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https://homeoflearning.in/congruent-triangles/
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math
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A triangle is a simple closed figure made up of three line segments.
A triangle has six parts or elements, namely: a) Three sides and b) Three angles.
The point of intersection of two adjacent sides of a triangle is called a vertex. A triangle has 3 vertices.
A median of a triangle is the line segment joining a vertex of the triangle to the mid-point of its opposite side. A triangle has three medians.
Congruent Triangles PDF
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http://augustabluescompany.com/315-66-homework-answer-key.php
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math
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Learn vocabulary, terms, and more with flashcards, games, and other study tools.
A trapezoid has one pair of parallel sides. If a quadrilateral is a kite, such as FGHJthen it has the following properties. Identify similar polygons.
You can create printable tests 6.6 homework answer key worksheets from these Grade 6 Quadrilaterals questions! The midsegment of a trapezoid is the segment that connects the midpoints of the legs.
Math properties and formulas 9th grade geometry. Prove that a favorite travel destination essay quadrilateral is a rectangle, rhombus, or square. Lesson 6. Perimeter of Quadrilaterals Worksheets. Trapezoid worksheets funtastic simon trapezoid. In this geometry lesson, 10th graders investigate the properties of trapezoids, isosceles trapezoids, and kites by measuring sides and angles in the figures and by Module 6: Midsegment of a Trapezoid.
The parallel sides are the. Trapezoids from trapezoid properties worksheet source: Trapezoid is a quadrilateral with exactly one pair of parallel sides. If a quadrilateral is a kite, then its diagonals are perpendicular.
Thanks for visiting our site, articleabove trapezoid properties worksheet published by admin. Trapezoid 2. To save a writing essay the time and expense of copying pages, you can give students the inexpensive Practice Your Skills Student Workbook, which does not have answers.
A is a quadrilateral with exactly one pair of parallel sides. Students will be thesis china to discover and prove properties of kites, trapezoids, and midsegments. Possible answer: Start studying 6.
Geometry worksheets quadrilaterals and polygons properties of trapezoids worksheets. L2 REF: Toggle navigation 6.
Perimeter of a trapezoid worksheets mathvine com download pdf. Is the trapezoid at the right isosceles?
Areas of Trapezoids, Rhombi application letter for funding sample Kites: Trapezoids can be deceptive. Notes on Trapezoids and Kites from Wyzant. Stalvey will post an answer key.
Properties of an Isosceles Triangle 2 If a trapezoid has one pair of congruent base angles, then the trapezoid is isosceles. What are the values of x and y?
Properties of trapezoids and kites wyzant resources, properties of trapezoids and kites now that we've seen several types of quadrilaterals that are parallelograms, let's learn about figures that do not have kiarostami homework dvd properties of parallelograms. Kite 3.
Opposite sides are congruent, so a rhombus is also a parallelogram and has all of the properties of a parallelogram. You may select between whole and decimal numbers, as well as whether the properties cinemark holdings case study have algebraic expressions to solve.
Geometry Worksheet Kites And Trapezoids Answers geometry worksheet kites and trapezoids answers kites and weather.
Day 1: Video on the properties of trapezoids and kites, watch after Test 7 Kites, Trapezoids, Rectangles, Rhombi, Squares, Parallelograms Answers may vary in how the venn diagram is designed to fit rhombi and squares within creative writing degree careers. Calculate the perimeter of various thesis china quadrilaterals like squares, rectangles, parallelograms, favorite travel destination essay, kites and trapezoids with this array of worksheets with dimensions presented as integers and decimals, comprehend the congruent all life is problem solving karl popper pdf and solve algebraic expressions to find the side length.
Trapezoid Find XY in each trapezoid. PS; QR 9. CC Kites, Trapezoids, Rectangles, Rhombi, Squares, Parallelograms Answers may vary in how the venn diagram is designed to fit rhombi and squares within kites.
Geometry Worksheet Kites and Trapezoids Answers Key Luxury 5th Grade Geometry, picture size x posted by at July 21, geometry notes g 9 8 4 8 5 rhombus rectangle square area notes and practice worksheet grade geometry in this worksheet students are to place the words trapezoid the triangle inequality theorem worksheets area of kites Worksheets are 6 properties of trapezoids, Midsegment of a triangle date period, Supplement name trapezoid midsegment period base t, Geometry work name kites and trapezoids period, Unit 4 syllabus properties of triangles quadrilaterals, Geometry, For problems 1 to 7 find the measure of each labeled, Answers work properties 6.6 homework answer key trapezoids.
Use the Pythagorean Properties of an Isosceles Triangle 1 If a quadrilateral is an isosceles triangle, then each pair of base angles are congruent. Repeat parts a and b for several other kites.
In the figure below, a glide reflection of the first trapezoid results in two congruent trapezoids that fit together to form a base height base parallelogram. What kind of segment is TS?
Prove certain triangles are Quadrilaterals - Properties of Quadrilaterals Color-By-Number Wintery WorksheetThis color-by-number worksheet covers creative writing courses in navi mumbai Properties of Quadrilaterals by asking students to problems involving parallelograms, rectangles, squares, rhombi, kites and isosceles trapezoids.
Show"FLIP"is"a"trapezoid"with"exactly" coordinate plane is m&a internship cover letter kite. Is the quadrilateral formed by connecting the midpoints of the trapezoid a parallelogram, rhombus, rectangle, or square? Students who took this test also took: Sign up to view the full version.
Nome ars lli Geometry worksheet kites and trap livinghealthybulletin Math practice worksheets Kites and trapezoids worksheet 29 beautiful pics of geometry Geometry worksheet kites and trap livinghealthybulletin Geometry worksheet kites and trapezoids worksheets for all Math practice worksheets within top 15 much more kites and Nome ars lli This preview has intentionally blurred sections.
The following theorems state the properties of an isosceles trapezoid. Proof of a theorem may require the use of an Section Use properties of kites to solve problems.
An business plan franchise pdf on each property is provided. What are some properties of trapezoids and kites? Geometry Unit 7 — Polygons and Quadrilaterals. Kites and Trapezoids Worksheet ; Chap: This is a simplistic model but there are others available on the internet which are more 6.
Practice Worksheet - Midsegments of Trapezoids Solve.
Module 6 Part 1 - Mr. E's Webpage Some of the worksheets for this concept are 6 properties of trapezoids, Geometry work name kites and trapezoids period, Area of trapezoids, Answers work properties of trapezoids, Performance based learning and assessment task properties, Name, Kites and trapezoids work answers, Name notes properties of trapezoids and kites.
Learn the properties of trapezoids and kites to solve problems. The Trapezoid Midsegment Theorem is similar to it.
Section 5. Students must use the definitions and properties of trapezoids and kites including medians of trapezoids. Section 6. Use properties of rectangles, rhombuses, and squares to solve problems.
Determine the sum of both the interior and exterior angle measures of a polygon Geometry Worksheet Name: TLW identify special quadrilaterals based on limited information, and will use both formal proof and coordinate geometry to prove that a quadrilateral is a special quadrilateral.
Key Words. Find GH in each trapezoid. The line containing the first diagonal bisects the Properties of Kites and Trapezoids.
Diagonals only bisect the vertex angles. Its measure is equal to one-half the sum of the lengths of the bases. Kite 47 9. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Properties of Kites and Trapezoids 1.
If a quadrilateral creative writing degree careers an isosceles favorite travel destination essay, then each pair of base angles is congruent. Geometry quadrilaterals test Geo. Create the worksheets you need with Infinite Geometry.
Geometry help: Answers for Geometry homework problems - Hotmath Kite 8. If AX and BY intersect at.
Find the area of the free-throw lane. Determine the most precise name for each quadrilateral.
Lucy is framing a kite with wooden dowels. Practice finding the areas of trapezoids. Use Properties of Trapezoids and Kites .
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https://www.physicsforums.com/threads/heat-equation-boundary-value-problem.585161/
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math
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http://img821.imageshack.us/img821/7901/heatp.png [Broken] Uploaded with ImageShack.us I'm having difficulty with the boundary conditions on this problem. I don't need a solution or a step by step. I've just never solved a boundary condition like this. Its the u(pi,t) = cos(t) that is giving me difficulty I tried getting a steady state solution for this. However, I end up with v(x) = (x/pi)*Cos(t) which makes no sense because v(x) should not be dependent on 't.' I can't make it homogeneous in order to solve it by separation of variables. Any advice would be greatly appreciated.
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https://www.schlockmercenary.com/blog/schlock-mercenary-aceo-series-one-roll-call/
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math
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Per my last post (where you'll find a definition of ACEO) there are ACEO "Sketch Cards" available in the Schlock Mercenary Store. The backs of these cards clearly read "Series One" which naturally implies that there will be a "Series Two." These are numbered cards, and there will be 250 of them in Series One. They take a lot longer to do than sketch edition books -- usually between 10 and 20 times as long.
After much hand-wringing, we've decided to limit Series One to seven (well, eight if you count the attorney drone) characters:
- Der Trihs
- Attorney Drone
Now that I've come out and said it, you completionists know exactly what you need.
I'm not planning to do equal numbers of each, mostly because I know that not everybody wants all seven. I plan on doing 50 each of Schlock, Tagon, and Kevyn, and 20 each of the remaining five characters. Of course, I've only drawn 65 cards total as of this writing, so there's still some wiggle-room. If there's a sudden run on "Brad" or "Attorney Drone" I'll adjust accordingly.
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https://homesearchend.com/how-to-find/how-to-find-eigen-vectors.html
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math
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How do you find the eigenvectors?
Finding Eigenvalues and Eigenvectors : 2 x 2 Matrix Example
How do you find eigenvectors and eigenvectors 3×3?
Find Eigenvalues of 3×3 Matrix –
How do you find eigenvalues and eigenvectors?
Given the Eigenvector Find Eigenvalues – Linear Algebra –
How do you find orthonormal eigenvectors?
Find Eigenvalues, Orthonormal eigenvectors , Diagonazible – Linear
What is the fastest way to find eigenvalues?
How to find eigenvalues quick and easy – Linear algebra explained
What do eigenvectors tell us?
Short Answer. Eigenvectors make understanding linear transformations easy. They are the “axes” (directions) along which a linear transformation acts simply by “stretching/compressing” and/or “flipping”; eigenvalues give you the factors by which this compression occurs.
What is the shortcut to find eigenvalues?
Easy method to find Eigen Values of matrices -Find within 10
How do you find the determinant of a 3×3?
Find the determinant of a 3×3 matrix the fast way –
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http://www.maa.org/publications/maa-reviews/sur-les-conjectures-de-gross-et-prasad-i?device=mobile
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math
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This is the first of two Astérisque volumes dedicated to a well-known conjecture of B. H. Gross and D. Prasad. (The second volume is here.) Formulated some twenty years ago, the conjecture has to do with representations of classical groups; specifically, the original conjecture described what should happen when one took an irreducible admissible representation of SO(n) and restricted it to a subgroup isomorphic to SO(n–1). The conjecture was recently proved by Waldspurger and Mœglin.
The two Astérisque volumes include articles by Gross and Prasad explaining (and generalizing) their conjecture and giving many examples, followed by a series of articles (mostly by Waldspurger) that culminate in the complete proof. Together, the two volumes make up a high-level expository account of the conjecture and proof.
I really like the idea of presenting a major new result in this form. It does not, of course, render it any less technical and difficult, but it does gather all the crucial ideas in one place. Kudos to the authors for having done it.
Fernando Q. Gouvéa is Carter Professor of Mathematics at Colby College and the Editor of MAA Reviews.
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https://research-information.bris.ac.uk/en/publications/stripping-torques-in-human-bone-can-be-reliably-predicted-prior-t/datasets/
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math
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Dataset for "Stripping torques in human bone can be reliably predicted prior to screw insertion with optimum tightness being found between 70% and 80% of the maximum"
Fletcher, J. W. A. (Creator), Zderic, I. (Creator), Gueorguiev, B. (Creator), Richards, R. G. (Creator), Whitehouse, M. R. (Contributor), Gill, R. (Creator) & Preatoni, E. (Creator), University of Bath, 21 Aug 2020
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http://mathhelpforum.com/algebra/220462-question-about-geometric-series-roots-unity.html
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math
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Hello. I am working on my own through the AQA FP2 Mathematics textbook (A2-Level) and I have come across a question I am stuck on. I am given the following equation, where is a complex number, and asked to solve the equation, in other words find the roots of the equation. My answers should be in the form .
Initially I noticed that this is a geometric series and can therefore be expressed as a summation:
But using the related formula for a geometric series I get:
note: n=4 because the final term of a sum of a geometric series is always n-1 (3).
However, when I substitute a random number in for , say 2, the two expressions do not equate. So somewhere, my geometric series expression is incorrect, but I cannot see where. I have not done geometric series in a while.
Continuing anyway, I would then go on to say that the roots of the initial equation must be three of the roots of or . But when finding the four solutions to this equation (disregarding the solution , as this would make the fraction indefinable), I do not get the right solutions.
To help, the actual solutions are .
If anyone could help me see what I have done wrong with the geometric series, I would be very grateful.
Thanks in advance.
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https://www.arxiv-vanity.com/papers/2102.07494/
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math
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Masking singularities in Weyl gravity and Ricci flows
Within vacuum Weyl gravity, we obtain a solution by which, using different choices of the conformal factor, we derive metrics describing (i) a bounce of the universe; (ii) toroidal and spherical wormholes; and (iii) a change in metric signature. It is demonstrated that singularities occurring in these systems are “masked”. We give a simple explanation of the possibility of masking the singularities within Weyl gravity. It is shown that in the first and third cases the three-dimensional metrics form Ricci flows. The question of the possible applicability of conformal Weyl gravity as some phenomenological theory in an approximate description of quantum gravity is discussed.
Weyl gravity is a conformally invariant theory of gravity where classes of conformally equivalent metrics serve as a physical object Weyl:gravity . For a host of reasons, such theory is not a fundamental theory of gravity, at least on cosmological and solar system scales. Nevertheless, there is a point of view that such theory can be useful in studying various gravitational effects near singularities. For example, in Refs. Gurzadyan:2010da ; Penrose:1 ; Penrose:2 ; Penrose:3 , the idea was proposed according to which classes of conformally equivalent metrics become a physically important object near a singularity. If this is the case, then such classes may contain both singular and regular metrics. By the singular metric we mean a metric which makes singular such scalar invariants like the scalar curvature and the squares of the Ricci and Riemann tensors. The existence both of regular and singular metrics within one class of conformally equivalent metrics results in the fact that, despite the divergence of the aforementioned scalar invariants, the conformally invariant tensors, and hence their squares, will be regular. For instance, we will demonstrate below that, within vacuum Weyl gravity, there exist solutions possessing such properties. It must be mentioned here that there exists an opposite point of view Mannheim:1988dj ; OBrien:2011vks , according to which Weyl gravity theory can be regarded as a viable macroscopic theory of gravitational phenomena. Moreover, there is a fact that some solutions of Einsteinian gravity are also exact solutions of Weyl gravity Li:2015bqa . Also, in Weyl gravity, there is an interesting approach to explain the nature of dark matter and dark energy Flanagan:2006ra .
In Refs. Hooft:2015mzk ; THooft:2015jcw , G. ’t Hooft considers the idea that “the conformal symmetry could be as fundamental as Lorentz invariance, and guide us towards a complete understanding of physics at the Planck scale.” In other words, the conformal invariance may be very important in quantum gravity in describing quantum gravity effects in high curvature regions. This means that Weyl gravity may serve as a phenomenological theory that approximately describes quantum gravity effects in high curvature regions, just as the Ginzburg-Landau theory is a phenomenological theory of superconductivity.
In Ref. Amelino-Camelia:2015dqa , the idea is proposed that gravity is responsible for breaking the fundamental conformal invariance. This can be treated as follows: in high curvature regions, the conformal invariance is not violated, but it is violated in going to low enough curvature regions.
In Refs. Maldacena:2011mk ; Anastasiou:2016jix , the connection between general relativity and Weyl gravity is under investigation. It is shown in Ref. Maldacena:2011mk that four-dimensional conformal gravity with a Neumann boundary condition is classically equivalent to ordinary four-dimensional Einstein gravity with a cosmological constant. Ref. Anastasiou:2016jix continues studies in this direction and provides a generic argument on the equivalence between Einstein gravity with a cosmological constant and conformal gravity for Bach-flat spacetimes. In Ref. Salvio:2017qkx , it is shown that agravity can be rewritten as conformal gravity plus two extra scalars with an SO(1,1) symmetry.
In Refs. Modesto:2014lga ; Modesto:2018def , the idea is discussed according to which consistent quantum gravity theory must be conformally invariant. In Ref. Modesto:2018def , on quantum conformal gravity grounds, an approach to resolving the problem of black hole singularity is even suggested. In Ref. narlikar:1977nf , the idea was pioneered that singularities arising in general relativity can be eliminated by an appropriate choice of conformal transformation. The papers Bambi:2016wdn ; Rachwal:2018gwu continue studying this subject and suggest an approach to eliminate singularities in Schwarzschild and Kerr solutions.
Concerning Ricci flows, in differential geometry, they are used in studying the topology of differentiable manifolds. Ricci flows define the occurrence of singular points on a manifold; this would lead us to expect that they can be used in gravitational theories when studying such singularities. For example, in Ref. Graf:2006mm , the field equations are postulated in the form of the Ricci flow equations, and Einstein’s theory is included as the limiting case where the flow is absent. In Ref. Herrera-Aguilar:2017hzc , the solutions to the equations for Ricci flows are given, and it is shown that these solutions contain metrics describing a change in metric signature. In Ref. Frenkel:2020dic , Ricci flows are under investigation, and the connection with a path integral in quantum gravity is demonstrated. In Ref. Dzhunushaliev:2008cz , the idea that the occurrence of quantum wormholes in spacetime foam can be described using Ricci flows is discussed. In Ref. Lashkari:2010iy , Ricci flows are used to study transitions between the AdS and warped AdS vacuum geometries within Topologically Massive Gravity.
Summarizing the above ideas, one can suppose that Weyl gravity may serve as some phenomenological theory that approximately describes quantum gravity effects in high curvature regions, and in low curvature regions the conformal invariance is violated. In the present paper, we would like to demonstrate that, within Weyl gravity, there exists such an interesting feature like masking of some singularities, that in fact is a consequence of quantum gravity, but in the case under consideration this effect is approximately described by Weyl gravity. By the “masking of singularities” we mean the fact that, in Weyl gravity, there can exist the following unusual situation: some tensors (and hence the corresponding scalar invariants) are singular (for example, the Ricci and Riemann tensors), while at the same time there are tensors which are not singular at the same points. As such tensors, there can be, for example, the Weyl and Bach tensors. From the mathematical point of view, this means that the Weyl tensor is constructed so that the combination of the Riemann and Ricci tensors and of the metric is such that the corresponding singularities eliminate each other. From the physical point of view, this means that Weyl gravity can be treated as an approximate description of quantum gravity effects describing the behavior of spacetime near some singularities. In this connection it may be also noted that it was shown in Ref. Hohmann:2018shl that the singularity of a black hole might be removed and replaced by the throat of a wormhole. It is noteworthy that in modified gravities, something similar can also exist: it was shown in Ref. Bahamonde:2016wmz that in ordinary gravity a singular cosmology in one frame might be nonsingular in the other frame.
The paper is organized as follows. In Sec. II, we introduce the Lagrangian and show the corresponding field equations, as well as the conformally invariant class of metrics which are the solution in Weyl gravity. For such class of metrics, in Sec. III, we discuss a cosmological bounce solution, singularities, and Ricci flows; in Sec. IV, we demonstrate the existence of a solution describing a change in metric signature, discuss the corresponding singularities, and show that the three-dimensional spatial metric is a Ricci flow; in Sec. V, we obtain toroidal, , and spherical, , wormholes and study the corresponding singularities. Finally, in Sec. VI, we discuss and summarize the results obtained.
Ii Weyl gravity
In this section we introduce the Lagrangian and write down the corresponding field equations in Weyl gravity. The action can be written in the form [hereafter we work in natural units and the metric signature is ]
where is a dimensionless constant, is the Weyl tensor.
The action (1) and hence the corresponding theory are invariant under the conformal transformations
where the function is arbitrary.
The corresponding set of equations in Weyl gravity is
where is the Bach tensor.
In what follows we will work with the solutions obtained in Ref. Dzhunushaliev:2020dom and consider in detail how such solutions can describe the structure of spacetime near singularities. To do this, let us consider the following metric:
Here, is the Hopf metric on the unit sphere; is an arbitrary function; is a constant; and . The Bach tensor for such metric is zero; this means that this metric is a solution of Eq. (2) for Weyl gravity.
In the following sections we will consider some interpretations of this solution and give the analysis of its singular points.
Iii Cosmological bounce solution and inflation
Consider the case with the conformal factor
Introducing the new time coordinate , we have the following expression for the metric (3):
If the function is chosen so that is an even function and , the metric (4) will describe a universe with a bounce at time . If, in addition, one chooses so that asymptotically , one will have a cosmological bounce solution with a subsequent inflationary expansion. This can be done by choosing, for example, , as was done in Ref. Dzhunushaliev:2020dom .
Consider the behavior of the metric at the bounce point . Since the function is assumed to be even, it can be expanded in a Taylor series as
where is an even function. We are interested in the behavior of the metric (4) at the bounce point , where the scalar invariants have the following expansions:
If one chooses the function so that , an interesting situation takes place at this point: there is a singularity, since the scalar invariants and diverge here. Nevertheless, the Weyl and Bach tensors are constructed so that they are just equal to zero, and hence the corresponding invariants and are also zero. This means that, in Weyl gravity, such singularities are masked! This is very interesting result, and one might reasonably suppose that it happens due to the fact that Weyl gravity is an approximate description of quantum gravity effects in high curvature regions, i.e., near some singularities.
Thus, the result obtained enables us to say that for small (notice that this parameter corresponds to the size of the universe at the bounce point) the transition from a contraction stage to expansion is a quantum gravity effect, and it may be approximately described by Weyl gravity.
Next, when the size of the universe increases, the spacetime becomes less curved; correspondingly, quantum gravity effects become negligible and the dynamics of the universe is no longer adequately described by Weyl gravity; that is, the spacetime becomes classical and it should be described by general relativity.
There is a very simple explanation why Weyl gravity “does not see” the singularity: the reason is that the singularity arises because of the factor in the spatial part of the metric (4). Since this factor tends to zero as , the volume of the space also goes to zero, thereby the singularities in the scalar invariants , and appear. But since Weyl gravity is conformally invariant, it “does not see” this change in the volume.
Consider now a sequence of spatial metrics from (4),
for . This sequence describes the occurrence of the singularity where the invariants (6) go to infinity. In differential geometry, this process is described by Ricci flows,
where is some parameter. The spatial metric tensor from (7) is defined as
where is the metric on the unit three-dimensional sphere in the Hopf coordinates ; is the corresponding three-dimensional Ricci tensor with the spatial indices ; the conformal factor also depends on the parameter .
The Ricci tensor for the metric (7) takes the form
with the solution
where is an integration constant. This means that the parameter , starting from the value , reaches zero value when . For this value, there appears a singularity for the scalar invariants and .
Thus, in this section, we have shown that there are cosmological bounce solutions in Weyl gravity. When the size of the universe decreases, at the bounce point, the scalar invariants , and diverge but the Weyl invariants and remain finite and equal to zero. This enables us to say that, in Weyl gravity, there is a kind of masking of singularities. It is also shown that the family of such solutions numbered by the size of the universe at the bounce time form the Ricci flow.
The idea that near singularities a conformal invariance and conformal transformations may be important has been considered in Refs. Gurzadyan:2010da ; Penrose:1 ; Penrose:2 ; Penrose:3 . The main idea of those papers is that near a singularity the conformal, but not the metric, structure of spacetime is of importance. And if there exists a conformal factor transferring a metric with a singularity into a metric without a singularity, from the physical point of view, there is no singularity in such a spacetime. From our point of view, this means that quantum gravity comes into play, and Weyl gravity is just an approximate description of quantum gravity effects in such a situation.
Iv Passing a singularity with a change in metric signature
Another interesting example of ignoring a singularity in Weyl gravity is its masking with a change in metric signature. To demonstrate this, consider the metric
In (11c), we have introduced for and for . We choose the function so that it changes its sign at some :
Thus, at , the metric signature is Lorentzian, , and at it is Euclidean, . To satisfy the conditions (12) and simplify calculations, let us choose the function in the form
where is an even function.
At the point , we have the following Taylor expansions for the scalar invariants:
Analogous to what was done in the previous section, the Weyl and Bach tensors are nonsingular when passing the point , while the invariants , and are singular. This means that, as well as in the previous section, at the point , where the metric changes its signature, the singularity is masked.
The spatial part of the metric from (11a) is
with the corresponding Ricci tensor
In contrast to the cosmological bounce solution considered in Sec. III where the quantity serves as a parameter in the Ricci flow, here the time coordinate may serve as such a parameter. This enables us to write the equation for Ricci flows (8) in the form
which gives us the following solution for a Ricci flow:
Thus, in this section, we have shown that, in Weyl gravity, there are solutions describing a change in metric signature. At the transition point, the scalar invariants , and go to infinity, but the Weyl invariants and remain finite and equal to zero. As in the case of the solution of Sec. III, this effect may be called the masking of singularities in Weyl gravity. Another interesting feature of the solutions with a change in metric signature obtained here is that there exists a special solution which is also the Ricci flow for the corresponding spatial part of the metric. Notice that the fact of a change in metric signature in passing through a singular point has also been pointed out in Ref. Herrera-Aguilar:2017hzc .
In the previous sections we have considered a choice of the conformal factor leading to metrics describing a bounce of the universe and a change in metric signature. In this section we examine metrics describing wormholes possessing different cross-sections: a torus and a sphere. In both cases we will obtain the corresponding Ricci flows.
v.1 Toroidal wormhole
In this subsection we consider the case where the conformal factor depends only on the spatial coordinate ,
Let us introduce the new spatial coordinate so that the function would have a minimum at and tend to as :
In this case we get the following metric:
The area of a torus spanned on the coordinates is defined by the determinant of the two-dimensional metric in the square brackets in Eq. (15):
Consistent with the conditions (14) for the function , it is seen that the area of the torus has a minimum at and goes to infinity as . Also, we require that as . Then this will mean that we have a toroidal wormhole.
Taking into account that and using the condition
we have the following solution for the function , see Ref. Dzhunushaliev:2020dom :
Then the metric (15) takes the form
and the coordinate covers the range .
Consider now Ricci flows for this case. In Secs. III and IV, we have considered three-dimensional Ricci flows for the spatial parts of four-dimensional metrics. The argument was that the singularities occurred because of the fact that the spatial volume vanishes. In this subsection we consider the case where the area of the wormhole throat goes to zero; therefore, we will consider two-dimensional Ricci flows defined on two-dimensional tori which are cross-sections of a wormhole.
Two-dimensional metric for the spacetime metric (17) is
and Ricci flows should be examined precisely for this two-dimensional metric. In this case a Ricci flow is written as
where the indices are two-dimensional indices defined on a two-dimensional torus with the metric (18). Since the Ricci tensor for the metric (18) is identically zero, , this means that the two-dimensional metric (18) is unchanged in the Ricci flow, as is obvious if we note that if the condition (16) is satisfied, we have only one solution – the metric (17).
One can ignore the condition (16) and consider wormholes without using it. In that case the even function has the following Taylor expansion near :
As , there will occur a singularity but, apparently, in this case the factor before the term with in (15) will also go to zero. This means that we will have a spherical wormhole which will be considered in the next subsection.
Thus, in this subsection, we have considered a toroidal wormhole and shown that for it the Ricci flow is stationary, and thereby singularities are absent in this case.
v.2 Spherical wormhole
In the above discussion, we have considered the spacetime with the spatial cross-section in the form of a three-dimensional sphere on which the Hopf coordinates are introduced. In particular, in the previous subsection, we have shown that, for some special choice of the conformal factor, it is possible to obtain a toroidal wormhole. Here, we will demonstrate that, by choosing the standard spherical coordinates on a three-dimensional sphere, it is possible to get a spherical wormhole with a cross-section in the form of a two-dimensional sphere .
Using the usual spherical coordinates, the spacetime metric can be written in the form
where are the angular coordinates on a three-dimensional sphere.
Let us define the conformal factor as . Then, introducing the new coordinate , we have from (19):
where we have used the function , which gives with . It is evident that this is the metric of a wormhole with the throat radius .
For simplicity, we will consider below a symmetric wormhole. This assumes that after introducing the new coordinate [see Eq. (20) above], the function will be even. Then the radius of the two-dimensional sphere in the metric (19) can be expanded in a Taylor series in the vicinity of as follows:
The parameter defines the area of a two-dimensional sphere at the center of the wormhole (that is, at the throat). Then the metric (19) takes the form
Let us keep track of the behavior of the scalar invariants when a cross-sectional area of the wormhole under consideration goes to zero:
It is seen from these expressions that the scalar invariants associated with the conformal tensors remain equal to zero, while the scalar invariants , and diverge. This means that, in Weyl gravity, when the cross-section of the wormhole decreases, nothing special happens, since the corresponding invariants do not diverge. From the physical point of view, this process of decrease (or of increase) of the cross-section can be interpreted as an annihilation (or creation) process of a quantum wormhole in spacetime foam.
As in the case of the toroidal wormhole from Sec. V.1, here, we will consider two-dimensional Ricci flows, defined now not on two-dimensional tori but on two-dimensional spheres, which are cross-sections of the wormhole under consideration. The corresponding two-dimensional metric follows from the spacetime metric (22),
For it, a Ricci flow is
where the indices are defined on a two-dimensional sphere. The Ricci tensor for the metric (23) is
with the solution
Thus, in this subsection, we have demonstrated that, in Weyl gravity, there is a family of solutions describing wormholes parameterized by the throat size . It is shown that when goes to zero, there occur singularities for such invariants like , and . At the same time, the scalar invariants associated with the conformally invariant tensors like and remain regular. This means that, in Weyl gravity, such singularities are masked. It is also shown that for the wormholes under investigation there are the Ricci flows whose presence can be physically interpreted as the description of the creation/annihilation process of quantum wormholes in spacetime foam.
Vi Discussion and conclusions
The main purpose of the present paper is to demonstrate that, in Weyl gravity, there is an interesting phenomenon – the masking of singularities. This means that there are solutions for which the scalar invariants , and are singular but the tensors employed in Weyl gravity (the Weyl and Bach tensors) remain regular. Perhaps this happens because Weyl gravity can be actually treated as an approximate theory describing quantum gravity effects near singularities, just as the Ginzburg-Landau theory is a phenomenological theory of superconductivity. In that case, in the region of strong gravitational fields, quantum gravity is approximately described by Weyl gravity or by some other modified gravity. By going to a low-curvature region, the conformal invariance violates. Similar ideas concerning the violation of the conformal invariance in quantum Weal gravity have been considered in Refs. Jizba:2020hre ; Jizba:2019oaf .
For better clarity, we would like to emphasize the distinction between the approaches suggested in Refs. Modesto:2014lga ; Modesto:2018def ; Rachwal:2018gwu and that of stated here. In the works Modesto:2014lga ; Modesto:2018def ; Rachwal:2018gwu and other similar papers, it is suggested to quantize some conformally invariant theory of gravitation, and general relativity follows from it as a classical limit. According to the idea suggested in the present paper, the primary theory is quantized general relativity, and Weyl gravity arises as an approximate description of some physical system; in the case under consideration, this is a gravitational field near singularities under discussion. Apparently, consistent quantum gravity will smooth out any singularities and, as it seemed to us, such a process can be approximately described using modified theories of gravity: Weyl gravity, as in the case considered by us (when the Weyl and Bach tensors vanish), or some other modified theories for black hole singularities (when the Weyl and Bach tensors are nonzero), for example, modified gravities. Notice also an interesting connection between the solution obtained here within Weyl gravity and “the Weyl curvature hypothesis ” proposed in Ref. Penrose:3 : in both cases, the Weyl tensor is equal to zero.
An unexpected result of the present study is that we have found the connection between the solutions obtained within Weyl gravity and Ricci flows. We have shown that for the cosmological bounce solution there is the family of solutions indexed by the size of the universe at the bounce time. The element of the family is the metric (the spatial part of the four-dimensional metric) which is the solution of the gravitational Weyl equations. In any such family, the metrics are a Ricci flow with the Ricci parameter . The solution found in Sec. IV, which describes a change in metric signature, is a Ricci flow where the Ricci parameter coincides with the time coordinate .
Another interesting result is that all the solutions discussed here refer to one conformally equivalent class of metrics, where both singular and regular metrics are present. There, they describe different physical situations: a bounce of the universe from a singularity with a possible subsequent exponential expansion, toroidal, , and spherical, , wormholes, and a change in metric signature.
A possible physical explanation of the fact that the metrics under discussion mask the singularities is that Weyl gravity is a phenomenological approximation for microscopic quantum gravity, just as the Ginzburg-Landau theory is a phenomenological description of superconductivity.
Thus, summarizing the results obtained:
Within Weyl gravity, there are obtained four types of solutions which are conformally equivalent each other but describe different physical situations.
It is shown that for all these solutions the singularities are masked in the sense that, even though such scalar invariants like the scalar curvature and the squares of the Ricci and Riemann tensors are singular, the squares of the Weyl and Bach tensors (which are employed in Weyl gravity) remain regular.
It is shown that for these solutions the three/two-dimensional spatial metrics are simultaneously Ricci flows.
A possible interpretation of Weyl gravity as a phenomenological theory which approximately describes quantum gravity effects is discussed.
We gratefully acknowledge support provided by the Program No. BR10965191 of the Ministry of Education and Science of the Republic of Kazakhstan. We are also grateful to the Research Group Linkage Programme of the Alexander von Humboldt Foundation for the support of this research.
- (1) H. Weyl, “Gravitation und Elekrizität”, Sitzungsberichte der Königlich Preusischen Akademie der Wissenschaften zu Berlin, 1918, pp. 465-480; English translation, “Gravitation and Electricity,” pp. 24-37 in O’Raifeartaigh’s book.
- (2) V. G. Gurzadyan and R. Penrose, Sir, “Concentric circles in WMAP data may provide evidence of violent pre-Big-Bang activity,” arXiv:1011.3706 [astro-ph.CO].
- (3) R. Penrose, Causality, quantum theory and cosmology (In On Space and Time ed. Shahn Majid, Cambridge University Press, Cambridge, 2008) pp. 141-195.
- (4) R. Penrose, The Basic Ideas of Conformal Cyclic Cosmology (In Death And Anti-Death, Volume 6: Thirty Years After Kurt Gödel (1906-1978), Chapter 7, pp. 223-242. Ed. Charles Tandy, Ria University Press, Stanford, Palo Alto, Calif., 2009).
- (5) R. Penrose, Cycles of Time: An Extraordinary New View of the Universe (Bodley Head, London, 2010).
- (6) P. D. Mannheim and D. Kazanas, “Exact Vacuum Solution to Conformal Weyl Gravity and Galactic Rotation Curves,” Astrophys. J. 342, 635 (1989).
- (7) J. G. O’Brien and P. D. Mannheim, “Fitting dwarf galaxy rotation curves with conformal gravity,” Mon. Not. Roy. Astron. Soc. 421, 1273 (2012).
- (8) Y. D. Li, L. Modesto and L. Rachwał, “Exact solutions and spacetime singularities in nonlocal gravity,” JHEP 1512, 173 (2015).
- (9) E. E. Flanagan, “Fourth order Weyl gravity,” Phys. Rev. D 74, 023002 (2006).
- (10) G. ’t Hooft, “Spontaneous breakdown of local conformal invariance in quantum gravity,” Les Houches Lect. Notes 97, 209 (2015).
- (11) G. ’t Hooft, “Local conformal symmetry: The missing symmetry component for space and time,” Int. J. Mod. Phys. D 24, no. 12, 1543001 (2015).
- (12) G. Amelino-Camelia, M. Arzano, G. Gubitosi, and J. Magueijo, “Gravity as the breakdown of conformal invariance,” Int. J. Mod. Phys. D 24, no. 12, 1543002 (2015).
