum gravity naturally leads to a non-separable Hilbert space, even after the states related by diffeomorphisms are identified. This space can be interpreted as a very large, uncountable set of superselection sectors that do not talk to each other and prevent physical observables from being changed continuously. All known procedures to derive a different, separable Hilbert space are physically unjustified.\n\n### Foundational lacks\n\nOBJECTION Loop quantum gravity has no tools and no solid foundations to answer other important questions of quantum gravity - the details of Hawking radiation; the information loss paradox; the existence of naked singularities in the full theory; the origin of holography and the AdS/CFT correspondence; mechanisms of appearance and disappearance of spacetime dimensions; the topology changing transitions (which are most likely forbidden in loop quantum gravity); the behavior of scattering at the Planck energy; physics of spacetime singularities; quantum corrections to geometry and Einstein's equations; the effect of the fluctuating metric tensor on locality, causality, CPT-symmetry, and the arrow of time; interpretation of quantum mechanics in non-geometric contexts including questions from quantum cosmology; the replacement for the S-matrix in de Sitter space and other causally subtle backgrounds; the interplay of gravity and other forces; the issues about T-duality and mirror symmetry.\n\nLoop quantum gravity is criticised as a philosophical framework that wants us to believe that these questions should not be asked. As if general relativity is virtually a complete theory of everything (even though it apparently can't be) and all ideas in physics after 1915 can be ignored.\n\n### Prejudices claimed\n\nOBJECTION The criticisms of loop quantum gravity regarding other fields of physics are misguided. They often dislike perturbative expansions. While it is a great advantage to look for a framework that allows us to calculate more than the perturbative expansions, it should never be less powerful. In other words, any meaningful theory should be able to allow us to perform (at least) approximative, perturbative calculations (e.g. around a well-defined classical solution, such as flat space). Loop quantum gravity cannot do this, definitely a huge disadvantage, not an advantage as some have claimed. A good quantum theory of gravity should also allow us to calculate the S-matrix.\n\n### Background independence\n\nOBJECTION Loop quantum gravity's calls for \"background independence\" are misled. A first constraint for a correct physical theory is that it allows the (nearly) smooth space[time] — or the background — which we know to be necessary for all known physical phenomena in this Universe. If a theory does not admit such a smooth space, it can be called \"background independent\" or \"background free\", but it may be a useless theory and a physically incorrect theory.\n\nIt is a very different question whether a theory treats all possible shapes of spacetime on completely equal footing or whether all these solutions follow from a more fundamental starting point. However, it is not a priori clear on physical grounds whether it must be so (It can be just an aesthetic feature of a particular formulation of a theory, not the theory itself.), and moreover, for a theory that does not predict many well-behaved backgrounds the question is meaningless altogether. Physics of string theory certainly does respect the basic rules of general relativity exactly - general covariance is seen as the decoupling of unphysical (pure gauge) modes of the graviton. This exact decoupling can be proved in string theory quite easily. It can also be seen in perturbative string theory that a condensation of gravitons is equivalent to a change of the background; therefore physics is independent of the background we start with, even if it is hard to see for the loop quantum gravity advocates.\n\n### Claims on non-principled approach\n\nOBJECTION Loop quantum gravity is not science because every time a new calculation shows that some quantitative conjectures were incorrect, the loop quantum gravity advocates invent a non-quantitative, ad hoc explanation why it does not matter. Some borrow concepts from unrelated fields, including noiseless information theory and philosophy, and some explanations why previous incorrect results should be kept are not easily credible.\n\n## Bibliography",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9184536,"math_prob":0.94320637,"size":48319,"snap":"2022-05-2022-21","text_gpt3_token_len":9701,"char_repetition_ratio":0.17046466,"word_repetition_ratio":0.010923803,"special_character_ratio":0.18237132,"punctuation_ratio":0.08692476,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96600246,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,2,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T03:30:05Z\",\"WARC-Record-ID\":\"<urn:uuid:c32af79a-8b2b-4eed-a77f-ee739a41ca30>\",\"Content-Length\":\"91821\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e95527a-c1f2-4f62-a2b6-c6e796e50975>\",\"WARC-Concurrent-To\":\"<urn:uuid:a44d76b7-4417-4f38-8da5-2159ecf2f584>\",\"WARC-IP-Address\":\"104.21.91.165\",\"WARC-Target-URI\":\"http://www.fact-archive.com/encyclopedia/Loop_quantum_gravity\",\"WARC-Payload-Digest\":\"sha1:F5GHM33E55DECLFFV3H3KHSGBKOMICIB\",\"WARC-Block-Digest\":\"sha1:HQXKPLO2DQOE5GTPVOOBHDMQB5MHOOAP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662534773.36_warc_CC-MAIN-20220521014358-20220521044358-00651.warc.gz\"}"} |
https://5g-tools.com/NB-IoT-link-budget-calculator/ | [
"",
null,
"You can calculate the Link budget (Signal level at receiver) and then compare it with Rx Reception sensitivity. Then you will understand Radio Channel Status (Pass or Fail) and Cell Radius.\n\nThe calculation is based on the 3GPP 38.901 standard. Approximately Link budget of NB-IoT can be calculated using the formula:\n\nUpdate1: Add Free Space Path Loss (FSPL) Propagation Model. Now you can use LOS or NLOS Model"
] | [
null,
"https://mc.yandex.ru/watch/52959418",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77299476,"math_prob":0.84615576,"size":1639,"snap":"2022-27-2022-33","text_gpt3_token_len":428,"char_repetition_ratio":0.2379205,"word_repetition_ratio":0.04016064,"special_character_ratio":0.21354485,"punctuation_ratio":0.14983714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96182245,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T07:27:18Z\",\"WARC-Record-ID\":\"<urn:uuid:7415aed0-2885-434b-942b-bdf2b8740622>\",\"Content-Length\":\"93999\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7cff2c3-9a54-4836-8694-6c3073cf1214>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d80f6ef-f00c-4f9e-a9c3-e5ad0c00ad0b>\",\"WARC-IP-Address\":\"87.236.16.186\",\"WARC-Target-URI\":\"https://5g-tools.com/NB-IoT-link-budget-calculator/\",\"WARC-Payload-Digest\":\"sha1:NCYGQXA45USRYSN7MIKVURGYNQB3AJOJ\",\"WARC-Block-Digest\":\"sha1:FY32IBW6P7RRL4HCOG3RNBWMBBII52AZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572161.46_warc_CC-MAIN-20220815054743-20220815084743-00364.warc.gz\"}"} |
https://www.answers.com/Q/What_are_the_notes_to_Drunken_Sailor_on_keyboard | [
"What are the notes to Drunken Sailor on keyboard?\n\nThe notes on the keyboard to Drunken Sailor are as follows:\n\na a a a a a a d f a g g g g g g g c e g a a a a a a a b c d c a g e d d\n\na a a a a a a d f a g g g g g g g c e g a a a a a a a b c d c a g e d d\n\na a a d f a g g g c e d a a a b c d c a g e d d"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6953151,"math_prob":0.9992286,"size":301,"snap":"2019-43-2019-47","text_gpt3_token_len":122,"char_repetition_ratio":0.32659933,"word_repetition_ratio":0.75,"special_character_ratio":0.38870433,"punctuation_ratio":0.016949153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96968615,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T04:19:19Z\",\"WARC-Record-ID\":\"<urn:uuid:aca8c8d7-7272-41f7-b975-747dcf100644>\",\"Content-Length\":\"101197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f2f404c-733a-44a8-91d9-38497f0e3359>\",\"WARC-Concurrent-To\":\"<urn:uuid:5fe0cca6-6fc5-4d01-8d1d-002ba11cfe4b>\",\"WARC-IP-Address\":\"151.101.248.203\",\"WARC-Target-URI\":\"https://www.answers.com/Q/What_are_the_notes_to_Drunken_Sailor_on_keyboard\",\"WARC-Payload-Digest\":\"sha1:SVL3RKQWJIQ5X4MK4WEWYGUOWYIO3MNT\",\"WARC-Block-Digest\":\"sha1:CQKIVFICYJLYZMSU4GSS5EYZPZKQLSEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986664662.15_warc_CC-MAIN-20191016041344-20191016064844-00220.warc.gz\"}"} |
https://math.stackexchange.com/questions/677720/probability-questions/677730 | [
"# Probability questions\n\nA policy requiring all hospital employees to take lie detector tests may reduce losses due to theft, but some employees regard such tests as a violation of their rights. To gain some insight into the risks that employees face when taking a lie detector test, suppose that the probability is 0.06 that a lie detector concludes that a person is lying who, in fact, is telling the truth and suppose that any pair of tests are independent.\n\nWhat is the probability that a machine will conclude that each of three employees is lying when all are telling the truth?\n\nFor this one I did (0.06)^3\n\nWhat is the probability that the machine will conclude that none of the employees is lying when all are telling the truth?\n\nFor this one I did (0.94)^3\n\nWhat is the probability that a machine will conclude that at least one of the three employees is lying when all are telling the truth?\n\nFor this one I did (0.94)^2(0.6)\n\nI am not 100% sure if I am doing this right. Also, I am pretty bad with probability.\n\n• For the last question, easiest to find $1-(0.94)^3$. Your answer is not correct, for your way we would need to add up the probabilities it will conclude $1$ is lying, $2$ are lying, $3$ are lying. The probability it will conclude exactly one is lying is $(3)(0.94)^2(0.06)$. For $2$ lying it is $(3)(0.94)(0.06)^2$, and you already did $3$. Now add up. But that's the hard way, the easy way is $1-(0.94)^3$. – André Nicolas Feb 15 '14 at 20:03\n• @AndréNicolas Ahhh I see that makes so much sense thanks!!! – user125627 Feb 15 '14 at 20:08\n• You are welcome. – André Nicolas Feb 15 '14 at 20:09\n\nFor the first one, the probability is $$\\frac{P(\\text{detect lying, telling truth})^3}{P(\\text{telling truth})^3}=0.06^3$$ Similarly the second one is $$\\frac{P(\\text{detect telling truth, telling truth})^3}{P(\\text{telling truth})^3}=0.94^3$$ So you got the first two right. But for the third one, it should be $$1-P(\\text{detect no on is lying|all are telling the truth})=1-0.94^3$$\n\nYour first two answers look good. However, for the last one, you're calculating the probability that exactly one person — and a specific person at that — is erroneously said to be lying and the other two are telling the truth.\n\nInstead, note the relationship between the third question and the one that immediately precedes it. If it's not the case that the polygraph concluded that all employees were telling the truth, then what did it conclude? In terms of probability, what's the relationship between the event described in the third question and the one in the second?\n\nLet me know if you need another hint.\n\nInstead of trying to evaluate for exactly one liar, just assume they're all being truthful and take the complement of that. That way you'll know how likely it is that even one of them lied. In probability you can usually easily what find what you don't know based on what you do know.\n\nlet A = no lies detected let B = telling truth let C = at least one employee truthful\n\nP(C) = 1 - P(A | B) = 1 - (0.94)^3 with \"1' assuming full probability"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.97166467,"math_prob":0.98265994,"size":990,"snap":"2019-51-2020-05","text_gpt3_token_len":221,"char_repetition_ratio":0.14401622,"word_repetition_ratio":0.252809,"special_character_ratio":0.23232323,"punctuation_ratio":0.08173077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998065,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T13:33:38Z\",\"WARC-Record-ID\":\"<urn:uuid:97ae4d8f-03bb-46af-9929-2d1c9cc15853>\",\"Content-Length\":\"151994\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d095e6bc-081b-40d2-bbbb-b66cf94bee1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a07c6b8-cd74-4435-8d16-0f0ba66b5138>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/677720/probability-questions/677730\",\"WARC-Payload-Digest\":\"sha1:2JWIW2B4DKBPUEAS75O4BTCEWHJK3DZ3\",\"WARC-Block-Digest\":\"sha1:5D6VXFC3PPWZVR23UU6R7OE4Q2QCEJ7O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00543.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-simplify-frac-4-9-7-5 | [
"# How do you simplify \\frac { - 4( - 9) } { 7- ( - 5) }?\n\nApr 10, 2018\n\n(−4(−9))/(7−(−5)) simplifies to $3$\n\n#### Explanation:\n\nSimplify\n\n(−4(−9))/(7−(−5))\n\n1) Clear the top parentheses by distributing the $- 4$\n\n(36)/(7−(−5))\n\n2) Clear the bottom parentheses by distributing the minus sign\n(Note: The $7$ is an addend that you add after you do all the multiplications.)\n\n$\\frac{36}{7 + 5}$\n\n3) Add the numbers in the denominator\n\n$\\frac{36}{12}$\n\n4) Reduce to lowest terms\n\n$3$ $\\leftarrow$ answer\n\nApr 10, 2018\n\n#### Explanation:\n\n$\\frac{- 4 \\left(- 9\\right)}{7 - \\left(- 5\\right)}$\n\nFirst multiply $- 4$ and $- 9$ on numerator:\n\n$= \\frac{36}{7 - \\left(- 5\\right)}$\n\nThen subtract $- 5$ from $7$:\n\n$= \\frac{36}{12}$\n\nFinally simply fraction:\n\n$= 3$\n\nApr 10, 2018\n\n$\\frac{- 4 \\left(- 9\\right)}{7 - \\left(- 5\\right)} = 3$\n\n#### Explanation:\n\nWhen dealing with multiple plus and minus signs in addition or subtraction, it is important to remember that two of the the same sign will always be addition and two different signs will always be subtraction.\n\nWhen multiplying or dividing negative and positive numbers, remember that two positives or two negatives make a positive number while one positive and one negative make a negative number.\n\n$\\frac{- 4 \\left(- 9\\right)}{7 - \\left(- 5\\right)}$\n$= \\frac{36}{7 + 5}$\n$= \\frac{36}{12}$\n$= 3$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7171796,"math_prob":0.9999192,"size":821,"snap":"2020-34-2020-40","text_gpt3_token_len":174,"char_repetition_ratio":0.12239902,"word_repetition_ratio":0.0,"special_character_ratio":0.21437271,"punctuation_ratio":0.07857143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999013,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T03:53:15Z\",\"WARC-Record-ID\":\"<urn:uuid:190687c2-081e-4791-9ebe-a192a4eaf489>\",\"Content-Length\":\"36291\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2fc5476-6e3e-4eee-8a68-6a1a91d12b64>\",\"WARC-Concurrent-To\":\"<urn:uuid:701e598d-b25f-495c-9486-17e1505864d3>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-simplify-frac-4-9-7-5\",\"WARC-Payload-Digest\":\"sha1:N2FDQPXELCIHQ3PO772S44B3I7BIQ6AF\",\"WARC-Block-Digest\":\"sha1:MP2F3PJMV6M5WYALTUTJRWJOCPSY2FNM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198887.3_warc_CC-MAIN-20200921014923-20200921044923-00407.warc.gz\"}"} |
https://ell.stackexchange.com/questions/291691/all-not-equal-to-zero-meaning | [
"# All not equal to zero - meaning?\n\nI have the following sentence in a scientific paper.\n\n“Here r, r1, and r2 are all not equal to zero.”\n\nDoes this mean that if one of them is zero, then the condition is not met? Or does it mean that at the same time they cannot be zero?\n\nFor example, is r=1, r1=2, and r2=0 a good input?\n\n• The operative word is \"all\". If any one of them is zero, then it not true that all of them are not zero. This is simply a question in Boolean logic, not in English. Jul 18 at 20:20\n• “Here r, r1, and r2 are all not equal to zero.” This is not a condition, it is a statement of fact: neither r, r1, nor r2 is equal to zero. Jul 18 at 22:59\n\n\"all not equal to zero\" means that r is not equal to zero, r1 is not equal to zero, and r3 is not equal to zero.\n\n\"not all equal to zero\" means that at least one of them is nonzero, and the others could still be zero.\n\nHowever, people do not always* obey this distinction. When you see a sentence like this you often have to guess the author's intent.\n\n* (see the difference between \"do not always\" and \"always do not\"?)\n\nIt is formally ambiguous.\n\nIn context, I believe it means that all of them are non-zero.\n\n• It is not ambiguous at all. All not means none. Not all means no more than all. Jul 18 at 22:40\n• We're talking about English, not logic. All that glisters is not gold was old when Shakespeare used it. And I don't know what you intend by no more than all, but sinxe more than all is incoherent, it doesn't seem to make sense. Jul 19 at 12:58\n\n“Here r, r1, and r2 are all not equal to zero.”\n\nUsually \"not equal to zero\" is found in math and science texts as \"non-zero\". As I read the sentence, it would be the same as and more clear as “Here r, r1, and r2 are all non-zero.”\n\nAll values are non-zero: r NOT EQUAL 0 r1 NOT EQUAL 0 r2 NOT EQUAL 0"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93262035,"math_prob":0.9624574,"size":284,"snap":"2021-43-2021-49","text_gpt3_token_len":81,"char_repetition_ratio":0.114285715,"word_repetition_ratio":0.0,"special_character_ratio":0.28521127,"punctuation_ratio":0.15068494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98487,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T22:08:07Z\",\"WARC-Record-ID\":\"<urn:uuid:52f248a1-d6df-4e22-95e1-d4751816fcc5>\",\"Content-Length\":\"134975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8e26e37-b5b5-4d29-ba23-ef37eef2143f>\",\"WARC-Concurrent-To\":\"<urn:uuid:78cf45c2-f953-47ff-9e67-c0f847a12a9c>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://ell.stackexchange.com/questions/291691/all-not-equal-to-zero-meaning\",\"WARC-Payload-Digest\":\"sha1:EEDOYLEYKHGHFB25EJGTFERSRWA2FP4V\",\"WARC-Block-Digest\":\"sha1:STMSQ5G4223PAZAUNERSPF4C53JK6OYK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363125.46_warc_CC-MAIN-20211204215252-20211205005252-00524.warc.gz\"}"} |
http://sciencewise.info/definitions/Benjamin-Ono_equation_by_Alexander_Abanov | [
"## Benjamin-Ono equation\n\nAlso available as: PDF, LaTeX.\nThe Benjamin-Ono equation is a nonlinear partial integro-differential equation. It was introduced by Benjamin (1967) and Ono (1975) as an equation describing one-dimensional internal waves in deep water.\nThe Benjamin-Ono equation has a form\nwhere $H$ is the Hilbert transform.\nIt is a soliton equation and can be solved by an inverse scattering transform.\nReferences:\n• T. Benjamin, Internal waves of permanent form in fluids of great depth, J. Fluid Mech. 29 (1967), 559-562.\n• H. Ono, Algebraic solitary waves in stratified fluids, J. Phys.Soc. Japan 39 (1975), 1082-1091.\n##### Ontology information:",
null,
""
] | [
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"http://sciencewise.info/media/graphviz/1432.160x160.png",
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https://costofcollegeeducation.net/flourless-chocolate-row/how-to-find-arc-length-90b80e | [
"# how to find arc length\n\nAn arc is a segment of a circle around the circumference. A circle is 360° all the way around; therefore, if you divide an arc’s degree measure by 360°, you find the fraction of the circle’s circumference that the arc makes up. {d} {y}\\right.}}} We say in geometry that an arc \"subtends\" an angle θ; literally, \"stretches under.\" Arc Length. The circumference is always the same distance from the centre - the radius. Calculate the arc length to 2 decimal places. How to Find Arc Length An arc is any portion of the circumference of a circle. So we need to find the fraction of the circle made by the central angle we know, then find the circumference of the total circle made by the radius we know. Our tips from experts and exam survivors will help you through. Rounded to 3 significant figures the arc length is 6.28cm. We will now look at some examples of computing arc lengths of curves. S =ba√1+ (f’ (x))2 dxThe Arc Length Formula. If you want really the length of the inner and the outer conture as shown by @mdavis22569 , then you can find it yourselfe: - create a new .idw - create a view at your model - mark the view (it must be selected) - add a new sketch - add the dimensions (choose the dimension type \"arc length\") => you will find the lenghts 6376.52 and 6219.77 … The smaller one is the sagitta as show in the diagram above. Solution : Given that l = 27.5 cm and Area = 618.75 cm 2 . The arc length is then given by: s = ∫ a b 1 + ( d y d x ) 2 d x . Enter the values you have into the boxes, and the arc length calculator will take care of it for you. And the ratio of the circumference to the diameter is the basis of radian measure. A circle is 360° all the way around; therefore, if you divide an arc’s degree measure by 360°, you find the fraction of the circle’s circumference that the arc makes up. If you need to determine the length of an arc or the area of a circle sector, then this is the tool for you. Circumference of a Circle. To find the arc length for an angle θ, multiply the result above by θ: 1 x θ corresponds to an arc length (2πR/360) x θ. An arc length is a portion of the circumference of a circle. Rounded to 3 significant figures the arc length is 6.28cm. A chord separates the circumference of a circle into two sections - the, It also separates the area into two segments - the, $$\\frac{1}{4} \\times \\pi \\times 8 = 2 \\pi$$. the latus rectum; this arc length is Then we just multiply them together. The circumference C of a circle is C = πd. $\\text{Arc length} = \\frac{144}{360} \\times \\pi \\times 7 = 8.8~\\text{cm}$. Let’s try an example where our central angle is 72° and our radius is 3 meters. The length of a curve can be determined by integrating the infinitesimal lengths of the curve over the given interval. To find arc length, start by dividing the arc’s central angle in degrees by 360. s = r θ. In this section we are going to look at computing the arc length of a function. Calculate the length of an arc with radius 10cm and the angle ssubtended by the radius is 50 degrees. Arc Length = θr. A chord separates the circumference of a circle into two sections - the major arc and the minor arc. Finally, multiply that number by 2 × pi to find the arc length. So finally, here’s the formula you’ve been waiting for. How do you find the radius of an arc? length = r = ∫ ab 1+ (dxdy)2 Find the radius (r) of that circle. Many real-world applications involve arc length. Now, arc length is given by (θ/360) ⋅ 2 Π r = l (θ/360) ⋅ 2 ⋅ (22/7) ⋅ 45 = 27.5. θ = 35 ° Example 3 : Find the radius of the sector of area 225 cm 2 and having an arc length of 15 cm. Since the arc is a portion of the circumference, if you know what portion of 360 degrees the arc’s central angle is, you can easily find the length of the arc. Finding an arc length requires knowing a bit about the geometry of a circle. or. Arc length = (Ø/ 360 o) x 2πr Here, r is the radius and Ø is the central angle. Circles are 2D shapes with one side and no corners. L = 234.9 million km. Find the radius, central angle and perimeter of a sector whose arc length and area are 27.5 cm and 618.75 cm 2 respectively. Calculate the sector’s area A = r² x Θ / 2 = 6² * π/4 / 2 = 14.14 inches-squared If you want to find out the central angle or radius of a circle, you can also use an arc length calculator. 90° is one quarter of the whole circle (360°). An arc length is just a fraction of the circumference of the entire circle. Arc Measure Definition. Online arc length calculator to find the arc length of a … Since the arc is a portion of the circumference, if you know what portion of 360 degrees the arc’s central angle is, you can easily find the length of the arc. First calculate what fraction of a full turn the angle is. Arc Length – Worksheet . Arc length is the distance from one endpoint of the arc to the other. In this section, we use definite integrals to find the arc length of a curve. You first get. In this lesson we learn how to find the intercepting arc lengths of two secant lines or two chords that intersect on the interior of a circle. Properties of Rhombuses, Rectangles, and Squares, Interior and Exterior Angles of a Polygon, Identifying the 45 – 45 – 90 Degree Triangle. Arc length is the distance from one endpoint of the arc to the other. How to find arc length using a measurement of Central angle in Degrees. You can find the length of the sagitta using the formula: s=r±√r2−l2where: Notice that there are two results due to the \"plus or minus\" in the formula. . Since the arc is a portion of the circumference, if you know what portion of 360 degrees the arc’s central angle is, you can easily find the length of the arc. 30The fraction is 110th110th the circumference. We say in geometry that an arc \"subtends\" an angle θ; literally, \"stretches under.\" I T IS CONVENTIONAL to let the letter s symbolize the length of an arc, which is called arc length. Figuring out the length of an arc on a graph works out differently than it would if you were trying to find the length of a segment of a circle. Round your answers to the nearest tenth. The length of the unsplitted pipe can be calculated easily by 6150*PI/2 -> 9660.4 (as shown in my image above). Curves with closed-form solutions for arc length include the catenary, circle, cycloid, logarithmic spiral, parabola, semicubical parabola and straight line. Calculate the major arc length to one decimal place. The arc length of the curve y = f (x) from x = a to x = b is given by: \\displaystyle\\text {length}= {r}= {\\int_ { {a}}^ { {b}}}\\sqrt { { {1}+ {\\left (\\frac { { {\\left. The angle measurement here is 40 degrees, which is theta. If you snip (a little more than) 1600 at each end, then the length of 6294.8 seems to be plausible. I didn't calculate it, but created a drawing and added the dimension of the lengthen of the arc (centerline of the pipe). Because, we will be armed with the power of circles, triangles, and radians, and will see how to use our skills and tools to some pretty amazing math problems. We can think of arc length as the distance you would travel if you were walking along the path of the curve. This step gives you. Use the central angle calculator to find arc length. Background is covered in brief before introducing the terms chord and secant. It also separates the area into two segments - the major segment and the minor segment. The result will be the radius. Where the length of a segment of a circle can be figured out with some simple knowledge of geometry (or trigonometry), finding the arc length of a function is a little more complicated. Example $$\\PageIndex{3}$$: Approximating arc length numerically. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. The derivation is much simpler for radians: By definition, 1 radian corresponds to an arc length R. Calculate the perimeter of a semicircle of radius 1. cm using the arc length formula. I T IS CONVENTIONAL to let the letter s symbolize the length of an arc, which is called arc length. And now suddenly we are in a much better place, we don't need to add up lots of slices, we can calculate an exact answer (if we can solve the differential and integral). Multiply 2πr2πr tim… This arc length calculator can help you to not only work out calculations in an instant, but educate you on how to do it by hand as well. Do you want to solve for. Some of the most common plane figures such as rectangles, squares, triangles, etc can be used to find the area and perimeter of any Complex figure. Sectors, segments, arcs and chords are different parts of a circle. If you want to learn how to calculate the arc length in radians, keep reading the article! Then, if you multiply the length all the way around the circle (the circle’s circumference) by that fraction, you get the length along the arc. d. X Research source For example, if the diameter of a circle is 14 cm, to find the radius, you would divide 14 by 2: 14÷2=7{\\displaystyle 14\\div 2=7}. Always verify your function before starting the problem. Area of sector = lr/2 ---(2) (1) = (2) The unit circle. Set up the formula for arc length. Its degree measure is 45° and the radius of the circle is 12, so here’s the math for its length. An arc measure is an angle the arc makes at the center of a circle, whereas the arc length is the span along the arc. Arc length is the distance from one endpoint to the arc of the other. Note: the integral also works with respect to y, useful if we happen to know x=g (y): S =. Then, multiply that number by the radius of the circle. Finding the Arc Length: The objective is to find the arc length of the curve by using the given function. Arc length is defined as the length along the arc, which is the part of the circumference of a circle or any curve. Arc Length. How do you find the angle with arc length and radius? The arc length is $$\\frac{1}{4}$$ of the full circumference. Arc Length of the Curve y = f(x) In previous applications of integration, we required the function $$f(x)$$ to be integrable, or at most continuous. Then, if you multiply the length all the way around the circle (the circle’s circumference) by that fraction, you get the length along the arc. The domain of the expression is all real numbers except where the expression is undefined. Keep in mind that different projections have different spatial properties and distortions. I can’t wait! The arc length formula An arc can be measured in degrees, but it can also be measured in units of length. Length of an arc. The length of an arc formed by 60° of a circle of radius “r” is 8.37 cm. Then, multiply that number by the radius of the circle. These can include arc lengths, the area and perimeter of sectors and the area of segments. Proof of the theorem. The radius is 10, which is r. Plug the known values into the formula. Example 1. For examples can be found on the Arc Length of Curves in Three-Dimensional Space Examples 2 page.. To find arc length, start by dividing the arc’s central angle in degrees by 360. Worksheet to calculate arc length and area of sector (radians). We can find the length of an arc by using the formula: $\\frac{\\texttheta}{360} \\times \\pi~\\text{d}$ $$\\texttheta$$ is the angle of … C = 2 πr. {d} {x}\\right. Arc length is the distance from one endpoint of the arc to the other. An arc is a segment of a circle around the circumference. Read about our approach to external linking. {\\displaystyle s=\\int _ {a}^ {b} {\\sqrt {1+\\left ( {\\frac {dy} {dx}}\\right)^ {2}}}dx.} You can also use the arc length calculator to find the central angle or the circle's radius. Rounded to 3 significant figures the arc length is 6.28cm. Enter the values into the formula (h/2) + (w^2/8h), where h is the arc height and w is the length of the chord. Things to Keep in Mind Image by Pixabay. The other is the longer sagitta that goes the other way across the larger part of the circle: where d is the diameter of the circle and r is the radius of the circle. Since the radius is half the diameter of a circle, to find the radius, simply divide the diameter by 2. The formulas for finding arc length utilize the circle’s radius. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. These two concepts are going to be so helpful when we get to calculus, and are asked to find the arc length and area of things other than circles. Example 1: Find the arc length if the radius of the arc is 8 cm and its central angle is 30 o. How to Find Arc Length An arc is any portion of the circumference of a circle. Even easier, this calculator can solve it for you. The formula to calculate the arc length is: $\\text{Arc length} = \\frac{\\text{angle}}{360^\\circ} \\times \\pi \\times d$ {d} {x}\\right.} An arc is a part of the circumference of a circle. Here is a step by step guide on how to find arc length. This expression for the parabola arc length becomes especially when the arc is extended from the apex to the end point (1 2 a, 1 4 a) of the parametre, i.e. Find the Arc Length, Check if is continuous. The circumference of a circle is the total length of the circle (the “distance around the circle”). The circumference of a circle is an arc length. Finally, multiply that number by 2 × pi to find the arc length. In this post, we will mainly focus on the circular shape and the various parts of the circle to find the area of sector and length of an Arc. The definition of radian measure. Arc length … Find the length of the sine curve from $$x=0$$ to $$x=\\pi$$. The arc length is $$\\frac{1}{4} \\times \\pi \\times 8 = 2 \\pi$$. The major sector has an angle of $$360 - 110 = 250^\\circ$$. }}}\\right)}^ {2}}} {\\left. It requires knowing a bit about the geometry of a circle. (π = 3.14)r = 24 (π = 3.14) r = 24 cm, θ = 60∘ θ = 60 ∘ An arc is part of a circle. Solution : Area of sector = 225 cm 2 ---(1) Arc length = 15 cm. This is somewhat of a mathematical curiosity; in Example 5.4.3 we found the area under one \"hump\" of the sine curve is 2 square units; now we are measuring its arc length. Again, when working with π, if we want an exact answer, we use π. Calculate the minor arc length to one decimal place. You can only calculate the area, length, or perimeter of features if the coordinate system being used is projected. If you want to learn how to calculate the arc length in radians, keep reading the article! To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps: Assign variable names to the values in the problem. Section 2-1 : Arc Length. Sign in, choose your GCSE subjects and see content that's tailored for you. Calculate the area of a sector: A = r² * θ / 2 = 15² * π/4 / 2 = 88.36 cm² . If you know the diameter of the circle, you can still find the arc length. An arc’s length means the same commonsense thing length always means — you know, like the length of a piece of string (with an arc, of course, it’d be a curved piece of string). L e n g t h = θ ° 360 ° 2 π r. The arc length formula is used to find the length of an arc of a circle. how to find the arc length of a 3-dimensional vector function, How to find arc length if a curve is given in parametric form, examples and step by step solutions, A series of … Below, find out what the arc length formula is, obtain instructions for its use, and get the equation for an arc’s sector. How to Find the Arc Length. Finding an arc length requires knowing a bit about the geometry of a circle. Finding an arc length requires knowing a bit about the geometry of a circle. Interval Notation: Set-Builder Notation: is continuous on . There are many geometric shapes we encounter in mathematics precisely. You can try the final calculation yourself by rearranging the formula as: L = θ * r. Then convert the central angle into radians: 90° = 1.57 rad, and solve the equation: L = 1.57 * 149.6 million km. { { {\\left. The arc length is the measure of the distance along the curved line making up the arc.It is longer than the straight line distance between its endpoints (which would be a chord) There is a shorthand way of writing the length of an arc: This is read as \"The length of the arc AB is 10\". arc length = [radius • central angle (radians)] arc length = circumference • [central angle (degrees) ÷ 360] where circumference = [2 • π • radius] Knowing two of these three variables, you can calculate the third. The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept. Make sure you don’t mix up arc length with the measure of an arc which is the degree size of its central angle. Arc Length and Sector Area – Example 1: Find the length of the arc. We can use the measure of the arc (in degrees) to find its length (in linear units). So arc length s for an angle θ is: s = (2π R /360) x θ = π θR /180. We also find the angle given the arc lengths. Calculate the arc length according to the formula above: L = r * θ = 15 * π/4 = 11.78 cm. So, the radius of the ci… Measure the length of the chord and the length of the bisecting line segment from the chord to the top of the arc. An arc measure is an angle the arc makes at the center of a circle, whereas the arc length is the span along the arc. Arc Measure Definition. Remember the circumference of a circle = $$\\pi d$$ and the diameter = $$2 \\times \\text{radius}$$. Then a formula is presented that we will use to meet this lesson's objectives. Solution. Therefore, the length of the arc is 8.728 cm. Simplify to solve the formula. The circumference of a circle is an arc length. where θ is the measure of the arc (or central angle) in radians and r is the radius of the circle. A circle is 360° all the way around; therefore, if you divide an arc’s degree measure by 360°, you find the fraction of the circle’s circumference that the arc makes up. In this case, there is no real number that makes the expression undefined. Then, if you multiply the length all the way around the circle (the circle’s circumference) by that fraction, you get the length along the arc. An angle of 1 radian. For a function f(x), the arc length is given by s = \\int_{a}^{b} \\sqrt{ 1 + (\\frac{dy}{dx})^2 } … The lower case L in the front is short for 'length'. Sometimes we need to know how to calculate values for specific sections of a circle. Finding the length of an arc is functionally not that different from finding the length of a diagonal line segment on a graph, although you have to use more complicated math to get the proper result. The formula to calculate the arc length is: $\\text{Arc length} = \\frac{\\text{angle}}{360^\\circ} \\times \\pi \\times d$. $\\text{Arc length} = \\frac{250}{360} \\times \\pi \\times 12 = 26.2~\\text{cm}$. Home Economics: Food and Nutrition (CCEA). Because it’s easy enough to derive the formulas that we’ll use in this section we will derive one of them and leave the other to you to derive. It for you to find arc length real numbers except where the expression is.! You were walking along the path of the chord and secant one quarter of entire... Then a formula is presented that we will now look at computing arc! The formulas for finding arc length utilize the circle 's radius circle ( 360° ) it also separates the C... Know x=g ( y ): s = end, then how to find arc length length a. Sector: a = r² * θ / 2 = 15² * π/4 / 2 88.36! Can be measured in degrees ) to \\ ( \\PageIndex { 3 } \\ ) of that.. Length, or perimeter of a circle s = that circle GCSE and. The circumference of a circle real number that makes the expression is all numbers. A bit about how to find arc length geometry of a function is presented that we will now at! { d } { 4 } \\ ) of that circle presented we... Try an Example where our central angle in degrees, which is theta about... Measurement here is a segment of a circle is C = πd, the length of 6294.8 seems to plausible. You would travel if you know the diameter by 2 × pi to find arc length area... S symbolize the length of the arc ( or central angle or the circle also works respect... This calculator can solve it for you, start by dividing the arc to the.! 4 } \\ ): s = ( Ø/ 360 o ) x θ = 15 * /... Online arc length an arc length to one decimal place walking along the path of circle! Formula is presented that we will now look at some examples of computing arc of... Y ): s = going to look at computing the arc length formula you travel! S symbolize the length of an arc subtends '' an angle θ ; literally, under. In degrees, which is r. Plug the known values into the formula above L. = 225 cm 2 -- - ( 1 ) arc length calculator to find arc length using measurement... With arc length = ( Ø/ 360 o ) x 2πr here r! Would travel if you snip ( a little more than ) 1600 at each end then... Home Economics: Food and Nutrition ( CCEA ) also find the arc you have the., stretches under. - 110 = 250^\\circ\\ ) circles are shapes! And the area, length, start by dividing the arc ’ s try an Example our. Can be found on the arc of the circumference to the diameter the! Or any curve a segment of a circle the distance from one endpoint to the diameter of the circle the. Solve it for you two sections - the major sector has an angle θ ; literally, stretches... You ’ ve been waiting for the math for its length ( in units.: a = r² * θ = π θR /180 geometry of a full turn the angle how to find arc length arc,! 360° ) were walking along the arc is any portion of the circumference of function... Radius ( r ) of the circle be measured in degrees length: the objective is find. The path of the circumference of a circle circle and r is the diameter of circumference! Is all real numbers except where the expression is all real numbers except where the expression is undefined defined... Length as the distance you would travel if you want to know x=g ( y ): =. Meet this lesson 's objectives if we happen to know x=g ( y:. Use the arc length of an arc length of a semicircle of radius 1. cm using given! By 360 one endpoint of the circle r ” is 8.37 cm ( \\PageIndex 3. At some examples of computing arc lengths of curves as show in the diagram above s. Presented that we will now look at computing the arc length s for an angle ;. Length in radians, keep reading the article = r * θ / 2 = *... Keep in mind that different projections have different spatial properties and distortions formula! Length an arc length of the bisecting line segment from the centre the... The other 3 significant figures the arc ’ s the formula for arc according. One is the distance you would travel if you were walking along the,! Called arc length = 88.36 cm² its degree measure is 45° and area... = 15 cm the measure of the circumference is always the same distance from one endpoint to the top the! End, then the length of an arc is 8.728 cm shapes we encounter in mathematics precisely Check if continuous. Covered in brief before introducing the terms chord and secant: given that L = *... Rounded to 3 significant figures the arc length can be measured in units of length examples can be in. Case, There is no real number that makes the expression is undefined 2D with. Can solve it for you to look at some examples of computing arc lengths of! Length as the distance from one endpoint of the circle sine curve from (. Radius of the circumference of a curve arc with radius 10cm and radius... R ” is 8.37 cm … There are many geometric shapes we encounter mathematics... You would travel if you want to learn how to find the arc length calculator to find length. With radius 10cm and the angle is 72° and our radius is degrees! Still find the central angle calculator to find the radius, simply divide the diameter a... Area and perimeter of features if the coordinate system being used is projected that! A semicircle of radius 1. cm using the given function take care it... Length in radians, keep reading the article to 3 significant figures the arc length a... L = 27.5 cm and area = 618.75 cm 2 walking along the path of the circle, is! In a couple of examples spatial properties and distortions ” is 8.37 cm chord the! = 88.36 cm² rocket is launched along a parabolic path, we use definite to... Called arc length of an arc, which is theta length requires knowing a bit about the geometry a... Arc to the top of the circumference of a circle angle ssubtended by the radius the... Segment from the centre - the radius of the circle is C = πd take care of for... To y, useful if we want an exact answer, we use definite integrals to the... Is length of the arc length formula - Example 1 Discuss the formula “ distance the. Radians and r is the distance from the chord to the other } ^ 2. -- - ( 1 ) arc length of the arc length is \\ ( \\frac { }! Angle calculator to find arc length requires knowing a bit about the geometry of a semicircle radius. Definite integrals to find the angle ssubtended by the radius of an arc length circumference to the diameter of full! Notation: is continuous on calculate the perimeter of sectors and the radius of the circumference a. Each end, then the length of the bisecting line segment from the chord to the diameter the... About the geometry of a semicircle of radius 1. cm using the given function, Check if continuous... Cm and area = 618.75 cm 2 -- - ( 1 ) arc length length utilize the circle, find... Arc with radius 10cm and the radius and Ø is the total length of arc. To meet this lesson 's objectives portion of the arc to the of. Let ’ s radius a parabolic path, we might want to know how far the rocket travels ”. Where d is the distance from one endpoint of the arc length is the is! One side and no corners that an arc can be measured in units of length at each end, the... Sectors, segments, arcs and chords are different parts of a around... } \\ ) of the circumference circumference C of a function is theta the front is short for 'length.. 4 } \\times \\pi \\times 8 = 2 \\pi\\ ) degrees, which is called arc length the..., useful if we want an exact answer, we use definite to... L in the diagram above be found on the arc length requires knowing a bit about the geometry a... = πd start by dividing the arc length fraction of a circle is an arc length formula:. Length of the circumference of a function now look at some examples computing. Calculator will take care of it for you introducing the terms chord and secant, then length... Is continuous you have into the formula measure of the circumference of a circle is the distance would! Introducing the terms chord and secant 8 = 2 \\pi\\ ) of features if the coordinate system being used projected... Geometric shapes we encounter in mathematics precisely length formula - Example 1: find angle... Front is short for 'length ' pi to find its length ( in,. Called arc length and area of segments: Approximating arc length ) in radians, keep the! Is always the same distance from one endpoint to the other is 10, which is arc. Angle ssubtended by how to find arc length radius Space examples 2 page geometry of a circle into two -."
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"# Casio FX-570 ES Plus Scientific Calculator\n\nAvailability:\n\n1 in stock\n\nEGP325.00 EGP350.00\n\n1 in stock\n\n## easy to teach and easy to learn!\n\n### New functions\n\n• New equation mode\n\n###",
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"• Random integers\n\n### Standard functions\n\n• Fraction calculations\n• Combination and permutation\n• Statistics (List-based STAT data editor, standard deviation, regression analysis)\n• 9 variables\n• Table function\n• Comes with new slide-on hard case\n\n### fx-82ES PLUS/85ES PLUS/350ES PLUS functions, in addition to:\n\n• Equation calculations\n• Integration/differential calculations\n• Matrix calculations\n• Vector calculations\n• Complex number calculations\n• CALC function\n• SOLVE function\n• Base-n calculation\n\n####",
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"",
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"SKU: 46685696 Category:\nBrand CASIO\n\n## Other features\n\n• One AAA-size(R03) battery\n• 417 Functions\n• Full Dot Display\n• Plastic Keys\n• Integration / Differential\n• Equation calculation\n• Matrix calculations / Vector calculations\n• Complex number calculation\n• CALC function / SOLVE function\n• Base-n calculation\n• Fraction calculation\n• Combination and Permutation\n• Logarithm log\n• List-based STAT data editor\n• Standard deviation\n• Paired-variable statistics regression analysis\n• Logical operations\n• Table function\n• 40 scientific constants\n• 40 metric conversions (20 conversion pairs)\n• 9 variable memories\n• Comes with new slide on hard case\n\nFor more Vist edu.casio.com\n\n## Based on 0 reviews\n\n0.0 overall\n0\n0\n0\n0\n0\n\nThere are no reviews yet."
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20295%20492'%3E%3C/svg%3E",
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"https://darkaram.com/wp-content/uploads/2019/10/img01_01.png",
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"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20440%20692'%3E%3C/svg%3E",
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"https://darkaram.com/wp-content/uploads/2019/10/Untisstled.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60787404,"math_prob":0.9620372,"size":659,"snap":"2019-51-2020-05","text_gpt3_token_len":147,"char_repetition_ratio":0.2259542,"word_repetition_ratio":0.0,"special_character_ratio":0.2124431,"punctuation_ratio":0.05319149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947246,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,2,null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T08:28:22Z\",\"WARC-Record-ID\":\"<urn:uuid:a2902f61-f5cf-4de7-8678-b3c51e2237a5>\",\"Content-Length\":\"190568\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05cf4e3d-1f17-46b8-bab3-4f67f58b76f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:a108a1db-3d20-4c43-a499-ba39ea6cd809>\",\"WARC-IP-Address\":\"104.27.166.175\",\"WARC-Target-URI\":\"https://darkaram.com/product/casio-fx-570-es-plus-scientific-calculator/\",\"WARC-Payload-Digest\":\"sha1:BV3ZCB3JJDFRAK2LCKAGT5JDMIJPAGH4\",\"WARC-Block-Digest\":\"sha1:3GDXZJIDR5YUK2QLZFMP7WWJROFT3GGD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592394.9_warc_CC-MAIN-20200118081234-20200118105234-00172.warc.gz\"}"} |
https://math.stackexchange.com/questions/3297902/square-integrable-eigenfunctions-of-the-schr%C3%B6dinger-operator-decay-pm-infty/3297996 | [
"# Square-integrable eigenfunctions of the Schrödinger operator decay $\\pm \\infty$ [closed]\n\nWhy is it necessary for an eigenfunction of $$H=\\frac{d^2}{dx^2}+u(x)$$ that is square-integrable that it tends to zero at $$\\pm \\infty$$?\n\n## closed as off-topic by cmk, RRL, mrtaurho, Xander Henderson, The CountJul 29 at 1:28\n\nThis question appears to be off-topic. The users who voted to close gave this specific reason:\n\n• \"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.\" – cmk, RRL, mrtaurho, Xander Henderson, The Count\nIf this question can be reworded to fit the rules in the help center, please edit the question.\n\n• What is the tail of integral? Could you give a detailed proof? – Lucas Pereiro Jul 19 at 16:26\n• What is $u$? $\\$ – cmk Jul 19 at 16:57\n• @cmk: $u$ is usually a potential function of some kind, like potential energy. – Adrian Keister Jul 19 at 18:35\n• @AdrianKeister You're right, of course, but I was encouraging them to add more context to their problem, like their assumptions on $u$. Otherwise, they're risking having their question closed. – cmk Jul 19 at 18:38\n• @AdrianKeister I don't disagree with you, but I'd say that if someone is looking at the problem statement and sees \"show $L^2$ eigenfunctions of $\\frac{d^2}{dx^2}+u$ vanish at infinity,\" they might like to know what assumptions are present on $u$ before providing a rigorous argument. FYI, I agree with your physical interpretation of the problem and +1'd. – cmk Jul 19 at 18:44\n\n## 1 Answer\n\nMainly because of the probability interpretation. If $$\\psi$$ is an eigenfunction of the Schrodinger equation, then the statistical interpretation says that $$\\int_a^b|\\psi(x)|^2\\,dx$$ gives the probability of finding the particle between $$a$$ and $$b$$. By the rules of probability, we must have $$\\int_{-\\infty}^{\\infty}|\\psi(x)|^2\\,dx=1<\\infty.$$ This certainly cannot happen unless the function itself decays to zero."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90945584,"math_prob":0.978227,"size":2203,"snap":"2019-35-2019-39","text_gpt3_token_len":594,"char_repetition_ratio":0.105957255,"word_repetition_ratio":0.036363635,"special_character_ratio":0.26373127,"punctuation_ratio":0.12832929,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921458,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T22:18:06Z\",\"WARC-Record-ID\":\"<urn:uuid:d0279508-4e68-4b14-b0c2-31f217ae1259>\",\"Content-Length\":\"125507\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:161a9cdb-254d-4deb-a90b-905a6a6475cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:d19edb70-95e9-4099-849b-6e21d9ab6dbc>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3297902/square-integrable-eigenfunctions-of-the-schr%C3%B6dinger-operator-decay-pm-infty/3297996\",\"WARC-Payload-Digest\":\"sha1:KEDAKHN62PM4MHHUST7BVR2XZNC44Y5I\",\"WARC-Block-Digest\":\"sha1:AHZGMUHCTFW5UJIIKT6ZAHISA43WADSM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027316549.78_warc_CC-MAIN-20190821220456-20190822002456-00104.warc.gz\"}"} |
https://gitlab.unige.ch/Conor.Mccoid/UNIGE/-/commit/0bf5ca77c2b1a2c09e63335d6fa8a1291e1b5468 | [
"### Extrap: equivalence bewteen MPE and GMRES, with both numerical and logical evidence\n\nparent eb72d8aa\n % Example - Extrapolation: fixed point iteration on linear function % solve the equation x = Ax + b for random A and b d = 10; n=2*d; d = 5; n=2*d; A = rand(d); b = rand(d,1); x_true = -(A - eye(d)) \\ b; X = b; for i=1:n ... ... @@ -25,7 +25,7 @@ for i=1:n res_MPE_v2(i) = norm(A*x_v2(:,i) + b - x_v2(:,i)); end [x_GMRES,~,~,~,res_GMRES]=gmres(A-eye(d),-b,d,0,n); [x_GMRES,~,~,~,res_GMRES]=gmres(A-eye(d),-b,d,0,n,eye(d),eye(d),X(:,1)); figure(1) semilogy(1:n,Error_v1,'b*--',1:n,Error_v2,'k.--',1:length(res_GMRES),res_GMRES,'ro') ... ...\n ... ... @@ -561,5 +561,47 @@ Table \\ref{tab:KrylovExtrap} gives the orthogonalization conditions and correspo \\end{table} The methods become Krylov methods most readily when the sequence to be accelerated may be expressed as $\\vec{x}_{n+1} = A \\vec{x}_n + \\vec{b}$. Under this sequence the function $\\fxi{n}=\\vec{x}_{n+1}-\\vec{x}_n$ satisfies \\begin{align*} \\fxi{n}= & \\vec{x}_{n+1} - \\vec{x}_n \\\\ = & A \\vec{x}_n + \\vec{b} - \\vec{x}_n \\\\ = & (A-I) \\vec{x}_n + \\vec{b}, \\\\ \\fxi{n+1} = & \\vec{x}_{n+2} - \\vec{x}_{n+1} \\\\ = & A \\vec{x}_{n+1} + \\vec{b} - A \\vec{x}_n - \\vec{b} \\\\ = & A (\\vec{x}_{n+1} - \\vec{x}_n) \\\\ = & A \\fxi{n}. \\end{align*} Suppose we wish to solve $(A-I) \\vec{x} = -\\vec{b}$. It is then the goal to minimize $\\norm{(A-I) \\hat{\\vec{x}} + \\vec{b}}$. We consider three types of solutions: \\begin{enumerate} \\item $\\hat{\\vec{x}} = X_{n,k} \\vec{u}_k$ such that $\\vec{1}^\\top \\vec{u}_k=1$; \\begin{align*} \\norm{(A-I) X_{n,k} \\vec{u}_k + \\vec{b}} = & \\norm{((A-I) X_{n,k} + \\vec{b} \\vec{1}^\\top) \\vec{u}_k} \\\\ = & \\norm{F_{n,k} \\vec{u}_k}. \\end{align*} \\item $\\hat{\\vec{x}} = \\vec{x}_n + X_{n,k} \\Delta \\tilde{\\vec{u}}_k$ where \\begin{equation*} \\Delta = \\begin{bmatrix} -1 \\\\ 1 & \\ddots \\\\ & \\ddots & -1 \\\\ & & 1 \\end{bmatrix}; \\end{equation*} \\begin{align*} \\norm{(A-I) \\vec{x}_n + (A-I) X_{n,k} \\Delta \\tilde{\\vec{u}}_k + \\vec{b}} = & \\norm{\\fxi{n} + F_{n,k} \\Delta \\tilde{\\vec{u}}_k} \\\\ = & \\norm{\\fxi{n} + (A-I) F_{n,k-1} \\tilde{\\vec{u}}_k}. \\end{align*} \\item $\\hat{\\vec{x}} =\\vec{x}_n + Q_k \\vec{y}_k$ where $Q_k$ is derived from the Arnoldi iteration on the Krylov subspace $\\mathcal{K}_{k-1}(A-I,\\fxi{n})$; \\begin{align*} \\norm{(A-I) \\vec{x}_n + (A-I) Q_k \\vec{y}_k + \\vec{b}} = & \\norm{\\fxi{n} + (A-I) Q_k \\vec{y}_k}. \\end{align*} \\end{enumerate} Method 1 is equivalent to MPE since minimizing this equation is equivalent to solving the normal equations. We have previously shown that methods 1 and 2 are equivalent. Method 3 is GMRES, and to show it is equivalent to method 2 it suffices to prove the Krylov subspace $\\mathcal{K}_{k-1}(A-I,\\fxi{n})$ is the same as the column space of $F_{n,k-1}$. This is trivial to show. If $Q_k$ is derived from another method but shares the column space of $F_{n,k-1}$ then there is still mathematical equivalence between methods 2 and 3. If one minimizes with respect to a different norm then the methods correspond to other methods discussed here. \\end{document} \\ No newline at end of file\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
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http://cocoa.dima.unige.it/download/CoCoAManual/html/cmdLexSegmentIdeal.html | [
"up previous next\n LexSegmentIdeal\n\nlex-segment ideal containing L, or with the same HilbertFn as I\n\n Syntax\n ```LexSegmentIdeal(L: LIST of power-products): IDEAL LexSegmentIdeal(I: IDEAL): IDEAL```\n\n Description\nIf the argument is a LIST of power-products L , this function returns the smallest lex-segment ideal containing the power-products in L .\n\nIf it is an IDEAL I , it returns the lex-segment ideal having the same Hilbert function as I .\n\n Example\n ```/**/ use R ::= QQ[x,y,z]; /**/ LexSegmentIdeal([y^3]); ideal(y^3, x*z^2, x*y*z, x*y^2, x^2*z, x^2*y, x^3) /**/ LexSegmentIdeal(ideal(y^3)); ideal(x^3) ```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.558826,"math_prob":0.89166385,"size":547,"snap":"2019-13-2019-22","text_gpt3_token_len":177,"char_repetition_ratio":0.160221,"word_repetition_ratio":0.0,"special_character_ratio":0.29981717,"punctuation_ratio":0.17322835,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9968218,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-22T20:59:40Z\",\"WARC-Record-ID\":\"<urn:uuid:eead4774-2ffd-4776-8197-8b4a633efa3f>\",\"Content-Length\":\"2322\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:94309378-d915-4f51-b1c7-9464e2d5346d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a901e10c-938f-4100-b9c5-87647ca99923>\",\"WARC-IP-Address\":\"130.251.60.30\",\"WARC-Target-URI\":\"http://cocoa.dima.unige.it/download/CoCoAManual/html/cmdLexSegmentIdeal.html\",\"WARC-Payload-Digest\":\"sha1:P5YIZG4WUZTLRBUNI4PUEDIBNLL3ETAM\",\"WARC-Block-Digest\":\"sha1:A4WZURY42BAZNMASHZWT73EIYMOHS7LG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202689.76_warc_CC-MAIN-20190322200215-20190322222215-00352.warc.gz\"}"} |
https://escholarship.org/uc/item/1618x8jb | [
"",
null,
"Open Access Publications from the University of California\n\n## Continuous data assimilation with stochastically noisy data\n\n• Author(s): Bessaih, H\n• Olson, E\n• Titi, ES\n• et al.\n\n## Published Web Location\n\nhttps://doi.org/10.1088/0951-7715/28/3/729\nAbstract\n\nWe analyse the performance of a data-assimilation algorithm based on a linear feedback control when used with observational data that contains measurement errors. Our model problem consists of dynamics governed by the two-dimensional incompressible Navier-Stokes equations, observational measurements given by finite volume elements or nodal points of the velocity field and measurement errors which are represented by stochastic noise. Under these assumptions, the data-assimilation algorithm consists of a system of stochastically forced Navier-Stokes equations. The main result of this paper provides explicit conditions on the observation density (resolution) which guarantee explicit asymptotic bounds, as the time tends to infinity, on the error between the approximate solution and the actual solutions which is corresponding to these measurements, in terms of the variance of the noise in the measurements. Specifically, such bounds are given for the limit supremum, as the time tends to infinity, of the expected value of the L2-norm and of the H1 Sobolev norm of the difference between the approximating solution and the actual solution. Moreover, results on the average time error in mean are stated.\n\nMany UC-authored scholarly publications are freely available on this site because of the UC's open access policies. Let us know how this access is important for you."
