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http://math.stackexchange.com/questions/366096/curvature-of-the-boundary-of-a-convex-set-is-positive | # curvature of the boundary of a convex set is positive
Let's consider $J\subset \mathbb R^2$ such that J is convex and such that it's boundary it's a curve $\gamma$. Let's suppose that $\gamma$ is anti-clockwise oriented, let's consider it signed curvature $k_s$. I want to prove the intuitive following fact:
$$\int\limits_\alpha {k_s } \left( s \right)ds \geqslant 0$$
For every sub-curve $\alpha \subset \gamma$.
And then prove that $k_s(s) \ge 0$
I have no idea how to attack this problem, intuitively I can see the result.
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Isn't $\kappa=||\frac{dT}{ds}||$ by definition? – user69810 Apr 19 '13 at 4:26
But here we are working with the signed curvature (it can be defined in the case of planar curves) – Miguel Apr 19 '13 at 4:31
Is the only difference the removal of the absolute value? – user69810 Apr 19 '13 at 4:33
If you put absolute values, is the same, but has an important geometric difference en.wikipedia.org/wiki/Curvature#Signed_curvature – Miguel Apr 19 '13 at 4:36
I see. Somehow we have to use the convexity to prove that the orientation doesn't change. – user69810 Apr 19 '13 at 4:40
If the curvature is negative, there must be a point with negative curvature. As you zoom up to that point, it looks more and more like the complement of a circle, which means that there are two points which are not connected by a straight line in the set.
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But I can prove it, in a more formal way? – Miguel Apr 19 '13 at 3:53
Yes; rotate the curve so the part with negative curvature is at the origin with the circle of curvature having its center on the y-axis. Then the difference between the circle and the curve can be given by a power series $a_3 x^3+...$. As $x\rightarrow 0$ from the right, this difference will become less then the height of the circle for some point; similarly for the left. The straight line between these points does not lie in the set, so the set is not convex. – Brian Rushton Apr 19 '13 at 12:57
I did not understand why that line does not live in the set – Miguel Apr 19 '13 at 14:20
Because the two points lie under the circle but above the x-axis. In particular, they have a non-zero y-coordinate, so the line between them intersects the y-axis at a point (0,y) which is not in the set. – Brian Rushton Apr 19 '13 at 17:06
This is a more formal version of Brian Rushton's answer. Suppose there is a point of negative curvature. Choose $xy$ coordinates so that this point is the origin $(0,0)$, the tangent direction is $x$-axis, and the $y$-axis points inside the convex set. Let $y=f(x)$ be the equation of a part of curve near $(0,0)$. (Implicit function theorem says you can solve for $y$ in terms of $x$.)
The curvature at $(0,0)$ is $f''(0)$, according to equation (14) here. Since $f''(0)<0$ and $f'(0)=0$, it follows that $f(x)<0$ for $0<|x|<\delta$, if $\delta$ is sufficiently small. This contradicts the convexity of the set: e.g., the line segment from $(x,f(x))$ to $(-x,f(-x))$ crosses the negative half of the $y$-axis.
If $s$ is arc-length, $T(s)$ is the unit tangent vector and $N(s)$ the counterclockwise unit normal, $\dfrac{d}{ds} T(s) = k(s) N(s)$. It's convenient to consider the plane as the complex plane, so $T(s) = e^{i\theta(s)}$ and $N(s) = i e^{i \theta(s)}$. Then we have $\dfrac{d\theta}{ds} = k(s)$. Now you want to show that $\theta(s)$ is nondecreasing... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917284369468689, "perplexity": 194.1415554159986}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999652955/warc/CC-MAIN-20140305060732-00089-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://kmr.csc.kth.se/wp/research/math-rehab/learning-object-repository/algebra/geomtric-algebra/historical-remarks/ | # Historical Remarks
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Related KMR-pages:
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Books:
John Stillwell (2016), Elements of Mathematics – From Euclid to Gödel,
Princeton University Press, ISBN 978-0-691-17854-7.
• John Stillwell (1999, (1989)), Mathematics and Its History, Springer, ISBN 0-387-96981-0.
• John Stillwell (1980), Classical Topology and Combinatorial Group Theory, Springer Verlag, ISBN 0-387-90516-2.
• Jeremy Gray (2007), Worlds Out of Nothing – A Course in the History of Geometry in the 19th Century, Springer, ISBN 1-84628-632-8.
Geometric Algebra for Computer Science – An Object-Oriented Approach To Geometry
///////
Other relevant sources of information:
Geometric Algebra (at Wikipedia)
• The Cauchy-Binét formula (which generalizes the product formula for determinants):
Det(AB) = Det(A) Det(B))
//////////////////////////////////////////////////////////////////////////////////////////
The rest of this page is based on Stillwell (2016, sections 5.10, 5.11 and 10.9)
//////////////////////////////////////////////////////////////////////////////////////////
/////// Quoting Stillwell (2016, section 5.10, p. 177):
Geometry / Historical Remarks
Euclid’s Elements is the most influential mathematical book of all time, and it gave mathematics a geometric slant that persisted until the twentieth century. Until quite late in the twentieth century, students were introduced to mathematical proof in the style of the Elements, and indeed it is hard to argue with a method that survived over 2000 years.
However, we know that Euclidean geometry has an algebraic description (vector space with an inner product), so we have another “eye” with which to view geometry, which we surely should use. To see how this new viewpoint came about, we review the history of elementary geometry.
Euclid’s geometry, illustrated in sections 5.2 to 5.4, has delighted thinkers through the ages with its combination of visual intuition, logic, and surprise. We can visualize what Euclid is talking about (points, lines, areas), we are impressed that so many theorems follow from so few axioms, and we are convinced by the proofs – even when the conclusion is unexpected. Perhaps the biggest surprise is the Pythagorean theorem. Who expected that the side lengths of a triangle would be related to their squares?
The Pythagorean theorem reverberates through Euclid’s geometry and all its decendants, changing from a theorem to a definition on the way (in a Euclidean space it holds virtually by definition of the inner product). For the ancient Greeks, the Pythagorean theorem led to $\, \sqrt{ 2 } \,$ and hence to the unwelcome discovery of irrational lengths. As mentioned in section 5.3, the Greeks drew the conclusion that lengths in general are not numbers and that they cannot be multiplied like numbers. Instead, the “product” of two lengths is a rectangle, and equality of products must be established by cutting and reassembling areas. This was complicated, though interesting and successful for the theory of area. However, as mentioned in section 5.3, one needs to cut volumes into infinitely many pieces for a satisfactory theory of volume.
It should be emphasized that the subject of the Elements is not just geometry, but also number theory (Books VII to IX on primes and divisibility) and an embryonic theory of real numbers (Book V), in which arbitrary lengths are compared by means of their rational approximations. Later advances in geometry, particularly Descartes (1637) and Hilbert (1899), work towards fusing all these subjects of the Elements into a unified whole.
The first major conceptual advance in geometry after Euclid came with the introduction of coordinates and algebra, by Fermat and Descartes in the 1620s. These two mathematicians seem to have arrived at the same idea independently, with very similar results. For example, they both discovered that the curves with equations of degree 2 are precisely the conic sections (ellipses, parabolas, and hyperbolas). Thus they achieved not only a unification of geometry and algebra, but also a unification of Euclid’s geometry with the Conics of Appolonius (written some decades after the Elements). Indeed, they set the stage for algebraic geometry, where curves of arbitrarily high degree could be considered. When calculus emerged, shortly thereafter, algebraic curves of degree 3 and higher provided test problems for the new techniques for finding tangents and areas.
However, algebra and calculus did not merely create new geometric objects; they also threw new light on Euclid’s geometry. One of Descartes’ first steps was to break the taboo on multiplying lengths, which he did with the similar triangle constructions for product, quotient, and square root of lengths described in sections 5.5 and 5.6. Thus, for all algebraic purposes, lengths could now be viewed as numbers. And by giving an algebraic description of numbers constructible by straightedge and compass, Descartes paved the way for the nineteenth-century proofs of non-constructibility, such as the one given in section 5.9.
Descartes did not intend to construct a new foundation for geometry, with “points” being ordered pairs $\, (x, y) \,$ of numbers, “lines” being point sets satisfying linear equations, and so on. He was content to take points and lines as Euclid described them, and mainly wished to solve geometric problems more simply with the help of algebra. (Sometimes, in fact, he tried to solve algebraic equations with the help of geometry.)
The question of new foundations for geometry came up only when Euclid’s geometry was challenged by non-Euclidean geometry in the 1820s. We say more about non-Euclidean geometry in the subsection below. The challenge of non-Euclidean geometry was not noticed at first, because the geometry was conjectural and not taken seriously by most mathematicians. This all changed when Beltrami (1868), building on some ideas of Gauss and Riemann, constructed models of non-Euclidean geometry, thereby showing that its axioms were just as consistent as those of Euclid.
Beltrami’s discovery was an earthquake that displaced Euclidean geometry from its long-held position at the foundation of mathematics. It strengthened the case for arithmetization: the problem of founding mathematics on the basis of arithmetic, including the real numbers, instead of geometry. Arithmetization was already under way in calculus and, thanks to Descartes, arithmetic was a ready-made foundation for Euclidean geometry. Beltrami’s models completed the triumph of arithmetization in geometry, because they too were founded on the real number and calculus.
By and large, geometry remains arithmetized today, with both Euclidean and non-Euclidean “spaces” situated among a great variety of manifolds that locally resemble $\, \mathbb{R}^n$, but with “curvature.” Among them, Euclidean geometry retains a somewhat privileged position as the one with zero Gaussian curvature
When a surface has a zero Gaussian curvature at all of its points, then it is a developable surface and the geometry of the surface is Euclidean geometry.
In this sense, Euclidean geometry is the simplest geometry, and the non-Euclidean spaces of Beltrami are among the next simplest, with constant (but negative) curvature at all points. In modern geometry of curved spaces, Euclidean spaces have a special place as tangent spaces. A curved manifold has a tangent space at each point, and one often works in the tangent space to take advantage of its simpler structures (particularly, its nature as a vector space).
It is all very well to base geometry on the theory of real numbers, but how well do we understand the real numbers? What is their foundation? Hilbert (1899) raised this question, and he had an interesting answer: the real numbers can be based on geometry! More precisely, they can be based on a “completed” version of Euclid’s geometry. Hilbert embarked on the project of completing Euclid’s axioms in the early 1890s, first with the aim of filling in some missing steps in Euclid’s proofs. As the project developed, he noticed that addition and multiplication arise from his axioms in a quite unexpected way, so that the field concept can be given a completely geometric foundation. Se section 5.11. Then, by adding an axiom guaranteeing that the line has no gaps, he was able to recover a complete “number line” with the usual properties of $\, \mathbb{R}$.
/////// End of Quote from Stillwell (2016)
/// Connect with the KMR page on Differential Geometry
/// Connect with the KMR pages on Plane Curves and Surfaces:
/////// Quoting Stillwell (2016, section 5.10, p. 180):
Non-Euclidean Geometry
In the 1820s, Janos Bolyai and Nikolai Lobachevsky independently developed a rival geometry to Euclid’s: a non-Euclidean geometry that satisfied all of Euclid’s axioms except the parallel axiom. The parallel axiom has a different character from the other axioms, which describe the outcome of finite constructions or “experiments” one can imagine carrying out:
1. Given two points, draw the line segment between them.
2. Extend a line segment for any given distance.
3. Draw a circle with given center and radius.
4. Any two right angles are equal (that is, one can be moved to coincide with the other).
The parallel axiom, on the other hand, requires an experiment that involves an indefinite wait:
5. Given two lines $\, l \,$ and $\, m$, crossed by another line $\, n \,$ making interior angles with $\, l \,$ and $\, m \,$ together less than two right angles, the lines $\, l \,$ and $\, m \,$ will meet, if produced indefinitely (figure 5.31).
Since the time of Euclid, mathematicians have been unhappy with the parallel axiom, and have tried to prove it from the other, more constructive, axioms. The most determined attempt was made by Saccheri (1733), who got as far as showing that, if non-diverging lines $\, l \,$ and $\, m \,$ did not meet, they would have a common perpendicular at infinity. This, Saccheri thought, was “repugnant to the nature of straight lines.” But it was not a contradiction, and in fact there is a geometry in which lines behave precisely in this fashion.
Bolyai and Lobachevsky worked out a large and coherent body of theorems that follow from Euclid’s axioms 1 to 4 together with the axiom:
5′. There exist two lines $\, l \,$ and $\, m$, which do not meet, although they are crossed by another line $\, n \,$ making interior angles with $\, l \,$ and $\, m$ together less than two right angles.
Their results were eventually published in Lobachevsky (1829) and Bolyai (1832) (the latter an appendix to a book by Bolyai’s father). They found no contradiction arising from this set of axioms, and Beltrami (1868) showed that no contradiction exists, because axioms 1, 2, 3, 4, and 5′ are satisfied under a suitable interpretation of the words “point,” “line,” and “angle”. (We sketch one such interpretation below.) Thus Euclid’s geometry had a rival, and deciding how to interpret the terms “point,” “line,” “distance,” and “angle” became an issue.
As we have seen, Euclid’s axioms had a ready-made interpretation in the coordinate geometry of Descartes: a “point” is an ordered pair $\, (x, y) \,$ of real numbers, a “line” consists of the points satisfying an equation $\, ax + by + c = 0$, and the “distance” between $\, (x_1, y_1) \,$ and $\, (x_2, y_2) \,$ equals $\, \sqrt{ {(x_2 - x_1)}^2 + {(y_2 - y_1)}^2 }$.
For Bolyai’s and Lobachevsky’s axioms, Beltrami found several elegant interpretations, admittedly with a somewhat complicated concept of “distance.” The simplest is probably the half-plane model, in which:
• “points” are points of the upper half-plane; that is, pairs $\, (x, y) \,$ with $\, y > 0$,
• “lines” are the open semicircles in the upper half-plane with their centers on the $\, x$-axis, and the open half-lines $\, \{ \, (x, y) \, : x = a \, , \, y > 0 \, \}$,
• the “distance” between “points” $\, P \,$ and $\, Q \,$ is the integral of $\, {\sqrt{ dx^2 + dy^2 }} / y \,$ over the “line” connecting $\, P \,$ and $\, Q$.
It turns out that “angle” in this model is just the ordinary angle between curves; that is, the angle between their tangents. This leads to some beautiful pictures of non-Euclidean geometric configurations, such as the one in figure 5.32.
/// Figure 5.32 GOES HERE
It shows a tiling of the half-plane by triangles which are “congruent” in the sense of non-Euclidean distance. In particular, they each have angles $\, \pi/2, \pi/3, \pi/7$. Knowing that they are congruent one can get a sense of non-Euclidean distance. One can see that the $\, x$-axis is infinitely far away – which explains why it is not included in the model – and perhaps also see that a “line” is the shortest path joining its endpoints, if one estimates distance by counting the number of triangles along a path.
It is also clear that the parallel axiom fails in the model. Take for example the “line” that goes straight up the center of the picture and the point on the far left with seven “lines” passing through it. Several of the latter “lines” do not meet the center line.
/////// End of Quote from Stillwell (2016)
/// Connect with the KMR pages on:
/////// Quoting Stillwell (2016, section 5.10, p. 182):
Vector Space Geometry:
Grassmann (1861), the Lehrbuch der Arithmetik mentioned in section 1.9, was not Grassmann’s only great contribution to the fundations of mathematics. The first was his Ausdehnungslehre (“Extention theory”), Grassmann (1844), in which he based Euclidean geometry on vector spaces. The Ausdehnungslehre, like the unfortunate Lehrbuch, was greeted at first with total incomprehension. The only person to review it was Grassmann himself, and its virtually unsold first edition was destroyed by the publisher. The full story of the Ausdehnungslehre, its genesis and aftermath, is in the Grassmann biography by Petsche (2009).
Grassmann was let down by an extremely obscure style, and terminology of his own invention, in attempting to explain an utterly new and complex idea: that of a real n-dimensional vector space with an outer product.
[Footnote 6: We will not define the concept of outer product, but it underlies the concept of determinant, then the center of what was called “determinant theory” and now at a less central position in today’s linear algebra]
The simpler concept of inner product was in Grassmann’s view an offshoot of the outer product – one he planned to expound in Ausdehnungslehre, volume two. Not surprisingly, the second volume was abandoned after the failure of the first.
Thus, Grassmann’s contributions to geometry might well have been lost – if not for a marvelous stroke of luck. In 1846, the Jablonowskian Society of Science in Leipzig offered an essay prize on a question only Grassmann was ready to answer: developing a sketchy idea of Leibniz about “symbolic geometry.” (The aim of the prize was to commemorate the 200th anniversary of Leibniz’s birth.) Grassmann (1847) duly won the prize with a revised version of his 1844 theory of vector spaces – one that put the inner product and its geometric interpretation at the center of the theory. He pointed out that his definition of inner product was motivated by the Pythagorean theorem, but that, once the definition was given, all geometric theorems follow from it by pure algebra.
Despite its greater clarity, Grassmann’s essay was not an overnight success. However, his ideas gathered enough momentum to justify a new version of the Ausdehnungslehre, Grassmann (1862), and they were gradually adopted by other mathematicians. Peano was among the first to appreciate Grassmann’s ideas, and was inspired by him to create the first axiom system for real vector spaces in Peano (1888), section 72. Klein (1909) brought Grassmann’s geometry to a wider audience by restricting it to three dimensions. Klein mentioned the inner product, but his version of Grassmann relied mainly on the determinant concept, which gives convenient formulas for areas and volumes.
/////// End of Quote from Stillwell (2016)
/// Connect with the Cauchy-Binét formula which generalizes the product formula for determinants.
/////// Quoting Stillwell (2016, section 5.11, p. 184):
Non-Euclidean Geometry
In this book I have made the judgement that non-Euclidean geometry is more advanced than Euclidean. There is ample historical reason to support this call, since non-Euclidean geometry was discovered more than 2000 years after Euclid. The “points” and “lines” of non-Euclidean geometry can be modeled by Euclidean concepts, so they are not advanced in themselves, but the concept of non-Euclidean distance surely is.
One way to see this is to map a portion of the non-Euclidean plane onto a piece of surface $\, S \,$ in $\, \mathbb{R}^3 \,$ in such a way that distance is preserved. Then ask: how simple is $\, S \,$? Well, the simplest possible $\, S \,$ is the trumpet-shaped surface shown in figure 5.33 and known as the pseudosphere.
Henry Segerman on the pseudosphere (YouTube):
The pseudosphere is obtained by rotating the tractrix curve with equation
$\, x \, = \, \ln {\dfrac{1 + \sqrt{1 - y^2}}{y}} - \sqrt{1 - y^2} \,$
about the $\, x$-axis.
The formula is complicated enough, but the conceptual complication is much greater. It is possible to compare only small pieces of the non-Euclidean plane with small pieces of a surface in $\, \mathbb{R}^3$, because a complete non-Euclidean plane does not “fit” smoothly in $\, \mathbb{R}^3$. This was proved by Hilbert (1901). The pseudosphere, for example, represents just a thin wedge of the non-Euclidean plane, the edges of which are two non-Euclidean lines that approach each other at infinity. These two edges are joined together to form the tapering tube shown in figure 5.33.
In contrast, Euclidean geometry is modeled by the simplest possible surface in – the plane!
[Footnote 7: It may be thought unfair to the hyperbolic plane to force it into the Euclidean straightjacket of $\, \mathbb{R}^3$. Might not the Euclidean plane look equally bad if forced to live in non-Euclidean space? Actually, this is not the case. Beltrami showed that the Euclidean plane fits beautifully into non-Euclidean space, where it is a “sphere with center at infinity.”]
/////// End of Quote from Stillwell (2016)
Tractrix and Catenary – Involute and Evolute of each other
The catenary is the evolute of the tractrix, and the tractrix is an involute of the catenary:
/// Connect with the Beltrami-Klein model
///////
/// Connect with Horocycles in Hyperbolic Geometry
/// Connect with Pseudospherical Surfaces
Concentric ‘ordinary’ circles in the Beltrami-Klein model
with a common center (inside the hyperbolic space) and varying radius:
Concentric horocycles in the Beltrami-Klein model
with a common center (at infinity)and varying radius:
Concentric hypercycles in the Beltrami-Klein model
with a common center (outside the hyperbolic space) and varying radius:
Illuminating hyperbolic geometry
(Henry Segerman and Saul Schleimer on YouTube):
Stereographic Projection
Stereographic projection of the Riemann sphere (by fsgm1):
Stereographic projection takes a circle on the sphere to a circle in the plane, but it does NOT map the center point of the original circle to the center point of the image circle. As shown in the movie below, the point that is mapped to the center point of the image circle is the vertex point of the cone that is tangent to the sphere along the original circle.
Stereographic projection of a circle on the sphere onto the plane (moving camera):
=======
/////// Quoting Stillwell (2016, section 5.11, p. 186):
Numbers and Geometry
Now let us return to the axioms of Hilbert (1899), and what they tell us about the relationship between numbers and geometry. Hilbert’s axioms probably capture Euclid’s concept of the line (with the finer structure explored in Book V of the Elements). So, given the Descartes “model” $\, \mathbb{R}^2 \,$ of Euclid’s axioms, Hilbert has shown Euclidean geometry to be essentially equivalent to the algebra of $\, \mathbb{R}$.
However, algebraists and logicians now prefer not to use the full set $\, \mathbb{R} \,$ in geometry. They point out that the set of constructible numbers suffices, because Euclid’s geometry “sees” only the points arising from straightedge and compass constructions. Thus
one can get by with an algebraically defined set of points, which is only a “potential” infinity, in contrast to the “actual” infinity $\, \mathbb{R}$.
Logicians also prefer the theory of constructible numbers because its “consistency strength” is less than that of the theory of $\, \mathbb{R}$. That is, it is easier to prove the consistency of constructible numbers (and hence the consistency of Euclid’s axioms) than it is to prove the consistence of the theory of $\, \mathbb{R} \,$ (and hence the consistency of Hilbert’s axioms).
/////// End of Quote from Stillwell (2016)
/////// Quoting Stillwell (2016, section 5.11, p. 186):
Geometry and “Reverse Mathematics”
In recent decades, mathematical logicians have developed a field called reverse mathematics, whose motivation was stated by Friedman (1975) as follows:
When the theorem is proved from the right axioms,
the axioms can be proved from the theorem
.
As logicians understand it, reverse mathematics is a technical field, concerned mainly with theorems about the real numbers (see section 9.9). However, if we understand reverse mathematics more broadly as the search for the “right axioms,” then reverse mathematics began with Euclid.
He saw that the parallel axiom is the right axiom to prove the Pythagorean theorem, and perhaps the reverse – that the Pythagorean theorem proves the parallel axiom (given his other axioms). The same is true of many other theorems of Euclidean geometry, such as the theorem of Thales and the theorem that the angle sum of a triangle is $\, \pi$. All of these theorems are equivalent to the parallel axiom, so it is the “right axiom” to prove them.
To formalize this and other investigations in reverse mathematics we choose a base theory containing the most basic and obvious assumptions about some area of mathematics. It is to be expected that the base theory will fail to prove certain interesting but less obvious theorems. We then seek the “right” axiom or axioms to prove these theorems, judging an axiom to be “right” if it implies the theorem, and conversely, using only assumptions from the base theory.
Euclid began with a base theory now known as neutral geometry. It contains basic assumptions about points, lines, and congruence of triangles but not the parallel axiom. He proved as many theorems as he could before introducing the parallel axiom – only when it was needed to prove theorems about the area of parallelograms and ultimately the Pythagorean theorem. He also needed the parallel axiom to prove the theorem of Thales and that the angle sum of a triangle is $\, \pi$. We now know, conversely, that all of these theorems imply the parallel axiom in neutral geometry, so the latter is the “right” axiom to prove them.
Neutral geometry is also a base theory for non-Euclidean geometry, because the latter is obtained by adding to neutral geometry the “non-Euclidean parallel axiom” stating that there is more than one parallel to a given line through a given point.
Grassmann’s theory of real vector spaces, as we have seen, can also be taken as a base theory for Euclidean geometry. It is quite different from the base theory of neutral geometry because the Euclidean parallel axiom holds in real vector spaces, and so does the theorem of Thales. Nevertheless, this new base theory is not strong enough to prove the Pythagorean theorem, or indeed to say anything about angles. Relative to the theory of real vector spaces, the “right” axiom to prove the Pythagorean theorem is existence of the inner product, because we can reverse the implication by using the Pythagorean theorem to define distance, hence angle and cosine, and then define the inner product by
$\, u \cdot v \, = \, |u| \cdot |v| \cos \theta$.
This raises the possibility of adding a different axiom to the theory of real vector spaces and obtaining a different kind of geometry, just as we obtain non-Euclidean geometry from neutral geometry by adding a different parallel axiom. Indeed we can, and simply by asserting the existence of a different kind of inner product. The inner product introduced by Grassmann is what we now call a positive-definite inner product, characterized by the property that $\, u \cdot u = 0 \,$ only if $\, u \,$ is the zero vector.
Non-positive-definite inner products also arise quite naturally. Probably the most famous one is the one on the vector space $\, \mathbb{R}^4 \,$ that defines the Minkowski space of Minkowski (1908). If we write the typical vector in $\, \mathbb{R}^4 \,$ as $\, u = (w, x, y, z) \,$ then the Minkowski inner product is defined by
$\, u_1 \cdot u_2 = -w_1 w_2 + x_1 x_2 + y_1 y_2 + z_1 z_2$.
In particular, the length $\, |u| \,$ of a vector in Minkowski space is given by
$\, {|u|}^2 = u \cdot u = -w^2 + x^2 + y^2 + z^2$,
so $\, |u| \,$ can certainly be zero when $\, u \,$ is not the zero vector.
Minkowski space is famous as the geometric model of Einstein’s special relativity theory. In this model, known as flat spacetime, $\, x, y, \,$ and $\, z \,$ are the coordinates of ordinary three-dimensional space and $\, w = c t$, where $\, t \,$ is the time coordinate and $\, c \,$ is the speed of light. As Minkowski (1908) said:
The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.
Undoubtedly, relativity theory put non-positive-definite inner products on the map, making them as real and important as the ancient concept of distance. But in fact such inner products had already been considered by mathematicians, and one of them is involved in a model of non-Euclidean geometry discovered by Poincaré (1881) – the so-called hyperboloid model.
To see where the hyperboloid comes from, consider the three-dimensional Minkowski space of vectors $\, u = (w, x, y) \,$ with one time coordinate $\, w \,$ and two space coordinates $\, x, y$. In this space, where $\, {|u|}^2 = -w^2 + x^2 + y^2$, we consider the “sphere of imaginary radius”
$\, \{ \, u \, : \, |u| = \sqrt{ -1 } \, \} = \{ \, (w, x, y) \, : \, -w^2 + x^2 + y^2 = -1 \, \}$.
This “sphere” consists of the points $\, (w, x, y) \,$ in $\, \mathbb{R}^3 \,$ such that
$\, w^2 - x^2 - y^2 = 1$,
so it is actually a hyperboloid; namely, the surface obtained by rotating the hyperbola $\, {|u|}^2 = w^2 - y^2 = 1$ in the $\, (w, y)$-plane about the $\, w$-axis. If we take “distance” on either sheet of the hyperboloid to be the Minkowski distance, it turns out to be a model of the non-Euclidean plane. As with the other models of the non-Euclidean plane, calculating distance is a little troublesome so I omit the details.
/////// Figure 5.34 GOES HERE
Instead, I offer figure 5.34 – a black-and-white version of a picture due to Konrad Polthier of the Freie Universität, Berlin – which shows what the triangle tesselation of figure 5.32 looks like in the hyperboloid model. (It also relates the hyperboloid model to another model – the conformal disk model – which is a kind of intermediary between the hyperboloid model and the half-plane model shown in figure 5.32.)
This elegant relationship between Minkowski space and the non-Euclidean plane has been used for some textbook treatments of non-Euclidean geometry, such as Ryan (1986). Just as the positive-definite inner product is the “right axiom” to develop Euclidean geometry over the base theory of real vector spaces, the Minkowski inner product is the “right axiom” to develop non-Euclidean geometry.
/////// End of Quote from Stillwell (2016)
/////// Quoting Stillwell (2016, section 5.11, p. 190):
Projective Geometry
Another wonderfully “right” axiom is the theorem discovered by Pappus a few hundred years after Euclid. Pappus viewed his theorem as part of Euclidean geometry, but it does not really belong there. It is unlike typical Euclidean theorems in making no mention of length or angle, so its home should be a geometry that does not involve these concepts. The statement of the theorem is the following, which refers to the configuration shown in figure 5.35.
/////// Figure 5.35 GOES HERE
Theorem of Pappus. If A, B, C, D, E, F are points of the plane lying alternately on two lines, then the intersections of the pairs of lines AB and DE, BC and EF, CD and FA, lie on a line.
The Pappus theorem has a Euclidean proof using the concept of length, and also a coordinate proof using linear equations to define lines. But it seems to have no proof using only concepts “appropriate” to its statement: points, lines, and the membership of points in lines. The appropriate setting for the Pappus theorem is the projective geometry of the plane, a geometry which tries to capture the behaviour of points and lines in a plane without regard to length and angle. In projective geometry, configurations of points and lines are considered the same if one can be projected onto the other. Projection can of course change lengths and angles, but the straightness of lines remains, as does the membership of points in lines.
If one seeks axioms for projective plane geometry, the following come easily to mind:
1. Any two points belong to a unique line.
2. Any two lines have a unique point in common.
3. There are four points, no three of which are in the same line.
The first axiom is also one of Euclid’s. The second disagrees with Euclid in the case of parallel lines, but projective geometry demands it, because even parallel lines can be projected so that they meet – on the “horizon.” The third axiom is there to ensure that we really have a “plane,” and not merely a line. However, these simple axioms are very weak, and it can be shown that they do not suffice to prove the Pappus theorem. They do, however, form a natural base theory to which other axioms about points and lines can be added.
What are the “right axioms” to prove the Pappus theorem? The answer is no less than the Pappus theorem itself, thanks to what the Pappus theorem implies; namely, that the abstract plane of “points” and “lines” can be given coordinates which form a field as defined in section 4.3. Thus geometry springs fully armed from the Pappus theorem! The Pappus axiom (as we should now call it) is the right axiom to prove coordinatization by a field, because such a coordinatization allows us to prove the Pappus axiom. As remarked above, this follows by using linear equations to define lines. The field properties then enable us to find intersections of lines by solving linear equations, and to verify that points of intersection lie on a line.
The idea of reversing the coordinate approach to geometry began with von Staudt (1847), who used the Pappus theorem to define addition and multiplication of points on a line. Hilbert (1899) extended this idea to prove that the coordinates form a field, but he had to assume another projective axiom, the so-called theorem of Desargues, which was discovered around 1640. (Roughly speaking, the Pappus axiom easily implies that addition and multiplication are commutative, while the Desargues axiom easily implies that they are associative.)
Quite remarkably, considering how long the Pappus and Desargues theorems had been around, Hessenberg (1905) discovered that the Pappus theorem implies the Desargues theorem. So the single Pappus axiom is in fact equivalent to coordinatization of the plane by a field.
This reversal of the Pappus theorem also tells us something remarkable about algebra: the nine field axioms follow from four geometric axioms – the three projective plane axioms plus Pappus!
/////// End of Quote from Stillwell (2016)
/// Connect with the KMR pages on:
/////// Quoting Stillwell (2016, ch 10.9 p. 384):
Groups and Geometry
Geometry is one of the most convincing cases where the group concept captures a phenomenon that mathematicians wish to study. At the same time, it is a sign of the depth of the group concept that its relationship with geometry was not uncovered until geometry had been studied for over 2000 years. One does not notice the group concept in geometry until several kinds of geometry have come to light – most importantly, projective geometry. It was projective geometry in particular that led Klein (1872) to notice the role of groups in geometry, and to define geometry as the study of groups and their invariants
The real projective line $\, \mathbb{R} \cup \{ \, \infty \, \} \,$ studied in section 10.4, involves perhaps the simplest example of an interesting group and an interesting invariant. The group is the group of linear fractional functions,
$\, f(x) = \dfrac{ax+b}{cx+d} \,$ where $\, a, b, c, d \in \mathbb{R} \,$ and $\, ad - bc \neq 0$,
under the operation of function composition. That is, given functions
$\, f_1(x) = \dfrac{a_1 x + b_1}{c_1 x + d_1} \,$ and $\, f_2(x) = \dfrac{a_2 x + b_2}{c_2 x + d_2}$,
we form the function $\, f_1(f_2(x))$, which corresponds to performing the projection corresponding to $\, f_2$, then the projection corresponding to $\, f_1$. This group is not commutative. For example if $\, f_1(x) = x + 1 \,$ and $\, f_2(x) = 2 x \,$
then
$\, f_1(f_2(x)) = 2 x + 1$, whereas $\, f_2(f_1(x)) = 2 (x + 1) = 2x + 2$,
so $\, f_1 f_2 \neq f_2 f_1$.
Nevertheless, we can find the invariant of the linear fractional transformations without knowing much about their group structure. It suffices to know that they are generated by the simple functions $\, x \, \mapsto \, x + b \, , \, x \, \mapsto \, a x \,$ for $\, a \neq 0$, and $\, x \, \mapsto \, 1/x$.
As we saw in section 10.4, traditional geometric quantities such as length, or the ratio of lengths, are not invariant under all linear fractional transformations. However, some simple computations with the generating transformations show the invariance of the cross-ratio
$\, [ \, p, q \, ; \, r, s \, ] \, = \, \dfrac{(r-p) (s-q)}{(r-q) (s-p)}$
for any four points on the real projective line. The invariance of the cross-ratio under projection was already known to Pappus, and was rediscovered by Desargues around 1640. However, its algebraic invariance had to wait for the identification of the appropriate group by Klein.
With hindsight, we can also see how length, and the ratio of lengths, are also algebraic invariants. The length of the line segment from $\, p \,$ to $\, q$, $\, | \, p - q \, |$, is the invariant of the group of translations $\, x \, \mapsto \, x + b \,$ of $\, \mathbb{R}$. We call $\, \mathbb{R} \,$ the Euclidean line when it is subject to these transformations, because they make any point “the same” as any other point, as Euclid intended.
The ratio of lengths, $\, \dfrac{p - r}{p - q}$, for any three points $\, p, q, r$, is the invariant group of similarities $\, x \, \mapsto \, a x + b \,$ where $\, a \neq 0$. These transformations are called affine, and when $\, \mathbb{R} \,$ is subject to these transformations it is called the affine line.
The deep difference between the Euclidean and affine lines and the projective line is of course the point at infinity $\, \infty$. The point $\, \infty \,$ arises on the projective line because the point $\, 0 \,$ otherwise has no place to go under the map $\, x \, \mapsto 1/x$. It is appropriate to call this point “infinity” because $\, 0 \,$ is the limit of $\, 1/n \,$ as $\, n \rightarrow \infty$, so the image of $\, 0 \,$ under $\, x \, \mapsto 1/x$ ought to be the “limit” of $\, n \,$ as $\, n \rightarrow \infty$. Nevertheless, it is also appropriate to view the projective line as a finite object; namely, the circle.
This is because $\, 0 \,$ is also the limit of $\, -1/n \,$ as $\, n \rightarrow \infty$, so the image of $\, 0 \,$ should be the limit of $\, -n \,$ as $\, n \rightarrow \infty$. Thus we “approach $\, \infty$” as we travel along the line in either direction. We can realize the common “limit” of $\, n \,$ and $\, -n \,$ by an actual point if we map $\, \mathbb{R}$ into a circle as shown in figure 10.15.
///// Figure 10.15 GOES HERE
The topmost point of the circle neatly corresponds to the “limit” of both $\, n \,$ and $\, -n$, because it is the actual limit of their images on the circle. So if we let the topmost point correspond to $\, \infty \,$ we can view the circle as a continuous and bijective image of the real projective line.
/////// End of Quote from Stillwell (2016)
/// Connect with the concept of a group acting on a set.
/////// Quoting Stillwell (2016, ch 10.9 p. 386):
Affine Geometry
The affine transformations of the line $\, \mathbb{R}$, mentioned above, extend to the projective line $\, \mathbb{R} \cup \{ \, \infty \, \} \,$ simply by sending the point $\, \infty \,$ to itself. Similarly, there are affine transformations of the plane, and of the projective plane. They are the projections that send finite points to finite points, and points at infinity to points at infinity. In particular, they send parallels to parallels. The affine geometry of the plane studies the images of the plane obtained by such projections. The term “affine” was introduced by Euler (1748b), motivated by the idea that images related by affine transformations have an “affinity” with one another.
Like projective geometry, affine geometry has an artistic counterpart. It may be seen in classical Japanese woodblock prints of the eighteenth and nineteenth centuries, such as Harunoby’s Evening Chime of the Clock shown in figure 10.16. This print, which dates from around 1766, clearly shows the preservation of parallel lines, since all parallel lines in the scene actually look parallel. This makes the scene look “flat,” though perfectly consistent. In fact, the picture shows how the scene would appear when viewed from infinitely far away, with infinite magnification.
///// Figure 10.16: Affine geometry in art. GOES HERE
Image courtesy of www.metmuseum.org
The art of representing parallel lines consistently demands some skill, as it is possible to go badly wrong. Figure 10.17 shows an example, The Birth of St Edmund. I have previously used this example, for example in Stillwell (2010, p. 128) – as a failure of projective geometry, because the parallel lines do not converge towards any horizon. But now I think it is better seen as a failure of affine geometry. The artist really wants the parallel lines to look parallel, but he has utterly failed to do so consistently.
Since affine maps pair finite points with finite points, an affine image cannot include the horizon. Most Japanese prints comply with this condition, but sometimes even an eminent artist slips up. Figure 10.18 shows the print Cat at Window by the nineteenth-century master Hiroshige. The interior is fine, but it is not compatible with the exterior, which includes the horizon.
///// Figure 10.18 Cat at Window – subtle failure of affine geometry GOES HERE
/////// End of Quote from Stillwell (2016)
/// Connect with the KMR pages on Oscar Reutersvärd and M.C. Escher.
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http://math.stackexchange.com/questions/356105/how-to-factor-x4-3x-2 | # How to factor $x^4 +3x -2$?
I have figured out there is two roots between $0$ and $1 ,-1$ and $-2$ for $x^4 +3x -2 = 0$.
Therefore there should be two factors $(x + a)$ and $(y - b)$ where $a,b \in R^+$. But how to find these $a$ and $b$?
When they found I can find the next factor in $ax^2+bx+c$ form and can check for further factors easily.
-
Try some general method like Descartes method or Ferrari(Check wikipedia or a google search if you are unfamiliar) – Moron plus plus Apr 9 '13 at 16:47
Wolfram alpha tells there are two real and two complex roots...wolframalpha.com/input/… – UrošSlovenija Apr 9 '13 at 16:48
## 3 Answers
Hints:
If it factors, you know the form will $(x^2 + bx \pm 1)(x^2 + cx \mp 2)$. You need a sum of $3$ and need for the cubic term to cancel out..
Now, can you use that and figure out the factors and find $b$ and $c$?
Result: $(x^2-x+2) (x^2+x-1)$
-
Actually he doesn't know that the form will be that... He might check the case when the quadratics are over $\mathbb Z$, to see if it works, but he doesn't know that that is the case.. He only knows that the roots are real... – N. S. Apr 9 '13 at 23:58
Very well put! +1 – amWhy Apr 10 '13 at 0:24
@amWhy: Thank you! – Amzoti Apr 10 '13 at 0:26
For it to have some "nice" linear factors, the roots must be one of $\pm 1,\pm 2$ (this is due to the rational root theorem). You can quickly check that these are not the roots. The next bet is quadratic factors, i.e., $$(x^4+3x-2) = (x^2+ax+b)(x^2+cx+d)$$ Expanding the right hand side gives us \begin{align} a+c & = 0\\ b + d + ac & = 0\\ ad+bc & = 3\\ bd & = - 2 \end{align} This gives us the factors to be $$(x^2-x+2) \text{ and } (x^2+x-1)$$
-
Over $\rm\,\Bbb Q\!:\:$ Rational Root Test excludes linear factors. Testing for quadratic factors is easy since the constant term is prime, which greatly constrains possible factors. We prove a more general case. Yours is $\rm\: p = 2,\:$ so $\rm\:3 = b = -a(p\!+\!1) = -3a\:\Rightarrow\:a = -1,\:$ so $\rm\:2 = p = 1\!+\!sa^2 = 1\!+\!s,\:$ so $\rm\:s = 1.$
Lemma $\$ If $\rm\ f(x) = x^4 + b x-p\$ has a quadratic factor $\rm\,\in \Bbb Q[x],\:$ and $\rm\:p\:$ is prime then
$$\rm f(x)\, =\, x^4\! -\! a(p\!+\!1)\, x\! -\! p\, =\, (s\,x^2\! +\! a\,x\!+\!p)(\color{#c00}s\,x^2\!-\!a\,x\!\color{#0a0}{-\!1}),\quad p = 1\!+\!sa^2,\ \ s =\pm1$$
Proof $\$ Invoking Gauss's Lemma, we may assume that it splits into monic quadratics $\rm\,\in\Bbb Z[x].\:$ Since $\rm\:p\:$ is prime, one of the factors has constant term $\rm\:\pm p,\:$ so scaling it by $\rm\,\pm1\,$ it will have the form $\rm\:s\,x^2\! +\! a\,x\!+\!p,\,\ s=\pm1 \:$. Comparing coef's, its cofactor must have leading coef $\rm\,= \color{#c00}s,\:$ constant coef $= \color{#0a0}{-1},\:$ and linear coef $\rm\ a' = - a,\:$ since the coef of $\rm\:x^3\:$ in the product $\rm\, = (a+a')s = 0.$ Finally, multiplying the two factors and comparing coef's yields the result. $\ \$ QED
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But $f(x) = x^4 + b x-p$ can always be factored as a product of two quadratics... Your Lemma only works if you factor it over $\mathbb Z$.... – N. S. Apr 9 '13 at 23:33
@N.S. But that is the OP's problem (as I read it). There is no indication that the OP seeks factorizations over $\,\Bbb C$ or algebraic number fields. – Math Gems Apr 9 '13 at 23:49
Nope, he only knows that the roots are real...And he knows that his Polynomial has a quadratic with two real roots, why is this over $\mathbb Z$? – N. S. Apr 9 '13 at 23:55
The above formula does split the OP's polynomial into quadratic factors, from which he may then easily conclude. It is not clear what the source of the problem is. The above completely characterizes all rational factorizations of this form, which may be of interest to the OP (ditto for the method used). – Math Gems Apr 10 '13 at 0:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8882566094398499, "perplexity": 474.5215801012813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507446525.24/warc/CC-MAIN-20141017005726-00244-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/u4-set.125235/ | # U4 set
1. Jul 4, 2006
### Taviii
I saw this problem in a book: calculate the sum of the elements of the U4 set.
The answer is 0, the elements of the sets being: 1, i, -1, -i.
My questions is: whats the U set?
Last edited: Jul 4, 2006
2. Jul 4, 2006
### maverick280857
In this terminology $U_{n}$ refers to the set of the n-th roots of unity, i.e. all roots of the equation
$$x^{n} - 1 = 0$$
So
$U_{3}$ = {1, $\omega$,$\omega^2$}
3. Jul 5, 2006
### Taviii
Thank you!
Your explanation was very helpful.
4. Jul 6, 2006
### HallsofIvy
If you multiply out (x-a1)(x-a2)...(x-an) the coefficient of x is easily seen to be -(a1+ a2+ ...+ an) so for any polynomial equation in which there is no "x" term, the sum of the roots must be 0. In particular, the sum of the roots of xn= 1 must be 0 for all n and so the sum of the elements of Un must be 0 for all n.
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Similar Discussions: U4 set | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9068698883056641, "perplexity": 1649.9335991172584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817999.51/warc/CC-MAIN-20180226025358-20180226045358-00581.warc.gz"} |
http://mathhelpforum.com/calculus/104612-don-t-get-pendelum-problem.html | Math Help - Don't get this pendelum problem??
1. Don't get this pendelum problem??
how the heck do i find the equation?? i dont get it!! i just need help with the equation then i can do the rest
2. Hello elpermic
Originally Posted by elpermic
how the heck do i find the equation?? i dont get it!! i just need help with the equation then i can do the rest
$d$ is a sinusoidal function of $t$; i.e. $d = a + b\sin(ct+d)$, for some constants $a, b, c, d$
$d_{max}=110\Rightarrow 110 = a+b$, since the maximum value of the sine function is $1$
$d_{min}=50 =a-b$, since the minimum value of the sine function is $-1$
$\Rightarrow a=80, b= 30$
So $d = 80+30\sin(ct+d)$
$\sin(ct+d)$ has period $\frac{2\pi}{c}\Rightarrow \frac{2\pi}{c}= 6$, since the pendulum performs one complete swing in $6$ seconds
$\Rightarrow c = \frac{\pi}{3}$
And $d = 110$ when $t = 1.3\Rightarrow \sin\Big(\frac{1.3\pi}{3}+d\Big) = 1$
$\Rightarrow \frac{1.3\pi}{3}+d = \frac{\pi}{2}$
$\Rightarrow d = \pi\Big(\frac{1}{2}-\frac{13}{30}\Big)=\frac{\pi}{15}$
$\Rightarrow d =80+30\sin\Big(\frac{\pi t}{3}+\frac{\pi}{15}\Big)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.964649498462677, "perplexity": 588.0445660917684}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375097512.42/warc/CC-MAIN-20150627031817-00018-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://testbook.com/objective-questions/mcq-on-evaluation-of-limits--5eea6a1439140f30f369f1a8 | What is $$\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}$$ equal to?
1. -1
2. Zero
3. -e
4. $$\rm -\dfrac{1}{e}$$
Option 1 : -1
Evaluation of Limits MCQ Question 1 Detailed Solution
Concept:
$$\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)$$
$$\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} = 1\\\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+x) }{x}$$
Calculation:
We have to find the value of $$\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}$$
As we know, $$\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)$$
$$\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} × \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1-x) }{x}$$
= 1 × $$\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{x}$$
= $$\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{-(-x)}$$
$$-1 × \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{(-x)}$$
= -1 × 1
= -1
What is $$\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}$$ equal to ?
1. 0
2. -1
3. 1
4. Limit does not exist
Option 1 : 0
Evaluation of Limits MCQ Question 2 Detailed Solution
Concept:
$$\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)$$
$$\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a$$
log mn = n log m
Calculation:
$$\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0$$
Find the value of $$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}$$
1. -2
2. 2
3. 1
4. - 1
Option 1 : -2
Evaluation of Limits MCQ Question 3 Detailed Solution
Concept:
$$\rm \sum n = \frac{n(n+1)}{2}$$
Calculation:
$$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}$$
$$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} { \left[\frac{n(n+1)}{2} \right ]}$$
$$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2(1-n^2)} {n^2+n}$$
Divide both numerator and denominator by n2
$$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1-n^2}{n^2} \right ]} {\left[\frac{n^2+n}{n^2} \right ]}$$
$$\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1}{n^2} - 1 \right ]} {1 + \frac{1}{n} }$$
$$\rm \frac{2(0-1)}{(1+0)}$$
= -2
Evaluate $$\lim_{x\rightarrow0} \frac {x\tan x}{1 - \cos x}$$
1. 1 / 2
2. -1 / 2
3. -2
4. 2
Option 4 : 2
Evaluation of Limits MCQ Question 4 Detailed Solution
Concept:
cos 2x = 1 - 2sin2x
$$\rm \lim_{x\rightarrow0} \frac {\tan x}{x}=1\\ \rm \lim_{x\rightarrow0} \frac {\sin x}{x}=1\\$$
Calculation:
Given, $$\lim_{x\rightarrow0} \frac {x\tan x}{1 - \cos x}$$
$$\Rightarrow \lim _{x \rightarrow 0} \frac{x\operatorname{tan} x}{2 \sin ^{2} \frac{x}{2}}$$
$$=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\tan x}{x} \times \frac{x.x}{\frac{\sin ^{2} \frac{x} {2}}{\left(\frac{x}{2}\right)^{2}} \times\left(\frac{x}{2}\right)^{2}}$$
$$= \frac{1}{2} \lim _{x \rightarrow 0} \frac{\tan x}{x} \times \frac{4}{\frac{\sin ^{2} \frac{x} {2}}{\left(\frac{x}{2}\right)^{2}}}$$
$$=\frac{1}{2}\times 1\times 4 =2$$
Hence, option (4) is correct.
Find the value of $$\rm \displaystyle \lim_{x \rightarrow \infty} x \sin \left(\frac{\pi} {x}\right)$$
1. $$\rm \frac {1}{π}$$
2. 0
3. π
4. 1
Option 3 : π
Evaluation of Limits MCQ Question 5 Detailed Solution
Concept:
$$\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$
Calculation:
$$\rm \displaystyle \lim_{x → ∞} x \sin \left(\frac{π} {x}\right)$$
$$\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{1}{x} \right )}$$
$$\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{π}{x} \right )} × π$$
Let $$\rm \frac {π}{x} = t$$
If x → ∞ then t → 0
$$\rm \displaystyle \lim_{t \rightarrow 0} \frac{\sin t}{t} × π$$
= 1 × π
= π
Evaluate $$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}$$
1. $$\frac{1}{{2\sqrt 2 }}$$
2. $$\frac{1}{{\sqrt 2 }}$$
3. 1/2
4. None of the above
Option 4 : None of the above
Evaluation of Limits MCQ Question 6 Detailed Solution
Calculation:
We have to find the value of $$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}$$
Rationalise the numerator, we get
$$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x} \times \frac{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }}{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }}$$
$$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm {a+x} - {\rm{\;}} {\rm (a - x )} }}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )}$$
$$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 2x}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )}$$
$$\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 1}{\rm 2({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )}$$
$$\frac{1}{{4 \rm \sqrt a }}$$
If $$\mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \frac{{\sin {\rm{x}}}}{{\rm{x}}} = {\rm{l}}$$ and $$\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{\cos {\rm{x}}}}{{\rm{x}}} = {\rm{m}}$$, then which one of the following is correct?
1. l = 1, m = 1
2. $${\rm{l}} = \frac{2}{{\rm{\pi }}},{\rm{\;m}} = \infty$$
3. $${\rm{l}} = \frac{2}{{\rm{\pi }}},{\rm{\;m}} = 0$$
4. l = 1, m = ∞
Option 3 : $${\rm{l}} = \frac{2}{{\rm{\pi }}},{\rm{\;m}} = 0$$
Evaluation of Limits MCQ Question 7 Detailed Solution
Concept:
Steps to solve the limit:
• Check the form by substituting the limit value in the function
• After substitution if indeterminate form (0/0, ∞/∞ etc.) obtain then solve it further
Calculation:
Given: $$\mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \frac{{\sin {\rm{x}}}}{{\rm{x}}} = {\rm{l}}$$ and $$\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{\cos {\rm{x}}}}{{\rm{x}}} = {\rm{m}}$$
$${\rm{l}} = \mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \frac{{\sin {\rm{x}}}}{{\rm{x}}}$$
$$= \frac{{\sin \left( {\frac{{\rm{\pi }}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2}}}$$ (put x = π/2)
$$= \frac{1}{{\frac{{\rm{\pi }}}{2}}}$$ (∵ sin(π/2) = 1)
$$\Rightarrow {\rm{l}} = \frac{2}{{\rm{\pi }}}$$
And $${\rm{m}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{\cos {\rm{x}}}}{{\rm{x}}}$$
$$= \frac{{\cos \left( \infty \right)}}{\infty }$$ (cos x lies between -1 to 1)
⇒ m = 0 (1/∞ = 0)
Hence, option (3) is correct.
Evaluate: $$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin \left( {ax} \right)}}{{\sin \left( {bx} \right)}}} \right] = ?$$
1. 1
2. $$\frac{a}{b}$$
3. $$\frac{b}{a}$$
4. 0
Option 2 : $$\frac{a}{b}$$
Evaluation of Limits MCQ Question 8 Detailed Solution
CONCEPT:
• L - Hospital's Rule: If $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = \frac{0}{0}\;or\frac{\infty }{\infty }$$ then we have to differentiate both the numerator and denominator with respect to x unless and until $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = l \ne \frac{0}{0}$$ where l is a finite value.
• $$\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$$
CALCULATION:
Here, we have to find the value of the limit $$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin \left( {ax} \right)}}{{\sin \left( {bx} \right)}}} \right]$$
⇒ $$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin \left( {ax} \right)}}{{\sin \left( {bx} \right)}}} \right] = \frac{0}{0}$$
Now, according to L-Hospital's rule if $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = \frac{0}{0}\;or\frac{\infty }{\infty }$$ then we have to differentiate both the numerator and denominator with respect to x unless and until $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = l \ne \frac{0}{0}$$ where l is a finite value.
⇒ $$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin \left( {ax} \right)}}{{\sin \left( {bx} \right)}}} \right] = \;\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{a\cos \left( {ax} \right)}}{{b\cos \left( {bx} \right)}}} \right] = \frac{a}{b}$$
Hence, option B is the correct answer.
What is $$\rm \underset{x\to 2}{\mathop{\lim }}\,\frac{x-2}{{{x}^{2}}-4}$$ equal to?
1. 0
2. 1 / 4
3. 1 / 2
4. 1
Option 2 : 1 / 4
Evaluation of Limits MCQ Question 9 Detailed Solution
Concept:
Indeterminate form of the type infinity divided by infinity (∞/∞) or zero divided by zero (0/0)
When it is possible then try to simplify the expression.
