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http://www.talkstats.com/showthread.php/11071-finding-p-value-when-F-value-and-df-s-are-known
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math
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Provided that any statpack gives you the p value for F-test, you could try to use Excel as I pointed out to you in my reply to a previous post of yours about chi-square (http://www.talkstats.com/showthread.php?t=11059).
1) click on any cell
2) type the following formula:
3) instead of the dots, put the F value and the degrees of freedom of your two samples, each separated by ";"
Excel should return the associated probability.
Please note that you have to put first the DF of the sample going in the numerator (that is the sample with the greater variance), and then the DF of the sample going into the denominator (that is the sample with the smallest variance).
Hope this helps
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CC-MAIN-2017-17
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https://brickipedia.fandom.com/wiki/Shu_Todoroki
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math
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Wiki Targeted (Games)
Do you like this video?
Shu Todoroki is a Japanese champion race car who competed in the World Grand Prix. He won fourth place in the first race, but was hit by Professor Zündapp's Death Ray in the second race.
- Oddly, the System variation of Shu Todoroki does not have printed sideplates, whereas the DUPLO variation has a dragon printing.
|view · talk · edit Cars Minifigures|
|System||Acer \ Broadside \ Crane \ Cruz Ramirez \ Fillmore \ Finn McMissile \ Francesco Bernoulli \ Grem \ Guido \ Holley Shiftwell \ Jackson Storm \ Jeff Gorvette \ Junior Moon \ Leland Turbo \ Lightning McQueen \ Luigi \ Mack \ Mater \ Max Schnell \ Miles Axlerod \ Miss Fritter \ Petrov Trunkov \ Pit Crew Helper \ Professor Z \ Raoul ÇaRoule \ Red \ Rod "Torque" Redline \ Sally Carrera \ Sarge \ Sheriff \ Shu Todoroki \ Smokey \ The Queen \ Tony Trihull|
|DUPLO Only||Doc Hudson \ Carla Veloso|
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s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00401.warc.gz
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https://rd.springer.com/article/10.1007/s10958-009-9679-5
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math
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It is known that under some conditions, a stationary random sequence admits a representation as a sum of two sequences: one of them is a martingale difference sequence, and another one is a so-called coboundary. Such a representation can be used for proving some limit theorems by means of the martingale approximation. A multivariate version of such a decomposition is presented in the paper for a class of random fields generated by several commuting, noninvertible, probability preserving transformations In this representation, summands of mixed type appear, which behave with respect to some group of directions of the parameter space as reversed rnultiparameter martingale differences (in the sense of one of several known definitions), while they look as coboundaries relative to other directions. Applications to limit theorems will be published elsewhere. Bibliography: 14 titles.
Russia Parameter Space Limit Theorem Random Field Random Sequence
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
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A. K. Basu and C. C. Y. Dorea, “On functional central limit theorem for stationary martingale random fields,” Acta Math. Acad. Sci. Hangar., 33, 307–316 (1979).MATHCrossRefMathSciNetGoogle Scholar
M. Gordin and M. Weber, “On the central limit theorem for a class of multivariate actions” (in preparation).Google Scholar
N. Maigret, “Théorème de limite centrale fonctionnel pour une chaîne de Markov récurrente au sens de Harris et positive,” Ann. Inst. H. Poincaré, Sect. B (N.S.), 14, 425–440 (1978).MATHMathSciNetGoogle Scholar
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CC-MAIN-2018-34
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https://linguix.com/english/misspelling/word/equilateral
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math
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1. An equilateral triangle
2. A side exactly equal to others.
3. A geometric figure having all sides equal.
4. A side exactly corresponding, or equal, to others; also, a figure of equal sides.
5. a figure whose sides are all equal
6. A figure having all its sides equal.
7. geometry Referring to a polygon all of whose sides are of equal length. Not necessarily a regular polygon since the angles can still differ (a regular polygon would also be equiangular).
8. Having all sides or faces equal.
9. (Geom.) one whose axes are equal.
10. (Zoöl.) one in which a transverse line drawn through the apex of the umbo bisects the valve, or divides it into two equal and symmetrical parts.
11. Having all the sides equal
12. applied to two figures, when every side of the one has its equal among the sides of the other.
13. having all sides or faces equal
14. In geometry, having all the sides equal: as, an equilateral triangle.
15. Having all the convolutions of the shell in one plane: said chiefly of foraminifers.
16. Having all the sides equal.
17. In zoology: Having the two sides equal: said of surfaces which can be divided into two parts of the same form by a longitudinal median line.
1. Ensure that the octagon is equilateral, meaning all sides on the ground have the same length.
2. a tiny mining company working claims in the remote "equilateral" cluster of asteroids far out in Jupiter's orbit.
3. For example, three Woolworths sites around Birmingham form an exact equilateral triangle (Wolverhampton, Lichfield and Birmingham stores) and if the base of the triangle is extended, it forms a 173.8 mile line linking the Conway and Luton stores.
4. From the moment that grand trine appeared in the sky—an equilateral triangle like the great pyramids—it was our turn to clean house.
5. Richardson, Taylor and Tomlinson are scattered, forming a sort of equilateral triangle.
6. How can anyone say that a scalene and an equilateral are really the same "kind"!
7. For example, if you enter 1, 3, 6, 10, 15, the server will tell you that is sequence A000217, the triangular numbers (these are the number of objects that can form an equilateral triangle like the standard arrangement of bowling pins.)
8. It happened that the Lancashire Queen, the shore at Turner's Shipyard, and the Solano Wharf were the corners of a big equilateral triangle.
9. Obtained by the rotation of 60 ° of two equilateral triangles superimposed and spaced in height, the form appears simple but at the same time full of spatial complexity.
10. ‘An equilateral triangle of side length one is called a unit triangle.’
11. ‘This is constructed by dividing a line into three equal parts and replacing the middle segment by the other two sides of an equilateral triangle constructed on the middle segment.’
12. ‘Then divide it into three equal pieces, and replace the middle piece with the other two sides of the equilateral triangle that has this middle piece as its base.’
13. ‘One way to draw a Reuleaux triangle is to start with an equilateral triangle, which has three sides of equal length.’
14. ‘But the image of the centroid of the equilateral triangle (which is, of course, the point of intersection of its medians) will be different, as indicated in Figure 6.’
15. ‘The vibration exciter and geophone were placed near the burrow with burrow opening, exciter and geophone occupying the three corners of an equilateral triangle having sides one meter in length.’
16. ‘If each edge has the same length and each face is an equilateral triangle, the result is a regular tetrahedron - one of the Platonic solids.’
17. ‘Now, both the original Asymmetric Propeller and Napoleon's theorem start with three equilateral triangles and discover the fourth one by construction.’
18. ‘As long as the three triangles are equilateral, the ‘midpoint ‘triangle is also equilateral.’’
19. ‘Any given triangle, either imagined or on paper, must either be acute, right, or obtuse, and either scalene, isosceles, or equilateral, and so any given triangle cannot represent all triangles.’
20. ‘The most interesting results show, with a very ingenious proof, that an equilateral triangle has a greater area than any isosceles or scalene triangle with the same perimeter.’
21. ‘The Von Koch snowflake is a fractal which is constructed from an equilateral triangle as follows.’
22. ‘For a given individual, the values of the three coefficients in the ancestry vector q are given by the distances to each of the three sides of the equilateral triangle.’
23. ‘The sides of this equilateral triangle will be five million kilometres long.’
24. ‘As an equilateral triangle rolls over one catenary, it ends up bumping into the next catenary’
25. ‘The theorem thus implies existence of the total of 18 equilateral triangles.’
26. ‘In 1772, J.L. Lagrange identified a periodic orbit in which three masses are at the corners of an equilateral triangle.’
27. ‘An equilateral triangle produces one of the observed crop-circle patterns; three isosceles triangles generate the other crop-circle geometries.’
28. ‘Naturally, the axes form equilateral triangles - 27 of them.’
29. ‘For example, it's possible to slice a square into four angular pieces that can be rearranged into an equilateral triangle.’
30. equilateral triangles are triangles of equal edge lengths
Other users have misspelling equilateral as:
1. equalatteral 33.33%
2. equilatral 33.33%
3. Other 33.34%
Use Linguix everywhere you write
Get your writing checked on millions of websites, including Gmail, Facebook, and Google Docs.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585215.14/warc/CC-MAIN-20211018221501-20211019011501-00413.warc.gz
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| 5,633 | 53 |
https://nyuscholars.nyu.edu/en/publications/spectral-gap-critical-exponent-for-glauber-dynamics-of-hierarchic
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math
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We develop a renormalisation group approach to deriving the asymptotics of the spectral gap of the generator of Glauber type dynamics of spin systems with strong correlations (at and near a critical point). In our approach, we derive a spectral gap inequality for the measure recursively in terms of spectral gap inequalities for a sequence of renormalised measures. We apply our method to hierarchical versions of the 4-dimensional n-component | φ| 4 model at the critical point and its approach from the high temperature side, and of the 2-dimensional Sine-Gordon and the Discrete Gaussian models in the rough phase (Kosterlitz–Thouless phase). For these models, we show that the spectral gap decays polynomially like the spectral gap of the dynamics of a free field (with a logarithmic correction for the | φ| 4 model), the scaling limit of these models in equilibrium.
ASJC Scopus subject areas
- Statistical and Nonlinear Physics
- Mathematical Physics
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00708.warc.gz
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CC-MAIN-2023-50
| 961 | 4 |
http://www.ingentaconnect.com/content/maa/cmj/2009/00000040/00000002/art00003
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math
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Automobile sunshades always fold into an odd number of loops. The explanation why involves elementary topology (braid theory and linking number, both explained in detail here with definitions and examples), and an elementary fact from algebra about symmetric group.
The College Mathematics Journal is designed to enhance classroom learning and stimulate thinking regarding undergraduate mathematics. CMJ publishes articles, short Classroom Capsules, problems, solutions, media reviews and other pieces. All are aimed at the college mathematics curriculum with emphasis on topics taught in the first two years.
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s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768529.27/warc/CC-MAIN-20141217075248-00073-ip-10-231-17-201.ec2.internal.warc.gz
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CC-MAIN-2014-52
| 609 | 2 |
https://dictionary.obspm.fr/?showAll=1&formSearchTextfield=transfer+function
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math
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modulation transfer function (MTF)
karyâ-ye tarâvaž-e degarâhangeš
Fr.: fonction de transfert de modulation
A measure of the ability of an optical system to reproduce (transfer) various levels of detail from the object to the image, as shown by the degree of contrast (modulation) in the image. → optical transfer function.
→ modulation; → transfer; → function.
optical transfer function (OTF)
karyâ-ye tarâvaž-e nuri
Fr.: fonction de transfert optique
The function that provides a full description of the imaging quality of an optical system. A combination of the → modulation transfer function (MTF) and the → phase transfer function (PTF) , the OTF describes the spatial (angular) variation as a function of spatial (angular) frequency.
phase transfer function (PTF)
karyâ-ye tarâvaž-e fâz
Fr.: fonction de transfert de phase
A measure of the relative phase in the image as function of frequency. It is the phase component of the → optical transfer function. A relative phase change of 180°, for example, results in an image with the black and white areas reversed.
Fr.: fonction de transfert
The mathematical relationship between the output of a control system and its input: for a linear system, it is the Laplace transform of the output divided by the Laplace transform of the input under conditions of zero initial-energy storage.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00729.warc.gz
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CC-MAIN-2023-14
| 1,366 | 15 |
http://miriamsharoni.de/index.php/kindle/c-algorithms-for-real-time-dsp
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math
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By Paul Embree
For electric engineers and computing device scientists.
Digital sign processing ideas became the tactic of selection in sign processing as electronic pcs have elevated in pace, comfort, and availability. whilst, the c program languageperiod is proving itself to be a necessary programming software for real-time computationally in depth software program initiatives. This publication is an entire advisor to electronic real-time sign processing suggestions within the C language.
Read or Download C Algorithms for real-time dsp PDF
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Extra info for C Algorithms for real-time dsp
In this very simple case, the number ik is the resolution bk . Assume also that 2 for each quantizer we can deduce (or measure) the quantization error power σQ (ik ) k that it generates. The number of bits that it requires is bk (ik ). We will see in Chapter 4 that an entropy coding can be carried out after a uniform quantization. In this case, we show that the necessary number of bits required to quantize the signal can be reduced, which explains the notation bk (ik ) and the fact that bk (ik ) can be a non-integer.
We write it as x ˆ0 (b = 0). If the number of vectors in the training data is L , the distortion is: 2 σQ (b = 0) = 1 1 L N L −1 2 ||x(m)||2 = σX m=0 since the signal is supposedly centered. – Next, we split this vector into two vectors written x ˆ0 (b = 1) and xˆ1 (b = 1) with 0 0 1 0 x ˆ (b = 1) = xˆ (b = 0) and x ˆ (b = 1) = xˆ (b = 0) + . Choosing the vector presents a problem. We choose “small” values. – Knowing that x ˆ0 (b = 1) and x ˆ1 (b = 1), we classify all the vectors in the training data relative to these two vectors (labeling all the vectors 0 or 1), and then calculate ˆ1 (b = 1) of the vectors labeled 0 and 1, the new centers of gravity x ˆ0 (b = 1) and x respectively.
From this, we deduce the M optimum quantizers. 3. 4. Optimum transform In a second step, we find among all the transformations T the one that minimizes 2 σQ after optimum allocation of the bM bits available for the transformed vector. 8], we need to find the transformation Topt that minimize σQ minimizes the geometric mean of the sub-band signal powers. We limit ourselves to the case of orthogonal transforms which already necessitate that N = M . Consider the covariance matrix of the vector X(m), which is an M × M dimensional Toeplitz matrix: ⎡ RX = 2 σX ⎢ ⎢ ⎢ ⎢ ⎣ 1 ρ1 ..
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https://vigoroushandwaving.wordpress.com/category/physics/general-relativity/black-holes/
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math
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The Cosmic Censorship Hypothesis (CSH) was put forward by Roger Penrose in 1969, and (roughly) states
“there are no naked singularities”.
The hypothesis proposes that whenever a singularity occurs, such as in the center of a black hole, it must occur behind an event horizon. A singularity outside of an event horizon is termed a naked singularity, and the CSH says that such singularites do not exist. Cosmic Censorship has pretty profound implications for fundamental physics. For instance, a failure of the Cosmic Censorship Hypothesis leads to a failure of determinism in classical physics, since one can’t predict the behaviour of spacetime in the causal future of a singularity. On the other hand, if the Cosmic Censorship Hypothesis holds then (outside the event horizon), the singularity does not affect determinism. In 1991, Stephen Hawking made a wager with physicists Kip Thorne and John Preskill that the Cosmic Censorship Hypothesis is true. Six years later in 1997, Hawking conceded the bet “on a technicality”, after computer calculations showed that naked singularities could exist, albeit only in exceptional (not physically realistic) circumstances. In this post, we’re going to be looking at extremal black holes, and their relationship to the Cosmic Censorship Hypothesis.
This is another post on my current theme of black holes. Recently, I wrote a blog post about light orbiting a Schwarzchild black hole. These so called Photon spheres are sufficiently interesting to wonder whether they exist for more complicated black holes. They do, and we’re going to find out about them. This post is based on a fantastic paper I read the other week. If you find this stuff interesting, and you’ve got an introductory level of GR you should definitely check out the paper here.
One of the coolest thing about black holes is you can describe any black hole with just three numbers: Mass , Charge , and Angular Momentum . Of course, if we’re being completely general we also need to describe where the black hole is and how fast it’s moving through space – but we can always perform a Lorentz transformation so that we’re in the rest-frame of the black hole and so we really only care about , and . The fact that you can describe black holes with only three numbers is one of the main reasons I find them so interesting. Consider for example, two black holes, A and B. Black hole A is made entirely from matter – protons, neutrons, electrons and the like. On the other hand, black hole B is made entirely from anti-matter – anti-protons, anti-neutrons, anti-electrons and the like. Further, suppose A and B have the same charge, angular momentum and mass. How can we tell A from B? That is, what experiment can we do to tell the difference between the matter black hole or the anti-matter black hole? The answer is, we can’t tell the difference. For all intents and purposes, there is no way of telling whether a black hole is made of matter or anti-matter. Indeed, a third black hole C, with the same mass, charge and angular momentum, but made entirely from light, would be similarly indistinguishable from the first two.
As promised in the title, we’re going to be looking at rotating black holes. A rotating black hole is, like the Schwarzchild black hole, uncharged (). Unlike the Schwarzchild solution however, the rotating black hole has a non-zero angular momentum . Rotating black holes are also called Kerr black holes, after Roy Kerr who was the first to write down an exact solution for one. We will consider an uncharged black hole which is rotating with constant angular momentum. For convenience, we define the angular momentum per unit mass . Since the black hole is rotating, we lose the spherical symmetry we have with the Schwarzchild metric, so we expect the solution to be a little more complicated. We do, however, still have axial-symmetry, i.e. rotational symmetry around the axis of rotation.
Ok, now that I have finished the preamble over here, I can finally start talking about what I originally intended to blog about: Photon spheres.
The defining feature of a black hole is that the gravitational attraction beneath the event horizon is so strong that even light can’t escape. That is, the curvature of spacetime in the vicinity of the black hole is so intense that there are no geodesics which are able to leave. This is what makes black holes so interesting, since anything (including light) which is dropped into a black hole is lost forever (we’re talking classical black holes at the moment, so no Hawking radiation). On the other hand, light (or matter) travelling near a black hole can escape, provided it stays outside of the event horizon.
General relativity predicts that light passing near a heavy object will be deflected by the gravitational field of that object. Equivalently, light will follow a geodesic in the curved spacetime around the heavy object.
This situation strongly resembles the situation we have in orbital mechanics, where a small object like a satellite or asteroid is moving near a large object, like a planet. Depending on the velocity of the object, it either escapes the planet, falls into the planet, or starts orbiting the planet. It seems like a natural question, then, to ask:
Can light orbit a black hole? Continue reading
I’ve been reading a lot about General Relativity this past week and so I thought I would do a couple of posts on Black Holes, since they are well and truly one of the most interesting things about General Relativity. This first post is really just a (very) brief introduction to General Relativity. My main goal is to write about some of the cool things I’ve come across lately, so think of this as setting the theme for the next few posts.
General Relativity (GR) is a geometric theory of gravitation proposed by Einstein. In GR, the flat background spacetime of special relativity is replaced by a curved spacetime. The curvature of spacetime is greater around objects with a higher mass, and particles (including light) travel along paths called geodesics (essentially the ‘shortest’ path between two points) in this curved spacetime.
The standard analogy here is bowling ball on a rubber sheet: If you take a big rubber sheet stretched flat and put a bowling ball in the middle, the sheet stretches and curves around the bowling ball, but as you get further away from the ball the sheet becomes less curved.
Now imagine a tiny little marble moving on the rubber sheet. As the marble gets closer to the bowling ball, the curvature of the sheet causes the marble to accelerate towards the bowling ball. The marble feels this acceleration as the gravitational force of the bowling ball – gravity is just the curvature of spacetime. The path that the marble traces out on the rubber sheet is called a geodesic. In flat space, a geodesic is simply a straight line, but on a curved surface geodesics can be more complicated. Archibald Wheeler described GR rather poetically, saying “Spacetime tells matter how to move, matter tells spacetime how to curve”
The rubber sheet analogy is all well and good, but let’s get into some honest mathematics!
General Relativity is described, rather succinctly, with the Einstein Field Equations (EFE).
The EFE relates the local curvature of spacetime with the local energy and momentum in that spacetime. Solutions to the EFE are metrics which describe the geometry of spacetime. It should be noted that the apparent simplicity of this equation is incredibly misleading. The Einstein Field Equations are actually a set of 10 coupled, non-linear, hyperbolic-elliptic, partial differential equations – solving them is a highly non-trivial task and exact solutions are only known for the simplest cases.
The EFE determine how a given distribution of matter/energy influences the geometry of spacetime. To describe the motion of freely falling matter through curved spacetime, we also need the geodesic equation.
The simplest solution to the Einstein Field Equations is the Minkowski metric – that is, flat space. Traditionally denoted by rather than , the Minkowski metric describes spacetime with no matter, no energy, no curvature – nothing. The metric has, in cartesian coordinates, the following form
The metric is related to the line element (spacetime interval) in the following way
where the repeated indices are implicitly summed over. For the Minkowski metric in cartesian coordinates , this becomes
A metric can either be specified by its components in a particular coordinate system , or the line element . Note that I’m using physicists terms like ‘line element’ and ‘spacetime interval’ in this article. There are two reasons for this. Firstly, I don’t want to go into the mathematics of differential geometry in this post – my goal is to describe some cool physics, and the clearest way to do that is to use physics notation. Secondly, this is the notation used in the literature.
I’m going to finish off this post by describing another solution to the Einstein Field Equations – the Schwarzchild metric. This metric describes spacetime around a spherically symmetric, uncharged, non-rotating massive object. In polar coordinates , the metric takes the form
where is the Schwarzchild radius, a constant related to the mass of the object.
If any spherically symmetric, uncharged, non-rotating massive object has a radius less than its Schwarzchild radius, the object forms a black hole and is called a Schwarzchild black hole. The physics of black holes is an incredibly cool area of General Relativity, and there are a number of surprising results that you can get. In the next blog post I will describe one of the lesser known results – the photon sphere.
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https://en.wikiquote.org/wiki/%C3%89variste_Galois
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math
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Évariste Galois (October 25, 1811 – May 31, 1832) was a French mathematician, who, while still in his teens, developed the well-known Galois theory. Galois theory is capable to determine whether a polynomial with rational coefficient can be solved by radicals and give a clear insight about what kind of length ratio can be constructed by compass and straightedge, thereby solving the long-standing problems of solving a polynomial by radicals. His life is considered to be one of the most romantic in all of mathematics because of the contributions he has made in such a short span of life.
|This article about a mathematician is a stub. You can help Wikiquote by expanding it.|
- ... un auteur ne nuit jamais tant à ses lecteurs que quand il dissimule une difficulté.
- ... an author never does more damage to his readers than when he hides a difficulty.
- in the preface of Deux mémoires d'Analyse pure, October 8, 1831, edited by Jules Tannery (1908). Manuscrits de Évariste Galois. Gauthier-Villars. p. 27.
- Ne pleure pas, Alfred ! J'ai besoin de tout mon courage pour mourir à vingt ans !
- Don't cry, Alfred! I need all my courage to die at twenty.
- Quoted in: Léopold Infeld (1978) Whom the gods love: the story of Évariste Galois. p. 299.
- [This] science is the work of the human mind, which is destined rather to study than to know, to seek the truth rather than to find it.
Following quotes are reproduced in The mathematical writings of Évariste Galois. (2011, edited by Peter M. Neumann).
- Nous avons transcrit textuellement la démonstration que nous avons donnée de ce lemme dans un mémoire présenté en 1830. Nous y joignons 154 IV The First Memoir comme document historique le note suivante qu’a cru ^devoir^ y apposer M. Poisson. On jugera. Note de l’auteur [We have faithfully transcribed the proof of this lemma that we have given in a memoir presented in 1830. We append as a historical document the following note which Mr Poisson believed he should add. Posterity will judge. Note by the author]
- Poisson, reading Galois' First Memoir, found the proof of Lemma III insufficient, and wrote in pencil the following comment. It angered Galois sufficiently that he wrote the above quote directly below.
- La démonstration de ce lemme n’est pas suffisante; mais il est vrai d’après le No . 100 du mémoire de Lagrange, Berlin, 1771. [The proof of this lemma is insufficient; but it is true according to No . 100 of the memoir by Lagrange, Berlin, 1771.]
- Mais je n’ai pas le temps et mes idées ne sont pas encore bien développées sur ce terrain qui est immense. [But I don't have the time and my ideas are not yet well developed on this immense terrain.]
- The Testamentary Letter of 29 May 1832
- Il parait après cela qu'il n'y a aucun fruit à tirer de la solution que nous proposons. [It seems there is no fruit to be drawn from the solution we offer.]
- Page 226, VI.4 Dossier 9: Preliminary discussion, folio 59a.
Quotes about Galois
- Langlands and Grothendieck are both (at least) Giants by any measure, and both were consciously successors of Galois.
- Michael Harris (18 January 2015). Mathematics without Apologies: Portrait of a Problematic Vocation. Princeton University Press. p. 24. ISBN 978-1-4008-5202-4.
- Since my mathematical youth, I have been under the spell of the classical theory of Galois. This charm has forced me to return to it again and again.
- Mario Livio (19 September 2005). The Equation that Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry. Simon and Schuster. p. 90. ISBN 978-0-7432-7462-3.
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| 3,611 | 22 |
https://dev.econometricsociety.org/publications/econometrica/2010/05/01/corrigendum-%E2%80%9Cgames-imperfectly-observable-actions-continuous
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math
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Econometrica: May 2010, Volume 78, Issue 3
Corrigendum to “Games With Imperfectly Observable Actions in Continuous Time”
Tadashi HashimotoSannikov (2007) investigated properties of perfect public equilibria in continuous‐time repeated games. This note points out that the proof of Lemma 6, required for the proof of the main theorem (Theorem 2), contains an error in computing a Hessian matrix. A correct proof of Lemma 6 is provided using an additional innocuous assumption and a generalized version of Lemma 5.
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| 546 | 4 |
http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.260819.html
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math
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SOLUTION: I need to factor 24-11w+w^2= ? Thanks so much for your help.
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https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2012-May/071836.html
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math
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[gmx-users] Calculating number density using g_density
adeyoung at andrew.cmu.edu
Sat May 26 18:00:34 CEST 2012
It is possible to compute the number density using g_density, with the
switch "-dens number". Do you know if this is the number density of
molecules? Or is it the number density of atoms?
Ideally, I would like to compute the number density of _molecules_.
Specifically, I would like to use the center of mass of each molecule to
represent that molecule's position. Then the center of mass of each
molecule should be used to calculate the number density of molecules. Do
you know if this is possible using any of the Gromacs utilities?
Thank you so very much for your time!
Carnegie Mellon University
More information about the gromacs.org_gmx-users
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| 760 | 14 |
https://betsiti-mobilnaja-versija.ru/solved-problems-on-fourier-transform-711.html
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math
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Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.Tags: Solve My Math ProblemBusiness Plan Checklist TemplateFor Euthanasia Essay IntroductionWalmart Swot Analysis EssayThesis Printing And Binding WaterfordAn Event That Changed My Life College Essay
\end$$ Boundary value problems of partial differential equations concerned with temperature as the unknown may be solved by a finite Fourier transform method.
The temperature at points other than the boundary, if they should be needed, can be obtained by summing the Fourier coefficients.
Similarly, partial differential equations are changed into ordinary differential equations by applying these transformations.
Two transformations which are particularly useful in solving boundary value problems are the finite Fourier sine and cosine transformations.
If you are unfamiliar with the Fourier Transform, check out my introduction here.
The method presented here works for initial value problems where the PDEs is solved with respect to some initial values known a priori.\end$$ $$\begin &\frac\int_^\sin nx\int_^F_(x-y) G_(y) \,dy\,dx -\frac \int_^\sin nx \int_ ^F_(x-y) G_(y) \,dy\,dx \ &\quad = \frac \int_^\sin nx \int_^F_(x-y) G_(y) \,dy\,dx \ &\quad = \frac \int_^\sin nx \int_^F_(x-y) G _(y) \,dy\,dx \ &\quad = \frac S\=S\biggl\ , \end$$ Since we will be concerned with functions of two independent variables, it is necessary to comment briefly on the characteristic of the finite Fourier transform of such a function.The finite Fourier sine transformation of Similar changes must be made in the other formulas in the applications which follow.Many other transforms exist which may be used to solve PDEs [4, 5].A feature which makes the finite transform a very economical method, is that the inverse transform may be solved only for regions of interest [6–8].In my previous post, PDEs using Fourier Analysis I, I investigated solving PDEs, in particular the wave equation, on a membrane using Fourier Series.In this article, I shall formulate a generalised method of solving PDEs, that is founded on the Fourier Transform.The particular transformation discussed in this paper is the finite Fourier transform, which is applicable to equations in which only the even order derivatives (of the function) with respect to transformed variable will be treated.The finite Fourier transform method is one of various analytical techniques in which exact solutions of boundary value problems can be constructed.The most widely used methods for the solution of boundary value problems are based on finite differences.These methods require certain assumptions about where the finite difference equals the derivative which by necessity have to be most loosely made on the boundaries.
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| 3,040 | 7 |
http://trhomeworkitjc.a3maal.me/am-modulation-paper-term.html
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math
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Open document below is an essay on amplitude modulation and demodulation from anti essays, your source for research papers, essays, and term paper examples. Chapter 3 amplitude modulation wireless information transmission system lab institute of communications en gggineering national sun national sun yatyat--sensen. Frequency modulation: theory, time domain (in contrast to amplitude modulation) phase modulation: theory, time domain, frequency domain. In byrnes, h ed 2006 advanced language learning the contribution of halliday and vygotsky london continuum pp 95-108 languaging, agency, and. What is overmodulation am over modulation causes the carrier wave to invert it's phase when the modulating signal has an amplitude that is above a certain level. Analog communication lab amplitude modulation date: amplitude modulation - paper amplitude modulation we can define another term k such that, k= am ac.
Pulse-amplitude modulation (pam) pulse-width modulation (pwm) pulse-position modulation (ppm) modulating signal chapter 7: pulse modulation. Application to automatic speech recognition (asr) is studied speech term “modulation” in this paper is speech production and amplitude modulation. Main difference between amplitude modulation and frequency modulation is, in amplitude modulation, the amplitude of the carrier wave is modified accordin.
Abstract: this paper presents a comparative theoretical study of amplitude, phase, and frequency modulation in amplitude modulation the modulation vector. Amplitude modulation objectives basic system pm may be an unfamiliar term but is commonly review paper of hybrid paper reduction technique for wavelet packet. Amplitude modulation (am) is a modulation technique used in electronic communication if we just look at the short-term spectrum of modulation.
Ece1352 analog integrated circuits i term paper university of under the name of automatic gain control (agc) circuits of amplitude modulation. See also amplitude modulation and frequency modulation quality process production backorder procedure fixed cost never miss another term.
Effects of frequency modulation (fm) transmitter microphone directivity on speech perception in noise. What is the basic difference between am and fm amplitude modulation (am) should i reject a request if i've already reviewed many papers from the same author. In telecommunications and signal processing, frequency modulation (fm) is the encoding of information in a carrier wave by varying the instantaneous frequency of the.
Time series of amplitude modulation of wind farm noise at noise receptor locations an improved version of the am assessment tool - featuring long term data. Amplitude modulation and demodulation of chaotic signals garded as a special case of the amplitude modulation tude modulation in this paper dsb case is. Read this essay on amplitude demodulation the term modulation is defined as the alteration of a carrier signal based on whether the amplitude modulation.
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| 2,988 | 6 |
https://www.briandunning.com/cf/200
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math
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FlexCeiling ( number ; precision )
An expanded "Ceiling" that accepts a precision argument
Average rating: 4.1 (37 votes) Log in to vote
Kevin Frank & Associates
FlexCeiling( 123.341 ; -2)
FlexCeiling( 123.341 ; -1)
FlexCeiling( 123.341 ; 0)
FlexCeiling( 123.341 ; 1)
FlexCeiling( 123.341 ; 2)
FlexCeiling( 123.341 ; 3)
Function definition: (Copy & paste into FileMaker's Edit Custom Function window)
While the Ceiling function by definition returns the smallest integer >= x, I found myself wishing it could accept a precision argument the way Round does. The FlexCeiling custom function allows this to happen.
Essentially, FlexCeiling is a "magic Round" that always makes a number larger -- regardless of whether that number is positive or negative -- whereas Rounding (via the Round function) a negative number "up" actually makes it smaller (more negative).
E.g., Round ( -4.7 ; 0 ) = -5 whereas FlexCeiling ( -4.7 ; 0 ) = -4
Note: see also the companion custom function, FlexFloor.
Note: these functions are not guaranteed or supported by BrianDunning.com. Please contact the individual developer with any questions or problems.
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https://fullpcsoftware.com/wolfram-mathematica/
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math
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Wolfram Mathematica activation key gives you the excellent computing environment. It creates many computing algorithms and visualization. This flawlessly joins a numeric and democratic computational engine, programming communications, documentation system, and illustrated the system. Wolfram Mathematica license server gives you powerful computing software with an easy, user-friendly interference. It includes symbolic and high-performance numeric total with 2D, 3D data visualization, board programming capabilities. This software includes notebook construction allows you to the produce entirely customization files which give the expert mathematical typesetting and publication-quality design.
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https://www.beselfsmarter.com/podcasts/page/3/
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Skip to content
podcast for leaders seeking to be better every day.
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https://m.scirp.org/papers/84750
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math
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System of almost periodic difference equations has been studied to describe phenomena of oscillations in the natural and social sciences. The investigation of almost periodic systems has been developed quite widely during the twentieth century, since relationships with the stability theory have been found. A main interest of the subject is the existence theorem for almost periodic solutions. Obviously an almost periodic solution is a bounded solution, but the existence of bounded solutions does not necessarily imply the existence of almost periodic solutions. Therefore, in order to prove the existence of almost periodic solutions, we need some additional conditions to the existence of bounded solutions. A main subject of the investigation has been to find such additional conditions, and up to now, many conditions have been considered (for example, in the linear system, J. Favard’s separation condition ).
In the Section 4, we consider the nonlinear almost periodic system of
where k is a positive integer, are almost periodic in n and satisfy
In the special case where are constant functions, system (1) is a mathematical model of gas dynamics and was treated by T. Carleman and R. D. Jenks . In the main theorem, we show that if the matrix is irreducible, then there exists a positive almost periodic solution which is unique and some stability. Moreover, we can see that this result gives R. D. Jenks’ result in the case where are constant functions. In the Section 5, we consider the linear almost periodic system with variable coefficients
where . Even in nonlinear problems, system (2) plays an important role, as their variational equations and moreover, it is requested to determine the uniformly asymptotic stability of the zero solution from the condition about . When is a constant matrix, it is well known that the stability is equivalent to the following condition (cf. );
“Absolute values of all eigenvalues of are less than one.”
However, it is not true in the case of variable coefficients, and hence we need additional conditions to (2). In the main theorem, we show that one of the such conditions is the diagonal dominance matrix condition on , that is, satisfies
This result improves a stability criterion based on results of F. Nakajima for differential equations.
We denote by Rm the real Euclidean m-space. Let and . Z is the set of integers, Z+ is the set of nonnegative integers. For , let be the Euclidean norm of x and be the i-th component. Let
We introduce an almost periodic function , where U is an open set in Rm.
Definition 1. is said to be almost periodic in n uniformly for , if for any and any compact set K in U there exists a positive integer such that any interval of length contains an integer τ for which
for all and all . Such a number τ in above inequality is called an ò-translation number of .
In order to formulate a property of almost periodic functions, which is equivalent to the above definition, we discuss the concept of the normality of almost periodic functions. Namely, let be almost periodic in n uniformly for . Then, for any sequence , there exist a subsequence of and a function such that
uniformly on as , where K is a compact set in U. There are many properties of the discrete almost periodic functions , which are corresponding properties of the continuous almost periodic functions [cf. ]. We denote by the function space consisting of all translates of f, that is, , where
Let denote the uniform closure of in the sense of (4). is called the hull of f. In particular, we denote by the set of all limit functions such that for some sequence , as and uniformly on for any compact subset S in Rm. Specially, for a function on Z with values in Rm, denotes the set of all function such that for some sequence ,
where the symbol “?” stands for the uniformly convergence on any compact set in Z (in short, “in Z”). Clearly, .
By (3), if is almost periodic in n uniformly for , so is a function in .
We define the irreducible matrix to need after.
Definition 2. An matrix is said to be irreducible if for any two nonempty disjoint subsets I and J of the set of m integers with , there exists an i in I and a j in J such that . In the case where is scalar, is said to be irreducible if . Otherwise, is said to be reducible, and we can assume that takes the form of
where * is square matrix, *' is matrix, is zero or a square irreducible matrix.
3. Linear Systems
We consider the system of linear difference equation
where and the matrix is bounded on Z and almost periodic function in n. We state discretization of Jenks and Nakajima' results for differential equations .
Now we define stability properties with respect to the subset K in Rm. Here, we denote by the solution of system (5) with initial condition .
Definition 3. The bounded solution of system (5) defined on Z is said to be;
i) uniformly stable (in short, U.S.) in K on Z+ if for any there exists a such that for all whenever and at some in Z+.
ii) uniformly asymptotically stable (in short, U.A.S.) in K on Z+ if it is U.S. in K on Z+ and if there exists a and, if for any there exists a such that for all whenever and at some in Z+.
iii) uniformly asymptotically stable (in short, U.A.S.) in the whole K on Z+ if it is U.S. in K on Z+ and if for any and there exists a such that for all whenever and , at some in Z+.
When Z+ in the definitions (i), (ii) and (iii) is replaced by Z, we say that is U.S. in K on Z, U.A.S. in K on Z and U.A.S. in the whole K on Z, respectively. Clearly Definition 3 agrees with the definitions of the usual stability properties in the case where .
Throughout this paper, we suppose the following conditions;
iii) each element in is irreducible.
First of all, we prove the following lemmas.
Lemma 1. Consider the m-equations , , where is continuous on second variable x in Rm, and assume that the initial value problem has a unique solution.
a) If , then the set Π is invariant.
b) If for and , then the set D is positively invariant, and in addition,
if , then the set Ω is positively invariant.
In the case of differential equations, the proof of the similar lemma is obvious (for instance, see ). We modify it to prove this lemma, but we omit it.
Lemma 2. If conditions (i) and (ii) are satisfied, then the trivial solution of system (5) is U.S. in P on Z and also it is U.S. on Z.
By modifying theorem in , we can easily prove Lemma 2 at same technique.
Lemma 3. If each element in is irreducible, then the each element in , we say and , has the property that for any two nonempty disjoint subsets I and J of the set of m integers with , there exists an and such that
Proof. Suppose not, Then there exists a in and two nonempty disjoint subsets I and J of with such that
Since is bounded on Z, there exists a subsequence as , such that
where . Clearly,
This show the reducibility of , which is a contradiction. This proves Lemma 3.
For system (5), we consider the system in of
Lemma 4. Assume that conditions (ii) and (iii) are satisfied for system (5), and let be a nontrivial solution of system (6) such that on Z. Then there exists a constant such that
Proof. Let be a solution of system (6) such that on Z. First of all, we show that if at some , then
Since satisfies the equation
where . Moreover, since , we have
Thus, we obtain
Because and on Z. Now suppose that Lemma 4 is not true. Then for some B in , the corresponding system (6) has a nontrivial solution , on Z, such that for some sequence ,
Set . Then, satisfies
Since the sequence is bounded, is uniformly bounded on any finite interval in Z, and hence there is a convergent subsequence of , which is again denoted by , such that
in Z for some function as .
We can also assume that
where and . Therefore, is the solution of the system
on Z and . Moreover (8) implies that . Thus, as was proved above, we have
For this , we define two subsets I and J of by for , where depends on and . Then , and since . By Lemma 3,
Now the -th equation of system (9) takes the form of
because of the definition of the set I. Since each term in the left hand side of (11) is nonnegative, all of them are equal to zero. Therefore
which implies, by (10),
This contradicts the definition of the set of J. The proof is completed.
The following proposition is an immediate result of Lemma 4.
Proposition 1. Under conditions (ii) and (iii), system (6) has no nontrivial solution such that
where for some .
We next consider a non-homogeneous system corresponding to system (5)
and assume that satisfies conditions (i), (ii) and (iii).
Lemma 5. If is bounded on Z with values in Rm and is bounded on Z+, then all solutions of system (12) are bounded on Z+.
Proof. It is sufficient to show that (12) has at least one bounded solution on Z+, because the trivial solution of (5) is U.S. by Lemma 2. We consider the system with real parameter ò
and show that for a sufficiently small ò, system (13) has a bounded solution on Z+, which implies the existence of a bounded solution on Z+ for system (12) by replacing x in (13) with . For a and for the m-vector e each of whose components is 1, let be a convex cone defined by
where denotes the inner product and . Clearly, . Every solution of (13) satisfies
because of condition (i). By replacing n with n-1 of the above both sides,
is sufficient small number and .
When , we have
Therefore, in order to show the boundedness of with in Ω, it is sufficient to prove that on Z+ if ò is sufficiently small. Now suppose that for each solution of (13) with in Ω, there exists an such that
We can assume that
where and denote the boundary and the closure of the set K, respectively. If we set , is a solution of the system
such that at and for . Thus, by (14),
The same argument in the proof of Lemma 4 enables us to assume that
in Z for some function as
in Z for some as
Therefore, satisfies and clearly, for ,
Moreover we have , which implies by Lemma 1 that
From this and (15) it follows that
Now we show that . In fact, if , we have
Thus because , and hence
which contradicts (16). Therefore (16) and (17) hold for . Moreover this enables us to assume that
Because is compact in the sense of the convergence. This contradicts the conclusion in Proposition 1. This proves that on Z+ if ò is sufficiently small. The proof is completed.
Lemma 6. Under the assumptions (i) and (ii), if for each B in , the trivial solution of the system
is U.S. on Z and U.A.S. on Z+, then the trivial solution of system (5) is U.A.S. on Z.
Proof. Let be the solution of (5). Since the trivial solution of (5) is U.S. on Z by Lemma 2, as is seen from (ii) in Definition 3, it is sufficient to show that for any there exists a such that
whenever and , where is the number in (i) of Definition 3.
Now suppose that there exists an and sequences in Z and in Rm such that and
Set . Then, satisfies
We can assume that
in Z+ for some function as
in Z for some as .
Therefore is a solution of the system
On the other hand, we have
because the trivial solution of (18) is U.A.S. on Z+. Therefore there arises a contradiction. Thus the proof is completed.
We show the following theorem, before we will mention a definition of the exponential dichotomy of a linear system;
System (5) is said to possess an exponential dichotomy if there exists a projection matrix P and positive constants and such that
where, I is a identical matrix and F is a fundamental matrix solution of system (5) (cf. ).
Theorem 1. Assume that system (5) satisfies conditions (i), (ii) and (iii), Then the trivial solution of system (5) is U.A.S. in P on Z.
Proof. On the set Π which is invariant for system (5), the system is written as the -system
where and is an matrix whose element is given by for . First of all, we can show that for each in , the system
has an exponential dichotomy on Z+ since (20) has at least one bounded solution, and as is well known (cf. ), it is equivalent to show the system
possesses at least one bounded solution on Z+ for any bounded function on Z+. For each in there corresponds some in such that the element of is equal to for . For , let be defined by
Obviously and are bounded on Z+. Applying Lemma 5 to the m-system
we obtain the bounded solution on Z+ with , and which yields
Hence we can verify that is a bounded solution on Z+ of system (21). The exponential dichotomy of (20) implies further that the trivial solution is U.A.S. on Z+, because the trivial solution is U.S. on Z by Lemma 2. Therefore it follows from Lemma 6 that the trivial solution of (19) is U.A.S. on Z, i.e., the trivial solution of (5) is U.A.S. in P on Z. The proof is completed.
4. Nonlinear Systems
We consider the nonlinear almost periodic system of
where is almost periodic function of n with conditions
In addition, assume that are continuously differentiable for , and for real number , where is the derivative of at u.
We first consider the linear system
and its perturbed system
where is an matrix function, almost periodic function in n, is continuous with respect to its second argument and uniformly for . Assume that the set Π is invariant for both system (23) and (24).
First of all, we can prove the following lemmas.
Lemma 7. If the trivial solution of system (23) is U.A.S. in Π on Z, then the trivial solution of system (24) has also the same stability property.
Proof. Let for . Then there are positive constants and such that
because . On the set Π, systems (23) and (24) are written as
respectively, where the element of , is given by for and uniformly for . Inequality (25) shows that the trivial solution of (23) is U.A.S. in P if and only if the trivial solution of (26) is U.A.S., and we have also the same equivalence between (24) and (27). As is well known, if the trivial solution of (26) is U.A.S., then the trivial solution of (27) has also the same stability property. Thus our assertion is proved.
The following lemma is obtained by the slight modification of the difference equation to Seifert’s result . Then, we will omit the proof (cf. ).
We consider the almost periodic nonlinear system
where is almost periodic in n uniformly for and for a constant , for and .
Lemma 8. Assume that the set Ω is positively invariant for system (28) and all solutions in Ω on Z are U.A.S. in Ω on Z. Then the set of such solutions is finite and consists of only almost periodic solutions which satisfy
on Z for and some constant .
Now we can show the following theorem. Since the last statements of the following theorem are alternative, under each assumption of these statements we can prove the existence of almost periodic solutions in Ω and the module containment.
Theorem 2. Under the assumptions (iv) and (v), system (22) has a nontrivial almost periodic solution in Ω whose module is contained in the module of . In addition to the above assumptions, if is irreducible, then the almost periodic solution of (22) is unique in Ω, which remains in on Z, and it is U.A.S. in the whole Ω on Z, where . Moreover, if is reducible, then at least one of the above almost periodic solutions satisfies that on Z, where .
Proof. First of all, we consider the case where is irreducible. Since system (22) satisfies the conditions of Lemma 1, the set Ω is positively invariant, namely, on Z+ for a solution of (22) with , and furthermore we can assume that
because of the almost periodicity of . We can show that this is U.A.S. in Ω on Z. If we set in system (22), then for x in Ω and
And Π is invariant for the above system. Considering the first approximation of system (29)
where is defined by , condition (iv) implies that Π is also invariant for (30). Then, by Lemma 6, if the trivial solution of (30) is shown to be U.A.S. in Π on Z, then the trivial solution of (29) has the same stability, and consequently is U.A.S. in Ω on Z. Therefore it is sufficient to show that the trivial solution of (30) is U.A.S. in Π on Z. Clearly is bounded and we have
because of conditions (iv) and (v), respectively. Thus satisfies conditions (i) and (ii). Condition (iii) will be verified in the following way. Applying the same argument as in the proof of Lemma 4 to system (22), we can see that there exists a constant such that
and hence there is a constant such that
Therefore, (31) implies
which guarantees that each element of is irreducible, because is irreducible and almost periodic. Thus it follows from Theorem 1 that the trivial solution of (30) is U.A.S. in Π on Z, i.e., all solutions of system (22) in Ω on Z are U.A.S. in Ω on Z. Therefore Lemma 8 concludes that system (22) possesses an almost periodic solution in Ω which remains in by (32), and the set of solutions in Ω on Z is finite and consists of only almost periodic solutions which satisfy
on Z for and some constant .