] | [
null,
"https://escholarship.org/images/logo_eschol-mobile.svg",
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https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-019-2272-4 | [
"Theory and Modern Applications\n\n# Comparison of fractional order techniques for measles dynamics\n\n## Abstract\n\nA mathematical model which is non-linear in nature with non-integer order ϕ, $$0 < \\phi \\leq 1$$ is presented for exploring the SIRV model with the rate of vaccination $$\\mu _{1}$$ and rate of treatment $$\\mu _{2}$$ to describe a measles model. Both the disease free $$\\mathcal{F}_{0}$$ and the endemic $$\\mathcal{F}^{*}$$ points have been calculated. The stability has also been argued for using the theorem of stability of non-integer order differential equations. $$\\mathcal{R} _{0}$$, the basic reproduction number exhibits an imperative role in the stability of the model. The disease free equilibrium point $$\\mathcal{F}_{0}$$ is an attractor when $$\\mathcal{R}_{0} < 1$$. For $$\\mathcal{R}_{0} > 1$$, $$\\mathcal{F}_{0}$$ is unstable, the endemic equilibrium $$\\mathcal{F}^{*}$$ subsists and it is an attractor. Numerical simulations of considerable model are also supported to study the behavior of the system.\n\n## 1 Introduction\n\nMeasles is a worldwide disease. Both epidemic and endemic occurrences are known. The highest incidence is in winter and spring. It is a respiratory disease (measles) triggered by a virus known as paramyxovirus. It is generally held that measles is not a good reason to consult a doctor. Instead, the parents prefer to visit a temple. The poorer community has the higher occurrence of infection at lower age . All this adds up to the problems of the child with measles. No specific treatment is available. General measures consist of isolation, cough sedatives, vasoconstrictor nasal drops, antipyretics, attention to eye and mucous membrane of mouth, antihistaminic for itching, and maintenance of proper fluid and dietary intake. In the case bacterial infection is superimposed, proper antibiotics should be given. Antiviral agents are not of proven value. Gamma globulins, hyper immune gamma globulins and steroids are of doubtful value . The symptoms of the measles appear 10 to 14 days after a person is infected with the measles virus. Worldwide, measles vaccination has been effective, reducing measles deaths by 78% from an estimated 562,400 deaths in 2000 to 122,000 in 2012. A few nonlinear models are given in [3,4,5,6,7,8,9,10,11,12,13,14,15,16, 36, 37].\n\nMeasles cases have continued to climb into 2019. Preliminary global data shows that reported cases rose by 300% in the first 3 months of 2019, compared to the same period in 2018. In 2017, it caused nearly 110,000 deaths . During 2000–2017, measles vaccination prevented an estimated 21.1 million deaths. Global measles death has decreased by 80% from an estimated 545,000 in 2000 to 110,000 in 2017 . The WHO estimated that 875,000 children died of measles in 1999 . This is 56% of all estimated deaths from vaccine-preventable diseases of childhood for that year, making measles the leading cause of vaccine-preventable child mortality.\n\nIn recent years’ fractional order derivatives have been introduced in compartment models, replacing ordinary derivatives with fractional derivatives. This has been motivated by the utility of fractional derivatives in incorporating history effects. The geometrical and physical significance of the fractional integral having a complex and real conjugate power-law exponent has been suggested. One somatic implication of the non-integer order in non-integer derivatives concerns the index of memory . Moreover, non-integer calculus displays a vivacious part of superdiffusive and subdiffusive measures, which make it an obliging instrument in the learning of syndrome transmission . As integer order differential equations cannot authoritatively depict the exploratory and field estimation information, as an alternative approach fractional order differential equations mock-ups are at the moment being widely applied . The fractional modeling is a beneficial approach which has been practiced to understand the comportment of syndromes; the non-integer derivative is a generalization of the ordinary derivative. The non-integer order is global in nature whereas the ordinary derivative is local. The fractional order may offer extra ‘freedom’ to amend the model to factual data from particular patients; specifically, that the fractional order index subsidizes positively to better fits of the patients’ information. The benefit of FDE systems over ODE frameworks is that they allow for more notable degrees of flexibility and they include a memory effect in the model. In other words, they give a superb manoeuvre for the portrayal of memory and traditional gears which were not considered in the established non-fractional order prototypical. Fractional calculus has heretofore been applied in epidemiological research . Also, fractional order models retain memory so a fractional order differential equation gives us a more realistic way to model measles. A significant role of modeling enterprises is that it can alert us to the paucities in our up-to-date understanding of the epidemiology of numerous infectious ailments, and propose critical queries for analysis and data that need to be collected. Lately, it has been applied to analyze dengue internal transmission model, bovine babesiosis disease and a tick population model, a HIV/AIDS epidemic model and the Lengyel–Epstein chemical reaction model [6, 7, 20, 21]. Although the operating of the non-integer approach is more complex than the out-dated one, there are numerical approaches for cracking systems of DEs which are nonlinear. Recently, a majority of the dynamical frameworks centered on the non-fractional order calculus have been altered into the non-integer order domain with regard of the additional degrees of opportunity and the malleability which can be utilized to resolutely fit the test information this being much superior over anything concerned with integer order molding.\n\nAs of not long ago, FC was considered as a fairly exclusive numerical hypothesis without applications, yet over the most recent couple of decades there has been a blast of research efforts on the use of FC to assorted scientific fields extending from the material science of dispersion and shift in weather conditions phenomena, to control frameworks to economics and financial matters. Without a doubt, at present, applications and/or efforts identified with FC have showed up in at least the following fields :\n\n• Non-integer control of engineering systems.\n\n• Development of calculus of variations and optimal control to non-integer dynamic structures.\n\n• Numerical and analytical tools and procedures.\n\n• Essential investigations of the electrical, mechanical, and thermal constitutive relations and other properties of numerous engineering materials, for instance foams, gels, animal tissues, and viscoelastic polymers, and their engineering and scientific applications.\n\n• Essential understanding of diffusion and wave phenomena, their measurements and verifications, comprising applications to plasma physics (such as diffusion in Tokamak).\n\n• Biomedical and bioengineering applications.\n\n• Thermal modeling of engineering structures for instance brakes and machine tools.\n\n• Signal and image processing.\n\nThis article contains five sections. The overview is the first section wherein we present some antiquities of non-integer calculus. In Sect. 2, we will present details of the concept of FDEs. In Sect. 3, we deliberate on the non-integer order model allied with the dynamics of measles model. Qualitative dynamics of the considerable system is resolute using elementary reproduction number. We provide the local stability analysis of the disease free equilibrium (DFE) and endemic equilibrium (EE) points. In Sect. 4, numerical replications are shown to confirm the core outcomes and the conclusion is in Sect. 5.\n\n## 2 Beginnings\n\nFor numerous ages, there have been many demarcations that address the idea of non-integer derivatives [6, 7]. In this section, the Caputo (C), Riemann Liouville (RL) and Grunwald–Letnikov (GL) fractional derivative (FD) demarcations are presented. Firstly, we present the definition of the RL non-integer integral,\n\n$$J^{\\varsigma } g ( z ) = \\bigl( \\varGamma (\\varsigma ) \\bigr) ^{-1} \\int _{0}^{z} \\frac{ g ( s )}{(z-s)^{1-\\gamma }}\\,ds,$$\n(1)\n\nwhere $$\\gamma >0$$, $$g\\in L^{1} ( R^{+} )$$, and $$\\varGamma (\\cdot)$$ is the Gamma function.\n\nThe RL derivative is\n\n\\begin{aligned} D_{R}^{\\gamma } g ( z ) =& \\frac{d^{m}}{d z^{m}} \\bigl[ J ^{m-\\gamma } g ( z ) \\bigr] \\\\ =& \\frac{1}{\\varGamma (m- \\gamma )} \\frac{d^{m}}{d z^{m}} \\int _{0}^{z} \\frac{g ( s )}{(z-s)^{1-m+ \\gamma }}\\,ds, \\quad m-1\\leq \\gamma < m. \\end{aligned}\n(2)\n\nThe Caputo fractional derivative is given by\n\n$$D_{C}^{\\gamma } g ( x ) = J^{M-\\varsigma } \\biggl[ \\frac{d ^{M}}{d x^{M}} g ( x ) \\biggr] = \\frac{1}{\\varGamma (M- \\varsigma )} \\int _{0}^{x} (x-s)^{M-\\varsigma -1} g^{(M)} ( s )\\,ds,$$\n(3)\n\nwhere $$M> \\gamma$$, $$\\forall M\\in Z^{+}$$.\n\nThe GL derivative is given by\n\n$${}_{a} D_{x_{k}}^{\\gamma } g ( x ) = \\lim _{h\\rightarrow 0} \\frac{1}{h^{\\gamma }} \\sum _{j=0}^{[ \\frac{x-a}{h} ]} (-1 )^{j} \\binom{ \\gamma }{j} g ( x-jh ),$$\n(4)\n\nwhere $$[\\,\\cdot\\,]$$ means the non-fractional quantity.\n\nThe Laplace transform of the Caputo FD is written\n\n$$\\mathcal{L} \\bigl[ D_{C}^{\\varphi } g ( x ) \\bigr] = s ^{\\varphi } G ( s ) - \\sum_{j=0}^{n-1} g^{(j)} ( 0 ) s^{\\varphi -j-1}.$$\n(5)\n\nThe Mittag-Leffler (ML) function is defined by using an infinite power series:\n\n$$E_{\\alpha ,\\beta } ( s ) = \\sum_{k=0}^{\\infty } \\frac{s ^{k}}{(\\alpha k+\\beta )}.$$\n(6)\n\nThe Laplace transform of the functions is\n\n$$\\mathcal{L} \\bigl[ t^{\\beta -1} E_{\\alpha ,\\beta } \\bigl( \\pm a t ^{\\alpha } \\bigr) \\bigr] = \\frac{s^{\\alpha -\\beta }}{s^{\\alpha } \\mp a}.$$\n(7)\n\nLet $$\\alpha , \\beta > 0$$ and $$z \\in \\mathbb{C}$$, and the Mittag-Leffler functions mollify the equality given by Theorem 4.2 in ,\n\n$$E_{\\alpha ,\\beta } ( z ) =z E_{\\alpha ,\\alpha +\\beta } ( z ) + \\frac{1}{\\varGamma (\\beta )}.$$\n(8)\n\n### Definition 1\n\n“A function F is Hölder continuous if there are non-negative amounts W, ν such that\n\n$$\\bigl\\Vert F ( p ) -F(\\tau ) \\bigr\\Vert \\leq W \\Vert p- \\tau \\Vert ^{\\nu },$$\n(9)\n\nfor all p, τ in the purview of F and ν is the Hölder exponent. We represent the space of Hölder-continuous functions by $$W^{0,\\nu }$$” .\n\nConsider the fractional order system:\n\n$$D_{C}^{\\varphi } \\varpi ( h ) =z(h,\\varpi ).$$\n(10)\n\nWe have the initial condition $$\\varpi ( 0 ) = \\varpi _{0}$$ where $$D_{C}^{\\varsigma } \\varpi ( h ) =( D_{C}^{\\varsigma } \\varpi _{1} ( h ), D_{C}^{\\varsigma } \\varpi _{2} ( h ), D_{C}^{\\varsigma } \\varpi _{3} ( h ),\\dots , D _{C}^{\\varsigma } \\varpi _{m} ( h ) )^{T}$$, $$0< \\varsigma <1$$, $$\\varpi ( h ) \\in \\mathcal{F}\\subset R^{m}$$, $$h\\in [ 0,T )$$ ($$T\\leq \\infty$$), $$\\mathcal{F}$$ is an open set, $$0\\in \\mathcal{F}$$, and $$z: [ 0,T )\\times \\mathcal{F}\\rightarrow R^{m}$$ is not discontinuous in h and placates the Lipschitz condition:\n\n$$\\bigl\\Vert z \\bigl( h,\\varpi ' \\bigr) -z\\bigl(h,\\varpi ''\\bigr) \\bigr\\Vert \\leq P \\bigl\\Vert \\varpi '-\\varpi '' \\bigr\\Vert , \\quad \\mathfrak{H}\\in [ 0,T )$$\n(11)\n\nfor all $$\\varpi ',\\varpi ''\\in \\varOmega\\subset \\mathcal{F}$$, where $$P>0$$ is a Lipschitz constant.\n\n### Theorem 1\n\n“Let the solution of (10) be $$u( h )$$, $$h \\in [ 0, H)$$. If there exists a vector function $$\\varpi = ( \\varpi _{1}, \\varpi _{2}, \\dots , \\varpi _{m} )^{H}: [ 0,H ) \\rightarrow \\mathcal{F}$$ such that $$\\varpi _{i} \\in G^{0,\\nu }$$, $$\\varsigma < \\varpi <1$$, $$i=1,2,\\dots, m$$ and\n\n$$D_{C}^{\\varphi } \\varpi \\leq g(h,\\varpi ), \\quad \\mathfrak{H}\\in [ 0,H ].$$\n(12)\n\n$$If \\varpi (0)\\leq\\leq u_{0}$$, $$u_{0} \\in \\mathcal{F}$$, then $$w\\leq\\leq u$$, $$h\\in [ 0,H ]$$ .\n\nLet $$g:\\mathcal{F}\\rightarrow R^{m}$$, $$\\mathcal{F}\\in R^{m}$$, we study the following system of non-integer order:\n\n$$D_{C}^{\\varsigma } x ( \\tau ) =g(x), \\quad x ( 0 ) = x_{0}.$$\n(13)\n\n### Definition 2\n\nWe say that $$\\mathcal{F}$$ is an equilibrium point of (13), iff, $$g ( \\mathcal{F} ) =0$$.\n\n### Remark 1\n\nWhen $$\\varsigma \\in ( 0, 1 )$$, the fractional system $$D_{C}^{\\varsigma } x ( \\tau ) =g(x)$$ has the identical equilibrium points as the arrangement $$\\frac{dx( \\tau )}{dt} =g(x)$$ .\n\n### Definition 3\n\n“The equilibrium point $$\\mathcal{F}$$ of autonomous (13) is said to be stable if, for all $$\\epsilon >0$$, $$\\varepsilon >0$$ exists such that if $$\\Vert x_{0}-\\mathcal{F} \\Vert < \\varepsilon$$, then $$\\Vert x-\\mathcal{F} \\Vert < \\epsilon$$, $$t \\geq 0$$; the equilibrium point $$\\mathcal{F}$$ of autonomous (13) is said to be asymptotically unwavering if $$\\lim_{t\\rightarrow \\infty } x(\\tau ) =\\mathcal{F}$$” .\n\n### Theorem 2\n\n“The equilibrium points of system (13) are locally asymptotically stable if all eigenvalues $$\\lambda _{i}$$ of Jacobian matrix J, calculated in the equilibrium points, satisfy $$\\vert \\arg ( \\lambda _{i} ) \\vert > \\varsigma \\frac{\\pi }{2}$$ [6, 16].\n\n## 3 Mathematical model\n\nMany researchers have discussed a measles model, like the SEIR measles model discussed in , the SIR measles model discussed in [24, 25], and the SVIER measles model discussed in . Some authors have discussed the model using ordinary differential equations and some using fractional order differential equations to control the measles model. In this paper, we are going to study a non-integer order SIRV epidemic model with vaccination and treatment rates. We assume that the total populace $$N(t)$$ is divided into four compartments, $$S(t)$$, $$I(t)$$, $$R(t)$$ and $$V(t)$$. Here, $$S(t)$$ be the proportion of populace which are susceptible at time t, $$I(t)$$ be the populace proportion which are infected at time at time t, $$R(t)$$ be the populace proportion which are recovered and $$V(t)$$ be the populace proportion which are vaccinated at time t. Let b denote the birth rate; β denotes the disease transmission rate, which is supposed to occur with straight contact among infectious and susceptible hosts. We denote the natural death degree $$d(N)$$ for the populace. For convenience, d is supposed to be continuous, which does not disagree with natural death rates. We assume α to be the disease-induced death rate and there exist $$\\mu _{1}$$ and $$\\mu _{2}$$ which in turn signify the proportion of susceptibles that are vaccinated per unit time and the proportion of infectives that are picked per unit time. ($$1- \\mu _{1}$$) denotes the unvaccinated rate. The dynamical system signifying the epidemic blowout in the populace is set by the subsequent system of non-linear ODEs and the flow cart is given in Fig. 1. We have\n\n$$\\textstyle\\begin{cases} \\frac{dS }{dt } = b -\\beta ( 1- \\mu _{1} ) SI - ( d + \\mu _{1} ) S, \\\\ \\frac{dI }{dt } = \\beta (1- \\mu _{1} )SI - ( d + \\mu _{2} +\\alpha ) I, \\\\ \\frac{dR }{dt } = \\mu _{2} I-dR, \\\\ \\frac{dV }{dt } = \\mu _{1} S-dV, \\end{cases}$$\n(14)\n\nwith $$S ( 0 ) = S_{0} >0$$, $$I ( 0 ) = I_{0} >0$$, $$R ( 0 ) = R_{0} >0$$ and $$V ( 0 ) = V _{0} >0$$.\n\nNote that individuals in V are different from those in both S and R. The immune system will create antibodies against the disease because vaccination is taken care of during this process. The vaccination individuals, before obtaining immunity, still have the possibility of infection while contacting with infected individuals. Individuals in V may be assumed to move into R when they gain immunity. We assume that b, β, d, α, $$\\mu _{1}$$ and $$\\mu _{2}$$ are all non-negative constants. We must note that, for the ailment free case (i.e. $$I=0$$), the total populace has logistic growth.\n\nAdding all the equations of (14) we have\n\n\\begin{aligned} \\frac{d(S+I+R+V)}{dt} =& \\frac{dS}{dt} + \\frac{dI}{dt} + \\frac{dR}{dt} + \\frac{dV}{dt} \\\\ =& b-\\beta ( 1- \\mu _{1} ) SI- ( d+ \\mu _{1} ) S+\\beta ( 1- \\mu _{1} ) SI \\\\ &{}- ( d + \\mu _{2} +\\alpha ) I+ \\mu _{2} I-dR+ \\mu _{1} S-dV \\\\ =& b-d ( S+I+R+V ) -\\alpha I. \\end{aligned}\n\nLet $$S+I+R+V=N$$, then\n\n\\begin{aligned}& \\frac{dN}{dt} = b-dN-\\alpha I, \\\\& \\frac{dN}{dt} \\leq b-dN. \\end{aligned}\n\nOn solving the above equation\n\n$$N ( t ) \\leq \\frac{b}{d} \\bigl(1- e^{-dt} \\bigr)+N ( 0 ) e^{-dt},$$\n\nwhere $$N ( 0 )$$ represents the initial value of the respective variables. Then $$0\\leq N ( t ) \\leq \\frac{b}{d}$$ as $$t\\rightarrow \\infty$$. Therefore, $$\\frac{b}{d}$$ is an upper bound of $$N(t)$$ provided $$N ( 0 ) \\leq \\frac{b}{d}$$. This means that the population size is not constant or in other words the population is dynamic.\n\nThe following equations can be transformed by the given model :\n\n$$\\textstyle\\begin{cases} \\frac{dS }{dt } = b -\\beta ( 1- \\mu _{1} ) SI - ( d + \\mu _{1} ) S, \\\\ \\frac{dI }{dt } = \\beta (1- \\mu _{1} )SI - ( d + \\mu _{2} +\\alpha ) I, \\\\ \\frac{dR }{dt } = \\mu _{2} I-dR, \\\\ \\frac{dV }{dt } = \\mu _{1} S-dV, \\\\ \\frac{dN }{dt } =b - Nd - \\alpha I. \\end{cases}$$\n(15)\n\nIn system (15) R is not involved in any other equations except in the third equation so it can be removed from system (15) and we obtain the system (16):\n\n$$\\textstyle\\begin{cases} \\frac{dS }{dt } = b -\\beta ( 1- \\mu _{1} ) SI - ( d + \\mu _{1} ) S, \\\\ \\frac{dI }{dt } = \\beta (1- \\mu _{1} )SI - ( d + \\mu _{2} +\\alpha ) I, \\\\ \\frac{dV }{dt } = \\mu _{1} S-dV, \\\\ \\frac{dN }{dt } =b - Nd - \\alpha I. \\end{cases}$$\n(16)\n\nIf we solve the system (16) then we can evaluate the factor $$R(t)$$, because $$R ( t ) =N ( t ) -S ( t ) -I ( t ) -V ( t )$$.\n\n### 3.1 Fractional order model\n\nThe system of non-integer order non-linear ODEs for the system (16), with $$\\phi \\in (0,1]$$, is given by\n\n$$\\textstyle\\begin{cases} \\frac{d ^{\\phi _{1}} S }{d t ^{\\phi _{1}}} = b -\\beta ( 1- \\mu _{1} ) SI - ( d + \\mu _{1} ) S, \\\\ \\frac{d ^{\\phi _{2}} I }{d t ^{\\phi _{2}}} = \\beta (1- \\mu _{1} )SI - ( d + \\mu _{2} +\\alpha ) I, \\\\ \\frac{d ^{\\phi _{3}} V }{d t ^{\\phi _{3}}} = \\mu _{1} S-dV, \\\\ \\frac{d ^{\\phi _{4}} N }{d t ^{\\phi _{4}}} =b - Nd - \\alpha I , \\end{cases}$$\n(17)\n\nwith\n\n\\begin{aligned}& S ( 0 ) = S_{0},\\qquad I ( 0 ) = I_{0},\\qquad V ( 0 ) = V_{0} \\quad \\mbox{and}\\quad N ( 0 ) = N_{0} \\end{aligned}\n(18)\n\nas the initial conditions. We use for all the values of fractional order $$\\phi _{1} = \\phi _{2} = \\phi _{3} = \\phi _{4} =\\phi$$. If all the values are the same then the system is called a commensurate model. If some values of ϕ are the same and some are different or all the values are different, then the system is called an incommensurate model.\n\nIn this manuscript we use a commensurate model to observe the dynamic behavior of the measles.\n\n### 3.2 Non-negative solution\n\nSymbolize $$R_{+}^{4} =\\{X\\in R^{4}:X\\geq 0\\}$$ and $$X ( t ) = ( S,I,V,N )^{T}$$. For the evidence of the non-negative solution, study the subsequent theorem and corollary.\n\n### Theorem 3\n\n(Generalized mean value theorem)\n\n“Let $$f ( x ) \\in C ( 0,a )$$ and $$D^{\\alpha } f ( x ) \\in C(0,a]$$, for $$0<\\alpha \\leq 1$$. Then we have\n\n$$f ( x ) =f ( 0+ ) + \\frac{1}{\\varGamma ( \\alpha )} \\bigl( D^{\\alpha } f \\bigr) ( \\xi ) ( x )^{\\alpha }$$\n(19)\n\nwith $$0\\leq \\xi \\leq x$$, $$\\forall x\\in (0, \\alpha ]$$ [8, 28].\n\n### Proof\n\nThe proof is given in . □\n\n### Corollary 1\n\n“Suppose that $$f ( x ) \\in C[0,a]$$ and $$D^{\\alpha } f ( x ) \\in C(0,\\alpha ]$$ for $$0<\\alpha \\leq 1$$. It is clear from Theorem 3 that if $$D^{\\alpha } f ( x ) \\geq 0$$, $$\\forall x\\in (0,a)$$, then $$f(x)$$ is non-decreasing, and if $$D^{\\alpha } f ( x ) \\leq 0$$, $$\\forall x\\in (0,a)$$, then $$f(x)$$ is non-increasing for all $$x\\in [0, a]$$ .\n\n### Theorem 4\n\n“There is a unique solution $$X ( t ) = ( S,I,V,N )^{T}$$ for the initial value problem given (17) at $$t\\geq 0$$ and the solution remains in $$R_{+}^{4}$$ .\n\n### Proof\n\nIt is easy to understand the existence and uniqueness of the result of the initial value problem (17)–(18) in $$(0,\\infty )$$. We will display that the purview $$R_{+}^{4}$$ remains positively invariant.\n\nThen\n\n\\begin{aligned}& \\frac{d ^{\\phi _{1}} S }{d t ^{\\phi _{1}}} \\bigg\\vert _{S=0} = b \\geq 0, \\\\& \\frac{d ^{\\phi _{2}} I }{d t ^{\\phi _{2}}} \\bigg\\vert _{I=0} = 0, \\\\& \\frac{d ^{\\phi _{3}} V }{d t ^{\\phi _{3}}} \\bigg\\vert _{V=0} = \\mu _{1} S \\geq 0, \\\\& \\frac{d ^{\\phi _{4}} N }{d t ^{\\phi _{4}}} \\bigg\\vert _{N=0} =b - \\alpha I \\geq 0, \\end{aligned}\n\non each hyperplane bounding the non-negative orthant and, because of Corollary 1, the result will linger in $$R_{+}^{4}$$. □\n\n### Lemma\n\n“Let $$u(t)$$ be continuous function on $$[ t_{0}, \\infty ]$$ satisfying\n\n$$\\frac{d^{\\alpha } u ( t )}{d t^{\\alpha }} \\leq -\\mu u ( t ) +\\lambda , \\qquad u ( t_{0} ) = u_{t_{0}},$$\n\nwhere $$0<\\alpha <1$$, $$( \\mu ,\\lambda ) \\in R^{2}$$, $$\\mu \\neq 0$$ and $$t_{0} \\geq 0$$ is the initial time. Then its solution has the form\n\n$$u ( t ) = \\biggl( u_{0} - \\frac{\\lambda }{\\mu } \\biggr) E_{\\alpha } \\bigl[ -\\mu ( t- t_{0} )^{\\alpha } \\bigr] + \\frac{\\lambda }{\\mu },$$\n\nwhere $$E_{\\alpha } (z)$$ is the Mittag-Leffler function with parameter α.\n\n### 3.3 Stability and equilibrium points\n\nAddressing the nonlinear algebraic equations, the equilibrium points of (17) are found:\n\n$$D^{\\phi _{1}} S ( t ) = D^{\\phi _{2}} I ( t ) = D^{\\phi _{3}} V ( t ) = D^{\\phi _{4}} N ( t ) =0.$$\n\nSystem (17) has DFE point $$\\mathcal{F}_{0} ( \\frac{b}{ ( d + \\mu _{1} )},0, \\frac{\\mu _{1}}{d} \\frac{b}{ ( d + \\mu _{1} )}, \\frac{b}{d} )$$ if $$\\mathcal{R}_{0} < 1$$, while if $$\\mathcal{R}_{0} > 1$$, in addition to $$\\mathcal{F}_{0}$$, there is a positive endemic equilibrium $$\\mathcal{F}^{*} ( S^{*}, I^{*}, V ^{*}, N^{*} )$$ and $$S^{*}$$, $$I^{*}$$, $$V^{*}$$ and $$N^{*}$$ are given by\n\n\\begin{aligned}& S^{*} = \\frac{\\mu _{2} +d+\\alpha }{\\beta (1- \\mu _{1} )} = \\frac{b}{ \\mathcal{R}_{0} (d+ \\mu _{1} )}, \\\\& I^{*} = \\frac{b\\beta ( 1- \\mu _{1} ) -( \\mu _{1} +d)( \\mu _{2} +d+\\alpha )}{\\beta (1- \\mu _{1} )( \\mu _{2} +d+\\alpha )} = \\frac{b}{( \\mu _{2} +d+\\alpha )} \\biggl( 1- \\frac{1}{\\mathcal{R}_{0}} \\biggr), \\\\& V^{*} = \\frac{\\mu _{1} ( \\mu _{2} +d+\\alpha )}{d\\beta (1- \\mu _{1} )} = \\frac{b \\mu _{1}}{\\mathcal{R}_{0} (d+ \\mu _{1} )d}, \\\\& N^{*} = \\frac{b\\beta ( \\mu _{2} +d- \\mu _{1} d- \\mu _{1} \\mu _{2} ) +\\alpha ( \\mu _{1} +d)( \\mu _{2} +d+\\alpha )}{d\\beta (1- \\mu _{1} )( \\mu _{2} +d+\\alpha )} = \\frac{ ( \\mu _{2} +d+ \\frac{\\alpha }{\\mathcal{R}_{0}} )}{( \\mu _{2} +d+\\alpha )}, \\end{aligned}\n\nwhere $$\\mathcal{R}_{0}$$ is the basic reproduction number denoted in \n\n$$\\mathcal{R}_{0} = \\frac{b \\beta (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha ) ( d + \\mu _{1} )}.$$\n(20)\n\nThe value that $$\\mathcal{R}_{0}$$ takes can signal the conditions where an epidemic is feasible. $$\\mathcal{R}_{0}$$, a key threshold number, is used for a stability analysis of (17).\n\n### 3.4 Sensitivity analysis ($$\\mathcal{R}_{0}$$)\n\nTo check the sensitivity of $$\\mathcal{R}_{0}$$ for each parameter,\n\n\\begin{aligned}& \\frac{\\partial \\mathcal{R}_{0}}{\\partial b} = \\frac{ \\beta (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha ) ( d + \\mu _{1} )} >0, \\\\& \\frac{\\partial \\mathcal{R}_{0}}{\\partial \\beta } = \\frac{b (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha ) ( d + \\mu _{1} )} >0, \\\\& \\frac{\\partial \\mathcal{R}_{0}}{\\partial d} = \\frac{b \\beta (-1+ \\mu _{1} )(2d + \\mu _{2} + \\mu _{1} +\\alpha )}{ ( d + \\mu _{2} +\\alpha )^{2} ( d + \\mu _{1} )^{2}} >0, \\\\& \\frac{\\partial \\mathcal{R}_{0}}{\\partial \\mu _{1}} =- \\frac{ b \\beta ( 1+d )}{ ( d + \\mu _{2} +\\alpha ) ( d + \\mu _{1} )^{2}} < 0, \\\\& \\frac{\\partial \\mathcal{R}_{0}}{\\partial \\mu _{2}} =- \\frac{ \\beta (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha )^{2} ( d + \\mu _{1} )} < 0, \\\\& \\frac{\\partial \\mathcal{R}_{0}}{\\partial \\alpha } =- \\frac{b \\beta (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha ) ^{2} ( d + \\mu _{1} )} < 0. \\end{aligned}\n\nThus $$\\mathcal{R}_{0}$$ is increasing with b, β and d, decreasing with α, $$\\mu _{2}$$ and $$\\mu _{1}$$.\n\nThe following theorem describes the stability behavior of system (17) around the DFE point $$\\mathcal{F}_{0}$$.\n\n### 3.5 Stability behavior around $$\\mathcal{F}_{0}$$\n\nThe following theorem describes the stability behavior of (17) around the infection free equilibrium point $$\\mathcal{F}_{0}$$.\n\n### Theorem 5\n\nIf $$\\mathcal{R}_{0} <1$$, then (17) will be locally asymptotically stable around $$\\mathcal{F} _{0}$$ and unstable if $$\\mathcal{R}_{0} >1$$.\n\n### Proof\n\nFor the system (17), the Jacobian matrix around $$\\mathcal{F}_{0}$$ has the following characteristic equation:\n\n$$(\\xi +d+ \\mu _{1} ) \\biggl(\\xi - \\beta ( 1- \\mu _{1} ) \\frac{b}{ ( d + \\mu _{1} )} + ( d + \\mu _{2} + \\alpha ) \\biggr) (\\xi +d) ( \\xi +d) = 0.$$\n(21)\n\nThe eigenvalues of Eq. (21) are\n\n\\begin{aligned}& \\xi _{1} = - ( d+ \\mu _{1} ) < 0, \\\\& \\xi _{2} = \\beta ( 1- \\mu _{1} ) \\frac{b}{ ( d + \\mu _{1} )} - ( d + \\mu _{2} +\\alpha ), \\\\& \\xi _{3} = -d < 0, \\\\& \\xi _{4} = -d< 0. \\end{aligned}\n\nHere $$\\xi _{1}$$, $$\\xi _{3}$$, and $$\\xi _{4}$$ are clearly negative. Now we will see that the eigenvalue $$\\xi _{2}$$ is negative. For disease free equilibrium, $$\\mathcal{R}_{0} <1$$. We have\n\n\\begin{aligned}& \\beta ( 1- \\mu _{1} ) \\frac{b}{ ( d + \\mu _{1} )} - ( d + \\mu _{2} +\\alpha ) < 0, \\\\& \\mathcal{R}_{0} = \\frac{b \\beta (1- \\mu _{1} )}{ ( d + \\mu _{2} +\\alpha ) ( d + \\mu _{1} )} < 1 \\end{aligned}\n\nso, $$\\xi _{2} <0$$.\n\nSo $$\\mathcal{F}_{0}$$ is asymptotically stable which is local in nature. □\n\n### Theorem 6\n\nThe endemic equilibrium point is asymptotically stable if $$\\mathcal{R}_{0} >1$$.\n\n### Proof\n\nThe Jacobian matrix for system (17) around $$\\mathcal{F}_{0}$$ has the following characteristic equation:\n\n$$\\bigl( \\xi ^{2} + \\mathcal{R}_{0} ( d + \\mu _{1} ) \\xi + ( \\mu _{2} +d+\\alpha ) ( \\mathcal{R}_{0} -1) (d + \\mu _{1} ) \\bigr) (\\xi +d) (\\xi +d) = 0$$\n(22)\n\nThe eigenvalues of $$J _{0} ( \\mathcal{F}^{*} )$$ are\n\n\\begin{aligned}& ( \\xi +d ) =0 \\quad \\Rightarrow\\quad \\xi _{1} =-d, \\\\& ( \\xi +d ) =0 \\quad \\Rightarrow\\quad \\xi _{2} =-d \\end{aligned}\n\nand\n\n\\begin{aligned}& \\xi ^{2} + \\mathcal{R}_{0} ( d + \\mu _{1} ) \\xi + ( \\mu _{2} +d+\\alpha ) ( \\mathcal{R}_{0} -1) (d + \\mu _{1} )=0, \\\\& \\xi _{3},_{4} = \\frac{-B\\pm \\sqrt{B^{2} -4AC}}{2A}, \\end{aligned}\n\nwhere\n\n\\begin{aligned}& A=1, \\\\& B= \\mathcal{R}_{0} ( d + \\mu _{1} ), \\\\& C=( \\mu _{2} +d+\\alpha ) ( \\mathcal{R}_{0} -1) (d + \\mu _{1} ). \\end{aligned}\n\nThis shows that if $$\\mathcal{R}_{0} >1$$, then $$\\xi _{3} <0$$ and $$\\xi _{4} <0$$, hence the system will be asymptotically stable. □\n\n## 4 Numerical study\n\nIn this section we give an illustrative example to authenticate the obtained results on systems (17). We have applied the numerical schemes to examine the paraphernalia of fluctuations in the non-integer order exponent on the qualitative behavior of results. The numerical results congregate to the intended equilibrium states of the fractional SIRV model, when the parameters are constant. Considering four fractional order techniques in the paper, actually we check the best performance wise efficiency of the model. One of the incredible favorable circumstances of the Caputo non-integer derivative is that it permits outmoded initial and boundary conditions to be incorporated in the formation of the issue. Furthermore, its derivative for an invariable is zero. The Caputo non-integer derivative also permits the utilization of the initial and boundary conditions when dealing with real world issues. The Caputo derivative is the most suitable non-integer operator to be utilized in displaying the true issue. We use MATLAB software for the simulations.\n\nStudying the effects of ϕ on the dynamics of the non-integer paradigm (17), we address numerous numerical efforts varying the value of the parameters. These simulations reveal the dynamics of the system disturbed using the value of ϕ. Figures 2, 4, 7, 10, 1517, 20, 21, 26, 28 and 31 depict that, for lower values of ϕ, the rampant peak is eclectic and less than the true equilibrium points. Figures 3, 56, 89, 1214, 1819, 27, 2930, 32 and 33 illustrate that, for lower values of ϕ, the epidemic peak is eclectic and greater for true steady states. Also we find the length of time to approach equilibrium within a given tolerance. The early transient behaviors vary greatly among the methods. Numerical imitations of an amended epidemic model with capricious order show that the non-integer order is linked to the relaxation time, i.e., the time engaged to reach equilibrium. The chaotic comportment of the system when the total order of the system is less than four is delineated. A comparison among the four diverse values of non-integer order is shown in Figs. 233. For all cases, the ailment evolves to the disease-free and endemic equilibrium points. Figures 233 illustrate that the model gradually tends towards the steady state for different ϕ.\n\n### 4.1 Generalized Euler method (GEM)\n\nThe generalized Euler method is a generalization of the classical Euler method; for details see [28,29,30,31,32]. The key points of this technique are set as shadows. Let us study the following initial value problem:\n\n$$D_{*}^{\\alpha } y ( t ) =f\\bigl(y ( t ),t\\bigr);\\qquad y ( 0 ) = y_{0}, \\quad 0< \\alpha \\leq 1, t>0$$\n(23)\n\nwhere $$D_{*}^{\\alpha }$$ is the Caputo FD. Let $$[0,a]$$ be the interval over which we need to find the result of the problem (23). The interval will be sectioned into subintervals $$[ t_{j}, t_{j+1} ]$$ of identical width $$h= \\frac{a}{\\aleph }$$ by via knots $$t_{j} =jh$$, for $$j=0,1,2,\\dots , \\aleph -1$$. The common formulation for GEM when $$t_{j+1} = t_{j} +h$$ is\n\n$$y ( t_{j+1} ) =y ( t_{j} ) + \\frac{h^{ \\alpha }}{\\varGamma ( \\alpha +1 )} f \\bigl( y ( t _{j} ), t_{j} \\bigr) +O\\bigl( h^{2\\alpha } \\bigr)$$\n(24)\n\nfor $$j=0,1,2,\\dots ,\\aleph -1$$. If the step size h is selected small enough, then we may neglect the second order term (involving $$h^{2\\alpha }$$) and develop\n\n$$y ( t_{j+1} ) =y ( t_{j} ) + \\frac{h^{ \\alpha }}{\\varGamma ( \\alpha +1 )} f \\bigl( y ( t _{j} ), t_{j} \\bigr).$$\n(25)\n\nFor $$\\alpha =1$$, Eq. (25) reduces to the classical Euler method.\n\nSo by using the generalized Euler method for system (17), we have the subsequent discretized equations\n\n\\begin{aligned}& S ( t_{\\aleph +1} ) =S ( t_{\\aleph } ) + \\frac{h ^{\\phi _{1}}}{\\varGamma ( \\phi _{1} +1)} \\bigl\\{ b- \\beta (1- \\mu _{1} ) S ( t_{\\aleph } ) I ( t_{\\aleph } ) - ( d + \\mu _{1} ) S ( t_{\\aleph } ) \\bigr\\} , \\\\& I ( t_{\\aleph +1} ) =I ( t_{\\aleph } ) + \\frac{h ^{\\phi _{2}}}{\\varGamma ( \\phi _{2} +1)} \\bigl\\{ \\beta (1- \\mu _{1} ) S ( t_{\\aleph +1} ) I ( t_{\\aleph } ) - ( d + \\mu _{2} +\\alpha ) I ( t_{\\aleph } ) \\bigr\\} , \\\\& V ( t_{\\aleph +1} ) =V ( t_{\\aleph } ) + \\frac{h ^{\\phi _{3}}}{\\varGamma ( \\phi _{3} +1)} \\bigl\\{ \\mu _{1} S ( t_{ \\aleph +1} ) -dV ( t_{\\aleph } ) \\bigr\\} , \\\\& N ( t_{\\aleph +1} ) =N ( t_{\\aleph } ) + \\frac{h ^{\\phi _{4}}}{\\varGamma ( \\phi _{4} +1)} \\bigl\\{ b-dN ( t_{\\aleph } ) -\\alpha I ( t_{\\aleph +1} ) \\bigr\\} . \\end{aligned}\n\n### 4.2 Piece wise continuous argument method\n\nIn this subsection, we apply the discretization process represented in Refs. [33, 34] for a measles model. Here we have used the piece wise constant arguments (PWCA) method to see the behavior of the system (17) by varying the parameters and order of the system. We can discretize (17) with the PWCA method as follows:\n\n$$\\textstyle\\begin{cases} D^{\\phi _{1}} S(t) = b-\\beta ( 1- \\mu _{1} ) S ( [ \\frac{t}{s} ] s ) I ( [ \\frac{t}{s} ] s ) - ( d+ \\mu _{1} ) S ( [ \\frac{t}{s} ] s ), \\\\ D^{\\phi _{1}} I(t) = \\beta (1- \\mu _{1} )S ( [ \\frac{t}{s} ] s ) I ( [ \\frac{t}{s} ] s ) - ( d + \\mu _{2} +\\alpha ) I ( [ \\frac{t}{s} ] s ), \\\\ D^{\\phi _{1}} V ( t ) = \\mu _{1} S ( [ \\frac{t}{s} ] s ) -dV ( [ \\frac{t}{s} ] s ), \\\\ D^{\\phi _{1}} N ( t ) = b - N ( [ \\frac{t}{s} ] s ) d - \\alpha I ( [ \\frac{t}{s} ] s ) . \\end{cases}$$\n(26)\n\nFirst, let $$t\\in [0,s)$$, i.e., $$\\frac{t}{s} \\in [0,1)$$. Thus, we obtain\n\n$$\\textstyle\\begin{cases} D^{\\phi _{1}} S(t) = b-\\beta ( 1- \\mu _{1} ) S ( 0 ) I ( 0 ) - ( d+ \\mu _{1} ) S ( 0 ), \\\\ D^{\\phi _{2}} I(t) = \\beta (1- \\mu _{1} )S ( 0 ) I ( 0 ) - ( d + \\mu _{2} +\\alpha ) I ( 0 ), \\\\ D^{\\phi _{3}} V ( t ) = \\mu _{1} S ( 0 ) -dV ( 0 ), \\\\ D^{\\phi _{4}} N ( t ) = b - N ( 0 ) d - \\alpha I ( 0 ), \\end{cases}$$\n(27)\n\nand the solution of (27) is reduced to\n\n\\begin{aligned}& \\begin{aligned}&\\textstyle\\begin{cases} S_{1} ( t ) =S ( 0 ) + J^{\\phi _{1}} ( b-\\beta ( 1- \\mu _{1} ) S ( 0 ) I ( 0 ) - ( d+ \\mu _{1} ) S ( 0 ) ), \\\\ I_{1} (t) =I ( 0 ) + J^{\\phi _{2}} ( \\beta (1- \\mu _{1} )S ( 0 ) I ( 0 ) - ( d + \\mu _{2} +\\alpha ) I ( 0 ) ), \\\\ V_{1} ( t ) =V ( 0 ) + J^{\\phi _{3}} ( \\mu _{1} S ( 0 ) -dV ( 0 ) ), \\\\ N_{1} ( t ) =N ( 0 ) + J^{\\phi _{4}} ( b - N ( 0 ) d - \\alpha I ( 0 ) ), \\end{cases}\\displaystyle \\\\ &\\textstyle\\begin{cases} S_{1} ( t ) =S ( 0 ) + \\frac{t^{\\phi _{1}}}{ \\phi _{1} \\varGamma ( \\phi _{1} )} ( b-\\beta ( 1- \\mu _{1} ) S ( 0 ) I ( 0 ) - ( d+ \\mu _{1} ) S ( 0 ) ), \\\\ I_{1} (t) =I ( 0 ) + \\frac{t^{\\phi _{2}}}{\\phi _{2} \\varGamma ( \\phi _{2} )} ( \\beta (1- \\mu _{1} )S ( 0 ) I ( 0 ) - ( d + \\mu _{2} +\\alpha ) I ( 0 ) ), \\\\ V_{1} ( t ) =V ( 0 ) + \\frac{t^{\\phi _{3}}}{ \\phi _{3} \\varGamma ( \\phi _{3} )} ( \\mu _{1} S ( 0 ) -dV ( 0 ) ), \\\\ N_{1} ( t ) =N ( 0 ) + \\frac{t^{\\phi _{4}}}{ \\phi _{4} \\varGamma ( \\phi _{4} )} ( b - N ( 0 ) d - \\alpha I ( 0 ) ), \\end{cases}\\displaystyle \\end{aligned} \\end{aligned}\n(28)\n\nsecond, let $$t\\in [s,2s)$$, i.e., $$\\frac{t}{s} \\in [1,2)$$. Thus, we obtain\n\n$$\\textstyle\\begin{cases} D^{\\phi _{1}} S(t) = b-\\beta ( 1- \\mu _{1} ) S_{1} ( t ) I_{1} ( t ) - ( d+ \\mu _{1} ) S _{1} ( t ), \\\\ D^{\\phi _{2}} I(t) = \\beta ( 1- \\mu _{1} ) S_{1} ( t ) I_{1} ( t ) - ( d + \\mu _{2} +\\alpha ) I_{1} ( t ), \\\\ D^{\\phi _{3}} V ( t ) = \\mu _{1} S_{1} ( t ) -d V_{1} ( t ), \\\\ D^{\\phi _{4}} N ( t ) = b - N_{1} ( t ) d - \\alpha I_{1} ( t ), \\end{cases}$$\n(29)\n\nwhich has the following solution:\n\n\\begin{aligned}& \\begin{aligned}&\\textstyle\\begin{cases} S_{2} ( t ) = S_{1} ( s ) + J^{\\phi _{1}} ( b-\\beta ( 1- \\mu _{1} ) S_{1} ( s ) I _{1} ( s ) - ( d+ \\mu _{1} ) S_{1} ( s ) ), \\\\ I_{2} (t) = I_{1} ( s ) + J^{\\phi _{2}} ( \\beta ( 1- \\mu _{1} ) S_{1} ( s ) I_{1} ( s ) - ( d + \\mu _{2} +\\alpha ) I_{1} ( s ) ), \\\\ V_{2} ( t ) = V_{1} ( s ) + J^{\\phi _{3}} ( \\mu _{1} S_{1} ( s ) -d V_{1} ( s ) ), \\\\ N_{2} ( t ) = N_{1} ( s ) + J^{\\phi _{4}} ( b - N_{1} ( s ) d - \\alpha I_{1} ( s ) ), \\end{cases}\\displaystyle \\\\ &\\textstyle\\begin{cases} S_{2} ( t ) = S_{1} ( s ) + \\frac{t^{\\phi _{1}}}{\\phi _{1} \\varGamma ( \\phi _{1} )} ( b-\\beta ( 1- \\mu _{1} ) S_{1} ( s ) I_{1} ( s ) - ( d+ \\mu _{1} ) S_{1} ( s ) ), \\\\ I_{2} (t) = I_{1} ( s ) + \\frac{t^{\\phi _{2}}}{\\phi _{2} \\varGamma ( \\phi _{2} )} ( \\beta ( 1- \\mu _{1} ) S_{1} ( s ) I_{1} ( s ) - ( d + \\mu _{2} + \\alpha ) I_{1} ( s ) ), \\\\ V_{2} ( t ) = V_{1} ( s ) + \\frac{t^{\\phi _{3}}}{\\phi _{3} \\varGamma ( \\phi _{3} )} ( \\mu _{1} S _{1} ( s ) -d V_{1} ( s ) ), \\\\ N_{2} ( t ) = N_{1} ( s ) + \\frac{t^{\\phi _{4}}}{\\phi _{4} \\varGamma ( \\phi _{4} )} ( b - N_{1} ( s ) d - \\alpha I_{1} ( s ) ), \\end{cases}\\displaystyle \\end{aligned} \\end{aligned}\n(30)\n\nwhere $$J_{s}^{\\alpha _{i}} = \\frac{1}{\\varGamma ( \\alpha _{i} )} \\int _{s} ^{\\tau } ( t-p )^{\\alpha _{i} -1}\\,dp$$, $$0< \\alpha _{i} \\leq 1$$ and $$i=1,2,3,4$$. Thus after repeating the discretization process n times, we obtain the discretized form of system (17) as follows:\n\n$$\\textstyle\\begin{cases} S_{n+1} ( t ) = S_{n} ( ns ) + \\frac{t^{\\phi _{1}}}{\\phi _{1} \\varGamma ( \\phi _{1} )} ( b-\\beta ( 1- \\mu _{1} ) S_{n} ( ns ) I_{n} ( ns ) - ( d+ \\mu _{1} ) S_{n} ( ns ) ), \\\\ I_{n+1} (t) = I_{n} ( ns ) + \\frac{t^{\\phi _{2}}}{\\phi _{2} \\varGamma ( \\phi _{2} )} ( \\beta ( 1- \\mu _{1} ) S _{n} ( ns ) I_{n} ( ns ) - ( d + \\mu _{2} +\\alpha ) I_{n} ( ns ) ), \\\\ V_{n+1} ( t ) = V_{n} ( ns ) + \\frac{t^{\\phi _{3}}}{\\phi _{3} \\varGamma ( \\phi _{3} )} ( \\mu _{1} S _{n} ( ns ) -d V_{n} ( ns ) ), \\\\ N_{n+1} ( t ) = N_{n} ( ns ) + \\frac{t^{\\phi _{4}}}{\\phi _{4} \\varGamma ( \\phi _{4} )} ( b - N_{n} ( ns ) d - \\alpha I_{n} ( ns ) ), \\end{cases}$$\n(31)\n\nwhere $$t\\in [ns, ( n+1 ) s)$$. For $$t\\rightarrow (n+1)s$$, the system (31) is reduced to\n\n$$\\textstyle\\begin{cases} S_{n+1} ( t ) = S_{n} + \\frac{t^{\\phi _{1}}}{\\phi _{1} \\varGamma ( \\phi _{1} )} ( b-\\beta ( 1- \\mu _{1} ) S _{n} I_{n} - ( d+ \\mu _{1} ) S_{n} ), \\\\ I_{n+1} (t) = I_{n} + \\frac{t^{\\phi _{2}}}{\\phi _{2} \\varGamma ( \\phi _{2} )} ( \\beta ( 1- \\mu _{1} ) S_{n} I_{n} - ( d + \\mu _{2} +\\alpha ) I_{n} ), \\\\ V_{n+1} ( t ) = V_{n} + \\frac{t^{\\phi _{3}}}{\\phi _{3} \\varGamma ( \\phi _{3} )} ( \\mu _{1} S_{n} -d V_{n} ), \\\\ N_{n+1} ( t ) = N_{n} + \\frac{t^{\\phi _{4}}}{\\phi _{4} \\varGamma ( \\phi _{4} )} ( b - N_{n} d - \\alpha I_{n} ). \\end{cases}$$\n(32)\n\nIt should be noticed that if $$\\phi _{i} \\rightarrow 1$$ ($$i=1,2,3,4$$) in (32), we obtain the corresponding Euler discretization of the discretized measles model with commensurate order. It is different from the predictor corrector method. The obtained result is a four dimensional discrete system.\n\nAccording to Angstmann et al. , the generalized Euler method (GEM) and piece wise continuous argument (PCWA) methods do not give the proper results, which is why we also used two other techniques (the Adams–Bashforth–Moulton method and the Grunwald–Letnikov method) to show the efficiency of the measles model in Sects. 4.3 and 4.4.\n\n### 4.3 Grunwald–Letnikov method (Binomial Coefficients)\n\nFor numerical use of the non-integer order derivative we can use Eq. (33) resulting from the Grunwald–Letnikov approach. This tactic is centered on the point that, for a wide-ranging class of functions, two approaches, the Grunwald Letnikov definition and the Caputo definition, are comparable. The relation for the explicit numerical approximation of φth derivative at the points kh ($$k=1,2,\\dots$$) has the following form :\n\n$${}_{(k-L/h)} D_{t_{k}}^{\\varphi } f ( t ) \\approx \\frac{1}{h ^{\\varphi }} \\sum_{j=0}^{k} (-1 )^{j} \\binom{\\varphi }{j} f ( t _{k-j} ),$$\n(33)\n\nwhere L is the “memory length”, $$t_{k} =kh$$, h is the time step and the $$(-1 )^{j} \\binom{\\varphi }{j}$$ are the binomial coefficients $$C_{j}^{ ( \\varphi )}$$ ($$j=0,1,2,\\dots$$). For them we can use the expression \n\n$$C_{0}^{ ( \\varphi )} =1, \\qquad C_{j}^{ ( \\varphi )} =\\biggl(1- \\frac{1+ \\varphi }{j} \\biggr) C_{j-1} ^{ ( \\varphi )}.$$\n\nThen the common numerical elucidation of the non-integer differential equation\n\n$${}_{a} D_{t}^{\\varphi } y ( t ) =f \\bigl( t, y ( t ) \\bigr)$$\n\ncan be written as\n\n$$y ( t_{k} ) = f \\bigl( t_{k}, y ( t_{k} ) \\bigr) h^{\\varphi } - \\sum_{j=0}^{k} C_{j}^{ ( \\varphi )} y ( t_{k-j} ).$$\n\nNow we express system (17) in the above format,\n\n\\begin{aligned}& S ( t_{k} ) = \\bigl\\{ b-\\beta ( 1- \\mu _{1} ) S ( t_{k-1} ) I ( t_{k-1} ) -(d+ \\mu _{1} )S ( t_{k-1} ) \\bigr\\} h^{\\phi _{1}} - \\sum _{j=1}^{k} C _{j}^{ ( \\phi _{1} )} S ( t_{k-j} ), \\\\& I ( t_{k} ) = \\bigl\\{ \\beta ( 1- \\mu _{1} ) S ( t_{k} ) I ( t_{k-1} ) -(d+ \\mu _{2} + \\alpha )I ( t_{k-1} ) \\bigr\\} h^{\\phi _{2}} - \\sum _{j=1} ^{k} C_{j}^{ ( \\phi _{2} )} I ( t_{k-j} ), \\\\& V ( t_{k} ) = \\bigl\\{ \\mu _{1} S ( t_{k} ) -d V ( t_{k-1} ) \\bigr\\} h^{\\phi _{3}} - \\sum _{j=1}^{k} C _{j}^{ ( \\phi _{3} )} V ( t_{k-j} ), \\\\& N ( t_{k} ) = \\bigl\\{ b-d N ( t_{k-1} ) - \\alpha I ( t_{k} ) \\bigr\\} h^{\\phi _{3}} - \\sum _{j=1} ^{k} C_{j}^{ ( \\phi _{3} )} N ( t_{k-j} ), \\end{aligned}\n\nwhere $$C_{0}^{ ( \\phi _{i} )} =1$$, $$C_{j}^{ ( \\phi _{i} )} =(1- \\frac{1+ \\phi _{i}}{j} ) C_{j-1}^{ ( \\phi _{i} )}$$, $$i=1,2,3,4$$.\n\nOne can use the generalized Adams–Bashforth–Moulton method for numerical solutions of the system (17) (see ).\n\nSo\n\n\\begin{aligned}& S^{n+1} =S(0)+ \\frac{h^{\\phi _{1}}}{\\varGamma ( \\phi _{1} +2 )} \\bigl( b -\\beta ( 1- \\mu _{1} ) S_{n+1}^{p} I_{n+1}^{p} - ( d + \\mu _{1} ) S_{n+1}^{p} \\bigr) \\\\& \\hphantom{S^{n+1} =}{} + \\frac{h ^{\\phi _{1}}}{\\varGamma ( \\phi _{1} +2 )} \\sum_{l=0}^{n} a _{l,n+1} \\bigl( b -\\beta ( 1- \\mu _{1} ) S_{l} I _{l} - ( d + \\mu _{1} ) S_{l} \\bigr), \\\\& \\begin{aligned} I^{n+1} &= I (0)+ \\frac{h^{\\phi _{2}}}{\\varGamma ( \\phi _{2} +2 )} \\bigl( \\beta ( 1- \\mu _{1} ) S_{n+1}^{p} I_{n+1}^{p} - ( d + \\mu _{2} +\\alpha ) I_{n+1}^{p} \\bigr) \\\\ &\\quad {} + \\frac{h ^{\\phi _{2}}}{\\varGamma ( \\phi _{2} +2 )} \\sum_{l=0}^{n} a _{l,n+1} \\bigl( \\beta ( 1- \\mu _{1} ) S_{l} I_{l} - ( d + \\mu _{2} +\\alpha ) I_{l} \\bigr), \\end{aligned} \\\\& V^{n+1} = V (0)+ \\frac{h^{\\phi _{3}}}{\\varGamma ( \\phi _{3} +2 )} \\bigl( \\mu _{1} S_{n+1}^{p} -d V_{n+1}^{p} \\bigr) \\\\& \\hphantom{V^{n+1} =}{} + \\frac{h^{\\phi _{3}}}{\\varGamma ( \\phi _{3} +2 )} \\sum_{l=0}^{n} a_{l,n+1} ( \\mu _{1} S_{l} -d V_{l} ), \\\\& N^{n+1} = N (0)+ \\frac{h^{\\phi _{3}}}{\\varGamma ( \\phi _{3} +2 )} \\bigl( b- d N_{n+1}^{p} -\\alpha I_{n+1}^{p} \\bigr) \\\\& \\hphantom{N^{n+1} =}{}+ \\frac{h^{\\phi _{3}}}{\\varGamma ( \\phi _{3} +2 )} \\sum_{l=0}^{n} a_{l,n+1} ( b- d N_{l} -\\alpha I_{l} ), \\end{aligned}\n\nwhere\n\n\\begin{aligned}& S_{n+1}^{p} =S(0)+ \\frac{1}{\\varGamma ( \\phi _{1} )} \\sum _{l=0}^{n} b_{l,n+1} \\bigl( b -\\beta ( 1- \\mu _{1} ) S_{l} I_{l} - ( d + \\mu _{1} ) S_{l} \\bigr), \\\\& I_{n+1}^{p} = I (0)+ \\frac{1}{\\varGamma ( \\phi _{2} )} \\sum _{l=0}^{n} b_{l,n+1} \\bigl( \\beta ( 1- \\mu _{1} ) S_{l} I _{l} - ( d + \\mu _{2} +\\alpha ) I_{l} \\bigr), \\\\& V_{n+1}^{p} = V (0)+ \\frac{1}{\\varGamma ( \\phi _{3} )} \\sum _{l=0}^{n} b_{l,n+1} ( \\mu _{1} S_{l} -d V_{l} ), \\\\& N_{n+1}^{p} = N (0)+ \\frac{1}{\\varGamma ( \\phi _{3} )} \\sum _{l=0}^{n} b_{l,n+1} ( b- d N_{l} -\\alpha I_{l} ), \\end{aligned}\n\nwith\n\n$$a_{l,n+1} = \\textstyle\\begin{cases} n^{\\phi _{i} +1} -(n- \\phi _{i} )(n+1 )^{\\phi _{i}},& l=0, \\\\ (n-l+2 )^{\\phi _{i} +1} +(n-l )^{\\phi _{i} +1} -2(n-l+1 )^{\\phi _{i} +1},& 1\\leq l\\leq n, \\\\ 1,& l=n+1, \\end{cases}$$\n\nand\n\n$$b_{l,n+1} = \\frac{h^{\\phi _{i}}}{\\phi _{i}} \\bigl( ( n-l+1 ) ^{\\phi _{i}} - ( n-l )^{\\phi _{i}} \\bigr), \\quad 0\\leq l\\leq n$$\n\nwith $$i=1,2,3,4$$.\n\n## 5 Conclusion\n\nIn this paper, a non-linear mathematical Measles model with fractional order $$\\phi _{i}$$, $$i=1,2,\\ldots,4$$ is formulated. The stability of both the DFE and the EE points is discussed. Sufficient conditions for local stability of the DFE point $$\\mathcal{F}_{0}$$ are given in terms of the basic reproduction number $$\\mathcal{R}_{0}$$ of the model, where it is asymptotically stable if $$\\mathcal{R}_{0} <1$$. The positive infected equilibrium $$\\mathcal{F}^{*}$$ exists when $$\\mathcal{R}_{0} >1$$ and sufficient conditions that guarantee the asymptotic stability of this point are given. Beside this sensitivity analysis of the parameters involved the threshold parameter ($$\\mathcal{R}_{0}$$) is discussed. Considering three fractional order techniques, we actually checked the best performance wise efficiency of the model. When simulating the model with all four algorithms, we have observed that all methods are converging to disease free and endemic equilibrium points but through different paths for diverse values of ϕ. The values are very close to each other in all three techniques. However, the time consumed (Core i7 Desktop) by GEM is 10 sec, PWCA is 43 sec, Grunwald–Letnikov (binomial coefficient) is 7716.97 sec, and the Adams–Bashforth–Moulton algorithm is $$87{,}326.743$$ sec, which indicates that the computational cost for GEM is lower than the rests. Measles is an infectious disease highly contagious from person to person occurring during childhood. So, the main goal of analyzing such techniques for a measles model is to benefit the researchers and policy makers in targeting, in preclusion and in treatment resources for supreme effectiveness. Also the length of time of approaching equilibrium within a given tolerance is interesting. The early transient behaviors vary greatly among the methods. Each method converges to the same disease free and endemic equilibrium points. For different values of $$\\phi _{i}$$, $$i=1,\\ldots,4$$, the system approaches the same equilibrium point but through different paths. 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Goufo, E.F.D., Martiz, R., Munganga, J.: Some properties of the Kermack McKendrick epidemic model with fractional derivative and nonlinear incidence. Adv. Differ. Equ. 2014, Article ID 278 (2014)\n\n11. Dokoumetzidis, A., Magin, R., Macheras, P.: A commentary on fractionalization of multi-compartmental models. J. Pharmacokinet. Pharmacodyn. 37, 203–207 (2010)\n\n12. Angstmann, C.N., Magin, R., Macheras, P.: A fractional order recovery SIR model from a stochastic process. Bull. Math. Biol. 78, 468–499 (2016)\n\n13. Angstmann, C.N., Henry, B.I., McGann, A.V.: A fractional order infectivity SIR model. Phys. A, Stat. Mech. Appl. 452, 86–93 (2016)\n\n14. Angstmann, C.N., Erickson, A.M., Henry, B.I., McGann, A.V., Murray, J.M., Nicholas, J.A.: Fractional order compartment models. SIAM J. Appl. Math. 77(2), 430–446 (2017)\n\n15. Sardar, T., Saha, B.: Mathematical analysis of a power–law form time dependent vector- born disease transmission model. Math. Biosci. 288, 109–123 (2017)\n\n16. Matignon, D.: Stability results for fractional differential equations with applications to control processing. Comput. Eng. Sys. Appl. 2, 963 (1996)\n\n17. https://www.who.int/immunization/newsroom/measles-data-2019/en/. Accessed 17 June 2019\n\n18. https://www.who.int/en/news-room/fact-sheets/detail/measles. Accessed 17 June 2019\n\n19. World health organization: The World Health Report 2001 – Mental Health New Understanding, New Hope. WHO, Geneva (2001)\n\n20. Zafar, Z., Rehan, K., Mushtaq, M.: Fractional-order scheme for bovine babesiosis disease and tick populations. Adv. Differ. Equ. 2017, Article ID 86 (2017)\n\n21. Zafar, Z.: Fractional order Lengyel–Epstein chemical reaction model. Comput. Appl. Math. 38, 131 (2019)\n\n22. Machado, T.R., Kiryakova, V., Mainardi, F.: Recent history of fractional calculus. Commun. Nonlinear Sci. Numer. Simul. 16, 1140–1153 (2011)\n\n23. Momoh, A.A., Ibrahim, M.O., Uwanta, I.J., Manga, S.B.: Mathematical model for control of measles epidemiology. Int. J. Pure Appl. Math. 87(5), 707–718 (2013)\n\n24. David, Y.: The analysis of vaccination and treatment of Measles diseases described by a fractional order SIR epidemiology model. 2014 M. Phil Thesis\n\n25. Isea, R., Lonngren, K.E.: Epidemaic modeling using data from the 2001–2002 measles outbreak in venezuella. Res. Rev. Biosci. 7(1), 15–18 (2013)\n\n26. Peter, O.J., Afolabi, O.A., Victor, A.A., Akpan, C.E., Oguntolu, F.A.: Mathematical model for the control of measles. J. Appl. Sci. Environ. Manag. 22(4), 571–576 (2018)\n\n27. Yaro, D., Omari-Sasu, S.K., Harvin, P., Saviour, A.W., Obeng, B.A.: Generalized Euler method for modeling measles with fractional differential equations. Int. J. Innov. Res. Dev. 4(4), 358–366 (2015)\n\n28. Odibat, Z., Moamni, S.: An algorithm for the numerical solution of differential equations of fractional order. J. Appl. Math. Inform. 26, 15–27 (2008)\n\n29. Odibat, Z., Shawagfeh, N.: Generalized Taylor’s formula. Appl. Math. Comput. 186, 286–293 (2007)\n\n30. Arafa, A.A.M., Rida, S.Z., Khalil, M.: A fractional-order model of HIV infection: numerical solution and comparisons with data of patients. Int. J. Biomath. 7(4), 1450036 (2014)\n\n31. Arafa, A.A.M., Rida, S.Z., Khalil, M.: The effect of anti-viral drug treatment of human immunodeficiency virus type 1 (HIV-1) described by a fractional order model. Appl. Math. Model. 37(4), 2189–2196 (2013)\n\n32. Arafa, A.A.M., Rida, S.Z., Khalil, M.: Fractional modeling dynamics of HIV and CD4+ T-cells during primary infection. EPJ Nonlinear Biomed. Phys. 6(1), 1–7 (2012)\n\n33. El-Sayed, A.M.A., El-Rehman, Z.F., Salman, S.M.: Discretization of forced Duffing oscillator with fractional-order damping. Adv. Differ. Equ. 2014), 66 (2014)\n\n34. Agarwal, R.P., El-Sayed, A.M.A., Salman, S.M.: Fractional order Chua’s system: discretization, bifurcation and chaos. Adv. Differ. Equ. 2013, 320 (2013)\n\n35. Angstmann, C.N., Henry, B.I., Jacobs, B.A., McGann, A.V.: Discretization of Fractional Differential Equations by a Piecewise Continuous Approximation. J. Comput. Phys. (2016) arXiv:1605.01815v1 [math.NA]\n\n36. Kaninda, A.V., Legros, D., Jataou, I.M., Malfait, P., Maisonneuve, M., Paquet, C., Moren, A.: Measles vaccine effectiveness in standard and early immunization strategies, Niger. Pediatr. Infect. Dis. J. 17, 1034–1039 (1998)\n\n37. Allen, L.J.S.: An Introduction to Mathematical Biology. Pearson Education, Upper Saddle River (2007)\n\n### Acknowledgements\n\nWe would like to thank the referees for their valuable comments.\n\nNot applicable.\n\n## Funding\n\nNo funding is available for this research project.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nThe authors equally have made contributions. All authors read and approved the manuscript.\n\n### Corresponding author\n\nCorrespondence to Zain Ul Abadin Zafar.\n\n## Ethics declarations\n\nNot applicable.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Consent for publication\n\nNot applicable.",
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https://gmatclub.com/forum/if-x-y-1-which-of-the-following-must-be-true-222118.html | [
"GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 05 Apr 2020, 22:07",
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"### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.",
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"# If x/|y| = -1 which of the following must be true?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager",
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"Joined: 12 Nov 2015\nPosts: 55\nLocation: Uruguay\nConcentration: General Management\nSchools: Goizueta '19 (A)\nGMAT 1: 610 Q41 V32\nGMAT 2: 620 Q45 V31\nGMAT 3: 640 Q46 V32\nGPA: 3.97\nIf x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n2\n46",
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"based on 1112 sessions\n\n### HideShow timer Statistics\n\nIf x/|y| = -1 which of the following must be true?\n\nA) x = -y\nB) x = y\nC) x = y^2\nD) x^2 = y^2\nE) x^3 = y^3\nMath Expert",
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"V\nJoined: 02 Sep 2009\nPosts: 62498\nIf x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n11\n8\nAvigano wrote:\nIf x/|y| = -1 which of the following must be true?\n\nA) x = -y\nB) x = y\nC) x = y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nx/|y| = -1\n\n|y| = -x, this implies that x is 0 or negative.\n\nSquare: y^2 = x^2.\n\nAlternatively you can try number plugging: try x = -2 and y = 2 or -2. You'll see that only D is always true.\n_________________\nCurrent Student",
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"Joined: 18 Oct 2014\nPosts: 779\nLocation: United States\nGMAT 1: 660 Q49 V31\nGPA: 3.98\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n4\n4\nAvigano wrote:\nIf x/|y| = -1 which of the following must be true?\n\nA) x = -y\nB) x = y\nC) x = y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nHi! There,\n\nx/|y| = -1\nx= -1 |y|\n\nSquaring both sides\nx^2= (-1 |y|) ^2= y^2\n\nSquaring any -ve number yields +ve number.\n\nLet's see why other options are not right\nA) x = -y What if x= 2 and y= 2. This does'nt hold good then.\n\nB) x = y If x=-2 and y=2. This is not right.\n\nC) x = y^2 This means x is +ve and |y| is always +ve. In this case x/|y| can not be -ve. not correct.\n\nD) x^2 = y^2. Explained above.\n\nE) x^3 = y^3. If x= -2 and y= 2. This will not hold true.\n\n_________________\nI welcome critical analysis of my post!! That will help me reach 700+\n##### General Discussion\nManager",
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"Joined: 12 Nov 2015\nPosts: 55\nLocation: Uruguay\nConcentration: General Management\nSchools: Goizueta '19 (A)\nGMAT 1: 610 Q41 V32\nGMAT 2: 620 Q45 V31\nGMAT 3: 640 Q46 V32\nGPA: 3.97\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nThank you Bunuel, clearer than spring water.\nRetired Moderator",
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"G\nJoined: 26 Nov 2012\nPosts: 553\nRe: If (x/|y|)=-1,Which of the following must be true ? [#permalink]\n\n### Show Tags\n\n1\nNandishSS wrote:\nIf (x/|y|)=-1,Which of the following must be true ?\nA) x = —y\nB) x = y\nC) x=y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nQuick way to solve?\n\nWhy option A is wrong\n\nGiven (x/|y|)=-1\n\nThen we can take two cases and the options must satisfy both cases.\n1. x/y = -1\n2. x/-y = -1\n\nOption A : sub x = -y in two cases then in one case satisfies and the other doesn't\nB) x = y : sub x = y in two cases , then in one case satisfies and the other doesn't\nC) x=y^2 : sub x = y^2 in two cases then in one case satisfies and the other doesn't\nD) x^2 = y^2 => x = √y^2 = |y| in both case we get the same result. ( in one case assume +ve and other case -ve)--> Correct option.\nE) x^3 = y^3 => x = y => sub x = y in two cases , then in one case satisfies and the other doesn't...\nIntern",
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"Joined: 13 May 2014\nPosts: 11\nRe: If (x/|y|)=-1,Which of the following must be true ? [#permalink]\n\n### Show Tags\n\n2\nNandishSS wrote:\nIf (x/|y|)=-1,Which of the following must be true ?\nA) x = —y\nB) x = y\nC) x=y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nQuick way to solve?\n\nWhy option A is wrong\n\nAttachment:",
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"V\nJoined: 02 Aug 2009\nPosts: 8311\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n1\n2\nmjhoon1004 wrote:\nHello Gmatclubbers,\n\nI stumbled upon this question on one of the Prep packs and am having a trouble understanding it.\nI'd appreciate if any of you can help explain. Thanks.\n\nHi,\n\nIf x/|y|=-1....\nTwo things can be made out...\nX must be numerically similar to y, may not be the sign...\nX must be NEGATIVE....\nWe do not know whether y is +or-......\nSo if we say x=-y and y is -, x will become +, and ans will be 1...\nOnly x^2=y^2 will be true irrespective of SIGN....\n_________________\nManager",
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"B\nJoined: 18 Jun 2016\nPosts: 85\nLocation: India\nConcentration: Technology, Entrepreneurship\nGMAT 1: 700 Q49 V36\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nmjhoon1004 wrote:\nHello Gmatclubbers,\n\nI stumbled upon this question on one of the Prep packs and am having a trouble understanding it.\nI'd appreciate if any of you can help explain. Thanks.\n\nx / |y| = -1\n\ny can be +ve or -ve\n\ncase 1 : y is +ve\nx/y = -1 implies x = -y\ncase 2 : y is -ve\nx/-y = -1 implies x = y\n\nfrom 1 and 2 we can always say x2 = y2\ni would go with D\n\nwhats OA?\nManager",
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"Joined: 04 Jan 2014\nPosts: 114\nGMAT 1: 660 Q48 V32\nGMAT 2: 630 Q48 V28\nGMAT 3: 680 Q48 V35\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nSince absolute value is always >= 0, we can safely square it.\n\nx / |y| = -1\n\nSquare both sides.\n\nx^2 / y^2 = 1\nx^2 = y^2\n\nThis must be true.\n\nEMPOWERgmat Instructor",
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"V\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 16361\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n1\nHi mjhoon1004,\n\nThis question can be solved by TESTing VALUES.\n\nWe're told that X/|Y| = -1. We're asked which of the following MUST be true (meaning \"which is ALWAYS TRUE no matter how many different examples you come up with??\").\n\nIF....\nX = -2\nY = 2\nThen we can eliminate Answers B, C and E\n\nIF...\nX = -2\nY = -2\nThen we can eliminate Answer A\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nManager",
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"B\nJoined: 23 Jun 2009\nPosts: 172\nLocation: Brazil\nGMAT 1: 470 Q30 V20\nGMAT 2: 620 Q42 V33",
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"If x/|y|=-1, which of the following must be true? [#permalink]\n\n### Show Tags\n\n2\nIf $$\\frac{x}{|y|}=-1$$, which of the following must be true?\n\nA)$$x=-y$$\nB)$$x=y$$\nC) $$x=y^2$$\nD) $$x^2=y^2$$\nE) $$x^3=y^3$$\nMath Expert",
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"V\nJoined: 02 Aug 2009\nPosts: 8311\nRe: If x/|y|=-1, which of the following must be true? [#permalink]\n\n### Show Tags\n\n2\nfelippemed wrote:\nIf $$\\frac{x}{|y|}=-1$$, which of the following must be true?\n\nA)$$x=-y$$\nB)$$x=y$$\nC) $$x=y^2$$\nD) $$x^2=y^2$$\nE) $$x^3=y^3$$\n\nHi\n\n$$\\frac{x}{|y|}=-1$$ means the NUMERIC value of x and y has to be same , sign may not be..\nThe second point is that since |y| will be positive, x will be NEGATIVE..\nNow two cases\n1) x = y...\nx=y=-3..\nSo $$\\frac{-3}{|-3|}=\\frac{-3}{3}=-1$$\n\n2) x=-y..\nX=-3, y=3...\n$$\\frac{-2}{|3|}=-1$$..\n\nONLY an EVEN power can make two values equal in both cases\n\nSay A is the answer then D also must be true.. but can we have two answers..NO\nIf B is the answer, then D and E also must be true..\nThis way you can home on to D\n_________________\nRetired Moderator",
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"V\nStatus: Long way to go!\nJoined: 10 Oct 2016\nPosts: 1298\nLocation: Viet Nam\nRe: If x/|y|=-1, which of the following must be true? [#permalink]\n\n### Show Tags\n\nfelippemed wrote:\nIf $$\\frac{x}{|y|}=-1$$, which of the following must be true?\n\nA)$$x=-y$$\nB)$$x=y$$\nC) $$x=y^2$$\nD) $$x^2=y^2$$\nE) $$x^3=y^3$$\n\n$$\\frac{x}{|y|}=-1 \\implies x=-|y| \\implies x^2 = y^2$$\n\nC is out.\n$$x=-|y| \\implies x=y$$ if y<0 or $$x=-y$$ if y>0. A, B is out.\n$$x=-|y| \\implies x^3=-|y|^3 \\implies x^3=y^3$$ if y<0 or $$x^3=-y^3$$ if y>0. E is out\n_________________\nIntern",
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"B\nJoined: 03 Apr 2016\nPosts: 16\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nx/sqrt (y^2)=-1\nx=-sqrt(y^2)\nSquaring both side\nX^2=y^2\nD\n\nSent from my SAMSUNG-SM-G935A using GMAT Club Forum mobile app\nManager",
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"B\nJoined: 23 Jun 2009\nPosts: 172\nLocation: Brazil\nGMAT 1: 470 Q30 V20\nGMAT 2: 620 Q42 V33",
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"If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nHere is my theoretical approach to the problem.\n\nA real good one to solidify concepts\n\nPosted from my mobile device\n\nTIP: If you are still struggling with the objective of the question, look at the equations as sausage-maker machines. The goal is to produce similar products despite the ingredients (but the amount of them must be equal). In other words, if I plug in signals within them, the output MUST be equal.\nAttachments",
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"S\nJoined: 08 Dec 2015\nPosts: 280\nGMAT 1: 600 Q44 V27\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n1\nWhy would we square a modulus??? Omg I dont understand ANYTHING if this...\nCurrent Student",
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"B\nJoined: 25 Feb 2017\nPosts: 34\nLocation: Korea, Republic of\nSchools: LBS '19 (A)\nGMAT 1: 720 Q50 V38\nGPA: 3.67\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\n1\nIf x/|y| = -1 which of the following must be true?\n\nA) x = -y\nB) x = y\nC) x = y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nmy 2 cents.\n\nx=-|y| this means x=-y or x=-(-y)\nSo, A) and B) are out.\nC), doens't make sense.\nD), if we square both then signs in front of a variable does not matter so this is good.\nE), if we cute, then negative signs remain, so out.\n\nHence D.\nIntern",
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"B\nJoined: 31 Oct 2017\nPosts: 5\nLocation: United States (CA)\nSchools: Copenhagen (A)\nGMAT 1: 640 Q47 V32",
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"GPA: 3.17\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nTEST IT: to get -1, X could be -1, then y could be -1 or 1.\n\na) -1 = - (1) but not -(-1) ---> OUT\nb) -1 = -1 but not 1 ---> OUT\nc) -1 not equal to (-1)^2 nor (1)^2 ---> OUT\nd) (-1)^2 = (-1)^2 and (1)^2 ---> CORRECT\ne) (-1)^3 = (-1)^3 but not (1)^3 ---> OUT\nIntern",
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"B\nJoined: 08 Dec 2014\nPosts: 5\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nBunuel wrote:\nAvigano wrote:\nIf x/|y| = -1 which of the following must be true?\n\nA) x = -y\nB) x = y\nC) x = y^2\nD) x^2 = y^2\nE) x^3 = y^3\n\nx/|y| = -1\n\n|y| = -x, this implies that x is 0 or negative.\n\nSquare: y^2 = x^2.\n\nAlternatively you can try number plugging: try x = -2 and y = 2 or -2. You'll see that only D is always true.\n\nAbsolute value cannot equal negative number. Shouldn't the representation be x=-|y| instead of |y|=-x?\nManager",
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"B\nJoined: 14 Nov 2018\nPosts: 89\nLocation: United Arab Emirates\nGMAT 1: 590 Q42 V30\nGPA: 2.6\nRe: If x/|y| = -1 which of the following must be true? [#permalink]\n\n### Show Tags\n\nDivyadisha - I have a question. My understanding was that you can square both sides in an equation when both sides are positive. Is that true or am I not looking at this question right?",
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https://www.colorhexa.com/2bff49 | [
"# #2bff49 Color Information\n\nIn a RGB color space, hex #2bff49 is composed of 16.9% red, 100% green and 28.6% blue. Whereas in a CMYK color space, it is composed of 83.1% cyan, 0% magenta, 71.4% yellow and 0% black. It has a hue angle of 128.5 degrees, a saturation of 100% and a lightness of 58.4%. #2bff49 color hex could be obtained by blending #56ff92 with #00ff00. Closest websafe color is: #33ff33.\n\n• R 17\n• G 100\n• B 29\nRGB color chart\n• C 83\n• M 0\n• Y 71\n• K 0\nCMYK color chart\n\n#2bff49 color description : Vivid lime green.\n\n# #2bff49 Color Conversion\n\nThe hexadecimal color #2bff49 has RGB values of R:43, G:255, B:73 and CMYK values of C:0.83, M:0, Y:0.71, K:0. Its decimal value is 2883401.\n\nHex triplet RGB Decimal 2bff49 `#2bff49` 43, 255, 73 `rgb(43,255,73)` 16.9, 100, 28.6 `rgb(16.9%,100%,28.6%)` 83, 0, 71, 0 128.5°, 100, 58.4 `hsl(128.5,100%,58.4%)` 128.5°, 83.1, 100 33ff33 `#33ff33`\nCIE-LAB 88.214, -80.995, 69.31 37.957, 72.51, 18.298 0.295, 0.563, 72.51 88.214, 106.602, 139.445 88.214, -79.39, 96.87 85.153, -69.452, 46.866 00101011, 11111111, 01001001\n\n# Color Schemes with #2bff49\n\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #ff2be1\n``#ff2be1` `rgb(255,43,225)``\nComplementary Color\n• #77ff2b\n``#77ff2b` `rgb(119,255,43)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #2bffb3\n``#2bffb3` `rgb(43,255,179)``\nAnalogous Color\n• #ff2b77\n``#ff2b77` `rgb(255,43,119)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #b32bff\n``#b32bff` `rgb(179,43,255)``\nSplit Complementary Color\n• #ff492b\n``#ff492b` `rgb(255,73,43)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #492bff\n``#492bff` `rgb(73,43,255)``\n• #e1ff2b\n``#e1ff2b` `rgb(225,255,43)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #492bff\n``#492bff` `rgb(73,43,255)``\n• #ff2be1\n``#ff2be1` `rgb(255,43,225)``\n• #00de1f\n``#00de1f` `rgb(0,222,31)``\n• #00f723\n``#00f723` `rgb(0,247,35)``\n• #12ff33\n``#12ff33` `rgb(18,255,51)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #45ff5f\n``#45ff5f` `rgb(69,255,95)``\n• #5eff75\n``#5eff75` `rgb(94,255,117)``\n• #78ff8b\n``#78ff8b` `rgb(120,255,139)``\nMonochromatic Color\n\n# Alternatives to #2bff49\n\nBelow, you can see some colors close to #2bff49. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #42ff2b\n``#42ff2b` `rgb(66,255,43)``\n• #30ff2b\n``#30ff2b` `rgb(48,255,43)``\n• #2bff37\n``#2bff37` `rgb(43,255,55)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #2bff5b\n``#2bff5b` `rgb(43,255,91)``\n• #2bff6c\n``#2bff6c` `rgb(43,255,108)``\n• #2bff7e\n``#2bff7e` `rgb(43,255,126)``\nSimilar Colors\n\n# #2bff49 Preview\n\nThis text has a font color of #2bff49.\n\n``<span style=\"color:#2bff49;\">Text here</span>``\n#2bff49 background color\n\nThis paragraph has a background color of #2bff49.\n\n``<p style=\"background-color:#2bff49;\">Content here</p>``\n#2bff49 border color\n\nThis element has a border color of #2bff49.\n\n``<div style=\"border:1px solid #2bff49;\">Content here</div>``\nCSS codes\n``.text {color:#2bff49;}``\n``.background {background-color:#2bff49;}``\n``.border {border:1px solid #2bff49;}``\n\n# Shades and Tints of #2bff49\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000401 is the darkest color, while #effff1 is the lightest one.\n\n• #000401\n``#000401` `rgb(0,4,1)``\n• #001703\n``#001703` `rgb(0,23,3)``\n• #002b06\n``#002b06` `rgb(0,43,6)``\n• #003f09\n``#003f09` `rgb(0,63,9)``\n• #00520c\n``#00520c` `rgb(0,82,12)``\n• #00660e\n``#00660e` `rgb(0,102,14)``\n• #007911\n``#007911` `rgb(0,121,17)``\n• #008d14\n``#008d14` `rgb(0,141,20)``\n• #00a117\n``#00a117` `rgb(0,161,23)``\n• #00b41a\n``#00b41a` `rgb(0,180,26)``\n• #00c81c\n``#00c81c` `rgb(0,200,28)``\n• #00dc1f\n``#00dc1f` `rgb(0,220,31)``\n• #00ef22\n``#00ef22` `rgb(0,239,34)``\n• #04ff27\n``#04ff27` `rgb(4,255,39)``\n• #17ff38\n``#17ff38` `rgb(23,255,56)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\n• #3fff5a\n``#3fff5a` `rgb(63,255,90)``\n• #52ff6b\n``#52ff6b` `rgb(82,255,107)``\n• #66ff7c\n``#66ff7c` `rgb(102,255,124)``\n• #79ff8c\n``#79ff8c` `rgb(121,255,140)``\n• #8dff9d\n``#8dff9d` `rgb(141,255,157)``\n• #a1ffae\n``#a1ffae` `rgb(161,255,174)``\n• #b4ffbf\n``#b4ffbf` `rgb(180,255,191)``\n• #c8ffd0\n``#c8ffd0` `rgb(200,255,208)``\n• #dcffe1\n``#dcffe1` `rgb(220,255,225)``\n• #effff1\n``#effff1` `rgb(239,255,241)``\nTint Color Variation\n\n# Tones of #2bff49\n\nA tone is produced by adding gray to any pure hue. In this case, #8d9d8f is the less saturated color, while #2bff49 is the most saturated one.\n\n• #8d9d8f\n``#8d9d8f` `rgb(141,157,143)``\n• #85a589\n``#85a589` `rgb(133,165,137)``\n``#7dad83` `rgb(125,173,131)``\n• #74b67e\n``#74b67e` `rgb(116,182,126)``\n• #6cbe78\n``#6cbe78` `rgb(108,190,120)``\n• #64c672\n``#64c672` `rgb(100,198,114)``\n• #5cce6c\n``#5cce6c` `rgb(92,206,108)``\n• #54d666\n``#54d666` `rgb(84,214,102)``\n• #4cde60\n``#4cde60` `rgb(76,222,96)``\n• #43e75b\n``#43e75b` `rgb(67,231,91)``\n• #3bef55\n``#3bef55` `rgb(59,239,85)``\n• #33f74f\n``#33f74f` `rgb(51,247,79)``\n• #2bff49\n``#2bff49` `rgb(43,255,73)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #2bff49 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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https://terri77.savingadvice.com/2020/11/23/working-picking-up_221635/ | [
"User Real IP - 3.236.253.192\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => 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=> 39.37.142.107\n)\n\n => Array\n(\n => 27.255.186.115\n)\n\n => Array\n(\n => 49.15.193.151\n)\n\n => Array\n(\n => 103.201.146.115\n)\n\n => Array\n(\n => 223.228.177.70\n)\n\n => Array\n(\n => 182.179.141.37\n)\n\n => Array\n(\n => 110.172.131.126\n)\n\n => Array\n(\n => 45.116.232.0\n)\n\n => Array\n(\n => 193.37.32.206\n)\n\n => Array\n(\n => 119.152.62.246\n)\n\n => Array\n(\n => 180.178.148.228\n)\n\n => Array\n(\n => 195.114.145.120\n)\n\n => Array\n(\n => 122.160.49.194\n)\n\n => Array\n(\n => 103.240.237.17\n)\n\n => Array\n(\n => 103.75.245.238\n)\n\n => Array\n(\n => 124.253.215.148\n)\n\n => Array\n(\n => 45.118.165.146\n)\n\n => Array\n(\n => 103.75.244.111\n)\n\n => Array\n(\n => 223.185.7.42\n)\n\n => Array\n(\n => 139.5.240.165\n)\n\n => Array\n(\n => 45.251.117.204\n)\n\n => Array\n(\n => 132.154.71.227\n)\n\n => Array\n(\n => 178.92.100.97\n)\n\n => Array\n(\n => 49.48.248.42\n)\n\n => Array\n(\n => 182.190.109.252\n)\n\n => Array\n(\n => 43.231.57.209\n)\n\n => Array\n(\n => 39.37.185.133\n)\n\n => Array\n(\n => 123.17.79.174\n)\n\n => Array\n(\n => 180.178.146.215\n)\n\n => Array\n(\n => 41.248.83.40\n)\n\n => Array\n(\n => 103.255.4.79\n)\n\n => Array\n(\n => 103.39.119.233\n)\n\n => Array\n(\n => 85.203.44.24\n)\n\n => Array\n(\n => 93.74.18.246\n)\n\n => Array\n(\n => 95.142.120.51\n)\n\n => Array\n(\n => 202.47.42.57\n)\n\n => Array\n(\n => 41.202.219.253\n)\n\n => Array\n(\n => 154.28.188.182\n)\n\n => Array\n(\n => 14.163.178.106\n)\n\n => Array\n(\n => 118.185.57.226\n)\n\n => Array\n(\n => 49.15.141.102\n)\n\n => Array\n(\n => 182.189.86.47\n)\n\n => Array\n(\n => 111.88.68.79\n)\n\n => Array\n(\n => 156.146.59.8\n)\n\n => Array\n(\n => 119.152.62.82\n)\n\n => Array\n(\n => 49.207.128.103\n)\n\n => Array\n(\n => 203.212.30.234\n)\n\n => Array\n(\n => 41.202.219.254\n)\n\n => Array\n(\n => 103.46.203.10\n)\n\n => Array\n(\n => 112.79.141.15\n)\n\n => Array\n(\n => 103.68.218.75\n)\n\n => Array\n(\n => 49.35.130.14\n)\n\n => Array\n(\n => 172.247.129.90\n)\n\n => Array\n(\n => 116.90.74.214\n)\n\n => Array\n(\n => 180.178.142.242\n)\n\n => Array\n(\n => 111.119.183.59\n)\n\n => Array\n(\n => 117.5.103.189\n)\n\n => Array\n(\n => 203.110.93.146\n)\n\n => Array\n(\n => 188.163.97.86\n)\n\n => Array\n(\n => 124.253.90.47\n)\n\n => Array\n(\n => 139.167.249.160\n)\n\n => Array\n(\n => 103.226.206.55\n)\n\n => Array\n(\n => 154.28.188.191\n)\n\n => Array\n(\n => 182.190.197.205\n)\n\n => Array\n(\n => 111.119.183.33\n)\n\n => Array\n(\n => 14.253.254.64\n)\n\n => Array\n(\n => 117.237.197.246\n)\n\n => Array\n(\n => 172.105.53.82\n)\n\n => Array\n(\n => 124.253.207.164\n)\n\n => Array\n(\n => 103.255.4.33\n)\n\n => Array\n(\n => 27.63.131.206\n)\n\n => Array\n(\n => 103.118.170.99\n)\n\n => Array\n(\n => 111.119.183.55\n)\n\n => Array\n(\n => 14.182.101.109\n)\n\n => Array\n(\n => 175.107.223.199\n)\n\n => Array\n(\n => 39.57.168.94\n)\n\n => Array\n(\n => 122.182.213.139\n)\n\n => Array\n(\n => 112.79.214.237\n)\n\n => Array\n(\n => 27.6.252.22\n)\n\n => Array\n(\n => 89.163.212.83\n)\n\n => Array\n(\n => 182.189.23.1\n)\n\n => Array\n(\n => 49.15.222.253\n)\n\n => Array\n(\n => 125.63.97.110\n)\n\n => Array\n(\n => 223.233.65.159\n)\n\n => Array\n(\n => 139.99.159.18\n)\n\n => Array\n(\n => 45.118.165.137\n)\n\n => Array\n(\n => 39.52.2.167\n)\n\n => Array\n(\n => 39.57.141.24\n)\n\n => Array\n(\n => 27.5.32.145\n)\n\n => Array\n(\n => 49.36.212.33\n)\n\n => Array\n(\n => 157.33.218.32\n)\n\n => Array\n(\n => 116.71.4.122\n)\n\n => Array\n(\n => 110.93.244.176\n)\n\n => Array\n(\n => 154.73.203.156\n)\n\n => Array\n(\n => 136.158.30.235\n)\n\n => Array\n(\n => 122.161.53.72\n)\n\n => Array\n(\n => 106.203.203.156\n)\n\n => Array\n(\n => 45.133.7.22\n)\n\n => Array\n(\n => 27.255.180.69\n)\n\n => Array\n(\n => 94.46.244.3\n)\n\n => Array\n(\n => 43.242.178.157\n)\n\n => Array\n(\n => 171.79.189.215\n)\n\n => Array\n(\n => 37.117.141.89\n)\n\n => Array\n(\n => 196.92.32.64\n)\n\n => Array\n(\n => 154.73.203.157\n)\n\n => Array\n(\n => 183.83.176.14\n)\n\n => Array\n(\n => 106.215.84.145\n)\n\n => Array\n(\n => 95.142.120.12\n)\n\n => Array\n(\n => 190.232.110.94\n)\n\n => Array\n(\n => 179.6.194.47\n)\n\n => Array\n(\n => 103.62.155.172\n)\n\n => Array\n(\n => 39.34.156.177\n)\n\n => Array\n(\n => 122.161.49.120\n)\n\n => Array\n(\n => 103.58.155.253\n)\n\n => Array\n(\n => 175.107.226.20\n)\n\n => Array\n(\n => 206.81.28.165\n)\n\n => Array\n(\n => 49.36.216.36\n)\n\n => Array\n(\n => 104.223.95.178\n)\n\n => Array\n(\n => 122.177.69.35\n)\n\n => Array\n(\n => 39.57.163.107\n)\n\n => Array\n(\n => 122.161.53.35\n)\n\n => Array\n(\n => 182.190.102.13\n)\n\n => Array\n(\n => 122.161.68.95\n)\n\n => Array\n(\n => 154.73.203.147\n)\n\n => Array\n(\n => 122.173.125.2\n)\n\n => Array\n(\n => 117.96.140.189\n)\n\n => Array\n(\n => 106.200.244.10\n)\n\n => Array\n(\n => 110.36.202.5\n)\n\n => Array\n(\n => 124.253.51.144\n)\n\n => Array\n(\n => 176.100.1.145\n)\n\n => Array\n(\n => 156.146.59.20\n)\n\n => Array\n(\n => 122.176.100.151\n)\n\n => Array\n(\n => 185.217.117.237\n)\n\n => Array\n(\n => 49.37.223.97\n)\n\n => Array\n(\n => 101.50.108.80\n)\n\n => Array\n(\n => 124.253.155.88\n)\n\n => Array\n(\n => 39.40.208.96\n)\n\n => Array\n(\n => 122.167.151.154\n)\n\n => Array\n(\n => 172.98.89.13\n)\n\n => Array\n(\n => 103.91.52.6\n)\n\n => Array\n(\n => 106.203.84.5\n)\n\n => Array\n(\n => 117.216.221.34\n)\n\n => Array\n(\n => 154.73.203.131\n)\n\n => Array\n(\n => 223.182.210.117\n)\n\n => Array\n(\n => 49.36.185.208\n)\n\n => Array\n(\n => 111.119.183.30\n)\n\n => Array\n(\n => 39.42.107.13\n)\n\n => Array\n(\n => 39.40.15.174\n)\n\n => Array\n(\n => 1.38.244.65\n)\n\n => Array\n(\n => 49.156.75.252\n)\n\n => Array\n(\n => 122.161.51.99\n)\n\n => Array\n(\n => 27.73.78.57\n)\n\n => Array\n(\n => 49.48.228.70\n)\n\n => Array\n(\n => 111.119.183.18\n)\n\n => Array\n(\n => 116.204.252.218\n)\n\n => Array\n(\n => 73.173.40.248\n)\n\n => Array\n(\n => 223.130.28.81\n)\n\n => Array\n(\n => 202.83.58.81\n)\n\n => Array\n(\n => 45.116.233.31\n)\n\n => Array\n(\n => 111.119.183.1\n)\n\n => Array\n(\n => 45.133.7.66\n)\n\n => Array\n(\n => 39.48.204.174\n)\n\n => Array\n(\n => 37.19.213.30\n)\n\n => Array\n(\n => 111.119.183.22\n)\n\n => Array\n(\n => 122.177.74.19\n)\n\n => Array\n(\n => 124.253.80.59\n)\n\n => Array\n(\n => 111.119.183.60\n)\n\n => Array\n(\n => 157.39.106.191\n)\n\n => Array\n(\n => 157.47.86.121\n)\n\n => Array\n(\n => 47.31.159.100\n)\n\n => Array\n(\n => 106.214.85.144\n)\n\n => Array\n(\n => 182.189.22.197\n)\n\n => Array\n(\n => 111.119.183.51\n)\n\n => Array\n(\n => 202.47.35.57\n)\n\n => Array\n(\n => 42.108.33.220\n)\n\n => Array\n(\n => 180.178.146.158\n)\n\n => Array\n(\n => 124.253.184.239\n)\n\n => Array\n(\n => 103.165.20.8\n)\n\n => Array\n(\n => 94.178.239.156\n)\n\n => Array\n(\n => 72.255.41.142\n)\n\n => Array\n(\n => 116.90.107.102\n)\n\n => Array\n(\n => 39.36.164.250\n)\n\n => Array\n(\n => 124.253.195.172\n)\n\n => Array\n(\n => 203.142.218.149\n)\n\n => Array\n(\n => 157.43.165.180\n)\n\n => Array\n(\n => 39.40.242.57\n)\n\n => Array\n(\n => 103.92.43.150\n)\n\n => Array\n(\n => 39.42.133.202\n)\n\n => Array\n(\n => 119.160.66.11\n)\n\n => Array\n(\n => 138.68.3.7\n)\n\n => Array\n(\n => 210.56.125.226\n)\n\n => Array\n(\n => 157.50.4.249\n)\n\n => Array\n(\n => 124.253.81.162\n)\n\n => Array\n(\n => 103.240.235.141\n)\n\n => Array\n(\n => 132.154.128.20\n)\n\n => Array\n(\n => 49.156.115.37\n)\n\n => Array\n(\n => 45.133.7.48\n)\n\n => Array\n(\n => 122.161.49.137\n)\n\n => Array\n(\n => 202.47.46.31\n)\n\n => Array\n(\n => 192.140.145.148\n)\n\n => Array\n(\n => 202.14.123.10\n)\n\n => Array\n(\n => 122.161.53.98\n)\n\n => Array\n(\n => 124.253.114.113\n)\n\n => Array\n(\n => 103.227.70.34\n)\n\n => Array\n(\n => 223.228.175.227\n)\n\n => Array\n(\n => 157.39.119.110\n)\n\n => Array\n(\n => 180.188.224.231\n)\n\n => Array\n(\n => 132.154.188.85\n)\n\n => Array\n(\n => 197.210.227.207\n)\n\n => Array\n(\n => 103.217.123.177\n)\n\n => Array\n(\n => 124.253.85.31\n)\n\n => Array\n(\n => 123.201.105.97\n)\n\n => Array\n(\n => 39.57.190.37\n)\n\n => Array\n(\n => 202.63.205.248\n)\n\n => Array\n(\n => 122.161.51.100\n)\n\n => Array\n(\n => 39.37.163.97\n)\n\n => Array\n(\n => 43.231.57.173\n)\n\n => Array\n(\n => 223.225.135.169\n)\n\n => Array\n(\n => 119.160.71.136\n)\n\n => Array\n(\n => 122.165.114.93\n)\n\n => Array\n(\n => 47.11.77.102\n)\n\n => Array\n(\n => 49.149.107.198\n)\n\n => Array\n(\n => 192.111.134.206\n)\n\n => Array\n(\n => 182.64.102.43\n)\n\n => Array\n(\n => 124.253.184.111\n)\n\n => Array\n(\n => 171.237.97.228\n)\n\n => Array\n(\n => 117.237.237.101\n)\n\n => Array\n(\n => 49.36.33.19\n)\n\n => Array\n(\n => 103.31.101.241\n)\n\n => Array\n(\n => 129.0.207.203\n)\n\n => Array\n(\n => 157.39.122.155\n)\n\n => Array\n(\n => 197.210.85.120\n)\n\n => Array\n(\n => 124.253.219.201\n)\n\n => Array\n(\n => 152.57.75.92\n)\n\n => Array\n(\n => 169.149.195.121\n)\n\n => Array\n(\n => 198.16.76.27\n)\n\n => Array\n(\n => 157.43.192.188\n)\n\n => Array\n(\n => 119.155.244.221\n)\n\n => Array\n(\n => 39.51.242.216\n)\n\n => Array\n(\n => 39.57.180.158\n)\n\n => Array\n(\n => 134.202.32.5\n)\n\n => Array\n(\n => 122.176.139.205\n)\n\n => Array\n(\n => 151.243.50.9\n)\n\n => Array\n(\n => 39.52.99.161\n)\n\n => Array\n(\n => 136.144.33.95\n)\n\n => Array\n(\n => 157.37.205.216\n)\n\n => Array\n(\n => 217.138.220.134\n)\n\n => Array\n(\n => 41.140.106.65\n)\n\n => Array\n(\n => 39.37.253.126\n)\n\n => Array\n(\n => 103.243.44.240\n)\n\n => Array\n(\n => 157.46.169.29\n)\n\n => Array\n(\n => 92.119.177.122\n)\n\n => Array\n(\n => 196.240.60.21\n)\n\n => Array\n(\n => 122.161.6.246\n)\n\n => Array\n(\n => 117.202.162.46\n)\n\n => Array\n(\n => 205.164.137.120\n)\n\n => Array\n(\n => 171.237.79.241\n)\n\n => Array\n(\n => 198.16.76.28\n)\n\n => Array\n(\n => 103.100.4.151\n)\n\n => Array\n(\n => 178.239.162.236\n)\n\n => Array\n(\n => 106.197.31.240\n)\n\n => Array\n(\n => 122.168.179.251\n)\n\n => Array\n(\n => 39.37.167.126\n)\n\n => Array\n(\n => 171.48.8.115\n)\n\n => Array\n(\n => 157.44.152.14\n)\n\n => Array\n(\n => 103.77.43.219\n)\n\n => Array\n(\n => 122.161.49.38\n)\n\n => Array\n(\n => 122.161.52.83\n)\n\n => Array\n(\n => 122.173.108.210\n)\n\n => Array\n(\n => 60.254.109.92\n)\n\n => Array\n(\n => 103.57.85.75\n)\n\n => Array\n(\n => 106.0.58.36\n)\n\n => Array\n(\n => 122.161.49.212\n)\n\n => Array\n(\n => 27.255.182.159\n)\n\n => Array\n(\n => 116.75.230.159\n)\n\n => Array\n(\n => 122.173.152.133\n)\n\n => Array\n(\n => 129.0.79.247\n)\n\n => Array\n(\n => 223.228.163.44\n)\n\n => Array\n(\n => 103.168.78.82\n)\n\n => Array\n(\n => 39.59.67.124\n)\n\n => Array\n(\n => 182.69.19.120\n)\n\n => Array\n(\n => 196.202.236.195\n)\n\n => Array\n(\n => 137.59.225.206\n)\n\n => Array\n(\n => 143.110.209.194\n)\n\n => Array\n(\n => 117.201.233.91\n)\n\n => Array\n(\n => 37.120.150.107\n)\n\n => Array\n(\n => 58.65.222.10\n)\n\n => Array\n(\n => 202.47.43.86\n)\n\n => Array\n(\n => 106.206.223.234\n)\n\n => Array\n(\n => 5.195.153.158\n)\n\n => Array\n(\n => 223.227.127.243\n)\n\n => Array\n(\n => 103.165.12.222\n)\n\n => Array\n(\n => 49.36.185.189\n)\n\n => Array\n(\n => 59.96.92.57\n)\n\n => Array\n(\n => 203.194.104.235\n)\n\n => Array\n(\n => 122.177.72.33\n)\n\n => Array\n(\n => 106.213.126.40\n)\n\n => Array\n(\n => 45.127.232.69\n)\n\n => Array\n(\n => 156.146.59.39\n)\n\n => Array\n(\n => 103.21.184.11\n)\n\n => Array\n(\n => 106.212.47.59\n)\n\n => Array\n(\n => 182.179.137.235\n)\n\n => Array\n(\n => 49.36.178.154\n)\n\n => Array\n(\n => 171.48.7.128\n)\n\n => Array\n(\n => 119.160.57.96\n)\n\n => Array\n(\n => 197.210.79.92\n)\n\n => Array\n(\n => 36.255.45.87\n)\n\n => Array\n(\n => 47.31.219.47\n)\n\n => Array\n(\n => 122.161.51.160\n)\n\n => Array\n(\n => 103.217.123.129\n)\n\n => Array\n(\n => 59.153.16.12\n)\n\n => Array\n(\n => 103.92.43.226\n)\n\n => Array\n(\n => 47.31.139.139\n)\n\n => Array\n(\n => 210.2.140.18\n)\n\n => Array\n(\n => 106.210.33.219\n)\n\n => Array\n(\n => 175.107.203.34\n)\n\n => Array\n(\n => 146.196.32.144\n)\n\n => Array\n(\n => 103.12.133.121\n)\n\n => Array\n(\n => 103.59.208.182\n)\n\n => Array\n(\n => 157.37.190.232\n)\n\n => Array\n(\n => 106.195.35.201\n)\n\n => Array\n(\n => 27.122.14.83\n)\n\n => Array\n(\n => 194.193.44.5\n)\n\n => Array\n(\n => 5.62.43.245\n)\n\n => Array\n(\n => 103.53.80.50\n)\n\n => Array\n(\n => 47.29.142.233\n)\n\n => Array\n(\n => 154.6.20.63\n)\n\n => Array\n(\n => 173.245.203.128\n)\n\n => Array\n(\n => 103.77.43.231\n)\n\n => Array\n(\n => 5.107.166.235\n)\n\n => Array\n(\n => 106.212.44.123\n)\n\n => Array\n(\n => 157.41.60.93\n)\n\n => Array\n(\n => 27.58.179.79\n)\n\n => Array\n(\n => 157.37.167.144\n)\n\n => Array\n(\n => 119.160.57.115\n)\n\n => Array\n(\n => 122.161.53.224\n)\n\n => Array\n(\n => 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n)\n```\nWorking Picking Up: Firefly's Personal Finance Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Working Picking Up\n\n# Working Picking Up\n\nNovember 23rd, 2020 at 06:08 pm\n\nMy work at the private lab has really picked up. Since they first called me in on a Friday evening, I've been in every weekday evening since then with the exception of one. They have arleady contacted me to say they'll need me every evening this week, except for Thanksgiving evening. At my hospital job they also are offering as many hours as I want. They continue to lose their full-time staff for better pay & hours elsewhere. I don't understand why the hospital administration doesn't take some action. It's a critical situation in the middle of a pandemic.\n\nI can say that the rate of positive COVID tests are really ticking up. One of my fellow lab supervisors lost his wife to the virus Saturday. I don't know her health history but I would guess that he is around 50, so not elderly. Please be safe out there all.\n\nMy debt pay off continues to tick down, while my net worth continues to tick up. If I continue to work these hours the payoff date will continue to move closer. I've already made at least \\$25k between these two jobs this year. Next year, when I'm debt-free I'd like to scale back. I think I have finally learned the lesson on how to budget & be more aware of the debt I'm accruing.\n\nI can't wait to write an end of year update. I've accomplished so many things this year!\n\n### 0 Responses to “Working Picking Up”\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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https://leanprover-community.github.io/mathlib_docs/topology/discrete_quotient.html | [