- (13) J. Maldacena, “Einstein Gravity from Conformal Gravity,” [arXiv:1105.5632 [hep-th]].
- (14) G. Anastasiou and R. Olea, “From conformal to Einstein Gravity,” Phys. Rev. D 94, no. 8, 086008 (2016).
- (15) A. Salvio and A. Strumia, “Agravity up to infinite energy,” Eur. Phys. J. C 78, no. 2, 124 (2018).
- (16) L. Modesto and L. Rachwał, “Super-renormalizable and finite gravitational theories,” Nucl. Phys. B 889, 228 (2014).
- (17) L. Modesto and L. Rachwał, “Finite conformal quantum gravity and spacetime singularities,” J. Phys. Conf. Ser. 942, no. 1, 012015 (2017).
- (18) J. V. Narlikar and A. K. Kembhavi, “Space-Time Singularities and Conformal Gravity,” Lett. Nuovo Cim. 19, 517 (1977).
- (19) C. Bambi, L. Modesto, and L. Rachwał, “Spacetime completeness of non-singular black holes in conformal gravity,” JCAP 1705, 003 (2017).
- (20) L. Rachwał, “Conformal Symmetry in Field Theory and in Quantum Gravity,” Universe 4, no. 11, 125 (2018).
- (21) W. Graf, “Ricci flow gravity,” PMC Phys. A 1, 3 (2007).
- (22) R. Cartas-Fuentevilla, A. Herrera-Aguilar, and J. A. Olvera-Santamaria, “Evolution and metric signature change of maximally symmetric spaces under the Ricci flow,” Eur. Phys. J. Plus 133, no. 6, 235 (2018).
- (23) A. Frenkel, P. Horava, and S. Randall, “Perelman’s Ricci Flow in Topological Quantum Gravity,” arXiv:2011.11914 [hep-th].
- (24) V. Dzhunushaliev, “Quantum wormhole as a Ricci flow,” Int. J. Geom. Meth. Mod. Phys. 6, 1033 (2009).
- (25) N. Lashkari and A. Maloney, “Topologically Massive Gravity and Ricci-Cotton Flow,” Class. Quant. Grav. 28, 105007 (2011).
- (26) M. Hohmann, C. Pfeifer, M. Raidal, and H. Veermäe, “Wormholes in conformal gravity,” JCAP 1810, 003 (2018).
- (27) S. Bahamonde, S. D. Odintsov, V. K. Oikonomou, and M. Wright, “Correspondence of Gravity Singularities in Jordan and Einstein Frames,” Annals Phys. 373, 96 (2016).
- (28) V. Dzhunushaliev and V. Folomeev, “Spinor field solutions in modified Weyl gravity,” Int. J. Mod. Phys. D 29, no. 13, 2050094 (2020).
- (29) P. Jizba, L. Rachwał, S. G. Giaccari and J. Kňap, “Dark side of Weyl gravity,” Universe 6 (2020) no.8, 123; [arXiv:2006.15596 [hep-th]].
- (30) P. Jizba, L. Rachwał and J. Kňap, “Infrared behavior of Weyl Gravity: Functional Renormalization Group approach,” Phys. Rev. D 101 (2020) no.4, 044050; [arXiv:1912.10271 [hep-th]].
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http://www.brothersoft.com/speq-mathematics-download-45772.html
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math
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https://secure.foodbankrockies.org/site/TR/OngoingVirtualFoodDrive/General?px=1540430&pg=personal&fr_id=1260
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Did you know:
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-1-foundations-for-algebra-1-1-variables-and-expressions-practice-and-problem-solving-exercises-page-7/30
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math
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The product of $12$ and $x$.
Work Step by Step
""The product of" can be used to denote the multiplication of two numbers. Therefore, $12x$ can be written as "the product of $12$ and $x$."
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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https://calculator.academy/pig-profit-calculator/
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math
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Enter the total revenue from selling the pig and the total cost to raise the pig into the calculator to determine the pig profit.
- All Profit Calculators
- Pig Weight Calculator
- AVI (Advanced Vegetation Index) Calculator
- Stocking Rate Calculator
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Pig Profit Formula
The following formula is used to calculate the estimated profit from selling a pig.
P = TR - TC
- Where P is the pig profit ($)
- TR is the total revenue generate from selling the pig ($)
- TC is the total cost to raise the pig ($)
The total revenue generated from a pig can be estimated by the weight of the pig. Pigs typically sell from $2.50 to $3.50 per pound, so a pig weighing 200 pounds would sell for $500-$700.
The total cost to raise the pig varies depending on what type of overhead is used in the calculation, but for the most part, the costs include the feed and land they use to raise the pig.
What is a pig profit?
A pig profit, just like any other profit, is a measure of the total revenue minus costs that one pig generates.
How to calculate pig profit?
The following example outlines the steps required to estimate the profit earned by one pig.
First, determine the weight of the pig. In this example, the weight of the pig is 400 lbs.
Next, determine the total selling price per pound of the pig. For this example, the selling price was $3.00 per pound.
Next, calculate the total revenue by multiplying the price per pound by weight. This yields a revenue of $1200.00.
Next, determine the total cost to raise the pig. In this case, it cost $800.00 total to raise the pig.
Finally, calculate the pig profit using the formula above:
P = TR – TC
P = 1200 – 800
P = $400 profit
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https://www.shaalaa.com/question-bank-solutions/select-the-correct-answer-from-the-option-given-below-and-rewrite-the-statement-a-company-issuing-______________-debenture-must-create-a-charge-on-the-assets-of-the-company-provisions-for-issue-of-debentures-as-per-companies-share-capital-and-debentures-rule-2014_158406
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math
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Select the correct answer from the option given below and rewrite the statement:
A company issuing ______________ debenture must create a charge on the assets of the company
A company issuing Secured debenture must create a charge on the assets of the company
Concept: Provisions for Issue of Debentures as per Companies (Share Capital and Debentures) Rule 2014
Is there an error in this question or solution?
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https://app-wiringdiagram.herokuapp.com/post/newtons-laws-of-motion-problems-and-solutions
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math
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NEWTONS LAWS OF MOTION PROBLEMS AND SOLUTIONS
DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion. Newton's First Law of Motion states that a body at rest will remain at rest unless an outside force acts on it,and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by an outside force.
The First and Second Laws of Motion - Glenn Research Center
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Newton's Laws of Motion - with Examples, Problems
An Introduction to Newton's Laws of Motion. Science originates by observing nature and making inferences from them followed by devising and doing experiments to verify or refute theories. The three laws of motion discovered by Newton govern the motion of every object in nature all the time but due to the presence of friction and air resistance, they are a little difficult to see.Physics · Simple Harmonic Motion
Newton's laws of motion – problems and solutions | Solved
Force of gravity and gravitational field – problems and solutions. 1. Two objects m1 and m2 each with a mass of 6 kg and 9 kg separated by a distance of 5.. Parabolic motion, work and kinetic energy, linear momentum, linear and angular motion – problems and solutions. 1.
Newton's second law of motion – problems and solutions
Solved problems in Newton's laws of motion - Newton's second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to
Mechanics: Newton's Laws of Motion - physicsclassroom
Newton's Laws of Motion: Problem Set Problem 1: An African elephant can reach heights of 13 feet and possess a mass of as much as 6000 kg. Determine the weight of an African elephant in Newtons and in pounds. (Given: 1 N = pounds) Audio Guided Solution
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Newton Second Law of Motion Example Problems with Answers
Newton Second Law of Motion Example Problems with Answers. It is the acceleration of an object produced by an action or force which is directly proportional to the magnitude of the net force in the same direction and inversely proportional to the object mass. Calculate net force, mass and acceleration of an object by referring the below Newton second law of motion example problems with answers.[PDF]
NEWTON’S LAWS PRACTICE PROBLEMS
NEWTON’S LAWS PRACTICE PROBLEMS Answer the following questions in your science notebook. Show all of your work for math problems (equation, plug-in numbers, box answer). Restate the question in your answer for answers that you explain in words. NET FORCE & NEWTON’S 1ST LAW OF MOTION 1. Describe the motion of the race car shown in the graphic[PDF]
CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION
SSM REASONING According to Newton's second law of motion, the net force applied to. the fist is equal to the mass of the fist multiplied by its acceleration. The data in the problem gives. the final velocity of the fist and the time it takes to acquire that velocity.
Solved Examples on Laws of Motion -Study Material for IIT
Solved Examples on Laws of Motion:-Problem 1:-. A 26-ton Navy jet as shown in the below figure requires an air speed of 280 ft/s for lift-off own engine develops a thrust of 24,000 lb. The jet is to take off from an aircraft carrier with a 300-ft flight deck.
All of Newton's laws of motion (practice) | Khan Academy
Practice all of newton's laws of motion with Khan Academy's free online exercises. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *atic and *ndbox are unblocked.
Newton's Laws of Motion - physicsclassroom
Newton's Laws of Motion: Problem Set Overview Mass versus Weight. This set of 30 problems targets your ability to distinguish between mass.. Newton's Second Law of Motion. Newton's second law of motion states that the acceleration ( a).. Free Body Diagrams. Free body diagrams represent the
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https://stason.org/TULARC/self-growth/puzzles/393-probability-amoeba-p.html
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math
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This article is from the Puzzles FAQ, by Chris Cole [email protected] and Matthew Daly [email protected] with numerous contributions by others.
A jar begins with one amoeba. Every minute, every amoeba
turns into 0, 1, 2, or 3 amoebae with probability 25%
for each case ( dies, does nothing, splits into 2, or splits
into 3). What is the probability that the amoeba population
eventually dies out?
If p is the probability that a single amoeba's descendants will die
out eventually, the probability that N amoebas' descendents will all
die out eventually must be p^N, since each amoeba is independent of
every other amoeba. Also, the probability that a single amoeba's
descendants will die out must be independent of time when averaged
over all the possibilities. At t=0, the probability is p, at t=1 the
probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be
equal. Extinction probability p is a root of f(p)=p. In this case,
p = sqrt(2)-1.
The generating function for the sequence P(n,i), which gives the
probability of i amoebas after n minutes, is f^n(x), where f^n(x) ==
f^(n-1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition
of f with itself.
Then f^n(0) gives the probability of 0 amoebas after n minutes, since
f^n(0) = P(n,0). We then note that:
f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4
so that if f^(n+1)(0) -> f^n(0) we can solve the equation.
The generating function also gives an expression for the expectation
value of the number of amoebas after n minutes. This is d/dx(f^n(x))
evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x))
and since f'(1) = 1.5 and f(1) = 1, we see that the result is just
1.5^n, as might be expected.
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| 1,720 | 28 |
https://edurev.in/course/quiz/attempt/-1_Test-Quantitative-Theory-of-The-PN-Diode-Currents/f8eb63e0-f34a-4b0c-85e6-233075853b26
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math
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|1 Crore+ students have signed up on EduRev. Have you?|
What is the thickness of ‘space charge region’ or ‘transition region’ in P-N junction diode?
The region of the junction is depleted by mobile charges, hence it is called space charge region or depletion region or transition region which is 10-4 cm = 10-6 m= 1 micron.
If what of the following is doped into a semiconductor say germanium a P-N junction is formed.
A P-N junction is formed only when a donor impurities and acceptor impurities are added to either side of a semiconductor like silicon and germanium.
Which of the factors doesn’t change the diode current.
I = Io [e(v/nVt) -1], as shown in this equation the diode current is dependent on temperature , voltage applied on the diode , Boltzmann’s constant but diode current is not dependent on resistance as it is independent of resistance.
The product of mobility of the charge carriers and applied Electric field intensity is known as
When the semiconductors like silicon and germanium is implied by an electric field the charge carriers when get drifted by certain velocity known as drift velocity. Drift velocity is product of mobility of charge carriers and field intensity.
The tendency of charge carriers to move from a region of heavily concentrated charges to region of less concentrated charge is known as.
In a semiconductor the charge will always have a tendency to move from higher concentrated area to less concentrated area to maintain equilibrium this movement of charges will result in diffusion current.
If the drift current is 100mA and diffusion current is 1A what is the total current in the semiconductor diode.
We know that the total current in a semiconductor is equal to sum of both drift current and diffusion current. Total current = 1A + 100mA =1.1A.
Which of the following is reverse biased?
The P-N junction diode is forward bias when the voltage applied to p type is greater than the n type and vice versa, since the voltage applied to p type is less in C) it is the answer.
The drift velocity is 5V and the applied electric field intensity 20v/m what will be the mobility of charge carriers.
We know that mobility of charge carriers is drift velocity divide by applied electric field intensity. Mobility = drift velocity / field intensity.
When there is an open circuit what will be the net hole current.
If there is any current present under open circuit there will be an indefinite growth of holes at one end of the semiconductor which is practically not possible hence zero amperes.
Rate of change of concentration per unit length in a semiconductor is called as.
In a semiconductor the holes as well as electrons which are the charge carriers is not equally concentrated on all regions of the semiconductor the change in their rate is referred as concentration gradient.
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| 2,834 | 21 |
https://outsidethehashes.com/?p=321
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math
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Continuing with my review of BCS computer rating systems, the 4th of the 6 systems in my series is Dr. Peter Wolfe’s ratings.
On his site, Wolfe only gives a brief explanation:
We rate all varsity teams of four year colleges that can be connected by mutual opponents, taking note of game locations….The method we use is called a maximum likelihood estimate. In it, each team i is assigned a rating value πi that is used in predicting the expected result between it and its opponent j, with the likelihood of i beating j given by:
πi / (πi + πj)
The probability P of all the results happening as they actually did is simply the product of multiplying together all the individual probabilities derived from each game. The rating values are chosen in such a way that the number P is as large as possible.
First thing to note is that Wolfe rates all teams from FBS through Division III and even NAIA. He includes all games between any two varsity teams at any level. Other systems, like Sagarin, only rate Division I teams. Some only rate the FBS teams. I am not sure any one method is more “right” than the others, but it is odd that the BCS allows different systems to rate different sets of teams.
On to the meat of the rating system, Wolfe uses a “maximum likelihood estimate”. But don’t fret, this is just a mathematically fancy way of saying that he is trying to find the ratings for each team that best explain the game results that we see. According to his formula, the likelihood of team i beating team j is calculated simply as the rating of team i divided by the sum of the two teams’ ratings. For example, let’s take Wolfe’s top two ranked teams: Oklahoma St. and Alabama. Oklahoma St. has a rating of 9.126, while Alabama comes in at 9.038. Thus, Oklahoma St. is 9.126 / (9.126 + 9.038) = 9.126 / 18.164 = 50.2% likely to beat Alabama. And against a weaker team like, say, Utah St. (rating of 3.749), Okie State would be 70.9% likely to win.
So what Wolfe’s ratings do is use some mathematical formulas to find the ratings that best explain the game results so far this season. This is a sound way of computing ratings. One thing that remains unclear, however, is how location of games is included. Wolfe does note that the system looks at all games “taking note of game locations”. However, he does not say how they do this.
One last note. Thanks to Austin Link who commented on my Anderson & Hester ranking and shared a link to something similar he did last year. On there, Austin notes with regard to the Wolfe ratings that “one problem is that under that method all undefeated teams should have infinitely high ratings. Since this doesn’t happen he presumably includes some limiting factor, but it’s likely sort of arbitrary, reducing the mathematical rigorousness.” Since undefeated teams are so important in BCS ratings–much more so than any other rating system I have seen (what other rating system is almost solely concerned with undefeated teams?!)–how Wolfe deals with this is actually very important.
Wolfe’s ratings are very solid. The only two questions are (1) how does he deal with game locations? and (2) how does he deal with undefeated teams, since all should theoretically have infinite ratings? Overall, Wolfe’s ratings seem to be a worth addition to the BCS computer ratings, though (as with most of these systems) it would be nice to see full methodology to provide a more rigorous review of their quality.
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CC-MAIN-2024-18
| 3,483 | 10 |
https://sites.rmit.edu.au/information-security-seminars/2019/04/05/seminar-embedding-partial-latins-squares/
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math
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Embedding partial Latin squares into sets of MOLS
SPEAKER: E. Sule Yazici, Koc University, Turkey.
Date: Wednesday 10th April
Time: 3:00pm–4:00pm (Talk & Q/A)
Venue: Building 8 Level 9 Room 66 (AGR) RMIT City campus
The seminar will be followed by snacks and drinks
All students, staff and visitors are welcome
ABSTRACT: In 1960 Evans proved that a partial Latin square of order n can always be embedded in some Latin square of order t for every t≥2n. In the same paper Evans asked if a pair of finite partial Latin squares which are orthogonal can be embedded in a pair of finite orthogonal Latin squares. It is known, that a pair of orthogonal Latin squares of order n can be embedded in a pair of orthogonal Latin squares of order t if t ≥3n, the bound of 3n being best possible. Jenkins considered embedding a single partial Latin square in a Latin square which has an orthogonal mate. His embedding was of order t2.
We showed that any partial Latin square of order n can be embedded in a Latin square of order at most 16n2 which has at least 2n-1 mutually orthogonal mates. We also showed that a pair of (partial) orthogonal Latin squares of order n can be embedded into a set of t mutually orthogonal Latin squares of order a polynomial with respect to n for any t≥2. Furthermore the constructions we provided, give a set of 9 MOLS(576).
Joint Work with Diane Donovan (UQ) and Mike Grannell (Open University UK)
BIOGRAPHY: Associate Professor Sule Yazuci is with the Department of Mathematics at Koc University in Turkey. Her research focuses on combinatorial designs.
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CC-MAIN-2022-33
| 1,582 | 11 |
http://yiessaykrev.skylinechurch.us/formulating-a-hypothesis.html
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math
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The null hypothesis is a hypothesis which the researcher tries to disprove, reject or nullify. Describes how to test the null hypothesis that some estimate is due to chance vs the alternative hypothesis that there is some statistically significant effect real statistics using excel i am not sure how to formulate my null hypothesis here. Video created by university of illinois at urbana-champaign for the course inferential and predictive statistics for business watch any infomercial and you hear many outrageous promises use this cream and your skin will look 80% firmer use. Planning phase step 4: formulate a hypothesis formulate a hypothesis worksheet your topic: _____ research/guiding question: _____ _____ purpose statement in one sentence state what the experiment will allow you to discover about your research/guiding question this is your purpose.
Research questions, hypotheses and objectives t tive to formulate a research question that is both clinically relevant and answerable researchhypothesis the primary research question should be driven by the hypothesis rather than the data1,2 that is. A significance test examines whether the null hypothesis provides a plausible explanation of the data the null hypothesis itself does not involve the data it is a statement about a parameter (a numerical characteristic of the population) these population values might be proportions or means or. The next stage of the scientific method is known as the hypothesis this word basically means a possible solution to a problem, based on knowledge and research the hypothesis is a simple statement that defines what you think the outcome of your experiment will be. Crafting good hypotheses for your startup is hard most people focus on solutions rather than problems that leads to a ton of products getting launched with zero traction the all-too-common solutions looking for problems a good hypothesis is important because it leads to good experimental. Join curt frye for an in-depth discussion in this video formulating a hypothesis, part of learning excel data-analysis. Form a hypothesis identify the independent and dependent variables as well as constants lesson: 91 formulating hypotheses and identifying variables rachel ryland boston collegiate charter school dorchester, ma 27158 views 125 formulating hypotheses.
Students write a hypothesis about how the length of a string affects the time it takes for a pendulum to complete one period, and conduct an experiment to test their hypothesis. Putting a certain hypothesis is important for the person making a certain experimentfor exampe,if s/he were a scientist,the hypothesis would help in knowing what experiments should be done(in another words,no hypothesis--no experiments will be done.
Conceptualizing nursing research problems, questions, and hypotheses in research reports introduction formulating answerable clinical questions you will click onto this and select one questions remember, a hypothesis can transform a research question into a quantitative prediction. After reading this material and performing the activities listed, you will be able to acquire knowledge about hypothesis and action hypothesis.
A hypothesis has classical been referred to as an educated guess in the context of the scientific method, this description is somewhat correct. A hypothesis is a guess about the outcome of an experiment due to research you basically say, i think the substance will dissolve when i add water. Learn what a hypothesis test is, three testing methods and how to interpret the results download excel add-in 30 day trial.
Advice on statistics research paper formulating a hypothesis: steps in formulating a hypothesis decide what you want to explain: choose a dependent variable your dependent variable must show variation run descriptives to see mean and dispersion statistics. Formulation of hypothesis hypothesis formulation once you have identified you research question, it is time to formulate your hypothesis while. Good hypothesis: poor hypothesis: when there is less oxygen in the water, rainbow trout suffer more lice kristin says: this hypothesis is good because it is testable, simple, written as a statement, and establishes the participants (trout), variables (oxygen in water, and numbers of lice), and predicts effect (as oxygen levels go down, the.
Define formulate: to reduce to or express in a formula to put into a systematized statement or expression devise — formulate in a sentence. Formulating the null hypothesis is not automated (though the calculations of significance testing usually are) sir david cox has said, how [the] translation from subject-matter problem to statistical model is done is often the most critical part of an analysis. This lesson will give the definition of a null hypothesis, as well as an alternative hypothesis examples will be given to clearly illustrate the. Reader approved how to write a hypothesis two parts: preparing to write a hypothesis formulating your hypothesis community q&a a hypothesis is a description of a pattern in nature or an explanation about some real-world phenomenon that can be tested through observation and experimentation.
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| 5,188 | 7 |
https://writemyresearchpaper.us/sincere-stationery-corporation-needs-to-raise-500000-to-improve-its-manufacturing-plant-it-has-decided-to-issue-a-1000-par-value-bond-with-a-14-percent-annual-coupon-rate-and/
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math
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Sincere Stationery Corporation needs to raise $500,000 to improve its manufacturing plant. It has decided to issue a $1,000 par value bond with a 14 percent annual coupon rate and a 10-year maturity. The investors require a 9 percent rate of return. a. Compute the market value of the bonds. b. will the net price be if flotation costs are 10.5 percent of the market price? c. How many bonds will the firm have to issue to receive the needed funds? d. is the firm’s after-tax cost of debt if its average tax rate is 25 percent and its marginal tax rate is 34 percent? Fijisawa Inc. is considering a major expansion of its product line and has estimated the following cash flows associated with such an expansion. The initial outlay would be $1,950,000, and the project would generate incremental free cash flows of $450,000 per year for 6 years. The appropriate required rate of return is 9 percent. a. Calculate the NPV. b. Calculate the PI. c. Calculate the IRR. d. Should this project be accepted? Spartan Stores is expanding operations with the introduction of a new distribution center. Not only will sales increase, but investment in inventory will decline due to increased efficiencies in getting inventory to showrooms. As a result of this new distribution center, Spartan expects a change in EBIT of $900,000. While inventory is expected to drop from $90,000 to $70,000, accounts receivables are expected to climb as a result of increased credit sales from $80,000 to $110,000. In addition, accounts payable are expected to increase from $65,000 to $80,000. This project will also produce $300,000 of depreciation per year and Spartan Stores is in the 34 percent marginal tax rate. is the project’s free cash flow in year 1?
https://writemyresearchpaper.us/wp-content/uploads/2021/08/whatsapp-logo-300x115.jpeg 0 0 Write my research paper https://writemyresearchpaper.us/wp-content/uploads/2021/08/whatsapp-logo-300x115.jpeg Write my research paper2019-07-07 23:14:462019-07-07 23:14:46Sincere Stationery Corporation needs to raise $500,000 to improve its manufacturing plant. It has decided to issue a $1,000 par value bond with a 14 percent annual coupon rate and
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| 2,178 | 2 |
http://www.vpr.net/news_detail/95555/bob-amos-returns-to-recording-studio/
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math
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Bob Amos Returns To The Recording Studio
08/16/12 12:50PM Matt Bushlow
| MP3 || Download MP3 |
Bluegrass musician Bob Amos moved to St. Johnsbury in 2005 after the death of a bandmate to take a break from music. VPR's Matt Bushlow reports that Amos is back performing and recording, with the recent album "Borrowed Time."
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| 325 | 4 |
http://library.iiitd.edu.in/cgi-bin/koha/opac-detail.pl?biblionumber=16190
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math
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Approximation, Randomization, and Combinatorial Optimization. Algorithms and Techniques [electronic resource] :7th International Workshop on Approximation Algorithms for Combinatorial Optimization Problems, APPROX 2004, and 8th International Workshop on Randomization and Computation, RANDOM 2004, Cambridge, MA, USA, August 22-24, 2004. Proceedings /
Contributor(s): Jansen, Klaus [editor.] | Khanna, Sanjeev [editor.] | Rolim, José D. P [editor.] | Ron, Dana [editor.] | SpringerLink (Online service).Material type: BookSeries: Lecture Notes in Computer Science: 3122Publisher: Berlin, Heidelberg : Springer Berlin Heidelberg, 2004.Description: X, 434 p. online resource.Content type: text Media type: computer Carrier type: online resourceISBN: 9783540278214.Subject(s): Computer science | Computers | Algorithms | Numerical analysis | Computer science -- Mathematics | Computer Science | Theory of Computation | Computer Science, general | Algorithm Analysis and Problem Complexity | Discrete Mathematics in Computer Science | Numeric Computing | AlgorithmsOnline resources: Click here to access online
Contributed Talks of APPROX -- Designing Networks with Existing Traffic to Support Fast Restoration -- Simultaneous Source Location -- Computationally-Feasible Truthful Auctions for Convex Bundles -- Randomized Approximation Algorithms for Set Multicover Problems with Applications to Reverse Engineering of Protein and Gene Networks -- On the Crossing Spanning Tree Problem -- A 3/4-Approximation Algorithm for Maximum ATSP with Weights Zero and One -- Maximum Coverage Problem with Group Budget Constraints and Applications -- The Greedy Algorithm for the Minimum Common String Partition Problem -- Approximating Additive Distortion of Embeddings into Line Metrics -- Polylogarithmic Inapproximability of the Radio Broadcast Problem -- On Systems of Linear Equations with Two Variables per Equation -- An Auction-Based Market Equilibrium Algorithm for the Separable Gross Substitutability Case -- Cost-Sharing Mechanisms for Network Design -- Approximating Max k CSP Using Random Restrictions -- Approximation Schemes for Broadcasting in Heterogenous Networks -- Centralized Deterministic Broadcasting in Undirected Multi-hop Radio Networks -- Convergence Issues in Competitive Games -- Cuts and Orderings: On Semidefinite Relaxations for the Linear Ordering Problem -- Min-Max Multiway Cut -- Contributed Talks of RANDOM -- The Chromatic Number of Random Regular Graphs -- Estimating the Distance to a Monotone Function -- Edge Coloring with Delays -- Small Pseudo-random Families of Matrices: Derandomizing Approximate Quantum Encryption -- The Sketching Complexity of Pattern Matching -- Non-Abelian Homomorphism Testing, and Distributions Close to Their Self-convolutions -- Robust Locally Testable Codes and Products of Codes -- A Stateful Implementation of a Random Function Supporting Parity Queries over Hypercubes -- Strong Refutation Heuristics for Random k-SAT -- Counting Connected Graphs and Hypergraphs via the Probabilistic Method -- Improved Randomness Extraction from Two Independent Sources -- The Diameter of Randomly Perturbed Digraphs and Some Applications -- Maximum Weight Independent Sets and Matchings in Sparse Random Graphs -- Estimating Frequency Moments of Data Streams Using Random Linear Combinations -- Fooling Parity Tests with Parity Gates -- Distribution-Free Connectivity Testing -- Testing the Independence Number of Hypergraphs -- A Note on Approximate Counting for k-DNF.
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| 3,520 | 3 |
https://staging.becomekinetic.com/blog/2020/02/11/what-is-changing-in-math-2/
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math
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What Is Changing In Math?
What Is Changing In Math?
What is changing in math? That is a very difficult problem for several who are searching for job opportunities by using a mathematics diploma.
It would seem that when a single graduates from the college, pupils are rankmywriter sometimes still left with this kind of trouble. Can we instruct math to anyone free of prior encounter? There seems to be nobody reply to.
The situation of instructing mathematics is 1 that has existed given that enough time with the Archimedes as well as Pythagoras. Then again, latest developments have generated educating mathematics less of a challenge than it was in the past. In fact, you can find job openings for those who are keen to teach math to virtually http://umich.edu/?cid=72& anyone. So, in the event you are prepared to show individuals mathematics, you’ll be able to possess a exceptionally worthwhile occupation.
For those that are interested in educating customers easy methods to examine, generate, and do the trick with figures, math will be the topic that features a amazing location within their coronary heart. It is the component within the area of examine from the Usa. While other countries have various ways of educating math, the united states is amongst the number of nations which locations as big precedence on training this science as they do on teaching other subjects.
People who are attracted to math are sometimes into science and math, but do not take pleasure in the very difficult sciences. This is where instructing math can assist them. With the Us, plenty of colleges offer majors in math and science, moreover to their usual college student majors.
For students who are worried about how to search out work in this particular discipline, learning exactly what does integers necessarily mean in math terms can really help them get ready to enter higher education. By discovering what does integers signify in math conditions, you are likely to learn more about how math functions and grow to be an improved man or woman in most cases.
If you’re a math key and so are involved about learning to work with figures and just how to write down sophisticated expressions, acquire some lessons this kind of as Linear Algebra and Integral Calculus. Integral Calculus is considered being quite possibly the most fundamental study course in algebra, and is also generally taken earliest in calculus programs. Additionally it is known as integral calculus. In linear algebra, you are taught tips on how to find matrices and how to perform operations on them, this sort of as addition and subtraction.
In the usa, several schools and universities give lessons in Indian Math which addresses enhanced math principles and enhanced mathematical procedures inside of a variety of disciplines. You will find courses like calculus, geometry, trigonometry, algebra, data, along with a number of other classes that will make it easier to find out how to work with quantities.
Take your number one fundamental training course in algebra to get ready you for the other programs. Algebra will be the branch of arithmetic which bargains with transforming one quantity into yet another to compute the result. Learners who be taught algebra in faculty should preferably look and feel at how it is utilized in their distinct job.
Online university courses may just be important for you personally. By using using the web classes, you will definitely realize math and science with little or no tension. Exactly the same applies to those who are wondering about taking math classes by means of classic colleges.
There are numerous factors why consumers that are seeking for work that has a mathematics diploma flip to on line classes. By using these classes, you certainly will understand math, and exactly how it may well use to any profession.
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| 3,845 | 13 |
http://www.mathemania.com/lesson/infimum-supremum-bounded-set/
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math
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A non-empty set is bounded from above if there exists such that
The number is called an upper bound of .
If a set is bounded from above, then it has infinitely many upper bounds, because every number greater then the upper bound is also an upper bound. Among all the upper bounds, we are interested in the smallest.
Let be bounded from above. A real number is called the supremum of the set if the following is valid:
(i) is an upper bound of :
(ii) is the least upper bound:
The supremum of we denote as
If , then we say that is a maximum of and we write
If the set it is not bounded from above, then we write .
Proposition 1. If the number is an upper bound for a set , then .
The question is, does every non- empty set bounded from above has a supremum? Consider the following example.
Example 1. Determine a supremum of the following set
The set is a subset of the set of rational numbers. According to the definition of a supremum, is the supremum of the given set. However, a set does not have a supremum, because is not a rational number. The example shows that in the set there are sets bounded from above that do not have a supremum, which is not the case in the set .
In a set of real numbers the completeness axiom is valid:
Every non-empty set of real numbers which is bounded from above has a supremum.
It is an axiom that distinguishes a set of real numbers from a set of rational numbers.
In a similar way we define terms related to sets which are bounded from below.
A non-empty set is bounded from below if there exists such that
The number is called a lower bound of .
Let be bounded from below. A real number is called the infimum of the set if the following is valid:
(i) is a lower bound:
(ii) is the greatest lower bound:
The infimum of we denote as
If , then we say that is the minimum and we write
If the set it is not bounded from below, then we write .
The existence of a infimum is given as a theorem.
Theorem. Every non-empty set of real numbers which is bounded from below has a infimum.
Proposition 2. Let such that . Then
Example 2. Determine , , and if
Solution. Firstly, we have to check what are the -s:
The inequality above will be less then zero if the numerator and denominator are both positive or both negative. We distinguish two cases:
1.) and , that is, and . It follows .
2.) and , that is, and . It follows .
From the proposition 2. follows that and .
The minimum and maximum do not exist ( because we have no limits of the interval).
Example 3. Determine , , and if
Firstly, we will write first few terms of :
We can assume that the smallest term is and there is no largest term, however, we can see that all terms do not exceed . That is, we assume , and do dot exists. Let’s prove it!
To prove that is the supremum of , we must first show that is an upper bound:
which is always valid. Therefore, is an upper bound. Now we must show that is the least upper bound. Let’s take some and show that then exists such that
and such surely exists. Therefore, .
However, is not the maximum. Namely, if , then such that
which is the contradiction. It follows that the maximum of does not exists.
Now we will prove that .
Since , it is enough to show that is a lower bound of . According to this, we have
which is valid for all . Therefore, .
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| 3,282 | 46 |
http://training.teamgupta.net/2023/02/rogue-suitcase-challenge.html
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math
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Suitcase challenge was held in 2021 and the top 3 were 590, 560 and 460 pounds. In 95th place the entry was 95 pounds and 50th place was 252.5 pounds. I can certainly manage 100 pounds each hand and could not manage over 150 pounds, but I am not sure where the sweet spot is.
UPDATE: 2/14/2022 1800 (about an hour later) - 95# resulted in an epic fail at about 20 seconds. 75# was fine and 80# felt near max effort. That would be a 160# total placing me at roughly 85 out of 95 in the 2021 leaderboard.
Balancing the bars was harder than I expected and a hook grip without tape really bit into my thumb.
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| 603 | 3 |
https://interpac-norwich.co.uk/product/pacplus-hot-air-shrink-gun/
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math
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Hot air shrink gun from Pacplus. 1200w electric heat gun.
• 1200W heat output
• Requires standard 230v 13amp power supply
• Ideal for shrink wrapping smaller packages
• May be stored on the HABSTD stand during use to prevent damage or injury
• Weight 0.86kg
|Dimensions||24 × 23.6 × 1 cm|
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| 300 | 7 |
https://www.physicsforums.com/threads/solubility-of-aniline-and-acetanilide-in-aqueous-hcl.304035/
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math
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Hi Everyone! Hopefully someone can help me out with this question....Why is aniline soluble in aqueous hydrochloric acid, but acetanilide is not? Is it because aniline has a lone pair on the nitrogen that is more available to react? The lone pairs in acetanilide are too tied up with resonance? Is that answer along the right track? Any help with this would be appreciated! Thank you!
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| 384 | 1 |
https://lists.boost.org/Archives/boost/2002/10/38711.php
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math
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From: David Bergman (davidb_at_[hidden])
Date: 2002-10-30 12:46:30
--- commenting on the virtues of DocBook vs LaTeX in a discussion with a
> DocBook can be as simple in two ways. First, if you use the SGML form
there are a lot of "minimization"
> ?techniques that are allowed. Your example could be coded in DocBook
as <emphasis/stuff to be emphasised/,
> which is practically identical. This input can then even be
translated to full XML compliant code, if there's
> a reason for it (such as allowing easy parsing, as discussed above).
Second, when using an XML editor you never
> really have to type the tags any way.
> However, I agree with Dave that it would be beneficial for all if we
had an even simpler solution (preferrably > one that could output
DocBook and/or LaTeX). So we need to keep our options open and do some
Has anybody looked at db2tex, or similar tools?
I am a LaTeX junkie myself, and consider it to be quite easy to get
running even on a Windows machine (on that platform I use MikTeX and
But, I must agree with Bill as to the more versatile transformation in
an XML (or SGML) solution such as DocBook. What TeX is extremely good
at, though, is typography, so if we could use DocBook and use a
XML->LaTeX tool for those of us enjoying the professional result of the
Then we could have (1) DocBook -(db2tex-similar tool)-> LaTeX
-(pdftex)-> PDF be the "paper" process and (2) DocBook -> XHTML be the
"web" process, thus using DocBook as the canonical representation.
What do you think (as long as we get a decent XML->LaTeX tool...) ?