Next we can show that there exists a such that each solution of (22) with satisfies that for some and the constant of (ii) in Definition 3,
because is U.A.S. in Ω. Suppose that this is not true. Then there exists a small constant less than β and sequences in Z and in Ω such that
Since is almost periodic in n uniformly for , we can choose a sequence , such that
If we set for and , these functions satisfy
because . Moreover,
We can assume that in Z for some function , as . Therefore are solutions of system (22), because in as , and
which shows that system (22) has distinct solutions in Ω on Z. This is contradiction. Therefore, is U.A.S. in the whole Ω on Z, if the uniqueness of is shown.
Now we will prove the uniqueness of . Suppose for and set
Then and are open sets in Ω, and moreover these sets are nonempty and disjoint, because on Z for . On the other hand , (33) shows that , which contradicts the connectedness of Ω. Thus the uniqueness of an almost periodic solution is proved, and moreover, as is seen from , this uniqueness guarantees the module containment of the almost periodic solution.
Now consider the case where is reducible. We can assume takes the form of
where is zero or a square irreducible matrix of order . If is zero, system (22) obviously has the constant solution in such that for and . In the latter case, if we set in system (22)
then system (22) is reduced to the lower dimensional system
where . Since is irreducible, the above system (34) has an almost periodic solution such that
and furthermore the module of is contained in the module of , i.e., of the module of . Thus, system (22) has an almost periodic solution in on Z such that for and for . The proof is completed.
Remark 1. As will be seen from the module containment, the above almost periodic solution is a critical point in the case where is a constant. Hence Theorem 2 is a discretization of Nakajimas’ result (Theorem 2 in ).
5. A Stability Criteria of Linear Systems
We consider a stability criterion for solutions of a linear system with coefficient matrix of diagonal dominance type.
We again consider a linear system (5).
Let be an matrix of functions for . We assume the following conditions;
where, denotes the determinant of matrix and
At first, we need the following lemmas for main theorem.
Lemma 9. If a square matrix A is irreducible and satisfies (38) and if for at least one j,
then A is nonsingular.
For the proof, see .
Lemma 10. If a nonsingular matrix satisfies (38), then all principal minors of A are nonsingular, namely,
Proof. Let be an principal minor of A. Then, for a permutation matrix Q,
where has rows and l columns and denotes the transposed matrix of Q. Moreover, from the definition of irreducibility, we can choose an permutation matrix such that
where is an irreducible matrix, , and has row and columns for . In particular, in the case where is irreducible, must be itself, and the matrices are not present. Setting for unit matrix I, where is the direct sum of and I, we have
where and has rows and columns. Since the diagonal dominance condition (38) is invariant under the permutation of indexes, B also satisfies (38). Hence, letting
for a fixed , we have
where the summations on j are taken along columns and for convenience. If or , then
and hence for this k,
by (39). Therefore it follows from Lemma 9 that
since is irreducible. If and , then we have the form of
which also implies (40), because . In any case, we have . Since these are true for all , it follows from (381) that
this proves Lemma 10.
Lemma 11. If system (5) satisfies conditions (36) and (38), then the norm of solution such that , is non-increasing, and consequently the zero solution is U.S..
For the proof, we can see (cf. ).
In the following theorem, we can prove that the zero solution is U.A.S., if is bounded on Z and if condition (36), (37) and (38) are satisfied.
Theorem 3. In system (5), let be bounded on Z. Assume that conditions (36) and (38) are satisfied for all and that there is a constant such that
Then the zero solution is U.A.S..
Proof. As is stated in Lemma 11, the zero solution is U.S., and hence it is sufficient to show that for any there exists a such that
whenever . Suppose that this is not true. Then there exists a constant , a sequence of solution of (5) and a sequence such that
Since is non-increasing, we have
and there exists a subinterval of such that
Set . Then, we obtain
Since is bounded, it follows from (41) and (42) that is uniformly bounded on any finite interval of Z, and thus, taking a subsequence, can be assumed to converge uniformly on any finite interval of Z. Defining by
it follows from (42) and (43) that there is a constant , such that
Since is defined on Z, we can choose an interval (for some ) such that
Here we note that because . Then
Then, we have
and there is a sequence such that
Moreover, since I is compact and is bounded on Z, we can assume that
Clearly B satisfies (36), (38) and . Taking a difference of both sides of (44) at and using relation (41), we find
where , since we have
By (45), we have
Since B satisfies (36) and (38),
Therefore each term of right hand side of (46) is non-positive, because we have
Then it follows from (46) that
since for . Thus we have
On the other hand, B satisfies (38) and , and thus it follows from Lemma 10 that all principal minors of B are nonsingular, which contradicts (461). This proves that the zero solution of system (5) is U.A.S..
Corollary 1. If system (5) is defined only for and all assumptions of Theorem 3 are satisfied for , then the zero solution is U.A.S. for .
Proof. We construct the system defined on Z by
Since system (47) satisfies all assumptions of Theorem 3 on Z, the zero solution is U.A.S. on Z, and furthermore, since system (5) coincides with system (47) for , this prove our conclusion.
Before Example 1, we state the following lemma is a special case of Theorem 3 in .
In the nonlinear system
let be almost periodic in n uniformly for and for any , let there exists a constant such that
Lemma 12. If is a bounded solution of (48) on and if for any solution of (48), is monotone decreasing to zero as , then is a unique almost periodic solution and its module is contained in the module of .
Example 1. Consider the variational linear difference equation
We now assume that is at least one bounded solution of (49) on and is any solution of (49) on , and , are some bounded functions on such that
for some positive constants and such that . We can verify that satisfies all assumptions in Corollary 1. First of all, is bounded in the future for , because and are bounded function on . It is clear that the diagonal elements of are negative and
The diagonal dominance condition (38) for requires that
which is equivalent to
and this is satisfied by
Therefore, by Theorem 3, the zero solution of (49) is U.A.S. and
where the convergence is monotone decreasing by Lemma 11. Thus, applying Lemma 12 to system (49), we find that there exists a unique almost periodic solution with the module contained in the module of .
In this paper, we obtain the existence and stability property of almost periodic solutions in discrete almost periodic systems. First, in Section 1, the research background is introduced. In Section 2, the fundamental concepts of the almost periodic solutions in discrete almost periodic systems is given. In Section 3, we are introduced to the several lemmas and have uniformly asymptotically stability theory of the linear system, and moreover, in Section 4, we consider the generalized gas almost periodic system, and if linear part is irreducible matrix, then we obtain the existence of almost periodic solutions of this system. Finally, in Sections 5 and 6, we consider and obtain an uniformly asymptotically stability criterion for solutions of a linear system with coefficient matrix of diagonal dominance conditions, and this result applies to meaningful example of a linear discrete system.
Yoshizawa, T. (1975) Stability Theory and the Existence of Periodic Solutions and Almost Periodic Solutions. Applied Mathematical Sciences 14, Springer-Verlag, Berlin.
Nakajima, F. (1976) Existence and Stability of Almost Periodic Solutions in Almost Periodic Systems. Publications of the Research Institute for Mathematical Sciences, Kyoto Univ., Kyoto, 12, 31-47.
Montandon, B. (1972) Almost Periodic Solutions and Integral Manifolds for Weakly Nonlinear Nonconservative Systems. Journal of Differential Equations, 12, 417-425.
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Dieses Lehrbuch ist der erste Band einer dreiteiligen Einf? hrung in die research. Es ist durch einen modernen und klaren Aufbau gepr? gt, der versucht den Blick auf das Wesentliche zu richten. Anders als in den ? blichen Lehrb? chern wird keine ok? nstliche Trennung zwischen der Theorie einer Variablen und derjenigen mehrerer Ver?
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Additional resources for Cours de mathematiques MP
E. area of regions) contained in the unit square, and are presented geometrically. The parameters of ~ are then extended onto the set of all equivalent pairs of distributions of X and Y. They are called grade parameters of any such pair. The so called normal concentration pattern is introduced as a reference model (which is also a grade parameter since, obviously, grade parameters can be numerical function - valued, etc). The normal concentration pattern enables easy visual comparison of the concentration curve for any pair (X; Y) with its normal counterpart.
Perhaps, it is the" expected," or familiar distribution where "things work," or where things are "in balance," or the distribution about which more is known . • Next, we calculate the concentrations (slopes) for each category in the distributions. The slopes are the ratios of the distribution proportions. 10. 15. Note: By convention, we put the reference distribution in the denominator. Each of the five corresponding categories has its own individual line segment and slope. 67 (Rounded). 22 F.
36 F. Ruland The diagonal represents the equal distribution line for gas amount and cylinder space. A "just to make sure" explanation: We call the diagonal curve the equal distribution curve. However, it is often referred to by the name uniform distribution curve. The" uniform" refers to the diagonal as being the graph of the cdf of the uniform distribution on interval (0,1). The "equal" refers to the equality between distributions, and not to equal (uniform) within distributions. " Let us keep the cylinder spaces in cells the same size, and vary the gas proportion amounts in each of the spaces.
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Presentation on theme: "CHAPTER 2 : CRYSTAL DIFFRACTION AND PG Govt College for Girls"— Presentation transcript:
1 CHAPTER 2 : CRYSTAL DIFFRACTION AND PG Govt College for Girls RECIPROCAL LATTICEProf. Harvinder KaurPG Govt College for GirlsSector-11, Chandigarh
2 Outline Reciprocal Lattice Reciprocal Lattice of various crystals Simple cubic latticeFace-centered cubic latticeBody-centered cubic latticeBragg law of DiffractionExperimental Diffraction methodsStructure and Form factor
3 Reciprocal LatticeThe reciprocal lattice of a Bravais Lattice is the set of all vectors K such thateiK.R = 1for all lattice point position vectors R. This reciprocal lattice is itself a Bravais lattice, and the reciprocal of the reciprocal lattice is the original lattice.For an infinite three dimensional lattice, defined by its primitive vectors, its reciprocal lattice can be determined by generating its three reciprocal primitive vectors, through the formulae
4 Properties Of Reciprocal Lattice Direct lattice is a lattice in ordinary space whereas the reciprocal lattice is a lattice in the Fourier space.The primitive vectors in reciprocal lattice has the dimensions of (length)-1 whereas the primitive vectors of the direct lattice have the dimensions of lengthA diffraction pattern of a crystal is a map of the reciprocal lattice of the crystal whereas a microscopic image is a map of direct latticeWhen we rotate a crystal, both direct and reciprocal lattice rotatesEach point in the reciprocal lattice represents a set of parallel planes of the crystal latticeIf the coordinates of reciprocal vector G have no common factor, then G is inversely proportional to the spacing of the lattice planes normal to GThe volume of unit cell of the reciprocal lattice is inversely proportional to the volume of unit cell of the direct latticeThe direct lattice is the reciprocal of its own reciprocal latticeThe unit cell of the reciprocal lattice need not be parallelopiped
6 Reciprocal Lattice of Simple Cubic Lattice The simple cubic Bravais, with cubic primitive cell of side a, has for its reciprocal a simple cubic lattice with a cubic primitive cell of side ( in the crystallographer's definition).The cubic lattice is therefore said to be self-dual, having the same symmetry in reciprocal space as in real space
7 Reciprocal lattice to Face-centered cubic The reciprocal lattice to an FCC lattice is the body-centered cubic (BCC) lattice.FCC in real spaceBCC in fourier space
8 Primitive Translation Vectors: Reciprocal lattice of Body-centered cubicThe reciprocal lattice to an BCC lattice is the face-centered cubic (FCC) lattice.Primitive Translation Vectors:FCC in real space
9 Real: FCC Reciprocal: BCC Real: BCC Reciprocal: FCC Brillouin zones for FCC,BCC, HCPReal: FCC Reciprocal: BCCReal: BCC Reciprocal: FCCHCP
10 Bragg law of Diffraction Prof. W.L.Bragg observed that X-rays can be reflected by the cleavage planes of the crystal as if these parallel planes were acting like mirrors to the light beam. The reflected beam lead to well defined diffraction patterns on a photographic plane. The cleavage planes are atomic planes of a crystal that are systematically stacked one over the other as a parallel set of planes. X-rays from the incident plane can penetrate deeper into a target crystal and get reflected. Thus, several beams of X-ray reflected from various planes are obtained. The final diffracted beam is the result of the superposition of these beams.Bragg law, dsin = n where, d is the interplanar spacing, is the wavelength of incident X rays and is the angle of reflection
11 Bragg law of Diffraction Significance of Bragg’s lawBragg’s law is a consequence of periodicity of the latticeThe law does not refer to the arrangement of atoms in the basis associated with each lattice pointThe composition of the basis determine the relative intensity of the various orders n of diffractionSince ~ 1Ao is inevitable, we can’t use visible light for such studies
12 Bragg law in Reciprocal Lattice (Ewald Construction)Chose a point according to the orientation of the specimen with respect to the incident beam.Draw a vector AO in the incident direction of length 2p/l terminating at the origin
13 Draw a vector AB to the point of the (Ewald Construction)Construct a circle of radius 2p/l with center at A. Note whether this circle passes through any point of the reciprocal lattice; if it does....Draw a vector AB to the point of theintersection
14 Draw a vector OB to the point of the intersection
15 Draw a line AE perpendicular to OB Complete the construction to all the intersection points in the same fashionDraw a line AE perpendicular to OB
16 By means of the Ewald construction we can write the Bragg law in vector form: Let G = OB and k = AO. For diffraction, it is necessarythat the vector k + G, that is, the vector AB, be equal in magnitude tothe vector k or(k + G)2 = k2or2k . G + G2 = 0 (1)If we call the scattered wave vector k*, thank* = k + G (2)so we can writek*2 = k2 (3)and k* - k = G, showing (2) that the scattering changes only the directionon k, and (3) that the scattered wave differs from the incident wave by areciprocal lattice vector G.Equations (2) and (3) are the momentum and energy conservation law forx-ray diffraction, which is an example of elastic scattering.Whether or not the k circle intersects a lattice vector, and hence reflects,depends on its magnitude and orientation. Using (1) we can construct in thereciprocal lattice the locus of all those waves that can produce Braggreflection. This locus represents a set of planes in three dimensions.The volume terminated by those planes is called Brillouin zone.
18 Experimental Diffraction Methods Laue’s Method : In this method a single crystal is held stationary in the path of a beam of e-m radiation (X-rays) or the Neutron radiation of continuous wavelengths. While is kept constant, the wavelength , is varied so that the Bragg law is satisfied. A plane film receives the diffracted beams. A developed film after its exposure shows a diffraction pattern that consists of series of spots.Laue spots in a diffraction pattern are actually map of the reciprocal lattice of the crystal under experiment.
19 Experimental Diffraction Methods Rotating Crystal Method : In this method , a single crystal is rotated about the fixed axis in a beam of monochromatic X-rays or neutrons. The angle is variable while the wavelength is kept constant. The variation of angle due to rotation of the crystal brings different atomic planes in the crystal into position for which Bragg’s reflection holds good. To record such reflections a film is mounted on a cylindrical holder that is concentric with a rotating spindle.
20 Experimental Diffraction Methods Powder Method : If a powdered specimen is used, instead of a single crystal, then there is no need to rotate the specimen, because there will always be some crystals at an orientation for which diffraction is permitted. Here a monochromatic X-ray beam is incident on a powdered or polycrystalline sample.This method is useful for samples that are difficult to obtain in single crystal form.The powder method is used to determine the value of the lattice parameters accurately. Lattice parameters are the magnitudes of the unit vectors a, b and c which define the unit cell for the crystal.For every set of crystal planes, by chance, one or more crystals will be in the correct orientation to give the correct Bragg angle to satisfy Bragg's equation. Every crystal plane is thus capable of diffraction. Each diffraction line is made up of a large number of small spots, each from a separate crystal. Each spot is so small as to give the appearance of a continuous line.
21 The Powder MethodIf a monochromatic x-ray beam is directed at a single crystal, then only one or two diffracted beams may result.
22 The Powder MethodIf the sample consists of some tens of randomly orientated single crystals, the diffracted beams are seen to lie on the surface of several cones. The cones may emerge in all directions, forwards and backwards.
23 The Powder MethodA sample of some hundreds of crystals (i.e. a powdered sample) show that the diffracted beams form continuous cones. A circle of film is used to record the diffraction pattern as shown. Each cone intersects the film giving diffraction lines. The lines are seen as arcs on the film.
24 Structure and Form Factor A crystal is a periodic arrangement of atoms in a particular pattern. Each of the atoms may scatter incident radiation such as X-rays, electrons and neutrons. Because of the periodic arrangement of the atoms, the interference of waves scattered from different atoms may cause a distinct pattern of constructive and destructive interference to form. This is the diffraction pattern caused by the crystal.In the kinematical approximation for diffraction, the intensity of a diffracted beam is given by:where is the wave function of a beam scattered a vector , and is the so called structure factor which is given by:Here, rj is the position of an atom j in the unit cell, and fj is the scattering power of the atom, also called the atomic form factor. The sum is over all atoms in the unit cell. It can be shown that in the ideal case, diffraction only occurs if the scattering vector is equal to a reciprocal lattice vector .
25 Structure Factor For Specific Lattice Types Body-centered cubic (BCC) : As a convention, the body-centered cubic system is described in terms of a simple cubic lattice with primitive vectors ax,ay,az with the basis consisting of ro = 0 and r1 = (a/2) (x+y+z). In a monoatomic crystal, all the form factors f are the same. The intensity of a diffracted beam scattered with a vector(2/a) (hx+ky+lz) by a crystal plane with Miller indices given by (h,k,l)
26 Structure Factor For Face Centered Cubic The face-centered cubic system is described as r0 =0, r1 = a/2(x+y), r2 = (a/2)(y+z)r3 = (a/2)(x+z) with indices given by (1/2,1/2,0), (0,1/2,1/2) and (1/2,0,1/2)
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https://www.homofaciens.de/technics-semiconductor-operational-amplifier_en.htm
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The video about op-amps
Integrated circuitsLike explained at the chapter about amplifying circuits, it is tricky to build circuits with a high gain. Many factors like temperature drift, crossover distortions or linearity have to be considered to minimize the distortions caused by the electronic circuits. Several components are required to build one ore more pre-amplifier(s) and a power circuit. Today, it is more easy to integrate a large number of transistors, capacitors or resistors into a small chip instead of assembling a circuit manually by using discrete electronic components. Amplifying circuits with different properties are available as cheap and tiny chips. A special group of amplifying circuits is called operational amplifiers or short op-amp. The drawing below, which is from Wikipedia, shows a component level diagram of the common 741 op-amp.
You would need a large board to build this circuit by using discrete components...
The drawing shows the circuit diagram symbol for an operational amplifier. The functionality of the pins is:
VS+ - positive power supply
VS- - negative power supply
V+ - non-inverting input
V- - inverting input
Vout - output
An op-amp amplifies the voltage drop between V+ and V-, which is called the differential input voltage. Operational amplifiers are often used with a symmetrical bipolar power supply, hence the referential potential (ground) is half the total voltage between the negative and the positive supply voltage. While the difference in potential at the two input clamps is zero (e. g. they are shortened), the resulting output voltage of an ideal operational amplifier is also zero volts which is sometimes called virtual ground meaning the output clamp of the op-amp is at a steady reference potential, without being connected directly to the reference potential.
Real devices normaly require a differencial DC voltage between the input clamps to make the output zero volts. This parameter is called input offset voltage (Vos) and it is usually around 0.1mV.
The gain of integrated circuit op-amps is typically 100000 or more, so a voltage drop of just +0.1V between the two input clamps would result in an output voltage of +10000V. Like mentioned at the chapter about amplifying circuits, the voltage drop at the output can't exceed the value of the supply voltage, so the output signal gets clipped. Situations in which the output voltage is equal or greater than the supply voltage are referred to as saturation of the amplifier. A sine curve with a peak voltage of just 1mV between the input clamps becomes a square wave signal at the output clamp.
Negative feedbackFigure 4:
If your intention is to amplify a sine wave signal, an op-amp with a gain of more than 100000 is nearly useless. By applying a portion of the output voltage to the inverting input, the total gain of the circuit can be reduced.
At the drawing, the signal source is attached between ground and non-inverting input of the op-amp. R1 and R2 are forming a voltage divider between output of the op-amp and ground. The inverting input is attached between R1 and R2. Let's assume the following values: The gain of the op-amp is 100000, R1 = 1kΩ, R2 = 10kΩ. VS- is attached to -12V, VS+ to +12V.
The situation is simple when attaching 0V to the non-inverting input: The output voltage is 0V and the resulting voltage drop at R1 and so the inverting input is 0V.
What's happening if the input voltage at the non-inverting input jumps up to 0.1V? The resulting differential input voltage is now 0.1V, the gain is 100000, hence the voltage output would climb up to 10000V, but it is clipped at +12V. Now that 12V are attached to the voltage divider, the resulting voltage drop at R1 is
(1) Vout = (V+ - V-) * β
(2) V- = Vout * R1 / (R1 + R2)
By inserting equation (2) in (1) we get:
Vout = (V+ - (Vout * R1 / (R1 + R2))) * β
Solving the equation for Vout:
V+ - Voltage drop between non-inverting input and ground (=input voltage)
V- - Voltage drop R1
Vout - Voltage output op-amp
R1, R2 - Resistors of the voltage divider
β - Voltage gain of op-amp
For very high values of β, the equation can be simplified to:
The voltage gain Vout / V+ of the op-amp with negative feedback is:
Adding a negative feedback via a voltage divider reduces the gain of an op-amp. The overall gain Vout / V+ is called closed-loop gain (ACL), because of the feedback provided by the resistors of the voltage divider. Without a feedback loop, the overall gain of the op-amp is called open-loop gain (AOL)
Without negative feedback, an op-amp acts like a comparator. At the drawing, the inverting input is connected to ground, hence the output will be maximum positive if the voltage applied to the non-inverting input is positive. If the input voltage becomes negative, the output will be maximum negative. The output of the op-amp can be either VS- or VS+ so it indicates if the input voltage is larger or lower than 0V.
By connecting the inverting input to a voltage divider, the voltage of the non-inverting input can be "compared" to any voltage level between VS- and VS+.
When connecting the non-inverting input to the voltage divider, the output signal is VS+ if the voltage at the inverting input is lower than that at the non-inverting input, respectively it becomes VS- if it climbs above that value. Now, the output signal is inverted.
Besides the negative feedback described above, we can also apply a positive feedback to an op-amp. Let's have a closer look at the behaviour of the feedback network of the drawing, assuming a resistance of 1kΩ for R1, 10kΩ for R2, a positive supply voltage of +12V and a negative supply voltage of -12V:
While connecting the circuit to a voltage source, Ui and V- and so the output voltage are zero, but just a slight random variation of V+ causes the output voltage to tilt either to the maximum or minimum value. Let's assume the output voltage is at it's maximum value and the input voltage is +0.1V. With the feedback via the voltage divider we get:
(3) V+ = Ui + (Vout - Ui) * R1 / (R1 + R2) = 1.18V
When the input voltage is decreasing to 0V we get (remember the output voltage is still +12V):
V+ = 0V + (12V - 0V) * 1000Ω / (1000Ω + 10000Ω) = 1.09V
Even at an input voltage of -0.1V we get:
V+ = -0.1V + (12V + 0.1V) * 1000Ω / (1000Ω + 10000Ω) = 1.00V
The op-amp will tilt to the minimum supply voltage, as soon as V+ drops slightly below 0V. Inserting V+ = 0 and solving equation (3) to Ui gives us:
Ui - Input voltage of the circuit
Vout - Voltage output of the op-amp
R1, R2 - Resistors of the voltage divider
To cause the op-amp to switch from +12V to -12V, the input voltage of the circuit must drop slightly below -1.2V.
Now that the output voltage is -12V, let's increase the input voltage to 0V:
V+ = 0V + (-12V - 0V) * 1000Ω / (1000Ω + 10000Ω) = -1.09V
The voltage output of the op-amp at an input level of 0V is now -12V! The input voltage of the circuit must rise above
-(-12V) * R1 / R2 = +1.2V
to cause the op-amp to switch back to an output voltage of +12V.
Input (blue) and corresponding output (red) signal of a Schmitt trigger:
On a normal comparator, the op-amp will switch at the same point of the rising and falling edge of the input signal. The output signal of the op-amp will start oscillating while the input signal is near the threshold, whenever the input voltage crosses the threshold because of noise.
On a Schmitt trigger, the situation is different:
The output signal is low, while the input signal is below a certain threshold and it is high if it is above a different (higher) threshold. The output signal retains it's value while the signal is between the two different input levels.
The dual threshold action is called hysteresis. The output voltage of the Schmitt trigger depends not only on the current input voltage, but also on that in the past. For example the output voltage at an input signal of 0.1V can be either +12V or -12V. It is +12V if the input signal was above +1.2V and is now falling to +0.1V. Vice versa the output signal is -12V if the input signal was below -1.2V in the past and is now rising to 0.1V.
Single supply voltage Schmitt-TriggerFigure 10:
This diagram shows a Schmitt trigger with a single supply voltage. Half the supply voltage is attached to the inverting input using a voltage divider composed of two identic resistors (R3 and R4).
Twilight switchFigure 11:
Just one practical circuit with a single supply Schmitt trigger:
The input voltage is provided by a voltage divider with one constant resistor (R3) and one PNP-phototransistor. If the phototransitor is exposed to light, the resistance decreases, hence the input voltage is decreasing, too. If the lower threshold is reached, the output voltage of the Schmitt-trigger drops down to 0V, hence a device (maybe a lamp) connected to the output clamp is turned off while the sun is shining. R4 is a potentiometer, operating as a voltage divider at the inverting input of the op-amp, by what the threshold can be adjusted, so that the light is switched off during twilight rather than in bright sunlight.
Ideal op-amp versus LM324N
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Explain the chain of effects of excess supply of a good on its equilibrium price
Chain effects of excess supply of a good on its equilibrium price
Consider DD to be the initial demand curve and SS to be the supply curve of the market. Market equilibrium is achieved at Point E, where the demand and supply curves intersect each other. Therefore, the equilibrium price is OP, and the equilibrium quantity demanded is OQ.
When there is the change in other factors than price, there will be the rise in the supply of goods. There will be a shift in the supply curve towards the right to SS1 with an increase in the supply, and the demand curve DD will remain the same. This implies that there will be a
a situation of excess supply at the equilibrium point
In the above diagram, there is an excess supply of OQ1 to OQ11 units of output at the initial price OP1. Thereby the producers will tend to reduce the price of the output to increase the sale in the market. A profit margin of the firm will come down and slowly some of the firms will tend to quit the market. Because of this, the market supply will decline to OQ2 level of output and the price of the output also gets reduced to the point OP2. Now, the new market equilibrium will be at Point E1, where the new supply curve SS1 intersects the demand curve DD.
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- Equilibrium Price
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s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178361849.27/warc/CC-MAIN-20210301030155-20210301060155-00137.warc.gz
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CC-MAIN-2021-10
| 1,366 | 8 |
http://forums.wolfram.com/mathgroup/archive/2007/Jun/msg00525.html
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math
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Are comments safe inside notebook cells in Mathematica 6?
- To: mathgroup at smc.vnet.net
- Subject: [mg77553] Are comments safe inside notebook cells in Mathematica 6?
- From: Andrew Moylan <andrew.j.moylan at gmail.com>
- Date: Wed, 13 Jun 2007 07:19:10 -0400 (EDT)
Prior to version 6, it has been considered unsafe to place comments (* like this *) inside e.g. Input cells, because it could lead to Mathematica corrupting the underlying notebook. Can anyone comment on whether this is still the case in version 6?
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00228.warc.gz
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CC-MAIN-2023-40
| 516 | 6 |
https://www.thestudentroom.co.uk/showthread.php?t=1177234
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math
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Hi guys, just wondering how would you solve this inequality:
6x-14 < 5x-2 ≤ 7x+2
Thanks a lot!
Turn on thread page Beta
- Thread Starter
- 08-02-2010 20:04
- 08-02-2010 20:31
Try splitting it into 2 separate inequalities, and then solve the 2 new inequalities by rearranging them. Should be okay from there. =]
- 08-02-2010 20:35
or solve the entire thing at once. i.e. add 2 to both sides so the middle fe diddle becomes 5x, then divinde the entire thing by 5
- 08-02-2010 20:36
Consider the problem as two separate inequalities.
Get two inequality solutions and use these to create a solution range for x.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221214538.44/warc/CC-MAIN-20180819012213-20180819032213-00101.warc.gz
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CC-MAIN-2018-34
| 609 | 13 |
https://bitcointalk.org/index.php?topic=56562.40
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math
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Juggling with numbers can lead to more interesting thoughts.
We know that in 2033, no additional BTC will be available. Let's assume that it will take up to 2033 to spread BTC:
*to the entire population of 6 billion people; or
*in exchange of the global total money supply of USD 60 trillion
To make our life simple, we assume that the annual factor and the base number (people, money supply) remain constant.
The graphs are inserted below.peoplemoney supply
The factor [people] is 3,6 (remarkable close to phi
), the factor [money supply] is 7,8. Of course, if the total time frame is less than 2009 to 2033, the factor to be used would have to be higher to meet the target.
An interesting aspect is the nature of these graphs, or better to say, the math behind it. The graphs are build up by assuming a constant factor that increases the total number, but halfway, the same factor is used to decrease the increase (=still growth, but less and less annually until it is a flat line). This is based on the assumption that, at first, there will be a low number of participants/money invested in BTC. Moving on, the numbers explode after which the increase will slow down again.
Since it follows from the formula, it is no surprise that the biggest increase is made around 2019-2020. It is clear that if you compress the total time frame to, let's say, 2022, the biggest increase would be round and about 2016. That's not the interesting part.
The interesting part (as far as I am concerned) is that, on the basis of these premises, the number will be approximately 10% of the total number to be reached just when the increase is at its maximum velocity. Above, I already presented the link to a study showing that, for an idea to be accepted, a critical mass
of 10% (or 13%) of all people should accept the idea. If the critical mass is reached, the rest will follow as a rule.spread of new idea
The curve used in that study aligns with the formula used in my numbers-juggling-exercise. First the slow start, then the increase, then the top and the following decrease.
To end this pick-your-number-and-base-your-argument-on-that-number
exercise, it seems fair to assume that the actual factor of the BTC spread (in terms of people or committed cash) is way higher than the numbers in my projection. At least, in this stage. It remains to be seen whether the actual factor remains very high over the course of the next years.
Someone else may calculate the actual
average factor to project the predicted date on which the critical mass of 10% will be reached. It could be a lot sooner than we think, prone to error as we humans are in understanding the exponential factor.
However, another thing to ponder is the known history of inventions, dating back to the 19th century. Usually, an invention lingers for about 10-15 years, before it suddenly gets adopted massively. Nowadays we see shorter time frames (Facebook, Youtube etc), but I think it is reasonable to assume that Bitcoin will stay in the dark for at least two more years before taking off.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917125841.92/warc/CC-MAIN-20170423031205-00248-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 3,050 | 18 |
https://coursesxrem.web.app/monachino18680la/math-help-online-free-answers-ruba.html
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math
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Socratic Math & Homework Help App Review Instant homework help with resources; watch for cheating. Read Common Sense Media's Socratic Math & Homework Help review, age rating, and parents guide. Homework help answers - Entertainment Box Free help services provide answers from math homework, 2016 - slader. Fingers and longicorn muffin sjambok their homework help.
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MATH Entrance This PDF book include australia electrical pre apprenticeship test answers information. To download free electrician pre-apprenticeship math entrance exam you Year : and Workplace Math 10 Year : Math 10 and Workplace Math 10.
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It is important to know how the answers came about in the first place. Plagiarism-free A plagiarism-free paper is a very important requirement of any reputable hw help site. Quizlets Live: Essay Writing Service with World-Class Skills Our writers provide outstanding writing help from scratch. We write for you top-grade and plagiarism-free essays, cost-efficient research, and term papers.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00634.warc.gz
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CC-MAIN-2021-21
| 3,182 | 10 |
https://en.cbmhealthcare.org/ostroslupHN7
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math
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Pyramid. Properties and patterns
A pyramid is a polyhedron with one base, and all side walls connect at one common point - the vertex. How do we divide the pyramids? What figure can be its basis? What is the formula for the volume and surface area of a pyramid?
See the movie: "How can you help your toddler find himself in a new environment?"
1. Definition and properties of a pyramid
The pyramid is a geometric solid, a three-dimensional figure, a polyhedron. All its vertices, except one, lie in one plane, on a polygon called the base. The sides of the pyramid are the edges of the base, and the plane is the plane of the base.
The point lying outside the plane of the pyramid is called the vertex, and the segments connecting it with the vertices of the base are the side edges. Each edge of the base, together with the vertex, forms an isosceles triangle we call the side wall.
The height of the pyramid is the distance from the vertex to the plane of the base. The point which is the perpendicular projection of the pyramid's apex onto the ground plane is defined as the bottom of the height.
The base of the pyramid can be any polygon. The pyramid with a triangle at its base is a tetrahedron. The total area of this solid is the sum of the areas of the pyramid's walls and the area of its base.
Triangles and their properties
A triangle, a polygon with three sides, is divided into sides and angles. How do they differ from ...read the article
2. How do we divide the pyramids?
The pyramids can be divided into:
- correct pyramid - its base is a regular polygon, hence all side edges of this solid are of equal length, all the angles of inclination of the side edges to the plane have equal measures, and the side walls are congruent isosceles triangles;
- simple pyramid - each side edge of the pyramid is of equal length;
- inclined pyramid - its side edges are not perpendicular to the base.
We also divide them due to the number of angles contained in the base and thus distinguish the pyramids:
Table of Contents...read the article
3. How to calculate the surface area of a pyramid?
The formula for the surface area of the pyramid looks like this:
Pc = Pp + Pb
- Pp is the area of the base of the pyramid;
- Pb is the sum of the areas of the side walls of the pyramid.
Prism. Characteristics and types
A prism is a polyhedron, the vertices of which are located on two parallel planes, and all ...read the article
4. How to calculate the volume of the pyramid?
The formula for the volume of the pyramid is:
V = ⅓ x Pp x H.
- Pp is the base area of the pyramid;
- H is the height of the pyramid.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00307.warc.gz
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CC-MAIN-2023-06
| 2,609 | 29 |
https://www.scientific.pictures/-/galleries/astronomy/-/medias/9b0b3e5f-6b6b-4cda-bcee-0b943060f0c2-geostationary-satellite/fs
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math
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Animation of three geostationary satellites orbiting the Earth. Satellites can be natural or artificial (man made). Satellites can be placed in various orbits. At a unique height (just under 35,800 km above sea level = an orbital radius of just under 42,200 km) their orbital speed can exactly match the rotation of the Earth and so the satellite effectively hovers above a fixed point. This is called the geostationary orbit and three silvery satellites are illustrated here in that orbit. The geostationary orbit is shown as a red ring in space. The geostationary orbital plane (pale pink plane) lies in the plane of the equator (red line around the middle of the Earth) and such satellites are often launched from sites close to the equator (by rockets or by the Space Shuttle). A great advantage of such an orbit is that receiving dishes or antennae (one is illustrated here at the latitude of London) can be trained on a satellite. Because the satellite does not move relative to the Earth, the dish does not have to change orientation to keep track the satellite. This makes geostationary satellites ideal as communications satellites and for satellite television broadcast (think of those fixed dishes on houses). Three such satellites arrayed around the Earth provide coverage of most of the planet (although in reality there are far more). In the graphic, cones of electromagnetic radiation (data) are shown emanating from the satellites to illustrate this point of global coverage.
WHITE BAND: poles to artic and antarctic circles
GREEN BAND: artic and antarctic circles to tropics
BLUE BAND: tropics to equator
RED BAND: equator
VERTICAL PURPLE LINES: represent lines of longitude at 30o intervals
YELLOW POLE: axis of rotation of Earth
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00482.warc.gz
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CC-MAIN-2024-18
| 1,747 | 7 |
https://www.hackmath.net/en/math-problems/rounding
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math
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Rounding - problems
- Rounding to ones
- Rounding to tens
- Rounding to hundreds
- Rounding to thousands
- Rounding to two decimal places
- Rounding to one decimal place
Allan keeps tropical fish. His aquarium is 4 feet long, 1 foot wide, and 2 feet tall. Each fish needs at least 0.5ft³ of water. What is the maximum number of fish that he can keep in the aquarium? Please show your solution. Please
In the mini-survey of our class about the popularity of individual subjects, it turned out that 11.1% of pupils like mathematics, 18.5% are enjoying languages, 30.4% of pupils like physical education, and the remaining 12 pupils have several popular subjec
- A tile
A tile setter is covering 5ft by 5ft square shower wall. Each tile covers 4 5/8in by 4 5/8in square. How many rows of tile are needed to reach 5ft? How many tiles are needed to cover 5ft by 5ft square
- Bricks pyramid
How many 50cm x 32cm x 30cm brick needed to built a 272m x 272m x 278m pyramid?
- Food weight
Stacie is a resident at the medical facility where you work. You are asked to chart the amount of solid food that she consumes. For the noon meal, today she ate 1/2 of a 3-ounce serving of meatloaf, 3/4 of her 3-ounce serving of mashed potatoes, and 1/3 of
- Find sales tax
What is 8.25% sales tax out of 1607.00?
- What is one third
What is 1/3 as a decimal? Give your answer rounded to 2 decimal places.
Mr. Billy calculated that excavation for a water connection dig for 12 days. His friend would take 10 days. Billy worked 3 days alone. Then his friend came to help and started on the other end. On what day since the beginning of excavation they met?
Calculate how many average minutes a year is the web server is unavailable, the availability is 99.99%.
- Homeless Dezider
Homeless Dežko has 9 coins in jacket: Calculate the value of its assets and calculate how many bottles of wine for 0.55 EUR can he buy.
Fire tank has cuboid shape with a rectangular floor measuring 13.7 m × 9.8 m. Water depth is 2.4 m. Water was pumped from the tank into barrels with a capacity of 2.7 hl. How many barrels were used, if the water level in the tank fallen 5 cm? Wr
Determine the number of all positive integers less than 4183444 if each is divisible by 29, 7, 17. What is its sum?
Write time in minutes rounded to one decimal place: 5 h 28 m 26 s.
How many steps you save if you go square estate for diagonal (crosswise), rather than circumvent on the two sides of its perimeter with 307 steps.
Camel owner wants to get out of the city into oasis. He bought 3000 bananas, which wants to sell in oasis. However oasis from the city divided 1,000 kilometers of desert. Camel can carry at one time up to 1000 bananas, and for every kilometer traveled, ea
- Salary raise
Monthly salary was 2390 Eur. During the year it was raised to 2617 Eur. Calculate the month from salary was increased that employee earned 29361 Eur during whole year.
From one ton of coal is produced 772 kg of coke for iron production. How many wagons of coal by 13 tonnes per day is needed for the blast furnace, which has a daily consumption of 1020 tons of coke?
Room is 40.1 m long, 25.2 dm wide and 369 cm high. How many people can simultaneously be in this room if for hygiene reasons is calculated 5100 dm3 of air per person?
- Civil protection
Students on civil protection exercise went 5800 m long trip. How many kilometers is it approximately?
- Report card
Ivor hit 5× grade 5 at the beginning of the school year. How many times must now catch grade 1 to get grade 2 on report card?
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s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583763839.28/warc/CC-MAIN-20190121070334-20190121092334-00576.warc.gz
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CC-MAIN-2019-04
| 3,539 | 36 |
https://projecteuclid.org/journals/annals-of-statistics/volume-13/issue-1/Universal-Domination-and-Stochastic-Domination--Estimation-Simultaneously-Under-a/10.1214/aos/1176346594.full
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math
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In Wald's statistical decision theory, the criterion of domination (or uniform betterness) is defined with respect to a specific loss. In practice, however, the exact form of a loss function is difficult to specify. Hence, it is important to study the domination criterion simultaneously under a class of loss functions. In this paper we focus on estimation problems. We mainly investigate the possibility of domination simultaneously under the class of loss functions $L(|\theta - \delta|)$, where $L$ is an arbitrary nondecreasing function. As usual, $\theta$ and $\delta$ (both in $p$-dimensional Euclidean space $R^p$) are, respectively, the unknown parameter of nature and the statistician's estimate. Domination simultaneously under this class of losses is called universal domination under Euclidean error. Several theoretical questions are resolved in this paper. In particular the criterion of universal domination is shown to be equivalent to the criterion of stochastic domination that compares the estimators by the stochastic ordering of their Euclidean distances from the estimators to the true parameter. Concrete results about universal domination relating to the usual estimator are also established. In particular when $X - \theta$ has a $p$-variate $t$ distribution, and $p = 1, 2$, there exists no estimator for $\theta$ that universally dominates $X$; however, for $p \geq 3$, estimators (of the type of James-Stein positive part estimators) that universally dominate $X$ are specified. When $X$ has a $p$-variate normal distribution with mean $\theta$ and identity covariance matrix, we show that for any dimension $p$, no James-Stein positive part estimators universally dominate $X$. However, under slightly smaller classes of losses, some James-Stein positive part estimators are shown to simultaneously dominate $X$. These hitherto unstudied losses are bounded and fairly practical.
"Universal Domination and Stochastic Domination: Estimation Simultaneously Under a Broad Class of Loss Functions." Ann. Statist. 13 (1) 295 - 314, March, 1985. https://doi.org/10.1214/aos/1176346594
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00631.warc.gz
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CC-MAIN-2022-05
| 2,107 | 2 |
https://science.lpnu.ua/keywords-paper/step-response
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math
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The current article describes the results of the study of the neural networks temperature prediction error dependence on measurement errors, which are random, nonlinear and multiplicative errors. It is noted applicability of the architecture of neural network for temperature prediction. The formula of temperature step response for ideal sensor is given.
Current article considers the results of the study of air and water flow temperature prediction error on the number of inputs in neural network. Authors guide the architecture of neural network for temperature prediction. The formula of temperature step response for real sensor is given. Also, the method for calculating the time constants for the temperature step response formula using real measurement data is considered.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00818.warc.gz
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CC-MAIN-2024-10
| 781 | 2 |
https://gradestack.com/UGC-NET-Papers-English-/-The-following-pie-chart/49-3524-3526-18365-sf
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math
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The following pie chart indicates the expenditure of a country on various sports during a particular year. Study the pie chart and answer the question.
If the total expenditure on sports during the year was Rs. 1,50,00,000 the expenditure on cricket and hockey together was :
A Rs. 60,00,000 B Rs. 50,00,000
C Rs. 37,50,000 D Rs. 25,00,000
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s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00327.warc.gz
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CC-MAIN-2017-30
| 339 | 4 |
https://docslib.org/doc/5788357/spacetime-thermodynamics-and-entanglement-entropy-the-einstein-eld-equations
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math
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Universitat de Barcelona
Master in Astrophysics, Particle Physics and Cosmology
Spacetime thermodynamics and entanglement entropy: the Einstein field equations
Author: Mart´ıBerenguer Mim´o NIUB: 17527646
Advisor: Roberto Emparan
September 2019 Ackonwledgements First of all, I would like to thank my tutor and advisor, Roberto Emparan, for all the help and dedication on supervising my thesis for the last months. He has always been there to assist me when needed. Thank you. I would also like to thank my family, for encouraging me during all these years of study, and to who I have stolen a lot of time. Finally, I would like to thank my master classmates. In particular, Oscar and Joseba, for all those sleepless nights working on the assignments, and with whom I have spent an amazing year.
1 Abstract In this master thesis, the posibility of a connection between spacetime dynamics (driven by the Einstein equations) and thermodynamics is discussed. Some known results, like the Raychaudhuri equation or the Unruh effect are reviewed in order to make the presentation self-contained. The Einstein equations are derived in two different ways from thermody- namic arguments. The first one (Section (3)) uses the thermodynamic relation δQ = T dS, together with the proporionality of entropy and horizon area. In the second derivation (Section (4)), the Einstein equations are derived from an hypothesis about entanglement entropy in a maximally symmetric spacetime. Some questions regarding the implications of this thermodynamic interpre- tation of spacetime are discussed as a conclusion of the thesis.
1 Introduction 3
2 Basic previous topics 6 2.1 Raychaudhuri equation ...... 6 2.2 Rindler space ...... 10 2.3 The Unruh effect ...... 12 2.4 Entanglement entropy ...... 18
3 First derivation. Equilibrium thermodynamics in Rindler space 22
4 Second derivation. Entanglement entropy 26
5 Comments and discussion 32
6 Conclusions 36
2 1 Introduction
One of the most surprising results of Albert Einstein’s general theory of relativity was the existence of black holes, regions of spacetime were gravity is so strong, that nothing, even light, could escape. The interior of a black hole is separated from the rest of the universe by an event horizon. This means that any particle, massive or massless, that is located inside the black hole, will never be able to escape, and is doomed to reach a singularity in its future, where its proper time suddenly ends and the known theories of physics stop to work. When Stephen Hawking studied black holes from a more mathematical point of view, he found an interesting result: the area of the event horizon never decreases with time and, in general, it will increase. This implies that, if two black holes collide and merge, the area of the final black hole will be larger than the sum of areas of the colliding black holes. This behavior is analogous to the behavior of entropy in thermodynamic systems, where the Second Law of Thermodynamics says that the entropy of a system can never decrease, and that the total entropy of a system is larger than the entropy of its subsystems:
Second Law of Black Hole Mechanics: δA ≥ 0
Second Law of Thermodynamics: δS ≥ 0
This analogy is more evident with the First Law of Black Hole Mechanics, which relates the change in mass of a black hole with the change in area of the horizon and the change in angular momentum and electric charge. From here one can see that if the area of the event horizon is analogous to the entropy, then the surface gravity κ is analogous to the temperature:
First Law of Black Hole Mechanics: κ δE = 8π δA + ΩδJ + ΦδQ First Law of Thermodynamics: δE = T δS + P δV
There is even a Zeroth Law of Black Hole Mechanics:
3 Zeroth Law of Black Hole Mechanics: κ is the same along the horizon in a time-independent black hole.
Zeroth Law of Thermodynamics: Temperature is the same in all points in a system in thermal equilibrium.