Calculation:
We have to find the value of $$\rm \lim_{x\rightarrow 2}\frac{x-2}{{{x}^{2}}-4}$$
$$\Rightarrow \rm \lim_{x\rightarrow 2}\frac{x-2}{{{x}^{2}}-4} = \lim_{x\rightarrow 2}\frac{x-2}{(x-2)(x+2)}$$
$$\rm = \lim_{x\rightarrow 2}\frac{1}{(x+2)}$$
= 1 / 4
ALTERNATE SOLUTION:
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}$$
II. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }$$
Then we can apply L-Hospital Rule ⇔ $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$
Note: We have to differentiate both the numerator and denominator with respect to x unless and until $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}$$ where l is a finite value.
Calculation:
We have to find the value of $$\rm \lim_{x\rightarrow 2}\frac{x-2}{{{x}^{2}}-4}$$
$$\rm \lim_{x\rightarrow 2}\frac{x-2}{{{x}^{2}}-4}$$ Form of limit is (0/0)
Applying L-Hospital Rule,
$$= \rm \lim_{x\rightarrow 2}\frac{1-0}{2x-0}$$
$$= \frac{1}{2\times 2} = \frac{1}{4}$$
The value of $$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{{x^3}}}$$ is:
1. $$\frac{-1}{6}$$
2. -∞
3. ∞
4. $$\frac{1}{6}$$
Option 3 : ∞
Evaluation of Limits MCQ Question 10 Detailed Solution
Concept:
If limiting expression $$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}}$$ leads to $$\frac{0}{0}$$ form, then apply L-Hospital’s rule to find the limiting value.. Applying L-Hospital’s rule we get,
$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}$$
Calculation:
$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{{x^3}}}$$
Now, e0 – 1 = 1 – 1 = 0 & 03 = 0
So, this is in the $$\frac{0}{0}$$ form, by applying L-Hospital rule, we get:
$$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left( {{e^x} - 1} \right)}}{{\frac{d}{{dx}}\left( {{x^3}} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{3{x^2}}}$$
$$\frac{{{e^0}}}{{3\left( 0 \right)}} = \frac{1}{0} = \infty$$
Note: 1/0 is not an indeterminate form.
Find out the value of x if logx 4 + logx 16 + logx 64 = 12
1. 1
2. 2
3. 7
4. 54
Option 2 : 2
Evaluation of Limits MCQ Question 11 Detailed Solution
Given:
logx 4 + logx 16 + logx 64 = 12
Formula Used:
If logx y = a, then xa = y
logx ab = b logx a
Calculation:
logx 4 + logx 16 + logx 64 = 12
⇒ logx 22 + logx 24 + logx 26 = 12
⇒ 2 logx 2 + 4 logx 2 + 6 logx 2 = 12
⇒ 12 logx 2 = 12
⇒ logx 2 = 1
⇒ 2 = x1
∴ x = 2
The value of $$\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}$$ is equal to:
1. 0
2. 1
3. $$\dfrac{1}{2}$$
4. $$\dfrac{1}{3}$$
Option 4 : $$\dfrac{1}{3}$$
Evaluation of Limits MCQ Question 12 Detailed Solution
Concept:
• $$\rm \displaystyle \lim_{x\to0}\dfrac{\tan x}{x}=1$$.
• $$\rm \dfrac{d}{dx}\tan x=\sec^2x$$.
• $$\rm \dfrac{d}{dx}\sec x=\tan x\sec x$$.
• $$\rm \dfrac{d}{dx}\left[f(x)\times g(x)\right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)$$.
Indeterminate Forms: Any expression whose value cannot be defined, like $$\dfrac00$$, $$\pm\dfrac{\infty}{\infty}$$, 00, ∞0 etc.
• For the indeterminate form $$\dfrac 0 0$$, first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the $$\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}$$, if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the $$\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)}$$ if it exists.
Calculation:
$$\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\dfrac00$$ is an indeterminate form. Let us simplify and use the L'Hospital's Rule.
$$\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\lim_{x\to 0}\left[\dfrac{\tan x - x}{x^3}\times\dfrac{x}{\tan x}\right]$$.
We know that $$\rm \displaystyle\lim_{x\to 0}\dfrac{x}{\tan x}=1$$, but $$\rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}$$ is still an indeterminate form, so we use L'Hospital's Rule:
$$\rm \displaystyle \rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}=\lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}$$, which is still an indeterminate form, so we use L'Hospital's Rule again:
$$\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}= \lim_{x\to 0}\dfrac{2\sec x(\sec x\tan x)}{6x}=\lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}$$, which is still an indeterminate form, so we use L'Hospital's Rule again:
$$\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}=\lim_{x\to 0}\dfrac{\sec^2 x\sec^2 x+\tan x[2\sec x(\sec x \tan x)]}{3}=\dfrac{1}{3}$$.
∴ $$\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=1\times\dfrac{1}{3}=\dfrac{1}{3}$$.
What is $$\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} {\rm{\;}}{{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}}$$ equal to?
1. 0
2. 1
3. -1
4. Limit does not exist
Option 1 : 0
Evaluation of Limits MCQ Question 13 Detailed Solution
Calculation:
We have to find the value of $$\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} {\rm{\;}}{{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}}$$
$$\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} {\rm{\;}}{{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}} = \;{e^{ - \frac{1}{0}}} = {e^{ - \infty }} = 0$$
Hence option 1st is correct answer.
What is $$\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x+x^2+x^3-3}{x-1}$$ equal to?
1. 1
2. 2
3. 3
4. 6
Option 4 : 6
Evaluation of Limits MCQ Question 14 Detailed Solution
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}$$
II. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }$$
Then we can apply L-Hospital Rule ⇔ $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$
Note: We have to differentiate both the numerator and denominator with respect to x unless and until $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}$$ where l is a finite value.
Calculation:
$$\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x+x^2+x^3-3}{x-1}$$ Form (0/0)
Apply L-Hospital Rule,
$$= \rm \displaystyle\lim_{x \rightarrow 1} \dfrac{\dfrac {d(x+x^2+x^3-3)}{dx}}{\dfrac {d(x-1)}{dx}}$$
$$=\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{1+2x+3x^2-0}{1-0}\\= \dfrac{1+2\times 1+3 \times 1^2-0}{1-0}\\= 6$$
Find the value of $$\rm \displaystyle \lim_{x \rightarrow 0} \cos x$$
1. 0
2. 1
3. -1
4. Not define
Option 2 : 1
Evaluation of Limits MCQ Question 15 Detailed Solution
Calculation:
$$\rm \displaystyle \lim_{x \rightarrow 0} \cos x$$
= cos 0
= 1
What is $$\rm \lim_{x \to 2} \frac{x^3 + x^2}{x^2 + 3x + 2}$$ equal to?
1. 0
2. 1
3. 2
4. 3
Option 2 : 1
Evaluation of Limits MCQ Question 16 Detailed Solution
Concept:
Continuity of a function:
• A function f(x) is said to be continuous at a point x = a in its domain, if $$\rm \lim_{x\to a}f(x)$$ exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ $$\rm \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)$$.
Calculation:
$$\rm \lim_{x \to 2} \frac{x^3 + x^2}{x^2 + 3x + 2}$$
$$\rm \frac{2^3 + 2^2}{2^2 + 3\cdot2 + 2}$$
$$\frac{12}{12}$$
= 1.
If $${\rm{F}}\left( {\rm{x}} \right) = \sqrt {9 - {{\rm{x}}^2}}$$, then what is $$\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F}}\left( {\rm{x}} \right) - {\rm{F}}\left( 1 \right)}}{{{\rm{x}} - 1}}$$ equal to?
1. $$- \frac{1}{{4\sqrt 2 }}$$
2. $$\frac{1}{8}$$
3. $$- \frac{1}{{2\sqrt 2 }}$$
4. $$\frac{1}{{2\sqrt 2 }}$$
Option 3 : $$- \frac{1}{{2\sqrt 2 }}$$
Evaluation of Limits MCQ Question 17 Detailed Solution
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}$$
II. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }$$
Then we can apply L-Hospital Rule $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$
Note: We have to differentiate both the numerator and denominator with respect to x unless and until $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}$$ where l is a finite value.
Calculation:
Let L = $$\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F}}\left( {\rm{x}} \right) - {\rm{F}}\left( 1 \right)}}{{{\rm{x}} - 1}}$$ (Form 0/0)
Apply L-Hospital rule,
$${\rm{L}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F'}}\left( {\rm{x}} \right){\rm{\;}} - 0)}}{{1 - 0}}$$
L = F’(1)
Given: $${\rm{F}}\left( {\rm{x}} \right) = \sqrt {9 - {{\rm{x}}^2}}$$
Differentiating with respect to x, we get
$$\Rightarrow {\rm{F'}}\left( {\rm{x}} \right) = {\rm{\;}}\frac{{1 \times \left( {0 - 2{\rm{x}}} \right)}}{{2\sqrt {9 - {{\rm{x}}^2}} }} = {\rm{\;}}\frac{{ - {\rm{x}}}}{{\sqrt {9 - {{\rm{x}}^2}} }}$$
Put x= 1
$$\Rightarrow {\rm{F'}}\left( 1 \right) = \frac{{ - 1}}{{\sqrt {9 - {1^2}} }} = \frac{{ - 1}}{{\sqrt 8 }} = - \frac{1}{{2\sqrt 2 }}$$
So, L $$= - \frac{1}{{2\sqrt 2 }}$$
If $$\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}$$, where k ≠ 0, then what is the value of k?
1. $$\dfrac{2}{3}$$
2. $$\dfrac{4}{3}$$
3. $$\dfrac{8}{3}$$
4. 4
Option 3 : $$\dfrac{8}{3}$$
Evaluation of Limits MCQ Question 18 Detailed Solution
Concept:
$$\rm a^2-b^2=(a-b)(a+b)$$
L-Hospital’s Rule: if we have an indeterminate form 0/0 or ∞/∞, all we need to do is differentiate the numerator and differentiate the denominator separately and then take the limit.
Calculation:
$$\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}$$
LHS =
$$\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} \\ =\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{(x^2-1)(x^2+1)}{x-1} \\ =\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{(x-1)(x+1)(x^2+1)}{x-1} \\ =\rm \displaystyle\lim_{x \rightarrow 1}(x+1)(x^2+1)\\ =(1+1)(1+1)\\ =4$$
RHS = $$\rm \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}$$
Here we have 0/0 form so apply L-Hospitals rule
$$\rm \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}=\rm \lim_{x\rightarrow k}\frac{3x^2}{2x}\\ =\rm \lim_{x\rightarrow k}\frac{3x}{2}\\ =\frac{3k}{2}$$
∴4 = 3k/2
⇒ 3k = 8
⇒ k = 8/3
Hence, option (3) is correct.
If $${\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}}$$, then $$\mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{f}}\left( {\rm{x}} \right)$$ is equal to
1. -2
2. -1
3. 0
4. 1
Option 4 : 1
Evaluation of Limits MCQ Question 19 Detailed Solution
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
$$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}$$ and $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }$$
Then we can apply L-Hospital Rule ⇔ $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$
Calculation:
Given:
$${\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}}$$
$$\mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}}$$
Putting x = 2 in f(x),
$$\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} \frac{{\sin \left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}} = \frac{{\sin \left( {{{\rm{e}}^{2 - 2}} - 1} \right)}}{{{\rm{In}}\left( {2 - 1} \right)}}$$
$$= \frac{{\sin \left( 0 \right)}}{{{\rm{In}}\left( 1 \right)}}$$
We get 0/0 form
$$\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{cos\;}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right){{\rm{e}}^{{\rm{x}} - 2}}}}{{\frac{1}{{\left( {{\rm{x}} - 1} \right)}}}}$$ (using L-Hospital rule)
$$\Rightarrow \frac{{\cos \left( {{{\rm{e}}^{2 - 2}} - 1} \right){{\rm{e}}^{2 - 2}}}}{{\frac{1}{{2 - 1}}}}$$
$$\Rightarrow \frac{{\cos \left( 0 \right)\left( 1 \right)}}{1}$$
⇒ 1
Hence, option (4) is correct.
If $${\rm{G}}\left( {\rm{x}} \right) = \sqrt {(25 - {{\rm{x}}^2}}$$ then what is $$\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{G}}\left( {\rm{x}} \right) - {\rm{G}}\left( 1 \right)}}{{{\rm{x}} - 1}}{\rm{\;}}$$equal to?
1. $$- \frac{1}{{2\sqrt 6 {\rm{\;\;}}}}$$
2. $$\frac{1}{5}$$
3. $$- \frac{1}{{\sqrt 6 }}$$
4. $$\frac{1}{{\sqrt 6 }}$$
Option 1 : $$- \frac{1}{{2\sqrt 6 {\rm{\;\;}}}}$$
Evaluation of Limits MCQ Question 20 Detailed Solution
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
1. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}$$
2. $$\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }$$
Then we can apply L-Hospital Rule ⇔ $$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$$
Calculation:
$$\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{G}}\left( {\rm{x}} \right) - {\rm{G}}\left( 1 \right)}}{{{\rm{x}} - 1}}{\rm{\;}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 1} \frac{{\sqrt {\left( {25 - {{\rm{x}}^2}} \right)} - \sqrt {\left( {25 - 1} \right)} {\rm{\;\;}}}}{{{\rm{x}} - 1}}$$
Here, we have 0/0 form, so applying L-Hospital Rule,
$$\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 1} \frac{{\sqrt {\left( {25 - {{\rm{x}}^2}} \right)} - \sqrt {\left( {25 - 1} \right)} {\rm{\;\;}}}}{{{\rm{x}} - 1}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 1} \frac{{\frac{{1\left( { - 2{\rm{x}}} \right)}}{{2\sqrt {\left( {25 - {{\rm{x}}^2}} \right)} }}}}{1}$$
$$= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 1} \frac{{ - {\rm{x}}}}{{\sqrt {\left( {25 - {{\rm{x}}^2}} \right)} }}$$
$$= - \frac{1}{{\sqrt {24} }}$$
$$= - \frac{1}{{2\sqrt 6 }}$$
Hence, option (a) is correct. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9921099543571472, "perplexity": 2261.7340232854363}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583087.95/warc/CC-MAIN-20211015222918-20211016012918-00074.warc.gz"} |
https://metric2011.wordpress.com/2014/06/19/notes-of-indira-chatterjis-rennes-lecture/ | ## Notes of Indira Chatterji’s Rennes lecture
${CAT(0)}$-cube complexes and the median class
Joint with Talia Fernos and Alessandra Iozzi.
1. Motivation
The following corollary.
Theorem 1 A cocompact, irreducible lattice in ${Sl(2,{\mathbb R})\times Sl(2,{\mathbb R})}$ is not cubical.
Completed by Fernos, Caprace, Lecureux in order to prove that anay such lattives, when acting isometrically on a ${CAT(0)}$ cube complex, must have a fixed point.
Consider lattices in semi-simple Lie groups ${G}$.
If ${G}$ has property (T), they can’t be cubical. If ${G=SO(3,1)}$, they are cubical (Bergeron-Wise, using Kahn-Markovic, unobvious). For ${SO(4,1)}$, some lattices are cubical, as Anne Giralt explained, but for the other ones we don’t know.
2. Main result
Theorem 2 Let ${\Gamma}$ act isometrically on a ${CAT(0)}$ cube complex ${X}$ in a non elementary manner (no fixed point on ${X}$ nor on the visual boundary ${\partial X}$). Then a certain bounded cohomology class ${m\in H_b^2(\Gamma,\pi)}$ vanishes.
Corollary 3 (Superrigidity) Let ${\Gamma}$ be a cocompact irreducible lattice in a product of locally compact groups ${G}$. Let ${\Gamma}$ act essentially and non-elementarily on a ${CAT(0)}$ cube complex. Then the action extends continuously to ${G}$, factoring via one of the factors.
The fact that this follows from the theorem is due to Shalom and Burger-Monod.
3. The median class
3.1. Case of trees
I explain the case when ${X}$ is a tree.
Let ${H}$ be the set of oriented paths of length 2 in the tree. Let ${\pi}$ be the obvious action of ${\Gamma}$ on ${\ell^2(H)}$. Let ${w:X\times X\rightarrow \ell^2(H)}$ be defined by
$\displaystyle \begin{array}{rcl} w(x,y)=1_{[[x,y]]}-1_{[[y,x]]}, \end{array}$
where ${[[x,y]]}$ denotes the set of ${a\in H}$ which are between ${x}$ and ${y}$. This is unbounded, but the coboundary ${dw}$ is bounded. Indeed, cancellations leave us only with paths that touch the median.
In the case of trees, this is not surprising (classical fact that generalizes to hyperbolic metric spaces).
3.2. Median metric spaces
In a metric, the side ${I(x,y)}$ is the set of points for wich the triangle inequality is an equality. A metric space is median if given 3 points, there is a unique common point to the 3 sides.
${CAT(0)}$ cube complexes equipped with the metric which is ${\ell^1}$ on cubes are median.
3.3. Case of CCC
${CAT(0)}$ cube complexes have half-spaces: start cutting s cube in equal parts and continue for ever in contiguous cubes. Say half-spaces ${h_1 \subset h_2}$ are tightly nested if any half-space that sits in between must be one of them. Define ${H}$ as the set of pairs of tightly nested half-spaces. The same formula defines a 1-cochain ${w}$. The same cancellations show that ${dw}$ only involves pairs touching the median point. Therefore it is bounded. However, the bound depends on the dimension of ${X}$. Pull-back ${dw}$ on ${\Gamma}$ via an orbit.
3.4. Non vanishing
Burger-Monod show that
$\displaystyle \begin{array}{rcl} H_b^2(\Gamma,\pi)\equiv ZL_{alt,*}^{\infty}(B,\pi)^{\Gamma}, \end{array}$
where ${B}$ is a Poisson boundary. We use the Roller compactification, defined as follows. ${X}$ embeds in the set of subsets of ${H}$ (a point is mapped to the set of half-space pairs that contain it). Take the closure of the image of that embedding. Then (Zimmer), there is an equivariant map of ${B}$ to the set of probability measures on ${\bar{X}}$. One shows that the image is in ${\partial X}$, this gives the image of ${dw}$ as a nonzero cocycle on ${B}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 50, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9801309108734131, "perplexity": 412.79321220268486}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00494.warc.gz"} |
https://arxiv-export-lb.library.cornell.edu/abs/2211.13602 | hep-ph
(what is this?)
# Title: Measuring the Quantum State of Dark Matter
Abstract: I demonstrate a simple example of how the time series obtained from searches for ultralight bosonic dark matter (DM), such as the axion, can be used to determine whether it is in a coherent or incoherent quantum state. The example is essentially trivial, but I hope that explicitly addressing it provokes experimental exploration. In the standard coherent state, $\mathcal{O}(1)$ oscillations in the number density occur over the coherence time, $\tau_c=h/m v^2$, where $m$ is the particle mass and $v$ is the galactic virial velocity, leading to a reduction in the constraining power of experiments operating on timescales $T<\tau_c$, due to the unknown global phase. On the other hand if the DM is incoherent then no such strong number oscillations occur, since the ensemble average over particles in different streams gives an effective phase average. If an experiment detects a signal then the coherent or incoherent nature of DM can be determined by time series analysis over the coherence time. This finding is observationally relevant for DM masses, $10^{-17}\text{ eV}\lesssim m\lesssim 10^{-11}\text{ eV}$ (corresponding to coherence times between a year and 100 seconds), and can be explored by experiments including CASPEr, DMRadio, and AION. Coherence may also be measurable at higher masses in the microwave regime, but I have not explored it.
Comments: 6 pages, 2 figures, perspective article submitted to special issue of Annalen der Physik for the "Wavy Dark Matter Summer" Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Experiment (hep-ex); Quantum Physics (quant-ph) Report number: KCL-PH-TH/2022-55 Cite as: arXiv:2211.13602 [hep-ph] (or arXiv:2211.13602v1 [hep-ph] for this version)
## Submission history
From: David Marsh [view email]
[v1] Thu, 24 Nov 2022 13:41:35 GMT (646kb,D)
Link back to: arXiv, form interface, contact. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8446351289749146, "perplexity": 1834.1758282078144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00506.warc.gz"} |
http://math.stackexchange.com/questions/133037/proof-involving-a-minimum-weight-spanning-tree | Proof involving a minimum weight spanning tree.
Let G be an undirected graph, $v: E\to R$ and $w: E\to R$ be two weight functions on the edges of $G$. Let $z: E\to R$ be defined as the sum of $v$ and $w$, i.e. for every edge $e$ of $G$ $z(e)=v(e) + w(e)$. Let $V$, $W$, and $Z$ be the total weights of the minimum weight spanning trees of $G$ with respect to weight functions $v$, $w$ and $z$, respectively. Prove or disprove: $Z \geq V+W$.
If this $Z \lt V + W$, then consider the the minimum weight spanning tree for which Z is attained. Let its weight according to v be V' and according to w be W'. $V' + W' \lt V + W$, so either V' < V or W' < W, which is not possible as V and W are the weights for the minimum spanning trees for v and w, respectively. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9882510304450989, "perplexity": 74.86926360479865}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988924.75/warc/CC-MAIN-20150728002308-00022-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://rd.springer.com/chapter/10.1007/978-1-4899-7405-1_5 | # Optimal Stopping Problems
• Xiaoqiang Cai
• Xianyi Wu
• Xian Zhou
Chapter
Part of the International Series in Operations Research & Management Science book series (ISOR, volume 207)
## Abstract
This chapter provides the foundations for the general theory of stochastic processes and optimal stopping problems. In Section 5.1, we elaborate on the concepts of s-algebras and information, probability spaces, uniform integrability, conditional expectations and essential supremum or infimum at an advanced level of probability theory. Then stochastic processes and the associated filtrations are introduced in Section 5.2, which intuitively explains the meaning of information flow. Section 5.3 deals with the concept of stopping times, with focus on the s-algebras at stopping times. Section 5.4 provides a concise introduction to the concept and fundamental results of martingales. The emphasis is focused on Doob’s stopping theorem and the convergence theorems of martingales, as well as their applications in studying the path properties of martingales. The materials in this chapter are essential in the context of stochastic controls, especially for derivation of dynamic policies.
## Keywords
Stochastic Process Probability Space Conditional Expectation Path Property Uniform Integrability
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.
## References
1. Dellacherie, C., & Meyer, P. (1978). Probabilities and potential A. Amsterdam/New York: North-Holland.Google Scholar
2. Dellacherie, C., & Meyer, P. (1982). Probabilities and potential B: Theory of martingales. Amsterdam/Oxford: North-Holland.Google Scholar
3. Karatzas, I., & Shreve, S. E. (1998). Methods of mathematical finance. New York: Springer.Google Scholar
4. Peskir, G., & Shiryaev, A. N. (2006). Optimal stopping and free-boundary problems (Lectures in mathematics). Basel: ETH Zürich/Birkhäuser.Google Scholar
5. Stoyanov, J. M. (1997). Counterexamples in probability (2nd ed.). Chichester/New York: Wiley.Google Scholar
## Authors and Affiliations
• Xiaoqiang Cai
• 1
• Xianyi Wu
• 2
• Xian Zhou
• 3
1. 1.Department of Systems Engineering and Engineering ManagementThe Chinese University of Hong KongShatin, N.T.Hong Kong SAR
2. 2.Department of Statistics and Actuarial ScienceEast China Normal UniversityShanghaiPeople’s Republic of China
3. 3.Department of Applied Finance and Actuarial StudiesMacquarie UniversityNorth RydeAustralia | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8455450534820557, "perplexity": 3065.9246193241233}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578526966.26/warc/CC-MAIN-20190419021416-20190419043416-00116.warc.gz"} |
http://mathhelpforum.com/differential-geometry/200302-question-about-proof-all-surfaces-zero-gaussian-curvature-developable.html | # Math Help - Question about proof that all surfaces with zero Gaussian curvature are developable
1. ## Question about proof that all surfaces with zero Gaussian curvature are developable
Hey guys, I need your help in understanding a proof given by Struijk on the claim that all surfaces with zero Gaussian curvature are developable surfaces. I am an engineering student rather than a mathematician, so excuse me if this is a complete beginners-question
Basically, the part of the proof I do understand is:
$K = 0$ means that the determinant of the second fundamental tensor must be zero $ef-g^2 = 0$.
We can rewrite this $ef - g^2 = (X_u \cdot N_u)(X_v \cdot N_v)-(X_v \cdot N_u)(X_u \cdot N_v) = (X_u \times X_v) \cdot (N_u \times N_v)$
We obtain the following requirement for zero gaussian curvature: $N \cdot (N_u \times N_v) = 0$.
Now we have two possibilities:
1. $N_u = 0$ or $N_v = 0$
2. $N_u$ is collinear with $N_v$, which basically implies that case 1 will be true if we change to different coordinates
So far so good, but now the part that I don't understand. If we look at case 1 only, how does for example $N_u = 0$ imply developability, meaning that the normal vector does not change along a straight line on the surface (generator)? Why must this be true and why can't the $u = constant$-curve be curved? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9388916492462158, "perplexity": 212.08391491270947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657127285.44/warc/CC-MAIN-20140914011207-00043-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.physicsforums.com/threads/taking-the-derivative-of-a-polynomial-fraction.639087/ | # Taking the derivative of a polynomial fraction?
1. Sep 26, 2012
### mooha
Taking the derivative of a polynomial fraction??
b]1. The problem statement, all variables and given/known data[/b]
Ok, so the question wants me to differentiate f(x)= (x)/(x+1). We are supposed to use the definition of the derivative f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h). We also have learned the power rule. I did the formula and had some issues, so I tried to check my answer with the power rule. However they didn't come out the same. I'm not sure where I made the mistake, the formula or the power rule!
2. Relevant equations[/b
f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h)
3. The attempt at a solution
Here's my work:
f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h
=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)
= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1)
= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)
= (lim h->0) h/(h)(x+hx+x^2+h+x)
=(lim h->0) 1/x^2+2x+hx+h
f'(x)= 1/x^2+2x
Power Rule:
y'= (x)/(x+1)
y'= (x)*(x+1)^-1
y'= -(x+1)^-2
y'= -1/(x+1)^2
Thank you so much!!
2. Sep 26, 2012
### SammyS
Staff Emeritus
Re: Taking the derivative of a polynomial fraction??
Hello mooha. Welcome to PF !
It's pretty difficult to read line after line of fractions in the form of
[(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1), etc,
especially when you don't always use proper parentheses, as in
1/x^2+2x+hx+h, which I assume you intended to be $\displaystyle \frac{1}{x^2+2x+hx+h}\ .$
You might make you life easier by writing f(x) as
$\displaystyle f(x)=1-\frac{1}{x+1}\ .$
3. Sep 26, 2012
### mooha
Re: Taking the derivative of a polynomial fraction??
Unfortunately my computer does not have an easy way to do that! Unless you are able to do that with the thread options? It would make my life a lot easier! It is hard to keep putting in those parenthesis! :) Thanks
4. Sep 26, 2012
### SammyS
Staff Emeritus
Re: Taking the derivative of a polynomial fraction??
I will look at it more carefully now.
5. Sep 26, 2012
### SammyS
Staff Emeritus
Re: Taking the derivative of a polynomial fraction??
You don't need to expand the denominator, & apparently that's where your mistake came from.
6. Sep 26, 2012
### mooha
Re: Taking the derivative of a polynomial fraction??
Ok thank you! So my use of the power rule is correct?
7. Sep 26, 2012
### mooha
Re: Taking the derivative of a polynomial fraction??
I fixed my mistake, but the answer from the power rule does not match up with the one from the equation.
WAIT! does the power rule go to zero?? because it is (x)(x+1)-1 and the x goes to zero? does the power rule not work at all for this kind of function?
8. Sep 26, 2012
### SammyS
Staff Emeritus
Re: Taking the derivative of a polynomial fraction??
Yes the power rule works for (x)(x+1)-1 ...
after you use the product rule !
Try the alternate expression for f(x): f(x) = 1 - (x+1)-1 .
9. Sep 26, 2012
### Ray Vickson
Re: Taking the derivative of a polynomial fraction??
It is hard to read such expressions without parentheses, so I don't bother trying.
RGV
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https://physics.stackexchange.com/questions/134615/modeling-magnetic-field-shielding | Modeling magnetic field shielding
I am wondering what is the best way to model magnetic and electric field shielding from a magnetic dipole in the near field?
For example, let's say you have a coil of current carrying wire in the x-y plane which creates an AC magnetic field along the z-axis. I am interested in a model in which I could put in material parameters (such as $\sigma$, $\epsilon$, $\mu$) for a shielding plate located distance $r$ from the source.
I tried using section 5.4 of an online electromagnetic book which talks about reflection, and transmission of of EM waves. I would estimate wave impedance as $Z=\omega\mu_0r$ which is the equation for near field magnetic source. I determine the intrinsic impedance of the shielding material as $\eta=\sqrt {\frac {j\omega\mu} {\sigma + j\omega\epsilon}}$. I want to try good conductors as well as ferrite so I'm not making any assumptions such as good conductor or loss-less medium.
I implemented this but the result was not what I expected. I would expect highest energy reflection with the ferrite because it has low reluctance to magnetic fields and would guide the flux through it and return to source. However, a good conductor (e.g aluminum) had almost 100% energy reflection, but I was expecting the shielding mechanism of conductors would be all energy lost as absorption from eddy currents.
In general, the eddy equation for "low" frequency is voltage based (dB/dt). I've recently been working on this issue from this side of the spectrum, where "shielding" is assumed to be negligible. In extending this assumption to higher orders, I found this assumption to break down. The shielding equation in my case is what I'm calling "shield factor" $S=\frac{B_{cur}}{B_{applied}}=k_{geom}\frac{\tau^2\mu f}{\rho}$ where $B_{cur}$ is flux density from eddy currents, $B_{applied}$ is flux density from external source, and $\tau$ is material thickness. The assumption that $\tau$ is small compared to other dimensions is also made. I'm not sure I kept all of the assumptions to extend this to $B_{cur} ~=B_{applied}$ where shielding is very effective, but for my case it seems to still be valid.
Selecting materials: One would want to select the highest conductivity first, then add some cheap permeability second. You'll see a lot of copper shielding for this reason, but also a lot of nickel shielding. Pure nickel is only 6x lower conductivity than copper, but you might get 10-100 for relative permeability in the MHz range. There is an added benefit of nickel that it can shield low frequency fields as well, so that can offset the cost. Copper will tend to function relatively better as the frequency increases, and nickel will tend to lose advantage at higher frequencies due to reduced permeability. This should give the first intuition into material selection. One might want to increase resistivity to reduce power loss in the shielding material, but $E^2/\sigma$ assumes a poor shield, or another way of saying $\int I\mu dA<<B$ at the frequency in question. For a shield, this is wrong, so you need to apply the field from the eddy currents back into the system, resolving the mesh for the given geometry (assuming this is FEA). http://www.public.iastate.edu/~nbowler/pdf%20final%20versions/conferences/QNDE2005Bowler.pdf
To get to the top equation, I used the assumptions and equations from NPTEL for eddy loss , which happened to be independent of permeability: $\frac{E^2}{\sigma} ->P_{eddy}=\frac{\pi^2f^2B^2_{max}\tau^2}{(6\rho)}(hL\tau)$ and $I=k_{geom}\frac{fB_{max}}{\rho}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212738275527954, "perplexity": 283.56797213522884}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00116.warc.gz"} |
https://www.physicsforums.com/threads/transverse-mechanical-waves-fundamentals.274210/ | # Transverse mechanical waves - fundamentals
1. Nov 22, 2008
### Mårten
Hello all!
I'm trying to understand and in detail analyse how transverse mechanical waves work, and have had troubles finding anything on the net, except for the link below. Let's begin with a water wave. The following questions then:
1) Firstly, I wonder if it's possible to find, buy or make some kind of liquid which exhibits a really slow wave speed, preferably something around 1 mm/s. This would have been wonderful to have to make experiments and really see how the wave slowly moves, like it was alive. I suppose it must be a pretty gelly liquid with high viscosity.
2) Someone who knows of an animated computer simulation which shows the various forces which build up the wave crest and the wave trough, and so forth?
3) Why, exactly, is the wave propagating? I've tried to study the following text (see link below). It says that if there is a depression in the water surface, then water on the sides flows down in this depression, building up a small hill (btw, how can this hill become higher than the original water level?). Then, where the water flowed down from the sides, a new depression is formed. But how can this new depression be as deep as the first depression? The water there should level out with the first depression?
http://www.lightandmatter.com/html_books/3vw/ch03/ch03.html#Chapter3 [Broken] (see figure and 2nd paragraph)
Last edited by a moderator: Apr 23, 2017 at 10:17 PM
2. Nov 23, 2008
### cesiumfrog
Right, that would be the point at which the force becomes zero, but because of inertia the medium can't instantly stop moving, not until it goes so far as to build up a force in the opposite direction..
3. Nov 24, 2008
### Mårten
I don't get that really. If the water molecules ran downhill this slope, they would run diagonally, not affecting the molecules right below them. So as the slope here becomes less and less steep, the molecules will have less and less energy, and when in equilibrium with the water in the center (i.e. the center of the original depression), no more water molecules will run down the slope (since there's no slope anymore).
Also, I've read that the water molecules should run in circles...
Last edited: Nov 24, 2008
4. Nov 24, 2008
### LURCH
Regarding your first question; do you have a video camera? You could record water waves, and then play them back at a lower speed.
This is actually two questions that have the same answer. Your first question, "how can this hill become higher than the original water level?", sounds like you're trying to figure out where the water gets the energy from to go past its original point of equilibrium. This energy is taken from whatever outside influence caused a depression in the water surface to begin with.
The answer to your second question, "how can the new depression be as deep as the first depression?", depends on the manner in which the waves are being formed. Again, the energy for this wave depression comes from the original event that disturbed the fluid surface in first place. If the surface is disturbed by a short-term event or impact (like a pebble being thrown in), then the second depression cannot be as deep as the first. The water's surface will descend to a point that is below its original level, but above the lowest point in the preceeding trough. If the surface is being disturbed by a continuous and ongoing external force (such as the wind), then this external force can continuously add energy so that it is even possible for the second depression to be deeper than the first.
But, in either case, some event has to put energy into the fluid.
5. Nov 24, 2008
### Mårten
No, I haven't... And it should be a high speed camera I guess. I found this formula here for the speed of a transverse wave, $v = \sqrt{F/\mu}$, and to get v = 0.001 m/s, yo got to have e.g. F = 1 N, and mu = 10^6 kg/m. Hm, that's pretty much...
Okey, thank's for a good explanation there regarding input of energy. I'm getting closer. Still, I cannot see the whole picture. I'll try to give a new, maybe more pure example.
Imagine a water surface (a glass of water may be good for experiment) where you carefully put down the end of a knife, just a millimetre or so, and let the knife be in that position. As soon as the knife touches the water surface, a wave pulse is seen emerging from where the knife touched the water surface. Imagine now the water surface close up to the knife, just after impact. There the water must be a little bit higher than the rest of the water, because the water where the knife now is, must go somewhere. Now, this higher water would like to come in equilibrium with the water outside, so the higher water flows outwards to the lower water.
Now, this had been perfectly okey for me, if what happened here had been a sort of pulse, where the higher water flowed outward, but as soon as that water was as high as the water outside no more flow would occur. So the pulse would not be like a hill going outwards from the impact place, more like a step function, that slowly decreased the height of the step as it moved further away from the impact. But now it is a hill - how come? And, as pointed out before, I cannot see that this higher water has momentum that makes it go below the equilibrium level, since the flow seems to be diagonal. And, it's not even diagonal, they say it's circular.
(Maybe, what I'm ultimately out for is the micro mechanics that's behind this circular motion of the water molecules.)
6. Nov 24, 2008
### Dblbond
7. Nov 24, 2008
### Staff: Mentor
The water molecules in a surface water wave do not run diagonally down the slope of the wave. Each molecule oscillates vertically. Neighboring molecules oscillate slightly out of step with each other, i.e. with slightly different phases, although still with the same frequency.
8. Nov 24, 2008
### cesiumfrog
Well, if the depth is finite, the water parcels do take "http://en.wikipedia.org/wiki/Airy_wave_theory" [Broken] paths. But I thought the OP's confusion might be better cleared up first with reference to simpler waves, such as of a rope.
Last edited by a moderator: Apr 23, 2017 at 10:19 PM
9. Nov 24, 2008
### Mårten
Okey, let's think of this vertical oscillation then. And my example above with the water that is higher just close up to the knifepoint that was just now inserted into the water surface. Now, what force drives these molecules vertically down?
I found this animation of water waves. It's nice, but I don't get the mechanics driving the points. http://www.kettering.edu/~drussell/Demos/waves/water.gif [Broken]
Btw, I suppose the water we are talking about here has a constant density, and that there is no kind of compression of the water anywhere. Correct me if this is an erroneous assumption.
I have indeed problems understanding the mechanics of a wave in a rope as well, but I thought I will take that issue later...
Last edited by a moderator: Apr 23, 2017 at 10:19 PM
10. Nov 24, 2008
### Staff: Mentor
You're right, I remember that now from somewhere. I agree that a stretched string or rope would be a better system to start with. I'm sure many classical mechanics textbooks derive the wave equation for a stretched string. I've seen it, but I don't remember the details and I don't have any of those books here at home.
For surface waves in deep water you probably have to deal with the surface tension, and I suspect that in some approximation you can consider the molecules as moving in very narrow vertical ellipses, i.e. practically in a vertical straight line. This would be for a propagating wave, relatively far from the source.
To take Mårten's example with the knife point touching the surface of the water to start the wave, the water next to the knife initially rises because of capillary action (I think that's what it's called) on the surface of the knife. This is what makes water rise by itself a bit in a narrow tube that you dip vertically into the water. The molecules next to the knife, in turn pull up their neighbors which then fall back a bit, and those neighbors pull up their neighbors, etc. I've never seen this "initial" process analyzed mathematically.
Last edited by a moderator: Apr 23, 2017 at 10:19 PM
11. Nov 25, 2008
### Mårten
Btw, I know of course there is gravity. But since the water is higher here, there is also a normal force from the underlying molecules, so the total force should be zero downwards (assuming that the water cannot be compressed). So the molecules ought not to go straight downwards. Like a steep pile of sand that you construct - the grains of sand will fall down more diagonally, not vertically, since there are other grains of sand in the way, just below them.
They do, but mine doesn't explain properly how the forces come about, at least so I can understand it...
I've been thinking of that, but I was thinking that you could disregard that fact. If you have dishwashing liquid instead (which I think doesn't have much of surface tension), and put down the knife point into the liquid, you will still have a wave, at least if you put down the knife point not too slowly. That was what I would expect at least...
Edit: Some reading makes me conclude that we cannot at all disregard surface tension and capillary effects, since what we are talking about here is actually called capillary waves, see:
http://en.wikipedia.org/wiki/Capillary_wave
Last edited: Nov 25, 2008
12. Nov 25, 2008
### Mårten
I've now started a new thread, dealing with the maybe more easily understood mechanics of the wave in a rope - see here: https://www.physicsforums.com/showthread.php?t=274913
I've now understood that this area is actually extremely complicated - even Feynman admits that... :yuck: (see link above in my previous post).
Still, I insist in that there ought to be some form of conceptual answer, not too much contaminated by ugly formulas, to why the water molecules behave like circles, like they do in this figure below, and why they ultimately spread their motion to neigboring molecules, so that the wave propagates, which is the main thing I haven't understood yet.
[Broken]
Last edited by a moderator: Apr 24, 2017 at 8:40 AM
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https://crossminds.ai/video/curse-of-dimensionality-on-randomized-smoothing-for-certifiable-robustness-606f431b072e523d7b7806fa/ | Curse of Dimensionality on Randomized Smoothing for Certifiable Robustness
# Curse of Dimensionality on Randomized Smoothing for Certifiable Robustness
Jul 12, 2020
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###### Details
Randomized smoothing, using just a simple isotropic Gaussian distribution, has been shown to produce good robustness guarantees against $\ell_2$-norm bounded adversaries. In this work, we show that extending the smoothing technique to defend against other attack models can be challenging, especially in the high-dimensional regime. In particular, for a vast class of i.i.d. smoothing distributions, we prove that the largest $\ell_p$-radius that can be certified decreases as $O(1/d^{\frac{1}{2} - \frac{1}{p}})$ with dimension $d$ for $p > 2$. Notably, for $p \geq 2$, this dependence on $d$ is no better than that of the $\ell_p$-radius that can be certified using isotropic Gaussian smoothing, essentially putting a matching lower bound on the robustness radius. When restricted to generalized Gaussian smoothing, these two bounds can be shown to be within a constant factor of each other in an asymptotic sense, establishing that Gaussian smoothing provides the best possible results, up to a constant factor, when $p \geq 2$. We present experimental results on CIFAR to validate our theory. For other smoothing distributions, such as, a uniform distribution within an $\ell_1$ or an $\ell_\infty$-norm ball, we show upper bounds of the form $O(1 / d)$ and $O(1 / d^{1 - \frac{1}{p}})$ respectively, which have an even worse dependence on $d$. Speakers: Aounon Kumar, Alexander Levine, Tom Goldstein, Soheil Feizi | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8465620875358582, "perplexity": 464.69603779476506}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337971.74/warc/CC-MAIN-20221007045521-20221007075521-00354.warc.gz"} |
https://www.physicsforums.com/threads/frictional-force-on-a-block-held-against-a-vertical-wall.674916/ | # Frictional force on a block held against a vertical wall
1. Feb 27, 2013
### freddyeddyjor
1. A 5kg block is held at rest against a vertical wall by a horizontal force of 100 newtons.
1-What is the frictional force exerted by the wall on the block
2-What is the minimum horizontal force needed to prevent the block from falling if the static coefficient of friction between the wall and the block is μs = .4
2. Relevant equations
MAx = Right forces- Left Forces ( This is a homemade equation, basically the forces pushing from the wall are subtracted from the forces pushing against the wall.
3. The attempt at a solution
I am unsure how to proceed with this problem. I know that for part 1, the frictional force should be about 49.1n, this comes from the equation shown above. If right forces- left forces= 0, then there is no movement, and therefore they must be equal. So 5kg*9.81m/s = 49.05N
The second part is what confuses me, I fail to see a way to incorporate the friction coefficient into the problem in a meaningful way.
Last edited: Feb 27, 2013
2. Feb 27, 2013
### Staff: Mentor
You don't need the coefficient of friction for the first part. Hint: The block is in equilibrium.
3. Feb 27, 2013
### haruspex
You know how large the frictional force needs to be from the first part. If the horizontal force is reduced, what does that do to the max frictional force available? At what point will it the available frictional force equal the required frictional force?
4. Feb 27, 2013
### KarlaQat
I think the part you're missing is that the frictional force is determined by:
Friction force = μs * Normal force
In this case you start out with the normal force pushing against the wall at 100N, but how low could it go before the block fell down?
5. Feb 28, 2013
### freddyeddyjor
So I should set μk*Fn= Frictional Force exerted by the wall.
.4*FN = 49.1
That actually makes a lot of sense, thank you all.
Draft saved Draft deleted
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http://physics.stackexchange.com/users/23752/belgi | # Belgi
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# 12 Questions
3 The lab have a constant electric field and a constant magnetic field, what is the electric and magnetic field inside a conductor and far from it 3 Using the image charges method to find the electric field 3 Why is the radial direction the preferred one in spherical symmetry? 2 A dielectric table is being inserted between a plate capacitor and $\triangle U<0$ how to deduce the table is attracted to the plates? 2 Why does the electric field perpendicular to every point on the surface of a conductor?
# 232 Reputation
+5 Why does the electric field perpendicular to every point on the surface of a conductor? +5 A dielectric table is being inserted between a plate capacitor and $\triangle U<0$ how to deduce the table is attracted to the plates? +15 The lab have a constant electric field and a constant magnetic field, what is the electric and magnetic field inside a conductor and far from it +5 What does it mean for a $2$ dimensional object to be charged with a charge distribution which is $1$ dimensional $\lambda$?
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# 19 Tags
0 electrostatics × 8 0 dielectric × 2 0 electric-fields × 4 0 coulombs-law × 2 0 homework-and-exercises × 4 0 gauss-law × 2 0 electromagnetism × 4 0 potential × 2 0 work × 3 0 forces × 2
# 13 Accounts
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http://mathhelpforum.com/discrete-math/165096-counting-problem.html | 1. ## Counting Problem
I simply don't know how to start.
In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.
I know that if the "difference between 2" wasn't there it would be C(100,10). But with the difference there how the heck would you set it up?
thanks
2. Originally Posted by bfpri
In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.
The idea is simple. We can say include $13~\&~15$ among the ten because $|13-15|\ge 2$ but not $15~\&~16$.
So how many ways can we pick ten of those numbers so no two are consecutive?
How many ways can you arrange a string of ten 1’s and ninety 0’s so no two 1’s are next each other?
3. Originally Posted by Plato
The idea is simple. We can say include $13~\&~15$ among the ten because $|13-15|\ge 2$ but not $15~\&~16$.
So how many ways can we pick ten of those numbers so no two are consecutive?
How many ways can you arrange a string of ten 1’s and ninety 0’s so no two 1’s are next each other?
That would be 90!/10!*80! if you split it into 10 "10"s and 90 "1"s.
4. Actually it would be $\dbinom{91}{10}$.
The ninety zeros create ninety-one places to place the ones.
5. Hmm, still don't quite understand why it's 91.
6. Simply using, $1111000000$, four ones and six zeros.
How many ways to arrange that string so no two ones are together?
Look at $\_\_0\_ \_0\_ \_0\_ \_0\_ \_0\_ \_0\_\_$.
Do you see there are seven places to put the four ones: $\dbinom{7}{4}$.
So ninety zeros create ninety one places: $\dbinom{91}{10}$.
7. Got it...thanks! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.870212733745575, "perplexity": 460.26868969195544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00076-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/24000/how-to-prove-the-mathematical-induction-is-true/24069 | # How to prove the mathematical induction is true?
I have no idea about the underlying theory from which the mathematical induction was derived.
How to prove the mathematical induction is true?
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good question... – anonymous Feb 27 '11 at 10:28
That should be "mathematical induction" not "mathematics induction" – Arjang Feb 27 '11 at 11:11
How complex an answer are you looking for? – barrycarter Feb 27 '11 at 16:26
I think a good interpretation to this question is "Is there a more obvious set of principles upon which all of the induction propositions are based?" It's a bit disingenuous to make something as not-obvious as induction into an axiom. – DanielV Mar 20 '14 at 1:27
A "proof" in mathematics always means a proof in some system/theory. You have to specify the system/theory that you want a proof for the induction axiom. (You should also formally specify what you mean by the induction axiom since there are various axioms that are called induction axiom.)
The induction axiom in an arithmetical theory (like Peano arithmetic) is an axiom, i.e. it is one of the axioms of the theory, and therefore the proof is just a single line stating the axiom.
In a set theory like $ZFC$ we can prove the induction axiom for the set of natural numbers using the fact that the set of natural numbers is defined as the smallest inductive set that contains zero, and the proof is almost trivial. (An inductive set means a set that contains the successor of $x$ whenever it contains $x$).
In high school or undergraduate courses, when one is asked to prove induction axiom, they are usually asked to derive the induction axiom from some other axioms like the least number principle for natural numbers.
Another possible question is what are the justifications for believing that the induction axiom is true (or for accepting it as an axiom), which is a question in philosophy of mathematics and might be more suitable for MathOverflow.
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The principle of induction (on the natural numbers) is equivalent to the axiom of well-foundedness of the natural numbers. Wikipedia gives (half of) a proof here.
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The induction axiom in Peano Arithmetic says that for any predicate (statement about numbers) $\phi$, if you can prove $\phi(0)$ is true and you can also prove that for any number $n$, $\phi(n) \implies \phi(n+1)$ then $\phi(n)$ is true for all $n$. So you have proven $\phi(0)$ and $\phi(0) \implies \phi(1)$ and $\dots$ $\phi(10) \implies \phi(11)$ and $\dots$. Intuitively, if there is an $m$ such that $\phi(m)$ is not true, the base case of $\phi(0)$ or one of these implications must have broken down.
The usual structure of an inductive proof is designed to take advantage of this, so following one carefully may help. Arturo Magidin gave an excellent discussion in his answer to this post
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If you are having difficulties accepting mathematical induction and other of Peano's Axioms, I find it helps to visualize the natural numbers as directed graph (a network of nodes connected by one-way arrows) as follows:
There is a node (call it 0) that has no arrows going into it (Axioms 1 and 3, below). Every node has an arrow leaving it (Axiom 2, below). Every node has at most one arrow going into it (Axiom 4). You can get to any node by starting at 0 and following the arrows; that is, there are no isolated portions of the graph (Axiom 5). You may also require that every node have at most one arrow leaving it (not explicitly covered in the axioms as presented below).
(Peano's Axioms from from Wolfram MathWorld http://mathworld.wolfram.com/PeanosAxioms.html)
1. Zero is a number.
2. If x is a number, the successor of x is a number.
3. Zero is not the successor of a number.
4. Two numbers of which the successors are equal are themselves equal.
5. If a set of numbers contains zero and also the successor of every number in , then every number is in .
The following is a better presentation of Peano's axioms (from http://www.ms.uky.edu/~lee/ma502/notes2/node7.html) It uses 1 instead of 0 as the first natural number. Unlike the above presentation, the fact that, graphically, there would be at most one arrow leaving each node is made explicit in Axiom 2, below.
Axiom 1: 1 is a natural number. That is, our set is not empty; it contains an object called 1 (read "one").
Axiom 2: For each x there exists exactly one natural number, called the successor of x, which will be denoted by x'.
Axiom 3: We always have ~x'=1. That is, there exists no number whose successor is 1. That is, there exists no number whose successor is 1.
Axiom 4: If x'=y' then x=y. That is, for any given number there exists either no number or exactly one number whose successor is the given number.
Axiom 5 (Axiom of Induction): Let there be given a set M of natural numbers, with the following properties: I. 1 belongs to M. II. If x belongs to M then so does x'. Then M contains all the natural numbers.
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Well, in principles induction is, having something holding for $0$, denoted as $h(0)$, and if $h(n)$ implies $h(n')$, where $n'$ is a successor of $n$, i.e. $n+1$, we have $h(n)$ for any natural number of $n$.
On the other hand, natural numbers are defined as the smallest set that contains 0 and closed to successor operation (i.e. if $n$ is natural so is $n'$).
The correctness of induction follows strictly from the fact the natural numbers are the the smallest set satisfying the same condition as induction.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9808005094528198, "perplexity": 267.1748793153346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375097710.32/warc/CC-MAIN-20150627031817-00009-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://eng.mephi.ru/news/120855 | MEPhI has developed an ultra-sensitive technique for detecting explosives
08.09.2022
MEPhI scientists have developed an ultra-sensitive technique for detecting trace amounts of explosives in the air. Its essence lies in the addition of impurities that stimulate ion formation to the composition of the gas sample.
According to the scientists of the Department of Physics of Micro and Nanosystems of the MEPhI, for the detection of low-volatile explosives in the gas phase at concentrations below 10-14 g/cm3, ion mobility spectrometry (IMS) and its subspecies — ion mobility increment spectrometry (IMS) get an advantage. The technique makes it possible to ionize the molecules present in the air, classify them according to their mobility, and thus detect impurities of nitro compounds.
“In order to improve the sensitivity of the SIP and SPIS methods, we proposed to introduce into the composition of the gas sample impurities that stimulate ion formation (dopants). So far, dopants have been actively used for traditional ionization sources, however, the effect of laser radiation parameters on the formation of ions of various types of nitro compounds, including the use of dopants, has not been studied,” said Vitaly Kostarev, research engineer at the Department of Physics of Micro and Nanosystems of the MEPhI.
The results of the study were published in the Talanta journal. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8418707251548767, "perplexity": 2500.5395555971377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335424.32/warc/CC-MAIN-20220930020521-20220930050521-00430.warc.gz"} |
http://mathhelpforum.com/discrete-math/29992-integer-arrangement.html | # Math Help - Integer Arrangement
1. ## Integer Arrangement
We want to arrange the integers 1,2,...,n (in a row) such that successive integers cannot be in adjacent positions. In how many ways can we do this for a general n?
Okay, let a_n denote the number of such arrangements. Then we have
a_2 = a_3 = a_4 = 0, a_5 = 2 (13524,14253). The only useful thing i could notice is that in forming the 5-integers arrangement, we can fix 1 and then add 3524 to 1 forwards and backwards. I think we can use this trick in any integer arrangement greater than 5. But how can we calculate the number of arrangements for larger numbers?
2. OK. Take you example of n=5.
We cannot have 13425
. But can we have 14325?
Do you see my question? In 14325 the “43” are consecutive.
3. Yes,i see what you mean but still i can't use it for determining the number of arrangements for larger integers. Can you explain it a little more for a specific case?
We cannot have 13425! We know that!
But can we have 14325?
5. No,we can't have 14325. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9299850463867188, "perplexity": 632.7384960731821}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049278887.13/warc/CC-MAIN-20160524002118-00211-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://likunxie.com/2020/06/ | # Calculate (co)limits as (co)equalisers (two examples)
There is a general formulation for constructing limits as equalisers: see Theorem 1 in Section V.2, Maclane. For the dual version, see Theorem A.2.1 in Appendix A written by me.
The constructions look like these (see the links above for details):
But in practice, these diagrams may not be helpful to see what the equalisers should be. Now I give proofs for the (co)equalisers in two examples: the connected component of a simplicial set and the sheaf condition.
### The connected components [Background]
For definitions and other backgrounds, see Subsection 00G5. For the record, see [P12, DLOR07] for the cosimplicial identities and Tag 000G for simplicial identities. (These identities are used in my proofs.)
# [Short Notes] Non-compactness of the closed unit ball in an infinite-dimensional Banach space
This is about an exercise in [Bass]:
Exercise 19.5. Prove that if $H$ is infinte-dimensional, that is, it has no finite basis, then the closed unit ball in $H$ is not compact.
Proof. Choose an orthonormal basis $\{x_i\}$, then $||x_i-x_j||^2=||x_i||^2+||x_j||^2=2$. This means the sequence is not Cauchy hence has no convergent subsequence.