Boost list run by bdawes at acm.org, gregod at cs.rpi.edu, cpdaniel at pacbell.net, john at johnmaddock.co.uk
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https://www.tutorpace.com/geometry/special-triangles-online-tutoring
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math
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In Geometry, there are two special triangles. The special triangles are the triangles in which the angles are either 30º-60º-90º or 45º-45º-90º. The triangles forming these angles are called special triangles because when their values are calculated using the six trigonometric functions, then exact values are produced instead of decimal numbers. Therefore using any of these angles and the length of any one side of the triangle, the lengths of other sides can be easily calculated using basic trigonometric functions such as sine, cosine or tan of the special angles.
Example 1: The side AC is the hypotenuse in the right triangle ABC. If the hypotenuse of this triangle is 4cm, then what is the measure of side BC if the measure of angle C is 60º?
Here triangle ABC is a special triangle because one of its angle measures 60º which implies that the angles
of this triangle are in the form of 30º-60º-90º.
The trigonometric function, cos(C) = (adjacent side)/ (hypotenuse) = BC/AC
This gives: cos(60)= BC/ 4==> 1/2= BC/ 4==> BC= 4 * 1/2
Therefore the measure of the side, BC= 2cm
Example 2: The side PR is the hypotenuse in the right triangle PQR. If the measure of side PQ is 9cm, then what is the measure of the side QR if measure of angle R is 45º?
Here triangle PQR is a special triangle because one of its angle measures 45º which implies that the angles of this triangle are in the form of 45º-45º-90º.
The trigonometric function, tan(R) = (opposite side)/ (adjacent side) = PQ/QR
This gives: tan(45)= 9/ QR==> 1= 9/ QR==> QR= 1* 9
Therefore the measure of the side, QR= 9cm
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| 1,599 | 12 |
https://www.coursehero.com/tutors-problems/Genetics/11639930-A-breeder-wishes-to-estimate-narrow-sense-heritability-for-adult-weigh/
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math
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A breeder wishes to estimate narrow sense heritability for adult weight in a population of Turkeys. The mean weight of the
parental population was 18 pounds. Turkeys with a mean weight of 26 pounds were used to produce the next generation,
which had a mean weight of 20 pounds. What is the narrow-sense heritability for these Turkeys. Would this estimate be
accurate if offspring and parents received a different diet? Explain.
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https://calendar.math.illinois.edu/?year=2021&month=09&day=17&interval=day
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math
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Abstract: The diffeomorphism group of a smooth manifold is really large, yet it is still possible to make sense of it as an infinite-dimensional Lie group. Things get more complicated when we look at subgroups preserving geometric structures. In this talk we will look at a few things: the smooth structure on the group of diffeomorphisms that makes it into a Lie group; subgroups preserving several geometric structures, such as metrics, complex structures, symplectic structures; how Lie groupoids can be used to describe the symmetries of several "intransitive" geometric structures, such as manifolds with a distinguished hypersurface (e.g. a boundary) and foliations. The main focus is on the geometry, so no technical knowlegde of infinite-dimensional manifolds is required. Also, Lie groupoids will only be used at the very end to provide context to some examples, so no worries if you have never seen them before. Hope to see many of you on Friday!
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| 956 | 1 |
https://justaaa.com/finance/364733-michelle-wishes-to-establish-a-university-fund
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math
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Michelle wishes to establish a university fund for her son who is currently 8 years old.
a. If her son will need a monthly income of $900, how much does he need to be in place at the start of his university life (ie start of first-year) so that the $900 per month is achievable? Assuming that the interest over the three years while her son is at university is 6%p.a. compounded monthly and he is paid the $900 at the start of the month for this present value annuity.
b. Using your answer from part a, how much does Michelle need to invest now as a lump sum (present value) for the next 10 years at 5%pa (compounded annually) so that there are sufficient funds to achieve the amount from part a.
SEE THE IMAGE. ANY DOUBTS, FEEL FREE TO ASK. THUMBS UP PLEASE
Get Answers For Free
Most questions answered within 1 hours.
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| 819 | 6 |
https://www.hackmath.net/en/word-math-problems/multiplication?tag_id=74
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math
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Multiplication + multiplication principle - math problems
Number of problems found: 47
There are 4 roads from city A to city B. There are 5 roads from city B to city C. How many different routes can we come from city A to city C via city B?
- Big numbers
How many natural numbers less than 10 to the sixth can be written in numbers: a) 9.8.7 b) 9.8.0
How many different ways can three people divide 7 pears and 5 apples?
- Cube construction
A 2×2×2 cube is to be constructed using 4 white and 4 black unit cube. How many different cubes can be constructed in this way? ( Two cubes are not different if one can be obtained by rotating the other. )
- Boys and girls
There are 20 boys and 10 girls in the class. How many different dance pairs can we make of them?
- Dice and coin
A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the probability that the number rolled is greater than 2 and the coin toss is head?
- Holidays with grandmam
We have packed three T-shirts - white, red, orange and five pants - blue, green, black, pink and yellow. How many days can we spend with the old mother if we put on a different combination of clothes every day?
- Bookshelf and books
How many ways can we place 7 books in a bookshelf?
How many ways can you place 20 pupils in a row when starting on practice?
- Combinations 7
A multi-course meal has 3 choices. 3 appetizers, 4 main courses, and 3 desserts. How many different combinations to choose from?
- How many 2
How many three lettered words can be formed from letters A B C D E G H if repeats are: a) not allowed b) allowed
- Dinning room
How many different combinations can we choose if there are 3 soups, 5 kinds of main dish and 2 desserts in the dining room?
- Pins 2
how many different possible 4 digits pins can be found on the 10-digit keypad?
How many ways can be rewarded 9 participants with the first, second and third prize in a sports competition?
Find all five-digit numbers that can be created from numbers 12345 so that the numbers are not repeated and then numbers with repeated digits. Give the calculation.
How many three-digit natural numbers is greater than 321 if no digit in number repeated?
- Friends in cinema
5 friends went to the cinema. How many possible ways can sit in a row, if one of them wants to sit in the middle and the remaining's place does not matter?
- The camp
At the end of the camp a 8 friends exchanged addresses. Any friend gave remaining 7 friends his card. How many addresses they exchanged?
- Three digits number
How many are three-digit integers such that in they no digit repeats?
- A three-digit numbers
Determine the total number of positive three-digit numbers that contain a digit 6.
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https://chess.stackexchange.com/questions/13020/dealing-with-s%C3%A9n%C3%A9chaud-countergambit-in-king-s-gambit/13828
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math
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How would you try to refute the Sénéchaud Countergambit in King's Gambit?
[FEN ""] 1.e4 e5 2.f4 Bc5 3.Nf3 g5
Chess Stack Exchange is a question and answer site for serious players and enthusiasts of chess. It only takes a minute to sign up.Sign up to join this community
Full disclosure: I don't play the King's Gambit. My first thought (and note that I am NOT a master-level player) was 4. Nxg5 exf4 5. Nf3 (which, in and of itself, isn't sufficient to be an answer).
I decided to just do some Googling to try to learn more about this, and stumbled across analysis.cpuchess.com, which allowed a position to be analysed by various chess engines.
Here's what they gave me:
- Nxg5 h6 5. Qh5 Qf6 6. Nf3 Qxf4 7. Qxe5+ Qxe5 8. Nxe5 d6 9. Nf3 Nc6 10. b3 Nge7 11. Bb2 Rg8 12. d4 Bb4+ 13. c3 Ba5 14. Nbd2 b5 15. g4 Bxg4 16. Bxb5 O-O-O
- Nxg5 Qe7 5. Bc4 exf4 6. Qh5 Nf6 7. Qxf7+ Qxf7 8. Bxf7+ Ke7 9. e5 Ng4
Seemed to be locking up mid-analysis.
- Nxg5 Nc6 5. Qh5 Qf6 6. Bc4 Nh6 7. Nc3 Qg6 8. Qxg6 hxg6 9. Nd5 Bb6 10. Bb3 d6 11. d3 f5 12. Rf1 Bd7 13. fxe5 dxe5 14. exf5 Nxf5
- Nxg5 Qe7 5. Nc3 exf4 6. Nf3 Bb4 7. e5 Bxc3 8. dxc3 d6 9. Qd5 Nc6 10. Bxf4 Nf6
- Nxg5 exf4 5. Qh5 Qe7 6. Bc4 Nf6 7. Qxf7+ Qxf7 8. Bxf7+ Ke7 9. Bb3 h6 10. Nf3 Nxe4 11. d4 Bd6 12. O-O Re8 13. Ne5 Nc6 14. Rxf4 Ng5 15. h4 Nxe5 16. hxg5
What I found most amusing about this was that all of them (except Fruit, of course) suggested 4. Nxg5 as the best move... but that most went different directions afterwards; 4. ... Qe7 on GarboChess JS and Toga2 being the only commonality, and even they diverged almost immediately.
... which leads me to conclude that the answer to "how would you try to refute" is 4. Nxg5 followed by "tactics."
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| 1,696 | 14 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=2645&CurriculumID=65&Num=3.19
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math
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kwizNET Subscribers, please login to turn off the Ads!
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3.19 Quadrilaterals Review Test
: In a trapezoid ABCD, AB is parallel to CD. If angle A = 50 degrees and angle B = 120 degrees, find angle C and angle D.
90 degrees and 60 degrees Respectively
60 degrees and 130 degrees Respectively
100 degrees and 70 degrees Respectively
: "In a square all angles are equal to 90
: A quadrilateral can have at most _____ obtuse angles.
: Name the quadrilateral which is both a rhombus and a rectangle?
: ABCD is a trapezoid. Using the mid segment theorem find the value of x.
: In a rectangle ABCD, the diagonal Ac = 8cm., then the diagonal BD = ____.
: A parallelogram which has equal diagonals is a ________.
: In a quadrilateral ABCD,
C = 180
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D. (write the number only).
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s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096504.12/warc/CC-MAIN-20150627031816-00242-ip-10-179-60-89.ec2.internal.warc.gz
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| 1,556 | 33 |
https://www.physicsforums.com/threads/capacitor-voltage-in-charge-pump.351403/
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math
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Hi all, I have the feeling I'm missing something big. Consider first a capacitor having +9V on both terminals, in this case nothing happens. Now I set the negative terminal to 0V, and make sure a current is able to flow; the capacitor will now charge up to 9V, right? Ok, now I make sure the capacitor is unable to discharge in any way; so it has its positive terminal always 9V higher than the lower, right? Here's the thing: If I now connect the negative terminal to a +9V potential again, then the positive terminal has +18V (of course, all relative to 0V ground). I understand this has to be the case in terms of the above boldface line, but I don't see how this works, physically. I have the feeling I'm almost there; in terms of energy stored per coulomb etc. The battery is still at a +9V relative to ground, but the same capacitor connected to it is now at +18V relative to ground..?
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s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00188.warc.gz
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CC-MAIN-2018-30
| 891 | 1 |
https://www.computer.org/csdl/trans/tc/1968/12/01687281-abs.html
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math
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Issue No. 12 - December (1968 vol. 17)
Abstract?Block-oriented simulation languages are a powerful tool for solving simulation problems. A problem of major importance in simulation, however, is the choice of the integration formula to be employed in the integration block. Users of early simulation languages had no difficulty selecting a formula since these languages provided only a single formula. Advanced languages have a number of formulas available, but leave the choice to the user who does not have much to go by except the good names of the formulas. Difficulties which were encountered in the use of multistep formulas made formulas of the Runge-Kutta type more popular. Experience has shown that they are less critical with respect to stability. However, the computing time for a Runge-Kutta formula is considerably larger and for this reason multistep formulas, which include the predictor- corrector scheme, are still appealing.
Index Terms?Digital simulation, error-stability trade-off, multistep integration formulas, numerical intergration, trapezoidal rule.
H. Grebe and W. Giloi, "Construction of Multistep Integration Formulas for Simulation Purposes," in IEEE Transactions on Computers, vol. 17, no. , pp. 1121-1131, 1968.
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CC-MAIN-2018-26
| 1,243 | 4 |
http://www.google.com/patents/US7245682?dq=6373188
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math
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US 7245682 B2
In some embodiments, a phase detector receives a set of sampling clock signals and a data signal and compares each of the clock signals to the data signal. A clock selector selects an optimal sampling clock signal from the set of sampling clock signals based on a trend of a relationship between the clock signals and the data signal. Other embodiments are described and claimed.
1. A method for determining a sampling clock comprising:
determining a trend in a relationship between clock signals and received data based on information included in the received data;
identifying a first clock as an optimal sampling clock for the data based on the trend;
storing information about characteristics of a previously optimal sampling clock;
identifying a change in the relationship between the clock signals and the data, in which the first clock is no longer an optimal sampling clock for the data;
identifying a second clock as an optimal sampling clock for the data;
determining the refresh rate time between the identification of the first clock as an optimal sampling clock and the identification of the second clock as an optimal sampling clock;
changing the sampling clock after a time period equal to the refresh rate time; and
programming an RC time constant value of a low pass filter based on the refresh rate time; and
using the low pass filter to change the optimal sampling clock based on the RC time constant value.
2. The method of
3. The method of
4. The method of
5. The method of
This description relates to determining an optimal sampling clock.
In order to sample and process an electronic signal, the phase of that signal is first determined. When a serial electronic device, such as a pager, receives a signal, the phase of that signal is not known. Thus there must be a mechanism for determining the phase before sampling and processing.
As shown in
An edge detector 32 analyzes the four sampled signals 28. The edge detector 32 detects transitions between digital logic levels (206). Hence it will detect that a sampled signal is the result of a sampling clock sampling the data when it was transitioning from low to high, or from high to low. The results of the edge detector 32 are sent to a decision matrix 34. The decision matrix 34 determines which clock has sampled a transition from low to high and which clock has sampled a transition from high to low (208). The clock whose phase is between the phases of those two clocks is sampling the data roughly at the midpoint of the data pulse (210). Thus, that clock is sampling the data better than any of the others. Data is then sampled by that clock.
The transmitter 22 and receiver 20, however, are independent of one another, and use different clocks. Even if those clocks are intended to be identical in frequency, subtle differences will materialize over time. Take, for example, an external device, such as a CD-ROM player, connected to a personal computer. The CD-ROM player transmits data to a receiver on the personal computer. The CD-ROM player's clock is independent from the personal computer's clock.
Because the decision matrix 34 also experiences drift and jitter, the decision matrix 34 may not be able to determine whether a sampling clock of increasing or decreasing phase is now the optimal sampling clock. Furthermore, as the speed of serial communication increases, the bit error rate (“BER”), or the rate at which data is incorrectly sampled, becomes a greater concern.
An optimal clock determination module 62 determines which clock is optimal, and sends this information to a sampling system 64. The sampling system 64 samples the data signal 54 using the optimal sampling clock. The clock determination module 62, the sampling system 64, and the receiver 58 can be controlled by a processor 66.
With reference to
The quadrature phase detector can include a capacitive circuit, for example a capacitor (not shown). The capacitor can be charged with a known base charge, for example by setting it to ground zero during a system reset. The capacitor can be charged while the clock signal is at a logic high value and the data signal is also at a logic high value, and discharged while the clock signal is at a logic high value and the data signal is at a logic low value. After the charge/discharge cycle, the capacitor may have a residual charge.
The residual charge represents how closely the clock signal is sampling the data signal. For example, if the clock signal is sampling the data signal at roughly the midpoint of the pulse, the charge and discharge times will be roughly equal, and the residual charge of the capacitor will be close to the base charge. If the clock signal is shifted left relative to the data signal, the capacitor will be charged for longer than it is discharged, and the residual charge on the capacitor will be greater than the base charge. Likewise, if the clock signal is shifted right relative to the input pulse, the capacitor will be charged for less time than it is discharged, and the residual charge will be less then the base charge.
The quadrature phase detector outputs 78 can be analog signals. Because the outputs may diminish over time, the quadrature phase detector output signals can be latched using D-type flip flops or latches 80, 82, to produce latched output signals 83.
Each of these latched output signals 83 can be analyzed and compared to one another with a quadrant determination module 86. The quadrant determination module 86 determines which of the output signals 78 has a time during which it is positive that is most equal to the time during which it is negative. This output signal represents the clock signal that is sampling most closely to the center of the input data pulse.
Each of the output signals 83 can be filtered with charge pump/low pass filter pairs 84 to remove noise. The absolute values of the residual capacitor charges for each of the latched output signals 83 are compared to one another using a voltage comparison module 88. In this example, the capacitor emitting the smallest analog voltage represents the clock/input signal pair that is the most optimal. Thus the voltage comparison module 88 compares the absolute values of the voltages of the analog signals from the discharging capacitors 78, or as in this example the latched output signals 83, to determine which is the smallest. An example of a voltage comparison module 88, a winner-take-all circuit 90, is shown in
The results of the voltage comparison module 88 can be sent to the decision matrix 34, shown in
Once the optimal sampling clock has been determined, the quadrature phase detectors 70, 72, shown in
The clock determination module 62 can also be used to determine a system time constant. The system time constant is the time it takes for the data signal to drift a certain amount. For example, the phase of the input data signal might drift by 45 degrees with respect to the VCO clock signals in 100 ms. The optimal sampling clock would then change every 100 ms. This is the optimal refresh rate. Once the time constant is determined, the sampling clock can be changed to a new sampling clock every 100 ms to reflect the change in which clock is optimal. The sampling clock can be automatically changed, for example by instruction from the processor 66. Alternately, the charge pump/low pass filter pairs 84, shown in
The optimal clock determination module 62 can be monitored to determine the system refresh rate. The output of the clock determination module 62 can be measured at two points in time, separated by a predetermined amount of time. The system refresh rate can thus be determined by measuring the amount of drift in the predetermine amount of time.
The optimal refresh rate can be determined from characteristics of the input signal and the VCO. For example, the refresh rate can be optimized based on the low frequency drift and jitter of the input signal and the VCO. The optimal refresh rate can be determined when the system resets and the transmitter sends an alignment sequence. The alignment sequence is a sequence of data that is typically used in serial communications, such as during the handshaking process between the transmitter 52 and the receiver 58, shown in
Although some implementations have been discussed, other embodiments are also within the scope of the invention.
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http://mathhelpforum.com/pre-calculus/175531-solving-equation-exponential.html
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math
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I seriously doubt that you will be able to find an exact value for . Just use a numerical method and/or some technology.
From the first line, multiply both sides by -3.
Then you'll be in a form where you can use...
Lambert W function - Wikipedia, the free encyclopedia
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| 268 | 4 |
https://www.physicsforums.com/threads/applied-probability-calculation-not-working-out-integrals.432525/
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math
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1. The problem statement, all variables and given/known data The joint density of X and Y is given by f(x,y) = [e-x/y)e-y ] / y x is between 0 and infinity; y is between 0 and infinity Show E[X |Y = y] = y. Jump to the last part of this post in bold if you just want to check my calculus calculations and skip the probability stuff. 2. Relevant equations 3. The attempt at a solution First I need to find fX|Y (x|y). To do this I need to take f(x,y) and divide by fY(y). I think it's the latter which is messing me up. My understanding is that to find it, I need to integrate f(x,y) with respect to x. The solutions manual does that, but there are no limits on their integral. Shouldn't the limits be from 0 to infinity? They have [(1/y)e-x/ye-y ] / [e-y ∫ (1/y)e(-x/y) dx ] = (1/y)e-x/y Which in my calculations doesn't work out! I get -1/y. What am I doing wrong?
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CC-MAIN-2018-51
| 867 | 1 |
http://collinsbooks.com.au/book/A-Concise-Introduction-to-Languages-and-Machines/9781848001213
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math
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Publisher: Springer London
Series: Undergraduate Topics in Computer Science
Publication Date: December 29, 2015
Binding: Kobo eBook
A Concise Introduction to Languages, Machines and Logic provides an accessible introduction to three key topics within computer science: formal languages, abstract machines and formal logic. Written in an easy-to-read, informal style, this textbook assumes only a basic knowledge of programming on the part of the reader.
The approach is deliberately non-mathematical, and features: - Clear explanations of formal notation and jargon, - Extensive use of examples to illustrate algorithms and proofs, - Pictorial representations of key concepts, - Chapter opening overviews providing an introduction and guidance to each topic, - End-of-chapter exercises and solutions, - Offers an intuitive approach to the topics.
This reader-friendly textbook has been written with undergraduates in mind and will be suitable for use on course covering formal languages, formal logic, computability and automata theory. It will also make an excellent supplementary text for courses on algorithm complexity and compilers.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00696.warc.gz
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CC-MAIN-2020-50
| 1,137 | 7 |
http://www.wired.co.uk/tags/Wolfram+Research
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math
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Get the digital magazine on your device!
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The founder of Wolfram Research, and a mathematician himself, Wolfram argues that there is currently a large void between maths in education and maths in the real world. It's time to change this once and for all »
Wolfram Research, the company behind "answer engine" Wolfram Alpha, has built an interactive document format for sharing complex data »
Conrad Wolfram is bringing maths to the masses with smart, searchable web apps »
If recent measurements of cosmic ray particles are correct, then we may have the first evidence that the universe as we know it is really a giant computer simulation »
Aristotle claimed that "happiness is the meaining and the purpose of life, the whole aim and end of human existence." And for good reason. »
Comparing Samsung's Galaxy S5, Sony's Xperia Z2 and HTC's One M8 - which is best? »
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s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00458-ip-10-147-4-33.ec2.internal.warc.gz
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CC-MAIN-2014-15
| 903 | 8 |
https://www.arxiv-vanity.com/papers/0810.0525/
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math
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The fate of black branes in Einstein-Gauss-Bonnet gravity
Black branes are studied in Einstein-Gauss-Bonnet (EGB) gravity. Evaporation drives black branes towards one of two singularities depending on the sign of , the Gauss-Bonnet coupling. For positive and sufficiently large ratio , where is the radius of compactification, black branes avoid the Gregory-Laflamme (GL) instability before reaching a critical state. No black branes with the radius of horizon smaller than the critical value can exist. Approaching the critical state branes have a nonzero Hawking temperature. For negative all black branes encounter the GL instability. No black branes may exist outside of the interval of the critical values, , where and is the radius of horizon of the black brane. The first order phase transition line of GL transitions ends in a second order phase transition point at .
pacs:04.50.+h, 04.70.Bw, 04.70.Dy
Gravity in higher dimensions has been in the forefront of research for a considerable time. Most of the studies start with the conventional Einstein action generalized to dimensions. In the low energy limit quantum theories of gravity like string theory yield additional higher order curvature correction terms to the Einstein action.callen Some higher order curvature terms when taken in isolation and quantized can potentially lead to problematic high energy behavior, such as the presence of negative norm states. The simplest term avoiding these problems is the Gauss-Bonnet (GB) term. In this letter we analyze the critical behavior of a class of classical solutions to Einstein gravity, augmented by the GB term. Such a term arises in the gravity action in string theory when corrections to zero slope limit are calculated nepomichie boulware . For simplicity, we restrict our discussion to five space-time dimension, but as we will point out later the results can be simply generalized to higher dimensions.
In Einstein gravity solutions, like a black hole can be trivially extended to a black string () or black brane solutions (). Solutions become more complicated if higher derivative terms are added to the Einstein action, such as the GB term. Though asymptotically Minkowski black hole solutions have been found in theories boulware , no exact black string (brane) solutions are known in EGB gravity. Yet the investigations of black strings (branes) is more important because with compact extra dimensions all but the lightest static objects must be black strings (branes).
Two alternative techniques have been used to investigate black brane solutions. Kobayashi and Tanaka kobayashi solved the 5 dimensional Einstein equations modified by the Lanczos tensor (variation of the Gauss-Bonnet term), numerically. Using an expansion around the event horizon they also found an exact lower bound for the black string mass, , where is the coupling constant of the GB term. For the solutions do not have a horizon, they represent naked singularities.
In a subsequent work chetiya , following kobayashi , we used a series expansion in to investigate black strings in EGB theory. We found that in every dimension, , a lower bound, similar to that in , exists for the mass of the black string. We also investigated the thermodynamics of the black string but the fifth order expansion employed in our paper was insufficient to get a definitive answer for the thermal behavior of the system when the mass of the black string approached the lower limit.
In this paper we also use improved numerical and expansion methods to investigate black string solutions including the neighborhood of critical points, and 3, where the dimensionless GB coupling is defined as
In (1) denotes the radius of the horizon. determines the critical radius. The critical radius has a one-to-one relationship with the critical mass, which is determined by the condition .
In Section 2 we will discuss the uniform black string, which should have a higher entropy than a black hole, when the mass is large enough. In fact, for sufficiently high mass it is the only static state of the system. We will use a horizon expansion to elucidate the nature of two critical points and 3. We will concentrate on but, except for numerical details theories are not different. Then we will use numerical calculations to investigate the properties of black string at large distances from the horizon. We discuss their thermodynamics and calculate their entropy.
In Section 3 we will investigate the extension of the Gregory-Laflamme instability gregory and the notion of the non-uniform black string, which is the preferred state if the radius of horizon goes under a critical multiple of the radius of compactification wiseman gubser harmark . We will also plot the phase diagram of black strings in the plane.
Section 4. summarizes our results. We will also discuss the possible directions for the continuation of this research.
Ii Uniform black string
In this section we will discuss the extension of a 4-dimensional solution to . If we assume that the metric is an asymptotically flat solution of the 4 dimensional Einstein equation then the trivial extension of this metric, obtained by defining and satisfies the dimensional Einstein equation. Here, and in what follows, . We tacitly assume that the coordinates, labeled by are compactified on circles, in which space we will maintain a symmetry throughout this paper. However, the trivially extended metric is not the solution of Einstein-Gauss-Bonnet equation
where is the coupling, the Lanczos tensor is
and is the Kretschmann scalar,
In an appropriate coordinate system the metric components of uniform black brane solutions depend only on a radial coordinate. Nonuniform black branes and the Gregory-Laflamme instability will be discussed later. The metric of nonuniform black branes is a periodic function of the compactified coordinate, , as well.
ii.1 Horizon expansion and scaling near critical points
Throughout this paper we will use dimensionless coordinates, scaled by the radius of the horizon, , of the uniform black string. In particular, we choose the horizon variable, defined by
to replace the radial coordinate, . When the GB coupling, , is also rescaled, by the introduction of the parameter, (See eq. 1), then all reference to the radius of the horizon is eliminated from the equations of motion. The only parameter remaining in the equations of motion is . True values of physical quantities are restored by multiplying by a power of , required by their dimensions.
Defining the metric of a uniform string, made dimensionless by factoring out , as
we can solve the equations of motion in a power series of (horizon expansion). The series coefficients depend only on . Displaying three leading orders only, the expansions have the following form:
The two constants, and , are not determined by the horizon expansion alone. They depend on only. They must be introduced in and to assure that the metric components are asymptotically Minkowski. They are determined in the process of numerical integration of the equations of motion.
The expansion coefficients have two singular points: and . Note that for . The physical domain is . The expansion coefficients are complex for and there are singularities outside the horizon for . Questions have been raised whether negative values of are admissible arkani . Nevertheless, for completeness we will also investigate that region, as well. Near the singular points the convergence radius of the expansion shrinks to zero. The leading singular terms of the expansion coefficients (subscript implies the coefficient of ) are the following:
Since the expansion coefficients become complex at the solution does not exist for negative , contradicting the assumption of the existence of a horizon or/and a regular expansion around the horizon kobayashi . This result can, however, be the result of a poor choice of the coordinates. Therefore, we also calculate scalars, like the Ricci scalar () and the Kretschmann scalar (), at the horizon. Being scalars, they must be independent of the scaling factors or which can be gauged away by rescaling the coordinates or . Indeed, we obtain
We can draw several conclusions from (9). First of all, and are non-analytic at and 3, ruling out mere coordinate singularities. Furthermore, both of these constants become complex for , therefore no uniform black string solution with regular horizon can exist in that region. implies and, as we will see later, , though not quite proportional, but is a regular and bounded function of the ADM mass. Thus, the bound imposes a lower limit on the mass of the black string.
Furthermore, while all curvature invariants are finite at they are infinite at . This points to the possibility of the existence of a critical solution at but not at . We will discuss the critical solution in the following section.
ii.2 The critical string
To investigate the critical solution we start with a rearrangement of the horizon expansion series. Keeping only the leading singular terms in every order of we are lead to functions of a scaling variable . Thus, one can define alternative expansions in series of scaling functions and . The scaling expansion is defined as
with similar expressions for and in terms of series of scaling functions, and . Substituting (10) into (2) we obtain differential equations for the scaling functions defined in (10). The equation for is
while and can be simply expressed by as and . Though we could not find the solution of (11) in analytic form the general behavior of can be analyzed fairly easily. Its power series in is provided by the most singular terms of horizon expansion (7). Analyzing an 80th order series expansion we were able to ascertain that has a singularity inside the horizon of the form
Eq. (12) suggests that the metric becomes singular inside the horizon, at and cannot be continued to smaller values of . Note that this singularity moves to at . We will indeed see below that the critical solution is singular at and cannot be continued to . There is no impediment, however to continue to positive values. In fact the analysis of Eq. (11) shows that the asymptotic behavior of is
The limit of higher order expansion terms lead to higher powers of , including all half-integer powers. This can be ascertained by repeating the horizon expansion (7) with half-integer powers included (starting from ). This leads to the following expansion of the critical black string () solution:
If one attempts an expansion with half-integer powers of for general then either all the half-integer powers vanish or . Conversely, no horizon expansion with integer powers exists for the critical black string.
The value of scalars and can also be calculated at using (15). One obtains and , precisely the values obtained from the limit of (9). In fact, one can show that all scalars are finite in the critical state. Correction to scalars, when one moves away from the horizon are of . Further evidence for the fact that the critical string is the limit of non-critical strings when comes from numerical integration of the equations of motion, to be discussed later.
The surface cannot really be called a horizon, just the limit of horizons when . While it is an infinite red shift null surface, the metric cannot be continued inside the surface. We believe that there is no way to facilitate such a continuation. Strictly speaking, though this is a string solution, it is not ”black.” It is however a limiting solution of a series of black string solutions in every quantity that we investigated. At all the horizon is completely regular and there is always a region inside the horizon, , where the solution can be continued without any impediment.
Time-like geodesics can be continued to the horizon both in the and cases. A test particle reaches the horizon in finite proper time, . However, while in the case the test particle passes through the horizon into a region from which it cannot escape any more, for geodesics have a pathological behavior. Choosing a geodesic for a particle falling towards the horizon, , and the time scale such that the radial coordinate turns complex for . Therefore, the role of the critical solution as a valid state of black strings is questionable. This question will be discussed later in more details, following the discussion of numerical integration of the equations of motion and of the Gregory-Laflamme instability.
ii.3 Numerical integration of the equations of motion
We build on the results of the previous sections to obtain the metric functions at arbitrary . Before starting the numerical integration from the horizon we calculate the value of the metric functions and their derivatives from the horizon expansion at . This step is necessary because is a singular point of the differential equations. is chosen to be small enough so that the last terms of the 30th order horizon expansion is smaller than . Then we integrate the equations numerically from using metric (6). The step size, , in the numerical integration is chosen such that .
Expansions (7) contain undetermined constants and . In an initial run these constants are chosen to be 1. Then, though the calculated metric functions tend to a finite value as , their asymptotic limit is not 1. Fitting a form
to , and a similar form to we define and to get, after a second round of numerical integrations, metric functions that are asymptotically 1, which corresponds to Minkowski space.
where is the Arnowit-Deser-Misner (ADM) mass, is the Hawking temperature of the black string, is the tension, and is circumference of the compact string.
where we restored the correct dimension of the surface gravity by reinstating a factor of . As we pointed out earlier and are determined in the process of numerical integration. They are functions of only. Combining (18) and (19) we can write the Hawking temperature as
Form factor , normalized to , is calculated in the process of our numerical integration procedure. It is a smooth function of between and . It tends to a finite value at . That limit, along with the limits of the other two form factors defined below, agree with the values obtained from the critical string at . in the limit implying that as . and the other two form factors are plotted in Fig.1,
The ADM mass and the tension are expressed simply by the long range components of and harmark . Namely, after renormalizing and so that and defining the leading, , asymptotic coefficients as and , then
where and are two additional form factors depending on only. They are both bounded on the interval . These form factors also converge smoothly to the corresponding values for the critical string at . They are plotted in Fig 1. along with . All three of the form factors are normalized to 1 at (vanishing GB coupling).
The three form factors are not independent from each other. The integrability condition
relates them. (23) implies
The calculated values of the form factors satisfy (24) to , except near , where the rapid variation of makes the evaluation of the numerical derivative difficult.
We can also calculate the entropy. We have
where is independent of . (25) can be rewritten by changing variables as
The function , a yet undetermined arbitrary function of , can be fixed if we take the derivative of (26) with respect to and use the integrability condition (24). Setting the lower bound of the integration at we obtain the final form of the entropy for as
An overall constant was fixed in (27) so that for . From our numerical integration we obtain .
For completeness we write down an equivalent expression for . Considering that diverges around the critical point the integral over is convergent at that point so we can write
implying that the entropy vanishes at the critical point, .
The limit of when can be calculated if we change the integration variable in (27) to . Then we obtain
Since and the form factors are equal to unity at , at that point (29) reduces to
which is the expression for the entropy of a black string in Einstein gravity.
The entropy can be represented in the form
at all . The function obtained from numerical integration is plotted in Fig.2. Note that and .
Iii Gregory-Laflamme instability and nonuniform black strings
iii.1 A simple picture of the life of a black string
We will investigate what happens to black strings as Hawking radiation reduces their mass. Consider the ADM mass as a function of and , (21). is a monotonic function of , because for the whole physical range of
With decreasing decreases if and increases if . Typical trajectories of black strings in the plane are shown in Fig. 3. According to (1) if . When hits the value the singular point, , is reached. In the critical state the black string has a finite Hawking temperature, but it cannot decay and stay in the black string state any more. We will return to the discussion of its further fate later.
If increases with decreasing . Eventually, it will hit the critical value , where . However, the Hawking temperature diverges at , so approaching this point the black string will completely evaporate.
Before turning to the investigation of instability we briefly discuss the fate of black branes in EGB theories. We will use the notation . The coefficients of the horizon expansion depend on only. Their singularity structure in is the same as that at , except has a more complicated expression
has a single zero for , which zero varies with from a minimum of at to at . For negative we have though is not a monotonic function of . Still, just like for the horizon expansion coefficients are singular at . Thus, aside for small numerical changes the fate of black branes is the same as that of black strings. They are driven to the critical state if and to if .
iii.2 Gregory-Laflamme instability
In the absence of the GB interaction term (), uniform black strings become unstable if gregory . At this point a nonuniform mode becomes marginally tachyonic, a zero mode. As it was shown by Gubser gubser , Harmark and Obers, Kol, Piran, and Sorkin harmark this point corresponds to a first order transition to a stable nonuniform black string state. Such a state is associated with a periodic dependence of the metric function on the compactified coordinates, in our case . It is fair to assume that such an instability also affects the EGB theory when . Therefore, the simplistic picture about the life of a black string which we depicted in the previous section may be misleading. The black string may hit an instability line before it reaches either of the singular points ( or 3). Therefore, we performed a stability analysis of EGB black strings using our numerical approximation to the metric components of the uniform string solution.
Following Refs. gubser and harmark we perturbed the metric with a periodic function with wave number of of the compact coordinate, . Considering the rescaling of coordinates with can be identified with . Applying the Landau-Ginzburg theory of phase transitions to the Einstein black string it was found gubser harmark that at the Gregory-Laflamme point gregory , , the system undergoes a first order phase transition to a nonuniform state with a finite non-uniformity. It is quite reasonable to expect that a first order transition exists at nonzero values of the GB coupling, as well.
where the exponents and are functions of both and . Though in (33) we gauged away a possible non-diagonal term still the gauge has not been fixed completely. A perturbation analysis is applied to (33), assuming a small deviation from the uniform string. It is also assumed that the smallest possible wavelength, dominates the perturbation. Higher order corrections, along with higher harmonics can be considered in a systematic perturbation analysis that we leave to future work. Accordingly, we use the following ansatz for the perturbed metric
We also introduce amplitudes , and for the perturbative form of coefficients and . Linearization of the equations of motion in leads to a set of equation for the amplitudes and . These equations still have a substantial gauge freedom. As we prefer to work with a single amplitude, we use the Kol-Sorkin gauge harmark to eliminate , and , using the gauge freedom.