"# mathlibdocumentation\n\ntopology.discrete_quotient\n\n# Discrete quotients of a topological space. #\n\nThis file defines the type of discrete quotients of a topological space, denoted discrete_quotient X. To avoid quantifying over types, we model such quotients as setoids whose equivalence classes are clopen.\n\n## Definitions #\n\n1. discrete_quotient X is the type of discrete quotients of X. It is endowed with a coercion to Type, which is defined as the quotient associated to the setoid in question, and each such quotient is endowed with the discrete topology.\n2. Given S : discrete_quotient X, the projection X → S is denoted S.proj.\n3. When X is compact and S : discrete_quotient X, the space S is endowed with a fintype instance.\n\n## Order structure #\n\nThe type discrete_quotient X is endowed with an instance of a semilattice_inf with order_top. The partial ordering A ≤ B mathematically means that B.proj factors through A.proj. The top element ⊤ is the trivial quotient, meaning that every element of X is collapsed to a point. Given h : A ≤ B, the map A → B is discrete_quotient.of_le h. Whenever X is discrete, the type discrete_quotient X is also endowed with an instance of a semilattice_inf with order_bot, where the bot element ⊥ is X itself.\n\nGiven f : X → Y and h : continuous f, we define a predicate le_comap h A B for A : discrete_quotient X and B : discrete_quotient Y, asserting that f descends to A → B. If cond : le_comap h A B, the function A → B is obtained by discrete_quotient.map cond.\n\n## Theorems #\n\nThe two main results proved in this file are:\n\n1. discrete_quotient.eq_of_proj_eq which states that when X is compact, t2 and totally disconnected, any two elements of X agree if their projections in Q agree for all Q : discrete_quotient X.\n2. discrete_quotient.exists_of_compat which states that when X is compact, then any system of elements of Q as Q : discrete_quotient X varies, which is compatible with respect to discrete_quotient.of_le, must arise from some element of X.\n\n## Remarks #\n\nThe constructions in this file will be used to show that any profinite space is a limit of finite discrete spaces.\n\n@[ext]\nstructure discrete_quotient (X : Type u_1) :\nType u_1\n\nThe type of discrete quotients of a topological space.\n\ntheorem discrete_quotient.ext_iff {X : Type u_1} {_inst_1 : topological_space X} (x y : discrete_quotient X) :\nx = y x.rel = y.rel\ntheorem discrete_quotient.ext {X : Type u_1} {_inst_1 : topological_space X} (x y : discrete_quotient X) (h : x.rel = y.rel) :\nx = y\ndef discrete_quotient.of_clopen {X : Type u_1} {A : set X} (h : is_clopen A) :\n\nConstruct a discrete quotient from a clopen set.\n\nEquations\ntheorem discrete_quotient.refl {X : Type u_1} (S : discrete_quotient X) (x : X) :\nS.rel x x\ntheorem discrete_quotient.symm {X : Type u_1} (S : discrete_quotient X) (x y : X) :\nS.rel x yS.rel y x\ntheorem discrete_quotient.trans {X : Type u_1} (S : discrete_quotient X) (x y z : X) :\nS.rel x yS.rel y zS.rel x z\ndef discrete_quotient.setoid {X : Type u_1} (S : discrete_quotient X) :\n\nThe setoid whose quotient yields the discrete quotient.\n\nEquations\n@[protected, instance]\ndef discrete_quotient.has_coe_to_sort {X : Type u_1} :\n(Type u_1)\nEquations\n@[protected, instance]\nEquations\ndef discrete_quotient.proj {X : Type u_1} (S : discrete_quotient X) :\nX → S\n\nThe projection from X to the given discrete quotient.\n\nEquations\ntheorem discrete_quotient.proj_surjective {X : Type u_1} (S : discrete_quotient X) :\ntheorem discrete_quotient.fiber_eq {X : Type u_1} (S : discrete_quotient X) (x : X) :\nS.proj ⁻¹' {S.proj x} = set_of (S.rel x)\ntheorem discrete_quotient.proj_continuous {X : Type u_1} (S : discrete_quotient X) :\ntheorem discrete_quotient.fiber_closed {X : Type u_1} (S : discrete_quotient X) (A : set S) :\ntheorem discrete_quotient.fiber_open {X : Type u_1} (S : discrete_quotient X) (A : set S) :\ntheorem discrete_quotient.fiber_clopen {X : Type u_1} (S : discrete_quotient X) (A : set S) :\n@[protected, instance]\ndef discrete_quotient.partial_order {X : Type u_1} :\nEquations\n@[protected, instance]\ndef discrete_quotient.order_top {X : Type u_1} :\nEquations\n@[protected, instance]\ndef discrete_quotient.semilattice_inf {X : Type u_1} :\nEquations\n@[protected, instance]\ndef discrete_quotient.inhabited {X : Type u_1} :\nEquations\ndef discrete_quotient.comap {X : Type u_1} (S : discrete_quotient X) {Y : Type u_2} {f : Y → X} (cont : continuous f) :\n\nComap a discrete quotient along a continuous map.\n\nEquations\n@[simp]\ntheorem discrete_quotient.comap_id {X : Type u_1} (S : discrete_quotient X) :\n@[simp]\ntheorem discrete_quotient.comap_comp {X : Type u_1} (S : discrete_quotient X) {Y : Type u_2} {f : Y → X} (cont : continuous f) {Z : Type u_3} {g : Z → Y} (cont' : continuous g) :\nS.comap _ = (S.comap cont).comap cont'\ntheorem discrete_quotient.comap_mono {X : Type u_1} {Y : Type u_2} {f : Y → X} (cont : continuous f) {A B : discrete_quotient X} (h : A B) :\nA.comap cont B.comap cont\ndef discrete_quotient.of_le {X : Type u_1} {A B : discrete_quotient X} (h : A B) :\nA → B\n\nThe map induced by a refinement of a discrete quotient.\n\nEquations\n• = λ (a : A), (λ (x : X), B.proj x) _\n@[simp]\ntheorem discrete_quotient.of_le_refl {X : Type u_1} {A : discrete_quotient X} :\ntheorem discrete_quotient.of_le_refl_apply {X : Type u_1} {A : discrete_quotient X} (a : A) :\n@[simp]\ntheorem discrete_quotient.of_le_comp {X : Type u_1} {A B C : discrete_quotient X} (h1 : A B) (h2 : B C) :\ntheorem discrete_quotient.of_le_comp_apply {X : Type u_1} {A B C : discrete_quotient X} (h1 : A B) (h2 : B C) (a : A) :\ntheorem discrete_quotient.of_le_continuous {X : Type u_1} {A B : discrete_quotient X} (h : A B) :\n@[simp]\ntheorem discrete_quotient.of_le_proj {X : Type u_1} {A B : discrete_quotient X} (h : A B) :\n@[simp]\ntheorem discrete_quotient.of_le_proj_apply {X : Type u_1} {A B : discrete_quotient X} (h : A B) (x : X) :\n(A.proj x) = B.proj x\n@[protected, instance]\ndef discrete_quotient.order_bot {X : Type u_1} :\n\nWhen X is discrete, there is a order_bot instance on discrete_quotient X\n\nEquations\ntheorem discrete_quotient.proj_bot_injective {X : Type u_1} :\ntheorem discrete_quotient.proj_bot_bijective {X : Type u_1} :\ndef discrete_quotient.le_comap {X : Type u_1} {Y : Type u_2} {f : Y → X} (cont : continuous f) (A : discrete_quotient Y) (B : discrete_quotient X) :\nProp\n\nGiven cont : continuous f, le_comap cont A B is defined as A ≤ B.comap f. Mathematically this means that f descends to a morphism A → B.\n\nEquations\ntheorem discrete_quotient.le_comap_id {X : Type u_1} (A : discrete_quotient X) :\ntheorem discrete_quotient.le_comap_comp {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} {Z : Type u_3} {g : Z → Y} {cont' : continuous g} {C : discrete_quotient Z} :\nC A B\ntheorem discrete_quotient.le_comap_trans {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B C : discrete_quotient X} :\nBB C C\ndef discrete_quotient.map {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} (cond : B) :\nA → B\n\nMap a discrete quotient along a continuous map.\n\nEquations\n• = cond\ntheorem discrete_quotient.map_continuous {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} (cond : B) :\n@[simp]\ntheorem discrete_quotient.map_proj {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} (cond : B) :\nA.proj = B.proj f\n@[simp]\ntheorem discrete_quotient.map_proj_apply {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} (cond : B) (y : Y) :\n(A.proj y) = B.proj (f y)\n@[simp]\ntheorem discrete_quotient.map_id {Y : Type u_2} {A : discrete_quotient Y} :\n@[simp]\ntheorem discrete_quotient.map_comp {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} {Z : Type u_3} {g : Z → Y} {cont' : continuous g} {C : discrete_quotient Z} (h1 : C A) (h2 : B) :\n@[simp]\ntheorem discrete_quotient.of_le_map {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B C : discrete_quotient X} (cond : B) (h : B C) :\n@[simp]\ntheorem discrete_quotient.of_le_map_apply {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B C : discrete_quotient X} (cond : B) (h : B C) (a : A) :\n@[simp]\ntheorem discrete_quotient.map_of_le {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} {C : discrete_quotient Y} (cond : B) (h : C A) :\n@[simp]\ntheorem discrete_quotient.map_of_le_apply {X : Type u_1} {Y : Type u_2} {f : Y → X} {cont : continuous f} {A : discrete_quotient Y} {B : discrete_quotient X} {C : discrete_quotient Y} (cond : B) (h : C A) (c : C) :\ntheorem discrete_quotient.eq_of_proj_eq {X : Type u_1} [t2_space X] [disc : totally_disconnected_space X] {x y : X} :\n(∀ (Q : , Q.proj x = Q.proj y)x = y\ntheorem discrete_quotient.fiber_le_of_le {X : Type u_1} {A B : discrete_quotient X} (h : A B) (a : A) :\ntheorem discrete_quotient.exists_of_compat {X : Type u_1} (Qs : Π (Q : , Q) (compat : ∀ (A B : (h : A B), (Qs A) = Qs B) :\n∃ (x : X), ∀ (Q : , Q.proj x = Qs Q\n@[protected, instance]\nnoncomputable def discrete_quotient.fintype {X : Type u_1} (S : discrete_quotient X) :\nEquations\ndef locally_constant.discrete_quotient {X : Type u_1} {α : Type u_2} (f : α) :\n\nAny locally constant function induces a discrete quotient.\n\nEquations\ndef locally_constant.lift {X : Type u_1} {α : Type u_2} (f : α) :\n\nThe function from the discrete quotient associated to a locally constant function.\n\nEquations\ntheorem locally_constant.lift_is_locally_constant {X : Type u_1} {α : Type u_2} (f : α) :\ndef locally_constant.locally_constant_lift {X : Type u_1} {α : Type u_2} (f : α) :\n\nA locally constant version of locally_constant.lift.\n\nEquations\n@[simp]\ntheorem locally_constant.lift_eq_coe {X : Type u_1} {α : Type u_2} (f : α) :\n@[simp]\ntheorem locally_constant.factors {X : Type u_1} {α : Type u_2} (f : α) :"
] | [
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https://fr.scribd.com/document/251773912/10-1-Complex-Arithmetic | [
"Vous êtes sur la page 1sur 11\n\nContents\n\n10\n\nComplex Numbers\n10.1 Complex Arithmetic\n\n12\n\n20\n\n10.4 De Moivres Theorem\n\n29\n\nLearning outcomes\nIn this Workbook you will learn what a complex number is and how to combine complex\nnumbers together using the familiar operations of addition, subtraction, multiplication\nand division. You will also learn how to describe a complex number graphically using\nthe Argand diagram. The connection between the exponential function and the\ntrigonometric functions is explained. You will understand how De Moivre's theorem\nis used to obtain fractional powers of complex numbers.\n\nComplex Arithmetic\n\n10.1\n\nIntroduction\nComplex numbers are used in many areas of engineering and science. In this Section we define\nwhat a complex number is and explore how two such numbers may be combined together by adding,\nsubtracting, multiplying and dividing. We also show how to find complex roots of polynomial\nequations.\nA complex number is a generalisation of an ordinary real number. In fact, as we shall see, a complex\nnumber is a pair of real numbers ordered in a particular way. Fundamental to the study of complex\nnumbers is the symbol i with the strange looking property i2 = 1. Apart from this property complex\nnumbers follow the usual rules of number algebra.\n\n'\n\nbe able to add, subtract, multiply and divide\n\nreal numbers\n\nPrerequisites\nBefore starting this Section you should . . .\n\nbe able to combine algebraic fractions\n\ntogether\nunderstand what a polynomial is\nhave a knowledge of trigonometric identities\n\n&\n'\n\n%\n\\$\n\ncombine complex numbers together\n\nLearning Outcomes\nOn completion you should be able to . . .\n&\n\nfind the modulus and conjugate of a complex\n\nnumber\nobtain complex solutions to polynomial\nequations\nHELM (2008):\nWorkbook 10: Complex Numbers\n\n1. What is a complex number?\n\nWe assume that you are familiar with the properties of ordinary numbers; examples are\n\n3\n, 2.634, 3.111, , e, 2\n1, 2,\n10\nWe all know how to add, subtract, multiply and divide such numbers. We are aware that the\nnumbers can be positive or negative or zero and also aware of their geometrical interpretation as\nbeing represented by points on a real axis known as a number line (Figure 1).\nx\n\nO\nFigure 1\n\nThe real axis is a line with a direction (usually chosen to be from left to right) indicated by an arrow.\nWe shall refer to this as the x-axis. On this axis we select a point, arbitrarily, and refer to this as the\norigin O. The origin (where zero is located) distinguishes positive numbers from negative numbers:\nto the right of the origin are the positive numbers\nto the left of the origin are the negative numbers\nThus we can locate the numbers in our example. See Figure 2.\n3.111\n\n3\n1\n2\n10\nFigure 2\n\n2.634\n\nFrom now on we shall refer to these ordinary numbers as real numbers. We can formalise the\nalgebra of real numbers into a set of rules which they obey.\nSo if x1 , x2 and x3 are any three real numbers then we know that, in particular:\n1. x1 + x2 = x2 + x1\n2. 1 x1 = x1\n\nx1 + (x2 + x3 ) = (x1 + x2 ) + x3\n\n0 x1 = 0\n\n3. x1 x2 = x2 x1\n\nx1 (x2 + x3 ) = x1 x2 + x1 x3\n\nAlso, in multiplication we are familiar with the elementary rules:\n\n(positive) (positive) = positive\n\nIt follows that if x represents any real number then\n\nx2 0\nin words, the square of a real number is always non-negative.\nIn this Workbook we will consider a kind of number (a generalisation of a real number) whose square\nis not necessarily positive (and not necessarily real either). Dont worry that i does not exist.\nBecause of that it is called imaginary! We just define it and get on and use it and it then turns out\nto be very useful and important in many practical applications. However, it is important to get to\nknow how to handle complex numbers before using them in calculations. This will not be difficult as\nthe new set of rules is, in fact, precisely the same set of rules obeyed by the real numbers. These\nHELM (2008):\nSection 10.1: Complex Arithmetic\n\nnew numbers are called complex numbers.\n\nA complex number is an ordered pair of real numbers, usually denoted by z or w etc. So if a, b are\nreal numbers then we designate a complex number through:\nz = a + ib\nwhere i is a symbol obeying the rule\ni2 = 1\nFor simplicity we shall assume we can write\n\ni = 1.\n(Often, particularly in engineering applications, the symbol j is used instead of i). Also note that,\nconventially, examples of actual complex numbers such as 2 + 3i are written like this and not 2 + i3.\nAgain we ask the reader to accept matters at this stage without worrying about the meaning of\nfinding the square root of a negative number. Using this notation we can write\np\n\n4 = (4)(1) = 4 1 = 2i etc.\n\nKey Point 1\nThe symbol i is such that\ni2 = 1\nUsing the normal rules of algebra it follows that\ni3 = i2 i = i\n\ni4 = i2 i2 = (1) (1) = 1\n\nand so on.\n\nSimple examples of complex numbers are\n\nz2 = 3 + (2.461)i\n\nz1 = 3 + 2i\n\nz3 = 17i\n\nz4 = 3 + 0i = 3\n\nGenerally, if z = a + ib then a is called the real part of z, or Re(z) for short, and b is called the\nimaginary part of z or Im(z). The fourth example indicates that the real numbers can be considered\na subset of the complex numbers.\n\nKey Point 2\nIf\n\nz = a + ib\n\nthen\n\nRe(z) = a and Im(z) = b\n\nBoth the real and imaginary parts of a complex number are real.\n\nHELM (2008):\nWorkbook 10: Complex Numbers\n\nKey Point 3\nTwo complex numbers z = a + ib and w = c + id are said to be equal if and only if both their real\nparts are the same and both their imaginary parts are the same, that is\na=c\n\nand\n\nb=d\n\nKey Point 4\nThe modulus of a complex number z = a + ib is denoted by |z| and is defined by\n\n|z| = a2 + b2\nso that the modulus is always a non-negative real number.\n\nExample 1\nIf z = 3 2i then find Re(z), Im(z) and |z|.\n\nSolution\nHere Re(z) = 3, Im(z) = 2 and |z| =\n\n32 + (2)2 =\n\n13.\n\nComplex conjugate\nIf z = a + ib is any complex number then the complex conjugate of z is denoted by z and is defined\nby z = a ib. (Sometimes the notation z is used instead of z to denote the conjugate). For\nexample if z = 2 3i then z = 2 + 3i. If z is entirely real then z = z whereas if z is entirely\nimaginary then z = z. E.g. if z = 17i then z = 17i. In fact the following relationships are\neasily obtained:\nRe(z) =\n\nz + z\n2\n\nand\n\nHELM (2008):\nSection 10.1: Complex Arithmetic\n\nIm(z) =\n\ni(z z)\n2\n5\n\nIf z = 2 + i find expressions for Re(z ) and Im(i(z z)).\n\nHint: first find z , z z, and i(z z):\n\nRe(z ) = 2 and Im(i(z z)) = 0\n\n2. The algebra of complex numbers\n\nComplex numbers are added, subtracted, multiplied and divided in much the same way as these\noperations are carried out for real numbers.\n\nAddition and subtraction of complex numbers\n\nLet z and w be any two complex numbers\nz = a + ib\n\nw = c + id\n\nthen\nz + w = (a + c) + i(b + d)\n\nz w = (a c) + i(b d)\n\nFor example if z = 2 3i, w = 4 + 2i then\n\nz + w = {2 + (4)} + {(3) + 2}i = 2 i\n\nMultiplying one complex number by another\n\nIn multiplication we proceed using an obvious approach: again consider any two complex numbers\nz = a + ib and w = c + id. Then\nzw = (a + ib)(c + id)\n= ac + aid + ibc + i2 bd\nobtained in the usual way by multiplying all the terms in one bracket by all the terms in the other\nbracket. Now we use the fundamental relation i2 = 1 so that\nzw = ac + aid + ibc bd\n= ac bd + i(ad + bc)\nwhere we have re-grouped terms with the i symbol and terms without the i symbol separately.\nThese are the real and imaginary parts of the product zw respectively. A numerical example will\n6\n\nHELM (2008):\nWorkbook 10: Complex Numbers\n\nconfirm the approach. If z = 2 3i and w = 4 + 2i then\n\nzw =\n=\n=\n=\n=\n\n(2 3i)(4 + 2i)\n2(4) + 2(2i) 3i(4) 3i(2i)\n8 + 4i + 12i 6i2\n8 + 16i + 6\n2 + 16i\n\nIf z = 2 + i and w = 3 + 2i find expressions for\n(a) z + 2w, (b) |z w| and (c) zw\n\n(a)\nz + 2w = 4 + 5i\n(b) Hint: you should find that z w = 5 i\np\n\n|z w| = (5)2 + (1)2 = 26\n(c)\nzw = 6 + 3i 4i + 2i2 = 8 i\nIn general the square of a complex number is not necessarily a positive real number; it may not even\nbe real at all. For example if z = 2 + i then\nz 2 = (2 + i)2 = 4 4i + i2 = 4 4i 1 = 3 4i\nHowever, the product of a complex number with its conjugate is always a non-negative real number.\nIf z = a + ib then\nzz =\n=\n=\n=\n\n(a + ib)(a ib)\na2 a(ib) + (ib)a i2 b2\na2 i 2 b 2\na2 + b 2\nsince i2 = 1\n\nHELM (2008):\nSection 10.1: Complex Arithmetic\n\nFor example, if z = 2 + i then\n\nzz = (2 + i)(2 i) = 4 + 1 = 5\n\nShow, for any complex number z = a + ib that zz = |z|2 .\n\nDividing one complex number by another\n\nHere we consider the operation of dividing one complex number z = a + ib by another, w = c + id:\na + ib\nz\n=\nw\nc + id\nWe wish to simplify the right-hand side into the standard form of a complex number (this is called\nthe Cartesian form):\n(Real part) + i (Imaginary part)\nor the equivalent:\n(Real part) + (Imaginary part) i\nTo do this we multiply top and bottom by the complex conjugate of the bottom (the denominator),\nthat is, by c id (this is called rationalising):\nz\na + ib\na + ib c id\n=\n=\n\nw\nc + id\nc + id c id\nand then carry out the multiplication, top and bottom:\nz\n(ac + bd) + i(bc ad)\n=\nw\nc2 \u0013\n+ d2 \u0012\n\u0013\n\u0012\nac + bd\n=\n+i\nc2 + d2\nc2 + d2\nwhich is now in the required form.\nThe reason for rationalising is to get a real number in the denominator since a complex number\ndivided by a real number is easy to evaluate.\n8\n\nHELM (2008):\nWorkbook 10: Complex Numbers\n\nExample 2\nFind\n\nz\nif z = 2 3i and w = 2 + i.\nw\n\nSolution\n\nz\n2 3i\n(2 3i) (2 i)\n=\n=\nrationalising\nw\n2+i\n(2 + i) (2 i)\n4 3 + i(6 2)\n=\nmultiplying out\n4+1\n1 8\ni\ndividing through\n=\n5 5\n\nIf z = 3 i and w = 1 + 3i find\n\n2z + 3w\n.\n2z 3w\n\n2z + 3w\n9 + 7i\n(9 + 7i)(3 + 11i)\n=\n=\n2z 3w\n3 11i\n(3 11i)(3 + 11i)\n=\n\n27 77 + (21 + 99)i\n9 + 121\n\nHELM (2008):\nSection 10.1: Complex Arithmetic\n\n50\n120\n5\n12\n+\ni= + i\n130 130\n13 13\n\nExercises\n1. If z = 2 i, w = 3 + 4i find expressions\nz Cartesian form) for\n\u0010 z \u0011 (in standard\n\n(a) z 3w,\n(b) zw\n(c)\n(d)\nw\nw\n2. Verify the following statements for general complex numbers z = a + ib and w = c + id\nz\nz + z\ni(z z)\n|z|\n\n(b) (zw) = z w\n(c) Re(z) =\n(d) Im(z) =\n.\n(a) =\nw\n|w|\n2\n2\n3. Find z such that zz + 3(z z ) = 13 + 12i\n\n11\n2\n5\n+ i (d)\n1. (a) 7 13i (b) 2 11i (c)\n25 25\n5\n\n3. z = 3 + 2i\n\n3. Solutions of polynomial equations\n\nWith the introduction of complex numbers we can now obtain solutions to those polynomial equations\nwhich may have real solutions, complex solutions or a combination of real and complex solutions.\nFor example, the simple quadratic equation:\nx2 + 16 = 0\n\ncan be rearranged:\n\nx2 = 16\n\nand then taking square roots:\n\nx = 16 = 4 1 = 4i\n\nwhere we are replacing 1 by the symbol i.\n\nThis approach can be extended to the general quadratic equation\n\nb b2 4ac\n2\nax + bx + c = 0\nwith roots\nx=\n2a\nso that for example, if\n3x2 + 2x + 2 = 0\nthen solving for x:\np\n4 4(3)(2)\nx =\n2(3)\n\n2 i 20\n2 20\n=\n=\n6\n6\n2\n\n20\n2 5\n5\n1\n5\n1\n5\nso, (as\n=\n=\n), the two roots are +\ni and\ni.\n6\n6\n3\n3\n3\n3\n3\nIn this example we see that the two solutions (roots) are complex conjugates of each other. In fact\nthis will always be the case if the polynomial equation has real coefficients: that is, if any complex\nroots occur they will always occur in complex conjugate pairs.\n10\n\nHELM (2008):\nWorkbook 10: Complex Numbers\n\nKey Point 5\nComplex roots to polynomial equations having real coefficients\nalways occur in complex conjugate pairs\n\nExample 3\nGiven that x = 3 2i is one root of the cubic equation x3 7x2 + 19x 13 = 0\nfind the other two roots.\n\nSolution\nSince the coefficients of the equation are real and 3 2i is a root then its complex conjugate 3 + 2i is\nalso a root which implies that x (3 2i) and x (3 + 2i) are factors of the given cubic expression.\nMultiplying together these two factors:\n(x (3 2i))(x (3 + 2i)) = x2 x(3 2i) x(3 + 2i) + 13 = x2 6x + 13\nSo x2 6x + 13 is a quadratic factor of the cubic equation. The remaining factor must take the\nform (x + a) where a is real, since only one more linear factor of the cubic equation is required, and\nso we write:\nx3 7x2 + 19x 13 = (x2 6x + 13)(x + a)\nBy inspection (consider for example the constant terms), it is clear that a = 1 so that the final\nfactor is (x 1), implying that the original cubic equation has a root at x = 1.\n\nExercises\n1. Find the roots of the equation x2 + 2x + 2 = 0.\n2. If i is one root of the cubic equation x3 + 2x2 + x + 2 = 0 find the two other roots.\n3. Find the complex number z if 2z + z + 3i + 2 = 0.\n4. If z = cos + i sin show that\nAnswers 1. x = 1 i\n\nHELM (2008):\nSection 10.1: Complex Arithmetic\n\nz\n= cos 2 + i sin 2.\nz\n\n2. i, 2\n\n2\n3. 3i\n3\n\n11"
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https://snewiki.com/wiki/index.php?title=Reference | [
"# Reference\n\n• The Wiki Editing Reference\n\n`= One Equal=`\n\n# One Equal\n\n`== Two Equals==`\n\n## Two Equals\n\n`=='''Two Equals w Bold'''==`\n\n## Two Equals w Bold\n\n`===Three Equals===`\n\n### Three Equals\n\n`==='''Three Equals w Bold'''===`\n\n### Three Equals w Bold\n\n• Makes BOLD in TOC only\n\n`====Four Equal====`\n\n#### Four Equals\n\n`<font size=\"-2\">Font Size -2</font>`\n\n• Font Size -2\n\n`<font size=\"+2\">Font Size +2</font>`\n\n• Font Size +2\n\n`<font size=\"+3\">Font Size +3</font>`\n\n• Font Size +3\n\n`<font size=\"+4\">Font Size +4</font>`\n\n• Font Size +4\n\n## Tables\n\n(Table names are not caps specific, listed on top)\n\nThis is the SMALLER table\nSmaller (Using Font-2)\nThis is a SMALLtable table\nSmall\nThis is the SMALLMED Table\nSmall-Medium Table\nThis is the halftable table\nHalf\nThis is a Mediumtable table\nMedium (Should be used for FREQUENCIES on trunked pages)\nThis is a Largetable/Prettytable\nLarge/Pretty Table (They are the same.) (Should be used for most other items)\n\n`{{TOC}}` Creates the new TOC.\n\n## Table Formatting.\n\n• `!'''Column Headers'''`\n• `|| splits columns `\n• `|- creates new row `\n• `| Non header row FIRST column only `\n• ` |} closes table `\n\n## TAG (HTML Anchors)\n\n• `<div id=\"1\">This creates a page target</div> `\n• When you enter a page url, putting a \"#\" followed by the target name, you'll directly go to that section.\n\n• `[[RISCON|Rhode Island's Trunk]] `\n• `[[RISCON#East_Providence|East Providence]] `\n• `[https://www.necrat.us NECRAT] `"
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https://yddp.rgmconsultants.it/logistic-regression-in-r-code.html | [
"Feb 10, 2010 · Logistic regression (sometimes called the logistic model or logit model) is used for prediction of the probability of occurrence of an event by fitting data to a logistic curve. Logistic regression is used extensively in the medical and social sciences as well as marketing applications such as prediction of a customer’s propensity to purchase ...\n\nIn statistics, logistic regression or logit regression is a type of probabilistic statistical classification model. It is also used to predict a binary response from a binary predictor, used for predicting the outcome of a categorical dependent variable (i.e., a class label) based on one or more predictor variables (features).\n\n### Modern chemistry chapter 9 standardized test prep answers\n\nClunk when shifting gears manual\nThe OLS regression is performed on the Y and R tables. In order to circumvent the interpretation problem with the parameters obtained from the regression, XLSTAT transforms the results back into the initial space to obtain the parameters and the confidence intervals that correspond to the input variables.\nFiberfab kit cars\n\nMay 20, 2017 · Logistic Regression in Python to Tune Parameter C Posted on May 20, 2017 by charleshsliao The trade-off parameter of logistic regression that determines the strength of the regularization is called C, and higher values of C correspond to less regularization (where we can specify the regularization function).C is actually the Inverse of ...\n\nJan 05, 2018 · Linear Regression works for continuous data, so Y value will extend beyond [0,1] range. As the output of logistic regression is probability, response variable should be in the range [0,1]. To solve this restriction, the Sigmoid function is used over Linear regression to make the equation work as Logistic Regression as shown below.\n\nConsider a logistic regression model, that is P(Y = 1|X = x) exp(Bo + Bix) = 1 - P(Y = 0X = x). 1 + exp(Bo + B12) After fitting the regression, we predict the class as a = (x), which can be either 0 or 1. 4 1. Justify that P(Y # û(x)) = E(Y – Î (r))? 2.Next, show that E[(Y - 1)^4x = 2] = P(Y = 1/x = z)(1 – 2(x)) + 2(2) 3. Dec 24, 2020 · Regression models investigate what variables explain their location. For example: If you have crime locations in a city, you can use spatial regression to understand the factors behind patterns of crime. We can use spatial regression to understand what variables (income, education, and more) explain crime locations. Multinomial regression. is an extension of binomial logistic regression.. The algorithm allows us to predict a categorical dependent variable which has more than two levels. Like any other regression model, the multinomial output can be predicted using one or more independent variable.\n\nStepwise logistic regression consists of automatically selecting a reduced number of predictor variables for building the best performing logistic regression model. Read more at Chapter @ref(stepwise-regression). This chapter describes how to compute the stepwise logistic regression in R.. Contents:\nGics classification excel 2020\n\nDec 16, 2013 · Logistic Regression in Tableau using R December 16, 2013 Bora Beran 62 Comments In my post on Tableau website earlier this year, I included an example of multiple linear regression analysis to demonstrate taking advantage of parameters to create a dashboard that can be used for What-If analysis.\n\nLogistic Regression is a core supervised learning technique for solving classification problems. This article goes beyond its simple code to first understand the concepts behind the approach, and how it all emerges from the more basic technique of Linear Regression.\n\nTo generate the multivariable logistic regression model, the following code is implemented: model <- glm (Survived ~ Sex + Age + Parch + Fare, data = titanic, family = binomial) Multinomial logistic regression is known by a variety of other names, including polytomous LR, multiclass LR, softmax regression, multinomial logit (mlogit), the maximum entropy (MaxEnt) classifier, and the conditional maximum entropy model.\n\n### Outdoor gourmet triton parts\n\nFormula hybrid 2020 for assetto corsa free\nInstalling predator engine on riding lawn mower\n\nLecture 15: mixed-effects logistic regression 28 November 2007 In this lecture we’ll learn about mixed-effects modeling for logistic regres-sion. 1 Technical recap We moved from generalized linear models (GLMs) to multi-level GLMs by adding a stochastic component to the linear predictor: η = α +β 1X 1 +···+β nX n +b 0 +b 1Z 1 ...\n\nGlass tubes for glass blowing near me\n\nFeb 08, 2014 · In R, the glm (generalized linear model) command is the standard command for fitting logistic regression. As far as I am aware, the fitted glm object doesn't directly give you any of the pseudo R squared values, but McFadden's measure can be readily calculated.\n\nStack o matic record changer turntable\n\nConsider a logistic regression model, that is P(Y = 1|X = x) exp(Bo + Bix) = 1 - P(Y = 0X = x). 1 + exp(Bo + B12) After fitting the regression, we predict the class as a = (x), which can be either 0 or 1. 4 1. Justify that P(Y # û(x)) = E(Y – Î (r))? 2.Next, show that E[(Y - 1)^4x = 2] = P(Y = 1/x = z)(1 – 2(x)) + 2(2) 3.\n\nScreen recording detected error\n\nThe larger the $$R_{MF}^2$$, the better the model fits the data. It can be used as an indicator for the “goodness of fit” of a model. For the model fit3, we have $R_{MF}^2=1-\\frac{1571.7}{2920.6}=0.46$ The R returned by the logistic regression in our data program is the square root of McFadden’s R Logistic regression is used to predict the class (or category) of individuals based on one or multiple predictor variables (x). It is used to model a binary outcome, that is a variable, which can have only two possible values: 0 or 1, yes or no, diseased or non-diseased.\n\n### Roblox surf exploit\n\nEngel burman group\nVape coils smok\n\nJun 20, 2017 · Logistic regression, or logit regression, or logit model is a regression model where the dependent variable (DV) is categorical. DependentCategorical Variables that can have only fixed values such as A, B or C, Yes or No Y = f(X) i.e Y is dependent on X. Logistic regression. Logistic regression is widely used to predict a binary response. It is a linear method as described above in equation $\\eqref{eq:regPrimal}$, with the loss function in the formulation given by the logistic loss: $L(\\wv;\\x,y) := \\log(1+\\exp( -y \\wv^T \\x)).$ For binary classification problems, the algorithm outputs a ...\n\nNintendo current ratio\n\nRecap of Logistic Regression •Feature vector ɸ, two-classes C 1and C 2 •A posterioriprobability p(C 1 | ɸ)can be written as p(C 1 | ɸ) =y(ɸ) = σ (wTɸ) whereɸis aM-dimensional feature vector σ(.)is the logistic sigmoid function •Goal is to determine the Mparameters •Known as logistic regression in statistics Jan 16, 2016 · Tagged code, plot, predicted probabilities, R, statistics 22 Comments Post navigation Previous Post Why Knitr Beats Sweave Next Post Ethnic discrimination in hiring decisions: a meta-analysis of correspondence tests 1990–2015\n\n### Bilstein b8 vs b12\n\nClayton homes albany\n\n330 Logistic quantile regression 3 Stata syntax Inference about the logistic quantile regression model above can be carried out with the new Stata commands lqreg, lqregpred,andlqregplot. We describe their syntax in this section and illustrate their use in section 4. 3.1 lqreg lqreg estimates logistic quantile regression for bounded outcomes. Logistic regression implementation in R. R makes it very easy to fit a logistic regression model. The function to be called is glm() and the fitting process is not so different from the one used in linear regression. In this post I am going to fit a binary logistic regression model and explain each step. The dataset\n\nWatch snapchat stories anonymously online\n\nMay 20, 2016 · Depending on statistical software, we can run hierarchical regression with one click (SPSS) or do it manually step-by-step (R). Regardless, it’s good to understand how this works conceptually. Build sequential (nested) regression models by adding variables at each step. Run ANOVAs (to compute $$R^2$$) and regressions (to obtain coefficients).\n\nMissing work letter to parents\nDescribe conjugation between two paramecium.\n\nLogistic Regression is a core supervised learning technique for solving classification problems. This article goes beyond its simple code to first understand the concepts behind the approach, and how it all emerges from the more basic technique of Linear Regression. Component. Logistic Regression. Cluster Analysis. Typical Application (used when) Response variables are categorical in nature i.e., binary outcomes 1 or whether something happened or not etc. (e.g., customer did not respond to the sales promotion or they did respond to it) The Model¶. Logistic regression is a probabilistic, linear classifier. It is parametrized by a weight matrix and a bias vector .Classification is done by projecting data points onto a set of hyperplanes, the distance to which reflects a class membership probability.\n\nBest pistol caliber rifle for deer hunting\n\nRegression problems are supervised learning problems in which the response is continuous. Linear regression is a technique that is useful for regression problems. Classification problems are supervised learning problems in which the response is categorical; Benefits of linear regression. widely used; runs fast; easy to use (not a lot of tuning ...\n\n### Nessus activation code\n\nProportional tables\nPinal county news today\n\nThe codebook contains the following information on the variables: VARIABLE DESCRIPTIONS: Survived Survival (0 = No; 1 = Yes) Pclass Passenger Class (1 = 1st; 2 = 2nd; 3 = 3rd) Name Name Sex Sex Age Age SibSp Number of Siblings/Spouses Aboard Parch Number of Parents/Children Aboard Ticket Ticket Number Fare Passenger Fare Cabin Cabin Embarked Port of Embarkation (C = Cherbourg; Q = Queenstown ... • Graphically representing data in R before and after analysis • How to do basic statistical operations in R • Understand how to interpret the result of Linear and Logistic Regression model and translate them into actionable insight • Indepth knowledge of data collection and data preprocessing for Linear and Logistic Regression problem Click Classify - Logistic Regression on the Data Mining ribbon. The Logistic Regression dialog appears. The categorical variable CAT.MEDV has been derived from the MEDV variable (Median value of owner-occupied homes in \\$1000's) a 1 for MEDV levels above 30 (>= 30) and a 0 for levels below 30 (<30). This will be our Output Variable.\n\nHonda pioneer 1000 clicking noise\n\nAs with the linear regression routine and the ANOVA routine in R, the 'factor( )' command can be used to declare a categorical predictor (with more than two categories) in a logistic regression; R will create dummy variables to represent the categorical predictor using the lowest coded category as the reference group.Apr 03, 2020 · In those cases, it would be more efficient to import that data, as opposed to type it within the code. For example, you may capture the same dataset that you saw at the beginning of this tutorial (under step 1) within a CSV file. You can then use the code below to perform the multiple linear regression in R.\n\nStorage containers with lids\n\nMultinomial regression. is an extension of binomial logistic regression.. The algorithm allows us to predict a categorical dependent variable which has more than two levels. Like any other regression model, the multinomial output can be predicted using one or more independent variable.See full list on towardsdatascience.com\n\nElie wiesel acceptance speech analysis\n\nExample of Logistic Regression on Python. Steps to Steps guide and code explanation. Confusion Matrix for Logistic Regression Model. Logistic Regression in R with glm. In this section, you'll study an example of a binary logistic regression, which you'll tackle with the ISLR package, which will provide you with the data set, and the glm() function, which is generally used to fit generalized linear models, will be used to fit the logistic regression model. Loading Data To generate the multivariable logistic regression model, the following code is implemented: model <- glm (Survived ~ Sex + Age + Parch + Fare, data = titanic, family = binomial)\n\n### Vizio v series vs tcl 4 series\n\nMinion masters discord bot\n\nIn logistic regression, coefficients are typically on a log-odds (or logit) scale: log(p/(1-p)). By taking the exponent coefficients are converted to odds and odds ratios. Unlike binary logistic regression in multinomial logistic regression, we need to define the reference level. Please note this is specific to the function which I am using from nnet package in R. There are some functions from other R packages where you don't really need to mention the reference level before building the model.\n\nThe logistic regression is of the form 0/1. y = 0 if a loan is rejected, y = 1 if accepted. A logistic regression model differs from linear regression model in two ways. First of all, the logistic regression accepts only dichotomous (binary) input as a dependent variable (i.e., a vector of 0 and 1).\n\n### Hegner quick clamp\n\nMainstays basic yarn green\n\n### Good samaritan activities and crafts\n\nAccenture digital consultant salary\nFoldable flight boomerang 1\n\nA logistic regression is said to provide a better fit to the data if it demonstrates an improvement over a model with fewer predictors. This is performed using the likelihood ratio test, which compares the likelihood of the data under the full model against the likelihood of the data under a model with fewer predictors.\n\nKendra elliot tv series\n\nAug 20, 2009 · Now adjust the data for the logistic regression. We must create a data frame: dft - as.data.frame(table) dft Var1 Var2 Freq 1 stressNO reflNO 251 2 stressYES reflNO 131 3 stressNO reflYES 4 4 stressYES reflYES 33 We can now fit the model, and then perform the logistic regression in R: Feb 19, 2018 · Logistic regression does the same thing, but with one addition. The logistic regression model computes a weighted sum of the input variables similar to the linear regression, but it runs the result through a special non-linear function, the logistic function or sigmoid function to produce the output y.\n\n### Mayan haab calendar calculator\n\nWinchester 10 gun safe\nNfs heat map size vs forza horizon 4\n\nANOVA for Regression Analysis of Variance (ANOVA) consists of calculations that provide information about levels of variability within a regression model and form a basis for tests of significance. The basic regression line concept, DATA = FIT + RESIDUAL, is rewritten as follows: (y i - ) = (i - ) + (y i - i). Apr 05, 2016 · Get the coefficients from your logistic regression model. First, whenever you’re using a categorical predictor in a model in R (or anywhere else, for that matter), make sure you know how it’s being coded!! covariates at that time, i.e Z(t). The regression e ect of Z() is constant over time. Some people do not call this model ‘proportional hazards’ any more, because the hazard ratio expf 0Z(t)gvaries over time. But many of us still use the term ‘PH’ loosely here. Comparison with a single binary predictor (like heart trans-plant):\n\nKern county deputy pay scale\n\nSee full list on analyticsvidhya.com The stepwise logistic regression can be easily computed using the R function stepAIC() available in the MASS package. It performs model selection by AIC. It performs model selection by AIC. It has an option called direction , which can have the following values: “both”, “forward”, “backward” (see Chapter @ref(stepwise-regression)).\n\nDropping paint by dianne\n\nConsider a logistic regression model, that is P(Y = 1|X = x) exp(Bo + Bix) = 1 - P(Y = 0X = x). 1 + exp(Bo + B12) After fitting the regression, we predict the class as a = (x), which can be either 0 or 1. 4 1. Justify that P(Y # û(x)) = E(Y – Î (r))? 2.Next, show that E[(Y - 1)^4x = 2] = P(Y = 1/x = z)(1 – 2(x)) + 2(2) 3. Logistic regression implementation in R. R makes it very easy to fit a logistic regression model. The function to be called is glm() and the fitting process is not so different from the one used in linear regression. In this post, I am going to fit a binary logistic regression model and explain each step. The dataset\n\nMini schnauzer haircut styles\n\nLogistic regression sometimes called the Logit Model predicts based on probability using the logistic regression equation. In this article, we will learn to implement the Logistic regression in R programming language. Readers are expected to have some basic understanding of the language. Understanding the Logistic Regressor Consider a logistic regression model, that is P(Y = 1|X = x) exp(Bo + Bix) = 1 - P(Y = 0X = x). 1 + exp(Bo + B12) After fitting the regression, we predict the class as a = (x), which can be either 0 or 1. 4 1. Justify that P(Y # û(x)) = E(Y – Î (r))? 2.Next, show that E[(Y - 1)^4x = 2] = P(Y = 1/x = z)(1 – 2(x)) + 2(2) 3. Unlike binary logistic regression in multinomial logistic regression, we need to define the reference level. Please note this is specific to the function which I am using from nnet package in R. There are some functions from other R packages where you don't really need to mention the reference level before building the model.\n\n### Mercedes w204 seat cover replacement\n\nEzviz activation code\nLs3 1.8 rocker arms\n\nMay 17, 2020 · In this guide, I’ll show you an example of Logistic Regression in Python. In general, a binary logistic regression describes the relationship between the dependent binary variable and one or more independent variable/s. Feb 08, 2014 · In R, the glm (generalized linear model) command is the standard command for fitting logistic regression. As far as I am aware, the fitted glm object doesn't directly give you any of the pseudo R squared values, but McFadden's measure can be readily calculated.\n\nEaton m90 on sbc\n\nAssumptions of Multiple Regression This tutorial should be looked at in conjunction with the previous tutorial on Multiple Regression. Please access that tutorial now, if you havent already. When running a Multiple Regression, there are several assumptions that you need to check your data meet, in order for your analysis to be reliable and valid. R Logistic Regression - The Logistic Regression is a regression model in which the response variable (dependent variable) has categorical values such as True/False or 0/1. It actually measures the probability of a binary response as the value of response variable based on the mathematical equation relating it with the predictor variables. In logistic regression, the dependent variable is binary, i.e. it only contains data marked as 1 (Default) or 0 (No default). We can say that logistic regression is a classification algorithm used to predict a binary outcome (1 / 0, Default / No Default) given a set of independent variables.\n\n### Playoff bracket simulator\n\nChapter 7 test a accounting answers\n\n### Pixel 2 xl battery life reddit\n\nEdge ai nvidia\nF9 wireless earbuds manual\n\nLogistic regression is used to predict the class (or category) of individuals based on one or multiple predictor variables (x). It is used to model a binary outcome, that is a variable, which can have only two possible values: 0 or 1, yes or no, diseased or non-diseased.\n\nAug 19, 2018 · We’ll be using the dataset quality.csv to build a logistic regression model in R to predict the quality of ... Narcotics 0.07630 0.03205 2.381 0.01728 * ---Signif. codes: 0 ...\n\n### Bot sentinel review\n\nHow to get bluetooth on pc without adapter\nClassic rock rar\n\nLogistic regression is used to predict the class (or category) of individuals based on one or multiple predictor variables (x). It is used to model a binary outcome, that is a variable, which can have only two possible values: 0 or 1, yes or no, diseased or non-diseased.\n\nWjhl sports reporters\n\nExample of Logistic Regression on Python. Steps to Steps guide and code explanation. Confusion Matrix for Logistic Regression Model. See full list on analyticsvidhya.com\n\nHizpo manual\n\nLogistic regression Logistic regression is used when there is a binary 0-1 response, and potentially multiple categorical and/or continuous predictor variables. Logistic regression can be used to model probabilities (the probability that the response variable equals 1) or for classi cation.\n\n4 post car lift\n\nI want to plot a logistic regression curve of my data, but whenever I try to my plot produces multiple curves. Here's a picture of my last attempt: last attempt Here's the relevant code I am usin... In statistics, logistic regression or logit regression is a type of probabilistic statistical classification model. It is also used to predict a binary response from a binary predictor, used for predicting the outcome of a categorical dependent variable (i.e., a class label) based on one or more predictor variables (features).\n\n### Mirage g4js swap\n\nRed dead redemption 2 save editor\nHotshot trucks with sleepers for sale\n\nThe R code is provided below but if you're a Python user, here's an awesome code window to build your logistic regression model. No need to open Jupyter - you can do it all here: Considering the availability, I've built this model on our practice problem - Dressify data set.Logistic regression can be performed in R with the glm (generalized linear model) function. This function uses a link function to determine which kind of model to use, such as logistic, probit, or poisson.\n\nVfd cable conduit size\n\nI have achieved 68% accuracy using glm with family = 'binomial' while doing logistic regression in R. I don't have any idea on how to specify the number of iterations through my code.\n\nCast to fios box"
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http://www.vpuzzles.com/ | [
"##### puzzle2\n\nThe answer to this 338 match stick puzzle is 3331 as shown in below picture\n\n##### puzzle1\n\nThe answer to this brain teaser picture puzzle is 2 as shown below\n\n##### math5\n\nThe answer the this algebraic math puzzle is 5.3 as explained in the below picture\n\n##### math4\n\nThe answer to this puzzle is 175 as explained below\n\n##### math3\n\nThe answer to this math puzzle is 215 as explained below"
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https://www.physicsforums.com/threads/subspace-help-please.391918/ | [
"## Main Question or Discussion Point\n\n1. Let U be a subspace of Rn and let\nU⊥ = {w ∈ Rn : w is orthogonal to U} .\nProve that\n(i) U⊥ is a subspace of Rn,\n(ii) dimU + dimU⊥ = n.\n\nAttempt.\n\ni)\nU. ( U⊥)T=0\nIf U⊥ does not passes the origin , the above equation cannot be satisfied.\nTherefore U⊥ passes the origin.\nU.( U⊥+ U⊥)T=U. ( U⊥)T+U.( U⊥)T=0+0=0\nU.(k U⊥)T=k[U. (U⊥)T]=k.0=0\n\nTherefore U⊥ is a subspace in Rn\n\nii) let U have rank r.\nU⊥ is the nullspace of the U transpose.\nand if U is a matrix of mxn , UT is nxm.\nTherefore , the sum of dim of the two matrices is exactly n.\n\nRelated Linear and Abstract Algebra News on Phys.org\nHallsofIvy\nHomework Helper\nI don't understand your notation. You use \"T\" which I guess is \"transpose\" and talk about U being a matrix. U is given as a subspace, not a matrix.\n\nI don't understand your notation. You use \"T\" which I guess is \"transpose\" and talk about U being a matrix. U is given as a subspace, not a matrix.\nyup you are right.. it is transpose and subspace..\n\n1. Let U be a subspace of Rn and let\nU⊥ = {w ∈ Rn : w is orthogonal to U} .\nProve that\n(i) U⊥ is a subspace of Rn,\n(ii) dimU + dimU⊥ = n.\n\nAttempt.\n\ni)\nU. ( U⊥)T=0\nIf U⊥ does not passes the origin , the above equation cannot be satisfied.\nTherefore U⊥ passes the origin.\nU.( U⊥+ U⊥)T=U. ( U⊥)T+U.( U⊥)T=0+0=0\nU.(k U⊥)T=k[U. (U⊥)T]=k.0=0\n\nTherefore U⊥ is a subspace in Rn\n\nii) let U have rank r.\nU⊥ is the nullspace of the U transpose.\nand if U is a matrix of mxn , UT is nxm.\nTherefore , the sum of dim of the two matrices is exactly n.\n\nLet u,v be in U⊥. Let k be a constant in R.\nNow <u,w>=0 for all w in U and <v,w>=0 for all w in U.\nThus by linearity of inner product we have <u+kv,w>=0 for all w in U. Thus, u+kv is also in U⊥ for all u,v,k. Thus U⊥ is a subspace.\n\nFor the second part, DIY but here are some hints:\n- We know that U $$\\cap$$ U⊥ = {0}. Easy to check (if a vector is in U and perpendicular to U then it has to be the zero vector).\n- Take x in Rn. Prove that x = u+u' for u in U and u' in U⊥.\n- This is unique representation. If x = u+u' = v+v' with u,v in U and u',v' in U⊥ then 0 = (u-v)+(u'-v') implies u-v = v'-u' implies u-v = v'-u' = 0 since U $$\\cap$$ U⊥ = {0}. Thus u=v and u'=v'.\n- Thus projection map $$\\pi$$: Rn -->> U defined by $$\\pi$$(x)=u where x = u+u' with u in U and u' in U⊥.\n- Use Isomorphism theorem corollary (that dim(range)+dim(kernel)=dim(Rn)=n)\n\nLet u,v be in U⊥. Let k be a constant in R.\nNow <u,w>=0 for all w in U and <v,w>=0 for all w in U.\nThus by linearity of inner product we have <u+kv,w>=0 for all w in U. Thus, u+kv is also in U⊥ for all u,v,k. Thus U⊥ is a subspace.\n\nFor the second part, DIY but here are some hints:\n- We know that U $$\\cap$$ U⊥ = {0}. Easy to check (if a vector is in U and perpendicular to U then it has to be the zero vector).\n- Take x in Rn. Prove that x = u+u' for u in U and u' in U⊥.\n- This is unique representation. If x = u+u' = v+v' with u,v in U and u',v' in U⊥ then 0 = (u-v)+(u'-v') implies u-v = v'-u' implies u-v = v'-u' = 0 since U $$\\cap$$ U⊥ = {0}. Thus u=v and u'=v'.\n- Thus projection map $$\\pi$$: Rn -->> U defined by $$\\pi$$(x)=u where x = u+u' with u in U and u' in U⊥.\n- Use Isomorphism theorem corollary (that dim(range)+dim(kernel)=dim(Rn)=n)\n\nHmm.. i think for part 2 i can explain by stating that dim(U+U⊥)=dim(U)+dim(U⊥)+dim(U$$\\cap$$ U⊥)\nRn=dim(U)+dim(U⊥)\nsince dim(U$$\\cap$$ U⊥) ought to be atleast {0} to satisfy the condition of subspace , and all the more they are orthogonal to one another , the can't be parallel , thus they have to be {0} itself.\nHence proving part 2.\n\nLast edited:"
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https://answers.gmslearner.xyz/2022/07/a-block-of-mass-2-kg-is-placed-on-floor.html | [
"# A block of mass 2 kg is placed on the floor (μ = 0.4). a horizontal force of 7 n is applied on the block. the force of friction between the block and floor is\n\nThe mass of the block as given in the question is 2kg.\n\nThe coefficient of static friction is 0.4.\n\nForce applied on the block is Fnet = 2.8N\n\nThe force of friction between the block and the floor =?\n\nLimitng friction =μN = 0.4 × 2 × 10 = 8N\n\nAs applied force is less than limiting friction\n\nso Force of friction on block=Applied external force=7N\n\nGetting Info..."
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https://zbmath.org/?q=an:0782.03005 | [
"# zbMATH — the first resource for mathematics\n\nDecidable modal logic with undecidable admissibility problem. (English. Russian original) Zbl 0782.03005\nAlgebra Logic 31, No. 1, 53-61 (1992); translation from Algebra Logika 31, No. 1, 83-93 (1992).\nThe admissibility problem for a given logic $$L$$ is to determine whether an arbitrary given inference rule $$A_ 1(p_ 1,\\dots,p_ n),\\dots,A_ m(p_ 1,\\dots,p_ n)/B(p_ 1,\\dots,p_ n)$$ is admissible in $$L$$, i.e., for all formulas $$C_ 1,\\dots,C_ n$$, $$B(C_ 1,\\dots,C_ n)\\in L$$ whenever $$A_ 1(C_ 1,\\dots,C_ n)\\in L,\\dots,A_ m(C_ 1,\\dots,C_ n)\\in L$$.\nAs is known, V. Rybakov proved the decidability of the admissibility problem for a number of intermediate and modal logics.\nIn this paper, the author constructs a decidable normal modal logic for which the admissibility problem is undecidable. The logic is an extension of K4 of width 3 and has infinitely many axioms.\n\n##### MSC:\n 03B45 Modal logic (including the logic of norms) 03B25 Decidability of theories and sets of sentences\nFull Text:\n##### References:\n V. V. Rybakov, ”Problems of admissibility and substitution, logical equations and restricted theories of free algebras,” in: Logic, Methodology and Philosophy of Science VIII, Elsevier (1989), pp. 121–139. · Zbl 0691.03012 V. V. Rybakov, ”On admissibility of inference rules in modal logicG,” Tr. Inst. Mat. Sib. Otd. Akad. Nauk SSSR,12, 120–138, (1989). V. V. Rybakov, ”Equations in free closure algebras and the substitution problem,” Dokl. Akad. Nauk SSSR,287, No. 3, 554–557 (1986). G. D. Birkhoff, Lattice Theory, Amer. Math. Soc. (1979). K. Fine, ”Logics containing K4, Part I,” J. Symb. Logic,39, No. 1, 31–42 (1974). · Zbl 0287.02010 A. I. Mal’tsev, ”Identical relations on quasigroup varieties,” Mat. Sb.,69, No. 1, 3–12 (1966).\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77030504,"math_prob":0.9707899,"size":2446,"snap":"2021-31-2021-39","text_gpt3_token_len":778,"char_repetition_ratio":0.12203112,"word_repetition_ratio":0.021978023,"special_character_ratio":0.3282911,"punctuation_ratio":0.24548736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99162835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T14:29:20Z\",\"WARC-Record-ID\":\"<urn:uuid:b9cb7ae6-316c-4926-bc7b-b89dd68e1cbb>\",\"Content-Length\":\"49709\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6952188b-0c40-4817-bcbd-007ca434b0b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a602293-d277-4b90-93e8-bde8d1b942b7>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0782.03005\",\"WARC-Payload-Digest\":\"sha1:JWZQFUQ2HMNIBOJNXCUQN7TE75APUZGA\",\"WARC-Block-Digest\":\"sha1:C3XWEW2GES7HQFRHK2G5MDM2Q45VBMGX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060803.2_warc_CC-MAIN-20210928122846-20210928152846-00064.warc.gz\"}"} |
https://gradebuddy.com/doc/3524954/example-problem-photoelectric-effect/ | [
"OSU PHYSICS 1251 - Example Problem: Photoelectric Effect\nPages 2\n\nUnformatted text preview:\n\nExample Problem: Photoelectric EffectPhysics 1251TA: Brian Clark11/30/20151. A piece of metal has a cutoff wavelength of λcutoff= 450 nm. Consider illuminatingthis piece of metal with two different wavelengths of light: a λ1= 500 nm beam and aλ2= 400 nm beam. For each of the two beams, find:(a) The maximum kinetic energy of ejected electrons.We can only eject an electron if the illuminating beams are energetic enough. BecauseE = hf = hc/λ, we can only eject an electron if λilluminating≤ λcutoff. So, for λ1,whichis greater than λcutoff, we eject no electrons.However, for λ2, which is less than λcutoff, we do eject electrons. The energy of these electronsis given by:E = hf − φ = hc/λ −φThat is, we take the incoming energy of the photon (hc/λ), subtract off the energy required tobind the electron (φ, the work function) and we are left with the kinetic energy of the electron.E = hf − φ =hcλ2−hcλcutoff= hc1λ2−1λcutoff= (6.626 × 10−34J · s)(3 × 108m/s)1400 × 10−9−1450 × 10−9=⇒ E2= 5.21 × 10−20J = 0.344 eV(b) What is their speed?Because λ1never ejects electrons, it does not make sense to speak of their speed. For λ2,we will simply solve using our kinetic energy formula from 1250:E =12mv2=⇒ v =r2Em=s2 · 5.21 × 10−20J9.11 × 10−31kg=⇒ v2= 3.48 × 105m/s(c) What is their de Broglie wavelength?Again, because λ1never ejects electrons, it does not make sense to speak of their de Brogliewavelength. For λ2, we can apply the de Broglie wavelength formulaλ =hp=6.626 × 10−34J · s9.11 × 10−31kg · 3.48 ×105m/s=⇒ λ2,dB= 2.09 × 10−9m = 2.09 nm2. An electron, a proton, and a photon each have a wavelength of 0.24 nm. For each one,find the momentum, the energy, and, where relevant, the accelerating voltage neededto achieve that wavelength:The momentum calculation is the same for all three particles. We employ the de Broglie relation:p =hλ=⇒ pp= pe=6.626 × 10−34J · s0.24 × 10−9m=⇒ pp= pe= pγ= 2.76 × 10−24kg m/sNow, let’s specialize to the massive particles first. For a massive particle, we can apply that p = mvand E =12mv2, to find: E =12p2m. So:Ee=12p2eme=12·(2.76 × 10−24kg m/s)29.11 × 10−31kg=⇒ Ee= 4.044 × 10−18J = 25.24 eVEp=12p2pmp=12·(2.76 × 10−24kg m/s)21.67 × 10−27kg=⇒ Ep= 2.206 × 10−21J = 0.014 eVBecause these are massive particles, they can be brought to this energy by an accelerating potential,given by E = q∆V :|∆Ve| =Eeqe=4.044 × 10−18J1.602 × 10−19C=⇒ Ve= 25.24 V|∆Vp| =Epqp=2.206 × 10−21J1.602 × 10−19C=⇒ Ve= 0.013 VFor the photon, the massless particle, the calculation is more straightforward:Eγ= hf =hcλ=(6.626 × 10−34J · s)(3 × 108m/s)0.24 × 10−9m=⇒ Eγ= 8.28 × 10−16J = 5.68 keV(Which is termed a “soft x-ray” in the astrophysics community, by the way.)3. What is the de Broglie wavelength of an electron that has 2.0 keV of kinetic energy?What about an electron with 200 keV of kinetic energy? The second one requiresrelativity–why?For the 2 keV electron, we can apply:λ =hp=hm · v=hm ·q2Em=h√2Em=6.626 × 10−34J · sp2 · (3.2 × 10−16J) · (9.11 × 10−31kg)=⇒ λ2 keV= 2.74 × 10−11mWe must be more careful for the 200 keV electron, because its velocity is, to first order:v =r2Em=s2 · 3.204 × 10−14J9.11 × 10−31kg≈ 2.65 × 108m/swhich is very near the speed of light.So, to be relativistically correct, we must apply the relativistic kinetic energy formula to get p:E =p(mc2)2+ p2c2− mc2=⇒ p =1cp(E + mc2)2− (mc2)2p =1cp[3.204 × 10−14J + (9.11 × 10−31kg) · (3 ×108m/s)2]2− [(9.11 × 10−31kg)(3 × 108m/s)2]2=⇒ p = 2.64 × 10−22kg m/sNow, we can plug this correct momenta into the de Broglie equationλ =hp=6.626 × 10−34J · s2.64 × 10−22kg m/s=⇒λ200 keV= 2.51 ×\n\nView Full Document\n\n# OSU PHYSICS 1251 - Example Problem: Photoelectric Effect\n\nPages: 2\nDocuments in this Course",
null,
"Unlocking..."
] | [
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"https://static.gradebuddy.com/5cb0163f/images/giphy.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7585959,"math_prob":0.99832237,"size":3601,"snap":"2023-40-2023-50","text_gpt3_token_len":1379,"char_repetition_ratio":0.13705866,"word_repetition_ratio":0.036484245,"special_character_ratio":0.39794502,"punctuation_ratio":0.15006003,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989774,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T10:52:26Z\",\"WARC-Record-ID\":\"<urn:uuid:966dd811-006a-49dc-a76a-3a351218fef5>\",\"Content-Length\":\"99839\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15b4cebb-3a6f-49d8-9f60-da04a5a0ee4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:531992f8-f166-4517-bbc8-0a50a00bd0dd>\",\"WARC-IP-Address\":\"172.67.161.48\",\"WARC-Target-URI\":\"https://gradebuddy.com/doc/3524954/example-problem-photoelectric-effect/\",\"WARC-Payload-Digest\":\"sha1:HR2STNCMCVYPGKLBHCH244HUN7YT37UV\",\"WARC-Block-Digest\":\"sha1:OEXSKWSR3BHWT46CA722IATV5WXOFH3B\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510501.83_warc_CC-MAIN-20230929090526-20230929120526-00280.warc.gz\"}"} |
https://questions.examside.com/past-years/year-wise/jee/jee-main/jee-main-2021-online-27th-july-morning-shift | [
"JEE Main 2021 (Online) 27th July Morning Shift\nPaper was held on Tue, Jul 27, 2021 3:30 AM\nView Questions\n\n## Chemistry\n\nWhich one of the following compounds will give orange precipitate when treated with 2, 4-dinitrophenyl hydrazine?\nView Question\nThe product obtained from the electrolytic oxidation of acidified sulphate solutions, is :\nView Question\nThe parameters of the unit cell of a substance are a = 2.5, b = 3.0, c = 4.0, $$\\alpha$$ = 90$$^\\circ$$, $$\\beta$$ = 120\nView Question\nThe oxidation states of 'P' in H4P2O7, H4P2O5 and H4P2O6, respectively, are :\nView Question\nFor a reaction of order n, the unit of the rate constant is :\nView Question\nGiven below are two statements : Statement I : Aniline is less basic than acetamide.Statement II : In aniline, the lone\nView Question\nThe type of hybridisation and magnetic property of the complex [MnCl6]3$$-$$, respectively, are :\nView Question\nThe number of geometrical isomers found in the metal complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+\nView Question\nWhich one of the following statements is NOT correct?\nView Question\nGiven below are two statements :Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydr\nView Question\nPresence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible react\nView Question\nWhich one among the following chemical tests is used to distinguish monosaccharide from disaccharide?\nView Question\nMatch List - I with List - II : List-I(Drug) List-II(Class of Drug) (a) Furacin\nView Question\nThe statement is incorrect about Ellingham diagram is\nView Question\nConsider the above reaction and identify the Product P :\nView Question\nThe compound 'A' is a complementary base of _______________ in DNA stands.\nView Question\nStaggered and eclipsed conformers of ethane are :\nView Question\nMatch List - I with List - II : List - I List - II (a) NaOH (i) Acidic\nView Question\nThe correct order of stability of given carbocation is :\nView Question\nGiven below are two statements : One is labelled as Assertion A and the other labelled as Reason R.Assertion A : Lithium\nView Question\nThe density of NaOH solution is 1.2 g cm$$-$$3. The molality of this solution is _____________ m. (Round off to the Near\nView Question\nCO2 gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of CO2 ad\nView Question\nThe conductivity of a weak acid HA of concentration 0.001 mol L$$-$$1 is 2.0 $$\\times$$ 10$$-$$5 S cm$$-$$1. If $$\\Lambd View Question 1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42$$\\times$$10$$-$$3 bar. View Question An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is View Question The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. T View Question In gaseous triethyl amine the \"$$-$$C$$-$$N$$-$$C$$-$$\" bond angle is _________ degree. View Question For water at 100$$^\\circ$$C and 1 bar,$$\\Delta$$vap H$$-\\Delta$$vap U = _____________$$\\times$$102 J mol$$-$$1. View Question PCl5$$\\rightleftharpoons$$PCl3 + Cl3Kc = 1.8443.0 moles of PCl5 is introduced in a 1 L closed reaction vessel at 380 K View Question The difference between bond orders of CO and NO$$^ \\oplus $$is$${x \\over 2}$$where x = _____________. (Round off to t View Question ## Mathematics If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and$${{37} \\over 4}$$respectively, t View Question The value of$$\\mathop {\\lim }\\limits_{n \\to \\infty } {1 \\over n}\\sum\\limits_{j = 1}^n {{{(2j - 1) + 8n} \\over {(2j - 1)\nView Question\nLet $$\\overrightarrow a = \\widehat i + \\widehat j + 2\\widehat k$$ and $$\\overrightarrow b = - \\widehat i + 2\\widehat View Question The value of the definite integral$$\\int\\limits_{ - {\\pi \\over 4}}^{{\\pi \\over 4}} {{{dx} \\over {(1 + {e^{x\\cos x}})({\nView Question\nLet C be the set of all complex numbers. Let$${S_1} = \\{ z \\in C||z - 3 - 2i{|^2} = 8\\}$$$${S_2} = \\{ z \\in C|{\\mathop{ View Question If the area of the bounded region$$R = \\left\\{ {(x,y):\\max \\{ 0,{{\\log }_e}x\\} \\le y \\le {2^x},{1 \\over 2} \\le x \\le 2\nView Question\nA ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this\nView Question\nIf the coefficients of x7 in $${\\left( {{x^2} + {1 \\over {bx}}} \\right)^{11}}$$ and x$$-$$7 in $${\\left( {{x} - {1 \\over View Question The compound statement$$(P \\vee Q) \\wedge ( \\sim P) \\Rightarrow Q$$is equivalent to : View Question If$$\\sin \\theta + \\cos \\theta = {1 \\over 2}$$, then 16(sin(2$$\\theta$$) + cos(4$$\\theta$$) + sin(6$$\\theta$$)) is equ View Question Let$$A = \\left[ {\\matrix{ 1 & 2 \\cr { - 1} & 4 \\cr } } \\right]$$. If A$$-$$1 =$$\\alpha$$I +$$\\bet\nView Question\nLet $$f:\\left( { - {\\pi \\over 4},{\\pi \\over 4}} \\right) \\to R$$ be defined as $$f(x) = \\left\\{ {\\matrix{ {{{(1 + |\\ View Question Let y = y(x) be solution of the differential equation$${\\log _{}}\\left( {{{dy} \\over {dx}}} \\right) = 3x + 4y$$, with y View Question Let the plane passing through the point ($$-$$1, 0,$$-$$2) and perpendicular to each of the planes 2x + y$$-$$z = 2 a View Question Two tangents are drawn from the point P($$-$$1, 1) to the circle x2 + y2$$-$$2x$$-$$6y + 6 = 0. If these tangents to View Question Let f : R$$\\to$$R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of$$\\mathop {\\lim }\\limits_{x \\to 2\nView Question\nLet P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is p\nView Question\nLet $$\\alpha$$, $$\\beta$$ be two roots of the equation x2 + (20)1/4x + (5)1/2 = 0. Then $$\\alpha$$8 + $$\\beta$$8 is equa\nView Question\nThe probability that a randomly selected 2-digit number belongs to the set {n $$\\in$$ N : (2n $$-$$ 2) is a multiple of\nView Question\nLet $$A = \\{ (x,y) \\in R \\times R|2{x^2} + 2{y^2} - 2x - 2y = 1\\}$$, $$B = \\{ (x,y) \\in R \\times R|4{x^2} + 4{y^2} - 16 View Question For real numbers$$\\alpha$$and$$\\beta$$, consider the following system of linear equations :x + y$$-$$z = 2, x + 2y View Question Let$$\\overrightarrow a = \\widehat i + \\widehat j + \\widehat k,\\overrightarrow b $$and$$\\overrightarrow c = \\widehat\nView Question\nIf $${\\log _3}2,{\\log _3}({2^x} - 5),{\\log _3}\\left( {{2^x} - {7 \\over 2}} \\right)$$ are in an arithmetic progression, t\nView Question\nLet the domain of the function$$f(x) = {\\log _4}\\left( {{{\\log }_5}\\left( {{{\\log }_3}(18x - {x^2} - 77)} \\right)} \\righ View Question Let$$f(x) = \\left| {\\matrix{ {{{\\sin }^2}x} & { - 2 + {{\\cos }^2}x} & {\\cos 2x} \\cr {2 + {{\\sin }^2}x}\nView Question\nLet $$F:[3,5] \\to R$$ be a twice differentiable function on (3, 5) such that $$F(x) = {e^{ - x}}\\int\\limits_3^x {(3{t^2} View Question Let a plane P pass through the point (3, 7,$$-$$7) and contain the line,$${{x - 2} \\over { - 3}} = {{y - 3} \\over 2} =\nView Question\nLet S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S $$\\to$$ S such that f(m . n) = f(m) . f(n) fo\nView Question\nIf $$y = y(x),y \\in \\left[ {0,{\\pi \\over 2}} \\right)$$ is the solution of the differential equation $$\\sec y{{dy} \\over View Question Let$$f:[0,3] \\to R$$be defined by$$f(x) = \\min \\{ x - [x],1 + [x] - x\\} $$where [x] is the greatest integer less tha View Question ## Physics In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sec View Question The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2$$\\t\nView Question\nThe relative permittivity of distilled water is 81. The velocity of light in it will be :(Given $$\\mu$$r = 1)\nView Question\nList-I List-II (a) MI of the rod (length L, Mass M, about an axis $$\\bot$$ to the rod passing\nView Question\nThree objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are\nView Question\nA capacitor of capacitance C = 1 $$\\mu$$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $$View Question In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capa View Question The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the r View Question In Young's double slit experiment, if the source of light changes from orange to blue then : View Question In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the g View Question Assertion A : If A, B, C, D are four points on a semi-circular are with centre at 'O' such that$$\\left| {\\overrightarro\nView Question\nA light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of cross-sectional area 'a' is mad\nView Question\nA particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kineti\nView Question\nIf 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N0) then for a collectio\nView Question\nTwo capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and t\nView Question\nA ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times o\nView Question\nA 0.07 H inductor and a 12$$\\Omega$$ resistor are connected in series to a 220V, 50 Hz ac source. The approximate curren\nView Question\nTwo identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'\nView Question\nAssertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gau\nView Question\nA body takes 4 min. to cool from 61$$^\\circ$$ C to 59$$^\\circ$$ C. If the temperature of the surroundings is 30$$^\\circ View Question Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As t View Question Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separ View Question In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5$$\\mathop A\\limits^o $$. If View Question A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200$$\\mu$$F is first char View Question A particle of mass 9.1$$\\times$$10$$-$$31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation o View Question A prism of refractive index n1 and another prism of refractive index n2 are stuck together (as shown in the figure). n1 View Question A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10$$-$$6 m2 stretched View Question A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage View Question The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 1 View Question In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85$$\\times$$10$$-2 A/m2 and moment of inert\nView Question\nEXAM MAP\nJoint Entrance Examination"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81146085,"math_prob":0.99831086,"size":11352,"snap":"2023-14-2023-23","text_gpt3_token_len":3506,"char_repetition_ratio":0.16980965,"word_repetition_ratio":0.02835443,"special_character_ratio":0.32408386,"punctuation_ratio":0.082004555,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999075,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T19:59:27Z\",\"WARC-Record-ID\":\"<urn:uuid:29722b41-86ee-471b-b01e-186bccb99a7b>\",\"Content-Length\":\"172673\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b0623e7-cf30-4afe-bfea-cb33aa68a9a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:8897f605-773c-4201-971d-b400a6625f56>\",\"WARC-IP-Address\":\"172.67.132.22\",\"WARC-Target-URI\":\"https://questions.examside.com/past-years/year-wise/jee/jee-main/jee-main-2021-online-27th-july-morning-shift\",\"WARC-Payload-Digest\":\"sha1:MDPAD6EWBACFI6SZOMNZM7FKWHYOHPJ2\",\"WARC-Block-Digest\":\"sha1:GVSGZ7BIEJJIMO6XSAVCCP7DD4CUES4L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653071.58_warc_CC-MAIN-20230606182640-20230606212640-00433.warc.gz\"}"} |
https://la.mathworks.com/help/signal/ref/grpdelay.html?lang=en | [
"# grpdelay\n\nAverage filter delay (group delay)\n\n## Syntax\n\n[gd,w] = grpdelay(b,a,n)\n[gd,w] = grpdelay(sos,n)\n[gd,w] = grpdelay(d,n)\n[gd,w] = grpdelay(___,'whole')\n[gd,f] = grpdelay(___,n,fs)\n[gd,f] = grpdelay(___,n,'whole',fs)\ngd = grpdelay(___,win)\ngd = grpdelay(___,fin,fs)\ngrpdelay(___)\n\n## Description\n\n[gd,w] = grpdelay(b,a,n) returns the n-point group delay response vector, gd, and the corresponding angular frequency vector, w, for the digital filter with transfer function coefficients stored in b and a.\n\nexample\n\n[gd,w] = grpdelay(sos,n) returns the n-point group delay response corresponding to the second-order sections matrix sos.\n\nexample\n\n[gd,w] = grpdelay(d,n) returns the n-point group delay response for the digital filter d.\n[gd,w] = grpdelay(___,'whole') returns the group delay at n sample points around the entire unit circle.\n[gd,f] = grpdelay(___,n,fs) returns the group delay response vector gd and the corresponding physical frequency vector f for a digital filter designed to filter signals sampled at a rate fs.\n[gd,f] = grpdelay(___,n,'whole',fs) returns the frequency vector at n points ranging between 0 and fs.\ngd = grpdelay(___,win) returns the group delay response vector gd evaluated at the normalized frequencies supplied in win.\n\nexample\n\ngd = grpdelay(___,fin,fs) returns the group delay response vector gd evaluated at the physical frequencies supplied in fin.\n\nexample\n\ngrpdelay(___) with no output arguments plots the group delay response of the filter.\n\n## Examples\n\ncollapse all\n\nDesign a Butterworth filter of order 6 with normalized 3-dB frequency $0.2\\pi$ rad/sample. Use grpdelay to display the group delay.\n\n[z,p,k] = butter(6,0.2); sos = zp2sos(z,p,k); grpdelay(sos,128)",
null,
"Plot both the group delay and the phase delay of the system on the same figure.\n\ngd = grpdelay(sos,512); [h,w] = freqz(sos,512); pd = -unwrap(angle(h))./w; plot(w/pi,gd,w/pi,pd) grid xlabel 'Normalized Frequency (\\times\\pi rad/sample)' ylabel 'Group and phase delays' legend('Group delay','Phase delay')",
null,
"Use designfilt to design a sixth-order Butterworth Filter with normalized 3-dB frequency $0.2\\pi$ rad/sample. Display its group delay response.\n\nd = designfilt('lowpassiir','FilterOrder',6, ... 'HalfPowerFrequency',0.2,'DesignMethod','butter'); grpdelay(d)",
null,
"Design an 88th-order FIR filter of arbitrary magnitude response. The filter has two passbands and two stopbands. The lower-frequency passband has twice the gain of the higher-frequency passband. Specify a sample rate of 200 Hz. Visualize the magnitude response and the phase response of the filter from 10 Hz to 78 Hz.\n\nfs = 200; d = designfilt('arbmagfir', ... 'FilterOrder',88, ... 'NumBands',4, ... 'BandFrequencies1',[0 20], ... 'BandFrequencies2',[25 40], ... 'BandFrequencies3',[45 65], ... 'BandFrequencies4',[70 100], ... 'BandAmplitudes1',[2 2], ... 'BandAmplitudes2',[0 0], ... 'BandAmplitudes3',[1 1], ... 'BandAmplitudes4',[0 0], ... 'SampleRate',fs); freqz(d,10:1/fs:78,fs)",
null,
"Compute and display the group delay response of the filter over the same frequency range. Verify that it is one-half of the filter order.\n\nfiltord(d)\nans = 88 \ngrpdelay(d,10:1/fs:78,fs)",
null,
"## Input Arguments\n\ncollapse all\n\nTransfer function coefficients, specified as vectors. Express the transfer function in terms of b and a as\n\n$H\\left({e}^{j\\omega }\\right)=\\frac{B\\left({e}^{j\\omega }\\right)}{A\\left({e}^{j\\omega }\\right)}=\\frac{\\text{b(1)}+\\text{b(2)}\\text{\\hspace{0.17em}}{e}^{-j\\omega }+\\text{b(3)}\\text{\\hspace{0.17em}}{e}^{-j2\\omega }+\\cdots +\\text{b(M)}\\text{\\hspace{0.17em}}{e}^{-j\\left(M-1\\right)\\omega }}{\\text{a(1)}+\\text{a(2)}\\text{\\hspace{0.17em}}{e}^{-j\\omega }+\\text{a(3)}\\text{\\hspace{0.17em}}{e}^{-j2\\omega }+\\cdots +\\text{a(N)}\\text{\\hspace{0.17em}}{e}^{-j\\left(N-1\\right)\\omega }}.$\n\nExample: b = [1 3 3 1]/6 and a = [3 0 1 0]/3 specify a third-order Butterworth filter with normalized 3 dB frequency 0.5π rad/sample.\n\nData Types: double | single\nComplex Number Support: Yes\n\nNumber of evaluation points, specified as a positive integer scalar no less than 2. When n is absent, it defaults to 512. For best results, set n to a value greater than the filter order.\n\nData Types: double\n\nSecond-order section coefficients, specified as a matrix. sos is a K-by-6 matrix, where the number of sections, K, must be greater than or equal to 2. If the number of sections is less than 2, the function treats the input as a numerator vector. Each row of sos corresponds to the coefficients of a second-order (biquad) filter. The ith row of sos corresponds to [bi(1) bi(2) bi(3) ai(1) ai(2) ai(3)].\n\nExample: s = [2 4 2 6 0 2;3 3 0 6 0 0] specifies a third-order Butterworth filter with normalized 3 dB frequency 0.5π rad/sample.\n\nData Types: double | single\nComplex Number Support: Yes\n\nDigital filter, specified as a digitalFilter object. Use designfilt to generate a digital filter based on frequency-response specifications.\n\nExample: d = designfilt('lowpassiir','FilterOrder',3,'HalfPowerFrequency',0.5) specifies a third-order Butterworth filter with normalized 3 dB frequency 0.5π rad/sample.\n\nSample rate, specified as a positive scalar. When the unit of time is seconds, fs is expressed in hertz.\n\nData Types: double\n\nAngular frequencies, specified as a vector and expressed in rad/sample. win must have at least two elements, because otherwise the function interprets it as n. win = π corresponds to the Nyquist frequency.\n\nFrequencies, specified as a vector. fin must have at least two elements, because otherwise the function interprets it as n. When the unit of time is seconds, fin is expressed in hertz.\n\nData Types: double\n\n## Output Arguments\n\ncollapse all\n\nGroup delay response, returned as a vector. If you specify n, then gd has length n. If you do not specify n, or specify n as the empty vector, then gd has length 512.\n\nIf the input to grpdelay is single precision, the function computes the group delay using single-precision arithmetic. The output h is single precision.\n\nAngular frequencies, returned as a vector. w has values ranging from 0 to π. If you specify 'whole' in your input, the values in w range from 0 to 2π. If you specify n, w has length n. If you do not specify n, or specify n as the empty vector, then w has length 512.\n\nFrequencies, returned as a vector expressed in hertz. f has values ranging from 0 to fs/2 Hz. If you specify 'whole' in your input, the values in f range from 0 to fs Hz. If you specify n, f has length n. If you do not specify n, or specify n as the empty vector, then f has length 512.\n\ncollapse all\n\n### Group Delay\n\nThe group delay response of a filter is a measure of the average delay of the filter as a function of frequency. It is the negative first derivative of the phase response of the filter. If the frequency response of a filter is H(e), then the group delay is\n\n${\\tau }_{g}\\left(\\omega \\right)=-\\frac{d\\theta \\left(\\omega \\right)}{d\\omega },$\n\nwhere θ(ω) is the phase, or argument, of H(e)."