For a Banach space, by Riesz’s lemma to find a non-Cauchy sequence.
[Bass] Bass, R. F. (2013). Real analysis for graduate students. Createspace Ind Pub. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 4, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.973597526550293, "perplexity": 984.9026906739994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00405.warc.gz"} |
https://almostsuremath.com/2011/05/26/predictable-stopping-times-2/?shared=email&msg=fail | # Predictable Stopping Times
Although this post is under the heading of the general theory of semimartingales’ it is not, strictly speaking, about semimartingales at all. Instead, I will be concerned with a characterization of predictable stopping times. The reason for including this now is twofold. First, the results are too advanced to have been proven in the earlier post on predictable stopping times, and reasonably efficient self-contained proofs can only be given now that we have already built up a certain amount of stochastic calculus theory. Secondly, the results stated here are indispensable to the further study of semimartingales. In particular, standard semimartingale decompositions require some knowledge of predictable processes and predictable stopping times.
Recall that a stopping time ${\tau}$ is said to be predictable if there exists a sequence of stopping times ${\tau_n\le\tau}$ increasing to ${\tau}$ and such that ${\tau_n < \tau}$ whenever ${\tau > 0}$. Also, the predictable sigma-algebra ${\mathcal{P}}$ is defined as the sigma-algebra generated by the left-continuous and adapted processes. Stated like this, these two concepts can appear quite different. However, as was previously shown, stochastic intervals of the form ${[\tau,\infty)}$ for predictable times ${\tau}$ are all in ${\mathcal{P}}$ and, in fact, generate the predictable sigma-algebra.
The main result (Theorem 1) of this post is to show that a converse statement holds, so that ${[\tau,\infty)}$ is in ${\mathcal{P}}$ if and only if the stopping time ${\tau}$ is predictable. This rather simple sounding result does have many far-reaching consequences. We can use it show that all cadlag predictable processes are locally bounded, local martingales are predictable if and only if they are continuous, and also give a characterization of cadlag predictable processes in terms of their jumps. Some very strong statements about stopping times also follow without much difficulty for certain special stochastic processes. For example, if the underlying filtration is generated by a Brownian motion then every stopping time is predictable. Actually, this is true whenever the filtration is generated by a continuous Feller process. It is also possible to give a surprisingly simple characterization of stopping times for filtrations generated by arbitrary non-continuous Feller processes. Precisely, a stopping time ${\tau}$ is predictable if the process is almost surely continuous at time ${\tau}$ and is totally inaccessible if the underlying Feller process is almost surely discontinuous at ${\tau}$.
As usual, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$. I now give a statement and proof of the main result of this post. Note that the equivalence of the four conditions below means that any of them can be used as alternative definitions of predictable stopping times. Often, the first condition below is used instead. Stopping times satisfying the definition used in these notes are sometimes called announceable, with the sequence ${\tau_n\uparrow\tau}$ said to announce ${\tau}$ (this terminology is used by, e.g., Rogers & Williams). Stopping times satisfying property 3 below, which is easily seen to be equivalent to 2, are sometimes called fair. Then, the following theorem says that the sets of predictable, fair and announceable stopping times all coincide.
Theorem 1 Let ${\tau}$ be a stopping time. Then, the following are equivalent.
1. ${[\tau]\in\mathcal{P}}$.
2. ${\Delta M_\tau1_{[\tau,\infty)}}$ is a local martingale for all local martingales M.
3. ${{\mathbb E}[1_{\{\tau < \infty\}}\Delta M_\tau]=0}$ for all cadlag bounded martingales M.
4. ${\tau}$ is predictable.
Before moving on to the proof of the theorem note that, for any stopping time ${\tau}$, ${1_{(\tau,\infty)}}$ is adapted and left-continuous, hence predictable. Since ${[\tau]=[\tau,\infty)\setminus(\tau,\infty)}$, the first statement of the theorem can equivalently by written as ${[\tau,\infty)\in\mathcal{P}}$.
Proof: 1 implies 2: As ${1_{[\tau]}}$ is a bounded predictable process, ${N\equiv\int1_{[\tau]}\,dM}$ is a local martingale. We need to show that ${N=\Delta M_\tau1_{[\tau,\infty)}}$. Although this might seem intuitively obvious, it does require some rather non-trivial properties of the stochastic integral (see the notes below). First,
$\displaystyle N^\tau=\int 1_{(0,\tau]}1_{[\tau]}\,dM=\int1_{[\tau]}\,dM = N,$
so N is constant on ${[\tau,\infty)}$. Secondly, ${\Delta N_t=1_{\{t=\tau\}}\Delta M_\tau}$, so ${N_t=N_{\tau-} + \Delta M_\tau}$ for all ${t\ge\tau}$. For any time ${t\ge0}$, note that ${1_{[\tau]}}$ vanishes over the interval ${[0,t]}$ whenever ${\tau > t}$. As previously shown, this means that ${N_t=\int_0^t1_{[\tau]}\,dM=0}$ whenever ${\tau > t}$. So, ${N=0}$ on ${[0,\tau)}$ and, hence, ${N_t=1_{\{t\ge\tau\}}\Delta M_\tau}$.
2 implies 3: If M is bounded, or just dominated in ${L^1}$, then ${N\equiv \Delta M_\tau1_{[\tau,\infty)}}$ is an ${L^1}$-dominated local martingale, and hence a proper martingale,
$\displaystyle {\mathbb E}\left[1_{\{\tau < \infty\}}\Delta M_\tau\right]={\mathbb E}\left[N_\infty\right]={\mathbb E}[N_0]=0.$
3 implies 4: To show that ${\tau}$ is predictable, we need to construct a sequence of stopping times ${\tau_n}$ increasing to ${\tau}$ such that ${\tau_n < \tau}$ whenever ${\tau > 0}$. The idea is simple enough. First define a right-continuous process giving roughly the expected time remaining before ${\tau}$ and use the debut theorem to construct ${\tau_n}$. To do this, start by choosing a continuous bounded and strictly increasing function ${f\colon{\mathbb R}_+\rightarrow{\mathbb R}_+}$. Then define the martingale
$\displaystyle M_t={\mathbb E}\left[f(\tau)\;\vert\mathcal{F}_t\right].$
Then, ${M_t-f(t)={\mathbb E}[f(\tau)-f(t)\;\vert\mathcal{F}_t]}$ tells us (roughly) how much longer we have to wait until time ${\tau}$. In order to be able to choose a cadlag version of M, it is necessary to assume that the filtration is right-continuous, so that ${\mathcal{F}_{t+}\equiv\bigcap_{s > t}\mathcal{F}_{s}}$ is equal to ${\mathcal{F}_t}$. The result does still hold without the assumption of right-continuity and, to be complete, the extension to the general case is given further below. Assuming that M is cadlag, define the following increasing sequence of stopping times,
$\displaystyle \tau_n=\inf\left\{t\ge0\colon M_t-f(t)\le1/n\right\}.$
As ${M_\tau=f(\tau)}$, we have that ${\tau_n\le\tau}$ for all n. Also, by optional sampling,
$\displaystyle {\mathbb E}[f(\tau)]={\mathbb E}[M_{\tau_n}]\le{\mathbb E}[f(\tau_n)+1/n]$
So, ${{\mathbb E}[f(\tau)-f(\tau_n)]\le1/n}$ decreases to zero, showing that ${f(\tau)-f(\tau_n)}$ tends to zero almost surely. As f was taken to be strictly increasing, this shows that ${\tau_n}$ increases to ${\tau}$. It only remains to show that ${\tau_n}$ is strictly less that ${\tau}$ whenever ${\tau > 0}$. The definition of M gives ${M_t\ge f(t)}$ for all times ${t\le\tau}$. So,
$\displaystyle \Delta M_\tau=f(\tau)-M_{\tau-}\le f(\tau)-f(\tau-)=0.$
Together with condition 3, which says that ${{\mathbb E}[\Delta M_\tau]=0}$, this implies that ${\Delta M_\tau}$ is almost surely zero. So, ${M_{\tau-}=f(\tau)}$. However, by construction,
$\displaystyle M_{\tau_n-}\ge f(\tau_n)+1/n\not=f(\tau_n)$
whenever ${\tau_n > 0}$, so, ${\tau_n\not=\tau}$ as required.
4 implies 1: As previously shown, ${[\tau,\infty)}$ is predictable for all predictable times ${\tau}$ and, therefore, so is ${[\tau]=[\tau,\infty)\setminus(\tau,\infty)}$. ⬜
An immediate consequence of this result is that the conclusion of the debut theorem can be strengthened in the case of right-continuous and predictable processes. This gives a significant generalization of the much simpler result that hitting times of continuous adapted processes are predictable.
Corollary 2 Let X be a right-continuous and predictable process. Then, for each constant K, the stopping time
$\displaystyle \tau=\inf\left\{t\ge0\colon X_t\ge K\right\}$
is predictable.
Proof: The left-continuous and adapted process ${1_{[0,\tau]}}$ is predictable. So,
$\displaystyle [\tau]=[0,\tau]\cap X^{-1}([K,\infty))$
is predictable and, by Theorem 1, ${\tau}$ is predictable. ⬜
A process X is locally bounded if there exists a sequence of stopping times ${\tau_n}$ increasing to infinity such that the stopped processes ${1_{\{\tau_n > 0\}}X^{\tau_n}}$ are each uniformly bounded. Continuous processes are easily seen to be locally bounded simply by stopping them as soon as they hit a level ${K > 0}$. This generalizes to cadlag predictable processes.
Lemma 3 All cadlag predictable processes are locally bounded.
Proof: Supposing that X is cadlag, define the sequence of stopping times
$\displaystyle \tau_m=\inf\left\{t\ge 0\colon\vert X_t\vert\ge m\right\}.$
These are increasing to infinity and, by Corollary 2, are predictable. So, for each m, there is a sequence ${\{\tau_{mn}\}_{n=1,2,\ldots}}$ of stopping times increasing to ${\tau_m}$ such that ${\tau_{mn} < \tau_m}$ whenever ${\tau_m > 0}$. Then, ${\sigma_n\equiv\tau_{1n}\vee\tau_{2n}\vee\cdots\vee\tau_{nn}}$ are stopping times increasing to infinity. Also, ${\sigma_n < \tau_n}$ whenever ${\tau_n > 0}$. So, ${1_{\{\sigma_n > 0\}}X^{\sigma_n}}$ is bounded by n and, hence, X is locally bounded. ⬜
Theorem 1 enables us to state several equivalent conditions for a cadlag adapted process X to be predictable. Note that the process ${X_-}$ of left-limits is automatically predictable, being left-continuous and adapted. For brevity, I write ${X_\tau}$ for the value of a process at a random time, even though this is not well defined when ${\tau=\infty}$. In that case, I take ${X_\tau}$ to be zero whenever ${\tau}$ is infinite, so ${X_\tau\equiv1_{\{\tau < \infty\}}X_\tau}$ (setting it to any ${\mathcal{F}_\infty}$-measurable value will not change any of the statements below). I also use ${\{\Delta X\not=0\}}$ as shorthand for the (progressively measurable and optional) set of times at which X is discontinuous which, more precisely, consists of the ${(t,\omega)\in{\mathbb R}_+\times\Omega}$ for which ${\Delta X_t(\omega)}$ is nonzero. The following notation will be used in the proofs. Given a stopping time ${\tau}$ and a set ${A\in\mathcal{F}_\tau}$, the random time ${\tau_A\colon\Omega\rightarrow\bar{\mathbb R}_+}$ is defined by
$\displaystyle \tau_A(\omega)\equiv\begin{cases} \tau(\omega),&\textrm{if }\omega\in A,\\ \infty,&\textrm{if }\omega\not\in A. \end{cases}$
It is clear that this defines a stopping time. The statement and proof of the equivalent conditions for a cadlag process to be predictable can now be given.
Lemma 4 If X is a cadlag adapted process then the following are equivalent.
1. X is predictable.
2. ${\Delta X}$ is predictable.
3. ${X_\tau}$ is ${\mathcal{F}_{\tau-}}$ measurable for all predictable stopping times ${\tau}$, and ${\Delta X_\tau=0}$ (almost surely) whenever ${\tau}$ is totally inaccessible.
4. there exists a sequence of predictable stopping times ${\{\tau_n\}_{n=1,2,\ldots}}$ such that ${\{\Delta X\not=0\}\subseteq\bigcup_n[\tau_n]}$ and ${X_{\tau_n}}$ is ${\mathcal{F}_{\tau_n-}}$-measurable for each n.
5. there exists a sequence of predictable stopping times ${\{\tau_n\}_{n=1,2,\ldots}}$ with disjoint graphs (${[\tau_m]\cap[\tau_n]=\emptyset}$ for ${m\not=n}$) such that ${\{\Delta X\not=0\}=\bigcup_n[\tau_n]}$ and ${X_{\tau_n}}$ is ${\mathcal{F}_{\tau_n-}}$-measurable for each n.
Proof: 1 implies 4: Let ${s_n,\epsilon_n}$ be a sequence running over the pairs of positive rational numbers, and define the stopping times
$\displaystyle \tau_n=\inf\left\{t\ge s_n\colon\vert X_t-X_{s_n}\vert\ge\epsilon_n\right\}.$ (1)
Corollary 2 implies that these are predictable. Looking at the individual sample paths of the process, consider a time t for which ${\Delta X_t\not=0}$. Then, ${\tau_n=t}$ whenever ${\epsilon_n < \vert\Delta X_t\vert}$ and ${s_n}$ approximates t closely enough from below. It follows that t is contained in the set of times ${\tau_n}$, so ${\{\Delta X\not=0\}\subseteq\bigcup_n[\tau_n]}$ as required. Also, as X is predictable, ${X_{\tau_n}}$ will be ${\mathcal{F}_{\tau_n-}}$-measurable.
4 implies 5: Let ${\tau_n}$ be stopping times satisfying condition 4. As these are predictable and ${\Delta X_{\tau_n}}$ is ${\mathcal{F}_{\tau_n-}}$-measurable, the processes
$\displaystyle Y^n\equiv\Delta X_{\tau_n}1_{[\tau_n]\setminus\bigcup_{m < n}[\tau_m]}$
are predictable. So, the set ${A_n\equiv\{Y^n_{\tau_n}\not=0\}}$ is ${\mathcal{F}_{\tau_n-}}$-measurable. We can therefore define new predictable stopping times ${\sigma_n=(\tau_n)_{A_n}}$. By construction, the graphs of ${\sigma_n}$ are disjoint and
$\displaystyle \bigcup_n[\sigma_n]=\bigcup_n[\tau_n]\cap\{\Delta X\not=0\}=\{\Delta X\not=0\}.$
5 implies 2: If 5 holds then ${\Delta X_{\tau_n}1_{[\tau_n]}}$ is predictable. So, the same is true of ${\Delta X=\sum_n\Delta X_{\tau_n}1_{[\tau_n]}}$.
2 implies 1: Simply write X as the sum ${X_-+\Delta X}$ of two predictable processes.
5 implies 3: We have already shown that 5 implies that X is predictable, so ${X_{\tau}}$ is ${\mathcal{F}_{\tau-}}$-measurable for any stopping time ${\tau}$. Also, for any totally inaccessible stopping time ${\tau}$, then ${{\mathbb P}(\tau_n=\tau)=0}$ by definition. So, ${\tau}$ is not in ${\bigcup_n[\tau_n]=\{\Delta X\not=0\}}$ (almost surely) and, therefore, ${\Delta X_\tau=0}$.
3 implies 4: Defining the sequence of stopping times ${\tau_n}$ by (1), we again have ${\{\Delta X\not=0\}\subseteq\bigcup_n[\tau_n]}$. By the decomposition of stopping times, there exists sets ${A_n\in\mathcal{F}_{\tau_n}}$ such that ${(\tau_n)_{A_n}}$ is accessible and ${(\tau_n)_{A_n^c}}$ is totally inaccessible. By condition 3, we have ${\Delta X_{(\tau_n)_{A_n^c}}=0}$ and, therefore, ${\{\Delta X\not=0\}}$ is almost surely contained in the graphs of ${(\tau_n)_{A_n}}$. By the definition of accessible stopping stopping times, this is contained in the union ${\bigcup_{m,n}[\tau_{nm}]}$ of predictable stopping times ${\tau_{nm}}$. Finally, ${X_{\tau_{nm}}}$ is ${\mathcal{F}_{\tau_{nm}-}}$-measurable by condition 3. ⬜
Applying this characterization to local martingales shows that, for such processes, predictability and continuity are equivalent. In the following, and throughout these notes, statements about the paths of processes are only intended in the almost sure sense. We do not care about what the sample paths look like on zero probability sets.
Lemma 5 A local martingale is predictable if and only if it is continuous.
Proof: As continuous processes are predictable by definition, only the converse needs to be shown. Suppose that M is a predictable local martingale and ${\tau}$ is a predictable stopping time. Then, for any ${A\in\mathcal{F}_{\tau-}}$, the stopping time ${\tau_A}$ is predictable so, by Theorem 1, the process ${Y\equiv1_A1_{[\tau,\infty)}\Delta M_\tau=1_{[\tau_A,\infty)}\Delta M_\tau}$ is a nonnegative local martingale. Taking ${A=\{\Delta M_\tau\ge0\}}$, so that Y is nonnegative and hence a supermartingale, we can take expectations to get
$\displaystyle {\mathbb E}[(\Delta M_\tau)_+]={\mathbb E}[Y_\tau]\le{\mathbb E}[Y_0]=0.$
So, ${\Delta M_\tau\le0}$ almost surely. Also applying this to ${-M}$ shows that ${\Delta M_\tau=0}$ for all predictable stopping times ${\tau}$. However, condition 4 of Lemma 4 shows that the jumps of M are contained in the graphs of a countable set of predictable stopping times, so M is almost surely continuous. ⬜
Note that for a filtration generated by a standard Brownian motion B, the martingale representation theorem implies that all local martingales are continuous, So, the third condition of Theorem 1 is trivially satisfied, giving the remarkable consequence that all stopping times are predictable. More generally, the property that all local martingales are continuous is equivalent to all stopping times being predictable.
Lemma 6 With respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge0},{\mathbb P})}$, the following are equivalent.
1. All stopping times are predictable.
3. All local martingales are continuous.
Proof: 1 implies 2: Let X be a cadlag adapted process. We can use the characterization of cadlag predictable processes given in the third condition of Lemma 4. It is immediate that ${\Delta X_\tau=0}$ (almost surely) for all totally inaccessible ${\tau}$ since, by assumption, all stopping times are predictable. Now suppose that ${\tau}$ is any stopping time. Then, for ${A\in\mathcal{F}_\tau}$, consider the stopping time ${\tau_A}$. By the condition, this is predictable, so the process ${Y\equiv1_A1_{[\tau,\infty)}=1_{[\tau_A,\infty)}}$ is predictable. Consequently ${Y_\tau=1_{A\cap\{\tau < \infty\}}}$ is ${\mathcal{F}_{\tau-}}$-measurable. So ${A\in\mathcal{F}_{\tau-}}$ and ${\mathcal{F}_{\tau-}=\mathcal{F}_\tau}$. So, ${X_\tau}$ is ${\mathcal{F}_{\tau-}}$-measurable for all stopping times ${\tau}$, and X is predictable.
2 implies 3: Any local martingale M is cadlag and adapted, hence is predictable. So, M is continuous by Lemma 5.
3 implies 1: If ${\tau}$ is a stopping time then any cadlag bounded martingale is continuous, so ${{\mathbb E}[\Delta M_\tau]=0}$. Theorem 1 says that ${\tau}$ is predictable. ⬜
As noted above, for the filtration generated by a Brownian motion, the martingale representation theorem has the consequence that all stopping times are predictable.
Corollary 7 Let ${\{\mathcal{F}_t\}_{t\ge0}}$ be the complete filtration generated by a d-dimensional Brownian motion B. Then, every ${\mathcal{F}_\cdot}$-stopping time is predictable.
Rather than using the martingale representation theorem, there is an alternative way to approach Corollary 7. Brownian motion is an example of a Feller process and, in fact, it can be shown that Corollary 7 extends to all continuous Feller processes.
#### Stopping Times of Feller Processes
For Feller processes it is possible to give a precise characterization of predictable and totally inaccessible stopping times. This follows from the following description of the times at which a local martingale can be discontinuous.
Lemma 8 Let X be a cadlag Feller process and ${\{\mathcal{F}_t\}_{t\ge0}}$ be its completed natural filtration, and suppose that M is a cadlag ${\mathcal{F}_\cdot}$-local martingale.
Then, with probability one, ${\Delta M_t=0}$ for all times t at which X is continuous.
Proof: By localization, it is enough to prove this result for all cadlag martingales of the form
$\displaystyle M_t={\mathbb E}[U\vert\mathcal{F}_t]$ (2)
where ${U}$ is an ${\mathcal{F}_\infty}$-measurable random variable. Let us use ${\mathcal{U}\subseteq L^1(\Omega,\mathcal{F}_\infty,{\mathbb P})}$ to denote the set of random variables U such that the cadlag martingale defined by (2) is almost-surely continuous wherever X is continuous. We need to show that ${\mathcal{U}}$ is equal to the whole of ${L^1(\mathcal{F}_\infty)}$. Clearly, ${\mathcal{U}}$ is closed under linear combinations. Furthermore, if ${U_n}$ is a sequence in ${\mathcal{U}}$ converging in ${L^1}$ to a limit U then the cadlag martingales ${M^n_t\equiv{\mathbb E}[U\vert\mathcal{F}_t]}$ converge in the ucp topology to a martingale M which will satisfy (2) and, by ucp convergence, is continuous at all times that ${M^n}$ are continuous. So, U is in ${\mathcal{U}}$. Therefore, ${\mathcal{U}}$ is a closed subspace of ${L^1(\mathcal{F}_\infty)}$.
Supposing that the Feller process X is defined by the transition function ${P_t}$ on the lccb space E, consider U of the form ${U=f(X_s)}$ for some ${f\in C_0(E)}$. Then, by the definition of Feller transition functions, ${(t,x)\mapsto P_tf(x)}$ defines a continuous real-valued function on ${{\mathbb R}_+\times E}$. Then,
$\displaystyle M_t\equiv P_{(s-t)_+}f(X_t)={\mathbb E}[U\vert\mathcal{F}_t]$
is a martingale which is continuous at all times that X is continuous. So ${U\in\mathcal{U}}$. Next, consider ${U=f_1(X_{s_1})f_2(X_{s_2})\cdots f_n(X_{s_n})}$ for a sequence of times ${0=s_0\le s_1\le\cdots s_n}$ and ${f_k\in C_0(E)}$, and let M be defined by (2). Then, for each ${k=1,\ldots,n}$, there exists a ${g_k\in C_0(E)}$ with
$\displaystyle M_t=f_1(X_{s_1})\cdots f_{k-1}(X_{s_{k-1}}){\mathbb E}\left[g_k(X_{s_k})\;\vert\mathcal{F}_t\right]$
for all ${s_{k-1}\le t\le s_k}$ (simply take ${g_n=f_n}$ and ${g_k=f_kP_{s_{k+1}-s_k}g_{k+1}}$ for ${k < n}$). As shown above, ${g_k(X_{s_k})\in\mathcal{U}}$, so M has a cadlag modification on ${(s_{k-1},s_k]}$ which is continuous wherever X is continuous. Therefore, ${U\in\mathcal{U}}$. Finally, by the monotone class theorem, the set of linear combinations of U of this form is dense in ${L^1(\mathcal{F}_\infty)}$, so ${\mathcal{U}=L^1(\mathcal{F}_\infty)}$ as required. ⬜
Applying Theorem 1 to this result, we obtain the promised characterization of predictable and totally inaccessible stopping times of a Feller process.
Theorem 9 Let X be a cadlag Feller process and ${\{\mathcal{F}_t\}_{t\ge0}}$ be its completed natural filtration. If ${\tau}$ is an ${\mathcal{F}_\cdot}$-stopping time then,
• ${\tau}$ is predictable if and only if ${X_{\tau-}=X_\tau}$ almost surely, whenever ${\tau < \infty}$.
• ${\tau}$ is totally inaccessible if and only if ${X_{\tau-}\not=X_{\tau}}$ almost surely, whenever ${\tau < \infty}$.
Proof: As cadlag Feller processes are quasi-left-continuous, if ${\tau}$ is predictable then ${X_{\tau-}=X_\tau}$ almost surely. Conversely, if ${X_{\tau-}=X_\tau}$ almost surely then, by Lemma 8, ${\Delta M_\tau=0}$ for any cadlag bounded martingale M. Then, ${{\mathbb E}[\Delta M_\tau]=0}$ and Theorem 1 says that ${\tau}$ is predictable.
For the second statement, suppose that ${X_{\tau-}\not=X_\tau}$ whenever ${\tau < \infty}$. Then, as shown above, ${X_{\sigma-}=X_\sigma}$ (almost surely) for any predictable stopping time ${\sigma}$ and consequently ${{\mathbb P}(\sigma=\tau < \infty)=0}$. So, ${\tau}$ is totally inaccessible. Conversely, suppose that ${\tau}$ is totally inaccessible and set ${A=\{X_{\tau-}=X_\tau\}\in\mathcal{F}_\tau}$. Then ${\tau_A}$ is a stopping time for which ${X_{\tau_A-}=X_{\tau_A}}$ and, hence, this is predictable so ${{\mathbb P}(A)={\mathbb P}(\tau_A=\tau < \infty)=0}$. Therefore, ${X_{\tau-}\not=X_\tau}$ whenever ${\tau < \infty}$ (almost surely). ⬜
One simple but surprising consequence of Theorem 9 is that, for Feller processes, the concepts of predictable and accessible stopping times actually coincide.
Corollary 10 Let ${\{\mathcal{F}_t\}_{t\ge0}}$ be the complete filtration generated by a Feller process. Then, an ${\mathcal{F}_\cdot}$-stopping time is predictable if and only if it is accessible.
Proof: Let X be a cadlag version of the Feller process. If ${\tau}$ is an accessible stopping time then the second statement of Theorem 9 says that ${X_{\tau-}=X_\tau}$ whenever ${\tau < \infty}$. So, by the first statement of Theorem 9, ${\tau}$ is predictable. ⬜
For continuous Feller processes, Theorem 9 simply states that, if X is a continuous process, then every stopping time is predictable. This gives the promised extension of Corollary 7 above to all continuous Feller processes.
Corollary 11 Let ${\{\mathcal{F}_t\}_{t\ge0}}$ be the complete filtration generated by a continuous Feller process. Then, every ${\mathcal{F}_\cdot}$-stopping time is predictable.
#### Non-Right-Continuous Filtrations
In Theorem 1 given above, the right-continuity of the filtration was required in the proof that the third condition implies the fourth. However, right-continuity is not required for the result to hold, and I will give an extension to the non-right-continuous case here. The idea is that we can apply Theorem 1 under the right-continuous filtration ${\mathcal{F}_{t+}\equiv\bigcap_{s > t}\mathcal{F}_s}$ and show that both the third and fourth statements of the theorem are unchanged under replacing ${\mathcal{F}_{t+}}$ by ${\mathcal{F}_t}$. First, we re-state the following simple lemma which was proven in the post on the Bichteler-Dellacherie theorem.
Lemma 12 Suppose that M is a cadlag and square-integrable ${\mathcal{F}_{t+}}$-martingale. Then, there exists a countable subset ${S\subset{\mathbb R}_+}$ such that ${{\mathbb P}(\Delta X_t\not=0)=0}$ for all ${t\in{\mathbb R}_+\setminus S}$. Furthermore, ${\int1_{{\mathbb R}_+\setminus S}\,dM}$ is an ${\mathcal{F}_t}$-martingale.
We can now give the proof that 3 implies 4 in Theorem 1 without the assumption that the filtration is right-continuous. So, suppose that 3 holds. Then, for any cadlag bounded ${\mathcal{F}_{\tau+}}$-martingale M, let S be the countable set of times at which ${{\mathbb P}(\Delta M_t\not=0) > 0}$. Letting ${N=\int1_{{\mathbb R}_+\setminus S}\,dM}$ then, as this is an ${\mathcal{F}_t}$-martingale, we have ${{\mathbb E}[\Delta N_\tau]=0}$. Next, for any fixed time t, if ${U}$ is an ${\mathcal{F}_t}$-measurable and bounded random variable, then ${\tilde N\equiv (U-{\mathbb E}[U\vert\mathcal{F}_{t-}])1_{[t,\infty)}}$ is easily seen to be a martingale. So, by assumption,
$\displaystyle {\mathbb E}\left[1_{\{\tau=t\}}(U-{\mathbb E}[U\vert\mathcal{F}_{t-}])\right]={\mathbb E}[\Delta\tilde N_\tau]=0.$
This implies that ${\{\tau=t\}}$ is in ${\mathcal{F}_{t-}}$. Hence, ${{\mathbb E}[1_{\{\tau=t\}}\Delta M_t]=0}$. Putting this together gives
$\displaystyle {\mathbb E}[\Delta M_\tau]={\mathbb E}[\Delta N_\tau]+\sum_{t\in S}{\mathbb E}[1_{\{\tau=t\}}\Delta M_t] =0.$
So, ${\tau}$ satisfies condition 3 with respect to the right-continuous filtration ${\mathcal{F}_{t+}}$. Applying Theorem 1 in this case shows that ${\tau}$ is predictable with respect to ${\mathcal{F}_{t+}}$. As shown previously, this implies that it is predictable with respect to the original filtration ${\mathcal{F}_t}$.
#### Notes on the Proof of Theorem 1
It is worth pausing here to consider the technical difficulties which had to be overcome in the proof of Theorem 1 above. The equivalence of statements 2 and 3 is easy to show without applying any advanced techniques, as is the fact that 4 implies 1. The proof that the third statement implies the fourth (fair stopping times are announceable) was a bit trickier to show but, still, constructing the sequence of stopping times announcing ${\tau}$ was achieved without too much difficulty, although extending the result to non-right-continuous filtrations gets a bit messy.
The proof that the first statement implies the second (predictable times are fair) can be the most technically demanding part of the proof of Theorem 1, so I will discuss some of the various approaches in this section. We managed to deal with this very efficiently in this post by making use of the identity
$\displaystyle \int1_{[\tau]}\,dM=\Delta M_\tau1_{[\tau,\infty)}$ (3)
and using the fact that stochastic integration preserves the local martingale property. Although (3) seems intuitively obvious by thinking about the integral in a pathwise sense, and is easy to prove for Riemann-Stieltjes integrals, it is much harder to show that the stochastic integral satisfies this identity. It does not follow easily from the defining properties of the stochastic integral — namely the bounded or dominated convergence theorem and the explicit expression for elementary integrands. Instead, we had to make use of the result that stochastic integrals coincide on any event for which the integrands coincide. This does seem like a simple enough statement, which we would expect to hold. However, the proof of this result required showing that semimartingales remain as semimartingales when the filtration is enlarged by adding a set to ${\mathcal{F}_0}$. This, in turn, required the characterization of semimartingales in terms of boundedness in probability of elementary integrals, which was rather demanding to prove, and was restated as part of the Bichteler-Dellacherie theorem. So, the seemingly simple statement that 1 implies 2 in Theorem 1 actually required some rather advanced stochastic calculus, and we would have been hard-pressed to give a short proof in these notes before the stochastic integral had been developed.
It is interesting to compare with the proof given in Rogers & Williams (Diffusions, Markov Processes, and Martingales, Volume 2, §VI.16.4-13), where the implication is denoted by P ⇒ F (predictable implies fair). They open with the following paragraph.
Proof that P ⇒ F. If you wish to understand the subject properly, you need to understand the proof of the section theorems, and this proof that P ⇒ F, much of which is based on §IV.76 of Dellacherie and Meyer, is a good introduction to the methods required. We did try for some time to find a quick proof that P ⇒ F, but though many proofs’ would immediately spring to the mind of anyone familiar with stochastic-integral theory, they all presuppose that P ⇒ A (although it sometimes takes a little thought to spot exactly where!).”
It seems likely that the proofs mentioned that would immediately spring to mind are those based on identity (3). If it is assumed that the stopping time ${\tau}$ is announceable, then this identity is easy to prove (hence, the presupposition that P ⇒ A). Fortunately, the approach that has been taken in these notes means that we were able to prove (3) without any such presupposition, so we could give a quick stochastic integration based proof that ${[\tau]\in\mathcal{P}}$ implies that ${\tau}$ is fair.
An alternative proof of the result is to apply the section theorems, as suggested by Rogers & Williams in the quote above. The section theorems are very powerful results on which much of the historical development of stochastic calculus depended, although their proofs are rather demanding and are based on predictive set theory and analytic sets. The predictable section theorem in particular implies that if ${[\tau]}$ is a predictable set then there exist predictable stopping times ${\tau_n}$ with ${{\mathbb P}(\tau_n\in [\tau])}$ tending to ${{\mathbb P}(\tau < \infty)}$. This implies that ${\tau}$ is predictable (i.e., that it is announceable), giving a proof that statement 1 implies 4 in Theorem 1. The proof given by Rogers & Williams does not use the section theorems themselves, but does involve ideas from their proofs, and is essentially based on the Choquet capacity theorem. However, in these notes I have taken a different approach to stochastic calculus, attempting to develop the stochastic integral in a more direct and slightly more intuitive way which avoids any use of the section theorems.
There do exist other methods of proving that stopping times satisfying ${[\tau]\in\mathcal{P}}$ are fair which avoid the use both of identity (3) and of the section theorems. For example, Metivier & Pellaumail (Stochastic Integration, 1980) give a relatively direct proof of the Doob-Meyer decomposition theorem without invoking any major machinery which, in particular, implies that every integrable increasing predictable process A and bounded cadlag martingale M with ${M_0=0}$ satisfies the identity
$\displaystyle {\mathbb E}\left[\int_0^t M_{s-}\,dA_s\right]={\mathbb E}[M_t A_t].$
Applying this to the predictable process ${A=1_{[\tau,\infty)}}$ implies that ${\tau}$ is fair. More details on this will be mentioned in a later post on compensators.
## 12 thoughts on “Predictable Stopping Times”
1. TheBridge says:
Hi,
So first,
In your proof of theorem 1, in the “2 implies 3” part, you mention the fact that if a local martinglae $M$ is $L^1$-dominated then it is a true martingale. As I couldn’t find in your notes this precise statement, would it be right to say the following ?
As $L^1$-domination entails that $M$ is locally integrable (by hypotheses $\exists X \in L^1$ such that $sup_{s \ge 0} M_s ), which in turn implies that $M$ is of class DL (see lemma 9 in your post "Localization"), and so the local martingale $M$ is a true martingale (by theorem 1 in your post on "Local Martingales"). Moreover here $N$ is uniformly integrable (or U.I.) as a $L^1$-dominated family of random variable is U.I.. So you can say that $N_\infty$ exists almost surely and use it in the equallity following the statement about the martingale property of $N$.
Maybe a more elementary proof exists but I missed it, otherwise if the result exists in your blog you consider hyperlinking it with the claim.
Best regards
1. I think you’ve more or less got the argument, except, for the locally integrable part. Local integrability is too week to imply that it is a true martingale. In fact, all local martingales are locally integrable anyway. Lemma 9 in the post mentioned only implies that it is locally of class (DL), not actually of class DL itself.
Instead, if M is L1-dominated (by X), then the set of random variables {XTT is a finite stopping time} must also be dominated by X, so uniformly integrable. Hence, M is of class (D) (so also of class DL). Now use Theorem 1 from the post on local martingales. And, class (D) implies that it is UI, so N exists by martingale convergence, and the martingale property E[Nt|Fs] = Ns also holds for t = ∞.
I tend not to spell out all the small steps once it gets down to just what is relatively standard manipulations that occur all over the place. Maybe in some places these steps are a bit big, depending on how familiar or not you are with this stuff. If I can make it easier to follow without making arguments which I think should be short and direct seem long and complicated, then I will (when I have time to go back and clean up posts).
1. TheBridge says:
Hi, first thank you for responding so fast,
Regarding spelling out small steps, well I think it depends on the size of one’s legs, and some kind of human factor is involved in the process.
I know there is always a trade-off between synthetized fancy style exposition of a mathematical proof and elementary exhaustive but tedious rigourous demonstration, as “élégance” is an determinant factor in the process.
At this game, your are particularly gifted in my opinion but as it is generally hard for high skilled people to get what is hard to get for not as smart guys, I only point out from time to time those steps which are sometimes too high for me.
Best regards
2. TheBridge says:
Hi,
Second,
I think there is a typo in “3 implies 4”. You wrote :
$\mathbb{E}[f(\tau_n)]=\mathbb{E}[M_{\tau_n}]\le \mathbb{E}[f(\tau)+1/n]$
Where I think it is :
$\mathbb{E}[f(\tau)]=\mathbb{E}[M_{\tau_n}] \le \mathbb{E}[f(\tau_n)+1/n]$
Shortly after that, using $M_t \ge f(t)$ you show that :
$\Delta M_\tau = f(\tau) - M_{\tau-} \le f(\tau) - f(\tau-)=0$ (by continuity of $f$)
Besides, I don’t see why then condition 3 is required to ensure that $\Delta M_\tau$ is almost surely 0.
Best regards
1. Yes, regarding the typo, you’re correct. I fixed this, thanks.
Condition 3 is certainly required to ensure that ΔM is almost surely zero, otherwise it would prove that every stopping time is predictable!
What happens if you try the same argument for a non-predictable stopping time is that ΔM can be strictly negative. Then, with positive probability, you have $M_\tau < M_{\tau-} - 1/n$ for some n, and then $\tau_m=\tau$ for m ≥ n, so the sequence $\tau_n$ does not announce $\tau$ strictly from below.
1. TheBridge says:
Hi
I got it thanks
Best regards
3. TheBridge says:
Hi,
For the proof in lemma 3, unless I missed something I think that the argument is a little too fast as $\sigma_n$ is non deacreasing but can be forced through (not so) appropriate choice of $\tau_{m,n}$ to make $\sigma_n$ to be “constant” (meaning that is the same random variable over and over),so a proper choice of the index $n$ (depending on $m$ and $\omega$) has to be done, or alternatively some condition over the construction of $\tau_{m,n}$ has to be imposed for every $m$, for example $\tau_{m,1}>\tau_{m-1}$ almost surely (I think this is possible with no arm).
In the proof of lemma 4, for the sake of exhaustivity you might also consider hyperlinking in the end of the “1 implies 4” argument, the sentence :
“Also, as X is predictable, $\Delta X_{\tau_n}$ will be $\mathcal{F}_{\tau_{n-}}$-measurable,” to lemma 1 of the post “Sigma Algebras at a Stopping Time” (appears also in the “5 implies 3” argument).
Hope this is not too much, but as I come almost at the end of my lecture of your notes the rythm of those comments should slow down soon dramatically.
Best regards
1. Regarding Lemma 3. Yes, there was a mistake, which I’ve fixed. The correct choice of $\sigma_n$ is actually much easier than you might expect when you first think about it (I think this is what I had in my head when I wrote this post, but it came out wrong).
And, no, its not too much at all! I like having the feedback on these notes. Any comments which help me improve them is much appreciated. Besides, I never normally know if anyone has actually read through the details of the more in-depth proofs. You’ve almost read through all of my notes? That’s pretty good going!
1. TheBridge says:
Hi,
Yes I must admit I did read almost all of them and had a great pleasure doing so, and as you don’t feel annoyed by my comments I will keep up making some when I feel I have to.
Best regards
4. Rudi says:
Hi,
I’ve got a question. I take a continuous process and it’s natural filtration. Is it possible to show, that all optional processes are predictable wrt to my filtration?
Best regards
5. Winfrida says:
hi, i am asking what is application of indicator function on stopping time and why also indicator function is used on writing jump process
1. I am not sure precisely what your question is, but $1_{[\tau,\infty)}$ is the process equal to zero before stopping time $\tau$ and equal to 1 at and after the stopping time, so is a very basic jump process. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 356, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9829162955284119, "perplexity": 184.96982626336396}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357935.29/warc/CC-MAIN-20210226175238-20210226205238-00303.warc.gz"} |
https://www.physicsforums.com/threads/photovoltaics-current.419042/ | Photovoltaics current
1. Jul 29, 2010
EvaristeG
Hello everyone !
I have a very basic question about phovoltaics: does a photovoltaic solar cell need to be current to work ? I mean it sounds stupid since we use solar cell to produce current and not the opposite... The fact is I do not really understand how the cell produces current, I was thinking that something like a external voltage is missing to create the current flow..
What I understood is that when you take the n-type silicon and the p-type silicon and put them together to form the pn junction a depletion zone is formed where you don't have any carrier... Due to the inbalance of the charges between the n side and the p side an electric field is formed (around 0.6-0.7 volts for Silicon)...and then ? Electrons from the n side would like to couple with the holes on the p side but they can't because of the depletion zone and therefore if you give them an external "track" they will flow through that circuit and is how current is produced ?
I don't know, I have the feeling I'm missing something..could anyone help me and explain me how it works ? I mean how you really get the current (the user could use) in the end ? Also is you need to supply the cell with power or not, which I know sounds weired but still, I'd like to know... I mean, you need to supply a photodiode don't you ? So what's the difference ?
Thank you very very much in advance for any help !
Evariste
2. Jul 29, 2010 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8534384369850159, "perplexity": 375.670483527341}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823009.19/warc/CC-MAIN-20181209185547-20181209211547-00134.warc.gz"} |
http://math.stackexchange.com/questions/216387/linear-algebra-problem-involving-the-characteristic-of-a-field/216399 | # Linear algebra problem involving the characteristic of a field
I'm having trouble with the following problem:
Let $\tau_A: F^2\times F^2 \rightarrow F$ be a symmetric bilinear form given by $\tau_A (v,w)=v^tAw$, $\forall v,w\in F^2$ and $A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$, where $F$ is a field.
Suppose the characteristic of $F$ is not equal to $2$. Prove that there exists a basis $\mathcal{B}=\{v_1,v_2\}$ of $F^2$ such that $\tau_A (v_1,v_1)=\tau_A (v_2,v_2)=0$
So far I've tried seeing if I could milk anything out of the non-degeneracy of $\tau_A$ (so that $(F^2,\ \tau_A)$ is an inner product space), but got stuck. I also split this problem into two cases: $Char(F)=0$ and $Char(F)=p$, but wasn't able to get anywhere. I have no experience dealing with the characteristic of a field, so I think conceptually I'm having a hard time understanding why it would matter in a problem like this.
Any tips or solutions (preferably as elementary as possible) would be appreciated! Thanks in advance!
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We can construct the vectors $v_1,v_2$ explicitly: $$v_1=(1,1),\qquad v_2=(1,-1).$$
It is clear that $\tau_A(v_1,v_1)= \tau_A(v_2, v_2)=0$. But when are they a basis?
They are a basis exactly when $1$ and $-1$ are different, i.e. everytime $\mathrm{char}(F)\neq 2$.
On the other hand if $\mathrm{char}(F)=2$ then $A=I$ the identity matrix. So $\tau_A((x,y),(x,y))=0$ implies $x^2+y^2=(x+y)^2=0$, so $x=-y=y$.
Then those vectors form a vector space of dimension $1$, so you cannot find a basis for your vector space of dimension $2$.
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Just proceed straightforwardly: if $v=(x,y)$ then $\tau_A(v)=0$ means $x^2=y^2$, so just take as basis $\{(1,1),(1,-1)\}$. Note that this is not a basis if the characteristic is 2.
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https://scipost.org/SciPostPhys.6.5.062 | Integrable Matrix Product States from boundary integrability
Balázs Pozsgay, Lorenzo Piroli, Eric Vernier
SciPost Phys. 6, 062 (2019) · published 27 May 2019
Abstract
We consider integrable Matrix Product States (MPS) in integrable spin chains and show that they correspond to "operator valued" solutions of the so-called twisted Boundary Yang-Baxter (or reflection) equation. We argue that the integrability condition is equivalent to a new linear intertwiner relation, which we call the "square root relation", because it involves half of the steps of the reflection equation. It is then shown that the square root relation leads to the full Boundary Yang-Baxter equations. We provide explicit solutions in a number of cases characterized by special symmetries. These correspond to the "symmetric pairs" $(SU(N),SO(N))$ and $(SO(N),SO(D)\otimes SO(N-D))$, where in each pair the first and second elements are the symmetry groups of the spin chain and the integrable state, respectively. These solutions can be considered as explicit representations of the corresponding twisted Yangians, that are new in a number of cases. Examples include certain concrete MPS relevant for the computation of one-point functions in defect AdS/CFT.
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Funders for the research work leading to this publication | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9348641633987427, "perplexity": 799.9353991783272}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875142603.80/warc/CC-MAIN-20200217145609-20200217175609-00168.warc.gz"} |
http://annals.math.princeton.edu/articles/11023 | Periodic complexes and group actions
Abstract
In this paper we show that the cohomology of a connected CW-complex is periodic if and only if it is the base space of a spherical fibration with total space that is homotopically finite dimensional. As applications we characterize those discrete groups that act freely and properly on $\mathbb{R}^n \times \mathbb{S}^m$; we construct nonstandard free actions of rank-two simple groups on finite complexes $Y\simeq \mathbb{S}^n \times \mathbb{S}^m$; and we prove that a finite $p$-group $P$ acts freely on such a complex if and only if it does not contain a subgroup isomorphic to $(\mathbb{Z}/p)^3$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8613739609718323, "perplexity": 101.58199573441856}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247481766.50/warc/CC-MAIN-20190217071448-20190217093448-00012.warc.gz"} |
http://mathoverflow.net/questions/18286/generalization-of-permanent-definition-based-on-number-of-permutation-cycles/18306 | # Generalization of permanent definition based on number of permutation cycles
Let $A$ be an $n$ by $n$ matrix and $x$ a free parameter. Define $$p(A,x)=\sum_{\pi \in S_n} x^{n(\pi)}A_{1\pi(1)}\ldots A_{n\pi(n)},$$ where $\pi$ ranges over the permutation group $S_n$ and $n(\pi)$ is the number of cycles in the cycle decomposition of $\pi$. Clearly, $p(A,1)=perm(A)$, the permanent. In general, $p(A,x)$ has properties in common with the permanent such as $p(PAQ,x)=p(A,x)$ for permutation matrices $P,Q$.
Is this a well-known structure in combinatorics and where might I find more information?
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$$Imm_\lambda(A) = \sum_{\pi \in S_m} \chi_\lambda(\pi) A_{i\pi(1)}...A_{n\pi(n)}$$
generalized the permanent and determinant and your $p(A,x)$, since we can write $x^{n(\pi)}$ as a linear combination of characters.
Continuing Doug's answer, when we write $x^{n(\pi)}$ as a linear combination of irreducible characters $\chi^\lambda$, then the coefficient of $\chi^\lambda$ is $\prod_u (x+c(u))/h(u)$, where $u$ runs over all squares in the diagram of $\lambda$, $c(u)$ is the content of $u$, and $h(u)$ is the hook length of $u$. – Richard Stanley Mar 15 '10 at 18:29
This exact version of a "generalized permanent" is called $\beta$-extension in this Foata-Zeilberger paper (see also my paper for the algebraic context and further non-commutative generalizations, Cartier-Foata style). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9875582456588745, "perplexity": 162.44148073611672}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095557.73/warc/CC-MAIN-20150627031815-00110-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/258001/problem-related-with-analytic-function | # Problem related with analytic function
I was thinking about the following problem:
Let $$S=\{0\}\cup \{ \frac{1}{4n+1}:n=1,2,3,4,\dots\}$$ Then what is the total number of analytic function which vanish only on $S$?
I was trying to use the fact that zeros of analytic functions are isolated. But I could not progress further. Am I going in the right direction? Please help. Thanks in advance for your time.
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Yes, you're on the right lines: is $0$ isolated in the set $\{ 0 \} \cup \left\{ \dfrac{1}{4n+1}\, :\, n \in \mathbb{N} \right\}$?
Given $\varepsilon > 0$, can you find an $n$ such that $\dfrac{1}{4n+1} < \varepsilon$?
No,sir.I do not think that here in the set $0$ is an isolated point. Here, we can choose $n$ to be arbitrarily large so that any nbd. of ${0}$ will contain points from the set ${1/(4n+1): n \in \mathbb N}$. Am i right? – user52976 Dec 13 '12 at 16:18
No... you were right in saying that there are no analytic functions vanishing on $S$. But this is because $S$ contains at least one non-isolated zero. – Clive Newstead Dec 13 '12 at 16:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9376862049102783, "perplexity": 86.30817647783901}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929561.80/warc/CC-MAIN-20150521113209-00199-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Borel_subgroup | # Borel subgroup
In the theory of algebraic groups, a Borel subgroup of an algebraic group G is a maximal Zariski closed and connected solvable algebraic subgroup. For example, in the group GLn (n x n invertible matrices), the subgroup of invertible upper triangular matrices is a Borel subgroup.
For groups realized over algebraically closed fields, there is a single conjugacy class of Borel subgroups.
Borel subgroups are one of the two key ingredients in understanding the structure of simple (more generally, reductive) algebraic groups, in Jacques Tits' theory of groups with a (B,N) pair. Here the group B is a Borel subgroup and N is the normalizer of a maximal torus contained in B.
The notion was introduced by Armand Borel, who played a leading role in the development of the theory of algebraic groups.
## Parabolic subgroups
Subgroups between a Borel subgroup B and the ambient group G are called parabolic subgroups. Parabolic subgroups P are also characterized, among algebraic subgroups, by the condition that G/P is a complete variety. Working over algebraically closed fields, the Borel subgroups turn out to be the minimal parabolic subgroups in this sense. Thus B is a Borel subgroup when the homogeneous space G/B is a complete variety which is "as large as possible".
For a simple algebraic group G, the set of conjugacy classes of parabolic subgroups is in bijection with the set of all subsets of nodes of the corresponding Dynkin diagram; the Borel subgroup corresponds to the empty set and G itself corresponding to the set of all nodes. (In general each node of the Dynkin diagram determines a simple negative root and thus a one-dimensional 'root group' of G---a subset of the nodes thus yields a parabolic subgroup, generated by B and the corresponding negative root groups. Moreover any parabolic subgroup is conjugate to such a parabolic subgroup.)
## Lie algebra
For the special case of a Lie algebra $\mathfrak{g}$ with a Cartan subalgebra $\mathfrak{h}$, given an ordering of $\mathfrak{h}$, the Borel subalgebra is the direct sum of $\mathfrak{h}$ and the weight spaces of $\mathfrak{g}$ with positive weight. A Lie subalgebra of $\mathfrak{g}$ containing a Borel subalgebra is called a parabolic Lie algebra. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9149653911590576, "perplexity": 229.18016585528287}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1433195036702.22/warc/CC-MAIN-20150601214356-00059-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://paradigms.oregonstate.edu/problem/786/ | ## Basic Calculus: Practice Exercises
• assignment Differentials of Two Variables
assignment Homework
##### Differentials of Two Variables
Static Fields 2023 (8 years) Find the total differential of the following functions:
1. $y=3u^2 + 4\cos 3v$
2. $y=3uv$
3. $y=3u^2\cos wv$
4. $y=u\cos(3v^2-2)$
• assignment Pressure of thermal radiation
assignment Homework
Thermal radiation Pressure Thermal and Statistical Physics 2020
(modified from K&K 4.6) We discussed in class that \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \end{align} Use this relationship to show that
1. \begin{align} p &= -\sum_j \langle n_j\rangle\hbar \left(\frac{d\omega_j}{dV}\right), \end{align} where $\langle n_j\rangle$ is the number of photons in the mode $j$;
2. Solve for the relationship between pressure and internal energy.
• assignment Mass of a Slab
assignment Homework
##### Mass of a Slab
Static Fields 2023 (6 years)
Determine the total mass of each of the slabs below.
1. A square slab of side length $L$ with thickness $h$, resting on a table top at $z=0$, whose mass density is given by $$\rho=A\pi\sin(\pi z/h).$$
2. A square slab of side length $L$ with thickness $h$, resting on a table top at $z=0$, whose mass density is given by $$\rho = 2A \Big( \Theta(z)-\Theta(z-h) \Big)$$
3. An infinitesimally thin square sheet of side length $L$, resting on a table top at $z=0$, whose surface density is given by $\sigma=2Ah$.
4. An infinitesimally thin square sheet of side length $L$, resting on a table top at $z=0$, whose mass density is given by $\rho=2Ah\,\delta(z)$.
5. What are the dimensions of $A$?
6. Write several sentences comparing your answers to the different cases above.
• assignment Coffees and Bagels and Net Worth
assignment Homework
##### Coffees and Bagels and Net Worth
Energy and Entropy 2021 (2 years)
In economics, the term utility is roughly related to overall happiness. Many things affect your happiness, including the amount of money you have and the amount of coffee you drink. We cannot directly measure your happiness, but we can measure how much money you are willing to give up in order to obtain coffee or bagels. If we assume you choose wisely, we can thus determine that your happiness increases when you decrease your amount of money by that amount in exchange for increasing your coffee consumption. Thus money is a (poor) measure of happiness or utility.
Money is also a nice quantity because it is conserved---just like energy! You may gain or lose money, but you always do so by a transaction. (There are some exceptions to the conservation of money, but they involve either the Fed, counterfeiters, or destruction of cash money, and we will ignore those issues.)
In this problem, we will assume that you have bought all the coffee and bagels you want (and no more), so that your happiness has been maximized. Thus you are in equilibrium with the coffee shop. We will assume further that you remain in equilibrium with the coffee shop at all times, and that you can sell coffee and bagels back to the coffee shop at cost.*
Thus your savings $S$ can be considered to be a function of your bagels $B$ and coffee $C$. In this problem we will also discuss the prices $P_B$ and $P_C$, which you may not assume are independent of $B$ and $C$. It may help to imagine that you could possibly buy out the local supply of coffee, and have to import it at higher costs.