Making the final gauge choice , can be solved for . Here and in what follows we use the notation for the components of the equation of motion. In the next step and are used to eliminate and its derivatives. Then and provide identical second order differential equations of the form
The coefficients , , , and are extremely complicated functions of the metric components and their derivatives 111The form of the coefficient functions can be found in the ’Mathematica’ file ”Coefficient functions” at the internet address: http://www.physics.uc.edu/suranyi/mathematicafiles. . An important feature of is its proportionality to . Consequently vanishes on the horizon making (35) singular at . The ratios , and when . Thus, (35) is a Schrodinger-like equation, with as a bound state eigenvalue. Our aim is to determine the single eigenvalue of the equation for every choice of , which is the only parameter entering (35).
Due to the singularity at the horizon we cannot integrate (35) starting from the horizon. However, we generated a 30th order Taylor series of around , using the known series expansion coefficients of the metric components of the uniform black string (, , ). This allows us to calculate and at , which is the same quantity which has also been used at the numerical integration of the equations of motion for the uniform black string.
We use the shooting method, starting from to find the eigenvalue. At each the coefficients of (35) are calculated using the stored values of functions and and their derivatives. We determine at a large number of values of between the two singular points, 0 and 3. The solid line of Fig. 3. represents the eigenvalue as a function of . While at reaches a finite value, , it diverges at . This implies that for black branes must undergo a Gregory-Laflamme transition before reaching the critical value and evaporate completely. The limit of as calculated from solutions agrees with calculated from the critical solution.
The time-lines of black strings are depicted by dash-dotted, dotted, and dashed lines in Fig. 3. These time-lines depend on only. At fixed different black strings may start their life at different points but they follow the same line. If
then black strings completely avoid the instability transition before reaching the critical state (dash-dotted line in Fig.2). However, if then every black string undergoes an instability transition and turns into a non-uniform black string before it could reach the critical line. The trajectory of black strings at such a value of is represented by the dotted line. Note that is completely determined by the geometry of space and the GB coupling. Furthermore, if then and the time-line approaches (dashed line). Finally, if then black strings follow the thin vertical dashed line at .
iii.3 Non-uniform black strings near
In contrast to uniform black strings non-uniform black strings require the solution of a two variable problem. Therefore, it is quite a bit more complicated to investigate the fate of non-unifom black strings. However, we can substantially simplify matters if we restrict ourselves to sufficiently small values of . Though we cannot quite follow non-uniform black strings through the region of instability, we assume that if they eventually also approach the line in the plot of Fig.2. Of course, it is quite possible that nonuniform black strings also undergo an instability transition, before they could reach the critical state. Still the following calculations, pertaining to the behavior of nonuniform stings near , certainly apply to a neighborhood of the critical point, , .
The simplest way to proceed is to investigate the horizon expansion (33) with -dependent coefficients. We write the following ansatz for the metric
Note that the dependence can be gauged away from and from the zeroth order contribution to . Expanding the equations of motion in , the lowest order contribution to implies that the periodic function is independent of . In the next step one can solve the lowest order contributions to and equations for and . Then the lowest order contributions to equations and are identical. They are second order equations for , containing and its second derivative. Solving the resulting equations for the discriminant of the equation must be positive for a real solution. This discriminant has the form
Now we know that for uniform black strings and for a small non-uniformity must have the form
Then expanding in and and keeping the leading order contributions only, we obtain
(40) implies that the discriminant is positive for all only if
First of all (41) implies that when we approach the critical state, . This makes our approximation of expanding in correct for small . Secondly, this implies that the non-uniform black string becomes a uniform black string again at the critical line. Thirdly, this implies the line of first order Gregory-Laflamme transition to the non-uniform state ends in a second order transition point at . This is a well-known feature of first order transitions.
We have studied uniform and non-uniform black strings in EGB gravity. We paid particular attention to the singularities at the endpoints of range, , , where is defined in (1). No black string solutions exists outside . In particular, we found a critical solution at which, in contrast to solutions at , has a horizon expansion that includes half-integer powers of the radial horizon variable, (5). Not only the metric components, but all scalars, including the Ricci scalar and the Kretschmann scalar, are also singular (though finite) at (9).
At the metric components are singular inside the horizon, at (12). It is always possible to continue the metric to a region inside the horizon. Introducing Eddington-Fonkelstein coordinates
the metric takes the form
regular everywhere including a neighborhood of the horizon.111Note that on the interval for the whole range, , of admissible values of . The singularity at moves to the horizon when . We have not been able to construct a metric inside this singular point. At , the critical metric cannot be continued inside the horizon. This can be shown if near the horizon we transform the metric to Kruskal coordinates. We define the Kruskal coordinates for the critical string such that the coefficient of is . They can be obtained in a asymptotic expansion from the Eddington-Finkelstein coordinates and
where has the following expression by the Kruskal coordinates and
where is a non-vanishing constant. It is obvious we cannot cross the light like surfaces or which represent the horizon. We also showed that a geodesic of a massive probe falling toward the horizon reaches it in finite proper time (at ), but it cannot pass the horizon because its radial coordinate turns complex at . Therefore, the critical solution does not represent a black string.
We used numerical techniques to calculate the solutions for the whole range . We calculated the ADM mass, tension, Hawking temperature, and entropy. These quantities are represented by form factors, depending on only, measuring the deviation from the (), Einstein gravity solutions. These form factors are defined in (20)(21)(22)(31).They are plotted in Figs 1. and 2. While the Hawking-temperature tends to a finite value at , it diverges at . The ADM mass and tension are modified only by a finite amount compared to pure Einstein gravity and bounded from above and below over the whole range, . We gave a closed expression for the entropy in terms of an integral over the form factors, which was evaluated. The entropy is a monotonically decreasing function of , vanishing in the limit . This is not surprising in view of the divergence of the Hawking temperature.
We paid particular attention to the life line of black strings decaying via Hawking radiation. We found that every black string follows a trajectory in the plane, approaching (depending on the sign of ) one of the singular lines in the -plane. This is depicted in Fig. 3. However, some of the black strings, depending whether , will encounter a line of first order singularities, which is the extension of the famed Gregory-Laflamme singularity. This line is the solid line shown in the phase diagram of Fig. 3. Under this line only non-uniform black string are stable. As we commented before, non-uniform black strings may become unstable at a subsequent phase transition line.
Finally, we showed, by investigating non-uniform string solutions near the horizon that the amplitude of non-uniformity vanishes near the line. This implies that non-uniform strings become uniform once again as they reach the line. As a corollary, we concluded that the line of first order phase transitions ends in a second order transition point at .
There are two unresolved problems concerning the fate of black strings. If they are driven toward the critical state at . That state, however is very problematic. Though has a linear zero and all scalars are finite at that zero, test particles that reach the point in finite time have nowhere to go. because the becomes complex as a function of proper time.
As finding solutions for black holes in compactified spaces is even more difficult than finding black string solutions we have no way of comparing the entropy and relative stability of these solutions. This is one of the reasons why we are not able to answer what happens with black strings that reach the end of their lives at the line. Since the limit of their Hawking temperature is
not zero, as a study based on expansions lead us to believe earlier, so they are certainly not frozen at that point. Their ADM mass is (21)
They cannot go into a black string state with a smaller ADM, mass, however, because such black string states do not exist. It seems that the only possibility is that some other, yet undiscovered, type of higher entropy static or possibly dynamic state exists, into which the black string can morph. Another possible solution, at least for , is that the inclusion of higher order Lovelock terms would cure this anomaly. The fate of black stings at and beyond is one of the most important questions we would like to investigate in the future. Other questions include a possible numerical study of black holes in compactified spaces.
This work used a Kaluza-Klein type compactification. It would be more difficult, but perhaps more interesting to repeat this work in a Randall-Sundrum type ADS space.
Acknowledgements.We thank Philip Argyres and Paul Esposito for discussions and Richard Gass for help in ”Mathematica”.. This work is supported in part by the DOE grant # DE-FG02-84ER40153.
- (1) C.Callen, E.Martinec, M.Perry and D.Friedan, Nuc. Phys. B262,593(1985).
- (2) Rafael I. Nepomechie Phys.Rev. D32:3201 (1985);B. Zwiebach Phys. Lett. 156 B, 315 (1985); B. Zumino, Phys. Rep. 137, 315 (1986).
- (3) D. Boulware, S. Deser Phys.Rev.Lett. .55, 2656 (1985);.
- (4) T. Kobayashi, T.Tanaka Phys.Rev. D 71 084005 (2005).
- (5) C. Sahabandu, P. Suranyi, C. Vaz, L.C.R. Wijewardhana Phys.Rev. D 73, 044009 (2006).
- (6) R. Gregory, R. Laflamme, Phys.Rev.Lett.70:2837-2840,1993.
- (7) Hideaki Kudoh and Toby Wiseman, Phys.Rev.Lett.94:161102, 2005.
- (8) S. S. Gubser, Class. Quant. Grav. 031601 (2004)
- (9) T. Harmark and N.A. Obers, Class. Quant. Grav. 21, 1709 (2004); B. Kol, E. Sorkin, and T. Piran, Phys.Rev. D 69, 064031 (2004), B. Kol and E. Sorkin, Class.Quant.Grav.23:4563-4592,2006.
- (10) Robert M. Wald, General Relativity, University of Chicago Press, Chicago, 1984.
- (11) R. Myers, J.. Simon Phys.Rev. D38, 2434 (1988); D.L. Wiltshire, Phys. Rev. 38, 2445 (1988).
- (12) Dmitri V. Fursaev, Sergei N. Solodukhin Published in Phys.Lett.B365:51-55,1996. e-Print: hep-th/9412020
- (13) Allan Adams, Nima Arkani-Hamed, Sergei Dubovsky, Alberto Nicolis, Riccardo Rattazzi, JHEP 0610:014,2006.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00043.warc.gz
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| 32,117 | 114 |
https://www.walmart.com/browse/auto-tires/nerf-bars/yellow/91083_1074784_4491859_2867839_5813602/Y29sb3I6WWVsbG93
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math
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CC-MAIN-2020-16
| 712 | 8 |
https://bmcinfectdis.biomedcentral.com/articles/10.1186/1471-2334-9-80/figures/3
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math
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PDE3B expression in a control kidney (A) and one treated with LPS (B). In the control group (A), immunohistochemical stain for PDE3B showed positive expression in the epithelial cells of the distal convoluted tubules (arrows). The glomerulus and proximal convoluted tubules showed no prominent expression for PDE3B. In the LPS group (B), both proximal and distal convoluted tubules showed strong expression for PDE3B (B). PCT: proximal convoluted tubule, DCT: distal convoluted tubule, G: glomerulus. Original magnification, ×200.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00663.warc.gz
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CC-MAIN-2022-33
| 531 | 1 |
http://openstudy.com/updates/566f24e1e4b0b9558262b66b
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math
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FREE MEDAL !!!!! She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0. is that correct ?
Stacey Warren - Expert brainly.com
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What's g(x)? We would need to know that equation
Three party-goers are in the corner of the ballroom having an intense argument. You walk over to settle the debate. They are discussing a function g(x). You take out your notepad and jot down their statements.
Professor McCoy: She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
Not the answer you are looking for? Search for more explanations.
there is no equation
She's wrong because assuming this is a polynomial the zero is -2
If a is a zero of f(x) then (x - a) will be a factor.
- and long division would give a zero reminder.
s=i will be back in a sec I am working it out
In order to know the answer quickly, you can use examples to help you understand. The simplest polynomial with 2 being a zero is definitely x−2.
Prof McCoy: x−2x+2 definitely doesn't have a remainder of 0
she is wrong because in order that 2 be a zero of g(x), it is (x-2) which must be a factor of g(x), not (x+2). (FACTOR THEOREM)
Ms. G is correct because if 2 is a zero of g(x), g(2) will be 0. (REMAINDER THEOREM)
Mr. R is also correct because if (x-2) is a factor of g(x), g(x) will have 2 as a zero. (FACTOR THEOREM)
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CC-MAIN-2018-05
| 2,247 | 21 |
https://www.imath.kiev.ua/~snmp2003/abstract2003/Daletskii.html
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math
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Institute of Mathematics of NAS of Ukraine,
3 Tereshchenkivs'ka Str.,
01601 Kyiv-4, UKRAINE
L2-Betti numbers of Poisson configuration spaces
The space GX of all locally finite configurations in a infinite covering X of a compact Riemannian manifold is considered. The deRham complex of square-integrable differential forms over GX, equipped with the Poisson measure, and the corresponding deRham cohomology and the spaces of harmonic forms are studied. We construct a natural von Neumann algebra which contains the projection onto the space of harmonic forms and obtain explicit formulae for the corresponding trace.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742968.18/warc/CC-MAIN-20181116004432-20181116030432-00469.warc.gz
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CC-MAIN-2018-47
| 616 | 5 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-6-rational-expressions-and-equations-6-1-rational-expressions-6-1-exercise-set-page-379/4
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math
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a factor equal to $1$
Work Step by Step
The numerator and the denominator of a rational expression will have no factors in common when the greatest common factor is equal to $1$. Thus, the final answer will be a factor equal to $1$.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00244.warc.gz
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CC-MAIN-2019-51
| 481 | 5 |
https://cheggwriters.com/roducing-hydrogen-atoms-from-h2-molecules-requires-436-kjmol-to-break-h/
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math
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roducing hydrogen atoms from H2 molecules requires 436 kJ/mol to break H-?H bonds. For the longest-?wavelength photon capable of breaking a single H-?H bond, calculate ?. Next, express the energy required to break a single H-?H bond as a percentage of the ground state energy of the hydrogen atom (with electron in the 1s-?orbital). [Hint: the numerical value of the constant RE = R·h·c is 2.18 x 10-?18 J/atom.] Use the Bohr energy En to calculate which n corresponds to the strength of the H-?H bond. [Hint: the n you calculate won’t be an integer.] Finally, round up n to the nearest integer. If an electron in energy level n (rounded) of the Hydrogen atom absorbs a photon with wavelength ? (calculated above), how much kinetic energy will it have after it escapes the atom?
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https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22
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math
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2021 AMC 10B Problems/Problem 22
Ang, Ben, and Jasmin each have blocks, colored red, blue, yellow, white, and green; and there are empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives blocks all of the same color is , where and are relatively prime positive integers. What is
Let our denominator be , so we consider all possible distributions.
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.
When we have at box with all balls the same color in that box, there are ways for the distributions to occur ( for selecting one of the five boxes for a uniform color, for choosing the color for that box, for each of the three people to place their remaining items).
However, we overcounted those distributions where two boxes had uniform color, and there are ways for the distributions to occur ( for selecting two of the five boxes for a uniform color, for choosing the color for those boxes, for each of the three people to place their remaining items).
Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth.
Our success by PIE is yielding an answer of .
As In Solution 1, the probability is Dividing by , we get Dividing by , we get Dividing by , we get .
Use complementary counting. Denote as the total number of ways to put colors to boxes by 3 people out of which ways are such that no box has uniform color. Notice . From this setup we see the question is asking for . To find we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have
In case b), there are ways to choose 2 boxes that have the same color, ways to choose the two colors, 2! ways to permute the 2 chosen colors, and ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is . Now, we just need to calculate , and .
We have , since we subtract the number of cases where the boxes contain uniform colors, which is 2.
In the same way, - again, we must subtract the number of ways at least 1 box has uniform color. There are ways if 3 boxes each contains uniform color. Two boxes each contains uniform color is the same as previous. If one box has the same color, there are ways to pick a box, and ways to pick a color for that box, 1! ways to permute the chosen color, and ways for the remaining 2 boxes to have non-uniform colors. Similarly,
Thus, the probability is and the answer is
Video Solution by OmegaLearn (Principle of Inclusion Exclusion)
Video Solution by Interstigation
~ Briefly went over Principal of Inclusion Exclusion using Venn Diagram
|2021 AMC 10B (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25|
|All AMC 10 Problems and Solutions|
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http://www.tropicalfishkeeping.com/beginner-freshwater-aquarium/where-find-square-fish-tank-7054/
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math
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I am not quite sure where to post this question, so here goes:
I have two goldfish in a teeny-tiny tank that need a bigger home. I have a solidly-constructed table that is built just like a fish tank stand and will hold a tremendous amount of weight, so I would like to find a fish tank that would fit on the table. It would have to be a square tank, either 26x26 or 27x27 in order to balance the weight correctly over the legs of the table. Can anyone tell me where I would find such a fish tank, or whether anyone makes such a tank? I would like it to be between 12 and 18" in height.
I am hoping not to have to custom order the thing or make it myself. In either of those cases, I will just get a whole new stand and tank and save time and funds. :)
I would also have to find or make a top for such a tank, too...
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http://kyhighperformance.org/mortgage-help/
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math
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http://www.veritasprep.com/community/viewtopic.php?f=20&t=11289
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math
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The number line strategy is much more useful when we don't have something with 64 potential steps spread over a range of only 5 percentage points...
I.e. - if we had 64 percentage points - where one was 20% nuts and the other was 84% nuts, then we'd be more able to apply the number line strategy here --
As it is, we have 5 percentage points.
If we had equal amounts of P and Q, (32 oz each) then we'd have 22.5% nuts in the final mix. Since we have 15/64 of the final mix almonds, which is roughly 23%. This means we must have more of the mix that contains 25% almonds than the mix that has 20% almonds.
Here's the concept:
The higher the proportion of the 25% mix we have, the higher the proportion of almonds in the final mix, and at the extreme end, if we have 100% of Brand Q, then we have 25% almonds in the final mixture. If, on the other hand, we have 100% of Brand P, then we have 20% almonds in the final mixture. Since our final mix is about 23%, we know that this is skewed closer to 25% than to 20%, telling us we have slightly more of Brand Q than of Brand P.
Hope this helps a bit -
Apologies for the typo - we don't use pre-written answers on these, so occasionally we make a typo in our calculations. We do our best to catch these, but one slips through from time to time. Hopefully, it doesn't detract from the explanation itself or the concepts presented.
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| 1,375 | 8 |
https://orac.amt.edu.au/cgi-bin/train/problem.pl?set=fario05&problemid=165
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math
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Want to try solving this problem? You can submit your code online if you log in or register.
Time Limit: 3 seconds
Memory Limit: 10 Mb
Your younger sister just got a mobile construction set for her birthday. The set consists of two different kinds of elements: rods and decorations, as illustrated below.
Each rod has three hooks attached via pieces of string. In the middle of each rod, a string rises upwards with a hook at the top. This allows the entire rod to be hung beneath some other object (either the ceiling or a higher decoration). One and only one rod must attach the mobile to the ceiling. At the ends of each rod, strings fall down with hooks at the bottom. This allows a decoration to be hung beneath the rod at each end.
There are many types of decorations, with differing shapes and weights. Each decoration has two rings attached to it, one at the top and one at the bottom. The top ring can be used to hang the decoration on a hook beneath the left or right end of a rod. The bottom ring can be used to hang another rod beneath the decoration.
When building a mobile with this set, no hook may be left free except for the very top hook that hangs the entire mobile from the ceiling. Every other hook must be attached to a decoration. Decorations are not so constrained; some decorations may have other rods hanging beneath them, whereas other decorations may have nothing attached to their bottom ring. An example of a completed mobile is illustrated below. The rods are numbered from 1 to 7, and the weight of each decoration is written beside it.
Your younger sister has finished making a beautiful mobile with her set, but she is complaining that it doesn't work! Indeed, the mobile is far from being well balanced: what is hanging on one side of a rod is sometimes much heavier than what is hanging on the other side.
You have offered your help, but your sister gets upset when you explain that you will have to move some things around. She doesn't want you to change anything in her beautiful construction — it's perfect the way it is, you just have to make it work! You manage to convince her that you need to change at least something to improve the balance, but she will only allow you to make a single change to her creation.
Your task is to make the entire mobile as balanced as possible, by making only one modification to it. This modification must involve detaching a rod from the decoration under which it is hanging, and then reattaching it beneath some other decoration. You don't need to worry about whether different parts of the mobile might bump into each other, since you can always fix this afterwards by changing the lengths of some strings. Making the mobile balanced is your only concern.
To measure how balanced a rod is, you must calculate the absolute value of the difference between the total weights hanging from each end (so a lower number is better). To measure how balanced the entire mobile is, you must add up these differences for every rod. Your goal is to reduce this total to its smallest possible value by making at most one change. Note that you are not required to make a change; you may choose to move nothing at all.
Note also that when moving a rod, you may not attach the rod to a decoration that already has another rod hanging from it. Rods, strings and hooks are so light that they can be considered as having no weight at all.
Furthermore, for 30% of the available marks, N <= 200.
The first line of input contains one integer: N, the number of rods used in the mobile. The rods are numbered from 1 to N, with rod 1 being attached to the ceiling.
Each of the following N lines describes a rod. These rods are given in order from 1 to N (so that the second input line describes rod 1, the third input line describes rod 2 and so on). Each of these lines is of the form W1 R1 W2 R2, where W1 and W2 are the weights of the decorations attached directly beneath the left and right ends of the rod, and R1 and R2 are the numbers of the rods hanging beneath these decorations (where rods R1 and R2 hang beneath the decorations of weight W1 and W2 respectively). If a decoration has no rod hanging beneath it, the corresponding rod number will be zero.
The output should contain one line, with a single integer: the minimum sum of weight differences that you can get by moving at most one rod of the mobile.
The following input scenario describes the large mobile illustrated earlier.
7 4 2 20 3 12 4 4 5 12 6 20 7 3 0 3 0 6 0 6 0 6 0 6 0 7 0 7 0
In the input scenario, the weight differences for rods 1,...,7 are 40, 2, 10, 0, 0, 0 and 0 respectively, coming to a total sum of 52. By moving rod 7 so that it hangs beneath one of the decorations attached to rod 5, these weight differences become 12, 12, 4, 0, 14, 0 and 0, coming to a total sum of 42. No smaller total sum can be obtained without moving more than one rod, and so 42 is the final answer.
The score for each input scenario will be 100% if the correct answer is output, and 0% otherwise.
© Australian Mathematics Trust 2001-2023
Page generated: 3 June 2023, 2:55pm AEST
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| 5,100 | 22 |
https://www.photonics.com/Article.aspx?AID=39714
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math
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The 320-page volume Signal Processing for Neuroscientists: An Introduction to the Analysis of Physiological Signals is targeted to neuroscientists and biomedical engineering students with backgrounds in mathematics, physics and computer programming. Focusing on averaging, Fourier analysis and filtering, it discusses techniques such as convolution, correlation, coherence and wavelet analysis, in the context of time- and frequency-domain analysis. It covers the full spectrum of signal analysis, from data acquisition to data processing, and from the mathematical background of the analysis to the application of processing algorithms. A goal is to enable the readers to construct their own analysis tools in an environment such as Matlab. Wim van Drongelen; Academic Press, an imprint of Elsevier, Burlington, Mass., 2006; $79.95, h66.95, paperbound.
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| 853 | 1 |
https://byjus.com/gate/control-systems-mcqs/
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math
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A Control system is a system of devices that are used to manage, command, direct, or regulate the other devices’ behaviours to achieve the desired result. A control system can achieve this with the help of control loops, which are a process designed to maintain a process variable at the desired set point.
MCQs on Control Systems
Solve Control Systems Multiple-Choice Questions to prepare better for GATE. Learn more about Control Systems and Control Systems MCQs by checking notes, mock tests, and previous years’ question papers. Gauge the pattern of MCQs on Control Systems by solving the ones that we have compiled below for your practice:
Control Systems Multiple-Choice Questions
1. In the block diagram given below, G2=10/s+1, G1=10/s, H2=1, and H1=s+3.
The overall transfer function here would be given by:
a. 100/11s² + 31s + 100
b. 10/11s² + 31s + 10
c. 100/11s² + 31s
d. 100/11s² + 31s +10
Answer: (a) 100/11s² + 31s +100
2. Any control system that has excessive noise would most likely suffer from:
c. Loss of gain
d. Saturation in the amplifying stages
Answer: (d) Saturation in the amplifying stages
3. Which of these is the most common hydraulic fluid?
c. Synthetic fluid
d. Mineral oil
Answer: (b) Water
4. A given control system yields a 0.20 steady-state error for a unit step input. We cascade a unit integrator to this system and then apply unit ramp input to this modified system. For this modified system, what would be the actual value of steady-state error?
Answer: (c) 0.25
5. With a feedback system, the transient response would:
a. Decays quickly
b. Decays slowly
c. Rises quickly
d. Rises slowly
Answer: (a) Decays quickly
6. We can apply the standard test signal to give output to:
a. Time-Invariant Systems
b. Time-Variant Systems
c. Non-Linear Systems
d. Linear Systems
Answer: (d) Linear Systems
7. What would be the intersection of the asymptotes of the root loci of a given system with a transfer function G(s)H(s) = K/s(s+1)(s+3) with an open-loop?
Answer: (c) -1.33
8. A system becomes ____________ as and when the polar plot would move toward the (-1, 0) point.
a. Marginally Stable
d. Conditionally Stable
Answer: (b) Unstable
9. In the Nyquist plot, the unit circle would transform into 0dB line of the Bode diagram’s amplitude plot at:
a. Any Frequency
b. High Frequency
c. Low Frequency
d. 0 Frequency
Answer: (a) Any Frequency
10. The steady-state error is specified (usually) in terms of:
a. Damping Factor
c. Error Constants
d. Speed of Response
Answer: (c) Error Constants
11. In a unity feedback system, the control transfer function with an open loop is given by:
In the closed-loop system, what should be the value of K that leads to sustained oscillations?
Answer: (d) 590
12. What would be the homogeneous solution of the following:
y(n) -9/16y(n-2) = x(n-1)
Answer: (b) C1(3/4)n+C2(3/4)-n
13. An LTI system’s transfer function is given as 1/(s+1). Here, what would be the steady-state value of any unit-impulse response?
Answer: (d) 0
14. What are the points at which the derivatives of all state variables happen to be zero?
a. Nonsingular Points
b. Singular Points
Answer: (b) Singular Points
15. An externally introduced signal that affects the controlled output is known as a:
c. Gain Control
Answer: (a) Stimulus
16. What are the mediums responsible for the transportation lag?
a. Belt Conveyors
d. All of the above
Answer: (d) All of the above
17. How are the robots used in automobile plants classified?
a. Mobile Robots
b. Perception Systems
c. Knowledge Robots
d. Industrial Robots
Answer: (d) Industrial Robots
18. In a servomechanism, the response is 1+0.2e-60t-1.2e-10t when we subject it to a unit step input. Here, what would be the damping ratio?
Answer: (c) 1.43
Keep learning and stay tuned to get the latest updates on the GATE Exam along with GATE MCQs, GATE Eligibility Criteria, GATE Syllabus for CSE (Computer Science Engineering), GATE Notes for CSE, GATE CSE Question Paper, and more.
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| 3,973 | 78 |
https://www.crackverbal.com/solutions/if-0-x-2-which-of-the-following-must-be-greatest/
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math
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Get detailed explanations to advanced GMAT questions.
If 0 < x < 2, which of the following must be greatest?
x + 2
Option D is the correct answer.
Plugging In for MUST-BE type Questions:
x = 1
A. x = 1
B. x^2 = 1^2 = 1
C. 2x = 2(1) = 2
D. x + 2 = 1 + 2 = 3 So this is the greatest. So eliminate all other answer choices.
E. 2^x = 2^1 = 2
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| 337 | 11 |
https://www.arxiv-vanity.com/papers/0905.0757/
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math
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Absorbing boundaries in the conserved Manna model
The conserved Manna model with a planar absorbing boundary is studied in various space dimensions. We present a heuristic argument that allows one to compute the surface critical exponent in one dimension analytically. Moreover, we discuss the mean field limit that is expected to be valid in space dimensions and demonstrate how the corresponding partial differential equations can be solved.
pacs:05.50.+q, 05.70.Ln, 64.60.Ht
Absorbing phase transitions are a particular class of nonequilibrium phase transitions (see [1, 2, 3, 4] for recent reviews). These transitions continue to attract attention since they have no equilibrium counterparts and occur even in one-dimensional systems. The stochastic process of directed percolation (DP) is recognized as a paradigmatic example of absorbing phase transitions. According to the well-known conjecture by Janssen and Grassberger[5, 6] a continuous phase transition should belong to the universality class of DP if the corresponding model has a scalar order parameter with a single absorbing state, evolves by short-range interactions without quenched disorder and has no unconventional symmetries and conservation laws. Non-DP behavior is expected to occur if one of these conditions is violated. In particular, particle conservation leads to an autonomous universality class of nonequilibrium phase transitions which is usually referred to as the Manna universality class. In the recent years, the scaling properties of the Manna universality class were investigated in several mostly numerical works (see and references therein). The same type of critical behavior occurs in various models such as the Manna sandpile model (which was introduced in the context of self-organized criticality), the conserved lattice gas (CLG) and the conserved threshold transfer process (CTTP) . In addition, the Maslov-Zhang sandpile and the Mohanty-Dhar sandpile , which were believed to belong to DP, were shown to be a member of the Manna universality class as well [12, 13, 14]. Moreover, the Manna universality class is of fundamental interest since it connects the critical behavior of absorbing phase transitions with the critical state of self-organized critical (SOC) systems . Actually, SOC sandpile models can be considered as driven-dissipative versions of (closed) systems exhibiting absorbing phase transitions . An intriguing consideration of the interplay between absorbing phase transitions and SOC in case of the CTTP can be found in .
As in equilibrium statistical mechanics, critical systems out of equilibrium respond strongly to the presence of boundaries. For systems in the DP class, the problem of boundary effects was investigated thoroughly in a series of papers, where numerical simulations, field theoretical approaches [16, 17, 18, 19], series expansions techniques [20, 21], as well as density matrix renormalization group methods have been applied. These studies focused on two different types of boundary conditions, namely, so-called absorbing walls, where activity is set to zero by removing particles, and active walls, where activity is artificially held at a high level. Usually absorbing walls generate a non-trivial surface-critical behavior which can be described in terms of an additional surface critical exponent . On the other hand, the impact of active walls can be explained in terms of the standard critical exponents only. In all cases the bulk critical behavior remains unaffected, i. e. the bulk scaling behavior still belongs to the ordinary DP universality class. Analogous results were obtained for the so-called parity-conserving universality class of absorbing phase transitions .
Such boundary effects may also be studied within the Manna universality class. Compared to DP, the implementation of an absorbing wall in the Manna class is more subtle since the dynamics at the boundary has to preserve the global number of particles. Furthermore, the existing knowledge about boundary effects in the Manna class is still limited and systematic studies started only recently . In the present paper we extend these studies, discussing the scaling behavior near the boundary, computing the surface exponent exactly in , and analyzing the mean-field limit which is expected to describe the system above the upper critical dimension .
2 Numerical results
The Manna model is defined on a lattice with sites, where each site is associated with an integer number representing the local number of particles. Sites with () grains are considered as inactive (active), where is a stability threshold, usually . The Manna model evolves by parallel updates by simultaneously redistributing all particles at active sites to randomly chosen nearest neighbor sites, which in turn may become active. Note that this update procedure conserves the total number of particles if periodic boundary conditions are applied. Each parallel update counts as a single time step. The Manna model is known to display a continuous phase transition from a fluctuating active phase into frozen configurations (lacking of active sites), where the particle density plays the role of a control parameter. The phase transition takes place at the critical threshold which depends on the spatial dimension, the lattice structure, as well as on the particular value of . In order to preserve the number of particles, the absorbing wall is implemented by removing all particles from active boundary sites and adding them to randomly chosen sites in the bulk of the system.
First we studied the spatio-temporal impact of absorbing boundaries in different space dimensions for systems starting with a homogeneously active state. Fig. 1 shows the decay of the density of active sites as a function of the perpendicular distance to the wall at different times. As expected, the absorbing wall creates a depletion zone that grows with time as . The density of active sites inside the depletion zone scales as
where denotes the distance to the boundary and is a new exponent. According to the standard scaling theory for surface-critical phenomena this suggest the scaling form
where is a scaling function with the asymptotic behavior
Our numerical simulations lead to the estimates
The asymptotic behavior of implies that the activity next to the boundary, e.g. at a site adjacent to the wall, decays as
which implies the scaling relation
Inserting the known estimates from Ref. this scaling relation yields the estimates
In one dimension our result is in very good agreement with the estimate reported in Ref. .
3 Heuristic proof for in one dimension
The numerical result suggests that might be exactly equal to . This conjecture can be supported as follows. It is argued in that the critical behavior of the Manna model can be described effectively by two mutually coupled Langevin equations of the form111In our opinion these Langevin equations should be considered as an educated guess. Derivations from microscopic models are available but they involve various approximations which are hard to verify.
Here describes the coarse-grained density of active sites while represents the coarse-grained background density of all particles. As usual, denotes an uncorrelated white Gaussian noise. Note that Eq. (9) differs from the Langevin equation for DP by an additional term which couples the two fields whereas Eq. (10) describes diffusive reordering of the background field during toppling. Furthermore note that in a system with periodic boundary conditions the structure of Eq. (10) ensures global conservation of the background field since .
In the following we show that the structure of these Langevin equations implies in one dimension. The idea is that the removal of active particles at the boundary and its redistribution in the (ideally infinite) bulk of the system depletes not only in the coarse-grained activity but also the background density . This is in fact the essential feature that makes boundary phenomena in the Manna class different from those of DP. Using the Langevin equation we compute the integrated loss of the background density and show that the results in one dimension would be inconsistent unless .
To this end we choose a particular site at distance and monitor the loss of the background density at this point. Since the depletion zone grows as , this density will begin to decrease at about a typical time which scales as . How much of the background density will be lost between and ? According to Eq. (10), the total loss is given by the integral
Inserting the scaling form (2) and substituting the scale-invariant ratio
where . Note that this expression is positive since has a negative curvature.
Since the total loss is a finite quantity, the integral (13) has to be finite as well. Obviously, the integral is finite if and only if the integrand diverges slower than as . In this limit the scaling function can be approximated by so that the integrand diverges as
For this reason the integral is finite if one of the following three conditions holds:
The first solution is unphysical because there would be no depletion zone, meaning that we are left with the other two possibilities. Which of them applies depends on the dimensionality of the system: In one dimension the exponent is clearly smaller than , hence the above integral is finite if and only if . In higher dimensions, however, is larger than so that can take arbitrary values. This explains why in one-dimensional systems only, reflecting particle conservation.
4 Mean field approximation
In dimensions, i.e. above the so-called upper critical dimension , the Manna model at criticality is expected to display mean field behavior on large scales which is usually characterized by simple values of the exponents. In the following we compute the surface exponent in the mean field limit, verify the validity of the scaling form (2), and show how the corresponding partial differential equations can be solved numerically.
As usual, we assume that the mean-field equations have the same form as the Langevin equations (9)-(10) if the noise term is neglected. Since the boundary breaks translational invariance, space-dependence has to be retained. Moreover, the time scale of the process may be fixed by choosing . Furthermore, the parameter can be absorbed in via rescaling of and thus we may set . The resulting set of mean field equations reads
If the boundary has the form of a -dimensional hyperplane it is clear that the mean field densities will only depend on the distance perpendicular to the wall. Denoting this distance by we may replace the gradient by an ordinary derivative perpendicular to the wall, leading to
where we omitted the arguments . Obviously the critical background field is . We iterated these partial differential equations numerically by means of standard methods (see Fig. 2). Because of the scaling form (2)
with the scale-invariant variable one expects a data collapse if is plotted against . As shown in the left panel of Fig. 2, a reasonable collaps is obtained for and , the well-known bulk exponents of DP in the mean field limit. Similarly, the background density should obey a scaling form
where the numerical integration (see right panel of Fig. 2) leads to the value . This is reasonable since the first and the third term in Eq. (17) should be equally relevant, hence in the mean field limit.