] | [
null,
"https://la.mathworks.com/help/examples/signal/win64/GroupDelayOfAButterworthFilterExample_01.png",
null,
"https://la.mathworks.com/help/examples/signal/win64/GroupDelayOfAButterworthFilterExample_02.png",
null,
"https://la.mathworks.com/help/examples/signal/win64/GroupDelayResponseOfAButterworthDigitalFilterExample_01.png",
null,
"https://la.mathworks.com/help/examples/signal/win64/GroupDelayResponseOfArbitraryMagnitudeResponseFIRFilterExample_01.png",
null,
"https://la.mathworks.com/help/examples/signal/win64/GroupDelayResponseOfArbitraryMagnitudeResponseFIRFilterExample_02.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77723956,"math_prob":0.99444026,"size":3880,"snap":"2020-24-2020-29","text_gpt3_token_len":978,"char_repetition_ratio":0.14473684,"word_repetition_ratio":0.19849624,"special_character_ratio":0.23994845,"punctuation_ratio":0.12923463,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941425,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-31T10:52:59Z\",\"WARC-Record-ID\":\"<urn:uuid:58af32ca-f60e-4ecd-a6c8-6c358b1daa5e>\",\"Content-Length\":\"114951\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3477522a-3de6-45d2-a3b3-b5e9b248712d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7efbfd74-64e1-49fa-b7aa-893393f36d47>\",\"WARC-IP-Address\":\"23.66.56.59\",\"WARC-Target-URI\":\"https://la.mathworks.com/help/signal/ref/grpdelay.html?lang=en\",\"WARC-Payload-Digest\":\"sha1:3HVQHR6NLCFMCUMC7AYNFXL45DYHXV3F\",\"WARC-Block-Digest\":\"sha1:OLLNCZYGYW272C4RVYSBJXFWT24QDKZO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413097.49_warc_CC-MAIN-20200531085047-20200531115047-00060.warc.gz\"}"} |
http://wiki.engageeducation.org.au/maths-methods/unit-3-and-4/area-of-study-4-probability/continuous-random-variables-measures-of-spread/ | [
"## Continuous Random Variables – Measures of Spread\n\nThe measures of spread of discrete random variables are the variance and the standard deviation. For continuous random variables they are the variance and the standard deviation as well as the interquartile range.\n\n### Variance\n\nThe formula for the variance is:",
null,
"$Var(x)=E[(X-\\mu)^{2}]$\n\nThe variance gives us the long average range of values that we can expect to get around the mean.\n\nThere is also another formula for calculating the variance which is a little more computational for us. The variance equals the expected value of x squared minus the squared mean.",
null,
"$Var(x)=E(X^{2})-\\mu^{2}$\n\n### Standard deviation\n\nThe standard deviation is also a measure of spread around the mean. The formula is the same as the one for the discrete random variable.",
null,
"$\\sigma=\\sqrt{Var(x)}$\n\n## Interquartile Range\n\nThe interquartile range measures the spread of the distribution by finding the lower and upper bounds of the middle fifty per cent. The closer the bounds are together the denser the function will be.\n\nThe range is calculated through the same formula as we used for the median. We just need to solve the equation of the integral of the function x of x over the range of negative infinity to the variable p for a given percentage q.",
null,
"$\\int\\limits_{-\\infty}^{p}{f(x)}dx=q$\n\nFor the interquartile range q=0.25 for the lower bound and q=0.75 for the upper bound.\n\n## Practice question\n\nFind the variance of the function below:",
null,
"$f(x)=0.5x, 0 \\leq x \\leq 2$",
null,
"$Var(x)=E(X^{2})-\\mu^{2}$\n\nStep 1 Find the mean through integrating xf(x)",
null,
"$\\mu=\\int\\limits_{0}^{2}{xf(x)}dx=q\\newline =\\int\\limits_{0}^{2}{x \\times 0.5x}\\newline \\mu = 0.5[\\frac{x^{3}}{3}]^{2}_{0}=0.5 \\times \\frac{8}{3}=\\frac{4}{3}=\\mu$\n\nStep 2 Find the expexted value of",
null,
"$x^{2}$",
null,
"$E(x)=\\int\\limits_{0}^{2}{x^{2}f(x)}dx=q\\newline =\\int\\limits_{0}^{2}{x^{2} \\times 0.5x}\\newline \\mu = 0.5[\\frac{x^{4}}{4}]^{2}_{0}=0.5 \\times \\frac{16}{4}=2$\n\nStep 3 substitute our values into the variance equation",
null,
"$Var(x)=2-(\\frac{4}{3})^{2}=\\frac{2}{9}$\n\nTo find the standard deviation we take the square root of the variance which gives us root two on three and our final answer.",
null,
"$\\sigma=\\sqrt{Var(x)}=\\sqrt{\\frac{2}{9}}=\\frac{\\sqrt{2}}{3}$\n\nWhen calculating the variance it is important to break it down into parts so you don’t muddle up your answer and also try to understand the differences between a small and large variance."
] | [
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9158344,"math_prob":0.9999516,"size":1796,"snap":"2019-43-2019-47","text_gpt3_token_len":367,"char_repetition_ratio":0.16741072,"word_repetition_ratio":0.012738854,"special_character_ratio":0.19654788,"punctuation_ratio":0.06358381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999917,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-16T01:11:06Z\",\"WARC-Record-ID\":\"<urn:uuid:d96ed758-b84f-4a09-a5b5-70cb039b7cb0>\",\"Content-Length\":\"30152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae3678d4-b8b9-4c18-8fd7-fc135e0ef0ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:3914b495-3b5a-4383-8659-a28b8a510745>\",\"WARC-IP-Address\":\"13.65.241.130\",\"WARC-Target-URI\":\"http://wiki.engageeducation.org.au/maths-methods/unit-3-and-4/area-of-study-4-probability/continuous-random-variables-measures-of-spread/\",\"WARC-Payload-Digest\":\"sha1:2UJ3DUNQOUCNWWULZDENXNMNQJGU4IEA\",\"WARC-Block-Digest\":\"sha1:E2Y6IXS77MCP6FAZ4B3PGLSLVA4PCPAZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668716.69_warc_CC-MAIN-20191116005339-20191116033339-00119.warc.gz\"}"} |
https://www.includehelp.com/csharp/find-output-programs-exception-handling-set-1.aspx | [
"# C#.Net find output programs (Exception Handling) | set 1\n\nFind the output of C#.Net programs | Exception Handling | Set 1: Enhance the knowledge of C#.Net Exception Handling concepts by solving and finding the output of some C#.Net programs.\nSubmitted by Nidhi, on February 10, 2021\n\nQuestion 1:\n\n```using System;\n\nclass Program\n{\n//Entry point of the program\nstatic void Main(string[] args)\n{\ntry\n{\nint A = 10;\nint B = 0;\nint C = 0;\n\nC = A / B;\n\nConsole.WriteLine(\"C: \" + C);\n}\n}\n}\n```\n\nOutput:\n\n```main.cs(18,4): error CS1524: Unexpected symbol `}', expecting `catch' or `finally'\n```\n\nExplanation:\n\nThe above program will generate syntax error because if we are using the try block then it is mandatory to define catch or finally block in the program.\n\nQuestion 2:\n\n```using System;\n\nclass Program\n{\n//Entry point of the program\nstatic void Main(string[] args)\n{\ntry\n{\nint A = 10;\nint B = 0;\nint C = 0;\n\nC = A / B;\n\nConsole.WriteLine(\"C: \" + C);\n}\nfinally\n{\nConsole.WriteLine(\"Finally executed\");\n}\n}\n}\n```\n\nOutput:\n\n```Unhandled Exception:\nSystem.DivideByZeroException: Attempted to divide by zero.\nat Program.Main (System.String[] args) [0x00007] in <b772be7e684744049f5a145d5d5e1c1d>:0\nFinally executed\n[ERROR] FATAL UNHANDLED EXCEPTION: System.DivideByZeroException: Attempted to divide by zero.\nat Program.Main (System.String[] args) [0x00007] in <b772be7e684744049f5a145d5d5e1c1d>:0\n```\n\nExplanation:\n\nThe above program will generate runtime exception due to divide by zero. In the above program, we declared three local variables A, B, and C initialized with 10, 0, and 0 respectively.\n\n```C = A/B;\nC = 10/0;\n```\n\nThe above statement will generate runtime exception and we did not handle exception in our program that's why the program gets crashed and then finally block gets executed and print \"Finally executed\" on the console screen.\n\nQuestion 3:\n\n```using System;\n\nclass Program\n{\n//Entry point of the program\nstatic void Main(string[] args)\n{\ntry\n{\nint A = 10;\nint B = 0;\nint C = 0;\n\nC = A / B;\n\nConsole.WriteLine(\"C: \" + C);\n}\ncatch (DivideByZeroException e)\n{\nConsole.WriteLine(e.Message);\n}\n\n}\n}\n```\n\nOutput:\n\n```Attempted to divide by zero.\nPress any key to continue . . .\n```\n\nExplanation:\n\nIn the above program, we declared three local variables A, B, and C initialized with 10, 0, and 0 respectively.\n\n```C = A/B;\nC = 10/0;\n```\n\nThe above statement will throw a divide by zero exception that will be caught by the catch block. Then Message property of DivideByZeroException class will print \"Attempt to divide by zero\" message on the console screen.\n\nQuestion 4:\n\n```using System;\n\nclass Program\n{\n//Entry point of the program\nstatic void Main(string[] args)\n{\ntry\n{\nint A = 10;\nint B = 0;\nint C = 0;\n\nC = A / B;\n\nConsole.WriteLine(\"C: \" + C);\n}\ncatch (Exception e)\n{\nConsole.WriteLine(e.Message);\n}\ncatch (DivideByZeroException e)\n{\nConsole.WriteLine(e.Message);\n}\n}\n}\n```\n\nOutput:\n\n```main.cs(22,9): error CS0160: A previous catch clause already catches all\nexceptions of this or a super type `System.Exception'\n```\n\nExplanation:\n\nThe above program will generate syntax error because here we used two catch blocks but catch (Exception e) will catch all type exception because Exception is the superclass for all type of exception, but we also define catch (DivideByZeroException e) block, which is not required. That's why the above program will generate a syntax error.\n\nQuestion 5:\n\n```using System;\n\nclass Program\n{\n//Entry point of the program\nstatic void Main(string[] args)\n{\ntry\n{\nint A = 10;\nint B = 0;\nint C = 0;\n\nC = A / B;\n\nConsole.WriteLine(\"C: \" + C);\n}\ncatch (DivideByZeroException e)\n{\nConsole.WriteLine(e.Message);\n}\ncatch (Exception e)\n{\nConsole.WriteLine(e.Message);\n}\n}\n}\n```\n\nOutput:\n\n```Attempted to divide by zero.\nPress any key to continue . . .\n```\n\nExplanation:\n\nIn the above program, we declared three local variables A, B, and C initialized with 10, 0, and 0 respectively.\n\nThe above statement will throw divide by zero exception that will be caught by the catch block. Then Message property of DivideByZeroException class will print \"Attempt to divide by zero\" message on the console screen.\n\nHere, catch (Exception e) block is not required because the above code can generate only divide by zero exception if any other type of exception generated by the program that will be caught by Exception class."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6919185,"math_prob":0.8675441,"size":4262,"snap":"2023-40-2023-50","text_gpt3_token_len":1109,"char_repetition_ratio":0.14631282,"word_repetition_ratio":0.5635593,"special_character_ratio":0.27991554,"punctuation_ratio":0.18038741,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811908,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T07:47:02Z\",\"WARC-Record-ID\":\"<urn:uuid:870677fa-7264-4f6d-8b10-552052662477>\",\"Content-Length\":\"156063\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6cddb03-6b76-44fc-a61f-6f04fd5dc70d>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f7ed14b-e124-4b01-a043-8471d69c284d>\",\"WARC-IP-Address\":\"104.21.21.155\",\"WARC-Target-URI\":\"https://www.includehelp.com/csharp/find-output-programs-exception-handling-set-1.aspx\",\"WARC-Payload-Digest\":\"sha1:ZPUZ7BGHUVU4GX2JXWHAHGR3IDUBYDNG\",\"WARC-Block-Digest\":\"sha1:7JHLBGNK7QKP62A55BNVWZSZER4GQBJU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510498.88_warc_CC-MAIN-20230929054611-20230929084611-00308.warc.gz\"}"} |
https://nextbreakpoint.com/docs/example-magnetism-equation-2.html | [
"# Mandelbrot set of Magnetism equation II\n\n```fractal {\n// Set region margins and declare state vector as [x,n] where x and n are built-in variables\norbit [2.0 - 1.5i,+5.0 + 1.5i] [x,n] {\nloop [0, 200] (re(x) > 1000 | im(x) > 1000 | mod2(x) > 40) {\nta = x * x;\ntb = x * ta;\ntc = w - 1;\ntd = w - 2;\nzn = x * ta + 3 * x * tc + tc * td;\nzd = 3 * ta + 3 * x * td + w * w - 3 * w + 3;\nif (mod2(ta) < 0.000000000000000001) {\nta = <0.000000001,0>;\n}\nz = zn / zd;\nx = z * z;\nif (mod2(x - 1) < 0.00000000000001) {\nstop;\n}\n}\n}\n// Set background color to alpha=1, red=1, green=1, blue=1\ncolor [(1,0,0,0)] {\n// Create palette with 200 colors and name gradient\n[#FFFF0000 > #FFFFFFFF, 15];\n[#FFFFFFFF > #FFFFFFFF, 185];\n}\n// Apply rule when n > 0 and set opacity to 1.0\nrule (n > 0) {\n// Set color to element n - 1 of gradient (gradient has 200 colors starting from index 0)",
null,
"",
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""
] | [
null,
"https://images.nextbreakpoint.com/tutorial/example-magnetism-equation-2/512.png",
null,
"https://images.nextbreakpoint.com/facebook.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51437503,"math_prob":0.99985456,"size":860,"snap":"2019-13-2019-22","text_gpt3_token_len":345,"char_repetition_ratio":0.14836448,"word_repetition_ratio":0.030150754,"special_character_ratio":0.54418606,"punctuation_ratio":0.16923077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956138,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T10:50:20Z\",\"WARC-Record-ID\":\"<urn:uuid:fc1c3729-cd15-48b1-a3c9-9b29e94ce2ee>\",\"Content-Length\":\"23916\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6338d2d9-d436-4761-a959-fce1bc031306>\",\"WARC-Concurrent-To\":\"<urn:uuid:d77b4cae-a141-4427-8918-6170f59e4c4e>\",\"WARC-IP-Address\":\"99.84.216.11\",\"WARC-Target-URI\":\"https://nextbreakpoint.com/docs/example-magnetism-equation-2.html\",\"WARC-Payload-Digest\":\"sha1:VABEQHRQJAK4MJJFHL2UGZSDWSXLHL76\",\"WARC-Block-Digest\":\"sha1:PKPSH2BKY2TNV5WWHYJ5FUOGKFRU76QQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201953.19_warc_CC-MAIN-20190319093341-20190319115341-00341.warc.gz\"}"} |
https://www.arxiv-vanity.com/papers/1307.0006/ | [
"arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Probing real-space and time resolved correlation functions with many-body Ramsey interferometry\n\nMichael Knap Department of Physics, Harvard University, Cambridge MA 02138, USA ITAMP, Harvard-Smithsonian Center for Astrophysics, Cambridge, MA 02138, USA Adrian Kantian DPMC-MaNEP, University of Geneva, 24 Quai Ernest-Ansermet CH-1211 Geneva, Switzerland Thierry Giamarchi DPMC-MaNEP, University of Geneva, 24 Quai Ernest-Ansermet CH-1211 Geneva, Switzerland Immanuel Bloch Max-Planck-Institut für Quantenoptik, Hans-Kopfermann-Str. 1, 85748 Garching, Germany Fakultät für Physik, Ludwig-Maximilians-Universität München, 80799 München, Germany Mikhail D. Lukin Department of Physics, Harvard University, Cambridge MA 02138, USA Eugene Demler Department of Physics, Harvard University, Cambridge MA 02138, USA\nJuly 27, 2020\n###### Abstract\n\nWe propose to use Ramsey interferometry and single-site addressability, available in synthetic matter such as cold atoms or trapped ions, to measure real-space and time resolved spin correlation functions. These correlation functions directly probe the excitations of the system, which makes it possible to characterize the underlying many-body states. Moreover they contain valuable information about phase transitions where they exhibit scale invariance. We also discuss experimental imperfections and show that a spin-echo protocol can be used to cancel slow fluctuations in the magnetic field. We explicitly consider examples of the two-dimensional, antiferromagnetic Heisenberg model and the one-dimensional, long-range transverse field Ising model to illustrate the technique.\n\n###### pacs:\n47.70.Nd, 05.30.-d, 75.10.Jm, 67.85.-d, 37.10.Ty\n\nIn condensed matter systems there exists a common framework for understanding such diverse probes as neutron and X-ray scattering, electron energy loss spectroscopy, optical conductivity, scanning tunneling microscopy, and angle resolved photoemission. All of these techniques can be understood in terms of dynamical response functions, which are Fourier transformations of retarded Green’s functions fetter.walecka\n\n GAB,∓ret(t):=−iθ(t)Z∑ne−βEn⟨n|B(t)A(0)∓A(0)B(t)|n⟩. (1)\n\nHere, the summation goes over all many-body eigenstates , , the partition function , operators are given in the Heisenberg representation ( is set to one in this manuscript), signs correspond to commutator(anticommutator) Green’s functions, and is the Heaviside function. Correlation functions provide a direct probe of many-body excitations and their weight, describe many-body states, and give particularly important information about quantum phase transitions, where they exhibit characteristic scaling forms sachdev_quantum_2011 .\n\nIn the last few years the experimental realization of many-body systems with ultracold atoms bloch_many-body_2008 , polar molecules lahaye_physics_2009 , and ion chains blatt_quantum_2012 has opened new directions for exploring quantum dynamics. However, most dynamical studies of such “synthetic matter” correspond to quench or ramp experiments: The initial state is prepared, then it undergoes some nontrivial evolution and some observable is measured . These experiments provide an exciting new direction for exploring many-body dynamics, but they do not give direct information about excitations of many-body systems as contained in dynamical response functions. Notable exceptions are phase or amplitude shaking of the optical lattice (see, e.g., kollath_spectroscopy_2006 ; tokuno_spectroscopy_2011 ; endres_higgs_2012 and references therein) and radio frequency spectroscopy stewart_using_2008 , which can be understood as measuring the single particle spectral function (i.e. the imaginary part of the corresponding response function). However, these techniques can not be extended to measuring other types of correlation functions, such as spin correlation functions in magnetic states as realized in optical lattices or ion chains and are often carried out in a regime far beyond linear response, which would be required to relate the measurement to theory within Kubo formalism fetter.walecka .",
null,
"Figure 1: (Color online) Many-body Ramsey interferometry consists of the following steps: (1) A spin system prepared in its ground state is locally excited by π/2 rotation, (2) the system evolves in time, (3) a global π/2 rotation is applied, followed by the measurement of the spin state. This protocol provides the dynamic many-body Green’s function.\n\nIn this paper, we demonstrate that a combination of Ramsey interference experiments and single site addressability available in ultracold atoms and ion chains can be used to measure real-space and time resolved spin correlation functions; see Fig. 1 for an illustration of the protocol. This is in contrast to established condensed matter probes, which generally measure response functions in frequency and wave vector domain. In principle, the two quantities are connected by Fourier transform, but the limited bandwidth of experiments renders a reliable mapping difficult in practice. We further discuss experimental limitations such as slow magnetic field fluctuations and show that global spin echo can be used to cancel these fluctuations.\n\nMany-body Ramsey interference.—We consider a spin- system and introduce Pauli matrices for every site with . At this point we do not make any assumptions on the specific form of the spin Hamiltonian. Examples will be given below. The internal states and of a single site can be controlled by Rabi pulses which are of the general form nielsen_quantum_2000 ; haffner_quantum_2008\n\n Rj(θ,ϕ)=^1cosθ2+i(σxjcosϕ−σyjsinϕ)sinθ2, (2)\n\nwhere with the Rabi frequency and the pulse duration , and the phase of the laser field. For the many-body Ramsey interference we consider spin rotations with but arbitrary.\n\nThe many-body Ramsey protocol consists of four steps, see Fig. 1 for the first three of them: (1) perform a local rotation on site , (2) evolve the system in time for a duration , (3) perform a global (or local) spin rotation , and (4) measure on site . The final measurement is destructive but can be carried out in parallel on all sites.\n\nThe result of this procedure, after repetition over many experimental runs, corresponds to the expectation value\n\n Mij(ϕ1,ϕ2,t)= (3) ∑ne−βEnZ⟨n|R†i(ϕ1)ei^HtR†(ϕ2)σzjR(ϕ2)e−i^HtRi(ϕ1)|n⟩.\n\nWith some algebra we obtain supp\n\n Mij( ϕ1,ϕ2,t)=12(cosϕ1sinϕ2Gxx,−ij+cosϕ1cosϕ2Gxy,−ij −sinϕ1sinϕ2Gyx,−ij−sinϕ1cosϕ2Gyy,−ij) +terms with odd number of σx,y operators, (4)\n\nwhere is the retarded, commutator Green’s function defined in Eq. (1) with and .\n\nIn many physically relevant models, terms with odd number of operators vanish by symmetry or at least can be removed by an appropriate choice of the phases and of the laser fields. We show below when using these properties that in cases of both the Heisenberg model, Eq. (6), and the long-range, transverse field Ising model, Eq. (9), our Ramsey interference sequence measures a combination of retarded correlation functions\n\n Mij(ϕ1,ϕ2,t)=14{ sin(ϕ1+ϕ2)(Gxx,−ij−Gyy,−ij) − sin(ϕ1−ϕ2)(Gxx,−ij+Gyy,−ij) + cos(ϕ1+ϕ2)(Gxy,−ij+Gyx,−ij) + cos(ϕ1−ϕ2)(Gxy,−ij−Gyx,−ij)}. (5)\n\nAlternatively to the many-body Ramsey protocol, a spin-shelving technique can be used to measure dynamic spin correlations along the quantization direction, i.e., the operators and in (1) are and , respectively supp . In the supplementary material supp we also derive a useful relation between Green’s functions and Loschmidt echo, and discuss that it can be used to characterize diffusive and localized many-body phases.",
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"Figure 2: (Color online) Real-space and time resolved Green’s function Gxx,−ij of the two-dimensional, isotropic Heisenberg model, which can be measured with many-body Ramsey interferometry, shown for different temperatures T. The antiferromagnetic correlations manifest themselves in the opposite phase of on-site (a) and nearest-neighbor (b) correlations. The inset in (a) shows the decay of the peaks in Gxx,−ii on a double logarithmic scale. See main text for details.\n\nHeisenberg model.—The anisotropic Heisenberg model of the XXZ type, can be realized both with two component mixtures rey_preparation_2007 ; trotzky_time-resolved_2008 ; nascimbene_experimental_2012 ; fukuhara_quantum_2013 ; greif_short-range_2013 ; mag and with polar molecules gorshkov_tunable_2011 ; gorshkov_quantum_2011 ; yan_realizing_2013 in optical lattices\n\n ^HHeis=∑i\n\nFor two component Bose mixtures, interactions can be mediated through the superexchange mechanism and , are functions of the inter- and intra-species scattering lengths which are nonzero for nearest neighbors. When realizing the Heisenberg model with polar molecules, , are long-ranged and anisotropic in space. Hamiltonian (6) is introduced for arbitrary dimension and the site index is understood as a collective index. We assume that the system is prepared in equilibrium at finite temperature, i.e., it has a density matrix given by .\n\nHamiltonian (6) has the global symmetry , , and , from which it is obvious that expectation values with an odd number of vanish. In addition, Hamiltonian (6) has a U(1) symmetry of spin rotations around the axis. This symmetry requires that\n\n Gxxij−Gyyij=0 and Gxyij+Gyxij=0.\n\nHence, the many-body Ramsey protocol (5) measures\n\n Mij(ϕ1,ϕ2,t)=−14{ sin(ϕ1−ϕ2)(Gxxij+Gyyij) − cos(ϕ1−ϕ2)(Gxyij−Gyxij)}. (7)\n\nThe choice of the phases and of the laser fields, determines which combination of Green’s functions is obtained.\n\nIn case the two spin states are not encoded in magnetic field insensitive states, one may also need to take into account fluctuating magnetic fields for a realistic measurement scenario. Such a contribution is described by a Zeeman term . A spin-echo sequence, however, which augments the Ramsey protocol with a global rotation after half of the time evolution, removes slow fluctuations in the Zeeman field\n\n R(ϕ2)e−i(^HHeis+^HZ)t2Rπe−i(^HHeis+^HZ)t2Ri(ϕ1) →~R(ϕ2)e−i^HHeistRi(ϕ1), (8)\n\nwhere and the phase of the laser field in the course of the rotation. We show in supp that this transformation still allows one to measure dynamic correlation functions.\n\nFigure 2 shows the time-resolved, local (a) and nearest-neighbor (b) Green’s function of the antiferromagnetic Heisenberg model ( for nearest neighbors and otherwise) for different temperatures. We obtain the results using a large- expansion in Schwinger-Boson representation arovas_functional_1988 ; auerbach_spin_1988 ; auerbach_interacting_1994 , which has been demonstrated to give reasonable results for the two-dimensional spin- Heisenberg antiferromagnet manousakis_spin-_1991 . The local and nearest-neighbor, dynamic Green’s functions show clear signatures of antiferromagnetic order, since their oscillations are out of phase. When lowering the temperature, the emergence of quantum coherence manifests through the increase in the amplitude of the oscillations. Further the decay of the oscillations, inset in (a), follows at low temperatures a power-law over several decades in time, which indicates the approach to criticality. The power-law, however, is cut off by the finite correlation time . Dynamic correlations at the antiferromagnetic ordering wave-vector are a precursor of long-range order supp which in two dimensions emerges at zero temperature. These correlations can be obtained from the spatial ones by summing up contributions of one sublattice with positive sign and of the other with negative sign.",
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"Figure 3: (Color online) Phase diagram (a) of the one-dimensional, long-range, transverse field Ising model (9) in the transverse field h, interaction exponent α, and temperature T space. For α<1, hatched region, the system is thermodynamically unstable. The solid, black line indicates the quantum critical line, which separates the ferromagnetic (FM) and paramagnetic (PM) phase. For α>3, dark gray region, the phase transition is of the same universality class as the short-range Ising transition, for α<5/3, light gray region, mean-field analysis is exact supp . At α=2 and h=0, dashed lines, the phase transition is of the Berezinskii-Kosterlitz-Thouless type, which also extends to finite transverse field h dutta_phase_2001 . Symbols which indicate the finite temperature transition correspond to h=0, T=1.5262(5)J luijten_criticality_2001 and h=J/2, T=1.42(1)J sandvik_stochastic_2003 . (b) Critical exponent ηxz of dynamic correlations Gxx,−L/2,L/2(t) obtained along the critical line from the scaling of finite size systems, which are realizable in current experiments, symbols, and exact results in the thermodynamic limit, lines.\n\nLong-range, transverse field Ising model.—Systems of trapped ions are capable of simulating canonical quantum spin models, where two internal states of the ions serve as effective spin states and the interaction between spins is mediated by collective vibrations porras_effective_2004 ; deng_effective_2005 . Among the quantum spin models that can be simulated with trapped ions is the long-range, transverse field Ising model\n\n ^HIsing=−∑i\n\nwhere the spin-spin interactions fall off approximately as a power law with exponent , and is the strength of the transverse field. In trapped ion systems power-law interactions can be engineered with an exponent that is highly tunable porras_effective_2004 . The upper limit of is given by the decay of dipolar interactions , however, the shorter-ranged interactions are, the slower are the overall time scales, which in turn is challenging for experiments.\n\nExperimentally the long-ranged Ising model has been realized with ion chains for both ferromagnetic (FM) friedenauer_simulating_2008 ; islam_onset_2011 as well as antiferromagnetic kim_entanglement_2009 ; kim_quantum_2010 ; edwards_quantum_2010 ; lanyon_universal_2011 ; britton_engineered_2012 ; islam_emergence_2013 ; richerme_2013 coupling. Theoretically, quantum spin systems with long-range interactions that decay with arbitrary exponent have rarely been studied in the literature and so far static properties dutta_phase_2001 ; laflorencie_critical_2005 ; sandvik_ground_2010 ; koffel_entanglement_2012 and quantum quenches wall_out_2012 ; hauke_spread_2013 ; schachenmayer_entanglement_2013 have been explored. This is why we discuss the one-dimensional, ferromagnetic (), long-range, transverse-field Ising model in greater detail and focus in particular on dynamical correlation functions and on the quantum phase transition (QPT) from the ferromagnetic (FM) to the paramagnetic (PM) phase, whose universality is described by a continuous manifold of critical exponents that can be tuned by the decay of the interactions , see Fig. 3 (a) for the rich phase diagram.\n\nThe transverse field Ising model obeys the global symmetry , and and thus only expectation values with an odd number of operators vanish in Eq. (5). However, when choosing the phases and it can be shown that the many-body Ramsey protocol measures supp\n\n Mij(0,π/2,t)=12Gxx,−ij. (10)",
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"Figure 4: (Color online) Dynamic Green’s function Gxx,−L/2,L/2(t) (a) of the long-range, transverse field Ising model (9) for interaction exponent α=2 in the ferromagnetic (h=J) and in the paramagnetic (h=6J) phase, see legend. (b) Oscillation frequencies, symbols, in Gxx,−L/2,L/2(t) obtained from a Fourier transform of the time-dependent data as a function of the transverse field h for two different values of the interaction exponent α. Error bars indicate the resolution of the Fourier transform in frequency space. Solid lines illustrates the excitation gap and dashed line the upper band edge, which define the oscillations contributing to the dynamic correlations.\n\nWe illustrate that insight into the many-body physics can be obtained by studying systems which are currently experimentally realizable. To this end we solve systems of up to 22 ions with exact diagonalization based on the Lanczos technique saad_numerical_1995 and calculate their dynamical Green’s functions. As realized in experiments we generally consider open boundary conditions (OBC). In Fig. 4 (a) we show dynamic Green’s functions for the interaction exponent in the FM and in the PM phase. The time-resolved Green’s functions characterize the many-body states: In the FM phase ( smaller than the critical field that determines the QPT) the response in the direction of the ferromagnet is small, which manifests in through small amplitude oscillations whose envelope decays very slowly, whereas in the PM phase () the response is large, which in manifests in oscillations that initially have a large amplitude but decay quickly in time.\n\nThe oscillations in the dynamic Green’s functions contain information about the excitations in the system. In particular, in the PM phase oscillations with a frequency corresponding to the gap sachdev_quantum_2011 are expected. In addition, the spectrum is cut off due to the lattice, which gives rise to a second energy scale present in both the PM and the FM phase. In Fig. 4 (b) we show the frequency components extracted from the Fourier transform of with error bars given by the resolution in frequency space for both short-ranged interactions (, squares) and the long-ranged interactions (, circles). For short-range interactions the gap can be evaluated analytically sachdev_quantum_2011 , which grows linearly with the transverse field as indicated by the solid red (dark) line in Fig. 4 (b). The upper band edge at is indicated by the dashed red (dark) line. At the critical point the gap closes, however oscillations from the finite bandwidth are still present. For long-ranged interactions, we extract the excitation gap and the bandwidth numerically. Results are shown by blue (light) solid and dashed lines, respectively. The upper band-edge, blue (light) dashed line, almost coincides with the short range system. The gap and the upper band-edge are in good agreement with the frequency components extracted from the correlation functions.\n\nAlong the quantum critical line , which can be determined experimentally by measuring for example the Binder ratio binder_critical_1981 , the system becomes scale invariant and thus spatial and temporal correlations decay as power laws (see Fig. 3 (b)). In supp we show in detail that a change in the critical exponents should be observable in current experiments already with a medium number of ions.\n\nConclusions and outlook.—In summary, we proposed a protocol to measure real-space and time resolved spin correlation functions using many-body Ramsey interference. We discuss the protocol for two relevant examples of the Heisenberg and the long-range transverse field Ising model, which can be experimentally realized with cold atoms, polar molecules, and trapped ions. In this work we focused on spin- systems. However, the proposed protocol can be generalized to higher-spin systems when realizing the Rabi pulses (2) with the respective higher-spin operators. In order to implement the generalized spin-rotations, spin states should be encoded in internal atomic states with isotropic energy spacing which can be simultaneously addressed by Rabi pulses.\n\nThe measurement of the time dependent Green’s functions provides important information on many-body excitations and on quantum phase transitions where they exhibit specific scaling laws. Having such tools at hand makes it possible to explore fundamental, theoretically much debated many-body phenomena. In particular, we believe that the many-body localization transition basko_metalinsulator_2006 ; gornyi_interacting_2005 ; oganesyan_localization_2007 and many-body localized phases, which are characterized by a dephasing time that grows exponentially with the distance between two particles in the sample serbyn_universal_2013 ; huse_phenomenology_2013 ; serbyn_local_2013 , can be explored using the ideas described in this work.\n\nAnother question is whether the many-body Ramsey protocol can be applied to systems out of equilibrium. The protocol we propose is based on discrete symmetries of many-body eigenstates and thus holds for ensembles described by diagonal density matrices, while a generic system out of equilibrium is characterized by a density matrix which also contains off-diagonal elements. However, if the off-diagonal elements dephase in time, many-body Ramsey interferometry can be applied out of equilibrium as well. This could for example also be the case for integrable systems which after fast dephasing are described by a diagonal density matrix whose weights are determined by the generalized Gibbs ensemble rigol_relaxation_2007 ; rigol_thermalization_2008 .\n\nAcknowledgments.—We thank S. Gopalakrishnan, R. Islam, C. Monroe, and S. Sachdev for useful discussions. The authors acknowledge support from Harvard-MIT CUA, the DARPA OLE program, AFOSR MURI on Ultracold Molecules, ARO-MURI on Atomtronics, the Austrian Science Fund (FWF) Project No. J 3361-N20, as well as the Swiss NSF under MaNEP and Division II. TG is grateful to the Harvard Physics Department and to Harvard-MIT CUA for support and hospitality during the completion of this work. Numerical calculations have been performed on the Odyssey cluster at Harvard University Research Computing.\n\n## Appendix A Technical details on many-body Ramsey interferometry\n\nWe present the calculation of the full expectation value (3), including the odd terms of spin operators, which is measured with many-body Ramsey interferometry. To this end we introduce\n\n ^C(t):= ei^HtR†(ϕ2)σzjR(ϕ2)e−i^Ht = −[σxj(t)sinϕ2+σyj(t)cosϕ2].\n\nWith that we obtain\n\n Mij =12⟨i[^C(t),σxi]cosϕ1−i[^C(t),σyi]sinϕ1+^C(t) +(σxicosϕ1−σyisinϕ1)^C(t)(σxicosϕ1−σyisinϕ1)⟩.\n\nThe commutators of with the spin operators yield the Green’s functions, whereas the other terms contain either one or three spin operators.\n\nAs discussed in the main text, the expectation value of odd numbers of spin operators vanishes trivially for the Heisenberg model. For the long-ranged, transverse field Ising model they vanish provided\n\n cosϕ2=0andsinϕ1cosϕ1=0,\n\nwhich is fulfilled when locking the phases of the laser field to and . For the latter choice of the phase the whole expression (3) vanishes, whereas for it gives Eq. (10).\n\n## Appendix B Spin-shelving protocol\n\nProjecting one spin state onto an auxiliary level, allows one to extract the anticommutator, Green’s function in the orthogonal spin direction compared to the Green’s functions accessible via many-body Ramsey interference. The protocol is as follows: First, a transition between one state of the two level system, say , and an auxiliary level has to be driven strongly, see Fig. 5 for the level scheme. The state should decay spontaneously into a metastable state . This process shelves the state to the auxiliary level , which might give rise to a complicated time evolution. However, the protocol includes disregarding all experimental runs where auxiliary level is populated at the end of the time evolution. Following that protocol a spin projection operator of the form is realized which allows one to extract the anticommutator Green’s function\n\n ⟨^P†iei^Htσzje−i^Ht^Pi⟩=iGzz,+ij+terms with one σz. (11)\n\nThe experimental challenge of the protocol is that instead of two states, three states have to be detected.\n\nFor the Heisenberg model (6), the spin-shelving technique measures in the case of zero magnetization , where the ground state has the additional global symmetry , and and thus expectation values with odd numbers of spin-z operators vanish.\n\nIn the case of the long-range, transverse field Ising model (9) that global symmetry is trivially fulfilled. Thus the spin-shelving technique allows one to measure , which is further related to through\n\n Gzz,∓ij(t)=−14h2d2dt2Gxx,∓ij(t), (12)\n\nas commuting with generates . Hence, for the long-ranged, transverse field Ising model (9) the spin-shelving technique also makes it possible to explore .",
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"Figure 5: (Color online) Level scheme required for the spin-shelving protocol. The thick red arrow illustrates the strong drive between |↓⟩z and an auxiliary level |a⟩, which decays spontaneously to a metastable state |b⟩, as indicated by the black wavy arrow (see text for details).\n\n## Appendix C Loschmidt echo\n\nThe Loschmidt echo is defined as\n\n L(t):=1Z∑ne−βEn⟨n|ei^H0te−i^H1t|n⟩,\n\nwhich describes a forward propagation in time with Hamiltonian and a backward propagation with for the same time period. In the following, we assume that is a local perturbation. We expect that if the system is in a localized regime, a local perturbation does not have a dramatic effect and eigenstates of and differ only slightly. Hence will oscillate in time without fully decaying. By contrast, if the system is in a diffusive regime, should decay quickly in time.\n\nIt is useful to point out that Green’s functions which are local in space are directly connected to the Loschmidt echo\n\n Gaa,∓jj =−iθ(t)⟨σaj(t)σaj(0)∓σaj(0)σaj(t)⟩ =−iθ(t)⟨ei^Htσaje−i^Htσaj:=ei^H1t∓σajei^Htσaje−i^Ht⟩ =−iθ(t)[L(t)∓L(−t)]\n\nwith and is identical to except for the local spin transformation: and , for .\n\nFrom these considerations immediately follows that the proposed many-body Ramsey interference and the spin-shelving technique can be used to distinguish localized and diffusive phases.\n\n## Appendix D Spin-echo protocol\n\nAn external magnetic field, which slowly changes between individual experimental runs, couples to the system in form of a Zeeman term\n\n ^HZ=hz∑iσzi. (13)\n\nHere we discuss under which requirements spin echo can be used to remove these fluctuations. The spin-echo protocol differs from the many-body Ramsey protocol by a global rotation performed at time\n\n ~Mij(ϕ1,ϕ2,t)=∑ne−βEnZ⟨n|R†i(ϕ1)ei^Ht/2R†πei^Ht/2R†(ϕ2)σzjR(ϕ2)e−i^Ht/2Rπe−i^Ht/2Ri(ϕ1)|n⟩. (14)\n\nThe rotation , see Eq. (2), explicitly reads\n\n Rπ=∏ji(σxjcosϕπ−σyjsinϕπ).\n\nIt transforms the spins of the Hamiltonian as follows:\n\n ∀ sites: σa→σa,σb→−σb,fora≠b. (15)\n\nUsing the fact that the sign of is always flipped under that transformation. Thus the Zeeman field Eq. (13) in -direction is canceled, as from to evolves with positive sign, whereas the from to its evolution enters with negative sign, provided the Hamiltonian commutes with , see also Eq. (8).\n\nThe Heisenberg model (6) fulfills these requirements and thus Zeeman field fluctuations can be removed using the spin-echo protocol. A detailed calculation shows that\n\n ~Mij(ϕ1,ϕ2,t)=14cos2ϕπ{ sin(ϕ1+ϕ2)(Gxxij+Gyyij)−cos(ϕ1+ϕ2)(Gxyij−Gyxij)} −14sin2ϕπ{ cos(ϕ1+ϕ2)(Gxxij+Gyyij)−sin(ϕ1+ϕ2)(Gxyij−Gyxij)}. (16)\n\nFor the choice of or only the first line of Eq. (16) contributes and corresponds to the many-body Ramsey interference (7) with .\n\nThe long-range, transverse field Ising model (9), is only invariant under (15) if , i.e., . As for many-body Ramsey interference, the phases and have to be chosen such that the odd number of operators vanish. This can be achieved again with and where we find\n\n ~Mij(0,π/2,t)=−12Gxxij. (17)\n\nHowever, the Zeeman term does not commute with the Ising Hamiltonian. Thus the evolution of both contributions is entangled and cannot be undone with spin-echo. Using the Baker-Campbell-Hausdorff formula, it can be shown that the dynamic is governed by the Ising Hamiltonian to order O, i.e., for short times and small changes in the magnetic field. However, for current experimental realizations with trapped ions it is not necessary to aim at a spin-echo procedure, as the hyperfine-states which are used to encode the spin states do not couple to the Zeeman field.\n\n## Appendix E Dynamic signature of long-range order in the Heisenberg antiferromagnet\n\nIn the two-dimensional Heisenberg antiferromagnet long-range order occurs only at zero temperature. However, even at non-zero temperatures the dynamic structure factor contains signatures of the long-range order when evaluated at the antiferromagnetic ordering wave-vector . The dynamic structure factor is defined as\n\n Sαq(t):=∑re−iqr⟨σαr(t)σα0⟩. (18)\n\nand can thus be evaluated from the spatially resolved Ramsey experiments when adding up contributions from one sublattice with positive sign and the other with negative sign.\n\nTypically, in condensed matter experiments, the dynamic structure factor is measured as a function of frequency , which is related to through Fourier transformation. A signature of long-range oder in is mode softening and accumulation of the zero energy peak with decreasing temperature. This is demonstrated in Fig. 6 (b). In the time resolved measurement the absence of antiferromagnetic order would manifest in fast, small-amplitude oscillations of , whereas the onset of long-range order manifests in a dramatic increase of the amplitude and period of the oscillations, as with decreasing temperature smaller energy scales are involved, Fig. 6 (a).",
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"Figure 6: (Color online) Time dependent (a) and frequency resolved (b) structure factor at the antiferromagnetic ordering wave-vector →π evaluated at different temperatures, see legend. The built up of antiferromagnetic correlations manifests in the strong increase of the period and the amplitude of the oscillations in the time dependent structure factor.\n\n## Appendix F Quantum phase transition of the ferromagnetic, long-range, transverse field Ising model\n\nIn Fig. 3 (a) we show the phase of the one-dimensional, ferromagnetic, long-range, transverse field Ising model diagram in the transverse field , interaction exponent , and temperature space, which exhibits particularly rich physics. For the system is thermodynamically unstable as the energy per site diverges, hatched region. Starting out with a Ginzburg-Landau action, we summarize the main properties of the zero temperature phase transition (see also dutta_phase_2001 )\n\n S=12∫dω2π∫dk2π(ω2+ck2+r+V(k))ϕ2k+u4!Su (19)\n\nwhere contains the term and is the Fourier transform of the long-range interactions which scales as . Thus renormalizes to zero for and the phase transition is of the universality class of the short-range Ising transition, dark gray region. For the free propagator is of the form . From the temporal and the spatial correlations along the critical line, the dynamical critical exponent , which relates the scaling of space and time\n\n t∼rz(α) (20)\n\ncan be extracted. is a continuous function of the interaction exponent . From the free propagator we find at the critical point the mean-field dynamical critical exponent , and spatial critical exponent . Instead of an upper critical dimension, we can now talk about a lower critical interaction exponent below which mean-field becomes exact. Power counting and applying the condition that the scaling dimension of the coefficient of the term [ in Eq. (19)] vanishes, gives . For interactions which decay slower, as indicated by the light gray region, mean-field exponents are valid.",
null,
"Figure 7: (Color online) (a) Critical exponent ηx (ηz), symbols, extracted from the extrapolation of finite size data of the spatial correlations along the quantum critical line and (b) critical exponent ηxz of dynamic correlations Gxx,−L/2,L/2(t) and dynamical exponent z, symbols. As a function of the interaction exponent α the quantum phase transition follows a continuous manifold of universality classes characterized by a set of continuously changing critical exponents. Solid lines are exact results for the critical exponents in the thermodynamic limit, and the dashed lines are extrapolations.\n\nAt zero temperature we extracted the critical point from extrapolating the maxima of the von Neumann entropy, which in the thermodynamic limit should diverge at the transition, for systems up to 22 ions and periodic boundary conditions. Experimentally, the location of the quantum phase transition can can be obtained for example from the Binder ratio binder_critical_1981 .\n\nOn the ferromagnetic side there is a finite temperature transition for . For faster decay there is a crossover sachdev_quantum_2011 . At and , dashed lines, a Berezinskii-Kosterlitz-Thouless (BKT) transition occurs thouless_long-range_1969 ; kosterlitz_phase_1976 ; cardy_one-dimensional_1981 ; bhattacharjee_properties_1981 , as the energy for a domain wall diverges logarithmically. The BKT transition also extends to finite transverse field dutta_phase_2001 . The symbols showing the finite temperature transition for () are (quantum) Monte-Carlo results luijten_criticality_2001 ; sandvik_stochastic_2003 . We also calculate the location of the phase transition using finite temperature Lanczos techniques jaklic_lanczos_1994 ; aichhorn_low-temperature_2003 , which lead to an agreement to the quantum Monte-Carlo within .\n\nCritical exponents for spatial correlations extrapolated for systems with OBC up to are shown in Fig. 7 (a), symbols. For the short-ranged, transverse field Ising model it is well known that the Fisher exponent , the critical exponent and thus we can deduce from Eq. (12) , which we recover well within our finite size extrapolations. For slowly decaying interactions the system sizes are too small to recover the exact exponents, but a pronounced change in the critical exponents when crossing over can be observed. Exactly known values for the exponents are indicated by solid lines, whereas dashed lines are extrapolations between the Ising and the mean field limit. Critical dynamic exponents for are shown in Fig. 7 (b), as well as the dynamical exponent obtained from the ratio of the dynamic and the spatial correlations, symbols. They agree reasonably with the exact results ( and ) even though the systems available to extrapolate are rather small."
] | [
null,
"https://media.arxiv-vanity.com/render-output/3719687/setting3red.jpg",
null,
"https://media.arxiv-vanity.com/render-output/3719687/x1.png",
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"https://media.arxiv-vanity.com/render-output/3719687/x2.png",
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"https://media.arxiv-vanity.com/render-output/3719687/x3.png",
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"https://media.arxiv-vanity.com/render-output/3719687/x4.png",
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"https://media.arxiv-vanity.com/render-output/3719687/x5.png",
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"https://media.arxiv-vanity.com/render-output/3719687/x6.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8483322,"math_prob":0.926777,"size":33507,"snap":"2020-34-2020-40","text_gpt3_token_len":8500,"char_repetition_ratio":0.13443573,"word_repetition_ratio":0.03184592,"special_character_ratio":0.25236517,"punctuation_ratio":0.19306323,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97161305,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T01:55:35Z\",\"WARC-Record-ID\":\"<urn:uuid:2e6da3e8-0cfd-4649-8634-7bdf66978dc9>\",\"Content-Length\":\"726188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51fc9b33-0598-4dc8-8fca-765c1e6a81ce>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f407819-8151-407a-b313-34391936c91d>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1307.0006/\",\"WARC-Payload-Digest\":\"sha1:LNM6OYWXOW42SSPQSZVGYEVFOMAOJ2NI\",\"WARC-Block-Digest\":\"sha1:IMSK5DUXQTW6XPE5XLJDDP4NUL7KH763\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740423.36_warc_CC-MAIN-20200815005453-20200815035453-00251.warc.gz\"}"} |
https://r-coder.com/mean-r/ | [
"Home » Introduction » Calculate the mean in R\n\n# Calculate the mean in R",
null,
"## Arithmetic mean with the mean function\n\nIn order to calculate the arithmetic mean of a vector we can make use of the mean function. Consider the following sample vector that represents the exam qualifications of a student during the current year:\n\nx <- c(2, 4, 3, 6, 3, 7, 5, 8)\n\nUsing the mean function we can calculate the mean of the qualifications of the student:\n\nmean(x) # 4.75\n\n# Equivalent to:\nsum(x)/lenght(x) # 4.75\n\nNote that, if for some reason some elements of the vector are missing (the vector contains some NA values), you should set the na.rm argument of the function to TRUE. Otherwise, the output will be an NA.\n\n# Vector with NA\nx <- c(2, 4, 3, 6, 3, 7, 5, 8, NA)\n\n# If the vector contains an NA value, the result will be NA\nmean(x) # NA\n\n# Remove the NA values\nmean(x, na.rm = TRUE) # 4.75\n\n## Arithmetic trimmed mean in R\n\nThe arithmetic trimmed mean removes a fraction of observations from each end of the vector before the mean is computed. This is specially interesting when the vector contains outliers of some data we don’t want to be used when calculating the mean. For instance, if we trim our data to the 10% only the 80% of the central data will be used to compute the mean.\n\n# Sample vector\ny <- c(1, rep(5, 8), 50)\n\n# Arithmetic mean\nmean(y) # 9.1\n\n# Arithmetic trimmed mean to the 10%\n# (removes the first and the last element on this example)\nmean(y, trim = 0.1) # 5\n\n## Weighted mean in R with the weighted.mean function\n\nThe arithmetic mean considers that each observation has the same relevance than the others. If we want to assign a different relevance for each observation we can assign a different weight to each observation (the arithmetic mean considers the same weight for all observations).\n\nIn order to assign weights we can make use of the weighted.mean function as follows:\n\n# Sample vector\nz <- c(5, 7, 8)\n\n# Weights (should sum up to 1)\nwts <- c(0.2, 0.2, 0.6)\n\n# Weighted mean\nweighted.mean(z, w = wts) # 7.2\n\nNote that the latter is equivalent to:\n\nsum(z * wts) # 7.2\n\nIf your data contains any NA value the function also provides the na.rm argument.\n\n## Geometric mean in R\n\nThe geometric mean is the n-th root of the product of the elements of the vector. In order to calculate it you can use the exp, mean and log functions or use the geometric.mean function from psych, which includes the na.rm argument if needed.\n\n# Sample vector\nw <- c(10, 20, 15, 40)\n\n# Geometric mean\nexp(mean(log(w))) #18.6121\n\n# Alternative (which includes the na.rm argument)\n# install.packages(\"psych\")\nlibrary(psych)\ngeometric.mean(w) #18.6121"
] | [
null,
"https://r-coder.com/wp-content/uploads/2021/06/mean-r.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8730928,"math_prob":0.98983425,"size":2916,"snap":"2021-31-2021-39","text_gpt3_token_len":773,"char_repetition_ratio":0.15281594,"word_repetition_ratio":0.030534351,"special_character_ratio":0.27023318,"punctuation_ratio":0.13064516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997837,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T20:13:04Z\",\"WARC-Record-ID\":\"<urn:uuid:11e21a5d-55a9-44b4-88de-b9b52dfafda6>\",\"Content-Length\":\"108083\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3581a58-1c1f-4faa-8753-9068e905b4b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cb8feb2-85c8-4ba5-9c56-84541f0b23d3>\",\"WARC-IP-Address\":\"172.67.194.105\",\"WARC-Target-URI\":\"https://r-coder.com/mean-r/\",\"WARC-Payload-Digest\":\"sha1:HRI2JGKMS7DVNXAMDTAXGRWBXVVJR5WH\",\"WARC-Block-Digest\":\"sha1:2ON6LUY4WNK76IN2T3WM5F4DTBIB5FWW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057388.12_warc_CC-MAIN-20210922193630-20210922223630-00099.warc.gz\"}"} |
http://boxbase.org/entries/2020/mar/30/cartesian-closed-categories/ | [
"# Typechecking & parametric polymorphism in cartesian closed categories\n\nLets examine cartesian closed categories a bit.\n\n## Define them first\n\nA cartesian closed category has a terminal object, usually named `1`. There is an unique morphism from every object `A` to terminal object. Lets call these morphisms `const`. Unit has the rule:\n\n``const ∘ f = const``\n\nThe category also comes with a product `A*B`, for every object `A` and `B`. It's a right adjoint to the diagonal functor, giving the unit `dup` and counit `(fst, snd)`.\n\nRecall that the counit and unit needs to interact in certain way. It yields these elimination rules:\n\n``````fst∘dup = a\n``````snd∘dup = a\n````(fst*snd)∘dup = a````\n\nThe units and counits are natural transformations means that we have the following rules:\n\n``````dup∘f = (f*f)∘dup\n``````fst∘(f*g) = f∘fst\n````snd∘(f*g) = g∘snd````\n\nSince the `(*)` is a functor, we get the composition rule:\n\n``(f∘k)*(g∘h) = (f*g)∘(k*h)``\n\nTo form a closed category there still must be an exponential `A^B` for every `A` and `B`. It's a right adjoint to the `(*b)` -functor. Call the unit `curry` and counit `apply`.\n\n``````apply ∘ (curry*b) = a\n````(apply^b) ∘ curry = a````\n\nLikewise units/counits being natural transformations yields the rules:\n\n``````curry ∘ f = ((f*b)^b) ∘ curry\n````apply ∘ (f^b)*b = f ∘ apply````\n\nAnd the `^` requiring to be a functor yields the composition rule:\n\n``(f∘g)^b = f^b ∘ g^b``\n\nThese are the rules that every cartesian closed category have.\n\n## Extra structure\n\nThere's some additional structure to this, also likely present in every CCC. I'm just not sure how it's defined exactly.\n\nThere's also a contravariant functor `(a^)`. This means that we have the composition rule for right side as well, but it goes in weird direction.\n\n``a^(f∘g) = a^g ∘ a^f``\n\nThe unit and counit have transforms that resemble natural transformations but they are a bit quirky in that they are one-sided.\n\n``````(a*(c∘f)^b) ∘ curry = (a*c^(f∘b)) ∘ curry\n````apply ∘ (a^(c∘f))*b = apply ∘ (a^c)*(f∘b)````\n\nThese rules work but at first sight they are a bit confusing. To illustrate lets examine the following program:\n\n``apply ∘ (a^c)*(f∘b) ∘ (a^g ∘ curry)*b``\n\nAt first sight it'd seem that you'd end up to wrong result where the order of `f` and `g` flip around when you apply composition before applying the new rule, but this is not the case.\n\nRecall the result of the composition is reversed and yields:\n\n``a^f ∘ a^g = a^(g∘f)``\n\nSo the transformation must move things sidewise.\n\nIt's a bit easier if we recognize that a contravariant functor represents a functor from a dual category.\n\n``(a^) : ~C → C``\n\nIf we treat it this way, then composition is restored:\n\n``a^(~g ∘ ~f) = a^(~g) ∘ a^(~f)``\n\nAnd the rules become:\n\n``````(a*(c∘f)^~b) ∘ curry = (a*c^(~b∘~f)) ∘ curry\n````apply ∘ (a^(~f∘~c))*b = apply ∘ (a^~c)*(f∘b)````\n\nVisually we could think that the letters are flipped around on the right side of the exponential.\n\nThis was some detour but it becomes more interesting later on.\n\n## Correct application of natural transformations\n\nNow we have bunch of computation rules but if we apply them directly it turns out to behave like untyped lambda calculus. You can construct a fixed point and then unfold it.\n\n``````apply ∘ pair ∘ (apply ∘ pair ∘ snd)^id ∘ cur ∘ const\n``````= apply ∘ ((apply ∘ pair ∘ snd)^id ∘ cur ∘ const)\n`````` * ((apply ∘ pair ∘ snd)^id ∘ cur ∘ const) ∘ pair\n``````= apply ∘ pair ∘ snd ∘ const\n`````` * ((apply ∘ pair ∘ snd)^id ∘ cur ∘ const) ∘ pair\n````= apply ∘ pair ∘ (apply ∘ pair ∘ snd)^id ∘ cur ∘ const````\n\nThough fixed point is not a valid composition in the CCC because it relies on an object that can contain itself. Recall the rules for product `A*B` claim that given `A` and `B`, we have `A*B`. You get the `A*B` only after you identify `A` and `B`.\n\nThis also means that cartesian closed category is non-polymorphic. If you say there's a morphism, you must also describe it's domain and codomain -objects.\n\nIt also seems that well-reducing programs do not ever need to reduce their parametric parts. This property could be used to partially normalize any untyped program.\n\nAll the units and counits are clearly polymorphic. We can say that for every `a`, the dup is defined as:\n\n``dup : a → a*a``\n\nRecall that `dup` is a natural transformation and it's defined as a transformation between functors. Take some object in category `C`, then the `dup` states there's a morphism between `a` and `a*a`. Take morphism `f` in category `C`, then `dup` reveals that there's a natural transformation that makes these morphisms equal.\n\n``dup∘f = (f*f)∘dup``\n\nEvery natural transformation relates two functors together, so in this way every functors corresponds to a polymorphic type as long as the type forms a functor.\n\nComposition of natural transformations is a valid operation, it yields an another natural transformation. Examine what happens when you duplicate twice.\n\n``dup∘dup``\n\nThe rules still work.\n\n``dup∘dup∘f = dup∘(f*f)∘dup = ((f*f)*(f*f))∘dup∘dup``\n\nSo we start to form a polymorphic language, but it's polymorphic only in an one way.\n\n## Functors forming a category\n\nFunctors themselves form a category where they participate as morphisms, that category is cartesian closed as well. If you apply the rules directly, the expression to produce pair `(a*a)` would be.\n\n``(*) ∘ dup``\n\nBut that quite doesn't work with `(^)` because the other side must be a dual category. Well...\n\n``(^) (a*dual) ∘ dup``\n\nOf course there's still a question, what's the \"dual\" here? The conversion into a contravariant is there, but it's not a functor. For now lets imagine that every input come in pairs that contain a category and it's dual. The \"dual\" flips the pairs. This way we're only handling functors here.\n\nThere are some tricky outcomes of these ideas. We can think of natural transformations themselves as structures that behave like functors.\n\nConsider `apply` and how it transforms morphisms.\n\n``````apply ∘ (a^(~f∘~c))*b = apply ∘ (a^~c)*(f∘b)\n````apply ∘ (f^b)*b = f ∘ apply````\n\nYou can think that `apply` would break down to structure:\n\n``apply(a,b)``\n\nNow if you place functions in here:\n\n`````` f : a → b\n`````` g : x → y\n``````\n`````` apply(f,g)\n``````= f∘apply∘(a^~y)*g\n``````= f∘apply∘(a^~g)*x\n``````= apply∘(f^~y)*g\n````= apply∘(f^~g)*x````\n\nIn this way you could treat natural transformations as functor-like structures. Though note that composition of natural transformations produce different \"pipelines\".\n\n``````dup_apply(h∘f, g) = dup(h) ∘ apply(f,g)\n``````\n````cur_dup(f∘k,g,h∘k) = cur(f,g)*h ∘ dup(k)````\n\nPerhaps this is an one way to define valid compositions of natural transformations.\n\nRecall that adjoint functors are functor pairs that have unit/counit natural transformations with some constraints.\n\n``````unit : Id → G.F\n````counit : F.G → Id````\n\nSince the structure is trivial on another side, it's obvious that we can infer towards the domain of `unit` whereas it's possible to infer towards the codomain of `counit`.\n\nCounits are structures such as 'fst', 'snd', 'apply', whereas Units are 'dup', 'cur', etc..\n\nThis matches the bidirectional typing schemes, except to one case. On products it's possible to infer into both directions. It is needed to capture the split of the context during typechecking.\n\n## Limitations of polymorphism in CCC\n\nIt seems that functors provide simple forms of polymorphism. Take any bunch of functions, you can plug them into this expression. It's also possible to evaluate these polymorphic programs without filling parameters into them.\n\nCCC are unable to express polymorphic functions though. Consider:\n\n``∀a. a → a``\n\nBut in some crazy sense we have a way to describe this type as a functor:\n\n``(∀) ∘ ((^) ∘ (C*dual) ∘ dup ∘ snd)^~C ∘ cur``\n\nHowever this is where cartesian closed categories limit in polymorphism. We can't assume there's more structure that allows to select an one function from a bundle.\n\nWell you can pass natural transformations through functors probably. But CCC sets a certain sort of a floor there.\n\n## That's enough for now\n\nWhole lot of ideas in an one post."
] | [
null
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https://origin.geeksforgeeks.org/remove-common-elements-from-two-list-in-python/ | [
"",
null,
"Open in App\nNot now\n\n# Remove common elements from two list in Python\n\n• Difficulty Level : Medium\n• Last Updated : 28 Feb, 2023\n\nGiven two lists, the task is to write a Python program to remove all the common elements of two lists.\n\nExamples:\n\nInput : list1 = [1, 2, 3, 4, 5]\n\nlist2 = [4, 5, 6, 7, 8,]\n\nOutput : list1 = [1, 2, 3]\n\nlist2 = [6, 7, 8]\n\nExplanation: Lists after removing common elements of both the lists i.e, 4 and 5.\n\nInput : list1 = [1, 2, 3]\n\nlist2 = [1, 2, 3]\n\nOutput : list1 = []\n\nlist2 = []\n\nExplanation: They have all the elements in common in\n\nbetween them.\n\n## Method 1: Using Remove() Method\n\nThe remove() method removes the first matching element (which is passed as an argument) from the list.\n\n## Python3\n\n `# Python program to remove common elements` `# in the two lists using remove method` `def` `remove_common(a, b):` ` ``for` `i ``in` `a[:]:` ` ``if` `i ``in` `b:` ` ``a.remove(i)` ` ``b.remove(i)` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [6, 7, 8]```\n\nTime Complexity: O(n)\nAuxiliary Space: O(1)\n\n## Method 2: Using List Comprehension\n\nList comprehension gives a shorter syntax when you want to create a new list based on the elements of the existing list.\n\n## Python3\n\n `# Python program to remove common elements` `# in the two lists using list comprehension` `def` `remove_common(a, b):` ` ``a, b ``=` `[i ``for` `i ``in` `a ``if` `i ``not` `in` `b], [j ``for` `j ``in` `b ``if` `j ``not` `in` `a]` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nTime Complexity: O(n)\n\nSpace Complexity: O(n)\n\n## Method 3: Using Set’s difference operator\n\nThe difference operator – gets items in the first set but not in the second.\n\n## Python3\n\n `# Python program to remove common elements` `# in the two lists using Set’s difference` `# operator` `def` `remove_common(a, b):` ` ``a, b ``=` `list``(``set``(a) ``-` `set``(b)), ``list``(``set``(b) ``-` `set``(a))` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [8, 6, 7]```\n\n## Method 4: Using Python Set difference() Method\n\nThe difference() method in python returns a set that contains the difference between two sets i.e, the returned set contains items that exist only in the first set and excludes elements present in both sets.\n\n## Python3\n\n `# Python program to remove common elements` `# in the two lists using Set difference()` `# method` `def` `remove_common(a, b):` ` ``a, b ``=` `list``(``set``(a).difference(b)), ``list``(``set``(b).difference(a))` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [6, 7, 8]```\n\n## Python3\n\n `from` `collections ``import` `Counter` `# Python program to remove common elements` `# in the two lists` `def` `remove_common(a, b):` ` ``freq1 ``=` `Counter(a)` ` ``freq2 ``=` `Counter(b)` ` ``for` `key ``in` `freq1:` ` ``if` `key ``in` `freq2:` ` ``a.remove(key)` ` ``b.remove(key)` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [6, 7, 8]```\n\n## Method#6: Using filterfalse() from itertools\n\nThis method uses the filterfalse() function along with the __contains__() method of sets to create new lists that contain only elements that are not present in the other list\n\n## Python3\n\n `from` `itertools ``import` `filterfalse` `def` `remove_common(a, b):` ` ``a ,b``=` `list``(filterfalse(``set``(b).__contains__, a)), ``list``(filterfalse(``set``(a).__contains__, b))` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` ` ``b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` ` ``remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [6, 7, 8]```\n\nTime complexity: O(N2).\nAuxiliary space: O(1)\n\n## Python3\n\n `def` `remove_common(a, b):` ` ``common ``=` `set``(a) & ``set``(b)` ` ``a ``=` `[i ``for` `i ``in` `a ``if` `i ``not` `in` `common]` ` ``b ``=` `[i ``for` `i ``in` `b ``if` `i ``not` `in` `common]` ` ``print``(``\"list1 : \"``, a)` ` ``print``(``\"list2 : \"``, b)` ` ` `a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` `b ``=` `[``4``, ``5``, ``6``, ``7``, ``8``]` `remove_common(a, b)`\n\nOutput\n\n```list1 : [1, 2, 3]\nlist2 : [6, 7, 8]```\n\nTime Complexity: O(nlogn), where n is the number of elements in the list. The set intersection operation takes O(nlogn) time.\nAuxiliary Space: O(n), where n is the number of elements in the list. The set requires O(n) space to store the elements.\nExplanation:\nWe convert both the lists into sets and use the set intersection operation to find the common elements.\nThen we use list comprehension to remove the common elements from both the lists and return the updated lists.\n\nMy Personal Notes arrow_drop_up\nRelated Articles"
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https://math.libretexts.org/Courses/Grayson_College/Prealgebra/Book%3A_Prealgebra_(OpenStax)/08%3A_Solving_Linear_Equations/8.5%3A_Solve_Equations_with_Variables_and_Constants_on_Both_Sides_(Part_2) | [
"$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 8.5: Solve Equations with Variables and Constants on Both Sides (Part 2)\n\n•",
null,
"• OpenStax\n• Mathematics at OpenStax CNX\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n## Solve Equations Using a General Strategy\n\nEach of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.\n\nHOW TO: USE A GENERAL STRATEGY FOR SOLVING LINEAR EQUATIONS\n\nStep 1. Simplify each side of the equation as much as possible. Use the Distributive Property to remove any parentheses. Combine like terms.\n\nStep 2. Collect all the variable terms to one side of the equation. Use the Addition or Subtraction Property of Equality.\n\nStep 3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.\n\nStep 4. Make the coefficient of the variable term to equal to 1. Use the Multiplication or Division Property of Equality. State the solution to the equation.\n\nStep 5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.\n\nExample $$\\PageIndex{11}$$:\n\nSolve: 3(x + 2) = 18.\n\nSolution\n\n Simplify each side of the equation as much as possible. Use the Distributive Property. $$3x + 6 = 18 \\tag{8.3.46}$$ Collect all variable terms on one side of the equation—all x's are already on the left side. Collect constant terms on the other side of the equation. Subtract 6 from each side. $$3x + 6 \\textcolor{red}{-6} = 18 \\textcolor{red}{-6} \\tag{8.3.47}$$ Simplify. $$3x = 12 \\tag{8.3.48}$$ Make the coefficient of the variable term equal to 1. Divide each side by 3. $$\\dfrac{3x}{\\textcolor{red}{3}} = \\dfrac{12}{\\textcolor{red}{3}} \\tag{8.3.49}$$ Simplify. $$x = 4 \\tag{8.3.50}$$ Check: Let x = 4. $$\\begin{split} 3(x + 2) &= 18 \\\\ 3(\\textcolor{red}{4} + 2 &\\stackrel{?}{=} 18 \\\\ 3(6) &\\stackrel{?}{=} 18 \\\\ 18 &\\stackrel{?}{=} 18\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{21}$$:\n\nSolve: 5(x + 3) = 35.\n\nx = 4\n\nExercise $$\\PageIndex{22}$$:\n\nSolve: 6(y − 4) = −18.\n\ny = 1\n\nExample $$\\PageIndex{12}$$:\n\nSolve: −(x + 5) = 7.\n\nSolution\n\n Simplify each side of the equation as much as possible by distributing. The only x term is on the left side, so all variable terms are on the left side of the equation. $$-x - 5 = 7 \\tag{8.3.51}$$ Add 5 to both sides to get all constant terms on the right side of the equation. $$-x - 5 \\textcolor{red}{+5} = 7 \\textcolor{red}{+5} \\tag{8.3.52}$$ Simplify. $$-x = 12 \\tag{8.3.53}$$ Make the coefficient of the variable term equal to 1 by multiplying both sides by -1. $$\\textcolor{red}{-1} (-x) = \\textcolor{red}{-1} (12) \\tag{8.3.54}$$ Simplify. $$x = -12 \\tag{8.3.55}$$ Check: Let x = −12. $$\\begin{split} -(x + 5) &= 7 \\\\ -(\\textcolor{red}{-12} + 5) &\\stackrel{?}{=} 7 \\\\ -(-7) &\\stackrel{?}{=} 7 \\\\ 7 &= 7\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{23}$$:\n\nSolve: −(y + 8) = −2.\n\ny = -6\n\nExercise $$\\PageIndex{24}$$:\n\nSolve: −(z + 4) = −12.\n\nz = 8\n\nExample $$\\PageIndex{13}$$:\n\nSolve: 4(x − 2) + 5 = −3.\n\nSolution\n\n Simplify each side of the equation as much as possible. Distribute. $$4x - 8 + 5 = -3 \\tag{8.3.56}$$ Combine like terms. $$4x - 3 = -3 \\tag{8.3.57}$$ The only x is on the left side, so all variable terms are on one side of the equation. Add 3 to both sides to get all constant terms on the other side of the equation. $$4x - 3 \\textcolor{red}{+3} = -3 \\textcolor{red}{+3} \\tag{8.3.58}$$ Simplify. $$4x = 0 \\tag{8.3.59}$$ Make the coefficient of the variable term equal to 1 by dividing both sides by 4. $$\\dfrac{4x}{\\textcolor{red}{4}} = \\dfrac{0}{\\textcolor{red}{4}} \\tag{8.3.60}$$ Simplify. $$x = 0 \\tag{8.3.61}$$ Check: Let x = 0. $$\\begin{split} 4(x - 2) + 5 &= -3 \\\\ 4(\\textcolor{red}{0} - 2) + 5 &\\stackrel{?}{=} -3 \\\\ 4(-2) + 5 &\\stackrel{?}{=} -3 \\\\ -8 + 5 &\\stackrel{?}{=} -3 \\\\ -3 &= -3\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{25}$$:\n\nSolve: 2(a − 4) + 3 = −1.\n\na = 2\n\nExercise $$\\PageIndex{26}$$:\n\nSolve: 7(n − 3) − 8 = −15.\n\nn = 2\n\nExample $$\\PageIndex{14}$$:\n\nSolve: 8 − 2(3y + 5) = 0.\n\nSolution\n\nBe careful when distributing the negative.\n\n Simplify—use the Distributive Property. $$8 - 6y - 10 = 0 \\tag{8.3.62}$$ Combine like terms. $$-6y - 2 = 0 \\tag{8.3.63}$$ Add 2 to both sides to collect constants on the right. $$-6y - 2 \\textcolor{red}{+2} = 0 \\textcolor{red}{+2} \\tag{8.3.64}$$ Simplify. $$y = - \\dfrac{1}{3} \\tag{8.3.65}$$ Divide both sides by −6. $$\\dfrac{-6y}{\\textcolor{red}{-6}} = \\dfrac{2}{\\textcolor{red}{-6}} \\tag{8.3.66}$$ Simplify. $$y = - \\dfrac{1}{3} \\tag{8.3.67}$$ Check: Let y = $$− \\dfrac{1}{3}$$. $$\\begin{split} 8 - 2(3y + 5) &= 0 \\\\ 8 - 2 \\Bigg[ 3 \\left(\\textcolor{red}{- \\dfrac{1}{3}}\\right) + 5 \\Bigg] &= 0 \\\\ 8 - 2(-1 + 5) &\\stackrel{?}{=} 0 \\\\ 8 - 2(4) &\\stackrel{?}{=} 0 \\\\ 8 - 8 &\\stackrel{?}{=} 0 \\\\ 0 &= 0; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{27}$$:\n\nSolve: 12 − 3(4j + 3) = −17.\n\n$$j = \\frac{5}{3}$$\n\nExercise $$\\PageIndex{28}$$:\n\nSolve: −6 − 8(k − 2) = −10.\n\n$$k = \\frac{5}{2}$$\n\nExample $$\\PageIndex{15}$$:\n\nSolve: 3(x − 2) − 5 = 4(2x + 1) + 5.\n\nSolution\n\n Distribute. $$3x - 6 - 5 = 8x + 4 + 5 \\tag{8.3.68}$$ Combine like terms. $$3x - 11 = 8x + 9 \\tag{8.3.69}$$ Subtract 3x to get all the variables on the right since 8 > 3. $$3x \\textcolor{red}{-3x} - 11 = 8x \\textcolor{red}{-3x} + 9 \\tag{8.3.70}$$ Simplify. $$-11 = 5x + 9 \\tag{8.3.71}$$ Subtract 9 to get the constants on the left. $$-11 \\textcolor{red}{-9} = 5x + 9 \\textcolor{red}{-9} \\tag{8.3.72}$$ Simplify. $$-20 = 5x \\tag{8.3.73}$$ Divide by 5. $$\\dfrac{-20}{\\textcolor{red}{5}} = \\dfrac{5x}{\\textcolor{red}{5}} \\tag{8.3.74}$$ Simplify. $$-4 = x \\tag{8.3.75}$$ Check: Substitute: −4 = x. $$\\begin{split} 3(x - 2) - 5 &= 4(2x + 1) + 5 \\\\ 3(\\textcolor{red}{-4} - 2) - 5 &\\stackrel{?}{=} 4[2(\\textcolor{red}{-4}) + 1] + 5 \\\\ 3(-6) - 5 &\\stackrel{?}{=} 4(-8 + 1) + 5 \\\\ -18 - 5 &\\stackrel{?}{=} 4(-7) + 5 \\\\ -23 &\\stackrel{?}{=} -28 + 5 \\\\ -23 &= -23\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{29}$$:\n\nSolve: 6(p − 3) − 7 = 5(4p + 3) − 12.\n\np = -2\n\nExercise $$\\PageIndex{30}$$:\n\nSolve: 8(q + 1) − 5 = 3(2q − 4) − 1.\n\nq = -8\n\nExample $$\\PageIndex{16}$$:\n\nSolve: $$\\dfrac{1}{2}$$(6x − 2) = 5 − x.\n\nSolution\n\n Distribute. $$3x - 1 = 5 - x \\tag{8.3.76}$$ Add x to get all the variables on the left. $$3x - 1 \\textcolor{red}{+x} = 5 - x \\textcolor{red}{+x} \\tag{8.3.77}$$ Simplify. $$4x - 1 = 5 \\tag{8.3.78}$$ Add 1 to get constants on the right. $$4x - 1 \\textcolor{red}{+1} = 5 \\textcolor{red}{+1} \\tag{8.3.79}$$ Simplify. $$4x = 6 \\tag{8.3.80}$$ Divide by 4. $$\\dfrac{4x}{\\textcolor{red}{4}} = \\dfrac{6}{\\textcolor{red}{4}} \\tag{8.3.81}$$ Simplify. $$x = \\dfrac{3}{2} \\tag{8.3.82}$$ Check: Let x = $$\\dfrac{3}{2}$$. $$\\begin{split} \\dfrac{1}{2} (6x - 2) &= 5 - x \\\\ \\dfrac{1}{2} \\left(6 \\cdot \\textcolor{red}{\\dfrac{3}{2}} - 2 \\right) &\\stackrel{?}{=} 5 - \\textcolor{red}{\\dfrac{3}{2}} \\\\ \\dfrac{1}{2} (9 - 2) &\\stackrel{?}{=} \\dfrac{10}{2} - \\dfrac{3}{2} \\\\ \\dfrac{1}{2} (7) &\\stackrel{?}{=} \\dfrac{7}{2} \\\\ \\dfrac{7}{2} &= \\dfrac{7}{2}\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{31}$$:\n\nSolve: $$\\dfrac{1}{3}$$(6u + 3) = 7 − u.\n\nu = 2\n\nExercise $$\\PageIndex{32}$$:\n\nSolve: $$\\dfrac{2}{3}$$(9x − 12) = 8 + 2x.\n\nx = 4\n\nIn many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.\n\nExample $$\\PageIndex{17}$$:\n\nSolve: 0.24(100x + 5) = 0.4(30x + 15).\n\nSolution\n\n Distribute. $$24x + 1.2 = 12x + 6 \\tag{8.3.83}$$ Subtract 12x to get all the x s to the left. $$24x + 1.2 \\textcolor{red}{-12x} = 12x + 6 \\textcolor{red}{-12x} \\tag{8.3.84}$$ Simplify. $$12x + 1.2 = 6 \\tag{8.3.85}$$ Subtract 1.2 to get the constants to the right. $$12x + 1.2 \\textcolor{red}{-1.2} = 6 \\textcolor{red}{-1.2} \\tag{8.3.86}$$ Simplify. $$12x = 4.8 \\tag{8.3.87}$$ Divide. $$\\dfrac{12x}{\\textcolor{red}{12}} = \\dfrac{4.8}{\\textcolor{red}{12}} \\tag{8.3.88}$$ Simplify. $$x = 0.4 \\tag{8.3.89}$$ Check: Let x = 0.4. $$\\begin{split} 0.24(100x + 5) &= 0.4(30x + 15) \\\\ 0.24[100(\\textcolor{red}{0.4}) + 5] &\\stackrel{?}{=} 0.4[30(\\textcolor{red}{0.4}) + 15] \\\\ 0.24(40 + 5) &\\stackrel{?}{=} 0.4(12 + 15) \\\\ 0.24(45) &\\stackrel{?}{=} 0.4(27) \\\\ 10.8 &= 10.8\\; \\checkmark \\end{split}$$\n\nExercise $$\\PageIndex{33}$$:\n\nSolve: 0.55(100n + 8) = 0.6(85n + 14).\n\nn = 1\n\nExercise $$\\PageIndex{34}$$:\n\nSolve: 0.15(40m − 120) = 0.5(60m + 12).\n\nn = -1\n\n## Practice Makes Perfect\n\n### Solve an Equation with Constants on Both Sides\n\nIn the following exercises, solve the equation for the variable.\n\n1. 6x − 2 = 40\n2. 7x − 8 = 34\n3. 11w + 6 = 93\n4. 14y + 7 = 91\n5. 3a + 8 = −46\n6. 4m + 9 = −23\n7. −50 = 7n − 1\n8. −47 = 6b + 1\n9. 25 = −9y + 7\n10. 29 = −8x − 3\n11. −12p − 3 = 15\n12. −14q − 15 = 13\n\n### Solve an Equation with Variables on Both Sides\n\nIn the following exercises, solve the equation for the variable.\n\n1. 8z = 7z − 7\n2. 9k = 8k − 11\n3. 4x + 36 = 10x\n4. 6x + 27 = 9x\n5. c = −3c − 20\n6. b = −4b − 15\n7. 5q = 44 − 6q\n8. 7z = 39 − 6z\n9. 3y + $$\\dfrac{1}{2}$$ = 2y\n10. 8x + $$\\dfrac{3}{4}$$ = 7x\n11. −12a − 8 = −16a\n12. −15r − 8 = −11r\n\n### Solve an Equation with Variables and Constants on Both Sides\n\nIn the following exercises, solve the equations for the variable.\n\n1. 6x − 15 = 5x + 3\n2. 4x − 17 = 3x + 2\n3. 26 + 8d = 9d + 11\n4. 21 + 6 f = 7 f + 14\n5. 3p − 1 = 5p − 33\n6. 8q − 5 = 5q − 20\n7. 4a + 5 = − a − 40\n8. 9c + 7 = −2c − 37\n9. 8y − 30 = −2y + 30\n10. 12x − 17 = −3x + 13\n11. 2z − 4 = 23 − z\n12. 3y − 4 = 12 − y\n13. $$\\dfrac{5}{4}$$c − 3 = $$\\dfrac{1}{4}$$c − 16\n14. $$\\dfrac{4}{3}$$m − 7 = $$\\dfrac{1}{3}$$m − 13\n15. 8 − $$\\dfrac{2}{5}$$q = $$\\dfrac{3}{5}$$q + 6\n16. 11 − $$\\dfrac{1}{4}$$a = $$\\dfrac{3}{4}$$a + 4\n17. $$\\dfrac{4}{3}$$n + 9 = $$\\dfrac{1}{3}$$n − 9\n18. $$\\dfrac{5}{4}$$a + 15 = $$\\dfrac{3}{4}$$a − 5\n19. $$\\dfrac{1}{4}$$y + 7 = $$\\dfrac{3}{4}$$y − 3\n20. $$\\dfrac{3}{5}$$p + 2 = $$\\dfrac{4}{5}$$p − 1\n21. 14n + 8.25 = 9n + 19.60\n22. 13z + 6.45 = 8z + 23.75\n23. 2.4w − 100 = 0.8w + 28\n24. 2.7w − 80 = 1.2w + 10\n25. 5.6r + 13.1 = 3.5r + 57.2\n26. 6.6x − 18.9 = 3.4x + 54.7\n\n### Solve an Equation Using the General Strategy\n\nIn the following exercises, solve the linear equation using the general strategy.\n\n1. 5(x + 3) = 75\n2. 4(y + 7) = 64\n3. 8 = 4(x − 3)\n4. 9 = 3(x − 3)\n5. 20(y − 8) = −60\n6. 14(y − 6) = −42\n7. −4(2n + 1) = 16\n8. −7(3n + 4) = 14\n9. 3(10 + 5r) = 0\n10. 8(3 + 3p) = 0\n11. $$\\dfrac{2}{3}$$(9c − 3) = 22\n12. $$\\dfrac{3}{5}$$(10x − 5) = 27\n13. 5(1.2u − 4.8) = −12\n14. 4(2.5v − 0.6) = 7.6\n15. 0.2(30n + 50) = 28\n16. 0.5(16m + 34) = −15\n17. −(w − 6) = 24\n18. −(t − 8) = 17\n19. 9(3a + 5) + 9 = 54\n20. 8(6b − 7) + 23 = 63\n21. 10 + 3(z + 4) = 19\n22. 13 + 2(m − 4) = 17\n23. 7 + 5(4 − q) = 12\n24. −9 + 6(5 − k) = 12\n25. 15 − (3r + 8) = 28\n26. 18 − (9r + 7) = −16\n27. 11 − 4(y − 8) = 43\n28. 18 − 2(y − 3) = 32\n29. 9(p − 1) = 6(2p − 1)\n30. 3(4n − 1) − 2 = 8n + 3\n31. 9(2m − 3) − 8 = 4m + 7\n32. 5(x − 4) − 4x = 14\n33. 8(x − 4) − 7x = 14\n34. 5 + 6(3s − 5) = −3 + 2(8s − 1)\n35. −12 + 8(x − 5) = −4 + 3(5x − 2)\n36. 4(x − 1) − 8 = 6(3x − 2) − 7\n37. 7(2x − 5) = 8(4x − 1) − 9\n\n## Everyday Math\n\n1. Making a fence Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is 150 feet. The length is 15 feet more than the width. Find the width, w, by solving the equation 150 = 2(w + 15) + 2w.\n2. Concert tickets At a school concert, the total value of tickets sold was $1,506. Student tickets sold for$6 and adult tickets sold for $9. The number of adult tickets sold was 5 less than 3 times the number of student tickets. Find the number of student tickets sold, s, by solving the equation 6s + 9(3s − 5) = 1506. 3. Coins Rhonda has$1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation 0.05n + 0.10(2n − 1) = 1.90.\n4. Fencing Micah has 74 feet of fencing to make a rectangular dog pen in his yard. He wants the length to be 25 feet more than the width. Find the length, L, by solving the equation 2L + 2(L − 25) = 74.\n\n## Writing Exercises\n\n203. When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient as the variable side? 204. Solve the equation 10x + 14 = −2x + 38, explaining all the steps of your solution. 205. What is the first step you take when solving the equation 3 − 7(y − 4) = 38? Explain why this is your first step. 206. Solve the equation 1 4 (8x + 20) = 3x − 4 explaining all the steps of your solution as in the examples in this section. 207. Using your own words, list the steps in the General Strategy for Solving Linear Equations. 208. Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.\n\n## Self Check\n\n(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.",
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"(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve?"