Because of these similarities, Bekenstein proposed that the entropy of a black hole should be proportional to the area of the event horizon, and proposed the Generalized Second Law : the sum of the entropy of the black hole and the entropy of the matter outside the black hole never decreases. The fact that black holes have temperature and entropy implies that they should radiate, but this was completely against the classical picture of black holes. One of the most important results in theoretical physics during the last century was the one obtained by Stephen Hawking in 1974, when he found that, when quantum effects around a black hole are considered, it radiates thermally at the so-called Hawking temperature : 1 T = (1.1) H 8πM Another important result is that this thermal behavior is not exclusive of black hole horizons. In 1976, W.G. Unruh demonstrated the following: the vacuum state, defined by inertial observers, has a thermal character for uniformly accelerated ob- servers with proper acceleration a (Rindler observers) at the Unruh temperature [3, 4]: a T = ~ (1.2) 2π This means that, around any event, in any spacetime, there is a class of observers that will perceive the spacetime as hot. This thermal character of spacetime (not only for black hole horizons) will be of great importance for the following sections. Moreover, this thermodynamic interpretation of spacetime invokes some questions about the structure of spacetime at smallest scales. From standard thermodynamics it is known that a macroscopic system like, for example, a gas, can be described with some thermodynamic variables, like the temperature or the entropy, but for a long time, the real meaning of these variables was unknown. It was Boltzmann who gave an explanation to these variables, essentially saying “if you can heat it, it has microscopic degrees of freedom”. Before that, it was considered that matter was continuous even at the smallest scales, and the concepts of heat and temperature were
4 added “by hand”. Boltzmann used the discrete interpretation of matter and found that the thermodynamic fenomena were related with the averages of the properties of these microscopic degrees of freedom. This is profound. It tells that the existence of microscopic degrees of freedom leaves a signature at macroscopic scales, in the form of temperature and heat. Then, if spacetime is seen as hot by some observers, what are the microscopic degrees of freedom that give raise to the temperature and the entropy? There are many ap- proaches that try to give an interpretation to these microscopic degrees of freedom [5, 6], but there is not a clear answer yet. What seems reasonable is that, if spacetime is, at its deepest level, a thermodynamic entity, we should be able to derive the equa- tions that drive its evolution (the Einstein equations) from a purely thermodynamic point of view. This is what we will do in Sections (3) and (4). In Section (3), the Einstein equations are derived from the thermodynamic relation δQ = T dS and the propor- tionality of the entropy and horizon area, working from the point of view of a Rindler observer in the neighbourhood of the causal horizon of the Rindler space. In Section (4), an alternative derivation of the Einstein equations will be given, based in the assumption that the entanglement entropy in a geodesic ball is maximal when the geometry and the quantum fields are varied from maximal symmetry. Section (2) includes some of the conceptual ideas and equations that will be necessary for the two derivations. In Section (5), some comments about the derivations and some questions about the implications of them are considered, while Section (6) contains the main conclu- sions of the thesis.
5 2 Basic previous topics
This section contains some topics that will be necessary to have in mind during the two derivations of the Einstein equations in the following sections. Here we will briefly talk about the Raychaudhuri equation, Rindler space, the Unruh effect, and entanglement entropy.
2.1 Raychaudhuri equation The Raychaudhuri equation is an evolution equation for what is called the expansion of a congruence of geodesics. In order to understand the meaning of the expansion (and two more quantities that appear in the equation, the shear and rotation), it is useful to think first about the kinematics of a deformable medium. Suppose, in a purely Newtonian context, a two-dimensional medium, with some internal motion whose dynamics are not of our interest. From a purely kinematic point of view, we can always write that, for an infnitesimal displacement ξa from a reference point O, dξa = Ba (t)ξb + O(ξ2) (2.1) dt b a for some tensor B b , which depends on the internal dynamics of the medium. For short intervals of time,
a a a ξ (t1) = ξ (t0) + ∆ξ (t0) (2.2) where
a a b 2 ∆ξ (t0) = B b (t0)ξ (t0)∆t + O(∆t ) (2.3) a and ∆t = t1 −t0. To describe the action of B b we will consider the situation that a ξ (t0) = r0(cos φ, sin φ); that is, a circle of radius r0 in the two-dimensional medium.
Expansion a Suppose that B b is a pure-trace matrix, i.e., proportional to the identity, with the form
1 a 2 θ 0 B b = 1 0 2 θ a 1 In this case, ∆ξ = 2 θr0∆t(cos φ, sin φ), which corresponds to a change in the 1 circle’s radius by an amount 2 θr0∆t. The correspoonding change in area is given by
6 2 ∆A = A1 − A0 = πr0θ∆t (2.4) This means that 1 ∆A θ = (2.5) A0 ∆t θ measures the fractional change of area per unit time, and is called the expansion parameter.
Shear a Suppose now that B b is symmetric and trace-free: a σ+ σ× B b = σ× −σ+ a In this case, ∆ξ = r0∆t(σ+ cos φ + σ× sin φ, −σ+ sin φ + σ× cos φ). If σ× = 0, we have an ellipse with the major axis oriented in the φ = 0 direction. If σ+ = 0, what we have is an ellipse oriented in the φ = π/4 direction. The general situation is an ellipse oriented along an arbitrary direction. The area of the figure is not affected by a the action of B b . What we have is a shearing of the figure, and the parameters σ× and σ+ are called the shear parameters.
Rotation a Finally, if B b is antisymmetric, 0 ω Ba = b −ω 0 a a 0 0 we have that ∆ξ = r0ω∆t(sin φ, − cos φ), and ξ (t1) = r0(cos φ , sin φ ), with φ0 = φ − ω∆t. This corresponds to an overall rotation of the original figure, keeping the area fixed. ω is called the rotation parameter. The most general decomposition of this tensor into algebraically irreducible com- ponents under rotations is
1 a 2 θ 0 σ+ σ× 0 ω B b = 1 + + 0 2 θ σ× −σ+ −ω 0 which can also be expressed as 1 B = θδ + σ + ω (2.6) ab 2 ab ab ab 7 a 1 where θ = B a (the expansion scalar) is the trace part of Bab , σab = B(a,b) − 2 θδab (the shear tensor) is the symmetric-tracefree part of Bab , and ωab = B[a,b] (the rotation tensor) is the antisymmetric part of Bab . For a three-dimensional medium, the decomposition is the same, but with a prefactor of 1/3 instead of 1/2 in the trace term, and the interpretation of the expansion, shear and rotation are the same, but changing the area by the volume. Once the classical 2-dimensional medium has been introduced, we can move now to the study of congruences of (for now, timelike) geodesics. Let O be an open region of spacetime. A congruence of geodesics in O is a family of geodesics such that through each point in O passes one and only one geodesic from this family. We will assume that the geodesics are timelike. We are interested in the evolution of the deviation vector ξa between two neighbouring geodesics in the congruence as a function of the proper time τ (see Figure (1)).
Figure 1: Two neighboring geodesics, with a deviation vector ξa as a function of τ.
Let ua be the (timelike) tangent vector to the geodesics. Then, the spacetime metric gab can be decomposed in a longitudinal part −uaub and a transverse part hab,
hab = gab + uaub (2.7)
The transverse metric hab is purely spatial, in the sense that it is orthogonal to ua. We introduce now the tensor field
Bab = ∇bua (2.8)
8 This tensor determines the evolution of the deviation vector ξa. To see this, note b a b a that from u ∇bξ = ξ ∇bu we obtain
b a a b u ∇bξ = B b ξ (2.9) a a That is, B b measures the failure of ξ to be parallel transported along the con- gruence. Equation (2.9) is analogous to (2.1), and therefore we can decompose the a tensor B b in the same way as before, with the same interpretation for the diffrent terms that appear: 1 B = θh + σ + ω (2.10) ab 3 ab ab ab In order to find the evolution equation for the expansion θ, we can start by finding an evolution equation for Bab :
c c u ∇cBab = u ∇c∇bua c d = u ∇b∇aua − Radbcu c c d = u ∇b∇cua − Radbcu u (2.11) c c c d = ∇b (u ∇cua) − (∇bu )(∇cua) − Radbcu u c c d = −B b Bac − Radbcu u Taking the trace of this equation, we obtain dθ = −BabB − R uaub (2.12) dτ ba ab ab 1 2 ab ab Now, from the definition of Bab , we find that Bab B = 3 θ + σ σab − ω ωab, so (2.12) becomes dθ 1 = − θ2 − σabσ + ωabω − R uaub (2.13) dτ 3 ab ab ab which is kown as the Raychaudhuri equation, and gives the evolution of the expansion parameter θ for a congruence of timelike geodesics. For the case of null geodesics, which is the one that will be of interest in the following sections, the line of argument is the same as for timelike geodesics, but the calculation is a bit more tedious because of the difficulty to precisely define the transverse spacetime. In the above case it was simply the spatial components, but in the case of null geodesics, if ka is the (null) tangent vector to the geodesics, the orthogonal space to ka includes ka because it is orthogonal to itself. Once this technical part is solved, the logic
9 of the derivation and the result are very similar. The Raychaudhuri equation for a congruence of null geodesics reads:
dθ 1 = − θ2 − σabσ + ωabω − R kakb (2.14) dλ 2 ab ab ab
2.2 Rindler space The Rindler space is introduced when one is interested in the motion of an accelerated observer in flat spacetime. This will be necessary in Section (3), where the whole argumentation line will be centered from the perspective of an accelerated observer in an approximately flat region of spacetime. For simplicity, let’s consider the 2-dimensional Minkowski space, whose metric, in the usual (t, x) coordinates is
ds2 = −dt2 + dx2 (2.15) An observer moving at a uniform acceleration α will follow the trajectory xµ(τ) given by 1 t(τ) = sinh(ατ) (2.16) α 1 x(τ) = cosh(ατ) (2.17) α This can be checked taking into account that the components of the 4-acceleration
D2xµ d2xµ aµ = = (2.18) dτ 2 dτ 2 where the covariant derivative is equal to the ordinary derivative because the Christoffel symbols vanish in these coordinates, are given by
at = α sinh(ατ) (2.19) ax = α cosh(ατ) (2.20) In this way, the magnitude of the acceleration is q p µ 2 2 2 2 aµa = −α sinh (ατ) + α cosh (ατ) = α (2.21) Thus, this trajectory corresponds to a uniformly accelerated observer. The tra- jectory of this observer obeys
10 1 x2(τ) = t2(τ) + (2.22) α2 which is an hyperboloid asymptoting to null paths x = −t in the past and x = t in the future (see Figure (2)).
Figure 2: Minkowski spacetime in Rindler coordinates. An observer with constant acceleration in the +x direction follows the hyperbolic trajectories dranw in region I. The patches H+ and H− act as horizons for this class of observers.
Notice from (2.22) that the larger the acceleration α, the closer the trajectory is to the patches x = −t and x = t. This fact will be important in Section (3). We can define new coordinates (η, ξ) in the following way: 1 1 t = eaξ sinh(aη), x = eaξ cosh(aη), (x > |t|) (2.23) a a which cover the wedge x > |t| (region I in Figure (2)). Although these are not the usual Rindler coordinates, they are the most appropiate for the derivation of the Unruh effect, which is the purpose of the next section. Notice that an accelerated observer with acceleration α = a follows a world line that is given by ξ = const and η = τ. In these coordinates, the metric is given by
11 ds2 = e2aξ −dη2 + dξ2 (2.24) Region I, with these coordinates, is known as Rindler space (although it is only a part of Minkowski space). A Rindler observer is an observer moving along a constant acceleration path (in the diagram, this corresponds to the hyperbolic trajectories).
Because the metric components are independent of η, the vector ∂η is a Killing vector. In the (t, x) coordinates, this Killing vector is
∂t ∂x ∂ = ∂ + ∂ η ∂η t ∂η x aξ (2.25) = e [cosh(aη)∂t + sinh(aη)∂x ]
= a (x∂t + t∂x )
Notice that the patches x = −t and x = t (H− and H+) act as Killing horizons for this vector field, because its norm vanishes (only) there:
∂ ∂ V = a x + t ⇒ V V µ = a2 t2 − x2 = a(t + x)(t − x) (2.26) ∂t ∂x µ
The surface gravity of this Killing horizon is
r 1 κ = − ∇µξν∇ ξ = a (2.27) 2 µ ν Although there is no gravitational field (we are in flat spacetime), the surface gravity characterizes the acceleration of the Rindler observers. It will be convenient for the Unruh effect to define coordinates (η, ξ) for the region IV, by flipping the signs of those defined in region I: 1 1 t = − eaξ sinh(aη), x = − eaξ cosh(aη), (x < |t|) (2.28) a a 2.3 The Unruh effect The basic statement of the Unruh effect is that an accelerating observer in flat space will observe the Minkowski vacuum as a thermal spectrum of particles. The basic idea of this result is the fact that observers with different notions of positive and negative frequency modes will disagree on the particle content of a given state.
12 In flat spacetime, this problem does not arise for non-accelerated (inertial) ob- servers. For inertial observers, we introduce a set of positive and negative frequency modes, and the fields are expressed as a combination of these modes, interpreting the operator coefficients as creation and annihilation operators. In flat spacetime we can choose a natural set of modes by demanding that they are positive-frequency modes with respect to the time coordinate. Obviously, the time coordinate is not unique, because we can perform Lorentz transformations, but the vacuum state and the number operators are invariant under these transformations. In curved spacetime (or accelerated observers) we can find a set of modes, but we can find many other sets that are equally good, and the notion of vacuum and number operators will be very sensitive to the set we choose. We can always find a set of orthonormal modes fi, and expand the fields in terms of these modes:
X † ∗ φ = aˆifi +a ˆi fi (2.29) i † where the operatorsa ˆi anda ˆi obey the usual commutation relations. We can define a vacuum state |0f i, which will be annihilated by all the annihilation operators,
aˆi |0f i = 0, ∀i (2.30) From this vacuum we can define an entire Fock basis, defining the excitations as † the states created by the action of ai . The number operator can be defined too,
† nˆfi =a ˆi aˆi (2.31) where the subscript f makes reference to the fact that this operator is defined with respect to the set of modes fi. But we can find another complete basis with respect to which expand the fields,
X ˆ ˆ† ∗ φ = bigi + bi gi (2.32) i ˆ ˆ† where, again, bi and bi obey the usual commutation relations. The vacuum state, ˆ the Fock basis, and the number operator for the bi operators are defined in the same way as for the operators of the fi modes:
ˆ ˆ†ˆ bi |0gi = 0 ∀i;n ˆgi = bi bi (2.33)
If one observer defines particles with respect to the set of modes fi and a different observer defines particles with respect to the set of modes gi, in general they will
13 disagree on the number of particles they observe. To see this, we can expand each set of modes in terms of the other:
X ∗ gi = αijfi + βijfj (2.34) j
X ∗ ∗ fi = αjigj − βjigj (2.35) j The transformation that allows to write one set of modes in terms of the other is called Bogoliubov transformation, and the coefficients αij, βij are called Bogoliubov coefficients, which satisfy the normalization conditions
X ∗ ∗ αijαjk − βikβjk = δij (2.36) j X (αikβjk − βikαjk) = 0 (2.37) j and can be used to relate not only the modes, but also the operators:
X ˆ ∗ ˆ† aˆi = αijbj + βjibj (2.38) j
ˆ X ∗ ∗ † bi = αijaˆj − βijaˆj (2.39) j The discrepancy on the number of particles can be seen from the following cal- culation: imagine that the system is in the f-vacuum (in which the observer using the fi modes would not see any particle). We want to know the number of particles that an observer using the g-modes will observe. Then, we compute the expectation value of the g number operator in the f-vacuum:
14 ˆ†ˆ h0f | nˆgi |0f i = h0f | bi bi |0f i X † ∗ † = h0f | αijaˆj − βijaˆj αikaˆk − βikaˆk |0f i jk X ∗ † = βijβik h0f | aˆjaˆk |0f i jk X ∗ † = βijβik h0f | aˆkaˆj + δjk |0f i jk (2.40) X ∗ = βijβikδjk h0f |0f i jk X ∗ = βijβij j X 2 = |βij| j
X 2 ⇒ h0f | nˆgi |0f i = |βij| (2.41) j In general, this coefficient does not vanish: an observer that defines particles with respect to the g-modes will detect particles where the observer that defines particles with respect to the f-modes will see the vacuum. This can be applied to the case of an accelerated observer in flat spacetime (Rindler observer). For simplicity, we will consider a massless Klein-Gordon field in 2 dimensions. The Klein-Gordon equation in Rindler coordinates takes the form
−2aξ 2 2 2φ = e −∂η + ∂ξ φ = 0 (2.42) which admits as solutions normalized plane waves of the form
1 −iωη+ikξ gk = √ e (2.43) 4πω with ω = |k|. Because the choice of coordinates for regions I and IV needed a difference of sign between them, we need to define two sets of modes, one for each of the two regions :
√ 1 e−iωη+ikξ I g(1) = 4πω k 0 IV
15 0 I g(2) = k √ 1 e+iωη+ikξ IV 4πω In this way, each set of modes is positive-frequency with respect to the corre- sponding future-directed timelike Killing vector,
(1) (1) ∂ηgk = −iωgk (2.44) (2) (2) ∂(−η)gk = −iωgk (2.45) Introducing the corresponding creation and annihilation operators for each region, the field can be expressed as Z ˆ(1) (1) ˆ(1)† (1)∗ ˆ(2) (2) ˆ(2)† (2)∗ φ = dk bk gk + bk gk + bk gk + bk gk (2.46) The modes to which we will compare them will be the usual Minkowski modes, which expand the field as Z † ∗ φ = dk aˆkfk +a ˆkfk (2.47)
The Minkowski vacuum state |0M i and the Rindler vacuum state |0Ri are defined as usual:
aˆk |0M i = 0 (2.48) ˆ(1) ˆ(2) bk |0Ri = bk |0Ri = 0 (2.49) The next step now is to compute the Bogoliubov coefficients relating both sets of modes, and compute the expectation value of the Rindler number operator in the Minkowski vacuum. This is a bit tedious, and the usual procedure is the following: (1) instead of using the previously defined Rindler modes, we will take a set of modes hk , (2) hk that share the same vacuum state as the Minkowski modes (but the excited states are different). The way to do this is to start with the Rindler modes, analytically extend them to the entire spacetime, and express them in terms of the orignal Rindler modes. Then, the field will be expanded as Z (1) (1) (1)† (1)∗ (2) (2) (2)† (2)∗ φ = dk cˆk hk +c ˆk hk +c ˆk hk +c ˆk hk (2.50)
The properly normalized version of these modes is
16 (1) 1 πω/2a (1) −πω/2a (2)∗ hk = e gk + e g−k (2.51) q πω 2 sinh a
(2) 1 πω/2a (2) −πω/2a (1)∗ hk = e gk + e g−k (2.52) q πω 2 sinh a Just like before, the Bogoliubov coefficients allow to relate also the creation and annihilation operators:
ˆ(1) 1 πω/2a (1) −πω/2a (2)† bk = e cˆk + e cˆ−k (2.53) q πω 2 sinh a
ˆ(2) 1 πω/2a (2) −πω/2a (1)† bk = e cˆk + e cˆ−k (2.54) q πω 2 sinh a In this way, the Rindler number operator in region I,
(1) ˆ(1)†ˆ(1) nˆR (k) = bk bk (2.55) (1,2) can be expressed in terms of the new operatorsc ˆk , and because they share the same vacuum state as the Minkowski modes, we have that
(1) (2) cˆk |0M i =c ˆk |0M i = 0 (2.56) The fact that the excited states do not coincide is not a problem, because we are only interested in what the Rindler observer sees when the state is the Minkowski vacuum. For a Rindler observer in region I, the expectation value of the number operator will be
(1) ˆ(1)†ˆ(1) h0M | nˆR (k) |0M i = h0M | bk bk |0M i 1 −πω/a (1) (1)† = πω h0M | e cˆ−kcˆ−k |0M i 2 sinh a −πω/a (2.57) e (1) (1)† = πω h0M | cˆ−kcˆ−k |0M i 2 sinh a 1 = e2πω/a − 1 This result corresponds to a Planck spectrum with temperature
17 a T = (2.58) 2π Thus, a uniformly accelerated observer through the Minkowski vacuum will detect a thermal flux of particles.
2.4 Entanglement entropy In order to conceptually understand entanglement entropy, it is useful to first take a look to the following discrete problem : imagine a lattice model, with discrete degrees of freedom located at the lattice sites, which are separated a distance (see Figure (3)). At each site (labeled by α) we have a finite-dimensional Hilbert space Hα (for instance, a qubit per site). A pure quantum state of the system can be written as
|Ψi ∈ ⊗αHα (2.59)
Figure 3: Discrete lattice system, with a Hilbert space at each place. The grey region is called A, while Ac is its complementary, separted by the boundary ∂A. The distance between places is .
We can divide the lattice system into two complementary subsystems, namely A and Ac, separated by the boundary ∂A, which we shall call the entangling surface,
18 as can be seen in Figure (3). The Hilbert space of the total system has been split into the direct product of two Hilbert spaces,
⊗α Hα = HA ⊗ HAc (2.60) Now, one can construct the reduced density matrix of the subsystem A, which is constructed by tracing out the degrees of freedom of Ac:
ρA = TrAc (|Ψi hΨ|) (2.61) If the state |Ψi is factorized when the system is split, then we will have a pure state in HA. However, if the state can not be written as a direct product of states from the two subsystems, the state is entangled and the density matrix gives the probabilities for the ocurrence of the states in HA. The amount of entanglement that exists in |Ψi when the system is split is quantified by the Von-Neumann entropy of the reduced density matrix, or entanglement entropy, which is given by
SA = − TrA(ρA log ρA) (2.62) In a discrete system, this can be computed diagonalizing the density matrix and obtaining its eigenvalues λi. Then, the entanglement entropy is simply X SA = − λi log λi (2.63) i Because |Ψi is a pure state, it can be decomposed via the Schmidt decomposition, P |Ψi = i λi |αiiA |βiiAc . This tells us that non-trivial eigenvalues of ρA are the same as those of Ac. Then, the traces are the same, and the entanglement entropies are also the same:
SA = SAc (2.64) The fact that the entropy is the same for both regions means that it can not depend on the size of each region, but only on the degrees of freedom shared by the two regions. That is, it must be proportional to the area of the boundary ∂A instead of being proportional to the volume of the regions, as it would be expected in classical thermodynamic systems. The continuum limit of this system can be defined as taking the limit → 0. When this is done, the result for the entropy is sensitive to the ultra-violet (UV) physics, as we should expect. For a d-dimensional free field theory, the entropy is a UV-divergent quantity, with the leading term being proportional to the area of the entangling surface :
19 Area(∂A) S = γ + ... (2.65) A d−2 where γ is a constant that depends on the model used. This quantity is divergent when → 0 unless there is some physical UV cutoff (presumably, of the order of 2 the Planck scale), with which the entropy would be finite and proportional to A/Lp, matching with the Bekenstein-Hawking entropy for black holes [9–12]. Thus we shall assume in all cases that due to the UV physics, the entanglement entropy is finite in small regions, with a leading term given by S=ηA. Another important quantity that will be useful is the relative entropy. Given two density matrices ρ and σ we can define the relative entropy,
S(ρ||σ) = Tr(ρ log ρ) − Tr(ρ log σ) (2.66) which gives information about the distinguishability between the two density matrices. An important property of the relative entropy is that it is always positive or equal to zero, being equal to zero only when the two density matrices are the same. We can define the modular hamiltonian as
Kρ = − log ρ (2.67) and rewrite the relative entropy as
S(ρ||σ) = Tr(ρ log ρ) − Tr(ρ log σ) + Tr(σ log σ) − Tr(σ log σ)
= −S(ρ) + Tr(ρKσ) − Tr(σKσ) + S(σ) (2.68) = ∆hKi − ∆S
where ∆S = S(ρ) − S(σ) is the entropy difference between the states, and ∆hKi = Tr(ρKσ)−Tr(σKσ) is the difference in the expectation values of the modular hamiltonian Kσ for ρ and σ. If we consider σ to be a reference state σ = ρ0, and ρ a state close to it, we can 2 expand the latter in a power series in a parameter λ, ρ(λ) = ρ0 + λρ1 + λ ρ2 + ... in such a way that ρ(0) = ρ0 = σ. The relative entropy can be expanded as
d 2 S(ρ(λ)||σ) = S(ρ(0)||σ) + S(ρ(λ)||σ) λ + O(λ ) (2.69) dλ λ=0 The first term is zero because of the definition of the relative entropy. The term of order λ is also zero, because the relative entropy is a monotonically increasing
20 function around σ. Thus, the relative entropy is at least quadratic in the deviation parameter. This means that, for first-order variations, we have that
δS = δhHσi (2.70) This is known as the first law of entanglement entropy.
21 3 First derivation. Equilibrium thermodynamics in Rindler space
In this approach, Einstein’s equations are derived from the proportionality of entropy and horizon area, together with the thermodynamic relation δQ = T dS, relating heat, temperature and entropy (and area, due to the relation entropy ∼ area). In standard thermodynamics, heat is defined as energy that flows from, or to a thermodynamic system. Here, we shall define heat as energy that flows across a causal horizon (not necessarily a black hole horizon). In the relation δQ = T dS, we associate δQ with an energy flux across the horizon, and we shall use that the entropy is proportional to the area of this horizon. It remains to identify the temperature T . Using Unruh’s results, we can take T to be the Unruh temperature if we consider that the observer is in accelerated motion. Then, for consistency, the heat flow must be defined as the energy flux that this observer measures. In order to apply local equilibrium thermodynamics, two conditions must be imposed in the construction of our system:
• We need the observer to be as near as possible to the horizon. In the limit that the accelerated worldine approaches the horizon, the acceleration diverges, and so do the temperature and the energy flux, but their ratio remains finite.
• In general, the horizon will be expanding, contracting or shearing. In order to impose equilibrium, we need the expansion, shear and rotation to be zero at first order in a neighbourhood of the horizon.
The introduction of an accelerated observer gives as a natural choice for the horizon the Rindler horizon associated to this accelerated observer. The key idea to be shown can be expressed as : “In order to satisfy the thermodynamic equilibrium relation δQ = T dS, in- trepreted in terms of the energy flux and area of local Rindler horizons, the grav- itational lensing by matter energy must distort the causal structure of spacetime in a way that the Einstein equation holds.” The next step is to define precisely this local causal horizon. It can be done as follows: By means of the equivalence principle, the neighbourhood of any point p can be thought as a piece of flat spacetime. Around p we consider a 2-dimensional surface P. As usual, this 2-surface will be represented as a point in the conformal diagram. The boundary of the past of P has two components, each of which is a null surface generated by a congruence of null generators ka orthogonal to P. The local causal
22 horizon is defined as one of these two components. We take λ as the affine parameter for ka, in such a way that λ vanishes at P and is negative to the past of P (see Figure (4)).
Figure 4: Rindler horizon H of a 2-sphere P. The accelerated observer follows the trajectory of the Killing vector χa. ka is the generator of the horizon.
In order to define the temperature and the heat, note that in the approximately flat region around p the usual Poincar´esymmetries hold. In particular, there is an approximate Killing field χa generating boosts orthogonal to P and vanishing at P. Because we are at very short distances, the Minkowski vacuum state (or any other state) is a thermal state with temperature T = ~a/2π with respect to the boost hamiltonian, where a is the acceleration of this orbit. The heat flow is then defined a through the boost-energy current of matter, Tab χ , where Tab is the stress-energy tensor. The Killing field defining the orbits of Rindler observers coincides at the null surface with the generators for sufficiently accelerated observers. Then, in the limit that the observer is sufficiently close to the horizon, the Killing field χa is parallel to the horizon generator ka, and, at first order, we have that χa = −κλka and dΣa = kadλdA, where dA is the area element on a cross section of the horizon [6, 13, 14]. Then, the heat flux is given by
23 Z Z a b a b δQ = Tab χ dΣ = −κ λTab k k dλdA (3.1) H H Assuming that the entropy is proportional to the area, we have that dS = ηδA, where δA is the area variation of a cross section of a pencil of generators of H. For now, the constant η is left undetermined. The area variation is given by Z δA = θdλdA (3.2) H where θ is the expansion of the horizon generators. The expression δQ = T dS ∝ δA is telling that the presence of the energy flux is associated with a focussing of the horizon generators. Then, the Raychaudhuri equation (2.14) enters in the game, because it tells precisely the rate of focussing of the generators. The stationarity conditions imposed above imply that, at p, both the expansion and the shear vanish1, and the Raychaudhuri equation simplifies to dθ = −R kakb (3.3) dλ ab where the θ2 and σ2 are higher-order contributions that can be neglected when integrating to find θ around P. For a small interval of λ, this integration is simply a b θ = −λRabk k . Then, Z a b δA = − λRabk k dλdA (3.4) H ~κ Now, from (3.4) and (3.1), we see that δQ = T dS = 2π ηδA is valid if η T kakb = ~ R kakb (3.5) ab 2π ab is valid for all null vectors ka. This is equivalent to the tensorial equation 2π Tab = Rab + fgab (3.6) ~η for some undetermined function f. The stress-energy momentum is divergence- free, which means that the rhs of (3.6) must also be divergence-free. This gives the R constraint f = − 2 + Λ for some undetermined constant Λ. Then, we find: 1It is always possible to find a 2-surface P so that both the expansion and shear vanish in a first order neighbourhood of p.
24 1 2π Rab − Rgab + Λgab = Tab (3.7) 2 ~η If η = 1 , as the Bekenstein-Hawking entropy formula tells, we recover the 4G~ Einstein equation:
1 R − Rg + Λg = 8πGT (3.8) ab 2 ab ab ab Thus, the Einstein equation appears from the relation δQ = T dS, from a purely thermodynamic point of view.
25 4 Second derivation. Entanglement entropy
In this derivation, the Einstein equation appears as a consequence of a maximal vacuum entanglement hypothesis in a small region of spacetime. The main hypothesis can be expressed as : “When the geometry and quantum fields are simultaneously varied from maximal symmetry, the entanglement entropy in a small geodesic ball is maximal at fixed volume.” The system to consider now can be defined as follows: Consider any point o of a spacetime of dimension d. If we choose a timelike unit vector ua, we can generate a (d − 1)−dimensional spacelike ball Σ of radius l if we consider all the geodesics of length l that leave p in all directions orthogonal to ua. The point p is located at the center of the ball, and we call the surface of the ball ∂Σ. The region causally connected to the sphere Σ is called the causal diamond (see Figure (5)). We will consider that the radius l of the ball is much smaller than the characteristic radius of curvature of the spacetime in that region: l Lcurv.
Figure 5: Causal diamond associated to a geodesic ball centered at o and geodesic radius l.
It is known that, at sufficiently short distances, all the fields look like the vacuum state. Moreover, if the condition l Lcurv is satisfied, the spacetime around p can be treated as flat. Then, when we perform the variations with respect to the geometry
26 and the quantum states, we will perform them with respect to flat space, and to the vacuum state. A way of interpreting geometrically the Einstein equation is the following: in classical vacuum (without any matter source), any small geodesic ball of given volume has the same area as in flat spacetime. However, when there is a source of matter or energy (given by some expectation value of the stress-energy tensor), curvature causes a spatial ball of given volume to have a smaller surface area than it would have in flat spacetime. This area deficit can be computed at fixed geodesic radius, or at fixed volume. The expressions are given by
d Ωd−2l δA = − R (4.1) l 6(d − 1) d Ωd−2l δA = − R (4.2) V 2(d2 − 1) ik where R = RikR is the spatial Ricci scalar at p. Note that
d + 1 δA = δA (4.3) l 3 V For convenience, as will be seen at the end, we will take the variations to be at fixed volume. To connect this expression with the Einstein equation, note that the spatial Ricci scalar can be related to the 00-component of the Einstein tensor as follows: 1 1 1 1 G = R − Rg = R − 2R 0 = R ik = R (4.4) 00 00 2 00 2 0 2 ik 2 Then, we can write
d Ωd−2l δA = − G (4.5) V d2 − 1 00 and, by virtue of the Einstein equation,
d 8πGΩd−2l δA = − T (4.6) V d2 − 1 00 Under a simultaneous variation of the geometry and the quantum fields, the variation of the entanglement entropy will have two contributions: a UV-contribution δSUV from the area change when the metric is varied with respect to flat spacetime (δgab), and an IR-contribution δSIR due to the variation of the fields (δ |ψi), so we can write
27 δS = δSUV + δSIR (4.7) We shall assume that the UV-part of the entanglement entropy is finite at leading order, and is proportional to the area variation computed above. That is, δSUV = ηδA. As in the previous derivation, the constant η is left undetermined until the end. In order to compute δSIR, we take into account that the vacuum state of any QFT, when restricted to the diamond, can be written as a thermal density matrix, 1 ρ = e−K/T (4.8) Z where T = ~/2π, and K is the modular hamiltonian. From this thermal density matrix, the entropy variation can be computed and it is given by δSIR = δhKi. In general, K is not a local operator, and there is not a general expression for it. However, in the case of the vacuum of a conformal field theory (CFT), the situation is different. The diamond has a conformal boost Killing vector generating it (see Figure (5)), given by 1 ζ = l2 − u2 ∂ + l2 − v2 ∂ (4.9) 2l u v in null coordinates u, v, or 1 ζ = l2 − r2 − t2 ∂ − 2rt∂ (4.10) 2l t r in the usual t, r coordinates. For the vacuum of a CFT, there is a conformal transformation relating the diamond to Rindler space and, in this case, K is equal to Hζ , the Hamiltonian generating the flow of the above Killing vector [9, 10, 16], which means that Z 2π ab Hζ = T ζb dΣa (4.11) ~ Σ With the previous Killing vector on the t = 0 surface, we obtain
Z Z 2 2 2π a b 2π d−1 l − r δhKi = δhTab iζ dΣ = d x δhT00 i (4.12) ~ ~ 2l
If we consider that δhT00 i is constant within the ball, it can be taken out of the integral, and we have
28 Z 2 2 2π d−1 l − r δhKi = δhT00 i d x ~ 2l Z Z l 2 2 2π d−2 l − r = δhT00 i dΩd−2 r dr (4.13) ~ 0 2l d 2π Ωd−2l = δhT00 i ~ d2 − 1 This result, together with the one providing the area variation at fixed volume, gives
δS V = ηδA + δhKi d Ωd−2l 2π (4.14) = −ηG00 + δhT00 i d2 − 1 ~ Now, imposing the assumption that the entanglement entropy is maximal at fixed volume (that is, δS V =0), we obtain the relation 2π G00 = δhT00 i (4.15) ~η If we require this variation to vanish at all points and with all timelike unit vectors, we obtain a tensor equation, 2π Gab = δhTab i (4.16) ~η This is the Einstein equation, provided we define the constant η to be η = 1 1 , 4 ~G which is the precise value required by the Bekenstein-Hawking entropy formula. For the non-CFT case, K is not given by (4.11), and some assumptions must be made in order to find an expression for δhKi. The main conjecture is to consider that δhKi is given by
Ω ld δhKi = d−2 (δhT i + δX) (4.17) d2 − 1 00
where δX is a spacetime scalar, maybe related to the trace of Tab . Calculations [17, 18] support this assumption, although it is still being investigated. When the maximization of entropy is considered, one obtains
29 2π Gab = (δhTab i − δhXigab) (4.18) ~η This result has a problem, because from the Bianchi identity, the lhs of (4.18) is divergence-free, and so is the term δhTab i because of energy-momentum conservation. This implies that ∇aδhXi = 0 and, if it is related to the trace of Tab , it is a too strong constraint. This problem can be solved if, instead of comparing it to the Minkowski vacuum, the variations are compared to some other maximally symmetric spacetime (MSS), because any MSS seems as good candidate for the vacuum as flat spacetime. The MSS Einstein tensor in a MSS of curvature scale λ is given by Gab = −λgab. When the area variation is compared to this MSS, the area variation at fixed volume is given MSS by the same expression as before, but replacing G00 by G00 − G00 . The variation of entropy reads now
2π δS = ηδA + δhKi V V ~ d (4.19) Ωd−2l 2π = −η (G00 + λg00) + (δhT00 i + δX) d2 − 1 ~ Again, when we consider that the variation vanishes at all points and with all timelike unit vectors, the equation becomes a tensorial equation, 2π Gab + λgab = (δhTab i − δXgab) (4.20) ~η Taking the divergence of this equation, the term of the Einstein tensor and the term of the stress-energy tensor vanish, because of the Bianchi identity and the con- servation of energy-momentum, respectively. Then, we obtain a constraint between λ and δX: 2π Λ = δX + λ (4.21) ~η where Λ is a spacetime constant. When this relation is plugged into (4.20), we obtain 2π Gab + Λgab = δhTab i (4.22) ~η
30 This is the Einstein equation with a cosmological constant Λ, provided that, again, η = 1 1 , in agreement with the Bekenstein-Hawking entropy. Thus, the 4 ~G Einstein equations have been derived from an entanglement entropy hypothesis.
31 5 Comments and discussion
This section contains a discussion about some issues related to the derivations, to- gether with some of the main questions that arise from this new interpretation of spacetime, and the possible answers (more or less satisfactory) that can be given with the current knowledge of physics.
• Why are the variations taken at fixed volume instead of at fixed geodesic radius?
In the derivation based on entanglement entropy, we have taken the variations at fixed volume “for convenience”. We argue here why this has been done. First of all, notice that we have obtained the desired result because the geometric d 2 term Ωd−2l /(d − 1) that appears as a prefactor in both variations is the same for δSUV and δSIR, and it can be factorized. Had we taken the variations at fixed geodesic radius instead of fixed volume, the terms would have not been the same. But there are other arguments to take the variations at fixed volume. The first law of causal diamonds is a variational identity, analogous to the first law of black hole mechanics, which relates variations, away from flat spacetime, of the area, volume, and cosmological constant inside the diamond. The first law reads κk V − δV + ζ δΛ = T δS (5.1) 8πG 8πG gen where κ is the surface gravity of the Killing horizon, k is the trace of the outward R extrinsic curvature of the boundary ∂Σ when embedded in Σ, and Vζ ≡ Σ |ζ|dV . Λ is the cosmological constant, T is minus the Hawking temperature, and Sgen is defined as the sum of the horizon entropy and the entanglement entropy of matter. At fixed volume and fixed cosmological constant, the first law of causal diamonds implies that the entropy is stationary when varied away from the vacuum, as it has been considered in the derivation.
• What is the best way to proceed if we want to find a quantum theory of gravity?
The fact that spacetime dynamics can be derived from thermodynamic arguments suggests the possibilty that gravity is not a fundamental force, but a macroscopic result of some microscopic degrees of freedom of spacetime [20–22]. These degrees of freedom have been called by some authors as “Atoms of Spacetime”, in analogy to the standard relation between thermodynamics and statistical mechanics.
32 If this is the case, it explains why the quantization of General Relativity has shown to be much more problematic than for other microscopic forces. In [21, 22], some properties that these atoms of spacetime should have are dis- cussed, and with a particular model of atoms of spacetime for the geometric part of the action, the Einstein’s equations are recovered from a purely thermodynamic argument. Other works [23, 24] have used particular models of microstructure to recover the Hawking temperature and entropy for black holes.
• Could we obtain higher-curvature corrections to the Einstein’s equa- tions with the thermodynamic interpretation?
The classical Einstein’s equations, 1 R − Rg = 8πGT (5.2) µν 2 µν µν are derived from the Einstein-Hilbert action, 1 Z √ S = −gR d4x (5.3) 16πG However, the Einstein-Hilbert gravity can be treated as a low-energy effective theory, so we should expect to have corrections to this action, of the form
1 Z √ S = −g R + α Λ + α R2 + α R Rµν + α R Rµνρσ + ... d4x (5.4) 16πG 1 2 3 µν 4 µνρσ and the field equations arising from this action would contain higher-curvature terms. These terms include higher derivatives of the metric, which correspond to terms with higher and higher curvature (and a lower and lower associated curvature radius). At some point, this curvature radius is of the order of the Planck length. Thus, in a theory of quantum gravity, we expect these terms to be important. The question that arises now is: we have obtained the classical Einstein’s equations from a thermodynamic point of view. If the spacetime is really a thermodynamic entity, should we be able to obtain these higher-curvature terms in the field equations with a similar argument? There is not a clear answer to this question. In the derivation of Section (4), in the expression for the area deficit we have neglected terms of order l/Lcurv, while the 2 next-higher-curvature correction to the field equations might be of order (l1/Lcurv) , with l1 a length scale appearing in the corresponding term in the action. To obtain 2 this next-order term in the field equations, we need l/Lcurv < (l1/Lcurv) ⇒ l/l1 <
33 l1/Lcurv. The rhs must be smaller than 1 (otherwise, the higher-order terms would dominate), which means that we need l < l1. That is, the diamond must be smaller than l1. If l1 is, for instance, the Planck length, the diamond should be smaller than the Planck length, and the classical geometry and quantum field theory used in the derivation would not work in that regime. There have been some attempts to find the field equations when these corrections are considered [25–27], but because of the presence of these terms, some technical difficulties appear and it is required to use non-equilibrium thermodynamics. However, an interesting result has been found in . There, they show that, for spherically symmetric systems with a horizon, the Einstein equations arising from the Einstein-Hilbert action can be put in the form of the relation T dS = dE + P dV , matching the entropy S and the energy E with the already know expressions. They go one step beyond, and do the same for the first correction to the Einstein-Hilbert action, the so-called Gauss-Bonnet correction:
1 Z √ S = −g (R + αL ) d4x, L = R2 − 4R Rµν + R Rµνρσ (5.5) 16πG GB GB µν µνρσ finding again that, once the field equations are written in the form T dS = dE + P dV , the entropy and the energy match with the expressions obtained by other authors. Finally, they generalize this result to the complete Lanczos-Lovelock action in D dimensions, matching again the results for S and E with independent calculations. These results suggest that the thermodynamic route to obtain the field equations also works for higher-curvature theories of gravity, and the quantum corrections to the Einstein-Hilbert action appear as quantum corrections for the entropy and the energy . However, the microscopic structure beyond this thermodynamics remains mysterious.
• Is it appropiate to consider the entanglement entropy to be finite at UV scales?
As we have seen, for a d-dimenional free field theory the entanglement entropy is a UV-divergent quantity, with the leading term being proportional to the area of the entangling surface, which we rewrite here for convenience:
Area(∂A) S = γ + ... (5.6) A d−2
34 where is the cutoff length that is sent to 0 in the continuum limit and originates the divergences. But, because of the fact that the fields that contribute more to the entanglement entropy are those of high energy, we expect to have modifications to the background geometry. This backreaction of spacetime may lead to a finite entropy. Susskind and Uglum show in that when these are considered, the divergences that appear are the same ones that appear in the renormalization of the gravitational constant G. When this renormalization process is done, one obtains a finite result for the entanglement entropy, with the leading term coinciding with the Bekenstein-Hawking entropy [30, 31]:
1 Area(∂A) SA = + ... (5.7) 4 GR~ where GR is the renormalized gravitational constant. Again, in order to un- derstand the microscopic origin of the entanglement entropy and its divergences, a microscopic understanding of the theory is needed. The great similarity between the entanglement entropy and the Bekenstein-Hawking entropy suggests that, maybe, black hole entropy can be originated purely from en- tanglement [5, 11]. This identification, however, can only be done once the diver- gences of entanglement entropy are properly solved, because the Bekenstein-Hawking entropy has no divergences.
35 6 Conclusions
After the realization of the thesis, some conclusions can be extracted:
• The analogy between thermodynamics and black holes found by Bekenstein and Hawking 50 years ago can be extended to spacetime itself. That is, spacetime is a thermodynamic entity.
• Given this thermodynamic nature of spacetime, one should be able to derive the Einstein equations from thermodynamic arguments. This has been done in two different ways: in Section (3), the Einstein equations have been derived from the thermodynamic relation δQ = T dS near a Rindler horizon, and in Section (4) they have been derived from an hypothesis about entanglement entropy.
• This thermodynamics suggests some kind of microstructure of spacetime at smallest scales. Although there is not fully satisfactory explanation of what this microstructure is, it seems reasonable that it will be important when trying to find a complete quantum theory of gravity.
J.D. Bekenstein. Generalized second law of thermodynamics in black-hole physics. Phys. Rev. D 9 3292 (1974). S.W. Hawking. Particle creation by black holes. Commun. Math. Phys. 43 199 (1975). W.G. Unruh. Notes on black-hole evaporation. Phys. Rev. D 14, 870 (1976). P.C.W. Davies. Scalar Production in Schwarzschild and Rindler Metrics. J. Phys. A.: Math. Gen. 8 609 (1975). J.D. Bekenstein. Do we understand black hole entropy? (1994). ArXiv: gr- qc/9409015. T. Jacobson; R. Parentani. Horizon entropy. Found. Phys. 33 323-348 (2003). ArXiv: gr-qc/0302099. S. Carroll. Spacetime and geometry. An introduction to General Relativity. Addison-Wesley (2004). T. Takayanagi M. Rangamani. Holographic entanglement entropy. (2017) ArXiv: hep-th/1609.01287. R. Myers D. Blanco; H. Casini; L. Hung. Relative entropy and holography. JHEP 2013 60 (2013). ArXiv: hep-th/1305.3182. H. Casini; M. Huerta; R. Myers. Towards a derivation of holographic entangle- ment entropy. JHEP 1105 036 (2011). ArXiv: hep-th/1102.0440. S. Solodukhin. Entanglement entropy of black holes. Living Rev. Rel. 14, 8 (2011). ArXiv: hep-th/1104.3712. M. Srednicki. Entropy and Area. Phys. Rev. Lett. 71 666-669 (1993). ArXiv: hep-th/9303048. T. Jacobson. Thermodynamics of Spacetime: the Einstein Equation of State. Phys. Rev. Lett. 75 1260 (1995). ArXiv: gr-qc/9504004. D. Kothawala. The thermodynamic structure of Einstein tensor. Phys. Rev. D 83, 024026 (2011). ArXiv: gr-qc/1010.2207. T. Jacobson. Entanglement Equilibrium and the Einstein Equation. Phys. Rev. Lett. 116 201101 (2016). ArXiv: gr-qc/1505.04753. D. Blanco; H. Casini; M. Leston; F. Rosso. Modular energy inequalities from relative entropy. JHEP 01 154 (2018) ArXiv: hep-th/1711.04816.