1. The prices of bagels and coffee $P_B$ and $P_C$ have derivative relationships between your savings and the quantity of coffee and bagels that you have. What are the units of these prices? What is the mathematical definition of $P_C$ and $P_B$?
2. Write down the total differential of your savings, in terms of $B$, $C$, $P_B$ and $P_C$.
3. Solve for the total differential of your net worth. Your net worth $W$ is the sum of your total savings plus the value of the coffee and bagels that you own. From the total differential, relate your amount of coffee and bagels to partial derivatives of your net worth.
• assignment Free Expansion
assignment Homework
##### Free Expansion
Energy and Entropy 2021 (2 years)
The internal energy is of any ideal gas can be written as \begin{align} U &= U(T,N) \end{align} meaning that the internal energy depends only on the number of particles and the temperature, but not the volume.*
The ideal gas law \begin{align} pV &= Nk_BT \end{align} defines the relationship between $p$, $V$ and $T$. You may take the number of molecules $N$ to be constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Free expansion” means that no work is done, but also that the process is also neither quasistatic nor reversible.
1. What is the change in entropy of the gas? How do you know this?
2. What is the change in temperature of the gas?
• assignment Bottle in a Bottle 2
assignment Homework
##### Bottle in a Bottle 2
heat entropy ideal gas Energy and Entropy 2021 (2 years)
Consider the bottle in a bottle problem in a previous problem set, summarized here.
A small bottle of helium is placed inside a large bottle, which otherwise contains vacuum. The inner bottle contains a slow leak, so that the helium leaks into the outer bottle. The inner bottle contains one tenth the volume of the outer bottle, which is insulated.
The volume of the small bottle is 0.001 m23 and the volume of the big bottle is 0.01 m3. The initial state of the gas in the small bottle was $p=106$ Pa and its temperature $T=300$ K. Approximate the helium gas as an ideal gas of equations of state $pV=Nk_BT$ and $U=\frac32 Nk_BT$.
1. How many molecules of gas does the large bottle contain? What is the final temperature of the gas?
2. Compute the integral $\int \frac{{\mathit{\unicode{273}}} Q}{T}$ and the change of entropy $\Delta S$ between the initial state (gas in the small bottle) and the final state (gas leaked in the big bottle).
• assignment Icecream Mass
assignment Homework
##### Icecream Mass
Static Fields 2023 (6 years)
Use integration to find the total mass of the icecream in a packed cone (both the cone and the hemisphere of icecream on top).
• face Energy and Entropy review
face Lecture
5 min.
##### Energy and Entropy review
Thermal and Statistical Physics 2020 (3 years)
This very quick lecture reviews the content taught in Energy and Entropy, and is the first content in Thermal and Statistical Physics.
assignment Homework
ideal gas internal energy engine Energy and Entropy 2020
A diesel engine requires no spark plug. Rather, the air in the cylinder is compressed so highly that the fuel ignites spontaneously when sprayed into the cylinder.
In this problem, you may treat air as an ideal gas, which satisfies the equation $pV = Nk_BT$. You may also use the property of an ideal gas that the internal energy depends only on the temperature $T$, i.e. the internal energy does not change for an isothermal process. For air at the relevant range of temperatures the heat capacity at fixed volume is given by $C_V=\frac52Nk_B$, which means the internal energy is given by $U=\frac52Nk_BT$.
Note: in this problem you are expected to use only the equations given and fundamental physics laws. Looking up the formula in a textbook is not considered a solution at this level.
1. If the air is initially at room temperature (taken as $20^{o}C$) and is then compressed adiabatically to $\frac1{15}$ of the original volume, what final temperature is attained (before fuel injection)?
2. By what factor does the pressure increase?
• assignment Flux through a Plane
assignment Homework
##### Flux through a Plane
Static Fields 2023 (4 years) Find the upward pointing flux of the vector field $\boldsymbol{\vec{H}}=2z\,\boldsymbol{\hat{x}} +\frac{1}{x^2+1}\boldsymbol{\hat{y}}+(3+2z)\boldsymbol{\hat{z}}$ through the rectangle $R$ with one edge along the $y$ axis and the other in the $xz$-plane along the line $z=x$, with $0\le y\le2$ and $0\le x\le3$.
• Static Fields 2023 (4 years) Determine the following derivatives and evaluate the following integrals, all by hand. You should also learn how to check these answers on Wolfram Alpha.
1. $\frac{d}{du}\left(u^2\sin u\right)$
2. $\frac{d}{dz}\left(\ln(z^2+1)\right)$
3. $\displaystyle\int v\cos(v^2)\,dv$
4. $\displaystyle\int v\cos v\,dv$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9374585151672363, "perplexity": 680.3820626680855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00058.warc.gz"} |
https://brilliant.org/problems/drunken-secertary/ | # Drunken Secretary
A secretary must send 4 emails to person A, B, C, and D.
However on a given day, suppose this incompetent secretary sends those 4 emails in a completely randomly order, let $\mathcal{P}$ denote the probability of each person received the wrong email.
What is the value of $24 \mathcal{P}$?
Details and Assumptions
• Each person A, B, C, D received exactly one email.
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http://www.html5freecode.com/MathML%E2%80%93_Underscript__Overscript_and_UnderOverscript.htm | Feel The Power
# Html Codes: MathML- Underscript, Overscript and UnderOverscript
In mathematical equations, sometimes it is needed to put a bar, hat, dot and etc. under or over the main characters. This feature called underscript and overscript, and have been defined in HTML5 MathML. You can also use both of them (UnderOverscript) at the same time.
The following example shows how to put some symbols above or below the main text. It is shown how to specify a script for a group of texts not just a single one and also how to use the combination of under and over scripts together.You can also turn on the accent option to make the scripts larger.
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https://emba.gnu.org/emacs/emacs/-/commit/1fb87c7784670dafa3dfb6b2326a5cb8b3f94137 | Commit 1fb87c77 by Karl Heuer
### Fix FSF address in comment.
parent 29bd7a8f
... ... @@ -16,8 +16,9 @@ rem MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the rem GNU General Public License for more details. rem You should have received a copy of the GNU General Public License rem along with GNU Emacs; see the file COPYING. If not, write to rem the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. rem along with GNU Emacs; see the file COPYING. If not, write to the rem Free Software Foundation, Inc., 59 Temple Place - Suite 330, rem Boston, MA 02111-1307, USA. rem ---------------------------------------------------------------------- rem YOU'LL NEED THE FOLLOWING UTILITIES TO MAKE EMACS: rem ... ...
... ... @@ -18,8 +18,9 @@ dnl MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the dnl GNU General Public License for more details. dnl dnl You should have received a copy of the GNU General Public License dnl along with GNU Emacs; see the file COPYING. If not, write to dnl the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. dnl along with GNU Emacs; see the file COPYING. If not, write to the dnl Free Software Foundation, Inc., 59 Temple Place - Suite 330, dnl Boston, MA 02111-1307, USA. AC_PREREQ(2.8)dnl AC_INIT(src/lisp.h) ... ...
... ... @@ -15,8 +15,9 @@ # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with GNU Emacs; see the file COPYING. If not, write to # the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ # along with GNU Emacs; see the file COPYING. If not, write to the # Free Software Foundation, Inc., 59 Temple Place - Suite 330, # Boston, MA 02111-1307, USA. # # ... ...
... ... @@ -13,9 +13,10 @@ FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. You should have received a copy of the GNU General Public License along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ ... ...
... ... @@ -12,7 +12,7 @@ # Author: Paul Eggert # \$Id: rcs2log,v 1.27 1996/01/15 01:17:03 eggert Exp \$ # \$Id: rcs2log,v 1.27 1996/01/15 01:17:56 eggert Exp kwzh \$ # Copyright 1992, 1993, 1994, 1995, 1996 Free Software Foundation, Inc. ... ... @@ -27,8 +27,9 @@ # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with this program; see the file COPYING. If not, write to # the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. # along with GNU Emacs; see the file COPYING. If not, write to the # Free Software Foundation, Inc., 59 Temple Place - Suite 330, # Boston, MA 02111-1307, USA. tab=' ' nl=' ... ...
... ... @@ -19,8 +19,9 @@ ;; GNU General Public License for more details. ;; You should have received a copy of the GNU General Public License ;; along with GNU Emacs; see the file COPYING. If not, write to ;; the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. ;; along with GNU Emacs; see the file COPYING. If not, write to the ;; Free Software Foundation, Inc., 59 Temple Place - Suite 330, ;; Boston, MA 02111-1307, USA. ;;; Commentary: ... ...
... ... @@ -24,8 +24,9 @@ ;; GNU General Public License for more details. ;; You should have received a copy of the GNU General Public License ;; along with GNU Emacs; see the file COPYING. If not, write to ;; the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. ;; along with GNU Emacs; see the file COPYING. If not, write to the ;; Free Software Foundation, Inc., 59 Temple Place - Suite 330, ;; Boston, MA 02111-1307, USA. ;;; Commentary: ... ...
... ... @@ -15,8 +15,9 @@ # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with GNU Emacs; see the file COPYING. If not, write to # the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ # along with GNU Emacs; see the file COPYING. If not, write to the # Free Software Foundation, Inc., 59 Temple Place - Suite 330, # Boston, MA 02111-1307, USA. # !include ..\nt\makefile.def ... ...
... ... @@ -18,8 +18,9 @@ ;; GNU General Public License for more details. ;; ;; You should have received a copy of the GNU General Public License ;; along with GNU Emacs; see the file COPYING. If not, write to ;; the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. ;; along with GNU Emacs; see the file COPYING. If not, write to the ;; Free Software Foundation, Inc., 59 Temple Place - Suite 330, ;; Boston, MA 02111-1307, USA. ;;; Quick Start: ;; ... ...
... ... @@ -14,8 +14,9 @@ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License Alongalong with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ /* Created by [email protected] */ ... ...
... ... @@ -21,8 +21,9 @@ # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with GNU Emacs; see the file COPYING. If not, write to # the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. # along with GNU Emacs; see the file COPYING. If not, write to the # Free Software Foundation, Inc., 59 Temple Place - Suite 330, # Boston, MA 02111-1307, USA. progname="\$0" ... ...
... ... @@ -14,8 +14,9 @@ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ /* No code in Emacs #includes config.h twice, but some of the code ... ...
... ... @@ -16,8 +16,9 @@ # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with GNU Emacs; see the file COPYING. If not, write to # the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ # along with GNU Emacs; see the file COPYING. If not, write to the # Free Software Foundation, Inc., 59 Temple Place - Suite 330, # Boston, MA 02111-1307, USA. # # ... ...
... ... @@ -14,8 +14,9 @@ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ /* Contributed by Mark Diekhans . */ ... ...
... ... @@ -14,8 +14,9 @@ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, 675 Mass Ave, Cambridge, MA 02139, USA. */ along with GNU Emacs; see the file COPYING. If not, write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */ /* Added by Kevin Gallo */ ... ...
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Finish editing this message first! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.99460369348526, "perplexity": 2591.0882462389905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178373241.51/warc/CC-MAIN-20210305183324-20210305213324-00256.warc.gz"} |
https://www.farlabs.edu.au/structure/engage-photoelectric-effect/ | Home Structure Engage: Photoelectric Effect
Engage: Photoelectric Effect
# Light: Wave or Particle?
A plane wave incident on two narrow slits.
In 1803, Thomas Young discovered that when light was passed between two very narrow slits, the light which emerged on the other side of the slits formed a series of light and dark bands (an interference pattern, as shown in the image on the right). This interference pattern could be explained by assuming that light consisted of waves that interfered with each other. This theory supporting the wave nature of light was further helped by James Clerk Maxwell around 1861 when he combined his four famous equations describing electricity and magnetism.
Toward the end of the 19th century, however, some problems emerged. Firstly, a wave theory of light could not correctly explain black-body radiation. Nor could it correctly explain phenomena, discovered in 1887 by Heinrich Hertz, in which light striking a metal surface was able to eject electrons, which we call the photoelectric effect. Specifically, the number and energy of these ejected electrons did not seem to fit with the wave theory of light.
It was Albert Einstein, in 1905, who successfully explained the photoelectric effect, not in terms of waves, but in terms of particles. He took an idea first proposed by Max Planck that light came in discrete amounts (called “quanta”) to explain black-body radiation and applied it to the data obtained by researchers investigating the photoelectric effect. The result was a simple explanation of both the energy and intensity of electrons ejected by light incident on some metals.
In this experiment, you will re-create the photoelectric effect, and gather evidence for the particle-like nature of light.
Next | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8002836108207703, "perplexity": 664.2719092137063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00243.warc.gz"} |
http://www.emathematics.net/circle.php?a=&figura=2 | User:
• Matrices
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Suma y resta Producto por escalar Producto Inversa
Monomials Polynomials Special products Equations Quadratic equations Radical expressions Systems of equations Sequences and series Inner product Exponential equations Matrices Determinants Inverse of a matrix Logarithmic equations Systems of 3 variables equations
2-D Shapes Areas Pythagorean Theorem Distances
Graphs Definition of slope Positive or negative slope Determine slope of a line Equation of a line Equation of a line (from graph) Quadratic function Parallel, coincident and intersecting lines Asymptotes Limits Distances Continuity and discontinuities
Sine Cosine Tangent Cosecant Secant Cotangent Trigonometric identities Law of cosines Law of sines
Equations of a straight line Parallel, coincident and intersecting lines Distances Angles in space Inner product
Perimeter
Circles: word problems
Ryan used a drafting compass to draw a circle on paper. The circle was 12 cm in radius. How many square centimeters were inside the circle?
Area = π × r2
Area = π x 122 = π x 144 = 144 π cm2
π ≈ 3.14, so the area is 3.14x144=452.16 inches squared
Robert drew a circle that has a diameter of 16 inches. What is the area of his circle?
Area = π × r2
You are told that the diameter is 16 inches, so the radius is one-half of this measure, or 8 inches. Substituting the information into the area formula gives you
Area = π ×82 = π × 64 = 64π
π ≈ 3.14, so the area is 3.14x64=200.96 inches squared.
Find the circumference of the following coin (Round to the nearest whole number):
1 Euro Coin Diameter=23.25 mm
C = π × d
Use 3.14 as an approximation for π.
C=(3.14)(23.25)≈73 mm.
Find the circumference of the following coin (Round to the nearest whole number):
Susan B.. Anthony Dollar Radius=0.52 in
C =2πr
Use 3.14 as an approximation for π.
C=(2)(3.14)(0.52)≈3.27 in. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9733688831329346, "perplexity": 3833.2380398540113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202589.68/warc/CC-MAIN-20190322014319-20190322040319-00065.warc.gz"} |
http://math.stackexchange.com/questions/46413/digit-in-the-tens-place-of-an-expression/46416 | # Digit in the ten's place of an expression
What is the digit in the ten's place of $23^{41}* 25^{40}$ ? How do you calculate this? The usual method for this kind of problem is using the Binomial theorem, but I couldn't solve it.
-
Note that $25^2=625\equiv 25\bmod 100$, so that in fact $25^n\equiv 25\bmod 100$ for any $n$. Because $\phi(100)=40$, by Euler's theorem we have that $a^{40}\equiv1\bmod 100$ for any $a$ relatively prime to 100 (as 23 is). Thus $23^{41}\equiv 23\bmod 100$. Now put these results together to find $$23^{41}\cdot 25^{40}\bmod 100.$$
-
$$(24+1)^{40} \times (24-1)^{41}$$ $$= 23\times (24^2-1)^{40}$$ $$=23\times (575)^{40}$$
You can extract a pattern for powers of $5$. The last digits will always be $5$ and for this case $575^n$ the last two digits will alternate between $25$ (even n) and $75$ for odd n.
So $$\cdots 25$$ $$\underline{\qquad \times 23}$$ $$\quad \cdots 75$$ $$\underline{\;\cdots 50\times}$$ $$=\cdots 75$$
I have tried to emulate the multiplication procedure to show it.
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More simply, $23^{41} 25^{40} = (23 \cdot 25)^{40} 23 = 575^{40} 23$. – Shai Covo Jun 20 '11 at 6:40
@shai you're right. When I saw the problem, somehow $24^2 =576$ came to my mind.. – kuch nahi Jun 20 '11 at 6:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9435509443283081, "perplexity": 201.2579256844536}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246657588.53/warc/CC-MAIN-20150417045737-00022-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.scribd.com/document/30812567/Font-Lectures-1-and-2 | # General relativistic hydrodynamics and MHD José A.
Font
Departamento de Astronomía y Astrofísica
December 7-11, 2009 Asia Pacific Center for Theoretical Physics, Seoul (Korea)
Outline
Lecture 1: General relativistic hydrodynamics equations. Lecture 2: General relativistic MHD equations. Formulations of Einstein’s equations. Lecture 3: Numerical methods for conservation laws. Lecture 4: Tests and applications in astrophysics. See also lectures by Dr. L. Baiotti.
Bibliography
A.M. Anile, “Relativistic fluids and magneto-fluids” Cambridge University Press (1989) J.M. Martí & E. Müller, “Numerical hydrodynamics in special relativity” Living Reviews in Relativity (2003) (www.livingreviews.org) J.A. Font, “Numerical hydrodynamics and magnetohydrodynamics in general relativity” Living Reviews in Relativity (2008) (www.livingreviews.org) F. Banyuls et al, “Numerical 3+1 general relativistic hydrodynamics: a local characteristic approach” Astrophysical Journal, 476, 221 (1997) L. Antón et al, “Numerical 3+1 general relativistic magneto-hydrodynamics: a local characteristic approach” Astrophysical Journal, 637, 296 (2006) R.J. LeVeque, “Numerical methods for conservation laws” Birkhäuser, Basel (1992) E.F. Toro, “Riemann solvers and numerical methods for fluid dynamics” Springer Verlag, Berlin (1997)
Lecture 1 General relativistic hydrodynamics
Astrophysical motivation
Natural domain of general relativistic hydrodynamics (GRHD) and magnetohydrodynamics (GRMHD) is the field of relativistic astrophysics. General relativity and relativistic (magneto-)hydrodynamics play a major role in the description of gravitational collapse leading to the formation of compact objects (neutron stars and black holes). Sources of Gravitational Radiation.
Astrophysical motivation
Natural domain of general relativistic hydrodynamics (GRHD) and magnetohydrodynamics (GRMHD) is the field of relativistic astrophysics. General relativity and relativistic (magneto-)hydrodynamics play a major role in the description of gravitational collapse leading to the formation of compact objects (neutron stars and black holes). Sources of Gravitational Radiation.
Time-dependent evolutions of fluid flow coupled to the spacetime geometry (Einstein’s equations) possible through accurate, large-scale numerical simulations. Some scenarios can be described in the test-fluid approximation: GRHD/GRMHD computations in curved backgrounds (highly mature, particularly GRHD case). GRHD/GRMHD equations are nonlinear hyperbolic systems. Solid mathematical foundations and accurate numerical methodology imported from CFD. A “preferred” choice: high-resolution shock-capturing schemes written in conservation form.
Motivation: intense work in recent years on formulating/solving the MHD equations in general relativistic spacetimes (either background or dynamical).
Pioneers: Wilson (1975), Sloan & Smarr (1985), Evans & Hawley (1988), Yokosawa (1993) More recently: Koide et al (1998 …), De Villiers & Hawley (2003 …), Baumgarte & Shapiro
(2003), Gammie et al (2003), Komissarov (2005), Duez et al (2005 …), Shibata & Sekiguchi (2005), Antón et al (2006), Neilsen et al (2006). Both, artificial viscosity and HRSC
schemes developed.
Most of the applications are in the field of black hole accretion and jet formation …
Development of the MRI in a magnetised torus around a Kerr black hole (Gammie, McKinney & Tóth 2003) Jet formation: the twisting of magnetic field lines around a Kerr black hole. The yellow surface is the ergosphere (Koide et al 2002)
… many others under way (you name it!)
The number of groups working in special relativistic MHD is even larger: Komissarov; Balsara; Koldoba et al; Del Zanna et al; Leisman et al; … Exact solution of the SRMHD Riemann problem found recently: Romero et al (2005) – particular case; Giacomazzo & Rezzolla (2005) – general case.
Fluid dynamics: introduction
The defining property of fluids (liquids and gases) lies in the ease with which they may be deformed. A “simple fluid” may be defined as a material such that the relative positions of its constituent elements change by a large amount when suitable forces, however small in magnitude, are applied to the material. The properties of solids and fluids are directly related to their molecular structure and to the nature of the forces between the molecules. For most simple molecules, stable equilibrium between two molecules is achieved when their separation d0 ~3-4x10-8 cm. The average spacing of the molecules in a gaseous phase at normal temperature and pressure is of the order of 10d0, while in liquids and solids is of the order of d0. Fluid dynamics deals with the behaviour of matter in the large (average quantities per unit volume), on a macroscopic scale large compared with the distance between molecules, l>>d0, not taking into account the molecular structure of fluids. Even when considering an infinitesimal volume element it must be assumed that it is much smaller than the dimensions of the fluid, L, but much larger than the distance between molecules, i.e. L>>l>>d0.
The macroscopic behaviour of fluids is assumed to be perfectly continuous in structure, and physical quantities such as mass, density, or momentum contained within a given small volume are regarded as uniformly spread over that volume. Hence, the quantities which characterize a fluid (in the continuum limit) are functions of time and position: density (scalar field) velocity (vector field) pressure tensor (tensor field)
Eulerian description: time variation of fluid properties in a fixed position in space. Lagrangian description: variation of properties of a “fluid particle” along its motion.
Both descriptions are equivalent: there exists a change of variables between them which is related to the Jacobian of the so-called “flux function” which describes the trajectories of fluid particles. Transport theorems: Scalar field: Vector field: Vt is a volume which moves with the fluid (Lagrangian description; image of V0 by the difeomorphism given by the flux function).
(Perfect) fluid dynamics: equations (1) Mass conservation (continuity equation)
Let Vt be a volume which moves with the fluid; its mass is given by: The principle of conservation of mass enclosed within that volume reads: Applying the so-called transport theorem for the density (scalar field) we get: where the convective derivative is defined as
Since above eq. must hold for any volume Vt, we obtain the continuity equation:
Corolary:
the variation of the mass enclosed in a fixed volume V is equal to the flux of mass across the surface at the boundary of the volume.
Incompressible fluid:
(Perfect) fluid dynamics: equations (2) Momentum balance (Euler’s equation)
“the variation of momentum of a given portion of fluid is equal to the net force (stresses plus external forces) exerted on it” (Newton’s 2nd law):
Applying the transport theorem on the l.h.s. of the above eq. we get
which must be valid for any volume Vt, hence:
After some algebra and using the equation of continuity we obtain Euler’s equation:
(Perfect) fluid dynamics: equations (3) Energy conservation
Let E be the total energy of the fluid, sum of its kinetic energy and internal energy:
Principle of energy conservation: “the variation in time of the total energy of a portion of fluid is equal to the work done per unit time over the system by the stresses (internal forces) and the external forces”.
After some algebra (transport theorem, divergence theorem) we obtain:
which, as must be satisfied for any given volume, implies:
Fluid dynamics equations as a hyperbolic system of conservation laws
The equations of perfect fluid dynamics are a nonlinear hyperbolic system of conservation laws: state vector fluxes sources
is a conservative external force field (e.g. the gravitational field): are source terms in the momentum and energy equations, respectively, due to coupling between matter and radiation (when transport phenomena are also taken into account). Hyperbolic equations have finite propagation speed: information can travel with speed at most that given by the largest characteristic curves of the system. The range of influence of the solution is bounded by the eigenvalues of the Jacobian matrix of the system.
A bit on viscous fluids
A perfect fluid can be defined as that for which the force across the surface separating two fluid particles is normal to that surface. Kinetic theory tells us that the existence of velocity gradients implies the appearance of a force tangent to the surface separating two fluid layers (across which there is molecular difussion). where is the pressure tensor which depends on the pressure and on the velocity gradients. where is the stress tensor given by:
distortion expansion
Using the pressure tensor in the previous derivation of the Euler equation and of the energy equation yields the viscous version of those equations:
shear and bulk viscosities
Navier-Stokes equation
Energy equation
General relativistic hydrodynamics equations
The general relativistic hydrodynamics equations are obtained from the local conservation laws of the stress-energy tensor, Tµν (the Bianchi identities), and of the matter current density Jµ (the continuity equation):
Equations of motion
As usual ∇µ stands for the covariant derivative associated with the four dimensional spacetime metric gµν. The density current is given by Jµ=ρuµ, uµ representing the fluid 4-velocity and ρ the rest-mass density in a locally inertial reference frame. The stress-energy tensor for a non-perfect fluid is defined as:
where ε is the rest-frame specific internal energy density of the fluid, p is the pressure, and hµν is the spatial projection tensor, hµν=uµuν+gµν. In addition, µ and ξ are the shear and bulk viscosity coefficients. The expansion, Θ, describing the divergence or convergence of the fluid world lines is defined as Θ=∇µuν. The symmetric, trace-free, and spatial shear tensor σµν is defined by:
Finally qµ is the energy flux vector.
In the following we will neglect non-adiabatic effects, such as viscosity or heat transfer, assuming the stress-energy tensor to be that of a perfect fluid:
where we have introduced the relativistic specific enthalpy, h, defined as:
Introducing an explicit coordinate chart the previous conservation equations read:
where the scalar x0 represents a foliation of the spacetime with hypersurfaces (coordinatised by xi). Additionally, is the volume element associated with the 4metric gµν, with g=det(gµν), and are the 4-dimensional Christoffel symbols. The system formed by the equations of motion and the continuity equation must be supplemented with an equation of state (EOS) relating the pressure to some fundamental thermodynamical quantities, e.g. • • Ideal fluid EOS: Polytropic EOS:
In the “test-fluid” approximation (fluid’s self-gravity neglected), the dynamics of the matter fields is fully described by the previous conservation laws and the EOS. When such approximation does not hold, the previous equations must be solved in conjunction with Einstein’s equations for the gravitational field which describe the evolution of a dynamical spacetime:
(Newtonian analogy: Euler’s equation + Poisson’s equation)
Einstein’s equations
The most widely used approach to solve Einstein’s equations in Numerical Relativity is the so-called Cauchy or 3+1 formulation (IVP). Spacetime is foliated with a set of non-intersecting spacelike hypersurfaces Σ. Within each surface distances are measured with the spatial 3-metric γij. There are two kinematical variables which describe the evolution between each hypersurface: the lapse function α, and the shift vector βi
3+1 GR Hydro equations: formulations
Equations of motion: local conservation laws of density current (continuity equation) and stress-energy (Bianchi identities) Perfect fluid stress-energy tensor Introducing an explicit coordinate chart: Different formulations exist depending on:
1. 2.
Choice of slicing: level surfaces of x0 can be spatial (3+1) or null (characteristic)
Choice of physical (primitive) variables (ρ, ε, ui …)
Wilson (1972) wrote the system as a set of advection equation within the 3+1 formalism. Non-conservative.
Conservative formulations well-adapted to numerical methodology are more recent:
• • • • Martí, Ibáñez & Miralles (1991): 1+1, general EOS Eulderink & Mellema (1995): covariant, perfect fluid Banyuls et al (1997): 3+1, general EOS Papadopoulos & Font (2000): covariant, general EOS
Wilson’s formulation (1972)
The use of Eulerian coordinates in multidimensional numerical hydrodynamics started with the seminal work of J. Wilson (1972). relativistic Introducing the basic dynamical variables D, Sµ, and E, i.e. the relativistic density, momenta, and energy, respectively, defined as:
the equations of motion in Wilson’s formulation are:
with the “transport velocity” given by Vµ=uµ/u0. Note that the momentum density equation is only solved for the three spatial components, Si, and S0 is obtained through the normalization condition uµuµ=-1.
A direct inspection of the system shows that the equations are written as a coupled set of advection equations.
Conservation of mass:
(linear advection eq.) This approach sidesteps an important guideline for the formulation of nonlinear hyperbolic systems of equations, namely the preservation of their conservation form. This is a necessary feature to guarantee correct evolution in regions of entropy generation (i.e. shocks). As a result, some amount of numerical dissipation (artificial viscosity) must be used to stabilize the numerical solution across discontinuities. Wilson’s formulation showed some limitations in dealing with situations involving ultrarelativistic flows, as first pointed out by Centrella & Wilson (1984). Norman & Winkler, in their 1986 paper “Why ultrarelativistic hydrodynamics is difficult?” performed a comprehensive numerical study of such formulation in the special relativistic limit.
Relativistic shock reflection
The relativistic shock reflection problem was among the 1D tests considered by Norman & Winkler (1986). This is a demanding test involving the heating of a cold gas which impacts at relativistic speed with a solid wall creating a shock which propagates off the wall.
1
Problem setup
shocked material
2
constant density cold gas flow velocity shock speed
high density high pressure zero velocity solid wall
shock front
Analytic solution:
(from Martí & Müller, 2003)
Limitations to handle ultrarelativistic flows (Centrella & Wilson 1984, Norman & Winkler 1986).
1
Problem setup
shocked material high density high pressure zero velocity
2
constant density cold gas flow velocity shock speed
Relativistic shock reflection test: Relative errors of the numerical solution as a function of the Lorentz factor W of the incoming gas. For W≈2 (v≈0.86c), errors are 5-7% (depending on the adiabatic index of the ideal fluid EOS), showing a linear growth with W.
shock front
solid wall
Analytic solution:
Presence of Lorentz factor in the convective terms of the hydrodynamic equations (and of the pressure in the specific enthalpy) make the relativistic equations much more coupled than their Newtonian counterparts. Norman & Winkler (1986) proposed the use of implicit schemes to capture more accurately such coupling. Limitation:W≈10 Ultrarelativistic flows could only be handled (with explicit schemes) once conservative formulations were adopted (Martí, Ibáñez & Miralles 1991;
Marquina et al 1992)
Norman & Winkler (1986) concluded that those errors were due to the way in which the artificial viscosity terms Q are included in the numerical scheme in Wilson’s formulation. These terms are added to the pressure terms only in some cases (at the pressure gradient in the source of the momentum equation and at the divergence of the velocity in the source of the energy equation), and not to all terms. However, Norman & Winkler (1986) proposed to add the viscosity terms Q globally, in order to consider the artificial viscosity as a real viscosity. Hence, the equations of motion should be rewritten for a modified stress-energy tensor of the form:
In this way, in flat spacetime, the Euler (momentum) equations take the form:
where
is the Lorentz factor,
being the lapse function.
In Wilson’s formulation is omitted in the two terms containing quantity . is in general a nonlinear function of the velocity and, hence, the quantity in the momentum density equation is a highly nonlinear function of the velocity and its derivatives.
Despite the nonconservative nature of the formulation and the limitations to handle ultrarelativistic flows, Wilson’s approach has been widely used by many groups in relativistic astrophysics and numerical relativity along the years, e.g.: 1. Axisymmetric stellar collapse: Wilson 1979, Dykema 1980, Nakamura et al 1980,
Nakamura 1981, Nakamura & Sato 1982, Bardeen & Piran 1983, Evans 1984, 1986, Stark & Piran 1985, Piran & Stark 1986, Shibata 2000, Shibata & Shapiro 2002.
2. Instabilities in rotating relativistic stars: Shibata, Baumgarte & Shapiro, 2000. 3. Numerical cosmology: Centrella & Wilson 1983, 1984, Anninos 1998. 4. Accretion on to black holes: Hawley, Smarr & Wilson 1984, Petrich et al 1989, Hawley
1991.
5. Heavy ion collisions (SR limit): Wilson & Mathews 1989. 6. Binary neutron star mergers: Wilson, Mathews & Marronetti 1995, 1996, 2000,
Nakamura & Oohara 1998, Shibata 1999, Shibata & Uryu 2000, 2002. Belodorov 1997, De Villiers & Hawley 2003, Hirose et al 2004.
7. GRMHD simulations of BH accretion disks: Yokosawa 1993, 1995, Igumenshchev & 8. etc. More recently, Anninos & Fragile (2003) have compared state-of-the-art AV schemes and high-order non-oscillatory central schemes using Wilson’s formulation and a conservative formulation. Ultrarelativistic flows could only be handled in the latter case. This highlights the importance of the conservative character of the formulation of the equations to the detriment of the particular numerical scheme employed.
Conservative formulations - Ibáñez et al (1991, 1997)
Numerically, the hyperbolic and conservative nature of the GRHD equations allows to design a solution procedure based on the characteristic speeds and fields of the system, translating to relativistic hydrodynamics existing tools of CFD. This procedure departs from earlier approaches, most notably in avoiding the need for artificial dissipation terms to handle discontinuous solutions as well as implicit schemes as proposed by Norman & Winkler (1986). Foliate spacetime with t=const spatial hypersurfaces
∑t
Let n be the unit timelike 4-vector orthogonal to ∑t such that
Eulerian observer: at rest in a given hypersurface, moves from ∑t to ∑t+∆t along the normal to the slice: Definitions: u : fluid’s 4-velocity, ε : specific internal energy density, p : isotropic pressure, ρ : rest-mass density e=ρ( 1+ε) : energy density
The extension of modern high-resolution shock-capturing (HRSC) schemes from classical fluid dynamics to relativistic hydrodynamics was accomplished in three steps: 1. 2. 3. Casting the GRHD equations as a system of conservation laws. Identifying the suitable vector of unknowns. Building up an approximate Riemann solver (or high-order symmetric scheme).
The associated numerical scheme had to meet a key prerequisite – being written in conservation form, as this automatically guarantees the correct propagation of discontinuities as well as the correct Rankine-Hugoniot (jump) conditions across discontinuities (the shock-capturing property). In 1991 Martí, Ibáñez, and Miralles presented a new formulation of the general relativistic hydrodynamics equations, in 1+1, aimed at taking advantage of their hyperbolic character. The corresponding 3+1 extension of the 1991 formulation was presented in Font et al (1994) in special relativity, and in Banyuls et al (1997) in general relativity. Replace the “primitive variables” in terms of the “conserved variables” :
Lorentz factor
specific enthalpy
Conservative formulations well-adapted to numerical methodology:
• Banyuls et al (1997); Font et al (2000): 3+1, general EOS
First-order flux-conservative hyperbolic system
Solved using HRSC schemes (either upwind or central)
Recovering special relativistic and Newtonian limits
Full GR
Newton Minkowski
HRSC schemes based on approximate Riemann solvers use the local characteristic structure of the hyperbolic system of equations. For the previous system, this information was presented in Banyuls et al (1997). The eigenvalues (characteristic speeds) are all real (but not distinct, one showing a threefold degeneracy), and a complete set of right-eigenvectors exists. The above system satisfies, hence, the definition of hiperbolicity.
Eigenvalues (along the x direction)
Right-eigenvectors
Special relativistic limit (along x-direction)
coupling with transversal components of the velocity (important difference with Newtonian case) Even in the purely 1D case:
For causal EOS the sound cone lies within the light cone Recall Newtonian (1D) case:
Black lines: SRHD Red lines: GRHD
Relativistic effects:
• Tangential component of the flow velocity (vt=0, 0.4, 0.8) • Gravitational field (r=1.5 Schwarzschild radius) Degeneracies in the eigenvalues occur as the flow speed reaches the speed of light and when the lapse function goes to zero (gauge effect), i.e. at the black hole horizon (sonic sphere).
Recovering primitive variables from state vector
A distinctive feature of the numerical solution of the relativistic hydrodynamics equations is that while the numerical algorithm updates the vector of conserved quantities, the numerical code makes extensive use of the primitive variables. Those would appear repeatedly in the solution procedure, e.g. in the characteristic fields, in the solution of the Riemann problem, and in the computation of the numerical fluxes. For spacelike foliations of the spacetime (3+1) the relation between the two sets of variables is implicit. Hence, iterative (root-finding) algorithms are required. Those have been developed for all existing formulations (Eulderink & Mellema 1995; Banyuls et al 1997, Papadopoulos & Font 2000) This feature, which is distinctive of the equations of general (and special) relativistic hydrodynamics (and also in GRMHD) – not existing in the Newtonian case – may lead to accuracy losses in regions of low density and small velocities, apart from being computationally inefficient. For null foliations of the spacetime, the procedure of connecting primitive and conserved variables is explicit for a perfect fluid EOS, a direct consequence of the particular form of the Bondi-Sachs metric.
Recovering primitive variables: expressions
Newtonian hydrodynamics: fully explicit to obtain “primitive” variables from state vector. 3+1 general relativistic hydrodynamics: root-finding procedure. The expressions relating the primitive variables to the state vector depend explicitely on the EOS p(ρ,ε). Simple expressions are only obtained for simple EOS, i.e. ideal gas. One can build a function of pressure whose zero represents the pressure in the physical state (other choices possible):
The root of the above function can be obtained by means of a nonlinear root-finder (e.g. a Newton-Raphson method).
Astrophysical applications of Banyuls et al formulation
1.
Simulations of relativistic jets (SR): Gamma-ray burst models:
Martí et al 1994, 1995, 1997, Gómez et al 1995, 1997, 1998, Aloy et al 1999, 2003, 2008 ..., Scheck et al 2002, Perucho et al 2008 ..., Mimica et al 2005, 2009 ... Aloy et al 2000, Aloy, Janka & Müller 2004, 2005 ...
2. 3.
Core collapse supernovae and GWs: QNM of rotating relativistic stars:
Dimmelmeier, Font & Müller 2001, 2002a,b, Dimmelmeier et al 2004, Cerdá-Durán et al 2004, 2008, Shibata & Sekiguchi 2004. Font, Stergioulas & Kokkotas 2000, Font et al 2001, 2002, Stergioulas & Font 2001, Stergioulas, Apostolatos & Font 2004, Shibata & Sekiguchi 2003, Dimmelmeier et al 2006, ...
4.
5. 6. 7. 8.
Neutron star collapse and black hole formation:
2007 ...
Baiotti et al 2004, 2005,
Accretion on to black holes: Disk accretion:
Font & Ibáñez 1998a,b, Font, Ibáñez & Papadopoulos 1999, Brandt et al 1998, Papadopoulos & Font 1998a,b, Nagar et al 2004, ... Font & Daigne 2002a,b, Daigne & Font 2004, Zanotti, Rezzolla & Font 2003, Rezzolla, Zanotti & Font 2003, ...
Binary neutron star mergers:
Miller, Suen & Tobias 2001, Shibata, Taniguchi & Uryu 2003, Evans et al 2003, Miller, Gressman & Suen 2004, Baiotti et al 2008, 2009.
list which probably needs updating ...
Eulderink & Mellema formulation (1995)
Eulderink & Mellema (1995) derived a covariant formulation of the GR hydrodynamic equations taking special care in the conservative form of the system. Additionally they developed a numerical method to solve them based on a generalisation of Roe’s approximate Riemann solver for the non relativistic Euler equations in Cartesian coordinates. Their procedure was specialized for a perfect fluid EOS. After the appropriate choice of the state vector variables, the conservation laws are rewritten in flux-conservative form. The flow variables are expressed in terms of a parameter vector ω as:
where The vector represents the state vector (the unknowns) and each vector corresponding flux in the coordinate direction is the
Eulderink and Mellema computed the appropriate “Roe matrix” for the above vector and obtained the corresponding spectral decomposition. This information is used to solve the system numerically using Roe’s generalised approximate Riemann solver. Formulation and numerical approach tested in 1D (shock tubes, spherical accretion onto a Schwarzschild black hole). In SR employed to study the confinement properties of relativistic jets. No astrophysical application in GR attempted.
Papadopoulos and Font (2000) derived another conservative formulation of the relativistic hydrodynamic equations form-invariant with respect to the nature of the spacetime foliation (spacelike, lightlike). In this formulation the spatial components of the 4-velocity, together with the restframe density and internal energy, provide a unique description of the state of the fluid and are taken as the primitive variables The initial value problem for the conservation laws is defined in terms of another vector in the same fluid state space, namely the conserved variables:
The flux vectors and the source terms (which depend only on the metric, its derivatives, and the undifferentiated stress energy tensor) are given by:
Characteristic structure of the above equations
Eigenvalues (along direction 1)
Right-eigenvectors
with the definitions:
Recovering primitive variables
For null foliations of the spacetime, the procedure of connecting primitive and conserved variables is explicit for a perfect fluid EOS, a direct consequence of the particular form of the Bondi-Sachs metric (g00=0). For the formulation of Papadopoulos & Font:
Lightcone hydrodynamics and characteristic numerical relativity
The formulation was tested in 1D relativistic flows (comparing with exact solutions in some cases) on null (lightlike) spacetime foliations (Papadopoulos & Font 1999): • Shock tube tests in Minkowski spacetime (advanced and retarded time).
• • • Perfect fluid accretion on to a Schwarzschild black hole (ingoing null EF coordinates). Dynamical spacetime evolutions of polytropes (TOV) sliced along the radial null cones. Accretion of self-gravitating matter on a dynamic black hole.
Existing applications in relativistic astrophysics include:
• Accreting dynamic black holes and quasi-normal modes
(Papadopoulos & Font, PRD, 63, 044016, 2001).
• Gravitational collapse of supermassive stars (GRB model)
(Linke et al, A&A, 376, 568, 2001).
• Interaction of scalar fields with relativistic stars (Siebel, Font
& Papadopoulos, PRD, 65, 024021, 2002).
• Nonlinear pulsations of axisymmetric neutron stars (Siebel,
Font, Müller & Papadopoulos, PRD, 65, 064038, 2002).
• Axisymmetric core collapse and gravitational radiation (Bondi news) (Siebel, Font, Müller & Papadopoulos, PRD, 67,
124018, 2003).
Winicour, Living Reviews in Relativity (2005)
Lecture 2 a) General relativistic magnetohydrodynamics
General Relativistic Magneto-Hydrodynamics (1)
GRMHD: Dynamics of relativistic, electrically conducting fluids in the presence of magnetic fields. Ideal GRMHD: Absence of viscosity effects and heat conduction in the limit of infinite conductivity (perfect conductor fluid). The stress-energy tensor includes contribution from the perfect fluid and from the magnetic field bµ measured by observer comoving with the fluid.
with the definitions:
Ideal MHD condition: electric four-current must be finite.
General Relativistic Magneto-Hydrodynamics (2)
Conservation of mass: Conservation of energy and momentum: Maxwell’s equations: • Divergence-free constraint: • Induction equation: Adding all up: first-order, flux-conservative, hyperbolic system + constraint
Antón et al. (2006)
RMHD: some issues on hyperbolic structure (I)
Wave structure classical MHD (Brio & Wu 1988): 7 physical waves
Anile & Pennisi (1987), Anile (1989) (see also van Putten 1991) have studied the characteristic structure of the equations (eigenvalues, right/left eigenvectors) in the space of covariant variables (uµ,bµ,p,s). Wave structure for relativistic MHD (Anile 1989): roots of the characteristic equation. Only entropic waves and Alfvén waves are explicit. Magnetosonic waves are given by the numerical solution of a quartic equation. Augmented system of equations: Unphysical eigenvalues/eigenvectors (entropy & Alfvén) which must be removed numerically (Anile 1989, Komissarov 1999, Balsara 2001, Koldoba et al 2002).
RMHD: some issues on hyperbolic structure (II)
Degeneracy in eigenvalues (wave speeds coincidence) as for classical MHD. The wave propagation velocity depends on the relative orientation of the magnetic field, θ. As in classical MHD there are two types of degeneracies: Degeneracy I: bµn=0 Degeneracy II: bµt=0 Implications: Physically two or more wave speeds become equal (compound waves). Numerically the spectral decomposition (needed in upwind HRSC schemes) blows up.
bµn bµt=0 (Deg II) bµt bµn=0 (Deg I)
Wavefront diagrams in the fluid rest frame (Jeffrey & Taniuti 1964).
Orientation of the magnetic field
Breakup of hyperbolicity (Komissarov 1999; degeneracy characterisation). Eigenvectors renormalization needed: Brio & Wu 1988 for classical MHD. Extended to RMHD for a Roe-type solver by Antón (PhD thesis, University of Valencia 2006).
Particular case: Minkowski + transverse B-field
Exact solutions of the Riemann problem in SRHD & SRMHD have recently been obtained: •SRHD: Martí & Müller 1996; Pons et al 2000; Rezzolla & Zanotti 2002; Rezzolla et al 2003. •SRMHD: Romero et al 2005, particular case; Giacomazzo & Rezzolla 2006, general orientations). Important tool for code validation.
Recovering primitive variables in RMHD
In RMHD the recovery of primitive variables more involved than in RHD. (see e.g. Noble et al 2006 for a comparison of methods.) Antón et al 2006 find the roots of an 8-th order polynomial using a 2d NewtonRaphson. See PhD thesis of L. Antón (2006), University of Valencia, for details.
Solution procedure of the GRMHD equations
The divergence-free constraint is not guaranteed to be satisfied numerically when updating the B-field with a HRSC scheme in conservation form.
• Same HRSC as for the GRHD equations • Wave structure information obtained • Primitive variable recovery more involved
Details: Antón, Zanotti, Miralles, Martí, Ibáñez, Font & Pons, ApJ (2006)
An ad-hoc scheme has to be used to update the magnetic field components.
Main physical implication of divergence constraint is that the magnetic flux through a closed surface is zero: essential to the constrained transport (CT) scheme (Evans & Hawley 1988, Tóth
2000).
For any given surface, the time variation of the magnetic flux across the surface is:
Stokes theorem
Induction equation
The magnetic flux through a surface can be calculated as the line integral of the electric field along its boundary.
Magnetic field evolution: flux-CT (1)
Numerical implementation (e.g. in axisymmetry): Assumption: B-field components constant at each cell surface E-field components constant along each cell edge
Evolution equations for the B-field (CT scheme) The polodial (r and θ) B-field components are defined at cell interfaces (staggered grid) The total magnetic flux through the cell interfaces is given by: (axisymmetry condition)
If the initial data satisfy the divergence constraint, it will be preserved during the evolution
Magnetic field evolution: flux-CT (2)
Discretisation:
Equations used by the code to update the B-field. The only remaining aspect is an explicit expression for the E-field.
The E-field components can be calculated from the numerical fluxes of the conservation equations for the B-field. Done solving Riemann problems at cell interfaces (characteristic information of the fluxvector Jacobians incorporated in the B-field evolution). This procedure is only valid for r and θ E-field components.
Balsara & Spicer (1999) proposed a practical way to compute the ϕ component of the Efield from the numerical fluxes in adjacent interfaces.
Resulting scheme flux-CT
Alternatives to the CT method
For AMR codes based on unstructured grids and multiple coordinate systems there are other schemes, such as projection methods or hyperbolic divergence cleaning, which appear more suitable than the CT method to enforce the magnetic field divergence-free constraint. The projection method involves solving an elliptic equation for a corrected magnetic field projected onto the subspace of zero divergence solutions by a linear operator. The magnetic field is decomposed into a curl and a gradient as whose divergence leads to an elliptic (Poisson) equation solved for Φ. The magnetic field is next corrected according to , which can be .
[An alternative projection scheme can be implemented by taking the curl of the above equation and solving for the vector potential.] The hyperbolic divergence cleaning approach is based on the introduction of an additional scalar field which is coupled to the magnetic field by a gradient term in the induction equation. The scalar field is calculated by adding an additional constraint hyperbolic equation given by Divergence errors are propagated off the grid in a wave-like manner with characteristic speed ch.
Lecture 2 b) (Some) formulations of Einstein’s equations
Numerical Relativity: dynamical spacetimes
Numerical Relativity is the field of research of General Relativity devoted to seeking numerical solutions of Einstein’s equations through (super) computer simulations.
Mathematical difficulties:
PDEs highly non-linear with hundreds of terms. Hyperbolic and elliptic character. Coordinates and gauge conditions (gauge freedom). Boundary conditions.
• Numerical difficulties:
Formulation of the equations. Development of numerical schemes. Stability and efficiency. Possible formation of curvature singularities (collapse to black hole).
• Huge computational resources needed (3D).
(Numerical) General Relativity: Which portion of spacetime shall we foliate?
(original formulation of the 3+1 equations)
Reformulating these equations to achieve numerical stability is one of the arts of numerical relativity.
(Numerical) General Relativity: Which portion of spacetime shall we foliate?
3+1 (Cauchy) formulation
Lichnerowicz (1944); Choquet-Bruhat (1962); Arnowitt, Deser & Misner (1962); York (1979)
Standard choice for most Numerical Relativity groups. Spatial hypersurfaces have a finite extension.
(original formulation of the 3+1 equations)
Reformulating these equations to achieve numerical stability is one of the arts of numerical relativity.
Conformal formulation: Spatial hypersurfaces have infinite extension (Friedrich et al). Characteristic formulation (Winicour
et al).
Hypersuperfaces are light cones (incoming/outgoing) with infinite extension.
Gravity: CFC metric equations
In the CFC approximation equations
(Isenberg 1985; Wilson & Mathews 1996)
reduce to a system of five coupled, nonlinear elliptic equations for the lapse function, conformal factor, and the shift vector:
CFC approximation
Gravity: CFC+ metric equations
Cerdá-Duran, Faye, Dimmelmeier, Font, Ibáñez, Müller & Schäfer (2005)
CFC+ metric:
The second post-Newtonian deviation from isotropy is given by:
Modified equations with respect to CFC
Gravity: CFC+ metric equations (cont’d)
The required intermediate potentials satisfy 16 elliptic linear equations: Linear solver: LU decomposition using standard LAPACK routines
Boundary conditions: Multipole development in compact-supported integrals
Fully Constrained Formulation (Meudon-Valencia) Bonazzola et al (2004), Cordero-Carrión et al (2009)
Hydrodynamical conserved variables
Fully Constrained Formulation (Meudon-Valencia) Bonazzola et al (2004), Cordero-Carrión et al (2009)
Spectral methods for gravity and HRSC methods for hydro
Gravity: BSSN metric equations (Free evolution)
Kojima, Nakamura & Oohara (1987); Shibata & Nakamura (1995); Baumgarte & Shapiro (1999)
Idea: Remove mixed second derivatives in the Ricci tensor by introducing auxiliary variables. Evolution equations start to look like wave equations for 3-metric and extrinsic curvature (idea goes back to De Donder 1921; ChoquetBruhat 1952; Fischer & Marsden 1972).
Conformal decomposition of the 3-metric:
BSSN evolution variables (trace of extrinsic curvature is a separate variable):
Introduce evolution variables (gauge source functions):
BSSN metric evolution equations
Kojima, Nakamura & Oohara (1987); Shibata & Nakamura (1995); Baumgarte & Shapiro (1999)
BSSN is currently the standard 3+1 formulation in Numerical Relativity. Long-term stable applications include strongly-gravitating systems such as neutron stars (isolated and binaries) and single and binary black holes! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.925705075263977, "perplexity": 2672.349599436055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886830.8/warc/CC-MAIN-20180117063030-20180117083030-00249.warc.gz"} |
https://www.physicsforums.com/threads/uncertainty-in-expectation-value.142554/ | # Uncertainty in expectation value.
1. Nov 6, 2006
### swain1
When trying to work out the uncerainty in position of the expectation value I have read that you have to find <r^2> aswell as <r>^2. I have worked out the value of 3a/2 for <r> but what do I have to do to find <r^2>. Do I just sqare the whole function before I integrate?
Also as I am integrating I found, in a book a table that had a general form of integrations between 0 and infinity. I used it without giving it much thought but how would you evaluate this? Thanks
2. Nov 6, 2006
### Tomsk
Yes to find <r^2> you integrate r^2 p(r) dr instead of rp(r)dr. Or for any function of r, it's just int[f(r)p(r)dr].
I'm guessing the table was for the gaussian interal int[exp(-ax^2)] a>0? The integral can't be evaluated directly, you have to do some weird substitutions, if you want to know how to do it I can show you. It's easiest to remember the the soln is sqrt(pi/a), from 0 to infinity is half this, and for x^2n exp(-ax^2), differentiate both sides wrt a.
3. Nov 7, 2006
### swain1
Ok thanks for that. I have done the inegral and worked out the uncertainty. The value I have got is sqrt(3)a/2. It seems like an awful lot to me. Anyway, do you know if this is right. I am doing this for the ground state of hydrogen. If I have got the correct value. Why is it so large?
4. Nov 8, 2006
### Tomsk
Not entirely sure, what exactly was the integral? And what is a? I haven't done any quantum, not sure what this is about exactly.
5. Nov 8, 2006
### stunner5000pt
$$\Delta r = \sqrt{<r^2> - <r>^2}$$
Similar Discussions: Uncertainty in expectation value. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.90811687707901, "perplexity": 838.8699243048417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121453.27/warc/CC-MAIN-20170423031201-00355-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://www.legisquebec.gouv.qc.ca/en/version/cs/H-2.1?code=se:2&history=20220622 | ### H-2.1 - Act respecting hours and days of admission to commercial establishments
2. Subject to sections 3 to 14.1, the public may be admitted to a commercial establishment only between the hours of
(1) 8:00 a.m. and 5:00 p.m. on Saturdays and Sundays, and 8:00 a.m. and 9:00 p.m. on the other days of the week;
(2) 8:00 a.m. and 5:00 p.m. on 24 and 31 December;
(3) 1:00 p.m. and 5:00 p.m. on 26 December where it falls on a Saturday or a Sunday, and 1:00 p.m. and 9:00 p.m. where it falls on another day of the week.
1990, c. 30, s. 2; 1992, c. 55, s. 1; 2006, c. 47, s. 1; 2019, c. 29, s. 81.
2. Subject to sections 3 to 14, the public may be admitted to a commercial establishment only between the hours of
(1) 8:00 a.m. and 5:00 p.m. on Saturdays and Sundays, and 8:00 a.m. and 9:00 p.m. on the other days of the week;
(2) 8:00 a.m. and 5:00 p.m. on 24 and 31 December;
(3) 1:00 p.m. and 5:00 p.m. on 26 December where it falls on a Saturday or a Sunday, and 1:00 p.m. and 9:00 p.m. where it falls on another day of the week.