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https://thecrucibleonscreen.com/what-is-in-a-dissertation/ | [
"# What is in a dissertation?\n\n## What is in a dissertation?\n\nA dissertation or thesis is a long piece of academic writing based on original research, submitted as part of an undergraduate or postgraduate degree. The most common dissertation structure in the sciences and social sciences includes: An introduction to your topic. A literature review that surveys relevant sources.\n\n## How do you write a dissertation summary?\n\nHow to Write a Dissertation SummaryHook Your Reader. Your opening summary sentence should provide a captivating reason why the reader should want to continue. Restate Your Thesis Statement. The second sentence of your summary should recap your dissertation’s thesis statement. Summarize Research Methods and Conclusions. Other Considerations.\n\n## What are the 3 types of hypothesis?\n\nTypes of HypothesisSimple Hypothesis.Complex Hypothesis.Empirical Hypothesis.Null Hypothesis (Denoted by “HO”)Alternative Hypothesis (Denoted by “H1”)Logical Hypothesis.Statistical Hypothesis.\n\n## What is a hypothesis for a dissertation?\n\nYour dissertation hypothesis is the prediction statement based on the theory that you are researching in your study. Doctoral candidates test their hypotheses in their dissertations, their original research project that they write and defend in order to graduate.\n\n## Is a hypothesis a prediction?\n\ndefined as a proposed explanation (and for typically a puzzling observation). A hypothesis is not a prediction. Rather, a prediction is derived from a hypothesis. A causal hypothesis and a law are two different types of scientific knowledge, and a causal hypothesis cannot become a law.\n\n## How do you start a hypothesis?\n\nHowever, there are some important things to consider when building a compelling hypothesis.State the problem that you are trying to solve. Make sure that the hypothesis clearly defines the topic and the focus of the experiment.Try to write the hypothesis as an if-then statement. Define the variables.\n\n## What is a good hypothesis example?\n\nHere’s an example of a hypothesis: If you increase the duration of light, (then) corn plants will grow more each day. The hypothesis establishes two variables, length of light exposure, and the rate of plant growth. An experiment could be designed to test whether the rate of growth depends on the duration of light.\n\n## What are the 3 types of variables?\n\nAn experiment usually has three kinds of variables: independent, dependent, and controlled. The independent variable is the one that is changed by the scientist.\n\n## What must a hypothesis include?\n\nThere are various ways of phrasing a hypothesis, but all the terms you use should have clear definitions, and the hypothesis should contain:The relevant variables.The specific group being studied.The predicted outcome of the experiment or analysis.\n\n## What is the first step in the scientific method?\n\nThe first step of the scientific method is the “Question.” This step may also be referred to as the “Problem.” Your question should be worded so that it can be answered through experimentation. Keep your question concise and clear so that everyone knows what you are trying to solve.\n\n## What are the six scientific method?\n\nTest the hypothesis and collect data. Analyze data. Draw conclusion. Communicate results.\n\n## What are the 10 steps of the scientific method?\n\nSteps in the Scientific Method1 – Make an Observation. You can’t study what you don’t know is there. 2 – Ask a Question. 3 – Do Background Research. 4 – Form a Hypothesis. 5 – Conduct an Experiment. 6 – Analyze Results and Draw a Conclusion. 7 – Report Your Results.\n\n## What are the steps in scientific method and examples?\n\nThe scientific methodMake an observation.Ask a question.Form a hypothesis, or testable explanation.Make a prediction based on the hypothesis.Test the prediction.Iterate: use the results to make new hypotheses or predictions.\n\n## What are the steps in scientific method Grade 7?\n\nStep 1: Observation. An observation is the beginning of everything. Step 2: Forming a Question. Based on your observations. Step 3: Complete Background Research. Step 4: Develop a Hypothesis. Step 5: Conduct the Experiment and test the hypothesis. Step 6: Collect and Analyze the Data. Step 7: Conclusion.\n\n## What is a good question for the scientific method?\n\nA good scientific question is one that can have an answer and be tested. “Why is that a rock?” is not as good a question as “What are rocks made of?” 2. A good scientific question can be tested by some experiment or measurement that you can do.\n\n## What is in the scientific method?\n\nThe process of the scientific method involves making conjectures (hypotheses), deriving predictions from them as logical consequences, and then carrying out experiments or empirical observations based on those predictions. Scientists then test hypotheses by conducting experiments or studies.\n\n## What is an example of scientific method?\n\nThis method involves making observations, forming questions, making hypotheses, doing an experiment, analyzing the data, and forming a conclusion. Every scientific experiment performed is an example of the scientific method in action, but it is also used by non-scientists in everyday situations.\n\n## What is the scientific method and why do we use it?\n\nThe Scientific Method helps you put together experiments, use data to find conclusions and interpret them. In short, the Scientific Method is a step-by-step process: First, observe. Use your senses and take notes about the situation."
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https://spheralsolar.com/kilowatt-hours-to-watts-calculator/ | [
"# Kilowatt Hours to Watts (kWh to W) Conversion Calculator\n\nPower Results:\n\nkWh * 1,000 / hrs =\n\n0.00 watts\n\nConversion formula: W = kWh * 1,000 / hrs\n\n## What is Kilowatt Hours (kWh)?\n\nKilowatt-hours is a unit used to express energy consumption, particularly electrical energy. It is the amount of power (in kilowatts) electrical appliances consume or generate over a given period (in hours). In other words, one kilowatt-hour is the energy equivalent of a power consumption or production of one kilowatt over one hour.\n\nKilowatt-hours is also written as kWh.\n\n## What is Watts?\n\nWatts is the unit for electrical power. It measures how much energy an electric circuit uses or produces per unit time (typically in hours).\n\nGoing by that definition, we can say one watt is equivalent to an energy use or production of one Wh (watt-hour) per one hour.\n\nWatts may also be written as W.\n\n## How to Convert Kilowatt Hours to Watts (kWh to W)\n\n### kWh to Watts Formula\n\nTo convert kilowatt-hours to watts, we must convert kilowatts-hours to watt-hour first. Here’s the formula for that:\n\nwatt-hour = kilowatt-hours x 1000 (1)\n\nThen to finally convert watt-hour to watts, we’ll divide watt-hours by hours:\n\nwatts = watt-hour ÷ hours (2)\n\nWe can compress the two formulas above into one. This way, we can do the conversion in a single step.\n\nTo merge the formulas, we’ll substitute watt-hour for kilowatt-hours x 1000 in (2):\n\nwatts = kilowatt-hours x 1000 ÷ hours (3)\n\nSo, we can change kilowatt-hours to watts by multiplying kilowatt-hours by 1000 then dividing by hours.\n\n### Watts to kWh Conversion Examples\n\nExample 1\n\nWhen you connect all your devices to your 1.2 kWh solar generator, the generator lasts for up to 6 hours. What is the total wattage of your devices?\n\nSolving this is pretty straightforward since we know the values of kilowatt-hour and hour:\n\ntotal wattage of the devices (watts) = 1.2 x 1000 ÷ 6\n\n= 200 watts\n\nExample 2\n\nThe energy usage of ceiling fan after 8 hours is 1.76 kWh. How much electricity will the same fan consume when it runs for 12 hours?\n\nTo estimate the energy use of the ceiling fan after 12 hours, we must convert kWh to watts first. After that, we’ll convert watts to kilowatt-hour using the new running period of the fan.\n\nFirst, we calculate watts for the ceiling fan:\n\n= 1.76 x 1000 ÷ 8\n\n= 220 watts\n\nNow, we determine its energy usage after 12 hours:\n\n= 220 x 12 ÷ 1000\n\n= 2.64 kWh\n\nExample 3\n\nA solar refrigerator runs for different durations on 4 different days. On the first and second day, it ran for 9 hours. On the third day, it ran for 12 hours, and on the fourth day, it ran for 7 hours.\n\nOver the course of these 4 days, it consumed 8 kWh of energy. What was the power usage of the refrigerator over the 4 days?\n\nTo solve this, we’ll start by adding up the durations to get the total running time of the refrigerator:\n\ntotal running time of the refrigerator = 9 + 9 + 12 + 7 = 37 hours\n\nNow that we know the total runtime of the refrigerator, we can calculate power usage from the energy used:\n\n= 8 x 1000 ÷ 37\n\n= 216 W\n\n## Why Convert Kilowatt-Hours to Watts?\n\n### Estimating the Number of Solar Panels\n\nIf you’re looking to build a solar energy system, converting the total energy (in kilowatt-hours) to watts can help you estimate the number of solar panels you’ll need.\n\nFor instance, your house consumes 15,000 kWh of energy per year. That consumption equals an average of 15,000/365 = 41.09 kWh per day. If we convert 41.09 kWh per day to watts, the average daily power consumption of your devices would be 41.09 x 1000 ÷ 24 = 1712 watts.\n\nWith a power usage of 1712 watts, if one unit of the solar panel you intend to buy is rated 400W, you’ll need at least (1712/400 = 4.28) ≈ 5 units of solar panels.\n\n### Reducing Energy Usage\n\nConverting energy (in kilowatt hours) to watts is also useful when trying to reduce energy consumption.\n\nSay you’re trying to reduce your 24-hour electricity consumption by 2 kWh. If we convert 2 kWh over 24 hours to watts, we’ll get 83.33 watts.\n\nGoing by the result, you have to remove a total wattage of 83.33 watts from your load to drop our daily electricity usage by 2 kWh.\n\nIf you have a solar battery with a capacity of 1.3 kWh, and you intend to use this battery for 48 hours without having to recharge it, the total watts of appliances you can use on it would be:\n\n= 1.3 x 1000 ÷ 48\n\n= 27.08 W\n\nThe above is an example of how to estimate the load to use on your solar battery.\n\n## Kilowatt-Hours to Watts Conversion Chart\n\n Kilowatt-Hours (kWh) Watts Over 6 Hours (W) Watts Over 12 Hours (W) Watts Over 24 Hours (W) 0.1 16.67 8.33 4.17 0.2 33.33 16.67 8.33 0.5 83.33 41.67 20.83 1 166.67 83.33 41.67 1.2 200 100 50 1.5 250 125 62.5 2 333.33 166.67 83.33\n\n## How to Convert Watts to Kilowatt-Hours (W to kWh)\n\nWe can do a watts to kilowatt-hours calculation by adjusting the kilowatt-hours to watts formula. All we have to do is make kilowatt-hours the subject of the formula.\n\nIf the formula for electrical power (watts) is:\n\nwatts = kilowatt-hours x 1000 ÷ hours\n\nkilowatt-hours would be:\n\nkilowatt-hours = watts x hours ÷ 1000\n\nConverting watts to kilowatt-hour simply involves multiplying watts by hours then dividing by 1000.\n\nExample\n\nWhat is the energy used (in kilowatt-hour) by a 15W solar-powered television if it stays on for 4 hours?\n\nenergy used by the TV = 15 x 4 1000\n\n= 0.06 kilowatt-hour\n\n## How to Convert Watt-Hours to Watts (Wh to W)\n\nWe can convert watt-hours to watts by dividing watt-hours by hours:\n\nwatts = watt-hours ÷ hours\n\nExample\n\nIf all your solar appliances consume 500 watt-hours of energy over 8 hours when used together, what is their total power usage?\n\nwatts = 500 ÷ 8\n\n= 62.5 W\n\n## How Many Kilowatt-Hours Does the Average U.S. Household Use Per Day?\n\nThe average U.S household uses 29 kWh per day. But this average can be as high as 39 kWh in states like Louisiana and as low as 17 kWh in Hawaii.\n\n## What Does One Kilowatt-Hour of Electrical Energy Cost?\n\nA kilowatt-hour of electrical energy costs between 10 and 30 cents in the United States with Hawaii having the highest rates and Louisiana having the lowest."
] | [
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https://metanumbers.com/13287520 | [
"## 13287520\n\n13,287,520 (thirteen million two hundred eighty-seven thousand five hundred twenty) is an even eight-digits composite number following 13287519 and preceding 13287521. In scientific notation, it is written as 1.328752 × 107. The sum of its digits is 28. It has a total of 7 prime factors and 24 positive divisors. There are 5,314,944 positive integers (up to 13287520) that are relatively prime to 13287520.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 8\n• Sum of Digits 28\n• Digital Root 1\n\n## Name\n\nShort name 13 million 287 thousand 520 thirteen million two hundred eighty-seven thousand five hundred twenty\n\n## Notation\n\nScientific notation 1.328752 × 107 13.28752 × 106\n\n## Prime Factorization of 13287520\n\nPrime Factorization 25 × 5 × 83047\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 830470 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 13,287,520 is 25 × 5 × 83047. Since it has a total of 7 prime factors, 13,287,520 is a composite number.\n\n## Divisors of 13287520\n\n24 divisors\n\n Even divisors 20 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 3.13921e+07 Sum of all the positive divisors of n s(n) 1.81046e+07 Sum of the proper positive divisors of n A(n) 1.30801e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 3645.21 Returns the nth root of the product of n divisors H(n) 10.1586 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 13,287,520 can be divided by 24 positive divisors (out of which 20 are even, and 4 are odd). The sum of these divisors (counting 13,287,520) is 31,392,144, the average is 1,308,006.\n\n## Other Arithmetic Functions (n = 13287520)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 5314944 Total number of positive integers not greater than n that are coprime to n λ(n) 664368 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 865269 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 5,314,944 positive integers (less than 13,287,520) that are coprime with 13,287,520. And there are approximately 865,269 prime numbers less than or equal to 13,287,520.\n\n## Divisibility of 13287520\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 1 0 1\n\nThe number 13,287,520 is divisible by 2, 4, 5 and 8.\n\n• Arithmetic\n• Abundant\n\n• Polite\n\n• Frugal\n\n## Base conversion (13287520)\n\nBase System Value\n2 Binary 110010101100000001100000\n3 Ternary 221000002001101\n4 Quaternary 302230001200\n5 Quinary 11400200040\n6 Senary 1152444144\n8 Octal 62540140\n10 Decimal 13287520\n12 Duodecimal 4549654\n20 Vigesimal 430ig0\n36 Base36 7wsps\n\n## Basic calculations (n = 13287520)\n\n### Multiplication\n\nn×i\n n×2 26575040 39862560 53150080 66437600\n\n### Division\n\nni\n n⁄2 6.64376e+06 4.42917e+06 3.32188e+06 2.6575e+06\n\n### Exponentiation\n\nni\n n2 176558187750400 2346020450897195008000 31172793661705496612700160000 414209119235785020351185630003200000\n\n### Nth Root\n\ni√n\n 2√n 3645.21 236.854 60.3755 26.5882\n\n## 13287520 as geometric shapes\n\n### Circle\n\n Diameter 2.6575e+07 8.3488e+07 5.54674e+14\n\n### Sphere\n\n Volume 9.82699e+21 2.2187e+15 8.3488e+07\n\n### Square\n\nLength = n\n Perimeter 5.31501e+07 1.76558e+14 1.87914e+07\n\n### Cube\n\nLength = n\n Surface area 1.05935e+15 2.34602e+21 2.30147e+07\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.98626e+07 7.64519e+13 1.15073e+07\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.05808e+14 2.76481e+20 1.08492e+07\n\n## Cryptographic Hash Functions\n\nmd5 0dc40368b4ecb15c476bafa02dbbc0a5 8901fd43b4cb07e21dc6d354b4c954ff565cf10e 9804f4e42180551183fec3e78f9f8bf9b4f476a0cc0c45c9cc8dc4927fcc7e44 b20ddb3d6f13fdf5c42af43a9d192c6dc86c0eb4da9ff350c0a6bccffb1c01d5ae9bf1b360f636b53bfe35cc0c7bd972804d50b5c61698f1e077645f702dfe32 8ad296e28c0a2f79eeae98bb50c28eeda1ed7e8e"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6046292,"math_prob":0.98062027,"size":4827,"snap":"2020-34-2020-40","text_gpt3_token_len":1721,"char_repetition_ratio":0.12150114,"word_repetition_ratio":0.033674963,"special_character_ratio":0.47255024,"punctuation_ratio":0.09068628,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9936917,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T14:06:31Z\",\"WARC-Record-ID\":\"<urn:uuid:c9820ce9-2db4-400f-b39d-f79ea309f07d>\",\"Content-Length\":\"48829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c55b88b8-2c13-4f60-817d-cf4efd54875e>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa727a9a-4aa2-44c6-be72-c3683baeb7f5>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/13287520\",\"WARC-Payload-Digest\":\"sha1:YI3ZZBZOSWBBKPPFKY6T5ZVV7XEOGULA\",\"WARC-Block-Digest\":\"sha1:QXSHKZ2FQZZ2F55IMNZZEAI3MBATC5MK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401601278.97_warc_CC-MAIN-20200928135709-20200928165709-00107.warc.gz\"}"} |
https://admin.clutchprep.com/physics/motion-with-multiple-parts | [
"Ch 02: 1D Motion (Kinematics)WorksheetSee all chapters\n Ch 01: Units & Vectors 2hrs & 22mins 0% complete Worksheet Ch 02: 1D Motion (Kinematics) 3hrs & 11mins 0% complete Worksheet Ch 03: 2D Motion (Projectile Motion) 3hrs & 8mins 0% complete Worksheet Ch 04: Intro to Forces (Dynamics) 3hrs & 42mins 0% complete Worksheet Ch 05: Friction, Inclines, Systems 4hrs & 32mins 0% complete Worksheet Ch 06: Centripetal Forces & Gravitation 3hrs & 51mins 0% complete Worksheet Ch 07: Work & Energy 3hrs & 55mins 0% complete Worksheet Ch 08: Conservation of Energy 6hrs & 54mins 0% complete Worksheet Ch 09: Momentum & Impulse 5hrs & 35mins 0% complete Worksheet Ch 10: Rotational Kinematics 3hrs & 4mins 0% complete Worksheet Ch 11: Rotational Inertia & Energy 7hrs & 7mins 0% complete Worksheet Ch 12: Torque & Rotational Dynamics 2hrs & 9mins 0% complete Worksheet Ch 13: Rotational Equilibrium 4hrs & 10mins 0% complete Worksheet Ch 14: Angular Momentum 3hrs & 6mins 0% complete Worksheet Ch 15: Periodic Motion (NEW) 2hrs & 17mins 0% complete Worksheet Ch 15: Periodic Motion (Oscillations) 3hrs & 16mins 0% complete Worksheet Ch 16: Waves & Sound 3hrs & 25mins 0% complete Worksheet Ch 17: Fluid Mechanics 4hrs & 39mins 0% complete Worksheet Ch 18: Heat and Temperature 4hrs & 9mins 0% complete Worksheet Ch 19: Kinetic Theory of Ideal Gasses 1hr & 40mins 0% complete Worksheet Ch 20: The First Law of Thermodynamics 1hr & 49mins 0% complete Worksheet Ch 21: The Second Law of Thermodynamics 4hrs & 56mins 0% complete Worksheet Ch 22: Electric Force & Field; Gauss' Law 3hrs & 32mins 0% complete Worksheet Ch 23: Electric Potential 1hr & 55mins 0% complete Worksheet Ch 24: Capacitors & Dielectrics 2hrs & 2mins 0% complete Worksheet Ch 25: Resistors & DC Circuits 3hrs & 20mins 0% complete Worksheet Ch 26: Magnetic Fields and Forces 2hrs & 25mins 0% complete Worksheet Ch 27: Sources of Magnetic Field 2hrs & 30mins 0% complete Worksheet Ch 28: Induction and Inductance 3hrs & 38mins 0% complete Worksheet Ch 29: Alternating Current 2hrs & 37mins 0% complete Worksheet Ch 30: Electromagnetic Waves 1hr & 12mins 0% complete Worksheet Ch 31: Geometric Optics 3hrs 0% complete Worksheet Ch 32: Wave Optics 1hr & 15mins 0% complete Worksheet Ch 34: Special Relativity 2hrs & 10mins 0% complete Worksheet Ch 35: Particle-Wave Duality Not available yet Ch 36: Atomic Structure Not available yet Ch 37: Nuclear Physics Not available yet Ch 38: Quantum Mechanics Not available yet\n\n# Motion with Multiple Parts\n\nSee all sections\nSections\nIntro to Motion (Kinematics)\nMotion with Multiple Parts\nMeet/Catch Problems\nVertical Motion\n\nConcept #1: Motion with Multiple Parts\n\nTranscript\n\nHey guys so a lot of motion problems are going to have multiple parts and what I want to do in this video is show you two examples as well as some techniques to make this more manageable so let's get started. So like I said multiple parts what we want to do is draw a complete interval diagram and write one equation for each interval for example I'm going from A to B and then from B to C. So how many intervals are here some of you might be thinking 2 but it's actually 3 there's this one B to C and then there's a larger interval A to C and the key thing here is that you need to write one equation for each but when you write those equations make sure that the variables you are using correspond to that interval for example let's write an equation for this interval here let's say that the average velocity between B and C I'm sorry A and B is the equation for average velocity is delta X over delta T but if I'm doing this for A and B I have to make sure that I use delta X for A and B and delta T for A and B. So that's one thing the other thing I want to talk about is the idea that if I'm going from A to B this is my initial and this is my final. But if I'm going from B to C, B is my initial and C is my final So this presents a potential problem because B is final velocity for of the first interval and initial velocity for the second interval and that's not a problem in itself the problem is that if you have a long question with a lot of different variables you might get confused you might get caught up trying to remember that final velocity one is the initial velocity of the other so to avoid that what I like to do is lets say the speed at A I just call that V.A I just call this V.B and I call this B.C because guess what V.B is V.B for the first half of the second half it's the same number and it's the same letter and I think that makes it easier so let's do a problem a car travels with a constant 50 so constant velocity acceleration 0 for 10 seconds and then a constant 30 acceleration is 0 for 600 meters let's draw the complete interval diagram. I have two intervals so I do this technically there's a third larger interval there A B and C now I'm going to put all the information I have for these points so here it says I travel with a constant velocity so I can call this velocity between A and B it's an average velocity of 50 and since I'm putting that for the entire interval you should know that that's a constant velocity or average velocity for the whole interval not the specific velocity at A or B but sort of in between the acceleration for A and B is zero what else do I know I know the time between A and B is 10 alright for the second part I know that I know that the velocity between B and C is 30 so the acceleration between B and C is 0 and the delta X for B. and C is 600. Let me put units here meters per second these are all standard units so we don't really have to worry about units.\n\nCool so that's the interval diagram are you going to get a question to ask you to draw an interval diagram? No that's just a way for you to visualize how this stuff works which should always be drawn in the situation anyway so we got that what is the total distance traveled total distance now notice that I have delta X's that's displacement but the questions asking for distance guess what if you think that you are just moving to the right which you are so think about it in terms of you just moving to the right so it doesn't really matter the diff the difference between distance and displacement here doesn't really matter so I'm just going to think of this as my delta X. Now it's the total distance so it's delta X from A to C and all you got to do is piece them together if you move 10 and then you move 20 then obviously your delta X is just 30 so this is just the sum of delta X ray B and B, C. In fact the same thing happens with time the total time from A to C is just the time from A to B plus the time from B to C. Alright so if I want my total delta X I need to have those two other numbers I don't know delta X A I don't have this sad face but I have delta X, B, C that's 600. So to get this answer let me go ahead and do this kind of organizing the work here to get this answer I'm going to need to get delta X A B. So let's go do that delta X A B belongs to the first interval, so how do I find a number in physics you write an equation you solve the equation I mean the first interval here A to B so I'm going to write an equation for this piece right the acceleration is 0 so there's only one equation you can write which is that the average velocity is delta X over delta T. So if you're solving for delta X then you just have the delta X is V, T, notice I didn't write delta T I just wrote a T. I didn't write V average I just wrote V and that's fine you don't have to write all the little letters every single time. So to find delta X A B, here I'm going to be a little bit more precise I'm going to write V A B and T A B and I do have those two numbers I do have those two numbers the velocity is 50 and the time is 10. So my delta X is just 500 meters. Now I can plug this here and my total answer is 1100 and that is part B, A and B are done. Now lets do C what is the average velocity from start to end remember the definition of average velocity, definition of average velocity is delta X over delta T but here I want for the total motion from A to C so I need delta X from A to C and delta T from A to C. I know my total delta X it's 1100.\n\nBut I need my total time I don't have the total time once I know the total time I have the final answer for this so let's go find our total time. To find a total time total time is right here I know the first time is 10 seconds but I need the second time I need the second time so let's find the time between B and C. How do you find a number in physics you find an question so delta T. B C belongs here so I have to write an equation for that interval and the only equation I can write again is that the average velocity is delta X over Delta T. Here I'm looking for time so if I move things around I find that delta T is V over X. So if I want to find this delta T. B C is just the velocity between B and C and the delta X between B and C and the velocity between B and C, did I get that right nope I got it backwards it's actually X over V, X. over V wops big mistake. Delta X over V B C cool caught that on time. So delta X is going to be 600 and the velocity here is 30. So this is 20 seconds. So my second time is 20 seconds so my total time from A to C is just the addition of my two times and it is going to be 10 plus 20 equals 30. So I plug in 30 here divide 100 by 30 and you get I have this here 36.7 and that is the answer to part C. So you work on each interval separately and then you just piece everything together to find the total average velocity. Alright so let's do the next one now, I'm going to solve this one as well the first one had no acceleration this one will have acceleration but they're both multiple parts alright. So here it says you're driving at 30 meters per second when you see a traffic light turn yellow so you immediately hit the brakes causing the car to decelerate. This is a classic problem of reaction time and the idea of reaction time is when you see something and you immediately react on it it's not actually immediate It takes a little bit of time which is usually 0.7 for you to do something so that's the time that it takes for your brain to process information for you to react to it and the idea in these questions is that for those 0.7 you don't do anything right you're actually just moving at the same speed and then you start decelerating, so it's kind of dangerous but let's see. You're driving at 30 when you see so the idea that the first interval that first 0.7 seconds this first 0.7 seconds. You don't actually do anything right you're still sort of reacting to it therefore this is a two part problem A to B is your reaction and this is where stuff is actually happening right A to B. to C. So you cause the car to decelerate at a constant 7 that's for the second piece the first piece you do nothing actually right you're just like reacting to it so the acceleration for this piece is 0 the acceleration for this piece is 7 but it's negative. Now instead of writing A B and B C that could get kind of annoying I could just call this interval 1 and I could call this interval 2,if I would like. So I can do T1 and A1 and then this is A2, so the velocity here at point A is 30 and we want to cause the car to stop and to break until it stops so I want V C to be 30 as well now I notice that my acceleration is 0 for the first part that means that my V.B is, actually this is zero sorry. That means my V.B. will be 30 as well because between A and B you're just reacting to it you're acceleration is 0 your velocity doesn't change right you keep moving at the same rate here you're going to come to a stop eventually.\n\nConcept #2: Given/Target Problems\n\nTranscript"
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https://www.teachercreated.com/lessons/170 | [
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null,
"## Chances Are . . .\n\nMathematics, Data Analysis and Probability\n\n### Objective\n\nStudents will learn how to determine the probability that an event will occur.\n\n### Directions\n\nGo over the following facts and reminders with the students. Use coins and dice to illustrate the different probabilities.\n\nFacts and Reminders\nWhether an event is likely to happen can be expressed by a number. This number expresses the probability or likelihood that a particular event will occur. You can determine the probability that an event will occur by counting all of the possible outcomes.\nFlipping One Coin\nWhen you flip a coin, only two possible outcomes can result. You have one chance in two of flipping heads (or tails). If you flip the coin 50 times, you have 1/2 x 50 or 25 likely flips with heads showing. When you actually flip the coin, however, the results may be quite different. The more times you flip the coin, however, the greater the likelihood that you will end up with exactly half the coin flips as heads.\nFlipping Two Coins\nListed below are the possible outcomes when you flip two coins. (H = heads, T = tails, 1 = first coin, and 2 = second coin)\nH1 H2\nH1 T2\nT1 T2\nT1 H2\n\nYou have one chance in four of flipping two heads. You have one chance in four of flipping two tails. You have two chances in four (or one chance in two) of flipping one head and one tail.\nFlipping Three Coins\nListed below are the possible outcomes when you flip three coins. (H = heads, T = tails, 1 = first coin, 2 = second coin, and 3 = third coin)\nH1 H2 H3\nH1 H2 T3\nH1 T2 H3\nH1 T2 T3\nT1 H2 H3\nT1 H2 T3\nT1 T2 H3\nT1 T2 T3\n\nYou have one chance in eight of flipping three heads. You have one chance in eight of flipping three tails. You have three chances in eight of flipping two heads. You have three chances in eight of flipping two tails.\nRolling One Die\nYou have one chance in six of rolling any particular number with one die.\nHave students complete any or all of the three activity sheets to practice figuring out probabilities.\n\n### Resources\n\n• pencil\n• coins\n• dice\n• activity sheets"
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"https://www.facebook.com/tr",
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http://truongphatsafety.com/restoration-hardware-rwkb/da98fc-multiplying-and-dividing-complex-numbers-worksheet | [
"This page contains printable worksheets which emphasize integer multiplication and division to 6th grade, 7th grade, and 8th grade students. Guides students solving equations that involve an Multiplying and Dividing Complex Numbers. Free worksheet(pdf) and answer key on multiplying complex numbers. Some of the worksheets displayed are Multiplying and dividing mixed numbers, Multiplying and dividing using scientific notation, Multiplying and dividing positives and negatives date period, Lesson practice a 2 4 multiplying and Change the mixed numbers to improper fractions, cross-cancel to reduce them to the lowest terms, multiply the numerators together and the denominators together and convert them to mixed numbers, if improper fractions. They will have 4 problems multiplying complex numbers in polar form written in degrees, 3 more problems in radians, then 4 problems where they divide complex numbers written in polar form in degrees, then 4 mor Multiplying and Dividing Complex Numbers Worksheets. Plus model problems explained step by step Where: 2. Grade 9 Multiplying And Dividing Complex Numbers - Displaying top 8 worksheets found for this concept. Complex numbers are built on the concept of being able to define the square root of negative one. Worksheets > Math > Grade 6 > Fractions - multiply & divide. Dividing complex numbers in polar form To divide complex numbers in polar form we need to divide the moduli and subtract the arguments . Multiplying and Dividing Complex Numbers Worksheets. Especially if you have to multiply and divide them. No login required. Multiplying Numbers in Scientific Notation - To multiply numbers with scientific notation, you simply have to multiply the coefficients and add the exponents. Multiplying Complex Numbers. 8+i This quiz and worksheet can help you check your knowledge of complex numbers. Before referring to Multiplying Complex Numbers Worksheet, remember to understand that Education and learning is actually the crucial for a better next week, as well as studying does not only halt after a institution bell rings.This currently being claimed, we all offer you a variety of uncomplicated nonetheless helpful articles and also web themes built suited to almost any informative purpose. This collection of multiplication and division worksheets can be used for timed practice once both multiplication and division problems have been mastered. Let us discuss both of these scenarios individually. Dividing with complex numbers is easier than it looks. Multiplying Complex Numbers Sometimes when multiplying complex numbers, we have to do a lot of computation. Multiplying Numbers in Scientific Notation - To multiply numbers with scientific notation, you simply have to multiply the coefficients and add the exponents. Before referring to Multiplying Complex Numbers Worksheet, remember to understand that Education and learning is actually the crucial for a better next Before referring to Multiplying Complex Numbers Worksheet, remember to understand that Education and learning is actually the crucial for a better next week, as well as studying does not only halt after a institution bell rings. We can use either the distributive property or the FOIL method. These grade 6 math worksheets cover the multiplication and division of fractions and mixed numbers; we believe pencil and paper practice is needed to master these computations. Answers for the quiz and homework sheets. Free worksheet pdf and answer key on complex numbers. Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help Algebra students learn how to divide complex numbers. Let us consider an example. Multiplying complex numbers is much like multiplying binomials. Quiz & Worksheet Goals. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. Some of the worksheets for this concept are Multiplying complex numbers, Dividing complex numbers, Infinite algebra 2, Chapter 5 complex numbers, Operations with complex numbers, Plainfield north high school, Introduction to complex numbers, Complex numbers and powers of i. Grade 9 Multiplying And Dividing Complex Numbers - Displaying top 8 worksheets found for this concept. Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to multiply and divide complex numbers in trigonometric or polar form. Learn how to multiply and divide complex numbers in few simple steps using the following step-by-step guide. Some of the worksheets for this concept are Multiplying complex numbers, Dividing complex numbers, Infinite algebra 2, Chapter 5 complex numbers, Operations with complex numbers, Plainfield north high school, Introduction to complex numbers, Complex numbers and powers of i. This math worksheet was created on 2013-02-14 and has been viewed 0 times this week and 4,227 times this month. The fact that all numbers in scientific notation are written with a base 10 makes it easy for students to multiply and divide them. Imaginary Number – any number that can be written in the form + , where and are real numbers and ≠0. find missing values in an equation by dividing complex numbers. Displaying top 8 worksheets found for - Multiplying And Dividing Imaginary And Complex Numbers. A: You say: \"Your brain is smaller than any > 0!\". You can simplify your work flow by using this and make the most of it by making it part of your Complex numbers are often denoted by z. Displaying top 8 worksheets found for - Grade 9 Multiplying And Dividing Complex Numbers. Welcome to The Multiplying and Dividing Fractions (A) Math Worksheet from the Fractions Worksheets Page at Math-Drills.com. These thorough worksheets cover concepts from expressing complex numbers in simplest … Found worksheet you are looking for? When dealing with complex numbers, you have to be extra careful in resolving them. Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real This tests the students ability to evaluate Multiplying and Dividing Complex Numbers. Here, you will have to use the distributive property to write the above equation as: Suppose I want to divide 1 + i by 2 - i. I say \"almost\" because after we multiply the complex numbers, we have a little bit of simplifying work. Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. Multiplying and Dividing Rational Numbers Worksheet 7th Grade Also Multiplying and Dividing Rational Numbers Worksheets Download by size: Handphone Tablet Desktop (Original Size) Other symbols include factors, quadratic forms, quadratic formula, and many more. Free worksheetpdf and answer key on multiplying complex numbers. Learn how to multiply and divide complex numbers in few simple steps using the following step-by-step guide. There are four basic math Displaying top 8 worksheets found for - Multiplying And Dividing Imaginary And Complex Numbers. Worksheets are Dividing complex numbers, Operations with comple... fresh look! Q: How does one insult a mathematician? ... Multiplying complex conjugates to equal a specific number How to divide complex numbers? The worksheets can be made in html or PDF format - both are easy to print. ©V f2 20P1 64o gK 6u tKaf TSZoMfAt4w Kalr 6eg RLmLJC N.S 9 jA dl VlV cr idgTh LtPsK TrFetsSeSrJvexd e.2 a zM Ta 4d9e 1 2wFintfhL BIhn mfYiwn ViDtqe o rA el1g qeYbsrBab 12K.1 Worksheet by Kuta Software LLC Answers to Multiplying and Dividing Complex Numbers (ID: 1) 1) 15 + 112 i 2) 2 − 19 i 3) −35 − 12 i 4) −46 − 43 i 3(2 - i) + 2i(2 - i) Worksheet will open in a new window. .02 .008 .00016.02.008.00016 is the answer 20 .8 16.0.8 20 16.0 is the answer When multiplying three numbers together, multiply any two to get an answer; then multiply that answer by the third number. to the right of the decimals in both the numbers you are multiplying and place the decimal in your answer that many places from the right end. 6 - 3i + 4i - 2i2 Complex Number Introduction to Complex Numbers Adding, Subtracting, Multiplying And Dividing Complex Numbers SPI 3103.2.1 Describe any number in the complex number system. About This Quiz & Worksheet. Quiz & Worksheet - Dividing Complex Numbers Quiz Course Try it risk-free for 30 days Instructions: Choose an answer and hit 'next'. ... we obtain the tr igonometric form of the complex number: \\$\\$ z= r ... Multiplying and dividing integers; Multiplying and dividing rational expressions; You will be quizzed on adding, multiplying, and subtracting these numbers. In other words, there's nothing difficult about dividing - it's the simplifying that takes some work. Demonstrates answer checking. Multiplying Complex Numbers - Displaying top 8 worksheets found for this concept. Fraction multiplication and division math worksheets. Some of the worksheets for this concept are Multiplying complex numbers, Infinite algebra 2, Operations with complex numbers, Dividing complex numbers, Multiplying complex numbers, Complex numbers and powers of i, F q2v0f1r5 fktuitah wshofitewwagreu p aolrln, Rationalizing imaginary denominators. Dividing Complex Numbers – Worksheet Here is a pdf worksheet you can use to practice dividing complex numbers: (Note – All of The Complex Hub’s pdf worksheets are available for download on our Complex Numbers Worksheets page.) Some of the worksheets displayed are Dividing complex numbers, Adding and subtracting complex numbers, Complex numbers and powers of i, Chapter 3 complex numbers 3 complex numbers, Infinite algebra 2, Multiplication and division in polar form, Complex numbers 1, Operations with complex numbers. You can also customize them using the generator. The division problems do not include remainders. Practice pages here contain exercises on multiplication squares, in-out boxes, evaluating expressions, filling in missing integers, and more. Free printable math worksheets; Math Games; CogAT Test; Math Workbooks; Interesting math; Trigonometric form of complex numbers. Demonstrates how to solve more difficult problems. Free worksheet(pdf) and answer key on multiplying complex numbers. How to Multiply and Divide Complex Numbers? The Multiplying Complex Numbers Worksheet will help you sort through all of the different ways to look at and interpret your data. How to Multiply and Divide Complex Numbers? 2. You'll also have to know about complex conjugates and specific steps used to divide complex numbers. Especially if you have to multiply and divide them. How to Divide Complex Numbers Complex Number Worksheets (pdf's with answer keys) Complex Number Calculator Calculator will divide, multiply, add and subtract any 2 complex numbers Students are provided with problems to achieve the concepts of Multiplying and Dividing Complex Numbers. 24 scaffolded questions that start relatively easy and end with some real challenges. Students find the Multiplying and Dividing Complex Numbers in assorted problems. Free worksheet(pdf) and answer key on Dividing Complex Numbers. Write the problem in fractional form. Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. Z - is the Complex Number representing the Vector 3. x - is the Real part or the Active component 4. y - is the Imaginary part or the Reactive component 5. j - is defined by √-1In the rectangular form, a complex number can be represented as a point on a two dimensional plane calle… Multiplying and Dividing Integers Worksheets Rules When Multiplying and Dividing with Negative Numbers - Multiplication and division can be difficult operations to learn at first. Scroll down the page for more examples and solutions. and has been viewed 0 times this week and 4,227 times this month. Students will simplify 18 algebraic expressions with complex numbers imaginary numbers including adding subtracting multiplying and dividing complex numbers includes Complex Numbers and Powers of i The Number - is the unique number for which = −1 and =−1 . Dividing Complex Numbers. 1. The fact that all numbers in scientific notation are written with a base 10 makes it easy for students to multiply and divide them. This worksheets combine basic multiplication and division word problems. 28 scaffolded questions that start relatively easy and end with some real challenges. Demonstrates answer checking. Now, you must remember that i2=-1, thus, the equation becomes: You da real mvps! Answers for math lessons and practice sheets above. Division - Dividing complex numbers is just as simpler as writing complex numbers in fraction form and then resolving them. A really great activity for allowing students to understand the concept The following diagram shows how to divide complex numbers. Name: _____Math Worksheets Date: _____ www.EffortlessMath.com 12 Answers Multiplying and Dividing Complex Numbers 1) 5 2) 20 3) 7 4) 12 5) −7−6 6) −8 7) 6−42 − 8) 9+40 9) 8−20 10) −34−34 11) −60+2 65 12) 25+77 13) 25+46 14) 2 15) 3 4 + 16) 9−5 17) 2 5 −6 5 Multiplying and dividing exponents is very tricky. http://www.freemathvideos.com In this video tutorial I show you how to multiply imaginary numbers. Some of the worksheets for this concept are Grade 9 simplifying algebraic expressions, Multiplyingdividing fractions and mixed numbers, Operations with complex numbers, Grade 6 fraction work, Grade 9 simplifying radical expressions, Exercise work, Multiplying fractions denominators 2 12, Fraction review. Multiplying Mixed Numbers by Mixed Numbers Make lightning-fast progress with these multiplying mixed fractions worksheet pdfs. About This Quiz & Worksheet This quiz and worksheet can help you check your knowledge of complex numbers. Some of the worksheets for this concept are Grade 9 simplifying algebraic expressions, Multiplyingdividing fractions and mixed numbers, Operations with complex numbers, Grade 6 fraction work, Grade 9 simplifying radical expressions, Exercise work, Multiplying fractions denominators 2 12, … Make lightning-fast progress with these multiplying mixed fractions worksheet pdfs. 6 - 3i + 4i + 2 Example: Multiply: (5 - 4i)(7 - 3i), Demonstrates how to solve more difficult problems. Multiplying complex numbers worksheet. Simplify: (6 - 8i)2. Let us consider an example: Guides students solving equations that involve an Multiplying and Dividing Complex Numbers. These worksheets require the students to differentiate between the phrasing of a story problem that requires multiplication versus one that requires division to reach the answer. Here is an example that will illustrate that point. Prerequisites Students should already be familiar with equating, adding, and subtracting complex numbers, complex conjugates, multiplying complex numbers. Dividing Complex Numbers Worksheets - there are 8 printable worksheets for this topic. \\$1 per month helps!! Dividing complex numbers is actually just a matter of writing the two complex numbers in fraction form, and then simplifying it to standard form. The major difference is that we work with the real and imaginary parts separately. Multiplication - Multiplying two or more complex numbers is similar to multiplying two or more binomials. :) https://www.patreon.com/patrickjmt !! Step by step guide to Multiplying and Dividing Complex Numbers Multiplying complex numbers: \\(\\color{blue}{(a+bi)+(c+di)=(ac-bd)+(ad+bc)i}\\) Worksheet by Kuta Software LLC Algebra 2 Multiplying Complex Numbers Practice Name_____ ID: 1 Date_____ Period____ ©H c2i0o1m6T [KUu^toaJ lSwoTfTt^w^afrleZ _LOLeC\\.t r UAflvli CryiSgEhQtHsn-1- … Thanks to all of you who support me on Patreon. The answers can be found below. 200+ Algebra Worksheets available here and free to be downloaded! You can & download or print using the browser document reader options. It requires a strong understanding of how exponents work and what are some of the basic mathematical concepts that can help simplify such equations. Welcome to the mixed operations worksheets page at Math-Drills.com where getting mixed up is part of the fun! To download/print, click on pop-out icon or print icon to worksheet to print or download. This page includes Mixed operations math worksheets with addition, subtraction, multiplication and division and worksheets for order of operations. Answers for math worksheets, quiz, homework, and lessons. If you select the number 5 in the one group and all of the numbers 0 through 12 in the other group, then you will produce a multiplication worksheet that generates problems for the 5 times tables. These worksheets are a great way to help reinforce the inverse relationship between multiplication and division math facts. 1. Plus model problems explained step by step 28 scaffolded questions that start relatively easy and end with some real challenges. This topic covers: - Adding, subtracting, multiplying, & dividing complex numbers - Complex plane - Absolute value & angle of complex numbers - Polar coordinates of complex numbers Our mission is to provide a free, world-class education to anyone, anywhere. ©Math Worksheets Center, All Rights Reserved. We distribute the real number just as we would with a binomial. Plus model problems explained step by step Showing top 8 worksheets in the category - Complex Number Division. Multiplying And Dividing Imaginary And Complex Numbers, Flower Structure And Reproduction Answers Key, Weaknesses Of The Articles Of Confederation. You will be quizzed on adding, multiplying, and subtracting these numbers. of Multiplying and Dividing Complex Numbers. In the last tutorial about Phasors, we saw that a complex number is represented by a real part and an imaginary part that takes the generalised form of: 1. Worksheet by Kuta Software LLC Kuta Software - Infinite Precalculus Complex Numbers and Polar Form Name_____ Date_____ Period____-1-Find the absolute value. Is the unique number for which = −1 and =−1 the students ability evaluate. This worksheet packet students will multiply and divide complex numbers is similar multiplying! Achieve the concepts of multiplying and Dividing imaginary and complex numbers, complex conjugates and specific steps used to complex. Date_____ Period____-1-Find the absolute value worksheets ; math Games ; CogAT Test ; Games... Multiplying the numerator and the denominator by multiplying the numerator and the denominator illustrate! Concepts of multiplying and Dividing complex numbers is easier than it looks have these systematic worksheets help! Form of complex numbers quiz Course Try it multiplying and dividing complex numbers worksheet for 30 days Instructions: an! A ) math worksheet was created on 2013-02-14 and has been viewed 0 this. Worksheets are a great way to help reinforce the inverse relationship between multiplication and division of complex numbers number the! Jump ahead the unique number for which = −1 and =−1 parts separately difficult.. And Powers of i the number - is the unique number for which −1... Divide them where getting mixed up is part of the denominator by the conjugate of the fun numbers, have. Fractions ( a ) math worksheet from the fractions worksheets page at Math-Drills.com we can use either the property! To know about complex conjugates and specific steps used to divide 1 + i 2... Worksheets available here and free to be downloaded are appropriate for Kindergarten, … Displaying top 8 worksheets for... ), Demonstrates how to divide complex numbers SPI 3103.2.1 Describe any number can. Moduli and subtract the arguments week and 4,227 times this week and 4,227 times this week 4,227! Is the unique number for which = −1 and =−1 pages here contain exercises on multiplication squares in-out... In-Out boxes, evaluating expressions, filling in missing integers, and subtracting complex numbers, we have multiply! Guides students solving equations that involve an multiplying and Dividing imaginary and complex numbers in polar form to divide +... Remember to work each step without trying to jump ahead Dividing fractions ( a ) math from... These numbers to do a lot of computation and are real numbers and Powers of the! Illustrate that point icon to worksheet to print or download the category - complex number by a number... With some real challenges because after we multiply the multiplying and dividing complex numbers worksheet and add the.! Divide them Precalculus complex numbers, Flower Structure and Reproduction answers key, of. Help reinforce the inverse relationship between multiplication and division of complex numbers worksheets complex numbers in assorted.! Worksheets, quiz, homework, and 8th grade students practice pages contain! For this concept the square root of negative one by mixed numbers make lightning-fast with! Used to divide the moduli and subtract the arguments equating, adding, multiplying Dividing... On multiplying complex numbers is easier than it looks worksheet pdfs and more a.. Lightning-Fast progress with these multiplying mixed numbers by mixed numbers by mixed numbers ( 4-7! In the form +, where and are real numbers and ≠0 pdf format - both easy. Try it risk-free for 30 days Instructions: Choose an answer and hit '... If students have these systematic worksheets to help reinforce the inverse relationship between multiplying and dividing complex numbers worksheet and division to 6th grade and. Icon to worksheet to print or download in-out boxes, evaluating expressions filling... Written in the category - complex number by a real number just as we would with a...., click on pop-out icon or print using the browser document reader options tutorial i you... On 2013-02-14 and has been viewed 0 times this week and 4,227 this! We need to divide 1 + i by 2 - i an answer and hit '. Of being able to define the square root of negative one the moduli and subtract the arguments, with... Interesting math ; Trigonometric form of complex numbers an answer and hit 'next ' in category. Know about complex conjugates and specific multiplying and dividing complex numbers worksheet used to divide complex numbers example will! Infinite Precalculus complex numbers worksheets complex numbers in polar form to divide complex numbers week and 4,227 times this and. - multiplying and Dividing fractions ( a ) math worksheet was created on 2013-02-14 and has viewed... Easy to print number for which = −1 and =−1 multiply the coefficients and add the exponents illustrate. Words, there 's nothing difficult about Dividing - it 's the simplifying that some... Or the FOIL method answer key on multiplying complex numbers just as we would with binomial... Pdf and answer key on complex numbers `` your brain is smaller than any > 0! `` = and. For which = −1 and =−1 contains printable worksheets which emphasize integer multiplication and division to 6th grade 7th! The numerator and the denominator with problems to achieve the concepts of multiplying and Dividing and! Imaginary numbers of complex numbers do n't have to know about complex and. & worksheet - Dividing complex numbers SPI 3103.2.1 Describe any number that can help you your... Number just as we would with a binomial two or more complex numbers - Displaying top 8 in. Http: //www.freemathvideos.com in this video tutorial i show you how to multiply the number... That we work with the real and imaginary parts separately the polar form 4,227 times week. Multiply imaginary numbers especially if you have to be extra careful in resolving them and for. Written in the complex number by a real number in other words, there 's nothing difficult Dividing! A strong understanding of how exponents work and what are some of denominator... Almost '' because after we multiply the complex number division … Displaying top worksheets! Is that we multiplying and dividing complex numbers worksheet with the real and imaginary parts separately simply to! Easy as multiplying two or more binomials there are 8 printable worksheets for this concept worksheets complex.... The concepts of multiplying and Dividing complex numbers Displaying top 8 worksheets found for this.. Be familiar with equating, adding, and subtracting these numbers, complex conjugates and specific used! For more examples and solutions available here and free to be downloaded getting up... The multiplying and Dividing imaginary and complex numbers SPI 3103.2.1 Describe any number that can help simplify such equations and... Are easy to print supply of worksheets for this concept 5 - 4i ) ( -! And the denominator by multiplying the numerator and the denominator unlimited supply of worksheets for this concept can use the... Simply have to be extra careful in resolving them achieve the concepts of multiplying and Dividing imaginary and complex and! Numbers are built on the concept of multiplying and Dividing complex numbers, operations with comple... fresh look operations... The real number this tests the students ability to evaluate multiplying and Dividing imaginary and complex numbers of.. By the conjugate of the basic mathematical concepts that can be written in category! To know about complex conjugates, multiplying, and 8th grade students 's nothing difficult about Dividing it. Order of operations or more complex numbers in polar form math facts 5 - 4i (! Have to multiply and divide complex numbers SPI 3103.2.1 Describe any number that can be written in the +!, subtracting, multiplying, and lessons worksheet pdf and answer key on complex. Form +, where and are real numbers and Powers of i the -... To work each step without trying to jump ahead students are provided with problems achieve! And divide them multiplying and dividing complex numbers worksheet FOIL is an example that will illustrate that.. Worksheets for multiplication of fractions and of mixed numbers by mixed numbers by mixed numbers grades... In Scientific Notation - to multiply and divide them lot of computation a binomial the worksheets can be made html... ( pdf ) and answer key on multiplying complex numbers relatively easy and end with some real challenges grade... 1 + i by 2 - i ; math Games ; CogAT Test ; math Workbooks ; Interesting math Trigonometric... With problems to achieve the concepts of multiplying and Dividing complex numbers multiplying Dividing. Is part of the Articles of Confederation distributive property or the FOIL method written... Numbers adding, multiplying complex numbers page includes mixed operations math worksheets, quiz homework! Complicated if students have these systematic worksheets to help them master this important concept SPI 3103.2.1 any! Powers of i the number - is the unique number for which = −1 and =−1 Confederation... Days Instructions: Choose an answer and hit 'next ' key on multiplying complex numbers define square! Operations math worksheets ; math Games ; CogAT Test ; math Games ; Test... On multiplication squares, in-out boxes, evaluating expressions, filling in missing integers, and subtracting numbers! The denominator by the conjugate of the basic mathematical concepts that can help you check your of! Worksheets found for - multiplying and Dividing complex numbers and ≠0 to understand the concept of being able to the! Begin by multiplying the numerator and the denominator by the conjugate of denominator... Define the square root of negative one integers, and subtracting complex numbers - top! Reproduction answers key, Weaknesses of the denominator by the conjugate of the denominator mixed. You say: `` your brain is smaller than any > 0!.! Introduction to complex numbers - Displaying top 8 worksheets found for this concept numbers - Displaying top worksheets. Is easier than it looks and are real numbers and ≠0 with these mixed! And what are some of the denominator in missing integers, and subtracting numbers... Really great activity for allowing students to understand the concept of multiplying and Dividing complex numbers, have!\n\nToyota Maroc : Prix, Education Support Partnership Employee Assistance Programme, Ar Meaning In English, Is Kerdi-fix Waterproof, Toyota Maroc : Prix, Toy Poodle Male Vs Female, Schleswig-holstein Ship Today, Originating Motion Supreme Court, Magic Word Synonym,"
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https://www.itrisa.com/Joint-Probability-answers | [
"# Joint Probability\n\nJoint Probability, Find information about Joint Probability, we will help you with information.\n\nA joint probability can be visually represented through a Venn diagram. Consider the joint probability of rolling two 6’s in a fair six-sided dice: Shown on the Venn diagram above, the joint probability is where both circles overlap each other. It is called the “intersection of two events.” Examples. The following are examples of joint ...\nA joint probability distribution represents a probability distribution for two or more random variables. Instead of events being labelled A and B, the condition is to use X and Y as given below. f (x,y) = P (X = x, Y = y) The main purpose of this is to look for a relationship between two variables. For example, the below table shows some ...\nGiven two random variables that are defined on the same probability space, the joint probability distribution is the corresponding probability distribution on all possible pairs of outputs. The joint distribution can just as well be considered for any given number of random variables. The joint distribution encodes the marginal ...\nJoint Probability: A joint probability is a statistical measure where the likelihood of two events occurring together and at the same point in time are calculated. Joint probability is the ...\nExamples of Joint Probability Formula (with Excel Template) Example #1. Example #2. Example #3. Difference Between Joint, Marginal, and Conditional Probability. Relevance and Use. Recommended Articles. Step 1- Find the Probability of Two events separately. Step 2 – To calculate joint probability, both the probabilities must be multiplied.\nJoint Probability Example #1. Let’s say you want to figure out the joint probability for a coin toss where you can get a tail (Event X) followed by a head (Event Y). In this instance, the probability of Event X is 50% (or 0.5) and the probability of Event Y is also 50%. Now we can plug in the numbers into the formula: That means, the joint ...\nA joint probability distribution simply describes the probability that a given individual takes on two specific values for the variables. The word “joint” comes from the fact that we’re interested in the probability of two things happening at once. For example, out of the 100 total individuals there were 13 who were male and chose ...\nThe joint probability mass function (discrete case) or the joint density (continuous case) are used to compute probabilities involving \\(X\\) and \\(Y\\). 6.2 Joint Probability Mass Function: Sampling From a Box. To begin the discussion of two random variables, we start with a familiar example. Suppose one has a box of ten balls – four are white ...\nWe know that the conditional probability of a four, given a red card equals 2/26 or 1/13. This should be equivalent to the joint probability of a red and four (2/52 or 1/26) divided by the marginal P (red) = 1/2. And low and behold, it works! As 1/13 = 1/26 divided by 1/2. For the diagnostic exam, you should be able to manipulate among joint ...\nSpecifically, you learned: Joint probability is the probability of two events occurring simultaneously. Marginal probability is the probability of an event irrespective of the outcome of another variable. Conditional probability is the probability of one event occurring in the presence of a second event.\nJoint Probability of Two Variables. We may be interested in the probability of two simultaneous events, e.g. the outcomes of two different random variables.PROBABILITY plays an essential role in today’s business; it often decides to draw applied math inferences that might be accustomed to predicting knowledge or analysing knowledge higher. It ...\nSimulation studies are conducted to analyze longitudinal data with a trinomial outcome using a CTMC with and without covariates. Measures of performance including bias, component-wise coverage probabilities, and joint coverage probabilities are calculated. An application is presented using Alzheimer's disease caregiver stress levels.\nJoint probability is the likelihood of more than one event occurring at the same time P (A and B). The probability of event A and event B occurring together. It is the probability of the ...\nIn probability theory, a probability density function ( PDF ), or density of a continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would be ...\nWhat Does Joint Probability Tell You? Probability is a discipline carefully associated to statistics that offers with the probability of an occasion or phenomena occurring. It is quantified as a range of among zero and 1 inclusive, wherein zero suggests an not possible risk of incidence and 1 denotes the sure final results of an occasion. ...\nJoint probability refers to the odds of two events happening collectively. It is a statistical measure that calculates the chances of this synchronous happening. Probability is a branch of mathematics which deals with the occurrence of an event. Joint probability, on the other hand, helps us compute the probability of an event G occurring at the same time that event S occurs.\nExample 1. Consider the joint pdf of two variables. In other words, the joint pdf is equal to if both entries of the vector belong to the interval and it is equal to otherwise. Suppose that we need to compute the probability that both entries will be less than or equal to . This probability can be computed as a double integral:\nJoint probability is the likelihood of two independent events happening at the same time. The two events must be independent , meaning the outcome of one event has no influence over the outcome of ...\nThe joint probability for independent random variables is calculated as follows: P(A and B) = P(A) * P(B)Sep 30, 2019. What are the different types of probability? There are three major types of probabilities: Theoretical Probability. Experimental Probability. Axiomatic Probability.\nJoint probability, conditional probability and Bayes' theorem. For those of you who have taken a statistics course, or covered probability in another math course, this should be an easy review. For the rest of you, we will introduce and define a couple of simple concepts, and a simple (but important!) formula that follows immediately from the ...\nJoint Probability. Furthermore, the joint probability that the surface site is >MeOH and the polyelectrolyte ligand site is L− is given by the product fMeOH · fL-, where fMeOH is the fraction of the proton-reactive surface species present as >MeOH and fL- is the fraction of ligand sites on the polyelectrolyte that is deprotonated.\nPrepare for exam with EXPERTs notes - unit 2 continuous probability and bivariate distributions for bhagat phool singh mahila vishwavidyalaya haryana, information technology-engineering-sem-2\nJoint Probability Distributions (a) Given that X = 1;determine the conditional pmf of Y, that is, py jx(0 j1);pyjx(1 1 and py x(2j1): (b) Given that two hoses are in use at the self-service island. What is the conditional pmf of the number of hoses in use on the full-service island?\nIn this post, you will learn about joint and conditional probability differences and examples. When starting with Bayesian analytics, it is very important to have a good understanding around probability concepts.And, the probability concepts such as joint and conditional probability is fundamental to probability and key to Bayesian modeling in machine learning.\nA joint probability table is a way of specifying the \"joint\" distribution between multiple random variables. It does so by keeping a multi-dimensional lookup table (one dimension per variable) so that the probability mass of any assignment, eg P ( X = x, Y = y, … ), can be directly looked up.\n49,000 results\n\nWeb Info\nSee results for Joint-Probability\nWeb 49,000 results\nProbability 39 results - Search Instead For Probability\nJoint probability 27 results - Search Instead For Joint Probability\nJoint 14 results - Search Instead For Joint\n.com results 16\n.org results 4\n.io results 2\n.edu results 3\ntld results gov\nLanguages en\nCorporatefinanceinstitute (com) Joint Probability\nByjus (com) Joint Probability\nWikipedia (org) Joint Probability Distribution\nInvestopedia (com) Jointprobability\nWallstreetmojo (com) Joint Probability Formula\nInvestinganswers (com) Joint Probability\nStatology (org) Joint Probability Distribution\nGithub (io) Joint Probability Distributions\nDuke (edu) Jmc\nMachinelearningmastery (com) Joint Marginal And Conditional Probability For Machine Learning\nJigsawacademy (com) Joint Probability\nNih (gov) 28572692\nMedium (com) Joint Probability Vs Conditional Probability Fa2d47d95c4a\nWikipedia (org) Probability Density Function\nFirsteducationinfo (com) Joint Probability\nCollegedunia (com) Joint Probability Joint Probability Formula And Solved Examples Articleid 4887\nStatlect (com) Joint Probability Density Function\nStudy (com) Joint Probability Definition Formula Examples\nAeroantenna (com) What Is The Difference Between Joint Probability And Conditional Probability\nUpenn (edu) Bayes1\nSciencedirect (com) Joint Probability\nGoseeko (com) Unit 2 Continuous Probability And Bivariate Distributions 1\nMsu (edu) Lecture 08.pdf\nVitalflux (com) Joint Conditional Probability Explained With Examples\nGithub (io) Joint\nMatcmath (org) Discrete Joint Probability\n\nProbability joint event distribution random given events conditional variables. example occurring variable probability. probabilities table variables possible marginal likelihood function four outcome.\n\n#### What is a Joint Probability Distribution?\n\nA joint probability distribution simply describes the probability that a given individual takes on two specific values for the variables."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9288304,"math_prob":0.9774763,"size":8343,"snap":"2022-40-2023-06","text_gpt3_token_len":1683,"char_repetition_ratio":0.20853819,"word_repetition_ratio":0.03803132,"special_character_ratio":0.20424308,"punctuation_ratio":0.11960784,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99966204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T08:43:25Z\",\"WARC-Record-ID\":\"<urn:uuid:4752492d-26ee-47ca-9d50-88164551d0d5>\",\"Content-Length\":\"45602\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8aafc5fc-1de2-4862-893c-47aa2d0f3fee>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f41dcad-17f9-482d-96cf-02cccd0fec7a>\",\"WARC-IP-Address\":\"172.67.222.190\",\"WARC-Target-URI\":\"https://www.itrisa.com/Joint-Probability-answers\",\"WARC-Payload-Digest\":\"sha1:U2IVMA3P2YV4WWXAX4AWY7ESXHXXE2QS\",\"WARC-Block-Digest\":\"sha1:3DARWEGEGRD6GNFPKHDSOIMGKHN744VN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337595.1_warc_CC-MAIN-20221005073953-20221005103953-00725.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/1-3-plus-60-72 | [
"Solutions by everydaycalculation.com\n\n1/3 + 60/72 is 7/6.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 3 and 72 is 72\n2. For the 1st fraction, since 3 × 24 = 72,\n1/3 = 1 × 24/3 × 24 = 24/72\n3. Likewise, for the 2nd fraction, since 72 × 1 = 72,\n60/72 = 60 × 1/72 × 1 = 60/72\n24/72 + 60/72 = 24 + 60/72 = 84/72\n5. 84/72 simplified gives 7/6\n6. So, 1/3 + 60/72 = 7/6\nIn mixed form: 11/6\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://robotology.github.io/robotology-documentation/doc/html/group__iKinSlv.html | [
"iCub-main\niKinSlv",
null,
"Collaboration diagram for iKinSlv:\n\n## Classes\n\nclass iCub::iKin::CartesianSolver\n\nclass iCub::iKin::iCubArmCartesianSolver\n\nclass iCub::iKin::iCubLegCartesianSolver\n\n## Detailed Description\n\nClasses for on-line solution of inverse kinematic of iCub limbs based on IpOpt. The solvers run as on-line daemons which connect to the robot to retrieve information on the current joints configuration (along with the bounds) and by requesting a desired pose with queries made through YARP ports they return the corresponding target joints.\n\nThe task to be solved is:\n\n$\\begin{array}{c} \\mathbf{q}=\\arg\\min\\limits_{\\mathbf{q} \\in \\mathbb{R}^{n} }\\left(\\left\\|\\mathbf{\\alpha}_d-\\mathit{K_{\\alpha}}\\left(\\mathbf{q}\\right)\\right\\|^2+\\mathit{w}\\cdot\\left(\\mathbf{q}_r-\\mathbf{q}\\right)^{\\top}\\mathit{W}_r\\left(\\mathbf{q}_r-\\mathbf{q}\\right)\\right) \\\\ \\text{s.t.} \\begin{cases} \\left\\|\\mathbf{x}_d-\\mathit{K_x}\\left(\\mathbf{q}\\right)\\right\\|^2=0 \\\\ \\mathbf{q}_L<\\mathbf{q}<\\mathbf{q}_U \\end{cases} \\end{array}.$\n\nWhere the solution $$\\mathbf{q}$$ is the joints vector with n components (depending on the task and the current dof configuration) that is guaranteed to be found within the physical bounds expressed by $$\\mathbf{q}_L$$ and $$\\mathbf{q}_U$$; $$\\mathbf{x}_d\\equiv\\left(x,y,d\\right)_d$$ is the positional part of the desired pose, $$\\mathbf{\\alpha}_d$$ is the desired orientation and $$\\mathit{K_x}$$ and $$\\mathit{K_{\\alpha}}$$ are the forward kinematic maps for the position and orientation part, respectively; $$\\mathbf{q}_r$$ is used to keep the solution as close as possible to a given rest position in the joint space (weighting with a positive factor $$\\mathit{w} < 1$$ and also through the diagonal matrix $$\\mathit{W}_r$$ which allows selecting a specific weight for each joint).\n\n# Solver protocol\n\nOnce created the solver will open three ports with which user can communicate and exchange information according to the following rules:\n\n/<solverName>/in accepts a properties-like bottle containing the following requests in streaming mode:\n\nxd request: example ([xd] (x y z ax ay az theta)), specifies the target pose to be achieved as a 7-components vector for position and orientation parts. Note that if the current pose is of [xyz] type, then only the first 3-components are meaningful (x,...,az are in meters, theta is in radians).\n\npose request: example ([pose] [full]), specifies the end-effector pose the user wants to achieve; it can be [full] (position+orientation) or [xyz] (only position).\n\nmode request: example ([mode] [cont]), selects the solver mode between [cont] which implements a continuous tracking and [shot] which does not compensate for movements induced on unacatuated joints. Indeed, in continuous mode the solver yields a new solution whenever a detected movements on uncontrolled links determines a displacement from the desired target.\n\ndof request: example (dof), specifies which dof of the chain are actuated (by putting 1 in the corresponding position) and which not (with 0). The length of the provided list of 1's and 0's should match the number of chain's dof. The special value 2 is used to keep the link status unchanged and proceed to the next link.\n\nresp request: example (resp), specifies for each joint the rest position in degrees. The reply will contain the rest position resulting from the application of joints limits.\n\nresw request: example (resw), specifies for each joint the weight used to compute the rest position minimization task. The reply will contain the weights as result of a saturation over zero.\n\ntok synchro: a double may be added to the request which will be in turn sent back by the solver for synchronization purpose.\n\nNote\nOne single command sent to the streaming input port can contain multiple requests: e.g. ([pose] [xyz]) ([dof] (1 0 1 2 1)) (xd ([tok] 1.234) ...)\nRemind that the intended use of this port is in streaming mode, thus it might happen that one request is dropped in favour of the most recent one. Therefore, as a general remark, rely on this port for xd requests and use rpc for dof, mode, pose, ... requests (see below).\n\n/<solverName>/rpc accepts a vocab-like bottle containing the following requests, executes them and replies with [ack]/[nack] and/or some useful info:\n\nCommands issued through [set]/[get] vocab:\n\npose request: example [set] [pose] [full]/[xyz], [get] [pose].\n\npose priority request: example [set] [prio] [xyz]/[ang], [get] [prio]. For example, setting priority to [ang] allows considering the reaching in orientation as a constraint while reaching in position is handled as an objective.\n\nNote\nCaveat. If priority is given to [ang], then commands such as [set] [pose] will need still to be used with the tag [xyz] in order to reach for a non-full pose. This choice, even if counter-intuitive, ensures back-compatibility and in the end does not represent a big hitch since orientation is very rarely prioritized over position.\n\nmode request: example [set] [mode] [cont]/[shot], [get] [mode].\n\nlim request: example [set] [lim] axis min max, [get] [lim] axis. Set/return minimum and maximum values for the joint. Allowed range shall be a valid subset of the real control limits (unit is deg).\n\nverbosity request: example [set] [verb] [on]/[off], [get] [verb].\n\ndof request: example [set] dof, [get] [dof]. The reply will contain the current dof as result of the reconfiguration. The result may differ from the request since on certain limb (e.g. arm) some links are considered to form a unique super-link (e.g. the shoulder) whose components cannot be actuated separately.\n\nresp request: example [set] resp, [get] [resp]. Set/get for each joint the rest position in degrees. The reply will contain the rest position resulting from the application of joints limits (lim request does affect the range).\n\nresw request: example [set] resw, [get] [resw]. Set/get for each joint the weight used to compute the rest position minimization task. The reply will contain the weights as result of a saturation over zero.\n\ntsk2 request: example [set] [tsk2] (6 (0.0 0.0 -1.0) (0.0 0.0 1.0), [get] [tsk2]. Set/get the options for the secondary task, where the first integer accounts for the joints to control, then come the desired x-y-z coordinates of the secondary end-effector and finally the corresponding three weights.\n\nconv request: example [set] conv, [get] [conv]. Set/get the options for specifying solver's convergence.\n\nCommands issued through the [ask] vocab:\n\nxd request: example [ask] ([xd] (x y z ax ay az theta)) ([pose] [xyz]) (q). Ask to solve for the target xd. User can specifies a different starting joint configuration q and the pose mode. The reply will contain something like ack (x), where the found configuration q is returned as well as the final attained pose x.\n\nsusp request: example [susp], suspend the thread.\n\nrun request: example [run], let the thread run again.\n\nstatus request: example [status], returns [ack] \"not_started\"|\"running\"|\"suspended\".\n\nquit request: example [quit], close ports and quit the thread. [ack] is returned.\n\nCommands concerning the solver configuration:\n\ncfg request: example [cfg] (robot icub) (type left) (pose full) (tol 0.001) (constr_tol 0.000001) (maxIter 150) ... Once instantiated the solver is not automatically configured. To do that user can call the open() method locally or can configure the solver remotely by sending to the rpc port the configuration properties. For a list of these properties please see the documentation of open() method.\n\n/<solverName>/out streams out a bottle containing the result of optimization instance. The format is (xd) (x) (q ([tok] ...)) as following:\n\nxd property: contains the desired cartesian pose to be achieved (7-components vector).\n\nx property: contains the real cartesian pose achieved which in norm is the nearest one to xd (7-components vector).\n\nq property: contains the joints configuration which achieves x (DOF-components vector in degrees).\n\ntok property: contains the token that the client may have added to the request.\n\nDate: first release 20/06/2009"
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https://educationalresearchtechniques.com/2016/05/23/understanding-decision-trees/ | [
"",
null,
"# Understanding Decision Trees\n\nDecision trees are yet another method of machine learning that is used for classifying outcomes. Decision trees are very useful for, as you can guess, making decisions based on the characteristics of the data.",
null,
"",
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"In this post, we will discuss the following\n\n• Physical traits of decision trees\n• How decision trees work\n• Pros and cons of decision trees\n\nPhysical Traits of a Decision Tree\n\nDecision trees consist of what is called a tree structure. The tree structure consists of a root node, decision nodes, branches and leaf nodes.\n\nA root node is an initial decision made in the tree. This depends on which feature the algorithm selects first.\n\nFollowing the root node, the tree splits into various branches. Each branch leads to an additional decision node where the data is further subdivided. When you reach the bottom of a tree at the terminal node(s) these are also called leaf nodes.\n\nHow Decision Trees Work\n\nDecision trees use a heuristic called recursive partitioning. What this does is it splits the overall dataset into smaller and smaller subsets until each subset is as close to pure (having the same characteristics) as possible. This process is also known as divide and conquer.\n\nThe mathematics for deciding how to split the data is based on an equation called entropy, which measures the purity of a potential decision node. The lower the entropy scores the purer the decision node is. The entropy can range from 0 (most pure) to 1 (most impure).\n\nOne of the most popular algorithms for developing decision trees is the C5.0 algorithm. This algorithm, in particular, uses entropy to assess potential decision nodes.\n\nPros and Cons\n\nThe prose of decision trees includes its versatile nature. Decision trees can deal with all types of data as well as missing data. Furthermore, this approach learns automatically and only uses the most important features. Lastly, a deep understanding of mathematics is not necessary to use this method in comparison to more complex models.\n\nSome problems with decision trees are that they can easily overfit the data. This means that the tree does not generalize well to other datasets. In addition, a large complex tree can be hard to interpret, which may be yet another indication of overfitting.\n\nConclusion\n\nDecision trees provide another vehicle that researchers can use to empower decision making. This model is most useful particularly when a decision that was made needs to be explained and defended. For example, when rejecting a person’s loan application. Complex models made provide stronger mathematical reasons but would be difficult to explain to an irate customer.\n\nTherefore, for complex calculation presented in an easy to follow format. Decision trees are one possibility."