37 A.J. Speranza. Entanglement entropy of excited states in conformal pertur- bation theory and the Einstein equation. JHEP 04 105 (2016). ArXiv: hep- th/1602.01380. H. Casini; D.A. Galante; R.C. Myers. Comments on Jacobson’s Entanglement equilibrium and the Einstein equation. JHEP 03 194 (2016). ArXiv: hep-th/1601.00528. T. Jacobson; M. Visser. Gravitational Thermodynamics of Causal Diamonds in (A)dS. (2019). ArXiv: hep-th/1812.01596. A. Granger. Thermodynamic Gravity and the emergence of space with geometry. Imperial College London (2010). T. Padmanabhan. Atoms of Spacetime and the Nature of Gravity. J. Phys. 701 012018 (2016). T. Padmanabhan. Atoms of Spacetime and the Nature of Gravity. Entropy 17 7420 (2015). ArXiv: gr-qc/1508.06286. J. M¨akel¨a. Partition Function of the Schwarzschild Black Hole. Entropy 2011, 13, 1324-1354. R. Raitio. A Statistical Model of Spacetime, Black Holes and Matter. Open Access Library Journal, 3: e2487. C. Elling; R. Guedens; T. Jacobson. Non-equilibrium Thermodynamics of Space- time. Phys. Rev. Lett. 96 121301 (2006). R. Guedens; T. Jacobson; S. Sarkar. Horizon entropy and higher curvature equations of state. Phys. Rev. D 85 064017 (2011). ArXiv: gr-qc/1112.6215. T. Jacobson. Gravitation and vacuum entanglement entropy. Int. J. Mod. Phys. D21 1242006 (2012). ArXiv: gr-qc/1204.6349. A. Paranjape; S. Sarkar; T. Jacobson. Thermodynamic route to field equations in Lanczos-Lovelock Gravity. Phys. Rev. D 74 104015 (2006). ArXiv: hep- th/0607240. L. Susskind; J. Uglum. Black hole entropy in canonical quantum gravity and superstring theory. Phys. Rev. D 50 2700-2711 (1994). ArXiv: hep-th/9401070. J.H. Cooperman; M.A. Luty. Renormalization of entanglement entropy and the gravitational effective action. J. High Energy Phys. 2014 45 (2014). ArXiv: hep-th/1302.1878. F. Larsen; F. Wilczek. Renormalization of black hole entropy and of the gravita- tional coupling constant. Nucl. Phys. B 458 249 (1996). ArXiv: hep-th/9506066.
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https://www.physicsforums.com/threads/applications-of-magnetic-fields.582369/
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An ion with a charge of +1e enters a region where the electric field produced by parallel plates at 630V separated by 7.0mm is at right angles ot a magnetic field B=450mT. The ion moves a straight line. Find the magnitude of the electric field and the speed of the ion.
Ampere's Law B=μ0I/2∏r
The Attempt at a Solution
Is this a mass spectrometer?
first of all, because the plates are parellel the force is maximized (sin90=1)
I am confused, isn't the ion supposed to curve in a radius defined by: r=mv/qB (mv2/r=qvB)
I just don't know what to do since the ion doesn't curve.
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http://homework.appleby.cumbria.sch.uk/?cat=7
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math
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Students to complete worksheet questions on Two Step Equations.
Complete the worksheet on Tree Diagrams
Complete the worksheet on Mode, Median and Range
Do Q 1 & 2 from sheet 4c. Extension Q 3 – 6
Do questions 1 -3 from sheet 11a. Extension Q 4
Students have a worksheet stuck in their maths exercise book on Percentages, which needs to be completed by Tuesday 23rd April.
If students need help, I am available at breaks and lunchtimes.
Complete the worksheet on Decimals and Fractions.
Complete the first three questions from either sheet 7a or 7b Linear equations
Extension Try Q 4 & 5
Complete both questions from the sheet 10e Plotting straight-line graphs.
Students to complete the worksheet, writing coordinates and plotting coordinates.
Remember ‘along the corridor, up or down the stairs’.
Remember to use brackets e.g. ( 5,6)
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https://algebra-answer.com/algebra-help/graphing-lines/simplifying-radical.html
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math
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Does someone here know anything regarding simplifying radical expressions ? I’m a little lost and I don’t know how to answer my math project about this topic. I tried researching all tutorials about it that could help me answer my math problems but I still can’t finish it. I’m having a hard time answering it especially the topics binomials, y-intercept and radicals. It will take me forever to answer my math homework if I can’t get any help. It would greatly help me if someone would suggest anything that can help me with my algebra homework.
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https://dspace.mit.edu/handle/1721.1/79372
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Saddle Point in the Minimax Converse for Channel Coding
MetadataShow full item record
A minimax metaconverse has recently been proposed as a simultaneous generalization of a number of classical results and a tool for the nonasymptotic analysis. In this paper, it is shown that the order of optimizing the input and output distributions can be interchanged without affecting the bound. In the course of the proof, a number of auxiliary results of separate interest are obtained. In particular, it is shown that the optimization problem is convex and can be solved in many cases by the symmetry considerations. As a consequence, it is demonstrated that in the latter cases, the (multiletter) input distribution in information-spectrum (Verdú-Han) converse bound can be taken to be a (memoryless) product of single-letter ones. A tight converse for the binary erasure channel is rederived by computing the optimal (nonproduct) output distribution. For discrete memoryless channels, a conjecture of Poor and Verdú regarding the tightness of the information spectrum bound on the error exponents is resolved in the negative. Concept of the channel symmetry group is established and relations with the definitions of symmetry by Gallager and Dobrushin are investigated.
DepartmentMassachusetts Institute of Technology. Department of Electrical Engineering and Computer Science
IEEE Transactions on Information Theory
Institute of Electrical and Electronics Engineers
Polyanskiy, Yury. Saddle Point in the Minimax Converse for Channel Coding. IEEE Transactions on Information Theory 59, no. 5 (May 2013): 2576-2595.
Author's final manuscript
INSPEC Accession Number: 13448776
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http://www.angara.com/jewelry-guide/about-birthstones.html
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math
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Know Your Birthstone And Birthstone Jewelry
Traditionally, birthstones or birthday stones are special gemstones associated with each month of the year. Each gemstone is attributed to various qualities symbolizing the specific month of birth in the Gregorian calendar.
The origin of birthstones is believed to date back to 1250 BC when Moses made the breastplate of the High Priest Aaron on the basis of the instructions he received during those 40 days that he spent in the mountains. According to the Bible, in Exodus 28 and 39 the original breastplate of the High Priest contained twelve gemstones representing the twelve tribes of Israel which were later linked with the signs of zodiac and then to the months in a year.
Throughout history, different cultures had their own rituals and beliefs linked with the use of a particular birthstone. Even many people believed that wearing all twelve birthstones together or in rotation is a way to ensure maximum benefit from these sacred crystals. Catherine de Medici wore a girdle set with twelve stones to derive utmost benefit. Also the kings and queens of different princely states in India wore jewels studded with twelve stones.
In 1912, the American National Association of Jewelers officially adopted a list which now works as the basis of modern birthstone list. This list was altered in 2002 when tanzanite was adopted as the birthstone for December.
These birthstones are usually worn in the form of jewelry, often as rings, pendants and amulets and are said to possess certain mystical powers. They are considered helpful in healing many ailments and known for radiating positive energy to the wearer.
These birthstones for centuries are loved by mankind for their mysterious appeal, rarity, and durability, play of light and divine colors. Each birthstone has some unique properties and is associated with a specific month. Do you know your birthstone? Would you like to get an idea of your lucky stone that might bring good fortune for you? Here are some interesting and alluring facts about birthstones. Check out which gem is special for you.
January Birthstone - Garnet
Garnet, the birthstone of January, signifies vitality, passion and trust. The gem comes virtually in all colors of spectrum ranging from red, pink and green to yellow, and orange. Garnet which is a group of minerals derived its name from Latin word granatus, meaning pomegranate seeds. The gem considered to heal ailments like blood deficiency, arthritis and rheumatism. Wearing garnet in form of rings, earrings and pendants helps in improving overall physical and mental wellbeing. This birthstone symbolizes faith, courage, loyalty, trust, constancy and fidelity. With a resinous luster and hardness of 6.5 to 7.5, garnet jewelry is a perfect present for those who love mystical charm and grace.
February Birthstone – Amethyst
Amethyst which is known as the gem of royalties is the official birthstone for February. It is one of the oldest stones known to human civilization. The history dates back the existence of amethyst as far as 25,000 years ago when prehistoric humans used it as a decorative piece. It was found in the remains of Neolithic age.
The gem represents power and symbolizes peace, love and happiness. It derived its name from Greek word 'amethystos' that means remedy against drunkenness. The name was given due to the belief that the gem has the powers to ward off the effect of intoxication and keeps the wearer clear headed and quick witted. Throughout history, many myths, lore and legends were associated with amethyst. Known to empower the minds and souls, this purple member of quartz family assists the wearer in attaining wisdom and courage. Amethysts are soft stones with hardness of 7 on moh's scale yet they are counted as a desiring gem for jewelry. Rings, earrings, pendants and amulets studded with amethyst are best said to ward off evil spirits and promote love and protection.
The gem is also said to have healing properties that work against headache, insomnia, arthritis and circulatory disorders.
March Birthstone – Aquamarine
Aquamarine is though officially the birthstone for March but it is also loved by people born in other months. This gem has derived its name from Latin word 'aqua' meaning water and 'marina' meaning sea so is known as 'Water of the sea'. It ranges from pale blue green to deep blue in color. As a stone of beryl family it is relatively a hard gem which is prized for its lavish beauty, mystical properties and symbolic association.
In ancient times aquamarines were carried by sailors as a guarantee of safe voyage. The gem signifies calmness, sympathy, trust, harmony, togetherness, peace, courage and serenity. It is said to have the mystical powers that reduce stress, sharpens the intellect, clears confusion, promotes self-expression and shields the aura. For ages, aquamarine jewelry specially rings and amulets have been associated with healing of many ailments like sore throats, swollen glands, thyroid problems, hormonal imbalance and immune system disorder. The best aquamarine stones come from Brazil but quality stones are also mined in Nigeria, Madagascar, Zambia and Mozambique.
April Birthstone – Diamond
People born in April are lucky to have diamonds as their official birthstone. This often colorless rock is an allotrope of carbon with tetrahedral geometry. Diamonds since inception are considered a traditional stone for engagement and wedding rings because the gem symbolizes unity, trust, fidelity, strength and devotion.
Known as a girl's best friend diamonds in the form of rings, earrings, pendants etc represent faith, love, innocence, beauty, clarity and purity.
Formed at the depth of 90 to 120 miles under the earth crust, this April birthstone is the hardest gem in the world which has the unique ability to refract and reflect light. Diamonds are examined and valued on the basis of color, clarity, cut and carat which are commonly known as the 4Cs of diamond.
Although colorless diamonds are most famous yet with the discoveries of different colored diamonds, the demand for fancy natural colored diamonds has tremendously increased. These colored diamonds make only 1% of the total diamonds that exist on the earth but their beauty and play of color add value to their worth. Fancy colored diamonds are formed due to any alteration in the diamond crystal structure. Pink, green, yellow, blue, orange, red, purple, brown, black, champagne etc are few of the many colors diamonds exhibit.
Diamonds are not only desired for their beauty but also cherished for their mysterious properties. They are said to balance the seventh Chakra and are known to aid creativity. Also the gem is said to enhance psychic ability, promotes harmony and balance between individual and ambience, removes toxins and heal the body while releasing stress.
May Birthstone – Emerald
Emerald the birthstone for May is known as the symbol of rebirth, spring and youth. It is believed to grant its owner foresight and good fortune.
This member of beryl family is one of the four precious stones. Its color ranges from light to dark green and it scores 8 on moh's scale of hardness. The gem is famous as the symbol of natural growth and harmony. It is very popular as a stone for fine jewelry.
The history of emeralds dates back to 1300 BC when Egypt was the only known source of the gem and it was mined for Cleopatra. The gem got its name from Greek word 'samaragdus', meaning green.
Emerald jewelry has been worn by kings and queens throughout the history. The stone in the form of rings, pendants, amulets etc is believed to promote romance, serenity, intelligence and communication. Emerald jewelry is believed to enhance unconditional love and provides domestic bliss.
The gem heals psychic ailments, clairvoyance and balance the heart chakra. It is also considered a good supporter in treating heart, lungs, spine and muscular disorders.
June Birthstone – Pearl and Alexandrite
As the birthstone of June, pearls, symbol of purity, are believed to grant its owner wisdom and innocence. It is associated with the third chakra and is known as the stone of sincerity and honesty. Pearls contrast to their fellow gemstones, has an organic origin. They are developed within the shell of oysters and clams. Pearls are mostly made of aragonite and are valued highly for their beauty and rarity. Wearing pearls in the form of jewelry is said to promote personal integrity, concentration and wisdom. It treats digestive disorders and relieves the conditions of bloating and biliousness.
Alexandrite is the second birthstone for the people born in month of June. This is relatively newer stone which was discovered in 1831 in Russia during the reign of Czar Alexander II. As it was discovered on his birthday, to honor him the stone was named alexandrite. This is an extremely rare chrysoberyl with chameleon like properties. The gem is an enchanting green in day light and fluorescence light which turns to deep red in incandescent light. As a rare and expensive gem, it enjoys a status of huge demand. Due to this the synthetic versions of alexandrite are also available in the market which possess the similar physical, chemical and optical properties as a natural gemstone. It is a new addition so got no time to be a part of myths and lore but people in Russia believe that it brings good fortune to the owner.
July Birthstone – Ruby
Also known as 'Ratanraj' or 'King of Gems' ruby is the symbol of love and passion. As the official birthstone for July it bestows the owner health, wisdom, wealth and success. It is a member of corundum family and is found only in shades of red. Large rubies are even rarer than sapphires and diamonds.
This gem of beauty is famous for its bright hues and mysterious properties. It is said to arouse the senses and energy levels. The stone is related to the base chakra and heart chakra. Since inception it is associated with feelings of happiness, devotion, love, care, romance and passion. Wearing ruby jewelry is believed to bring prosperity, power and leadership qualities. The gem had a long written history and it enjoyed a prominent position in the hearts and vaults of royals.
According to ancient Asian lore and legends the gem was self-luminous so called the 'glowing stone'. It was believed to guard against wicked thoughts, amorous desires and disputes. Ruby in rings, amulets, pendants etc. was told to be a good supporter in detoxifying body and treating fevers, infectious diseases and cures bleeding. It was famous for stimulating adrenals, kidneys, reproductive organs, spleen and pineal gland. The gem was also known to aid in retaining wealth and passion by clearing negative energy and imparting potency and vigor.
August Birthstone – Peridot
Peridot is associated with fame, dignity, protection, purity and rebirth. Being the birthstone for the month of august it strengthens its owner by enhancing patience, confidence and assertion. The member of olivine family achieves the hardness of 6.5 to 7 on moh's scale and exhibits yellowish green to golden green tints.
It was called 'the gem of the sun' by Egyptians as it shines brightly in the sun. The Greeks called it 'evening emerald' because the stone becomes deep green at night and glows in dark.
Peridot symbolizes prosperity and openness. It is believed to host the magical powers to protect against nightmares and negative energy. It reduces stress, banishes lethargy and cleanses the body. Benefits immune system, aids heat, lungs, gallbladder and treats ulcer and hypochondria. Peridot jewelry is used to treat anger, stress and jealousy. It is also a lavishing accessory to enhance looks and brings the spring glow to fashion statement.
September Birthstone – Sapphire
This 'wisdom stone' has a long written history. Sapphire unlike its sibling ruby comes in various colors from pink, green, yellow, orange, purple and white. But the most coveted of all is the blue sapphire ranging from light to intense hues of blue. Sapphire is one of the most powerful gems that possess enormous mysterious powers. Sapphires known as the gem of heavens for their beauty and brilliance and are believed to promote prosperity. Medieval clergy wore sapphire jewelry to represent strength, power and position. Sapphire engagement rings were also very famous among royals.
This birthstone of September symbolizes luck, love, generosity and loyalty. It is associated with creative expressions, spiritual development and metaphysical balance.
Sapphires are thought to release mental tensions and depression. It balances the body, mind and soul, brings serenity, peace and aligns the physical, mental and spiritual plans. Sapphires are also know to treat blood and cellular disorder and they regulate the glands and calm overactive body system.
October Birthstone – Opal and Tourmaline
Opal the birthstone of October is an emotional stone which reflects the mood of the wearer. The Greek word "Opallios" which means "to see a change" had given opal its name. From milky white to black with shades red, blue, green, yellow, and orange, opals have a range of colors. Its spherical arrangement is the reason of its radiance that adds extra beauty to its color play.
This non-crystalline silica gel product is associated with love, loyalty, peace, and faithfulness. Opals are thought to encourage freedom, enhance cosmic consciousness and induce mystical vision. Wearing opal jewelry especially rings stimulates creativity, releases anger and soothes stress. It is also believed to treat infections and fevers.
Tourmaline with a wide variety of colors is a favorite gemstone of designers and gem collectors. This magnificent stone with trigonal geometry comes in various colors from pink, red, green to blue, black, yellow and orange. In the world where fashion changes with every minute, pink tourmaline which is an alternative birthstone or people born in October holds spiritual, sentimental and styling attributes all together.
November Birthstone – Citrine
Citrine, the birthstone for November, is known as the 'healing quartz'. This golden gemstone is plentiful in nature and comes in the shades from yellow to brown.
Signifying prosperity, protection, strength and stability, the gem encourages hope, comfort, wealth and warmth. It was said to attract wealth and success so merchants in ancient times used to keep citrine in their vaults. This is the reason why it is also called as "Merchant's stone". Wearing citrine jewelry stimulates the brain by increasing concentration and releasing negative traits, depression and phobias. It imparts joy, pleasure, stability and delight, raises self-confidence and promotes motivation.
December Birthstone – Tanzanite
Discovered in 1960s and added in 2002 to the list of birthstones, tanzanite is officially the birthstone for the month of December. The stone has received praises for its exemplary beauty and alluring color changing property. Named after Tanzania, its place of origin, tanzanite comes in a range of colors from blue to violet. The stone helps in raising self-confidence and consciousness by dispelling lethargy and enhancing creativity. It is thought to heal heart, lungs and spleen disorders as it is a great detoxifying agent. The gem also promotes motivation, regenerates cells and strengthens immune system.
3 payments of $233.33
3 payments of $330.00
- United States
- United Kingdom
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s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696382185/warc/CC-MAIN-20130516092622-00080-ip-10-60-113-184.ec2.internal.warc.gz
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CC-MAIN-2013-20
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http://www.chegg.com/homework-help/questions-and-answers/carnot-cycle-heat-supplied-350-c-rejected-27-c-working-fluid-water-receiving-heat-evaporat-q4255208
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math
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In a Carnot cycle, heat is supplied at 350�C and rejected at 27�C. The working fluid is
water, which, while receiving heat, evaporates from liquid at 350�C to steam at 350�C.
The associated entropy change is 1.44 kJ/kg K.
If the cycle operates on a stationary mass of 1 kg of water, how much is the workdone per cycle, and how much is the heat supplied?
If the cycle operates in steady flow with a power output of 20 kW,what is thesteam flow rate?
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s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395166.84/warc/CC-MAIN-20160624154955-00153-ip-10-164-35-72.ec2.internal.warc.gz
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CC-MAIN-2016-26
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https://www.geogebra.org/m/NxFe4u2S
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math
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SSA, The Ambiguous Case for the Law of Sines
A visual explanation of the ambiguous case for the Law of Sines given Side Side Angle (SSA).[br]Sides [math]a[/math] and [math]b[/math] and [math]∠A[/math] are given and can be changed by the user using the sliders or directly inputting values.[br]The student should start by increasing the length of side [math]a[/math], going through all 3 cases.[br]Then change all values at will to see the effects.
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http://teachingmathematics.blog/background-information/aligning-instruction-with-purpose/
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math
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Whether you view mathematics as a tool for application to other investigations or as free-standing logical academic study, it is very important to realize that it is not a collection of hundreds or thousands of formulas, facts, and processes which are to be memorized.
Current mathematics instruction is such a mess that most students view it as a vast collection of disparate formulas and processes to solve a myriad of non-sensical problems.
In my blog postings, as in all previous writing and speaking, I wish to promote the philosophy that mathematics is about deductive reasoning using very general principles.
The purpose of high school and early college algebra courses is to introduce the student to the use of abstraction, generalization, problem solving techniques, deductive reasoning, and critical thinking while exploring the structure, patterns, and relationships of a variety of algebraic entities including, but not limited to, relations, unary operations, binary operations, and mathematics objects such as equations, inequalities, algebraic fractions, polynomials, and functions. Furthermore, I hope to instill a firm belief that it is better (and easier) to use a single mathematics principle to solve a multitude of problems rather than use what appears to be a random process with little apparent connection to mathematics properties for each of a multitude of problems.
I intend to use The Number Line, The Distributive Law, The Transitive Law, The Law of Trichotomy, and The Euclidean Algorithm to demonstrate how we can achieve the above stated goals in early algebra courses.
I strongly recommend that you read my essay “What is Wrong with Mathematics Instruction” found in the “Pages” section of this blog.
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CC-MAIN-2019-51
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https://www.varsitytutors.com/high_school_math-help/how-to-find-the-perimeter-of-an-acute-obtuse-isosceles-triangle
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math
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All High School Math Resources
Example Question #1 : Acute / Obtuse Isosceles Triangles
One side of an acute isosceles triangle is 15 feet. Another side is 5 feet. What is the perimeter of the triangle in feet?
Because this is an acute isosceles triangle, the third side must be the same as the longer of the sides that you were given. To find the perimeter, multiply the longer side by 2 and add the shorter side.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170864.16/warc/CC-MAIN-20170219104610-00085-ip-10-171-10-108.ec2.internal.warc.gz
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https://rprogramminghelp.xyz/cross-validation-aic-assignment-help-16056
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math
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Cross-Validation, AIC Assignment Help
The Akaike details requirement (AIC) is a procedure of the relative quality of analytical designs for a provided set of information. Provided a collection of designs for the information, AIC approximates the quality of each design, relative to each of the other designs.
AIC supplies a way for design choice. AIC wases established on details theory: it uses a relative quote of the info lost when a provided design is utilized to represent the procedure that creates the information. In doing so, it handles the compromise in between the goodness of fit of the design and the intricacy of the design.
Utilize the Akaike info requirement (AIC), the Bayes Information requirement (BIC) and cross-validation to choose an ideal worth of the regularization specification alpha of the Lasso estimator. Outcomes gotten with LassoLarsIC are based upon AIC/BIC requirements. Information-criterion based design choice is really quick, however it depends on a correct estimate of degrees of liberty, are obtained for big samples (asymptotic outcomes) and presume the design is appropriate, i.e. that the information are in fact produced by this design. When the issue is terribly conditioned (more functions than samples), they likewise tend to break.
For cross-validation, we utilize 20-fold with 2 algorithms to calculate the Lasso course: coordinate descent, as executed by the LassoCV class, and Lars (least angle regression) as executed by the LassoLarsCV class. Both algorithms offer approximately the exact same outcomes. They vary with concerns to their execution speed and sources of mathematical mistakes. It is able to calculate the complete course without setting any meta criterion. On the opposite, coordinate descent calculate the course points on a pre-specified grid (here we utilize the default). In terms of mathematical mistakes, for greatly associated variables, Lars will build up more mistakes, while the coordinate descent algorithm will just sample the course on a grid.
Akaike Information Criterion (AIC) is often utilized in the semiparametric setting of choice of copula designs, even though as a design choice tool it was established in a parametric setting. Just recently a Copula Information Criterion (CIC) has actually been particularly created for copula design choice. Among the primary factors for utilizing cross-validation rather of utilizing the standard validation (e.g. separating the information set into 2 sets of 70% for training and 30% for test) is that there is inadequate information readily available to partition it into different training and test sets without losing substantial modelling or screening ability. In these cases, a reasonable method to effectively approximate design forecast efficiency is to utilize cross-validation as an effective basic method.
The information set is separated into 2 sets, called the training set and the screening set. The mistakes it makes are collected as before to provide the mean outright test set mistake, which is utilized to examine the design. The examination might depend greatly on which information points end up in the training set and which end up in the test set, and hence the assessment might be substantially various depending on how the department is made. Each time, one of the k subsets is utilized as the test set and the other k-1 subsets are put together to form a training set. Every information point gets to be in a test set precisely when, and gets to be in a training set k-1 times. A version of this technique is to arbitrarily divide the information into a test and training set k various times.
Of the k subsamples, a single subsample is kept as the validation information for evaluating the design, and the staying k-1 subsamples are utilized as training information. The cross-validation procedure is then duplicated k times (the folds), with each of the k subsamples utilized precisely when as the validation information. The Akaike Information Criterion (AIC) is a method of choosing a design from a set of designs. The selected design is the one that lessens the Kullback-Leibler range in between the fact and the design.
In utilizing AIC to try to determine the relative quality of econometric designs for an offered information set, AIC supplies the scientist with a price quote of the info that would be lost if a specific design were to be used to show the procedure that produced the information. The AIC works to stabilize the compromises in between the intricacy of a provided design and its goodness of fit, which is the analytical term to explain how well the design "fits" the information or set of observations.
Provided a collection of designs for the information, AIC approximates the quality of each design, relative to each of the other designs. In a forecast issue, a design is generally provided a dataset of recognized information on which training is run (training dataset), and a dataset of unidentified information (or initially seen information) versus which the design is checked (screening dataset). The Akaike Information Criterion (AIC) is a method of picking a design from a set of designs. In utilizing AIC to try to determine the relative quality of econometric designs for a provided information set, AIC supplies the scientist with a quote of the info that would be lost if a specific design were to be utilized to show the procedure that produced the information. The AIC works to stabilize the compromises in between the intricacy of a provided design and its goodness of fit, which is the analytical term to explain how well the design "fits" the information or set of observations.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00051.warc.gz
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CC-MAIN-2022-33
| 5,668 | 10 |
https://www.physicsforums.com/threads/consecutive-number-problem.212346/
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math
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1. The problem statement, all variables and given/known data Given the integers from 1 to 500. How many integer sets of 15 are there? Do any of the integer sets contain consecutive numbers that are equal? 2. Relevant equations 3. The attempt at a solution I do not know what is meant by an integer set of 15. I do not know what is meant by consecutive numbers being equal.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00557.warc.gz
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CC-MAIN-2018-30
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https://dodgerocksgasmonkey.com/problemsolving-502
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math
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When b 2 − 4 a c roots. Quadratic Formula: The quadratic formula x = − b ± b 2 − 4 a c 2 a is used to solve quadratic equations where a ≠ 0 (polynomials with an order of 2) a x 2 + b x + c = 0 Examples using
Determine math tasks
To determine what the math problem is, you will need to look at the given information and figure out what is being asked. Once you know what the problem is, you can solve it using the given information.
Get detailed step-by-step solutions
Get detailed step-by-step solutions to math, science, and engineering problems with Wolfram
Doing homework can help you learn and understand the material covered in class.
Deal with math questions
Math is all about solving equations and finding the right answer.
Quadratic Equation Roots Well, the quadratic equation is all about finding the roots and the roots are basically the values of the variable x and y as the case may be. The roots are basically the solutions of the whole equation or
Mathematics is the study of numbers, shapes, and patterns. It is used to solve problems.
Enhancing your scholarly performance can be as simple as reading one extra book on the subject matter per week.
Doing math questions can be fun and engaging. It can also help improve your math skills.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00638.warc.gz
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CC-MAIN-2023-06
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https://www.worldtransitresearch.info/research/7572/
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math
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Optimal sectional fare and frequency settings for transit networks with elastic demand
economics - pricing, economics - profitability, place - asia, operations - frequency, planning - methods, planning - network design, ridership - elasticity
Fare optimization, Frequency setting, Transit assignment, Bilevel optimization, Transit network design, Sensitivity-based heuristic
Sectional fares have been used in transit services in practice but are rarely examined analytically and compared with flat and distance-based fares, especially under the considerations of path choice, elastic demand, service frequency, and profitability. This paper proposes a bilevel programming model to jointly determine the fare and frequency setting to maximize transit operator's profit. The preceding three fare structures can be incorporated into the bilevel model. To consider the path choice and elastic demand in the bilevel model, the existing approach-based stochastic user equilibrium transit assignment model for the fixed demand was extended to the elastic demand case and the resultant model was used in the lower level model. To solve the bilevel model, the sensitivity-based descent search method that takes into account the approach-based formulation for the elastic demand transit assignment is proposed, in which the approach-based formulation was solved by the cost-averaging self-regulated averaging method. Numerical studies and mathematical analyses were performed to examine the model properties and compare the three fare structures. The result of the Tin Shui Wai network instance is also provided to illustrate the performance of the solution method.
It is proven that when all passengers’ destinations are located at transit terminals, the sectional fare structure is always better than the other two fare structures in terms of profitability. For more general networks, the sectional fare structure is always better than the flat fare structure, but the choice between sectional and distance-based fare structures depends on the geometry of the network (e.g., the route structure and the distance between stops), the demand distribution, and the maximum allowable fares. It is also proven that the optimal profit (total vehicle mileage) is strictly monotonically decreasing (monotonically decreasing) with respect to the unit operating cost. Moreover, it is proven that the lower level approach-based assignment problem with elastic demand has exactly one solution. However, the bi-level problem can have multiple optimal solutions. Interestingly, it is found that from the operator's profitability point of view, providing better information to the passengers may not be good.
Permission to publish the abstract has been given by Elsevier, copyright remains with them.
Sun, S., & Szeto, W.Y. (2019). Optimal sectional fare and frequency settings for transit networks with elastic demand. Transportation Research Part B: Methodological, Vol. 127, pp. 147-177.
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CC-MAIN-2024-18
| 2,968 | 7 |
https://en.wikipedia.org/wiki/Hu_Hesheng
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math
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This article's use of external links may not follow Wikipedia's policies or guidelines. (January 2020) (Learn how and when to remove this template message)
Hu Hesheng (Chinese: 胡和生; pinyin: Hú Héshēng; born 20 June 1928) is a Chinese mathematician. She served as vice-president of Chinese Mathematical Society, president of the Shanghai Mathematical Society, and is an academician of Chinese Academy of Science. She held the Noether Lecture in 2002.
Education and career
Born in Shanghai, Hu studied mathematics at National Chiao Tung University (now Shanghai Jiaotong University) and Great China University. She received her master's degree in mathematics from Zhejiang University in 1952, under the supervision of Su Buqing. During 1952-1956, she was a researcher at the Institute of Mathematics, Chinese Academy of Sciences. 1956, she went to Fudan University in Shanghai, became a lecture in mathematics, then associate professor, and full-time professorship.
Service and recognition
Hu served as vice president of the Chinese Mathematical Society and president of the Shanghai Mathematical Society. In 1991, Hu was elected as an academician of the Chinese Academy of Sciences. 2002, She held the Emmy Noether Lecturer during the International Congress of Mathematicians in Beijing, China.
Her main academic interest is differential geometry. She led a research group at Fudan University during the 1980s and 1990s.
- Holeung Ho Lee Foundation Awardee of Mathematics & Mechanics Prize
- The Noether Lecturers Archived 2008-02-09 at the Wayback Machine Hesheng Hu, 2002* , Emmy Noether Lecture at International Congress of Mathematics
- "Hu Hesheng at Fudan University". Archived from the original on 2008-03-14. Retrieved 2008-01-18.
- Biographies of Women Mathematicians
- "Gu Chaohao & Hu Hesheng". Archived from the original on 2008-09-28. Retrieved 2008-01-18.
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CC-MAIN-2020-50
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https://coursezxyl.web.app/seabright63122jy/help-me-solve-a-math-problem-1954.html
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math
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Step-by-Step Calculator - Symbolab Solve problems from Pre Algebra to Calculus step-by-step . Learning math takes practice, lots of practice. You write down problems, solutions and notes to go How to solve Algebra Word Problems? (solutions, examples, videos) Many students find solving algebra word problems difficult. The best way to approach word problems is to “divide and conquer”. Break the problem down into smaller bits and solve each bit at a time. First, we need to translate the word problem into equation(s) with Math Homework Help - Answers to Math Problems - Hotmath
Help me solve a math problem | SQICM
Many students find solving algebra word problems difficult. The best way to approach word problems is to “divide and conquer”. Break the problem down into smaller bits and solve each bit at a time. First, we need to translate the word problem into equation(s) with Math Homework Help - Answers to Math Problems - Hotmath Math homework help. Hotmath explains math textbook homework problems with step-by-step math answers for algebra, geometry, and calculus. Online tutoring available for math help. Welcome to QuickMath Welcome to QuickMath Your step-by-step homework solver. To start working on your math problems, please click on the button below. If you need help entering a problem, click on the Help navigation button.
Algebra Calculator - MathPapa
Now, as it is still stated in the introductory paragraph of the article, the transformation problem is a mathematical one, concerning the existence of a set of functions with certain well-defined properties. Problem solving – Denise Gaskins' Let's Play Math Posts about Problem solving written by Denise Gaskins GitHub - math10/Solved-Problem Contribute to math10/Solved-Problem development by creating an account on GitHub. Problem Solving Strategies | PrivateWriting Here is an article on problem solving strategies. Problem solving strategies are a necessary part of college, work environment, and everyday life. The better you become at the different types of strategies and at determining which type to…
How to Solve Math Word Problems - HelpingWithMath.com
Solve Any Math Problem in Seconds - brighthubeducation.com An Introduction on How to Solve Any Math Problem in Seconds "Wow!" you scream. "This guy's going to teach me how to solve any math problem in seconds! That seems to good to be true. What's he really going to do?" Learning how to solve any math problem in seconds is a metaphorical combination of two societal phenomena: Help me solve word problems for math - juicedwithjess.com Factoring problem solving definition Factoring problem solving definition, solving volume problems lesson 9-5 answer key marketing strategies in business plan business plan for a nonprofit templates how to write a good college entrance essay professional help with writing a business plan english argument essay topics examples dissertation ... Help me with math problem solving problems Homework planning hseey solving exponential distribution problems conceptual framework in a dissertation sample of essay writing about environment go math homework book 3rd grade essay topics on greek mythology essay on life in a big city with quotations for 10th class boren scholarship sample essays essay topics on greek mythology how to write ... Math Problems - Math Practice for Kids - Math Blaster
How to solve Algebra Word Problems? (solutions, examples, videos)
Free Math Problem Solver - Basic mathematics Free math problem solver The free math problem solver below is a sophisticated tool that will solve any math problems you enter quickly and then show you the answer. I recommend that you use it only to check your own work because occasionally, it might generate Solve inequalities with Step-by-Step Math Problem Solver FIRST-DEGREE EQUATIONS AND INEQUALITIES. In this chapter, we will develop certain techniques that help solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For example, the stated problem "Find a number
Obviously, graphing isn't the most effective means to address a system of equations. Equations are useful to fix our everyday life issue. As stated earlier, Help me with math problem solving / excelwebtechnologies.com Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Minimum Maximum Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. Help Me Solve This Math Problem Help! - John Foster Tennis The Little-Known Secrets to Help Me Solve This Math Problem Therefore, as part of the international public good on learning, a string of tools to help countries at every stage of the assessment procedure is being developed.
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https://oneclass.com/class-notes/ca/utsg/soc/soc101y1/12274-research-methods-ii.en.html
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math
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April 6 , 2011
Research Methods II
• A sample is part of a group.
• A population is the entire group.
• A voluntary response sample is a group of people who chose themselves in
response to a general appeal.
• A representative sample is a group is a group of people chosen so their
characteristics closely match those of the population of interest.
• A convenience sample consists of people who are easiest to reach.
• If respondents are chosen at random and an individual’s chance of being chosen
is known and greater than zero, the respondents constitute a probability sample.
• A sampling frame is a list of all the people in the population of interest.
• A randomizing method is a way of ensuring every person in the sampling frame
has a known, equal, and non-zero chance of being selected.
• A mail questionnaire is a form containing questions is mailed to the respondent
and returned to the researcher through the mail system.
• The response rate
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https://photo.stackexchange.com/questions/9853/what-is-the-quantitative-relation-between-f-of-the-lens-aperture-exposure
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math
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Apertures work in steps of
sqrt(2) -- i.e., each time you increase the area of the aperture, you double the amount of light that can go through the lens in a given period of time.
For simplicity, let's start with, say, f/2 and f/5.6. The full stops in this case are f/2, f/2.8, f/4 and f/5.6. That means changing f/2 to f/5.6 decreases the exposure by three stops. To compensate for that, you need a change of 3 stops in the ISO or shutter speed (or a combination of the two).
ISO numbers and shutter speeds both work exponentially -- i.e., multiplying or dividing by a fixed factor changes the amount of light translated by a fixed number of stops. To use your example, starting from ISO 100, going to ISO 200 is a one-stop increase, to ISO 400 is a two-stop increase, and ISO 800 is a three-stop increase.
That leaves us with one minor detail to deal with: fractional f/stops, which are kind of a pain. f/4.5 is about 1/3rd of a stop slower than f/4. If you started from f/1.4 and went to f/4.5, that would be ~3 1/3rd stops, so you'd need to increase the ISO by ~3 1/3rd stops to compensate. Starting from ISO 100, that works out to ISO 1000 (200, 400, 800, 1000[3 1/3]). Technically, ISO 1000 isn't quite right, but it's close enough for any practical purpose and it's what your camera will (probably) provide.
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| 1,314 | 5 |
http://rspa.royalsocietypublishing.org/content/284/1398/325
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math
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An increase with time in resistivity occurs in plutonium at low temperatures as a result of radioactive self damage. The effect has been studied in experiments at liquid helium temperature lasting over 8000 h and has been found to exist in the $\alpha$, $\beta$ and $\delta$ phases of the metal. In all cases the resistivity approaches saturation, the rates of increase depending on the crystal structure and isotropic content of the sample. In spite of wide differences in these rates the functional increase is roughly the same. The temperature dependence of the accumulated resistivities in all three modifications deviates very strongly from a simple additive resistance obeying Matthiesen's rule. The resistivity of fully damaged $\alpha$ plutonium is almost temperature independent, becoming smaller than that of the undamaged specimen above about 50 $^\circ$K. Measurements on neptunium and uranium 233 have also shown resistivity increases with time at helium temperatures, but the effect is too small to decide whether Matthiesen's rule is obeyed in these metals. The observed effects have been discussed under the assumption that in plutonium scattering of electrons at low temperatures is mainly due to a co-operative phenomenon.
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CC-MAIN-2017-22
| 1,240 | 1 |
https://pwntestprep.com/2011/07/working-in-3-d-on-the-sat/
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math
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It’s not uncommon for a question or two involving three-dimensional shapes to appear on the SAT. Luckily, most of the time these questions either deal directly with the simple properties of three-dimensional shapes (like surface area and volume), or are just 2-D questions in disguise. It’s pretty rare to come across a truly difficult 3-D question — but you know I’m gonna give you some in this post because I care about you so.
Generally speaking, the SAT will give you every volume formula that you need, either in the beginning of the section (rectangular solid — V = lwh; right circular cylinder — V = πr2h) or in the question itself in the (exceedingly) rare case where you’ll have to deal with the volume of a different kind of solid. It’s worth mentioning, though, that the volume of any right prism* can be calculated by finding the area of its base, and multiplying that by its height.
For example, if you needed to calculate the volume of a prism with an equilateral triangle base, you’d find the area of an equilateral triangle:
And multiply that by the height of the prism:
You almost definitely won’t need this particular formula on the SAT, but it’s nice to know how to find the volume of a right prism in general: just find the area of the base, and multiply it by the height.
Most volume questions you’ll see on the SAT will require you to deftly maneuver between the volume of a solid and its dimensions. Let’s see an example (and showcase my fresh new drawing software):
- If the volume of the cube in the figure above is 27, what is the length of AF?
Remember that a cube is the special case of rectangular solid where all the sides are equal, so the volume of a cube is the length of one edge CUBED:
So far, so good, right? Now it’s time to do the thing that you’re going to find yourself doing for almost every single 3-D question you come across: work with one piece of the 3-D figure in 2-D.
The segment we’re interested in is the diagonal of the square base of the cube. If we look at it in 2 dimensions, it looks like this:
The diagonal of a square is the hypotenuse of an isosceles right triangle, so we can actually skip the Pythagorean Theorem here since we’re so attuned to special right triangles. AF = 3√2. That’s choice (B).
The surface area of a solid is simply the sum of the areas of each of its faces. Easy surface area problems are really easy. Trickier surface area problems will often also involve volume, like this example:
- If the volume of a cube is 8s3, which of the following is NOT a value of s for which the value of the surface area of the cube is greater than the value of the volume of the cube?
Yuuuuck. What to do? Well, to find the surface area of a solid, you need to know the areas of its faces. To find those areas, you need to know the lengths of the sides of the solid. Luckily for us, it’s pretty easy to find the lengths of the sides of this cube, since we know that the volume is 8s3. Take the cube root of the volume to find the length of one side of the cube:
If a side of the cube is 2s, then the area of one face of the cube is (2s)2, or 4s2. There are 6 sides on a cube, so the surface area of the cube is found thusly:
From here, it’s trivial to either backsolve, or solve the inequality spelled out in the question:
The answer must be (E), the one choice for which the inequality is NOT true.
* Right circular cylinders and rectangular solids are both special cases of right prisms — a right prism is any prism whose top lines up directly above its bottom.
Break it down.
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The last one seems a bit tough, even for a #20…
You’re right that it’s tough, but I don’t think it’s outside of the realm of possibility. See #19 on page 401 of the Blue Book for another 3-D question that actually requires a bit of 3-D reasoning and a similar number of steps. That one provides a diagram, but I actually think it’s a bit tougher than my #20.
There’s also pg 548, #16, which is eerily similar, considering I just made the question up without consulting the Blue Book and am only now looking for similar problems. :
Thank you for the references, I will look them up tonight.
I think the problem would be easier if you gave the volume and asked for the radius. Going the other way around seems to require more steps and more advanced reasoning. OTOH, I may have just gone the long way without realizing it.
pg 401 took me 3 minutes including rationalizing the denominator and then realizing the answer choice wasn’t there and having to go back and un-rationalize the denominator to find the answer (what’s up with that? very poor form). Figure not drawn to scale and the whole e=m thing are only there to obfuscate but it consumes time to digest these things. Great example of College Board using tricks that have nothing to do with math. What is up with question #20 on that page?? That is difficulty 2 at the most, but it is listed as a 5.
pg 548 is exactly what I suggested above! Of course this only took as long as it takes to read the question because I had just derived the whole x : x sqrt 2 : x sqrt 3 relationship between the three legs of the right triangle in question. But if I didn’t have the head start I think it still would have been much quicker than your example. You are given ‘boo/boo square root of 2’ and the only trick is to correctly visualize the right triangle inside the cube. In your example, you had to accomplish the visualization first and then work backwards into ‘boo/boo square root of 2’. It is tricky algebra (including exponents) on top of tricky visualization vs just tricky visualization in the CB example. Again, maybe I just missed a quicker way to do your example so I’m not 100% sure. (fyi I wasn’t timing but I think it took me almost 10 minutes to solve your problem)
Thanks again for the great problems!
Hmm…I’m not sure how you solved mine, but I think the tricky step exists in both directions: recognizing that the diameter of the sphere connects two opposite corners of the cube, and then determining the relationship between that diagonal and the length of the cube’s side.
What I was trying to get at with the question is that often, on the hardest 3-D questions, what you’re really doing is doubling up on the Pythagorean Theorem — using it once in the plane of a solid’s side, and then using the value found that way to form another right triangle that cuts through the solid.
Not too bad (although certainly not trivial) once you know to look for it, but brutal the first time you encounter it. I wonder, if you had encountered those two BB questions before mine, if mine still would have taken 5 mins.
As for #20 on pg 401, I agree. Pretty easy stuff. It’s important to remember that some questions make it to #20 because a lot of kids get them wrong, not because they’re hard. Symbol Functions are another great example of this.
Students are notoriously imprecise when it comes to the word “percent.” The questions is easy if you translate “k percent” to mean “k/100,” but I’ve seen countless kids get stymied by this kind of question because they’ve become so accustomed to “moving the decimal point” that they forget WHY that is done, and they don’t know how to move a decimal point on a variable.
You may be right, the way I did your problem was kinda weird.
First I figured out I was looking at a right triangle with the sides being – edge : diagonal of face : diagonal of cube (this took a little while, I had to draw a diagram which I never do)
But given the diagonal of cube = 8, I didn’t know how to figure out the length of the legs, so I used the only method I KNEW would work…
I set the side edge to 4 (arbitrarily), then realized the diagonal of the face would be 4 sqrt2 and the cube diagonal would be 4 sqrt3.
I then tested another edge size to make sure the x : x sqrt2 : x sqrt3 held.
Then I solved for x by setting the third formula equal to 8.
x sqrt3 = 8
x = (8 sqrt3) / 3
x = edge of the cube so the volume just equals (8 sqrt3)/3 cubed
That whole thing with setting the edge equal to 4 and testing the relationship between the legs is not necessary when the question is asked the other way around because you are starting out knowing the edge. I now see how I could have solved for x directly without the abstraction but this was not obvious at the time.
Could you explain number 20 please? I found this, but I still don’t understand:
http://answers.yahoo.com/question/index?qid=20080904045041AAQ2J0f I lose him at the, ”
the distance from the centre of a face to the corner = x/√2″ part… Thanks.
It’s hard to explain this one without diagrams, and it’s impossible to draw diagrams in the comments, so I’m going to attach an image of the solution I put in my book for this question. If this doesn’t help, let me know, and I’ll try to clarify the parts that are confusing.
Thank you so much. That was very helpful.
I lost you after the 45 45 90 triangle was shown. I’m not sure how it becomes 8 Sroot2 and S?
8, S, and S√2 are the sides of the right triangle that’s drawn in dotted lines in the first drawing. Does that help?
Oh! I see that now. We were trying to find the diagonal of the square. I was about to say I couldn’t see the 30 60 90, but I can’t argue with the math. Thank you very much!
Can you please explain number 18?
Like so many 3-D questions, this one is really a right triangle question. First, picture the two ants as far apart as they can possibly be from each other. They need to be on opposite rims of the cylinder, first of all. They also need to be on opposite sides of the tube…like if it’s held horizontally then one is all the way top right, and the other is all the way bottom left. Check out the picture below, since I’m clearly having a hard time explaining this with only words.
Once you’ve got the picture in your head, you’re ready to solve. The distance between the ants is the hypotenuse of a right triangle with legs of 12 and 5. If you know your Pythagorean Triples, you know the hypotenuse of that triangle has to be 13. If you don’t, just do the theorem! 🙂
here’s that pic I forgot to attach above…
Wow! You’re a life saver! 😀
Wouldn’t you account for the curve as well? Like when measuring distance on the earth, you don’t draw a line thru it from the US to china but you take a portion of its circumference. Wouldn’t the same apply here as well?
Making the equation 12^2 + (2.5pi)^2 = c^2
No. Here, you just cut right through. It’s a cardboard tube! You’re not burrowing through earth.
I was about to ask how can you get the answer C. But, I realised I didn’t carefully look at “Prism’s base”.. I always do such mistakes and end up with 1 or 2 mistakes..Hopefully I’ll get 800 this time.. and, 18 was easy..for 20, If one can imagine that figure in the head, and little bit of geometry basics, the job is done..:D
I’ve been staring at 19 for about 45 minutes, and I know it’s got to be something much simpler than the paths I’ve tried. I really need to work on the “be nimble” philosophy. Anyway, can you put me in the right direction?