1990, c. 30, s. 2; 1992, c. 55, s. 1; 2006, c. 47, s. 1.
2. Subject to sections 5 to 14, the public may be admitted to a commercial establishment only between the hours of
(1) 8:00 a.m. and 5:00 p.m. on Saturdays and Sundays, and 8:00 a.m. and 9:00 p.m. on the other days of the week;
(2) 8:00 a.m. and 5:00 p.m. on 24 and 31 December;
(3) 1:00 p.m. and 5:00 p.m. on 26 December where it falls on a Saturday or a Sunday, and 1:00 p.m. and 9:00 p.m. where it falls on another day of the week.
1990, c. 30, s. 2; 1992, c. 55, s. 1.
2. Subject to sections 4 to 14, the public may be admitted to a commercial establishment only between the hours of
(1) 8:00 a.m. and 7:00 p.m. on Mondays and Tuesdays;
(2) 8:00 a.m. and 9:00 p.m. on Wednesdays, Thursdays and Fridays;
(3) 8:00 a.m. and 5:00 p.m. on Saturdays;
(4) 8:00 a.m. and 9:00 p.m. on Mondays and Tuesdays in December before 25 December;
(5) 8:00 a.m. and 5:00 p.m. on Sundays in December before 25 December;
(6) 8:00 a.m. and 5:00 p.m. on 24 and 31 December where they fall on any day other than Sunday;
(7) 1:00 p.m. and 7:00 p.m. on 26 December where it falls on a Monday or a Tuesday, 1:00 p.m. and 9:00 p.m. where it falls on a Wednesday, a Thursday or a Friday, and 1:00 p.m. and 5:00 p.m. where it falls on a Saturday.
1990, c. 30, s. 2. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8227025866508484, "perplexity": 1946.641088431455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00134.warc.gz"} |
https://blancosilva.wordpress.com/teaching/past-sections/ma242-section-3/third-midterm-practice-test/ | Third Midterm—Practice Test
In the exam, we will be using Table 6.2.1 from Boyce-DiPrima as aid in our problems involving Laplace transforms. Feel free to use that table in these problems as well.
1. Show that $y_1=x^3$ and $y_2 = \lvert x^3 \rvert$ are linearly independent solutions on the real line of the equation $xy'' - 3xy' +3y =0.$ Verify that the Wronskian $W(y1,y2)$ is identically zero. Why do these facts not contradict Theorem 3 in page 154?
2. A second-order Euler equation is one of the form
$ax^2 y''+bxy' +cy = 0$
where $a,b,c$ are constants. Show that for non-negative values of $x$ the substitution $v=\ln x$ transforms the previous equation into the constant-coefficient linear equation
$a \dfrac{d^2y}{dv^2} + (b-a) \dfrac{dy}{dv} + cy =0$
with independent variable $v.$ If the roots $r_1$ and $r_2$ of the characteristic equation of the previous differential equation are real and distinct, conclude that a general solution of the Euler equation is $y=c_1 x^{r_1} + c_2 x^{r_2}.$
3. Find a function $y(x)$ such that $y^{(4)}(x)=y^{(3)}(x)$ for all $x,$ and $y(0)=18, y'(0)=12, y''(0)=13,$ and $y^{(3)}(0)=7.$
4. Find a particular solution of the equation $2y''+4y'+7y=x^2$ using both the method of variation of parameters, and undetermined coefficients.
5. Find the solution of the equation $y^{(4)}-y^{(3)}-y''-y'-2y=8x^5$ that satisfies the initial conditions $y(0)=y'(0)=y''(0)=y^{(3)}(0)=0.$ Use both undetermined coefficients and methods based on the Laplace transform.
6. Find the Laplace transform of the function $f(x) = \sin 3x \cos 3x.$
7. Find the inverse Laplace transform of the function $F(s)= \dfrac{1}{(s^2+b^2)^2}.$
8. Find the inverse Laplace transform of the function $F(s) = \dfrac{s^2+1}{s^3-2s^2-8s}.$
9. Solve the initial value problem $y^{(4)}+2y''+y=4xe^x$ with $y(0)=y'(0)=y''(0)=y^{(3)}(0)=0$ using exclusively methods based on the Laplace transform.
10. Find the Laplace transform of the functions $f(x)=\dfrac{\sin x}{x}$ and $g(x)=xe^{2x}\cos 3x.$
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=Finance/GammaProcess | Finance - Maple Programming Help
Home : Support : Online Help : Mathematics : Finance : Stochastic Processes : Finance/GammaProcess
Finance
GammaProcess
create new Gamma process
Calling Sequence GammaProcess(mu, sigma)
Parameters
mu - real constant; mean parameter sigma - real constant; variance parameter
Description
• The GammaProcess command creates a Gamma process with the specified parameters. The Gamma process $G\left(t\right)$ with mean parameter mu and variance parameter sigma is a continuous-time process with stationary, independent gamma increments such that for any $0, $G\left(t+h\right)-G\left(t\right)$ has a Gamma distribution with shape parameter $\frac{{\mathrm{\mu }}^{2}h}{\mathrm{\sigma }}$ and scale parameter $\frac{\mathrm{\sigma }}{\mathrm{\mu }}$.
• The parameter mu is the mean. The parameter sigma is the variance.
Examples
> $\mathrm{with}\left(\mathrm{Finance}\right):$
> $\mathrm{μ}≔1:$$\mathrm{σ}≔3:$
> $G≔\mathrm{GammaProcess}\left(\mathrm{μ},\mathrm{σ}\right):$
> $\mathrm{PathPlot}\left(G\left(t\right),t=0..3,\mathrm{timesteps}=100,\mathrm{replications}=10,\mathrm{thickness}=3,\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)$
> $\mathrm{ExpectedValue}\left(G\left(3\right),\mathrm{replications}={10}^{4}\right)$
$\left[{\mathrm{value}}{=}{2.376349494}{,}{\mathrm{standarderror}}{=}{0.01792180394}\right]$ (1)
> $S≔\mathrm{SampleValues}\left(G\left(2\right)-G\left(1.98\right),\mathrm{timesteps}={10}^{2},\mathrm{replications}={10}^{3}\right)$
${S}{:=}\left[\begin{array}{c}{\mathrm{1 .. 1000}}{\mathrm{Array}}\\ {\mathrm{Data Type:}}{{\mathrm{float}}}_{{8}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{C_order}}\end{array}\right]$ (2)
The variance gamma process, introduced by Madan and Seneta, is the difference of two independent gamma processes representing the up and down movements of the underlying asset.
> $\mathrm{Xu}≔\mathrm{GammaProcess}\left(1,3\right):$
> $\mathrm{Xd}≔\mathrm{GammaProcess}\left(0.9,3\right):$
> $X≔t→\mathrm{Xu}\left(t\right)-\mathrm{Xd}\left(t\right)$
${X}{:=}{t}{→}{\mathrm{Xu}}{}\left({t}\right){-}{\mathrm{Xd}}{}\left({t}\right)$ (3)
> $\mathrm{PathPlot}\left(X\left(t\right),t=0..3,\mathrm{timesteps}=20,\mathrm{replications}=5,\mathrm{thickness}=2,\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)$
References
Glasserman, P., Monte Carlo Methods in Financial Engineering. New York: Springer-Verlag, 2004.
Compatibility
• The Finance[GammaProcess] command was introduced in Maple 15. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.992520809173584, "perplexity": 3214.2950317500763}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719155.26/warc/CC-MAIN-20161020183839-00301-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://projecteuclid.org/euclid.aoms/1177699260 | ## The Annals of Mathematical Statistics
### Some Concepts of Dependence
E. L. Lehmann
#### Abstract
Problems involving dependent pairs of variables $(X, Y)$ have been studied most intensively in the case of bivariate normal distributions and of $2 \times 2$ tables. This is due primarily to the importance of these cases but perhaps partly also to the fact that they exhibit only a particularly simple form of dependence. (See Examples 9(i) and 10 in Section 7.) Studies involving the general case center mainly around two problems: (i) tests of independence; (ii) definition and estimation of measures of association. In most treatments of these problems, there occurs implicitly a concept which is of importance also in other contexts (for example, the evaluation of the performance of certain multiple decision procedures), the concept of positive (or negative) dependence or association. Tests of independence, for example those based on rank correlation, Kendall's $t$-statistic, or normal scores, are usually not omnibus tests (for a discussion of such tests see [4], [15] and [17], but designed to detect rather specific types of alternatives, namely those for which large values of $Y$ tend to be associated with large values of $X$ and small values of $Y$ with small values of $X$ (positive dependence) or the opposite case of negative dependence in which large values of one variable tend to be associated with small values of the other. Similarly, measures of association are typically designed to measure the degree of this kind of association. The purpose of the present paper is to give three successively stronger definitions of positive dependence, to investigate their consequences, explore the strength of each definition through a number of examples, and to give some statistical applications.
#### Article information
Source
Ann. Math. Statist. Volume 37, Number 5 (1966), 1137-1153.
Dates
First available in Project Euclid: 27 April 2007
https://projecteuclid.org/euclid.aoms/1177699260
Digital Object Identifier
doi:10.1214/aoms/1177699260
Mathematical Reviews number (MathSciNet)
MR202228
Zentralblatt MATH identifier
0146.40601
JSTOR | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8190862536430359, "perplexity": 625.9658266461884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690376.61/warc/CC-MAIN-20170925074036-20170925094036-00643.warc.gz"} |
http://math.stackexchange.com/questions/184982/polynomial-of-degree-infty | # Polynomial of degree $-\infty$?
I'm reading E.J Barbeau Polynomials. I'm in a page where he asks a polynomial of degree $-\infty$. Then I thought about $77x^{-\infty}+1$, but when I went for the answers, the answer to this question was zero.
Then I thought about making $n^{-\infty}$ on Mathematica and it outputed $Indeterminate$ as a result.
I thought the problem was in my understanding of exponantiation, then I tried to "algebrize" it. (I guess that's the name of the procedure)
Then I thought:
$2^3=\overbrace{2\cdot 2\cdot 2}^{\text{3 times}}$
$a^b=\overbrace{a\cdot a\cdot a\cdot ...}^{\text{b times}}$
And in this case:
$a^{-\infty}=\overbrace{a\cdot a\cdot a\cdot ...}^{{-\infty}\text{ times}}$
But this gave me no insight of what could be done to better understand this. I can't see why $n^{-\infty}=0$ so clearly.
With the last example, I'm thinking that there will be no $a$'s to multiply, can you help me?
I thought about some other thing:
$$2^{-8}=\frac{1}{256}=\frac{1}{2^8}$$
Then considering this example, I would get: $$a^{-\infty}=\frac{1}{\infty}=0$$ Right?
-
Well, negative powers are defined as $x^{-k} := \frac{1}{x^k}$. Thus, you can define $$x^{-\infty} := \lim\limits_{k\to+\infty}\frac{1}{x^k}$$ and since for $|x|>1$ the absolute value of $x^k$ grows unboundedly with $k\to\infty$, you obtain that $x^{-\infty} = 0$ whenever $|x|>1$. – Ilya Aug 21 '12 at 10:49
Consider $$\lim_{k\to\infty}\frac1{a^k}$$ – Ｊ. Ｍ. Aug 21 '12 at 10:50
Surely the point is that we (some of us, rather) conventionally define the polynomial $0$ to have degree $-\infty$, so that formulae like $\deg(pq) = \deg(p)+\deg(q)$ continue to hold. – Sean Eberhard Aug 21 '12 at 10:52
@Gustavo: would you please give a hint, are you looking for the reasons for $0$ to be a polynomial of the degree $-\infty$, or your question is different? It is not that clear from your post. – Ilya Aug 21 '12 at 13:15
It sounds like the question in the book was "What polynomial should we define to have degree $-\infty$ ?" As it's not a mathematical question and has a "come up with the same opinion I already have" feel, it's a rather poor question in my opinion. – MartianInvader Aug 21 '12 at 17:24
IMO it comes down to conventions. We say the zero polynomial has degree $-\infty$. Let's see why this is a good convention:
Usually the degree is the highest power with a non-vanishing coefficient. Following this logic it is not really clear what the degree of the zero-polynomial should be. We could just say it has no degree, or we could say it is just a special case of a degree $0$ polynomial (i.e. a constant polynomial), or maybe it's something different?
What properties does the degree have? More specifically what happens if I add or multiply two polynomials $P$ and $Q$ of degree, say, $n$ and $m$?
You can check that the degree of the sum of $P$ and $Q$ will be smaller or equal to the maximum of the degrees of $P$ and $Q$, while the product will have degree $m+n$.
In particular if we multiply any polynomial $P$ with the zero polynomial we want:
$$\deg 0=\deg P\cdot 0=\deg P+ \deg 0$$
To make sense of this equation $\deg 0$ has to be $\pm \infty$ but $+\infty$ doesn't agree with the property for sums. So $-\infty$ remains as the only sensible choice.
-
Yes. I imagined this: $ax^2+bx^1+c^0$ which is a second degree polynomial if at least $a\neq 0$. If $(a=0)$ then it's a first degree polynomial, if $(a=b=0)$ it's a zero degree polynomial, if $(a=b=c=0)$ we have a polynomial of degree $-\infty$. – Vÿska Aug 21 '12 at 23:02
I got one doubt with this: If my suggested polynomial is going to be evaluated, I'm will meet some trouble on the $c\,$ term, since $c^0=1$ (and I know that the $c\,$ term could be $\neq 1$). I guess that $ax^2+bx^1+cd^0$ would be more adequate, isn't it? – Vÿska Aug 21 '12 at 23:37
It's $ax^2+bx^1+cx^0=ax^2+bx+c$ – Simon Markett Aug 22 '12 at 8:20
Turn back to the beginning of the first chapter, a page or so before the problem you're attempting, and you should find Barbeau's definition of degree. It contains the words "a nonzero constant polynomial has degree $0$, but, by convention, the zero polynomial (all coefficients vanishing) has degree $-\infty$." The question then becomes rather easy.
-
Then 0 is a polynomial? – Vÿska Aug 21 '12 at 11:04
@GustavoBandeira Maybe the question is, what is $0$? Is the zero polynomial the same thing as the natural number $0$, as the real number $0$,...? – Simon Markett Aug 21 '12 at 11:07
This is one of the reasons which led me to think about that a polynomial is similar to a number – Vÿska Aug 21 '12 at 11:13
Gustavo: Sorry but a polynomial is not a number (nor a function, by the way), but a sequence $(a_n)_{n\geqslant0}$ with values in a given ring, with the property that the set $\{n\mid a_n\ne0\}$ is finite. And this sequence, when viewed as a polynomial, is often denoted by $\sum\limits_na_nX^n$ by convention. – Did Aug 21 '12 at 12:37
@SimonMarkett set-wise: $\Bbb{N} \subset \Bbb{N}[x].$ – user2468 Aug 21 '12 at 13:35
You can write $n^{-\infty}=(\frac{1}{n})^{\infty}$ and if $|n|>1$ then you will get
$$\lim_{k\rightarrow -\infty}n^{k}=0$$ As $|n|<1$ then $$\lim_{k\rightarrow -\infty}n^{k}=\infty$$ thats why $77x^{-\infty}+1$ is undefined.
-
You might mean $|n|\gt1$ instead of $|n|\lt1$, whatever the relevance of your remark to the question asked. – Did Aug 21 '12 at 13:07
@did: it's not clear from the OP whether it asks about reasons for $0$ to be the polynomial of the degree $-\infty$, or why $n^{-\infty} = \mathsf{Indeterminate}$, or why $n^{-\infty} = 0$ (whatever the value of $n$ is meant in the latter expression). Moreover, I would say that the former meaning seems to be less probable given the text of OP (especially when updated). – Ilya Aug 21 '12 at 13:12
@Seyhmus: you may want to mention that $n^{-\infty} = \infty$ for $n\in (0,1)$ which is one of the reasons, why $77x^{-\infty}$ is undefined. – Ilya Aug 21 '12 at 13:13
@Ilya thanks for the comment. – Seyhmus Güngören Aug 21 '12 at 13:21
In the division algorithm for polynomials you want to divide $f$ by a non-zero polynomial $g$ and get a remainder $r$ of smaller degree than tat of $g$: $f=qg+r$ where $q, r$ are polynomials and $\deg(r)<\deg(g)$. In case $\deg(g)=0$, i.e. $g$ is a non-zero constant then $r=0$, $\deg(r)=\deg(0)=-\infty$ makes this all work out nicely.
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https://www.physicsforums.com/threads/gravity-problems-help.21649/ | # Gravity Problems Help
1. Apr 20, 2004
### Dingo Jellybean
I'm stuck on 2 problems.
T5.11.96) One big question is planetary science is whether each of the rings of Saturn is solid or composed of many smaller satellites. There is a simple observation that can now be made to resolve this issue. Measure the velocity of the inner and outer portion of the ring: if the inner portion of the ring moves more slowly than the outer portion, then the ring is solid; if the opposite is true, then it is composed of many seperate chunks. a) If the thickness of the ring is r, the average distance of the ring from the center of Saturn is R, and the average velocity of the ring is v, show that v(out)-v(in) ~ rv/R if the ring is solid. Here, v(out) is the speed of the outmost portion of the ring, v(in) is the speed of the outermost portion, and v is the average velocity of the ring. b) If, however, the ring is composed of many small chunks, show that v(out)-v(in) ~ -.5(rv)/R. (Assume that r<<R.)
I tried integration, which didn't work. I tried using linear equations of motion to solve, but that got me nowhere either. This problem is a level 3 question, which is the toughest in the book. I've tried...but to no avail. I know that the rings closer to Saturn have a higher velocity...but that didn't help me much either.
T5.11.101.) United Federation Spaceship Excelsior is dropping two robot probes to the surface of a neutron star for exploration. The mass of the star is the same as that of the sun, but the star's diameter is only 10 km. The robot probes are linked together by a 1-m-long steel cord, and are dropped vertically(that is, one always above the other). a) Explain why there seems to be a "force" trying to pull the robots apart. b) How close will the robots be to the surface of the star before the cord breaks? Assume that the cord has a breaking tension of 25000N and that the robots each have a mass of 1kg.
I know this question seems stupid because a neutron star would crush these probes in a matter of seconds. I'd figure since there is so much massin such a small area that the force of gravity would pull down on them more. But I also know that the force of gravity increases with a shorter radius...so wouldn't an object farther out be less affected?
b) For part b, do I just plug in the equation for gravitational force? I know the answer is 220km...but I'm unsure how to get it. I thought it was more like setting the force to 25000N and then solving for "r^2"...but somehow it went wrong. Am I doing something wrong here or maybe it's just calculations error? Thanks.
2. Apr 21, 2004
### HallsofIvy
Staff Emeritus
Actually I thought it was pretty well known that the rings of Saturn are NOT solid! :)
You don't need any fancy calculus for this. The inner edge of the ring is distance R from the center, the outer edge, distance R+r. A point on the outer ring goes around a total distance of 2π(R+r) in one complete revolution while a point on the inner ring only goes 2πR. If the rings were solid, they would take the same time. That is:
VOT= 2π(R+r) and VIT= 2πR. Dividing one equation by the other VO/VI= (R+ r)/R= 1+ r/R.
Number 2 is what is called a "tidal" problem- the question is not just the strength of the gravity but how it changes with distance.
F= -GMm/r2 but there are two different "r"s! Assume the nearer of the two masses is distance r meters from the center of the star. The other would be distance r+ 1 meters. The force on the nearer would be -GMm/r2 and the on the farther -GMm/(r+1)2. What is the difference between the two forces? When will that difference be 25000 N. (You will need to get "-GM" from the fact that the mass of this star is the same as the sun.)
3. Apr 24, 2004
### Dingo Jellybean
I got the first part of problem 1, but the 2nd part of problem 1 is bugging me. How can you actually know what the velocity(in symbolic form) is?
You said for part b of the 2nd problem was like, set 25000 = -GmM/r^2...but I got some gigantic number(about 73000km) instead of 220. I swapped both F and r^2, but no cigar. Thanks for being so helpful though. These 2 problems have been bugging me all week.
4. Apr 25, 2004
### HallsofIvy
Staff Emeritus
I didn't say the forces would be 25000, I said the difference in forces would be 25000!
(By the way, what are you getting for GM?)
5. Apr 25, 2004
### Dingo Jellybean
I got G = 6.67E-11, and M=2E30 kg and just multiplied GM together.
But I think I might get it now...it's the first problem of part b that's bugging me. Thanks. :)
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https://www.studyadda.com/sample-papers/neet-sample-test-paper-21_q22/210/276985 | • # question_answer A spherical hollow cavity is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through the center. What is the shift in the center of mass of the lead sphere due to hollowing? A) $\frac{R}{7}$ B) $\frac{R}{14}$ C) $\frac{R}{2}$ D) $R$
Let $\rho$by the density of lead. Then M$M=\frac{4}{3}\pi {{r}^{3}}p=$mass of total sphere ${{m}_{1}}=\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}\rho =$ Mass of removed part $=\frac{M}{8}$ ${{m}_{2}}=M-\frac{M}{8}=\frac{7M}{8}=$Mass of remaining sphere choosing the centre of the sphere as the origin. ${{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$or, $o=\frac{\frac{M}{8}\times \frac{R}{2}+\frac{7M}{8}{{x}_{2}}}{M}$or ${{x}_{2}}=-\frac{R}{14}\,\,\therefore$Shift in centre of mass$=R/14.$ Hence, the correction option is (b). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9500346183776855, "perplexity": 2002.9493119327974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487600396.21/warc/CC-MAIN-20210613041713-20210613071713-00128.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-1-introduction-to-algebraic-expressions-1-4-positive-and-negative-real-numbers-1-4-exercise-set-page-35/36 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
We can write a fraction as a decimal by dividing the numerator by the denominator. $\frac{-1}{9}=-1\div9=-.11111.....$ In this case, the 1 repeats after the tenths digit, so we can write $\frac{-1}{9}$ as -.11. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794304966926575, "perplexity": 688.2206238257069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00243.warc.gz"} |
https://www.physicsforums.com/threads/algebra-problem.153094/ | # Algebra Problem
1. Jan 25, 2007
### complexPHILOSOPHY
I am working through this algebra book and some of the problems. The chapter this comes out of is General Algebraic Systems and the section is Isomorphisms. I am new to proofs and maths higher than calculus I so I am not sure if I am following the text or not. There aren't any solutions and this book is really old, out of print and not on the internet (temporary until I can purchase one of the texts already suggested to me).
Here is the problem:
Prove that the mapping $$a+bi\rightarrow a-bi$$ of each complex number onto its conjugate is an automorphism of the additive group of C.
$$(a+bi)\alpha=a-bi$$
$$(c+di)\alpha=c-di$$
$$[(a+bi)+(c+di)]\alpha=$$
$$[(a+c)+(b+d)i]\alpha=$$
$$(a+c)-(b+d)i=$$
$$(a-bi)+(c-di)=$$
$$(a+bi)\alpha+(c+di)\alpha$$
Since $$a-bi=c-di$$
Then $$(a-c)+(d-b)i=0$$
$$a=c$$ and $$b=d$$
$$a+bi=c+di$$
This is one-to-one right?
I think this is correct but I am not sure if it is even complete or if I am on the right track. If someone could help me, that would be great.
Last edited: Jan 25, 2007
2. Jan 25, 2007
### Kreizhn
Everything looks fine.
You've shown that the mapping preserves the binary operation under each respective group.
You've shown injectivity with
$$a=c$$ and $$b=d$$
Technically, you only need to show surjectivity, though it's a trivial exercise.
Other than that, everything looks perfectly fine.
3. Jan 26, 2007
### complexPHILOSOPHY
An automorphism is an isomorphism from a set of elements onto itself, right? So, by demonstrating surjection as well, would I be illustrating a bijection? Do I need to show surjection for proof or were you suggesting it to help improve my understanding?
I appreciate your response. Also, since my book seems to be the only copy that was printed on the earth, I don't know how difficult the problems are considered. This problem seemed almost too simple (considering I have no idea what I am doing) so I am wondering if my book is very easy? Does this problem look pretty average or would you suggest more difficult problems to get a better intuitive feel for groups? I am a complete algebra newbie. This is my third day reading through the text, so bear with me.
Also, one further question. How long does the average person spend on each chapter when working through maths? I am completely self-taught all the way up to Calculus 1 and I have no experience learning math from anyone except books, so I am not sure what the pace of average students is compared to me.
I am curious because I will be taking Calculus II this semester and I am wondering if I will be slower than them.
Last edited: Jan 26, 2007
4. Jan 26, 2007
### Kreizhn
Well, from what I recall, the definition for a group isomorphism is
Definition The map $$\phi : G \rightarrow H$$ is an isomorphism if
1) $$\phi$$ is a homomorphism
2) $$\phi$$ is a bijection
where the group homomorphism preserves the binary operation mapped from G to H, and a bijective function is both injective and surjective. With that in mind, it's necessary, although relatively trivial to show that the complex conjugate is a one-to-one correspondence (i.e. Injective and Surjective).
It isn't usually until somewhat higher level courses that one is introduced to this kind of algebra (except possibly in an advanced classical algebra class), and so it's outstanding that you'd be learning this on your own.
As for this question in particular, it does indeed seem fairly rudimentary, but the difficulty of question you might want depends greatly on your background. Have you been introduced to the formal idea of proofs for abstract algebra? Is that the kind of thing you'd be interested in?
It's also somewhat hard to gauge what exactly can be meant by a chapter, and I suppose that also depends on the text that you're using. I too learned Calculus I and II via the text book, and I found that doing a "chapter" daily was very sufficient. However, note that a university student doing Calc II typically only has this lecture 3 times a week (at least where I come from) and so a chapter daily would actually be an advanced pace.
I would imagine that this also depends on the amount of theoretical material presented in the course. When you studied Calc I were you introduced to the $$\epsilon\delta$$ notion of continuity? Cauchy Sequences? The notion of compact sets? Banach Spaces? Lipschitz, Topological, or metric definitions of continuity? I don't believe these are taught outside of advanced Calculus classes, but if this is the kind of thing you are learning, it wouldn't hurt to take a bit more time to ensure that you fully understand the material.
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http://motls.blogspot.com/2011/10/any-use-for-f4-in-hep-th.html | Saturday, October 08, 2011 ... /////
Any use for $F_4$ in hep-th?
Yuji asked: In high energy physics, the use of the classical Lie groups are common place, and in the Grand Unification the use of $E_{6,7,8}$ is also common place.
In string theory $G_2$ is sometimes utilized, e.g. the $G_2$-holonomy manifolds are used to get 4d $\mathcal{N}=1$ SUSY from M-theory.
That leaves $F_4$ from the list of simple Lie groups. Is there any place $F_4$ is used in any essential way?
Of course there are papers where the dynamics of $d=4$ $\mathcal{N}=1$ susy gauge theory with $F_4$ are studied, as part of the study of all possible gauge groups, but I'm not asking those.
LM answered: $F_4$ is the centralizer of $G_2$ inside an $E_8$. In other words, $E_8$ contains an $F_4\times G_2$ maximal subgroup; the decomposition of the fundamental (=adjoint) representation under the group is simple:
${\bf 248} = ({\bf 52},{\bf 1}) \oplus ({\bf 1},{\bf 14}) \oplus ({\bf 26},{\bf 7})$
It's a direct sum of the two subgroups' adjoint representations and the tensor product of their fundamental representations. A reason I wrote it is to remind you of the 26-dimensional fundamental and 52-dimensional adjoint representation of $F_4$. Incidentally, one also sees that the dimension of the adjoint representation of $G_2$, 14, is also twice the dimension of its fundamental representation, 7.
The existence of this maximal subgroup is why by embedding the spin connection into the $E_8\times E_8$ heterotic gauge connection on $G_2$ holonomy manifolds, one obtains an $F_4$ gauge symmetry. See, for example,
http://arxiv.org/abs/hep-th/0108219
Gauge theories and string theory with $F_4$ gauge groups, e.g. in this paper
http://arxiv.org/abs/hep-th/9902186
depend on the fact that $F_4$ may be obtained from $E_6$ by a projection related to the nontrivial ${\mathbb Z}_2$ automorphism of $E_6$ which you may see as the left-right symmetry of the $E_6$ Dynkin diagram.
$E_6$ has six simple roots (nodes). If you identify two pairs of nodes by the ${\mathbb Z}_2$ symmetry, you obtain two new nodes which are $\sqrt{2}$-times shorter (from the pairs), aside from the two original "unpaired" nodes on the $E_6$ axis. In other words, you obtain the $F_4$ Dynkin diagram. The $G_2$ Dynkin diagram may similarly be obtained from the $SO(8)$ Dynkin diagram by the $S_3$-identification based on triality of $SO(8)$, an unusually large outer automorphism group.
This automorphism may be realized as a nontrivial monodromy which may break the initial $E_6$ gauge group to an $F_4$ as in
http://arxiv.org/abs/hep-th/9611119
Because of similar constructions, gauge groups including $F_4$ factors (sometimes many of them) are common in F-theory:
http://arxiv.org/abs/hep-th/9701129
More speculatively (and outside established string theory), a decade ago, Pierre Ramond had a dream
http://arxiv.org/abs/hep-th/0112261
http://arxiv.org/abs/hep-th/0301050
that the 16-dimensional Cayley plane, the $F_4/SO(9)$ coset (note that $F_4$ may be built from $SO(9)$ by adding a 16-spinor of generators, analogously to $E_8$ which may be built by adding a 128-dimensional chiral real spinor to the adjoint of $SO(16)$; the only other – third – example of the same thing is getting $SO(9)$ from $SO(8)$ by adding a spinor, an operation that differs from the normal "vectorial" extension of $SO(8)$ to $SO(9)$ by triality), may be used to define all of M-theory.
As far as I can say, it hasn't quite worked but it is interesting. Sati and others recently conjectured that M-theory may be formulated as having a secret $F_4/SO(9)$ fiber at each point:
TRF: Is M-theory hiding Cayley plane fibers?
There's one funny episode from the history of maths concerning $F_4$ in the discussions of holonomy. Aside from 7-dimensional $G_2$-holonomy manifolds and 8-dimensional $spin(7)$-holonomy manifolds, Berger originally conjectured that there were also 16-dimensional $spin(9)$-holonomy manifolds. But they were later proved to be trivial – either locally flat or locally isomorphic to the Cayley plane, $F_4/spin(9)$. So $F_4$ was the only new, non-trivial external "killer" that made people stop talking about the $spin(9)$ holonomy. ;-)
Less speculatively, the noncompact version $F_{4(4)}$ of the $F_4$ exceptional group is also the isometry of a quaternion manifold relevant for the maximal ${\mathcal N}=2$ matter-Einstein supergravity, see
http://arxiv.org/abs/hep-th/9708025
In that paper, you may also find cosets of the $E_6/F_4$ type and some role is also being played by the fact that $F_4$ is the symmetry group of a $3\times 3$ matrix Jordan algebra of octonions.
Recall that while $G_2$ is the automorphism group of the octonions ${\mathbb O}$, the group $F_4$ may similarly be defined as the automorphism group of the algebra $J=A_3({\mathbb O})$ of $3\times 3$ "Hermitian" octonion matrices (Hermitian conjugation involves both transposition and the sign flip for all 7 imaginary units) with the bilinear operation
$\hat A \diamond \hat B \equiv \hat A\cdot \hat B +\hat B \cdot \hat A$
Note that the matrices in the algebra have 3 real numbers on the diagonal and 3 inequivalent octonions above the diagonal (copied below it, with the complex conjugation), making the total number of real components $3+3\times 8=27$. However, the identity matrix obviously has to be mapped onto itself under automorphisms; the remaining 26 components form the (previously advertised) fundamental representation of $F_4$.
If you have a feeling that $F_4$ only arises in applications that are related to string theory or at least supergravity, you're right: that's where you encounter exceptional mathematical structures in physics. If there were no string theory and not even supergravity, physicists wouldn't be forced to learn exceptional groups.
Well, the $E_6$ grand unified theories are the only exception but they would still probably remain ignorant about the other four. $E_6$ is unique among the exceptional groups when it comes to grand unified theory model building because it's the only one that has complex representations, representations inequivalent to their complex conjugates (which is needed to get chiral fermions and P- and CP-violation). This fact is actually equivalent to the existence of a symmetry of the $E_6$ Dynkin diagram – and this symmetry of the diagram, as a white picture above showed, may be "orbifolded by" to get nothing else than the Dynkin diagram of $F_4$, the main hero of this blog entry.
As you can see, the relationships between the exceptional groups and similar structures are diverse, rich, and sometimes unexpected. To describe some basic features of $F_4$, I needed an $E_8$ as well as $G_2$, the centralizer of $F_4$ in $E_8$, as well as $E_6$ from which $F_4$ could have been built by the orbifolding of the Dynkin diagram.
Only $E_7$ hasn't been mentioned as an important partner of $F_4$ so far. Let me just assure you it was a random omission: $E_7$ is exactly what you get by the Kantor-Koecher-Tits construction applied to the very algebra $J=A_3({\mathbb O})$ whose automorphism group was said to be $F_4$ above.
Many of these relationships have a physical interpretation we have already found; the other interesting relationships have a physical/stringy explanation, interpretation, or visualization that we will find in the future. ;-) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8361371755599976, "perplexity": 415.37148806048066}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701145751.1/warc/CC-MAIN-20160205193905-00330-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Simple_Harmonic_Motion | # A-level Physics (Advancing Physics)/Simple Harmonic Motion
Simple harmonic motion occurs when the force on an object is proportional and in the opposite direction to the displacement of the object. Examples include masses on springs and pendula, which 'bounce' back and forth repeatedly. Mathematically, this can be written:
${\displaystyle F=-kx}$,
Graph of displacement against time in simple harmonic motion.
where F is force, x is displacement, and k is a positive constant. This is exactly the same as Hooke's Law, which states that the force F on an object at the end of a spring equals -kx, where k is the spring constant. Since F = ma, and acceleration is the second derivative of displacement with respect to time t:
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}=-kx}$
${\displaystyle {\frac {d^{2}x}{dt^{2}}}={\frac {-kx}{m}}}$
The solution of this second order differential equation is:
${\displaystyle x=A\cos {\omega t}}$,
where A is the maximum displacement, and ω is the 'angular velocity' of the object. The derivation is given here, since it will seem very scary to those who haven't met complex numbers before. It should be noted that this solution, if given different starting conditions, becomes:
${\displaystyle x=A\sin {\omega t}}$,
## Angular Velocity
Angular velocity in circular motion is the rate of change of angle. It is measured in radians per. second. Since 2π radians is equivalent to one complete rotation in time period T:
${\displaystyle \omega ={\frac {2\pi }{T}}=2\pi f}$
If we substitute this into the equation for displacement in simple harmonic motion:
${\displaystyle x=A\cos {2\pi ft}}$
The reason the equation includes angular velocity is that simple harmonic motion is very similar to circular motion. If you look at an object going round in a circle side-on, it looks exactly like simple harmonic motion. We have already noted that a mass on a spring undergoes simple harmonic motion. The following diagram shows the similarity between circular motion and simple harmonic motion:
## Time Period
The time period of an oscillation is the time taken to repeat the pattern of motion once. In general:
${\displaystyle T={\frac {2\pi }{\omega }}}$
However, depending on the type of oscillation, the value of ω changes. For a mass on a spring:
${\displaystyle \omega ={\sqrt {\frac {k}{m}}}}$
For a pendulum:
${\displaystyle \omega ={\sqrt {\frac {g}{l}}}}$,
where g is the gravitational field strength, and l is the length of the string. By substitution, we may gain the following table:
Type of Oscillation Spring Pendulum
Angular Velocity ${\displaystyle {\sqrt {\frac {k}{m}}}}$ ${\displaystyle {\sqrt {\frac {g}{l}}}}$
Time Period ${\displaystyle 2\pi {\sqrt {\frac {m}{k}}}}$ ${\displaystyle 2\pi {\sqrt {\frac {l}{g}}}}$
## Velocity and Acceleration
The displacement of a simple harmonic oscillator is:
${\displaystyle x=A\cos {\omega t}}$
Velocity is the rate of change of displacement, so:
${\displaystyle v={\frac {dx}{dt}}=-A\omega \sin {\omega t}}$
Acceleration is the rate of change of velocity, so:
${\displaystyle a={\frac {dv}{dt}}={\frac {d^{2}x}{dt^{2}}}=-A\omega ^{2}cos{\omega t}=-\omega ^{2}x}$
## Questions
1. A 10N weight extends a spring by 5cm. Another 10N weight is added, and the spring extends another 5cm. What is the spring constant of the spring?
2. The spring is taken into outer space, and is stretched 10cm with the two weights attached. What is the time period of its oscillation?
3. What force is acting on the spring after 1 second? In what direction?
4. A pendulum oscillates with a frequency of 0.5 Hz. What is the length of the pendulum?
5. The following graph shows the displacement of a simple harmonic oscillator. Draw graphs of its velocity, momentum, acceleration and the force acting on it.
6. A pendulum can only be modelled as a simple harmonic oscillator if the angle over which it oscillates is small. Why is this?
Worked Solutions | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 17, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.945282518863678, "perplexity": 419.20464673538817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719566.74/warc/CC-MAIN-20161020183839-00554-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://cstheory.stackexchange.com/questions/16344/how-can-i-find-all-numbers-for-which-the-xor-sum-is-0 | # How can I find all numbers for which the XOR-sum is 0?
Given a list of integers $[a_1, a_2, \dots a_n]$, I want to find the number of $n$-tuples $(x_1,\dots,x_n)$ of integers such that the following three conditions are satisfied:
1. $x_1 \oplus x_2 \oplus \dots \oplus x_n = 0$.
2. $x_k \in [0,a_k]$ for each $k$.
3. $x_k=a_k$ for at least one $k$.
For instance, if the given list is [1,2,3], the answer is 4, and the specific solutions are
• [0,2,2],
• [1,0,1],
• [1,1,0], and
• [1,2,3].
A brute-force approach would just compute the Cartesian product of all $x_k$s or simply count through the whole number space starting with [0,0,0]. But how can I tackle this problem with a more sophisticated algorithm? What paradigm is a good choice? Dynamic programming? If I only consider condition 2, then I could think of denoting the sub-problem by
$$E[n,i]$$
Where $n$ defines the number of integers and $i$ the current slot. Also given the caps $[m_1, m_2, \dots,m_n]$ then the number of combinations could be computed with:
\begin{align*} E[0,0] & = 0 \\ E[0,i] & = 1 + m_i \\ E[n,i] & = (1 + m_i) \cdot E[n, i-1] \\ E[n,i] & = \text{invalid if } i + 1 > n \end{align*}
But how can I integrate the XOR constraint here?
On the other hand it seems this is a classical Constraint Satisfaction Problem which could be backtracked.
-
I wonder if [0, 0, 0] and [0, 1, 1] are also solutions of your example. – Yoshio Okamoto Feb 3 '13 at 12:37
@YoshioOkamoto No, sorry for making a false impression. But look at the last paragraph starting with Another constraint. [0, 0, 0] and [0, 1, 1], both, do not have one number slot, which was kept untouched. There is no $x_k$ in it, which sill has the original value $a_k$. – Mahoni Feb 3 '13 at 12:40
@Mahoni: Following up on Yoshio's comment: why is $[2, 2, 0]$ not a solution? Similarly, why is $[3, 0, 3]$ not a solution? – Rachit Feb 3 '13 at 12:52
This is easy to solve by dynamic programming in $O(n A^2)$ time, where $A = \max_i a_i$. Let $F(j, t)$ denote the number of tuples $(x_1, x_2, \dots, x_j)$ where (1) $0\le x_i \le a_i$ for every index $i\le j$, (2) $x_i = a_i$ for some index $i\le j$, and (3) $x_1\oplus x_2\oplus\cdots\oplus x_j = t$. There are $O(nA)$ subproblems, each of which can be evaluated in $O(A)$ time. – JɛffE Feb 3 '13 at 21:08
Can you solve the decision problem in (strongly) polynomial time? – JɛffE Feb 3 '13 at 21:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794296622276306, "perplexity": 456.8334878269586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988922.24/warc/CC-MAIN-20150728002308-00046-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://www.contrib.andrew.cmu.edu/~ryanod/?tag=derandomization | ## §6.4: Applications in learning and testing
In this section we describe some applications of our study of pseudorandomness.
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https://dsp.stackexchange.com/questions/61391/where-should-i-set-my-anti-aliasing-filter-corner-frequency-for-this-signal | # Where should I set my anti-aliasing filter corner frequency for this signal?
I have a Fourier transform of an arbitrary digital signal. The transform has For 1024 frequency bins. The magnitude plot of the first 100 is below, with no surprises in the unplotted 924 bins.
Now, suppose that I want to design a single-pole anti-aliasing filter for this signal. How would one approach selecting a corner frequency for that filter?
• The anti-alias filter really doesn't have anything to do with the signal you are showing (the FFT signal as shown could have any shape but would not drive the answer): The anti-alias filter design is dependent on your analog spectrum, the sampling rate, and the Nyquist zone that you want to capture digitally. Keep in mind the purpose of the anti-alias filter is to reject signals in other Nyquist zones that would otherwise fold into the one of interest (folding/aliasing due to the sampling process). – Dan Boschen Oct 21 '19 at 1:42
• Thanks for being patient, I'm an EE, but my formal DSP is a little weak. I think I know most of those words and understand the order they're in! I think your point is that, since this is an FFT of a digital signal without an anti-aliasing filter, we can't use this to infer the frequency content of the original signal. Entirely true! But I think we can assume it to be comparable, for the purposes of this question. – Stephen Collings Oct 21 '19 at 2:28
• Yes and the concern with the design of the anti-alias signal is BEFORE you sample it, so you would want to be showing a plot of the analog spectrum and indicating what part of that spectrum you want to maintain digitally (usually but not always it would be the lowest frequency portion of that spectrum, but there are applications where we "bandpass" sample as well so my wording was more generic). – Dan Boschen Oct 21 '19 at 12:11
• But also digitally you may be interested in "re-sampling" such as changing to a lower sampling rate, and then you would also have concern with the design of the anti-alias filter. In either case (from the analog to digital or from the digital to digital at another rate) you would need to specify what the target sampling rate is – Dan Boschen Oct 21 '19 at 12:13
Suppose if your sampling rate is fs, then the corner frequency of the Anti Aliasing filter would be fs/2
• Hi Yeshwanth- thanks for providing an answer. Consider that anti-alias filters cannot be "brick-wall" filters, so if the filter had a corner frequency as fs/2, this is the frequency where the filter starts to attenuate higher frequency signals that can fold in. One must define the frequency band they are interested in, and how much attenuation they want from the alias frequencies in order to design the filter. See my answer for more details on that. – Dan Boschen Oct 21 '19 at 12:15
The purpose of an anti-alias filter is to remove signal and noise from frequency bands that would otherwise "fold" or alias into the band of interest during the sampling process. This can occur when going from an analog signal to digital (and thus would be the filter just prior to the A/D converter), or when resampling a digital signal such as done in multi-rate digital signal processing where a sampling rate is changed completely in the digital domain.
That said, the design of the anti-alias filter depends on the signal bandwidth of interest and how much of the Nyquist zone in occupies (see further description below on Nyquist zone). In order to realized a practical filter design (a brick-wall filter cannot be realized), the bandwidth of the signal of interest must always be sufficiently less than the Nyquist zone. A rejection requirement is established knowing the worst case interference (noise and other signals) that exist in the other Nyquist bands; note that all the spectrum that passes through to the A/D Converter in other Nyquist bands will alias multiple times on top of the digital spectrum. Ultimately the A/D itself will have analog input bandwidth acting as a low pass filter to keep the otherwise infinite folding from occurring. An initial misconception many make when first approaching this subject is to dictate the rejection based on now allowing any aliasing into the digital spectrum and thus requiring full rejection right at the Nyquist boundary. This often over-constrains the filter design, as signals just beyond the Nyquist boundary will indeed alias in, but they will not land within the signal bandwidth of interest and therefore can be further filtered out digitally later downstream where such filtering can be done much more effectively than in the analog domain. This is clearer in the plot below that shows that the Nyquist boundary is placed mid-way in the transition bandwidth of the filter; full rejection is applied at the point where an interfering signal if present would alias into the signal bandwidth of interest. The plot below shows the filter as it would appear in the digital spectrum: if this was a signal from the first Nyquist zone in the analog domain then this would be an analog low pass filter centered on DC (f=0). However this could also be a bandpass filter centered on any arbitrary frequency at a higher Nyquist zone, which also shows how good frequency planning (of signal of interest and sampling rate) is to place the signal, when real, in the center of a Nyquist zone, thus optimizing the filter realization by maximizing the allowable transition band.
This approach is also done in digital resampling applications; for example, a digital anti-alias filter must precede a decimator (throwing away samples to reduce the sampling rate). Multi=band filters are often used in this application, again applying rejection only at the frequencies that would fold into the signal bandwidth of interest and ignoring the rest to be further filtered downstream at the lower sampling rate where such filtering can be done more efficiently.
The bottom three diagrams help illustrate the aliasing mechanism and the idea of "Nyquist Zones". Once the aliasing mechanism is understood, then the design considerations for the anti-alias filter are clearer for signals of interest in any Nyquist zone.
All three plots show the frequency spectrum of the analog signal of interest on the top line, the digital sampling spectrum on the middle line (sampling is multiplying by a stream of impulses in the time domain, and the Fourier Transform of a stream of impulses in the time domain is a stream of impulses in the frequency domain, spaced by the sampling rate). The bottom line is the convolution of the two, resulting in the digital spectrum. The digital spectrum is unique from $$-f_s/2$$ to $$+f_s/2$$ where $$f_s$$ is the sampling rate, thus it is typically only shown over that domain (I find it helpful in many applications to extend it to $$+/-\infty$$ as I do below, showing the repetition over the $$f_s$$ interval.
The first plot show the case where the signal of interest is in the "first Nyquist zone" which is the most common form of sampling an analog signal-- the analog signal is entirely below $$+/-f_s/2$$ and therefore the frequency of the signal in the analog domain maps directly to the frequency of the signal in the digital domain.
Note that I will explain this all with complex signals: $$cos(\omega t)$$ consists of a positive and negative frequency consisting of $$\frac{1}{2}(e^{j\omega t}+e^{-j\omega t})$$. Once you see the cosine wave as two impulses in frequency each being of the form of \$e^{j\omega t}, the rest becomes a lot more intuitive and covers all cases.
As a analog signal begins to cross the Nyquist boundary, it's content will roll around (alias) to the other boundary. For example if a signal is sampled at 10 Hz, the Nyquist boundary is +/-5 Hz. If the analog signal was $$e^{j4\pi t}$$ which is +4 Hz, it will be at +4 Hz in the digital domain (and at ...-16 Hz, -6 Hz, +14 Hz, +24 Hz,... etc). If this signal frequency was increased to +4.5 Hz it would digitally be at (...-15.5, -5.5, 4.5, 14.5....). Notice the signal approaching the -5 Hz boundary and what occurs when we increase the signal to +5.5 Hz: (...-14.5, -4.5, 5.5, 10.5....) Aliasing!
This is what causes a "spectral inversion" for signals sampled in the second Nyquist Zone as shown in the plot below:
As the frequency of the signal extends higher into an odd Nyquist Zone, the same digital spectrum is produced!
Thus we see how all higher frequencies will fold or alias into the digital spectrum. Therefore we must pre-filter the analog spectrum PRIOR to sampling to reject any signals (and noise) that will exist in these unwanted bands so that only our band of interest maps to the digital spectrum. (Note this could be the first Nyquist zone as in traditional sampling where we say the sampling rate must be greater than twice the highest frequency of interest, or undersampling applications where the anti-alias filter is a bandpass filter about a signal of interest in a higher Nyquist zone. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8022019863128662, "perplexity": 821.6504088468098}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00418.warc.gz"} |
http://ibmaths.gr/ib-%CE%BC%CE%B1%CE%B8%CE%B7%CE%BC%CE%B1%CF%84%CE%B9%CE%BA%CE%AC-sl-differentiation/ | # IB μαθηματικά SL – Differentiation
##### Differentiation, Derivatives, gradient, tangent – IB Mathematics SL
Find the equation of the tangent to the curve with equation
$$f(x)=2x^3+5x^2-20x+1$$ at the point $$(0,1)$$
Solution
$$f(x)=2x^3+5x^2-20x+1 \Rightarrow$$ $$f’(x)=6x^2+10x-20$$
The gradient of the curve at $$(0,1)$$ is
$$f’(0)=6(0)^2+10(0)-20=-20$$
The equation of the tangent on this point is given by the equation
$$y-y_{0}=m_{t}(x-x_{0})$$
where $$m_{t}=-20$$ is the gradient of the tangent
and the touch point is $$(x_{0},y_{0})=(0,1)$$.
So, the equation of the tangent at the given point is
$$y-1=-20(x-0) \Rightarrow y=-20x+1$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8936514854431152, "perplexity": 398.933681216742}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00568.warc.gz"} |
https://std.iec.ch/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/79a4c16d37619fd0c12584cd0049cf4e?OpenDocument | IEVref: 113-05-16 ID: Language: en Status: Standard Term: elementary electric charge Synonym1: elementary charge [Preferred] Synonym2: Synonym3: Symbol: e Definition: fundamental physical constant equal to the quantum of electric charge, the value of which is set to exactly 1,602 176 634 × 10−19 C Note 1 to entry: The elementary electric charge is equal to the charge of the proton and opposite to the charge of the electron. Note 2 to entry: The elementary electric charge is used in the SI to define the ampere. Publication date: 2020-12 Source: SI Brochure, 9th edition, 2019, 2.2, modified – The full form of the term, including the adjective "electric", has been specified as the preferred term to differentiate clearly this quantity from other charges such as "color charge". The term specified in the SI Brochure has been given as a synonym. The definition in the SI Brochure has been revised to comply with the IEV rules. Replaces: 113-05-16:2011-04 Internal notes: CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9861701130867004, "perplexity": 2099.3823825533045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00134.warc.gz"} |
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Contributions to High–Dimensional Analysis under Kolmogorov Condition
2015 (English)Doctoral thesis, comprehensive summary (Other academic)
##### Abstract [en]
This thesis is about high–dimensional problems considered under the so{called Kolmogorov condition. Hence, we consider research questions related to random matrices with p rows (corresponding to the parameters) and n columns (corresponding to the sample size), where p > n, assuming that the ratio $\small\frac{p}{n}$ converges when the number of parameters and the sample size increase.
We focus on the eigenvalue distribution of the considered matrices, since it is a well–known information–carrying object. The spectral distribution with compact support is fully characterized by its moments, i.e., by the normalized expectation of the trace of powers of the matrices. Moreover, such an expectation can be seen as a free moment in the non–commutative space of random matrices of size p x p equipped with the functional $\small \frac{1}{p}E[Tr\{\cdot\}]$. Here, the connections with free probability theory arise. In the relation to that eld we investigate the closed form of the asymptotic spectral distribution for the sum of the quadratic forms. Moreover, we put a free cumulant–moment relation formula that is based on the summation over partitions of the number. This formula is an alternative to the free cumulant{moment relation given through non{crossing partitions ofthe set.
Furthermore, we investigate the normalized $\small E[\prod_{i=1}^k Tr\{W^{m_i}\}]$ and derive, using the dierentiation with respect to some symmetric matrix, a recursive formula for that expectation. That allows us to re–establish moments of the Marcenko–Pastur distribution, and hence the recursive relation for the Catalan numbers.
In this thesis we also prove that the $\small \prod_{i=1}^k Tr\{W^{m_i}\}$, where $\small W\sim\mathcal{W}_p(I_p,n)$, is a consistent estimator of the $\small E[\prod_{i=1}^k Tr\{W^{m_i}\}]$. We consider
$\small Y_t=\sqrt{np}\big(\frac{1}{p}Tr\big\{\big(\frac{1}{n}W\big)^t\big\}-m^{(t)}_1 (n,p)\big),$,
where $\small m^{(t)}_1 (n,p)=E\big[\frac{1}{p}Tr\big\{\big(\frac{1}{n}W\big)^t\big\}\big]$, which is proven to be normally distributed. Moreover, we propose, based on these random variables, a test for the identity of the covariance matrix using a goodness{of{t approach. The test performs very well regarding the power of the test compared to some presented alternatives for both the high–dimensional data (p > n) and the multivariate data (p ≤ n).
##### Place, publisher, year, edition, pages
Linköping: Linköping University Electronic Press, 2015. , p. 61
##### Series
Linköping Studies in Science and Technology. Dissertations, ISSN 0345-7524 ; 1724
##### Keyword [en]
Eigenvalue distribution, free moments, free Poisson law, Marchenko-Pastur law, random matrices, spectral distribution, Wishart matrix
Mathematics
##### Research subject
Natural Science, Mathematics
##### Identifiers
ISBN: 978-91-7685-899-8 (print)OAI: oai:DiVA.org:lnu-58164DiVA, id: diva2:1047433
##### Public defence
2015-12-11, Visionen, ingång 27, B-huset, 13:15 (English)
##### Supervisors
Available from: 2016-11-18 Created: 2016-11-17 Last updated: 2016-11-21Bibliographically approved
##### List of papers
1. Closed Form of the Asymptotic Spectral Distribution of Random Matrices Using Free Independence
Open this publication in new window or tab >>Closed Form of the Asymptotic Spectral Distribution of Random Matrices Using Free Independence
2015 (English)Report (Other academic)
##### Abstract [en]
The spectral distribution function of random matrices is an information-carrying object widely studied within Random matrix theory. Random matrix theory is the main eld placing its research interest in the diverse properties of matrices, with a particular emphasis placed on eigenvalue distribution. The aim of this article is to point out some classes of matrices, which have closed form expressions for the asymptotic spectral distribution function. We consider matrices, later denoted by $\mathcal{Q}$, which can be decomposed into the sum of asymptotically free independent summands.