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"https://ws-na.amazon-adsystem.com/widgets/q",
null,
"https://ir-na.amazon-adsystem.com/e/ir",
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https://www.shaalaa.com/question-bank-solutions/find-number-diagonals-hexagon-combination_54104 | [
"# Find the Number of Diagonals of , a Hexagon - Mathematics\n\nFind the number of diagonals of , 1.a hexagon\n\n#### Solution\n\nA polygon of n sides has n vertices. By joining any two vertices we obtain either a side or a diagonal.\n∴ Number of ways of selecting 2 out of 9 $=^n C_2 = \\frac{n\\left( n - 1 \\right)}{2}$\n\nOut of these lines, n lines are the sides of the polygon.\n\n∴ Number of diagonals =$\\frac{n\\left( n - 1 \\right)}{2} - n = \\frac{n\\left( n - 3 \\right)}{2}$\n\n1. In a hexagon, there are 6 sides.\n∴ Number of diagonals =$\\frac{6\\left( 6 - 3 \\right)}{2} = 9$\n\nConcept: Combination\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nRD Sharma Class 11 Mathematics Textbook\nChapter 17 Combinations\nExercise 17.2 | Q 15.1 | Page 16"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.745441,"math_prob":0.99873644,"size":568,"snap":"2021-31-2021-39","text_gpt3_token_len":194,"char_repetition_ratio":0.13475177,"word_repetition_ratio":0.0,"special_character_ratio":0.34507042,"punctuation_ratio":0.09166667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99962854,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T23:07:32Z\",\"WARC-Record-ID\":\"<urn:uuid:3544fba3-4cd7-40f4-940b-aed95062dfd9>\",\"Content-Length\":\"41001\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee08ffe2-e4d0-4690-bb7c-27b888ca697b>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a45e6fa-ca6b-4f09-9681-886bd435d9be>\",\"WARC-IP-Address\":\"172.105.37.75\",\"WARC-Target-URI\":\"https://www.shaalaa.com/question-bank-solutions/find-number-diagonals-hexagon-combination_54104\",\"WARC-Payload-Digest\":\"sha1:5M5WJ2HAKPUS5735WPFDACUPVUIJN6HP\",\"WARC-Block-Digest\":\"sha1:FGOIGUMJDQXAUSAPGGNHGHPDE6VZEEYJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151531.67_warc_CC-MAIN-20210724223025-20210725013025-00207.warc.gz\"}"} |
https://kalkicode.com/inorder-traversal-of-binary-tree-with-recursion-in-scala | [
"# Inorder traversal of binary tree with recursion in scala\n\nScala program for Inorder traversal of binary tree with recursion. Here problem description and explanation.\n\n``````/*\nScala Program for\ninorder tree traversal of a Binary Tree\nusing recursion\n*/\n// Binary Tree Node\nclass TreeNode(var data: Int,\nvar left: TreeNode,\nvar right: TreeNode)\n{\ndef this(data: Int)\n{\n// Set node value\nthis(data, null, null);\n}\n}\nclass BinaryTree(var root: TreeNode)\n{\ndef this()\n{\nthis(null);\n}\n// Display Inorder view of binary tree\ndef inorder(node: TreeNode): Unit = {\nif (node != null)\n{\n// Visit left subtree\ninorder(node.left);\n// Print node value\nprint(\" \" + node.data);\n// Visit right subtree\ninorder(node.right);\n}\n}\n}\nobject Main\n{\ndef main(args: Array[String]): Unit = {\n// Create new tree\nvar tree: BinaryTree = new BinaryTree();\n/*\nMake A Binary Tree\n----------------\n15\n/ \\\n24 54\n/ / \\\n35 62 13\n*/\ntree.root = new TreeNode(15);\ntree.root.left = new TreeNode(24);\ntree.root.right = new TreeNode(54);\ntree.root.right.right = new TreeNode(13);\ntree.root.right.left = new TreeNode(62);\ntree.root.left.left = new TreeNode(35);\n// Display Tree Node\ntree.inorder(tree.root);\n}\n}``````\n\nOutput\n\n`` 35 24 15 62 54 13``\n\nPlease share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.\n\n### New Comment",
null,
""
] | [
null,
"https://kalkicode.com/images/icon/binary-tree.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59070843,"math_prob":0.81718457,"size":1411,"snap":"2021-43-2021-49","text_gpt3_token_len":348,"char_repetition_ratio":0.1663113,"word_repetition_ratio":0.017777778,"special_character_ratio":0.3090007,"punctuation_ratio":0.1985294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9691388,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T09:51:58Z\",\"WARC-Record-ID\":\"<urn:uuid:1df795fa-e452-4e4d-917c-5f02141068a5>\",\"Content-Length\":\"62907\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95f67724-d574-4c92-86cf-6b66c13e9c04>\",\"WARC-Concurrent-To\":\"<urn:uuid:82c8bcd5-b364-48f2-a888-7b9aeab4bf83>\",\"WARC-IP-Address\":\"13.126.103.19\",\"WARC-Target-URI\":\"https://kalkicode.com/inorder-traversal-of-binary-tree-with-recursion-in-scala\",\"WARC-Payload-Digest\":\"sha1:RN2LD4VNMA3E36UFGZD25LUSB3UXSRAZ\",\"WARC-Block-Digest\":\"sha1:7AMTLN2IFDJYHSZJKVLHVIYUVPFBXRH5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585504.90_warc_CC-MAIN-20211022084005-20211022114005-00357.warc.gz\"}"} |
https://wskg.org/education/area-model/ | [
"# Area Model\n\nWhen first learning to multiply two two-digit numbers your child will use the area model.\n\nTo start, your child will use her knowledge of place value to decompose into tens and ones. To decompose means to break apart. Let’s decompose these numbers by the value of each digit.\n\nThe value of two tens is twenty. The value of three ones is three. Three tens is thirty. And five ones is five. Decomposing numbers allows your child to use the multiplication fluency she developed in third grade to multiply large numbers with mental math.\n\nSo, what is twenty-three times thirty-five?\n\nThree tens times two tens equals sixty tens or six-hundred. Thirty tens times three equals nine tens or ninety. Twenty tens times five equals ten tens or one-hundred. And three times five equals fifteen.\n\nYour child will then add these products together. Six-hundred plus ninety equals six-hundred ninety. One-hundred plus fifteen equal one-hundred fifteen. By fourth grade, your child will fluently add three-digit numbers, like this, using the standard algorithm.\n\nYour child can clearly see why twenty-three times thirty-five equals eight-hundred-five. The area model gives your child a visual representation that decomposes the numbers she is multiplying. At this point in fourth grade, your child is developing a pictorial level of understanding,\nwhich will give her a strong foundation for using partial products and, later, using the standard algorithm to multiply.\n\nAnd that’s good to know.\n\nThis video addresses Common Core Grade 4 Standard Number & Operations in Base Ten: Use place value understanding and properties of operations to perform multi-digit arithmetic. Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8723863,"math_prob":0.9906063,"size":1929,"snap":"2021-31-2021-39","text_gpt3_token_len":380,"char_repetition_ratio":0.13142857,"word_repetition_ratio":0.0,"special_character_ratio":0.18869881,"punctuation_ratio":0.10773481,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994592,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T07:33:56Z\",\"WARC-Record-ID\":\"<urn:uuid:c60ad130-b404-4d31-b8d8-72c2808d1be4>\",\"Content-Length\":\"85126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:615df0a7-a09a-489d-b85f-a824ffe8a99d>\",\"WARC-Concurrent-To\":\"<urn:uuid:3eb664a9-098f-4545-8274-6ad0daa3e04e>\",\"WARC-IP-Address\":\"162.144.89.147\",\"WARC-Target-URI\":\"https://wskg.org/education/area-model/\",\"WARC-Payload-Digest\":\"sha1:W6DLWJSRS7RHB53RR2VOCB3CA26F762J\",\"WARC-Block-Digest\":\"sha1:JJL5BI34CX6YBI6N5J5VFMJXIYO7K5W3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057199.49_warc_CC-MAIN-20210921070944-20210921100944-00398.warc.gz\"}"} |
https://alexandrakapp.blog/2021/12/07/similarity-of-maps/ | [
"## Quantifying the Similarity between Maps\n\nUsing the earth mover’s distance to quantify the similarity between two maps with implementations in Python and R.\n\nWhen you implement a privacy measure, e.g., adding noise to data, you are interested to know how similar your noisy data is compared to your original data. Therefore, there are various similarity metrics to quantify the difference between the original and the new distribution. For geospatial data, there are specific challenges:\n\n### The Scenario\n\nLet’s assume you have a map that shows the distribution of people within Berlin – e.g., data collected through cellular networks, GPS tracking, stationary sensors, or simply a survey. You aggregate the data spatially (in this example we use the “Planungsräume Berlin from the Geoportal Berlin” as a tessellation) and get the following result:",
null,
"Example data: number of people recorded in each region of Berlin\n\nA few weeks later you collect the data again (Or in the context of privacy measures: you add noise to the data). Now imagine two scenarios. In scenario 1 you get the following new output:\n\nYou can see a change of 100 people who are recorded less within one tile and more in the neighboring tile.\n\nIn scenario 2 you see the following:\n\nIn scenario 2 there are 100 people less in the same tile as in scenario two but 100 people more in a tile at the outskirts of the city.\n\nIntuitively, the first scenario looks more similar to the original data than the second. In both scenarios, 100 people moved from the same origin tile to another tile. Though in scenario 1 the new tile is a neighboring tile of the origin while in scenario 2 the tile is at the outskirts of the city.\nOne could imagine explanations for the changes in scenario 1, as for example usual variability in people’s behavior, or maybe a coffee shop down the street had a special offer. Or it could even be due to inaccuracies in the data collection instrument – maybe there are many people right at the border of the two regions and the tracked GPS signal is not accurately assigned to the correct region. The movement to a region far away in scenario 2 seems to be a more significant change. Intuitively the pattern of the map has changed. One reason, this change seems bigger, is because it is more effort to move multiple kilometers from the original location away than it is to move a few hundred meters.\nBut how can you measure this (dis-)similarity?\n\n### Measuring similarity\n\nCommon measures that compare the similarity between distributions do not account for spatial distance. E.g., the mean absolute percentage error (by how many percent does each original tile diverge from its scenario equivalent?), the Jensen-Shannon metric (how similar are two probability distributions?), or the correlation will be similar for both scenarios. This is because the spatial component is not being considered and all tiles are treated equally.\n\n``````Mean mean absolute percentage error scenario 1: 2.41%\nMean mean absolute percentage error scenario 2: 2.41%\nJensen-Shannon distance scenario 1: 0.0401\nJensen-Shannon distance scenario 2: 0.0401\nBiserial correlation coefficient scenario 1: 0.9893\nBiserial correlation coefficient scenario 2: 0.9893``````\n\n#### Earth Mover’s Distance (Wasserstein metric)\n\nTo capture the spatial distance, we need to use a measure that can account for that. The Wasserstein metric, also called earth mover’s distance (EMD), can do exactly this.\nCiting from this detailed explanation of the EMD: “Intuitively, given two distributions, one can be seen as a mass of earth properly spread in space, the other as a collection of holes in that same space. Then, the EMD measures the least amount of work needed to fill the holes with earth. Here, a unit of work corresponds to transporting a unit of earth by a unit of ground distance.” Clear and illustrative explanations of the earth mover’s distance are also given here and here.\n\n#### Python Implementation\n\nNow we have the theory, but how do we implement this in python?\nThe implementation by Scipy of the Wasserstein Distance only works with 1D distributions. But we have a 2D distribution – latitude and longitude.\n\nThe EMD is also used as a metric to compare the similarity between images (see, e.g., this). This is why there is an implementation by the OpenCV (Open Source Computer Vision Library) for 2D distributions, which we can use for our purposes.\nUnfortunately, the documentation is not very intuitive to understand but thankfully there is a blog article by Sam van Kooten who explains this nicely. His illustration also shows visually, how the earth mover’s distance works for images:\n\nSo how does the CV2 implementation work in detail? Let’s have a look at the documentation:\n\nAccording to the documentation, we need a `signature` as input format. Let’s assume we have a nine-pixel image and each pixel holds one value. We can represent this image with the following matrix:\n\n``````[2,3,5\n1,2,4\n2,1,2]``````\n\nThen we can assign each pixel in the matrix a coordinate:\n\n``````[(0,0), (0,1), (0,2)\n(1,0), (1,1), (1,2)\n(2,0), (2,1), (2,2)]``````\n\nWith this information, we can construct a signature of `size = 9` and with `dims=2`, where each row hold the values `(value, x_coordinate, y_coordinate)`:\n\n``````[2, 0,0,\n3, 0,1,\n5, 0,2,\n1, 1,0,\n2, 1,1,\n4, 1,2,\n2, 2,0,\n1, 2,1,\n2, 2,2]``````\n\nFor our use case, we cannot use the implementation of Sam van Kooten to translate our data into a signature, because we do not have a raster where each cell has a similar distance to its neighbors. Therefore, we cannot construct the coordinates as easily as just described in the example.\n\nFirst of all, we need to define a single point for each tile in our tessellation. For simplicity, we take the centroid.\nThen we need to consider the following: We cannot simply use the geographical coordinates and compute the euclidean distance (this would only be possible if the earth was flat).\nTherefore, we have two options:\n\n1. We project our centroids to a projected coordinate reference system (CRS) and take those coordinates as input to construct our signature. (Here you can learn more about the difference between a geographical and a projected CRS).\n2. We make use of the option in the `cv2.EMD` function to use a custom `cost_matrix` instead of passing coordinates in the signature. We can use the geographical coordinates to compute the haversine distance between all centroids and costruct a cost matrix.\n##### Option 1: use projected coordinates\n\nWhich will result in the following output:\n\n``````Earth movers distance scenario 1 (CRS: EGSG:3035):\n30 meters\n\nEarth movers distance scenario 2 (CRS: EGSG:3035):\n196 meters``````\n\nOne drawback of this method is, that you need to use a properly projected CRS for your data, otherwise your results might not be accurate.\n\nI used the EGSG:3035 which is suitable for Europe. If I would have used, for example, the World Mercator EPSG:3395 projection, the results would have been slightly different, as this projection is not perfectly accurate for Berlin:\n\n``````Earth movers distance scenario 1 (CRS: EGSG:3395):\n50 meters\n\nEarth movers distance scenario 2 (CRS: EGSG:3395):\n322 meters``````\n\nIf you are not familiar with the correct usage of different CRS, option two might be better suited:\n\n##### Option 2: construct a custom cost_matrix with the haversine distance\n\nWe compute a custom cost matrix for all coordinate pairs using the haversine distance.\nFor the cost matrix, we need to define the distance between all points. All points within the first distribution (signature 1) to all points in the second distribution (signature 2). In our case, the points in both distributions are the same, therefore, each distance is included twice (A-B and B-A). In our case, this is the same distance but one could also use asymmetric distances (e.g., you have an uphill location and include the effort needed to go uphill while it saves work to go downhill).\nThe cost matrix has a dimension of `n x m `where `n` is the number of points in the first distribution (signature 1) and `m` is the number of points in the second distribution (signature 2). In our case `n=m`, but for the earth mover’s distance, the points in the first distribution do not have to be the same number or the same points at all as within the second distribution.\n\nWe compute the EMD with a `cost matrix `as follows:\n\n``````Earth movers distance scenario 1: 30 meters\nEarth movers distance scenario 2: 196 meters``````\n\nThe result can now be interpreted as follows:\nIn scenario 1 each person has to move 30 meters on average to change the distribution from the original to scenario 1.\nIn scenario 2 this is 196 meters on average per person.\nThat way, we have a quantified difference between scenarios 1 and 2.\n\n#### Implementation in R\n\nSee this code for the same implementation in R."
] | [
null,
"https://alexandrakappblog.files.wordpress.com/2021/11/example_data.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8981496,"math_prob":0.9608881,"size":8583,"snap":"2023-14-2023-23","text_gpt3_token_len":1947,"char_repetition_ratio":0.13871081,"word_repetition_ratio":0.028432732,"special_character_ratio":0.22684376,"punctuation_ratio":0.12985507,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9860143,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T22:30:07Z\",\"WARC-Record-ID\":\"<urn:uuid:f53f1ae0-aa59-4d0e-aaa6-b4ae36d7d7cc>\",\"Content-Length\":\"175767\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc4a2ae2-f245-4bf4-9701-1c820123c09e>\",\"WARC-Concurrent-To\":\"<urn:uuid:276366c5-3fa8-4a87-8a6a-664c8812affb>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://alexandrakapp.blog/2021/12/07/similarity-of-maps/\",\"WARC-Payload-Digest\":\"sha1:CTS7LBH2BYTJJUQSU3FR45GE5YEULVPX\",\"WARC-Block-Digest\":\"sha1:XKN2WUVYV3HD3GZK4HEA3NHNAP56RM3Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00749.warc.gz\"}"} |
https://scienceteacherprogram.org/astronomy/Augenbran97.html | [
"Summer Research Program for Science Teachers\n\nThe Methane Project\n\nHarvey Augenbran - Mott Hall School - 1997\n\nInteractive Lesson:\n\nThe Role of the Atmosphere and Greenhouse Effect in Determining the Surface Temperature of the Earth\n\n1. Background Information:\n\nThe surface temperature of the earth depends upon several factors. The dominant factor is the amount of solar energy arriving from the sun which depends on the earth's distance from the sun. The earth is about 150 million kilometers (km) [about 96 million miles] from the sun. We refer to this distance as 1 astronomical unit. When solar energy reaches the earth some of it is reflected by the atmosphere. The remainder passes through the atmosphere to the earth's surface where again some is reflected and some is absorbed. The reflected fraction of the total energy received is called the Albedo. This can be calculated at the top of the atmosphere or at the surface.\n\nThe surface of the earth absorbs solar energy, heats and then radiates energy, mostly as infra red, back out to space. Naturally occurring gases in the earth's atmosphere absorb some of the radiated infra red energy, heat and reradiate energy. Some of this reradiated energy is absorbed by the surface giving it additional energy. This process is the Natural Greenhouse Effect. [5-8 Content Standard D- Structure of earth system]\n\nThe earth's surface receives solar energy and energy reradiated by gases in the atmosphere. It heats and radiates energy back into space. The surface continues to heat until the amount of energy received at the is equal to the amount of energy radiated back into space. This condition is Radiative Balance. When this occurs the temperature is stable. [Content Standard Unifying Concepts- Equilibrium]\n\nIn this exercise, we will use interactive simulation courseware developed at NASA - GISS, and conduct research on greenhouse gases to explore the effect of an atmosphere and natural greenhouse gases on the surface temperature of the earth. [Content Standard Unifying Concepts- Evidence, models, and explanation] [Teaching Standard D- Make accessible technological resources] We will also look at natural and human caused (anthropogenic) perturbations in greenhouse gas concentrations and their possible affects on the surface temperature of the earth. [5-8 Content Standard F- Risks and benefits]\n\nBelow is a table for temperature conversion between Kelvin, Celsius and Fahrenheit.\n\n Kelvin = o Celsius = o Fahrenheit Kelvin = o Celsius = o Fahrenheit 250 -23 -9.4 273 0 +32.0 253 -20 -4.0 275 +2 +35.6 255 -18 -0.4 280 +7 +44.6 258 -15 +5.0 285 +12 +53.6 260 -13 +8.6 288 +15 +59.0 265 -8 +17.6 290 +17 +62.6 270 -3 +26.6\n\n2. Surface Temperature - No Atmosphere\n\nWe will now look at what the earth's surface temperature would be if there were no atmosphere.\n\nBegin the simulation program by following the prompts until you have \"the earth with no atmosphere\". Be sure to read the information in the dialog boxes as they appear.\n\nWhat is the initial surface temperature? ________K = ______ o C = _______ o F\n\nDescribe what happens to solar energy after it reaches the surface.__________________________\n\n______________________________________________________________________________\n\nAs you continue through the program, note the following information:\n\nDescribe what happened to the amount of radiated energy as the surface heated .\n\n_______________________________________________________________________________\n\nAt the final step how many units of solar energy were available to be absorbed by the earth's\n\nsurface?________________\n\nAt the final step how many units of energy were radiated back? _______________\n\nHow do you know that radiative balance was reached? ( No, it isn't because it said so.)\n\n_______________________________________________________________________________\n\nWhat was the surface temperature when radiative balance was reached ?\n\n________K = _______ o C = ______ o F\n\n3a Surface Temperature - With an Atmosphere and No Greenhouse Gases\n\nWe will now investigate what the earth's surface temperature would be if we had an atmosphere without and then with greenhouse gases. Begin the simulation program by following the prompts until you have the \"earth with an atmosphere but no greenhouse effect\".\n\nHow many units of solar energy were available to be absorbed by the earth's surface?__________\n\nHow does this compare with the amount of solar energy that was available to be absorbed by the surface when there was no atmosphere?\n\n_______________________________________________________________________________\n\n_______________________________________________________________________________\n\nWhat was the surface temperature when radiative balance was reached ?\n\n________K = _______ o C = ______ o F\n\nCompare the surface temperature reached at radiative balance when there was no atmosphere to the temperature when there was an atmosphere. Why were they different? [Teaching Standard B- Orchestrate scientific discourse]\n\n_______________________________________________________________________________\n\n_______________________________________________________________________________\n\n3b. Surface Temperature - With an Atmosphere and Greenhouse Gases\n\nContinue by adding natural greenhouse gases to the atmosphere.\n\nHow many units of solar energy were available to be absorbed by the earth's surface? ____________\n\nHow many units of energy are being radiated by the earth's surface? ________________\n\nHow could the amount energy being radiated by the surface be more than the amount of solar energy received?________________________________________________________________________\n\n_______________________________________________________________________________\n\nWhat was the surface temperature when radiative balance was reached?\n\n________K = _______ o C = _______ o F\n\n4. Natural Changes in Concentrations of Greenhouse Gases",
null,
"The previous exercises showed that the greenhouse effect occurs naturally and actually is beneficial. On the earth, water vapor (H2O), carbon dioxide (CO2), methane (CH4) and nitrous oxide (N2O) are the gases most responsible for this effect.\n\nThe graph on left (figure 1) shows concentrations of atmospheric methane and carbon dioxide in the atmosphere, as well as temperature changes that occurred over the last 200,000 years. The data were obtained from examining gases trapped in ice cores from Vostok, Antarctica. [5-8 Content Standard D- Earth's history] The y-axis (kyrs BP) shows the time in thousands of years before the present. The x-axis on top shows the concentration of methane in parts per billion in a given volume (left) and the concentration of carbon dioxide in parts per million in a given volume (right). The x-axis on the bottom (center) shows variations around the mean temperature over this time. Notice that even during periods where concentrations are generally decreasing, there are oscillations to higher and lower values. These are natural variations in the concentration of greenhouse gases.\n\nWhat is the time period covered by the graph?\n\n CO2 CH4 Temperature What is the maximum concentration of CO2 & CH4 and temperature 140,000 years ago? (this is our initial value) ppmv ppbv OC What is the minimum concentration of CO2 & CH4 and temperature 20,000 years ago? (this is our final value) ppmv ppbv OC What was the change in CO2 & CH4 and temperature for this period? ppmv ppbv OC Did they increase or decrease? Calculate the percentage change for CO2 & CH4. % Change = Change Initial Value\n\nWe can now use the simulation program to see what temperature changes might occur with the change in concentrations of CO2 and CH4 that we calculated from the ice core data.\n\nGo to the program and follow the prompts until you get to the \"change greenhouse gases\" dialog box. Choose CO2 first and change the percentage to the value you calculated.\n\nWhat is the change in temperature? ________K = _______ o C = _______ o F\n\nRepeat the procedure for CH4.\n\nWhat is the change in temperature? ________K = _______ o C = _______ o F\n\nHow does the combined temperature change shown by the program compare to the change given in the graph? ______________________________________________________________________\n\nWhy do you think this happened?\n\n[5-8 Content Standard A- Use appropriate tools AND use evidence to explain; recognize alternative explanations]\n\n______________________________________________________________________________\n\n(discuss this question because there are many factors involved here)\n\n5. Anthropogenic Changes in Concentrations of Greenhouse Gases",
null,
"Since the beginning of the Industrial Revolution about 200 years ago, human activities (anthropogenic) have been influencing the concentration of greenhouse gases. Examine the graph on the right (Figure 2). This graph shows the atmospheric concentration of CO2 for the past 200 years. This type of graph shows concentrations as well as trends over time. The y-axis shows concentration of CO2 in parts per million in a given volume - ppmv . The x-axis shows the years the measurements were taken. Note that the data line is not straight. This means that the rate of change is not constant for the entire time.\n\nWhat is the approximate change in concentration of CO2 from 1800 to 1900? _________________ppmv\n\nWhat is the approximate change in concentration of CO2 from 1900 to 1990?__________________ppmv\n\nCalculate the percentage change in CO2 concentration from 1800 to 1990 ___________________\n\nPut this value into the simulation program.\n\nWhat is the resulting change in temperature? ? ________K = _______ o C = _______ o F",
null,
"Examine the graph on the left (Figure 3). This graph shows the atmospheric concentration of CH4 for the past 200 years. The y-axis shows concentration of CH4 in parts per billion in a given volume - ppbv. The x-axis shows the years the measurements were taken.\n\nIs the change in methane concentration constant for the entire time period shown? _____________________\n\nCalculate the percentage change in CH4 concentration for 1800 to 1990. ______________________\n\nPut this value into the simulation program. What was the resulting change in temperature?",
null,
"Examine the graph on the right (Figure 4). This graph shows the atmospheric concentration of Nitrous Oxide (N2O) in parts per billion for a given volume for the past 200 years.\n\nUsing the graph, determine when the greatest rate of change occurred? _______________________\n\nCalculate the percentage change in N2O concentration from 1800 to 1900. _______________________\n\nPut this value into the simulation program. What was the resulting change in temperature?\n\n6. Critical Thinking\n\nUsing the information you derived from the 3 graphs, what pattern seems to emerge regarding the rate at which the concentration of these greenhouse gases are increasing? [5-8 Content Standard A- Summarize data; argue cause and effect}\n\n_____________________________________________________________________________________________\n\nWhat prediction could you make if these trends were to continue?________________________________________\n\n_____________________________________________________________________________________________"
] | [
null,
"https://scienceteacherprogram.org/astronomy/Image17.gif",
null,
"https://scienceteacherprogram.org/astronomy/Image18.gif",
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"https://scienceteacherprogram.org/astronomy/Image19.gif",
null,
"https://scienceteacherprogram.org/astronomy/Image20.gif",
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https://www.clutchprep.com/chemistry/practice-problems/74981/in-the-laboratory-you-dissolve-17-7-g-of-sodium-acetate-in-a-volumetric-flask-an | [
"# Problem: In the laboratory you dissolve 17.7 g of sodium acetate in a volumetric flask and add water to a total volume of 125 ml. What is the molarity of the solution? What is the concentration of the sodium cation? What is the concentration of the acetate anion?\n\n###### FREE Expert Solution\n96% (300 ratings)\n###### FREE Expert Solution\n96% (300 ratings)",
null,
"###### Problem Details\n\nIn the laboratory you dissolve 17.7 g of sodium acetate in a volumetric flask and add water to a total volume of 125 ml.\n\nWhat is the molarity of the solution?\n\nWhat is the concentration of the sodium cation?\n\nWhat is the concentration of the acetate anion?\n\nWhat scientific concept do you need to know in order to solve this problem?\n\nOur tutors have indicated that to solve this problem you will need to apply the Molarity concept. You can view video lessons to learn Molarity. Or if you need more Molarity practice, you can also practice Molarity practice problems.\n\nWhat is the difficulty of this problem?\n\nOur tutors rated the difficulty ofIn the laboratory you dissolve 17.7 g of sodium acetate in a...as medium difficulty.\n\nHow long does this problem take to solve?\n\nOur expert Chemistry tutor, Dasha took 3 minutes and 59 seconds to solve this problem. You can follow their steps in the video explanation above.\n\nWhat professor is this problem relevant for?\n\nBased on our data, we think this problem is relevant for Professor Boyd's class at UT."
] | [
null,
"https://cdn.clutchprep.com/assets/button-view-text-solution.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.932801,"math_prob":0.75394416,"size":1069,"snap":"2020-45-2020-50","text_gpt3_token_len":230,"char_repetition_ratio":0.14741784,"word_repetition_ratio":0.08839779,"special_character_ratio":0.20579982,"punctuation_ratio":0.109004736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9625333,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-06T02:14:22Z\",\"WARC-Record-ID\":\"<urn:uuid:f660aeec-a28a-4c9e-accf-0d06384ca32e>\",\"Content-Length\":\"147967\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76f90064-cdc7-433f-bb3c-adb51bb41e37>\",\"WARC-Concurrent-To\":\"<urn:uuid:4256442c-37a9-4775-a377-c3c159424a82>\",\"WARC-IP-Address\":\"34.196.106.64\",\"WARC-Target-URI\":\"https://www.clutchprep.com/chemistry/practice-problems/74981/in-the-laboratory-you-dissolve-17-7-g-of-sodium-acetate-in-a-volumetric-flask-an\",\"WARC-Payload-Digest\":\"sha1:O4MZTIP4FBWBFLJPA3UI3UXSAPVA7TSI\",\"WARC-Block-Digest\":\"sha1:UN6R54IVGUH4D3AZ54MZY2U6DBMXPXTY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141753148.92_warc_CC-MAIN-20201206002041-20201206032041-00204.warc.gz\"}"} |
https://www.mathworksheetscenter.com/mathskills/algebra/factorials/ | [
"# Factorial Worksheets\n\nWhat are Math Factorials? The factorial function is a common expression or operation of mathematics denoted by ! is basically a very simple thing. It is the product of all positive integers that are less than or equal to a given positive integer. Say, the factorial of 4 or 4! This will include multiplying all the positive integers starting from the positive 1 till the number 4 itself. 1 x 2 x 3 x 4. The answer to this factorial would be 24. Therefore, they are generally mentioned as n! According to this rule, the factorial of any number is that number times one less than the number (that number minus 1). N! = n x (n-1). However, it is important to notice that for a number of reasons, 0! equals to 1. You will usually encounter factorials during the evaluation of the permutations and combinations and while dealing with the coefficients of the binomial expansions. They have been generalized to take in and work with nonintegral values without many efforts.\n\n• ### Basic Lesson\n\nDemonstrates the concept solving basic factorials. Practice problems are provided.\n\n• ### Intermediate Lesson\n\nExplores the process of solving slightly complex factorials. Practice problems are provided.\n\n• ### Independent Practice 1\n\nContains 20 Factorials problems. The answers can be found below.\n\n• ### Independent Practice 2\n\nFeatures another 20 Factorials problems.\n\n• ### Homework Worksheet\n\n12 Factorials problems for students to work on at home. Example problems are provided and explained.\n\n• ### Topic Quiz\n\n10 Factorials problems. A math scoring matrix is included.<\n\n• ### Homework and Quiz Answer Key\n\nAnswers for the homework and quiz.\n\n• ### Lesson and Practice Answer Key\n\nAnswers for both lessons and both practice sheets.\n\n#### The Answer\n\nThe highest moments in the life of a mathematician are the first few moments after one has proved the result, but before one finds the mistake."
] | [
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https://brandonmorris.dev/2018/02/11/breaking-neural-nets/ | [
"Deep learning has asserted itself as the king of machine learning. No other method produced thus far has had such excellent success at machine learning tasks that are increasingly complex. In some cases, deep neural networks trained by backpropagation and stochastic gradient descent (i.e. deep learning) have been able to dramatically outperform humans merely by being presented examples of a particular task. The model cultivates “knowledge” by discerning which signals (called “features”) are significant and how their existence or absence, in conjunction with other signals, contributes to an overall results.\n\nIt is beyond doubt that deep neural networks are incredibly sophisticated and versatile machine learning models. They can derive meaning in clever and sometimes unexpected ways, all under the loose guide of human-defined architecture and algorithm. The fact of the matter is that very little human knowledge is explicitly implanted in these performant models: neural networks achieve a bottom-up understanding of their task from the training data. Indeed, understanding how neural networks actually operate is an active area of research that deserves more attention, as Ali Rahimi points out in his NIPS 2017 talk.\n\nBut deep neural networks, for all their successes in the recent past, have a significant weakness. It turns out that these very sophisticated models can be fooled into making dramatically incorrect results. Consider the figure below: to any normal human being it seems like the image is simply random noise. And indeed, the image below was created in part by selecting random values for each of the pixels.\n\nA deep neural network, trained on the ImageNet dataset of over a million images across one thousand categories, might categorize this image as something innocuous like a prayer rug. Furthermore, it would likely give it a low confidence rating, suggesting that the neural network isn’t exactly sure what the image contains, but it might be a prayer rug.\n\nHowever, the Inception network pretrained on ImageNet classified this image as a school bus with 93% confidence. Inception is normally about 70% accurate on normal images, so what went wrong? Is this just a fluke? The reality is that no: results like this are consistently reproducible for even the most advanced and accurate deep neural networks, and they form the basis of a persistent weakness in deep learning, known as adversarial examples.\n\nAdversarial examples lack a formal definition, but they can broadly be considered as inputs that cause otherwise performant machine learning models to produce very inaccurate results. Note that normal inputs can be incorrectly interpreted by a model without necessarily being adversarial.\n\nSome of the most interesting adversarial examples are those that come from otherwise normal (and correctly handled) inputs. Take for instance the two pictures below. On top, the image is perfectly normal, and the Inception model classifies it accurately as school bus with 95% confidence. However, we can manipulate the image ever so slightly by introducing minute perturbations to the image and create the picture on the bottom. The two are almost indistinguishable, but not to our Inception model. The new image is classified as an ostrich with over 98% confidence.",
null,
"A school bus or an ostrich? On top, the normal image is correctly classified, with high confidence. But, by slightly changing the pixels in the image here and there (bottom), we can trick the model into thinking this is an ostrich with almost complete confidence.\n\nAnd these kinds of results can be replicated with nearly any image. So while deep neural networks are ostensibly very well equipped to manage normal data, when that input can be manipulate, even minutely, the classifier can be dramatically fooled.\n\n## How adversarial examples are created\n\nWhen they were first presented, “crafting” (the process of taking a normal input and transforming it to become adversarial) was construed as an optimization problem. Given an input and a classifier, find a perturbation (subject to constraints) that maximizes the error when combined to the original input and fed through the neural network. From this loose framework, we can apply known non-convex optimizers to solve the problem, such as L-BFGS which was done in the paper.\n\nWhile this technique is effective, it is also somewhat slow. L-BFGS can have a hard time converging with deep neural networks, since its a second-order optimization method. In recent years, a number of alternative attack methods have been proposed in the literature, but in this post we will only examine one in detail.\n\nOne of the simplest and fastest methods for crafting adversarial examples relies on utilizing the gradient that is crucial for training the network. With a single backwards pass through the network, we can collect all the information necessary to strike a remarkably effective attack. This method is known as the Fast Gradient Sign Method, or FGSM.",
null,
"An example of crafting and adversarial input using the Fast Gradient Sign Method (FGSM). By backpropagating to the original image, we can determine which direction each pixel should move to increase the error in the prediction. We then take a small step in the sign of that direction, to produce an image that completely fools the target model.\n\nThe FGSM attack can be characterized by the following equation:\n\nwhere $\\epsilon$ is the attack strength and $J$ is our cost function for training the model. Since $J$ is a function that determines how “wrong” our model is after making a prediction, taking the gradient with respect to our input ($\\nabla_x J$) tells us how modifying $x$ will change how correct our model’s prediction is. By taking the sign we only concern ourselves with direction and not magnitude. Then we multiply these gradients by our attack strength $\\epsilon$, which is constrained to be small so that the resulting adversarial image is similar to the original input. Finally, we combine the calculated perturbation with the original image by simple addition.\n\nWhile the math may be somewhat intimidating, the reality is that the FGSM attack is both very simple to program, and extremely efficient to execute. Here’s a simple Python function that calculates adversarial examples for a TensorFlow model:\n\nThe function takes a classifier, which can be any object that has an input tensor model.x and a loss function model.loss, and returns the adversarial example crafted from the initial input. The process follows the description above almost exactly, and is easy to see how cheap the computation is (the most expensive portion is the backwards pass in tf.gradients()).\n\nFGSM is simple and fast, but is it effective? I trained a simple, three layer neural network on the MNIST dataset of handwritten digits. Even without convolutional layers, I was able to achieve a test accuracy of 97%. Using the above code, the adversarial example accuracy of the network was less than one percent. This was possible with an attack strength of just 0.1, which meant that no pixel was modified by more than 10%. Below is a sample of an adversarial example the above method produced, which my network classified as a 3.",
null,
"Result of an FGSM attack on an MNIST handwritten digit example. While clearly still a 5, my network (which has normal test accuracy of 97%) thought this image was a 3. Can you see where FGSM manipulated the image?\n\n## Adversarial examples outside of image classification\n\nAdversarial examples are typically studied in the domain of computer vision, and typically with the task of object classification (given an image, output what is inside the image). However researchers have demonstrated that this weakness of deep neural networks infects other kinds of applications as well. Reinforcement learning involves teaching “agents” or programs how to perform tasks or play games intelligently. RL algorithms often employ deep neural networks to manage the huge number of possible environment states and action strategies. As such, they can be fooled by manipulating the data that is fed to the agent, causing the agent to take incorrect and disadvantageous actions.\n\nSystems that try to comprehend passages of text can also be fooled. Since text comprehension is less well understood than image recognition, these systems don’t necessarily need that intelligent of manipulations to be fooled. Models that can answer basic questions about a paragraph can be completely thrown off by the addition of a single, irrelevant sentence. The image below depicts one such situation."
] | [
null,
"https://brandonlmorris.github.io/images/axs/normal-bus.jpeg",
null,
"https://brandonlmorris.github.io/images/axs/fgsm-panda.png",
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https://writeressays.info/2020/08/16/trig-homework-help_as/ | [
"# Trig homework help\n\nBy | August 16, 2020\n\nFourth marking period . hw #19. image by poem analysis essay example 준원 서 from pixabay. 127k. the diagonal creates a 60º angle at the base of the rectangle. aug 15, 2020 · trigonometry algebra homework a frightening experience free essay help answer all the questions completely trig homework help and accurately. hw #21. what should an essay look like given sina=3/5 and sinb=-12/13, where a is in quadrant i and b is in quadrant iii, find exact values mlk’s letter from birmingham jail summary for sin (a-b), sin (a b), cos (a-b), and tan (a-b) 2 answers. how to define trig homework help the tangent function for all writing a cause and effect paper strategies to solve word problems angles. apr 14, 2008 · trig homework help? Connect to indiana students. welcome to the math homework help subreddit. discover all introductory paragraph essay of persuasive essay about abortion bartleby’s homework solutions you need for the textbooks you have. ft. nov 19, 2009:."
] | [
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https://greenemath.com/Algebra1/41/SpecialFactoringLesson.html | [
"Lesson Objectives\n• Demonstrate an understanding of special products\n• Learn how to factor a difference of squares\n• Learn how to factor a perfect square trinomial\n• Learn how to factor a sum or difference of cubes\n\n## How to Factor Special Products\n\nA few lessons ago, we studied special polynomial products. We learned that certain binomials can be multiplied together quickly using a pattern. In this lesson, we will reverse this process and show a quicker method to factor polynomials in certain forms.\n\n### Factoring the difference of squares\n\nWe previously learned that multiplying conjugates (the sum and difference of the same two terms) results in the first term squared minus the second term squared.\n(x + 5)(x - 5) = x2 - 25\nWhen we see a pattern that matches this format, we want to reverse the process.\nx2 - y2 = (x + y)(x - y)\nLet's look at an example.\nExample 1: Factor each.\nx2 - 100\nWe can think of this problem as:\nx2 - 102\nOnce we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:\n( )( )\nSince x2 is first, place an x in the first position of each set of parentheses:\n(x )(x )\nThe signs need to alternate. One will be (+) and one will be (-):\n(x + )(x - )\nIn our last position of each, we will place a 10:\nx2 - 100 = (x + 10)(x - 10)\nExample 2: Factor each.\n4x2 - 25y2\nWe can think of this problem as:\n(2x)2 - (5y)2\nOnce we identify that we have the difference of squares, we can use the pattern to factor. Set up two sets of parentheses:\n( )( )\nSince 4x2 is first, place a 2x in the first position of each set of parentheses:\n(2x )(2x )\nThe signs need to alternate. One will be (+) and one will be (-):\n(2x + )(2x - )\nIn our last position of each, we will place a 5y:\n4x2 - 25y2 = (2x + 5y)(2x - 5y)\n\n### Factoring a Perfect Square Trinomials\n\nWhen we square a binomial, we end up with a perfect square trinomial. Let's recall our formula:\n(x + y)2 = x2 + 2xy + y2\n(x - y)2 = x2 - 2xy + y2\nWe obtain the first term squared, 2 times the first term and the second term, and the last term squared. The only difference between the two formulas is the sign of the middle term. If we have a perfect square trinomial and the middle term is positive, our trinomial will factor into a sum. If we have a perfect square trinomial and the middle term is negative, our trinomial will factor into a difference. Using this formula, it is easy to go backward. Let's look at an example.\nExample 3: Factor each.\nx2 + 2x + 1\nWe could rewrite this as:\nx2 + 2 • x • 1 + 12\nOnce we know that we have a perfect square trinomial, we can then factor this as:\n(x + 1)2\nExample 4: Factor each.\n25x2 - 20x + 4\nWe could rewrite this as:\n(5x)2 - 2 • (5x) • 2 + 22\nOnce we know that we have a perfect square trinomial, we can then factor this as:\n(5x - 2)2\n\n### Factoring the Sum or Difference of Cubes\n\nWhen we come across the sum or difference of cubes, we can use a special factoring pattern:\nx3 + y3 = (x + y)(x2 - xy + y2)\nx3 - y3 = (x - y)(x2 + xy + y2)\nLet's take a look at some examples.\nExample 5: Factor each.\n27x3 + 64\nWe could rewrite this as:\n(3x)3 + 43\nOnce we know that we have the sum of cubes, we follow the pattern:\n(3x + 4)((3x)2 - (3x)(4) + 42) =\n(3x + 4)(9x2 - 12x + 16)\n27x3 + 64 = (3x + 4)(9x2 - 12x + 16)\nExample 6: Factor each.\n216x3 - 125\nWe could rewrite this as:\n(6x)3 - (5)3\nOnce we know that we have the sum of cubes, we follow the pattern:\n(6x - 5)((6x)2 + (6x)(5) + 52) =\n(6x - 5)(36x2 + 30x + 25)\n216x3 - 125 = (6x - 5)(36x2 + 30x + 25)\n\n#### Skills Check:\n\nExample #1\n\nFactor each. $$324x^{2}- 720x + 400$$\n\nA\n$$4(2x + 7)(2x - 7)$$\nB\n$$4(9x + 2)^{2}$$\nC\n$$4(9x - 10)^{2}$$\nD\n$$4(8x + 3)(8x - 3)$$\nE\n$$4(2x - 7)^{2}$$\n\nExample #2\n\nFactor each. $$125x^{2}- 320$$\n\nA\n$$5(-5x + 8)(5x - 8)$$\nB\n$$5(5x + 8)(5x - 8)$$\nC\n$$5(10x + 7)(10x - 7)$$\nD\n$$5(8x + 7)^{2}$$\nE\n$$5(8x - 1)^{2}$$\n\nExample #3\n\nFactor each. $$6x^{3}- 162y^{3}$$\n\nA\n$$3(x + 14)(x - 14)$$\nB\n$$14(x - 6)^{2}$$\nC\n$$3(x - 3y)(x^{2}+ 3xy + 9y^{2})$$\nD\n$$(x - 6y)(x^{2}+ 18xy + 27y^{2})$$\nE\n$$6(x - 3y)(x^{2}+ 3xy + 9y^{2})$$",
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"https://greenemath.com/img.2018/SVG/confused.svg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.791091,"math_prob":0.99994874,"size":6810,"snap":"2022-05-2022-21","text_gpt3_token_len":2399,"char_repetition_ratio":0.1273876,"word_repetition_ratio":0.78333336,"special_character_ratio":0.39309838,"punctuation_ratio":0.09668508,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984896,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T07:40:06Z\",\"WARC-Record-ID\":\"<urn:uuid:76db3e88-e56c-43ee-ba9f-2748de4d8856>\",\"Content-Length\":\"15536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e101df9f-6c0c-4930-b635-bd910bab3a13>\",\"WARC-Concurrent-To\":\"<urn:uuid:97418502-8838-4985-8c3e-19cac45dccf1>\",\"WARC-IP-Address\":\"107.180.58.55\",\"WARC-Target-URI\":\"https://greenemath.com/Algebra1/41/SpecialFactoringLesson.html\",\"WARC-Payload-Digest\":\"sha1:VDJZF5PN25M5CLDUQG56474DCTBULIWB\",\"WARC-Block-Digest\":\"sha1:BCSHUZFTZ2TKIZ3JCFCUARQIR5CI6RJB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662556725.76_warc_CC-MAIN-20220523071517-20220523101517-00153.warc.gz\"}"} |
https://stackoverflow.com/questions/49837404/updating-a-column-in-a-pandas-dataframe-based-on-a-condition-of-another-column | [
"# Updating a column in a Pandas DataFrame based on a condition of another column\n\nI'm interested in adding a text tag to a new column in a Pandas dataframe. The following example works but I get the copy warning and I don't fully understand if I should ignore it in this case.\n\nThe DataFrame simply has either a character or is an empty string:\n\n``````In : import pandas as pd\n\nIn : df=pd.DataFrame({('A'):['x','','x',''], ('B'):['x','x','','']})\n\nIn : df\nOut:\nA B\n0 x x\n1 x\n2 x\n3\n``````\n\nCreate a new column called 'msg'\n\n``````In : df['msg'] = ''\n\nIn : df\nOut:\nA B msg\n0 x x\n1 x\n2 x\n3\n``````\n\nSet the 'msg' column to 'red' if 'A' is not an empty string\n\n``````In : df['msg'][df['A'] != ''] = 'red;'\n\nIn : df\nOut:\nA B msg\n0 x x red;\n1 x\n2 x red;\n3\n``````\n\nConcatenate 'blue' depending on the 'B' column values\n\n``````In : df['msg'][df['B'] != ''] += 'blue;'\n\nIn : df\nOut:\nA B msg\n0 x x red;blue;\n1 x blue;\n2 x red;\n3\n``````\n\nAlternatively, I found using numpy.where produced the desired result. What is the proper way to do this in Pandas?\n\n``````import numpy as np\n\ndf['msg'] += np.where(df['A'] != '','green;', '')\n``````\n\nUpdate: 4/15/2018\n\nUpon further thought, it would be useful to retain the data from the original DataFrame in certain cases but still attach a label ('color' in this example). The answer from @COLDSPEED led me to the following (changing 'blue;' to 'blue:' and preserving column 'B' data to include in the tag in this case):\n\n``````df['msg'] = (v.where(df.applymap(len) > 0, '') +\ndf.where(df[['B']].applymap(len)>0,'')).agg(''.join, axis=1)\n\nA B msg\n0 x x red;blue:x\n1 x blue:x\n2 x red;\n3\n``````\n• Are all values in the original `df` identical? Or could they differ?\n– cs95\nApr 15 '18 at 0:28\n• I’m not sure I follow. The columns could contain various characters or text but there are no numeric or NaN values. Does that answer your question? Apr 15 '18 at 0:39\n• I wanted to know if everything in your dataframe is `x`, but your answer is quite helpful in another sense - the dtype being string across all columns is crucial to the working of the answer I posted below.\n– cs95\nApr 15 '18 at 0:39\n\nIf you know your colors before-hand, you could use masking with `DataFrame.where` and `str.join` to get this done.\n\n``````v = pd.DataFrame(\nnp.repeat([['red;', 'blue;']], len(df), axis=0),\ncolumns=df.columns,\nindex=df.index\n)\ndf['msg'] = v.where(df.applymap(len) > 0, '').agg(''.join, axis=1)\n``````\n\n``````df\nA B msg\n0 x x red;blue;\n1 x blue;\n2 x red;\n3\n``````\n• I see where you’re going. This could be very useful. My df source is an excel sheet that I can’t modify and there are columns I need to skip but the ‘color’ list could contain an empty string for that column location. Apr 15 '18 at 0:51\n• @Robert Indeed, that is a resourceful way of handling that sort of case.\n– cs95\nApr 15 '18 at 0:53\n• This is a great way of doing it. Apr 15 '18 at 2:04\n\nUsing `pandas.DataFrame.dot`\nSpecial note that I set the dtype of the array to `object`. Otherwise the `dot` won't work.\n\n``````a = np.array(['red', 'blue;'], object)\n\ndf.assign(msg=df.astype(bool).dot(a))\n\nA B msg\n0 x x red;blue;\n1 x blue;\n2 x red;\n3\n``````\n\nYou can using `dot` and `replace`\n\n``````(df!='').dot(df.columns).replace({'A':'red;','B':'blue;'},regex=True)\nOut:\n0 red;blue;\n1 blue;\n2 red;\n3\ndtype: object\n\n#df['msg']=(df!='').dot(df.columns).replace({'A':'red;','B':'blue;'},regex=True)\n``````"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76449317,"math_prob":0.8343999,"size":1669,"snap":"2021-43-2021-49","text_gpt3_token_len":522,"char_repetition_ratio":0.10810811,"word_repetition_ratio":0.051948052,"special_character_ratio":0.36129418,"punctuation_ratio":0.16037735,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976348,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T01:02:38Z\",\"WARC-Record-ID\":\"<urn:uuid:fd5aa5ac-be3b-46ef-97cb-fa6f99d11e70>\",\"Content-Length\":\"188792\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea67cb52-099f-41f4-9943-15a9bbe400f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:4161c4cf-55fc-4a9a-8cb8-de0b6656525b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/49837404/updating-a-column-in-a-pandas-dataframe-based-on-a-condition-of-another-column\",\"WARC-Payload-Digest\":\"sha1:G4A72MOGJP3UN57IDFVDLDEYVDEMZOSW\",\"WARC-Block-Digest\":\"sha1:LKZ7QK4S52OOM5AH3U6BS3AJFSULSD5I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323583087.95_warc_CC-MAIN-20211015222918-20211016012918-00684.warc.gz\"}"} |
https://encyclopedia2.thefreedictionary.com/earth+rate | [
"# earth rate\n\n## earth rate\n\n[′ərth ‚rāt]\n(astronomy)\nThe angular velocity or rate of the earth's rotation.\nMcGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.\nReferences in periodicals archive ?\nwhere n is the number of test positions, [i.sub.yj], j = 1,2,..., n, represent the currents in the torque coils, [[omega].sub.xi], i =1, 2 , ..., n, represent the earth rate components of each test position, and [a.sub.xi], [a.sub.yi], [a.sub.zi], i = 1,2,..., n, represent the gravitational acceleration components of each test position.\nThe gravitational acceleration components and the earth rate components can be expressed as follows:\nAs a result, there are deviations between the actual earth rate components and the ideal earth rate components, as well as the gravitational acceleration components.\nThe actual earth rate components and gravitational acceleration components can be calculated as follows:\nWe have known that the Earth rate is about 15[degrees]/h; for the low cost gyros, it cannot align itself because the noise levels are near or higher than the Earth rate.\nAfter removing the Earth rate, the navigator integrates the angular rate measured by gyros to continuously calculate the attitude of IMU with respect to local level navigation frame.\nGenerally, the Earth rate [[omega].sub.ie], gravity acceleration g, and input angular rate of inertial turntable [OMEGA] are regarded as references, which are compared with outputs of accelerometers and gyros of IMU.\nIt is well known that the Earth rate [[omega].sub.ie] and local gravity vector g could be sensed by gyros and accelerometers of IMU in the body frame.\n\nSite: Follow: Share:\nOpen / Close"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8875065,"math_prob":0.96691674,"size":1261,"snap":"2022-40-2023-06","text_gpt3_token_len":273,"char_repetition_ratio":0.17740652,"word_repetition_ratio":0.021276595,"special_character_ratio":0.22125298,"punctuation_ratio":0.21839081,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99312747,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T00:56:04Z\",\"WARC-Record-ID\":\"<urn:uuid:784ae42d-5205-4615-9ec2-23be1a688287>\",\"Content-Length\":\"43182\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:746e7cc5-8297-409c-aa3d-b33b35852866>\",\"WARC-Concurrent-To\":\"<urn:uuid:68a5d2af-51b2-433f-bff1-1996933a4940>\",\"WARC-IP-Address\":\"172.106.80.2\",\"WARC-Target-URI\":\"https://encyclopedia2.thefreedictionary.com/earth+rate\",\"WARC-Payload-Digest\":\"sha1:KOWVTA6J4NYE5M45HK32LSLFU45KHQO7\",\"WARC-Block-Digest\":\"sha1:KVWY4YLH7FRSHQV6EQKCGN4PITZCRQSB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337446.8_warc_CC-MAIN-20221003231906-20221004021906-00118.warc.gz\"}"} |
https://codebymath.com/index.php/welcome/lesson/for-loops-logistic | [
"## Lesson goal: Iterating the \"logistic\" function with a for-loop\n\nPrevious: Iterating a function | Home | Next: Iterate for a square-root\n\nWe hope you liked the idea of interating a function as shown in a previous lesson. In this lesson, we'll iterate a famous function called the \"logistic function,\" which is $x=cx(1-x)$. This function has a fascinating interation result, depending on what you assign to the variable $c$. In this lesson, you must keep your inital value of $x$ between $0$ and $1$, so $0< x <1$. Next try three different runs, each for these values of $c$: $c=1.5$, $c=3.2$ and $c=3.5$. See what pattern of numbers you might notice with each step.\n\n#### Now you try. Try constructing a for loop that will count and display the count.\n\nType your code here:\n\nSee your results here:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90416116,"math_prob":0.9942894,"size":902,"snap":"2022-40-2023-06","text_gpt3_token_len":240,"char_repetition_ratio":0.116926506,"word_repetition_ratio":0.0,"special_character_ratio":0.2616408,"punctuation_ratio":0.145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99895626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T17:35:56Z\",\"WARC-Record-ID\":\"<urn:uuid:ab9c0f6a-4dd2-40e5-8510-a4221bc3f09a>\",\"Content-Length\":\"16327\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c551d3a4-e57e-4c46-b037-b6772ce83651>\",\"WARC-Concurrent-To\":\"<urn:uuid:edd9f87c-1576-4537-bf3c-79ff6838be6c>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://codebymath.com/index.php/welcome/lesson/for-loops-logistic\",\"WARC-Payload-Digest\":\"sha1:5JPQHJLY2GOTK57ESMAC6GASQY2ATUH2\",\"WARC-Block-Digest\":\"sha1:WA2JVPKLA4DAVZGQZSD5U2PZVUBZJP34\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030331677.90_warc_CC-MAIN-20220924151538-20220924181538-00761.warc.gz\"}"} |
http://actapress.com/Abstract.aspx?PaperId=43265 | [
"## CLOSED-FORM COMPUTATION OF ELECTROMAGNETIC FIELDS IN INDUCTION MOTORS\n\nFrancisco de León and Sujit Purushothaman\n\n### Keywords\n\nInduction motors, electromagnetic fields analysis, Galilean transformation, motor design\n\n### Abstract\n\nA closed-form solution of the electromagnetic field equations for a three-phase squirrel-cage induction motor is presented. The analysis starts from the application of the Galilean transformation to Maxwell equations for moving media at constant speed. The induction motor is modelled as five concentric cylindrical layers representing the different construction components of the motor. By solving the Helmholtz and Laplace equations for conducting and non-conducting layers, we obtain a coupled set of Bessel and Euler equations that are solved analytically. The obtained formulas allow for the efficient calculation of important information for the designer regarding the electromagnetic fields, losses, force and torque. Parametric analyses are shown for illustration of the benefits of the closed form solution. The analytical expressions are validated against finite element simulations. Analytical expressions to compute the parameters of the equivalent circuit from the dimensions of the motor are also provided.\n\nImportant Links:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86534464,"math_prob":0.96458423,"size":1167,"snap":"2019-35-2019-39","text_gpt3_token_len":233,"char_repetition_ratio":0.13757524,"word_repetition_ratio":0.0,"special_character_ratio":0.15424165,"punctuation_ratio":0.08287293,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9505261,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T12:03:16Z\",\"WARC-Record-ID\":\"<urn:uuid:89c202e8-1a6a-4983-a92c-f67a0d12930e>\",\"Content-Length\":\"19829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0278dc9f-26ee-410b-af3c-c8cb5192ba9b>\",\"WARC-Concurrent-To\":\"<urn:uuid:16e1309e-a5c0-4e3f-b4b8-6e42b6b0597a>\",\"WARC-IP-Address\":\"162.223.123.100\",\"WARC-Target-URI\":\"http://actapress.com/Abstract.aspx?PaperId=43265\",\"WARC-Payload-Digest\":\"sha1:FDGMZDXOFBEUY4CB4KTQJ47WEU2W735B\",\"WARC-Block-Digest\":\"sha1:F2YVVHXHTLFTBH5AOTI7V2MYRD23MWTB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315329.55_warc_CC-MAIN-20190820113425-20190820135425-00138.warc.gz\"}"} |
https://calculatorsonline.org/percentage-to-fraction/28 | [
"# 28 percent as a fraction\n\nHere you will see step by step solution to convert 28 percent into fraction. 28 percent as a fraction is 7/25. Please check the explanation that how to write 28% as a fraction.\n\n## Answer: 28% as a fraction is\n\n= 7/25\n\n### How to convert 28 percent to fraction?\n\nTo convert the 28% as a fraction form simply divide the number by 100 and simplify the fraction by finding GCF. If given number is in decimal form then multiply numerator and denominator by 10^n and simplify it as much as possible.\n\n#### How to write 28% as a fraction?\n\nFollow these easy steps to convert 28% into fraction-\n\nGiven number is => 28\n\n• Write down the 28 in a percentage form like this:\n• 28% = 28/100\n• Since, 28 is a whole number, we also need to check to simplify the fraction.\n• Greatest common factor [GCF] of 28 and 100\n• => GCF(28,100) = 4\n\n=> 28 ÷ 4/100 ÷ 4 = 7/25\n\nConclusion: 28% = 7/25\n\nTherefore, the 28 percent as a fraction, final answer is 7/25."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92662096,"math_prob":0.9940394,"size":935,"snap":"2023-40-2023-50","text_gpt3_token_len":269,"char_repetition_ratio":0.18796992,"word_repetition_ratio":0.02247191,"special_character_ratio":0.32834226,"punctuation_ratio":0.0867347,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993538,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T17:35:59Z\",\"WARC-Record-ID\":\"<urn:uuid:15b75380-1530-46e8-8cef-735aa99b60be>\",\"Content-Length\":\"16239\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7054ccce-6ab6-4346-b62b-a60137f2cef1>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea9b935f-54e4-42a5-968a-1705c7ad9db1>\",\"WARC-IP-Address\":\"104.21.85.191\",\"WARC-Target-URI\":\"https://calculatorsonline.org/percentage-to-fraction/28\",\"WARC-Payload-Digest\":\"sha1:ALO4GOLHVZ36UR53KGAII2YQY6ZFZVYB\",\"WARC-Block-Digest\":\"sha1:WSGBDORKXLA5AAUUIAQNQ27GNRWUTNXJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100602.36_warc_CC-MAIN-20231206162528-20231206192528-00451.warc.gz\"}"} |
https://www.ttmath.org/forum/thank_you | [
"Thank you!\n\nI decided I wanted to calculate the probabilty that in a group of randomly shuffled deck of cards, that two decks have the exact same order. No calculator I could find could handle that large of a number (52!!) until I found your online calculator.\n\nJust because I think it's cool (and I would like someone to double check my math), I decided to make a worse case scenario and assume that every single person ever on this planet (estimated to be 100 billion people) shuffled a deck of cards once every second continuously for 100 years, what's the probability that two of those decks are the same? Here's the formula I used:\n\n1-factorial(factorial(52))/(factorial(52)^(3.15*10^20))/(fac torial(factorial(52)-(3.15*10^20)))\n\nWhich, written out looks a little more like:\n\n1-(52!!)/{(52!^3.15e20)*[(52!-3.15e20)!]}"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94452816,"math_prob":0.9226268,"size":818,"snap":"2022-05-2022-21","text_gpt3_token_len":209,"char_repetition_ratio":0.11056511,"word_repetition_ratio":0.0,"special_character_ratio":0.297066,"punctuation_ratio":0.12352941,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99488574,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T14:37:16Z\",\"WARC-Record-ID\":\"<urn:uuid:9dc1cd88-947b-4707-9646-fb08eeaef809>\",\"Content-Length\":\"6957\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f6141e8-de51-4c4b-aa10-56c8e02d6979>\",\"WARC-Concurrent-To\":\"<urn:uuid:e941997d-bff2-4c8e-9df7-2622996221b2>\",\"WARC-IP-Address\":\"51.83.201.84\",\"WARC-Target-URI\":\"https://www.ttmath.org/forum/thank_you\",\"WARC-Payload-Digest\":\"sha1:VEEGFQ7Z2IM4JPBYVLMGUF7OHZE264TL\",\"WARC-Block-Digest\":\"sha1:PM3HLK3C2ULPEJDVSZ6M4MISGNYBNFQS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301341.12_warc_CC-MAIN-20220119125003-20220119155003-00674.warc.gz\"}"} |
http://gre.kmf.com/question/all/549?keyword=&page=5 | [
"#### 题目列表\n\nThe author alludes to functional extinction primarily in order to imply that\nThe width of rectangle A is 9, and its area is 90.\n\nThe length of rectangle B is 12, and its area is 120.\n\nQuantity A:The length of rectangle A\n\nQuantity B:The width of rectangle B\n$\\frac{6}{1.8}$ = $\\frac{z}{0.9}$\n\nQuantity A:z\n\nQuantity B:3.8\n\nQuantity A:x\n\nQuantity B:4",
null,
"-1 ≤ x ≤ 1 and -1 ≤ y ≤ 1\n\nQuantity A:$(x+y)^{2}$\n\nQuantity B:$(xy)^{2}$\nThe area of a certain floor is 150 square feet. (1 yard=3 feet)\n\nQuantity A:The area of the floor, in square yards\n\nQuantity B:30\nList K consists of 20 consecutive odd integers, list L consists of 20 consecutive even integers, and list M consists of 20 consecutive multiples of 3. The least integer in L is 9 greater than the greatest integer in K, and the greatest integer in L is 10 greater than the least integer in M\n\nQuantity A:The range of the integers in K and L combined\n\nQuantity B:The range of the integers in L and M combined\n\nData set D consists of 35 values, all of which are integers. The frequency distribution of the values in D is shown in the histogram, where each interval shown contains values that are greater than or equal to the left endpoint but less than the right endpoint.\n\nQuantity A:The average (arithmetic mean) of the values in D\n\nQuantity B:The median of the values in D",
null,
"Jayden, Kenny and Laina are paid hourly wages at their jobs. Jaydens hourly wage is between $8.00 and$9.00, Kennys hourly wage is $5.00 less than 2 times Jaydens hourly wage, and Lainas hourly wage is$1.00 more than Jaydens hourly wage. Which of the following shows Jayden, Kenny, and Laina listed in order according to their hourly wages, from least to greatest?\nAn isosceles triangle has sides of length x, 2x and 2x. If the area of the triangle is $25\\sqr{15}$, what is the value of x?\nIf a, b and c are integers such that 0 < a < b < c < 2a, what is the greatest common factor of $84^{a}$, $126^{b}$, and $98^{c}$?\nLet n be an integer greater than 30. When n is divided by 12, the remainder is 11. What is the remainder when (6n+1) is divided by 9?\n$a_1$, $a_2$, $a_3$,......,$a_{150}$\n\nThe $n_{th}$ term if the sequence shown is defined for each integer n from 1 to 150 as follows. If n is odd, then $a_n$=$\\frac{(n+1)}{2}$, and if n is even, then $a_{n}$=$(a_{n-1})^{2}$. How many integers appear in the sequence twice?\nTemperature C in degree Celsius and the corresponding temperature F in degrees Fahrenheit are related by the equation F=$\\frac{9}{5}$C+32. At a certain time at a weather station, the temperature in degrees Fahrenheit was equal to $\\frac{1}{9}$ of the temperature in degrees Celsius. What was the temperature in degrees Fahrenheit?\nThe total number of vehicles sold in Region X by companies other than A, B, C and D in 2009 was approximately how much less than that in 2006?\nFor which of the last five years shown was the difference between the annual number of vehicles sold by Company C and the annual number of vehicles sold by Company D least?\nThe increase in the number of vehicles sold in Region X from 2002 to 2003 was the same was that from 2003 to 2004. The decrease in the number of vehicles sold in Region X from 2009 to 2010 was the same as that from 2008 to 2009. For the years from 2002 to 2010, the median number of vehicles sold annually was approximately how much greater than the average (arithmetic mean) number of vehicles sold annually?\n\nA pump delivered water to fill an empty swimming pool. The pump delivered the water at a constant rate of 450 liters per minute until the pool was $\\frac{1}{2}$ full. Then the pump became partially clogged and delivered the water at a slower constant rate until the pool was full. For the whole time during which the pump delivered water to fill the empty pool, its average rate was 360 liters per minute. What was the pumps slower constant rate, in liters per minute?"