This is one of my favorite questions! Here’s a clue: the surface area is base + base + side + side + side. But the expression the question gives you only adds 4 things…
It seems like ab has to be bh or hb, which makes sense because the whole base of one triangle times the height of one triangle would give you the area for both. I don’t know where to go from here though, as I can’t figure out which leg is the base, or understand how the height could equal one of the legs. I must be on the wrong track 🙁
Another hint: since d appears in 3 of the terms, it must be the height. So a, b, and c, are the sides of the triangular bases.
I finally got it! I hope I used a reliable method. I ended up looking at the 30-60-90 special right triangle to get the answer. This may seem like a silly question and I’m sure I just need to brush up on my basic geometry, but how was I supposed to know there was a 90 degree angle in there? Visually it doesn’t seem like any of the sides form a right angle to me. Do right prisms always have right angles somewhere in their bases?
Awesome! The way you figure out it’s a 90º angle is that only a right triangle’s area is half of the product of two of its sides. Any other triangle, you’d need another constant in the expression. There’d be the sides of the triangle, and then the “height” of it. Since you ONLY have the sides of the triangle in an expression for its area, it must be a right triangle.
This is, admittedly, a difficult leap, and you can safely assume the SAT won’t make you do something QUITE as hard on test day. But I love this question too much to make it any easier. 🙂
I didnt get question no. 19. Can you pleasee explain ?
This is a pretty tough problem. The first step is to realize that the surface area should be base + base + side + side + side, but the expression you’re given, ab + ad + bd + cd, only has 4 terms. So there’s something funny going on here.
Since 3 terms contain a d, you can conclude that d is your height—it’s multiplied by each side of the base to create the three vertical sides.
Now you’re left with the fact that ab has to represent both the top and bottom face. Remember that the area of a triangle is ½bh, and you’ve got 2 congruent triangles’ areas represented by ab. That must mean each of them is ½ab, making the base a right triangle with legs a and b. So the hypotenuse, which is what we want, must be c.
For the longest time i was wondering why it can’t be side D and then i realized its asking for a side of the prism’s base.
i want to improve my score from 580 to 650 🙁 any advice? for num20#
8=square root of x^2+x^2+x^2
then use the volume equations
To improve from 580 to 650, concentrate on getting all the easy and medium difficulty questions right. You can score a 650 without answering many of the hard questions. For more information, see this post.
your website is so helpful. huge thanks for maintaining this blog, i realise how much is it of a hard work. having my sat in 3 days, hopefully i’ll do well…
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https://www.mun.ca/computerscience/departmental-events/seminar-random-cnf-formulas-are-hard-to-refute-in/
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math
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Seminar: Random CNF formulas are hard to refute in Cutting Planes
University of Toronto
Random CNF formulas are hard to refute in Cutting Planes
Department of Computer Science
Friday, August 3, 2018, 12:00 pm, Room EN 2022
Random SAT formulas form one of the most important testbeds of hard examples for AI and SAT algorithms. The conjectured hardness of certifying the unsatisfiability of random SAT formulas has been connected to many areas in theoretical computer science. In 1988, Chvátal and Szeméredi proved that for k>=3, with high probability, a randomly chosen k-CNF formula requires an
exponential number of lines to refute in Resolution. Since then, it has been an open problem to show the conjectured hardness of refuting a random formula.
This work makes progress towards resolving the question of refuting random k-CNF formulas for the Cutting Planes proof system. We prove that random O(log(n))-CNF formulas require an exponential number of lines to refute in Cutting Planes. This result was proved independently as well by Pudlak and Hrubeš.
I aim to make this talk self contained, although an understanding of NP-completeness will be useful. The first half of the talk will be an introduction to proof complexity, the study of short certificates of unsatisfiability, with a focus on the Cutting Planes proof system. In the second half, I will sketch how to obtain the lower bound on the length of Cutting Planes refutations of random
formulas, by proving an equivalence between strengthening of Cutting Planes proofs and monotone circuits.
The work presented here follows the paper Random Θ(log n)-CNFs are hard for Cutting Planes, and is joint work with Denis Pankratov, Toniann Pitassi, and Robert Robere.
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https://slogix.in/cloud-computing/a-pareto-based-approach-for-cpu-provisioning-of-scientific-workflows-on-clouds/
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math
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Research Area: Cloud Computing
As of recently, cloud providers have started offering CPU resources that can be selected from a wide range of different CPU frequencies. CPU resources at higher frequencies have a higher price than CPU resources at lower frequencies that are available at a lower price. When executing applications, such as large scientific workflow applications, multiple CPU resources are required. In this case, this new pricing scheme allows users to choose from a large number of possible CPU configurations that may include relatively fast and relatively slow CPUs. However, such an option raises the problem of how to select appropriate CPU frequency configurations that strike a good balance between cost and execution time performance. As the search space is large with a wide range of choices that have different trade-offs, the problem becomes how to choose Pareto-efficient solutions with respect to execution time and (monetary) cost to use the (CPU) resources. This paper proposes an algorithm to efficiently explore alternative CPU configurations for a given number of resources and identify Pareto-efficient solutions for cost and execution time trade-offs. The algorithm is evaluated through simulation using three different pricing models to charge for CPU provisioning according to the allocated CPU frequency and four widely used scientific workflow applications.
Author(s) Name: Ilia Pietri and Rizos Sakellariou
Journal name: Future Generation Computer Systems
Publisher name: ELSEVIER
Volume Information: Volume 94, May 2019, Pages 479-487
Paper Link: https://www.sciencedirect.com/science/article/abs/pii/S0167739X18310264
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CC-MAIN-2023-50
| 1,679 | 7 |
https://jacobbradjohnson.com/find-quadratic-formula-343
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math
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We'll provide some tips to help you choose the best Functions solver for your needs. Math can be a challenging subject for many students.
The Best Functions solver
Apps can be a great way to help learners with their math. Let's try the best Functions solver. Quadratic equations are a common type of algebraic equation that can be difficult to solve. However, there are a number of Quadratic equation solvers that can help to make the process easier. These solvers will typically provide a step-by-step solution, making it easy to see how to solve the equation. In addition, some Quadratic equation solvers will also provide a visual representation of the solution, which can be helpful in understanding the concept. There are a number of Quadratic equation solvers available online, and many of them are free to use.
How to solve perfect square trinomial. First, identify a, b, and c. Second, determine if a is positive or negative. Third, find two factors of ac that add to b. Fourth, write as the square of a binomial. Fifth, expand the binomial. Sixth, simplify the perfect square trinomial 7 eighth, graph the function to check for extraneous solutions. How to solve perfect square trinomial is an algebraic way to set up and solve equations that end in a squared term. The steps are simple and easy to follow so that you will be able to confidently solve equations on your own!
There are a variety of websites that offer help with math word problems. Some of these sites provide step-by-step solutions, while others simply give the answer. However, there are a few things to keep in mind when using these websites. First, make sure that the site you're using is reputable. There are many fake sites out there that will give you incorrect answers. Second, be sure to read the instructions carefully. Many sites require you to input specific information, such as the type of problem and the variables involved. Finally, take your time and double-check your work. With a little patience and effort, you should be able to find a website that will help you solve even the most difficult math word problem.
3 equation solver is a mathematical tool that can be used to solve three simultaneous equations. It is a simple and straightforward tool that can be used by anyone with a basic understanding of algebra. The 3 equation solver can be used to solve equations that have one unknown variable, two unknown variables, or three unknown variables. It can also be used to solve equations that are linear or nonlinear. 3 equation solver is a valuable tool for students, teachers, and professionals who need to solve three simultaneous equations.
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| 2,641 | 6 |
https://xanadu.math.utah.edu/graduate/mathed/mathteaching-resources.php
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math
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MATHEMATICS TEACHING RESOURCES
To teach mathematics well one must have a real interest in the subject along with a strong grasp of mathematical ideas well beyond the level at which one teaches. In addition, one must have a good understanding of how young people learn and think about mathematics, what their questions and problems are at different stages of growth. It is our goal to provide you with these skills in our program.
If you are interested in attending graduate school, you may want to consider the Master of Science in Mathematics Teaching.
PROFESSIONAL DEVELOPMENT AT TH
- Math Teachers' Circle
- Summer Workshop for Teachers
RESOURCES FOR TEACHING MATHEMATICS
The internet is a wonderful resource for information related to teaching m athematics. The level of information provided varies over all grades K-12, through college or university, and beyond. There are sites that present topics in mathematics, pedagogical issues and questions, lesson plans, and software reviews. The sites listed below have been selected for their depth of coverage, clear and concise format, and the usefulness of the links provided by the site. The list has been kept small on purpose but should offer you a good starting point in your search for information on teaching mathematics.
Shodor: interactive activities and instruction materials. The Math Forum is an excellent site for all grade levels. UtahLink provides information for teachers on classroom materials, lesson plans, the Utah State Core Curriculum, professional development opportunities as well as links to individual school web pages. The MacTutor History of Mathematics Archive is one of the best collections on the history of mathematics. Math Archives offers information on topics in mathematics as well as a database of teaching materials and software. NASA Internet Educational Resources and MathMol is a large and well-organized site with on-line lessons and tutorials on topics illustrating applications of mathematics in the areas of biology, chemistry, and physics.The Gateway provides a large searchable collection of mathematics materials. PBS MATHLINE
Mathematics Teaching in the Middle School is an official journal of the NCTM intended as a resource for middle school students, teachers, and teacher educators. The Mathematics Teacher The journal's focus is the improvement of mathematics instruction in grade 8 through two-year and teacher-education colleges. The Prime Pages is a wonderful site for information about prime numbers. Sieve of Eratosthenes, Prime Machine, Platonic Solids are some of the interesting applets along with useful background information on each of these topics on this site developed by Peter Alfeld, Mathematics Department, University of Utah. There is also a nice study guide for students, Understanding Mathematics andCHANCE is a wonderful site for anyone teaching statistics and probability. The Geometry of the Sphere is a site which examines the geometry of a sphere in contrast to plane geometry.
Creative Publications (K-8), Summit Learning (K-12), Tessallations: Creative Combinations of Math, Art, and Fun (K-12), Key Curriculum Press (9-12), Sunburst Technology (K-12), Geogebra is a freeware useful for teaching algebra, geometry and calculus. The site has Wiki pages with lesson plans and other ideas. Online courses is a collection of various courses: open source.
If you are interested in attending graduate school, you may want to consider the Department of Mathematics' Master of Science in Mathematics Teaching.
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| 3,533 | 12 |
https://chathamtownfc.net/how-to-find-the-height-of-a-flagpole-using-its-shadow/
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math
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You are watching: How to find the height of a flagpole using its shadow
I"m confused about the provided problem over because ns can"t determine yet the length of the shadow, and likewise I"m grounding at the angle of depression, in i beg your pardon I claimed that it has to be the angle of elevation instead.
enter image description here
edited january 22 "18 at 22:08
asked january 22 "18 at 21:42
344 bronze badges
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First, we draw a flagpole, $x$m long. Climate we attract a shadow that is $x-3$m long. Climate the sunlight ray i m sorry casts the shadow connects the optimal of the pole and TOP of the shadow. This means the angle created by the hypotenuse (the sunlight ray) and the zero is $58^o$. Currently we have $ an 58^o=fracxx-3$, offering us $x an 58^o-3 an 58^o=x$. We lug $x$ to one side, having $xleft( an 58^o-1 ight)=3 an 58^o$. $x=7.997...=8.00left(3sf ight)$.
edited Feb 19 "20 in ~ 8:48
93744 silver- badges1313 bronze title
answered january 22 "18 at 21:52
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https://rd.springer.com/chapter/10.1007/978-1-4899-0004-3_12
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In this final chapter we touch on a variety of topics of special interest The Kaiman recursions have had a profound impact in time series analysis and in many related areas. In Section 12.1 the basic recursions are derived and applied to ARMA processes with observational noise. A similar analysis is used in Sections 12.3 and 12.7 to deal with data having unequally spaced (or missing) values. In Section 12.2 we consider transfer function models, designed to exploit, for predictive purposes, the relationship between two time series when one leads the other. Section 12.4 deals with long-memory models, characterized by very slow convergence to zero of the autocorrelations ρ(h) as h → ∞. Such models are suggested by numerous observed series in hydrology and economics. In Section 12.5 we examine linear time-series models with infinite variance and in Section 12.6 we briefly consider non-linear models and their applications.
KeywordsThreshold Model ARMA Model Transfer Function Model Infinite Variance ARMA Process
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Can someone please help me out with this practice test and provide explanations?
So for number 1 a, when r<R, I take +Ze as q and plug it into the equation E = kq/r2 and get E = kZe/R2. Not sure if that's correct
Then for number 1 b when r>R, I assume that E is zero since you set a Gaussian surface that encloses the entire thing so you take the enclosed charge which would be zero since it is +Ze and -Ze which cancels. So when you plug that in to solve for E, you would get zero. Please lmk if this is correct.
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Journal of Medical Engineering
Volume 2013 (2013), Article ID 193578, 15 pages
A MATLAB-Based Boundary Data Simulator for Studying the Resistivity Reconstruction Using Neighbouring Current Pattern
Department of Instrumentation and Applied Physics, Indian Institute of Science Bangalore, Bangalore, Karnataka 560012, India
Received 30 August 2012; Revised 14 December 2012; Accepted 28 December 2012
Academic Editor: Nicusor Iftimia
Copyright © 2013 Tushar Kanti Bera and J. Nagaraju. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Phantoms are essentially required to generate boundary data for studying the inverse solver performance in electrical impedance tomography (EIT). A MATLAB-based boundary data simulator (BDS) is developed to generate accurate boundary data using neighbouring current pattern for assessing the EIT inverse solvers. Domain diameter, inhomogeneity number, inhomogeneity geometry (shape, size, and position), background conductivity, and inhomogeneity conductivity are all set as BDS input variables. Different sets of boundary data are generated by changing the input variables of the BDS, and resistivity images are reconstructed using electrical impedance tomography and diffuse optical tomography reconstruction software (EIDORS). Results show that the BDS generates accurate boundary data for different types of single or multiple objects which are efficient enough to reconstruct the resistivity images for assessing the inverse solver. It is noticed that for the BDS with 2048 elements, the boundary data for all inhomogeneities with a diameter larger than 13.3% of that of the phantom are accurate enough to reconstruct the resistivity images in EIDORS-2D. By comparing the reconstructed image with an original geometry made in BDS, it would be easier to study the inverse solver performance and the origin of the boundary data error can be identified.
Electrical impedance tomography (EIT) [1, 2] reconstructs the spatial distribution of electrical conductivity or resistivity of a closed conducting domain () from the surface potentials developed by a constant current injection through the surface electrodes surrounding the domain to be imaged. Before carrying out the practical measurements on patients, it is advised to test an EIT system with a tissue mimicking model of known properties called practical phantoms [4–10]. Hence, phantoms are often required to assess the performance of EIT systems for their validation, calibration, and comparison purposes. Two-dimensional (2D) EIT (2D-EIT) assumes that the electrical current flows in a 2D space which is actually three-dimensional inside real volume conductors. Hence, the development of a perfect 2D practical phantom is a great challenge as the real electrodes always have a definite surface area, and hence the injected current signal cannot be confined in a 2D plane in bathing solution . Researchers have developed a number of practical phantoms which are three-dimensional objects, and those phantoms are designed and developed, generally, for their own EIT systems. Practical phantoms containing electrolyte (or other conducting medium) [4–10] are three-dimensional in shape and hence they will have some data error due to the three dimensional current conduction. Also, the phantoms containing electrolytes (e.g., NaCl solution or saline) [5, 7, 8] are difficult to transport and are prone to errors since the evaporation of the water gives rise to changes in conductivity . In addition, temperature variations have a marked effect on the conductivity because the temperature coefficient is large . Therefore, the practical phantoms will have a poor stability and a gradually increasing data error over time. Network or mesh phantoms [12, 13] are compact, more stable, rugged, portable, easy to move, consistent over time, and less temperature dependent. But these phantoms need a huge number of identical electronic components properly designed in a mesh mimicking the conductivity distribution of a practical biological tissue. Furthermore, for a large tissue structure, a mesh phantom requires a huge number of very precision components. The reproduction of these kinds of phantoms having different properties is often time-consuming . The option for changing the position and property of an inhomogeneity is limited by the phantom structure and the number of elements in mesh phantom but the practical phantoms allow us to put several types of object in different positions in the bathing solution, but they produce several errors contributing to the poor signal to noise ratio (SNR) in boundary data.
Reconstructed image quality in impedance tomography depends on the errors associated with practical phantom, electronic hardware, and inverse solver performance. Image quality is largely affected by the practical phantom design parameters such as phantom geometry, electrode geometry, electrode materials, and the nature and behavior of the inhomogeneity and bathing solution. SNR is also reduced by the error contributed by current injector, data acquisition system, and signal conditioner circuits. In practical phantoms, the voltage data developed by a three-dimensional current conduction are collected form surface electrodes connected to an analog instrumentation. Therefore, it is quite confusing to identify the source of the errors responsible for poor image quality in a 2D-EIT system. In order to overcome the difficulties and limitations of practical and mesh phantoms, a MATLAB-based boundary data simulator (BDS) is developed to generate accurate 2D boundary data for assessing the EIT inverse solvers. BDS is an absolute 2D data simulator which is required to generate the errorless 2D boundary data to study and modify the inverse solver of a 2D EIT system. As the BDS is a computer program, it is free from the instrumentation errors and allows us to generate voltage profile with different types of phantom geometry, inhomogeneity and background conductivity profile, and inhomogeneity geometry (shape, size, and position). Moreover, it is absolutely stable, compact, easy to use, and easy to handle and modify for further development. Boundary data for different phantom geometries are generated in BDS, and resistivity images are reconstructed in standard reconstruction algorithm. BDS is studied to conform its suitability to use for boundary data generation with different phantom configurations which are required to assess the EIT inverse solvers.
2.1. Mathematical Modelling of EIT
EIT image reconstruction is a nonlinear inverse problem in which the electrical conductivity distribution of a closed domain () in a volume conductor is reconstructed from the surface potential data developed at the boundary by injecting a constant current signal. A low frequency and low magnitude constant sinusoidal current is injected through an array of electrodes attached to the boundary, and the boundary potentials are measured using a data acquisition system. The voltage data collected from surface electrodes are then used by an image reconstruction algorithm which reconstructs the conductivity distribution of the domain under test (DUT). The reconstruction algorithm computes the boundary potential for a known current injection and known conductivity values and tries to compute the conductivity distribution for which the difference between the measured boundary potential () and the calculated is minimum. The reconstruction algorithm is developed with two parts: forward solver (FS) [5, 15–17] and inverse solver (IS) [15–17]. Forward solver calculates the boundary potential data for a known current injection and known conductivity values. Inverse solver computes the conductivity distribution for which the boundary voltage difference () becomes minimum.
The DUT will have the distinct conductivity values at each points defined by their corresponding coordinates (). Due to a constant current injection, a potential profile is developed within DUT, and its potential profile without any internal energy sources depends on the conductivity profile. Hence, a relationship, called EIT governing equation, between the electrical conductivity () of the points within the DUT and their corresponding potential values () can be established. The governing equation in EIT [1, 2] can be derived from the Maxwell’s equation and can be represented as
To calculate the domain potential developed for a constant current injected to the DUT with a known conductivity distribution, the above equation is essentially to be solved. As the EIT governing equation is a nonlinear partial differential equation, the direct or analytical technique fails to solve it. Therefore, to calculate the domain potential, the equation is solved by developing a mathematical model called “forward model” which is derived from (1) using a numerical technique like finite element method (FEM) .
The EIT governing equation has an infinite number of solutions, and hence the FEM formulation of the EIT technique is essentially required to be provided by some boundary conditions [18–20] to restrict its solutions space. The boundary conditions are imposed into the FEM formulation of EIT by specifying the value of certain parameters (voltage or current). The parameters defining the boundary conditions may be either the potentials at the surface or the current density crossing the boundary or mixed conditions.
The boundary conditions, in which the parameters are the potential at the surface, are called the Dirichlet boundary conditions and are represented as [1, 5, 19, 20] where are the measured potentials on the electrodes.
where is the boundary, and is the outward unit normal vector on an electrode surface.
In EIT, the FEM technique is used to derive the forward model from the governing equation in the form of a matrix equation establishing the relationship between the injected current and the developed potential within a DUT. The relationship can be assumed as the transfer function of the system which is mathematically represented as a matrix called global stiffness matrix (GSM) or transformation matrix constructed with the elemental conductivities and nodal coordinates (). In EIT, FEM discretizes the DUT by a finite element mesh containing finite number of elements of defined geometry and finite number of node. FEM applied on the governing equation to derive the forward model of a DUT in the form of a matrix equation using the and nodal coordinates. In the EIT forward model, the relationship established between the current injection matrix (matrix of the applied signal) and the nodal potential matrix (matrix of the developed signal) through the transformation matrix is mathematically represented as
Now, in FEM formulation in EIT, when the current matrices and are known, and the nodal potential matrix is unknown, the forward model or the mathematical problem is termed as the “forward problem”. The procedure of calculating the by solving the forward problem (3) with known and known is termed as “forward solution”. In EIT, the forward solver first computes the potential distribution with the assumed initial conductivity distribution () with a known constant current simulation, and then the inverse solver reconstructs the conductivity distribution from the measured boundary potential data for a same constant current injection through surface electrodes. The EIT reconstruction algorithm tries to mathematically find the elemental conductivity values (conductivity distribution) for which the difference between the estimated nodal potentials () computed in the FS and the potentials measured () on the surface electrodes (for a same current injection values) becomes minimum.
The inverse solver of the EIT reconstruction algorithm is developed with a mathematical minimization algorithm (MMA) [19–22] such as Gauss-Newton-based mathematical minimization algorithm (GN-MMA). In GN-MMA, the conductivity update vector () is calculated and the boundary data mismatch vector () is minimized by an iteration technique like the modified Newton-Raphson iteration technique (NRIT) [19–22]. The matrix is the desired variation in the elemental conductivity values in matrix for which the forward solver calculates the boundary potentials more similar to the measured value in next iteration using NRIT. Therefore, the algorithm starts with an initial elemental conductivity vector (), and it is then updated to () in the next iteration. Using this , FS calculates a new potential distribution in DUT and a new voltage mismatch vector is thus obtained and compared with the previous voltage mismatch vector . If the is not found as the minimum, the iteration process is continued till the kth iteration using the conductivity update vector () developed by GN-MMA. Using, NRIT the matrix is iteratively updated to and repetitively tries to find out the minimum value of .
Hence, in the EIT inverse solver, it is understood that the desired elemental conductivity matrix is obtained by a minimization algorithm (MMA) which is composed of Gauss-Newton method and Newton-Raphson iteration in which the technique iteratively tries to find out an optimum conductivity distribution for which the voltage mismatch vector is minimized . At a particular iteration in this MMA, the elemental conductivity matrix is calculated when the current matrices and or are known. This process is logically an opposite process to the forward problem. Thus, when the current matrices and are known, and the elemental conductivity matrix is unknown, the model or the problem is called the “inverse problem.” The procedure of calculating the or using with known and the known is termed as “inverse solution.”
2.2. Image Reconstruction with GN-MMA and NRIT
Electrical conductivity imaging is a highly nonlinear and ill-posed inverse problem [19–22]. In EIT, a minimization algorithm is used to obtain an optimized elemental conductivity value for which the voltage mismatch vector becomes minimum. In the image reconstruction process, the minimization algorithm [17, 18] first defines an objective function () from the computational predicted data and the experimental measured data and runs iteratively to minimize it. Generally, in the EIT image reconstruction algorithm, the inverse solver searches for a least square solution of the minimized object the function () using by a Gauss-Newton method and the NRIT-based iterative approximation techniques.
If is a function mapping a t-dimensional (t is the number of element in the FEM mesh) impedance distribution into a set of M (number of the experimental measurement data () available) approximate measured voltages, then the Gauss-Newton-method-based minimization algorithm [19–26] tries to find a least square solution of the minimized object function (s) [19–26] which is defined as:
Now, differentiating (4) with respect to the conductivity , it reduces to where the matrix is known as Jacobin matrix [19–22], which may be calculated by a method as described in [19, 22] or by the adjoint method represented by (6) where is the forward solution for a particular source location, and is the forward solution for the adjoint source location (source at the detector location and detector at the source location).
Differentiating (5) with respect to again, the equation reduces to
By Gauss-Newton method, the conductivity update vector is given by
In general, using NRIT method, the conductivity update vector expressed as in (10) can be represented for kth iteration (where is a positive integer) as where and are the voltage mismatch matrix and Jacobian matrix, respectively.
The matrix in (11) is always ill conditioned [19–24], and hence small measurement errors will make the solution of (11) changes greatly. In order to make the system well posed, the regularization method [19–26] is incorporated into the reconstruction algorithm by redefining the object function [19–26] with regularization parameters as where is the constrained least-square error of the regularized reconstructions, is the regularization operator, and (the positive scalar) is called the regularization coefficient [19–26]
Differentiating the inject function in (12) with respect to the elemental conductivity: the following relations are obtained
Now, using Gauss-Newton- (GN-) method-based minimization process, the conductivity update vector is obtained as
Replacing by and by (identity matrix) (21) reduces to
where the matrix is the Jacobin as stated earlier.
Thus, the conductivity update vector () is found as
Sometimes, the last term () is neglected , and the conductivity update vector is calculated as
In general, the EIT image reconstruction algorithm provides a solution of the conductivity distribution of the DUT for the kth iteration as
The EIT algorithm starts with the solution of FP obtained from the EIT governing equation, and the is calculated for a known current injection matrix and an initial guess (known or assumed) conductivity matrix . The voltage mismatch matrix is estimated, and then it is used to calculate the conductivity update matrix using GN-MMA and is added to the initial conductivity matrix () to update it to a new conductivity matrix using NRIT. New update matrix is used in forward solver to obtain a new calculated boundary data matrix which provides a new voltage mismatch matrix . Therefore, the NRIT algorithms iteratively calculate the using GN-MMA to find out an optimized matrix for which the reaches its minimum value. Thus, the EIT reconstruction algorithm is found to work in the following sequences:(1)forward solver calculates the boundary potential matrix for a known current injection matrix and an initial guess (known) conductivity matrix ,(2)measured voltage data matrix is compared with to estimate the as ,(3)Jacobian () is computed,(4)conductivity update vector is calculated by Gauss-Newton-based minimization algorithm,(5) matrix is updated to a new conductivity matrix by adding to using Newton-Raphson iteration technique (NRIT),(6)new update matrix is used in forward solver to calculate the new voltage mismatch matrix ,(7)check whether the is minimum or not or compare the with a specified error limit () if provided,(8)stop the algorithm if condition is achieved, otherwise repeat the steps 1 to 7 until the specified stopping criteria () is achieved.
2.3. Boundary Data Simulator (BDS)
A two-dimensional boundary data simulator (BDS) is developed in MATLAB R2010a using finite element method (FEM) to generate accurate boundary data for studying the EIT reconstruction algorithms. The MATLAB-based BDS is developed as an absolute 2D data simulator for EIT image reconstruction studies, and it is used suitably to generate the errorless 2D boundary data to study and modify the inverse solver of a 2D EIT system. As BDS is developed in a computer software, it is found free from errors produced by the EIT instrumentation and phantom. BDS also allows us to generate boundary potential data for different type of phantom geometry, inhomogeneity geometry (shape, size, and position), inhomogeneity conductivity profiles, and background conductivity profiles. Moreover, it is developed as a compact, absolutely stable, and easy to use and handle for EIT studies. It is developed in such a way that it can be modified for further modifications.
BDS is developed with MATLAB-based computer program consisting of four-part imaging domain simulator (IDS), EIT model developer (EMD), current injection simulator (CIS), and boundary data calculator (BDC). Imaging domain simulator (IDS) in BDS simulates a domain with inhomogeneity with their corresponding conductivity distributions. EIT model developer (EMD) derives a mathematical model of the forward solver by applying FEM on the governing equation of the DUT in the form of a matrix equation. Current injection simulator (CIS) simulates a constant current injection through the definite points at the domain boundary with neighbouring current injection protocol [1, 2, 28–30]. The boundary data calculator (BDC) solves the governing equation by solving the forward model and calculates the potentials at all electrodes at the domain boundary.
Imaging domain simulator (IDS) first defines a DUT with a desired area defined by a required diameter and defined with a particular coordinate system. Imaging domain simulator applies the FEM to discretize the domain with a 2D finite element mesh containing finite element of triangular elements () and finite number of nodes (). In IDS, a circular domain () to be imaged is defined with a required radius () using the Cartesian coordinate system (Figure 1(a)), and the domain is discretized with a finite element (FE) mesh (Figure 1(b)). The mesh is symmetrically composed of the first-order triangular elements with linear shape functions [18, 31]. The FE mesh is generated with the pdetool of MATLAB R2010a in such a way that it can be refined further to increase the number of elements as per the requirement. All the coordinates and parameters assigned to the finite elements and the nodes are stored in corresponding matrices. Boundary nodes are identified, and the sixteen nodes among the boundary nodes are assigned as the electrodes called the electrode nodes. Inside the domain one (or more) smaller region (regions) is (are) defined as the inhomogeneity (inhomogeneities) positioned at a particular place. The center point () of the inhomogeneity with the required shape and size is positioned inside the phantom domain by defining its center with a polar coordinate ( as shown in Figure 1(a). Single or multiple inhomogeneities are defined with their desired areas () inside the DUT, and elements within the inhomogeneity and the background are identified. The background area is defined as the area of the domain surrounding the inhomogeneity (), and the elements within the background area () are identified. The elements within the inhomogeneity are assigned with a particular conductivity called inhomogeneity conductivity, while the rest of the elements are assigned with a different conductivity called background conductivity () as shown in Figure 1(b). The assigned conductivity values of all the elements are assumed to be featured at their corresponding centroids.
EIT model developer (EMD) develops the mathematical model of the forward solver by applying FEM on the governing equation and derive the forward model of a DUT in the form of a matrix equation (3) using the elemental conductivities and nodal coordinates. The EMD establishes a relationship between the current injection matrix, (matrix of the applied signal), and the nodal potential matrix, (matrix of the developed signal), through the transformation matrix which is mathematically represented by (3). The global stiffness matrix in EIT is actually an admittance matrix that is formed using the nodal coordinates of all the elements with their corresponding conductivities. Thus, the inforward model represents the transfer function of the EIT system obtained from the governing equation by FEM formulation .
The current injection simulator (CIS) is used to simulate a constant current injection through the sixteen nodes called simulated electrodes (SE) on the domain boundary with neighbouring current injection protocol. The CIS works in a “for” loop to execute all the projections [1, 28, 30, 32] of current injection process. In BDS, a constant current injection is simulated into the DUT surrounded by the sixteen simulated current electrodes () with all the possible combination of SEI pairs, and the potential data are calculated on all the electrodes called voltage electrodes () in BDC. The current injection through a particular current electrode pair (say and ) and corresponding voltage data collection from all the possible voltage electrodes (, , , , , , , , , , , , , and ) is known as a simulated current projection (SCP). Hence, in an N-electrode EIT system, there will be N-different current projections each of which will inject current through a particular current electrode pair and collect m voltage (differential/grounded) data where m may be either equal to N or less than N depending on the EIT data collection strategy called the current pattern [1, 28, 30, 32]. Therefore, a complete scan (containing all the current projections) conducted on the DUT yields voltage data. As the BDS is studied for sixteen electrode system, the CIS runs for sixteen times and provides sixteen current projections (, , , , , , , , , , , , , , , and ). Therefore, a complete data collection procedure (called a complete scan) in the BDS collects m voltage data from the voltage electrodes or voltage electrode pairs in all the sixteen current projections and computes voltage data.
Boundary data calculator (BDC) calculates the potentials (developed for a constant current injection by CIS) at all electrode points (electrode nodes) at the domain boundary in each current projection for a particular current pattern. The current injection matrix is formed in CIS using the Neumann type boundary conditions, and the potential matrix is calculated from (3) using the matrix inversion technique working on L-U factorization process. The BDS is developed to run in an another “for” loop for m times to calculate the m electrode potentials from voltage electrodes or voltage electrode pairs at each of the steps of the loop. This second “for” loop runs within the first “for” loop for m times and collects m voltage data for each step of first “for” loop and hence collects voltage data as first “for” loop runs for sixteen times. Moreover, as the EIT reconstruction process needs a complete scan, the BDS runs in each current projection and computes sixteen electrode potentials at each projection. The domain potential is calculated from the forward model (3), and the potential values of all the nodes are stored in a nodal potential matrix [33, 34] denoted by . Boundary potential data are separated from and stored in a different matrix called boundary potential matrix . The electrode potential data are extracted from the nodal potential matrix and are stored in a separate matrix called electrode potential matrix . In sixteen electrode EIT system, the is formed as a column matrix and contains the electrode potentials (differential or grounded) obtained for all the projections.
2.4. Neighbouring or Adjacent Current Injection Method
In neighbouring or adjacent current injection method, first reported by Brown and Segar , the current is applied through two neighbouring or adjacent electrodes, and the differential voltages is measured successively from all other adjacent electrode pairs excluding the pairs containing one or both of the current electrodes. For a sixteen electrode EIT system with domain under test surrounded by equally spaced sixteen electrodes (, , , , , , , , , , , , , , , and ), the neighbouring method injects current through the current electrode pairs for sixteen current projections (Figure 2), and the differential voltages are measured across the voltage electrode pairs using four electrode method in each projection.
As shown in Figure 2(a) in the first current projection () of adjacent method, the current is injected through electrode 1 () and electrode 2 (), and the thirteen differential voltage data () are measured successively between the thirteen electrode pairs -, -, and -, respectively (Figure 2(a)). As reported by Brown and Segar, in neighbouring current injection method, the current density within the DUT is found highest between the current electrodes ( and for ); the current density then decreases rapidly as a function of distance . Similarly, in current projection 2 (), the current signal is injected through electrodes 2 () and 3 (), and an another set of thirteen differential voltage data are collected between the thirteen electrode pairs -, -, -, and so on. Lastly, in the current projection 16 (), the last set of thirteen differential voltage data are collected between the thirteen-electrode pairs -, -, and - by injecting the current through the electrodes and . Thus, the neighbouring current injection method in a sixteen electrode EIT system data collection procedure consists of sixteen current projections (, and ), and each of the current projection yields thirteen differential voltage data . Therefore, a complete data collection scan with the neighbouring current injection method in a sixteen electrode EIT system yields voltage measurements.
Though in neighbouring method, EIT boundary data are not collected across the electrode pairs containing one or two current electrode for contact impedance problem , but sometimes it is advantageous to collect the boundary data from all the electrodes including the current electrodes to obtain the greatest sensitivity to the resistivity changes in the domain as reported by Cheng et al. . In the present study, the boundary potentials are calculated at all the electrodes (Figure 2(b)) with respect to a virtual ground point selected within the DUT. Hence, in a complete data collection scan, the potentials on all the electrodes are collected in all the sixteen current projection and are stored in . Therefore, the is found as a column matrix containing voltage data all collected with respect to the virtual ground point of the DUT. Hence, in the present study, with neighbouring current injection method, the is found as a matrix containing 256 electrode potentials. In the present study, 1 mA current injection is simulated through the electrodes of the simulated domain containing sixteen nodal electrodes using adjacent or neighboring current injection protocol (Figure 2(b)). The potentials on all the sixteen electrodes are calculated using boundary data calculator (BDC) for all the current projections, and the electrode potential matrix is used as the calculate boundary potential matrix to reconstruct the conductivity distribution of DUT.
The BDS is designed in such a way that a huge number of voltage data sets can be generated using different types of phantoms with their different design parameters. Boundary potential data are generated for different type of phantom configurations, and the boundary data have been tested with electrical impedance tomography and diffuse optical tomography reconstruction software (EIDORS) [37, 38] for 2D-EIT. A large number of data sets are generated by changing the values of one or more phantom parameters like: phantom diameter (), inhomogeneity radius (), inhomogeneity geometry (shape, size, and position), inhomogeneity number (), bathing solution conductivity (), and inhomogeneity conductivity (). 1 mA current injection is simulated to the domain boundary, and corresponding boundary data sets are used for image reconstruction in EIDORS. Data generation in BDS and image reconstruction in EIDORS are studied for different inhomogeneity geometries in DUT. Reconstruction is also studied for different iterations and for multiple inhomogeneity reconstruction to evaluate the BDS.
3. Results and Discussion
Image reconstruction quality in EIT depends on the boundary data accuracy which is dependent on the geometric accuracy of the inhomogeneity developed in BDS. Dimensional accuracy of the inhomogeneity depends on the number of finite elements in the FE mesh or mesh refinement number () as shown in Figure 3. As the increases, the number of elements in the FE mesh is increased, and hence the geometric accuracy of the inhomogeneity increases which gives more accurate boundary data and better image reconstruction (Figure 3). But the BDS with a highly refined mesh needs a high PC memory and large computation time. In this paper, the mesh refinement is found suitable as as per the configuration of the PC (2.4 GHz/1.5 GBRAM/ P-IV) used. It is observed that the FE mesh with (containing 2048 elements and 1089 nodes) gives almost an accurate geometry (Figure 3) to the desired inhomogeneity and generates a reconstructible data set in less than 10 seconds. EIDORS reconstructs the resistivity images from the BDS data sets using regularized image reconstruction technique.
Results show that the resistivity or conductivity can be successfully reconstructed from the boundary data generated by our BDS using a circular domain () with a circular inhomogeneity ( mm, , , , and ) in the 9th iteration (Figure 4). It is also observed that the reconstructed shape of the inhomogeneity is similar to that of the original one (Figure 4(a)), and the reconstructed conductivity profile in Figure 4(b) is almost similar to that of the original object in Figure 4(a).
Iteration studies shows that in different reconstruction steps called iterations (Figure 5), the reconstructed images become more localized from iteration to iteration and the reconstruction errors (appeared by the red color at phantom periphery) are gradually reduced (Figure 5).
It is observed that the resistivity is successfully reconstructed from the boundary data in the 9th iteration (Figures 5(i) and 5(j)), though the shape of all the reconstructed images in 9th–12th iterations is almost similar to that of the original one (shown by dotted circles in Figure 5). As the reconstructed resistivity profile similar to that of the original is obtained only in the 9th iteration, the 9th iteration is taken as the optimum reconstruction. In 13th and 14th iterations, the resistivity is overestimated, and the images are lost. The optimum iteration number depends on the data accuracy and reconstruction algorithm, and hence the BDS can be used to generate the boundary data sets required for assessing the inverse solver in EIT.
Voltage data are also generated for a domain () with the circular inhomogeneities (, , and ) positioned at different places using the BDS (Figure 6). It is observed that the reconstructed image is more circular for an inhomogeneity positioned at the phantom centre where and (Figure 6(a)). On the other hand, for , that is, for the inhomogeneities near domain boundary (Figure 6(b)), reconstructed images are not perfectly circular because of the comparatively less accurate shape of the original object obtained for . For a less number of mesh refinements, the geometry of the original side objects is not exactly circular itself (Figure 4), and hence the corresponding boundary data have lower accuracy. An FE mesh with large can easily produce an accurate geometry for the boundary objects (objects near domain boundary) with proper shape, which gives a boundary data without geometric error and automatically improves the image shape.
Boundary data sets are also generated with a circular domain ( and ) with a circular inhomogeneity () with different diameters () and all positioned at the phantom center (). The boundary data are calculated and used for reconstructing the resistivity images. Results show that for the domain discretized with , the data sets, generated with a diameter larger than 13.3% of the phantom diameter, are accurate enough (Figures 7(a)–7(f)) to reconstruct the resistivity images in EIDORS-2D. It is clearly observed that for , the triangular elements within the inhomogeneity with smaller are unable to shape themselves into a proper circle (Figure 7(g)). Hence, the data obtained for the inhomogeneity with a diameter of 20 mm has low accuracy (Figure 7(g)), and hence the resistivity image (Figure 7(h)) is found with low resolution showed and some reconstruction error (appeared in the red color at phantom periphery). Increasing the FE elements in BDS, the boundary data error can be minimized, and the improved resistivity image can be achieved even for smaller inhomogeneities with a diameter less than 13.3% of .
Boundary potential data are also generated for domains () containing multiple circular inhomogeneities (, , , and ) placed at different positions inside the domain (Figure 8). Figure 8(a) shows a domain with two circular inhomogeneities ( apart from each other) which are placed at a central distance () of 37.5 mm. Similarly, another domain with three circular inhomogeneities (120° apart from each other) placed inside the phantom domain is shown in Figure 8(c). All the inhomogeneities in both the domains are positioned at a central distance () of 37.5 mm. 1 mA current is simulated with the neighbouring current pattern, and the boundary data are collected for resistivity reconstruction. It is noticed that the resistivity images (Figures 8(b) and 8(d)) of inhomogeneities in both the domains are reconstructed successfully.
Results show that the boundary data simulator can be efficiently used to generate boundary potential data for a huge number of phantom configurations in less than 10 seconds. BDS is software-based virtual EIT phantom, and hence it has a number of advantages over the practical and mesh phantoms. The literatures [39–41] presenting the phantom simulations are limited, and they only discuss the software phantoms developed for their own systems. BDS is a software-based versatile boundary data simulator which generates boundary data suitable for studying the reconstruction algorithm required for several EIT systems, and hence it is better suited for assessing the performance of the inverse solver of 2D electrical impedance tomography.
A MATLAB boundary data simulator (BDS) is developed for studying the resistivity reconstruction in inverse solvers of 2D-EIT. BDS is developed with four parts: imaging domain simulator (IDS), EIT model developer (EMD), current injection simulator (CIS), and boundary data calculator (BDC). Imaging domain simulator (IDS) simulates a domain with single or multiple inhomogeneities of different geometries defined with their corresponding conductivity distributions, whereas the EIT model developer (EMD) derives a forward model using FEM to solve the governing equation of the DUT. Current injection simulator (CIS) simulates a constant current injection through the simulated electrodes positioned at the domain boundary with the neighbouring current injection protocol. The boundary data calculator (BDC) solves the forward model to solve the governing equation and calculates the potentials at all the simulated electrodes. Boundary data are generated with different type of domains simulated in BDS by changing its input parameters. Resistivity images are reconstructed from the boundary data using standard EIT reconstruction software called EIDORS, and the BDS is evaluated. It is observed that the BDS with FE mesh with 2048 elements can simulate an inhomogeneity of desired geometry with suitable accuracy. The BDS with 2048 elements suitably generates the boundary data for simulated domains containing the objects with different geometries which are found efficient for image reconstruction in EIDORS. Results also show that the conductivity or resistivity profiles of the domains simulated in BDS are successfully reconstructed from their corresponding boundary data generated for different type of single and multiple inhomogeneities. By changing the inhomogeneity position, diameter, and number in BDS, boundary data are successfully generated as well as the resistivity images are reconstructed successfully. Multiple inhomogeneity imaging shows that the BDS suitably generates boundary data with the desired accuracy, and the boundary data are found efficient for resistivity reconstruction in EIDORS. Results also show that for the simulated domains discretized with , the boundary data sets generated for circular inhomogeneity with a diameter larger than 13.3% of the phantom diameter are accurate enough to reconstruct the resistivity images in EIDORS. Increasing the FE elements in BDS, the boundary data error can further be minimized, and the improved resistivity image reconstruction can be obtained even for smaller inhomogeneities. Hence, it is concluded that the BDS generated a number of boundary data sets which can suitably be used for inverse solver assessment in EIT.
- J. G. Webster, Electrical Impedance Tomography, Adam Hilger Series of Biomedical Engineering, Adam Hilger, New York, NY, USA, 1990.
- D. S. Holder, Electrical Impedance Tomography: Methods, History and Applications, Series in Medical Physics and Biomedical Engineering, Institute of Physics Publishing Ltd., Bristol, UK, 1st edition, 2005.
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- T. K. Bera and J. Nagaraju, “A reconfigurable practical phantom for studying the 2D Electrical Impedance Tomography (EIT) using a FEM based forward solver,” in Proceedings of the 10th International Conference on Biomedical Applications of Electrical Impedance Tomography (EIT '09), School of Mathematics, The University of Manchester, Manchester, UK, June 2009.
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- T. K. Bera and J. Nagaraju, “A stainless steel electrode phantom to study the forward problem of Electrical Impedance Tomography (EIT),” Sensors & Transducers Journal, vol. 104, no. 5, pp. 33–40, 2009.
- H. Griffiths, “A phantom for electrical impedance tomography,” Clinical Physics and Physiological Measurement, vol. 9, supplement A, pp. 15–20, 1988.
- T. K. Bera and J. Nagaraju, “Resistivity imaging of a reconfigurable phantom with circular inhomogeneities in 2D-electrical impedance tomography,” Measurement, vol. 44, no. 3, pp. 518–526, 2011.
- H. Griffiths, Z. Zhang, and M. Watts, “A constant-perturbation saline phantom for electrical impedance tomography,” Physics in Medicine and Biology, vol. 34, no. 8, pp. 1063–1071, 1989.
- H. Gagnon, A. E. Hartinger, A. Adler, and R. Guardo, “A resistive mesh phantom for assessing the performance of EIT systems,” in Proceedings of the International Conference on Biomedical Applications of Electrical Impedance Tomography (EIT '09), Manchester, UK, 2009.
- H. Gagnon, Y. Sigmen, A. E. Hartinger, and R. Guardo, “An active phantom to assess the robustness of EIT systems to electrode contact impedance variations,” in Proceedings of the International Conference on Biomedical Applications of Electrical Impedance Tomography (EIT '09), Manchester, UK, 2009.
- G. Hahn, A. Just, and G. Hellige, “Determination of the dynamic measurement error of EIT systems,” in Proceedings of the 13th International Conference on Electrical Bioimpedance and the 8th Conference on Electrical Impedance Tomography (ICEBI '07), IFMBE Proceedings 17, pp. 320–323, September 2007.
- W. R. B. Lionheart, “EIT reconstruction algorithms: pitfalls, challenges and recent developments,” Physiological Measurement, vol. 25, pp. 125–142, 2004.
- T. K. Bera, S. K. Biswas, K. Rajan, and J. Nagaraju, “Improving image quality in Electrical Impedance Tomography (EIT) using projection error propagation-based regularization (PEPR) technique: a simulation study,” Journal of Electrical Bioimpedance, vol. 2, pp. 2–12, 2011.
- T. K. Bera, S. K. Biswas, K. Rajan, and J. Nagaraju, “Improving conductivity image quality using block matrix-based multiple regularization (BMMR) technique in EIT: a simulation study,” Journal of Electrical Bioimpedance, vol. 2, pp. 33–47, 2011.
- J. N. Reddy, An Introduction to the Finite Element Method, TATA McGraw-Hill Publishing Company Ltd., New Delhi, India, 3rd edition, 2006.
- T. J. Yorkey, Comparing reconstruction methods for electrical impedance tomography [Ph.D. thesis], University of Wisconsin at Madison, Madison, Wis, USA, 1986.