Let $(\Omega,\mathcal{F},P)$ be a probability space. We consider the particular example of a non-commutative space$(RM_p(\mathbb{C}),\tau)$, where $RM_p(\mathbb{C})$ denotes the set of all $p \times p$ random matrices, with entries which are com-plex random variables with finite moments of any order and $\tau$ is tracial functional. In particular, explicit calculations are performed in order to generalize the theorem given in [15] and illustrate the use of asymptotic free independence to obtain the asymptotic spectral distribution for a particular form of matrix$Q\in\mathcal{Q}$.
Finally, the main result is a new theorem pointing out classes of the matrix $Q$ which leads to a closed formula for the asymptotic spectral distribution. Formulation of results for matrices with inverse Stieltjes transforms, with respect to the composition, given by a ratio of 1st and 2nd degree polynomials, is provided.
##### Place, publisher, year, edition, pages
Linköping: Linköping University Electronic Press, 2015. p. 25
##### Series
LiTH-MAT-R, ISSN 0348-2960 ; 2015:12
##### Keyword
Closed form solutions, Free probability, Spectral distribution, Asymptotic, Random matrices, Free independence
Mathematics
##### Identifiers
urn:nbn:se:lnu:diva-58165 (URN)
Available from: 2016-11-17 Created: 2016-11-17 Last updated: 2018-04-20Bibliographically approved
2. Cumulant-moment relation in free probability theory
Open this publication in new window or tab >>Cumulant-moment relation in free probability theory
2014 (English)In: Acta et Commentationes Universitatis Tartuensis de Mathematica, ISSN 1406-2283, E-ISSN 2228-4699, Vol. 18, no 2, p. 265-278Article in journal (Refereed) Published
##### Abstract [en]
The goal of this paper is to present and prove a cumulant-moment recurrent relation formula in free probability theory. It is convenient tool to determine underlying compactly supported distribution function. The existing recurrent relations between these objects require the combinatorial understanding of the idea of non-crossing partitions, which has been considered by Speicher and Nica. Furthermore, some formulations are given with additional use of the Möbius function. The recursive result derived in this paper does not require introducing any of those concepts. Similarly like the non-recursive formulation of Mottelson our formula demands only summing over partitions of the set. The proof of non-recurrent result is given with use of Lagrange inversion formula, while in our proof the calculations of the Stieltjes transform of the underlying measure are essential.
##### Place, publisher, year, edition, pages
Tartu University Press, 2014
##### Keyword
R-transform, Free cumulants, Moments, Free probability, Non-commutative probability space, Stieltjes transform, Random matrices
##### National Category
Probability Theory and Statistics Other Mathematics
##### Research subject
Natural Science, Mathematics
##### Identifiers
urn:nbn:se:lnu:diva-58167 (URN)10.12697/ACUTM.2014.18.22 (DOI)
Available from: 2016-11-17 Created: 2016-11-17 Last updated: 2018-04-20Bibliographically approved
3. On E\big[\prod_{i=0}^k Tr\{W^{m_i}\} \big], where $W\sim\mathcal{W}_p(I,n) Open this publication in new window or tab >>On E\big[\prod_{i=0}^k Tr\{W^{m_i}\} \big], where$W\sim\mathcal{W}_p(I,n)
2017 (English)In: Communications in Statistics - Theory and Methods, ISSN 0361-0926, E-ISSN 1532-415X, Vol. 46, no 6, p. 2990-3005Article in journal (Refereed) Published
##### Abstract [en]
In this paper, we give a general recursive formula for $\small E[\prod_{i=0}^k Tr\{W^{m_i}\}]$, where $\small W \sim \mathscr{W}_p(I,n)$ denotes a real Wishart matrix. Formulas for fixed n, p are presented as well as asymptotic versions when $\small \frac{n}{p}\overset{n,p\rightarrow\infty}{\rightarrow}c$i.e. when the so called Kolmogorov condition holds. Finally, we show application of the asymptotic moment relation when deriving moments for the Marchenko-Pastur distribution (free Poisson law). A numerical illustration using implementation of the main result is also performed.
##### Place, publisher, year, edition, pages
Taylor & Francis, 2017
##### Keyword
Eigenvalue distribution, Free moments, Free Poisson law, Marchenko– Pastur law, Random matrices, Spectral distribution, Wishart matrix
Mathematics
##### Research subject
Natural Science, Mathematics
##### Identifiers
urn:nbn:se:lnu:diva-58169 (URN)10.1080/03610926.2015.1053942 (DOI)000390425800031 ()
Available from: 2015-11-12 Created: 2016-11-17 Last updated: 2018-04-20Bibliographically approved
4. On p/n-asymptoticsapplied to traces of 1st and 2nd order powers of Wishart matrices with application to goodness-of-fit testing
Open this publication in new window or tab >>On p/n-asymptoticsapplied to traces of 1st and 2nd order powers of Wishart matrices with application to goodness-of-fit testing
(English)Manuscript (preprint) (Other academic)
##### Abstract [en]
The distribution of the vector of the normalized traces of $\small \frac{1}{n}XX'$ and of $\small \Big(\frac{1}{n}XX'\Big)^2$, where the matrix $\small X:p\times n$ follows a matrix normal distribution and is proved, under the Kolmogorov condition $\small \frac{{n}}{p}\overset{n,p\rightarrow\infty}{\rightarrow}c>0$, to be multivariate normally distributed. Asymptotic moments and cumulants are obtained using a recursive formula derived in Pielaszkiewicz et al. (2015). We use this result to test for identity of the covariance matrix using a goodness–of–fit approach. The test performs well regarding the power compared to presented alternatives, for both c < 1 or c ≥ 1.
##### Keyword
goodness–of–fit test, covariance matrix, Wishart matrix, multivariate normal distribution
Mathematics
##### Identifiers
urn:nbn:se:lnu:diva-58170 (URN)
Available from: 2015-11-12 Created: 2016-11-17 Last updated: 2018-04-20Bibliographically approved
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v. 2.32.0
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https://planetmath.org/IntegralMeanValueTheorem | # integral mean value theorem
###### The Integral Mean Value Theorem.
If $f$ and $g$ are continuous real functions on an interval $[a,b]$, and $g$ is additionally non-negative on $(a,b)$, then there exists a $\zeta\in(a,b)$ such that
$\int_{a}^{b}\!{f(x)g(x)}\,\mathrm{d}{x}=f(\zeta)\int_{a}^{b}\!{g(x)}\,\mathrm{% d}{x}.$
###### Proof.
Since $f$ is continuous on a closed bounded set, $f$ is bounded and attains its bounds, say $f\left(x_{0}\right)\leq f(x)\leq f\left(x_{1}\right)$ for all $x\in[a,b]$. Thus, since $g$ is non-negative for all $x\in[a,b]$
$f\left(x_{0}\right)g(x)\leq f(x)g(x)\leq f\left(x_{1}\right)g(x).$
Integrating both sides gives
$f\left(x_{0}\right)\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}\leq\int_{a}^{b}\!{f(x)g% (x)}\,\mathrm{d}{x}\leq f\left(x_{1}\right)\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}.$
If $\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}=0$, then $g(x)$ is identically zero, and the result follows trivially. Otherwise,
$f\left(x_{0}\right)\leq\frac{\int_{a}^{b}\!{f(x)g(x)}\,\mathrm{d}{x}}{\int_{a}% ^{b}\!{g(x)}\,\mathrm{d}{x}}\leq f\left(x_{1}\right),$
and the result follows from the intermediate value theorem. ∎
Title integral mean value theorem IntegralMeanValueTheorem 2013-03-22 17:15:56 2013-03-22 17:15:56 me_and (17092) me_and (17092) 9 me_and (17092) Theorem msc 26A06 EstimatingTheoremOfContourIntegral | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 18, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989626407623291, "perplexity": 310.55261041618667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744750.80/warc/CC-MAIN-20181118221818-20181119003818-00117.warc.gz"} |
https://online.stat.psu.edu/stat414/book/export/html/683 | # 7.5 - More Examples
7.5 - More Examples
## Example 7-8
A lake contains 600 fish, eighty (80) of which have been tagged by scientists. A researcher randomly catches 15 fish from the lake. Find a formula for the probability mass function of $$X$$, the number of fish in the researcher's sample which are tagged.
#### Solution
This problem is very similar to the example on the previous page in which we were interested in finding the p.m.f. of $$X$$, the number of defective bulbs selected in a sample of 4 bulbs. Here, we are interested in finding $$X$$, the number of tagged fish selected in a sample of 15 fish. That is, $$X$$ is a hypergeometric random variable with $$m = 80$$, $$N = 600$$, and $$n = 15$$. Therefore, the p.m.f. of $$X$$ is:
for the support $$x=0, 1, 2, \ldots, 15$$.
## Example 7-9
Let the random variable $$X$$ denote the number of aces in a five-card hand dealt from a standard 52-card deck. Find a formula for the probability mass function of $$X$$.
#### Solution
The random variable $$X$$ here also follows the hypergeometric distribution. Here, there are $$N=52$$ total cards, $$n=5$$ cards sampled, and $$m=4$$ aces. Therefore, the p.m.f. of $$X$$ is:
$$f(x)=\dfrac{\dbinom{4}{x} \dbinom{48}{5-x}}{\dbinom{52}{5}}$$
for the support $$x=0, 1, 2, 3, 4$$.
## Example 7-10
Suppose that 5 people, including you and a friend, line up at random. Let the random variable $$X$$ denote the number of people standing between you and a friend. Determine the probability mass function of $$X$$ in tabular form. Also, verify that the p.m.f. is a valid p.m.f.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8482000827789307, "perplexity": 414.156535614951}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00386.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-approach-4th-edition/chapter-3-exercises-page-131/35 | ## Chemistry: Molecular Approach (4th Edition)
a. $Ca(OH)_2$ b. $CaCrO_4$ c. $Ca_3(PO_4)_2$ d. $Ca(CN)_2$
Use table 3.4 (Page 100) and figure 3.7 (Page 98) to see the charges and the formula of the ions. - Calcium has a charge of $2+$, and its symbol is $Ca$: $Ca^{2+}$ ------- a. Hydroxide ion is: $OH^-$ (See table 3.4). Calcium: $Ca^{2+}$ Put these two together, where the subscript number will be the charge of the other ion. Calcium will receive 1, and hydroxide will receive 2. $Ca_1(OH)_2$: $Ca(OH)_2$ b. Chromate ion is: $CrO_4^{2-}$ (See table 3.4). Calcium: $Ca^{2+}$ Put these two together, where the subscript number will be the charge of the other ion. Calcium will receive 2, and chromate will receive 2. $Ca_2(CrO_4)_2$: $CaCrO_4$ c. Phosphate ion is: $PO_4^{3-}$ (See table 3.4). Calcium: $Ca^{2+}$ Put these two together, where the subscript number will be the charge of the other ion. Calcium will receive 3, and phosphate will receive 2. $Ca_3(PO_4)_2$ c. Cyanide ion is: $CN^{-}$ (See table 3.4). Calcium: $Ca^{2+}$ Put these two together, where the subscript number will be the charge of the other ion. Calcium will receive 1, and cyanide will receive 2. $Ca_1(CN)_2$ : $Ca(CN)_2$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8995336890220642, "perplexity": 1323.0273357221276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00158.warc.gz"} |
http://math.stackexchange.com/questions/8312/discriminant-of-a-monic-irreducible-integer-polynomial-vs-discriminant-of-its-s | # Discriminant of a monic irreducible integer polynomial vs. discriminant of its splitting field
Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't?
-
The discriminant of $K$ is equal to the gcd of $disc(f)$ as $f$ ranges over irred. polys. of all algebraic integers in $K$ which are primitive elements (i.e. which generate $K$ over $\mathbb Q$). – Matt E Oct 30 '10 at 3:18
Edit: Every time I say "splitting field" in the following answer I really mean $\mathbb{Q}[x]/f(x)$.
The two are the same if the roots of $f$ form an integral basis of the ring of integers of the splitting field (e.g. if $f$ is a cyclotomic polynomial) because then, well, they're defined by the same determinant (see Wikipedia), but in general they don't. In the general case $\mathbb{Z}[\alpha_1, ... \alpha_n]$ is an order in $\mathcal{O}_K$ so one can write the $\alpha_i$ as an integer linear combination of an integral basis, so the matrices whose determinants define the two discriminants should be related by the square of a matrix with integral entries, hence integral determinant.
In fact if I'm not totally mistaken, the quotient of the two discriminants should be precisely the index of $\mathbb{Z}[\alpha_1, ... \alpha_n]$ in $\mathcal{O}_K$ as lattices, or maybe its square...?
In any case, since the discriminant of the field is defined in terms of $\mathcal{O}_K$ it is the "right" choice for carrying information about, for example, ramification. One can see this even in the quadratic case: if $d \equiv 1 \bmod 4$ then the discriminant of $x^2 - d$ is $4d$ but the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$, and the latter is the "right" choice because $2$ doesn't ramify in $\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$.
-
It hardly ever happens when the roots of $f(x)$ forms the the integral basis of the splitting field. For example, take $f(x)=x^3-2$, then the splitting field has degree $6$ over $\mathbb{Q}$, but $f(x)$ has only 3 roots. – Jiangwei Xue May 7 '11 at 17:19
I think there was some confusion about the splitting field and the field $\mathbb{Q}[x]/(f(x))$, which is isomorphic to the field generated by one root of $f(x)$. (We always assume that $f(x)$ is monic irreducible.)
Let $\alpha$ be a root of $f(x)$, and $L=\mathbb{Q}(\alpha)$ be the field generated by $\alpha$, $\mathbb{Z}[\alpha]$ be the subring of $\mathcal{O}_L$ generated by $\alpha$. The the discriminant of $f(x)$ is the discriminant of the lattice $\mathbb{Z}[\alpha]$. So $\mathrm{Disc}(f)/\mathrm{Disc}(\mathcal{O}_L)$ is the square of $[\mathcal{O}_L: \mathbb{Z}[\alpha]]$. (See III.3 of Lang's "Algebraic number theory").
However, for splitting fields, these things hardly compares. For example, take $f(x)=x^4-x+1$, then the discriminant of $f(x)$ is 229 (a prime, which coincides with the discriminant of the field $L$ in this case) , but the discriminant of the splitting field of $f(x)$ is $229^{12}$ (calculated using Pari), which has 28-digits. (Well, it is not hard to show the discriminant of the splitting field of $f(x)$ share the same prime divisors as the field $L$.)
Sorry about bring up a really old question. It is just I asked myself the same thing today.
-
Thanks for the correction, Jiangwei. – Qiaochu Yuan May 7 '11 at 19:26
$Disc(f)/Disc(\mathcal{O}_K)$ is the square of the index of $\mathbb{Z}[ \alpha _1, \ldots , \alpha _n ]$ in $\mathcal{O}_K$. The index itself is the determinant of the change of basis matrix from $(\alpha _1, \ldots , \alpha _n )$ to an integral basis for $\mathcal{O}_K$. This matrix is squared when taking the discriminant. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754934310913086, "perplexity": 121.52221032658838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999655040/warc/CC-MAIN-20140305060735-00073-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.jiskha.com/questions/549210/if-freddy-tosses-a-coin-three-times-what-is-the-probability-that-the-coin-will-turn | # math
if freddy tosses a coin three times, what is the probability that the coin will turn up hesds,tails,heads, tails in that order? I came up with 1/2 correct?
also the ratio length to the width of the alamo isabout 5to 3 if the width of the alamo is 63ft about how long is the alamo? would that be 37.4ft?
THANK YOU.
1. 👍 0
2. 👎 0
3. 👁 73
1. It would help if you proofread your questions before you posted them. You indicate three tosses with four outcomes.
The probability of any one toss = 1/2
The probability of all/both events occurring = product of the individual events.
If ratio of length to width = 5/3 and width = 63, the length will be > 63.
5/3 = x/63
Solve for x.
1. 👍 0
2. 👎 0
posted by PsyDAG
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More Similar Questions | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8342925906181335, "perplexity": 855.137695216164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00016.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-4-section-4-1-solving-systems-of-linear-equations-by-graphing-exercise-set-page-292/52 | ## Algebra: A Combined Approach (4th Edition)
a. $2x + y = 0$ and $y = -2x + 1$ are parallel lines. We can put the second line equation in standard form ($Ax+By=C$) as the first one: $2x+y = 1$. Their slopes are calculated as $m=-\frac{A}{B}=-\frac{2}{1}=-2$. Since they have the same slope, they are parallel lines. b. Since, they do not intersect, the system does not have a solution. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9176594614982605, "perplexity": 211.25461898715594}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00051.warc.gz"} |
http://mathhelpforum.com/statistics/141498-basic-standard-deviation.html | ## Basic Standard Deviation
Hi all,
I have a simple question really. I know that standard devaition is given by the following formula:
However, does anyone know if there is a special name just for the portion:
(Xμ)2
Many thanks. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9558926224708557, "perplexity": 749.595415461836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00144-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.arxiv-vanity.com/papers/nucl-th/9805017/ | # On the Flow of Kaons Produced in Relativistic Heavy Ion Collisions
C. David, C. Hartnack and J. Aichelin
SUBATECH
Université de Nantes, EMN, IN2P3/CNRS
4, Rue Alfred Kastler, BP 20722, 44307 Nantes Cedex 3, France
###### Abstract
Abstract: We investigate the different contributions to the in-plane flow of mesons observed recently by the FOPI collaboration in the reaction Ni (1.93 A GeV) + Ni. Due to the kinematics of the three body phase space decay the flow of the kaons produced in baryon-baryon interactions is smaller than that of the baryons in the entrance channel. On the contrary, in N interactions the flow of the sources and of the kaons are identical. Therefore the total kaon flow depends on the relative number of and reactions and hence on the lifetime of the , in addition to the already known dependence on the potential interaction of the kaons with the nuclear environment.
## I introduction
The production of K mesons in heavy ion collisions is presently one of the most challenging topics in nuclear physics. At beam energies below or close to the threshold (in NN collisions GeV) we observe a strong enhancement of the kaon production as compared to the extrapolation of pp collisions. Detailed investigations have shown that most of the K’s are created in two step processes via an intermediate or and are produced at a density well above the normal nuclear matter density [1, 2]. This triggered the conjecture that K’s may be of use as a messenger of the high density zone.
What makes a straight forward analysis complicated are several problems:
• The nucleons as well as the K’s and interact with each other via static and momentum dependent interactions. Hence particle properties like the particle mass are modified.
• This modification of the particle properties changes the thresholds of the relevant strange particle production cross sections , , as well as of the the rescattering reactions.
• Most of the K’s are produced in a collision in which a nuclear resonance or its decay product is involved. Hence the lifetime of a resonance in nuclear matter becomes important. If the life time is short as compared to , where is the density dependent mean free path of the nuclear resonance and is its average velocity, the resonance decays before it encounters another nucleon for creating a kaon. Then only the decay product, i.e. the can create a kaon. If the opposite is the case, the dominant production channel will be Resonance + N .
It is the purpose of this article to investigate the influence of these in medium properties on the production of K’s, especially on the observed in-plane flow which has been proposed as a signature of the relative strength of the scalar and vector part of the interaction of the K’s with its hadronic environment [3].
For this purpose we employ simulations of the heavy ion reaction with the Quantum Molecular Dynamics (QMD) approach. This approach is described in ref [4]. In addition we have implemented the kaon producing cross sections and with and . We have displayed in fig. 1 the important cross section with a in the final state. Similar fits to the (fewer) available data for the reaction with a in the final state have also been made and are included in the calculation. The energy range of importance is shown in fig. 2 where we display the number of elementary collisions as a function of in the simulation. The relevance of the different channels is given by the number of collisions times the kaon production cross section. We see that , N and NN collision contribute all to the kaon production in a non negligible way. We see as well that the deviations of our parameterization from data at high is not of importance for the beam energy considered here.
After being produced the K’s move in a potential created by the nuclear environment which has the form [5]:
uKopt=√(→k−gv→Σv)2+m2K+mKgsΣs+gvΣ0v−√k2+m2K (1)
and rescatter with the nucleons. The cross section of the rescattering of the produced kaons is a parametrisation of the results presented in [6].
## Ii Kaon’s: messenger from the high density zone?
Naively one expects that K’s are created in the high density zone of the reaction. Because we are at subthreshold energies the kaon is produced easiest if there is a nuclear resonance in the entrance channel, which - due to its higher mass as compared to a nucleon - needs only a smaller relative momentum with respect to its scattering partner to overcome the threshold. The shorter its mean free path, the more probable the resonance encounters a nucleon before it disintegrates. Because the kaon production via resonances appears preferably at high densities. When created, the kaon is still surrounded by nuclear matter and may encounter a collision with nucleons while the system disintegrates. The cross section for rescattering of is small but nonnegligable. Such a collisions may destroy the information about the high density zone which is carried by the K’s.
Fig 2. displays the distribution of nuclear densities at the places where the K’s are produced. We see that even for a system as small as Ni+Ni 80% of the K’s are produced at a density above normal nuclear matter density and 60% do not rescatter at a density lower than normal nuclear matter density. Hence K’s can really be regarded as messenger from the high compressed phase of the reaction. It is the only meson which has this property at the energy considered in this article.
## Iii In-plane flow of the K’s
Recently it has been speculated [3, 7] that the in-plane flow of the K’s is directly related to the relative strength of the scalar and vector potential of K’s in nuclear matter. Therefore it is interesting to discuss in detail the origin of the in-plane flow of the K’s, which is much lower than the in-plane flow of the nucleons measured in the same reaction.
Before discussing the kaon flow it is, however, necessary to verify that the simulations reproduce the nucleon flow although, as we will see later, both are only very loosely connected. The experimental nucleon in-plane flow as compared to the simulation is displayed in fig. 4. We see that the soft momentum dependent interaction [4] reproduces the experimental flow as well as the less realistic static soft equation of state.
Fig.5 displays the time evolution of the in-plane flow of different classes of protons. The average directed in-plane flow is defined by
pdirx=1NN∑i=1pix⋅sign(yicm) (2)
where is the rapidity of the nucleon i in the nucleus-nucleus center of mass system and is the momentum of the nucleon i in the direction of the impact parameter. The calculation has been done at an impact parameter of 2 fm.
The averaged directed in-plane flow of all nucleons is given by a line. We see that it reaches asymptotically 80 MeV/c. For the dotted line we have counted only those nucleons which have passed a density of . As already found in ref. [8] the in-plane flow of nucleons which have passed the high density zone is considerably smaller than that of all nucleons. If we include in the calculation of only those nucleons, which have been involved in a collision, in which a kaon has been produced (dashed line), we observe about the same in-plane flow as for those nucleons which have passed the high density zone. This agreement is compatible with the fact that K’s are produced in the high density zone as shown in fig.1. More interesting than the in-plane flow of the protons is the in-plane flow of the sources in which the K’s are produced according to phase space, i.e. isotropically:
pdirx=1MM∑i=1(p1ix+p2ix)⋅sign(y1icm+y2icm). (3)
Here the sum runs over all collisions in which a kaon is produced. 1 and 2, respectively, mark the two hadrons which scatter in these collisions. The result is displayed as the dashed-dotted line. This in-plane flow has about the same size as that for the nucleons involved in a kaon producing collision and is not twice as large as one may think naively. Hence the in-plane directed source velocity / mass is about half as large as the directed in-plane velocity of the participating nucleons.
The reason for this astonishing result is displayed on the left hand side of fig. 6 where we show the rapidity distribution of the kaon producing sources as well as that of the and of the K’s at the moment of their creation (i.e. before rescattering) in the reaction baryon+baryon . We see that - due to kinematical reasons - most of the sources are centered very close around midrapidity in the nucleus-nucleus center of mass system. Hence it happens quite often that one of the reaction partners has a negative and the other a positive rapidity. In this case the average of the both nucleons points into opposite directions and the vector sum becomes small. This explains the low value of the in-plane flow of the source velocity. The three body phase space decay broadens the rapidity distribution of the sources considerably. This lowers the in-plane flow of the K’s a second time. At a given rapidity we find K’s from sources with quite different rapidities and consequently with quite different in-plane source velocities. Hence the in-plane velocity of those K’s is an average value of the in-plane velocities of the different sources. Because the in-plane velocity changes sign at mid rapidity, the average value is very small. This is seen in fig. 6 (left) where the rapidity distribution and the directed in-plane velocity of sources, ’s and K’s are displayed. Fig. 6 (right) displays the same quantities for the channel . Here the in-plane velocity of the sources is of about the same size as that for the 3-body channel. However, having only two particles in the exit channel, this reaction does not spread the in-plane velocity as far in rapidity as a three body decay. Hence finally K’s coming from a collision have a larger in-plane flow than those from a baryon baryon collision.
In conclusion we see that the small value of the in-plane flow of the K’s as compared to that of the nucleons is expected even if one neglects completely any potential or collisional interaction of the K’s with the nuclear environment. It is caused by two processes: nucleons which pass the high density zone of the reaction have a much smaller average flow than the average nucleon and the production of K’s according to phase space transports the K’s far away in rapidity space with respect to the rapidity of the center of mass of the collision in which it is produced. Thus the in-plane velocity is smeared out.
## Iv The role of the lifetime of the Δ resonance
Most of the collisions in which a kaon is produced involve either directly a or the produced by its disintegration. The relative fraction depends on the lifetime of the . For a long time it has been thought that the life time of a with a given mass depends on the phase space for decay by.
τ=1Γ;Γ∝∫d3pπd3pN||2δ4(pi−pf) (4)
Usually one assumed that the matrix element is independent of the mass of the resonance and therefore the mass dependence is given by phase space. Only recently it has been realized [10] that this formula is only valid for a well prepared resonance which has the Breit-Wigner distribution of their mass. In simulations we are confronted with a different situation. The variance of the energy of the incoming particles is zero in BUU and small as compared to the variance of the resonance in QMD simulations. Therefore the center of mass energy - which corresponds to the resonance mass - is determined with a very good precision and its variance is much smaller than the variance of the Breit Wigner distribution which characterizes the resonance. In this case the lifetime of is given by the more general formula [10]
τ=dδdE=Γ/2(E−E0)2+Γ2/4. (5)
Whereas in the former case (eq.4) the lifetime increases with decreasing mass of the due to the smaller phase space, in the latter case (eq.5) the life time has its maximum at the center of the resonance and approaches zero at energies well above or below. Fig.7 compares the lifetimes presently used in the different numerical programs with that calculated with the above equation. The influence of the different assumptions on the lifetime on the kaon flow will become increasingly important if one lowers the beam energy because more and more will be produced close to the threshold due to the limited available energy. At the reaction considered here the correct equation increases the average lifetime of the ’s as compared to the former approaches because due to the sufficient energy the ’s are created preferably at the center of the resonance.
Fig.8 displays the influence of the different lifetimes of the resonance on the kaon in-plane flow. We compare the flow calculated by using the -lifetime as derived by Huber et al. [11] () with that obtained by eq.6 () and that obtained by the parameterization of () from data of [12]. For these calculations we have set . We see that the differences for the in-plane flow are considerable what offers the possibility to study the -lifetime in medium.
How our calculation compares with experiment is seen in fig.9 where we display the data of the FOPI collaboration [13] in comparison with different calculations. For this purpose we have filtered our data. The limitation to particles with a transverse momentum larger than 250 MeV/c decreases the statistics by a large factor. For we display calculations in which we set and , respectively, to zero and compare them with the result obtained by including and excluding the total static potential interaction. A more realistic -lifetime as compared to [7] and the addition of the momentum dependent part of the vector potential balance each other at that energy. Thus finally we come close to the results of Ko [7]. Finally, we display in fig. 10 the influence of the vector part of the vector potential on the in-plane flow of kaons. The temporal vector potential represents the case when we take only the term of the potential into account (i.e. ). The additional term increases the flow of kaons. This is easy to understand: We have seen (fig.3) that all kaons come from the high density fireball region where due to the symmetry of the system the momentum distribution of the baryons is symmetric with respect to the nucleus-nucleus center of mass system. Assuming that is proportional to the baryon current the scalar product in the optical potential, , is zero. Therefore for symmetric systems the optical potential is reduced to
uKopt=√k2+(gv→Σv)2+m2K+mKgsΣs+gvΣ0v−√k2+m2K (6)
The term has the same sign as and depends in the same way on the local baryon density. It enhances therefore the action of term as seen in fig. 10. We would like to mention that in ref. [14] an opposite result has been found.
In conclusion we have found that the low value of the in-plane flow of the kaons as compared to nucleons has little to do with the interaction of the kaons after their creation. It is a consequence of the kinematics before and at the time point of the creation. The potential interaction of the kaons with the nuclear environment as well as the life time of the nuclear resonances and the rescattering modifies the directed in-plane kaon momentum. In addition, the value obtained in the simulation is modified by the experimental cuts.
Therefore it is premature to interpret the kaon flow as a signal for strength of one or the other of these processes. Because each of the above mentioned processes is interesting in itself exciting physics lie ahead of us on the way to interpret the data on the kaon flow.
## References
• [1] J. Aichelin and C.M. Ko Phys. Rev. Lett. 55 (1985) 2661
• [2] C.Hartnack et al., Nucl. Phys. A580 (1994) 643
• [3] G.Q. Li, C.M. Ko and B.A. Li, Phys. Rev. Lett 74. (1995) 235
• [4] J. Aichelin, Phys. Rep. 202, 233 (1991), and references therein.
• [5] J. Schaffner et al, Phys. Lett B334 (1994) 268 J. Schaffner-Bielich et al., Nucl. Phys. A625 (1997) 325
• [6] C.B. Dover and G.E. Walker, Phys. Rep. 89, 1 (1982)
• [7] G.Q. Li et al., Phys. Rev C57 (1998) 434
• [8] C. Hartnack, H. Stöcker and W. Greiner, Proc. on the NASI on the Nuclear Equation of State, Peniscola (Spain), NASI Series B, Vol. 216A, Plenum, New York, p. 239.
J. Jänicke and J. Aichelin, Nucl. Phys. A 547 (1992) 542
• [9] J. Ritman and al., Z. Phys. A352 (1995) 355
• [10] J. Aichelin, Proceedings of Hirschegg 1997, ed. by H. Feldmeier and W. Nörenberg, and references therein. P. Danielewisz, Proceedings of Hirschegg 1997, ed. by H. Feldmeier and W. Nörenberg, and references therein.
• [11] S. Huber and J. Aichelin, Nucl. Phys. A573 (1994) 587
• [12] W. Lock and D. Measday, Intermediate energy nuclear physics, ed. Nederlandse Boekdruk Industrie n.v., 1970.
• [13] D.Best et al., Nucl. Phys. A625 (1997) 307
• [14] C. Fuchs, nucl-th 980148 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9604402184486389, "perplexity": 688.3598290958246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00354.warc.gz"} |
https://forums.tornbanner.com/topic/4976/solved-udk-exe-issue | # [SOLVED] UDK.exe Issue.
• Hello there,
Ive been having issues with starting the game through steam as I start the applications I get the appcrash…If I try to manually run it through the C:ProgramFiles/Steam/Steamapps/Common/Chivalry/Binaries/Win32 I get an error saying " This application has failed to start because D3DCOMPILER_42.dll was not found.Re-installing the application may fix this problem". Now I did try to re-install it and still doesnt work.In an other thread a possible fix was that I should delete the Config folder then run a Verify cache files.I also tried that to no avail.Anybody had this problem before?I would be very grateful for a reply
thank you
btw here’s my dxdiag.txt
• If that file is accurate, you should update your video drivers. You appear to be running 8.X where 12.X is the latest.
Also, have you tried reinstalling DirectX as a whole? Downloading and installing loose .dll files is never a good idea. They tend to add malicious code, but even if they’re safe, most .dlls don’t work unless they have a matching registry key. Which might be missing if you are only placing the dll file, as opposed to using an installer.
• Ill try reinstalling my DirectX
• Ill try reinstalling my DirectX
Try running DXSETUP.exe from C:\Program Files (x86)\Steam\steamapps\common\chivalrymedievalwarfare\Binaries\RedistChiv\DXRedistCutdown
• Ill try reinstalling my DirectX
Try running DXSETUP.exe from C:\Program Files (x86)\Steam\steamapps\common\chivalrymedievalwarfare\Binaries\RedistChiv\DXRedistCutdown
• Wow thank you very much it worked for me :)
• Wow thank you very much it worked for me :)
Yay! I’m glad that it works for you! See you on the battlefield. :k1:
• Wow thank you very much it worked for me :)
That’s excellent news. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9054681658744812, "perplexity": 4486.376232081016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655896374.33/warc/CC-MAIN-20200708031342-20200708061342-00195.warc.gz"} |
https://tex.stackexchange.com/questions/118244/what-is-the-difference-between-unicode-math-and-mathspec/118254 | # What is the difference between unicode-math and mathspec?
I'm currently typesetting my lecture notes for classical electrodynamics using the tufte-book class and XeLaTeX.
This is my first time using XeLaTeX and so I stumbled upon the two packages unicode-math and mathspec, when it came to include an OpenType math font (here TeX Gyre Pagelle Math).
I browsed the documentation of the two packages a little bit, but wasn't able to find out, which is the "better practice" to use.
Question: Should I use unicode-math or mathspec for my XeLaTeX documents?
## unicode-math
The unicode-math and mathspec packages have very different goals. The unicode-math package is designed to map math markup into unicode characters as supplied by real OpenType math fonts such as the Latin Modern Math, STIX, Asana Math. It also allows (as much as makes sense/is possible) Unicode input of math, as well as output so that you can enter unicode math characters in the source and have them be interpreted correctly as math by TeX.
For example, you could create the following document:
% !TEX TS-program = XeLaTeX
\documentclass[12pt]{article}
\usepackage{unicode-math}
\begin{document}
This is some math text entered with math in the source:
$∀X [ ∅ ∉ X ⇒ ∃f:X ⟶ ⋃ X\ ∀A ∈ X (f(A) ∈ A ) ]$
This is some math text entered with regular markup
$\forall X [\emptyset \not\in X \Rightarrow \exists f:X \rightarrow \bigcup X\ \forall A \in X (f(A) \in A ) ]$
\end{document}
which would produce
The unicode-math package is compatible with both XeTeX and LuaTeX.
## mathspec
The mathspec package is not designed to allow you to use unicode characters in your source math input. Instead it is designed to allow you to use open-type fonts within math so that you can, for example, match the text font of your math with the text font in your document.
So for example you can use:
% !TEX TS-program = XeLaTeX
\documentclass[12pt]{article}
\usepackage{mathspec}
\setmainfont{Linux Libertine O}
\setmathfont(Digits,Latin)[Scale=MatchLowercase]{Linux Libertine O}
\begin{document}
This is some text in Libertine \emph{X(f(A))}
$\forall X [\emptyset \not\in X \Rightarrow \exists f:X \rightarrow \bigcup X\ \forall A \in X (f(A) \in A ) ]$
\end{document}
which will produce:
The mathspec package can only be used with XeTeX; it cannot be used with LuaTeX.
## Which should you use
If you want to use one of the available OpenType math fonts, then using unicode-math makes sense.
If you want to match a non-math font with the math font then mathspec can help, but because TeX does not apply kerning between characters from different fonts, using mathspec has its drawbacks, since you may find places in which the switch from the text font to the math font produces bad spacing, as in the example given.
• @HenriMenke Yes, that's true. mathspec is not compatible with LuaTeX. (The extra \setmathfont was a typo). – Alan Munn Jun 8 '13 at 14:33
• (+1) You could use ⋃ (U+22C3) instead of \bigcup to have an a no-markup example. – David Carlisle Jun 8 '13 at 14:47
• @DavidCarlisle Thanks. I've fixed the example, but it reinforces the point about how painful it can be to enter unicode source: I was didn't immediately find that character with my character viewer; also in the source the n-ary union is almost indistinguishable from the regular union character, which is kind of defeats the purpose of having more readable input. – Alan Munn Jun 8 '13 at 15:29
• @AlanMunn There is no need to use Unicode input with unicode-math; it works perfectly with traditional input. – egreg Jun 8 '13 at 17:09
• I don’t think the description of unicode-math is right, the Unicode input is merely a side effect of the main goal; supporting Unicode math fonts, and you can still use csnames for math inout, which is actually what most people do. – Khaled Hosny Jun 8 '13 at 17:27
mathspec is a clever attempt to make it possible to use math in XeLaTeX documents, with system fonts for the letters and symbols from standard math fonts.
With unicode-math one can basically use only specially tailored OpenType math fonts such as Latin Modern Math, TeX Gyre Termes Math, TeX Gyre Pagella Math, XITS Math, Asana Math (among the free ones) or Cambria Math and Lucida Bright Math (not free).
## mathspec
With this package you can do limited math with the main system font. Just a silly example:
\documentclass{article}
\usepackage{mathspec}
\setmainfont[Ligatures=TeX]{Old Standard}
\setmathsfont(Digits,Latin,Greek){Old Standard}
\begin{document}
Some text and a formula $a+b=\int_{\xi}^{\theta} f(x)\,dx$.
\end{document}
The emulated math is not bad, and the package has features for manually correcting the possible bad spacings, which happen because the fonts have really no support for math. For instance, inputting the formula as
$a+b=\int_{\xi}^{\theta} "f(x)\,dx$
would give the better result
The " means ”add some space on both sides of the following letter”.
Digits, Latin and Greek letters can be taken from different fonts.
## unicode-math
The unicode-math is compatible both with XeLaTeX and LuaLaTeX. Here's the same example with Lucida Bright:
\documentclass{article}
\usepackage{unicode-math}
\setmainfont[Ligatures=TeX,Scale=0.85]{Lucida Bright OT}
\setmathfont[Scale=0.85]{Lucida Bright Math OT}
\begin{document}
Some text and a formula $a+b=\int_{\xi}^{\theta} f(x)\,dx$.
\end{document}
Spacing is correct without manual intervention, because the font used for math is a real math font. Here's the same with TeX Gyre Pagella:
\documentclass{article}
\usepackage{unicode-math}
\setmainfont[Ligatures=TeX]{TeX Gyre Pagella}
\setmathfont{TeX Gyre Pagella Math}
\begin{document}
Some text and a formula $a+b=\int_{\xi}^{\theta} f(x)\,dx$.
\end{document}
## Unicode input
One can input math with Unicode symbols, but this is not mandatory. It's possible both with mathspec and unicode-math. For instance, the integral with mathspec could be input as
$∫_ξ^θ "f(x)\,dx$
provided that ∫ is made known to the environment, for instance with
\usepackage{newunicodechar}
\newunicodechar{∫}{\int}
Some symbols (in particular letters) are already known.
Nothing particular is needed for inputting the integral as
$∫_ξ^θ f(x)\,dx$
with unicode-math. However the traditional syntax is understood by the package.
• Great answer, thanks! It is very helpful to know, how to define additional unicode symbols as math operators. – Henri Menke Jun 8 '13 at 18:42
• I would not call it 'clever' tbh. that's a standard feature of some programming language. to not have it by default in a language dedicated to typesetting is laughable – nicolas Apr 25 '16 at 7:20
• @egreg Is it possible to somehow use sequences of unicode symbols without having to load unicode-math? I am thinking in particular about mapping ⁻¹ to ^{-1}. – Gaussler May 24 '16 at 8:35
• @Gaussler It may be possible; whether it's convenient is another matter – egreg May 24 '16 at 8:49
• Is it better to use TeX-Gyre Pagella Math, NewPX, Neo-Euler or Asana?. I'm looking for some opentype font with nice math, and with text looking like Palatino (because of university restrictions). – skan Nov 25 '16 at 13:45
unicode-math is mainly concerned about facilitating the use of using Unicode math fonts, namely OpenType math fonts in LaTeX. While mathspec is about allowing to take math alphabets from text fonts in the absence of matching math font.
Unicode defines a large set of math symbols and many fonts include them, but proper math typesetting requires many font parameters to aid the typesetting engine and those have to be provided by the font. Traditionally TeX fonts would provide the needed parameters in the font dimension of TFM files, which meant that non-TFM fonts can not be used to typeset math in XeTeX (and later LuaTeX). When Microsoft started adding TeX-like math typesetting in Word, they extended OpenType with a special MATH table to hold the needed parameters for math typesetting and XeTeX implemented it (and later LuaTeX) and unicode-math was written to provide macro support for using the new fonts. So with unicode-math you get proper math typesetting with proper OpenType math fonts, like TeX Gyre Pagelle Math, but you are limited to those specially prepared fonts.
• Thank you for answer. I like espacially, that you provided additional information on the history. – Henri Menke Jun 8 '13 at 18:39
• So the reason that using mathspec leads to the differences shown in the other answers is that it tries to be compatible to any font that defines the Unicode math characters, not only those that provide the additional parameters you mentioned? Put differently, it will ignore those parameters even if they are available? – balu Jan 8 '15 at 19:48
• @balu The fonts don't necessarily need to define the unicode maths characters. mathspec lets you use just e.g. the letters for a calligraphic alphabet or the letters from your text font for operator and variable names etc. so that text and maths use the same ABC ... abc ... or whatever. You can substitute just the bits you want - pick 'n mix. – cfr Aug 2 '16 at 22:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9803345799446106, "perplexity": 3740.601538016492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00462.warc.gz"} |
http://pipingdesigner.co/index.php/properties/dimensionless-numbers/156-lewis-number | # Lewis Number
Written by Jerry Ratzlaff on . Posted in Dimensionless Numbers
Lewis number, abbreviated as Le, is a dimensionless number which is the ratio of thermal diffusivity to mass diffusivity. It is used to characterize fluid flows where there is simultaneous heat and mass transfer.
$$\large{ Le = \frac{ \alpha }{ D_m } }$$ Where: $$\large{ Le }$$ = Lewis number $$\large{ D_m }$$ = mass diffusivity $$\large{ \alpha }$$ (Greek symbol alpha) = thermal diffusivity Solve for: $$\large{ \alpha = Le \; D_m }$$ $$\large{ D_m = \frac{ \alpha }{ Le } }$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9881856441497803, "perplexity": 4241.277236121205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00265.warc.gz"} |
http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/pete1.html | SEARCH HOME
Math Central Quandaries & Queries
Question from Pete, a student: Hi, How do you express ³√h-4 in exponential form. I am having a lot of trouble with this one. thanks Pete
Hi Pete. It looks like you wrote this:
The radical sign with a 3 in the crook means cube root. That means the same as the exponent of 1/3.
Remember that with rules of exponents, you multiply the exponents together to simplify from this form:
Hope this helps,
Stephen La Rocque.
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.801223874092102, "perplexity": 1093.0104944835575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934809392.94/warc/CC-MAIN-20171125032456-20171125052456-00082.warc.gz"} |
https://www.vedantu.com/maths/equation-of-a-plane-in-normal-form | Equation Of Plane In Normal Form Easy Method
The three-dimensional geometry is a nightmare for some students who are appearing for their board exams. Nothing can be scarier for a student than to know a question, which is a sure shot will come in an exam, and they don’t know how to solve it. What makes geometry so tricky is the concept of planes, which is an integral part of 3D geometry. Today we are going to slay this behemoth named equation of the normal form of a plane and make it easy for students to understand it. Please make no mistake even we have to churn the gears of our mind to understand it fully, so we know how difficult this concept is for students to learn.
There are two ways to find out the general equation of a plane, the first one is by using the standard form, and the other way of doing it is by using the Cartesian form. The first method is quite essential, and once you get to know how to solve the normal vector of a plane with it, you can derive the Cartesian form on your own. So in this article, we are covering the first method only.
Theorem For Equation Of Plane In Normal Form
First, you need to think about a plane that is perpendicular to the origin, and its distance is D. In addition to this, D is not equal to 0.
Here D ≠ 0
$\overrightarrow{ON}$ ---> Normal from the origin
$\widehat{n}$ ---> is the unit vector, which is normal along $\overrightarrow{ON}$
Now we have $\overrightarrow{ON}$ = d$\widehat{n}$
Step 2:- Now here, you need to take any point (P) on the plane. Thus making vector $\overrightarrow{NP}$ perpendicular to $\overrightarrow{ON}$
As a result, the dot product of these two vectors will come out to be zero, meaning $\overrightarrow{NP}$ . $\overrightarrow{ON}$ = 0
Step 3:- Here we will take a line from the origin (0) and connects it with point (P)
This will be our position vector $\overrightarrow{r}$ of point (P).
Step 4:- From here, you take out the triangle ∆OPN, and we have the given equation.
$\overrightarrow{NP}$ + $\overrightarrow{ON}$ = $\overrightarrow{OP}$
As a result, we can say $\overrightarrow{NP}$ = $\overrightarrow{OP}$ - $\overrightarrow{ON}$
We know $\overrightarrow{OP}$ = $\overrightarrow{r}$ and$\overrightarrow{ON}$ = d$\widehat{n}$ .
Now we will put these values in our given equation i.e $\overrightarrow{NP}$ = $\overrightarrow{OP}$ - $\overrightarrow{ON}$
$\overrightarrow{NP}$ = $\overrightarrow{OP}$ - $\overrightarrow{ON}$
$\overrightarrow{NP}$ = $\overrightarrow{r}$ - d$\widehat{n}$
Now here comes a twist, you need to remember the very first equation that we showed you, that product of two products will be zero.
i.e $\overrightarrow{NP}$ . $\overrightarrow{ON}$ = 0
nowhere, in this equation, we will put the value of $\overrightarrow{NP}$ = $\overrightarrow{r}$ - d$\widehat{n}$
now it looks like this,
( $\overrightarrow{r}$ - D$\widehat{n}$ ) . d$\widehat{n}$ = 0
Step 5:- In this step, we are just going to simplify the equation to reach the final answer.
$\overrightarrow{r}$ - d$\widehat{n}$ - $d^{2}$ ( $\widehat{n}$ . $\widehat{n}$ ) = 0
d [ $\overrightarrow{r}$ . $\widehat{n}$ - d ] = 0
[ $\overrightarrow{r}$ . $\widehat{n}$ - d ] = 0/d
$\overrightarrow{r}$ . $\widehat{n}$ - d = 0
$\overrightarrow{r}$ . $\widehat{n}$ = dThis is your equation of Plane in normal form.
Now for solving problems, you need to know about the Cartesian form, which is Ax + By + Cz = d. Where (A, B, C) are direction cosines of n and (x,y,z) is the distance of point P from the origin.
It might look difficult at the start, but once you start solving it, the answer will come on its own.
Solved Example
Now we know how to get to the equation; let’s try to apply it by solving some problems so you can better understand its usage.
Question. The distance of a given plane from the origin O is $\frac{10}{\sqrt{36}}$, the normal vector given to us is 4$\widehat{i}$ + 3$\widehat{j}$ - 2$\widehat{k}$ You have to find out the vector equation for the plane?
Answer. First, you need to find out the unit vector of a normal vector of a plane.
$\widehat{n}$ = $\overrightarrow{n}$ / | $\overrightarrow{n}$ |
Thus, putting in the values we have
$\widehat{n}$ = $\frac{4\widehat{i} + 3\widehat{j} - 2\widehat{k}}{\sqrt{16+9+4}}$
= $\frac{4\widehat{i} + 3\widehat{j} - 2\widehat{k}}{\sqrt{29}}$
Now, you need to substitute the vector equation in order to find out the required equation of the plane.
$\overrightarrow{r}$ . $\widehat{n}$ = d
$\overrightarrow{r}$ . $(\frac{5}{\sqrt{29}}\widehat{i}\; + \;\frac{3}{\sqrt{29}}\widehat{j}\; + \;\frac{-2}{\sqrt{29}}\widehat{k}\;)$ = $\frac{10}{\sqrt{36}}$ your final answer
There you have it, the equation of a plane in normal form with its solved theorem and example. Now, you might have gone through this article and think you have learned it, well, that’s where you are wrong. Mathematics is not something that you can learn without solving. Hence, you must practice and try to solve some of these problems on your own, and if you have any issues, we are here to help you out. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8279014229774475, "perplexity": 363.7978694884143}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655929376.49/warc/CC-MAIN-20200711095334-20200711125334-00032.warc.gz"} |
http://proceedings.mlr.press/v80/moreau18a.html | # DICOD: Distributed Convolutional Coordinate Descent for Convolutional Sparse Coding
Thomas Moreau, Laurent Oudre, Nicolas Vayatis ;
Proceedings of the 35th International Conference on Machine Learning, PMLR 80:3626-3634, 2018.
#### Abstract
In this paper, we introduce DICOD, a convolutional sparse coding algorithm which builds shift invariant representations for long signals. This algorithm is designed to run in a distributed setting, with local message passing, making it communication efficient. It is based on coordinate descent and uses locally greedy updates which accelerate the resolution compared to greedy coordinate selection. We prove the convergence of this algorithm and highlight its computational speed-up which is super-linear in the number of cores used. We also provide empirical evidence for the acceleration properties of our algorithm compared to state-of-the-art methods. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8899027705192566, "perplexity": 1336.247098640833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232258453.85/warc/CC-MAIN-20190525224929-20190526010929-00448.warc.gz"} |
https://arxiv.org/abs/1010.1203 | math.GR
(what is this?)
Title: First cohomology for finite groups of Lie type: simple modules with small dominant weights
Abstract: Let $k$ be an algebraically closed field of characteristic $p > 0$, and let $G$ be a simple, simply connected algebraic group defined over $\mathbb{F}_p$. Given $r \geq 1$, set $q=p^r$, and let $G(\mathbb{F}_q)$ be the corresponding finite Chevalley group. In this paper we investigate the structure of the first cohomology group $H^1(G(\mathbb{F}_q),L(\lambda))$ where $L(\lambda)$ is the simple $G$-module of highest weight $\lambda$. Under certain very mild conditions on $p$ and $q$, we are able to completely describe the first cohomology group when $\lambda$ is less than or equal to a fundamental dominant weight. In particular, in the cases we consider, we show that the first cohomology group has dimension at most one. Our calculations significantly extend, and provide new proofs for, earlier results of Cline, Parshall, Scott, and Jones, who considered the special case when $\lambda$ is a minimal nonzero dominant weight.
Comments: 24 pages, 5 figures, 6 tables. Typos corrected and some proofs streamlined over previous version Subjects: Group Theory (math.GR); Representation Theory (math.RT) MSC classes: 20G10 (Primary), 20G05 (Secondary) Journal reference: Trans. Amer. Math Soc. 365 (2013), 1025-1050 DOI: 10.1090/S0002-9947-2012-05664-9 Cite as: arXiv:1010.1203 [math.GR] (or arXiv:1010.1203v3 [math.GR] for this version)
Submission history
From: Christopher Drupieski [view email]
[v1] Wed, 6 Oct 2010 16:54:23 GMT (26kb)
[v2] Thu, 7 Oct 2010 13:52:29 GMT (26kb)
[v3] Sun, 24 Jul 2011 15:14:51 GMT (28kb) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8787301778793335, "perplexity": 596.3758679446065}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00162.warc.gz"} |
http://cogsci.stackexchange.com/questions/7921/what-to-do-when-you-have-a-speed-accuracy-trade-off-in-your-data | # What to do when you have a speed-accuracy trade-off in your data?
Many studies report that a speed-accuracy trade-off (SATO) did not occur in the data since there is a positive correlation between RTs and error rates. In other words, people took longer to respond for the trials that were more difficult, i.e. the same that elicited more errors on average, rather than trading speed for accuracy or vice-versa, a participant behaviour seen as undesirable. The absence of a SATO is then strongly indicative of a participant having given their best during the task, i.e. correctly following the usual instructions to give fast but accurate responses. A certain level of skill is then said to characterise a particular SATO curve, and this parameter of the curve can change as people become better at the task.
However, rarely if ever have I seen papers that DO report a SATO, i.e. a negative correlation betewen RTs and error rates, or, in other words, participants that remained at a relatively constant level of skill but simply moved "up and down" their SATO curve. Clearly this is not a sign of "good data", but still, what is the correct analysis to perform in such situations, assuming that you can't just proceed to do analyses on RTs and error rates the way you normally do (in the absence of a SATO)?
Also, the above two paragraphs refer to the cases where a RT-errors correlation exists and is positive (1st para) or negative (2nd para). But what if there isn't a significant correlation at all?
A good paper discussing this phenomenon is Pachella's 1974 paper "The Interpretation of Reaction Time in Information Processing Research". However, I don't believe it addresses my questions above.
-
– Jeromy Anglim Jul 30 '14 at 2:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8246700167655945, "perplexity": 1090.829441494505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246651471.95/warc/CC-MAIN-20150417045731-00057-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://groundpotential.org/forum/viewtopic.php?f=8&t=40&sid=50e7c6be46f172634f8b2134617e866e | It is currently Tue Aug 21, 2018 8:17 pm
## Galaxy Rotation Curves
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Steven Sesselmann
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Everyone is now familiar with the problem of galaxy rotation curves, and how it has given rise to speculations about galaxies containing dark matter. The problem was first announced by American astronomers Vera Rubin and Kent Ford in 1975, who collaborated to show that galaxies displayed a flat rotation curve and did not exhibit the expected Keplerian motion.
Unable to explain such flat rotation curves, theoreticians proposed that there had to be additional invisible matter in the galaxy in order to account for the flat rotation curve, and it was coined "dark matter".
Keplerian orbital velocity follows the function;
$v = \sqrt{\frac{GM}{r}}$
We see when plotting the keplerian function for increasing radius, we get a velocity curve with exponential decay as in the yellow scetch below. (Google "galaxy rotation curves" to see excamples of real plots)
galaxy rotation curve
Blackboard_rot_curve.png (42.86 KiB) Viewed 5575 times
Keplers law describes planetary motions with great accuracy, but somehow fails to describe orbital velocities of stars in galaxies, why is this?
Confident that Ground Potential would solve this problem, I started thinking about this and soon realised how a flat rotation curve was perfectly normal, it was instead Keplers law which was an anomaly.
According to GPT, velocity of orbiting bodies ought to increase as radius increases, because $$\Delta v$$ is proportional to $$\Delta \phi$$ so if GPT is correct, then was Kepler wrong?