Although the data collapses become gradually better for large , they are not really perfect, raising the question whether slightly different critical exponents would lead to a better data collapse. However, this would be not consistent with the structure of the mean field equations (17) and (18). To see this we eliminate the parameter by a shift of , solve the first differential equation for and insert it into the second one, arriving at a single partial differential equation for the density of active sites
Inserting the scaling form (19) one obtains an autonomous ordinary differential equation for (meaning that it does no longer depend explicitely on and ) if and only if and . This leads to a non-linear third-order differential equation of the form
The trivial solution is unphysical while the solution describes the homogeneous case without boundary. With an absorbing wall one has to impose a Dirichlet boundary condition together with the bulk limit . The second term in the second line of Eq. (4) requires that either or vanishes, hence the solution is either linear or quadratic at the origin. The linear solution can be excluded because it would imply that the loss of the background density according to the second Langevin equation (10) would be zero. Therefore, the solution behaves as for small , hence we obtain the exponent .
In the present paper we have investigated the properties of an absorbing hyperplane in a critical Manna model. All results are consistent with earlier findings by Bonachela et al [23, 14] in 1d, who used such boundary effects as a criterion to discriminate between directed-percolation and Manna scaling. We have extended these numerical studies to higher dimensions and have confirmed that the observed scaling properties are consistent with the mean field limit. Moreover we have shown that in one spatial dimension, the structure of the Langevin equations, which are believed to describe Manna-type critical behavior on a coarse-grained scale, allows one to calculate the surface-critical exponents by means of a scaling relation.
- J. Marro and R. Dickman, Nonequilibrium phase transitions in lattice models (Cambridge University Press, Cambridge, UK, 1999).
- H. Hinrichsen, Adv. Phys. 49, 815 (2000), [cond-mat/0001070].
- G. Ódor, Rev. Mod. Phys. 76, 663 (2004).
- S. Lübeck, Int. J. Mod. Phys. B 18, 3977 (2004).
- H.-K. Janssen, Z. Phys. B42, 151 (1981).
- P. Grassberger, Z. Phys. B 47, 365 (1982).
- S. Manna, J. Phys. A Math. Gen. 24, L363 (1991).
- M. Rossi, R. Pastor-Satorras, and A. Vespignani, Phys. Rev. Lett. 85, 1803 (2000).
- J. F. F. Mendes, R. Dickman, M. Henkel, and M. C. Marques, J. Phys. A 27, 3019 (1994).
- S. Maslov and Y.-C. Zhang, Physica A 223, 1 (1996).
- P. K. Mohanty and D. Dhar, Phys. Rev. Lett. 89, 104303 (2002).
- J. A. Bonachela, J. J. Ramasco, H. Chaté, I. Dornic, and M. A. Muñoz, Phys. Rev. E 74, 050102 (2006).
- P. K. Mohanty and D. Dhar, Physica A 384, 34 (2007).
- J. A. Bonachela and M. A. Muñoz, Phys. Rev. E 78, 041102 (2008).
- P. Bak, C. Tang, and K. Wiesenfeld, Phys. Rev. Lett. 59, 381 (1987).
- H. K. Janssen, B. Schaub, and B. Schmittmann, Z. Phys. B 72, 111 (1988).
- K. B. Lauritsen, P. Fröjdh, and M. Howard, Phys. Rev. Lett. 81, 2104 (1998).
- P. Fröjdh, M. Howard, and K. Lauritsen, J. Phys. A 31, 2311 (1998).
- P. Fröjdh, M. Howard, and K. Lauritsen, Int. J. Mod. Phys. B15, 1761 (2001).
- J. W. Essam, A. J. Guttmann, I. Jensen, and D. TanlaKishani, J. Phys. A 29, 1619 (1996).
- I. Jensen, J. Phys. A 32, 6055 (1999).
- E. Carlon, M. Henkel, and U. Schollwöck, Eur. Phys. J. B 12, 99 (1999).
- J. A. Bonachela and M. A. Muñoz, Physica A 384, 89 (2007).
- A. Vespignani, R. Dickman, M. A. Muñoz, and S. Zapperi, Phys. Rev. Lett. 81, 5676 (1998).
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| 15,237 | 61 |
https://www.enotes.com/homework-help/what-creates-greatest-torque-301769
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math
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What creates the greatest torque?
Torque is a force applied at right angles, moving through some distance over a lever arm.
torque = Fd
So to have the maximum torque you want the lever arm (the distance about which you are rotating) to be as long as possible, and at the same time to apply the maximum possible force. Generally, if course, the opposite is true. A long lever arm is used so a smaller amount of force is required to generate the desired torque.
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CC-MAIN-2018-05
| 459 | 4 |
https://nursingpaperwriting.com/topic5dq12-docx/
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math
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Topic 5 DQ 1
A doctoral learner has decided to do a case study for his/her proposed dissertation research study topic because it is believed to be the best approach to address the research questions. The researcher's choices of data sources for this particular study are to conduct interviews, to conduct observations, and to conduct focus groups. Will these data sources generate the breadth and depth of the data necessary for this design? Why or why not? What challenges might the researcher encounter in collecting data from these sources? Explain.
Topic 5 DQ 2
Based on the sources of data identified in the first discussion question in this topic, how would the researcher demonstrate "triangulation" of the data across the sources to address each of the research questions? What are the assumptions, limitations, or delimitations related to the data? Explain.
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https://slideplayer.com/slide/229177/
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math
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Jacques Charles: 1746-1823 Balloonist, noticed relation between temperature and volume At constant P: V=kT or V/T=k More commonly V 1 /T 1 =V 2 /T 2
Charles noticed that no matter what the starting volume was, if P is constant and he increased T, the V also increased
Graphing the relationship between V and T produces a straight line This is a direct proportion Note that if you extend the lines, you can reach a temperature at which volume is ZERO!
1824-1907 Extended work of Joule, Charles and others to use gas as thermometer Gas volume changes 1/273 for every degree change Kelvin scale: based on T at which volume of gas is zero
To use temperature in gas law equations, you MUST convert the temperature to Kelvins Otherwise, you may end up calculating a negative volume! 0 K = -273.15 o C To go from o C to K, add 273.15 To go from K to o C, subtract 273.15
A 2.0 L sample of air is collected at 298 K then cooled to 278 K. The pressure is held constant at 1.0 atmo. What is the new volume of the air? A sample of gas at 15 o C has a volume of 2.58 L. The temperature is then raised to 38 o C. What is the new volume? A child blows a soap bubble at 28 o C with a volume of 1.0 L. As the bubble rises, it encounters a pocket of cold (18 o C air). What is the new volume of the bubble?
Gas volume has been used as a way to measure temperature (gas thermometer). If a gas has a volume of 0.675 L at 35 o C and 1 atmo P, what is the temperature of a room if the gas has a volume of 0.535 L in the room (at 1 atmo P)?
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| 1,518 | 7 |
http://www.chegg.com/homework-help/questions-and-answers/4-jet-passes-airstrip-40-000-feet-belowit-flying-nw-45-degrees-north-west-speed-600mph-air-q217481
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math
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#4. Jet A passes over an airstrip that is about 40,000 feet belowit and is flying NW (45 degrees north of west) at a speed of 600MPH. At the airstrip a twin engine plane (B) is prepared fortakeoff. It takes off at a 40 degree angle with respect to theground. The plane is headed due east. When it is 200 feet above theground, it is moving at a speed of 200 MPH. Find the relativevelocity of the jet to the plane (VA/B) and the plane tothe jet (VA/B). HINT: Do not worryabout using the jet's height while doing the problem. Be concernedabout the velocities.
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s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736679145.29/warc/CC-MAIN-20151001215759-00088-ip-10-137-6-227.ec2.internal.warc.gz
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https://www.deepdyve.com/lp/springer_journal/a-quasilinear-hierarchical-size-structured-model-well-posedness-and-faDBy692pQ
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math
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Appl Math Optim 51:35–59 (2005)
2004 Springer Science+Business Media, Inc.
A Quasilinear Hierarchical Size-Structured Model:
Well-Posedness and Approximation
Azmy S. Ackleh, Keng Deng, and Shuhua Hu
Department of Mathematics, University of Louisiana at Lafayette,
Lafayette, LA 70504, USA
Abstract. A finite difference approximation to a hierarchical size-structured model
with nonlinear growth, mortality and reproduction rates is developed. Existence-
uniqueness of the weak solution to the model is established and convergence of
the finite-difference approximation is proved. Simulations indicate that the mono-
tonicity assumption on the growth rate is crucial for the global existence of weak
solutions. Numerical results testing the efficiency of this method in approximating
the long-time behavior of the model are presented.
Key Words. Hierarchical size-structured model, Existence-uniqueness, Finite-
AMS Classification. 35L60, 65M06, 65M12, 92D25.
In this paper we consider the following initial-boundary value problem which models
the evolution of a hierarchical size-structured population:
+ (g(x , Q(x, t))u)
+ m(x, Q(x, t))u = 0,(x, t) ∈ (0, L] × (0, T ],
g(0, Q(0, t))u(0, t) = C(t ) +
β(x, Q(x, t ))u(x, t) dx, t ∈ (0, T ],
u(x, 0) = u
(x), x ∈ [0, L].
The work of ASA and SH was supported in part by the National Science Foundation under Grant
No. DMS-0311969. The work of KD was supported in part by the National Science Foundation under Grant
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https://fratstock.eu/Test-Bank-Calculus-Scientists-Engineers-1st-Edition-Briggs
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math
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|<< Solutions Manual for Calculus for Scientists and Engineers Early Transcendentals 1st Edition by Briggs||Solutions Manual for Calculus for Scientists and Engineers 1st Edition by Briggs >>|
Download FREE Sample Here for Test Bank for Calculus for Scientists and Engineers 1st Edition by Briggs. Note : this is not a text book. File Format : PDF or WordProduct Description Complete downloadable Test Bank for Calculus for Scientists and Engineers 1st Edition by Briggs. INSTRUCTOR RESOURCE INFORMATION TITLE: Calculus for Scientists and Engineers RESOURCE: Test Bank EDITION: 1st Edition AUTHOR: Briggs, Cochran, Gillett PUBLISHER: Pearson PREVIEW PDF SAMPLE Test-Bank-Calculus-Scientists-Engineers-1st-Edition-Briggs Table of Contents 1. Functions 2. Limits 3. Derivatives 4. Applications of the Derivative 5. Integration 6. Applications of Integration 7. Logarithmic and Exponential Functions 8. Integration Techniques 9. Differential Equations 10. Sequences and Infinite Series 11. Power Series 12. Parametric and Polar Curves 13. Vectors and Vector-Valued Functions 14. Functions of Several Variables 15. Multiple Integration 16. Vector Calculus DOWNLOAD SAMPLES
Once the order is placed, the order will be delivered to your email less than 24 hours, mostly within 4 hours.
If you have questions, you can contact us here
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s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00362.warc.gz
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http://people.brandeis.edu/~bernardi/everytopic/everytopic.html
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math
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Everytopic Seminar - Spring 2015
Room 226, Goldsmith building
Organizer: Olivier Bernardi
The Everytopic Seminar is a seminar aimed at a broad audience of mathematicians.
The talks are 80 minutes long. The first half the talk should be given in a
colloquium style, and should be accessible to faculty and graduate students
from any field of mathematics. It should allow everyone to grasp the content and
significance of the results being discussed. The second half of the talk can be in
the style of a regular research seminar, providing the audience with a deeper understanding
of the results as well as some details of the proofs.
Tuesday February 3
Speaker: Siu-Cheong Lau (Harvard)
Title: A constructive and functorial approach to mirror symmetry
Abstract: Homological mirror symmetry conjecture asserts an equivalence between the category of Lagrangian submanifolds and the category of coherent sheaves of the mirror. The conjecture has been verified in several interesting cases by computing generators and relations of the categories. However such a computational approach hardly explains why homological mirror symmetry exists in general. In this talk I will introduce a mirror construction which always comes with a functor between the two categories. It produces mirrors which are not reachable by the SYZ construction, and is naturally connected with non-commutative geometry.
Tuesday February 10
Speaker: Alison Miller (Harvard)
Title: Classical knot invariants, arithmetic invariant theory, and counting simple (4q+1)-knots
Certain knot invariants coming from the Seifert matrix and Alexander polynomial of a knot have interesting number-theoretic structure. I'll explain how a classical construction of knot invariants that are ideal classes of rings fits into the context of Bhargava and Gross's arithmetic invariant theory.
I'll then discuss the following related asymptotic counting question: on the set of knots with squarefree Alexander polynomial of fixed degree and bounded height, our ideal class invariant can take on only finitely many values -- how many? This is a quantitative version of the question "how much additional information is contained in the Seifert matrix that is not captured by the Alexander polynomial?" It also has consequences in higher-dimensional knot theory, where Kearton, Levine, and Trotter identified an important class of knots, the "simple knots", which are entirely classified by higher-dimensional analogues of our invariants.
Tuesday February 24
Speaker: Dylan Rupel (Northeastern)
Tuesday March 3
Speaker: Arunima Ray (Brandeis)
Tuesday March 10
Speaker: Paul Monsky (Brandeis)
Title: Some Characteristic 2 Hecke algebras. A theme of Nicolas and Serre, with variations
Abstract: For each odd prime p, there is a formal Hecke operator Tp: Z/2[[x]] -> Z/2[[x]]. The mod 2 reductions of modular forms give rise to certain Tp stable subspsaces of Z/2[[x]]. Nicolas and Serre have analyzed the action of the Tp on a space attached to level 1 forms. Their space is spanned by the odd powers of x+x^9+x^25+x^49+... . They show that the Hecke algebra arising is a power series ring in T3 and T5.
I'll sketch an alternative proof and present related results on some spaces attached to level 3 forms. One of the Hecke algebras I get is a power series ring in T7 and T13 with an element of square 0 adjoined. My arguments are not technical, and the talk should be accessible.
Tuesday March 17
Speaker: Anna Medvedovsky (Brandeis)
Title: The nilpotence method for obtaining lower bounds for dimensions of
I will present a new method for obtaining a lower bound on the
dimension of a Hecke algebra acting on modular forms modulo a prime p.
Along the way, I will discuss linear recurrence sequences in
characteristic p. The method is elementary and the talk should be
accessible to a wide audience.
Tuesday March 24
Speaker: Jonathan Novak (MIT)
Tuesday March 31
Speaker: Charles Smart (MIT)
Here are some information about how to reach Brandeis Mathematic department.
For further information please contact:
Olivier Bernardi (bernardi at brandeis dot edu).
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https://mail.scipy.org/pipermail/scipy-user/2005-June/004661.html
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math
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[SciPy-user] do complex numbers default to double precision?
ryanfedora at comcast.net
Thu Jun 30 07:56:26 CDT 2005
The matrix is currently 4x4 but will grow to probably 6x6. It is
definitely nonlinear. The matrix contains sinh, cosh, sin, and cos. I
am using the transfer matrix method to analyze structures. When you say
two-parameter, do you mean the real and imaginary part of the
independent variable? I guess you are right that I don't necessarily
need to use the determinant. In order to satisfy the boundary
conditions of the problem this 4x4 or 6x6 matrix (which is really a
submatrix of an 8x8 or 12x12) must have a null space. So, what would be
the better thing to look for? An eignevalue that approaches 0?
Nils Wagner wrote:
> Ryan Krauss wrote:
>> I have a matrix that is a function of a complex valued input. I am
>> trying to find that value of that input that drives the determinant
>> of the matrix to zero. I am searching for this value using fmin. The
>> error I am trying to minimize is the abs(det(complex matrix)).
> It's not a good idea to use the determinant directly since det(A) is a
> rapidly varying function. As far as I understand your problem,
> you are interested in the solution of a two-parameter nonlinear
> eigenvalue problem. Is that correct ? How about the size of your
> complex matrix A ?
>> I don't seem to be able to drive this error lower that roughly 9e-17,
>> regardless of the values for ftol and xtol I use.
>> Am I hitting some internal limitation? Are complex values by default
>> single or double precision?
>> SciPy-user mailing list
>> SciPy-user at scipy.net
> SciPy-user mailing list
> SciPy-user at scipy.net
More information about the SciPy-user
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https://www.scribd.com/document/4026812/Counters
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math
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This action might not be possible to undo. Are you sure you want to continue?
Lecturer : Room Email : : Yongsheng Gao Tech - 3.25 [email protected] 6 lectures 1 Tutorial 1 Laboratory (5%) 1 Test (20%) Floyd, Digital Fundamental Chapter 9 Chapter 10 Asynchronous Counters, Synchronous Counters, Design of Synchronous Counters, Shift Registers, Johnson & ring counters, Applications etc.
Structure : Assessment: Textbook : Chapters : Synopsis :
Aim: In the previous lectures, you have learnt D, S-R, J-K flip-flops. These flip-flops can be connected together to perform certain operations. In the following lectures, we will focus on a variety of sequential circuits used mostly as storage elements: registers, counters. We look at how to construct the different types of counter. We will also look at how to build registers for storage and shifting from simple D-type flip-flops sharing common clock lines.
Review of Flip-Flops
Flip-flops are synchronous bistable devices. The term synchronous means the output changes state only when the clock input is triggered. That is, changes in the output occur in synchronization with the clock.
Positive edge-triggered (without bubble at Clock input): S-R, J-K, and D.
Negative edge-triggered (with bubble at Clock input): S-R, J-K, and D.
An edge-triggered flip-flop changes states either at the positive edge (rising edge) or at the negative edge (falling edge) of the clock pulse on the control input. The three basic types are introduced here: S-R, J-K and D. The S-R, J-K and D inputs are called synchronous inputs because data on these inputs are transferred to the flip-flop's output only on the triggering edge of the clock pulse. On the other hand, the direct set (SET) and clear (CLR) inputs are called asynchronous inputs, as they are inputs that affect the state of the flip-flop independent of the clock. For the synchronous operations to work properly, these asynchronous inputs must both be kept LOW.
Edge-triggered S-R flip-flop
The basic operation is illustrated below, along with the truth table for this type of flip-flop. The operation and truth table for a negative edgetriggered flip-flop are the same as those for a positive except that the falling edge of the clock pulse is the triggering edge.
Note that the S and R inputs can be changed at any time when the clock input is LOW or HIGH (except for a very short interval around the triggering transition of the clock) without affecting the output.
Edge-triggered J-K flip-flop
The J-K flip-flop works very similar to S-R flip-flop. The only difference is that this flip-flop has NO invalid state. The outputs toggle (change to the opposite state) when both J and K inputs are HIGH.
Edge-triggered D flip-flop
The operations of a D flip-flop is much more simpler. It has only one input addition to the clock. It is very useful when a single data bit (0 or 1) is to be stored. If there is a HIGH on the D input when a clock pulse is applied, the flip-flop SETs and stores a 1. If there is a LOW on the D input when a clock pulse is applied, the flip-flop RESETs and stores a 0. The truth table below summarize the operations of the positive edgetriggered D flip-flop. As before, the negative edge-triggered flip-flop works the same except that the falling edge of the clock pulse is the triggering edge.
A counter is a sequential machine that produces a specified count sequence. The count changes whenever the input clock is asserted. There is a great variety of counter based on its construction. 1. 2. 3. 4. 5. Clock: Synchronous or Asynchronous Clock Trigger: Positive edged or Negative edged Counts: Binary, Decade Count Direction: Up, Down, or Up/Down Flip-flops: JK or T or D
• A counter can be constructed by a synchronous circuit or by an asynchronous circuit. With a synchronous circuit, all the bits in the count change synchronously with the assertion of the clock. With an asynchronous circuit, all the bits in the count do not all change at the same time. • A counter may count up or count down or count up and down depending on the input control. • Because of limited word length, the count sequence is limited. For an n-bit counter, the range of the count is [0, 2n-1]. The count sequence usually repeats itself. When counting up, the count sequence goes in this manner: 0, 1, 2, … 2n-2, 2n-1, 0, 1, …etc. When counting down the count sequence goes in the same manner: 2n-1, 2n-2, … 2, 1, 0, 2n1, 2n-2, … etc. Example: 3-bit Up Counter 000 001 010 011 100 101 110 111 3-bit Down Counter 000 111 110 101 100 011 010 001
• The complement of the count sequence counts in reverse direction. If the uncomplemented output counts up, the complemented output counts down. If the uncomplemented output counts down, the complemented output counts up. Example: 3-bit Up Counter 000 001 010 011 100 101 110 111 Complement of the Count 111 110 101 100 011 010 001 000
• The natural count sequence is to run through all possible combinations of the bit patterns before repeating itself. External logic can be used to arbitrary cause the counter to start at any count and terminate at any count. • A binary counter produces a count sequence similar to the binary numbers. A decade counter counts from 0 to 9, thus making it suitable for human interface. Other counters count to 12 making them suitable for clocks.
Uses of Counters
The most typical uses of counters are • To count the number of times that a certain event takes place; the occurrence of event to be counted is represented by the input signal to the counter (Fig. 1.1a) • To control a fixed sequence of actions in a digital system (Fig. 1.1b) • To generate timing signals (Fig. 1.2a) • To generate clocks of different frequencies (Fig. 1.2b)
Two Classes of Counters
• Counters are classified into two categories: • Asynchronous Counters (Ripple counters) • Synchronous Counters Asynchronous & Synchronous Asynchronous: The events do not have a fixed time relationship with each other and do not occur at the same time. Synchronous: The events have a fixed time relationship with each other and do occur at the same time. • Counters are classified according to the way they are clocked. In asynchronous counters, the first flip-flop is clocked by the external clock pulse and then each successive flip-flop is by clocked the output of the preceding flip-flop. In synchronous counters, the clock input is connected to all of the flip-flop so that they are clocked simultaneously.
2.0 Asynchronous Counters
An asynchronous counter is one in which the flip-flop within the counter do not change states at exactly the same time because they do not have a common clock pulse. • 2 Bit asynchronous binary counter • 3 Bit asynchronous binary counter • 4 Bit asynchronous binary counter The main characteristic of an asynchronous counter is each flip-flop derives its own clock from other flip-flops and is therefore independent of the input clock. Consequently, the output of each flip-flop may change at different time, hence the term asynchronous. From the asynchronous counter diagram above, we observed that the output of the first flip-flop becomes the clock input for the second flip-flop, and the output of the second flip-flop becomes the clock input for the third flip-flop etc. For the first flip-flop, the output changes whenever there is a negative transition in the clock input. This means that the output of the first flipflop produces a series of square waves that is half the frequency of the clock input. Since the output of the first flip-flop becomes the clock of the second flip-flop, the output of the second flip-flop is half the frequency of its clock, i.e. the output of the first flip-flop that in turn is half the frequency of the clock input. This behaviour, in essence is captured by the binary bit pattern in the counting sequence
2-Bit Asynchronous Binary Counter
Figure 1.3a: two-bit asynchronous counter • A two-bit asynchronous counter is shown on the left. The external clock is connected to the clock input of the first flip-flop (FF0) only. So, FF0 changes state at the falling edge of each clock pulse, but FF1 changes only when triggered by the falling edge of the Q output of FF0. Because of the inherent propagation delay through a flip-flop, the transition of the input clock pulse and a transition of the Q output of FF0 can never occur at exactly the same time. Therefore, the flip-flops cannot be triggered simultaneously, producing an asynchronous operation. • Note that for simplicity, the transitions of Q0, Q1 and CLK in the timing diagram above are shown as simultaneous even though this is an asynchronous counter. Actually, there is some small delay between the CLK, Q0 and Q1 transitions. • Usually, all the CLEAR inputs are connected together, so that a single pulse can clear all the flip-flops before counting starts. The clock pulse fed into FF0 is rippled through the other counters after propagation delays, like a ripple on water, hence the name Ripple Counter. • The 2-bit ripple counter circuit above has four different states, each one corresponding to a count value. Similarly, a counter with n flip-
flops can have 2 to the power n states. The number of states in a counter is known as its mod (modulo) number. Thus a 2-bit counter is a mod-4 counter. • A mod-n counter may also be described as a divide-by-n counter. This is because the most significant flip-flop (the furthest flip-flop from the original clock pulse) produces one pulse for every n pulses at the clock input of the least significant flip-flop (the one triggers by the clock pulse). Thus, the above counter is an example of a divide-by-4 counter. Example 2:
Figure 1.3b: Two-bit asynchronous binary counter, timing diagram, binary state sequence
3-Bit Asynchronous Binary Counter
The following is a three-bit asynchronous binary counter and its timing diagram for one cycle. It works exactly the same way as a two-bit asynchronous binary counter mentioned above, except it has eight states due to the third flip-flop.
Figure 1.4a: Three-bit asynchronous binary counter, timing diagram, binary state sequence
Figure 1.4b: Propagation Delay in a 3-bit asynchronous binary counter Asynchronous counters are commonly referred to as ripple counters for the following reason: The effect of the input clock pulse is first “felt” by FFO. This effect cannot get to FF1 immediately because of the propagation delay through FF0. Then there is the propagation delay through FF1 before FF2 can be triggered. Thus, the effect of an input clock pulse “ripples” through the counter, taking some time, due to propagation delays, to reach the last flip-flop.
2.3 4 Bit Asynchronous Binary Counter
The following is a 4-bit asynchronous binary counter and its timing diagram for one cycle. It works exactly the same way as a 2-bit or 3 bit asynchronous binary counter mentioned above, except it has 16 states due to the fourth flip-flop.
Figure 1.5: Four-bit asynchronous binary counter, timing diagram [Floyd]
2.4 Asynchronous Decade Counters
• The binary counters previously introduced have two to the power n states. But counters with states less than this number are also possible. They are designed to have the number of states in their sequences, which are called truncated sequences. These sequences are achieved by forcing the counter to recycle before going through all of its normal states. • A common modulus for counters with truncated sequences is ten. A counter with ten states in its sequence is called a decade counter.
The circuit below is an implementation of a decade counter.
Figure 1.6: Asynchronous decade counter, timing diagram [Floyd]
• Once the counter counts to ten (1010), all the flip-flops are being cleared. Notice that only Q1 and Q3 are used to decode the count of ten. This is called partial decoding, as none of the other states (zero to nine) have both Q1 and Q3 HIGH at the same time. • The sequence of the decade counter is shown in the table below:
1 1 0
Recycles 0 (normal next state)
Glitch: Notice that there is a glitch on the Q1 waveform. The reason for this glitch is that Q1 must first go HIGH before the count of ten can be decoded. Not until several nanoseconds after the counter goes to the count of ten does the output of the decoding gate go LOW (both inputs are HIGH). Therefore, the counter is in the 1010 state for a short time before it is reset to 0000, thus producing the glitch on Q1 and the resulting glitch on the CLR line that resets the counter.
Example: Modulus Twelve Asynchronous Counter An Asynchronous counter can be implemented having a modulus of 12 with a straight binary sequence from 0000 through 1011.
Figure 1.7: Asynchronous modulus-12 counter & timing diagram.
2.5 Asynchronous Up-Down Counters
In certain applications a counter must be able to count both up and down. The circuit below is a 3-bit up-down counter. It counts up or down depending on the status of the control signals UP and DOWN. When the UP input is at 1 and the DOWN input is at 0, the NAND network between FF0 and FF1 will gate the non-inverted output (Q) of FF0 into the clock input of FF1. Similarly, Q of FF1 will be gated through the other NAND network into the clock input of FF2. Thus the counter will count up.
Figure 1.8: 3-bit up-down counter • When the control input UP is at 0 and DOWN is at 1, the inverted outputs of FF0 and FF1 are gated into the clock inputs of FF1 and FF2 respectively. If the flip-flops are initially reset to 0's, then the counter will go through the following sequence as input pulses are applied.
• Notice that an asynchronous up-down counter is slower than an up counter or a down counter because of the additional propagation delay introduced by the NAND networks.
2.6 Commercially Available Asynchronous Counters
Example 1: The 74LS93 Asynchronous Binary Counter
Figure 1.9: The 74LS93A 4-bit asynchronous binary counter logic diagram Three configurations of the 74LS293 asynchronous counter:
RO(1), R0(2) are the gated reset inputs. If both of these inputs are HIGH, the counter is reset to the 0000 state by CLR .
3.0 Synchronous Counters
In synchronous counters, the clock inputs of all the flip-flops are connected together and are triggered by the input pulses. Thus, all the flip-flops change state simultaneously (in parallel).
3.1 2-Bit Synchronous Binary Counter
Figure 3.1: Two-bit synchronous binary counter, timing diagram
3.2 3-Bit Synchronous Binary Counter
The circuit below is a 3-bit synchronous counter. The J and K inputs of FF0 are connected to HIGH. FF1 has its J and K inputs connected to the output of FF0, and the J and K inputs of FF2 are connected to the output of an AND gate that is fed by the outputs of FF0 and FF1.
Figure 3.3a: A 3-bit synchronous binary counter Pay attention to what happens after the 3rd clock pulse. Both outputs of FF0 and FF1 are HIGH. The positive edge of the 4th clock pulse will cause FF2 to change its state due to the AND gate.
Figure 3.3b: Timing diagram
(recycles) Figure 3.3c: Binary state sequence
The count sequence for the 3-bit counter is shown in Figure 3.3c. The most important advantage of synchronous counters is that there is no cumulative time delay because all flip-flops are triggered in parallel. Thus, the maximum operating frequency for this counter will be significantly higher than for the corresponding ripple counter.
3.3 4-Bit Synchronous Binary Counter
Figure 3.4(a) shows a 4-bit synchronous binary counter and Figure 3.4(b) reveals its timing diagram.
Figure 3.4: Four-bit synchronous binary counter, timing diagram
3.4 Synchronous Decade Counters
Similar to an asynchronous decade counter, a synchronous decade counter counts from 0 to 9 and then recycles to 0 again. This is done by forcing the 1010 state back to the 0000 state. This so called truncated sequence can be constructed by the following circuit.
Figure 3.5a: A synchronous BCD decade counter
Figure 3.5b: States of a BCD decade
Figure 3.5c: Timing diagram for the BCD decade counter (Q0 is the LSB)
From the sequence in the Figure 3.5b, we notice that: • • • • Q0 toggles on each clock pulse. Q1 changes on the next clock pulse each time Q0=1 and Q3=0. Q2 changes on the next clock pulse each time Q0=Q1=1. Q3 changes on the next clock pulse each time Q0=1, Q1=1 and Q2=1 (count 7), or then Q0=1 and Q3=1 (count 9).
Flip-flop 2 (Q2) changes on the next clock pulse each time both Q0=1 and Q1=1. Thus we must have J2 = K2 = Q0Q1 Flip-flop 3 (Q3) changes to the opposite state on the next clock pulse each time Q0=1, Q1=1, and Q2=1 (state 7), or when Q0=1 and Q3=1 (state 9). Thus we must have J3 = K3 = Q0Q1Q2 + Q0Q3 These characteristics are implemented with the AND/OR logic connected as shown in the logic diagram (Figure 3.5b).
3.5 Up-Down Synchronous Counters
A circuit of a 3-bit synchronous up-down counter and a table of its sequence are shown in Figure 3.6. Similar to an asynchronous up-down counter, a synchronous up-down counter also has an up-down control input. It is used to control the direction of the counter through a certain sequence.
Figure 3.6: A basic 3-bit up/down synchronous counter and its up/down sequence An examination of the sequence table shows: • for both the UP and DOWN sequences, Q0 toggles on each clock pulse. • for the UP sequence, Q1 changes state on the next clock pulse when Q0=1. • for the DOWN sequence, Q1 changes state on the next clock pulse when Q0=0. • for the UP sequence, Q2 changes state on the next clock pulse when Q0=Q1=1. • for the DOWN sequence, Q2 changes state on the next clock pulse when Q0=Q1=0.
These characteristics are implemented with the AND, OR & NOT logic connected as shown in the Figure 3.6 Example: 4-bit synchronous up-down counter
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s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720941.32/warc/CC-MAIN-20161020183840-00150-ip-10-171-6-4.ec2.internal.warc.gz
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CC-MAIN-2016-44
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http://www3.math.tu-berlin.de/dm/radopreis2012.html
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math
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Eine "Honorable Mention" ging an Juliane Dunkel (MIT) für ihre Promotion über den Chvatal-Gormory-Abschluss in der ganzzahligen Optimierung. Juliane Dunkel kommt ursprünglich von der TU Berlin, hat am MIT bei Andreas Schulz (ebenfalls vorm. TU Berlin) promoviert.
Laudatio von Michel Goemans zur Preisverleihung an Heidi Gebauer und Juliane Dunkel:
Maker-Breaker games are two-person combinatorial games in which Maker tries to build a subgraph with certain properties by fixing edges while Breaker tries to prevent it from happening by deleting edges. In her dissertation, Heidi Gebauer tackles several of the most humiliating open problems in the area, an adjective coined by Joszef Beck in his reference book on the subject. Among several other results, she resolves the long-standing threshold question for the biased connectivity game in the affirmative, shows the counterintuitive result that Maker can do better in the clique game if both players can select 6 or more edges at a time, and refutes the strongest form of the Neighborhood conjecture with consequences to the satisfiability of Boolean formulas. For these contributions to combinatorial game theory, Heidi Gebauer is eminently deserving of the 2012 Richard Rado prize awarded by the Discrete Mathematics section of the German mathematical Society.
Cutting planes constitute one of the most fundamental tools in integer programming. In particular, the Gomory-Chvátal closure provides an automatic strengthening of any polyhedron towards its integer hull. In her dissertation, Juliane Dunkel proves that the closure of a possibly non-rational polytope is always a rational polytope. This answers in the affirmative a long-standing question left open by Lex Schrijver in a seminal paper over 30 years ago, in which he showed that the closure of any rational polyhedron is always rational. She also introduces a strengthening of the Gomory-Chvátal closure in her dissertation. For these contributions to the theory of cutting planes, Juliane Dunkel is eminently deserving of an Honorable Mention for the 2012 Richard Rado prize.
Der Richard-Rado-Preis wurde von Fachgruppe Diskrete Mathematik der DMV zum achten Mal für hervorragende Dissertationen in der Diskreten Mathematik vergeben. Der Preis wird vom Springer Verlag gestiftet und ist in diesem Jahr mit EUR 1000,- dotiert.
Text: Thomas Vogt; Foto: Jannik Matuschke
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CC-MAIN-2018-17
| 2,391 | 6 |
https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/4/lesson/4.1.4/problem/4-59
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math
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Michelle has the problem and is thinking of expanding the terms and then solving. Erin suggests that she let then solve . Use Erin’s suggestion to solve for and then use your solutions to find the values of .
Solve for the values of first. Then substitute the values of into . Finally, solve for . You should get two answers.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00472.warc.gz
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CC-MAIN-2022-21
| 327 | 2 |
http://www.residentadvisor.net/feed-item.aspx?id=60749
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math
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Fri, 08 Mar 2013 / RAOut-there techno from the Brooklyn-based up-and-comer.
Fri, 08 Mar 2013 / Bill_LeeOm'mas Keith guests on the first single from the trio's upcoming album on !K7
Fri, 08 Mar 2013 / terrencefullerPitchfork also have a statement from the duo explaining the video's implied commentary on gender roles.
Fri, 08 Mar 2013 / Max_CherryJust in case you were wondering.
Thu, 07 Mar 2013 / brianpkHead to self-titled to hear the collaborative effort, which will appear on Blake's upcoming Overgrown LP.
Thu, 07 Mar 2013 / Bill_LeeSavas Pascalidis, Mad Alba and Andre Galluzzi will also play live-streamed sets at the Berlin record shop between now and 9 PM GMT.
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s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164567400/warc/CC-MAIN-20131204134247-00009-ip-10-33-133-15.ec2.internal.warc.gz
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CC-MAIN-2013-48
| 694 | 6 |
http://picsdownloadz.com/puzzles/flower-puzzle-60-30-viral-flower-math-puzzle-with-answer/
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math
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Solve this viral flower puzzle image! Viral math puzzle image for facebook with answers! Only for genius puzzles!
Hello, This one is most popular puzzle image over the internet, knows as flower puzzle. Solve this tricky or confusing brainteasers puzzle and share your answer via comments. All you have to do, just find the values of these flowers and put into the last equation and find the answer. Simple right?