- Vauhkonen Marko Electrical Impedance Tomography and Prior Information, Kuopio University Publications, Natural and Environmental Sciences, 1997.
- T. J. Yorkey, J. G. Webster, and W. J. Tompkins, “Comparing reconstruction algorithms for electrical impedance tomography,” IEEE Transactions on Biomedical Engineering, vol. 34, no. 11, pp. 843–852, 1987.
- C. J. Grootveld, Measuring and modeling of concentrated settling suspensions using electrical impedance tomography [Ph.D. thesis], Delft University of Technology, Delft, The Netherlands, 1996.
- B. M. Graham, Enhancements in Electrical Impedance Tomography (EIT) image reconstruction for 3D lung imaging [Ph.D. thesis], University of Ottawa, 2007.
- M. C. Kim, K. Y. Kim, S. Kim, H. J. Lee, and Y. J. Lee, “Electrical impedance tomography technique for the visualization of the phase distribution in an annular tube,” Journal of Industrial and Engineering Chemistry, vol. 8, no. 2, pp. 168–172, 2002.
- T. K. Bera, S. K. Biswas, K. Rajan, and J. Nagaraju, “Improving the image reconstruction in Electrical Impedance Tomography (EIT) with block matrix-based Multiple Regularization (BMMR): a practical phantom study,” in Proceedings of the IEEE World Congress on Information and Communication Technologies, pp. 1346–1351, Mumbai, India, 2011.
- T. K. Bera, S. K. Biswas, K. Rajan, and J. Nagaraju, “Image reconstruction in Electrical Impedance Tomography (EIT) with projection error propagation-based regularization (PEPR): a practical phantom study,” in Advanced Computing, Networking and Security, vol. 7135 of Lecture Notes in Computer Science, pp. 95–105, Springer, 2012.
- MATLAB: The Language of Technical Computing, Version R2010a, The MathWorks, Inc., Natick, Mass, USA, 2010.
- J. Malmivuo and R. Plonsey, Bioelectromagnetism: Principles and Applications of Bioelectric and Biomagnetic Fields, chapter 26, section 26.2.1, Oxford University Press, New York, NY, USA, 1995.
- T. K. Bera and J. Nagaraju, “Studying the resistivity imaging of chicken tissue phantoms with different current patterns in Electrical Impedance Tomography (EIT),” Measurement, vol. 45, pp. 663–682, 2012.
- T. K. Bera and J. Nagaraju, “Studying the 2D resistivity reconstruction of stainless steel electrode phantoms using different current patterns of Electrical Impedance Tomography (EIT), biomedical engineering,” in Proceeding of the International Conference on Biomedical Engineering (ICBME '11), pp. 163–169, Narosa Publishing House, Manipal, India, 2011.
- T. K. Bera and J. Nagaraju, “A FEM-based forward solver for studying the forward problem of Electrical Impedance Tomography (EIT) with a practical biological phantom,” in Proceedings of the IEEE International Advance Computing Conference (IACC '09), pp. 1375–1381, Patiala, India, March 2009.
- T. K. Bera and J. Nagaraju, “A study of practical biological phantoms with simple instrumentation for Electrical Impedance Tomography (EIT),” in Proceedings of the IEEE Instrumentation and Measurement Technology Conference (I2MTC '09), pp. 511–516, Singapore, May 2009.
- E. Kreyszig, Advanced Engineering Mathematics, chapter 18, section 18.2, John Wiley & Sons, 8th edition, 1999.
- T. K. Bera and J. Nagaraju, “Studying the boundary data profile of a practical phantom for medical electrical impedance tomography with different electrode geometries,” in Proceedings of the World Congress on Medical Physics and Biomedical Engineering, IFMBE Proceedings 25/II, pp. 925–929, Munich, Germany, September 2009.
- B. H. Brown and A. D. Segar, “The Sheffield data collection system,” Clinical Physics and Physiological Measurement, vol. 8, supplement A, pp. 91–97, 1987.
- K. S. Cheng, S. J. Simske, D. Isaacson, J. C. Newell, and D. G. Gisser, “Errors due to measuring voltage on current-carrying electrodes in electric current computed tomography,” IEEE Transactions on Biomedical Engineering, vol. 37, no. 1, pp. 60–65, 1990.
- N. Polydorides and W. R. B. Lionheart, “A Matlab toolkit for three-dimensional electrical impedance tomography: a contribution to the Electrical Impedance and Diffuse Optical Reconstruction Software project,” Measurement Science and Technology, vol. 13, no. 12, pp. 1871–1883, 2002.
- M. Vauhkonen, W. R. B. Lionheart, L. M. Heikkinen, P. J. Vauhkonen, and J. P. Kaipio, “A MATLAB package for the EIDORS project to reconstruct two-dimensional EIT images,” Physiological Measurement, vol. 22, no. 1, pp. 107–111, 2001.
- M. Kuzuoplu, M. Moh'dSaid, and Y. Z. Ider, “Analysis of three-dimensional software EIT (electrical impedance tomography) phantoms by the finite element method,” Clinical Physics and Physiological Measurement, vol. 13, supplement A, pp. 135–138, 1992.
- R. P. Patterson and J. Zhang, “Evaluation of an EIT reconstruction algorithm using finite difference human thorax models as phantoms,” Physiological Measurement, vol. 24, no. 2, pp. 467–475, 2003.
- R. Davalos and B. Rubinsky, “Electrical impedance tomography of cell viability in tissue with application to cryosurgery,” Journal of Biomechanical Engineering, vol. 126, no. 2, pp. 305–309, 2004.
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https://mcq.jobsandhan.com/look-at-the-series-3-7-24-11-__-19-19-what-number-should-fill-the-blank/
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math
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Look at the series : 3, 7, 24, 11, __, 19, 19,… What number should fill the blank ? January 24, 2020 by mcq A) 15 B) 13 C) 21 D) 16 View AnswerOption – A. More QuestionsWhich city is located on the banks of the river Mula-Mutha?1.12.91 is the first Tuesday. Which is the fourth Thursday of December 91 ?We can still post credit memos for AUC after it has been fully capitalized if we allow negative APC.Spyware can result in all of the following exceptIf price of an article decreases from Rs 40 to Rs 30, quantity demanded increases from Q1 units to 7500 units. If point elasticity of demand is -1 find Q1?If the positions of the third and tenth letters of the word ‘DOCUMENTATION’ are interchanged, and likewise the position of the fourth and seventh letters, the second and sixth letters, is also interchanged, which of the following will be eleventh letter from the right end ?R enter into a partnership in the ratio 2 : 3 : 5 . After 2 months, P increases his share 20% and Q by 10%. If the total profit at the end of one year is Rs 1,90,500, what amount will ‘R’ receive at the end of year as share in profit ?From a bunch of flowers having 16 red roses and 14 white roses, four flowers have to be selected. In how many different ways can they be selected such that at least one red rose is selected?Name another word for friend?What is the role of the Emergency Change Advisory Board (ECAB)?
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573540.20/warc/CC-MAIN-20220819005802-20220819035802-00020.warc.gz
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CC-MAIN-2022-33
| 1,410 | 1 |
https://www.oreilly.com/library/view/handbook-of-computer/9780444511041/xhtml/B9780444511041500046.htm
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math
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Geometries for CAGD
Helmut Pottmann and Stefan Leopoldseder
Chapter 2 describes the fundamental geometric setting for 3D modeling and addresses Euclidean, affine and projective geometry, as well as differential geometry. In the present chapter, the discussions will be continued with a focus on geometric concepts which are less widely known. These are projective differential geometric methods, sphere geometries, line geometry, and non-Euclidean geometries. In all cases, we outline and illustrate applications of the respective geometries in geometric modeling.
Special emphasis is put on a general important principle, namely the simplification of a geometric problem by application of an appropriate geometric transformation. For example, ...
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s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00055.warc.gz
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CC-MAIN-2020-24
| 747 | 4 |
https://www.helpwithassignment.com/Solution-Library/ways-of-reducing-gender-bias-in-sampling
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math
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You want to conduct a survey of social work students here at the University. You have determined that a random sample is best and have been provided with a list of all students’ names. Given that there are less male than female students, you are concerned that a simple random sample would produce a sample of mostly female students. Describe another sampling strategy that you could use to address your concern and describe the procedure that you would follow to obtain your sample.
The question belongs to Statistics and it is about the alternatives for arriving at a sample which has equal number of participants from both gender. The solution discusses this further.
Total Word Count 130
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CC-MAIN-2019-22
| 693 | 3 |
https://www.physicsforums.com/threads/finding-velocity-and-momentum-of-a-bullet-and-recoiling-gun.866823/
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math
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A bullet leaves the barrel of a gun with a kinetic energy of 90 J. The barrel of the gun is 50 cm long. The gun has a mass of 4 kg, the bullet 10 g. (a) Find the bullets final velocity. (b) Find the bullets final momentum. (c) Find the momentum of the recoiling gun. (d) Find the kinetic energy of the recoiling gun.
KE = 1/2mv^2
p = mv
The Attempt at a Solution
(a) Since I know the kinetic energy of the bullet as it leaves the barrel, I attempted to find the velocity with this information.
90 J = (1/2)(.01 kg)v^2
which gives v = 134.16 m/s
(b) To find momentum of the bullet I plugged this value for v into p= mv, this gives the result p = 1.3416 kg m/s.
for (c) I used conservation of momentum, since the initial momentum of both the bullet and gun are zero.
0 = (1.3416 kg m/s) + (4 kg)v
solving this give v = -.3354 m/s
(d) KE =(1/2)mv^2 so the kinetic energy of the gun is KE = (1/2)(4 kg)(-.3354 m/s)^2 = .225 J
I'm not entirely sure if I solved this correctly. My main problem is that it gives the length of the barrel of the gun and I did not use this in the solution. Is this unnecessary information? or did I do this incorrectly? I thought that the 50 cm could be used to find the work done on the gun, but 50 cm is how far the bullet moved, not how far the gun moved.
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| 1,282 | 13 |
http://www.sadlier-oxford.com/math/mc_prerequisite.cfm?sp=student&grade=8&id=1807
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math
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Chapter 6, Lesson 7
6-7.A identify the slope and y-intercept of a line from an equation
Which of these lines have a slope of 3 and a y-intercept of 2?
I 2y = 3x + 2
II y = 3x — 2
III y — 2 = 3x
(a) I and II only (b) I and III only (c) I only (d) III only
6-7.B graph an equation in slope-intercept form
Graph the line whose equation is .
6-7.C write an equation of a line using slope and y-intercept
Write, in standard form, the equation of the line whose slope = and whose y-intercept = 3.
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https://www.hackmath.net/en/examples?list=1&page_num=15
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math
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Latest problems - page 15
- Ratio of two unknown numbers
Two numbers are given. Their sum is 30. We calculate one-sixth of a larger number and add to both numbers. So we get new numbers whose ratio is 5:7. Which two numbers were given?
- Mason with assistant
Mason had to complete the work in 10 days. Two and a half days later his assistant arrived. Together they completed the work in next three and a half days. How many days would he need an assistant for the same work?
- Direct route
From two different places A and B connected by a direct route, Adam (from city A) and Bohus (from city B) started at a constant speed. As Adam continued to go from A to B, Bohus turned around at the time of their meeting and at the same speed he returned t
- Cars speeds
Distance from city A to city B is 108 km. Two cars were simultaneously started from both places. The speed of a car coming from city A was 2 km/h higher than the second car. What was the speed of each car if they met in 54 minutes?
- Odd numbers
The sum of four consecutive odd numbers is 1048. Find those numbers ...
- Two numbers
The sum of the two numbers is 1. Find both numbers if you know that half of the first is equal to one seventh of the second number.
Students of high school have 10 points for each good solved task. The wrong answer is deducted by 5 points. After solving 20 tasks, student Michael had 80 points. How many tasks did he solve correctly and how many wrong?
- How old
The student who asked how many years he answered: "After 10 years I will be twice as old than as I was four years ago. How old is student?
- Cars 6
At 9:00 am two cars started from the same town and traveled at a rate of 35 miles per hour and the other car traveled at a rate of 40 miles per hour. After how many hours will the cars be 30 miles apart?
- Boys and girls 2
The ratio of boys to girls in math club is 4:3 . After 8 more girls joined the Club, the ratio become 1:1. How many members are there in the club now?
- Money 6
Lita had some money. She spent 1/2 of it on a hand bag and 1/3 of the remainder of the blouse. She had 260.00. How much money did she have at start?
- Two cuboids
Find the volume of cuboidal box whose one edge is: a) 1.4m and b) 2.1dm
- Bathroom 2
A bathroom is 2.4 meters long and 1.8 meters wide. How many square tiles 1 dm on each side are to be used to cover it?
- Pipe cross section
The pipe has an outside diameter 1100 mm and the pipe wall is 100 mm thick. Calculate the cross section of this pipe.
- Doctor 2
A doctor noted the Diastolic Blood Pressure (DBP) of a large number of patients. Later, he scrambled the data to keep the privacy of his patients. Based on the scrambled dataset, he finds that the lower inner fence is equal to 50 and the upper inner fenc
- Unknown number 24
f we add 20, we get 50% of its triple. What is this unknown number?
- Six terms
Find the first six terms of the sequence a1 = -3, an = 2 * an-1
- Parcel 4
To send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=6 pounds?
- Mail train
The speed of mail train is 1370 meter per minute. Express it in miles per hour correct to three significant figure . Given that 1 meter =39.37 inches.
There are two laths in the garage opposite one another: one 2 meters long and the second 3 meters long. They fall against each other and stay against the opposite walls of the garage and both laths cross 70 cm above the garage floor. How wide is the garag
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http://www.e-booksdirectory.com/details.php?ebook=7090
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math
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Recent Progress on the Random Conductance Model
by Marek Biskup
Publisher: arXiv 2012
Number of pages: 80
Recent progress on the understanding of the Random Conductance Model is reviewed and commented. A particular emphasis is on the results on the scaling limit of the random walk among random conductances for almost every realization of the environment, observations on the behavior of the effective resistance as well as the scaling limit of certain models of gradient fields with non-convex interactions.
Home page url
Download or read it online for free here:
by Curtis T. McMullen - Harvard University
Contents: The Sample Space; Elements of Combinatorial Analysis; Random Walks; Combinations of Events; Conditional Probability; The Binomial and Poisson Distributions; Normal Approximation; Unlimited Sequences of Bernoulli Trials; etc.
by Pierre Simon Laplace - Chapman & Hall
Classic book on probability theory. It demonstrates, without the use of higher mathematics, the application of probability to games of chance, physics, reliability of witnesses, astronomy, insurance, democratic government, and many other areas.
by Douglas Kennedy - Trinity College
This material was made available for the course Probability of the Mathematical Tripos. Contents: Basic Concepts; Axiomatic Probability; Discrete Random Variables; Continuous Random Variables; Inequalities, Limit Theorems and Geometric Probability.
by Rick Durrett - Cambridge University Press
An introduction to probability theory covering laws of large numbers, central limit theorems, random walks, martingales, Markov chains, ergodic theorems, and Brownian motion. It concentrates on the results that are the most useful for applications.
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https://elitewritings.net/create-a-chart-in-excel-that-is-a-scatterplot-of-mileage-x-axis-and-price-y-axis/
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math
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Create a chart in Excel that is a scatterplot of mileage (x-axis) and price (y-axis). Add the regression line, the regression equation, and the r-squared (in Excel, R2) value to the chart. Copy it from Excel and paste it into Word. Write a brief summary (1 paragraph) of what the chart indicates about the relationship between the two variables. Identify any outliers that you see.
V. Calculate a 95% confidence interval (CI) for both of the two variables (price and mileage). In your report, state the 95% CI for both variables and provide a brief written summary (1 paragraph) interpreting the CI’s.
VI. Copy the results of the simple linear regression hypothesis test that you conducted in Excel and paste them into Word.
VII. Write a detailed summary (3 – 4 paragraphs) of the results explaining what they mean and how they answer the question listed in part I. State whether you rejected the null hypothesis in favor of the alternative hypothesis or whether you failed to reject the null hypothesis, including the F-value and p-value for your test and your interpretation of these values. Report the Pearson correlation coefficient (r) and the goodness of fit (r2) and interpret what they tell you about the relationship in terms of strength of the correlation between the variables and the fit of the line to the data, respectively. State the linear regression equation y = a + bx + e, using price as y and mileage as x (b is the regression coefficient for mileage, which comes from the Excel output).
VIII. Conclusion: Briefly summarize (1 paragraph) what you did for this project and what you determined about the relationship between mileage and price of a used car of a particular make, model, and year. Make sure that your summary clearly answers the question asked in part I.
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https://www.thefreelibrary.com/Towards+the+equation+of+state+for+neutral+(%5BC.sub.2%5D%5BH.sub.4%5D)%2C+polar...-a0439363597
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math
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Towards the equation of state for neutral ([C.sub.2][H.sub.4]), polar ([H.sub.2]O), and ionic ([bmim][[BF.sub.4]], [bmim][[PF.sub.6]], [pmmim][[Tf.sub.2]N]) liquids.
Behavior of low-melting organic salts or ionic liquids (ILs) [1-6] in the region of phase transitions is qualitatively similar to that either for high-temperature nonorganic molten salts or long-hydrocarbon-chain organic solvents and, even, for polymer systems. Such characteristic features as negligible vapor pressure [P.sub.[sigma]](T), undefined critical parameters [P.sub.c], [[rho].sub.c], [T.sub.c] for vapor-liquid (v, l)-transition, split of liquid-solid (l,s)-boundary onto melting [P.sub.m](T) and freezing [P.sub.f](T) branches, existence of glassy states make the problem of metastability to be quite complex but vital for many potential uses of ILs. In particular, thermodynamic modeling and computer simulation of the phase behavior in mixtures formed by ILs with water and low-molecular organic solvents such as ethylene can be of great importance for the further tuning of their operational parameters. If one proceeds from a pure to a mixed fluid, it is especially advantageous to develop the same format of reference equation of state (EOS) and the common format of reference pair potential (RPP) for each component and mixture.
As a first step toward consistent modeling of the phase behavior of IL and its solution we demonstrate in this work how the fluctuational-thermodynamic (FT) EOS [7-12] and the relevant finite-range Len-nard-Jones (LJ) RPP can be applied to model the underlying structure and properties of low-molecular ([C.sub.2][H.sub.4], [H.sub.2]O) and imidazolium-based (1-butyl-3-methylimidazolium tetrafluorob orate ([bmim][[BF.sub.4]]), 1-butyl-3-methylimida-zolium hexafluorophosphate ([bmim][[PF.sub.6]]), 2,3-dimethyl-1-propylimidazolium bis(trifluoromethylsulfonyl)imide ([pmmim][[Tf.sub.2]N])) solvents. For any pure component FT-model is based either on the measurable coexistence-curve input data [P.sub.[sigma]](T), [[rho].sub.v](T), [[rho].sub.l](T) (if they are achievable as for [C.sub.2][H.sub.4] and [H.sub.2]O) or on the also measurable one-phase density of liquid at atmospheric pressure [rho] ([P.sub.0] [approximately equal to] 0,1 MPa, T) for ILs. This methodology becomes purely predictive for density [rho](P,T) in any one-phase v,l,s-regions including their metastable extensions. Only the measurable isobaric heat capacity data [C.sub.P]([P.sub.0], T) have to be added to the set of input data for prediction of other caloric properties (isochoric heat capacity [C.sub.V](P,T), speed of sound W(P,T), and Gruneisen parameter Gr(P, T)) at higher pressures P > [P.sub.0] and lower T < [T.sub.m] or higher T > [T.sub.b] temperatures where [T.sub.b] is the hypothesized normal boiling temperature [T.sub.b]([P.sub.0]). Its existence itself is a debatable question because the thermal decomposition [T.sub.d] maybe former [T.sub.d] < [T.sub.b].
Such approach was proposed recently [7,8] to reconstruct the hypothetical (V, l)-diagram of any ILs in its stable and metastable regions on the base of only standard reference data on density [rho](T) at [P.sub.0] [1-4] and one free parameter, an a priori unknown value of the excluded volume [b.sub.0]. To our knowledge this is first attempt to predict simultaneously the whole set of one-phase and two-phase properties for ILs without the fit at any other pressures including the negative ones. It was argued that the particular low-temperature variant of the most general FT-EOS [9-12] should be used to obtain the consistent prediction of volumetric properties and the standard response functions [a.sub.P], [[beta].sub.T], [[gamma].sub.[rho]] = [a.sub.P]/[[beta].sub.T] by the following equations:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)
where [b.sub.0] is the excluded molecular volume and a(T) is the T-dependent effective cohesive energy. The derivative da/dT affects the thermal expansion [[alpha].sub.P] and the thermal-pressure coefficient [[gamma].sub.[rho]] while the isothermal compressibility [[beta].sub.T] depends only on [b.sub.0]-value at the given pressure. The changeable sign of two thermal derivatives [[alpha].sub.P], [[gamma].sub.[rho]] offers a possibility to predict the properties of anomalous low-temperature substances (such as water, for example) too [7,8].
Fortunately we have obtained now [13-19] a possibility to test our predictions not only by the direct experimental one-phase data [14, 16, 18, 19] on [rho](P,T)- and W(P,T)-surfaces. Another possibility is offered by comparison of the predictions obtained by FT-EOS for the critical parameters of ILs ([bmim][[BF.sub.4]]: [T.sub.c] = 962,3 K, [P.sub.c] = 3503,9 kPa, [[rho].sub.c] = 438,565 [kg.m.sup.-3] with those predicted here by the Sanchez-Lacombe EOS for lattice fluid (LF) : [T.sub.c] = 885,01 K, [P.sub.c] = 2829 kPa, [[rho].sub.c] = 248,565 [kg.m.sup.-3] as well as with those simulated by GEMC-methodology : [T.sub.c] = 1252 K, [P.sub.c] = 390 kPa, [[rho].sub.c] = 181 [kg.m.sup.-3]. It seems that the relatively close location of ([T.sub.c],[P.sub.c])-parameters predicted by both EOSs is some guarantee of their reliability while [T.sub.c] and [P.sub.c] from are significantly overestimated and underestimated, respectively. Interestingly, the known descriptive factor of compressibility [r.sub.t] = [P.sub.c]/([[rho].sub.t][RT.sub.c]) estimated by Guggenheim in the vicinity of triple point [T.sub.t] for argon as [r.sub.t] = 0,108 is equal to close values [r.sub.t] = 0,082 for FT-EOS and [r.sub.t] = 0,072 for LF-EOS but only to very small value [r.sub.t] = 0,007 for result of GEMC-simulations if the common realistic estimate (see below) [[rho].sub.t] [approximately equal to] [[rho].sub.t] = 5,350646 [mol*dm.sup.-3] at T = 290 K is used. Moreover, it will be shown that the characteristic dimensional parameters [P.sup.*.sub.c], [T.sup.*.sub.c], [[rho].sup.*.sub.c] and another compressibility factor [r.sup.*.sub.c] = [P.sup.*.sub.c]/([[rho].aup.*.sub.c ][RT.aup.*.sub.c]) obtained by Machida et al. by the fit to (P, [rho], T)-experimental data for [bmim][[BF.sub.4]] and [bmim][[PF.sub.6]] provide the structural estimates of hard-core volume, number of lattice sites in a cluster, and energy of near-neighbor pair interactions which are surprisingly close to ones independently predicted by the FT-model of a continuum substance.
Taking into account the compatibility of above results it is important to consider the presumable similarity between the square-well fluid (which may be thought of as a continuum analogue of the lattice-gas (LG) or lattice-fluid (LF) systems) on the one hand and the LJ-fluid of finite-range interactions (RPP) on the other. This conceptual analogy has been pointed out long ago for the critical region by Widom who suggested that it is the propagation of attractive correlations in the LG which determines the peculiarities of criticality However, such unphysical LG-predictions at low temperatures of the ([rho], T)-plane as the nonexistence of a (l, s)-transition suggest that repulsive forces are not being treated properly by this RPP-model. In contrast with the discrete LG-model, it seems that both attractive and repulsive forces are being dealt with properly in the square-well continuum fluid because it exhibits both (v, l)- and (l, s)-transitions. The serious restriction of latter is however evident since any singularities of RPP imply an artificial jump of pair-distribution isotropic function g(r) at the point of cutoff radius [r.sub.c] for attractive interactions.
In this context only the shifted and smoothed at [r.sub.c]-point LJ-potential [5, 6] seems to be appropriate as RPP for a continuum system. Of course, the algebraic form of the respective reference EOS is essential too. In accordance with the statistical-mechanical arguments presented by Widom there are the set of alternative forms including the original vdW-EOS and the LG-EOS in the well-known Bragg-Williams approximation which share the common restrictive feature. One may suppose that the probability of finding some prescribed value of the potential energy [TEXT NOT REPRODUCIBLE IN ASCII] at an arbitrary point in the fluid is independent of T at fixed [TEXT NOT REPRODUCIBLE IN ASCII]. Another simplifying assumption is that such EOS supposes only two types of fluid structure, one of the excluded (or hardcore) volume [Nv.sub.0] where the singular hard-sphere branch of potential is infinite and one of free volume (V - [Nv.sub.0]) where the potential is uniform, weak, and unrestricted (an infinite-range rectilinear well). It should be directly proportional to density e = U/N = -ap where U is the total configurational energy and a is the constant vdW-coefficient. These historical notes are important to explain how one can go beyond the above restriction of T-independency by adoption of linear [rho]-dependence for a generalized specific or molar energy (see also (8) below). Consider
a (T) = - [([partial derivative]e/[partial derivative][rho]).sub.T] (5)
Another aim of the developed FT-EOS follows from the possibility to estimate the effective LJ-parameters without any fit. Indeed, their general T-dependent values,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6a)
[epsilon](T)/k = T (1- [Z.sub.l]) (6b)
are determined simply in the low-temperature range of all ILs where [b.sub.0] is constant in ((1)-(4)) while the compressibility factor of saturated liquid [Z.sub.t] = [P.sub.[sigma]]/([[rho].sub.l]RT) becomes negligible as well as the vapor pressure [P.sub.[sigma]](T) trends to zero. Taking into account this asymptotic behavior it is especially important to study the possible correlations of these parameters in the RPP-model of an effective LJ-potential for ILs as the functions of total molecular weight M. This concept is unusual for the conventional consideration of a separate influence of the anions [M.sub.a] and cations [M.sub.c] components. It may provide, in principle, the useful insight the nature of (v,l)-transition in ILs by effective capturing underlying pair interactions.
The distinction of both FT-EOS and LF-EOS from the conventional hard-sphere reference EOS is crucial to provide the quantitative description of one-phase liquid. The formers include the quadratic in density contribution, which is dominating at high pressures along the isotherms. The latter considers this term as a small vdW-perturbation for the hard-sphere EOS. Such perturbation approach is not directly applicable to associating fluids such as water and alcohols for which presence of hydrogen bonding, anisotropic dipolar 1/[r.sup.3], or coulombic 1/r interactions in addition to isotropic dispersive 1/[r.sup.6] attractions is inconsistent with the main assumption of the perturbation methodology that the structure of a liquid is dominated by repulsive forces .
The FT-model promotes the more flexible approach in which the above factors of attraction and clustering can be effectively accounted by the a(T)-dependence. It was firstly confirmed by Longuet-Higgins and Widom and, then, by many authors that a combination of Carnahan-Starling EOS, for example, with the vdW-perturbation a[[rho].sup.2] is a reasonable approximation for the l- and s-phases but not the v-phase. Guggenheim has concluded its applicability only to a liquid when large clusters are more important than small clusters (i.e., at low temperatures [T.sub.m] < T < [T.sub.b]). In contrast with this observation, the general FT-EOS provides the adequate representation of entire subcritical range [T.sub.m] < T < [T.sub.c] including the critical region and (v, l)-phase transition [9-12]. It will be shown below by FT-model without undue complexity of calculations.
2. Universal FT-EOS for Any Low-Themperature Fluids
2.1. General Form of FT-EOS for Subcritical Themperatures. It is often claimed that the original van der Waals (vdW)-EOS with two constant coefficients a, b determined by the actual critical-point properties [[rho].sub.c], [T.sub.c] is only an approximation at best and cannot provide more than qualitative agreement with experiment even for spherical molecules. However, it was proved recently [9-12] that the general FT-EOS with three T-dependent coefficients,
p = [[rho]RT [1 - c(T)]/1 - b(T) [rho]] - a(T) [[rho].sup.2], (7)
is applicable to any types of fluids including ILs. The measurable volumetric data of coexistence curve (CXC) have been used for evaluation of T-dependences without any fit. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (8)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (9)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (10)
where the reduced slope [A.sub.[sigma]](T) of [P.sub.[sigma]](T)-function is defined by the thermodynamic Clapeyron's equation:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (11)
This fundamental ratio of the (v,l)-latent heat to the thermodynamic work of (v, l)-expansion is the main parameter of FT-coefficients determined by ((8)-(9)). It should be calculated separately in each of high-temperature ([T.sub.b] [less than or equal to] T [less than or equal to] [T.sub.c]) [9-12] v- and l-phases to obtain the reasonable quantitative prediction of one-phase thermophysical properties. The general FT-EOS is applicable to the entire subcritical range ([T.sub.t][less than or equal to] T [less than or equal to] [T.sub.c]) but it can be essentially simplified to the form of (1) if ([T.sub.t ][less than or equal to] T [less than or equal to] [T.sub.b]).
2.2. Particular Form of FT-EOS for Low Themperatures. An absence of input CXC-data [P.sub.[sigma]](T), [[rho].sub.g](T), [[rho].sub.l](T) for ILs is the serious reason to develop the alternative method for the evaluation of T-dependent FT-coefficients. The thermodynamically-consistent approach has been proposed in [7, 8] for the particular form of FT-EOS (1) applicable in the low-temperature range from the triple [T.sub.t] (or melting [T.sub.m]) point up to the [T.sub.b]-point. Former one is usually known for ILs while the latter one is, as a rule, more than temperature of thermal decomposition [T.sub.b] > [T.sub.d] ~ 650 K. The methodology was tested on two low-molecular-weight substances ([C.sub.2][H.sub.4], [H.sub.2]O) and two imidazolium-based ILs ([bmim][[PF.sub.6]], [pmmim][[Tf.sub.2]N]) with the promising accuracy of predictions even for the isothermal compressibility [[beta].sub.T] up to the pressure P = 200 MPa.
To illuminate the distinction between the particular (reference) and general form of FT-EOS let us discuss in brief the main steps of the proposed procedure. Its detailed analysis can be found elsewhere [7, 8]. The algorithm is as follows. Step 1. At the chosen free parameter [T.sub.b] one determines the orthobaric molar densities [[rho].sub.l]([T.sub.b]) = [rho]([T.sub.b],[P.sub.0]); [[rho].sub.g]([T.sub.b]) = [P.sub.0]/[RT.sub.b] to solve the transcendent equation:
[[rho].sub.g]/[[rho].sub.l] = 1 + y (x) [e.sup.-x]/ 1 + y (x) [e.sup.x], (12)
for the reduced entropy (disorder) parameter x([T.sub.b]) and the respective molar heat [r.sub.[sigma]]([T.sub.b]) of vaporization. Consider
x = [([s.sub.g] - [s.sub.l])/2R], (13a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (13b)
Step 2. The universal CXC-function y(x) in (12) is determined by equalities
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (14)
and it provides the possibility to estimate a preliminary value of [b.sub.0],
[b.sub.0] = [x([T.sub.b]) - 1]/[[rho].sub.l]([T.sub.b]) [x([T.sub.b]) - 1/2], (15)
as well as to evaluate the orthobaric densities at any T if the function x(T) is known. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (16a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (16b)
Step 3 (A-variant of x(T)-prediction [7, 8]). To calculate its values one must obtain two densities [[rho].sup.[+ or -]] (at the assumption [P.sub.[sigma]] [approximately equal to] 0) from equation
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (17)
where [rho] = [rho]([T,P.sub.0]), [Z.sub.0] = [P.sub.0]/[rho]RT and the [[rho].sup.+](T)-function provides the preliminary estimate of x(T) for the low-temperature range (at the consistent assumption: [[rho].sup.+] >> [[rho].sub.g] [approximately equal to] 0). Consider
x(T) = 1 - [b.sub.0][[rho].sup.+]/2/1- [b.sub.0][[rho].sup.+]. (18)
Step 4. One substitutes x(T) from (18) in ((13a), (13b), (16a), (16b)) to calculate [r.sub.[sigma]](T), [[rho].sub.g](T), [[rho].sub.l](T), respectively. Step 5. A preliminary value of [a.sub.0](T) may be estimated then by the more restrictive assumption [P.sub.0] [approximately equal to] 0 (used also in the famous Flory-Orwoll-Vrij EOS developed for heavy n-alkanes). Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (19)
Step 6 (B-variant of x(T)-prediction). To control the consistency of methodology one may use instead of Step 3 (A-variant) the same equation (17) with the approximate equality [[rho].sub.l](T) [approximately equal to] [[rho].sup.+](T) to solve (16b) at the a priori chosen [b.sub.0]-value for determination of alternative x(T), and so forth, (Steps 4 and 5). Just this approach (B-variant) has been used below in the low-temperature range of [bmim][[BF.sub.4]].
Step 7. The self-consistent prediction of a hypothetical (v, l)-diagram requires the equilibration of CXC-pressures [P.sub.[sigma]](T, [[rho].sub.g]) = [P.sub.[sigma]](T, [[rho].sub.l]) by FT-EOS (1) with the necessary final change in [a.sub.0](T)-value from (19) to satisfy the equality
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (20)
Only in the low-temperature range T [less than or equal to] [T.sub.b] the distinction between the preliminary definition (18) and its final form (19) for a(T)-values is not essential at the prediction of vapor pressure [P.sub.[sigma]](T,[[rho].sub.g]).
3. Reference Equation of State, Effective Pair Potential, and Hypothetical Phase Diagram
To demonstrate universality of approach and for convenience of reader we have collected the coefficients of FT-EOS (1) for neutral ([C.sub.2][H.sub.4]) and polar ([H.sub.2]O) fluids [7, 8] in Table 1 and added in Table 2 to other ILs ([bmim][[PF.sub.6]], [pmmim][[Tf.sub.2]N] [7, 8]) the data for [bmim][[BF.sub.4]] obtained in this work (Table 3). When temperature is low [T.sub.m] < T < [T.sub.b] FT-model follows a two-parameter ([eplison](T), [[sigma].sub.0]) correlation of principle of corresponding states (PCS) on molecular level as well as a two-parameter (a(T), [b.sub.0]) correlation of PCS on macroscopic level.
One the most impressed results of FT-methodology is shown in Figure 1 where the comparison between such different high- and low-molecular substances as ILs and [C.sub.2][H.sub.4], [H.sub.2]O is represented. The results based on the coefficients of Tables 1 and 2 demonstrate that the proposed low-temperature model provides the symmetric two-value representation of vapor pressure [+ or -][P.sub.s](T) similar to that observed for the ferromagnetic transition in weak external fields.
To estimate the appropriate excluded molar volume [b.sub.0] (M = 225,82 g/mol) of FT-model we consider that it belongs to the range [v.sub.0] = M/[[rho].sub.0] [approximately equal to] 162, [v.sub.l] = M/[[rho].sub.l] [approximately equal to] 187[cm.sup.3]/mol]. The extrapolated to zero temperature T = 0 K "cold" volume [v.sub.0] = 162 cm /mol follows from (27). The fixed value: [b.sub.0] = 178 [cm.sup.3]/mol ([b.sub.0] [approximately equal to] 1,[1v.sub.0]) has been used in this work to demonstrate the main results of the proposed methodology. Such choice for [bmim][[BF.sub.4]] on the ad hoc basis is in a good correspondence with the respective values: [b.sub.0] = 195,3 [cm.sup.3]/mol for [bmim][[PF.sub.6]] and [b.sub.0] = 271,1 [cm.sup.3]/mol for [pmmim][[Tf.sub.2]N] where the empirical relationship [b.sub.0] [approximately equal to] 1,[1v.sub.0] was also observed [7, 8]. Our estimates of the effective LJ-diameters by (6a) for ILs: [sigma]([bmim][[BF.sub.4]]) = 5,208 A, [sigma]([bmim][[PF.sub.6]]) = 5,371 A, and [sigma]([pmmim][[Tf.sub.2]N]) = 5,992 A can be tested by comparison with the independently determined values for anions [[sigma].sub.a]([[BF.sub.4]]) = 4,51 [Angstrom]; [[sigma].sub.a]([[PF.sub.6]]) = 5,06 [Angstrom]. We have verified Berthelots combining rule for spherical molecular ions (21a) and van der Waals' combining rule for chain molecules (21b) usually considered by van der Waals'-type of EOS for mixtures . Consider
[sigma] = [[sigma].sub.c] + [[sigma].sub.a]/2, (21a)
[b.sub.0] = [b.sub.c] + [b.sub.a]/2. (21b)
The predicted by former rule of LJ-diameter for the same [bmim]-cation were close but still different, 5,906 [Angstrom] and 5,682 [Angstrom]. For the latter rule their values and distinction become even smaller, 5,757 [Angstrom] and 5,651 [Angstrom]. As a result, the chain rule (21b) seems preferable for ILs and its average value for [[sigma].sub.c][bmim] = 5,704 [Angstrom] can be used to estimate the LJ-diameter of [[Tf.sub.2]N]-anion: [[sigma].sub.a][[Tf.sub.2]N] = 6,254 [Angstrom] taking into account the equality: [M.sub.c] [bmim] = [M.sub.c] [pmmim] = 139 g/mol. The collected in Table 4 effective LJ-diameters are linear functions of [M.sub.a] in the set of ILs with different anions and cations if the molecular weight of latters [M.sub.c] is the same one.
Since the low-temperature compressibility factor [Z.sub.l](T) is very small for all discussed liquids their dispersive energies [eplison](T) (molecular attractions parameters) are comparable in accordance with (6b). However, the differences in cohesive energies a(T) (collective attraction's parameters) between the low-molecular substances ([C.sub.2][H.sub.4], [H.sub.2]O) and ILs are striking as it follows from Tables 1 and 2. The physical nature of such distinction can be, at the first glance, attributed to omitted in the reference LJ-potential influence of intramolecular force-field parameters and anisotropic (dipole-dipole and coulombic) interactions. At the same time, one must account the collective macroscopic nature of a(T)-parameter. It corresponds to the scales which are compatible or larger than the thermodynamic correlation length [xi]([rho], T). FT-model [9-12] provides an elegant and simple estimation of this effective parameter based on the concept of comparability between energetic and geometric characteristic of force field determined by the given RPP. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (22)
Taking into account the above results and the coefficients from Tables 1-3 we have used (22) at T = 300 K (T* = [k.sub.B]T/[eplison] [approximately equal to] 1) to compare the thermodynamic correlation length predicted for [bmim][[BF.sub.4]] (a = 8900,9 J*d[m.sup.3] /[mol.sup.2]; [b.sub.0] = 178 [cm.sup.3] /mol; [rho] = 5,322294 mol/d[m.sup.3]) and at T = 298,15 K for water (a = 548,27 J*d[m.sup.3]/[mol.sup.2]; [b.sub.0] = 16,58 cm /mol; [rho] = 55,444 mol/d[m.sup.3]) . The dimensional and reduced ([xi]* = [xi]/[sigma]) values for former are, respectively, [xi] = 177,7 [Angstrom], [xi]* = 34,12 while for latter [xi] = 69,86 [Angstrom], [xi]* = 29,45. No more need be said to confirm the universality of FT-model.
One may note that our estimates of correlation length are significantly larger than those usually adopted for the dimensional or reduced cutof radius ([r.sub.c] or [r*.sub.c] = [r.sub.c]/[sigma]) of direct interactions at computer simulations. As a result, the standard assumption [[xi]*.sub.c] [approximately equal to] [r*.sub.c] may become questionable in the comparatively small (mesoscopic) volumes of simulation [L.sup.3] < [xi][([rho],T).sup.3]. At this condition the simulated properties are mesoscopic although their lifetime may be essentially larger than its simulated counterpart. The key point here is the same as one near a critical point where the problem of consistency between the correlation length for statics and the correlation time for dynamics becomes crucial. In any case, the computer study of possible nongaussian nature of local fluctuations within the thermodynamic correlation volume [[xi].sup.3] may be quite useful. The relevant inhomogeneities in the steady spacial distributions of density and enthalpy can affect, first of all, the simulated values of volumetric ([[alpha].sub.P], [[beta].sub.T]) and caloric ([C.sub.p],[C.sub.v]) derived quantities. Simultaneously, an account of internal degrees of freedom and anisotropy by the perturbed RPP may change the correlation length itself.
The above described by ((12)-(20)) FT-methodology has been used to reconstruct the hypothetical phase diagram (HPD) for [bmim][[BF.sub.4]] shown in Figures 2, 3, and 4 and represented in Table 3. Both (T, [rho]) (Figure 2) and (P, T) (Figure 4) projections contain also the branches of classical spinodal calculated by the LF (Sanchez-Lacombe)-EOS obtained in . Its top is the location of a respective critical point. It seems that the relatively close ([P.sub.c],[T.sub.c])-parameters predicted independently by FT-EOS and by LF-EOS (see Section 1) are reasonable.
The FT-model provides a possibility to estimate, separately, the coordination numbers of LJ-particles in the orthobaric liquid [[rho].sub.l](T)- and vapor [[rho].sub.g](T)-phases. An ability to form the respective "friable" ([N.sub.l,g] + 1)-clusters is defined by the ratio of effective cohesive and dispersive molar energies at any subcritical temperature. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (23)
The term "friable" is used here to distinguish the clusters formed by the unbounded LJ-particles at the characteristic distance l* = l/[sigma] [approximately equal to] [cube root of 2] > 1 from the conventional "compact" ones with the bonding distance l* < 1 studied, in particular, by the GEMC-methodology to model of molecular association. It is straightforwardly to obtain the low-temperature estimates based on the assumptions.
[Z.sub.l] << [Z.sub.g] [approximately equal to] 1,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (24a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (24b)
and to find the critical asymptotics based on the difference of classical ([a.sup.0] ,[b.sup.0] ,[c.sup.0]) and nonclassical (a, b, c) T-dependent FT-EOS' coefficients [9-12]. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25b)
The crucial influence of excluded-volume in (24a) and its relative irrelevance in (24b) for [N.sub.l,g,] -predictions are illustrated by Figure 5 where [N.sub.l](T) function is shown also for the entire l-branch based on the evaluated in the present work HPD. For comparison, the low-temperature ability to form the ([N.sub.l] + 1)-clusters in liquid water [7, 24] is represented in Figure 5 too.
In according with ((25a), (25b)) the "friable" clusters can exist only as dimers in the classical critical liquid phase ([N.sub.l.sup.c] [approximately equal to] 1). It is not universal property in the meaning of scaling theory but it corresponds to the PCS-concept of similarity between two substances ([H.sub.2]O and [C.sub.4]mim [[BF.sub.4]], e.g.) if their Z.sub.c]-values are close. On the other side, the scaling hypothesis of universality is confirmed by the FT-EOS' estimates in the nonclassical critical vapor phase. For the set of low-molecular-weight substances studied in (Ar,[C.sub.2][H.sub.4], [CO.sub.2], [H.sub.2]O); for example, one obtains by (25b) the common estimate ([N.sup.c].sub.g] [approximately equal to] 2,5) which shows a significant associative near-mean-field behavior.
It is worthwhile to note here the correspondence of some FT-EOS'-estimates with the set of GEMC-simulated results. One may use the approximate estimate of critical slope [A.sub.c] [approximately equal to] 7,86 for [bmim][[BF.sub.4]] based on the similarity of its [Z.sub.c]-value with that for [H.sub.2]O . In such case, the respective critical excluded volume [b.sub.c] [approximately equal to] 220 [cm.sup.3]/mol becomes much more than vdW-value 1/3[[rho].sub.c] = [b.sub.c.sup.0] [approximately equal to] [b.sub.0] [approximately equal to] 178[cm.sup.3]/mol. Another observation seems also interesting. Authors have calculated (see Figure 3 in ) for the "compact" clusters at [l.sup.*] = 0,7; 0,5; 0,45; the ([T.sup.*], [[rho].sup.*])-diagram of simple fluids. One may note that only the value [l.sup.*] = 0,7 corresponds to the shape of strongly-curved diameter shown in Figure 2 for the HPD while the smaller values: l* = 0,5; 0,45 give the shape of HPD and the nearly rectilinear diameter strongly resembling those obtained by the GEMC-simulations for the complex ILs force-field. If this correspondence between the "friable" and "compact" clustering is not accidental one obtains the unique possibility to connect the measurable thermophysical properties with the both characteristics of molecular structure in the framework of FT-EOS.
4. Comparison with the Empirical Tait EOS and Semiempirical Sanchez-Lacombe EOS
The empirical Tait EOS is based on the observation that the reciprocal of isothermal compressibility [[beta].sup.-.sub.T] for many liquids is nearly linear in pressure at very high pressures. Consider
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (26)
where some authors [14, 19] omit the T -dependence in coefficient C and ignore the value [P.sub.0] [approximately equal to] 0 . Such restrictions transform the Tait EOS into the empirical form of two-parameter (B(T),C) PCS because the sets of C-values for different ILs become close one to another. For example, Machida et al. have found the sets C = 0,09710 for [bmim][[PF.sub.6]], C = 0,09358 for [bmim][[BF.sub.4]], and C = 0,08961 for [bmim][OcSO.sub.4] which is rather close to the set obtained by Matkowska and Hofman C = 0,088136 for [bmim][[BF.sub.4]] and C = 0,0841547 for [bmim][MeS[O.sub.4]]. At the same time, Gu and Brennecke have reported the much larger T-dependent values C(298,2 K) = 0,1829 and C(323,2 K) = 0,1630 for the same [bmim][[PF.sub.6]].
Two other reasons of discrepancies in the Tait methodology is the different approximations chosen by authors for the reference input data [rho]([P.sub.0], T) and for the compound-dependent function B(T). Some authors [4, 14, 18] prefer to fit the atmospheric isobars [rho]([P.sub.0],T) and[C.sub.p]([P.sub.0],T) with a second-order or even third-order polynomial equation while the others [1, 2, 16, 19] use a
linear function for this aim.. As a result, the extrapolation ability to lower and higher temperatures of different approximations becomes restricted.