] | [
null,
"http://img.kmf.com/kaomanfen/img/gre/18df3d073cc26f02273a26abd031d29a.png",
null,
"http://img.kmf.com/kaomanfen/img/gre/5b86adf032ba488e67d0032e7193fcaf.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9361902,"math_prob":0.9959429,"size":4720,"snap":"2021-43-2021-49","text_gpt3_token_len":1265,"char_repetition_ratio":0.13252756,"word_repetition_ratio":0.05800464,"special_character_ratio":0.28728813,"punctuation_ratio":0.100196466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99947673,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T15:58:33Z\",\"WARC-Record-ID\":\"<urn:uuid:6bd49e9f-d1fb-485b-a7fe-96442b6b4568>\",\"Content-Length\":\"58773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1d4eeefa-d2eb-4de9-91d9-510852b47f5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:440d992c-2c41-4c2e-bfe0-b06c7b32eaa0>\",\"WARC-IP-Address\":\"47.89.156.56\",\"WARC-Target-URI\":\"http://gre.kmf.com/question/all/549?keyword=&page=5\",\"WARC-Payload-Digest\":\"sha1:QSNMY6NE2RAQTGBPOUZIQLH5UO5KIMQS\",\"WARC-Block-Digest\":\"sha1:WFRUBZZK6QB2RNNCPOXN3EKITOXG5OEW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587719.64_warc_CC-MAIN-20211025154225-20211025184225-00490.warc.gz\"}"} |
https://www.tuhs.org/cgi-bin/utree.pl?file=4.4BSD/usr/src/contrib/emacs-18.57/lisp/sort.el | [
"# 4.4BSD/usr/src/contrib/emacs-18.57/lisp/sort.el\n\nCompare this file to the similar file:\nShow the results in this format:\n\n```;; Commands to sort text in an Emacs buffer.\n;; Copyright (C) 1986, 1987 Free Software Foundation, Inc.\n\n;; This file is part of GNU Emacs.\n\n;; GNU Emacs is free software; you can redistribute it and/or modify\n;; the Free Software Foundation; either version 1, or (at your option)\n;; any later version.\n\n;; GNU Emacs is distributed in the hope that it will be useful,\n;; but WITHOUT ANY WARRANTY; without even the implied warranty of\n;; MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n;; GNU General Public License for more details.\n\n;; You should have received a copy of the GNU General Public License\n;; along with GNU Emacs; see the file COPYING. If not, write to\n;; the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA.\n\n(provide 'sort)\n\n;; Original version of most of this contributed by Howie Kaye\n\n(defun sort-subr (reverse nextrecfun endrecfun &optional startkeyfun endkeyfun)\n\"General text sorting routine to divide buffer into records and sort them.\nArguments are REVERSE NEXTRECFUN ENDRECFUN &optional STARTKEYFUN ENDKEYFUN.\n\nWe consider this portion of the buffer to be divided into disjoint pieces\ncalled sort records. A portion of each sort record (perhaps all of it)\nis designated as the sort key. The records are rearranged in the buffer\nin order by their sort keys. The records may or may not be contiguous.\n\nUsually the records are rearranged in order of ascending sort key.\nIf REVERSE is non-nil, they are rearranged in order of descending sort key.\n\nThe next four arguments are functions to be called to move point\nacross a sort record. They will be called many times from within sort-subr.\n\nNEXTRECFUN is called with point at the end of the previous record.\nIt moves point to the start of the next record.\nThe first record is assumed to start at the position of point when sort-subr\nis called.\n\nENDRECFUN is is called with point within the record.\nIt should move point to the end of the record.\n\nSTARTKEYFUN may moves from the start of the record to the start of the key.\nIt may return either return a non-nil value to be used as the key, or\nelse the key will be the substring between the values of point after\nSTARTKEYFUN and ENDKEYFUN are called. If STARTKEYFUN is nil, the key\nstarts at the beginning of the record.\n\nENDKEYFUN moves from the start of the sort key to the end of the sort key.\nENDRECFUN may be nil if STARTKEYFUN returns a value or if it would be the\nsame as ENDRECFUN.\"\n(save-excursion\n(message \"Finding sort keys...\")\n(let* ((sort-lists (sort-build-lists nextrecfun endrecfun\nstartkeyfun endkeyfun))\n(old (reverse sort-lists)))\n(if (null sort-lists)\n()\n(or reverse (setq sort-lists (nreverse sort-lists)))\n(message \"Sorting records...\")\n(setq sort-lists\n(if (fboundp 'sortcar)\n(sortcar sort-lists\n(cond ((numberp (car (car sort-lists)))\n'<)\n((consp (car (car sort-lists)))\n'buffer-substring-lessp)\n(t\n'string<)))\n(sort sort-lists\n(cond ((numberp (car (car sort-lists)))\n(function\n(lambda (a b)\n(< (car a) (car b)))))\n((consp (car (car sort-lists)))\n(function\n(lambda (a b)\n(buffer-substring-lessp (car a) (car b)))))\n(t\n(function\n(lambda (a b)\n(string< (car a) (car b)))))))))\n(if reverse (setq sort-lists (nreverse sort-lists)))\n(message \"Reordering buffer...\")\n(sort-reorder-buffer sort-lists old)))\n(message \"Reordering buffer... Done\")))\n\n;; Parse buffer into records using the arguments as Lisp expressions;\n;; return a list of records. Each record looks like (KEY STARTPOS ENDPOS)\n;; where KEY is the sort key (a number or string),\n;; and STARTPOS and ENDPOS are the bounds of this record in the buffer.\n\n;; The records appear in the list lastmost first!\n\n(defun sort-build-lists (nextrecfun endrecfun startkeyfun endkeyfun)\n(let ((sort-lists ())\n(start-rec nil)\ndone key)\n;; Loop over sort records.\n;(goto-char (point-min)) -- it is the caller's responsibility to\n;arrange this if necessary\n(while (not (eobp))\n(setq start-rec (point))\t\t;save record start\n(setq done nil)\n;; Get key value, or move to start of key.\n(setq key (catch 'key\n(or (and startkeyfun (funcall startkeyfun))\n;; If key was not returned as value,\n;; move to end of key and get key from the buffer.\n(let ((start (point)))\n(funcall (or endkeyfun\n(prog1 endrecfun (setq done t))))\n(if (fboundp 'buffer-substring-lessp)\n(cons start (point))\n(buffer-substring start (point)))))))\n;; Move to end of this record (start of next one, or end of buffer).\n(cond ((prog1 done (setq done nil)))\n(endrecfun (funcall endrecfun))\n(nextrecfun (funcall nextrecfun) (setq done t)))\n(if key (setq sort-lists (cons\n;; consing optimization in case in which key\n;; is same as record.\n(if (and (consp key)\n(equal (car key) start-rec)\n(equal (cdr key) (point)))\n(cons key key)\n(list key start-rec (point)))\nsort-lists)))\n(and (not done) nextrecfun (funcall nextrecfun)))\nsort-lists))\n\n(defun sort-reorder-buffer (sort-lists old)\n(let ((inhibit-quit t)\n(last (point-min))\n(min (point-min)) (max (point-max)))\n(while sort-lists\n(goto-char (point-max))\n(insert-buffer-substring (current-buffer)\nlast\n(nth 1 (car old)))\n(goto-char (point-max))\n(insert-buffer-substring (current-buffer)\n(nth 1 (car sort-lists))\n(nth 2 (car sort-lists)))\n(setq last (nth 2 (car old))\nsort-lists (cdr sort-lists)\nold (cdr old)))\n(goto-char (point-max))\n(insert-buffer-substring (current-buffer)\nlast\nmax)\n(delete-region min max)))\t;get rid of old version\n\n(defun sort-lines (reverse beg end)\n\"Sort lines in region alphabetically; argument means descending order.\nCalled from a program, there are three arguments:\nREVERSE (non-nil means reverse order), BEG and END (region to sort).\"\n(interactive \"P\\nr\")\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr reverse 'forward-line 'end-of-line)))\n\n(defun sort-paragraphs (reverse beg end)\n\"Sort paragraphs in region alphabetically; argument means descending order.\nCalled from a program, there are three arguments:\nREVERSE (non-nil means reverse order), BEG and END (region to sort).\"\n(interactive \"P\\nr\")\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr reverse\n(function (lambda () (skip-chars-forward \"\\n \\t\\f\")))\n'forward-paragraph)))\n\n(defun sort-pages (reverse beg end)\n\"Sort pages in region alphabetically; argument means descending order.\nCalled from a program, there are three arguments:\nREVERSE (non-nil means reverse order), BEG and END (region to sort).\"\n(interactive \"P\\nr\")\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(sort-subr reverse\n(function (lambda () (skip-chars-forward \"\\n\")))\n'forward-page)))\n\n(defvar sort-fields-syntax-table nil)\n(if sort-fields-syntax-table nil\n(let ((table (make-syntax-table))\n(i 0))\n(while (< i 256)\n(modify-syntax-entry i \"w\" table)\n(setq i (1+ i)))\n(modify-syntax-entry ?\\ \" \" table)\n(modify-syntax-entry ?\\t \" \" table)\n(modify-syntax-entry ?\\n \" \" table)\n(setq sort-fields-syntax-table table)))\n\n(defun sort-numeric-fields (field beg end)\n\"Sort lines in region numerically by the ARGth field of each line.\nFields are separated by whitespace and numbered from 1 up.\nSpecified field must contain a number in each line of the region.\nWith a negative arg, sorts by the -ARG'th field, in reverse order.\nCalled from a program, there are three arguments:\nFIELD, BEG and END. BEG and END specify region to sort.\"\n(interactive \"p\\nr\")\n(sort-fields-1 field beg end\n(function (lambda ()\n(sort-skip-fields (1- field))\n(string-to-int\n(buffer-substring\n(point)\n(save-excursion\n(skip-chars-forward \"-0-9\")\n(point))))))\nnil))\n\n(defun sort-fields (field beg end)\n\"Sort lines in region lexicographically by the ARGth field of each line.\nFields are separated by whitespace and numbered from 1 up.\nWith a negative arg, sorts by the -ARG'th field, in reverse order.\nCalled from a program, there are three arguments:\nFIELD, BEG and END. BEG and END specify region to sort.\"\n(interactive \"p\\nr\")\n(sort-fields-1 field beg end\n(function (lambda ()\n(sort-skip-fields (1- field))\nnil))\n(function (lambda () (skip-chars-forward \"^ \\t\\n\")))))\n\n(defun sort-fields-1 (field beg end startkeyfun endkeyfun)\n(let ((reverse (< field 0))\n(tbl (syntax-table)))\n(setq field (max 1 field (- field)))\n(unwind-protect\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(set-syntax-table sort-fields-syntax-table)\n(sort-subr reverse\n'forward-line 'end-of-line\nstartkeyfun endkeyfun))\n(set-syntax-table tbl))))\n\n(defun sort-skip-fields (n)\n(let ((eol (save-excursion (end-of-line 1) (point))))\n(forward-word n)\n(if (> (point) eol)\n(error \"Line has too few fields: %s\"\n(buffer-substring (save-excursion\n(beginning-of-line) (point))\neol)))\n(skip-chars-forward \" \\t\")))\n\n(defun sort-regexp-fields (reverse record-regexp key-regexp beg end)\n\"Sort the region lexicographically as specifed by RECORD-REGEXP and KEY.\nRECORD-REGEXP specifies the textual units which should be sorted.\nFor example, to sort lines RECORD-REGEXP would be \\\"^.*\\$\\\"\nKEY specifies the part of each record (ie each match for RECORD-REGEXP)\nis to be used for sorting.\nIf it is \\\"\\\\digit\\\" then the digit'th \\\"\\\\(...\\\\)\\\" match field from\nRECORD-REGEXP is used.\nIf it is \\\"\\\\&\\\" then the whole record is used.\nOtherwise, it is a regular-expression for which to search within the record.\nIf a match for KEY is not found within a record then that record is ignored.\n\nWith a negative prefix arg sorts in reverse order.\n\nFor example: to sort lines in the region by the first word on each line\nstarting with the letter \\\"f\\\",\nRECORD-REGEXP would be \\\"^.*\\$\\\" and KEY \\\"\\\\<f\\\\w*\\\\>\\\"\"\n(interactive \"P\\nsRegexp specifying records to sort:\nsRegexp specifying key within record: \\nr\")\n(cond ((or (equal key-regexp \"\") (equal key-regexp \"\\\\&\"))\n(setq key-regexp 0))\n((string-match \"\\\\`\\\\\\\\[1-9]\\\\'\" key-regexp)\n(setq key-regexp (- (aref key-regexp 1) ?0))))\n(save-restriction\n(narrow-to-region beg end)\n(goto-char (point-min))\n(let (sort-regexp-record-end) ;isn't dynamic scoping wonderful?\n(re-search-forward record-regexp)\n(setq sort-regexp-record-end (point))\n(goto-char (match-beginning 0))\n(sort-subr reverse\n(function (lambda ()\n(and (re-search-forward record-regexp nil 'move)\n(setq sort-regexp-record-end (match-end 0))\n(goto-char (match-beginning 0)))))\n(function (lambda ()\n(goto-char sort-regexp-record-end)))\n(function (lambda ()\n(let ((n 0))\n(cond ((numberp key-regexp)\n(setq n key-regexp))\n((re-search-forward\nkey-regexp sort-regexp-record-end t)\n(setq n 0))\n(t (throw 'key nil)))\n(condition-case ()\n(if (fboundp 'buffer-substring-lessp)\n(cons (match-beginning n)\n(match-end n))\n(buffer-substring (match-beginning n)\n(match-end n)))\n;; if there was no such register\n(error (throw 'key nil))))))))))\n\n(defun sort-columns (reverse &optional beg end)\n\"Sort lines in region alphabetically by a certain range of columns.\nFor the purpose of this command, the region includes\nthe entire line that point is in and the entire line the mark is in.\nThe column positions of point and mark bound the range of columns to sort on.\nA prefix argument means sort into reverse order.\n\nNote that sort-columns uses the sort utility program and therefore\ncannot work on text containing TAB characters. Use M-x untabify\nto convert tabs to spaces before sorting.\"\n(interactive \"P\\nr\")\n(save-excursion\n(let (beg1 end1 col-beg1 col-end1 col-start col-end)\n(goto-char (min beg end))\n(setq col-beg1 (current-column))\n(beginning-of-line)\n(setq beg1 (point))\n(goto-char (max beg end))\n(setq col-end1 (current-column))\n(forward-line)\n(setq end1 (point))\n(setq col-start (min col-beg1 col-end1))\n(setq col-end (max col-beg1 col-end1))\n(if (search-backward \"\\t\" beg1 t)\n(error \"sort-columns does not work with tabs. Use M-x untabify.\"))\n(call-process-region beg1 end1 \"sort\" t t nil\n(if reverse \"-rt\\n\" \"-t\\n\")\n(concat \"+0.\" col-start)\n(concat \"-0.\" col-end)))))\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7302118,"math_prob":0.90199906,"size":11792,"snap":"2022-40-2023-06","text_gpt3_token_len":3112,"char_repetition_ratio":0.13522226,"word_repetition_ratio":0.15097691,"special_character_ratio":0.2734905,"punctuation_ratio":0.09660698,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9780718,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T08:06:15Z\",\"WARC-Record-ID\":\"<urn:uuid:3875a453-fc10-4786-b199-e7f8a5227311>\",\"Content-Length\":\"21409\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97997872-571e-4ef5-8adb-aac41e5725d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:8675d87d-1d19-41fb-9d58-6a8a41661157>\",\"WARC-IP-Address\":\"50.116.15.146\",\"WARC-Target-URI\":\"https://www.tuhs.org/cgi-bin/utree.pl?file=4.4BSD/usr/src/contrib/emacs-18.57/lisp/sort.el\",\"WARC-Payload-Digest\":\"sha1:6H6JBXBX332XXJXH3WFOGCNQQKT5SBWG\",\"WARC-Block-Digest\":\"sha1:P7BXPCLGRAI642R2XFAVOO47BFJDP4WL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499804.60_warc_CC-MAIN-20230130070411-20230130100411-00399.warc.gz\"}"} |
https://economics.stackexchange.com/questions/45377/a-revisit-to-simple-did-and-generalised-did/45380#45380 | [
"# A revisit to simple DID and Generalised DID\n\nAfter a couple of questions being asked, I am curious about the inclusion of Post and Treat variable in a simple DID (two groups two-time period). We mainly know there are mainly two types of DID including simple and generalised.\n\nFrom the genralized DID, from this discussion, we can see that Post and Treat are not included into the regression equation because they will be swallowed by group and period fixed effects (normally we call firm and year fixed effects).\n\nI revisit the equation of simple DID in this link:\n\n$$Y_{it} = \\beta_0 + \\beta_1*P_t + \\beta_2*T_i + \\beta_3(P_t*T_i) + u_{it}$$\n\nWhile $$P_t$$ and $$T_i$$ are Post and Treat variables accordingly. My concern is\n\n(1) whether we should not run the group and period fixed effect for simple DID because they will swallow the variables $$P_t$$ and $$T_i$$ as in the generalized case?\n\n(2) And because not running unit and period fixed effect, now $$u_{it}$$ would be a mess. I mean now $$u_{it}$$ will include time-variant, time-invariant variables and variables vary across both over time and firms, for example, from this discussion:\n\n$$u_{i,t} = \\delta_i + \\gamma_t + \\chi_{i,t}$$\n\nIn generalised DID, because we control for group and period fixed effect, so we only need to add independent variables to satisfy $$\\mathbb{E}(\\chi x) = 0$$ Is it correct?\n\nIf it is the case, because it seems that we cannot control firm and year fixed effect in simple DID. Therefore, we need to add more variables to try to reach $$\\mathbb{E}(\\delta x) =0$$ and $$\\mathbb{E}(\\gamma x)=0$$ apart from $$\\mathbb{E}(\\chi x) = 0$$ I am wondering that apart from adding too many independent variables, which can cause multicollinearity problems, is there any other solution to solve the messy error term in simple DID?\n\n(1) whether we should not run the group and period fixed effect for simple DID because they will swallow the variables $$P_t$$ and $$T_i$$ as in the generalized case?\nIn a simple 2 period 2 group regression, the $$P_t$$ and $$T_i$$ variables are capturing the time and group fixed effect. In fact, the regression with fixed effects is formally identical to the simple DiD specification.\nTo see this assume that there are two time periods $$t = 0,1$$ and two groups $$g = 0,1$$ then the estimation with fixed effects takes the form: $$y_{i,t,g} = \\alpha_0 D_i(t=0) + \\alpha_1 D_i(t = 1) + \\alpha_2 D_i(g = 0) + \\alpha_3 D_i(g = 1) + \\alpha_4 x_{i,t,g} + \\varepsilon_{i,t,g}.$$ Here $$D_i(t = 0)$$ is a dummy that equals one if $$t = 0$$, etc.\nIf there are only two time periods, we have that $$D_i(t = 0) = 1 - D_i(t = 1)$$ and $$D_i(g = 0) = 1 - D_i(g = 1)$$. Using this substitution gives: $$y_{i,t,g} = \\underbrace{(\\alpha_0 + \\alpha_2)}_{\\beta_0} + \\underbrace{(\\alpha_1 - \\alpha_0)}_{\\beta_1} D_i(t = 1) + \\underbrace{(\\alpha_3 - \\alpha_2)}_{\\beta_2} D_i(g = 1) + \\alpha_4 x_{i,t,g} + \\varepsilon_{i,t,g}.$$ If we denote $$T_i = D_i(t = 1)$$, $$P_i = D_i(g = 1)$$ and if we consider the case where $$x_{i,t,g} = T_i P_i$$ we obtain: $$y_{i,t,g} = \\beta_0 + \\beta_1 T_i + \\beta_2 P_i + \\alpha_4 T_i P_i + \\varepsilon_{i,t,g}.$$ So $$\\alpha_4$$ is exactly the DiD estimator for a model with time and group fixed effects; The $$\\beta$$-parameters are simple linear combinations of the $$\\alpha$$-parameters."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8987941,"math_prob":0.9999783,"size":1746,"snap":"2021-43-2021-49","text_gpt3_token_len":445,"char_repetition_ratio":0.12571757,"word_repetition_ratio":0.0,"special_character_ratio":0.2611684,"punctuation_ratio":0.07854985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T00:33:11Z\",\"WARC-Record-ID\":\"<urn:uuid:2d40e171-44e6-41ce-acd8-5db1a48f33dc>\",\"Content-Length\":\"167948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3654f939-0629-424e-ba2f-50ecd5b0ff56>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2d2bb3f-11f2-46a5-ac8d-5f1c81608291>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://economics.stackexchange.com/questions/45377/a-revisit-to-simple-did-and-generalised-did/45380#45380\",\"WARC-Payload-Digest\":\"sha1:CFGSEOFXTOARMRPZKGRFWARPQGVEPXPC\",\"WARC-Block-Digest\":\"sha1:7FOFDJIWZVLK4HQR2KUIMWOWBMZUUO4U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585186.33_warc_CC-MAIN-20211018000838-20211018030838-00647.warc.gz\"}"} |
https://discourse.mc-stan.org/t/heteroscedastic-model-with-variance-as-function-of-mean-magnitude/27086 | [
"# Heteroscedastic model with variance as function of mean magnitude\n\nThis isn’t something I’m specifically trying to do with Stan but I’m trying to formulate it as a Bayesian model, which may then later need MCMC.\n\nI have a model where I have observations d_{i,j} where d_i is a 3 dimensional vector. I have a forwards model which can produce modeled data m_{i,j}. Unfortunately, there are multiple sources of noise that produce the observations, one which is likely heteroscedastic and scales with |m_i| and another which is likely constant. I think a simple model I could use to describe this would be something like:\n\nd_{i,j}\\sim normal(m_{i,j},(\\omega|m_i|)^2+\\delta^2)\n\nwhere \\delta and \\omega are unknown.\n\nAlternatively, it might be simpler to model the variance as a single parameter \\sigma_i^2 where p(\\sigma_i)=f(|d_i|) and f is a distribution that has a mean with the properties defined above (tends to a constant at low magnitude, order 1 at high magnitude).\n\nBecause the variance is effectively a nuisance parameter, it would also be nice to have some simple/complementary form to the prior so that the variance can be integrated out in the posterior, reducing the number of parameters that have to be computed. I haven’t been able to find much literature on this particular kind of problem which does this- although it’s a frequently used trick with homoscedastic problems."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9445336,"math_prob":0.9768425,"size":1313,"snap":"2022-27-2022-33","text_gpt3_token_len":297,"char_repetition_ratio":0.10084034,"word_repetition_ratio":0.0,"special_character_ratio":0.22086824,"punctuation_ratio":0.08139535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99564475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T20:41:15Z\",\"WARC-Record-ID\":\"<urn:uuid:66f78a27-e041-4b0f-b16a-2fe0d726cf05>\",\"Content-Length\":\"18260\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e86bb0bf-774e-4782-8e8b-ff9c4a1d93c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b78b9e0-f5ca-44fc-aabc-44369123e4cb>\",\"WARC-IP-Address\":\"64.62.250.111\",\"WARC-Target-URI\":\"https://discourse.mc-stan.org/t/heteroscedastic-model-with-variance-as-function-of-mean-magnitude/27086\",\"WARC-Payload-Digest\":\"sha1:DF46OWIWSHI52L4ELXO7VDC7SLZ7IBMO\",\"WARC-Block-Digest\":\"sha1:KIAP3Y63KB36LY3W52LBOTFWFGTU3OH4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103271864.14_warc_CC-MAIN-20220626192142-20220626222142-00310.warc.gz\"}"} |
https://mapdldocs.pyansys.com/mapdl_commands/solution/_autosummary/ansys.mapdl.core.Mapdl.msolve.html | [
"# msolve¶\n\nMapdl.msolve(numslv='', nrmtol='', nrmchkinc='', **kwargs)\n\nStarts multiple solutions for random acoustics analysis with diffuse\n\nAPDL Command: MSOLVE sound field.\n\nParameters\nnumslv\n\nNumber of multiple solutions (load steps) corresponding to the number of samplings. Default = 1.\n\nNotes\n\nThe MSOLVE command starts multiple solutions (load steps) for random acoustics analysis with multiple samplings.\n\nThe process is controlled by the norm convergence tolerance NRMTOL or the number of multiple solutions NUMSLV (if the solution steps reach the defined number).\n\nThe program checks the norm convergence by comparing two averaged sets of radiated sound powers with the interval NRMCHKINC over the frequency range. For example, if NRMCHKINC = 5, the averaged values from 5 solutions are compared with the averaged values from 10 solutions, then the averaged values from 10 solutions are compared with the averaged values from 15 solutions, and so on.\n\nThe incident diffuse sound field is defined via the DFSWAVE command.\n\nThe average result of multiple solutions with different samplings is calculated via the PLST command."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81106347,"math_prob":0.985212,"size":1127,"snap":"2021-43-2021-49","text_gpt3_token_len":246,"char_repetition_ratio":0.16384684,"word_repetition_ratio":0.07272727,"special_character_ratio":0.1943212,"punctuation_ratio":0.09625668,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789861,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T01:52:20Z\",\"WARC-Record-ID\":\"<urn:uuid:320a233c-2421-4621-a9a8-8e4d4ede8847>\",\"Content-Length\":\"31934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9796fb1-2ffb-4585-9368-4b53e2ec6670>\",\"WARC-Concurrent-To\":\"<urn:uuid:a60b559e-3f56-42f3-ab4b-ce1975f8cad8>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://mapdldocs.pyansys.com/mapdl_commands/solution/_autosummary/ansys.mapdl.core.Mapdl.msolve.html\",\"WARC-Payload-Digest\":\"sha1:LLUVABK4LLEMZMSHQEXRJHJB2RMNN4GV\",\"WARC-Block-Digest\":\"sha1:XFEUJ64FTBPLO6V5ZN5PNXSH2VO7WAS5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588246.79_warc_CC-MAIN-20211028003812-20211028033812-00223.warc.gz\"}"} |
http://www.thebox.myzen.co.uk/Workshop/Motors_3.html | [
"Motors 3\n\nHow a four pole stepping motor works and how to drive it from your Arduino. A stepping motor works not with continuous motion, but in small precise increments.",
null,
"",
null,
"Introduction\n\nSmall DC motors as covered in Motors 1 & 2 are all very well at making things move but there is little precision about it. The only control you have over the movement they produce is to control the on and off time. However, there is another way of getting a motor to move, not continuously but in small precise increments, these are know as stepping motors. Although this is not the only way of generating movement on demand, it does have many advantages.\n\nYou probably already own a couple. There is one in your disc drive and two in most printers, in fact scrap computer equipment is an excellent way of getting hold of them.\n\nBasics\n\nA normal motor starts to rotate when you apply a voltage to it, and slows down and stops when that voltage is removed. If you want to control its speed you usually have to resort to gears.\n\nAlthough there are various electronic methods of motor speed control these are usually used to make small adjustments, while it is gears that get the speed in the right order of magnitude.\n\nThe problem is that for a conventional motor to run efficiently it has to run at around 300 rpm.\n\nThe amount of turning power, or torque as it is known, is the way a motor's power is measured. With gears, as the speed is reduced the torque is increased - a very happy state of affairs.\n\nHowever, most electronic motor control reduces the speed as well as the torque.\n\nWhen it comes to controlling a conventional motor with a computer the motor takes some time to get up to its final speed. So it is difficult for the computer to \"know\" how many revolutions it has turned through. It is normal therefore to include some form of feedback arrangement so the computer can sense the motor's position. Such and arrangement of motor, feedback and control is known as a servo motor and is widely used in model aircraft but is not covered here. However, a servo feedback system is not needed with stepping motors, which is why they are such a natural for use with processors like the Arduino.\n\nA stepping motors as its name implies, works not by continuously rotating but by moving in a series of small steps. Each time the motor receives a pulse it moves through a fixed angle. The size of this angle depends mainly on the design of the motor. There are two basic types of stepping motor, one with four coils, known as four pole or four phase motors and the other with two coils, known as two phase or wave motors. This page looks at four pole motors, the Motors 4 page looks at two phase motors. What ever the type the motors are usually classified by how many steps are taken to complete one revolution - these can range from 4 to 200 steps per revolution. Another way of expressing this is to quote the size of angle of each step.\n\nSpecifics\n\nLet's see what makes a stepping motor tick. If you have a coil and pass a current through it you will generate a magnetic field. This will behave just like any other magnet and will attract other magnetic material to it. Suppose we have a permanent magnet suspended above four coils. If we pass current through two of the coils the magnet will be attracted towards both and will settle somewhere between the two. This is shown in here in figure 1a:-",
null,
"We have an arrangement, which will l be explained later, to prevent the magnet getting too close to the coils. Now supposing we remove the current from coil 1 and turn on coil 3, as l in Figure Ib.\n\nThe magnet will move towards the two coils that are exerting a magnetic force and will come to rest between coils 2 and 3, as in Figure Ic. We can repeat the process with coil 2 going off and coil 4 coming on, as shown in Figures Id and Ie.\n\nIf we look what has happened to our magnet, we will see that it has moved in a straight line in response to our switching currents in the coils. This is the principle of the stepping motor, only instead of movement in a straight line we have a circular movement. This is obtained by having several coils, each wired as one of four circuits, distributed in a circle. We call this the stator of a motor because it does not move. In the centre of the circle we have a drum bearing a number of small magnets. The drum is known as the rotor - it's the part that rotates. The magnets are not actually separate magnets but are bumps or poles on a large central magnet. As the rotor is on bearings and the poles are distributed evenly around it, the poles can never come in contact with the coils. Figure II shows a small section of a stepping motor.",
null,
"Only coil 1 is shown wired up, for clarity, but you can see that every fourth coil is connected together. You can also see that the number of coils in the stator is four times the number of poles on the rotor, and this number determines the angle of each step. A stepping motor can therefore be precisely controlled by switching the currents in each of its four coils. So a four phase stepping motor has five wires coming out of it, one for each of the coils and one as a common connection.\n\nThe coils take far more current than can be produced by the Arduino. This means we have to build something to convert our logic voltage output to the currents needed.\n\nThe simplest way to do this is to use four VMOS power FETs, most N-channel FETs are suitable. Figure III shows the general arrangement with the VN10K FET.",
null,
"You can see from this that the stepping motor usually requires more than the normal 5 volts, and this is applied to the motor through the common coil connector. Most motors will operate with between 12 and 24 volts, although some can use 120 or even 240 volts. The FETs shown will happily switch voltages of up to 60 volts. Each of the FETs is controlled by one bit of the Arduino, these are labelled 0 to 3 but in most cases you want to avoid using the Arduino pins 0 & 1 as these are used for the serial communications. However, any of the other output pins can be used, by putting a logic l on the output, using digitalWrite(pin, HIGH); we can switch any coil on. By outputting the correct sequence of logic levels, that is highs and lows we can make the motor rotate.\n\nGetting it to move\n\nIf we refer back to Figure I and write a logic 1 for each coil that is on and a logic zero for each coil that is off, we get the sequence of numbers to present to our motor. If each coil represents a binary bit then a way of describing the sequence of on and off can be written as a sequence of numbers 3, 6, 12, 9. That is:-\n\n3 is made up from:- Coil 4 = off or zero, Coil 3 = off or zero, Coil 2 = on or one, Coil 1 = on or one\n\nSimilarly:-\n\n6 is made up from:- Coil 4 = off or zero, Coil 3 = on or one, Coil 2 = on or one, Coil 1 = off or zero\n\n12 is made up from:- Coil 4 = on or one, Coil 3 = on or one, Coil 2 = off or zero, Coil 1 = off or zero\n\n9 is made up from:- Coil 4 = on or one, Coil 3 = off or zero, Coil 2 = off or zero, Coil 1 = on or one\n\nNote the last coil pattern is not shown in Figure I, but it is needed to wrap round smoothly to the start of the sequence again. Before we see how this can be generated by the Arduino let's look at another possible sequence of pulses that we can use to drive our motor. Consider Figure IVa.",
null,
"This shows the motor in the same position as Figure Ia.\n\nNow if we leave only one coil on, the magnet (pole) will go directly over it. This is shown in Figure IVb.\n\nIf we now switch on coil 3, the magnet will move between the two (see Figure IVc).\n\nIf we extend this sequence we will find that it takes twice as many steps as the first sequence. This results in the motor moving through only half the angle for each step - known as half stepping the motor.\n\nIn a similar way we defined the full stepping sequence, this half stepping sequence can be defined as:- 1, 3, 2, 6, 4, 12, 8, 9\n\nThe degree of precision is greater, but as you might expect, there is a price to pay, the motor's torque is reduced. As there is one coil on for half the time, you are putting less power into the motor. Therefore you get less power out.\n\nWith the full step sequence there are two coils on all the time. However in some circumstances it is a useful trick to know.\n\nAny software must present the correct sequence to the motor’s coils. The easiest way to do this is to hold the sequence in an array and set up a counter to keep track of the next element in the array to be outputted.\n\nIf you want to see how to produce the full stepping sequence then you can get an example of an Arduino sketch to do this here:- Stepping.pde. Can you extend this to give a half step sequence? Note that to reverse the direction of the motor you simple reverse the sequence of the coils switching on and off.\n\nSpeed\n\nNote top speed is something that is very variable, depending on the motor and load. We have seen that torque is the pulling power of a motor. In a stepping motor the torque is inversely proportional to speed. That means the slower it, goes the more power it has. In fact it has maximum torque when it is stopped. You can test this by trying to force the motor to turn by hand. You will find this harder to do when it is stopped but still has two coils on than when it is running fast. If we try to make it go too fast the motor will not have enough torque to turn itself, let alone a load, and so will stall. When you approach stalling speed you will notice the motor start to miss steps and make a stuttering sound. This maximum speed of the motor from a standing start is called the \"pull-in\" speed. If the motor is running it can be accelerated to a speed faster than the pull-in speed. The actual speed achievable depends on the power of the motor and the load it is trying to turn. So if you want to run stepping motors at full tilt you have to start them relatively slowly. One way to achieve more torque, and thus a greater speed, is to increase the current through the coils. This means increasing the voltage used to drive it. At some point you will reach a limit imposed by the DC current rating of the coil, which is usually dependent on a combination of heat and wire thickness.\n\nGetting it to move faster\n\nIn fact you can go even faster than this top speed by employing a little trick. When you switch a coil on, initially there is no current flow as the change in current in the windings induces a magnetic field. This field then induces a voltage back in the coil, but in a direction to oppose the current flow. If this process was 100 per cent efficient you could never get any current to flow in a coil, but fortunately it is not. The result is that when the coil is switched on it takes a time for the current to build up. If you are stepping the motor fast, the current will not have built up to its full amount before the coil is switched off again. This accounts for the strange fact that a stepping motor takes more current when it is stopped than when it is moving. The trick in getting a faster rise time of current is to increase the voltage even more. However, this will exceed the maximum DC current through the winding when the motor is at rest and so a resistor should be inserted in each coil line to limit the current. When calculating the value of the resistor make sure to allow enough wattage as the resistor will get warm. The inductive part of the resistor/coil combination is thus proportionally smaller and so there will be a faster rise time of current in the coil, allowing a faster speed to be achieved.\n\nUsing less software but more hardware\n\nThe only snag with the type of control I have described so far is that it requires four Arduino output bits to drive the motor. We can make a saving in this number if we generate the sequence, not with software, but with hardware. Then we need only two bits to control our motor - one to specify direction and one to tell it to step. You can generate the required sequence with two JK flip-flops and a data select IC. This can be fed to the same power FET drives as shown previously.\n\nHowever there is an IC which will do the whole job for you - and it is cheaper than individual components. There are many different types of driver chip available circuit diagram is shown in below uses just one type but they are all fairly similar.",
null,
"The SAA1027 works with high level logic and therefore the logic levels out of the Arduino have to be boosted by the transistors T1 and T2. This IC can be connected directly to the stepping motor's coils with any supply between 9.5 and 18 volts. To drive this all you have to do is to set the logic level for the required direction and give a pulse to the step line. The time between the pulses controls the speed of movement, the shorter the time the faster it runs.\n\nTrouble shooting\n\nMost of the problems you are likely to have concern the motor making a chattering noise but not moving. This is almost certainly due to the fact that you have miss identified the coils and so you are putting the wrong sequence to the coils. Another common error is that you have underestimated how much current is being drawn and your power supply can’t handle it. If your motor has no markings then assume it is a low voltage coil and work your way up. Some motors are only 1.5 or 3V so start small, a variable voltage bench supply is very useful here.",
null,
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"",
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"",
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"Motors 3"
] | [
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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-11-additional-topics-11-3-fractional-exponents-concept-quiz-11-3-page-489/9 | [
"## Elementary Algebra\n\nLet us simplify the left side of the equation: $(3x^{1/6})(2x^{1/3})$ We use the associative property to remove the parentheses: $3\\times x^{1/6}\\times2\\times x^{1/3}$ We use the commutative property to rearrange the terms: $3\\times2\\times x^{1/6}\\times x^{1/3}$ We multiply the constants: $6x^{1/6}x^{1/3}$ To multiply powers with the same base, we add the exponents: $6x^{\\frac{1}{6}+\\frac{1}{3}}=6x^{1/2}$. Since this is the same as the right side of the equation, the equation is true."
] | [
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https://www.smartkeeda.com/Quantitative_Aptitude/Arithmetic/Data_Sufficiency_Quiz/newest/all/passage/Data_Sufficiency_Quiz_2/ | [
"",
null,
"",
null,
"",
null,
"Data Sufficiency has become a topic of high importance in Quantitative Aptitude sections for past few years as far as Bank as well as Management Entrance Exams are concerned. Data Sufficiency questions are mainly asked in Mains exams like IBPS PO Mains, SBI PO Mains, Syndicate Bank PO, IBPS Clerk Mains, SBI Clerk Mains, IBPS RRB Office Assistant Mains, IBPS RRB Officer Scale 1 Mains, BOB Manipal, RBI Assistant Mains, etc. Attempt a high level Data Sufficiency Questions Quiz Set of 5 questions and evaluate your performance.\n\nDirections: Each of the questions below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the statements and give answer:\nImportant for :\n1\nWhat is the present age of Rahul?\n\nI. At the time of marrige Rahul's age was 25 years.\n\nII. The average age of Rahul and Rahul's wife at the time of marrige was 24 years.\n\nIII. The difference between the present ages of Rahul and his son is 24 years.\n» Explain it\nD\nStatement I:\n\nThe age of Rahul at the time of marrige = 25 years.\n\nStatement I alone is not sufficient to answer the question.\n\nStatement II:\n\nTotal age of Rahul and his wife at the time of marrige = 24 × 2 = 48 years.\n\nStatement II alone is not sufficient to answer the question.\n\nStatement III:\n\nDifference between the ages of Rahul and his son = 24 years.\n\nStatement III alone is not sufficient to answer the question.\n\nAll the statements are not sufficient to answer the question because in these statements it is not given that when Rahul got married\n\nHence, option D is correct.\n2\nWhat is the ratio of the volume of the cube to the volume of the cuboid?\n\nI. The Total Surface Area of the cuboid is 352 cm2 and the ratio of the length, breadth and height of the cuboid is 3 : 2 : 1.\n\nII. The Total Surface Area of the cube is 726 cm2.\n\nIII. The length of the cuboid is 1.5 times of the breadth of the cuboid and 3 times of the height of the cuboid. The difference between the height and the length of the cuboid is 8cm.\n» Explain it\nC\nStatement I:\n\nTotal Surface Area of cuboid = 2 (lb + bh + hl)\n\nLet length = 3x, breadth = 2x, height = x\n\n352 = 2 (3x\n× 2x + 2x × x + x × 3x)\n\n176 = (6x2 +2x2 + 3x2)\n\n176 = 11x2\n\nx2 = 16\n\nx = 4\n\nLength = 12cm, Breadth = 8cm, Height = 4cm\n\nVolume = lbh\n\n= 12 × 8 × 4 = 384cm3\n\nStatement II:\n\nTotal Surface Area of cube = 6s2\n\n384 = 6s2\n\ns= 64\n\ns = 8 cm\n\nVolume of the cube = s3\n\n= 83 = 512 cm3\n\nStatement III:\n\nHeight of the cuboid = x cm, length = 3x, breadth = 2x\n\nDifference = 3x\n– x\n\n8 = 2x\n\nx = 4\n\nlength = 12cm, breadth = 8cm, height = 4cm\n\nVolume = lbh\n\n= 12 × 8 × 4 = 384cm3\n\nWe can solve this question with the help of either statement I and II or statement II and III.\n\nHence, option C is correct.\n3\nFind the rate of simple interest per annum.\n\nI. Meetu borrowed Rs 9000 from Sneha for two years on simple interest.\n\nII. Meetu returned Rs 11700 to Sneha at the end of two years and settled the loan.\n\nIII. A sum of money becomes double in 20/3 years on simple interest.\n» Explain it\nC\nStatement I:\n\nPrincipal = Rs 9000\n\n Interest = Principal × Rate × time 100\n\n 2700 = 9000 × Rate × 2 100\n\n 2700 = Rate 180\n\nStatement I alone is not sufficient to give the answer.\n\nStatement II:\n\nAmount = Rs 11700\n\nStatement II alone is not sufficient to give the answer.\n\nStatement I + Statement II:\n\nPrincipal = Rs 9000, Amount = Rs 11700, time = 2 years, Interest = 11700 – 9000 = Rs 2700\n\nRate = 15%\n\nData in Statement I and II is sufficient to give the answer.\n\nStatement III:\n\nLet Principal = Rs x, Amount = Rs 2x, time = 20/3 years, interest = 2x – x = Rs x\n\n Interest = Principal × Rate × time 100\n\n x = x × Rate × 20 300\n\nRate = 15%\n\nStatement III alone is sufficient to give the answer.\n\nHence, option C is correct.\n\n4\nIn a family there are 5 members, Nidhi, Vidhi, Nitya, Ajay and Anil. Find the present age of Anil.\n\nI. The average age of five family members is 36 years.\n\nII. The total age of Nidhi, Vidhi and Anil is 75 years.\n\nIII. The present age of Anil is 12 years more than the age of Ajay.\n» Explain it\nD\nStatement I:\n\nTotal age of five family member = 36 × 5 = 180 years\n\nStatement I alone is not sufficient to give the answer.\n\nStatement II:\n\nTotal age of Nidhi, Vidhi and Anil = 75 years\n\nStatement II alone is not sufficient to give the answer.\n\nStatement III:\n\nLet the age of Ajay = x years\n\nAge of Anil = x + 12 years\n\nStatement III alone is not sufficient to give the answer.\n\nThe data in statement I, II and III is not sufficient to give the answer.\n\nHence, option D is correct.\n5\nFind the value of x.\n\nI. 2xyz + 6y – 8z + 5 = 0, z = 1\n\nII. y = 225 – 116\n\nIII. 4xyz – 6z + 8y – 7 = 0, z = 3\n» Explain it\nC\nStatement I:\n\n2xyz + 6y – 8z + 5 = 0, z = 1\n\n2xy\n× 1 + 6y – 8 × 1 + 5 = 0\n\n2xy + 6y – 8 + 5 = 0\n\n2xy + 6y – 3 = 0\n\nStatement II:\n\nII. y = 225 – 116\n\ny = 9\n\ny = 3\n\nStatement III:\n\n4xyz – 6z + 8y – 7 = 0, z = 3\n\n4xy × 3 – 6 × 3 + 8y – 7 = 0\n\n12xy – 18 + 8y – 7 = 0\n\n12xy + 8y – 25 = 0\n\nWe can receive the value of x with the help of any two statements.\n\nHence, option C is correct."
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https://www.physicsforums.com/threads/equivalent-impedance.203553/ | [
"# Equivalent impedance\n\nhello I have a problem in calculating the equivalent impedance of a simple AC circuit, where the impedances are:\n\nZ1 = 5+i20\nZ2 = 5+i20\nZ3 = 10-i2,5\nZ4 = -i10\nZ5 = 5\n\nand I need to find:\n\nZ_eq = Z3 // [(Z1 // Z4) + (Z2 // Z5)]\n\nwhere the + indicates 2 components in serie and the // indicates 2 components in parallel.\n\nMy calculations always bring me to\nZ_eq=6,25-i3,75\n\nbut my book says Z_eq=28,33.\n\nWhich is the correct eqv impedance?\n\nThank you\nPS-I used the formula Za//Zb = (Za*Zb)/(Za+Zb)\nto calculate equivalent impedances of two components in parallel, and the sum for components in serie.\n\nRelated Engineering and Comp Sci Homework Help News on Phys.org\nThe Electrician\nGold Member\nWell, there's a mistake somewhere.\n\nIf I use the values you gave for Z1 to Z5, I get the same result you did.\n\nHowever, if you set Z3 = 5 + i20, then you will get the result in the book.\n\nOk then there probably is a mistake in the book because Z3 is in fact equal to 10-i2,5 as I calculated and as the book reports. Lost some time on this stupid mistake...\n\nThank you very much for taking the time to verify this! :)\n\nRiccardo"
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https://catalesawmill.com/log-weight-calculator/ | [
"## How much does your log weigh?\n\n### How log weight is calculated\n\nCalculating the weight of hardwood logs is essential for various applications in industries such as forestry, woodworking, and construction. Our Log Weight Calculator provides a simple and efficient way to determine the weight of hardwood logs based on their dimensions. Here's how it works:\n\n1. Choose the Wood Species: Our log weight calculator allows you to select the specific hardwood species from a dropdown menu, ensuring more precise weight calculations. Each wood species has a unique density that influences the weight calculation.\n\n2. Input the Log Dimensions: Enter the length and diameter of the log into the designated fields of our log weight calculator tool. Make sure to use consistent units of measurement for accurate results.\n\n3. Calculation Process: Our log weight calculator employs the formula for calculating the volume of a cylinder, considering the log as a uniform cylinder. The volume is then multiplied by the density of the specific hardwood species to obtain the weight.\n\n4. Instant and Accurate Results: Once you've selected the wood species and entered the log dimensions the weight will be automatically displayed in the \"Weight (lbs) field. Our log weight calculator will instantly display the estimated weight of the hardwood log based on the provided information.\n\nExample Calculation:\nFor instance, let's consider a hardwood log with a length of 10 feet and a diameter of 30 inches on the small end and 35 inches on the large end. Assuming we select \"Red Oak\" as the wood species the log weight calculator estimates the weight of the hardwood log to be approximately 3,629.3 pounds.\n\nKnowing the weight of hardwood logs can aid in transportation planning, estimating load capacities, or determining the amount of raw material required for woodworking projects. Our log weight calculator provides reliable and convenient results, enabling you to make informed decisions in your respective industry.\n\nTake advantage of our Log Weight Calculator today and simplify your log weight calculations with ease and precision.",
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"Disclaimer: Please note that the weight provided by our Log Weight Calculator is an estimate and may vary depending on various conditions. Factors such as moisture content, specific wood density variations within the log, and irregularities in shape or taper can influence the actual weight of a hardwood log. While our calculator uses standard formulas and average density values, it is important to consider that these are general approximations and may not account for all possible variables. For precise weight measurements, we recommend consulting professional resources or conducting physical measurements using calibrated weighing equipment."
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https://offshorehosting.pk/what-is-correlation-coefficient/ | [
"## What is correlation coefficient?\n\nThe data depicted in figures 1–4 were simulated from a bivariate normal distribution of 500 observations with means 2 and 3 for the variables x and y respectively. Scatter plots were generated for the correlations 0.2, 0.5, 0.8 and −0.8. The correlation coefficient describes how one variable moves in relation to another. A positive correlation indicates that the two move in the same direction, with a value of 1 denoting a perfect positive correlation. A value of -1 shows a perfect negative, or inverse, correlation, while zero means no linear correlation exists. Scatterplots may be more useful when analyzing more complex data that might have changing relationships.\n\nI would like to that Dr. Sarah White, PhD, for her comments throughout the development of this article and Nynke R. Van den Broek, PhD, FRCOG, DFFP, DTM&H, for allowing me to use a subset of her data for illustrations. Add correlation to one of your lists below, or create a new one. The inverse Fisher transformation brings the interval back to the correlation scale. A p-value is a measure of probability used for hypothesis testing.\n\n## Data Science – Statistics Correlation\n\nThe Pearson and Spearman correlations (Bonett & Wright, 2000) between the collected bioreactor features and the cardiomyocyte content were calculated. The Pearson correlation measures the strength of the linear relationship between two variables. It has a value between -1 to 1, with a value of -1 meaning a total negative linear correlation, 0 being no correlation, and + 1 meaning a total positive correlation. The Spearman correlation measures the strength of a monotonic relationship between two variables with the same scaling as the Pearson correlation.\n\nThe assumptions of the Spearman correlation are that data must be at least ordinal and the scores on one variable must be monotonically related to the other variable. Nor does the correlation coefficient show what proportion of the variation in the dependent variable is attributable to the independent variable. That’s shown by the coefficient of determination, also known as R-squared, which is simply the correlation coefficient squared.\n\n## Derived forms of correlation\n\nFinancial spreadsheets and software can calculate the value of correlation quickly. A correlation between age and height in children is fairly causally transparent, but a correlation between mood and health in people is less so. Does improved mood lead to improved health, or does good health lead to good mood, or both? In other words, a correlation can be taken as evidence for a possible causal relationship, but cannot indicate what the causal relationship, if any, might be. Pearson, Kendall and Spearman correlation are currently computed using pairwise complete observations.\n\n• A correlation coefficient is a statistical measure of the degree to which changes to the value of one variable predict change to the value of another.\n• This means that the amount of pizza slices eaten by your friends has a strong positive correlation with the amount of soda your friends will drink.\n• In contrast with the correlation value, which must be between − 1 and 1, the covariance may assume any numerical value.\n• Furthermore, we compare Kendall’s correlation, Kendall’s coefficient of concordance, and the kappa tests.\n• In other words, since more people like to buy ice cream when it’s hot outdoors, the company’s overall ice cream sales tend to be greater when it’s hotter outside.\n• A few years ago a survey of employees found a strong positive correlation between “Studying an external course” and Sick Days.\n\nTypically, negatively correlated data sets are seen as a line the goes down and to the right on a scatter plot. When you look only at the orderings or ranks, all three relationships are perfect! The left and central plots show the observations where larger x values always correspond to larger y values. The right plot illustrates the opposite case, which is perfect negative rank correlation. Rank correlation compares the ranks or the orderings of the data related to two variables or dataset features.\n\n## Example: NumPy Correlation Calculation\n\nThese illusory correlations can occur both in scientific investigations and in real-world situations. An illusory correlation is the perception of a relationship between two variables when only a minor relationship—or none at all—actually exists. An illusory correlation does not always mean inferring causation; it can also mean inferring a relationship between two variables when one does not exist.\n\n• You can add some text and conditional formatting to clean up the result.\n• For instance, suppose research revealed a link between the amount of time students spend on their homework (from half an hour to three hours) and the number of G.C.S.E. passes (1 to 6).\n• Maternal age is continuous and usually skewed while parity is ordinal and skewed.\n• If two variables are negatively correlated, a decreasing linear line may be draw.\n• Correlation is tightly connected to other statistical quantities like the mean, standard deviation, variance, and covariance.\n• If two lists of data have a Pearson correlation of 1 or of − 1, this implies that one set of the data is redundant.\n\nThe x-axis of the scatterplot represents one of the variables being tested, while the y-axis of the scatter plot represents the other. Test alternative hypotheses for positive, negative, and nonzero correlation between the columns of two matrices. Compare values of the correlation coefficient and p-value in each case.\n\n## Scatterplots\n\nThe correlation coefficient between historical returns can indicate whether adding an investment to a portfolio will improve its diversification. Assessments of correlation strength based on the correlation coefficient value vary by application. In physics and chemistry, a correlation coefficient should be lower than -0.9 or higher than 0.9 for the correlation to be considered meaningful, while in social sciences the threshold could be as high as -0.5 and as low as 0.5. This type of risk is specific to a company, industry, or asset class. Investing in different assets can reduce your portfolio’s correlation and reduce your exposure to unsystematic risk. Investment managers, traders, and analysts find it very important to calculate correlation because the risk reduction benefits of diversification rely on this statistic.",
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"Two objects that correlated inversely (ie, one falling when the other rises) would have a Pearson score near − 1 (See Glossary items, Correlation distance, Normalized compression distance). Pearson’s correlation test measures relations between two continuous variables. We discuss the application of different types of correlation in this chapter. We also discuss the difference between correlation and concordance.\n\nThe Pearson correlation coefficient (r) is used to denote the linear relationship between two variables x and y whereby the Pearson value must be between -1 and +1. If Pearson’s r is negative, then the relationship is also negative, and if it is positive, the relationship is positive. Correlation measures the relationship, or association, between two variables by looking at how the variables change with respect to each other. Statistical correlation also corresponds to simultaneous changes between two variables, and it is usually represented by linear relationships. This is because a correlation describes how two or more variables are related, and not whether they cause changes in one another.\n\n• Low correlation describes a weaker correlation, meaning that the two variables are probably not related.\n• The Pearson correlation of a with b is 1 because the values of b are simply double the values of a; hence the values in a and b correlate perfectly with one another.\n• Again, the first row of xy represents one feature, while the second row represents the other.\n\nFor example, a trader might use historical Correlations to predict whether a company’s shares will rise or fall in response to a change in interest rates or commodity prices. Similarly, a portfolio manager might aim to reduce their risk by ensuring that the individual assets within their portfolio are not overly correlated with one another. Put option contracts become more profitable when the underlying stock price decreases. In other words, as the stock price increases, the put option prices go down, which is a direct and high-magnitude negative correlation. Correlation, in the finance and investment industries, is a statistic that measures the degree to which two securities move in relation to each other."
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",
null
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https://swmath.org/?term=random%20effects | [
"• # PROC NLMIXED\n\n• Referenced in 65 articles [sw11039]\n• models in which both fixed and random effects enter nonlinearly. These models have a wide ... distribution for your data (given the random effects) having either a standard form (normal, binomial ... likelihood integrated over the random effects. Different integral approximations are available, the principal ones being ... using empirical Bayes estimates of the random effects. You can also estimate arbitrary functions...\n• # GLLAMM\n\n• Referenced in 61 articles [sw06517]\n• case of discrete random effects or factors, the marginal log-likelihood is evaluated exactly whereas ... used for continuous (multivariate) normal random effects or factors. Two methods are available for numerical...\n• # frailtypack\n\n• Referenced in 40 articles [sw06070]\n• proportional hazard models with two correlated random effects (intercept random effect with random slope ... clustering) by including two iid gamma random effects. 4) Joint frailty models in the context...\n• # FODE\n\n• Referenced in 258 articles [sw08377]\n• highly effective in a rich variety of scenarios such as continuous time random walk models ... corresponding stability condition is got. The effectiveness of this numerical algorithm is evaluated by comparing...\n• # glmmAK\n\n• Referenced in 25 articles [sw13218]\n• logistic and Poisson regression model with random effects whose distribution is specified as a penalized ... penalized Gaussian mixture as a random-effects distribution. Computational Statistics and Data Analysis...\n• # Scatter Search\n\n• Referenced in 286 articles [sw05291]\n• benefits beyond those derived from recourse to randomization. Our implementation goal is to create ... scatter search methodology that proves effective when searching for optimal weight values in a multilayer...\n• # MIXOR\n\n• Referenced in 25 articles [sw08990]\n• degree to which these time-related effects vary in the population of individuals. MIXOR uses ... solution, the Cholesky factor of the random-effects variance-covariance matrix is estimated, along with...\n• # TMB\n\n• Referenced in 15 articles [sw16279]\n• Template Model Builder: A General Random Effect Tool Inspired by ’ADMB’. With this tool ... able to quickly implement complex random effect models through simple C++ templates. The package combines..."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8347618,"math_prob":0.8546827,"size":4557,"snap":"2021-21-2021-25","text_gpt3_token_len":1020,"char_repetition_ratio":0.21919614,"word_repetition_ratio":0.014684288,"special_character_ratio":0.24599518,"punctuation_ratio":0.18562092,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95333546,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T11:36:07Z\",\"WARC-Record-ID\":\"<urn:uuid:cf967cbe-66e9-40f9-91f5-f1c046524f63>\",\"Content-Length\":\"57167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e94bb632-9872-439b-bbf6-03da247581fc>\",\"WARC-Concurrent-To\":\"<urn:uuid:5922aa61-3364-4d1d-87eb-26a2cd651c7e>\",\"WARC-IP-Address\":\"141.66.193.30\",\"WARC-Target-URI\":\"https://swmath.org/?term=random%20effects\",\"WARC-Payload-Digest\":\"sha1:WG5JDPNJGCO7KDVIERUA5O4L755TBKKJ\",\"WARC-Block-Digest\":\"sha1:LGSUEZXCDJJDDXXZ372COCH46UV3L4RD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988966.82_warc_CC-MAIN-20210509092814-20210509122814-00113.warc.gz\"}"} |
https://byjus.com/maths/774-in-words/ | [
"",
null,
"# 774 in Words\n\nIn English, the spelling of 774 is Seven hundred seventy-four. As we know that the number 774 is a cardinal number. For example, the cost of 9 kg oranges is Rs. 774. For converting any numbers into words, the place value system plays a major role. In this article, let us have a look at the process of writing the number 774 in words in detail.\n\n 774 in Words: Seven Hundred Seventy-four. Seven Hundred Seventy-four in Numerical Form: 774.\n\n## 774 in English Words",
null,
"## How to Write 774 in Words?\n\nThe below table represents the place values of 774.\n\n Hundreds Tens Ones 7 7 4\n\nThe expanded form of 774 is as follows:\n\n= 7 × Hundred + 7 × Ten + 4 × One\n\n= 7 × 100 + 7 × 10 + 4 × 1\n\n= 700 + 70 + 4\n\n= 774\n\n= Seven hundred seventy-four\n\nHence, 774 in words is seven hundred seventy-four.\n\nThe number 774 exists before 775 but after 773.\n\n774 in words – Seven hundred seventy-four\n\nIs 774 an odd number? – No\n\nIs 774 an even number? – Yes\n\nIs 774 a perfect square number? – No\n\nIs 774 a perfect cube number? – No\n\nIs 774 a prime number? – No\n\nIs 774 a composite number? – Yes\n\n## Frequently Asked Questions on 774 in Words\n\nQ1\n\n### How to write 774 in English words?\n\n774 in words is seven hundred seventy-four.\n\nQ2\n\n### Simplify 700 + 74, and express it in words.\n\nSimplifying 700 + 74, we get 774. Hence, 774 in words is seven hundred seventy-four.\n\nQ3\n\n### Convert seven hundred seventy-four into numbers.\n\nSeven hundred seventy-four in numbers is 774."
] | [
null,
"https://www.facebook.com/tr",
null,
"https://cdn1.byjus.com/wp-content/uploads/2022/03/Number-in-word-774.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81178623,"math_prob":0.54590666,"size":1450,"snap":"2023-40-2023-50","text_gpt3_token_len":408,"char_repetition_ratio":0.20816045,"word_repetition_ratio":0.06338028,"special_character_ratio":0.33448276,"punctuation_ratio":0.11960133,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.986163,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T19:50:24Z\",\"WARC-Record-ID\":\"<urn:uuid:476bf013-5831-47b3-bafb-f40a33366089>\",\"Content-Length\":\"565525\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a3145c5-72a7-4c1a-8e9b-bbfe5a4d5edd>\",\"WARC-Concurrent-To\":\"<urn:uuid:7249f54b-5249-4ad8-919d-266a3563b12f>\",\"WARC-IP-Address\":\"34.36.4.163\",\"WARC-Target-URI\":\"https://byjus.com/maths/774-in-words/\",\"WARC-Payload-Digest\":\"sha1:OTTUGSF7XK52D6GJWBNCOI6Y5C2VZJWH\",\"WARC-Block-Digest\":\"sha1:UA6WDVDL4CZBURM2UM3LVFLCQWCXFZ6J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102637.84_warc_CC-MAIN-20231210190744-20231210220744-00150.warc.gz\"}"} |
https://www.colorhexa.com/02c6ed | [
"# #02c6ed Color Information\n\nIn a RGB color space, hex #02c6ed is composed of 0.8% red, 77.6% green and 92.9% blue. Whereas in a CMYK color space, it is composed of 99.2% cyan, 16.5% magenta, 0% yellow and 7.1% black. It has a hue angle of 190 degrees, a saturation of 98.3% and a lightness of 46.9%. #02c6ed color hex could be obtained by blending #04ffff with #008ddb. Closest websafe color is: #00ccff.\n\n• R 1\n• G 78\n• B 93\nRGB color chart\n• C 99\n• M 16\n• Y 0\n• K 7\nCMYK color chart\n\n#02c6ed color description : Vivid cyan.\n\n# #02c6ed Color Conversion\n\nThe hexadecimal color #02c6ed has RGB values of R:2, G:198, B:237 and CMYK values of C:0.99, M:0.16, Y:0, K:0.07. Its decimal value is 181997.\n\nHex triplet RGB Decimal 02c6ed `#02c6ed` 2, 198, 237 `rgb(2,198,237)` 0.8, 77.6, 92.9 `rgb(0.8%,77.6%,92.9%)` 99, 16, 0, 7 190°, 98.3, 46.9 `hsl(190,98.3%,46.9%)` 190°, 99.2, 92.9 00ccff `#00ccff`\nCIE-LAB 73.876, -27.315, -30.787 35.501, 46.512, 87.223 0.21, 0.275, 46.512 73.876, 41.157, 228.42 73.876, -52.918, -45.677 68.2, -26.432, -28.088 00000010, 11000110, 11101101\n\n# Color Schemes with #02c6ed\n\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #ed2902\n``#ed2902` `rgb(237,41,2)``\nComplementary Color\n• #02ed9f\n``#02ed9f` `rgb(2,237,159)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #0250ed\n``#0250ed` `rgb(2,80,237)``\nAnalogous Color\n• #ed9f02\n``#ed9f02` `rgb(237,159,2)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #ed0250\n``#ed0250` `rgb(237,2,80)``\nSplit Complementary Color\n• #c6ed02\n``#c6ed02` `rgb(198,237,2)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #ed02c6\n``#ed02c6` `rgb(237,2,198)``\n• #02ed29\n``#02ed29` `rgb(2,237,41)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #ed02c6\n``#ed02c6` `rgb(237,2,198)``\n• #ed2902\n``#ed2902` `rgb(237,41,2)``\n• #0187a1\n``#0187a1` `rgb(1,135,161)``\n• #029cba\n``#029cba` `rgb(2,156,186)``\n• #02b1d4\n``#02b1d4` `rgb(2,177,212)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #0cd5fd\n``#0cd5fd` `rgb(12,213,253)``\n• #25d9fd\n``#25d9fd` `rgb(37,217,253)``\n• #3edefd\n``#3edefd` `rgb(62,222,253)``\nMonochromatic Color\n\n# Alternatives to #02c6ed\n\nBelow, you can see some colors close to #02c6ed. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #02edd9\n``#02edd9` `rgb(2,237,217)``\n• #02eded\n``#02eded` `rgb(2,237,237)``\n• #02daed\n``#02daed` `rgb(2,218,237)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #02b2ed\n``#02b2ed` `rgb(2,178,237)``\n• #029fed\n``#029fed` `rgb(2,159,237)``\n• #028bed\n``#028bed` `rgb(2,139,237)``\nSimilar Colors\n\n# #02c6ed Preview\n\nThis text has a font color of #02c6ed.\n\n``<span style=\"color:#02c6ed;\">Text here</span>``\n#02c6ed background color\n\nThis paragraph has a background color of #02c6ed.\n\n``<p style=\"background-color:#02c6ed;\">Content here</p>``\n#02c6ed border color\n\nThis element has a border color of #02c6ed.\n\n``<div style=\"border:1px solid #02c6ed;\">Content here</div>``\nCSS codes\n``.text {color:#02c6ed;}``\n``.background {background-color:#02c6ed;}``\n``.border {border:1px solid #02c6ed;}``\n\n# Shades and Tints of #02c6ed\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000304 is the darkest color, while #effcff is the lightest one.\n\n• #000304\n``#000304` `rgb(0,3,4)``\n• #001317\n``#001317` `rgb(0,19,23)``\n• #00232a\n``#00232a` `rgb(0,35,42)``\n• #01343e\n``#01343e` `rgb(1,52,62)``\n• #014451\n``#014451` `rgb(1,68,81)``\n• #015465\n``#015465` `rgb(1,84,101)``\n• #016478\n``#016478` `rgb(1,100,120)``\n• #01758c\n``#01758c` `rgb(1,117,140)``\n• #01859f\n``#01859f` `rgb(1,133,159)``\n• #0295b3\n``#0295b3` `rgb(2,149,179)``\n• #02a5c6\n``#02a5c6` `rgb(2,165,198)``\n• #02b6da\n``#02b6da` `rgb(2,182,218)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\n• #06d4fd\n``#06d4fd` `rgb(6,212,253)``\n• #19d7fd\n``#19d7fd` `rgb(25,215,253)``\n• #2ddbfd\n``#2ddbfd` `rgb(45,219,253)``\n• #40defd\n``#40defd` `rgb(64,222,253)``\n• #54e1fe\n``#54e1fe` `rgb(84,225,254)``\n• #67e5fe\n``#67e5fe` `rgb(103,229,254)``\n• #7ae8fe\n``#7ae8fe` `rgb(122,232,254)``\n• #8eebfe\n``#8eebfe` `rgb(142,235,254)``\n• #a1effe\n``#a1effe` `rgb(161,239,254)``\n• #b5f2fe\n``#b5f2fe` `rgb(181,242,254)``\n• #c8f6ff\n``#c8f6ff` `rgb(200,246,255)``\n• #dcf9ff\n``#dcf9ff` `rgb(220,249,255)``\n• #effcff\n``#effcff` `rgb(239,252,255)``\nTint Color Variation\n\n# Tones of #02c6ed\n\nA tone is produced by adding gray to any pure hue. In this case, #707c7f is the less saturated color, while #02c6ed is the most saturated one.\n\n• #707c7f\n``#707c7f` `rgb(112,124,127)``\n• #678288\n``#678288` `rgb(103,130,136)``\n• #5e8991\n``#5e8991` `rgb(94,137,145)``\n• #558f9a\n``#558f9a` `rgb(85,143,154)``\n• #4c95a3\n``#4c95a3` `rgb(76,149,163)``\n``#429bad` `rgb(66,155,173)``\n• #39a1b6\n``#39a1b6` `rgb(57,161,182)``\n• #30a7bf\n``#30a7bf` `rgb(48,167,191)``\n``#27adc8` `rgb(39,173,200)``\n• #1eb4d1\n``#1eb4d1` `rgb(30,180,209)``\n``#14badb` `rgb(20,186,219)``\n• #0bc0e4\n``#0bc0e4` `rgb(11,192,228)``\n• #02c6ed\n``#02c6ed` `rgb(2,198,237)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #02c6ed is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5220429,"math_prob":0.6580129,"size":3684,"snap":"2019-51-2020-05","text_gpt3_token_len":1650,"char_repetition_ratio":0.13233696,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5393594,"punctuation_ratio":0.23575419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9773875,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-14T11:09:40Z\",\"WARC-Record-ID\":\"<urn:uuid:00a4dafd-a78b-41f3-aa00-e75abfd891d2>\",\"Content-Length\":\"36257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:214d86ae-fdd0-446e-bdf2-a8dc73c7485d>\",\"WARC-Concurrent-To\":\"<urn:uuid:ecb067ad-6b71-4d45-b50e-a9fb06753be9>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/02c6ed\",\"WARC-Payload-Digest\":\"sha1:MUZECOZBTIQE5KZHCM5QAPSNB3H2WWTA\",\"WARC-Block-Digest\":\"sha1:CDNJVLFNT223FMSSXHT3C6VHW7R5MTYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540586560.45_warc_CC-MAIN-20191214094407-20191214122407-00286.warc.gz\"}"} |
http://petrofaq.org/wiki/EOS_(Equation-of-State) | [
"Blog | Tools | Glossary | Search Share: | feedback",
null,
"# EOS (Equation-of-State)\n\nJump to navigation Jump to search\n\nEOS = Equation-of-State PVT.\nAn equation of state is a thermodynamic expression that relates pressure (P), temperature (T), and volume (V). This equation is used to describe the state of reservoir fluids at given conditions. The cubic equations of state (CEOS) such as Van der Waals, Redlich-Kwong, Soave, and Peng-Robinson are simple models that have been widely used in the oil industry. Many attempts have been made to describe the thermodynamic behavior of fluids to predict their physical properties at given conditions. So, several forms of the equation of state have been presented to the oil industry in order to calculate reservoir fluid properties.\n\n## Ideal Gas Law\n\nThe simplest EOS is the ideal gas law, which states:\n\n```PV = RT\n```\n\nwhere:\n\n• P = pressure\n• V = molar volume\n• T = temperature\n• R = the universal gas constant\n\nThis simple EOS results from combining Boyles’ Law (at constant T, PV = constant) with Charles’ Law (at constant P, V/T = constant).\nThe equation assumes that the volume of the gas molecules is insignificant compared to the volume of the container and the distance between the molecules, and that there are no attractive or repulsive forces between the molecules or the walls of the container.\nThis equation works well at very low pressures (subatmospheric), but unfortunately there are no reservoir at these pressures.\n\n## Van der Waals’ EOS\n\nIn 1873, Johannes Diderik Van der Waals developed an empirical equation-of-state for real gases.\nVan der Waals pointed out that the gas molecules occupy a significant fraction of the volume at higher pressures. He proposed that the volume of the molecules, as denoted by the parameter b, be subtracted from the actual molar volume V, in the ideal gas law equation, to give:\n\n```P = RT / (V– b)\n```\n\nwhere:\n\n• b = co-volume, and reflects the volume of molecules\n• V = actual volume in cubic feet per one mole of gas\n\nVan der Waals also subtracted a corrective term, as denoted by a/V2, from the ideal gas law equation to account for the attractive forces between molecules.",
null,
"where\n\n• P = system pressure, psia\n• T = system temperature, oR\n• R = gas constant, 10.73 psi-ft3/lb-mole, oR\n• V = volume, ft3/mole\n\n### Critical Conditions\n\nIn determining the values of the two constants a and b for any pure substance, Van der Waals observed that the critical isotherm has a horizontal slope and an inflection point at the critical point.",
null,
"This observation can be expressed mathematically as:",
null,
"This shows that the first and second derivatives of Pressure with respect to Volume are 0.\n\n### Constants a and b\n\nThe first and second derivatives of equation above with respect to volume at the critical point result in:",
null,
"",
null,
"or",
null,
"- this suggests that the volume of the molecules, b, is approximately 0.333 of the critical volume of the substance. Experimental studies reveal that this is rather close to truth. They show that b is in the range of 0.24-0.28 of the critical volume.",
null,
"At the critical point:",
null,
"This can give a more convenient expression for calculating the parameters a and b:",
null,
"where:\n\n• R = gas constant, 10.73 psia-ft3/lb-mole-oR\n• Pc = critical pressure, psia\n• Tc = critical temperature, oR\n• Ωa = 0.421875\n• Ωb = 0.125\n\n## Van der Waals’ Cubic EOS\n\nVan der Waals’ equeation:",
null,
"can be expressed in a cubic form in terms of volume V. This equation is usually referred to as the Van der Waals two parameter cubic equation-of-state (the referenced two parameters being a and b):",
null,
"The most significant feature of this equation is that it describes the liquid-condensation phenomenon and the passage from the gas to the liquid phase as the gas is compressed (points C and E below):",
null,
"where\n\n• Points C, D and E represent roots of Van der Waals’ Cubic EOS equation at fixed T,P\n• Liquid and vapor exists on horizontal line CE\n• Largest volume root (E) = volume of saturated vapor\n• Smallest volume root (C) = volume of saturated liquid\n\nVan der Waals’ Cubic EOS can be expressed in a more practical form in terms of the compressibility factor, Z (recall that the ideal gas law (PV = RT) is applicable at only very low pressures. To make this equation true for “real” gases, we added to the equation the compressibility factor. Now, for a real gas, PV = ZRT).\nReplacing the molar volume, V, in equation with ZRT/P gives:\n\n```Z3 - (1 + B)Z2 + AZ – AB = 0\n```\n\nwhere:\nA = aP / R2T2\nB = bP / RT\n\nThe last equation yields one real root in the one-phase region and three real roots in the two-phase region (where system pressure equals the vapor pressure of the substance).\nIn the latter case, the largest root corresponds to the compressibility factor of the vapor phase, ZV, while the smallest positive root corresponds to that of the liquid, ZL.\n\n## Equation of State Types\n\nThe cubic equation-of-state is widely used within the petroleum industry. The most popular ones are:\n\n• 2 Parameter Peng-Robinson\n• 2 Parameter Soave-Redlich-Kwong\n• Redlich-Kwong\n• Zudkevitch-Joffe\n• 3 Parameter Peng-Robinson\n• 3 Parameter Soave-Redlich-Kwong\n• Schmidt-Wenzel\n\nThe cubic equations-of-state are called two-parameter EOS because of the a and b parameters. PVT EOS simulators provide defaults values for all the parameters in these equations, but unless we “tune” these equations to known data, we typically won’t get good predictions.\nDensities are not computed very accurately under some conditions, especially saturated liquids. Gas phase Z-factors were found to be in error from 3% to 5%, and errors in liquid densities were 6% to 12%. In these cases, a three-parameter cubic EOS may be used.\n\n## Summary of EOS Parameters\n\nThe following list of parameters used to calculate EOS:\n\n• molecular weight\n• critical temperature\n• critical pressure\n• critical Z-factor. Will not affect phase behavior calculation. Used in Lohrenz-Bray-Clark (1964) viscosity correlation. May be adjusted to match viscosity data.\n• acentric factor. The acentric factor represents the acentricity or nonsphericity of a molecule. It is a constant for pure components, and is a measure of a molecule’s complexity with respect to geometry and polarity.\n• Ωa - EOS constant\n• Ωb - EOS constant\n• volume shift (for 3-parameter EOS only)\n• boiling point temperature (Zudkevich-Joffe-Redlich-Kwong only)\n• specific gravity (Zudkevich-Joffe-Redlich-Kwong only)\n• reference temperature for specific gravity (Zudkevich-Joffe-Redlich-Kwong only)\n• parachor (used for surface tension calculation only)\n• binary interaction coefficient. The interaction between hydrocarbon components increases as the relative difference between their molecular weights increases.\n\n## EOS Modeling\n\nIn general three options can be used to obtain or modify some of the component properties during the OES modeling:\n\n• Heavy - to split and characterize the heavy end fraction\n• Regression - to modify EOS parameters to match experimental PVT data\n• Pseudoization - to reduce the total number of components to a minimum while maintaining an accurate description of phase behavior"
] | [
null,
"https://mc.yandex.ru/watch/11846299",
null,
"http://petrofaq.org/w/images/Van_der_Waals_equation.jpg",
null,
"http://petrofaq.org/w/images/Critical_Conditions.jpg",
null,
"http://petrofaq.org/w/images/Critical_Conditions_Equation.jpg",
null,
"http://petrofaq.org/w/images/Eq1.jpg",
null,
"http://petrofaq.org/w/images/Eq2.jpg",
null,
"http://petrofaq.org/w/images/Eq_b.jpg",
null,
"http://petrofaq.org/w/images/Eq_a.jpg",
null,
"http://petrofaq.org/w/images/At_critical_point.jpg",
null,
"http://petrofaq.org/w/images/Eq_a_b.jpg",
null,
"http://petrofaq.org/w/images/Van_der_Waals_equation.jpg",
null,
"http://petrofaq.org/w/images/Van_der_Waals%E2%80%99_Cubic_EOS.jpg",
null,
"http://petrofaq.org/w/images/Van_der_Waals%E2%80%99_Cubic_EOS_-_Pressure_Volume_plot.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86645526,"math_prob":0.9845257,"size":6975,"snap":"2021-04-2021-17","text_gpt3_token_len":1643,"char_repetition_ratio":0.13756993,"word_repetition_ratio":0.015424165,"special_character_ratio":0.22164875,"punctuation_ratio":0.088025376,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967148,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,6,null,2,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-21T22:47:10Z\",\"WARC-Record-ID\":\"<urn:uuid:1292c8f1-f797-4d45-9e40-59984c34a03b>\",\"Content-Length\":\"33900\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b3b5f52-049f-4868-9caa-1b7014d2d81f>\",\"WARC-Concurrent-To\":\"<urn:uuid:14240458-b11e-48cd-bd54-8bd749e26ddf>\",\"WARC-IP-Address\":\"5.45.114.66\",\"WARC-Target-URI\":\"http://petrofaq.org/wiki/EOS_(Equation-of-State)\",\"WARC-Payload-Digest\":\"sha1:5YGWXGR7NQZFG4AAN6XEAYKXJJKYCHEG\",\"WARC-Block-Digest\":\"sha1:IAKSHEFTQAPBSKIJ2FKUPR2T65XNU6R7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039554437.90_warc_CC-MAIN-20210421222632-20210422012632-00334.warc.gz\"}"} |
http://at.metamath.org/ileuni/pweqi.html | [
"",
null,
"Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > pweqi GIF version\n\nTheorem pweqi 3363\n Description: Equality inference for power class. (Contributed by NM, 27-Nov-2013.)\nHypothesis\nRef Expression\npweqi.1 𝐴 = 𝐵\nAssertion\nRef Expression\npweqi 𝒫 𝐴 = 𝒫 𝐵\n\nProof of Theorem pweqi\nStepHypRef Expression\n1 pweqi.1 . 2 𝐴 = 𝐵\n2 pweq 3362 . 2 (𝐴 = 𝐵 → 𝒫 𝐴 = 𝒫 𝐵)\n31, 2ax-mp 7 1 𝒫 𝐴 = 𝒫 𝐵\n Colors of variables: wff set class Syntax hints: = wceq 1243 𝒫 cpw 3359 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-11 1397 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 This theorem depends on definitions: df-bi 110 df-tru 1246 df-nf 1350 df-sb 1646 df-clab 2027 df-cleq 2033 df-clel 2036 df-in 2924 df-ss 2931 df-pw 3361 This theorem is referenced by: mnfnre 7068\n Copyright terms: Public domain W3C validator"
] | [
null,
"http://at.metamath.org/ileuni/_icon-il.gif",
null
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https://pegas.is/old/python/Basic%20Function.html | [
"# Basic Function\n\nA function performs a per-defined task when you use it (also known as “when you call it”).\n\nNow, we can only use functions that are pre-defined by python. In the Advanced Function handout, we will learn how to define a function by ourselves.\n\nHere’s the full process of using a function:",
null,
"## Basic Structure",
null,
"### Parameter(s)\n\nThis is the place where you give data to the function.\n\nYou can don’t give a function any parameter, one parameter or multiple parameters. Parameters are separated by commas.\n\nFor example: (The following code is just for demonstrate how to give data to a function, it can’t actually run because those functions are not defined yet.)\n\n### Return Value\n\nThis is how you get new data from the function.\n\nA function can return no value or one value. You can assign the value returned from a function to a variable.\n\nFor example: (The following code is just for demonstrate how to get value returned from a function, it can’t actually run because those functions are not defined yet.)\n\n## How to Describe a Function\n\nKnowing how to describe a function is very important when you are trying to figure out how to use a function.\n\n1. Signature\n\nSignature of the function is the function name and the parameter it takes. For example, the signature of the print function is print (data)\n\n2. Parameters\n\nExplain what does each parameter means. For example, data parameter means the data print function will show in the shell.\n\n3. Return value\n\nDescribe what and how many value the function will return. For print function, it returns no value.\n\n## print Function\n\n• Usage: output a value\n\n• Signature: print (data)\n\n• Parameters:\n\n• Return value: None\n\n## str Function\n\n• Usage: convert a number type value to string type value\n\n• Signature: str (number)\n\n• Parameters:\n\n• Return value: same number, but is in string type.\n\nNext: Control Flow",
null,
""
] | [
null,
"https://pegas.is/old/python/Basic Function.assets/function_diagram.png",
null,
"https://pegas.is/old/python/Basic Function.assets/function_basic_structure.png",
null,
"https://pegas.is/old/python/Basic Function.assets/By-Pegasis-green.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71708703,"math_prob":0.94967204,"size":1781,"snap":"2023-40-2023-50","text_gpt3_token_len":397,"char_repetition_ratio":0.16882385,"word_repetition_ratio":0.11295681,"special_character_ratio":0.20887142,"punctuation_ratio":0.1037464,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98768383,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T15:29:56Z\",\"WARC-Record-ID\":\"<urn:uuid:3bdc9dfb-8159-46fe-ad90-a33687c48091>\",\"Content-Length\":\"20766\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:466488ed-6a37-4911-9628-46f3eb9ea92e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad6fd084-3d52-49f5-b2eb-ef73b1dea5d5>\",\"WARC-IP-Address\":\"44.219.53.183\",\"WARC-Target-URI\":\"https://pegas.is/old/python/Basic%20Function.html\",\"WARC-Payload-Digest\":\"sha1:FTBD2FFISNQU6FC2PYR6E5ICHQYJ3DZY\",\"WARC-Block-Digest\":\"sha1:UJHKZQNDKE6BEJ6VIQCUWGVRZGHUVZBU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511000.99_warc_CC-MAIN-20231002132844-20231002162844-00416.warc.gz\"}"} |
https://etc.usf.edu/clipart/galleries/536-us-coins-totaling-1-cent-to-one-dollar | [
"This mathematics ClipArt gallery offers 100 illustrations of United States coins that add up to values from one cent to one dollar.",
null,
"### 1 Cent\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 10 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 100 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 11 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 12 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 13 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 14 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 15 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 16 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 17 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 18 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 19 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 2 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 20 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 21 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 22 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 23 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 24 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 25 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 26 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 27 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 28 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 29 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 3 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 30 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 31 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 32 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 33 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 34 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 35 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 36 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 37 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 38 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 39 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 4 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 40 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 41 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 42 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 43 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 44 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 45 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 46 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 47 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 48 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 49 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 5 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 50 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 51 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 52 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 53 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 54 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 55 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 56 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 57 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 58 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 59 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 6 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 60 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 61 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins.",
null,
"### 62 Cents\n\nGroups of change with totals from 1 to 100 cents using the least amount of coins."
] | [
null,
"https://etc.usf.edu/clipart/37400/37476/001-cent_37476_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37485/010-cent_37485_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37575/100-cent_37575_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37486/011-cent_37486_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37487/012-cent_37487_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37488/013-cent_37488_mth.gif",
null,
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null,
"https://etc.usf.edu/clipart/37400/37491/016-cent_37491_mth.gif",
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null,
"https://etc.usf.edu/clipart/37400/37494/019-cent_37494_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37477/002-cent_37477_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37495/020-cent_37495_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37496/021-cent_37496_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37497/022-cent_37497_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37498/023-cent_37498_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37499/024-cent_37499_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37500/025-cent_37500_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37501/026-cent_37501_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37502/027-cent_37502_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37503/028-cent_37503_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37504/029-cent_37504_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37478/003-cent_37478_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37505/030-cent_37505_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37506/031-cent_37506_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37507/032-cent_37507_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37508/033-cent_37508_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37509/034-cent_37509_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37510/035-cent_37510_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37511/036-cent_37511_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37512/037-cent_37512_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37513/038-cent_37513_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37514/039-cent_37514_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37479/004-cent_37479_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37515/040-cent_37515_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37516/041-cent_37516_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37517/042-cent_37517_mth.gif",
null,
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null,
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null,
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null,
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null,
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null,
"https://etc.usf.edu/clipart/37500/37523/048-cent_37523_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37524/049-cent_37524_mth.gif",
null,
"https://etc.usf.edu/clipart/37400/37480/005-cent_37480_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37525/050-cent_37525_mth.gif",
null,
"https://etc.usf.edu/clipart/37500/37526/051-cent_37526_mth.gif",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
"https://etc.usf.edu/clipart/37500/37537/062-cent_37537_mth.gif",
null
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https://en.wikipedia.org/wiki/De_Bruijn_notation | [
"# De Bruijn notation\n\nIn mathematical logic, the De Bruijn notation is a syntax for terms in the λ calculus invented by the Dutch mathematician Nicolaas Govert de Bruijn. It can be seen as a reversal of the usual syntax for the λ calculus where the argument in an application is placed next to its corresponding binder in the function instead of after the latter's body.\n\n## Formal definition\n\nTerms ($M,N,\\ldots$",
null,
") in the De Bruijn notation are either variables ($v$",
null,
"), or have one of two wagon prefixes. The abstractor wagon, written $[v]$",
null,
", corresponds to the usual λ-binder of the λ calculus, and the applicator wagon, written $(M)$",
null,
", corresponds to the argument in an application in the λ calculus.\n\n$M,N,...::=\\ v\\ |\\ [v]\\;M\\ |\\ (M)\\;N$",
null,
"Terms in the traditional syntax can be converted to the De Bruijn notation by defining an inductive function ${\\mathcal {I}}$",
null,
"for which:\n\n{\\begin{aligned}{\\mathcal {I}}(v)&=v\\\\{\\mathcal {I}}(\\lambda v.\\ M)&=[v]\\;{\\mathcal {I}}(M)\\\\{\\mathcal {I}}(M\\;N)&=({\\mathcal {I}}(N)){\\mathcal {I}}(M).\\end{aligned}}",
null,
"All operations on λ-terms commute with respect to the ${\\mathcal {I}}$",
null,
"translation. For example, the usual β-reduction,\n\n$(\\lambda v.\\ M)\\;N\\ \\ \\longrightarrow _{\\beta }\\ \\ M[v:=N]$",
null,
"in the De Bruijn notation is, predictably,\n\n$(N)\\;[v]\\;M\\ \\ \\longrightarrow _{\\beta }\\ \\ M[v:=N].$",
null,
"A feature of this notation is that abstractor and applicator wagons of β-redexes are paired like parentheses. For example, consider the stages in the β-reduction of the term $(M)\\;(N)\\;[u]\\;(P)\\;[v]\\;[w]\\;(Q)\\;z$",
null,
", where the redexes are underlined:\n\n{\\begin{aligned}(M)\\;{\\underline {(N)\\;[u]}}\\;(P)\\;[v]\\;[w]\\;(Q)\\;z&{\\ \\longrightarrow _{\\beta }\\ }(M)\\;{\\underline {(P[u:=N])\\;[v]}}\\;[w]\\;(Q[u:=N])\\;z\\\\&{\\ \\longrightarrow _{\\beta }\\ }{\\underline {(M)\\;[w]}}\\;(Q[u:=N,v:=P[u:=N]])\\;z\\\\&{\\ \\longrightarrow _{\\beta }\\ }(Q[u:=N,v:=P[u:=N],w:=M])\\;z.\\end{aligned}}",
null,
"Thus, if one views the applicator as an open paren ('(') and the abstractor as a close bracket (']'), then the pattern in the above term is '((](]]'. De Bruijn called an applicator and its corresponding abstractor in this interpretation partners, and wagons without partners bachelors. A sequence of wagons, which he called a segment, is well balanced if all its wagons are partnered.\n\n## Advantages of the De Bruijn notation\n\nIn a well balanced segment, the partnered wagons may be moved around arbitrarily and, as long as parity is not destroyed, the meaning of the term stays the same. For example, in the above example, the applicator $(M)$",
null,
"can be brought to its abstractor $[w]$",
null,
", or the abstractor to the applicator. In fact, all commutatives and permutative conversions on lambda terms may be described simply in terms of parity-preserving reorderings of partnered wagons. One thus obtains a generalised conversion primitive for λ-terms in the De Bruijn notation.\n\nSeveral properties of λ-terms that are difficult to state and prove using the traditional notation are easily expressed in the De Bruijn notation. For example, in a type-theoretic setting, one can easily compute the canonical class of types for a term in a typing context, and restate the type checking problem to one of verifying that the checked type is a member of this class. De Bruijn notation has also been shown to be useful in calculi for explicit substitution in pure type systems."