It appears Kepler made a rather naive assumption, namely that planets move forward in time which turns out to be wrong, at least when looking up from a lower potential to a higher potential.
If we take the sun to be the reference point, the arrow of time points towards the centre of gravity i.e. the past is radially outwards, therefore an observer on the Sun is temporally ahead of the planets which indeed move backwards relative to the sun, so the velocities are subsequently negative. Therefore the sum of the negative velocities from consecutive Kepler orbits will result in a real velocity increase.
Planet Orbital Velocity
data-table.png (119.79 KiB) Viewed 5448 times
In the table above we can see how the forward velocity assumption differs from the retro temporal motion clearly changing the velocity curve as seen in the chart below.
Decreasing negative values resulting in increasing velocity overall.
Planet Rotation Curve
planet-plot.png (67.33 KiB) Viewed 5448 times
In the following plot I have lifted the negative velocity up into the positive number line by summing the sign reversed negative velocities.
Combined Rotation Curve
combined.png (36.99 KiB) Viewed 5448 times
So I have deliberately plotted the rotation curve in the positive quadrant to show the similarity between my plot and those measured by astronomers, like this one below. It should however appear in the bottom quadrant of the graph.
M33 Galaxy
M33_rotation_curve_HI.png (245.15 KiB) Viewed 5575 times
My conclusion is, that the temporal direction of an orbiting body depends on the potential of the observer, so for an observer looking down into a gravitational well with orbiting bodies, these bodies will appear to move according to Keplers law, ie faster the further down the well they orbit, but for an observer standing at the bottom of a potential well, looking up, the orbiting bodies move backwards at increasing velocities as the radius increases (non Keplerian motion).
When we observe a galaxy from Earth, we are looking into the past, GPT states that potential falls over time, so we should expect to se galaxies follow non Keplerian velocity curves, which indeed we do.
According to GPT there is no need to postulate any additional dark matter to explain galaxy rotation curves, they appear more or less excactly as they should.
Steven
PS: If you agree with my conclusion above, help me like, share & tweet the good news, so all the scientists trying to find dark matter can do something more useful :)
Last edited by Steven Sesselmann on Sat Apr 11, 2015 4:12 pm, edited 8 times in total.
Reason: Sorry, had to amend the velocity function to reflect the sum of differences rather than the sum of velocities - SS
Steven Sesselmann
Only a person mad enough to think he can change the world, can actually do it...
Steven Sesselmann
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In this sketch I show how the backwards moving planets give the appearance of forward motion. The apparent forward motion is an optical illusion and would have been less obvious if there were more planets in orbits. the fact that we only have a few planets means that their relative forward motion appears more obvious than their backwards motions.
Planetary Orbits
revolution.png (59.7 KiB) Viewed 5484 times
Keplers law describe the forward motion of planets, which gives us the incorrect sum for the velocity.
This discovery is a direct prediction of GPT, as Keplers law of diminishing velocity with increasing potential was simply incompatible with GPT.
In GPT the relative velocity between two bodies follows $$\Delta V = c*(\frac{\Delta \phi}{\Phi})$$ therefore velocities in consecutive orbits should/must increase with increasing radii.
Pretty revolutionary isn't it ?
Steven
Steven Sesselmann
Only a person mad enough to think he can change the world, can actually do it...
Steven Sesselmann
• Posts: 104
• Joined: Thu Jul 17, 2014 9:41 pm
• Location: Sydney - Australia
Here is a nice animation on Wikipedia showing prograde and retrograde motion of the planets.
The problem here is I don't think the interpretation is correct, because the planets currently believed to be rotating in a prograde fashion are retrograde and vise versa.
Steven Sesselmann
Only a person mad enough to think he can change the world, can actually do it...
Steven Sesselmann
• Posts: 104
• Joined: Thu Jul 17, 2014 9:41 pm
• Location: Sydney - Australia
As an ultimate test to see weather relative motion is positive or negative we can imagine having a very long string on a spool, then tying one end of the string to the ground and place the spool on the moving object.
Spool
spool.jpg (20.71 KiB) Viewed 5337 times
If the spool is unwinding, then the relative motion is negative, ie. it is receeding, which confirms it has higher potential than ground potential,on the other hand if the string is winding up on the spool, the velocity is positive, confirming that the object is approaching, and therefore of lower potential than ground.
This may seem like an obvious argument, but when it comes to planetary motion, Kepler's law contradicts the obvious and should therefore be wrong. If one were to tie one end of a string to the ground on earth and place the spool on the moon, it is obvious that the spool would be unwinding, therefore the moon is receding in its orbit, and not approaching.
Now take this concept one step further, hammer a peg into Mercury, and tie a string to Venus, then continue the string to Earth, Mars, Jupiter and Saturn, and it will immediately become obvious that these planets are moving in a retrograde fashion with increasing velocity, and not with prograde motion as currently believed to be the case.
Solar System
solarsys_poster.jpg (153.58 KiB) Viewed 5337 times
Why does it matter which way the planets rotate?
It does because, retrograde motion produces the same flat rotation curve as we see in Galaxies, which means there is no need to postulate additional dark matter in order to explain galaxy rotation rates. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8880114555358887, "perplexity": 1448.4984745570705}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221218101.95/warc/CC-MAIN-20180821092915-20180821112915-00057.warc.gz"} |
http://www.linuxtoday.com/infrastructure/2000093001006NWBRLD | # Kuro5hin.org: Review of Linux Network Administrators Guide
Sep 30, 2000, 18:11 (1 Talkback[s])
(Other stories by Paul Dunne)
[ Thanks to Paul Dunne for this link. ]
"The book is 24 chapters straight-through; but it seems to me that, like Gaul, the whole thing falls into three parts. In the first, the basics of TCP/IP networking are dealt with, taking us from an introduction to the protocols through to configuring networking hardware and getting the network acutally up and running. In the second, we turn to basic low-level features of our working network. In the third, we consider network services, what the network actually exists to do from day to day. I like this structure, though I suppose it's an obvious one. We follow a path from a newbie install through to a working network providing services. Because of the length of the book, I haven't done a blow-by-blow account of each chapter. Instead, here's the TOC, split up as I think it should be, and with remarks on points that particularly held my notice."
"Earlier in this series, and before "NAG II" was available, I reviewed TCP/IP Network Administration by Craig Hunt, another O'Reilly offering. Hunt assumes a lot; NAG goes through everything step by step. NAG is great on actually getting TCP/IP up and running on your machine -- perhaps this is overkill when today's distributions do so much for you, but it is nice to have nevertheless. The first eight chapters of NAG, pp.1-124, are all about this. Hunt, on the other hand, has a tendency to say, or rather imply, RTFM! Hunt is easier to read straight through. NAG is more a work of reference. ... NAG is very up-to-date: BIND 8, nsswitch.conf, all three varieties of Linux firewall admin., for example. Hunt, dating from 1996 for the second edition, is already sadly out-of-date in these areas. ... In conclusion, the 2nd edition of Linux Network Administrator's Guide is indeed so much improved on the first edition that it now makes a well-nigh indispensable companion to Hunt. In fact, the relationship is now reversed; if there is *one* must-get book, it's now NAG rather than Hunt. But I still recommend both."
Complete Story
Related Stories: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8535114526748657, "perplexity": 2382.047595000173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00032-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/help-on-series.6755/ | Help on Series.
1. Oct 5, 2003
mrnoll
hi.
how can I prove that if A is a converges and B diverges
that the Sum of these series (A +B) diverges..
( A = a1 + a2 + a3 + ...
B = b1 + b2 + b3 + ...) if the series start from n=1
2. Oct 5, 2003
Kalimaa23
Limit of a sum is the sum of the limits.
The convergence of the series is determined by the row of partial sums.
Therefore, if both rows converge, the sum of both will too.
3. Oct 5, 2003
HallsofIvy
In order to apply Dimitri Terryn's suggestion, you will also have to note that if C=(A+ B), then Sum(B)= Sum(C- A). Assume both Sum(C) and Sum(A) converge. What does that tell you about Sum(B)?
4. Oct 5, 2003
mrnoll
Hi,
I dont think it´s that easy actually.
Because what I forgot to say was that these series go from
n=1 to n=infinity.
And in infinity normal rules dont always apply like SumA + SumB = Sum(A+B)
I need a proof... :)
5. Oct 6, 2003
phoenixthoth
it is that easy because what the infinite series converges to is defined to be the limit of the sequence of partial sums.
6. Oct 8, 2003
HallsofIvy
In this case they do. If Sum(A) and Sum(B) exist then
Sum(A+ B) exists and is equal to Sum(A)+ Sum(B).
I had assumed you were talking about infinite series since otherwise it wouldn't make sense to talk about "convergent" and "divergent". | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9579747915267944, "perplexity": 1705.0725348769186}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513441.66/warc/CC-MAIN-20181020205254-20181020230754-00553.warc.gz"} |
http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-p?answertab=votes | # Geometric interpretation of the exact sequence for the cotangent bundle of the projective space
Edit: As Dan Petersen pointed out, this question is a duplicate of a previous one. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).
I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field): $$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$ The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$.
But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?
-
I am definitely not an expert, but what happens if you realize the conormal bundle as a sub bundle of the cotangent bundle? Don't you find that the cotangent bundle of the embedded variety gets realized as a piece of the conormal? – Filippo Alberto Edoardo Jun 2 '12 at 4:53
There is a quotient map $\pi:\mathbb A^{n+1} \setminus \{0\} \to \mathbb P^n$, and so we can pull back a differential form $\omega$ on $U \subset \mathbb P^n$ to a differential form $\pi^* \omega$ on $\pi^{-1}(U)$. The sheaf of differential forms on $\mathbb A^{n+1} \setminus \{0\}$ is globally free, with a basis $dx_0, \ldots, dx_n$. If you think about it, you can then interpret the pull-back map as an embedding $\Omega_{\mathbb P^n} \hookrightarrow \mathcal O(-1)^{n+1}$. Not every section of $\mathcal O(-1)^{n+1}$ will give rise to a differential form, though --- when you interpret such ... – Emerton Jun 2 '12 at 4:58
... a section as a differential form on $\mathbb A^{n+1}\setminus \{0\}$, to come from $\Omega_{\mathbb P^n}$ is has to be invariant under the scaling action of $\mathbb G_m$. This can be tested by pairing with the Euler class $x_0\partial_{x_0} + \cdots + x_n\partial_{x_n}$, and this gives the map $\mathcal O(-1)^{n+1} \to \mathcal O$ which appears in the exact sequence. – Emerton Jun 2 '12 at 5:00
You might be interested to know that Ravi Vakil has a short but sweet exposition of this exact sequence and its various generalizations in section 23.3 of his notes (math.stanford.edu/~vakil/216blog/FOAGmay1612public.pdf) – Sam Lichtenstein Jun 2 '12 at 5:58
This is an exact duplicate of mathoverflow.net/questions/5211 – Dan Petersen Jun 2 '12 at 12:17
By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*)$$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$
So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea inspired by differential geometry.
The "curve" $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [algebraic geometers consider very short curves!] gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}$$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to a colinear vector $(a_0',...,a_n')$, you also have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2
-
Thanks! This is pretty close to what I wanted. – auniket Jun 2 '12 at 19:38
Here is another (unknown?) way of optaining the Euler sequence (though not really geometric): Since $\Omega^1_{\mathbb{P}}$ is a coherent sheaf; by Serre it has a "twisted presentation". For that one has to find some $k > 0$ such that $\Omega^1_{\mathbb{P}}(k)$ is generated by global sections. You will find that $k=2$ suffices, namely there is an epimorphism $\bigoplus_{u<v} \mathcal{O}(-2) \twoheadrightarrow \Omega^1$, which is given on $D_+(x_i)$ by mapping
$$x_i^{-2} e_{uv} \mapsto \dfrac{x_u}{x_i} \cdot d\left(\dfrac{x_v}{x_i}\right)- \dfrac{x_v}{x_i} \cdot d\left(\dfrac{x_u}{x_i}\right).$$
You can also compute the relations between these elements and arrive at the exact sequence
$$\bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \Omega^1 \to 0.$$
But now the (graded) Koszul resolution of $R[x_0,\dotsc,x_n]/(x_0,\dotsc,x_n)$ (here $R$ is an arbitrary base ring; it doesn't have to be an algebraically closed field) yields the long exact sequence
$$\dotsc \to \bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0.$$
These combine to the Euler sequence $0 \to \Omega^1 \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9642526507377625, "perplexity": 263.2174037940512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768957.83/warc/CC-MAIN-20141217075248-00130-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/integral-problem-please-help/ | ×
What is the value of:
integral of ln|sin x| dx
(by using partial method)
Note by Dina Andini Sri Hardina
3 years, 11 months ago
Sort by:
http://www.wolframalpha.com/input/?i=integral+ln%7Csin%28x%29%7C I think you might've made a typo. · 3 years, 11 months ago
what typo? · 3 years, 11 months ago
Well the function you gave doesn't have an elementary antiderivative. · 3 years, 11 months ago
If you can't see the link, Wolfram Alpha says this evaluates to something nasty that involves imaginaries. · 3 years, 11 months ago
Integration by parts formula states that $$\int u_{}dv = uv-\int v_{}du$$
In this case we have $$\int \ln\lvert\sin x\rvert dx$$
comparing this to the integration by parts formula
$$u = \ln\lvert\sin x\rvert\Rightarrow du = \cos x\ln\lvert\sin x\rvert dx$$
$$dv = dx \Rightarrow v = x$$
Now we can apply the integration by parts formula
$$\int \ln\lvert\sin x\rvert dx = x\ln\lvert\sin x\rvert-\int x\cos x\ln\lvert\sin x\rvert dx$$
now we have an integral $$\int x\cos x\ln\lvert \sin x\rvert dx$$. though this may look more complex, it is actually more simple. we can apply integration by parts again. The requirement for integration by parts is that one variable must be easy to differentiate, in this case it is $$x$$, and one variable must be easy to integrate, in this case $$\cos x\ln\lvert\sin x\rvert dx$$. We know this is easy to integrate as we can see there is a $$\cos x$$ and a $$\sin x$$, so we can use a simple u-substitution. To make things simple, i shall removed the modulus sigh, but u can factor it back in and u will get 2 answers for the integral
Applying integration by parts:
$$u = x\Rightarrow du = dx$$
$$dv = \cos x\ln(sin x) dx \Rightarrow v = \sin x\ln(\sin x)-\sin x$$ <-- u-substitution
$$\int x\cos x\ln( \sin x) dx = x(\sin x\ln(\sin x) - \sin x)- \int (\sin x\ln(\sin x)-\sin x)dx$$
this last integral $$\int (\sin x\ln(\sin x)-\sin x)dx$$is essentially $$\int\sin x\ln(\sin x)dx - \int \sin x dx$$
$$\int\sin x\ln(\sin x) dx$$ can be found by integration by parts. after finding that, sub everything back into the orginal equation for your final answer · 3 years, 10 months ago
itu integralnya buat cari rumus apa? kecepatan, percepatan atau posisi? atau yang lain? · 3 years, 11 months ago
By partial method you mean by the method of partial sums? · 3 years, 11 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9773613810539246, "perplexity": 1223.9284942885297}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00571.warc.gz"} |
https://www.komal.hu/verseny/feladat.cgi?a=honap&h=201701&t=fiz&l=en | Magyar Information Contest Journal Articles
# KöMaL Problems in Physics, January 2017
Please read the rules of the competition.
Show/hide problems of signs:
## Problems with sign 'M'
Deadline expired on 10 February 2017.
M. 365. Measure the depth of water, at which a table-tennis ball is to be released in order that
$\displaystyle a)$ the whole ball emerges from the water;
$\displaystyle b)$ the ball emerges to the highest level above the water surface.
(6 pont)
statistics
## Problems with sign 'G'
Deadline expired on 10 February 2017.
G. 590. The vapour trail of air planes is formed not exactly at the planes, but some distance behind them, and follows the plane as if it was pulled by a rope. Why does this happen?
(3 pont)
solution (in Hungarian), statistics
G. 591. A journey from $\displaystyle A$ to $\displaystyle B$ takes 8 hours for a plane flying into a headwind, whilst the backward journey from $\displaystyle B$ to $\displaystyle A$ with tailwind takes 7 hours. If the wind speed is considered to be constant, by what factor is the speed of the air plane (without wind) greater than the wind speed?
(3 pont)
solution (in Hungarian), statistics
G. 592. A ball is dropped from a height of $\displaystyle H$ and it collides totally elastically with a slant wall at a height of $\displaystyle h<H$. The angle between the wall and the horizontal is $\displaystyle 45^\circ$.
$\displaystyle a)$ What should the height $\displaystyle h$ be in order that the range (horizontal displacement) of the ball is to be the greatest?
$\displaystyle b)$ What is this maximum range?
(3 pont)
solution (in Hungarian), statistics
## Problems with sign 'P'
Deadline expired on 10 February 2017.
P. 4894. Foucault, a French physicist, carried out his famous pendulum experiment in the Pantheon in Paris to demonstrate that the Earth is rotating, by means of a pendulum of length 67 m and of mass 28 kg.
$\displaystyle a)$ Why was it necessary to use such a long wire and such a heavy pendulum bob?
$\displaystyle b)$ What was the period of the pendulum?
(4 pont)
solution (in Hungarian), statistics
P. 4895. A rod of length $\displaystyle \ell$, and of mass $\displaystyle M$, and another rod of length $\displaystyle 2\ell$, and mass $\displaystyle 2M$ are arranged as shown in the figure. What is the direction and the magnitude of the gravitational force exerted on the point-like object of mass $\displaystyle m$? (Look for an elementary solution.)
(5 pont)
solution (in Hungarian), statistics
P. 4896. A small sport plane makes a trip, first flying into a headwind, from $\displaystyle A$ to $\displaystyle B$ towards the north in 3 hours, whilst the backward trip from $\displaystyle B$ to $\displaystyle A$ flying with tailwind takes 2 hours. How long would the full trip (i.e., forth and back) take if the wind blew constantly from the northeast? (The speed of the wind can be considered as constant.)
(4 pont)
solution (in Hungarian), statistics
P. 4897. Water flows out through a tap at the bottom of a vertical cylindrical container, open at its top. The water level in the cylinder decreases at a certain speed. How does this speed change? If half of the water flows out in time $\displaystyle T$, how long will it take to empty the full container?
(5 pont)
solution (in Hungarian), statistics
P. 4898. The arms of a U-shaped tube are vertical. The arm on the right side is closed and the other arm is closed by a moveable piston. There is mercury in the tube and initially the level of the mercury in the arms is the same. Above the mercury, there is an air column of height $\displaystyle h$ in each arm, and the initial air pressure is the same as the atmospheric pressure in both arms. How much does the mercury level in each arm move if the piston is slowly pushed down by a distance of $\displaystyle h/2$?
Data: $\displaystyle h=30$ cm, and the atmospheric pressure is the same as the pressure of a mercury column of height $\displaystyle H=76$ cm.
(4 pont)
solution (in Hungarian), statistics
P. 4899. The electric field lines of a vast region of uniform electric field are horizontal. The electric field strength is $\displaystyle E=10^4$ N/C. At one point of this region a metal ball of mass $\displaystyle m=4$ g is projected vertically upward at a speed of $\displaystyle v_0=2$ m/s. Initially the metal ball was charged positively to a charge of $\displaystyle q=3\cdot10^{-6}$ C.
$\displaystyle a)$ What is the displacement of the ball when its speed becomes the same as its initial speed was?
$\displaystyle b)$ How much time elapses until this instant?
$\displaystyle c)$ What is the minimum speed of the ball during its motion?
$\displaystyle d)$ Where is the ball when it is the slowest?
(4 pont)
solution (in Hungarian), statistics
P. 4900. An interesting optical toy consists of two opposite concave spherical mirrors, having the same radius of curvature, the one at the top having a circular hole of diameter of a few centimetres at its centre. The distance between the mirrors is set in a way, that if a small object (e.g. a piece of candy) is placed at the centre of the mirror at the bottom, then its image is formed at the centre of the mirror with the hole in it. The light beam forms the image after a reflection first in the top, then in the bottom mirror.
$\displaystyle a)$ What is the distance between the centers of the two mirrors?
$\displaystyle b)$ Is the image upright or inverted? Is it a virtual or a real image? What is the magnification?
(5 pont)
solution (in Hungarian), statistics
P. 4901. Joseph Fraunhofer, a German physicist, started his measurements in 1814 in order to investigate the spectrum of the Sun, and found 570 dark lines in the spectrum, which he denoted with letters (or sometimes letters with numbered indices). A particular diffraction grating has 500 gratings in 1 mm. Two images (formed symmetrically about the zeroth-order maximum) of one of the Fraunhofer lines are formed at a distance of 196.6 cm on the screen, which is at a distance of 3.6 m from the diffraction grating.
What is the wavelength of this spectral line and which Fraunhofer line is it?
(4 pont)
solution (in Hungarian), statistics
P. 4902. An initially neutral solid red copper ball of radius 5 cm, attached to an insulating handle, was charged by the negative terminal of a 5 kV voltage supply. (The positive terminal was earthed.) By what percent does the number of electrons of the copper ball increase?
(4 pont)
solution (in Hungarian), statistics
P. 4903. A small ball of mass $\displaystyle m$ and of charge $\displaystyle Q$ is attached to the bottom end of a piece of negligible-mass thread of length $\displaystyle L$, whose top end is fixed. The system formed by the thread and the ball is in uniform magnetic field, $\displaystyle \boldsymbol B$, which is perpendicular to the plane of the figure and points into the paper.
The ball is started in a direction perpendicular both to the magnetic induction and to the direction of the thread. At a certain initial speed the ball moves along a circular path such that the thread remains tight during the whole motion.
$\displaystyle a)$ What is the magnitude of the magnetic induction $\displaystyle B$ if the minimum initial speed at which the described motion of the ball occurs is $\displaystyle v_0=\frac 12 \sqrt{17Lg}\,$?
$\displaystyle b)$ By what factor is the force acting on the thread at point $\displaystyle A$ greater than that at point $\displaystyle C$, when the initial speed of the ball is the above stated one?
(5 pont)
solution (in Hungarian), statistics
P. 4904. The initial voltage across the capacitor of capacitance $\displaystyle C$, connected as shown in the figure is $\displaystyle 2U_0$, whilst the capacitor of capacitance $\displaystyle 2C$ is neutral.
How much heat is dissipated at the resistor of resistance $\displaystyle R$ after closing the switch?
(6 pont)
solution (in Hungarian), statistics
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http://www.math-only-math.com/properties-of-arithmetic-mean.html | # Properties of Arithmetic Mean
To solve different types of problems on average we need to follow the properties of arithmetic mean.
Here we will learn about all the properties and proof the arithmetic mean showing the step-by-step explanation.
What are the properties of arithmetic mean?
The properties are explained below with suitable illustration.
Property 1:
If x is the arithmetic mean of n observations x1, x2, x3, . . xn; then
(x1 - x) + (x2 - x) + (x3 - x) + ... + (xn - x) = 0.
Now we will proof the Property 1:
We know that
x = (x1 + x2 + x3 + . . . + xn)/n
⇒ (x1 + x2 + x3 + . . . + xn) = nx. ………………….. (A)
Therefore, (x1 - x) + (x2 - x) + (x3 - x) + ... + (xn - x)
= (x1 + x2 + x3 + . . . + xn) - nx
= (nx - nx), [using (A)].
= 0.
Hence, (x1 - x) + (x2 - x) + (x3 - x) + ... + (xn - x) = 0.
Property 2:
The mean of n observations x1, x2, . . ., xn is x. If each observation is increased by p, the mean of the new observations is (x + p).
Now we will proof the Property 2:
x = (x1 + x2 +. . . + xn)/n
⇒ x1 + x2 + . . . + xn) = nx …………. (A)
Mean of (x1 + p), (x2 + p), ..., (xn + p)
= {(x1 + p) + (x2 + p) + ... + (x1 + p)}/n
= {(x1 + x2 + …… + xn) + np}/n
= (nx + np)/n, [using (A)].
= {n(x + p)}/n
= (x + p).
Hence, the mean of the new observations is (x + p).
Property 3:
The mean of n observations x1, x2, . . ., xn is x. If each observation is decreased by p, the mean of the new observations is (x - p).
Now we will proof the Property 3:
x = (x1 + x2 +. . . + xn)/n
⇒ x1 + x2 + . . . + xn) = nx …………. (A)
Mean of (x1 - p), (x2 - p), ...., (xn - p)
= {(x1 - p) + (x2 - p) + ... + (x1 - p)}/n
= {(x1 + x2 + …. + xn) - np}/n
= (nx - np)/n, [using (A)].
= {n(x - p)}/n
= (x - p).
Hence, the mean of the new observations is (x + p).
Property 4:
The mean of n observations x1, x2, . . .,xn is x. If each observation is multiplied by a nonzero number p, the mean of the new observations is px.
Now we will proof the Property 4:
x = (x1 + x2 + . . . + xn)/n
⇒ x1 + x2 + . . . + xn = nx …………… (A)
Mean of px1, px2, . . ., pxn,
= (px1 + px2 + ... + pxn)/n
= {p(x1 + x2 + ... + xn)}/n
= {p(nx)}/n, [using (A)].
= px.
Hence, the mean of the new observations is px.
Property 5:
The mean of n observations x1, x2, . . ., xn is x. If each observation is divided by a nonzero number p, the mean of the new observations is (x/p).
Now we will proof the Property 5:
x = (x1 + x2 + ... + xn)/n
⇒ x1 + x2 + ... + xn) = nx …………… (A)
Mean of (x1/p), (x2/p), . . ., (xn/p)
= (1/n) ∙ (x1/p + x2/p + …. xn/p)
= (x1 + x2 + ... + xn)/np
= (nx)/(np), [using (A)].
= (x/p).
To get more ideas students can follow the below links to understand how to solve various types of problems using the properties of arithmetic mean. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9565385580062866, "perplexity": 1486.2844568470034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721555.36/warc/CC-MAIN-20161020183841-00335-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://www.openproblemgarden.org/op/approximation_ratio_for_maximum_edge_disjoint_paths_problem | # Approximation Ratio for Maximum Edge Disjoint Paths problem
Importance: Medium ✭✭
Author(s): Bentz, Cedric
Subject: Graph Theory
Keywords: approximation algorithms Disjoint paths planar graph polynomial algorithm
Posted by: jcmeyer on: April 18th, 2010
Conjecture Can the approximation ratio be improved for the Maximum Edge Disjoint Paths problem (MaxEDP) in planar graphs or can an inapproximability result stronger than -hardness?
Assume a flow graph with vertices and edges. Each edge has a capacity function (Flow network). The graph contains a list of terminal vertices called sources () and sinks (). Each pair () defines a net or commodity.
A Multiflow is a way of routing commodities from their sources to the respective sinks while ensuring that the flow of each commodity is conserved at each non-terminal vertex and that the sum of the flows of all commodities through an edge does not exceed the capacity of the edge.
The Maximum Integer Multiflow problem (MaxIMF) seeks to maximize the number of flow units routed between the nets in the graph. The Maximum Edge Disjoint Paths (MaxEDP) problem seeks to find the maximum number of disjoint paths between the sources and sinks. When the capacities for all edges are set to one, MaxIMF simplifies into the MaxEDP problem.
Bentz provides an algorithm to find the MaxEDP with a proven approximation ratio (Approximation and integrality gap) of . Can the approximation ratio be improved for MaxEDP in planar graphs, or can an inapproximability result stronger than -hardness be proved for this problem? And what about the general graphs?
## Bibliography
Cédric Bentz, Edge disjoint paths and max integral muliflow/min multicut theorems in planar graphs, Electronic Notes in Discrete Mathematics 22 (2005), 55–60
* indicates original appearance(s) of problem. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8909696936607361, "perplexity": 1495.5363574603493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578650225.76/warc/CC-MAIN-20190424154437-20190424180437-00261.warc.gz"} |
http://www.openproblemgarden.org/category/fixed_point | # fixed point
## Nonseparating planar continuum ★★
Author(s):
\begin{conjecture} Does any path-connected, compact set in the plane which does not separate the plane have the fixed point property?
A set has the fixed point property if every continuous map from it into itself has a fixed point. \end{conjecture}
Keywords: fixed point | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8990148901939392, "perplexity": 572.4452202226959}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886792.7/warc/CC-MAIN-20180117003801-20180117023801-00683.warc.gz"} |
http://lbc.oa-roma.inaf.it/Documentation/Science/node32.html | Next: The Clustering of Faint Up: Extremely Red Galaxies Previous: Scientific Justification
### Survey Characteristics and Observation Strategy
It follows from the above discussion that the nature of the extremely red galaxies should be intensively investigated in order to understand their role in galaxy evolution. The reddest galaxies (candidate to be the most dusty) have a poorly determined sky surface density of 0.010- 0.015 arcmin-2 at (Hu & Ridgway 1994), and they are very faint in the optical.
Because of their extremely steep and red spectra, these galaxies can be found with (1) moderately deep J-band imaging, (2) with very deep optical imaging in BVRI, and bf (3) covering a relatively large area of the sky.
The availability of a J filter for the LBT Wide Field Imager provides a unique possibility to perform a very deep and wide-field survey for extremely red galaxies. Adopting a conservative sky surface density of 0.01 arcmin-2 (Hu & Ridgway 1994), we will need to image about 3 square degrees in order to select a statistically significant and complete sample of about 100 objects with . Such a sky area can be covered with the LBT Wide Field Imager (FOV arcmin = 600 arcmin2) observing totally 18 fields at moderately deep magnitude limit in J and at very deep limits in BVRI. The sample will be selected in J, but the key point in such a survey will be to reach extremely deep magnitudes in BVRI in order to be able to : (1) reliably find and classify the extremely red galaxies, (2) characterize their SEDs, i.e. detecting them in at least one of the BVRI bands. For example, if the limiting (completeness) magnitude in J will be 23.0, we will need to detect objects with magnitudes , and even fainter in BVR. These limits will be reachable with integration times in Iof the order of >10000 seconds. The great advantage of the LBT will be the possibility to image simultaneously the same field with the blue and the red channels.
The use of at least 3 filters (e.g. RIJ) will be also very useful in order to produce colour-colour plots (R-K) vs. (R-I) where it is easier to investigate, to isolate and to characterize the various populations of objects present in the field (stars, brown dwarfs, galaxies at low- and high- redshifts).
The proposed survey will immediately constrain the surface density of these galaxies, and it will provide samples for additional photometric follow-up (in order to estimate the photometric redshifts) and for spectroscopic optical and/or near-IR spectroscopy with 8-10m class telescopes (when the object will be not too faint).
In addition, the JCMT+SCUBA and IRAM 30m telescopes will be used to do photometry at 450,850,1250m in order to measure the dust content and the far-IR luminosity of these systems. We then can address the following issues related to these galaxies: (1) what is their space density ? (2) what fraction of distant galaxies is hidden by dust ? (3) how much do they contribute to the total star formation rate per comoving volume in the Universe, and does it really peak at ? (4) what fraction of stars are being formed in major starburst as opposed to the more quiescent star-forming rates as inferred for spirals ? (5) is it a likely scenario that a large fraction of massive elliptical went through a dusty phase in which most of their stars were formed ? (6) Are the extremely red galaxies a major contributor to the cosmic far-IR-mm background ? (7) As a byproduct, we will investigate also the galaxies which will turn out to have no dust thermal emission and whose colours are therefore intrinsically red because of an evolved (>1 Gyr) stellar population, thus being cosmologically important because they would represent the oldest envelope of the distant galaxy population, and whose ages would in principle constrain the earliest epoch of galaxy formation, H0 and q0.
Finally, we would like to stress a point which is valid for our two programmes, and also probably for several others: the LBT-WF imager will be a necessary complement of the other LBT instruments (spectrographs and high resolution imagers), since it takes a very efficient and wide field imager like the LBT-WF to find interesting targets at a sufficient rate to feed the other LBT instruments.
Next: The Clustering of Faint Up: Extremely Red Galaxies Previous: Scientific Justification
Guido Buscema
1999-01-29 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8128540515899658, "perplexity": 1943.2452700898407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637899132.1/warc/CC-MAIN-20141030025819-00112-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://ashwinnarayan.blogspot.com/2010/08/i-for-i.html | ## Saturday, 31 July 2010
### An 'i' for an 'i' !!!
I'm learning about complex and imaginary numbers right now. Now, the previous sentence probably throws my less mathematically inclined readers into a state of utter confusion. I'm afraid it just can't be helped because this little mathematical curiosity is so interesting that I find myself unable to delay the moment of revelation even to give a short overview about what complex and imaginary numbers are. If you are very keen on understanding this post I suggest you brush up by clicking here and here.
As is usual when I start learning about a topic I started to fiddle around with the concept. A little bit of manipulation here and there goes a long way. Very soon I stumbled across a mystery which I managed to explain to myself. We all know that
i 2 = −1
$i^{4n} = 1\,$
$i^{4n+1} = i\,$
$i^{4n+2} = -1\,$
and
$i^{4n+3} = -i.\,$
So I took this simple problem: i^5 and decided that it could be represented as (i^4)^(5/4). Since i^4 = 1 the expression can be simplified to 1^(5/4). 1^(5/4) = 1.
I was mystified. I could use this technique to prove that any power of i is equal to one. Had I discovered some disrepancy, some fallacy in mathematics? Not likely, I thought. So I set out to analyse my results.
I wish to prove that i^(5) is i and not 1.
Very soon I found the error in my calculations. I was just beginning to understand how quirky imaginary numbers can be. My mistake lay in assuming that 1^(5/4) = 1. After some thinking, I figured out that it was not. Here's what I did. I reasoned that
1^(5/4) = (1^5)^(1/4)
= (1^(1/2))^(1/2) Now since 1^(1/2) = +1 or -1
= 1^(1/2) or -1^(1/2)
= -1 or 1 or i or -i
So in the end I concluded that by breaking up the exponent of 5 to 4 and 5/4 I was actually introducing 3 new solutions into the problem. It is similar to a situation in which you have a linear equation and by squaring the equation you introduce new spurious solutions which were never part of the original problem. Since (i^4)^(5/4) is not equal to i^5, i^5 remains equal to i and only i. Q.E.D
Interesting isn't it? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8375369906425476, "perplexity": 644.7470843400997}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818695439.96/warc/CC-MAIN-20170926103944-20170926123944-00453.warc.gz"} |
https://www.physicsforums.com/threads/physics-12-kinematics-question.243393/ | # Physics 12 Kinematics Question
1. Jul 4, 2008
### jameo15
1. The problem statement, all variables and given/known data
A ball rolls off the top of a 12.5m building with a speed of 3.4m/s. How long does it take to hit the ground? What is the range from the base of the building?
2. Relevant equations
V = V0 + at
Avg V = (V = V0) /2
d = V0 + 1/2 at2
3. The attempt at a solution
All I've got was a sketch drawn.. Please help me. Well I timed 12.5 by 3.4 to get how long it took to hit the ground but I'm not sure if it is right. I'm guessing to find the range you must need an angle of some sort?
Last edited: Jul 4, 2008
2. Jul 4, 2008
### Hootenanny
Staff Emeritus
Welcome to PF jameo15,
You need to consider the horizontal and vertical components of motion separately. When the ball rolls off the building, just at the point where it leaves the surface, what is it's vertical speed?
3. Jul 4, 2008
### jameo15
Would it's vertical speed be the acceleration due to gravity? But at the point where it leaves the surface, it should be zero right?
4. Jul 4, 2008
### Hootenanny
Staff Emeritus
No, but that would be it's acceleration.
Correct. So using this information, can you determine how long it takes to hit the ground (just considering vertical motion).
Similar Discussions: Physics 12 Kinematics Question | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8314412832260132, "perplexity": 967.7394902669934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948604248.93/warc/CC-MAIN-20171218025050-20171218051050-00340.warc.gz"} |
http://mathoverflow.net/questions/64517/faithful-actions-of-finite-groups-on-topological-spaces | # Faithful actions of finite groups on topological spaces
Suppose that $G$ is a finite group acting faithfully on a topological space $X$. In the smooth setting, one can deduce that for each $x$ in $M$, the induced map $$G_x \to Diff_x\left(M\right)$$ from the isotropy group of $x$ to the group of germs of locally defined diffeomorphisms is a monomorphism. Does this result hold true in the topological setting? (Replacing diffeomorphisms with homeomorphisms)
In the smooth setting, the result follows from the following standard lemma:
Lemma: *Let $M$ be a manifold and $G$ a finite subgroup of $\mathit{Diff}\left(M\right)$. Then for any smooth map $f:V \to M$ defined on a non-empty open connected submanifold of $M,$ such that $f\left(x\right) \in G \cdot x$ for all $x,$ there exists a unique element $g \in G$ such that $f=g|_V.$ *
I'm guessing the answer is NO for general topological spaces. However, a counter-example would be nice.
-
The answer is no. Consider your space to be a wedge of a bunch of circles -- and don't use common wedge points. Then an involution can fix a neighbourhood of a point and still flip some of the circle wedge summands. – Ryan Budney May 10 '11 at 17:47
I suspect the answer is yes for connected topological manifolds but I don't have an argument off the top of my head. These kinds of questions were investigated in the 60's as far as I know. – Ryan Budney May 10 '11 at 18:57
@Ryan: Thanks. I'm glad to see that there is such a simple counter-example. – David Carchedi May 10 '11 at 23:03
maybe Ryan or David can repost this as an answer and close the question ? – HenrikRüping May 11 '11 at 12:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8321219682693481, "perplexity": 158.52505974011135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00369-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.enotes.com/homework-help/how-do-you-solve-this-polynomial-331296 | # How do you solve this polynomial: (3x + 1 )(x2 – 4x + 2)
You need to perform multiplication of two polynomials, hence you need to multiply each term from first brackets to each term from the second brackets such that:
`(3x + 1 )(x^2 – 4x + 2) = 3x*x^2 + 3x*(-4x) + 3x*2 + 1*x^2 + 1*(-4x) + 1*2`
`(3x + 1 )(x^2 – 4x + 2) = 3x^3 - 12x^2 + 6x + x^2 - 4x + 2`
You need to add or subtract the coefficients of like powers such that:
`(3x + 1 )(x^2 – 4x + 2) = 3x^3 + x^2(-12 + 1) + x(6 - 4) + 2`
`(3x + 1 )(x^2 – 4x + 2) = 3x^3 - 11x^2 + 2x ` `+ 2`
Hence, performing the multiplication of polynomials yields `(3x + 1 )(x^2 – 4x + 2) = 3x^3 - 11x^2 + 2x + 2.`
Approved by eNotes Editorial Team
Every term in the first binomial must be multiplied by every term in the trinomial.
Therefore, when multiplying 3x by every term in the second set of parentheses, you get:
3x^3 - 12x^2 + 6x
and multiplying 1 by every term in the second set of parentheses gives you:
x^2 - 4x + 2
Combine the two, and you get:
3x^3 - 12x^2 + 6x + x^2 - 4x + 2
Now, combine like terms and write your answer in descending order of the variable:
3x^3 - 11x^2 + 2x + 2
Approved by eNotes Editorial Team | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9612109065055847, "perplexity": 826.9110305189599}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00735.warc.gz"} |
https://www.physicsforums.com/threads/exponentiation-multiplication-addition.169807/ | # ?, exponentiation, multiplication, addition, ?
1. May 10, 2007
### Loren Booda
What simple operations, if any, precede or succeed the series ". . . exponentiation, multiplication, addition. . ."?
2. May 10, 2007
### Moo Of Doom
..., hexation, pentation, tetration, exponentiation, multiplication, addition, ...
Maybe succession is a good candidate for the next operation in the sequence. I don't think there's much after that, though.
3. May 11, 2007
### chroot
Staff Emeritus
Are you trying to remember PEMDAS?
- Warren
4. May 11, 2007
### Loren Booda
chroot,
More like: multiplication represents repeated additions, and exponentiation represents repeated multiplications, etc. (Thanks for the PEMDAS blast to my algebraic past, though.)
5. May 11, 2007
### Moo Of Doom
Yep, that's the list I gave you.
Tetration ($\uparrow\uparrow$) is repeated exponentiation, right associated:
$$a\uparrow\uparrow b = a^{a^{...^{a^a}}}$$
where there are $b$ $a$'s on the left side.
$$2\uparrow\uparrow 2 = 2^2 = 4$$
$$2\uparrow\uparrow 3 = 2^{2^2} = 2^4 = 16$$
$$2\uparrow\uparrow 4 = 2^{2^{2^2}} = 2^{16} = 65536$$
Pentation ($\uparrow\uparrow\uparrow$) is repeated tetration, also right associated, etc.
Succession is adding 1, so you might say that adding a and b is like adding 1 to a, b times.
EDIT: I use Knuth uparrow notation for the higher operators. This is not strictly speaking a universally agreed upon thing, so you might have to explain it to pretty much anyone you show it to.
Last edited: May 11, 2007
6. May 11, 2007
### Loren Booda
Moo,
That's what I sought. "Succession" as you define it seems reasonable to me. My guess is that the next step involves fractals. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8873884081840515, "perplexity": 3758.469902041053}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00183.warc.gz"} |
https://tex.stackexchange.com/questions/455375/what-does-aka-backslash-exclamation-mark-do | What does “\!” aka backslash exclamation mark do?
I just came across this ! (backslash exclamation mark) in the code for a table in somebody else's tex and I was wondering what it does.
\begin{table}[hbt]
\centering
\resizebox{\columnwidth}{!}{%
\setlength\extrarowheight{1pt}
\begin{tabular}{l | c c c | c r}
\toprule
\textbf{Thing} & $a$ & $b$ & $c$ & $d$ & $e$\\
\midrule
Stuff & \!\!100\!\! & \!\!$10^{6}$\!\! & \!\!$\pi_0$\!\! & $\sigma$ & $2^{30}$\\
\bottomrule
\end{tabular}%
}
\end{table}
My tex editor says unrecognized command but it compiles just fine. I've tried on overleaf and it seems to tighten the table somehow. Does anybody know more?
• A \! is a negative space, see tex.stackexchange.com/questions/9091/… for more information. – BambOo Oct 16 '18 at 9:13
• It doesn't compile fine! – egreg Oct 16 '18 at 9:42
• @egreg I'm not getting any errors for the overleaf example. What do you mean? – Elias Oct 16 '18 at 16:56
• @Elias I get 11 errors, after adding a suitable minimal preamble. – egreg Oct 16 '18 at 17:14
• Hm..strange. Compiling the overleaf example with pdflatex I don't get any. – Elias Oct 16 '18 at 17:18
I don't know what the commands \! are supposed to do in the code you show, other than raising many errors.
The command \! is only allowed in math mode (unless redefined, which I'd discourage). It's purpose is to insert a negative thin space, which is useful in several places. For instance
\biggl(\frac{121}{12}-1\biggr)^{\!2}
is an improvement over ^{2}, because it moves the exponent towards the parenthesis and takes care of its bending.
Another place where it is helpful is in 2/\!\log x, because for technical reasons / is an ordinary symbol and TeX would add a thin space between it and the “log” operator.
Here's a picture: left the output with \!, right without.
The negative spacing of \! exactly matches the positive one by \, (which is automatically inserted in some places) in math mode. Note that \, can be used outside of math mode, where it inserts a sixth of a quad of space.
If you want to tighten a table, reduce the size of \tabcolsep. Using explicit negative spaces all over the place is not the correct way.
And never use \resizebox on a table.
• Huh...I always thought this kind of stuff was automatic in tex. I have to fix such spaces manually? – Elias Oct 16 '18 at 16:57
• @Elias There are more things in heaven and math formulas than are dreamt of by your TeX engine. – egreg Oct 16 '18 at 17:12
• In any case, good to know, although I have no intention of fiddling with this kind of thing. :) – Elias Oct 16 '18 at 17:15
This symbol is usually defined as
\! negative thin space (normally 1/6 of a quad)
It is commonly used in tables, and formulas among other ways to manipulate spaces in LaTeX
And before you do post a question, please do run a search with similar keywords on this site. Hope this helps.
• Commonly used in tables? Please explain. I hardly never see it used other than in a few math constructions. – daleif Oct 16 '18 at 9:34
• @daleif I am not sure if I can provide examples that are relevant here, since it is a matter of usage, and remember coming across it in tables. – GermanShepherd Oct 16 '18 at 9:38
• @GermanShepherd I tried searching for it but got no useful results. The search can't handle the special symbols, I guess. I didn't know it was called a negative thin space. ;) – Elias Oct 16 '18 at 16:54
• @daleif Well, it can be used also for more bizarre goals as draw a donkey :-) – Fran Oct 16 '18 at 19:19
• @Fran That was a good one :-) – GermanShepherd Oct 23 '18 at 3:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.897862434387207, "perplexity": 1317.3020078931802}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00011.warc.gz"} |
https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book%3A_Statistical_Mechanics_(Styer)/06%3A_Quantal_Ideal_Gases/6.07%3A_Bose-Einstein_Statistics | # 6.7: Bose-Einstein Statistics
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Perhaps you thought the Fermi-Dirac results were strange: Non-interacting particles forming a collection as hard as steel. . . room temperature being effectively zero. Wait until you see the Bose-Einstein results.
## 6.7.1 Theory
For independent bosons, whether free or subject to an external potential, the mean occupation number function is
$b(\mathcal{E})=\frac{1}{e^{\beta(\mathcal{E}-\mu)}-1}.$
To begin to understand this function, note that
$\begin{array}{cc}{\text { when: }} & {\text { we have: }} \\ \hline \mathcal{E}<\mu & {b(\mathcal{E}) \text { negative }} \\ {\mathcal{E}=\mu} & {b(\mathcal{E})=\infty} \\ {\mathcal{E}>\mu} & {b(\mathcal{E}) \text { positive }}\end{array}$
Thus this function has the general character sketched below:
Although we have written $$b ( \mathcal{E})$$ as a function of the continuous variable $$\mathcal{E}$$, we will in fact have occasion to evaluate it only at the energy eigenvalues $$\epsilon_1,\epsilon_2, \epsilon_3,. . . , \epsilon_r,. . .$$, eigenvalues which of course differ for different external potentials. It seems bizarre that b(E) can be negative, and indeed this is only a mathematical artifact: Recall that in our derivation of the Bose function we needed to assume that µ < $$\varepsilon_1$$ in order to insure convergence (see equation 6.35). Evaluating $$b ( \mathcal{E})$$ at any eigenvalue r will always result in a positive mean occupation number.
The character of the Bose function is dominated by the singularity at $$\mathcal{E}$$ = µ, so in trying to understand the function and its physical implications one must first locate the chemical potential. This section will provide a tour of Bose-Einstein behavior with decreasing temperature, throughout which the chemical potential shifts to the right. (When we investigated Fermi-Dirac behavior, we started at T = 0 and toured with increasing temperature, so the chemical potential shifted left.) This rightward shift presents a potential problem, because as the temperature decreases µ might shift right all the way to $$\epsilon_1$$, and we know that µ < 1. We will just have to go ahead and take the tour, being wary because we know that a µ = 1 road block might pop up right in front of us as we view the countryside. With any luck µ will not yet have reached $$\epsilon_1$$ when our tour halts at T = 0.
For the case of free and independent bosons (subject to periodic boundary conditions), the ground level energy is $$\epsilon_1$$ = 0. The natural first step is to find µ(T, V, N) by demanding that
$N=\int_{0}^{\infty} G(\mathcal{E}) b(\mathcal{E}) d \mathcal{E}.$
Natural though this may be, caution is in order. Remember that the integral above is an approximation to the sum over discrete energy levels
$N=\sum_{r} b\left(\epsilon_{r}\right).$
It is legitimate to replace the sum with the integral when the value of $$b ( \mathcal{E})$$ changes little from one energy level to the next. We saw in section 6.5 that in the thermodynamic limit, the free-particle level spacings approach zero, so for the most part this approximation is excellent. . . even exact in the thermodynamic limit. But there is one exception: At $$\epsilon$$ = µ, the Bose function $$b ( \mathcal{E})$$ is infinite, so b(µ) = ∞ is very different from b(µ + δ), no matter how small the positive number δ is. Usually we can ignore this caution, because µ < 1 = 0. But if µ approaches 0 then we expect the approximation (6.71) to fail.
In summary, the integral (6.71) is a good approximation for the sum (6.72) as long as the integrand varies slowly. Now for any value of $$\mathcal{E}$$ greater than µ, you can make b($$\mathcal{E}$$ + δ) very close to b(E) simply by choosing δ > 0 small enough. This is what happens in the thermodynamic limit. But for $$\mathcal{E}$$ = µ, there is always a large difference between $$b ( \mathcal{E})$$ = ∞ and b($$\mathcal{E}$$ + δ), which is finite. Thus the integral approximation will be a good one as long as we avoid $$\mathcal{E}$$ = µ.
In situations where the sum can be legitimately replaced with the integral, we have
$N=\int_{0}^{\infty} G(\mathcal{E}) b(\mathcal{E}) d \mathcal{E}$
$=V\left[\frac{\sqrt{2 m^{3}}}{2 \pi^{2} \hbar^{3}}\right] \int_{0}^{\infty} \sqrt{\mathcal{E}} \frac{1}{e^{\beta(\mathcal{E}-\mu)}-1} d \mathcal{E}.$
Use of the obvious substitution $$x = \beta \mathcal{E}$$ gives
$N=V\left[\frac{\sqrt{2 m^{3}}}{2 \pi^{2} \hbar^{3}}\right]\left(k_{B} T\right)^{3 / 2} \int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x} e^{-\beta \mu}-1} d x,$
and remembering the definition (5.4) of thermal de Broglie wavelength results in a more convenient expression
$N=\frac{2}{\sqrt{\pi}} \frac{V}{\lambda^{3}(T)} \int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x} e^{-\beta \mu}-1} d x.$
Note that the definite integral above is not “just a number”. . . it is a function of the product βµ. There is no closed-form expression for the integral, although it is readily evaluated numerically and can be found in tabulations of functions. However it is easy to find an analytic upper bound: Because µ < 0, we have e−βµ > 1 whence
$\int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x} e^{-\beta \mu}-1} d x<\int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x}-1} d x.$
The expression on the right is just a number, and a little bit of work (see problem 6.26) shows that it is the number
$\zeta\left(\frac{3}{2}\right) \frac{\sqrt{\pi}}{2},$
where the Riemann zeta function is defined by
$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}} \quad \text { and } \quad \zeta\left(\frac{3}{2}\right)=2.612375348 \dots$
So, how does this upper bound help us? It shows that
$N=\frac{2}{\sqrt{\pi}} \frac{V}{\lambda^{3}(T)} \int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x} e^{-\beta \mu}-1} d x<\frac{V}{\lambda^{3}(T)} \zeta\left(\frac{3}{2}\right),$
but what is the significance of this result? Remember that the upper bound is just the value of the function when µ = 0, which is exactly where we expect a road block due to the breakdown of the integral approximation (6.71). Our hopes that we could avoid the issue have been dashed. The breakdown occurs at the temperature T0 that satisfies
$N=\frac{V}{\lambda^{3}\left(T_{0}\right)} \zeta\left(\frac{3}{2}\right) \quad \text { or } \quad \lambda^{3}\left(T_{0}\right)=\zeta\left(\frac{3}{2}\right) / \rho$
or
$T_{0}(\rho)=\frac{h^{2}}{2 \pi m k_{B} \zeta^{2 / 3}\left(\frac{3}{2}\right)} \rho^{2 / 3}.$
For temperatures above T0, the chemical potential shifts right as the temperature falls, and the integral approximation (6.71) is legitimate (indeed, exact in the thermodynamic limit). But as the temperature decreases below T0, the chemical potential sticks at µ = 0 and the integral approximation (6.71) must be modified.
What is the proper modification? The function $$b(\mathcal{E})$$ is slowly varying for all values of $$\mathcal{E}$$ except $$\mathcal{E}$$ = µ, so the integral approximation is legitimate for all the energy levels except the ground level at 1 = 0 = µ. Only for the ground level is some other result needed, so we just add in the ground level occupancy by hand:
$N=\left\langle n_{1}\right\rangle+\int_{0}^{\infty} G(\mathcal{E}) b(\mathcal{E}) d \mathcal{E}.$
Now, we have already seen that when µ = 0—the condition for validity of this equation—the integral can be evaluated exactly and we have
$N=\left\langle n_{1}\right\rangle+\frac{V}{\lambda^{3}(T)} \zeta\left(\frac{3}{2}\right).$
Note that $$\langle n_1 \rangle$$ here is not given by its traditional formula (6.45), because
$\frac{1}{e^{\beta(\mathcal{E}-\mu)}-1}=\frac{1}{e^{0}-1}=\infty.$
Instead, equation (6.84) is the formula for $$\langle n_1 \rangle$$ when T < T0. The mean ground level occupancy $$\langle n_1 \rangle$$ is an intensive quantity when T > T0 but an extensive quantity when T < T0.
In summary, the correct normalization equation breaks into two parts, namely
$N=\left\{\begin{array}{ll}{\frac{2}{\sqrt{\pi}} \frac{V}{\lambda^{3}(T)} \int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x} e^{-\beta \mu}-1} d x} & {\text { for } T>T_{0}(\rho)} \\ {\left\langle n_{1}\right\rangle+\frac{V}{\lambda^{3}(T)} \zeta\left(\frac{3}{2}\right)} & {\text { for } T<T_{0}(\rho)}\end{array}\right.$
We should expect that each part will behave quite differently, i.e. we expect a sudden change of behavior as the temperature drops through T0.
What does all this mean physically? A naive reading of equation (6.80) suggests an upper bound on the number of particles that can be placed into the volume V. This would be sensible if the particles were marbles with hard-core repulsions. But these are non-interacting particles! Surely we can add more particles just by throwing them into the container. Indeed we can do so, and the associated excess mean occupation number is due to the level that particles like best at low temperatures, namely the ground level. The ground level thus has a much higher mean occupancy than the first excited level, and this rapid variation of mean occupancy with energy renders the approximation of sum by integral invalid. The inequality (6.80) does not limit the number of particles in the system: instead it shows the domain within which it is legitimate to approximate the sum (6.72) by the integral (6.71).