In this work we have used for [bmim] [[BF.sub.4]] the simplest linear approximation of both density and heat capacity,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (27)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (28)
taken from . The extrapolated to zero of temperature value [[rho].sub.0](0 K) = 1394,65 [kg*m.sup.-3] is in a good correspondence with that from [[rho].sub.0](0 K) = 1393,92 [kg*m.sup.-3], in reasonable correspondence with that from [[rho].sub.0](0 K) = 1416,03 [kg.m.sup.-3] and that from [[rho].sub.0](0 K) = 1429[kg*m.sup.-3] but its distinction from value [[rho].sub.0](0 K) = 1476,277 [kg*m.sup.-3] reported by authors is rather large. The similar large discrepancy is observable between [C.sup.0.sub.p] (0 K) = 273,65 J/mol*K from and [C.sup.0.sub.p] (0 K) = 464,466 J/mol*K from .
The different choices of an approximation function for B(T) (so authors have used the exponential form while authors have preferred the linear form) may distort the derivatives [[beta].sub.T] and [[alpha].sub.P] calculated by the Tait EOS (26). The problem of their uncertainties becomes even more complex if one takes into account the often existence of systematic distinctions of as much as 0,5% between the densities measured by different investigators even for the simplest argon . Machida et al. , for example, pointed out the systematic deviations measured densities from those reported by the de Azevedo et al. and Fredlake et al. for both [bmim][[BF.sub.4]] and [bmim][[PF.sub.6]]. Matkowska and Hofman concluded that the discrepancies between the different sets of calculated [[beta].sub.T-] and [[alpha].sub.P-]derivatives increase with increasing of T and decreasing of P due not only to experimental differences in density values but also result from the fitting equation used. The resultant situation is that the expansivity [[alpha].sub.P] of ILs reported in literature was either nearly independent of T or noticeably dependent of T [3,19].
We can add to these observations that the linear in molar (or specific) volume Tait Eos (26) is inadequate in representing the curvature of the isotherm P([rho]) at low pressures. It fails completely in description of (v, l)-transition where the more flexible function of volume is desirable. However, this has been clearly stated and explained by Streett for liquid argon that the adjustable T-dependence of empirical EOS becomes the crucial factor in representing the expansivity [[alpha].sub.P] and, especially, heat capacities [C.sub.P],[C.sub.v] at high pressures even if the reliable input data of sound velocity W(P, T) were used.
From such a viewpoint, one may suppose that the linear in temperature LF-EOS proposed by Sanchez and Lacombe,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (29)
is restricted to achieve the above goal but can be used as any unified classical EOS common for both phases to predict the region of their coexistence. Such conjecture is confirmed by the comparison of FT-EOS with LF-EOS presented in Figures 2 and 4 and discussed below. The obvious advantage of former is the more flexible T-dependence expressed via the cohesive-energy coefficient a(T). On the other hand, the LF-EOS is typical form of EOS (see Section 1) in which the constraint of T-independent potential energy [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is inherent .
One may consider it as the generalized variant of the well-known Bragg-Williams approximation for the ordinary LG presented here in the dimensional form
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (30)
Such generalization provides the accurate map of phenomenological characteristic parameters T*, P*, [rho]* which determine the constant effective number of lattice sites [N.sub.l] ]occupied by a complex molecule,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (31)
into the following set of molecular characteristic parameters for a simple molecule ([N.sub.l] = 1):
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (32)
where [v.sub.0] is the volume of cell and z is the coordination number of lattice in which the negative [eplison] is the energy of attraction for a near-neighbor pair of sites. In the polymer terminology [eplison]* from (31) is the segment interaction energy and v* is the segment volume which determines the characteristic hard core per mole M/[rho]* (excluded volume b in the vdW-terminology).
Another variant of described approach is the known perturbed hard-sphere-chain (PHSC) EOS proposed by Song et al. for normal fluids and polymers
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (33)
where g(d) = (1 - [eta]/2)/[(1 - [eta]).sup.3] is the pair radial distribution function of nonbonded hard spheres at contact and the term with ([N.sub.l] - 1) reflects chain connectivity while the last term is the small perturbation contribution. Though the PHSC-EOS has the same constraint of the potential energy field [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] authors have introduced two universal adjustable [[PHI].sub.a](T)- and [[PHI].sub.b](T)-functions to improve the consistency with experiment. The vdW-type coefficients were rescaled as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (34a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (34b)
where s([N.sub.l]) is the additional scaling function for T* = [eplison]/[k.sub.B]. It provides the interconnection of molecular LJ-type parameters ([eplison], [sigma]) with the phenomenological vdW-ones (a, b). The resultant reduced form of PHSC is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (35)
where the following characteristic and reduced variables are used:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (36)
It was compared with the simpler form of LF-EOS (29). Their predictions of the low-temperature density at saturation [[rho].sub.l](T) are comparable but, unfortunately, inaccurate (overestimated) even for neutral low-molecular liquids. The respective predictions of the vapor pressure [P.sub.[sigma]](T) are reasonable excepting the region of critical point for both EOSs. Our estimates based on the LF-EOS shown in Figures 2 and 4 are consistent with these conclusions.
The comparison of volumetric measurements and derived properties [14,18] with the purely predictive (by the FT-EOS) and empirical (by the Tait EOS and LF-EOS) methodologies used for [bmim][[BF.sub.4]] is shown in Figures 6-9. Evidently, that former methodology is quite promising. Machida et al. have reported two correlations of the same (P, [rho], T)-data measured for [bmim][[BF.sub.4]] at temperatures from 313 to 473 and pressures up to 200 MPa. To examine the trends in properties of ILs with the common cation [bmim] the Tait empirical EOS was preliminarily fitted as the more appropriate model. The estimate of its extrapolation capatibilities for (P,[rho], T)-surface in the working range (290 < T/K < 350) follows from the compatibility of experimental points (where those measured by de Azevedo et al. in the range of temperature 298 < T/K < 333 and pressure (0,1 < P/MPa < 60) were also included) with the thick curves in Figure 6. It is noticeable, for example, that the extrapolated Tait's isotherm T = 290 K coincides practically with isotherm T = 298,34 K from because the measured densities of latter source are systematically higher than those from . Density data of Fredlake et al. for [bmim][[BF.sub.4]] (not shown in Figure 6) are also systematically shifted from measurements .
The consequence of such discrepancies is also typical for any simple liquids (Ar, Kr, Xe) at moderate and high pressures. It is impossible to reveal an actual T-dependence of volumetric (mechanical) derived functions [[alpha].sub.P], [[gamma].sub.[rho]] due to systematic deviations between the data of different investigators. In such situation an attempt "to take the bull by the horns" and to claim the preferable variant of EOS based exclusively on volumetric data maybe erroneous. Indeed, since the Tait EOS is explicit in density while the LF-EOS--in temperature the direct calculation of [[alpha].sub.P], [[beta].sub.T]-derivatives for former and [[alpha].sub.P], [[gamma].sub.[rho]]-derivatives for latter are motivated. To illustrate the results of these alternative calculations we have used in Figures 6-9 the coefficients of LF-EOS reported by Machida et al. for the restricted range of moderate pressures 0,1 < P/MPa < 50. The thick dashed curves represent the boundaries of working range where the extrapolation to T = 290 K is again assumed. One may notice the qualitative similarity of FT-EOS (the thin curves) and LF-EOS which can be hypothesized as an existence of certain model substance at the extrapolation to higher pressures P/MPa > 50. It demonstrates the smaller compressibility [[beta].sub.T] (Figure 8) and expansivity [[alpha].sub.P] (Figure 7) than those predicted by the Tait EOS while the value of thermal-pressure coefficient [[gamma].sub.[rho]] for FT-EOS (Figure 9) becomes larger. It determines the distinctions in the calculated internal pressure. The choice of the FT-models substance as a reference system for the perturbation methodology provides the set of advantages in comparison with the LF-EOS.
It follows from Figure 6 that at moderate pressures P/MPa < 50 the predictive FT-EOS is more accurate than the fitted semiempirical LF-EOS although the discrepancies of both with the empirical Tait EOS become significant at the lowest (extrapolated) temperature T = 290 K. The Tait's liquid has no trend to (v, l)-transition (as well as polymers) in opposite to the clear trends demonstrated by FT-EOS and LF-EOS. One may suppose a competition between vaporization of IL (primarily driven by the isotropic dispersive attraction 1/[r.sup.6] in RPP) and chain formation (driven mainly by the anisotropic dipolar interactions 1/[r.sup.3]) reflected by the Tait EOS fitted to the experimental data. Of course, such conjecture must be, at least, confirmed by the computer simulations and FT-model provides this possibility by the consistent estimate of RPP-parameters ([eplison], [sigma]) at each temperature.
The differences of calculated expansivity [[alpha].sub.P] in Figure 7 are especially interesting. FT-EOS predicts even less variation of it with temperature than that for the Tait EOS. This result and crossing of [[alpha].sub.P](P)-isotherms are qualitatively similar to those obtained by de Azevedo et al. although the pressure dependence of all mechanical ([[alpha].sub.P], [[beta].sub.T], [[gamma].sub.[rho]]) and caloric ([C.sub.P],[C.sub.v]) derivatives (see Figures 10,11, and 12) is always more significant for the FT-EOS predictions. It seems that the curvature of the [rho](T)-dependence following from the LF-EOS (29) is not sufficient to predict the [[alpha].sub.P](P) behavior (Figure 7) correctly. The strong influence of the chosen input [rho]([P.sub.0], T)-dependence is obvious from Figures 7-9.
The prediction of caloric derivatives ([C.sub.P],[C.sub.v], [C.sub.P]/[C.sub.v], Gr) is the most stringent test for any thermal (P, [rho], T) EOS. It should be usually controlled by the experimental (W, [rho], T)-surface to use the thermodynamical identities,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (37)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (38)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (39)
in addition to the chosen input [C.sub.P]([P.sub.0], T)-dependence. de Azevedo et al. applied this strategy to comprise the approximated by the Pade-technique measured speed of sound data for [bmim][[PF.sub.6]] and [bmim][[BF.sub.4]] (Figure 13) with the evaluated at high pressures heat capacities.
Our predictive strategy is based on the differentiation of a(T)-dependence to evaluate directly the most subtle ([C.sub.V], P, T)-surface in a low-temperature liquid
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (40)
where the influence of the consistence for the chosen input [rho]([P.sub.0], T)- and[C.sub.p]([P.sub.0], T)-dependences (via (37) used for estimate of [C.sub.V]([P.sub.0], T) at the atmospheric pressure [P.sub.0]) becomes crucial. The use of first derivative da/dT (even by its rough approximation in terms of finite differences: [DELTA]a/[DELTA]T) to calculate simultaneously by ((3), (4), (37), (40)) all volumetric and caloric derivatives is the important advantage over the standard integration of thermodynamic identities:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (41a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (41b)
To illustrate such statement it is worthwhile to remind the situation described by the Streett for liquid argon. Since isotherms of [[alpha].sub.P](P) cross over for many simple liquids (Ar, Kr, Xe), this author concludes that the sign of [([[partial derivative].sup.2]V/[[partial derivative]T.sup.2]).sub.P] changes also from positive to negative at the respective pressure. This conclusion is not valid because to change the sign of derivative [([partial derivative][[alpha].sub.P]/[partial derivative]T).sub.P] it is enough to account for the exact equality
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (42)
in which [([[partial derivative].sup.2]V/[[partial derivative]T.sup.2]).sub.P] can be always positive. In this case one would expect the monotonous decrease of [C.sub.p] with increasing P in accordance to ((41a), (41b)) while the presence of extremum (minimum or maximum of [C.sub.p](P)-dependence) seems to be artificial.
There is the variety of pressures reported by different investigators as a presumable cross-point for the same ILs. Machida et al. have estimated it to be about 10 MPa on the base of Tait EOS for [bmim][[PF.sub.6]] but have not found it (Figure 7) for [bmim][[BF.sub.4]]. For latter our estimate by the FT-EOS is: P = 20,6 MPa. de Azevedo et al. have reported the mild decrease of [[alpha].sub.P](T)-dependence and the sharp decrease [[alpha].sub.P](P)-dependence while a presumable cross-point is located between about 100 and 120 MPa for [bmim][[BF.sub.4]]. Taking into account the above distinction in the evaluated ([[alpha].sub.P], P, T)-surface it is interesting to consider their consequences for caloric ([C.sub.v],P,T)-([C.sub.p],P,T)- and[C.sub.v]/[C.sub.p]-surfaces shown in Figures 10-12.
The remarkable qualitative and even quantitative (<8%) correspondence between the predicted by FT-EOS[C.sub.v]-values and those reported by de Azevedo et al. follows from Figure 10. At the same time, although the discrepancies between[C.sub.p]-values and those predicted by the FT-EOS are again within acceptable limits (<10%) the formers demonstrate the weak maximum and very small pressure dependence for [bmim][[BF.sub.4]] (for [bmim][[PF.sub.6]] this[C.sub.p](P)-dependence is monotonous as well as that predicted by the FT-EOS). It seems that the resultant ratio of heat capacity [C.sub.p]/[C.sub.v] shown in Figure 12 which demonstrates the irregular crossing of isotherms is questionable. It suggests that their pressure dependence either needs the more accurate approximation or reflects the realistic distinction of reference FT-EOS from the actual behavior of [bmim][[BF.sub.4]].
The lock of noticeable variations in pressure is the common feature of integration methodology based on the given (W, P, T)- and ([rho], P, T)-surfaces. The unavoidable accumulation of uncertainties at each stage of calculations in the set W - [C.sub.p] - [C.sub.v] may cause the unplausible behavior of adiabatic exponent [C.sub.p]/[C.sub.v] in liquid. The same is true for the set [C.sub.v ] - [C.sub.p] - W used in the FT-methodology. It is the most appropriate explanation of significant discrepancies for W(P)-dependence shown in Figure 13. Let us remind also that the precise mechanical measurements of speed velocity in the very viscous IL cannot be attributed exactly to the condition of constant entropy.
Thus, strictly speaking, the measured (W, P, T)-surface reflects the strong dispersive properties of media and must be less than its thermodynamic counterpart in the ideal (without a viscosity) liquid.
There are the structure-forming factors related to the above-discussed thermodynamic characteristic. Despite the certain discrepancies between the predicted and derived properties for FT-EOS and LF-EOS, both ones provide the close estimates of structure factors represented in Table 5.
Our aim here is to show that the thermodynamically-consistent predictions of thermodynamical properties by the FT-EOS yields also the molecular-based parameters which are, at least, realistic (see also Table 3). The estimate of average T-dependent well-depth [bar.[eplison] by (6b) as well as estimate of average value [[bar.N].sub.l] by (24a) is related to the middle of temperature range: T = 320 K. The distinction of [bar.[eplison] from the respective [eplison]*-parameters of LF-EOS can be attributed to the difference between nonbonded interactions in the discrete (LF-EOS) and continuum (FT-EOS) models of fluid. Our estimate of cohesive-energy density [[eplison].sub.coh] by equality,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (43)
represented in Table 6 seems also physically plausible. Mag-inn et al. [5, 6] have determined it within the framework of GEMC-simulations by the knowledge of [rho]([P.sub.0], T) and the internal energy difference between an ideal-gas ion pair and the average internal energy of an ion pair in the liquid state.
Such definition indicates that cohesive energy densities of many ILs are on the order of 500-550 J/[cm.sup.3] (see, for comparison, Table 6) and demonstrate a slight decrease as temperature increases.
Another relevant characteristic is the internal pressure determined by the derivative of molar (or specific) internal energy,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (44)
which is compared to ones calculated by different authors [14, 16] for [bmim][[BF.sub.4]] in Table 6. As in the other cases, the FT-EOS predicts the much faster change of both cohesive energy density [[eplison].sub.coh] and internal pressure [([partial derivative]e/[partial derivative]v).sub.T] as temperature increases.
One should collect a large number of precise experimental measurements to reconstruct the thermodynamic surface of a substance. FT-methodology provides a possibility of preliminary reliable estimates of relevant macroscopic and molecular-based correlations. Its thermodynamic consistency provides the serious advantage in comparison with the purely empiric treatment of any volumetric measurements at the description of derived heat capacities.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Department of Physics, Odessa State Academy of Refrigeration, Dvoryanskaya Street 1/3, Odessa 65082, Ukraine
Correspondence should be addressed to Vitaly B. Rogankov; [email protected]
Received 5 August 2014; Accepted 4 November 2014; Published 16 December 2014
Academic Editor: Pedro Jorge Martins Coelho
Copyright [c] 2014 V. B. Rogankov and V. I. Levchenko. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Table 1: Coefficients of FT-EOS (1) for neutral ([C.sub.2][H.sub.4]) and polar ([H.sub.2]O) substances. [b.sub.0] [[dm.sup.3]/mol], a [[J.dm.sup.3]/[mol.sup.2]], T [K] [C.sub.2][H.sub.4] [H.sub.2]O [b.sub.0] = 0,04181 [b.sub.0] = 0,01658 T a T a 103,99 1557,12 273,16 517,162 105 1459,76 278,15 527,171 110 1127,30 283,15 534,941 115 935,490 288,15 540,955 120 812,151 293,15 545,767 125 725,634 298,15 548,270 130 661,817 303,15 549,648 135 613,602 308,15 549,424 140 575,528 313,15 548,755 145 545,191 318,15 547,172 150 520,785 323,15 544,760 155 500,788 328,15 541,609 160 484,652 333,15 537,810 165 471,466 338,15 533,455 169,35 462,006 343,15 529,044 348,15 524,605 353,15 519,789 358,15 514,676 363,15 509,674 368,15 504,480 373,15 499,159 Table 2: Coefficients of FT-EOS (1) for ILs: [bmim][[PF.sub.6]], [pmmim] [[Tf.sub.2]N], and [bmim][[BF.sub.4]]. [b.sub.0] [[dm.sup.3]/mol], a [[J.dm.sup.3]/[mol.sup.2]], T [K] [bmim][[PF.sub.6]] [pmmim][[Tf.sub.2]N] [bmim][[BF.sub.4]] [b.sub.0] = 0,1953 [b.sub.0] = 0,2711 [b.sub.0] = 0,178 T a T a T a 285 8433,52 288,15 11724,8 290 9470,0 290 8214,24 295,15 12224,5 300 8900,9 295 8015,04 304,25 12279,0 310 8435,0 300 7833,44 313,65 11728,9 320 8047,5 305 7667,34 324,35 11308,4 330 7720,8 310 7514,97 335,15 10672,3 340 7442,2 315 7374,83 344,65 10051,1 350 7202,6 320 7245,64 350,15 9709,50 325 7126,31 330 7015,89 335 6913,55 340 6818,59 345 6730,37 350 6648,36 Table 3: Predicted hypothetical (v, l)-transition in the low-temperature range for FT-model of [bmim][[BF.sub.4]] based on the experimental data [1, 2] treated by FT-EOS (B-variant of x(T)-prediction). T (K) [[rho].sub.l] (mol/[dm.sup.3]) [[rho].sub.g] (mol/[dm.sup.3]) 290 5,350646 3,09E - 08 300 5,322170 2,1E - 07 310 5,293693 1,01E - 06 320 5,265215 3,74E - 06 330 5,236735 1,13E - 05 340 5,208254 2,92E - 05 350 5,179771 6,61E - 05 360 5,151287 0,000135 370 5,122802 0,000253 380 5,094315 0,000441 390 5,065828 0,000725 400 5,037338 0,001131 410 5,008848 0,001689 420 4,980355 0,002428 430 4,951862 0,003379 440 4,923367 0,004569 450 4,894871 0,006027 460 4,866373 0,007778 470 4,837873 0,009843 480 4,809372 0,012244 490 4,780869 0,014997 500 4,752365 0,018119 510 4,723859 0,021622 511 4,721008 0,021993 512 4,718157 0,022369 513 4,715307 0,022748 514 4,712456 0,023132 515 4,709605 0,023519 516 4,706754 0,023910 517 4,703903 0,024306 T (K) [P.sub.[sigma]] (kPa) [r.sub.[sigma]] (J/mol) 290 7,45E - 05 53082 300 5,24E - 04 49866 310 0,002 47230 320 0,009 45032 330 0,031 43175 340 0,082 41587 350 0,192 40216 360 0,404 39022 370 0,778 37974 380 1,393 37048 390 2,347 36225 400 3,754 35488 410 5,740 34826 420 8,446 34228 430 12,017 33684 440 16,604 33188 450 22,359 32733 460 29,432 32314 470 37,968 31926 480 48,105 31566 490 59,977 31230 500 73,704 30914 510 89,400 30618 511 91,082 30589 512 92,785 30561 513 94,509 30532 514 96,254 30504 515 98,020 30476 516 99,806 30448 517 101,615 30420 T (K) a (J.[dm.sup.3]/[mol.sup.2]) [eplison]/k (K) 290 9470,0 290 300 8900,9 300 310 8435,0 309,99 320 8047,5 319,99 330 7720,8 329,99 340 7442,2 339,99 350 7202,6 349,99 360 6994,6 359,99 370 6813,0 369,98 380 6653,5 379,96 390 6512,7 389,94 400 6387,8 399,91 410 6276,6 409,86 420 6177,4 419,79 430 6088,7 429,70 440 6009,1 439,59 450 5937,6 449,45 460 5873,2 459,27 470 5815,4 469,05 480 5763,2 478,79 490 5716,3 488,49 500 5674,1 498,13 510 5636,1 507,72 511 5632,5 508,67 512 5629,0 509,63 513 5625,4 510,58 514 5622,0 511,54 515 5618,6 512,49 516 5615,2 513,44 517 5611,8 514,40 Table 4: Effective LJ-diameters of FT-model for ILs determined by (6a), (6b), and (21b) on the base of estimates [7,13] and the choice [b.sub.0] = 178 [cm.sup.3]/mol for [bmim][[BF.sub.4]] in this work. IL M (g/mol) [sigma]([Angstrom]) [bmim][[BF.sub.4]] 225,82 5,208 [bmim][[PF.sub.6]] 284 5,371 [pmmim][[Tf.sub.2]N] 419,1 5,992 IL [M.sub.c]/[M.sub.c] [bmim][[BF.sub.4]] 139/86,82 [bmim][[PF.sub.6]] 139/145 [pmmim][[Tf.sub.2]N] 139/280,1 IL [[sigma].sub.c]/[[sigma].sub.a] [bmim][[BF.sub.4]] 5,757/4,51 [bmim][[PF.sub.6]] 5,651/5,06 [pmmim][[Tf.sub.2]N] 5,704/6,254 Table 5: Comparison of excluded volumes (M/[rho]* and [b.sub.0]), characteristic interaction energy ([eplison]* and [bar.[eplison]]), and efffective number of bonded units (MP*/RT*[rho]* [14,15] and [[bar.N].sub.l]) (see text). Compound M/[rho]* [b.sub.0] [eplison]* ([cm.sup.3]/mol) ([cm.sup.3]/mol) (J/mol) [bmim][[BF.sub.4]] 175,3 178 5642 [bmim][[PF.sub.6]] 196,2 195,3 5658 Compound [bar.[eplison]] MP*/RT*[rho]* [[bar.N].sup.l] (J/mol) [bmim][[BF.sub.4]] 2661 17,6 17,0 [bmim][[PF.sub.6]] 2661 18,8 18,1 Table 6: Comparison of internal pressure ([partial derivative]e/[[partial derivative]v).sub.T] for [bmim][[BF.sub.4]] based on the LF-EOS and FT-EOS (this work) with the values estimated by experimental data on speed of sound W, density [rho], and isobaric heat capacity [C.sub.p]. I ([partial derivative]e/[[partial derivative]v).sub.T] (K) (MPa) 313,1 482,78 332,6 471,50 352,6 460,35 372,7 450,11 392,8 440,00 412,9 429,23 432,6 419,35 452,3 408,84 472,2 399,19 I T ([partial derivative]e/[[partial derivative]v).sub.T] (K) (K) (MPa) 313,1 283,15 459,91 332,6 288,15 459,31 352,6 293,15 459,20 372,7 298,15 457,79 392,8 303,15 455,76 412,9 308,15 453,73 432,6 313,15 451,20 452,3 318,15 448,06 472,2 323,15 446,64 328,15 444,61 333,15 443,09 338,15 441,68 343,15 439,75 I T [FT] ([partial derivative]e/[[partial derivative]v).sub.T] (K) (K) (MPa) 313,1 290 743,76 332,6 310 573,16 352,6 330 463,91 372,7 350 388,61 392,8 370 333,84 412,9 390 292,27 432,6 410 259,66 452,3 430 233,40 472,2 450 211,77 470 193,64 490 178,19 510 166,76 517 162,41 I [[eplison].sub.coh] (K) (J/[cm.sup.3]) 313,1 555,15 332,6 486,41 352,6 437,85 372,7 401,57 392,8 373,35 412,9 350,67 432,6 331,94 452,3 316,13 472,2 302,52 290,60 280,01 270,46 267,31
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|Title Annotation:||Research Article|
|Author:||B. Rogankov, Vitaly; I. Levchenko, Valeriy|
|Publication:||Journal of Thermodynamics|
|Date:||Jan 1, 2014|
|Previous Article:||Second law analysis for third-grade fluid with variable properties.|
|Next Article:||Thermal performance and economic analysis of 210 MWe coal-fired power plant.|
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s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00035.warc.gz
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CC-MAIN-2021-31
| 61,994 | 179 |
https://brainmass.com/engineering/electronic-engineering/circuit-analysis-calculating-voltage-531867
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math
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Please refer to the attachment questions and diagrams. Questions are asking to calculate overall resistance, voltage, and finding the values for input impedance and output impedance.
l. Assume the transistor you will use has a beta of 155 and V_BE(on)=0.65 V. For the circuit in Fig. 2. (attached) calculate the overall resistance (R_B and pot) such that I_C = 1 mA.
2. Compute the voltage V_c.
3. Find the values for the input impedance R_i, output impedance R_o, and small signal transcouductauce g_m.
4. Find the value of the small signal voltage gain.
5. Construct the circuit in Fig. 2 in Pspice and find the small signal parameters g_m and R_ce. Compare SPICE results and hand calculations. Explain the differences, if any.
This posting contains the solution to the given problem.
Circuit analysis with LEDs, BJTs, Relays, Opto Couplers
(b) The diagram of FIGURE 2 shows a bi-directional opto coupler input interface circuit. When a supply voltage of 20 V is applied the LED carries a current and 2 V is dropped across it. Calculate the value of the LED current and the value of current through the 3 k resistance.
Separate supply used for inputs.
Note: Either polarity.
(c) The circuit shown in FIGURE 3 is part of the interface of a relay output module. Ib is 1 mA and VCC is 9 V. The relay requires a
minimum of 50 mA to energise.
Complete the values of the assumptions listed below in order to calculate:
• voltage across R1
• value of R1
• voltage across the relay coil
• voltage across R2
• value of R2
• collector of current Ic.
Assumptions: Logic '1' = V
Logic '0' = V
Transistor forward current gain hfe = LED current = 10 mA LED voltage drop at 10 mA = V Base/emitter voltage = V
Collector emitter voltage when transistor is on = 1 V
FIG. 3View Full Posting Details
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s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515029.82/warc/CC-MAIN-20181022092330-20181022113830-00349.warc.gz
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CC-MAIN-2018-43
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https://www.wikiind.com/2011/01/chain-rule/
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math
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1. If 15 toys cost Rs.234, what do 35 toys cost ?
A. 456 B. 534
C. 546 D. 528
2)If 36 men can do a piece of work in 25hours, in how many hours will 15men do it?
A. 60 B. 55
C. 62 D. 58
3. If 9 engines consume 24metric tonnes of coal, when each is working 8 hours a day, how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?
A. 23 B. 26
C. 22 D. 24
4. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
A. 83 B. 86
C. 81 D. 84
5. A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that?
A. 53 B. 50
C. 58 D. 49
6. A contractor undertook to do a certain piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?
A. 5 B. 6
C. 3 D. 4
7. 12 men and 18 boys,working 7 ½ hors a day, can do a piece if work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working 9 hours a day?
A. 44 B. 42
C. 52 D. 39
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https://www.jiskha.com/questions/833219/A-Quadratic-Word-Problem-If-one-side-of-a-square-is-increased-by-2-inches-and-an
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math
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A Quadratic Word Problem
If one side of a square is increased by 2 inches and an adjacent side is decreased by 2 inches, the area of the resulting rectangle is 32 square inches. Find the length of one side of the square.
If each edge of a cube is increased by 2 inches, the A. volume is increased by 8 cubic inches B. area of each face is increased by 4 square inches C. diagonal of each face is increased by 2 inches D. sum of the edges is increased
The side of a square is Three raised to the five halves power inches. Using the area formula A = s2, determine the area of the square. This is the side of the square in fraction(3^5/2) A = 9 square inches A = 15 square inches A =
Jim is installing ceramic tile behind his kitchen sink. Each tile is a square with side length 4 1/4 inches. For each tile, he allows an additional 1/8 inch per side for the grout between tiles. How many tiles should he buy if the
Orlando is making a design for a logo. He begins with a square measuring 20 inches on each side. The second square has a side length of 16 inches, and the third square has a side length of 12.8 inches. Which square will be the
okay guys, just one more word problem. and i don't think this one has any typos in it: The perimeter of a rectangle is 27 cm and its area is 35 square centimeters. Find the length and the wide of the rectangle. i just need help
a square has a length of 5 inches. Then it is increased by 20%. Then the same square has a width of 10 inches and it is increased by 10%. How Many Percents has the area of the square been increased By? Thanx 4 your Help
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https://www.economicsdiscussion.net/monopoly/inefficiency-of-monopoly-markets/23840
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math
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Each firm in a competitive industry operates at a point where its MC becomes equal to the (exogenously given) price of the product. That is why the short-run supply curve of a competitive industry is the horizontal summation of the short-run supply curves (or the short-run MC curves on and above the minimum AVC) of individual firms.
Therefore, a competitive industry operates at a point where price equals marginal cost. A monopolized industry, on the other hand, operates where price is greater than MC. Thus, in general, price will be higher and output lower under monopoly than under competition.
We may illustrate the point with the help of Fig, 11.20. In this figure, DD is the demand curve for the product of a competitive industry, and EMC is the industry’s supply curve. Therefore, the market (demand = supply) equilibrium point will be Ec, which is the point of intersection between the DD and the IMC curves.
At this point, the price of the product will be pc = opc and the quantity demanded and supplied will be qc = oqc. All the firms would operate at their respective p = MC points. Let us now assume that the competitive industry with all its firms becomes monopolized under a single owner, i.e., the competitive industry now becomes a multi-plant monopoly.
The demand curve DD for the product now becomes the AR curve of the monopolist and its corresponding marginal curve is given by MR in Fig. 11.20.
Now, in order to minimise the cost of production of any particular quantity of output, the monopolist would have to distribute the production of that quantity over the different plants in such a way that MC in each plant may become the same, and this MC would be the MC of that particular quantity of output.
Understood in this way, the MC of any particular quantity of output produced by the multi-plant monopolist would be given by the curve EMC of Fig. 11.20, which is the horizontal summation of the MC curves of individual plants.
The profit-maximising equilibrium point of the monopolist in this case would be the point of intersection (Em) between the MR and the ∑MC curves. At this point, the output of the monopolist would be qm = Oqm which would be less than the competitive output qc = Oqc and the price charged by the monopolist would be pm = Opm which is greater than MC (= Em qm) and which is greater than the competitive price (pc).
Since the price under monopoly is greater and output lower than under competition, the consumers would always be worse off in a monopolistic industry than in one under perfect competition. But a portion of the consumers’ loss would be converted into the gain of the industry.
This can be seen in Fig. 11.20. Here, as we move from competition to monopoly, i.e., as the equilibrium (p, q) combination moves from (pc, qc) to (pm, qm), the consumers’ surplus diminishes by □pcpmAEc (from □TpcE,. to □TpmA) and of this, □pcpmAB is converted into the producer’s surplus under monopoly.
Producer’s surplus, □SpcEc, decreases by □BEmEc and increases by □pcpmAB. Taking both the industry’s gain and the consumers’ loss into consideration, it is not clear which would be the better arrangement for the society—competition or monopoly, unless we make a value judgement about the relative welfare of the consumers and of the owners of the industry.
However, we may argue against monopoly on grounds of efficiency alone. An economic arrangement is Pareto-efficient if there is no way to make anyone better off without making somebody else worse off. We shall now see that the level of output under monopoly is not Pareto-efficient.
Let us remember that at any point on the demand curve of a monopoly firm like DD in Fig. 11.20, the price p stands for how much people are willing to pay for the marginal unit of the good.
Since p is greater than MC for all output levels between qm and qc, there is a whole range of output (i.e., between qm and qc) where people are willing to pay more for the marginal unit of output than it costs to produce it. Clearly, there is a case for a Pareto improvement here.
For example, let us consider the situation at the monopoly level of output qm. Since p(qm) > MC (qm), we know that there is someone who is willing to pay more for an extra unit of output than it costs to produce that extra unit. Let us suppose that the firm produces an extra unit and sells it to this person at any price p where p(qm) > p > MC(qm).
Then this consumer is made better off because he was just willing to pay p(qm) for that unit of consumption, and it was sold for p < p (qm). Similarly, it cost the monopolist MC(qm) to produce that extra unit of output and he sold it for p > MC(qm).
If all the other units of output are being sold for the same price as before, then from the sale of the extra unit, both sides of the market—the buyers’ side and the sellers’ side— are better off and no one else is made worse off. Therefore, we have found here a Pareto improvement. That is, the usual monopoly solution (pm, qm) is Pareto-ineflicient.
The reason for this inefficiency of monopoly is this. In the case of competition, price is constant irrespective of output, making MR at any output a constant and equal top. So the firm’s profit maximising p = MR = MC point is also the Pareto-efficient p = MC point.
On the other hand, since the AR curve of the monopolist is downward sloping, the extra revenue (MR) obtained from selling the marginal unit of output is not equal to, but less than, the price of that unit (i.e., MR < p). That is why, in monopoly, the firm’s profit maximising MR = MC point is a Pareto-inefficient p > MC point.
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CC-MAIN-2024-18
| 5,647 | 19 |
https://itectec.com/spec/5-1-3-sample-rate-conversion-to-12-8-khz/
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math
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26.4453GPPCodec for Enhanced Voice Services (EVS)Detailed algorithmic descriptionRelease 15TS
The linear predictive (LP) analysis, the long-term prediction (LTP), the VAD algorithm and signal are performed at the 12.8 kHz sampling rate. The HP-filtered input signal is therefore converted from the input sampling frequency to 12.8 kHz.
18.104.22.168 Conversion of 16, 32 and 48 kHz signals to 12.8 kHz
For 16, 32 and 48 kHz signals, the sampling conversion is performed by first up‑sampling the signal to 192 kHz, then filtering the output through a low-pass FIR filter that has the cut‑off frequency at 6.4 kHz. Then, the signal is down-sampled to 12.8 kHz. The filtering delay is 15 samples at 16 kHz sampling frequency which corresponds to 0.9375 ms.
The up-sampling is performed by inserting 11, 5 or 3 (for 16, 32 or 48 kHz, respectively) zero-valued samples between each 2 samples for each 20-ms frame of 320 samples (at 16 kHz sampling frequency)
where is the signal at 192 kHz sampling frequency and is the up-sampling factor equal to 12 for a 16 kHz input, 6 for a 32 kHz input and 4 for a 48 kHz input. Then, the signal is filtered through the LP filter and decimated by 15 by keeping one out of 15 samples. The filter is a 361-tap linear phase FIR filter having a cut-off frequency of 6.4 kHz in the 192 kHz up-sampled domain. The filtering and decimation can be done using the relation
where is the impulse response of . The operations in equations (3) and (4) can be implemented in one step by using only a part of the filter coefficients at a time with an initial phase related to the sampling instant n. That is
In case the encoder is externally forced to narrow-band processing of the input signal, the cut-off frequency of the LP filter is changed from 6.4 kHz to 4 kHz.
22.214.171.124 Conversion of 8 kHz signals to 12.8 kHz
For 8 to 12.8 kHz resampling a sharper resampling filter is beneficial. Double length low-pass FIR filter is used in this case. The doubling of the impulse response length is compensated by a low delay resampling method. The filter is a 241-tap linear phase FIR filter having a cut-off frequency of 3.9 kHz and is applied in the up-sampled domain which is 64 kHz. Direct FIR filtering with this filter would yield a delay of 120/64 = 1.875 ms. In order to reduce this delay to 0.9375 ms, future samples are determined at 8 kHz by adaptive linear prediction. The exact number of future samples is found based on the difference between the actual delay (1.875 ms) and the desired delay (0.9375 ms) at 8 kHz. Therefore future samples are predicted. These predicted samples are concatenated at the end of the current frame to form a support vector. Then, the sample rate conversion of is performed in a similar way as for the other sampling rates, i.e. is first up-sampled to 64 kHz, the output is filtered through the low-pass FIR filter and the resulting signal is down-sampled to 12.8 kHz. The final filtering delay is aligned with that of the other resampling configurations, i.e 12 samples at 12.8 kHz sampling frequency which corresponds to 0.9375 ms.
To determine the future samples, linear prediction coefficients of order 16 are computed in the pre-emphasized domain in the following way. The last Lss =120 samples of the input frame at 8 kHz are windowed by an asymmetrical analysis window winss_120:
and a first order autocorrelation analysis is made on the windowed signal . The pre-emphasis coefficient ss is obtained by
where rw(0) and rw(1) are the autocorrelation coefficients
The last 120 samples of the signal are pre-emphasized using the adaptive filter
to obtain the pre-emphasized signal of Lss =120 samples. Then is windowed by the asymmetrical analysis window winss_120 and a 16th order autocorrelation analysis is made on the windowed signal
These autocorrelation coefficients are lag-windowed by
where wlag8k(k) is defined as
Based on the autocorrelation coefficients rpwl(k), the linear prediction coefficients ass(k) are computed by the Levinson-Durbin algorithm. The future samples in the pre-emphasized domain are predicted by zero input filtering through the 1/Ass(z) synthesis filter
Finally, the concatenated signal is de-emphasized through the filter . Note that only the last 7 predicted samples need to be de-emphasized. These 7 de-emphasized samples are concatenated to (at positions n = 160,…,166) to form the support vector.
The up-sampling of is then performed by inserting 7 zero-valued samples between each 2 samples for each 20-ms frame of 160 samples (at 8 kHz sampling frequency) completed by 7 predicted future samples (167 in total)
where is the signal at 64 kHz sampling frequency. Then, the signal is filtered through the LP filter and decimated by 5 by keeping one out of 5 samples. The filtering and decimation can be done using the relation
where is the impulse response of and assures that the index of s64 is never higher than the highest available index for (which is 1335). Indeed, it corresponds to the delay of this filtering at 64 kHz. To reduce complexity, the operations in equations (14) and (15) can be implemented in one step by using only a part of the filter coefficients at a time with an initial phase related to the sampling instant n. This polyphase implementation of the resampling filter is applied on the concatenated support vector. That is
where is derived from the delay of this filtering at 8 kHz. It assures that the index of sHPC is never higher than the highest available index (which is 166).
126.96.36.199 Conversion of input signals to 16, 25.6 and 32 kHz
If ACELP core is selected for WB, SWB or FB signals at bitrates higher than 13.2 kbps (see subclause 5.1.16), its internal sampling rate is set to 16 kHz rather than 12.8 kHz. If the input signal is sampled at 8 kHz, there is no conversion needed because for NB signals, ACELP core is always operated at 12.8 kHz. If the input signal is sampled at 16 kHz, no conversion is needed either and the input signal is only delayed by 15 samples which corresponds to 0.9375 ms. This is to keep all pre-processed signals aligned regardless of the bitrate or bandwidth. Thus, the input signal is resampled to 16 kHz only if its sampling frequency is 32 or 48 kHz.
The resampling operation is done in the same way as for the case of 12.8 kHz (see subclause 188.8.131.52), i.e. by means of FIR filtering. The coefficients of the LP filter are different but the filtering delay is still the same, i.e. 0.9375 ms.
The resampled signal is denoted where n=0,..,319.
The input signal is converted to 25.6 kHz at 48 kbps and to 32 kHz at 96 or 128kbps but only for SWB and FB signals. The sampling conversion is again similar as in the case of 12.8 kHz with differences in LP filter coefficients. The resampled signals are denoted and , respectively.
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http://physics.stackexchange.com/questions/tagged/geometry+quantum-mechanics
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math
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What would be good introductory and follow-up references to understand the close ties between physics and geometry. I'm a retired engineer with the math background to handle Shankar's Principles of ...
Can the electroweak and strong forces be written as geometric theories? - Why and why not? Can quantum mechanics in general? For example, the Kaluza-Klein theory explains the electromagnetic field ...
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s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345766127/warc/CC-MAIN-20131218054926-00096-ip-10-33-133-15.ec2.internal.warc.gz
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http://www.ipam.ucla.edu/abstract/?tid=10404&pcode=sar2012
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math
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We discuss two inverse problems related to anisotropic media for Maxwell's equations. The first one is the inverse scattering problem of determining the anisotropic surface impedance of a bounded obstacle from a knowledge of electromagnetic scattered field due to incident plane waves. Such an anisotropic boundary condition can arise from surfaces covered with patterns of conducting and insulating patches. We show that the anisotropic impedance is uniquely determined if sufficient data is available, and characterize the non-uniqueness presence if a single incoming wave is used. We derive an integral equation for the surface impedance in terms of solutions of a certain interior impedance boundary value problem. These solutions can be reconstructed from far field data using the Herglotz theory underlying the Linear Sampling Method. The second problem is to obtain information about matrix index of refraction of an anisotropic media. This problem plays a special role in inverse scattering theory due to the fact that the (matrix) index of refraction is not in general uniquely determined from scattering data. Our imaging tool is a new class of eigenvalues associated with the scattering by inhomogeneous media, known as transmission eigenvalues. In this presentation we describe how transmission eigenvalues can be determined from scattering data and be used to obtain upper and lower bounds on the norm of the index of refraction. For both problems the support of the scatterer can be determined by mean of the Linear Sampling Method. Preliminary numerical results will be shown for both problems.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159744.52/warc/CC-MAIN-20180923193039-20180923213439-00453.warc.gz
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CC-MAIN-2018-39
| 1,609 | 1 |
https://mcqzs.com/online-practice-paper/applied-physics-energy-mcqs/630
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math
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Online Practice Paper
Applied Physics: Energy MCQs Online Test
Applied physics MCQs, applied physics multiple choice questions, and answers for distance learning online courses. Applied physics quiz questions and answers to practice physics tests, online learning for colleges and universities courses. Free applied physics MCQs with answers, online competitive exam prep questions. Applied physics topics for distance learning courses areas:
Total questions: 5
Set 1 - The test contains 5 questions and there is no time limit. You will get 1 point for each correct answer. You will get your online test score after finishing the complete test.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00712.warc.gz
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CC-MAIN-2021-31
| 644 | 5 |
https://www.superphysics.org/research/bohr/quantum/intro/
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math
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In an attempt to develop certain outlines of a theory of line-spectra based on a suitable application of the fundamental ideas introduced by Planck in his theory of temperatureradiation to the theory of the nucleus atom of Sir Ernest Rutherford, the writer has shown that it is possible in this way to obtain a simple interpretation of some of the main laws governing the line-spectra of the elements, and especially to obtain a deduction of the well known Balmer formula for the hydrogen spectrum1)
The theory in the form given allowed of a detailed discussion only in the case of periodic systems, and obviously was not able to account in detail for the characteristic difference between the hydrogen spectrum and the spectra of other elements, or for the characteristic effects on the hydrogen spectrum of external electric and magnetic fields.
Recently, however, a way out of this difficulty has been opened by Sommerfeld2 ) who, by introducing a suitable generalisation of the theory to a simple type of non-periodic motions and by taking the small variation of the mass of the electron with its velocity into account, obtained an explanation of the fine-structure of the hydrogen lines which was found to be in brilliant conformity with the measurements. Already in his first paper on
) N. Bohr, Phil. Mag., XXVI, pp. 1, 476, 857 (1913), XXVII, p. 506 (1914), XXIX. p. 332 (1915), XXX. p. 394 (1915). 2 ) A. Sommerfeld, Ber. Akad. M¨unchen, 1915, pp. 425, 459, 1916, p. 131. 1917. p. 83. Ann. de Phys., LI. p. 1 (1916). 2 this subject, Sommerfeld pointed out that his theory evidently offered a clue to the interpretation of the more intricate structure of the spectra of other elements. Briefly afterwards Epstein1 ) and Schwarzschild, 2 ) independent of each other, by adapting Sommerfeld’s ideas to the treatment of a more extended class of non-periodic systems obtained a detailed explanation of the characteristic effect of an electric field on the hydrogen spectrum discovered by Stark.
Subsequently Sommerfeld3 ) himself and Debye4 ) have on the same lines indicated an interpretation of the effect of a magnetic field on the hydrogen spectrum which, although no complete explanation of the observations was obtained, undoubtedly represents an important step towards a detailed understanding of this phenomenon.
In spite of the great progress involved in these investigations many difficulties of fundamental nature remained unsolved, not only as regards the limited applicability of the methods used in calculating the frequencies of the spectrum of a given system, but especially as regards the question of the polarisation and intensity of the emitted spectral lines.
These difficulties are intimately connected with the radical departure from the ordinary ideas of mechanics and electro1 ) P. Epstein, Phys. Zeitschr. XVII, p. 148 (1916), Ann. d. Phys. L, p. 489. LI. p. 168 (1916). 2 ) K. Schwarzschild, Ber. Akad. Berlin, 1916, p. 548. 3 ) A. Sommerfeld, Phys. Zeitschr. XVII, p. 491 (1916). 4 ) P. Debye, Nachr. K. Ges. d. Wiss. G¨ottingen, 1916, Phys. Zeitschr. XVII, p. 507 (1916). 3 dynamics involved in the main principles of the quantum theory, and with the fact that it has not been possible hitherto to replace these ideas by others forming an equally consistent and developed structure. Also in this respect, however, great progress has recently been obtained by the work of Einstein1 ) and Ehrenfest. 2 ) On this state of the theory it might therefore be of interest to make an attempt to discuss the different applications from a uniform point of view, and especially to consider the underlying assumptions in their relations to ordinary mechanics and electrodynamics. Such an attempt has been made in the present paper, and it will be shown that it seems possible to throw some light on the outstanding difficulties by trying to trace the analogy between the quantum theory and the ordinary theory of radiation as closely as possible. The paper is divided into four parts. Part I contains a brief discussion of the general principles of the theory and deals with the application of the general theory to periodic systems of one degree of freedom and to the class of non-periodic systems referred to above. Part II contains a detailed discussion of the theory of the hydrogen spectrum in order to illustrate the general 1 ) A. Einstein, Verh. d. D. phys. Ges. XVIII, p. 318 (1916), Phys. Zeitschr. XVIII, p. 121 (1917). 2 ) P. Ehrenfest, Proc. Acad. Amsterdam, XVI. p. 591 (1914), Phys. Zeitschr. XV. p. 657 (1914), Ann. d. Phys. LI. p. 327 (1916), Phil. Mag. XXXIII. p. 500 (1917). 4 considerations. Part III contains a discussion of the questions arising in connection with the explanation of the spectra of other elements. Part IV contains a general discussion of the theory of the constitution of atoms and molecules based on the application of the quantum theory to the nucleus atom. Copenhagen, November 1917.