Flower Math Puzzle Image:
No need to be in hurry take your time. Check the details of the puzzle and try to solve the puzzle. Don’t forget to comment before checking the answer! Answer is given at the end of the post.
Flower Puzzles Image, Math Puzzles Image with Answer, Brain Teasers Puzzles Image, Only for Genius Puzzles, Logic Puzzles Pictures, Difficult Puzzle Image, Confusing Math Puzzles, Hard Puzzles, Mind Puzzle Game, Confusing Riddles, Math Puzzles Only for Geniuses, Maths Picture Puzzles with answers, Math Question Image, Maths Puzzle Images, Maths Picture Puzzles, Only for Geniuses Answer, Puzzles for Facebook, Puzzles for Whatsapp, Only for Geniuses, Math Question Image, Math Puzzle Images, Maths Picture Puzzles, Genius Puzzles, Math Game, Solve this if you are a Genius,
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s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948597295.74/warc/CC-MAIN-20171217171653-20171217193653-00683.warc.gz
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CC-MAIN-2017-51
| 1,208 | 5 |
https://www.tutoraspire.com/mann-whitney-u-test-stata/
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math
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A Mann-Whitney U test (sometimes called the Wilcoxon rank-sum test) is used to compare the differences between two samples when the sample distributions are not normally distributed and the sample sizes are small (n two sample t-test.
This tutorial explains how to perform a Mann-Whitney U test in Stata.
Example: Mann-Whitney U Test in Stata
Researchers want to know if a fuel treatment leads to a change in the average mpg of a car. To test this, they conduct an experiment in which they measure the mpg of 12 cars with the fuel treatment and 12 cars without it.
Because the sample sizes are small and they suspect that the sample distributions are not normally distributed, they decided to perform a Mann-Whitney U test to determine if there is a statistically significant difference in mpg between the two groups.
Perform the following steps to conduct a Mann-Whitney U test in Stata.
Step 1: Load the data.
First, load the data by typing use http://www.stata-press.com/data/r13/fuel2 in the command box and clicking Enter.
Step 2: View the raw data.
Before we perform the Mann Whitney U test, let’s first view the raw data. Along the top menu bar, go to Data > Data Editor > Data Editor (Browse). The first column, mpg, shows the mpg for a given car while the second column, treat, shows whether or not the car received fuel treatment.
Step 3: Perform a Mann-Whitney U test.
Along the top menu bar, go to Statistics > Summaries, tables, and tests > Nonparametric tests of hypotheses > Wilcoxon rank-sum test.
For Variable, choose mpg. For Grouping variable, choose treat. Check both of the boxes below to display a p-value for an exact test and display an estimate of the probability that the variable for the first group is larger than the variable for the second group. Lastly, click OK.
The results of the test will be displayed:
The values that we are primarily interested in are z = -1.279 and Prob > |z| = 0.2010.
Since the p-value of the test (0.2010) is not smaller than our significance level of 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the true mean mpg is different between the two groups.
Step 5: Report the results.
Lastly, we will report the results of our Mann-Whitney U test. Here is an example of how to do so:
A Mann-Whitney U test was conducted on 24 cars to determine if a new fuel treatment lead to a difference in mean miles per gallon. Each group had 12 cars.
Results showed that the mean mpg was not statistically significantly different between the two groups (z = -1.279, p = .2010) at a significance level of 0.05.
Based on these results, the new fuel treatment does not have a significant impact on the miles per gallon of cars.
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CC-MAIN-2024-18
| 2,714 | 21 |
https://cubo.ufro.cl/ojs/index.php/cubo/article/view/1486
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math
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Simple Fixed Point Theorems on Linear Continua
A simple fixed point theorem is formulated for multivalued maps with a connected graph on closed intervals of linear continua. These intervals either cover themselves or are concerned with self-maps. We discuss a question when the original map can possess a fixed point, provided the same assumptions are satisfied only for some of its iterate. We are particularly interested in a situation on noncompact connected linearly ordered spaces. Many illustrating examples are supplied.
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CC-MAIN-2023-50
| 527 | 2 |
http://flightstlouis.blogspot.com/2011/04/plane-takes-off-at-st-louis-flies.html
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math
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A plane takes off at St. Louis, flies straight to Denver, and then returns the same way.
The plane flies at the same speed with respect to the ground during the entire flight, and there are no head winds or tail winds. Since the earth revolves around its axis once a day, you might expect that the times for the outbound trip and the return trip differ, depending on whether the plane flies against the earth's rotation or with it. However, under the conditions given, the two flight times are identical. WHY?
Physics - 1 Answers
Random Answers, Critics, Comments, Opinions :
Here is an important point to understand: EARTH's ORBIT DOES NOT AFFECT FLIGHT TIME The Earth's orbit has no direct effect on flight time. The planes do not actually fly through space but they fly through the air. For example Earth moves at about 5000m/s but a balloon moving in the air does not travel at 5000m/s. The atmosphere rotates with the air where the balloon moves through this air. You can imagine 2 different things moving, the Earth moving with its atmospher moving on top. though the earth's moves quickly as seen by the rapid changes of the continents if you were on the moon. the atmosphere may not even change when it is moving. the plane would be moving through this shell of atmosphere which isn't moving much at all so expect flight speeds to be similar
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s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578527720.37/warc/CC-MAIN-20190419121234-20190419143234-00553.warc.gz
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CC-MAIN-2019-18
| 1,349 | 5 |
https://tkart.it/en/magazine/tech-talk/ackermann-steering-geometry/
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math
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Kart steering is far more complex than a simple system for turning the front wheels on a bend. Its operation uses the Ackermann system: a revolutionary solution invented and patented in 1817 to optimise the direction of the front tyres and, consequently, driving around a bend
The starting point for explaining the Ackermann steering geometry is obvious: while on a bend, the outer front wheel has a wider trajectory (that is, a wider curve) than the inner wheel (which has a narrower curve). It should also be considered that a turning kart must have a rotation centre around which to do so, and around which the front wheels can rotate.
That being so, it is evident that if the front wheels are perfectly parallel during the bend, no rotation centre would be created because the axles of the wheels would remain parallel and without a common rotation point (which would be obtained from the intersection of the two axles of the wheels). In this way, the front wheels would slide, generating friction between tread and asphalt, tyre wear and loss of performance.
Ackermann’s steering geometry serves to offset the different direction covered by the front wheels along the curved trajectory to avoid this happening.
How? By causing the front wheels to rotate in a non-linear direction when turning the steering wheel. In doing so, a theoretical point is created (“theoretical” because, with the slipping of a 4-wheeled kart, the point actually varies) in which the 2 axes of the front wheels and the axis of the rear axle intersect.
Because, among other things, because of the caster angle, the bend geometry of a kart’s steering will lower the inner front wheel and lift the outer front wheel, with the transfer of much of the load to the front. The grip on the front is accentuated and the wheel travel direction is even more decisive.
Steering rods, stub axles and steering columns: these are the elements that determine Ackermann’s steering geometry
To ensure that the Ackermann steering geometry is more than zero and therefore the front wheels turn in a non-linear manner, the front stub axles are made with tie-rods directed towards the inside of the chassis, the ends of which have the holes the steering rods are hooked onto. In this way, a system is created in which the distance between the rotation axis of the stub axles is greater than that between the points of attachment of the steering columns with the steering rods. The steering geometry thus generated is, in fact, Ackermann’s steering geometry.
However, this is not enough. In fact, if it stopped there, the variation of the two angles of the front wheels with respect to the axis of the front stub axle would not be in optimal motion.
Therefore, in order to give greater progression to the variation of the two steering geometries of the front wheels (i.e. at the Ackermann angle) as the steering wheel rotates, the tie rods are connected to two different points on the plate of the steering system on the steering column instead of the same position.
This determines the non-linear variation of the front wheel steering angle, which will adapt in the best possible way to any steering geometry.
In order to have a single theoretical rotation centre there are equations that relate the front wheels (X), the kart’s pitch (Y) and the two angles, alpha and beta, of the two front wheels with the axis that joins the stub axles.
X/Y = cotg (alpha) – cotg (beta)
However, it needs to be said that, on the one hand Ackermann’s solution is simple and economical to achieve, but at the same time this system does not always ensure a perfect theoretical rotation centre: there is almost always a misalignment that does not allow the exact definition of a common point between the two front axle rotation axes and the axis of rotation of the rear axle.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00463.warc.gz
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CC-MAIN-2023-40
| 3,965 | 15 |
https://circuitglobe.com/single-line-diagram-of-power-system.html
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math
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Definition: Single line diagram is the representation of a power system using the simple symbol for each component. The single line diagram of a power system is the network which shows the main connections and arrangement of the system components along with their data (such as output rating, voltage, resistance and reactance, etc.).
It is not necessary to show all the components of the system on a single line diagram, e.g., circuit breaker need not be shown in the load flow study but are the must for a protection study. In the single line diagram, the system component is usually drawn in the form of their symbols. Generator and transformer connections, star, delta and neutral earthing are indicated by symbols drawn by the side of the representation of these elements.
Circuit breakers are represented by rectangular blocks. The figure shown below represents the single line diagram of a typical block system. It is difficult to draw the line diagram of the few components. So for simplification, the impedance diagram is used for representing the power system components.
Impedance Diagram for the Power System
In impedance diagram, each component is represented by its equivalent circuit, e.g., the synchronous generator at the generating station by a voltage source in series with the resistance and reactance, the transformer by a nominal ∏-equivalent circuit. The load is assumed to be passive and are represented by a resistive and inductive reactance in the series. Neutral earthing impedance does not appear in the diagram as the balanced condition is assumed.
The diagram shown below is the balanced 3-phase diagram. It is also called positive sequence diagram. Three separate diagrams are also used for representing the positive, negative and zero sequence networks. The three separate impedance diagrams are used in the short circuit for the studies of unsymmetrical fault.
The impedance diagram can further be simplified by making certain assumptions and reduced to simplified reactance. Reactance diagram is drawn by neglecting the effective resistance of generator armature, transformer winding resistance, transmission line resistance line charging and the magnetising circuit of transformers. Reactance diagram of the power system is shown below.
Reactance Diagram for the Power System
The reactance diagram gives an accurate result for many power system studies, such as short-circuit studies, etc. The winding resistance, including the line resistance, are quite small in comparison with leakage reactance and shunt path which includes line charging and transformer magnetising circuit provide a very high parallel impedance with fault.
It is considered that if the resistance is less than one-third of the reactance, and resistance is ignored, then the error introduced will be not more than 5 %. If the resistance and reactance ignored errors up to 12% may be introduced. The errors mean their calculation gives a higher value than the actual value.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100551.17/warc/CC-MAIN-20231205105136-20231205135136-00194.warc.gz
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CC-MAIN-2023-50
| 2,980 | 10 |
http://www.49erswebzone.com/forum/nfl-draft/175323-sammy-watkins-wr-clemson/page19/
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math
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Originally posted by HarbaughzDeep:
I know the consensus (from me and most here) is that Watkins is the top rated WR. Do you think any GM's in the league have Evans ahead of Watkins? Just asking because I heard Polian say that Matthews was the best WR in this class. Other scouting people have said that the gap between Watkins and Evans is very slim. I don't agree but what are your thoughts on the GM's boards?
Well, let's make the comparison line by line to find out who is better...
Rushing: How many rushing yards do they each have? How many rushing TD's do they each have? How many times have they each broken a long run for a touchdown? How many times have they each broken a tackle?
Kickoff returns: How many return yards do they each have? How many return TD's do they each have?
Screens: How many times have they been used on screens? How many yards do they each have on screen plays? How many touchdowns do each of them have on screen plays?
Passing: How many passing yards do they each have? How many touchdown passes?
Receiving: How many receiving yards do they each have? How many touchdown catches? How many receptions? How often do they break tackles after the catch?
I know it's very close, but after you compare these things, you will see that one of them has a slight edge.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122174.32/warc/CC-MAIN-20170423031202-00295-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 1,292 | 9 |
http://asvao.biemdas.com/archives/1387
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math
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Volume 4, Issue 1, 1 April 2022, Pages 73-82
Abstract. We propose a projection algorithm for solving strongly monotone equilibrium problems. An error bound for the algorithm is derived, and the application to a strongly multivalued monotone variational inequality is discussed. Some computational results are reported to show the efficiency and behavior of the proposed algorithm.
How to Cite this Article:
Le Dung Muu, Nguyen Van Quy, A new algorithm for solving strongly monotone equilibrium and multivalued variational inequality problems, Appl. Set-Valued Anal. Optim. 4 (2022), 73-82.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00445.warc.gz
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CC-MAIN-2022-21
| 589 | 4 |
http://www.phy.ntnu.edu.tw/ntnujava/msg.php?id=7473
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math
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There is nothing wrong with the code.
It is because you have too large value for k1,k2
The resistor force is assume to be and the mass is assumed to be 1.
K1 and K2 has to be much smaller than 1.
for k2=0.1 it is not an object moving in air. It is an object moving in densed fluid.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00127.warc.gz
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CC-MAIN-2020-45
| 281 | 5 |
https://www.interviewkiller.com/question/741-the-cost-of-type-1-rice.html
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math
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The cost of Type 1 rice is Rs. 15 per kg a...
The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:
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- Job Interview Questions and Answers - Recruitment Tips
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00846.warc.gz
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CC-MAIN-2022-49
| 742 | 19 |
https://ocw.mit.edu/ans7870/16/16.unified/thermoF03/mud/T10mud03.html
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math
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I reviewed the material we discussed in the last two lectures starting with the steady flow energy equation. This is a form of the first law written for an open system (with the same mass flow coming in and out). The only assumption I invoked was that I neglected changes in potential energy. That got me to and . The common difficulty with these two equations is understanding the physical difference between flow work (p2v2-p1v1) and shaft (or other external) work.
We then looked at the special case of an adiabatic process with no external work (q=0 and ws=0). This is an excellent model for the acceleration or deceleration of many fluids. For this case the stagnation enthalpy of the flow (hsubT = cpTsubT for an ideal gas) is a constant. If the flow is accelerated (c increasing), the enthalpy decreases (and the internal energy and the temperature). And vice versa if the flow is decelerated. If we deecelerate the flow to zero speed (c=0) then the temperature = the stagnation temperature and the enthalpy = the stagnation enthalpy. We also noted that if the process was also quasi-static, then pv^g=constant and we can define a stagnation pressure as the pressure one would reach if the flow were decelerated to zero speed in a particular reference frame via an adiabatic, quasi-static process with no external work. The stagnation quantities are dependent on the speed of the reference frame in which we assume the flow is stagnating ( a faster moving vehicle has a higher skin temperature). Also note that you can only apply the steady flow energy equation in a reference frame that is STEADY. So you must put yourself on the moving blade or on the moving aircraft (so you have a steady flow coming towards you), then determine what the speed of the flow is that is moving towards you, allow all of this kinetic energy (c^2/2) to be converted to enthapy (cpT) via an adiabatic process with no external work , arriving at the stagnation enthalpy (=cpTsubT for an ideal gas).(Add quasi-static as a requirement for the process if you want to determine the stagnation pressure).
We reviewed the PRS question we ended the last lecture with, and introduced one new PRS question.
Responses to 'Muddiest Part of the Lecture Cards'
(20 respondents out of 66 students)
1)Unclear on the frame dependence of stagnation temperature and related questions. (16 students) All of the mud cards focused on the same thing. So I will make a few general comments rather than answering each separately. First, we will have more time to go over this in the recitation on Tuesday. Also, you will have practice on the homework.
Now, I will review the rotor blade and supersonic airplane examples (they are the same case--instead of blades spinning around, think of tiny supersonic airplanes spinning around in the engine). There are two ways to think about the problems. First, in the reference frame of the blade or the airplane. They see flow coming at them with kinetic energy associated with the moving flow, and with an internal energy (or more appropriately enthalpy) associated with its temperature far away in the atmosphere. When the flow is stagnated on the blade, both of these sum into the stagnation enthalpy. Thus the temperature is higher than the ambient temperature. The second way to look at it is in the fixed reference frame (not moving with respect to the atmosphere). In this case the flow has some stagnation temperature as it comes in the inlet which is equal to the ambient atmospheric temperature. Then the blade or airplane comes along and the fluid sticks to the blade or airplane (work is done on the fluid particle) gaining kinetic energy. Thus the total or stagnation temperature on the surface of the blade is greater than the stagnation temperature the flow had when coming in the inlet. But this is an unsteady problem (the blades and the airplane are moving), so you can only think about it this way conceptually -- you can't apply the steady flow energy equation in this frame. You need to apply the steady flow energy equation in a reference frame where the flow is steady.
Now the engine sitting on the ground example The flow starts out stagnant someplace (in the atmosphere far away) and moves to a new location with no heat or external work where it stagnates again on a surface -- the wall of the engine inlet. The two reference frames are the same with no relative motion, therefore the stagnation temperature is the same. Any time the flow is moving relative to the fixed reference frame (like in the inlet of the engine), the (static) temperature is lower. With the stagnation temperature constant, enthalpy (cpT) is just traded for kinetic energy (c^2/2) as the flow accelerates and decelerates. So as the flow moves faster, its static temperature drops. As the flow moves slower, its static temperature increases.
2) No mud (4 students). Good.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103556871.29/warc/CC-MAIN-20220628142305-20220628172305-00796.warc.gz
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CC-MAIN-2022-27
| 4,877 | 9 |
https://www.larsen.no/en/puzzles/AR7-subtra-puzzle
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math
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AR7 - Subtra-Puzzle
Maxi (36.5x28.5 cm) with 58 pieces
Learn subtraction. There is one math problem on each piece. Solve the problem and place the piece by its correct answer. The piece will not fit if the answer is wrong.
AR7 - Subtra-Puzzle is available in the following versions:
- Neutral (EAN/GTIN: 7023852115725)
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s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00343.warc.gz
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CC-MAIN-2020-05
| 318 | 5 |
https://www.teacherspayteachers.com/Product/Find-Slope-Write-Linear-Equations-Given-Two-Points-Placemat-Activity-2894329
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math
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Two placemat activities include practice finding slope given two ordered pairs and practice writing linear equations given two ordered pairs.
With a partner, students solve all four equations on the placemat. After both students agree on the answers or solutions for all problems on the page, they then add the answers together to find the “Sum”. The teacher easily and quickly checks the “Sum” to evaluate the students’ understanding which allows the students to know if they made a mistake.
CCSS: Use functions to model relationships between quantities.
8.F.B.4 Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
CCSS: Construct and compare linear, quadratic, and exponential models and solve problems.
HSF.LE.A.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).
CCSS: Build a function that models a relationship between two quantities.
HSF.BF.A.1 Write a function that describes a relationship between two quantities.
CCSS: Interpret functions that arise in applications in terms of the context.
HSF.IF.B.6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
CCSS: Create equations that describe numbers or relationships.
HSA.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
Print several copies of the placemat on one color of paper and slide them into plastic page protectors. The students can use dry erase markers on the plastic and easily wipe off their work when finished. Sometimes I print two different placemats with similar topics, each on their own color of paper, and put them back-to-back into the plastic page protectors.
This activity works well at the beginning of a lesson when students are actively learning a new skill and can also be used as a review.
The instructions are clearly marked on the students’ page and the answer key is included.
Included in the package:
Two placemats activities, each with four problems and space to write the “Sum” of the solutions.
-Find Slope Given Two Points
-Write Equations Given Two Points
Answer Keys for Teacher
***THIS PRODUCT IS PART OF A MONEY-SAVING BUNDLE***
Write Linear Equations Activities Bundle 1 (5 Activities)
You might also like:
• Find Slope & Y-Intercept Sum It Up Activity (7 stations)
• Rewrite Linear Equations Into Slope-Intercept Form & Standard Form Placemat Activities (2 activities)
• Point-Slope, Slope-Intercept & Standard Form Sum It Up Activity (8 stations)
• Parallel & Perpendicular Lines Sum It Up Activity (6 stations)
• Write Linear Equations From Context Problem Pass Activity (6 rounds)
• Write Linear Equations In Context & Mixed Review Placemat Activities (2 activities)
This purchase is for one teacher only.
This resource is not to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in a site license, please contact me for a quote at [email protected]. This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00763.warc.gz
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CC-MAIN-2018-09
| 3,774 | 31 |
http://www.chegg.com/homework-help/problems-section-212-solved-assuming-material-elastoplastic-chapter-2.12-problem-10p-solution-9781285225784-exc
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math
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The problems for Section 2.12 are to be solved assuming that the material is elastoplastic with yield stress σY, yield strain εY, and modulus of elasticity E in the linearly elastic region (see Fig. 2-72).
Two cables, each having a length L of approximately 40 m, support a loaded container of weight W (see figure). The cables, which have effective cross-sectional area A = 48.0 mm2 and effective modulus of elasticity E = 160 GPa, are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d = 100 mm. The cables are made of steel having an elastoplastic stressstrain diagram with σY = 500 MPa. Assume that the weight W is initially zero and is slowly increased by the addition of material to the container.
(a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine the corresponding elongation δY of the shorter cable.
(b) Determine the weight WP that produces yielding of both cables. Also, determine the elongation δP of the shorter cable when the weight W just reaches the value WP.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation δ of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region )
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s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660931.25/warc/CC-MAIN-20160924173740-00128-ip-10-143-35-109.ec2.internal.warc.gz
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CC-MAIN-2016-40
| 1,322 | 5 |
https://www.stromtrooper.com/threads/how-many-chain-links.424361/
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math
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Im doing chain and sprockets this next week on my bike and im going up two teeth in the rear from 17/41 to 17/43. I know the stock chain is 112 links but i have read here and on other forums it may or may not fit. Was wondering how many links i should expect to use? Is their a formula for this? Or am i just going to have to measure the new chain on the sprockets and go from there?
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s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00448.warc.gz
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CC-MAIN-2020-10
| 383 | 1 |
https://pdfslide.us/documents/introductory-lectures-on-fluid-dynamics.html
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math
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These notes are intended to provide a survey of basic concepts in fluid dynamicsas a preliminary to the study of dynamical meteorology. They are based on a moreextensive course of lectures prepared by Professor B. R. Morton of Monash University,Australia.
1.1 Description of fluid flow
The description of a fluid flow requires a specification or determination of the velocityfield, i.e. a specification of the fluid velocity at every point in the region. In general,this will define a vector field of position and time, u = u(x, t).
Steady flow occurs when u is independent of time (i.e., ∂u/∂t ≡ 0). Otherwisethe flow is unsteady.
Streamlines are lines which at a given instant are everywhere in the directionof the velocity (analogous to electric or magnetic field lines). In steady flow thestreamlines are independent of time, but the velocity can vary in magnitude along astreamline (as in flow through a constriction in a pipe) - see Fig. 1.1.
Figure 1.1: Schematic diagram of flow through a constriction in a pipe.
CHAPTER 1. INTRODUCTION 4
Particle paths are lines traced out by “marked” particles as time evolves. Insteady flow particle paths are identical to streamlines; in unsteady flow they aredifferent, and sometimes very different. Particle paths are visualized in the laboratoryusing small floating particles of the same density as the fluid. Sometimes they arereferred to as trajectories.
Filament lines or streaklines are traced out over time by all particles passingthrough a given point; they may be visualized, for example, using a hypodermicneedle and releasing a slow stream of dye. In steady flow these are streamlines; inunsteady flow they are neither streamlines nor particle paths.
It should be emphasized that streamlines represent the velocity field at a specificinstant of time, whereas particle paths and streaklines provide a representation ofthe velocity field over a finite period of time. In the laboratory we can obtaina record of streamlines photographically by seeding the fluid with small neutrallybuoyant particles that move with the flow and taking a short exposure (e.g. 0.1 sec),long enough for each particle to trace out a short segment of line; the eye readilylinks these segments into continuous streamlines. Particle paths and streaklines areobtained from a time exposure long enough for the particle or dye trace to traversethe region of observation.
1.2 Equations for streamlines
The streamline through the point P , say (x, y, z), has the direction of u = (u, v, w).
Figure 1.2: Velocity vector and streamline
Let Q be the neighbouring point (x+ δx, y+ δy, z+ δz) on the streamline. Thenδx ≈ uδt, δy ≈ vδt, δz ≈ wδt and as δt→ 0, we obtain the differential relationship
between the displacement dx along a streamline and the velocity components. Equa-tion (1.1) gives two differential equations (why?). Alternatively, we can represent thestreamline parameterically (with time as parameter) as
CHAPTER 1. INTRODUCTION 5
Find the streamlines for the velocity field u = (−Ωy, Ωx, 0), where Ω is a constant.
Eq. (1.1) gives
The first pair of ratios give ∫Ω (x dx+ y dy) = 0
orx2 + y2 = Γ(z),
where Γ is an arbitrary function of z. The second pair give∫dz = 0 or z = constant.
Hence the streamlines are circles x2 + y2 = c2 in planes z = constant (we havereplaced Γ(z), a constant when z is constant, by c2).
Note that the velocity at P with position vector x can be expressed as u = Ωk∧xand corresponds with solid body rotation about the k axis with angular velocity Ω.
1.3 Distinctive properties of fluids
Although fluids are molecular in nature, they can be treated as continuous mediafor most practical purposes, the exception being rarefied gases. Real fluids generallyshow some compressibility defined as
change in density per unit change in pressure
but at normal atmospheric flow speed, the compressibility of air is a relative bysmall effect and for liquids it is generally negligible. Note that sound waves owe theirexistence to compressibility effects as do “supersonic bangs” produced by aircraftflying faster than sound. For many purposes it is accurate to assume that fluids are
CHAPTER 1. INTRODUCTION 6
incompressible, i.e. they suffer no change in density with pressure. For the presentwe shall assume also that they are homogeneous, i.e., density ρ = constant.
When one solid body slides over another, frictional forces act between them toreduce the relative motion. Friction acts also when layers of fluid flow over oneanother. When two solid bodies are in contact (more precisely when there is anormal force acting between them) at rest, there is a threshold tangential force belowwhich relative motion will not occur. It is called the limiting friction. An exampleis a solid body resting on a flat surface under the action of gravity (see Fig. 1.3).
Figure 1.3: Forces acting on a rigid body at rest.
As T is increased from zero, F = T until T = μN , where μ is the so-calledcoefficient of limiting friction which depends on the degree of roughness between thesurface. For T > μN , the body will overcome the frictional force and accelerate.A distinguishing characteristic of most fluids in their inability to support tangentialstresses between layers without motion occurring; i.e. there is no analogue of limitingfriction. Exceptions are certain types of so-called visco-elastic fluids such as paint.
Fluid friction is characterized by viscosity which is a measure of the magnitudeof tangential frictional forces in flows with velocity gradients. Viscous forces areimportant in many flows, but least important in flow past “streamlined” bodies. Weshall be concerned mainly with inviscid flows where friction is not important, but itis essential to acquire some idea of the sort of flow in which friction may be neglectedwithout completely misrepresenting the behaviour. The total neglect of friction isrisky!
To begin with we shall be concerned mainly with homogeneous, incompressibleinviscid flows.
1.4 Incompressible flows
Consider an element of fluid bounded by a “tube of streamlines”, known as a streamtube. In steady flow, no fluid can cross the walls of the stream tube (as they areeverywhere in the direction of flow).
Hence for incompressible fluids the mass flux ( = mass flow per unit time) acrosssection 1 (= ρv1S1) is equal to that across section 2 (= ρv2S2), as there can be no
CHAPTER 1. INTRODUCTION 7
accumulation of fluid between these sections. Hence vS = constant and in the limit,for stream tubes of small cross-section, vS = constant along an elementary streamtube.
vS = constant along an elementary stream tube.
It follows that, where streamlines contract the velocity increases, where they ex-pand it decreases. Clearly, the streamline pattern contains a great deal of informationabout the velocity distribution.
All vector fields with the property that
(vector magnitude) × (area of tube)
remains constant along a tube are called solenoidal. The velocity field for an incom-pressible fluid is solenoidal.
1.5 Conservation of mass: the continuity equation
Apply the divergence theorem∫V
∇ · u dV =
u · n ds
to an arbitrarily chosen volume V with closed surface S (Fig. 1.5).Let n be a unitoutward normal to an element of the surface ds u.c. If the fluid is incompressibleand there are no mass sources or sinks within S, then there can be neither continuingaccumulation of fluid within V nor continuing loss. It follows that the net flux offluid across the surface S must be zero, i.e.,∫
u · n dS = 0,
whereupon∫V∇ · u dV = 0. This holds for an arbitrary volume V , and therefore
∇ · u = 0 throughout an incompressible flow without mass sources or sinks. This isthe continuity equation for a homogeneous, incompressible fluid. It corresponds withmass conservation.
CHAPTER 1. INTRODUCTION 8
Equation of motion: somepreliminaries
The equation of motion is an expression of Newtons second law of motion:
mass × acceleration = force.
To apply this law we must focus our attention on a particular element of fluid,say the small rectangular element which at time t has vertex at P [= (x, y, z)] andedges of length δx, δy, δz. The mass of this element is ρ δx δy δz, where ρ is thefluid density (or mass per unit volume), which we shall assume to be constant.
Figure 2.1: Configuration of a small rectangular element of fluid.
The velocity in the fluid, u = u(x, y, z, t) is a function both of position (x, y, z)and time t, and from this we must derive a formula for the acceleration of the elementof fluid which is changing its position with time. Consider, for example, steady flowthrough a constriction in a pipe (see Fig. 1.1). Elements of fluid must accelerateinto the constriction as the streamlines close in and decelerate beyond as they openout again. Thus, in general, the acceleration of an element (i.e., the rate-of-changeof u with time for that element) includes a rate-of-change at a fixed position ∂u/∂t
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 10
and in addition a change associated with its change of position with time. We derivean expression for the latter in section 2.1.
The forces acting on the fluid element consist of:
(i) body forces, which are forces per unit mass acting throughout the fluid becauseof external causes, such as the gravitational weight, and
(ii) contact forces acting across the surface of the element from adjacent elements.
These are discussed further in section 2.2.
2.1 Rate-of-change moving with the fluid
We consider first the rate-of-change of a scalar property, for example the temperatureof a fluid, following a fluid element. The temperature of a fluid, T = T (x, y, z, t),comprises a scalar field in which T will vary, in general, both with the position andwith time (as in the water in a kettle which is on the boil). Suppose that an elementof fluid moves from the point P [= (x, y, z)] at time t to the neighbouring point Q attime t+ Δt. Note that if we stay at a particular point (x0, y0, z0), then T (x0, y0, z0)is effectively a function of t only, but that if we move with the fluid, T is a functionboth of position (x, y, z) and time t. It follows that the total change in T betweenP and Q in time Δt is
ΔT = TQ − TP = T (x+ Δx, y + Δy, z + Δz, t+ Δt) − T (x, y, z, t),
and hence the total rate-of-change of T moving with the fluid is
T (x+ Δx, y + Δy, z + Δz, t+ Δt) − T (x, y, z, t)
For small increments Δx,Δy,Δz,Δt, we may use a Taylor expansion
T (x+ Δx, y + Δy, z + +Δz, t+ Δt) − T (x, y, z, t) +
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 11
Δz + higher order terms in Δx, Δy, Δz, Δt .
Hence the rate-of-change moving with the fluid element
since higher order terms → 0 and u = dx/dt, v = dy/dt, w = dz/dt, where r = r(t) =(x(t), y(t), z(t)) is the coordinate vector of the moving fluid element. To emphasizethat we mean the total rate-of-change moving with the fluid we write
Here, ∂T/∂t is the local rate-of-change with time at a fixed position (x, y, z), while
∂z= u · ∇T
is the advective rate-of-change associated with the movement of the fluid element.Imagine that one is flying in an aeroplane that is moving with velocity c(t) =(dx/dt, dy/dt, dz/dt) and that one is measuring the air temperature with a ther-mometer mounted on the aeroplane. According to (2.1), if the air temperaturechanges both with space and time, the rate-of-change of temperature that we wouldmeasure from the aeroplane would be
∂t+ c · ∇T. (2.2)
The first term on the right-hand-side of (2.2) is just the rate at which the tem-perature is varying locally ; i.e., at a fixed point in space. The second term is therate-of-change that we observe on account of our motion through a spatially-varyingtemperature field. Suppose that we move through the air at a speed exactly equal tothe local flow speed u, i.e., we move with an air parcel. Then the rate-of-change ofany quantity related to the air parcel, for example its temperature or its x-componentof velocity, is given by
∂t+ u · ∇ (2.3)
operating on the quantity in question. We call this the total derivative and oftenuse the notation D/Dt for the differential operator (2.3). Thus the x-component ofacceleration of the fluid parcel is
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 12
∂ t+ u · ∇u, (2.4)
while the rate at which its potential temperature changes is expressed by
∂ t+ u · ∇θ. (2.5)
Consider, for example, the case of potential temperature. In many situations,this is conserved following a fluid parcel, i.e.,
Dt= 0. (2.6)
In this case it follows from (2.5) and (2.6) that
∂t= −u · ∇θ. (2.7)
This equation tells us that the rate-of-change of potential temperature at a point isdue entirely to advection, i.e., it occurs solely because fluid parcels arriving at thepoint come from a place where the potential temperature is different.
For example, suppose that there is a uniform temperature gradient of −1◦ C/100km between Munich and Frankfurt, i.e., the air temperature in Frankfurt is cooler. Ifthe wind is blowing directly from Frankfurt to Munich, the air temperature in Munichwill fall steadily at a rate proportional to the wind speed and to the temperaturegradient. If the air temperature in Frankfurt is higher than in Munich, then thetemperature in Munich will rise. The former case is one of cold air advection (coldair moving towards a point); the latter is one of warm air advection.
∂T+ (u · ∇) F
represents the total rate-of-change of any vector field F moving with the fluid velocity(velocity field u), and in particular that the acceleration (or total change in u movingwith the fluid) is
∂t+ (u · ∇) u.
The previous result for the rate-of-change of a scalar field can be applied to each ofthe component of F, or to each of the velocity components (u, v, w) and these resultsfollow at once.
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 13
∂t+ (u · ∇) r = 0 +
)(x, y, z) = (u, v, w)
as x, y, z, t are independent variables.
2.2 Internal forces in a fluid
An element of fluid experiences “contact” or internal forces across its surface due tothe action of adjacent elements. These are in many respects similar to the normalreaction and tangential friction forces exerted between two rigid bodies, except, asnoted earlier, friction in fluids is found to act only when the fluid is in non-uniformmotion.
Figure 2.3: Forces on small surface element δS in a fluid.
Consider a region of fluid divided into two parts by the (imaginary) surface S,and let δS be a small element of S containing the point P and with region 1 belowand region 2 above S. Let (δX, δY, δZ) denote the force exerted on fluid in region 1by fluid region 2 across δS.
This elementary force is the resultant (vector sum) of a set of contact forces actingacross δS, in general it will not act through P ; alternatively, resolution of the forces
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 14
will yield a force (δX, δY, δZ) acting through P together with an elementary couple
with moment of magnitude on the order of (δS)1/2 (δX2 + δY 2 + δZ2)1/2
.The main force per unit area exerted by fluid 2 on fluid 1 across δS,[
]is called the mean stress. The limit as δS → 0 in such a way that it always containsP , if it exists, is the stress at P across S. Stress is a force per unit area. The stressF is generally inclined to the normal n to S at P , and varies both in magnitude anddirection as the orientation n of S is varied about the fixed point P .
The stress F may be resolved into a normal reaction N , or tension, acting normalto S and shearing stress T , tangential to S, each per unit area.
Figure 2.4: The stress on a surface element δS can be resolved into normal andtangential components.
Note that in the limit δS → 0 there is no resultant bending moment as
provided that the stress is bounded.The stress and its reaction (exerted by fluid in region 1 on fluid in region 2) are
equal and opposite. This follows by considering the equilibrium of an infinitesimalslice at P ; see Fig. 2.5.