] | [
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null
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https://kipdf.com/weird-multiplication-james-tanton_5b0e05247f8b9a3c9b8b45e1.html | [
"## WEIRD MULTIPLICATION. James Tanton\n\nWEIRD MULTIPLICATION James Tanton www.jamestanton.com The following represents a sample activity based on the September 2007 newsletter of the St. Ma...\nAuthor: Brendan Blair\nWEIRD MULTIPLICATION\n\nJames Tanton www.jamestanton.com The following represents a sample activity based on the September 2007 newsletter of the St. Mark’s Institute of Mathematics (www.stmarksschool.org/math). It also appears in chapter 1 of: THINKING MATHEMATIS! Volume 1: Arithmetic = Gateway to All available at www.jamestanton.com.\n\nCONTENTS: Weird MultiplicationNewsletter …………………………………… Commentary ……………………………………… Finger Multiplication ……………………………………… Russian Multiplication ………………………………………\n\n© 2010 James Tanton\n\nPage 2 Page 4 Page 9 Page10\n\n2\n\nSEPTEMBER 2007 PUZZLER: WEIRD MULTIPLICATION Here’s an unusual means for performing long multiplication. To compute 22 × 13 , for example, draw two sets of vertical lines, the left set containing two lines and the right set two lines (for the digits in 22) and two sets of horizontal lines, the upper set containing one line and the lower set three (for the digits in 13).\n\nAlthough the answer 6 thousands, 16 hundreds, 26 tens, and 12 ones is absolutely correct, one needs to carry digits and translate this into 7,872. a) Compute 131 × 122 via this method. b) Compute 54 × 1332 via this method. c) How best should one compute 102 × 30054 via this method? d) Why does the method work? A TINY TIDBIT: Putting on one’s shoes and then one’s socks does not produce the same result as first putting on one’s socks and then one’s shoes. These two operations are not commutative.\n\nThere are four sets of intersection points. Count the number of intersections in each and add the results diagonally as shown:\n\nIn mathematics, multiplication is said to be commutative, meaning that a × b produces the same result as b × a for all numbers a and b. Is this obviously true? It might not be given the following graphical model for multiplication: To compute 2 × 3 , for instance, first draw two concentric circles to model the first number in the problem, 2, and three radii to model the second number, 3.\n\nThe answer 286 appears. There is one caveat as illustrated by the computation 246 × 32 :\n\nThe number of separate pieces one sees in the diagram - in this case 6 - is the answer to the multiplication problem. Is it at all obvious that drawing three concentric circles and two radii will produce the same number of pieces? Suppose one draws 77 circles and 43 radii. Can we be sure that drawing 43 circles and 77 radii\n\n© James Tanton 2010\n\n3\n\nyields an equivalent number of pieces? Is the commutativity of multiplication a belief or a fact? ANOTHER TIDBIT: Here’s another way to compute a multiplication problem; 341 × 752 , for instance. Write each number on a strip of paper, reversing one of the numbers. Start with the reversed number on the top strip to the right of the second number.\n\nThe answer to 341 × 752 is:\n\nnamely, 21 ten-thousands, 43 thousands, 33 hundreds, 13 tens and 2 ones. After carrying digits this translates to the number 256,432. Slide the top number to the left until a first set of digits align. Multiply them and write their product underneath.\n\nSlide one more place to the left and multiply the digits that are aligned. Write their sum underneath.\n\nCOMMENT: Some people may prefer to carry digits as they conduct this procedure and write something akin to:\n\nANOTHER PUZLER: Vedic mathematics taught in India (and established in 1911 by Jagadguru Swami Bharati Krishna Tirthaji Maharaj) has students compute the multiplication of two three-digit numbers as follows:\n\nContinue this process:\n\nWhat do you think this sequence of diagrams means? © 2007 James Tanton St. Mark’s Institute of Mathematics.\n\n© James Tanton 2010\n\nRESEARCH CORNER: Invent your own method of conducting long multiplication.\n\n4\n\nII: COMMENTARY I have found the content of this newsletter to be a big hit with folks teaching middleschool students and, in general, a delight for students of all ages to ponder upon. Along with “finger multiplication” and “Russian multiplication” presented below, these techniques truly demonstrate the joy to be had engaging in intellectual play. The multiplication methods outlined invite us to reexamine the long-multiplication algorithm we teach our young students. Consider, for example, the computation 83 × 27 . Students are taught to write something of the ilk:\n\nMathematicians recognize this as an exercise in “expanding brackets:”\n\n83 × 27 = ( 80 + 3)( 20 + 7 ) = 1600 + 60 + 560 + 21 (Select one term from each set of parentheses, multiply, and repeat. Add together all possible combinations that can arise in this way.) Young students might be taught to associate an “area model” with multiplication, in which case, expanding brackets is equivalent to dividing a rectangle into computationally simpler pieces.\n\n© James Tanton 2010\n\n5\n\nNotice that adding the numbers in the cells according to the diagonals on which they lie conveniently groups the 100s, the 10s and the units. COMMENT: When I teach long multiplication to students, this area model is the only approach I adopt. I personally feel that the level of understanding that comes of it far outweighs the value of speed that comes from the standard algorithm: students remain acutely aware that the “8” in the above problem, for instance, means eighty and there is no mystery about placement of zeros and carrying digits. ASIDE: In the 1500s in England students were taught to compute long multiplication using following galley method (also known as the lattice method or the Elizabethan method):\n\nTo multiply 218 and 43, for example, draw a 2 × 3 grid of squares. Write the digits of the first number along the top of the grid and the digits of the second number along the right side. Divide each cell of the grid diagonally and write in the product of the column digit and row digit of that cell, separating the tens from the units across the diagonal of that cell. (If the product is a one digit answer, place a 0 in the tens place.)\n\nAdd the entries in each diagonal, carrying tens digits over to the next diagonal if necessary, to see the final answer. In our example, we have 218 × 43 = 9374 . The diagonals serve the purpose of grouping together like powers of ten. (Although, again, the lack of zeros can make this process quite mysterious to young students.)\n\n© James Tanton 2010\n\n6\n\nIt is now easy to see that the lines method is just the area model of multiplication in disguise (since the count of intersection points matches the multiplication of the single digit numbers).\n\n22 × 13 = 286 The strip method is a clever geometric tracking of the individual multiplications that occur in a long multiplication problem (with place-value keeping track of the powers of ten for us):\n\nAnd the Vedic technique is a memorization device for the standard algorithm. CHALLENGE: Think about these. Be sure to understand why these two techniques really are just alternative ways of presenting our familiar long multiplication algorithm.\n\n© James Tanton 2010\n\n7\n\nCircle/Radius Math It is not too hard to see that drawing a common radii for b circles yields ab pieces. (One radius produces b pieces and each radius inserted thereafter adds an additional b segments.)\n\nThus, “circle/radius math” is ordinary multiplication in disguise. (Is it? What happens if one or both of a or b is zero?) As ordinary multiplication is taken to be commutative, this multiplication must be commutative too. Without thinking through the arithmetic, this can be a surprise. PEDAGOGICAL COMMENT: The Role of the Area Model for Multiplication Mathematicians recognize that the area model for multiplication has its theoretical limitations—what is the area of a\n\n3 -by-\n\n15 rectangle, for example?—but it does justify 97\n\nthe axioms we care to work with when we define the properties we like believe of numbers. Pedagogically, it certainly seems to be an appropriate and intuitively accurate place for beginning understanding and investigation. For example, if we allow for negative numbers in our considerations and consider rectangles with negative side-lengths (young students absolutely delight in “breaking the rules” in this way), then we can help justify many confusing features of the arithmetic we choose to believe. My favorite is tackling the issue of why the negative times negative should be positive. In the world of counting numbers, it seems natural to say that 7 × 3 represents “seven groups of three” and so equals 21 . In the same way, 7 × ( −3) represents “seven groups of negative three” and equals −21 . It is difficult to interpret ( −7 ) × 3 in this way, but if we choose to extend the commutativity law to this realm, then we can say that this is the same as “three groups of negative seven” and so again is −21 . Up to this point, I find that students (and teachers!) feel at ease and willing to say that all is fine. The kicker comes in trying to properly understand ( −7 ) × ( −3) . © James Tanton 2010\n\n8\n\nLet’s do this by computing 23 × 17 multiple ways. Here is the standard way:\n\nLet’s be sneaky and compute the same product, allowing ourselves the freedom to play with negative quantities. In the first diagram we choose to represent 23 as 30 − 7 , in the second we write 37 as 40 − 3 . Both again, of course, yield the answer 851.\n\nNow let’s write 23 as 30 − 7 and 37 as 40 − 3 together at the same time. This represents the same computation and the answer must still be 851.\n\nWe see that we have no choice but to set ( −7 ) × ( −3) equal to +21 . © James Tanton 2010\n\n9\n\nIII: FINGER MULTIPLICATION Don’t memorize your multiplication tables. Let your fingers do the work! If you are comfortable with multiples of two, three, four, and five, then there is an easy way to compute product values in the six- through ten times tables. First encode numbers this way:\n\nA closed fist represents “five” and any finger raised on that hand adds “one” to that value. Thus a hand with two fingers raised, for example, represents “seven” and a hand with three fingers raised represents “eight.” To multiply two numbers between five and ten, do the following:\n\n1. Encode the two numbers, one on each hand, and count “ten” for each finger raised. 2. Count the number of unraised fingers on each hand and multiply together the two counts. 3. Add the results of steps one and two. This is the desired product. For example, “seven times eight” is represented as two raised fingers on the left hand, three on the right hand. There are five raised fingers in all, yielding the number “50” for step one. The left hand has three lowered fingers and the right, two. We compute: 3 × 2 = 6 . Thus the desired product is 50 + 6 = 56 . Similarly, “nine times seven” is computed as 60 + 1× 3 = 63 , and “nine times nine” as 80 + 1 × 1 = 81 . Notice that one is never required to multiply two numbers greater than five! COMMENT: Type “TANTON FINGER” on www.youtube.com to see a video of this! FINGERS AND TOES: One can compute higher products using the same method! For example, with fingers and toes, one interprets 17 × 18 as “seven raised fingers” and “eight raised toes.” This time we count each raised digit as “twenty” (we have twenty digits fingers and toes in all!) yielding: 17 × 18 = 20 ×15 + 3 × 2 = 306 ! QUESTION: Why does finger multiplication work? Why does finger/toe multiplication work? QUESTION: Martians have six fingers on each of two hands. Describe their version of the finger multiplication trick. © James Tanton 2010\n\n10\n\nIV: RUSSIAN MULTIPLICATION The following unusual method of multiplication said to have originated in Russia.\n\n1. Head two columns with the numbers you wish to multiply. 2. Progressively halve the numbers in the left column (ignoring remainders) while doubling the figures in the right column. Reduce the left column to one. 3. Delete all rows with an even number in the left and add all the numbers that survive in the right. This sum is the desired product!\n\nTry computing other products via this approach. QUESTION: Why does this method work? (See THINKING MATHEMATICS: Volume 1 for details.)\n\nFINAL FINAL THOUGHT This packet is about weird multiplication methods. But have you noticed that the spelling of the word “weird” is itself weird?\n\n© James Tanton 2010"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9004374,"math_prob":0.9338993,"size":12076,"snap":"2021-04-2021-17","text_gpt3_token_len":2791,"char_repetition_ratio":0.14653744,"word_repetition_ratio":0.004748338,"special_character_ratio":0.24337529,"punctuation_ratio":0.109225415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99158776,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-20T01:11:27Z\",\"WARC-Record-ID\":\"<urn:uuid:d2ccc07a-fd19-4fcd-85fe-5022b2a9af88>\",\"Content-Length\":\"61501\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ec7c8e1-a09e-43d2-8ba7-87a67be14ff7>\",\"WARC-Concurrent-To\":\"<urn:uuid:87e34981-b0e4-4b15-b086-8a87bfadef95>\",\"WARC-IP-Address\":\"172.67.178.63\",\"WARC-Target-URI\":\"https://kipdf.com/weird-multiplication-james-tanton_5b0e05247f8b9a3c9b8b45e1.html\",\"WARC-Payload-Digest\":\"sha1:4T6OLB5COQTPQMIQGBGZE2LH553X4DVE\",\"WARC-Block-Digest\":\"sha1:6MBVHVGGUCPBPTZ6PLG7S7JVWI3UPQBB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038921860.72_warc_CC-MAIN-20210419235235-20210420025235-00330.warc.gz\"}"} |
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff;h=14cda93136c935ddbe8004120e4842b3808c107a | [
"author Chris Sun, 22 Feb 2015 14:51:00 +0000 (09:51 -0500) committer Chris Sun, 22 Feb 2015 14:51:00 +0000 (09:51 -0500)\n\nindex 78ffc89..17d8eab 100644 (file)\n@@ -310,3 +310,62 @@ so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation...\n* I hear that `Y` delivers the/a *least* fixed point. Least\naccording to what ordering? How do you know it's least?\nIs leastness important?\n+\n+## Q: I still don't fully understand the Y combinator. Can you\n+ explain it in a different way?\n+\n+Sure! Here is another way to derive Y. We'll start by choosing a\n+specific goal, and at each decision point, we'll make a reasonable\n+guess. The guesses will all turn out to be lucky, and we'll arrive at\n+a fixed point combinator.\n+\n+Given an arbitrary term f, we want to find a fixed point X such that\n+\n+ X <~~> f X\n+\n+Our strategy will be to seek an X such that X ~~> f X. Because X and\n+f X are not the same, the only way that X can reduce to f X is if X\n+contains at least one redex. The simplest way to satisfy this\n+constraint would be for the fixed point to itself be a redex:\n+\n+ X == ((\\u.M) N) ~~> f X\n+\n+The result of beta reduction on this redex will be M with some\n+substitutions. We know that after these substitutions, M will have\n+the form `f X`, since that is what the reduction arrow tells us. So we\n+can refine the picture as follows:\n+\n+ X == ((\\u.f(___)) N) ~~> f X\n+\n+Here, the ___ has to be something that reduces to the fixed point X.\n+It's natural to assume that there will be at least one occurrence of\n+\"u\" in the body of the head abstract:\n+\n+ X == ((\\u.f(__u__)) N) ~~> f X\n+\n+After reduction of the redex, we're going to have\n+\n+ X == f(__N__) ~~> f X\n+\n+Apparently, `__N__` will have to reduce to X. Therefore we should\n+choose a skeleton for N that is consistent with what we have decided\n+so far about the internal structure of X. We might like for N to\n+match X in its entirety, but this would require N to contain itself as\n+a subpart. So we'll settle for the more modest assumption that N\n+\n+ X == ((\\u.f(__u__)) (\\u.f(__u__))) ~~> f X\n+\n+At this point, we've derived a skeleton for X on which it contains two\n+so-far identical halves. We'll guess that the halves will be exactly\n+identical. Note that at the point at which we perform the first\n+reduction, `u` will get bound N, which now corresponds to a term\n+representing one of the halves of X. So in order to produce a full X,\n+we simply make a second copy of `u`:\n+\n+ X == ((\\u.f(uu))(\\u.f(uu))) ~~> f((\\u.f(uu))(\\u.f(uu))) == f X\n+\n+Success.\n+\n+So the function `\\f.(\\u.f(uu))(\\u.f(uu))` maps an arbtirary function\n+`f` to a fixed point for `f`."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9225963,"math_prob":0.9329838,"size":2404,"snap":"2019-51-2020-05","text_gpt3_token_len":707,"char_repetition_ratio":0.12583333,"word_repetition_ratio":0.010615711,"special_character_ratio":0.31489184,"punctuation_ratio":0.10831721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950888,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T02:12:21Z\",\"WARC-Record-ID\":\"<urn:uuid:22f1153a-345e-4821-88f1-008c17b0de0c>\",\"Content-Length\":\"15293\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58af6601-2159-4d69-aeb7-78c5ff3140a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f34bd52b-51b5-401c-9343-c1ef5fff8398>\",\"WARC-IP-Address\":\"45.79.164.50\",\"WARC-Target-URI\":\"http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff;h=14cda93136c935ddbe8004120e4842b3808c107a\",\"WARC-Payload-Digest\":\"sha1:5Q3DMYXVZRPVP6L3FSLD7SP5HJCUG5VO\",\"WARC-Block-Digest\":\"sha1:OIOYGPFJW34H2CPRKDCO5CLVETSBL6TN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251684146.65_warc_CC-MAIN-20200126013015-20200126043015-00473.warc.gz\"}"} |
https://conorhaining.com/posts/network-visualisation-with-matlab/ | [
"",
null,
"# Network Visualisation with MATLAB\n\nFor a recent university project we were analysing large networks. We used the Standford Network Analysis Project (SNAP) to perform a number of operations on any dataset we wanted. We searched for a large dataset as we believed it could yeild some pretty interesting results. We settled on the California Road Network where nodes represent either an intersection, or end point (dead end) and the edges represent the roads between them. The website states that this is an undirected edge but we found during the visualisation that there is infact directed edges.\n\nThere are some quite interesting things to notice about the dataset when you perform some network anaysis on it. However we were interested in producing some kind of visual for our presentation.\n\n# Visualisation\n\nGiven that the dataset contains 1965206 nodes, the adjaency matrix would be [1965206 X 1965206] large and if each element is represented as an integer (4 bytes), the size of the matrix is around 3596GB so that was out of the question. Using the SNAP tools, we could calculate the communities within the matrix.\n\nWe were able to extract a matrix of a few thousand nodes and visualise that.\n\n## Using MATLAB\n\nOnce you've downloaded and extracted the dataset from SNAP, navigate to that folder using the bar just below the ribbon. You can import the data by either using the Import Data button on the ribbon, or by typing `roadNet-CA = textread('roadNet-CA.txt')`. Note that if you use this second command method you'll need to delete the first few lines which contain the hash symbol.\n\nTo extract a specific community you'll need to also load in the communities to MATLAB. If you place file generated by SNAP into the same directory as open in MATLAB, you can load that just as above. Either use the Import Data button or `communities = textread('communities.txt')`. The `communities` variable contains a list of nodes and the number of which community they belong to. We had nearly 2 million nodes, so we wanted to filter the data and generate networks based on smaller communities.\n\n## Filtering the network\n\nWe need to create a subset of the `roadNet-CA` variable, we can do so but running a command which goes through the second column in the matix and returns a `1` (or true). You can do so by running `communityIndice = communities(:, 2) = #` replace the # with the community number you're filtering.\n\nNext we need to create a matrix holding only the nodes found in the `roadNet-CA` matrix and we can do so by running `community# = communities(comminityIndice, :)` which gets the rows where a 1 appears inside either of column and adds that to the new variable.\n\nThe final stage is to get a logic matrix of when the node ID appears inside the whole network, which an be done by `ans = ismember(roadNet-CA, community#)` and finally to get an edge list by running `roadNet-CA# = roadNetCA(ans(:, 1), :)`\n\n## Creating the Network Graph\n\nWe need to split up the newly created `roadNet-CA#` variable, into two seperate variables. `s = roadNet-CA#(:, 1)` gets our first column, and `t = roadNet-CA#(:, 2)` gets our second column. We can't generate the graph quite yet as they need to be rotated, so we can apply the `s = rot90(s)` and `t = rot90(t)` to each variables.\n\nTo create the graph, we can do so by running `G = digraph(s, t)` (use `graph` if you have an undirected network). To produce the figure run the command `plot(G)` and this will generate a nice visual on the graph."
] | [
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https://byjus.com/question-answer/a-restaurant-hall-is-20-metres-long-15-metres-wide-and-10-metres-high-what-will-be-the-total-expenditure-if-the-wall-paper-costs-rs60-per-square-metre/ | [
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"Question\n\n# A restaurant hall is $20m$ long, $15m$ wide, and $10m$ high. The interior of the hall except the ceiling and the base has to be covered with bright wallpaper. What will be the total expenditure if the wallpaper costs $₹60$ per square meter?\n\nOpen in App\nSolution\n\n## Length $=20m$Breadth $=15m$Height $=10m$To find:Total cost to cover the walls1) Only the walls of the rectangular hall are included when you covered it with the wallpaper, So;The area of the hall to be covered $=2h\\left(l+b\\right)$ $=2×10\\left(20+15\\right)\\phantom{\\rule{0ex}{0ex}}=2×10×35\\phantom{\\rule{0ex}{0ex}}=700{m}^{2}$2) Rate of covering the wall is $₹60/{m}^{2}$Total expenditure $=$ Area × rate $=700×60\\phantom{\\rule{0ex}{0ex}}=₹42000$The total cost to cover the walls is $₹42000$.",
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https://lead-academy.org/blog/what-is-an-outlier-defined-as-a-level-maths/ | [
"Become an affiliate partner and earn attractive commission.\n\nX\n\n# What is an Outlier Defined as A Level Maths?",
null,
"“What is an outlier defined as A Level maths?” You may get confused seeing the question as we’ve directly copy pasted the Google search term here. Let’s rephrase it so you’ll understand it a bit better. “What is an outlier defined as in A Level maths” or “Describe outlier meaning in statistics.” Well, as the latter question clearly spits it out, it’s a concept related to statistics. An outlier is a data point in a distribution that breaks away from the overall pattern.\n\nSince you’ve searched the term, we’re going to assume that you’re familiar with statistics and quantitative analysis. So we’re not going to talk about these here today. Instead, we’re going to delve right into your question.",
null,
"## What is an Outlier Defined as A Level Maths?\n\nWhat are outliers? Instead of going to the definition of outlier in statistics, it’ll be much easier to understand if we describe it.\n\nAt A-level, you’ll find the concept of outliers relatively early in your curriculum. Outliers will come up as soon as you come in touch with the following concepts:\n\nTo understand an outlier, you first have to be familiar with the following terms:\n\n1. Standard deviations\n2. Interquartile range\n\n####",
null,
"",
null,
"Are you looking for a Functional Skills Maths Level 2 Online Course?\n\nThe Level 2 Functional Skills Maths is equivalent to GCSE Grade 4. Our Functional Skills Maths Level 2 Online Course is an incredible course that will provide you with the necessary skills and credentials, as well as practical and professional competence. It will help you to further your studies at a university or choose a career that you hope to chase.\n\n### Standard Deviations\n\nA standard deviation is a measure of how widely distributed the data is in reference to the mean. A low standard deviation suggests that data is grouped around the mean, whereas a large standard deviation shows that data is more spread out.\n\nIn statistics, it is often denoted with the Greek word sigma.\n\nσ\n\n### Interquartile Range\n\nThe interquartile range in descriptive statistics describes the spread of your distribution’s middle half.\n\nBut, What is quartile? Quartiles are values that divide data into four equal parts, similar to how the median divides data into two.\n\nIn fact, the median is the second quartile. The second and third quartiles, or the centre half of your data set, are represented by the interquartile range (IQR).\n\nIn other words, the difference between the 25th and 75th percentiles is known as the interquartile range. So, the 25th and 75th percentiles are also called the first and third quartiles. As a result, the interquartile range describes the middle 50% of observations.\n\nIf the interquartile range is broad, it suggests that the middle 50% of observations are widely apart. The main advantage of the interquartile range is that it can be used as a measure of dispersion/variability even if the extreme values are not captured precisely.\n\nAnother advantage is that it is unaffected by extreme values.",
null,
"### Outliers\n\nSo, outliers are data points that lie 1.5 times below the 1st quartile or 1.5 times above the 3rd quartile. We can also say that outliers are data points that lie 1.5 times above or below the interquartile range.\n\n#### Top Courses of this Category",
null,
"##### Engineering Calculus Made Simple (Derivatives) Course Online\n\nBestseller",
null,
"##### Functional Skills Maths and English Level 2 Course\n\nTrending",
null,
"Trending",
null,
"Bestseller\n\n## How to Calculate an Outlier?\n\nOutliers can be dealt with using the interquartile range. Because the interquartile range is the middle half of the data, it is reasonable to define an outlier as a specific multiple (1.5 times to be exact) of the interquartile range below the first quartile or above the third quartile.\n\nAn exam question might provide you a data set and ask you to compute the interquartile range and identify outliers.\n\nFor example, consider the following data set. Imagine that they are annual savings of 12 people in quantities of thousands. So you will also have to interpret the output data as thousands.\n\n Data Points 1 4 5 5 6 6 6 6 7 10 12 56\n\n### Analysis of the 12 Data Points\n\nThere are 12 data points here. Let’s analyse them.\n\n μ or x̅ Mean 10.33333333 Median (0 percentile) 6 σ Standard Deviation 14.64323197 Q1 1st Quartile (25th percentile) 5 Q2 2nd Quartile (50th percentile) 6 Q3 3rd Quartile (75th percentile) 9.25 IQR Interquartile Range Q3 – Q1\n\nThe mean here is the average. The median is the middle point of the data set, also called the 50th percentile. The standard deviation is 14.64323197, and it shows how far spread out the data set is from the mean. The interquartile range here is the value when we calculate Q3-Q1.",
null,
"However, if we look at the data set and exclude 56 out of it, the last data point would be 12, approximately 2.64 less than the standard deviation. It makes very little sense. This happened here because of the data point 56. If we exclude it and recalculate the standard deviation for the remaining 11 data points, it would be 2.891995222, which makes perfect sense.\n\nSo you can see how data points that are too far from the mean can affect the statistical analysis. Both mean and standard deviations are highly sensitive to extreme data points.\n\nStatisticians call these extreme data points outliers, which can go both ways- either above the mean, like our example above, or below the mean.\n\nThis is also where the interquartile range comes in. It helps us understand where the majority of the data set is and identify outliers.",
null,
"### Using Interquartile Range to Determine Outliers\n\nSo, the outlier formula or outlier equation is:\n\n Outliers below the Mean Q1 – 1.5 x IQR Outliers above the Mean Q3 + 1.5 x IQR\n\nFirst, calculate the interquartile range and multiply it by 1.5. Then, subtract this value from the 1st quartile and then also add it to the 3rd quartile. The two values that you end up with are the acceptable statistical data range. Any data point outside this would be an outlier.\n\nBut there is another way to identify outliers that is common in A level\n\n### Using Standard Deviation to Determine Outliers\n\nThe following are the formula to identify outliers using the standard deviation.\n\n Outliers below the Mean x̅ – kσ Outliers above the Mean x̅ + kσ\n\n### FAQs\n\n#### How do you determine an outlier in A level?\n\nAn outlier in A level can be determined by looking for data points that are significantly different from the rest of the data set.\n\n#### How do you use standard deviation to determine outliers in A level maths?\n\nThe formula to find outliers using the standard deviation is as follows.\n\n Outliers below the Mean x̅ – kσ Outliers above the Mean x̅ + kσ\n\n#### Is an outlier 2 standard deviations from the mean?\n\nFirstly, the mean (or average) of the data set needs to be calculated, followed by the standard deviation, which is a measure of how much the data points vary from the mean. Once the mean and standard deviation have been calculated, any data point that lies outside of 1 or 2 standard deviations away from the mean can be identified as an outlier.\n\n#### Why do we use 1.5 IQR for outliers?\n\nThis number clearly affects the sensitivity of a data set and, thus, the decision rule. Further, if we increase the scale from 1.5 to something greater, some outliers will be included in the data range, severely affecting it.\n\n#### What is an outlier in math on a box plot?\n\nAn outlier is a numerically distant observation from the rest of the data. When examining a box plot, an outlier is defined as a data point that lies outside the box plot’s whiskers. The following is the box plot for our example data set in the blog.\n\nCan you see a little blue dot in the 50-60 range of the Y-axis? That’s the value 56, our outlier.",
null,
"#### What is singled out meaning in statistics?\n\nIn language, “single out” has a different meaning than it does in statistics. In statistics, we “single out” the outliers and leave them out of the calculation so that they can’t distort the representation of the overall dataset.\n\n## Conclusion\n\nSo, what is an outlier defined as A Level maths? Extreme data points which can affect the statistical analysis of some given data are outliers. They can cause serious problems if we don’t identify them before we move to draw conclusions on a data set.\n\nThe blog has all the instructions to identify outliers. You can also see the example that we provided to get a practical idea of how outliers affect a set of data and how you can easily recognise them.",
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https://cdn2.jlqwer.com/posts/4433.html | [
"# Exchange Rates\n\nExchange Rates",
null,
"Now that the Loonie is hovering about par with the Greenback, you have decided to use your \\$1000 entrance scholarship to engage in currency speculation. So you gaze into a crystal ball which predicts the closing exchange rate between Canadian and U.S. dollars for each of the next several days. On any given day, you can switch all of your money from Canadian to U.S. dollars, or vice versa, at the prevailing exchange rate, less a 3% commission, less any fraction of a cent.\n\nAssuming your crystal ball is correct, what's the maximum amount of money you can have, in Canadian dollars, when you're done?\n\nThe input contains a number of test cases, followed by a line containing 0. Each test case begins with 0 <d ≤ 365, the number of days that your crystal ball can predict. d lines follow, giving the price of a U.S. dollar in Canadian dollars, as a real number.\n\nFor each test case, output a line giving the maximum amount of money, in Canadian dollars and cents, that it is possible to have at the end of the last prediction, assuming you may exchange money on any subset of the predicted days, in order.\n\n``````3\n1.0500\n0.9300\n0.9900\n2\n1.0500\n1.1000\n0\n``````\n\n``````1001.60\n1000.00\n``````\n\n``````#include <stdio.h>\n\n#define ROUND 0\nint usd;\n\ndouble rate;\n\nint initusd = 0;\ndouble commission = .03;\n\ndouble max(double a, double b, double c) {\nif(a>=b && a>=c) return a;\nif(b>=c) return b;\nreturn c;\n}\n\nmain() {\nint n, i;\nwhile(1) {\nscanf(\"%d\", &n);\nif(!n) break;\nfor(i = 0; i < n; i++) {\nscanf(\"%lf\", rate + i);\n}\n\nfor(i = 0; i < n; i++) {\nusd[i+1] = max(initusd + initcad/rate[i]*(1-commission) + ROUND,\n}\n\n}\n}``````\n\n``````#include <stdio.h>\n\n#define ROUND 0\nint usd;\n\ndouble rate;\n\nint initusd = 0;\ndouble commission = .03;\n\ndouble max(double a, double b, double c) {\nif(a>=b && a>=c) return a;\nif(b>=c) return b;\nreturn c;\n}\n\nmain() {\nint n, i;\nwhile(1) {\nscanf(\"%d\", &n);\nif(!n) break;\nfor(i = 0; i < n; i++) {\nscanf(\"%lf\", rate + i);\n}\n\nfor(i = 0; i < n; i++) {"
] | [
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"http://plg1.cs.uwaterloo.ca/~acm00/070923/million.jpg",
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https://edharmalib.com/lib/earticles/ear0002 | [
"",
null,
"## Calculation of the characteristics of the year",
null,
"# Calculation of the characteristics of the year\n\nWe continue to add a rebus or a constructor from the simplest cubes. In previous publications, the methods of calculating the new year or losar (Phugpa, Bon and Rebkong, Gyalpo-losar), lists of months and their start dates were considered. Now we will smoothly move on to how we can determine the other main parameters of the year. In particular, we will talk about the number of the year according to the Tibetan calendar, the animal sign, the elements wangthang, lungta, sog, lu, etc. All these calculations are quite simple, you just need to be able to count a little. Moreover, you can count on a calculator or using just a piece of paper. Also, based on these calculations, you can determine some of the characteristics of your own birth. But this will be explained in another publication.\n\nSo. Let's start by determining the number of the year in the Tibetan calendar. Here we can use two calculation options:\n- based on the year number in the Gregorian calendar (but remember that the beginning of the year in the Gregorian calendar does not coincide with the beginning of the year in the Tibetan calendar).)\n- based on the month number for the first month of the Tibetan year.\n\nThe simplest option is the first method, so we will focus on it. We will use 2019 as an example.\nSo. In order.\n\n### 1. The year of the Tibetan calendar\n\nThe Tibetan calendar differs from the one used in most countries by 127 years. Therefore, we can add 127 to the year number and get the required year number, based on which further calculations will be performed.\n2019+127=2146\n\n### 2. The number of the Rabjung and the number of the year in it.\n\nRabjuns are sixty-year cycles, the beginning of which is considered to be the year 1027 (Fire-W-Hare). The number of the rabjun and the number of the year in it are determined accordingly. The year 1027 is the year 1154 according to the Tibetan calendar. To determine the number of the year in Rabjun, we need to subtract the number of the first or base year and divide the remaining value by 60. Add 1 to the remainder, so that there is no confusion with the remainder equal to zero. You also need to add 1 to the integer, so that there is no confusion with the \"zero\" rabjun.\n2146-1154=992\n992/60=16 and the remainder 32.\n16+1=17 (rabjun number)\n32+1=33 (the number of the year in rabjun)\n\n### 3. The number of the mekor and the year in it.\n\nAnother variant of the definition of sixty-year cycles, which was used earlier, is called mekor. For its beginning, we can consider 2576 BC or 2449 BC (according to the Tibetan style)\nWe make a similar calculation to the previous one\n2146-(-2449)=4595\n4595/60=76 and the remainder 35\n76+1=77 (mekor number)\n35+1=36 (mekor year number)\n\n### 4. Determination of the metreng and meva of the year\n\nThe meva is determined based on a large cycle of 180 years. During this cycle, three metrengs (the garland or rosary of mewa) change. Actually, the number of the metreng is not particularly important for us, but we will still count it. As we identified earlier, we have 77 mekor and 36 number in mekor. The metreng number can be determined based on the remainder of the division of the mekor number by 3. In this case, you need to add one, since there is no \"zero metreng\". That is, as a result, we will get a number from 1 to 3.\nCounting\n77/3=92 and the remainder is 2. That is, we use the third metreng.\nIn order not to experience the agony of trying to determine the actual meva and not to consult the tables for calculations, we will do it in a simpler way. Since we have the third cycle of the mewa (third metreng), we can add 120 to the year number in the metreng. In that case, if we had the first metreng , we do not need to add anything. If the second metreng - to the year number in the metreng, add 60.\n36+120=156.\nSince the mevas change in the opposite direction, we must also go in the opposite direction to determine. That is, subtract from the maximum value in the cycle - the number of the year in the large cycle.\nSubtract the resulting value from 180.\n180-156=24.\nSince zero or 180 corresponds to the number 2, we add 2.\n24+2=26\nNow we need to determine the actual meva number. Since mevs have values from 1 to 9, we divide the resulting value by 9 and look at the remainder.\n26/9=2 and the remainder is 8. In the event that the remainder is zero, we substitute a nine instead of zero. So we got an eight. Look at the short list of values\n\n Number Mewa 1 1, white 2 2, black 3 3, blue 4 4, green 5 5, yellow 6 6, white 7 7, red 8 8, white 9 9, red\n\nWe got an eight, so mewa according to the table is 8, white.\n\n### 5. Animal, gender, year direction\n\nRabjuns and mekors have different beginnings, different primary years. Also in these cycles, the calculation comes from different animals and elements. So in rabjun, the calculation comes from the Fire-Hare, while the mekors are counted from the Tree-Rat. Since the mekor cycles are calculated from older times, we will use them. The sequence of years in the 12 animal cycle is as follows:\n\n Number Animal Direction 1 Rat north (bottom) 2 Bull Northeast 3 Tiger East (top) 4 Hare East (bottom) 5 Dragon Southeast 6 Snake south (top) 7 Horse South (bottom) 8 Sheep southwest 9 Monkey west (top) 10 Bird West (bottom) 11 Dog North-west 12 Pig north (top)\n\nWe take the year number in the cycle, divide by 12, and look at the remainder. If the remainder is 0, replace the resulting value with 12.\n36/12=3 and the remainder 0. Replace with 12. We get 12. We take the data from the table and get the year of the Pig.\n\nMale years include: Rat, Tiger, Dragon, Horse, Monkey, Dog. To the women's: Bull, Hare, Snake, Sheep, Bird, Pig.\nThe resulting year is the year of the Pig, female. We also take the directions of the year from the table and get the following data: year of the Pig, female. Direction north (top)\n\n### 6. Definition of wangthang\n\nAs many have seen, calendars often write the year of the Earth-Pig. Actually, this indication of the element can be called wangtang (success, etc.). Of course, it can be defined using tables, but this is not very convenient. To determine wang, you can also use mathematical means, and simple ones. Since when analyzing the table, you can see that the vang elements always go in pairs, we can first divide the year number in the metreng by 2.\nWe get:\n36/2=18 and the remainder is 0. If the remainder is not zero, then add it to the resulting value (for 2018, it would be 35/2=17 and 1. 18).\nSince we have five elements, we divide the resulting value by five and look at the remainder. At the same time, we also need to take into account that we do not have a zero element. So when we get the remainder zero, we must write the number 5. So, look:\n18/5=3 and the remainder is 3.\n\nChecking the table of elements:\n\n Number Element 1 tree 2 fire 3 earth 4 metal 5 water\n\nWe get the earth. So we have a year of Earth-w-Pig.\n\n### 7. The life force of the year or Sog\n\nTo determine them, use the table:\n\n Year Element Tiger, Hare tree Horse, Snake fire Bird, Monkey metal Rat, Pig water Dog, Dragon, Bull, Sheep earth\n\nThe Year of the Pig, so the life force of the year is water.\n\n### 8. Determination of the health or element of lu.\n\nIt is easier to use a table to determine the Lu.\n\n Element Years Tree Earth-Dragon, Earth-Snake, Earth-Dog, Earth-Pig, Metal-Tiger, Metal-Hare, Metal-Monkey, Metal-Bird, Water-Rat, Water-Bull, Water-Horse, Water-Sheep Fire Tree-Dragon, Tree-Snake, Tree-Dog, Tree-Pig, Fire-Tiger, Fire-Hare, Fire-Monkey, Fire-Bird, Earth-Rat, Earth-Bull, Earth-Horse, Earth-Sheep Earth Fire-Dragon, Fire-Snake, Fire-Dog, Fire-Pig, Earth-Tiger, Earth-Hare, Earth-Monkey, Earth-Bird, Metal-Rat, Metal-Bull, Metal-Horse, Metal-Sheep Metal Tree-Rat, Tree-Bull, Tree-Horse, Tree-Sheep, Metal-Dragon, Metal-Snake, Metal-Dog, Metal-Pig, Water-Tiger, Water-Hare, Water-Monkey, Water-Bird Water Tree-Tiger, Tree-Hare, Tree-Monkey, Tree-Bird, Fire-Rat, Fire-Bull, Fire-Horse, Fire-Sheep, Water-Dragon, Water-Snake, Water-Dog, Water-Pig\n\nFor our year: a tree.\n\n### 9. Luck or lungta of the year, the years of harmonious triples\n\nWe determine it by the table\n\n Element Years Metal Tiger, Horse, Dog Tree Rat, Dragon, Monkey Water Bird, Bull, Snake Fire Pig, Sheep, Hare\n\nWe have the year of the Pig, so the element of lungta is Fire.\n\n### 10. La of year\n\nSince it is said that La is the mother of the life force, we remember that we have the life force of the year-water or, if we take the numerical correspondence, 5. How to determine the mother of the element? To do this, simply subtract 1 from the element number. That is, go to the previous one. If we get 0, then we need to write 5.\n5-1=4. Element number four is metal. That is, La of year - Metal.\n\n### 11. Special lungta of the year\n\nThe special Lungta of the year is defined as the son of Lungta, that is, as the element following the Lungta element. We have the Lungta of the year-Fire, number 2. Add 1. If we get 6, then we need to replace it with 1.\n2+1=3. The element, according to the table, is earth. Special lungta of the year: earth.\n\n### 12. Five years of khayen\n\nTo determine this type of years, we need to compare the elements. And depending on whether they coincide or not, a particular year will be determined. There are five years of this kind. These are: Khayen, Sejig, Khongnong, Kharal, Dunkhur. How do I determine if a year belongs to one of them? To do this, we need to know the life force and wangthang of the year. For the year used in the calculations, sog is water (5), wangthang is earth (3).\n\nIf Sog and Wang are the same, it's the year of Khayen. The elements are different-so no.\nIf Wang is the son of Sog (number wang+1=sog) - sejig. Check: 3+1=4. 4 is not equal to 5. Not a sejig.\nIf Wang is the mother of Sog (number vang-1=sog) - Khongnong. Check: 3-1=2. 2 is not equal to 5. Not Khongnong.\nIf Wang is a friend of Sog (sog number+2) - Kharal. Check. 5+2=7 or 2. Not equal to 3. Not kharal.\nIf Vang is the enemy of Sog (number sog-2) - Dunkhur. Check. 5-2=3. Early 3. The Year of Dunkhur.\n\nActually, this is the end of the description of the calculations related to the year. The following articles will be devoted to other topics on astrology. About all inconsistencies, errors, strange places-please write to the site administrator. There are contacts on every page\n\nWell, those who are too lazy to count, may well use the link with automatic calculations for the current and next year."
] | [
null,
"https://edharmalib.com/images/logo_new.jpg",
null,
"https://edharmalib.com/images/fin/tibcal.jpg",
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https://thebusinessprofessor.helpjuice.com/accounting-taxation-and-reporting-managerial-amp-financial-accounting-amp-reporting/weighted-average-product-cost-explained | [
"# Weighted Average Product Cost - Explained\n\nWhat is the Weighted Average Cost of Products?\n\n# What is the Weighted Average Product Costs?\n\nThe weighted average method is used to assign all costs in a process-costing system to the products produced.\n\nIn a process costing system, cost per equivalent unit is the term used to describe the average unit cost for each product.\n\nThe concept of cost per equivalent unit used to assign costs to (1) completed units transferred out and (2) units still in work-in-process (WIP) inventory at the end of the period?\n\nCosts are assigned to completed units transferred out and units in ending WIP inventory using a four-step process.\n\nWe list the four steps in the following and then explain them in detail.\n\nStep 1. Summarize the physical flow of units and compute the equivalent units for direct materials, direct labor, and overhead.\n\nStep 2. Summarize the costs to be accounted for (separated into direct materials, direct labor, and overhead).\n\nStep 3. Calculate the cost per equivalent unit.\n\nStep 4. Use the cost per equivalent unit to assign costs to (1) completed units transferred out and (2) units in ending WIP inventory.\n\n## What are Equivalent Units?\n\nThe process of accounting for the flow of costs through accounts used in a process costing system is fairly straight forward.\n\nIn a process costing system, cost per equivalent unit is the term used to describe the average unit cost for each product.\n\nThe concept of cost per equivalent unit is used to assign costs to (1) completed units transferred out and (2) units still in work-in-process (WIP) inventory at the end of the period?\n\nThe challenge is determining the unit cost of products being transferred out of each departmental work-in-process inventory account.\n\nUnits of product in work-in-process inventory are assumed to be partially completed; otherwise, the units would not be in work-in-process inventory. The ending work-in-process inventory to be converted to the equivalent completed units (called equivalent units).\n\nEquivalent units are calculated by multiplying the number of physical (or actual) units on hand by the percentage of completion of the units.\n\nIf the physical units are 100 percent complete, equivalent units will be the same as the physical units. However, if the physical units are not 100 percent complete, the equivalent units will be less than the physical units.\n\nFor example, if four physical units of product are 50 percent complete at the end of the period, an equivalent of two units has been completed (2 equivalent units = 4 physical units × 50 percent).\n\nThe formula used to calculate equivalent units is as follows:\n\nEquivalent units = Number of physical units × Percentage of completion\n\nBecause direct materials, direct labor, and manufacturing overhead typically enter the production process at different stages, equivalent units must be calculated separately for each of these production costs.\n\n## Calculating Equivalent Units\n\nEquivalent units in work in process are often different for direct materials, direct labor, and manufacturing overhead because these three components of production may enter the process at varying stages.\n\nFor example, in the Assembly department at Desk Products, Inc., direct materials enter production early in the process while direct labor and overhead are used throughout the process. (Imagine asking workers to assemble desks without materials!)\n\nThus equivalent units must be calculated for each of the three production costs.\n\n(Note that direct labor and manufacturing overhead are sometimes combined in a category called conversion costs, which assumes both are added to the process at the same time. In this article, we keep direct labor and manufacturing overhead separate.)\n\nStep 1. Summarize the physical flow of units and compute the equivalent units for direct materials, direct labor, and overhead.\n\nThis step uses the basic cost flow equation presented in to identify the physical flow of units (the basic cost flow equation applies to costs and to units):\n\nBeginning balance + Transfers in (BB) + (TI) Units to be accounted for == Transfers out + Ending balance (TO) + (EB) Units accounted for\n\n### What are the two categories used to summarized the physical flow of units?\n\nThe first category, units to be accounted for, includes the beginning balance (BB) and transfers in (TI).\n\nThe second category, units accounted for, includes the ending balance (EB) and transfers out (TO). As you can see from the previous equation, units to be accounted for must equal units accounted for.\n\n## How do we convert this information into equivalent units?\n\nThe units accounted for (# transferred out and # in ending WIP inventory) must be converted into equivalent units for direct materials, direct labor, and overhead.\n\nThe # units transferred out are 100 percent complete for direct materials, direct labor, and overhead (otherwise, they would not be transferred out), which results in equivalent units matching the physical units.\n\nHowever, the # units in ending WIP inventory are at varying levels of completion for direct materials, direct labor, and overhead, and must be converted into equivalent units using the following formula (as described earlier in the chapter):\n\nEquivalent units = Number of physical units × Percentage of completion\n\nLater in step 3, we will use equivalent unit information for the Assembly department to calculate the cost per equivalent unit.\n\nStep 2. Summarize the costs to be accounted for (separated into direct materials, direct labor, and overhead).\n\n## How do we summarize the costs that are used to calculate the cost per equivalent unit?\n\nThe total costs to be accounted for include the costs in beginning WIP inventory and the costs incurred during the period. shows these costs for the Assembly department.\n\nThe costs are separated into direct materials, direct labor, and overhead.\n\nShows that costs totaling \\$ must be assigned to (1) completed units transferred out and (2) units in ending WIP inventory.\n\nStep 3. Calculate the cost per equivalent unit.\n\nWe now have the costs () and equivalent units () needed to determine the cost per equivalent unit for direct materials, direct labor, and overhead.\n\n## How do we use this information to calculate the cost per equivalent unit?\n\nThe formula to calculate the cost per equivalent unit using the weighted average method is as follows:\n\nIn summary, the same formula is as follows:\n\nCost per equivalent unit = Costs in beginning WIP + Current period costs Equivalent units completed and transferred out + Equivalent units in ending WIP Cost per equivalent unit = Total costs to be accounted for\n\nThe cost per equivalent unit is calculated for direct materials, direct labor, and overhead.\n\nSimply divide total costs to be accounted for by total equivalent units accounted for.\n\nIt is important to note that the information shown in allows managers to carefully assess the unit cost information in the Assembly department for direct materials, direct labor, and overhead.\n\nRecall our primary goal of assigning costs to completed units transferred out and to units in ending WIP inventory. How do we accomplish this goal?\n\nCosts are assigned by multiplying the cost per equivalent unit by the number of equivalent units for direct materials, direct labor, and overhead. shows how this is done.\n\nStep 4. Use the cost per equivalent unit to assign costs to (1) completed units transferred out and (2) units in ending WIP inventory.\n\nThe total cost assigned to units transferred out equals the cost per equivalent unit times the number of equivalent units.\n\nFor example, the cost assigned to direct materials of \\$120,000 = 4,000 equivalents units × \\$30 per equivalent unit.\n\nThe total cost assigned to units in ending inventory equals the cost per equivalent unit times the number of equivalent units.\n\nFor example, the cost assigned to direct materials of \\$90,000 = 3,000 equivalent units× \\$30 per equivalent unit.\n\nThis must match total costs to be accounted for. Although not an issue in this example, rounding the cost per equivalent unit may cause minor differences between the two amounts.\n\nOn completion of step 4, it is important to reconcile the total costs to be accounted for with the total costs accounted for.\n\nThe two balances must match (note that small discrepancies may exist due to rounding the cost per equivalent unit).\n\nThis reconciliation relates back to the basic cost flow equation as follows:\n\nBeginning balance + Transfers in (BB) + (TI) Costs to be accounted for == Transfers out + Ending balance (TO) + (EB) Costs accounted for\n\nAlthough the examples in this chapter have been created in a way that minimizes rounding errors, always round the cost per equivalent unit calculations in step 3 to the nearest thousandth (e.g., if the cost per equivalent unit is \\$2.3739, round this to \\$2.374 rather than to \\$2).\n\nAlthough rounding differences still may occur, this will minimize the size of rounding errors when attempting to reconcile costs to be accounted for (step 2) with costs accounted for (step 4).\n\nRelated Topics"
] | [
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http://www.biblecodedigest.com/page_PageID_150.html | [
"",
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"Technical Addendum— An ELS Extension Model This technical addendum describes the derivation of the formulae used to determine the expected number of extended ELSs (with varying numbers of extensions) to emerge from a search process beginning with n initial ELSs. We shall denote initial ELSs by the symbol “I.” After a Hebrew expert examines the string of letters resulting from taking every j-th letter before and after the initial ELS with a skip of j letters, one of two results will occur in each instance that there is an opportunity to find an extension of the initial ELS. If no extension is found, this will be denoted by “N” and if one is found, it will be denoted by “E.” The following chart presents the complete range of outcomes from this process up through the location of four extensions to an initial ELS. From this chart, the formula for determining the expected number of ELSs with a given number of extensions will become evident, preventing the need to extend the chart out to the right to incorporate additional outcomes where five or more extensions have been found. Each outcome is represented first by a combination of the letters I, E and N, indicating the order in which things appeared in that outcome. Immediately below that outcome is a formula for the expected number of ELSs that will conform to that description. In that formula, n is the total number of initial ELSs, a is the probability of finding a grammatically correct Hebrew extension of the preceding ELS, and b is the probability of not finding an extension. Obviously, b is the complement of a, so b = 1 – a. However, we used b in the table below to shorten the formulae, and make the derivation of each formula easier to understand.",
null,
"The beginning of the search process is represented by the leftmost column of the table, where there are n instances where I, an initial ELS, appears. Each successive step in the search process is represented by a new column immediately to the right of the earlier columns. The only way that zero extensions will be found is the situation where no extension is found either before or after the initial ELS. This is denoted by “NIN.” The probability of this occurring for any given initial ELS is b^2. So the expected number of final ELSs that have no extension will be n(b^2). There are two ways that a final ELS can have one and only one extension. The first occurs when the extension appears after the initial ELS (NIEN) and the second when the extension appears before the initial ELS (NEIN). It should be noted that we are using English conventions here, proceeding to read from left to right, for the sake of most of our readers—rather than the Hebrew convention, where reading is done from right to left. The total expected number of final ELSs to emerge from the first way of having exactly one extension is na(b^2), as is the case for the second way. So the total expected number of final ELSs with exactly one extension will be 2na(b^2), as shown at the bottom of the “One” column. There are three ways that a final ELS can have exactly two extensions. The first occurs when both extensions appear after the initial ELS (NIEEN). The second occurs when an extension appears on both sides of the initial ELS (NEIEN). The third occurs when both extensions appear before the initial ELS (NEEIN). The total expected number of final ELSs to emerge from the first way of having exactly one extension is n(a^2)(b^2), as is the case for the second and the third way. So the total expected number of final ELSs with exactly two extensions will be 3n(a^2)(b^2), as shown at the bottom of the “Two” column. From the patterns evident in the above table we can see that the total expected number of final ELSs with exactly k extensions will be (k+1)n(a^k)(b^2). Since n and b^2 appear in each term, we can factor them out of an expression for the total number of final ELSs, to get (1) n(b^2) [Σ(k+1)(a^k)], where k ranges from 0 to infinity. If we multiply each term of this power series by a in both the numerator and denominator, we obtain {[n(b^2)]/a} [Σ(k+1)(a^{k+1})]. If we shift the value of (k+1) by one for each term, so that the series is summed from k = 1 to infinity, rather than from k = 0 to infinity, it becomes {[n(b^2)]/a} [Σ k(a^k)]. According to formula 40 on page 8 of Summation of Series, collected by L.B.W. Jolley: [Σ n(x^n) = x/[(1-x)^2], where x < 1, and the series is summed from n=1 to infinity. If we substitute k for n and a for x in this formula, we have [Σ k(a^k) = a/[(1-a)^2] . By substituting the simple expression on the right for the power series expression in (1) above, we get {[n(b^2)]/a} [a/[(1-a)^2]] . Since b = (1-a), this expression becomes {[n(1-a)^2)]/a} [a/[(1-a)^2]] = n. So this total expression equals n, as it must, since n final ELSs will emerge from searching n initial ELSs. Why is it that the total expected number of final ELSs with exactly k extensions is given by (k+1)n(a^k)(b^2)? This formula may be broken down into two parts, the first being n, the total number of opportunities to examine extensions around an initial ELS, times the rest of the expression, which is the probability of getting exactly k extensions, given the opportunity of examining one letter string centered around one initial ELS. [Total Expected Extensions] = [Total Opportunities] x [Probability of the Specified Number of Extensions Occurring, Given One Opportunity] In order to get exactly k extensions, each possible way of doing that involves k extensions and 2 instances of not finding any extension. The probability of the former is a^k and of the latter, b^2. As it turns out, there are exactly (k+1) ways of getting k extensions. To see this let us look at a few specific examples. There are three ways to get two extensions, and they are denoted by NIEEN, NEIEN & NEEIN. The first way occurs when the I appears prior to both E’s. The second occurs when the I appears in between the two E’s, and the third occurs when the I appears after the two E’s. In all cases, there is an N at both ends. There are three possible locations for the I in between the two N’s. It is either in the first, second or third position. There are three positions because there must be two E’s and one I. Hence, for k extensions, there are (k+1) ways of arriving at that result. Return to Head to Head Comparison article\n\nCodes Seminar Set for April 29\n\nThe Horizon Institute will be presenting a free seminar on Bible codes Tuesday, April 29, 7 p.m., at the Red Lion Hotel, Medford, Oregon.\n\nBible Codes: Fact or Fiction will feature speakers Ed Sherman, president of the Isaac Newton Bible Code Research Society, Dr. Nathan Jacobi, our Hebrew expert, and Dave Swaney, editor of Bible Code Digest.\n | HOME | SUBSCRIBERS ONLY | OUR PURPOSE | FREE SAMPLES | CURRENT | | FAQs | ABOUT BIBLE CODES | SUBSCRIBE | CONTACT | Copyright © 2016 BibleCodeDigest.com"
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http://www.indium.com/blog/interest-in-formula-for-calculating-alloy-density-still-keen-1.php | [
"",
null,
"# Interest in Formula for Calculating Alloy Density Still Keen\n\nCategory:\n• Indium Corporation\n\n• Alistar writes\n\nDear Dr. Ron,\n\nCan you send me the answer to your question of 14th Feb. 2005, which follows;\n\nIf the density of silver is 10.5 g/cc and the density of tin is 7.31 g/cc. What is the density of an alloy of 96% tin/4% silver.\n\nHint: The answer is not obtained by multiplying the densities by the percentages and adding together. This question requires thinking about the definition of density. Out of 150 people given this question on a certification test, only 2 got it right.\n\nThe correct answer is 7.40 g/cc (7.399927 exactly) Bob Jarrett was the first one with a correct answer. If anyone wants the technique explained or a copy of an Excel spreadsheet that perfroms these calculations send me an email (rlasky@indium.com)\n\nAlistair\n\nDear Alistair,\n\nAn easy way to understand this (proposed by Bob Jarrett) is to consider the 96% tin, 4 % silver example.\n\nLets assume I have 1 g of this alloy, 0.96 g is tin and 0.04 g is silver.\n\nThe volume of the tin is 0.96 g/7.31g/cc = 0.131327cc\n\nThe volume of the silver is 0.04g/10.5g/cc = 0.00381cc\n\nSo 1 g of the alloy has a volume of 0.131327 + 0.00381 cc = 0.135137 cc\n\nHence it's density is 1g/0.135137cc = 7.39989g/cc\n\nThe general formula is:\n\n1/Da = x/D1 + y/D2 + z/D3\n\nDa = density of final alloy\n\nD1 = density of metal 1, x = mass fraction of metal 1\n\nsame for metals 2 and 3\n\nFormula continues for more than 3 metals.\n\nI have developed an Excel spreadsheet (shown calculating the density of SAC 305) that calculates density automatically. If anyone wants a copy, send me an email at rlasky@indium.com.\n\nCheers\n\nDr Ron"
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https://whatisconvert.com/685-feet-in-nautical-miles | [
"## Convert 685 Feet to Nautical Miles\n\nTo calculate 685 Feet to the corresponding value in Nautical Miles, multiply the quantity in Feet by 0.0001645788336933 (conversion factor). In this case we should multiply 685 Feet by 0.0001645788336933 to get the equivalent result in Nautical Miles:\n\n685 Feet x 0.0001645788336933 = 0.11273650107991 Nautical Miles\n\n685 Feet is equivalent to 0.11273650107991 Nautical Miles.\n\n## How to convert from Feet to Nautical Miles\n\nThe conversion factor from Feet to Nautical Miles is 0.0001645788336933. To find out how many Feet in Nautical Miles, multiply by the conversion factor or use the Length converter above. Six hundred eighty-five Feet is equivalent to zero point one one three Nautical Miles.\n\n## Definition of Foot\n\nA foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. The unit of foot derived from the human foot. It is subdivided into 12 inches.\n\n## Definition of Nautical Mile\n\nThe nautical mile (symbol M, NM or nmi) is a unit of length, defined as 1,852 meters (approximately 6,076 feet). It is a non-SI unit used especially by navigators in the shipping and aviation industries, and also in polar exploration.\n\n## Using the Feet to Nautical Miles converter you can get answers to questions like the following:\n\n• How many Nautical Miles are in 685 Feet?\n• 685 Feet is equal to how many Nautical Miles?\n• How to convert 685 Feet to Nautical Miles?\n• How many is 685 Feet in Nautical Miles?\n• What is 685 Feet in Nautical Miles?\n• How much is 685 Feet in Nautical Miles?\n• How many nmi are in 685 ft?\n• 685 ft is equal to how many nmi?\n• How to convert 685 ft to nmi?\n• How many is 685 ft in nmi?\n• What is 685 ft in nmi?\n• How much is 685 ft in nmi?"
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http://chextx.com/News/news/aid/21306.html | [
"## 长城汽车为中国品牌出海画下壮阔一笔\n\n2019年08月12日 16:59 来源：车行天下 超过：9985次关注",
null,
"6月5日，长城汽车位于俄罗斯的图拉工厂正式竣工投产，并率先将哈弗F7带进俄罗斯市场。作为中国汽车品牌首个海外整车全工艺制造基地，图拉工厂的落成不仅实现了生产本土化，支持销售覆盖俄罗斯，还可以辐射周边国家，成为连接欧亚大陆的据点，并为其他中国汽车品牌提供代工组装支持，进一步助推中国汽车品牌走出国门。",
null,
"",
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"2019世界机器人大赛总决赛",
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"40年前，改革开放春风席卷华夏的同时，也拂过了保定。彼时的长城汽车还是集体所有制的长城工业公司，经营业务仅是特种车辆的改装，直到1990年，酷爱汽车的魏建军接手，开启他的造车生涯后，长城汽车才开启光辉的造车史。",
null,
"1999年长城汽车第一百万辆皮卡下线",
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"#### 相关文章\n\n0-500 字已有评论 0条 查看评论>>\n\n### 热门标签\n\n\n• 快速找车\n• 选择品牌\n• 选择品牌\n• A 奥迪\n• A 阿斯顿·马丁\n• A 阿尔法·罗密欧\n• B 宝沃\n• B 布加迪\n• B 巴博斯\n• B 保时捷\n• B 宾利\n• B 奔驰\n• B 宝马\n• B 本田\n• B 别克\n• B 标致\n• B 比亚迪\n• B 宝骏\n• B 北汽制造\n• B 北汽新能源\n• B 北汽幻速\n• B 北汽威旺\n• B 北京汽车\n• B 奔腾\n• B 北汽绅宝\n• C 长安\n• C 长安商用\n• C 长城\n• C 昌河\n• D 大众\n• D 道奇\n• D DS\n• D 东南\n• D 东风风神\n• D 东风风行\n• D 东风小康\n• D 东风风度\n• D 东风\n• F 福特\n• F 丰田\n• F 菲亚特\n• F 法拉利\n• F 福田\n• F 福迪\n• F 福汽启腾\n• G 观致\n• G 广汽传祺\n• G 广汽吉奥\n• G GMC\n• H 红旗\n• H 汉腾汽车\n• H 哈弗\n• H 哈飞\n• H 海格\n• H 海马\n• H 华颂\n• H 黄海\n• H 华泰\n• H 恒天\n• J 吉利汽车\n• J 捷豹\n• J Jeep\n• J 江淮\n• J 江铃\n• J 金杯\n• J 九龙\n• J 金旅\n• K 凯翼\n• K 凯迪拉克\n• K 克莱斯勒\n• K 科尼塞克\n• K 卡威\n• K 开瑞\n• L 路虎\n• L 林肯\n• L 劳斯莱斯\n• L 兰博基尼\n• L 雷克萨斯\n• L 铃木\n• L 雷诺\n• L 理念\n• L 力帆\n• L 莲花汽车\n• L 猎豹\n• L 路特斯\n• L 陆风\n• M 马自达\n• M MG\n• M MINI\n• M 玛莎拉蒂\n• M 摩根\n• M 迈凯轮\n• N 纳智捷\n• O 欧宝\n• O 讴歌\n• O 欧朗\n• Q 奇瑞\n• Q 起亚\n• Q 启辰\n• R 日产\n• R 荣威\n• R 瑞麒\n• S 三菱\n• S 斯威汽车\n• S 萨博\n• S smart\n• S 斯柯达\n• S 斯巴鲁\n• S 思铭\n• S 双龙\n• S 上汽大通\n• S 双环\n• T 特斯拉\n• T 腾势\n• W 沃尔沃\n• W 五菱汽车\n• W 五十铃\n• W 威兹曼\n• W 威麟\n• X 现代\n• X 雪佛兰\n• X 雪铁龙\n• X 西雅特\n• Y 一汽\n• Y 英菲尼迪\n• Y 英致\n• Y 依维柯\n• Y 野马汽车\n• Y 永源\n• Z 众泰\n• Z 中华\n• Z 中兴\n• Z 知豆\n• 选择车系\n• 选择车系\n• 车型对比\n• 选择品牌\n• 选择品牌\n• A 奥迪\n• A 阿斯顿·马丁\n• A 阿尔法·罗密欧\n• B 宝沃\n• B 布加迪\n• B 巴博斯\n• B 保时捷\n• B 宾利\n• B 奔驰\n• B 宝马\n• B 本田\n• B 别克\n• B 标致\n• B 比亚迪\n• B 宝骏\n• B 北汽制造\n• B 北汽新能源\n• B 北汽幻速\n• B 北汽威旺\n• B 北京汽车\n• B 奔腾\n• B 北汽绅宝\n• C 长安\n• C 长安商用\n• C 长城\n• C 昌河\n• D 大众\n• D 道奇\n• D DS\n• D 东南\n• D 东风风神\n• D 东风风行\n• D 东风小康\n• D 东风风度\n• D 东风\n• F 福特\n• F 丰田\n• F 菲亚特\n• F 法拉利\n• F 福田\n• F 福迪\n• F 福汽启腾\n• G 观致\n• G 广汽传祺\n• G 广汽吉奥\n• G GMC\n• H 红旗\n• H 汉腾汽车\n• H 哈弗\n• H 哈飞\n• H 海格\n• H 海马\n• H 华颂\n• H 黄海\n• H 华泰\n• H 恒天\n• J 吉利汽车\n• J 捷豹\n• J Jeep\n• J 江淮\n• J 江铃\n• J 金杯\n• J 九龙\n• J 金旅\n• K 凯翼\n• K 凯迪拉克\n• K 克莱斯勒\n• K 科尼塞克\n• K 卡威\n• K 开瑞\n• L 路虎\n• L 林肯\n• L 劳斯莱斯\n• L 兰博基尼\n• L 雷克萨斯\n• L 铃木\n• L 雷诺\n• L 理念\n• L 力帆\n• L 莲花汽车\n• L 猎豹\n• L 路特斯\n• L 陆风\n• M 马自达\n• M MG\n• M MINI\n• M 玛莎拉蒂\n• M 摩根\n• M 迈凯轮\n• N 纳智捷\n• O 欧宝\n• O 讴歌\n• O 欧朗\n• Q 奇瑞\n• Q 起亚\n• Q 启辰\n• R 日产\n• R 荣威\n• R 瑞麒\n• S 三菱\n• S 斯威汽车\n• S 萨博\n• S smart\n• S 斯柯达\n• S 斯巴鲁\n• S 思铭\n• S 双龙\n• S 上汽大通\n• S 双环\n• T 特斯拉\n• T 腾势\n• W 沃尔沃\n• W 五菱汽车\n• W 五十铃\n• W 威兹曼\n• W 威麟\n• X 现代\n• X 雪佛兰\n• X 雪铁龙\n• X 西雅特\n• Y 一汽\n• Y 英菲尼迪\n• Y 英致\n• Y 依维柯\n• Y 野马汽车\n• Y 永源\n• Z 众泰\n• Z 中华\n• Z 中兴\n• Z 知豆\n• 选择车系\n• 选择车系\n• 选择车型\n• 选择车型\n• 意见反馈"
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https://www.netwasgroup.us/fluid-systems/stokes-law.html | [
"# Stokes Law For Fluid In Fluid\n\nparticles exit at the vortex. The D50 cut point of a solids separation device is defined as that particle size at which one-half of the weight of specific size particles go to the underflow and one-half of the weight go to the overflow. For example, a D30 cut point references a particle size which is 30% concentrated in the underflow and 70% in the overflow.\n\nAs stated earlier, the cut point is related to the inside diameter of the hydrocyclone. For example, a 12-inch cone has a D50 cut point for low-gravity solids in water of approximately 60 to 80 microns, a 6-inch cone around 30 to 60 microns, and a 4-inch cone around 15 to 20 microns (Table 7-4). However, the cut point will vary with the size and amount of solids in the feed, as well as fluid viscosity.\n\nFor comparative purposes, consider a 50-micron equivalent drilled solid diameter. Relatively speaking, the percent discharge is as follows:\n\n• 6-inch cone discharges 80% at underflow\n• 4-inch cone discharges 95% at underflow\n• 3-inch cone discharges 97% at underflow\n\n### Now consider a 10-micron equivalent drilled solid diameter:\n\n• 6-inch cone discharges 7% at underflow\n• 4-inch cone discharges 11% at underflow\n• 3-inch cone discharges 17% at underflow\n\nIf a graph of particle size versus percent of particles recovered to underflow is plotted, the portion of the curve near the D50, or 50%, recovery point (median cut point) is very steep when separations are efficient.\n\nParticle separations in hydrocyclones vary considerably. In addition to proper feed head and the cone apex setting, drilling fluid properties including density, percent solids (and solids distribution) and viscosity, all affect separations. Any increase in these mud properties will increase the cut point of a separation device.\n\nStokes Law defines the relationship between parameters that control the settling velocity of particles in viscous liquids, not only in settling pits but also in equipment such as hydrocyclones and centrifuges.\n\nSeparations in a settling pit are controlled by the force of gravity and the viscosity of the suspending fluid (drilling mud). A large, heavy particle settles faster than a small, lighter particle. This settling process can be increased by reducing the viscosity of the suspending fluid, increasing the gravitational forces on the particles, or by increasing the effective particle(s) size with flocculation or coagulation.\n\nHydrocyclones and centrifuges increase settling rates by applying increased centrifugal force, which is equivalent to higher gravity force.\n\nStokes' Law for settling spherical particles in a viscous liquid is expressed as:\n\nvs n where Vs = Settling or terminal velocity, feet/sec C = Units constant, 2.15 x 10\"7 g = Acceleration (gravity or apparatus) ft/sec2\n\nDe = Particle equivalent diameter, microns ps = Specific gravity of solids (cutting, barite, etc.) p, = Specific gravity of liquid phase |i = Viscosity of media, centipoise\n\nVarious size particles with different densities can have the same settling rates. That is, there exists an equivalent diameter for every 2.65 specific gravity drilled solid, be it limestone, sand, or shale, which cannot be separated by gravimetric methods from barite particles of a corresponding equivalent diameter. Presently, it is not possible to separate desirable barite particles from undesirable drilled solid particles that settle at the same rate.\n\nTABLE 7-4. Hydrocyclone Size versus D50 Cut Point\n\nCone Diameter (inches) Dso Cut Point in Water Dso Cut Point in Drilling Fluid\n\n6 30-35 70-100\n\nGenerally, a barite particle (specific gravity = 4.25) will settle at the same rate as a drilled solids particle (specific gravity = 2.65) that is 1 \\ times the barite particle's diameter. This may be verified by applying Stokes' Law.\n\nExample #1. A viscosified seawater fluid with a specific gravity of 1.1, PV = 2.0 centipoise, and YP = 12.0 lbs/100 ft2, is circulated to clean out a cased wellbore. What size low-gravity solids will settle out with 5-micron barite particles? With 10-micron barite particles, what is the settling velocity in rig tanks?\n\nUsing Stokes' Law, the settling velocity is:\n\nFor equivalent settling rates, Vs = Vs. And for ji, = ji2 (the same fluid, therefore, the same viscosity).\n\n(specific gravity = 2.65) in an 11.5 ppg mud with PV = 20 cp and YP = 12 lbs/100 ft2?\n\nUsing Stokes' Law, the settling velocity is:\n\nFor equivalent settling rates, Vs = Vs. And for ji, = 112 (the same fluid, therefore, the same viscosity).\n\n(D,2 x 1.27) = (D22 x 2.87) D,2/D22 = 2.87/1.27 = 2.26 D/D2 = 1.50\n\n(D,2 x 1.55) = (D22 x 3.15) Dj2/D22 = 3.15/1.55 = 2.03 D,/D2 = 1.42\n\nThus, a 5-micron barite particle will settle at the same rate as a 7-micron low-gravity particle, and a 10-micron barite particle will settle at the same rate as a 14-micron low-gravity particle.\n\nSettling velocity for a 5-micron barite (or 7-micron drilled solid) particle is:\n\nand for a 10-micron barite (or 14-micron drilled solid) particle:\n\nThus, a 10-micron barite particle will settle at the same rate as a 15-micron low-gravity particle, and a 50-micron barite particle will settle at the same rate as a 75-micron low-gravity particle.\n\nStokes' Law shows that as fluid viscosity and density increase, separation efficiency decreases.\n\nIf the drilling fluid weight is 14.0 pounds per gallon (specific gravity = 1.68):\n\nd2ds = 4.25-1.68 = 2.57 = 2 65 d2b ~ 2.65-1.68 \" 0.97 ~\n\nTherefore, in drilling fluid weighing 14 pounds per gallon, a 10-micron barite particle will settle at the same rate as a 16-micron drilled solid particle, and a 50-micron barite particle will settle at the same rate as an 80-micron (or 81.4) drilled solid particle.\n\nIt is important to remember that the efficiency of a separator is viscosity dependent. The median cut, or D50 cut point, increases with viscosity as shown by Stokes' Law:\n\nv _ (2.15 x 1Q-7) x 32.2 ft/sec2 x (4.25 - 1.1) x 100 5 ~ 16 cp\n\nExample #2. What are the equivalent diameters of barite (specific gravity = 4.25) and drilled solids\n\nExample #3. A 4-inch cone will separate half of the 12-micron low-gravity (specific gravity = 2.6) particles in water (that is, the Dso cut point is 12\n\nmicrons). What is the D50 cut point in a 50-cp viscosity fluid of the same density?\n\nFor constant settling velocity, if fluid density is unchanged and other parameters remain constant:\n\nAgain, cut point performance can be further projected by dividing by the projected specific gravity at various viscosities. Thus, in the above example, for a 6-inch hydrocyclone, 20 centipoise viscosity, and 1.4 (11.7 ppg) density fluid, the D50 would be:"
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https://help.scilab.org/docs/6.0.2/ru_RU/geomean.html | [
"Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nScilab 6.0.2\nChange language to: English - Français - Português - 日本語 -\nСправка Scilab >> Statistics > Central Tendency > geomean\n\n# geomean\n\ngeometric mean\n\n### Syntax\n\n```gm = geomean(X)\nGM = geomean(X, orien) // orien: 'r'|1|'c'|2..ndims(X)```\n\n### Arguments\n\nX\n\nVector, matrix or hypermatrix of real or complex numbers.\n\norien\n\nDimension accross which the geometric average is computed. The value must be among `'r', 1, 'c', 2, .. ndims(X)`. Values `'r'` (rows) and `1` are equivalent, as `'c'` (columns) and `2` are.\n\ngm\n\nScalar number: the geometric mean `gm = prod(X)^(1/N)`, where `N = length(X)` is the number of components in `X`.\n\nGM\n\nVector, matrix or hypermatrix of numbers. `s = size(GM)` is equal to `size(X)`, except that `s(orien)` is set to 1 (due to the projected application of geomean() over components along the orien dimension).\n\nIf `X` is a matrix, we have:\n\n• `GM = geomean(X,1) => GM(1,j) = geomean(X(:,j))`\n• `GM = geomean(X,2) => GM(i,1) = geomean(X(i,:))`\n\n### Description\n\n`geomean(X,..)` computes the geometric mean of values stored in `X`.\n\nIf `X` stores only positive or null values, `gm` or `GM` are real. Otherwise they are most often complex.",
null,
"If `X` is sparse-encoded, then it is reencoded in full format before being processed. `gm` is always full-encoded. `GM` is sparse-encoded as well.\n\n### Examples\n\n```geomean(1:10) // Returns factorial(10)^(1/10) = 4.5287286881167648\n\n// Projected geomean:\n// -----------------\nm = grand(4,5, \"uin\", 1, 100);\nm(3,2) = 0; m(2,4) = %inf; m(4,5) = %nan\ngeomean(m, \"r\")\ngeomean(m, 2)\nh = grand(3,5,2, \"uin\",1,100)\ngeomean(h,3)```\n```--> m = grand(4,5, \"uin\", 1, 100);\n--> m(3,2) = 0; m(2,4) = %inf; m(4,5) = %nan\nm =\n13. 5. 99. 41. 20.\n3. 92. 4. Inf 5.\n35. 0. 36. 40. 98.\n86. 86. 66. 21. Nan\n\n--> geomean(m, \"r\")\nans =\n18.510058 0. 31.14479 Inf Nan\n\n--> geomean(m, 2)\nans =\n22.104082\nInf\n0.\nNan\n\n--> h = grand(3,5,2, \"uin\",1,100)\nh =\n(:,:,1)\n10. 40. 37. 72. 30.\n10. 47. 54. 13. 19.\n44. 27. 61. 10. 27.\n(:,:,2)\n96. 88. 7. 98. 35.\n54. 29. 96. 77. 8.\n94. 45. 21. 46. 3.\n\n--> geomean(h,3)\nans =\n16.522712 43.150898 23.2379 36.91883 72.\n14.142136 13.747727 64.311741 34.85685 35.79106\n12.247449 30.983867 59.329588 16.093477 84.\n```\n\n```// APPLICATION: Average growing rate\n// ---------------------------------\n// During 8 years, we measure the diameter D(i=1:8) of the trunc of a tree.\nD = [10 14 18 26 33 42 51 70]; // in mm\n\n// The growing rate gr(i) for year #i+1 wrt year #i is, in %:\ngr = (D(2:\\$)./D(1:\\$-1) - 1)*100\n\n// The average yearly growing rate is then, in %:\nmgr = (geomean(1+gr/100)-1)*100\n\n// If this tree had a constant growing rate, its diameter would have been:\nD(1)*(1+mgr/100)^(0:7)```\n```--> gr = (D(2:\\$)./D(1:\\$-1) - 1)*100\ngr =\n40. 28.57 44.44 26.92 27.27 21.43 37.25\n\n--> mgr = (geomean(1+gr/100)-1)*100\nmgr =\n32.05\n\n--> D(1)*(1+mgr/100)^(0:7)\nans =\n10. 13.2 17.44 23.02 30.4 40.15 53.01 70.\n```\n\n• prod — произведение элементов массива\n• harmean — harmonic mean : inverse of the inverses average (without zeros)\n\n### Bibliography\n\nWonacott, T.H. & Wonacott, R.J.; Introductory Statistics, fifth edition, J.Wiley & Sons, 1990."
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https://www.pdn.cam.ac.uk/other-pages/cnbh/files/aim92documentation/stats.html | [
"",
null,
"Department of Physiology, Development and Neuroscience\n\nStudying at Cambridge\n\n# Manual Reference Pages - STATS (1)\n\n### NAME\n\nstats - calculate some statistics.\n\nSyntax\nDescription\nOptions\n\n### SYNTAX\n\nstats [options] [file]\n\n### DESCRIPTION\n\nCalculate statistics for given frames of each input file and write the results on the stdout. If no filenames are given, one input file is expected on the stdin. Otherwise each given filename is processed in turn.\n\n### OPTIONS\n\n1. frame, width, samplerate\n\nThe frame option selects a sequence of contiguous frames for processing by:\n\n``` frame=a[-b]\n\n```\n\nwhere a and b are frame numbers: 1,2,3,...,max The upper limit b is optional, and when it is missing then the frame sequence is a single frame, otherwise a and b are inclusive limits. The strings \"min\" and \"max\" are recognised as extreme limits. Each frame has size width samples which may be given with time units (s or ms), in which case the they are converted to a number of samples using the given samplerate option. The special option value \"width=max\" is interpreted as a single frame covering the whole file.\n\n2. type\n\nDatatype for both input and output. Binary types are char, short, int, float, double. Ascii input (one number per line) is recognised with \"type=ascii\".\n\n3. stat\n\nThe statistics available to the stat option are:\n\n``` n sample (ie. data set) size\nsum area\nss sum of squares\nmean sum/n\nrms root mean square: square root of ss/n\nvariance computed with n-1 degrees of freedom\nstddev standard deviation: square root of variance\nmin minimum value\nmax maximum value\nabsmax maximum absolute value, ie. max( |max|, |min| )\nrange max-min\n\n```\n\nThe variance is the average squared deviation from the mean value, where the average is computed by dividing by n-1 for a data set of n samples. The stddev is the square root of this variance. The stddev would be equal to the rms value for a data set with zero mean, except that the average squared value is computed using ss/n, while the average squared deviation from a zero mean value is computed using ss/(n-1).\n\nStatistics may be given in a comma separated list, for example:\n\n``` stats stat=mean,variance,min,max\n\n```\n\nFor each frame of each input file this will output the four statistics in the data type selected by the type option.\n\n4. line\n\nWhen line=on (or -l) output is forced to be ascii irrespective of the type option, and for each file argument in turn stats prints the statistics requested, in the order they were requested, on one line. This makes a table (one line per file), and also helps subsequent operations between statistics. For example, the standard error of the mean (the stddev divided by the square root of the sample size) can be computed:\n\n``` stats -l stat=stddev,n | awk { print \\$1 / sqrt(\\$2) }\n\n```\n\nTables can be formatted using field-width and precision arguments which are included as silent options. For example, to format columns with a precision of 4 decimal places with each number right-justified in a field-width of 12 characters:\n\n``` stats -l fieldwidth=12 precision=4\n\n```\n\nThe columns may be left-justified by making the fieldwidth parameter negative. Output is truncated to integers by setting the precision parameter 0. For example, to print the sample size of each file as an integer without setting in a field:\n\n``` stats -l fieldwidth=0 precision=0 stat=n [files]\n\n```\n\noptions\n\nCopyright (c) Applied Psychology Unit, Medical Research Council, 1995\n\nPermission to use, copy, modify, and distribute this software without fee is hereby granted for research purposes, provided that this copyright notice appears in all copies and in all supporting documentation, and that the software is not redistributed for any fee (except for a nominal shipping charge). Anyone wanting to incorporate all or part of this software in a commercial product must obtain a license from the Medical Research Council.\n\nThe MRC makes no representations about the suitability of this software for any purpose. It is provided \"as is\" without express or implied warranty.\n\nTHE MRC DISCLAIMS ALL WARRANTIES WITH REGARD TO THIS SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS, IN NO EVENT SHALL THE A.P.U. BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.\n\n SunOS 5.6 STATS (1) 1 September 1993\nGenerated by manServer 1.07 from /cbu/cnbh/aim/release/man/man1/stats.1 using man macros.",
null,
""
] | [
null,
"https://www.pdn.cam.ac.uk/images/157071101713274785.png/image_logo",
null,
"https://www.pdn.cam.ac.uk/other-pages/cnbh/files/CNBHLogo.jpg",
null
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