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CC-MAIN-2023-14
| 4,949 | 7 |
http://link.springer.com/article/10.1007%2FBF02803517
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math
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We consider an Abel equation (*)y’=p(x)y2 +q(x)y3 withp(x), q(x) polynomials inx. A center condition for (*) (closely related to the classical center condition for polynomial vector fields on the plane) is thaty0=y(0)≡y(1) for any solutiony(x) of (*).
Folowing , we consider a parametric version of this condition: an equation (**)y’=p(x)y2 +εq(x)y3p, q as above, ε ∈ ℂ, is said to have a parametric center, if for any ɛ and for any solutiony(ɛ,x) of (**)y(ɛ, 0)≡y(ɛ, 1)..
We give another proof of the fact, shown in , that the parametric center condition implies vanishing of all the momentsmk (1), wheremk(x)=∫0xpk(t)q(t)(dt),P(x)=∫0xp(t)dt. We investigate the structure of zeroes ofmk(x) and generalize a “canonical representation” ofmk(x) given in . On this base we prove in some additional cases a composition conjecture, stated in [6, 7] for a parametric center problem.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607960.64/warc/CC-MAIN-20170525010046-20170525030046-00460.warc.gz
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CC-MAIN-2017-22
| 902 | 3 |
http://mxplx.com/meme/2999/
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math
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...Carlsen is demonstrating one of his most feared qualities, namely his “nettlesomeness,” to use a term coined for this purpose by Ken Regan. Using computer analysis, you can measure which players do the most to cause their opponents to make mistakes. Carlsen has the highest nettlesomeness score by this metric, because his creative moves pressure the other player and open up a lot of room for mistakes. In contrast, a player such as Kramnik plays a high percentage of very accurate moves, and of course he is very strong, but those moves are in some way calmer and they are less likely to induce mistakes in response.
A characteristic of chess players. A measure of how often they make moves that cause their opponent to make mistakes.
highest nettlesomeness score (0.915154 (neutral:0.000000)), accurate moves (0.820447 (positive:0.335608)), creative moves (0.805012 (negative:-0.457740)), mistakes (0.801825 (negative:-0.561495)), feared qualities (0.775708 (negative:-0.367253)), Ken Regan (0.754812 (negative:-0.323302)), chess players (0.724334 (neutral:0.000000)), high percentage (0.717507 (positive:0.335608)), Carlsen (0.643772 (negative:-0.367253)), Kramnik (0.509423 (positive:0.335608)), characteristic (0.500286 (neutral:0.000000)), measure (0.496627 (negative:-0.822382)), opponent (0.489997 (negative:-0.822382)), opponents (0.482629 (negative:-0.581551)), contrast (0.480027 (neutral:0.000000)), term (0.458177 (negative:-0.323302)), purpose (0.457999 (negative:-0.323302)), analysis (0.457302 (neutral:0.000000)), course (0.454694 (positive:0.404666)), way (0.454161 (negative:-0.384309))
Player (0.920162): dbpedia
Chess (0.845293): dbpedia | freebase | opencyc
Fundamental physics concepts (0.713884): dbpedia
Topological space (0.676746): dbpedia | freebase | opencyc
Players (0.644805): dbpedia
Algorithm (0.614640): dbpedia | freebase | opencyc
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CC-MAIN-2017-17
| 1,874 | 9 |
http://www.allindiaexams.in/online-test/cat-free-online-test-cat-mock-test-/382
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math
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Time Left: 00:50:00
1. A number consists of two digits. If 3/5 of 1/5 of the number is 9. Find the sum of its two digits?
2. The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:10. If P invested the money for 5 months, find for how much time
did Q invest the money?
3. If cost of sugar increases by 25%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed?
4. A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then, A's speed is equal to?
5. The sum of four consecutive even numbers is 292. What would be the smallest number?
6. Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
7. The average weight of a group of boys is 30 kg. After a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
8. An amount of Rs. 3000 becomes Rs. 3600 in four years at simple interest. If the rate of interest was 1% more, then what was be the total amount?
9. In a 1000 m race, A beats B by 200 meters or 25 seconds. Find the speed of B?
10. In how many ways can a committee consisting of three men and four women be formed from a group of six men and seven women?
11. Three men start together to travel the same way around a circular track of 11 kilometers in circumference. Their speeds are 4, 5 and 8 kilometers per hour respectively. When will they meet at a starting point?
12. A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after
8 months. If the annual gain be Rs.18000. A's share is?
13. A thief steals at a car at 2.30 p.m. and drives it at 60 km/hr. The theft is discovered at 3 p.m. and the owner sets off in another car at 75 km/hr. When will he overtake the thief?
14. If A:B = 1/2: 1/3 B:C = 1/2:1/3 then A:B:C?
15. The average of 1st 3 of 4 numbers is 16 and of the last 3 are 15. If the sum of the first and the last number is 13. What is the last numbers?
16. A certain sum becomes Rs. 20720 in four years and 24080 in six years at simple interest. Find sum and rate of interest?
17. (562 - 242) * 1/32 + ?% of 1200 = 146
18. A train is 360 meter long is running at a speed of 45 km/hour. In what time will it pass a bridge of 140 meter length?
19. In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?
20. Anand and Deepak started a business investing Rs. 22,500 and Rs. 35,000 respectively. Out of a total profit of Rs. 13,800, Deepak's share is:
21. How long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length?
22. There is 60% increase in an amount in 6 years at S.I. What will be the C.I. of Rs. 12,000 after 3 years at the same rate?
23. 32% of 425 - ?% of 250 = 36
24. The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream?
25. Anil invested a sum of money at a certain rate of simple interest for a period of five years. Had he invested the sum for a period of eight years for the same rate, the total intrest earned by him would have been sixty percent more than the earlier interest amount. Find the rate of interest p.a.
26. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?
27. The tax on a commodity is diminished by 20% and its consumption increased by 15%. The effect on revenue is?
28. 25 * 25 / 25 + 15 * 40 = ?
29. Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?
30. A and B invests Rs.3000 and Rs.4000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit?
31. Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?
32. 4500 * ? = 3375
33. The least number of four digits which is divisible by 4, 6, 8 and 10 is?
34. A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it
takes him 1 hour to row to a place and black. What is the total distance traveled by the man?
35. A and B can do a piece of work in 7 days. With the help of C they finish the work in 5 days. C alone can do that piece of work in?
36. The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream?
37. A bag contains equal number of Rs.5, Rs.2 and Re.1 coins. If the total amount in the bag is Rs.1152, find the number of coins of each kind?
38. If selling price is doubled, the profit triples. Find the profit percent:
39. (8x9 / 27y-6)-2/3 = ?
40. Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
41. The probability that A speaks truth is 3/5 and that of B speaking truth is 4/7. What is the probability that they agree in stating the same fact?
42. A father said to his son, "I was as old as you are at present at the time of your birth." If the father's age is 38 years now, the son's age five years back was:
43. If (461 + 462 + 463 + 464) is divisible by ?, then ? =
44. Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately?
45. On dividing 2272 as well as 875 by 3-digit number N, we get the same remainder. The sum of the digits of N is:
46. By how much is 3/5th of 875 greater than 2/3 of 333?
47. The value of a machine depreciates at 20% per annum. If its present value is Rs. 1,50,000, at what price should it be sold after two years such that a profit of Rs. 24,000 is made?
48. There are two positive numbers in the ratio 5:8. If the larger number exceeds the smaller by 15, then find the smaller number?
49. In how much time will a train of length 100 m, moving at 36 kmph cross an electric pole?
50. Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What
time will they take to cross each other if they are running in opposite directions?
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s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670601.75/warc/CC-MAIN-20191120185646-20191120213646-00051.warc.gz
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CC-MAIN-2019-47
| 6,747 | 55 |
https://rebab.net/the-bending-of-light-is-called/
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math
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Refraction is the change in the direction of a tide passing indigenous one tool to another. A light ray refracts whenever it travels at an angle right into a tool of various refractive index. This readjust in speed outcomes in a adjust in direction.
The ratio of the rate of irradiate in a vacuum come its rate in a specific medium is referred to as refractive index. Refractive index is also referred to as refraction table of contents or index of refraction. The rate of irradiate in a medium depends top top the properties of the medium. In electromagnetic waves, the rate is dependent on the optical thickness of the medium.
You are watching: The bending of light is called
Why recently hydroelectricity is desired over heat power?The break down in a turning back biased p-n junction diode is more likely to occur due to.Consider a wedge of mass 2m and also a cube of massive m . Between the cube and wedge, there is no friction....n the same cubes each of fixed m and also edge L space lying top top a floor. If the cubes space to be arranged as...A close up door coil is composed of 500 transforms on a rectangular frame of the area 4.0cm2 and also has a resistance o...Current sensitivity the a galvanometer can be increased by decreasing?A parallel bowl capacitor stores a charge Q in ~ voltage V. Expect the area that the capacitor and also the...A boat, which has actually a speed of 5km/h in quiet water, crosses a river of width 1km follow me the shortest p...For an item projected from the floor with speed u, horizontal selection is 2 times the preferably heig...A particle of fee q and also mass m move in a one orbit of radius r through angular speed ω. The r...The resistance in the 2 arms that the meter bridge room 5 Ω and R Ω , respectively. Once the resist...−2 . Find the angular acceleration the the ring and also the angular velocity at that instant. Radius the the ring is 2m">A ring rotates about the z axis as presented in the figure. The aircraft of rotation is x-y. At a details i... In a LR circuit, R=10Ωand L=2H. If an alternate voltage the 120V and also 60Hz is associated in this cir...A copper wire of diameter 1.6mm carries a present of 20A. Find the maximum size of the magnetic...Moment that inertia go not rely upon A. Angular velocity that the human body B. Shape and size C. Mass D. ...2 . At the prompt when that is velocity is 4m/s a ball is projected indigenous the floor the the lift through a speed of 4m/s , relative to the floor in ~ an key of 300 . The time taken by the round to return the floor is">A an extremely broad elevator is going up vertically through a constant acceleration that 2m/s2 . At t...An electron is revolving roughly a proton in a circular orbit of diameter 1A∘. If the produces a mag...In a collection RLC circuit the is operating above the resonant frequency, the current A. Lags the appl...−1. What will be the present necessity to create a magnetic ar of 20mT inside the solenoid (μ04π=10−7metreAmpere−1)">A lengthy solenoid is formed by winding 20 turns cm−1. What will certainly be the present necessity ...A taxi leaves the terminal X for terminal Y every 10 minutes. Simultaneously, a taxi leaves terminal Y ...
See more: How Many Words In 10 Minute Speech ? How Many Words Should A Speech Contain
Grade/ExamClass 1Class 2Class 3Class 4Class 5Class 6Class 7Class 8Class 9Class 10Class 11Class 12IASCATBank ExamGATE
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00572.warc.gz
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CC-MAIN-2022-33
| 3,411 | 6 |
https://cutiumum.net/graphing-linear-equations-t-chart-worksheet/
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math
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Table of Contents :
Top Suggestions Graphing Linear Equations T Chart Worksheet :
Graphing Linear Equations T Chart Worksheet And while slide rules use logarithmic scales nomograms can use either linear or logarithmic scales depending on the requirements of the equation the bar graph and the pie chart and pursue Many different equations used in the analysis of electric circuits may be graphed take for instance ohm s law for a 1 k resistor plot this graph following ohm s law then plot another graph I will be mapping j t linear model while the current catcher value sits at around 7 5 million war it is projected to rise above the 8 million mark during realmuto s extension years the.
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Graphing A Linear Equation Using A T Chart
Let S Review The Steps For Graphing With A T Chart Or Table Of Values 1 Make A T Chart Or Table Of Values To Show Three Or More Possible Solutions 2 If You Need To Write The Solutions As Coordinate Pairs In A Third Column 3 Plot The Coordinate Pairs On A Graph 4 Connect The Points Using A Line With Arrows 5 Write The Equation
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Graphing Linear Inequalities Worksheet
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Graphing Linear Equation Type 3 To Graph A Linear Equation First Make A Table Of Values Assume Your Own Values For X For All Worksheets Provided Here Substitute The X Values Of The Equation To Find The Values Of Y Complete The Tables Plot The Points And Graph The LinesEquations Tables And Graphs 6th Grade
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Graphing Linear Equations Homework Help Homework Help
Overview Graphing Linear Equations One Way To Express Linear Equations Is To Do Homework To Help Tutors Graph On The Coordinate Plane In Order To Graph The Line Accurately It Is Important To Know What The Slope Of The Line Is As An Aid To The Homework Of Graphing Linear Equations Homework Help The Persian War Sentences Phrases Homework Quest Homework Help Physics Helps You Graphically
Linear Graphs Worksheets Ks3 Gcse Teaching Resources
Two Worksheets For Ks3 Gcse On Linear Graphs Starting With The Basics Walk Pupils Through The Basics Of Finding A Table Of Values And Plotting Linear Graphs As Well As The Key Terms And Properties Of The Graphs Read More Free Loading Save For Later Preview And Details Files Included 2 Pptx 93 Kb Linear Graphs Algebra Worksheet 1 Pptx 86 Kb Linear Graphs Algebra Worksheet
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Tags: #linear equation algebra 1 worksheets#finding slope from a graph worksheet#linear equation practice worksheets#solve by graphing worksheet#graphing linear equations fun worksheets#linear equations worksheets 7th grade#point slope form worksheet
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s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735963.64/warc/CC-MAIN-20200805153603-20200805183603-00479.warc.gz
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CC-MAIN-2020-34
| 10,554 | 32 |
https://www.dealzon.com/computers/desktops
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math
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Cheap deals and best prices on desktop computers from HP, Dell, Sony, Lenovo, Apple, and many more manufacturers. From the latest Intel Haswell 4th Gen equipped computer desktops, to space-saving All-in-One computers - you'll find the latest coupons and deals curated by Dealzon's editors.
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$1,299.00$599.00at B&H PhotoEXPIRED Feb 12
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s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00228.warc.gz
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CC-MAIN-2018-09
| 528 | 7 |
https://groupprops.subwiki.org/wiki/Orbit-counting_theorem
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math
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- 1 Name
- 2 Statement
- 3 Related facts
- 4 Facts used
- 5 Proof
This result is called the orbit-counting theorem, orbit-counting lemma, Burnside's lemma, Burnside's counting theorem, and the Cauchy-Frobenius lemma.
In terms of group actions
Suppose a finite group has a group action on a set . For , denote by the stabilizer of in . Let be the set of orbits in under the action of . Further, for , let be the set of elements of fixed by . Then:
In terms of linear representations
Suppose a finite group has a permutation representation on a set . When viewed as a linear representation, this has character . Then, the number of orbits of under the action of is the inner product where is the trivial (principal) character of .
Related facts about group actions
Related counting techniques
- Polya's theorem is a more sophisticated version of this that involves a group acting on combinatorial configurations on a set.
Related ideas in other disciplines
The formula of the orbit-counting theorem, which in this case counts the number of orbits, gives an effective measure of the size of a quotient of a set by a group action. This formula can be generalized to a groupoid acting on a set.
Applications to conjugacy class-representation duality
The orbit-counting theorem has important applications to conjugacy class-representation duality. In particular, it is used to prove statements of the form for a given Galois automorphism group or group automorphism group, the number of orbits on the set of conjugacy classes is the same as the number of orbits on the set of irreducible representations. These statements hold even in cases where the actual orbit structures differ. (For cyclic groups of automorphisms/Galois automorphisms, the orbit structures must also coincide, by Brauer's permutation lemma).
Here are some applications:
- Number of irreducible representations over rationals equals number of equivalence classes under rational conjugacy
- Number of orbits of irreducible representations equals number of orbits of conjugacy classes under any subgroup of automorphism group, of which a special case is that number of orbits of irreducible representations equals number of orbits under automorphism group
We first prove equality of the first two sides, and then prove equality of the second and third side.
Equality of the first two sides
Given: A group acting on a set .
To prove: .
|No.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||For any orbit and any , we have||Fact (1)||acts on||Fact-direct|
|2||For any orbit and any , we have||Fact (2)||Step (1)||[SHOW MORE]|
|3||For any orbit and any , we have .||Step (2)||Algebraic rearrangement|
|4||For any orbit , we have||Step (3)||[SHOW MORE]|
|5||We get .||Step (4)||[SHOW MORE]|
|6||We get||Step (5)||Algebraic rearrangement|
Equality of the second and third side
To prove: .
Proof: Note that we can ignore the on both sides for the proof. The proof essentially relies on the observation that both sides are equal to the cardinality of the set:
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| 3,058 | 34 |
https://onlinessaywriting.com/the-construction-of-parallel-lines-using-a-point-and-a-line-makes-use-of-the-pro/
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math
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The construction of parallel lines using a point and a line makes use of the property “____________________________”
A.Parallel lines do not intersect.
B.Many transversals can be drawn through a set of parallel lines.
C.Transversals do not intersect.
D.A transversal makes equal angles of inclination with the parallel lines.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00484.warc.gz
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CC-MAIN-2022-49
| 329 | 5 |
https://zahedi.info/advanced-mathematical-economics-rakesh-vohra-55/
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math
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Detailed yet student-friendly, Vohra’s book contains chapters in, amongst as well as postgraduate students in mathematical economics will find this book. “In this volume, Rakesh V. Vohra sets out the basic concepts of mathematics as they relate Print version Vohra, Rakesh V. Advanced mathematical economics. Buy Advanced Mathematical Economics (Routledge Advanced Texts in Economics and Finance) 1 by Rakesh V. Vohra (ISBN: ) from Amazon’s.
|Published (Last):||27 July 2016|
|PDF File Size:||20.22 Mb|
|ePub File Size:||5.18 Mb|
|Price:||Free* [*Free Regsitration Required]|
Development Finance Stephen Spratt. Paperbackpages. Other links Inhaltsverzeichnis at http: Kevin Zollman marked it as to-read Nov 02, Selected pages Title Page. Mthematical marked it as to-read Jan 11, Mathematicsl online Borrow Buy Freely available Show 0 more links Lattices and supermodularity 8. Thanks for telling us about the problem. The University of Melbourne Library.
Includes bibliographical references and index. Skip to content Skip to search.
My library Help Advanced Book Search. Behavioral Economics Edward Cartwright. Strategic Entrepreneurial Finance Darek Klonowski.
Higher level undergraduates matyematical well as postgraduate students in mathematical economics will find this book extremely useful in their development as economists.
Advanced Mathematical Economics by Rakesh V.
Advanced mathematical economics / Rakesh Vohra. – Version details – Trove
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This concise textbook presents students with all they need for advancing in mathematical economics. The Best Books of Home All editions This editionEnglish, Book edition: There are no discussion topics on this book yet. Add a tag Cancel Be the first to add a tag for this edition. Detailed yet student-friendly, Vohra’s book contains chapters in, amongst others:. Of particular salience to mathmatical economic thought are sections on lattices, supermodularity, matroids and their applications.
Advanced Mathematical Economics by Rakesh V. Regional Economics Roberta Capello. Advanced Mathematical Economics Rakesh V. University of Western Australia Library.
Visit our Beautiful Books page and find lovely books for kids, photography lovers and more. The University of Melbourne. Just a moment while we sign you in to your Goodreads account. Separate different tags with a comma.
Advanced Mathematical Economics
These 3 locations in Western Australia: Looking for beautiful books? Federation University Australia Library. Higher rakexh undergraduates as well as postgraduate students in mathematical economics will find this book extremely useful in their development as economists.
Higher level undergraduates as well as postgraduate students in mathematical economics will find this book extremely useful in their development as economists.
Game Theory and Exercises Gisele Umbhauer. Vohra Limited preview – Refresh and try again. Estimated delivery business days. Tags What are tags? With access conditions MyiLibrary at http: University of the Sunshine Coast Library.
Common terms and phrases agent allocation assignment bidder Bolzano-Weierstrass theorem called choose columns concave function cone cone A constraint contradiction convex combination convex hull convex set defined Definition denoted equation equilibrium example exists extreme points Farkas lemma feasible region feasible solution finite number fixed point function f graph Hence hyperplane implies inequality Let f linear program local maximum m x n matrix mathematical matroid maximum mixed strategy Nash equilibrium non-decreasing non-empty objective function objective function value optimal objective function optimal solution optimization problem pair payoff polyhedral polyhedron price vector primal prove pure strategies rank function real numbers S C R satisfies semi-matching separating hyperplane theorem sequence short player span A Sperner’s lemma submodular subset sufficiently small supermodular theorem triangle utility function valued function vertex vertices white soma zero.
Public Private login e. We were unable to find this edition in any bookshop we are able to search. Mathematical Finance Nikolai Dokuchaev.
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https://www.giancolianswers.com/giancoli-physics-7th-edition-solutions/chapter-3/problem-48
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math
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A boat, whose speed in still water is 2.50 m/s, must cross a 285-m-wide river and arrive at a point 118 m upstream from where it starts (Fig. 3-45). To do so, the pilot must head the boat at a $45.0^\circ$ upstream angle. What is the speed of the river's current?
Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.
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CC-MAIN-2020-16
| 351 | 2 |
https://www.quesba.com/questions/1-a-fixed-charge-coverage-a-is-a-balance-sheet-indication-of-debt-carrying--53
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math
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1. A fixed charge coverage: a. is a balance sheet indication of debt carrying ability. b. is an income statement indication of debt carrying ability. c. is a liquidity ratio. d. frequently includes research and development. e computation is standard from firm to firm. If a firm has substantial capital or financing leases disclosed in the notes but not capitalized in the financial statements, then: a. the times interest earned ratio will be overstated, based upon the financial statements. b. the fixed charge ratio will be overstated, based upon the financial statements. c. the debt ratio will be understated. d. the working capital will be understated. e. None of the answers are correct. 3. Return on assets cannot fall under which of the following circumstances? 1. II. III. IV. Net Profit Margin decline rise rise decline Total Asset Turnover rise decline rise decline a. I b. II c. III d. IV e. The ratio could fall under all of the answers. 4. Which of the following would most likely cause a rise in net profit margin? a. Increased sales b. Decreased preferred dividends c. Increased cost of sales d. Decreased operating expenses e. Decreased earnings per share I Interest expense creates magnification of earnings through financial leverage because: 5. a. the interest rate is variable. b. interest accompanies debt financing. c. the use of interest causes higher earnings. d. interest costs are cheaper than the required rate of return to equity owners. e. while earnings available to pay interest rise, earnings to residual owners rise faster.
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CC-MAIN-2021-21
| 1,558 | 1 |
http://fqpaperqbck.cyclingjersey.us/find-the-answer-to-a-math-problem.html
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math
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Math mystery picture worksheets for addition, subtraction, division, and multiplication math solve the math problems students will have to find the answers to basic division problems all problems have a divisor of 6 3rd and 4th grades. Math problems with mistakes can you find the mistake in each problem menu and widgets for more math mistakes, see mathmistakesorg the correct answer to the problem on the first line above is. Yhomework - math solver 50,003 math underground education not only does it give an answer to the problem but it shows it step by step so you could get it enter your math problem, and let cymath solve it for you step-by-step. Discover how to use online math solver to find the right answer without we have introduced our free online math problem solver section where you can find answers to your problems what it means is that unlike other sites where their free math solvers only show the final answer. Introduction to math 40 solving problems: in mathematics, numeration is one of the main sources describing about numerals such as number system.
Answers to questions from people who know at ask experience project find the answer to your question. How to solve math problems want to compare your answers to the estimates you came up with for each step as well as your overall estimate for the problem's answer ask yourself, do my answers match or closely resemble my estimates. Solve math problems to decode the answers to funny riddles over 30 worksheets, covering addition, subtraction, multiplication, division, place value, rounding, and more. 'correct answers, but let's find the 'friendly' numbers'' a second grader's answers to a common core this common core math problem asks kids to write the 'friendly' answer, instead of the correct one robby soave reporter 10:06 pm 03/25/2014. Here's the answer to the cheryl's birthday math problem it's not too often a math problem goes viral huffpost impact news us news world news business environment health social justice entertainment culture & arts media.
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First you do have to distrubute the negative, this allows for removal of the parenthesis this gives 21 + -k + 3 = 16 + -3k + 4 now the variable, k, needs to be isolated on one side of the equation. Solving a math problem results in a numerical answer of equal value to the problem's equation most math problems require the solver to find the value of an unknown variable in the algebraic math. Developed by mit graduates, mathscore provides online math practice for simple interest and hundreds of other types of math problems. Ask any math question and get an answer from our subject experts in as little as 2 hours step-by-step solutions to problems over 22,000 isbns find textbook solutions math questions and answers from chegg.
The verb solve is often used in mathematics, and it means to answer a math problem you can solve other, non-mathematical problems too like the question of what to wear when you've run out of clean clothes (your sibling's clothes. This page includes very detailed explanations of the math practice problems the actual rectangle, so you can compare the proportions and find the correct answer the width of the scale drawing is 12 cm the. Finding answers to your math questions trying to do your math homework, but keep getting stuck on a tricky problem read on to learn about the resources available to help you with your toughest math assignments.
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| 5,210 | 7 |
https://19january2021snapshot.epa.gov/chp/methods-calculating-chp-efficiency_.html
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math
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Methods for Calculating CHP Efficiency
Every CHP application involves the recovery of heat that would otherwise be wasted. In this way, CHP increases fuel-use efficiency.
Two measures are commonly used to quantify the efficiency of a CHP system: total system efficiency and effective electric efficiency.
- Total system efficiency is the measure used to compare the efficiency of a CHP system to that of conventional supplies (the combination of grid-supplied electricity and useful thermal energy produced in a conventional on-site boiler). If the objective is to compare CHP system energy efficiency to the efficiency of a site's conventional supplies, then the total system efficiency measure is likely the right choice.
- Effective electric efficiency is the measure used to compare CHP-generated electricity to electricity generated by power plants, which is how most electricity is produced in the United States. If CHP electrical efficiency is needed to compare CHP to conventional electricity production (i.e., grid-supplied electricity), then the effective electric efficiency metric is likely the right choice.
Certain assumptions are implicit in each methodology that are not appropriate in all cases. Consequently, the measure employed should be selected carefully and the results interpreted with caution.
Key Terms Used in Calculating CHP Efficiency
Calculating a CHP system's efficiency requires an understanding of several key terms:
- CHP system. The CHP system includes the prime mover (e.g., combustion turbine, engine, microturbine), the electric generator, and the heat recovery unit that transforms otherwise wasted heat to useful thermal energy.
- Total fuel energy input (QFUEL). The heating value of the total fuel input. Total fuel input is the sum of all the fuel used by the CHP system. The total fuel energy input is often determined by multiplying the quantity of fuel consumed by the heating value of the fuel.
Commonly accepted heating values for natural gas, coal, and diesel fuel are:
- 1020 Btu per cubic foot of natural gas
- 10,157 Btu per pound of coal
- 138,000 Btu per gallon of diesel fuel
- Net useful electric output (WE). The gross electric output of the generator minus any parasitic electric losses. In other words, the net useful electric output is the total electric output from the CHP system that is put to a useful purpose.
- Gross electric output is the total electric output of the generator.
- Parasitic electric losses are the electrical power consumed by the CHP system; for example, the electricity used to compress natural gas before it is used as fuel in a combustion turbine.
- Net useful thermal output (∑QTH). The gross thermal output of the CHP system minus any thermal output that is not put to a useful purpose. In other words, the net useful thermal output is the total thermal output from the CHP system that is put to a useful purpose.
- In the case of a CHP system that produces 10,000 pounds of steam per hour, with 90 percent of the steam used for space heating and the remaining 10 percent exhausted in a cooling tower, the energy content of the 9,000 pounds of steam per hour is the net useful thermal output.
- Gross thermal output is the total thermal output of the CHP system.
Total System Efficiency
The total system efficiency (ηo) of a CHP system is the sum of the net useful electric output (WE) and net useful thermal output (∑QTH) divided by the total fuel energy input (QFUEL), as shown below:
The calculation of total system efficiency evaluates the combined CHP outputs (i.e., electricity and useful thermal output) based on the fuel consumed. CHP systems typically achieve total system efficiencies of 60 to 80 percent.
Note that this measure does not differentiate between the value of the electric output and the thermal output; instead, it treats electric output and thermal output as having the same value which allows them to be added (kWh can be converted to Btu using a standard conversion factor). In reality, electricity is considered a more valuable form of energy because of its unique properties.
Effective Electric Efficiency
Effective electric efficiency (ℰEE) can be calculated using the equation below, where WE is the net useful electric output, ∑QTH is the sum of the net useful thermal output, QFUEL is the total fuel energy input, and α equals the efficiency of the conventional technology that would be used to produce the useful thermal energy output if the CHP system did not exist:
For example, if a CHP system is natural gas-fired and produces steam, then α represents the efficiency of a conventional natural gas-fired boiler. Typical boiler efficiencies are 80 percent for natural gas-fired boilers, 75 percent for biomass-fired boilers, and 83 percent for coal-fired boilers.
The calculation of effective electric efficiency is the CHP net electric output divided by the additional fuel the CHP system consumes over and above what would have been used by a boiler to produce the thermal output of the CHP system.
Typical effective electric efficiencies for combustion turbine-based CHP systems range from 50 to 70 percent. Typical effective electric efficiencies for reciprocating engine-based CHP systems range from 70 to 85 percent.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00710.warc.gz
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CC-MAIN-2023-40
| 5,259 | 29 |
https://www.physicsforums.com/threads/trajectory-does-it-clear-a-certain-height-problem-help.317906/
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math
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Trajectory, "does it clear a certain height" problem help! 1. The problem statement, all variables and given/known data A place kicker must kick a football from a point 45.5 m from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 19.0 m/s at an angle of 48.0° to the horizontal. By how much does the ball clear or fall short of clearing the crossbar? 2. Relevant equations Dy = distance in y = Voy *t -1/2gt^2 Dx = ~~~ x = Vox * t Vox = initial velocity in x Voy = ~~~~ y 3. The attempt at a solution I tried to express Y in terms of x... since x = Vox * t .... t = x / Vox then Dy = Voy(x) / Vox - 4.9 (x^2/Vox^2) assume gravity's 9.8 Plug in 45.5 m for x Voy = 14.12 m/s Vox = 12.71 m/s Dy = -12.23 m So when the ball is 45.5 m long, its 12.23m underground. which means the ball is 12.23 - 3.05 = 9.18m short of crossing the bar. Idk... seems simple enough but apprently the answer's wrong. Thanks alot for your help!
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CC-MAIN-2018-30
| 997 | 1 |
https://onepalate.biz/significant-figures-pdf-free-download/
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math
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Significant Figures Rules: 1) All nonzero digits are significant. 2) Zeros between significant digits are significant. 3) Zeros to the left of nonzero digits are not significant. 4) Zeroes at the end of a number are significant only if they are to the right of the decimal point. Significant Figures Examples Number Significant Figures 70.2 3 0. PDF Significant Figures Download ebook full free. Significant Figures available for download and read online in pdf, epub, mobi.
Author: Ian Stewart
Publisher: Basic Books
ISBN: 9780465096121 The inpaint online.
Significant Figures Pdf Free Download Free
Significant Figures Pdf Free Download Adobe Reader For Windows 10
Significant Figures Pdf Free Download Windows 10
In Significant Figures, acclaimed math popularizer Ian Stewart introduces the visionaries of mathematics throughout history. Delving into the lives of twenty-five great mathematicians, Stewart examines the roles they played in creating, inventing, and discovering the mathematics we use today. Through these short biographies, we get acquainted with the history of mathematics from Archimedes to Benoit Mandelbrot, and learn about those too often left out of the cannon, such as Muhammad ibn Musa al-Khwarizmi (c. 780-850), the creator of algebra, and Augusta Ada King (1815-1852), Countess of Lovelace, the world's first computer programmer. Tracing the evolution of mathematics over the course of two millennia, Significant Figures will educate and delight aspiring mathematicians and experts alike.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585460.87/warc/CC-MAIN-20211022052742-20211022082742-00230.warc.gz
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CC-MAIN-2021-43
| 1,517 | 8 |
https://sharemylesson.com/standards/virginia-doe/4.4.10-0
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math
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In order to develop and use strategies to learn the multiplication facts through the twelves table, students should use concrete materials, a hundreds chart, and mental mathematics. Strategies to learn the multiplication facts include an understanding of multiples, properties of zero and one as factors, commutative property, and related facts. Investigating arithmetic operations with whole numbers helps students learn about the different properties of arithmetic relationships. These relationships remain true regardless of the whole numbers.
Lessons for this standard
Resources cannot be aligned to this standard, browse sub-standards to find lessons.
Similar standards in other grades
Apply strategies, including place value and the properties of multiplication and/or addition when multiplying and dividing whole numbers. (a, b, c, d)
Create practical problems to represent a multiplication or division fact. (b)
As students work to solve multiplication and division problems, they naturally tend to utilize strategies that involve place value understanding and properties of the operations. Applying the commutative property of multiplication (e.g., 5 x 8 = 8 x 5) reduces in half the number of multiplication facts that students must learn. The distributive property of multiplication allows students to find the answer to a problem such as 6 x 7 by decomposing 7 into 3 and 4 (e.g., 6 x 7= 6 x (3 + 4)) allowing them to think about (6 x 3) + (6 x 4) = 18 + 24 = 42.
dividends do not exceed four digits.
The student will a) represent multiplication and division through 10 × 10, using a variety of approaches and models; b) create and solve single-step practical problems that involve multiplication and division through 10 × 10; c) demonstrate fluency with multiplication facts of 0, 1, 2, 5, and 10; and d) solve single-step practical problems involving multiplication of whole numbers, where one factor is 99 or less and the second factor is 5 or less.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00337.warc.gz
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CC-MAIN-2020-05
| 1,966 | 9 |
https://brilliant.org/problems/double-rainbow-all-the-way-across-the-sky/
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math
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In a marathon, there are 14 runners. There are exactly of two runners that are wearing the same color shirt for each color of the rainbow.
At the finish line, their configuration is as follows:
- 1 runner between the red pairs,
- 2 runners between the orange pairs,
- 3 runners between the yellow pairs,
- 4 runners between the green pairs,
- 5 runners between the blue pairs,
- 6 runners between the indigo pairs,
- 7 runners between the violet pairs.
If we know that the first runner wore a red shirt, what is the total number of possible configuration(s) of all the runners (from fastest to slowest)?
As an explicit example, if the runners were arranged as ROYGBIVROYGBIV, then there are 6 runners between all of the colored pairs.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00168.warc.gz
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CC-MAIN-2021-49
| 734 | 11 |
https://www.jiskha.com/display.cgi?id=1314641466
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math
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posted by renee
At a certain time a particle had a speed of 91 m/s in the positive x direction, and 4.7 s later its speed was 73 m/s in the opposite direction. What was the average acceleration of the particle during this 4.7 s interval?
a = (Vf - Vo) / t.
a = (-73 - 91) / 4.7 = -34.9m/s^2.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00381.warc.gz
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CC-MAIN-2018-30
| 291 | 4 |
https://forums.unrealengine.com/t/suggestion-anchor-reroute/76586
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math
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What you guys think about this idea?
GIV3_UP.zip (126 KB)
I don`t know if I can put this type of message here in this forum, but I do not make idea where I can put. is just a suggestion, but I think this can help (or not ) so much in visualization of bluprints. What you guys think?
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00506.warc.gz
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CC-MAIN-2021-17
| 282 | 3 |
https://amp.doubtnut.com/ncert-solutions/class-10-maths-1
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math
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Doubtnut.com presents complete NCERT solutions for class 10 maths in video tutorial format. To make it easier for you to learn and understand maths the video tutorials are prepared by our esteemed mathematicians from the renowned IITs of India. These are some of the best online lectures on maths, where our experts have discussed a wide array of class 10 maths topics. Class 10th Maths as a subject is vast; therefore, we’ve listed every important topic by segregating its chapters into subsequent exercises. In our video tutorials, we've discussed Real Numbers, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Coordinate Geometry, Trigonometry, Some Applications of Trigonometry, Circles, Constructions, Area Related to Circles, Surface Areas and Volumes, Statistics and Probability and more…
Cengage Chapterwise Maths Solutions
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
INTRODUCTION TO TRIGONOMETRY
SOME APPLICATIONS OF TRIGONOMETRY
AREAS RELATED TO CIRCLES
SURFACE AREAS AND VOLUMES
Key Highlights of NCERT Solutions for Class 10 Maths:
1. Strictly as per the CBSE examination guidelines.
2. Fully solved exercises of NCERT Math textbook for Class 10.
3. The questions have been answered by the IITians and best mathematicians of the country.
4. Solutions are as per the CBSE marking scheme.
5. Videos are full of tips and tricks to help you gain the competitive edge to ace the board examination.
Diagrams are given to help the students in visualizing the solutions.
All the questions from each chapter are covered.
Videos are enriched with expert advice, math's tips and step-by-step maths solutions to give a proper justification to every math's problem. It would provide students detailed insights on every important topic. The NCERT solutions for class 10 maths in Hindi are prepared to keep the student's need in mind. You can easily learn either using your mobile or personal computer. The given video tutorials provide students detailed insights on every important topic.
Chapter 1: Real Numbers
Do you know natural numbers, whole numbers, integers, fractions, rational numbers and irrational numbers are all real numbers? Any real number that can be found on the number line is known as the real number. The which we use and apply in the real world calculations are the real numbers... In this chapter, you'll be learning some interesting concepts related to a real number and their useful applications. The video tutorial covers every important topic related to real numbers and gives some great examples. The exercise wise NCERT solution class 10 maths book are listed below: Exercise 1.1 Introduction Exercise 1.2 Euclid’s Division Lemma Exercise 1.3 The Fundamental Theorem of Arithmetic Exercise 1.4 Revisiting Irrational Numbers Exercise 1.5 Revisiting Rational Numbers and Their Decimal Expansions Exercise 1.6 Summary
Chapter 2: Polynomials
In this chapter, you'll be learning about the polynomials in-depth. Starting from the exercise 2.1 Introduction to polynomials, you’ll be learning how to simplify and evaluate polynomials. Then you’ll learn about the zeros or roots of polynomials, roots of quadratic equations and cubic equation and it’s coefficient. In the final exercise 2.5, you'll learn about how to do long division of the polynomials. Exercise-wise solutions we’ve covered include: Exercise 2.1 Introduction Exercise 2.2 Geometrical Meaning Of The Zeroes Of A Polynomial Exercise 2.3 Relationship Between Zeroes And Coefficients Of A Polynomial Exercise 2.4 Division Algorithm For Polynomials Exercise 2.5 Polynomials Exercise 2.6 Summary
Chapter 3: Pair of Linear Equations in Two Variables
In NCERT class 10 Chapter 3, you'll be learning about the interpretation of linear equations in two variables. With given NCERT solutions for class 10, maths students can easily model the linear equations into real-world problems. You’ll enjoy learning to solve the linear equation problems both algebraically and graphically. In the algebraic method, you will learn some important concepts like the elimination method, substitution method, and cross-multiplication method. Then finally we will move on to equations reducible to a pair of linear equations in two variables. The exercise-wise NCERT solutions we’ve covered include: Exercise 3.1 Introduction Exercise 3.2 Pair of Linear Equations in Two Variables Exercise 3.3 Graphical Method of Solution of a Pair of Linear Equations Exercise 3.4 Algebraic Methods of Solving a Pair of Linear Equations Exercise 3.4.1 Substitution Method Exercise 3.4.2 Elimination Method Exercise 3.4.3 Cross-Multiplication Method Exercise 3.5 Equations Reducible To a Pair of Linear Equations in Two Variables Exercise 3.6 Summary
Chapter 4: Quadratic Equations
So far you have learned about the linear equations and linear equations in two variables. In chapter 4, you'll be learning some new concepts, which includes variables raise to the second power and how to take square roots on both sides. You'll also learn to solve the factored equation like (x-1) (x+3) = 0 and using the factorization method, solution of quadratic equation by completing the square and finally the nature of roots. Listed below are the exercise wise NCERT solutions for chapter 4 quadratic equations: Exercise 4.1 Introduction Exercise 4.2 Quadratic Equations Exercise 4.3 Solution of a Quadratic Equation by Factorization Exercise 4.4 Solution of a Quadratic Equation by Completing the Square Exercise 4.5 Nature of Roots Exercise 4.6 Summary
Chapter 5: Arithmetic Progressions
Find here detailed and in-depth answers to Class 10 Maths Chapter 5 Arithmetic Progressions (AP). If the difference between two consecutive terms is constant in a progression, it is known as the arithmetic progression. In this lesson five, you’ll be introduced important concepts of arithmetic progression and you’ll be studying about constructing an arithmetic progression (A.P). After learning the basics of AP, you can learn to calculate the Nth term of an AP. Then you will see how to find the sum of n terms of an AP. It is interesting and fun to learn the concept of mathematics. The exercise wise NCERT solutions for chapter 5 Arithmetic Progressions we’ve covered are: Exercise 5.1 Introduction Exercise 5.2 Arithmetic Progressions Exercise 5.3 Nth Term of An AP Exercise 5.4 Sum Of First N Terms Of An AP Exercise 5.5 Summary
Chapter 6: Triangles
Find here complete study material on class 10 chapter 6 triangles. Triangle is an interesting 3 cornered shape that has some unique properties to explore. In this chapter 6 triangles, you'll be studying all about triangles, similarity criterion of triangles and their properties. Itemized below are the exercises that we’ve covered here: Exercise 6.1 Introduction Exercise 6.2 Similar Figures Exercise 6.3 Similarity Of Triangles Exercise 6.4 Criteria for Similarity of Triangles Exercise 6.5 Areas of Similar Triangles Exercise 6.6 Pythagoras Theorem Exercise 6.7 Summary
Chapter 7: Coordinate Geometry
Get Free NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry. In lesson 7 you’ll learn about finding distances between the two points whose coordinates are given. You’ll also learn how to find the coordinates of the point using the distance formula, section formula, area of triangle etc. Given NCERT Solutions were prepared by experienced mathematicians to helps students build a stronger foundation in mathematics. Detailed exercise wise answers to Chapter 7 Maths Class 10 Coordinate Geometry are provided below: Exercise 7.1 Introduction Exercise 7.2 Distance Formula Exercise 7.3 Section Formula Exercise 7.4 Area of a Triangle Exercise 7.5 Summary
Chapter 8: Introduction to Trigonometry
In chapter eight introductions to trigonometry, you'll learn about some important concepts like trigonometric ratios, trigonometric ratios of some specific angles, Trigonometric Ratios of Complementary Angles and finally Trigonometric Identities. This lesson is restricted to the acute angles only, however, ratios can be extended to other angles as well. We will also define and calculate trigonometric ratios and some simple identities involving these ratios, called trigonometric identities. The exercise wise NCERT solution for Class 10 Maths Chapter 8 Introduction to Trigonometry we’ve covered includes: Exercise 8.1 Introduction Exercise 8.2 Trigonometric Ratios Exercise 8.3 Trigonometric Ratios Of Some Specific Angles Exercise 8.4 Trigonometric Ratios Of Complementary Angles Exercise 8.5 Trigonometric Identities Exercise 8.6 Summary
Chapter 9: Some Applications of Trigonometry
In this lesson, you’ll be learning about the practical application of trigonometry. The NECRT solutions will teach about calculating the heights and distances of various objects, without actually quantifying them. The exercise covered includes: Exercise 9.1 Introduction Exercise 9.2 Heights and Distances Exercise 9.3 Summary
Chapter 10: Circles
With NCERT Class 10 chapter 10 Circles you’ll be learning various types of situations that can arise when a circle and a line are given in a plane. You'll also learn the concept of a tangent and number of tangents from a point on a circle. The exercise wise NCERT solution for Class 10 Maths Chapter 10 Circles that we’ve covered includes: Exercise 10.1 Introduction Exercise 10.2 Tangent to a circle Exercise 10.3 Number of tangents from a point on a circle Exercise 10.4 Summary
Chapter 11: Constructions
Class 10 Chapter 11 constructions are an important part of the geometry that has some useful real-world applications. In this chapter, you be learning about various types of constructions in geometry. The exercise wise NCERT solutions we’ve covered are listed below: Exercise 11.1 Introduction Exercise 11.2 Division of A Line Segment Exercise 11.3 Construction Of Tangents To A Circle Exercise 11.4 Summary
Chapter 12: Area Related to Circles
Chapter 12 will begin with the concepts of circumference and area of a circle. Then you'll be shifted to the area of sector and segment of a circle and apply this you’ll be extending your knowledge to sector and segment of circles. In the last exercise, you are to learn about the areas of combinations of plane figures. Here are the exercise wise solutions that we’ve included: Exercise 12.1 Introduction Exercise 12.2 Perimeter and area of a circle – a review Exercise 12.3 Areas of sector and segment of a circle Exercise 12.4 Areas of combinations of plane figures Exercise 12.5 Summary
Chapter 13: Surface Areas and Volumes
NCERT chapter 13 surface areas and volumes are an important part of geometry in which you'll be learning to calculate the surface area, volume, or perimeter of a variety of geometrical solid shapes such as circles, rectangle, pyramid, cube, triangle etc. Each has its own specific formulas and method to meet the solution. Listed below are the exercise wise NCERT solutions for chapter 13 surface areas and volumes: Exercise 13.1 Introduction Exercise 13.2 Surface Area of a Combination Of Solids Exercise 13.3 Volume of a Combination Of Solids Exercise 13.4 Conversion of Solid from One Shape to Another Exercise 13.5 Frustum of a Cone Exercise 13.6 Summary
Chapter 14: Statistics
In this lesson, you’ll be learning about the grouped data such as mean, median and mode and the concept of cumulative frequency. The NCERT solutions will provide a comprehensive and in-depth study of the statistics. Find here detailed answers to Chapter 14 Maths Class 10 Statistics: Exercise 14.1 Introduction Exercise 14.2 Mean of Grouped Data Exercise 14.3 Mode of Grouped Data Exercise 14.4 Median of Grouped Data Exercise 14.5 Graphical Representation Of Cumulative Frequency Distribution Exercise 14.6 Summary
Chapter 15: Probability
Probability is the final chapter of NCERT class 10th maths books. In this chapter, you'll be introduced to theoretical, also known as the classical probability of an event. The video tutorials talk about some of the very simple problems based on the concept of probability. Listed below are the exercise wise solutions that we’ve covered: Exercise 15.1 Introduction Exercise 15.2 A Theoretical Approach Exercise 15.3 Summary
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