2.2.1 Fluid and solids: pressure
If the stress in a material at rest is always normal to the measuring surface forall points P and surfaces S, the material is termed a fluid ; otherwise it is a solid.Solids at rest sustain tangential stresses because of their elasticity, but simple fluidsdo not possess this property. By assuming the material to be at rest we eliminatethe shearing stress due to internal friction. Many real fluids conform closely to this
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 15
Figure 2.5: The stress and its reaction are equal and opposite.
definition including air and water, although there are more complex fluids possessingboth viscosity and elasticity. A fluid can be defined also as a material offering noinitial resistance to shear stress, although it is important to realize that frictionalshearing stresses appear as soon as motion begins, and even the smallest force willinitiate motion in a fluid in time. The property of internal friction in a fluid is knownas viscosity.
Although the term tension is usual in the theory of elasticity, in fluid dynamicsthe term pressure is used to denote the hydrostatic stress, reversed in sign. In a fluidat rest the stress acts normally outwards from a surface, whereas the pressure actsnormally inwards from the fluid towards the surface.
2.2.2 Isotropy of pressure
The pressure at a point P in a continuous fluid is isotropic; i.e., it is the same forall directions n. This is proved by considering the equilibrium of a small tetrahedralelement of fluid with three faces normal to the coordinate axes and one slant face.The proof may be found in any text on fluid mechanics.
2.2.3 Pressure gradient forces in a fluid in macroscopic equi-
Pressure is independent of direction at a point, but may vary from point to point ina fluid. Consider the equilibrium of a thin cylindrical element of fluid PQ of lengthδs and cross-section A, and with its ends normal to PQ. Resolve the forces in thedirection P for the fluid at rest. Then pressure acts normally inwards on the curved
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 16
cylindrical surface and has no component in the direction of PQ (2.6). Thus theonly contributions are from the plane ends.
Figure 2.6: Pressure forces on a cylindrical element of fluid.
The net force in the direction PQ due to the pressure thrusts on the surface ofthe element is
pA− (p+ δp)A = −∂p∂sA δs = −∂p
where dV is the volume of the cylinder. In the limit δs→ 0, A→ 0, the net pressurethrust → − (∂p/∂s) dV, or − ∂p/∂s = −s · ∇p per unit volume of fluid (s being aunit vector in the direction PQ). It follows that −∇p is the pressure gradient forceper unit volume of fluid, and −n · ∇p is the component of pressure gradient forceper unit volume in the direction n.
Figure 2.7: A horizontal cylindrical element of fluid in equilibrium.
2.2.4 Equilibrium of a horizontal element
The cylindrical element shown in Fig. 2.7 is in equilibrium under the action ofthe pressure over its surface and its weight. Resolving in the direction PQ, thex-direction, the only force arises from pressure acting on the ends
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 17
pA− (p+ δs) A = −Aδpδxδx = −δp
and hence in equilibrium in the limit δV → 0,
Alternatively, the horizontal component of pressure gradient force per unit volumeis −i · ∇p = −∂p/∂x = 0, from the assumption of equilibrium.
Thus p is independent of horizontal distance x, and is similarly independent ofhorizontal distance y. It follows that
p = p(z)
and surfaces of equal pressure (isobaric surfaces) are horizontal in a fluid at rest.
2.2.5 Equilibrium of a vertical element
For a vertical cylindrical element at rest in equilibrium under the action of pressurethrusts and the weight of fluid
−k · ∇p δV + ρg δV = 0, where k = (0, 0, 1).
Thus 1 dpdz
= ρg, per unit volume, since p = p(z) only (otherwise we would write
∂p/∂z!). Hence ρ = 1gdpdz
is a function of z at most, i.e., ρ = ρ(z).
2.2.6 Liquids and gases
Liquids undergo little change in volume with pressure over a very large range ofpressures and it is frequently a good assumption to assume that ρ = constant. Inthat case, the foregoing equation integrates to give
p = p0 + ρgz,
where p = p0 at the level z = 0.Ideal gases are such that pressure, density and temperature are related through
the ideal gas equation, p = ρRT , where T is the absolute temperature and R isthe specific gas constant. If a certain volume of gas is isothermal (i.e., has constanttemperature), then pressure and density vary exponentially with depth with a so-called e-folding scale H = RT/g (see Ex. 4).
1Here z measures downwards so that sgn(δz) = sgn(δp). Normally we take z upwards whereupondp/dz = −ρg.
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 18
Figure 2.8: Equilibrium forces on a vertical cylindrical element of fluid at rest.
2.2.7 Archimedes Principle
In a fluid at rest the net pressure gradient force per unit volume acts verticallyupwards and is equal to −dp/dz (when z points upwards) and the gravitational forceper unit volume is ρg. Hence, for equilibrium, dp/dz = −ρg. Consider the vertically-oriented cylindrical element P1P2 of an immersed body which intersects the surfaceof the body to form surface elements δS1 and δS2. These surface elements havenormals n1, n2 inclined at angles θ1, θ2 to the vertical.
The net upward thrust on these small surfaces
= p2 cos θ2 δS2 − p1 cos θ1 δS1 = (p2 − p1) δS,
where δS1cosθ1 = δS2cosθ2 = δS is the horizontal cross-sectional area of the cylinder.Since
p1 − p2 = −∫ z1
ρg dz ,
the net upward thrust
= The weight of liquid displaced by the cylindrical element.
If this integration is now continued over the whole body we have Archimedes Principlewhich states that the resultant thrust on an immersed body has a magnitude equal
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 19
Figure 2.9: Pressure forces on an immersed body or fluid volume.
to the weight of fluid displaced and acts upward through the centre of mass of thedisplaced fluid (provided that the gravitational field is uniform).
1. If you suck a drink up through a straw it is clear that you must acceleratefluid particles and therefore must be creating forces on the fluid particles nearthe bottom of the straw by the action of sucking. Give a concise, but carefuldiscussion of the forces acting on an element of fluid just below the open endof the straw.
2. Show that the pressure at a point in a fluid at rest is the same in all directions.
3. Show that the force per unit volume in the interior of homogeneous fluid is−∇p, and explain how to obtain from this the force in any specific direction.
4. Show that, in hydrostatic equilibrium, the pressure and density in an isothermalatmosphere vary with height according to the formulae
p(z) = p(0) exp(−z/HS) ,ρ(z) = ρ(0) exp(−z/HS) ,
where Hs = RT/g and z points vertically upwards. Show that for realisticvalues of T in the troposphere, the e-folding height scale is on the order of 8km.
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 20
5. A factory releases smoke continuously from a chimney and we suppose that thesmoke plume can be detected far down wind. On a particular day the windis initially from the south at 0900 h and then veers (turns clockwise) steadilyuntil it is from the west at 1100 h. Draw initial and final streamlines at 0900and 1100 h, a particle path from 0900 h to 1100 h, and filament line from 0900to 1100 h.
6. Show that the streamline through the origin in the flow with uniform velocity(U, V,W ) is a straight line and find its direction cosines.
7. Find streamlines for the velocity field u = (αx,−αy, 0), where α is constant,and sketch them for the case α > 0.
8. Show that the equation for a particle path in steady flow is determined by thedifferential relationship
where u = (u, v, w) is the velocity at the point (x, y, z). What does thisrelationship represent in unsteady flow?
9. A stream is broad and shallow with width 8 m, mean depth 0.5 m, and meanspeed 1ms−1. What is its volume flux (rate of flow per second) in m3 s−1? Itenters a pool of mean depth 3 m and width 6 m: what then is its mean speed?It continues over a waterfall in a single column with mean speed 10ms−1 at itsbase: what is the mean diameter of this column at the base of the waterfall?Will the diameter of the water column at the top of the waterfall be greater,equal to, or less at its base? Why?
10. Under what condition is the advective rate-of-change equal to the total rate-of-change?
11. Express u·∇ and ∇·u in Cartesian form and show that they are quite different,one being a scalar function and one a scalar differential operator.
12. Some books use the expression df/dt. Would you identify this with Df/Dt or∂f/∂t in a field f(x, y, z, t)?
13. The vector differential operator del (or nabla) is defined as
in rectangular Cartesian coordinates. Express in full Cartesian form the quan-tities: ∇ · u, ∇∧ u, u · ∇, ∇ · ∇ and identify each.
CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 21
14. Are the two x-components in rectangular Cartesian coordinates,
(u · ∇u)x and(∇1
the same or different? Note that (u ·∇)u = ∇(12u)2 −u∧ω, where ω = ∇∧u.
Equations of motion for an inviscidfluid
The equation of motion for a fluid follows from Newtons second law, i.e.,
mass × acceleration = force.
If we apply the equation to a unit volume of fluid:
(i) the mass of the element is ρ kg m−3;
(ii) the acceleration must be that following the fluid element to take account bothof the change in velocity with time at a fixed point and of the change in positionwithin the velocity field at a fixed time,
∂t+ (u · ∇) u =
(iii) the total force acting on the element (neglecting viscosity or fluid friction)comprises the contact force acting across the surface of the element −∇p perunit volume, which is a pressure gradient force arising from the difference inpressure across the element, and any body forces F, acting throughout the fluidincluding especially the gravitational weight per unit volume, −gk.
The resulting equation of motion or momentum equation for inviscid fluid flow,
Dt= −∇p + ρF , per unit volume,
∂ t+ (u · ∇) u = −1
ρ∇p+ F , per unit mass,
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 23
is known as Euler’s equation. In rectangular Cartesian coordinates (x, y, z) withvelocity components (u, v, w) the component equations are
∂y+ Y ,
∂z+ Z ,
where F = (X, Y, Z) is the external force per unit mass (or body force). These arethree partial differential equations in the four dependent variables u, v, w, p and fourindependent variables x, y, z, t. For a complete system we require four equations inthe four variables, and the extra equation is the conservation of mass or continuityequation which for an incompressible fluid has the form
∇ · u = 0, or∂ u
∂ x+∂ v
∂ y+∂ w
∂ z= 0.
3.1 Equations of motion for an incompressible vis-
It can be shown that the viscous (frictional) forces in a fluid may be expressed asμ∇2u = ρν∇2u where μ the coefficient of viscosity and ν = μ/ρ the kinematicviscosity provide a measure of the magnitude of the frictional forces in particularfluid, i.e., μ and ν are properties of the fluid and are relatively small in air or waterand large in glycerine or heavy oil. In a viscous fluid the equation of motion for unitmass,
∂t+ (u · ∇) u = − 1
ρ∇p + F + ν∇2u
body force viscousforce
is known as the Navier-Stokes equation. We require also the continuity equation,
∇ · u = 0,
to close the system of four differential equations in four dependent variables. Thereis no equivalent to the continuity equation in either particle or rigid body mechanics,because in general mass is permanently associated with bodies. In fluids, however,we must ensure that holes do not appear or that fluid does not double up, and wedo this by requiring that ∇ · u = 0, which implies that in the absence of sources
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 24
or sinks there can be no net flow either into or out of any closed surface. We mayregard this as a geometric condition on the flow of an incompressible fluid. It is not,of course, satisfied by a compressible fluid (c.f. a bicycle pump). We say that anyincompressible flow satisfying the continuity equation ∇ · u = 0 is a kinematicallypossible motion.
The Navier-Stokes equation plus continuity equation are extremely important,but extremely difficult to solve. With possible further force terms on the right, theyrepresent the behaviour of gaseous stars, the flow of oceans and atmosphere, themotion of the earth’s mantle, blood flow, air flow in the lungs, many processes ofchemistry and chemical engineering, the flow of water in rivers and in the permeableearth, aerodynamics of aeroplanes, and so forth....
The difficulty of solution, and there are probably no more than a dozen or sosolutions known for very simple geometries, arises from:
(i) the non-linear term (u ·∇)u as a result of which, if u1 and u2 are two solutionsof the equation, c1u1 + c2u2 (where c1 and c2 are constants) is in general not asolution, so that we lose one of our main methods of solution;
(ii) the viscous term, which is small relative to other terms except close to bound-aries, yet it contains the highest order derivatives(
∂2u/∂ x2 , ∂2u/∂ y2 , ∂2u/∂ z2),
and hence determines the number of spatial boundary conditions that must beimposed to determine a solution.
The Navier-Stokes equation is too difficult for us to handle at present and weshall concentrate on Euler’s equation from which we can learn much about fluidflow. Euler’s equation is still non-linear, but there are clever methods to bypass thisdifficulty.
Find the velocity field u = (−Ωy,Ωx, 0) for Ω constant as a possible flow of anincompressible liquid in a uniform gravitational field F ≡ g = (0, 0, −g).
(i) This is a kinematically-possible steady incompressible flow, as u satisfies thecontinuity equation
∇ · u =∂ u
∂ x+∂ v
∂ y+∂ w
∂ z= 0 + 0 + 0 = 0.
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 25
(ii) We find the corresponding pressure field from Euler’s equation.
u · ∇u = −1
If the given velocity field is substituted in the Euler’s equation and it is rear-ranged in component form,
∂ x= ρΩ2x,
∂z= −ρ g.
We may now solve these three equations as follows.
= ρΩ2x ⇒ p =1
2ρΩ2x2 + constant
where “constant” can include arbitrary functions of both y and z (Check:∂ p/∂ x = ρΩ2x + 0). We continue in like manner with the other componentequations:
∂ y= ρΩ2y ⇒ p =
2ρΩ2y2 + g(z, x),
∂ z= −ρ g ⇒ p = −ρ gz + h(x, y),
where f(y, z), g(z, x) and h(x, y) are arbitrary functions. By comparison of thethree solutions we see that f(y, z) must incorporate 1
2ρΩ2y2 and −ρgz and so
forth. Hence the full solution is
2ρΩ2(x2 + y2) − ρ gz + constant,
and we find that this solution does in fact satisfy each of the component Eulerequations. On a free surface containing the origin O(x = 0, y = 0, z = 0), p =po ⇒ the constant = po, where po is atmospheric pressure, and r2 = x2 + y2,
p = po +1
2ρΩ2r2 − ρ gz.
(iii) The equation for the free surface is now given by p = p0 over the whole liquidsurface, which therefore has equation
2ρ gr2 =
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 26
(iv) Streamlines in the flow are given by
yielding two relations
∫Ωy dy = 0 ⇒ x2 + y2 = constant∫
dz = 0 ⇒ z = constant
and streamlines are circles about the z-axis in planes z = constant. The velocityfield represents rigid body rotation of fluid with angular velocity Ω about theaxis Oz (imagine a tin of water on turntable!).
3.2 Equations of motion in cylindrical polars
Take the cylindrical polars (r, θ, z) and velocity (vr, vθ, vz). These are more com-plicated than rectangular Cartesians as vr, vθ change in direction with P (in factOP rotates about Oz with angular velocity vθ/r). Suppose that r, n, z are the unitvectors at P in the radial, azimuthal and axial directions, as sketched in Fig. 3.1.Then zis fixed in direction (and, of course, magnitude) but r and n rotate in theplane z = 0 as P moves, and it follows that dz/dt = 0, but that
dt= (−r)θ = −rθ
Hence, as θ = vθ/r,
v = (vrr + vθn + vzz) ,
v = vr r + vrdr
dt+ vθn + vθ
dt+ vzz = (vr − vrθ/r) r + (vθ + vrvθ/r) n + vzz.
Recalling also that d/dt must be interpreted here as D/Dt, the acceleration is[DvrDt
If we now write (u, v, w) in place of (vr, vθ, vz), Euler’s equations in cylindricalpolar coordinates take the form
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 27
Figure 3.1: Velocity vectors and coordinate axes in cylindrical polar coordinates.
∂ θ+∂ ω
∂ z= 0
∂ t+ u
∂ θ+ w
∂ z− v2
∂ r+ Fr ,
∂ t+ u
∂ θ+ w
r= − 1
∂ θ+ Fθ ,
∂ t+ u
∂ θ+ w
∂ z= − 1
∂ z+ Fz .
3.3 Dynamic pressure (or perturbation pressure)
If in Euler’s equation for an incompressible fluid,
ρ∇p+ g, (3.1)
we put u = 0 to represent the equilibrium or rest state,
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 28
0 = −1
ρ∇p0 + g (3.2)
This is merely the hydrostatic equation
∇p0 = ρg or∂ p0
∂ x= 0,
∂ y= 0,
∂ z= −ρ g,
where p0 is the hydrostatic pressure. Subtracting (3.1)-(3.2) we obtain
ρ∇(p− p0) = −1
where pd = p−p0 = (total pressure) - (hydrostatic pressure) is known as the dynamicpressure (or sometimes, especially in dynamical meteorology, the perturbation pres-sure). The dynamic pressure is the excess of total pressure over hydrostatic pressure,and is the only part of the pressure field associated with motion.
We shall usually omit the suffix d since it is fairly clear that if g is included weare using total pressure, and if no g appears we are using the dynamic pressure,
∂ t+ (u · ∇) u = −1
3.4 Boundary conditions for fluid flow
(i) Solid boundaries : there can be no normal component of velocity through theboundary. If friction is neglected there may be free slip along the boundary,but friction has the effect of slowing down fluid near the boundary and itis observed experimentally that there is no relative motion at the boundary,either normal or tangential to the boundary. In fluids with low viscosity, thistangential slowing down occurs in a thin boundary layer, and in a number ofimportant applications this boundary layer is so thin that it can be neglectedand we can say approximately that the fluid slips at the surface; in many othercases the entire boundary layer separates from the boundary and the inviscidmodel is a very poor approximation. Thus, in an inviscid flow (also called theflow of an ideal fluid) the fluid velocity must be tangential at a rigid body, and:
for a surface at rest n · u = 0;for a surface with velocity us n · (u − us) = 0.
(ii) Free boundaries: at an interface between two fluids (of which one might bewater and one air) the pressure must be continuous, or else there would bea finite force on an infinitesimally small element of fluid causing unboundedacceleration; and the component of velocity normal to the interface must becontinuous. If viscosity is neglected the two fluids may slip over each other. Ifthere is liquid under air, we may take p = p0 = atmospheric pressure at theinterface, where p0 is taken as constant. If surface tension is important theremay be a pressure difference across the curved interface.
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 29
3.4.1 An alternative boundary condition
As the velocity at a boundary of an inviscid fluid must be wholly tangential, itfollows that a fluid particle once at the surface must always remain at the surface.Hence for a surface or boundary with equation
F (x, y, z, t) = 0,
if the coordinates of a fluid particle satisfy this equation at one instant, they mustsatisfy it always. Hence, moving with the fluid at the boundary,
∂ t+ u · ∇F = 0,
as F must remain zero for all time for each particle at the surface.
3.1 Describe briefly the physical significance of each term in the Euler equation forthe motion of an incompressible, inviscid fluid,
∂ t+ ρ (u · ∇)u = −∇p + ρg,
explaining clearly why the two terms on the left are needed to express the massacceleration fully. To what amount of fluid does this equation apply?
3.2 The velocity components in an incompressible fluid are
u = − 2xyz
(x2 + y2)2 , v =(x2 − y2) z
(x2 + y2)2 , w =y
x2 + y2.
Show that this velocity represents a kinematically possible flow (that is, thatthe equation of continuity is satisfied).
3.3 Find the pressure field in the inviscid, incompressible flow with velocity field
u = (nx,−ny, 0).
3.4 If r, n are the unit radial and azimuthal vectors in cylindrical polars (r, θ, z)show that
CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 30
3.5 State the boundary conditions for velocity in an inviscid fluid at (a) a stationaryrigid boundary bisecting the 0x, 0y axes; (b) a rigid boundary moving withvelocity V j in the direction of the y axis.
3.6 Write down Euler’s equation for the motion of an inviscid fluid in a gravitationaluniform field: (i) in terms of the total pressure p, and (ii) in terms of thedynamic pressure pd. Relate p and pd.
3.7 Explain briefly why DF/Dt = 0 provides an alternative form of the bound-ary condition for flow in a region of inviscid fluid bounded by the surfaceF (x, y, z, t) = 0. Find the boundary condition on velocity at a fixed planey + mx = 0 and show that the equation y = m(x + y − Ut) represents a cer-tain inclined plane moving with the speed U in a certain direction. Find thisdirection and obtain the boundary condition at this plane.
For steady inviscid flow under external forces which have a potential Ω such thatF = −∇Ω Euler’s equation reduces to
u · ∇u = −1
and for an incompressible fluid
u · ∇u +1
ρ∇(p+ ρΩ) = 0.
We may regard p + pΩ as a more general dynamic pressure; but for the particularcase of gravitation potential, Ω = gzand F = −∇Ω = −(0, 0, g) = −gk.
We note that
u · (u · ∇)u = u(u · ∇) u+ v(u · ∇) v + w(u · ∇)w
= u · ∇12
(u2 + v2 + w2
)= (u · ∇) 1
using the fact that u · ∇ is a scalar differential operator. Hence,
u · [u · ∇u + ∇ (p/ρ+ Ω)] = u · ∇ [12u2 + p/ρ+ Ω
and it follows that(
12u2 + p/ρ+ Ω
)is constant along each streamline (as u · ∇ is
proportional to the rate-of-change in the direction u of streamlines). Thus for steady,incompressible, inviscid flow
(12u2 + p/ρ+ Ω
)is a constant on a streamline, although
the constant will generally be different on each different streamline.
CHAPTER 4. BERNOULLI’S EQUATION 32
4.1 Application of Bernoulli’s equation
(i) Draining a reservoir through a small hole
If the draining opening is of much smaller cross-section than the reservoir (Fig.4.1), the water surface in the tank will fall very slowly and the flow may beregarded as approximately steady. We may take the outflow speed uA as ap-proximately uniform across the jet and the pressure pA uniform across the jetand equal to the atmospheric pressure p0 outside the jet (for, if this were notso, there would be a difference in pressure across the surface of the jet, andthis would accelerate the jet surface radially, which is not observed, althoughthe jet is accelerated downwards by its weight). Hence, on the streamline AB,
12u2A + p0/ρ = 1
2u2B + p0/ρ+ gh,
and as uB << uA uA =√
Figure 4.1: Draining of a reservoir.
This is known as Toricelli’s theorem. Note that the outflow speed is that offree fall from B under gravity; this clearly neglects any viscous dissipation ofenergy.
(ii) Bluff body in a stream; Pitot tube
Suppose that a stream has uniform speed U0 and pressure p0 far from anyobstacle, and that it then flows round a bluff body (Fig. 4.2). The flow mustbe slowed down in front of the body and there must be one dividing streamlineseparating fluid which follows past one side of the body or the other. Thisdividing streamline must end on the body at a stagnation point at which thevelocity is zero and the pressure
p = p0 +1
CHAPTER 4. BERNOULLI’S EQUATION 33
Figure 4.2: Flow round a bluff body in this case a cylinder.
This provides the basis for the Pitot tube in which a pressure measurement isused to obtain the free stream velocity U0. The pressure p = p0 + 1
0 isthe total or Pitot pressure (also known as the total head) of the free stream,and differs from the static pressure p0 by the dynamic pressure 1
0 . The
Figure 4.3: Principal of a Pitot tube.
Pitot tube consists of a tube directed into the stream with a small centralhole connected to a manometer for measuring pressure difference p− p0 (Fig.4.3). At equilibrium there is no flow through the tube, and hence the left handpressure on the manometer is the total pressure p0 + 1
0 . The static pressurep0 can be obtained from a static tube which is normal to the flow.
The Pitot-static tube combines a Pitot tube and a static tube in a single head(Fig. (4.4). The difference between Pitot pressure (p0 + 1
0 ) and static pres-sure (p0) is the dynamic pressure 1
0 , and the manometer reading thereforeprovides a measure of the free stream velocity U0. The Pitot-static tube canalso be flown in an aeroplane and used to determine the speed of the aeroplanethrough the air.
(iii) Venturi tube
This is a device for measuring fluid velocity and discharge (Fig. 4.5). Supposethat there is a restriction of cross-sections in a pipe of cross-section S, with
CHAPTER 4. BERNOULLI’S EQUATION 34
Figure 4.4: A Pitot-static tube.
Figure 4.5: A Venturi tube.
velocities v, V and pressures p, P in the two sections, respectively, the pipebeing horizontal. Then
v2 − V 2 =2
ρ(P − p) =
ρρm gh = 2gh
CHAPTER 4. BERNOULLI’S EQUATION 35
The dischargeQ = vs = V S,
and substitution gives [Q
Q =sS√S2 − s2
s√S2 − s2
1. Hold two sheets of paper at A and B with a finger between the two at top andbottom, and blow between the sheets as illustrated in Fig. 4.6. The trailingedges of the sheets will not move apart as you might have anticipated, buttogether. Explain this in terms of Bernoulli’s equation, assuming the flow tobe steady.
2. Explain why there is an increase in pressure on the side of a building facingthe wind.
3. A uniform straight open rectangular channel carries a water flow of mean speedU and depth h. The channel has a constriction which reduces its width by halfand it is observed that the depth of water in the constriction is only 1
applying Bernoulli’s theorem to a surface streamline find U in terms of g andh.
4. Using Bernoulli’s equation (often referred to as Bernoulli’s theorem):
(i) show that air from a balloon at excess pressure p1 above atmospheric willemerge with approximate speed
CHAPTER 4. BERNOULLI’S EQUATION 36
(ii) find the depth of water in the steady state in which a vessel, with a wastepipe of length 0.01 m and cross-sectional area 2 × 10−5 m2 protrudingvertically below its base, is filled at the constant rate 3 × 10−5 m3 s−1.
5. A vertical round post stands in a river, and it is observed that the water level atthe upstream face of the post is slightly higher than the level at some distanceto either side. Explain why this is so, and find the increase in the height interms of the surface stream speed U and acceleration of gravity g. Estimatethe increase in height for a stream with undisturbed surface speed 1 ms−1.
The vorticity field
The vector ω = ∇× u ≡ curl u is called the vorticity (from Latin for a whirlpool).The vorticity vector ω(x, t) defines a vector field, just like the velocity field u(x, t).In the case of the velocity, we can define streamlines that are everywhere in thedirection of the velocity vector at a given time. Similarly we can define vortex linesthat are everywhere in the direction of the vorticity vector at a given time. We willshow that the vorticity is twice the local angular velocity in the flow.
(i) Bundles of vortex lines make up vortex tubes.
(ii) Thin vortex tubes, such that their constituent vortex lines are approximatelyparallel to the tube axis, are called vortex filaments (see below).
(iii) The vorticity field is solenoidal, i.e. ∇ · ω = 0. This very important resultresult is proved as follows:
∇ · ω = ∇ · (∇× u)
CHAPTER 5. THE VORTICITY FIELD 38
∂ y− ∂ v
∂ z− ∂ w
∂ x− ∂ u
From the divergence theorem, for any volume V with boundary surface S∫S
ω · n ds =
∇ · ω dV = 0,
and there is zero net flux of vorticity (or vortex tubes) out of any volume:hence there can be no sources of vorticity in the interior of a fluid (cf. sourcesof mass can exist in a velocity field!).
(iv) Consider a length P1P2 of vortex tube. From the divergence theorem∫S
ω · n ds =
∫∇ · ω dV = 0.
We can divide the surface of the length P1P2 into cross-sections and the tubewall,
S = S1 + S2 + Swall,
ω · n ds =
ω · n ds+
ω · n ds+
ω · n ds = 0. (5.1)
CHAPTER 5. THE VORTICITY FIELD 39
However, the contribution from the wall (where ω ⊥n) is zero, and hence∫S2
ω · n ds =
ω · (−n) ds
where the positive sense for normals is that of increasing distance along thetube from the origin. Hence ∫
ω · n ds
measured over a cross-section of the vortex tube with n taken in the same senseis constant, and taken as the strength of the vortex tube.
In a thin vortex tube, we have approximately:∫S
ω · n dS ≈ ω · n
dS = ωS
and ω × area = constant along tube (a property of all solenoidal fields). Here,ω = |ω|.
u · drFrom Stokes’ theorem ∫
(∇× u) · n dS =
u · dr
Hence the line integral of the velocity field in any circuit C that passes onceround a vortex tube is equal to the total vorticity cutting any cap S on C, andis therefore equal to the strength of the vortex tube. We measure the strengthof a vortex tube by calculating
u · dr around any circuit C enclosing thetube once only. The quantity
u · dr is termed the circulation.
Vorticity may be regarded as circulation per unit area, and the component inany direction of ω is
u · dr
where C is a loop of area S perpendicular to the direction specified.
CHAPTER 5. THE VORTICITY FIELD 40
Show that 12u2 + p/ρ+ Ω = constant along a vortex line for steady, incompressible,
inviscid flow under conservative external forces.
As beforeu · ∇u + ∇ (p/ρ+ Ω) = 0,
u · ∇u = ∇(
)− u × (∇× u) = ∇
)− u × ω.
Henceu× ω = ∇ [
12u2 + p/ρ+ Ω
u. u · (u × ω) ≡ 0 = u · ∇ [12u2 + p/ρ+ Ω
ω ω · (u× ω) ≡ 0 = ω · ∇ [12u2 + p/ρ+ Ω
From Eq.(5.2) 12u2 + p/ρ + Ω = constant along a streamline, and from Eq.(5.3)
12u2 + p/ρ + Ω = constant along a vortex line. Thus we have a Bernoulli equation
for vortex lines as well as for streamlines.
CHAPTER 5. THE VORTICITY FIELD 41
1. Define the circulation round a closed circuit C and show that it is equal to thenet vorticity cutting any cap on that circuit.
2. Show that vorticity may be interpreted as circulation per unit area of section.
3. Does fluid with velocity
[z − 2x
r, 2y − 3z − 2y
r, x− 3y − 2z
possess vorticity (where u = (u, v, w) is the velocity in the Cartesian framer = (x, y, z) and r2 = x2 + y2 + z2)? What is the circulation in the circlex2 + y2 = 9, z = 0? Is this flow incompressible?
4. Find the vorticity passing through the circuit x2+y2 = a2, z = 0 in the velocityfield u = U(z, x, y)/a.
5.1 The Helmholtz equation for vorticity
From Euler’s equation for an incompressible fluid in a conservative force field.
as ω is always solenoidal and u is solenoidal in an incompressible fluid; we obtain
∂t+ (u · ∇) ω = (ω · ∇)u,
which is the Helmholtz vorticity equation.
CHAPTER 5. THE VORTICITY FIELD 42
5.1.1 Physical significance of the term (ω · ∇)u
We can understand the significance of the term (ω · ∇)u in the Helmholtz equationby recalling that ∇ is a directional derivative and (ω · ∇)u is proportional to thederivative in the direction of ω along the vortex line (see example 7).
Dt= (ω · ∇)u = |ω| ω · ∇u = ω
where δsω is the length of an element of vortex tube. We now resolve u into compo-nents uω parallel to ω and u⊥ at right angles to ω and hence to δsω. Then
∂sω(uω + u⊥) δsω
≈ [uω (r + δsω) − uω(r)]︸ ︷︷ ︸rate of stretching of element
+ [u⊥ (r + δsω) − u⊥(r)]︸ ︷︷ ︸rate of turning of element
CHAPTER 5. THE VORTICITY FIELD 43
• stretching along the length of the filament causes relative amplification of thevorticity field;
• turning away from the line of the filament causes a reduction of the vorticityin that direction, but an increase in the new direction.
Discuss properties of the directional derivative.
Suppose that P is a point on the level surface φ of a scalar function, and that N andP are points on the neighbouring surface φ + δφ in the direction of the normal atP (n) and a specified curve (s). Then
∇φ = n ∂φ/∂n is the largest of the directional derivatives at P (as δn is the minimumseparation distance between the surfaces, φ, φ + δφ) and has the direction n of theoutward normal at P . Then
s · ∇φ = s · n∂φ∂n
∂ncos θ =
and ω · ∇u = |ω| ω · ∇u = ω ∂u∂sω
where sω is distance along the vortex line.
CHAPTER 5. THE VORTICITY FIELD 44
5.2 Kelvin’s Theorem
The ideas of vorticity and circulation are important because of the permanence ofcirculation under deformation of the flow due to pressure forces. We next look at therate-of-change of circulation round a circuit moving with an incompressible, inviscidfluid:
∮u · dr =
Dt(u · dr)
Dt· dr +
∮u · D
The first integral on the right may be written∮ (
)· dr, and the
u · DDtdr =
∮u · du (see Example 7). Hence∮
Dt· dr =
∮u · dr =
]· dr +
∮u · du
ρdp− dΩ + d
(−pρ− Ω +
as −p/ρ − Ω + 12u2 returns to its initial value after one circuit since it is a single
u · DDtdr =
∮u · du.
Suppose that the elementary vector P �Q = δr at t is advected with the flow toP′�Q′ = δr (t+ δt) at t+ δt. Then
δr (t+ δt) ≈ −u (r) δt+ δr(t) + u (r + δr) δt,
orδr (t+ δt) − δr(t) ≈ u (r + δr) δt− u (r) δt,
δr (t+ δt) − δr(t)
u (r + δr) − u (r)
Dt(δr) ≈ ∂u
∂sδs ≈ δu
CHAPTER 5. THE VORTICITY FIELD 45
in a fixed reference frame 0xyz, where |δr| = δs and s is arc length along the pathP. In the limit as δr → dr , δu → du,
Dt(dr) = du.
5.2.1 Results following from Kelvins Theorem
(i) Helmholtz theorem: vortex lines move with the fluid
Consider a tube of particles T which at the instant t forms a vortex tube ofstrength k. At that time the circulation round any circuit C ′ lying in the tubewall, but not linking (i.e. embracing) the tube is zero, while that in an circuitC linking the tube once is k. These circulations suffer no change moving withthe fluid: hence the circulation in C ′ remains zero and that in C remains k,i.e. the fluid comprising the vortex tube at T continues to comprise a vortextube (as the vorticity component normal to the tube wall - measured in C ′- isalways zero), and the strength of the vortex remains constant. A vortex lineis the limiting case of a small vortex tube: hence vortex lines move with (arefrozen into) inviscid fluids.
(ii) A flow which is initially irrotational remains irrotational Circulation isadvected with the fluid in inviscid flows, and vorticity is “circulation per unitarea”. If initially for all closed circuits in some region of flow, it must remainso for all subsequent times. Motion started from rest is initially irrotational(free from vorticity) and will therefore remain irrotational provided that it isinviscid.
(iii) The direction of the vorticity turns as the vortex line turns, and itsmagnitude increases as the vortex line is stretched.
CHAPTER 5. THE VORTICITY FIELD 46
The circulation round a thin vortex tube remains the same; as it stretches thearea of section decreases and
increases in proportion to the stretch.
1. Explain the physical significance of each term in the Helmholtz equation forvorticity in inviscid incompressible flow.
2. Show that in two-dimensional flow, with u = (u(x, y), v(x, y), 0) vorticity isnecessarily normal to the xy-plane, ω = (0, 0, ζ). Hence show that in two-dimensional inviscid incompressible flow the Helmholtz vorticity equation re-duces to the form
so that if the distribution of vorticity is initially uniform it must remain so,and if the motion is initially irrotational (free from vorticity) it must remainso.
3. Explain the statement that in inviscid flows vorticity is “frozen into the fluid”.
4. Show that the circulation in any circuit embracing a vortex tube (i.e. passingonce round it) in otherwise irrotational fluid is equal to the strength of thevortex tube ∮
ω · n dS
CHAPTER 5. THE VORTICITY FIELD 47
taken over any section of tube. Hence, or otherwise, show that a vortex tubecannot terminate in the interior of a fluid region.
5.3 Rotational and irrotational flow
Flow in which the vorticity is everywhere zero (∇ × u = 0) is called irrotational.Other terms in use are vortex free; ideal ; perfect. Much of fluid dynamics used tobe concerned with analysing irrotational flows and deciding where these give a goodrepresentation of real flows, and where they are quite wrong.
We have neglected compressibility and viscosity. It can be shown that the neglectof compressibility is not very serious even at moderately high speeds, but the effectof neglecting viscosity can be disastrous. Viscosity diffuses the vorticity (much asconductivity diffuses heat) and progressively blurs the results derived above, theerrors increasing with time.
There is no term in the Helmholtz equation
Dt= (ω · ∇)u
corresponding to the generation of vorticity: the term ω · ∇u represents processingby stretching and turning of vorticity already present. It follows, therefore, that inhomogeneous fluids all vorticity must be generated at boundaries. In real (viscous) flu-ids, this vorticity is carried away from the boundary by diffusion and is then advectedinto the body of the flow. But in inviscid flow vorticity cannot leave the surface bydiffusion, nor can it leave by advection with the fluid as no fluid particles can leavethe surface. It is this inability of inviscid flows to model the diffusion/advection ofvorticity generated at boundaries out into the body of the flow that causes most ofthe failures of the model.