The abrupt transition at T0(ρ) is called Bose-Einstein condensation and the material at temperatures below T0(ρ) is called the Bose condensate. These terms are unfortunate: they conjure images of a gas condensing into a liquid, in which circumstance the atoms separate into two different classes: those in the liquid and those remaining in the vapor. This suggests that in Bose-Einstein condensation too there are two classes of particles: those in the ground level and those in the excited levels. This picture is totally false. It is incorrect to say “one particle is in the ground level, another is in the fourth excited level”. In truth the individual particles are not in individual levels at all: instead the whole system is in a state produced by multiplying together the individual level wavefunctions (“building blocks”) and then symmetrizing them. The literature of Bose-Einstein statistics is full of statements like “at temperatures below T0, any particle added to the system goes into the ground level.” Such statements are wrong. They should be replaced with statements like “at temperatures below T0, any increase in particle number occurs through an increase in $$\langle n_1 \rangle$$, the mean occupancy of the ground level.” Or alternatively, “at temperatures below T0, it is very likely that many of the building blocks from which the system wavefunction is built are the ground level.” Or again, to be absolutely precise, “at temperatures below T0, if the energy is measured then it is very likely that many of the building blocks from which the resulting energy eigenfunction is built are the ground level.” Read again the previous paragraph—the one that begins “What does all this mean physically?”. Notice that I never need to say that a particle “is in” or “goes into” a given level.
## 6.7.2 Experiment
References: M.H. Anderson, J.R. Ensher, M.R. Matthews, C.E. Wieman, E.A. Cornell, “Observation of Bose-Einstein condensation in a dilute atomic vapor”, Science, 269 (14 July 1995) 198–201; Randall G. Hullet at Rice; Malcolm W. Browne, “Two groups of physicists produce matter that Einstein postulated”, New York Times, 14 July 1995, page 1.
## 6.7.3 Problems
6.23 Character of the Bose function
What are the limits of the Bose function $$b( \mathcal{E})$$ (equation 6.70) as $$\mathcal{E}$$ → ±∞? Is the curvature of the function greater when the temperature is high or when it is low?
6.24 Thermodynamics of the Bose condensate
For temperatures less than the Bose condensation temperature T0, find the energy, heat capacity, and entropy of an ideal gas of spin-zero bosons confined to a volume V. Write your answers in terms of the dimensionless integral
$I=\int_{0}^{\infty} \frac{x^{3 / 2}}{e^{x}-1} d x$
but don't bother to evaluate it. Show that
$C_{V}=\frac{5}{2} \frac{E}{T} \quad \text { and } \quad S=\frac{5}{3} \frac{E}{T}.$
6.25 More thermodynamics of the Bose condensate
For the system of the previous problem, show that
$F=-\frac{2}{3} E \quad \text { and } \quad p=\frac{2}{3} \frac{E}{V}.$
From this show that at low temperatures, the pressure of a collection of free and independent bosons goes like pT5/2. (This is always less than the classical pressure pT.)
6.26 An integral important for Bose condensation
Show that
$\int_{0}^{\infty} \frac{x^{1 / 2}}{e^{x}-1} d x=\zeta\left(\frac{3}{2}\right) \frac{\sqrt{\pi}}{2}$
where
$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}}.$
Clue:
$\frac{1}{e^{x}-1}=\frac{1}{e^{x}\left(1-e^{-x}\right)}=e^{-x} \sum_{n=0}^{\infty} e^{-n x} \quad \text { for } \quad x>0.$
6.7: Bose-Einstein Statistics is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9451172351837158, "perplexity": 541.0456518455866}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337803.86/warc/CC-MAIN-20221006092601-20221006122601-00742.warc.gz"} |
http://mathhelpforum.com/calculus/40666-integral-ln-x-2-a.html | # Math Help - integral (ln(x))^2
1. ## integral (ln(x))^2
Man I'm having problems.
Question is
$\int_1^2 {(\ln (x))^2 } dx \\$
I get this
$= \int_1^2 {\ln (x)\ln (x)dx} \\$
$= \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\$
$= \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\$
$= \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\$
$= \frac{1}{x}\ln (x) + \frac{1}{x} \\$
$= [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\$
$= \frac{1}{2}\ln (2) - \frac{1}{2} \\
$
But apparently it should be
$= 2(\ln (2))^2 - 4\ln (2) + 2
$
2. Originally Posted by Craka
Man I'm having problems.
Question is
$\int_1^2 {(\ln (x))^2 } dx \\$
I get this
$= \int_1^2 {\ln (x)\ln (x)dx} \\$
$= \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\$
$= \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\$
$= \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\$
$= \frac{1}{x}\ln (x) + \frac{1}{x} \\$
$= [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\$
$= \frac{1}{2}\ln (2) - \frac{1}{2} \\
$
But apparently it should be
$= 2(\ln (2))^2 - 4\ln (2) + 2
$
You have not applied the integration by parts forumla correctly. Go back and check what it says.
There are many ways of doing it. The best way for you I think is to make the substitution $u = \ln x$. Then the integral becomes $\int_{0}^{\ln 2} u^2 \, e^u \, du$. Now use integration by parts - twice.
3. Originally Posted by Craka
Man I'm having problems.
Question is
$\int_1^2 {(\ln (x))^2 } dx \\$
I get this
$= \int_1^2 {\ln (x)\ln (x)dx} \\$
$= \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\$
$= \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\$
$= \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\$
$= \frac{1}{x}\ln (x) + \frac{1}{x} \\$
$= [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\$
$= \frac{1}{2}\ln (2) - \frac{1}{2} \\
$
But apparently it should be
$= 2(\ln (2))^2 - 4\ln (2) + 2
$
See here
personally, i like my way the best (this is usually not the case)
...oh, i didn't show the full working... oh well, i gave a small outline. try it. Mr F's method is there too
There should be another thread around where this was done completely...you can do a search if you want
4. Originally Posted by Craka
Man I'm having problems.
Question is
$\int_1^2 {(\ln (x))^2 } dx \\$
I get this
$= \int_1^2 {\ln (x)\ln (x)dx} \\$
$= \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\$
$= \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\$
$= \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\$
$= \frac{1}{x}\ln (x) + \frac{1}{x} \\$
$= [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\$
$= \frac{1}{2}\ln (2) - \frac{1}{2} \\
$
But apparently it should be
$= 2(\ln (2))^2 - 4\ln (2) + 2
$
$\int\ln^2(x)dx$
Let $\ln(x)=u\Rightarrow{x=e^u}$
So $dx=e^u$
Giving
$\int{e^uu^2du}$
Parts gives
$\int{e^uu^2du}=u^2e^u-2\int{e^u\cdot{u}du}=u^2e^u-2\bigg[e^uu-\int{e^u}\bigg]=u^2e^u-2ue^u+2e^u=$ $(u^2-2u+2)e^u$
Now back subbing we get
$\int\ln^2(x)=(\ln^2(x)-2\ln(x)+2)x+C$
EDIT: Sorry Mr. F! Didn't even see that you had that there! Well, at least the poster sees what you meant
5. Originally Posted by Craka
Man I'm having problems.
Question is
$\int_1^2 {(\ln (x))^2 } dx \\$
I get this
$= \int_1^2 {\ln (x)\ln (x)dx} \\$
$= \frac{1}{x}\ln (x) - \int {\frac{1}{x}} \times \frac{1}{x}dx \\$
$= \frac{1}{x}\ln (x) - \int {x^{ - 2} } dx \\$
$= \frac{1}{x}\ln (x) - [ - x^{ - 1} ] \\$
$= \frac{1}{x}\ln (x) + \frac{1}{x} \\$
$= [\frac{1}{2}\ln (2) + \frac{1}{2}] - [\frac{1}{1}\ln (1) + \frac{1}{1}] \\$
$= \frac{1}{2}\ln (2) - \frac{1}{2} \\
$
But apparently it should be
$= 2(\ln (2))^2 - 4\ln (2) + 2
$
But if you were to do it your way
let $u=\ln(x)$
and math $dv=\ln(x)dx$
So $du=\frac{dx}{x}$
and $v=\int\ln(x)dx=x\ln(x)-x=x(\ln(x)-1)$
Giving us
$\int\ln^2(x)=\ln(x)x(\ln(x)-1)-\int\frac{x(\ln(x)-1)}{x}dx$
So for the second integral we get
$\int\frac{x(\ln(x)-1)}{x}dx=\int\ln(x)-1dx=x(\ln(x)-2)$
Giving us a grand finale of
$\int\ln^2(x)=x\ln(x)(\ln(x)-1)-x(\ln(x)-2)=x(\ln^2(x)-2\ln(x)+2)+C$
An easier method that might have been mentioned elsewhere would be as if you were just calculating $\int\ln(x)dx$
So for this case let
$u=\ln^2(x)$
and $dv=dx$
so then
$du=\frac{2\ln(x)}{x}dx$
and $v=x$
So once again using our parts formula we get
$\int\ln^2(x)dx=\ln^2(x)\cdot{x}-\int\frac{2\ln(x)\cdot{x}}{x}$
Now for the second one we apply parts again and once again let dv=dx to get
$\int\ln^2(x)dx=\ln^2(x)\cdot{x}-2(\ln(x)\cdot{x}-x)=(\ln^2(x)-2\ln(x)+2)x+C$
as you can see your method is by far the messiest, there are very few integrals where you have $\int{u^2(x)dx}$
and let $u=u(x)$ and $dv=u(x)$
There are some I think, but they are uncommon | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 71, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9693480730056763, "perplexity": 1639.3788070268138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776401658.57/warc/CC-MAIN-20140707234001-00097-ip-10-180-212-248.ec2.internal.warc.gz"} |
https://brilliant.org/problems/carbon-potential/ | # Carbon Potential
If the electric potential inside a carbon atom can be expressed as
$V(r) = \frac{Ze}{a\pi {\epsilon}_{0}} \left(\frac{1}{r} - \frac{b}{cR} + \frac{{r}^{2}}{d{R}^{3}}\right)$
where $$R$$ is the radius of the carbon atom, calculate $$Z+a+b+c+d$$.
Notes
1) $$e$$ is the charge of a proton
2) $${\epsilon}_{0}$$ is the permittivity of free space
3) $$V(r)$$ should not be confused with the potential energy, which would be of the form $U(r) = -\frac{Z{e}^{2}}{a\pi {\epsilon}_{0}} \left(\frac{1}{r} - \frac{b}{cR} + \frac{{r}^{2}}{d{R}^{3}}\right)$
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http://hpcourseworkaybn.alisher.info/rate-of-reaction-sodium-thiosulphate-and-hydrochloric-acid-concentration.html | # Rate of reaction sodium thiosulphate and hydrochloric acid concentration
Hydrochloric acid: hcl sodium thiosulfate:na2s2o3 aim the aim of this experiment is to see how the reaction rate is effected by temperature, specifically heated and chilled sodium thiosulfate, and also concentration of hydrochloric acid, 1 mole and 2 moles to see how the speed of the chemical reaction is affected. Sodium thiosulphate and hydrochloric acid and the effect of different -concentration on reaction rate: the rate of reaction measures the frequency of successful collisions documents similar to rate of reaction between sodium thiosulphate courswork chemistry idocx. The rate of reaction between hydrochloric acid and sodium thiosulphate chemistry coursework how concentration affects a reaction aim: the aim of this experiment is to find out how concentration affects the rate of reaction between hydrochloric acid and sodium thiosulphate. When hydrochloric acid and sodium thiosulphate react, the solution turns cloudy you can measure the rate of the reaction by altering the concentration of sodium thiosulphate and measuring the time it takes for the solution to turn fully opaque. How reaction rate varies with sodium thiosulphate concentration background information sodium thiosulphate and hydrochloric acid are both colourless liquids, when the two reactants are reacted together they produce sulphur the sulphur that is produced from the reaction changes the solution to yellow and cloudy, this is a precipitation reaction, where a two solutions react and a solid forms.
Published: mon, 5 dec 2016 the aim of this experiment is to study the rate of reaction and the different parameters that affect it in this experiment, we will be investigating the effect of temperature on the reaction between sodium thiosulphate and hydrochloric acid. Reaction of sodium thiosulphate with hydrochloric acid safety: good laboratory practice should be followedconcentrations of hydrochloric acid stronger than 2m but weaker than 65m should be labelled. Aim my aim is to investigate how changing the concentration of reactants can change the rate of reaction between hydrochloric acid and sodium thiosulphate i will be timing and watching: how quickly the reactants are used up how quickly a product is formed the rate of reaction (ror) describes how fast a reaction takes place.
Factors affecting rate of reaction between sodium thiosulphate and hydrochloric acid results in sulphur being precipitated slowly thiosulphate is one because the graph of rate of reaction versus concentration of sodium thiosulphate shows the linear relationship. Solutions provided sodium thiosulphate 40g/l hydrochloric acid 73g/l planning how will we time the reaction we will put the solution on top of the cross and add the chemicals together. Sodium thiosulfate solution is reacted with acid – a sulfur precipitate forms the time taken for a certain amount of sulfur to form can be used to indicate the rate of the reaction hands-on practical activities or ideas for front of class demonstrations hands-on practical activities or ideas for. The effect of concentration of reactants on rate of a reaction can be studied easily by the reaction between sodium thiosulphate and hydrochloric acid na 2 s 2 0 3 + 2hcl ——– s(s) + 2nacl(aq) + so 2 (g) + h 2 o (l.
Learning outcomes: 1 to investigate the effect of concentration of sodium thiosulphate in a reaction with hydrochloric acid 2to investigate the effect of temperature of sodium thiosulphate in a reaction with hydrochloric acid 3a. The effect of concentration of the reactant on the rate of a reaction can be studied easily by the reaction between sodium thiosulphate and hydrochloric acid sodium thiosulphate reacts with dilute acid to produce sulphur dioxide, sulphur and water. Learning outcomes: what affects the rate of reactions experiments to find out how, temperature, concentration and particle size affect the rate of reaction using sodium thiosulphate and hydrochloric acid.
- sodium thiosulphate and hydrochloric acid reaction investigation aim to see the effects of concentration on the rate of a reaction between sodium thiosulphate and hydrochloric acid background information the collision theory briefly: for a reaction to occur particles have to collide with each other. 80 experiment starter sheet - investigating the rate of reaction between sodium thiosulphate and hydrochloric acid here is a suggested method to investigate the effect of varying the concentration. Example \(\pageindex{1}\): the reaction between sodium thiosulphate solution and hydrochloric acid this is a reaction which is often used to explore the relationship between concentration and rate of reaction in introductory courses. In this experiment the effect of temperature on the rate of reaction between sodium thiosulfate and hydrochloric acid is investigated hands-on practical activities or ideas for front of class demonstrations hands-on practical activities or ideas for front of class demonstrations.
Decomposition of sodium thiosulphate aim: the aim or purpose of this investigation is to explain how concentration affects the rate of reaction (the decomposition of sodium thiosulphate in reaction with hydrochloric. Gcse chemistry coursework: investigating the rate of a reaction your task is to plan, and carry out, an experiment to discover how to change the rate the reaction between sodium thiosulphate solution and dilute hydrochloric acid. 2) measure out 15ml of 10g/l (concentration) sodium thiosulphate using a measuring cylinder and place into a clean labelled boiling tube 3) measure out 15ml of 1 mol/l (concentration) hydrochloric acid using a 25ml measuring cylinder and transfer into a clean labelled boiling tube. In this experiment i will be seeing how the concentration of sodium thiosulphate, na2s2o3, and hydrochloric acid, hcl, affects the rate of the reaction in this reaction the solution turns milky yellow as sulphate is displaced and forms a solid precipitate.
The purpose of this demonstration is to investigate the effect of sodium thiosulfate concentration on the rate of reaction of sodium thiosulfate with hydrochloric acid the reaction, which produces solid sulfur, will be followed by measuring the. Experiment: to study the effect of concentration on the rate of reaction between sodium thiosulphate and hydrochloric acid how to perform experiment take three clean and dry conical flasks and mark them 1, 2 and 3. Both sodium thiosulphate and diluted hydrochloric acid are colorless solution sulphur dioxide is a very soluble gas and dissolves completely in the aqueous solution. Reactions that do not take long to make a 3 | page antonio motta-marques dr fowler rates of reaction chemistry sodium thiosulphate and hydrochloric acid reaction rates product have a high rate of reaction, and reactions that take a longer time have a low rate of reaction.
The purpose of this demonstration is to investigate the effect of sodium thiosulfate concentration on the rate of reaction of sodium thiosulfate with hydrochloric acid. The rate of a chemical reaction will be increased by increasing the concentration this can be seen in the reaction between dilute hydrochloric acid and sodium thiosulfate solution. The rate of reaction between sodium thiosulphate and hydrochloric acid candidate name: yeo jin kim (kimberly kim) candidate number: 000791-021 teacher: yitao duan 10 aim the aim of this experiment is to investigate the changing of concentration effect the rate of reaction between sodium thiosulphate and hydrochloride.
Rate of reaction sodium thiosulphate and hydrochloric acid concentration
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2018. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9089049696922302, "perplexity": 2088.1636907741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513508.42/warc/CC-MAIN-20181020225938-20181021011438-00165.warc.gz"} |
https://www.emathhelp.net/notes/calculus-1/sequence-theorems/squeeze-sandwich-theorem-for-sequences/ | Squeeze (Sandwich) Theorem for Sequences
Related Calculator: Limit Calculator
Consider two sequences x_n and y_n. When we write that x_n=y_n we mean that corresponding values are equal, i.e. x_1=y_1, x_2=y_2 etc.
Fact 1. If two sequences x_n and y_n are equal: x_n=y_n, and each of them has limit (finite or infinite): lim x_n=a and lim y_n=b then a=b.
This fact is used in limiting process: from x_n=y_n we conclude that lim x_n=lim y_n.
Fact 2. If for two sequences x_n and y_n we have that x_n>=y_n, and each of them has limit (finite or infinite): lim x_n=a and lim y_n=b then a>=b.
This fact is used in limiting process: from x_n>=y_n we conclude that lim x_n>=lim y_n.
Of course sign >= can be replaced by sign <=, i.e. from x_n<=y_n we conclude that lim x_n<=lim y_n.
CAUTION! Inequality x_n>y_n we can't conclude that lim x_n>lim y_n, we can only conclude that lim x_n>=lim y_n.
For example, consider two sequences x_n=1/n and y_n=-1/n. Clearly, x_n>y_n but lim 1/n=lim -1/n=0.
Fact 3 (Squeeze Theorem for Sequences). Consider three sequences x_n,y_n,z_n. If we have that x_n<=y_n<=z_n, and sequence x_n and z_n have same limit (finite or infinite), i.e. lim x_n=lim z_n=a, then lim y_n=a.
This theorem tells us folowing: if there are three sequences, two of which have same limit and third is "squeezed" between them, then third will have same limit as first two.
Consider figure to the right. Let sequence drawn in green and sequence drawn in blue have same limit a. Since sequence drawn in pink is "squeezed" between them, then it will also have limit a.
From this theorem it follows that if for all n a<=y_n<=z_n and z_n->a then y_n->a. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9890183806419373, "perplexity": 4907.9054405500165}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00147.warc.gz"} |
http://mbi.osu.edu/publications?view=keywords&id=63&item=Cancer | ## MBI Publications
### MBI Publications for Cancer (1)
• A. Friedman and Y. Kim
Tumor cells proliferation and migration under the influence of their microenvironment
Mathematical Biosciences and EngineeringVol. 8 No. 2 (2011) pp. 373-385
#### Abstract
It is well known that tumor microenvironment affects tumor growth and metastasis: Tumor cells may proliferate at different rates and migrate in different patterns depending on the microenvironment in which they are embedded. There is a huge literature that deals with mathematical models of tumor growth and proliferation, in both the avascular and vascular phases. In particular, a review of the literature of avascular tumor growth (up to 2006) can be found in Lolas [8] (G. Lolas, Lecture Notes in Mathematics, Springer Berlin / Heidelberg, 1872, 77 (2006)). In this article we report on some of our recent work. We consider two aspects, proliferation and of migration, and de- scribe mathematical models based on in vitro experiments. Simulations of the models are in agreement with experimental results. The models can be used to generate hypotheses regarding the development of drugs which will confine tumor growth. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8145537376403809, "perplexity": 1461.530979859197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443438.42/warc/CC-MAIN-20141017005723-00126-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://finpis.mi.sanu.ac.rs/i0qds/5e7ee0-hasse-diagram-examples-pdf | ... diagram (or Hasse diagram) if y is higher in the plane than x whenever y covers x in P. Order Diagrams and Cover Graphs Order Diagram Cover Graph. (Posets) has_bottom() Return True if the poset has a unique minimal element. /D [58 0 R /XYZ 114.52 517.826 null] (Modular Lattices) The poset $$(\{1,5,25,125\},\mid)$$ is also a totally ordered set. 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Q�ٛ���.�g������}��}� H�g���YP���YxI �y���ǖ�D��>k��K(�������Ď��HS�ǟXZ�CӁ�"Z?�!M���]�����s�x(4�4H� ���.C��Rapv�l��iQ�E�dP_��5�D��/�k�Q����,�p�9#ENݑ���(���NqL ��:1L1������8ap���݅�*вI�4,i3M���Y^ �d���ݝ;s������9t0��v��η���7�6��ڑ�+b�F��9z�?��Zu���A3���ͣ�Td�� Q�"r�a'�-�G!��!5�1���o��BH�Ya�m]��������ѧt@�d� These graphs are called Hasse diagrams after the twentieth-century German number theorist Helmut Hasse. /D [58 0 R /XYZ 114.52 565.15 null] 1. hasse (data, labels = c (), parameters = list ()) Arguments. Description. 32 0 obj Hasse diagrams in the design and analysis of experiments R. A. Bailey University of St Andrews / QMUL (emerita) In memory of Paul Darius, Leuven Statistics Days, 5 December 2014 1/42. has_top() Return True if the poset contains a unique maximal element, and False otherwise. Therefore, it is also called an ordering diagram. Dies gilt weltweit. Hasse Diagram. ��@�5�wX��48v����Ng+����d�i�m� endobj %���� endobj 36 0 obj /D [58 0 R /XYZ 114.52 463.997 null] /D [58 0 R /XYZ 114.52 479.148 null] %PDF-1.4 /MediaBox [0 0 595.276 841.89] labels: Vector containing labels of elements. endstream >> endobj endobj 8 0 obj Skip to content. Ganesha 10 Bandung 40132, Indonesia [email protected] Abstrak—Himpunan, relasi, dan graf merupakan tiga dari sekian banyak topik yang dibahas pada Matematika Diskrit. << /S /GoTo /D (section.3.1) >> (The M\366bius Inversion Formula) Menu. endobj Download scientific. The prerequisite for Hasse Diagram is to know how to represent relations using graphs. Example $$\PageIndex{5}\label{eg:ordering-05}$$ The poset $$(\mathbb{N},\leq)$$ is a totally ordered set. For example, the definition of an equivalence relation requires it to be symmetric. (3.7) De nition The Hasse diagram of a nite poset Pis the graph with vertices x2Pand if x> endobj Die Richtung der Kante wird dadurch zum Ausdruck gebracht, dass sich der Knoten b oberhalb von a befindet. 3 0 obj endobj /Filter /FlateDecode is_bounded() (The M\366bius Function of Lattices) (Hierbei ist a < b als a b und a b zu verstehen.) Although hasse diagrams are simple as well as intuitive tools for dealing with finite posets it turns out to be rather difficult to draw good diagramsthe reason is that there will in general be many possible ways to draw a hasse diagram for a given poset. << /S /GoTo /D (section.3.9) >> << /S /GoTo /D [58 0 R /Fit ] >> This function draws Hasse diagram – visualization of transitive reduction of a finite partially ordered set. endobj How to draw a Hasse Diagram in LATEX For example, the Hasse diagram of set f2, 4, 5, 10, 12, 20, 25g with divisibility condition is given by 12 20 4 | | | | | | | | 10 25 Weitere Auflösungen: 170 × 240 Pixel ... Logical_connectives_Hasse_diagram.svg: Quelle: Eigenes Werk: Urheber: Watchduck (a.k.a. Download scientific << /S /GoTo /D (section.3.5) >> 2) Eliminate all loops 3) Eliminate all arcs that are redundant because of transitivity 4) eliminate the arrows at the ends of arcs since everything points up. Return an iterator over greedy linear extensions of the Hasse diagram. >> endobj Hasse Diagram — from Wolfram MathWorld. 1. hasse (data, labels = c (), parameters = list ()) Arguments. u4��(�q��Lq�uN��yQLX�����uj�#FY��uR��C��Hg�d�Kjc�QR4��T��j0�M� m�����,��Հ6d3�]�ݯ�����|s��U���g�]�1��'� #�~2m��c�!܇�?���I�PPH�������e=��j9}a�a�M��HP��~e���,#s���1�WsP�y����l]�&���yއ��2�ƣ��,��[��=d\�ƂV� (%� 5O���n*�~ۇٜa=��� ���ɟ� Its Hasse diagram is shown below. �zyCߤ%�>��Mrm��pf���(���^�Z̄Q���Z؛���sm�$ׅ�t�aP�i���O�m�uw;D^;�Z �~�_������?��N 28 0 obj endobj View 4.pdf from AA 1Assignment 4 : Relations - Solutions Assignment 4 : Relations 1. The randomization diagrams and the Hasse diagrams are graphic tools that help to construct the analysis of variance table. /D [58 0 R /XYZ 114.52 743.442 null] Tilman Piesk) Genehmigung (Weiternutzung dieser Datei) Public domain Public domain false false: Ich, der Urheberrechtsinhaber dieses Werkes, veröffentliche es als gemeinfrei. endobj Solch ein… 61 0 obj << DIAGRAMAS HASSE PDF - 29 set. [vA�Q� �E�.5,��|�B,�����������l3��\���{ee6�,�����]#�lD���Çɸχ������{� �=�ys��p6b�.9� �߲�?ClJ?�DA ��u| Here is an exercise for you to practice. The Hasse diagram of P Q is the Cartesian product of the Hasse diagrams of P and Q. Example When P is a collection of sets, set x ≤ y in P when x is a subset of y. >> endobj /D [58 0 R /XYZ 114.52 330.994 null] /D [58 0 R /XYZ 114.52 579.237 null] 60 0 obj << >> endobj >> 1�����N�����-���J��w�z���8����d����b~Dx�LL����Z�gi�Z�6 � ��H./!��=��i��Z��0����{n>�G��H�|6�. 79 0 obj << The randomization diagrams and the Hasse diagrams are graphic tools that help to construct the analysis of variance table. Download scientific For some reason, TikZ doesn't want to cooperate with my installation. These diagrams show relationships between factors. >> endobj (Distributive Lattices) *k^���@�ĭ��4�:zy��^��0��/1&k�����;���2�lI�ི���AuY:��*k��Z4lڈ�����BB����&��\d�D0� (����ÃU�� 9��A�q�:qy3�|�A>Z�C?��j}�,7�88��M���#FH�\d�MOy���٬����6x��KЃ�Pv4��mUg�5������6�6�5�3��b���/��RT[v�1��.n�8�d �Pd�@$�-��MN��y����I �[����I�u*S�4��K�L�ń�|�ΥY�X�:P�dP�N���V�5�'[S�,��=�:a&�tJ!�E�hp�(�\�z�n�NS�Sf���v2��$��u�)?�yV�)��b�F���А�0����:���*S/�Jh�����X���:;�'�>Y�/� ��sM�$�왭��!��t� �ѶG�ta���.����Ζ��W�X�B� q�"���D���S�����!���L˘����������3TpD"�4��FR*�����>���%|�S��#=��w5Β�^G����Y�:�II�e�Jx��HU��g����bFE�;�������[&�diL�{r���g,�{c������0M�Z��4�Q��p�8�BXq�1R2��!bU�@F=km��)�����W�VeD02ؐ In particular the attached Moebius function is used to compute the so-called homogenous weight in Coding Theory. Relasi Pengurutan Parsial, Poset, dan Diagram Hasse Hasna Nur Karimah - 13514106 Program Studi Teknik Informatika Sekolah Teknik Elektro dan Informatika Institut Teknologi Bandung, Jl. endobj In hasseDiagram: Drawing Hasse Diagram. Other tutorial like answers here which might be useful for new tikz users include: Recently, I also had to draw Hasse diagrams, so I post a few examples to get you going. (Applications of the M\366bius Inversion Formula) Length of C0= fx 0 x 1 x ngis n. Hence, ‘(C) = 7 1 = 6. x��ZIo#���W�����8�0y��!b�fr���݀,9�2���S${o�"��y̖�Ū�W�pt��3�����웏TED!J�.W� Recently, I also had to draw Hasse diagrams, so I post a few examples to get you going. %�����/�w��u�|�xY�C;,�ֽ�f�-�'�*��c;,!�Ώ�8�����(�l�=�|"5��}���w����-���v5cY�_/��d���1S�K�G�=���d�!U�I��6;Ms�jⰞ�{��j�e���qqw�?���e�O�2�9��d��Q��!1��ś�Oe��Y?��5��A��L!�:91(�[^D4d�Za�.4�l����4DD��0O���+C������7&A�!�R2�=���o=���s�Z�َ�p�_�a�o��~�A6ڑ�kH����Cj�̉��ۃ�g�)=��c�p|3�p�$�OM= ��5��D�J��7Ur:Ϋ)#�m����fߩD�X"}.�3�ă7_I� /Length 2548 /Type /Page 44 0 obj endobj (Incidence Algebra of a Poset) �����-��S!�$��1. endobj d��=�o4�� 12 0 obj z EXAMPLE 7.1-4 Diagram the following posets: a) The poset of Example 3b: the divisors of 36 ordered by m|n. In order to visualize finite posets, we often draw a picture called its Hasse diagram. This video explain how can we construct a Hasse Diagram directly for a given POSET. Die Einschränkung auf solche a , b nennt man transitive Reduktion der Halbordnung. endobj /D [58 0 R /XYZ 114.52 256.595 null] DIAGRAMAS HASSE PDF - 29 set. 48 0 obj 76 0 obj << Therefore, while drawing a Hasse diagram following points must be … This function draws Hasse diagram – visualization of transitive reduction of a finite partially ordered set. Distinct sets on the same horizontal level are incomparable with each other. 10 0 obj << Das Hasse-Diagramm für eine Halbordnung ( M , ) ergibt sich als Darstellung eines gerichteten Graphen, wobei die Elemente von M die Knoten bilden. 16 0 obj Posets are Everywhere!! labels: Vector containing labels of elements. (Partially Ordered Sets $$Posets$$) For the Hasse diagram given below; find maximal, minimal, greatest, least, 1. (Lattices) In this poset {2, 5} < {2, 5, 7, 8} and {5, 8, 9} ≥ {5, 8, 9}. Some other pairs, such as {x} and {y,z}, are also incomparable. A number of results on upward planarity ciagramas on crossing-free Hasse diagram construction are known:. 29 0 obj /Length 2474 41 0 obj (Young's lattice) Example 1.4 Let R;S X be two relations on X, with corresponding incidence matrices M and N. Replace each star of the matrices M and N by the Boolean 1. data : n x n matrix, which represents partial order of n elements in set. << /S /GoTo /D (section.3.2) >> endobj The Hasse diagram for (P({a,b,c}),⊆) is {a,b,c} Zwei Knoten a und b werden durch eine Kante verbunden, wenn a < b gilt und es keinen Knoten c gibt mit a < c < b . Each cell [i, j] has value TRUE iff i-th element precedes j-th element. 21 0 obj endobj 53 0 obj >> endobj In order to keep our picture from getting too complicated, we will only show the minimal set of relations needed to deduce all of the relations in the poset. 6 0 obj << Example-1: Draw Hasse diagram for ({3, 4, 12, 24, 48, 72}, /) Explanation – According to above given question first, we have to find the poset for the divisibility. << /S /GoTo /D (section.3.6) >> x��[[s�6~��У4��H���n�6�N'���}�%��Ԗ���y�߾$�T�fӇD ���|�;���jRL~8����w/����B����D���RXOΗ����������ٜR2�h6��N_��>{=Stz����?���!��~~�������oN�߸ۿ�x���Wx���O'�_�w/�h��b��̩B� This diagram is from a great discussion on using TikZ to draw Hasse diagrams, but I'd prefer to use xypic. << /pgfprgb [/Pattern /DeviceRGB] >> /ProcSet [ /PDF /Text ] It is clear that the Hasse diagram of any totally ordered set will look like the one displayed above. 37 0 obj 77 0 obj << >> endobj 80 0 obj << Here's an example of the kind of thing I'm looking to do: Note the equal space between the four entries of the second row. endobj Abstract I Paul Darius was a great advocate of the use of Hasse diagrams as an aid to thinking about factors in a designed experiment. 52 0 obj Let A be a poset, A = { 2, 4, 6, 8 } and the relation a | b is ‘a divides b. /D [58 0 R /XYZ 114.52 309.966 null] endobj 65 0 obj << /Filter /FlateDecode /Length 3340 That is, for all sets U and V in ({a, b, c}), Construct the Hasse diagram for this relation. endobj (The Zeta Function) >> endobj is_antichain_of_poset() Return True if elms is an antichain of the Hasse diagram and False otherwise. We give examples for Hasse diagrams of the entire moduli space of theories with arXiv:2004.01675v2 [hep-th] 27 Aug 2020 enhanced Coulomb branches. ARPTIALLY ORDERED SETS (POSETS) (3.25) De nition An antichain of a poset P is a subset A P such that every pair of elements of Aare incomparable. 58 0 obj << 13 0 obj Solution: Draw the directed graph of the relation in such a way that all arrows except loops point upward. As Hasse diagrams are the visualization of a mathematical concept, namely of partial order, one has to go back until the end of the nineteenth century, where Dedekind and Vogt (see Rival, 1985 ) made the first important … 58. DIAGRAMAS HASSE PDF - 29 set. Example 7 – Constructing a Hasse Diagram Consider the “subset” relation, ⊆, on the set ({a, b, c}). Observe that if a> _____ Example: Construct the Hasse diagram of (P({a, b, c}), ⊆ ). It is a useful tool, which completely describes the associated partial order. ��xdr�Ѵ��UHBj�ſ�>�a}2Q8�Ң��M��c�q��c�|$�|Q&J�:�:�&�.��IC�ug[���|A,%���e��bsH����]r��"�!��}� That is, for all sets U and V in ({a, b, c}), Construct the Hasse diagram for this relation. << /S /GoTo /D (section.3.8) >> ���me~cy{,�H����C�(Z�G1D���͉�>��h{&sf#̏���,� ���9FB�z���$*���Q����J�D����IkY�7�j�x1#r��ڗH�����'�QA�],�n�:��"H�5 data : n x n matrix, which represents partial order of n elements in set. stream �(Q�����!�]����ߩ�kK�on:ꂊ�"g= ��}Ũj�덪�Z6U�K�Џa!4����U�TQ܍���"K�#�%���$g��ś��B�(Mm�$OB��&���έ��*�C�_�!��v�h��]��0��m-�. /Font << /F27 63 0 R /F48 64 0 R /F51 66 0 R /F26 68 0 R /F16 69 0 R /F52 70 0 R /F54 71 0 R /F15 72 0 R /F30 73 0 R /F33 74 0 R /F31 75 0 R /F42 81 0 R >> endobj 67 0 obj << >> endobj Hasse diagrams got the name from the German mathematician H. Hasse, who lived from 1898 to 1979 and who used them to represent algebraic structures (Hasse, 1967). endobj endobj In ring theory the Hasse diagram of ideals ordered by inclusion is used often. Also it represents (2) of Example 3.5 if we /Contents 60 0 R stream Typical examples include Aristotelian diagrams, Hasse diagrams and duality diagrams [6{13]. >> endobj 45 0 obj Recently, I also had to draw Hasse diagrams, so I post a few examples … LB, For the diagram given below; << /S /GoTo /D (section.3.4) >> Wt X��v~��ʀ�y�b�6�ܗ���.sk�vO*�]& ����4hIֲ5���4��Fh�7FX���sFؔ%�����=����]_�拓~��lD,^�7�zl(
hR0}������^��. Tutorial in how to draw the Hasse diagram from a graph, and determine the minimal elements. 20 0 obj edges upward are left implicit instead of cluttering up the diagram. /D [58 0 R /XYZ 114.52 768.348 null] Hasse Diagrams •Since partial orderings is a binary relation, it can be represented by a directed graph •However, many edges can be omitted, because such an ordering must be reflexive and transitive •Also, we may order the vertices in the graph in a ‘vertical’ manner, … 57 0 obj stream 2) Eliminate all loops 3) Eliminate all arcs that are redundant because of transitivity 4) eliminate the arrows at the ends of arcs since everything points up. The randomization diagrams and the Hasse diagrams are graphic tools that help to construct the analysis of variance table. 9 0 obj The Hasse diagram of a poset is a simpler version of the digraph representing the partial order relation. The Hasse diagram of the set of all subsets of a three-element set {x, y, z}, ordered by inclusion. Example 4.10.1. ��x�>F���*��\�����?�%녳�Mnu� �Hs}��{��_��M�sJ52D�QU1P��q���r�I���_�x�����R1x?Y���E�^95�տ�H�9�-�����1ĵ5qph�ӡ��:yL���c����V]\$9o�41 /Resources 59 0 R (Linear extensions) It is very easy to convert a directed graph of a relation on a set A to an equivalent Hasse diagram. As pointed out by Paul Gaborit, the out and in options are really only for the to directive so some might prefer a syntax that more explicitly places those options for the to as in:. It means that there is a set of elements in which certain element are ordered sequenced or arranged in some way. >> endobj Example 9 B n ’2|{z 2} n times Proof: De ne a candidate isomorphism f : 2 2 !B n (b 1; ;b n) 7!fi 2[n] jb i = 2g: It’s easy to show that f is bijective. (3.8) Example If P= fa;b;c;d;e;fgand a> 87 0 obj << CHAPTER 3. endobj (3.24) Example Poset Pwith Hasse diagram: e 0 0 0 0 0 0 0 ~ ~ ~ ~ c d b @ @ @ @ a C= fa;b;b;b;c;c;egis a multichain. Solution: Draw the directed graph of the relation in such a way that all arrows except loops point upward. /D [58 0 R /XYZ 114.52 172.773 null] A natural example of an ordering which is not total is provided by the subset ordering. endobj >> endobj (Rank-selection) endobj In these cases, one diagram contains several formulas, and diagrammatic reasoning. �J���p�����j �=�� /�о�y���)z����Ļ��E[3�����*��S��W~�bP�k�����x�Js��9��[�>b�C�HD��Y,!����d��sY�������0O�e^���? 5 0 obj 62 0 obj << >> endobj Hasse diagram for classical Higgs branches which are more general symplectic singularities. Description Usage Arguments Examples. Hasse or Poset Diagrams To construct a Hasse diagram: 1) Construct a digraph representation of the poset (A, R) so that all arcs point up (except the loops). 83 0 obj << Since maximal and minimal are unique, they are also the greatest and least element of the poset. 49 0 obj �=h�E���ƅ�í���c�l����ed��>�J��aFU|���y�ݬ��Cz����d�RH?ݮ�]zHB��P& ���o���a����ғޮBt�AJ�'狲7_D����6Ha�j��l���':H().��]h fې�2h��B;���p�Lڈ���w�lu����G�Or�g*���)�Q�5���[��-~n�{HaY��Jug��8�q�N������%H���b����Taā�T���� E�Pua�Bx��hu� >> 59 0 obj << Example 7 – Constructing a Hasse Diagram Consider the “subset” relation, ⊆, on the set ({a, b, c}). /Filter /FlateDecode 78 0 obj << >> endobj 24 0 obj 25 0 obj Usage. Hasse Diagrams. All definitions tacitly require transitivity and reflexivity. endobj From a graph, and False otherwise with my installation P is simpler! An equivalence relation requires it to be symmetric partial order relation iterator over greedy extensions! 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http://www.vallis.org/blogspace/preprints/1207.3088.html | ## [1207.3088] Towards models of gravitational waveforms from generic binaries: A simple approximate mapping between precessing and non-precessing inspiral signals
Authors: Patricia Schmidt, Mark Hannam, Sascha Husa
Date: 12 Jul 2012
Abstract: One of the greatest theoretical challenges in the build-up to the era of second-generation gravitational-wave detectors is the modeling of generic binary waveforms. We introduce an approximation that has the potential to significantly simplify this problem. We show that generic precessing-binary inspiral waveforms (covering a seven-dimensional parameter space) can be mapped to only a two-dimensional space of non-precessing binaries, characterized by the mass ratio and a single effective total spin. The mapping consists of a time-dependent rotation of the waveforms into the quadrupole-aligned frame, and is extremely accurate (matches $> 0.99$ with parameter biases in the total spin of $\Delta \chi \leq 0.04$), even in the case of transitional precession. In addition, we demonstrate a simple method to construct hybrid post-Newtonian--numerical-relativity precessing-binary waveforms in the quadrupole-aligned frame, and provide evidence that our approximate mapping can be used all the way to the merger. Finally, based on these results, we outline a general proposal for the construction of generic waveform models, which will be the focus of future work.
#### Jul 16, 2012
1207.3088 (/preprints)
2012-07-16, 00:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8460111618041992, "perplexity": 1458.9848632690814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00067.warc.gz"} |
http://mathoverflow.net/feeds/question/81384 | Degree conjecture and automorphic L-functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:14:13Z http://mathoverflow.net/feeds/question/81384 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81384/degree-conjecture-and-automorphic-l-functions Degree conjecture and automorphic L-functions Sylvain JULIEN 2011-11-19T23:10:23Z 2011-11-21T22:36:35Z <p>Hello,</p> <p>this may be a very naive question, but has the degree conjecture (namely "the degree of any function of the Selberg class is a non negative integer") been proven for automorphic L-functions? Thank you in advance.</p> http://mathoverflow.net/questions/81384/degree-conjecture-and-automorphic-l-functions/81387#81387 Answer by Stefano V. for Degree conjecture and automorphic L-functions Stefano V. 2011-11-20T00:14:58Z 2011-11-20T07:12:28Z <p>To the best of my knowledge, we are very far from proving such a conjecture. If $\mathcal S_d$ is the subclass of the Selberg class $\mathcal S$ consisting of the functions of degree $d$ then it is known that</p> <p>1) $\mathcal S_0={1}$ (Conrey-Ghosh, 1993);</p> <p>2) $\mathcal S_d=\emptyset$ for $0\lt d\lt1$ (Richert, 1957 and others);</p> <p>3) $\mathcal S_1$ consists of the Riemann zeta function $\zeta(s)$ and the shifted Dirichlet $L$-functions $L(s+i\tau,\chi)$ with $\tau\in\mathbb R$ and $\chi$ a primitive character (Kaczorowski-Perelli, 1999);</p> <p>4) $\mathcal S_d=\emptyset$ for $1\lt d\lt2$ (Kaczorowski-Perelli, 2002 and 2011).</p> <p>Apart from these results, I think that nothing has been established in general. A nice survey of the results obtained so far can be found in the introduction to</p> <p>J. Kaczorowski, A. Perelli, "On the structure of the Selberg class, VII: $1\lt d\lt2$", <em>Ann. of Math. (2)</em> <strong>173</strong> (2011), 1397-1441.</p> <p>Note that, in fact, Kaczorowski and Perelli prove their results for functions in the so-called <em>extended Selberg class</em> $\mathcal S^\sharp$, whose elements are not required to satisfy the Ramanujan hypothesis and the Euler product property. </p> http://mathoverflow.net/questions/81384/degree-conjecture-and-automorphic-l-functions/81401#81401 Answer by BR for Degree conjecture and automorphic L-functions BR 2011-11-20T06:53:39Z 2011-11-21T22:36:35Z <p>Going off <a href="http://en.wikipedia.org/wiki/Selberg_class#Basic_properties" rel="nofollow">wikipedia</a>, it is true that automorphic $L$-functions for $GL_n$ over a number field have non-negative integral degree, where by degree I mean the number $2\sum_{i=1}^k \omega_i$, where the $\omega_i$ are the coefficients of $s$ appearing in the gamma factor, which is more-or-less $$L_\infty(s,F)=Q^s\prod_{i=1}^k\Gamma(\omega_i s+\mu_i)$$ We know that for $GL_n(\mathbb R)$ and $GL_n(\mathbb C)$ the $\omega_i$ are either $1/2$ or $1$ (see, e.g., Knapp's "Local Langlands Correspondence: The Archimedean case", in Motives, vol 2), so twice the sum will always be an integer (of course, only a few of these $L$-functions are known to be in Selberg's class).</p> <p>For general $G$, it depends on whether someone has written done the $L$-factors for general real reductive groups. I don't know if this has been done, or if it is technically known but difficult to write out, or if it is not known.</p> | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9332007169723511, "perplexity": 999.6777025599133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368699675907/warc/CC-MAIN-20130516102115-00048-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://blog.computationalcomplexity.org/2020/07/erdos-turan-for-k3-is-true.html | ## Sunday, July 19, 2020
### Erdos-Turan for k=3 is True!
(All of the math in this post is summarized (without proofs) in a writeup by Erik Metz and myself which you can find here. It is a pdf file so you can click on links in it to get to the papers it refers to. There have been posts on this topic by Gil Kalai and Luca Trevisan. If you know of others then let me know so I can add them to this post.)
This is a sequel to A BREAKTHROUGH result on density and 3-AP's and Big news on W(3,r)!
For this post N is large, and all inequalites have a big-O or a big-Omega.
For this post [N] is {1,...,N}
Let
r(N) be the least w such that if A is a subset of [N] and |A| > w, then A has a 3-AP.
There has been a long sequence of results getting smaller and smaller upper bounds on r(N).
The motivation for getting these results is that if r(N) is < N/(log N)^{1+\delta} with delta>0 then the following holds:
If sum_{x\in A} 1/x diverges then A has a 3-AP.
This is the k=3 case of one of the Erdos-Turan Conjectures.
Bloom and Sisack HAVE gotten N/(log N)^{1+delta} so they HAVE gotten ET k=3. Wow!
1) I am NOT surprised that its true.
2) I am SHOCKED and DELIGHTED that it was proven. Shocked because the results leading up to it (see the write up referenced at the beginning of this post) seemed Zeno-like, approaching the result needed got but not getting there. Delighted because... uh, as the kids say, just cause.
I've heard that k=4 really is much harder (see my comments and Gil's response on his blog post, pointed to at the beginning of this post) and it is true that there has been far less progress on that case (the write up I pointed to at the beginning of this post says what is known). Hence I will again be shocked if it is proven. So, unlike The Who (see here) I CAN be fooled again. That's okay--- I will be delighted. (ADDED LATER- there are more comments no Gil's website, from Thomas Bloom and Ben Green about what is likely to happen in the next 10 years.)
Erdos offered a prize of $3000 for a proof that A has, for all k, a k-AP. The prize is now$5000. After Erdos passed away Ronald Graham became the Erdos-Bank and paid out the money when people solved a problem Erdos put a bounty on. What happens now? (If I have the facts wrong and/or if you know the answer, please leave a polite and enlightening comment.) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8331515192985535, "perplexity": 1353.3224851994034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735989.10/warc/CC-MAIN-20200805212258-20200806002258-00038.warc.gz"} |
http://mathoverflow.net/questions/21782/generalized-chinese-remainder-theorem/21801 | # Generalized Chinese Remainder Theorem
Let $U,V$ be submodules of a $R$-module $M$. Then the diagonal induces an isomorphism
$M/(U \cap V) \to M/U \times_{M/(U+V)} M/V.$
This is a (useful!) generalization of the Chinese Remainder Theorem and the proof is very easy. But I'm interested what happens when we take finitely many submodules $U_1,...,U_n$. How can we relate $M/(U_1 \cap ... \cap U_n)$ with the $M/U_i$? I think the case $n=2$ can not be used for an induction, there are more compatiblities to check for an element in $\prod_i M/U_i$ to come from $M$. I wonder if there is a nice description.
For $M=R$, this question asks for a sort of sheaf condition for sections on closed subschemes.
-
– Gjergji Zaimi Apr 18 '10 at 23:10
you may post this as an answer ;) – Martin Brandenburg Apr 19 '10 at 7:42
So this is what's in Kleinert's paper "Some remarks on the Chinese Remainder Theorem" that I mentioned in the comments.
If $\mathcal F=\{U_1,U_2,\dots,U_n\}$ is a family of submodules of the $R$-module M, then there is an embedding $\phi(\mathcal F)$ of $M/U_1\cap \cdots \cap U_n$ into $$M(\mathcal F):= \{(u_i)\in \prod M/U_i \quad \rvert u_i\equiv u_j \mod (U_i+U_j),\forall i,j\}.$$ Let the cokernel of $\phi$ be $$O(\mathcal F)=M(\mathcal F)/\phi(M/U_1\cap \cdots \cap U_n).$$ $O(F)$ is thought of as the obstruction against the ability to solve simultaneous congruences, and so we say that the generalized Chinese Remainder Theorem holds if $O(\mathcal F)=0$. He proceeds to the following sheaf-theoretical interpretation of the problem:
Let $X$ be the discrete topological space $\{1,2,\dots,n\}$, and define the presheaf $\mathcal P(\mathcal F)$ on $X$ by $\mathcal P(V)=M/\sum_{i\notin V}U_i$, for $V\subset X$. If $V\subset W$ the restriction map is given by the residue map $$\mathcal P(W)=M/\sum_{i\notin W}U_i\to M/\sum_{i\notin V}U_i=\mathcal P(V).$$ Now let $\mathcal U$ be the covering $\{X/\{i\}\}$. It follows that $M(\mathcal F)$ is the set of cocycles $C^0(\mathcal U,\mathcal P)$ and that $O(\mathcal F)=0$ iff $\mathcal P$ satisfies the second sheaf axiom with respect to $\mathcal U$. He also makes the remark that when $n=2$ , which you described in the question, this is always the case and so the generalized Chinese Remainder Theorem always holds, even though it doesn't always in the general case.
-
Thanks. The paper studies the obstruction from the image to the "naive" image (pairwise compatibility). I'm interested in the image of $M$ in $\prod_i M/U_i$ itsself. – Martin Brandenburg Apr 19 '10 at 11:07
In the remark about the surjectivity of q, I think that (2) is wrong and should be replaced by $b_i \equiv a_i$ mod $A_i + A_{n+1}$. Other opinions? – Martin Brandenburg Apr 19 '10 at 11:12
If we consider submodules generated by applying ideals to $M$, ie. $I \ M$, where $I \subseteq R$ is an ideal, we can generalize the two submodule case to any finite number of modules by induction from n = 2. All we need to check is that if $A_{i} \ M = U_{i}$ and the $A_{i}$'s are pairwise comaximal, then $A_{1} ... A_{n-1}$ and $A_{n}$ are comaximal.
-
I don't assume any comaximality-condition. Please read the question carefully. – Martin Brandenburg Apr 19 '10 at 6:31
To be honest, I wasn't sure exactly what you were assuming. Do you mean R to be commutative and unital? – Ryan Thorngren Apr 19 '10 at 6:48
this is not important. – Martin Brandenburg Apr 19 '10 at 7:33
Sorry, no clue but I will take a wild guess. Your Chinese theorem can be stated as exactness of the sequence $$0 \rightarrow M/A\cap B \rightarrow M/A \times M/B \rightarrow M/A+B \rightarrow 0$$ The third arrow is $(x+A,y+B)\mapsto (x-y)+(A+B)$. I can envision that you may be able to produce a long exact sequence starting with $$0 \rightarrow M/\cap_i A_i \rightarrow$$ as soon as the lattice generated by $A_i$-s is distributive. Notice that distributivity will ensure that you have $2^n$ submodules to work with...
Sorry if I misunderstood anything or said something completely ridiculous...
-
Assuming $A_1 \cap (A_2 + A_3) = A_1 \cap A_2 + A_1 \cap A_3$ etc. makes an induction easy. But I think this is almost never the case? – Martin Brandenburg Apr 19 '10 at 13:41
What do you mean? Off course, it is rare but it holds for PID-s (where you have usual Chinese RT holds) and for your case as 2 subs generate a distributive lattice. – Bugs Bunny Apr 28 '10 at 14:31
No the equality is also wrong in PIDs. – Martin Brandenburg Oct 6 '10 at 17:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9391036629676819, "perplexity": 195.4483537506105}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095270.70/warc/CC-MAIN-20150627031815-00182-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.h2knowledgecentre.com/content/journal2147 | 1900
# Aging Effects on Modelling and Operation of a Photovoltaic System with Hydrogen Storage
### Abstract
In this work, the aging effects on modelling and operation of a photovoltaic system with hydrogen storage in terms of energy production decrease and demand for additional hydrogen during 10 years of the system operation was analysed for the entire energy system for the first time. The analyses were performed with the support of experimental data for the renewable energy system composed of photovoltaic modules, fuel cell, electrolysers, hydrogen storage and hydrogen backup.
It has been found that the total degradation of the analysed system can be described by the proposed parameter – unit additional hydrogen consumption ratio. The results reveal a 33.2–36.2% increase of the unit fuel requirement from an external source after 10 years in reference to the initial condition. Degradation of the components can, on the other hand, be well described with the unit hydrogen consumption ratio by fuel cell for electricity or the unit electricity consumption ratio by electrolyser for hydrogen production, which has been found to vary for the electrolyser in the range of 4.6–4.9% and for the fuel cell stack in the range of 13.4–15.1% during the 10 years of the system operation. The analyses indicate that this value depends on the load profile and PV module types and the system performance decline is non-linear."
Funding source: Polish Ministry of Science (Grant AGH No. 16.16.210.476 and grant number 0711/SBAD/4514).
Keywords:
Related subjects:
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