In inviscid flows we are left with a free slip velocity at the boundaries which wemay interpret as a thin vortex sheet wrapped around the boundary.
5.3.1 Vortex sheets
Consider a thin layer of thickness δ in which the vorticity is large and is directedalong the layer (parallel to 0y), as sketched. The vorticity is
where ∂u/∂z is large (but not ∂w/∂x, which would lead to very large w). We cansuppose that within the vortex layer
u = u0 + ωz
changing from u0 to u0 + ωδ between z = 0 and δ, with mean vorticity
η =(u0 + ωδ) − u0
CHAPTER 5. THE VORTICITY FIELD 48
This vortex layer provides a sort of roller action, though it is not of course rigid, andit also suffers high rate-of-strain.
If we idealize this vortex layer by taking the limit δ → 0, ω → ∞, such that ωδremains finite, we obtain a vortex sheet, which is manifest only through the free slipvelocity. Such vortex sheets follow the contours of the boundary and clearly maybe curved. They are infinitely thin sheets of vorticity with infinite magnitude acrosswhich there is finite difference in tangential velocity.
5.3.2 Line vortices
We can represent approximately also strong thin vortex tubes (e.g. tornadoes, wa-terspouts, draining vortices) by vortex lines without thickness. The circulation in acircuit round the tube tends to a definite non-zero limit as the circuit area (S) →zero. If the flow outside the vortex is irrotational then all circuits round the vortexhave the same circulation, the strength κ of the vortex:∮
u · dr → κ as C → 0.
As a consequence, the velocity → ∞ as the line vortex is approached, like κ ∝(distance)−1.
The effect of viscosity is to thicken vortex sheets and line vortices by diffusion;however, the effect of diffusion is often slow relative to that of advection by the flow,and as a result large regions of flow will often remain free from vorticity. Vortexsheets at surfaces diffuse to form boundary layers in contact with the surfaces; orif free they often break up into line vortices. Boundary layers on bluff bodies oftenseparate or break away from the body, forming a wake of rotational, retarded flowbehind the body, and it is these wakes that are associated with the drag on the body.
CHAPTER 5. THE VORTICITY FIELD 49
5.3.3 Motion started from rest impulsively
Viscosity (which is really just distributed internal fluid friction) is responsible forretarding or damping forces which cannot begin to act until the motion has started;i.e. take time to act. Hence any flow will be initially irrotational everywhere except atactual boundaries. Within increasing time, vorticity will be diffused form boundariesand advected and diffused out into the flow.
Motion started from rest by an instantaneous impulse must be irrotational. For,if we integrate the Euler equation over the time interval (t, t+ δt)∫ t+δt
F dt−∫ t+δt
F dt− 1
p dt .
In the limit δt→ 0 for start-up by an instantaneous impulse, the impulse of the bodyforce → 0 (as the body force is unaffected by the impulsive nature of the start) and
u − u0 = −1
where the fluid responds instantaneously with the impulsive pressure field P =∫ δtp dt, and the impulse on a fluid element is −∇P per unit volume, producing
a velocity from rest of
u0 = −1
This is irrotational as
∇× v = −1
ρ∇× (∇P ) ≡ 0.
Two dimensional flow of ahomogeneous, incompressible,inviscid fluid
In two (x, z) dimensions, the Euler equations of motion are
and the continuity equation is
∂z= 0. (6.3)
The vorticity ω has only one non-zero component, the y-component, i.e., ω =(0, η, 0), where
Taking (∂/∂z) (6.1) −(∂/∂x) (6.2) and using the continuity equation we can showthat
Dt= 0. (6.5)
This equation states that fluid particles conserve their vorticity as they movearound. This is a powerful and useful constraint. In some problems, η = 0 for allparticles. Such flows are called irrotational.
Consider, for example, the problem of a steady, uniform flow U past a cylinderof radius a. All fluid particles originate from far upstream (x → −∞) where u = 0,
CHAPTER 6. TWO DIMENSIONAL FLOW OF A HOMOGENEOUS, INCOMPRESSIBLE, INVIS
w = 0, and therefore η = 0. It follows that fluid particles have zero vorticity for alltime.
The inviscid flow problem can be solved as follows. Note that the continuityequation (6.3) suggests that we introduce a streamfunction ψ, defined by the equa-tions
∂z, w = −∂ψ
Then Eq. (6.3) is automatically satisfied and it follows from (6.4) that
In the case of irrotational flow, η = 0 and ψ satisfies Laplaces equation:
∂z2= 0. (6.8)
Appropriate boundary conditions are found using (6.6). For example, on a solidboundary, the normal velocity must be zero, i.e., u · n = 0 on the boundary. Ifn = (n1, 0, n3), it follows using (6.6) that n1
= 0, or n ∧ ∇ψ = 0 on theboundary. We deduce that ∇ψ is in the direction of n, whereupon ψ is a constanton the boundary itself.
Let us return to the example of uniform flow past a cylinder of radius a: seediagram below.
The problem is to solve Eq. (6.8) in the region outside the cylinder (i.e. r > a)subject to the boundary condition that
CHAPTER 6. TWO DIMENSIONAL FLOW OF A HOMOGENEOUS, INCOMPRESSIBLE, INVIS
∂z, 0, −∂ψ
)→ (U, 0, 0) as r → ∞, (6.9)
u · n = 0 on r = a, (6.10)
where r = (x2 + y2)1/2
. For this problem it turns out to be easier to work in cylin-drical polar coordinates centred on the cylinder.
It is easy to check that the solution of (6.8) satisfying (6.9) and (6.10) is
ψ = U
(r − a2
)sin θ. (6.11)
Note that for large r, ψ ∼ Ur sin θ = Uz, whereupon u = ∂ψ/∂z ∼ U as required.
and z = r sin θ ⇒ ∂r/∂z = 1/ sin θ and 1 = r cos θ ∂/∂z, whereupon
r cos θ
and x = r cos θ ⇒ ∂r/∂x = 1/cosθ and 1 = −r sin θ ∂/∂x, whereupon
r sin θ
∂ θ. (6.13)
The boundary condition on the cylinder expressed by (6.10) requires that
∂zcos θ − ∂ψ
∂xsin θ = 0
CHAPTER 6. TWO DIMENSIONAL FLOW OF A HOMOGENEOUS, INCOMPRESSIBLE, INVIS
at r = a and for all θ and, using (6.12) and (6.13), this reduces to
∂ θ= 0 at r = a. (6.14)
This equation implies that ψ is a constant on the cylinder; i.e., the surface of thecylinder must be a streamline. Substitution of (6.11) into (6.14) confirms that ψ ≡ 0on the cylinder.
It remains to show that ψ satisfies (6.8). To do this one can use (6.12) and (6.13)to transform (6.8) to cylindrical polar coordinates; i.e.,
∂ z2= 0. (6.15)
It is now easy to verify that (6.11) satisfies (6.15) and is therefore the solution forsteady irrotational flow past a cylinder. Note that the solution for ψ is unique onlyto within a constant value; if we add any constant to it, it will satisfy equation (6.8)or (6.15), but the velocity field would be unchanged.
It is important to note that we have obtained a solution without reference tothe pressure field, but the pressure distribution determines the force field that drivesthe flow! We seem, therefore, to have by-passed Newton’s second law, and haveobviously avoided dealing with the nonlinear nature of the momentum equations(6.1) and (6.2). Looking back we will see that the trick was to use the vorticityequation, a derivative of the momentum equations. For a homogeneous fluid, thevorticity equation does not involve the pressure since ∇∧∇p ≡ 0. We infer from thevorticity constraint [Eq. (6.7)] that the flow must be irrotational everywhere and usethis, together with the continuity constraint (which is automatically satisfied whenwe introduce the streamfunction) to infer the flow field. If desired, the pressure fieldcan be determined, for example, by integrating Eqs. (6.1) and (6.2), or by usingBernoulli’s equation along streamlines.
Now the solution itself. The streamline corresponding with (11) are sketched inthe figure overleaf. Note that they are symmetrical around the cylinder. ApplyingBernoulli’s equation to the streamline around the cylinder we find that the pressuredistribution is symmetrical also so that the total pressure force on the upstream side
CHAPTER 6. TWO DIMENSIONAL FLOW OF A HOMOGENEOUS, INCOMPRESSIBLE, INVIS
of the cylinder is exactly equal to the pressure on the downwind side. In other words,the net pressure force on the cylinder is zero! This result, which in fact is a generalone for irrotational inviscid flow past a body of any shape, is known as d’Alembert’sParadox. It is not in accord with our experience as you know full well when you tryto cycle against a strong wind. What then is wrong with the theory? Indeed, whatdoes the flow round a cylinder look like in reality? The reasons for the breakdownof the theory help us to understand the limitations of inviscid flow theory in generaland help us to see the circumstances under which it may be applied with confidence.First, let us return to the viscous theory.
The Navier-Stokes’ equation is the statement of Newton’s second law of motionfor a viscous fluid. It reads
ρ∇p+ ν∇2u. (6.16)
The quantity of ν is called the kinematic viscosity. For air, ν = 1.5×10−5 m2 s−1;for water ν = 1.0 × 10−6 m2 s−1. The relative importance of viscous effect ischaracterized by the Reynolds’ number Re, a nondimensional number defined by
where U and L are typical velocity and length scales, respectively. The Reynolds’number is a measure of the ratio of the acceleration term to the viscous term in(6.16). For many flows of interest, Re >> 1 and viscous effects are relatively unim-portant. However, these effects are always important near boundaries, even if only ina thin “boundary-layer” adjacent to the boundary. Moreover, the dynamics of thisboundary layer may be crucial to the flow in the main body of fluid under certaincircumstances. For example, in flow past a circular cylinder it has important con-sequences for the flow downstream. The observed streamline pattern in this case atlarge Reynolds numbers is sketched in the figure overleaf. Upstream of the cylinderthe flow is similar to that predicted by the inviscid theory, except in a thin viscous
CHAPTER 6. TWO DIMENSIONAL FLOW OF A HOMOGENEOUS, INCOMPRESSIBLE, INVIS
boundary-layer adjacent to the cylinder. At points on the downstream side of thecylinder the flow separates and there is an unsteady turbulent wake behind it. Theexistence of the wake destroys the symmetry in the pressure field predicted by theinviscid theory and there is net pressure force or form drag acting on the cylinder.Viscous stresses at the boundary itself cause additional drag on the body.
Boundary layers in nonrotatingfluids
We consider the boundary layer on a flat plate at normal incidence to a uniformstream U as shown.
The Navier Stokes’ equations for steady two-dimensional flow with typical scaleswritten below each component are:
and the continuity equation is
CHAPTER 7. BOUNDARY LAYERS IN NONROTATING FLUIDS 57
From the continuity equation we infer that since |∂u/∂x| = |∂w/∂z|, W ∼ UH/Land hence the two advection terms on the left hand sides of (7.1) and (7.3) are thesame order of magnitude: U2/L in (7.2) and (U2/L)(H/L) in (7.4). Now, for a thinboundary layer, H/L << 1 so that the derivatives ∂2/∂x2 in (7.1) and (7.3) can beneglected compared with ∂2/∂z2. Then in (7.1), assuming that the pressure gradientterm is not larger than both inertial or friction terms1, we have
The first two terms imply that
H ∼ L Re−1/2
where Re = UL/ν has the form of a Reynolds’ number. Alternatively, this ex-pression implies that the boundary thickness increases downstream like x1/2 [i.e.,H ∼ L1/2(ν/U)1/2]. Now from (7.4) we find that
ν= Re >> 1.
But if both the inertia terms and friction terms in (7.3) are much less than thepressure gradient term, the equation must be accurately approximated by
This implies that the perturbation pressure is constant across the boundary layer.It follows that the horizontal pressure gradient in the boundary layer is equal to thatin free stream.
Collecting these results together we find that an approximate form of the Navier-Stokes’ equations for the boundary layer to be
1Note that if this were not true, steady flow would not be possible as the large pressure gradientwould accelerate the flow further.
CHAPTER 7. BOUNDARY LAYERS IN NONROTATING FLUIDS 58
∂z= 0, (7.7)
and U = U(x) being the (possible variable) free stream velocity above the boundarylayer, Equations (7.6) and (7.7) are called the boundary layer equations.
7.1 Blasius solution (U = constant)
Equation (7.6) reduces to
and we look for a solution satisfying the boundary conditions u = 0, w = 0 at z = 0,u → U as z → ∞ and u = U at x = 0. Equation (7.7) suggests that we introduce astreamfunction ψ such that
whereupon ψ must satisfy the conditions ψ = constant, ∂/∂z = 0 at z = 0,ψ ∼ Uz as z → ∞ and ψ = Uz at x = 0. It is easy to verify that a solutionsatisfying these conditions is
∂z, w = −∂ψ
χ = (U/2νx)1/2 z, (7.10)
CHAPTER 7. BOUNDARY LAYERS IN NONROTATING FLUIDS 59
f(χ) satisfies the ordinary differential equation
f ′′ + ff ′ = 0, (7.11)
subject to the boundary conditions
f(0) = f ′(0) = 0; f ′(∞) = 1. (7.12)
Here, a prime denotes differentiation with respect to χ. It is easy to solve Eq.(7.11) subject to (7.12) numerically (see e.g. Rosenhead, 1966, Laminar BoundaryLayers, p. 222-224). The profile of f ′ which characterizes the variation of u acrossthe boundary layer thickness is proportional to χ and we might take χ = 4 ascorresponding with the edge of the boundary layer. Then (7.10) shows that thedimensional boundary thickness δ(x) = 4(2νx/U)1/2; i.e., increases like the squareroot of the distance from the leading edge of the plate. We can understand thethickening of the boundary layers as due to the progressive retardation of more andmore fluid as the fictional force acts over a progressively longer distance downstream.
Often the boundary layer is relatively thin. Consider for example the boundarylayer in an aeroplane wing. Assuming the wing to have a span of 3 m and that theaeroplane flies at 200 ms−1, the boundary layer at the trailing edge of the wing (as-
suming the wing to be a flat plate) would have thickness of 4 (2 × 1.5 × 10−5 × 3/200)1/2
=2.7×10−3 m, using the value ν = 1.5×10−5m2s−1 for the viscosity of air. The calcu-lation assumes that the boundary layer remains laminar; if it becomes turbulent, therandom eddies in the turbulence have a much larger effect on the lateral momentumtransfer than do random molecular motions, thereby increasing the effective valueof ν, possibly by an order of magnitude or more, and hence the boundary layerthickness.
Note that the boundary layer is rotational since ω = (0, η, 0), where η = ∂u∂z
,or approximately just −∂u/∂z.
7.2 Further reading
Acheson, D. J., 1990, Elementary Fluid Dynamics, Oxford University Press, pp406.
Morton, B. R., 1984: The generation and decay of vorticity. Geophys. Astrophys.Fluid Dynamics, 28, 277-308.
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math
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Definition of Carrying Amount
Examples of Carrying Amount
Here are some examples when the term carrying amount or carrying value is used:
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Adaptive confidence balls are constructed for individual resolution levels as well as the entire mean vector in a multiresolution framework. Finite sample lower bounds are given for the minimum expected squared radius for confidence balls with a prespecified confidence level. The confidence balls are centered on adaptive estimators
based on special local block thresholding rules. The radius is derived from an analysis of the loss of this adaptive estimator. In addition adaptive honest confidence balls are constructed which have guaranteed coverage probability over all of $\RR^N$ and expected squared radius adapting over a maximum range of Besov bodies.
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very difficult to understand, and it is not easy to learn the concepts from the book alone. All are available for free download in PDF after free registration. Look for a question similar to yours Under each topic you'll find a listing of questions related to the topic. It solved everything I trigonometry isosceles triangle calculator.
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CC-MAIN-2018-26
| 18,359 | 33 |
http://downloadingnowstarted.top/2018/04/08/statistics-for-beginners-pdf-download-22/
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math
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To test a statistical hypothesis, you take a sample, collect data, form a statistic, standardize it to form a test statistic so it can be interpreted on a standard scaleand decide whether the test statistic refutes the claim. When designing a study, the sample size is an important consideration because the larger the sample size, the more data you have, and the more precise your results will be assuming high-quality data.
Statistics For Dummies Cheat Sheet. Quantitative Analysis is written by an experienced mathematics teacher, this e-book is presented in tutorial fashion as if a tutor was sitting next to you.
Are you using one of our books in a class? This series consists of six book on the elementary part of the theory of real functions in one variable. This eBook explains statistical concepts. Please share with your friends, let’s read it!! This first fog contains over solved beginhers and exercises.
You use hypothesis tests to challenge whether some claim about a population is true for example, statistkcs claim that 40 percent of Americans own a cellphone. In this book we present a collection of examples of applications of the theory of Fourier series. This series consists of six books on the elementary part of Linear Algebra.
There are many books concerned with statistical theory. In every case, the files were especially designed to demonstrate some specific capability of SPSS. Think Stats emphasizes simple techniques you can use to explore real data sets and answer interesting questions.
This is not one of them. tor
Medical Statistics: For Beginners – Free eBooks Download
This book, together with the linked YouTube videos, reviews a first course on differential equations. How spread out is the data? Understanding Formulas for Common Statistics After data has been collected, the first step in analyzing it is to crunch out some descriptive statistics to get a statistica for the data.
The book is an introduction to advanced demographic techniques and has been designed for researchers and students of demography and statistics as well as social scientists and health workers. Some mathematicians including the author of this book believe that counterexamples will guide the reader towards a better understanding of the underlying theory.
This is an introduction to linear algebra. Essential Group Theory is an undergraduate mathematics text book introducing the theory of groups.
Introductory Algebra is a primer for students considering an entrance level college algebra course. We have detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading.
This book introduces you to the basics of Matlab without requiring any previous experience of programming. Topics progress from the beginnrrs of mathematical proof, to groups, fields, and then rings. If you know nothing you get a basic knowledge, if you know some statistical methods you get a better understanding of the ideas behind them. Finally, a technical note: Medical Statistics Author by: A book with a collection of examples of how to solve linear differential equations with polynomial coefficients by the method of power series.
Stability, Riemann Surfaces, Conformal Mappings is one of the great eBooks available to download from our website.
This consists of the elementary aspects of linear algebra which depend mainly on row operations involving elementary manipulations of matrices. This book contains a brief description of general descent methods and a detailed study of Newton’s method and the important class of so-called self-concordant functions.
Statistics For Dummies Cheat Sheet
The monograph presents a generalization of the well-known Lyapunov function method and related concepts to the matrix function case within systematic stability analysis of dynamical systems. By taking advantage of the PMF and CDF libraries, it is possible for beginners to learn the concepts and solve challenging problems.
For your convenience, we have put all the books in this category into a zip file which you can download in one go. Calculus 4c-1, Systems of differential systems.
This book is an introduction to basic mathematics and is intended for students who need to reach the minimum level of mathematics required for their sciences, engineering and business studies. The goal of this textbook is to be a source for a first undergraduate course in abstract algebra.
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CC-MAIN-2018-17
| 4,437 | 18 |
https://academyofengineers.com/entropy-actually-mean/
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math
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Entropy : What this actually mean ??
Let we discuss today on this topic —
“Entropy is a function of a quantity of heat which shows the possibility of conversion of that heat into work. The increase in entropy is small when heat is added at a high temperature and is greater when heat addition is made at a lower temperature. Thus for maximum entropy, there is minimum availability for conversion into work and for minimum entropy there is maximum availability for conversion into work.”
In heat engine theory, the term entropy plays a vital role and leads to important results.
All heat is not equally valuable for converting into work. Heat that is supplied to a substance at high temperature has a greater possibility of conversion into work than heat supplied to a substance at a lower temperature.
Let us consider a system undergoing a reversible process from state 1 to state 2 along path L and then from state 2 to the original state 1 along path M. Applying the Clausius theorem to this reversible cyclic process, we have –
∮R dQ / T = 0
want to know in detail about Iron and Ferrous Materials
Let S1 = E n tropy at the initial state 1, and
S2 = E n tropy at the final state 2.
Then, the change in entropy of a system, as it undergoes a change from state 1 to 2, becomes
S2 – S1 = ∫21 (dQ / T)R
The third law of thermodynamics states “When a system is at zero absolute temperature, the entropy of system is zero”.
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CC-MAIN-2021-25
| 1,438 | 13 |
http://www.uwlax.edu/faculty/Okamoto/seminar.html
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math
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Sheldon Lee (Viterbo University)|
Random Walks and Continuum Coupling
A random walk can be thought of as a discrete version of the diffusion equation. We will begin with a basic overview of random walks, and their connection to the diffusion equation. Next, we discuss basic deterministic coupling, along with stability and convergence issues. Finally, we discuss how a random walk region can be coupled to a continuous region.
Rainbow Connectivity of Graphs -- An Introduction --
Isometries on the Bloch Space
Let X be a Banach space (a complete normed linear space). The problem of characterizing the isometries from X to X, that is the linear operators on X which preserve the norm, is open for most spaces. In this talk, I will discuss what is known of the isometries when X is taken to be a classical space, and in particular, when X is the Bloch space. Also, I will discuss current research in the characterization of isometries amongst specific types of operators. This talk will be accessible to a general audience.
Statistical Methods for Modeling 3-Dimensional Orientation Data
Orientation data are common in areas such as materials science and human kinematics. While some attention has been paid to methods for modeling orientations in the statistical literature, these methods suffer many limitations and are often difficult to apply in real-world problems. Motivated by a materials science problem, a new, useful class of distributions on orientations in 3 dimensions is developed. Statistical inference for this class is explored and application is made to the motivating problem. This talk will be accessible to a general audience.
Hamiltonian Line Graphs and Related Problems
Turing Instabilities in Reaction-Diffusion Equations: A Model for the Formation of Mammalian Coat Patterns
In this talk, I will discuss the analysis of the bifurcation structure of a reaction-diffusion equation with Thomas non-linearities, and the spatial patterns produced by what are called Turing instabilities. We use AUTO to construct the bifurcation structure for our system. We will discuss pattern formation in one and two dimensions. Also, we will present a visualization system developed to enhance AUTO. This is joint work with Evelyn Sander, Richard Tatum and Thomas Wanner.
Computing the Modular Chromatic Number for Trees
Michael A. Wodzak (Viterbo University)|
The Bard the Laird and the Lender
Shakespeare's play "The Merchant of Venice" was written at about the same time as John Napier was creating the Logarithm. We will look at the mathematical development of logarithms and see that those mathematical themes are actually echoed throughout the play.
|4/16 ||No meeting (MAA Wisconsin Annual Meeting at UW Oshkosh)
Chris Malone (Winona State University)|
Resequencing Topics in an Introductory Applied Statistics Course
The introductory applied statistics course taken by many thousands of undergraduate students has undergone a transformation over the past 25 years. Changes in what we teach, how we teach, and how we assess have impacted introductory statistics courses at institutions worldwide. In this article we shift focus from what we teach and how we teach to when we teach. We propose changes to the sequence in which core statistical concepts are presented in an introductory applied statistics course. The proposed ordering of topics repeats the sequence of descriptive summaries - probability theory - statistical inference several times throughout the course in various contexts.
|4/30 || TBA
|5/7 || TBA
Fall 2009 schedule
|Go to...||UW-L Math Dept Home|
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| 3,594 | 23 |
http://sst2012-s109maths.blogspot.com/2012/07/chap-6-error-analysis-homework-2-ex-62_3352.html
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math
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he cannot times the 6 with -8
There is a mistake at Line 2. At Line 2, RHS, y - 48 should be 2y - 96. Because we must multiply at everything by 12 as it is the LCM of the three denominators - 3, 4, 6.
There is a mistake at line 2 . 48 should be changed to 96 as the common denominator is 12.
There is a mistake from line 1 to 2, the error is in 12(y/6+-8)=y-48 it should be 12(y/6+-8)=2y-96 because both sides must be multiplied by the same amount
In Line 2, the LHS cannot be cross multiplied if the sign is and addition sign. He should change the denominators all to 12 first.
In line 2, y - 48 should be 2y - 96.
In line 2, y-48 should have been 2y-96 as he cross multiplied the LHS
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| 685 | 7 |
https://socratic.org/questions/how-do-you-find-the-sum-of-the-infinite-series-sigma-1-10-k-from-k-1-to-oo#493709
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math
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How do you find the sum of the infinite series #Sigma(1/10)^k# from k=1 to #oo#?
Find the common ratio, by calculating the first three terms.
This could have been seen from the summation expression.
The sum of a geometric series is:
This could also have been arrived at using the limit to
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| 288 | 5 |
https://internetoracle.org/digest.cgi?N=921
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math
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921-09 (cevEj dist, 3.3 mean)
Selected-By: "Leo L. Schwab" <[email protected]>
The Internet Oracle has pondered your question deeply.
Your question was:
And in response, thus spake the Oracle:
> [Beavis and Butthead enter the information age]
> Butthead: Hey Beavis -- check it out! I stole this e-mail address
> from Mr. Van Driessen's computer. It's for some guy named, like, the
> Oracle, or something. He can answer all of our questions about
> whatever we want. For free.
> Beavis: So?
> Butthead: Now all we need to do is ask him a question.
> Beavis: Let's ask him about chicks... like, how soon before I get
> Butthead: Naw. We already know you aren't gonna get any. We
> need a more difficult question.
> Beavis: Ummmm... heh-heh ... ummmmmmmm ... heh-heh ...
> Butthead: Uh, I've got one. Huh-huh-huh.
> Beavis: No way, buttknocker. I've got one.
> Butthead: Tough, @sswipe. I'm at the keyboard.
> [Butthead begins to type:]
> H ... O ... W M ... U ... C ... H W ... O ... O ... D ...
> Beavis: Heh-heh. Heh-heh. You said "wood".
> Butthead: Oh yeah, I did. Huh. Huh-huh-huh.
> Beavis: Do you think, like, this Oracle dude gets morning wood?
> Butthead: Uuuuuhh... I dunno. Depends on his, uh, relationship ...
> with, like, Lisa, or Michelle, or sumthin'.
> Beavis: Oh yeah. Heh-heh-heh.
> Butthead: He's the MAN.
> [Butthead continues typing:]
> ... W ... O ... U ... L ... D ...
> Beavis: Look Buttmunch, that's not how you spell "wood". I think
> it's, like, shorter.
> Butthead: Shut up Dilweed. It's not THAT "wood".
> [Typing again:]
> ... A W ... O ... O ... D ...
> Beavis: LOOK! LOOK! You said "wood" again! Heh-heh-heh-heh-heh.
> Butthead: Beavis, if you don't shut up, I'm gonna have to kick you in
> the nads.
> Beavis: No way, Butthead. I'm gonna kick YOU in the nads.
> [Typing again:]
> ... C ... H ... U ... C ... K C ... H ... U ... C ... K I ... F
> A W ... O ... O ... D ...
> Beavis: LOOK! LOOK! You did it again! Heh-heh-heh-heh-heh-heh!!!
> [Butthead kicks Beavis in the groin:] <CRACK!> [Beavis finally shuts
> [Typing again:]
> ... C ... H ... U ... C ... K C ... O ... U ... L ... D
> C ... H ... U ... C ... K W ... O ... O ... D?
> Beavis [wheezing]: Hhhhh-hhh-hhh. "Wood". Hhhhh-hhh-hhh. Cool.
} [Bart turns the TV off:] *click*
} Bart: I thought "Itchy and Scratchy" was on!
} Lisa: Bart, you *know* that's only on Thursdays. Besides, aren't
} you curious about this mystical Oracle being?
} Bart: No way, man. I mean, Beavis and Butthead need to get a life.
} And who cares about woodchucks?
} Lisa: Bart, the woodchuck, also known as the groundhog or
} Marmota monax, is a fascinating rodent closely related
} to the squirrel. It is also a burrowing animal,
} hibernating for several--
} Bart: Dad, make Lisa shut up!
} Homer: Lisa, shut up!
} Marge: Homer!
} Homer: Doh!
} [Doorbell rings, and Lisa opens the door:]
} Ned: Hey-dilly dey-dilly, neighbors!
} Bart and Homer: Doh!
} Ned: What'cha doin?
} Lisa: I was telling Bart about woodchucks.
} Ned: Woodchucks? What's a woodchuck?
} Lisa: The woodchuck, also known as the groundhog or Marmota monax,
} is a fascinating rodent--
} Marge: Lisa, shut up!
} Bart (to Homer): Yeah, man!
} [Krusty enters through the kitchen:]
} Krusty (muttering and stumbling): I can't find any more beer...
} Homer: [belches]. Not anymore. Hey! Why are you looking for my
} Krusty: Wha-- This isn't my house!
} [Krusty leaves through the front door.]
} Lisa: As I was saying, the woodchuck is a burrowing animal. In
} fact, the volume of an average woodchuck's burrow is 6860
} cubic inches, meaning that if a woodchuck could chuck as
} much wood as dirt for the burrow, a woodchuck could chuck--
} [Oracle turns the TV off:] *click*
} You owe the Oracle a better cable TV package.
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CC-MAIN-2021-21
| 3,770 | 83 |
http://visit-holstebro.com/unit-7-polygons-and-quadrilaterals-test-answer-key/
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math
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Unit 7 Polygons And Quadrilaterals Test Answer Key. Interior and exterior angles worksheet with answers pdf. Interior and exterior angles add up to 180 degree.
Ixl review properties of quadrilaterals geometry Polygons & quadrilaterals homework 5: Unit 7 polygons and quadrilaterals test answer key we looked at interior and exterior angle sums along with.
Unit 7 Polygons And Quadrilaterals Test Answer Key We Looked At Interior And Exterior Angle Sums Along With.
Unit 7 test study guide. What is the formula to find the measure of each Unit 7 test study guide (polygons & qpadrilaterals) topic 1:
Unit 7 Polygons And Quadrilaterals Test Answer Key.
Uv 8 and wx 5 2. A polygon with 4 sides. Unit 7 polygons and quadrilaterals homework 7 kites answer key trestans 10 march, 2021 edit unit 7 polid and quadrilaterals answers all things to lgebra / unit 7 polids and quadrilaterals homework 3 key response:
This Unit 7 Polygons And Quadrilaterals Homework 1 Angles Of Polygons Answer Key Class 12 Mcq Test Contains Random 30 Objective Questions.
At the end of this unit you will be able to: If the sum of the interior angles of a polygon name: No, because some policies may have more than 4 sides all policies 4 sides it's quadrilotes.
Read And Learn The Chapter
Unit 4 test study guide congruent triangles answer key pdf the chapter 15 of class 9 maths concise selina textbook helps students in learning the construction of different polygons. Geometry unit 2 test review answer key. What is the formula for the sum of the interior angles of a polygon?
Module 5 Answer Key By Nrweg3 22959.
From quadrilaterals to parallelograms, rectangles, squares, trapezium etc. A polygon is a shape that has no curves. Geometry chapter 6 practice test wordpress com.
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http://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Krull's_principal_ideal_theorem
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Krull's principal ideal theorem
In commutative algebra, Krull's principal ideal theorem, named after Wolfgang Krull (1899–1971), gives a bound on the height of a principal ideal in a commutative Noetherian ring. The theorem is sometimes referred to by its German name, Krulls Hauptidealsatz (Satz meaning "proposition" or "theorem").
Precisely, if R is a Noetherian ring and I is a principal, proper ideal of R, then each minimal prime ideal over I has height at most one.
This theorem can be generalized to ideals that are not principal, and the result is often called Krull's height theorem. This says that if R is a Noetherian ring and I is a proper ideal generated by n elements of R, then each minimal prime over I has height at most n.
The principal ideal theorem and the generalization, the height theorem, both follow from the fundamental theorem of dimension theory in commutative algebra (see also below for the direct proofs). Bourbaki's Commutative Algebra gives a direct proof. Kaplansky's Commutative ring includes a proof due to David Rees.
Proof of the principal ideal theorem
Let be a Noetherian ring, x an element of it and a minimal prime over x. Replacing A by the localization , we can assume is local with the maximal ideal . Let be a strictly smaller prime ideal and let , which is a -primary ideal called the n-th symbolic power of . It forms a descending chain of ideals . Thus, there is the descending chain of ideals in the ring . Now, the radical is the intersection of all minimal prime ideals containg ; is among them. But is a unique maximal ideal and thus . Since contains some power of its radical, it follows that is an Artinian ring and thus the chain stabilizes and so there is some n such that . It implies:
from the fact is -primary (if is in , then with and . Since is minimal over , and so implies is in .) Now, quotienting out both sides by yields . Then, by Nakayama's lemma, letting , we get that both sides are zero and , thus . Using Nakayama's lemma again, and is an Artinian ring; thus, the height of is zero.
Proof of the height theorem
Krull’s height theorem can be proved as a consequence of the principal ideal theorem by induction on the number of elements. Let be elements in , a minimal prime over and a prime ideal such that there is no prime strictly between them. Replacing by the localization we can assume is a local ring; note we then have . By minimality, cannot contain all the ; relabeling the subscripts, say, . Since every prime ideal containing is between and , and thus we can write for each ,
with and . Now we consider the ring and the corresponding chain in it. If is a minimal prime over , then contains and thus ; that is to say, is a minimal prime over and so, by Krull’s principal ideal theorem, is a minimal prime (over zero); is a minimal prime over . By inductive hypothesis, and thus .
- Matsumura, Hideyuki (1970), Commutative Algebra, New York: Benjamin, see in particular section (12.I), p. 77
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https://www.24houranswers.com/college-homework-library/Business/Finance/8953
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math
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This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.FACTS
(1) The call option covers 100 shares
(2) The option price is $310.25
(3) Stock price rises to $45
(4) Current stock price is $22.50
(5) You would pay$25 for the stock should you exercise the option
To calculate the pre-tax profit, you compute the profit you would make per share, and then deduct the amount you paid for the call option. Since you bought the call option when the price was $22.50, and you would pay $25 per share, and the stock price has now risen to $45, your profit per share is:
Profit per share = $45 - $25 = $20...
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http://wildaboutmath.com/2011/09/
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math
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After a hiatus of several months Dr. Tanton is making videos again. Here are two new ones.
Lulu has two children. You are told that at least one of her children is a boy who was born on a Tuesday. What is the probability that her other child is also a boy?
The answer will surprise you!
Here is a cute geometry puzzle: Imagine you are an archeologist and have come across just a small section of a rim of an ancient wheel. What size wheel did it come from?
This is a great puzzle to give to geometry students too. Hand out a picture of an arc of a circle and ask if is possible to find the measure of that arc using only basic tools - and them have students actually do it.
Math joke from Anna, the bartender and civil engineering student: an infinite number of mathematicians walk into a bar. The first one tells the bartender he wants a beer. The second one says he wants half a beer. The third one says he wants a fourth of a beer. The bartender puts two beers on the bar and says “You guys need to learn your limits.”
Hat tip to Algut Runeman at MathFuture.
Here's a great video by Vi Hart that shows a couple of proofs of the Pythagorean Theorem. Nothing new in terms of the proofs; they're ones I've seen often. What's clever about Vi's approach is that she uses paper-folding to create the elements of the proof. I like this kinesthetic approach to a couple of familiar proofs.
As always with Vi's exuberant videos, it would be nice to be able to watch them at half speed!
Review: Number-Crunching: Taming Unruly Computational Problems from Mathematical Physics to Science Fiction
Great stories. Interesting and challenging problems. Instructive MATLAB code. Lots of physics. That's my in-a-nutshell assessment of Princeton University Press's hot-off-the-press Number-Crunching: Taming Unruly Computational Problems from Mathematical Physics to Science Fiction.
Paul Nahin is a great story teller. Some of you might recall my review of an earlier book of Nahin's An Imaginary Tale where I noted Nahin's enjoyable writing style. Nahin has, in fact, written quite a number of books.
Nahin takes on the subject of using computers to solve difficult problems, many in physics, that couldn't be solved before computers. The publisher's page introduces some of the problems.
How do technicians repair broken communications cables at the bottom of the ocean without actually seeing them? What's the likelihood of plucking a needle out of a haystack the size of the Earth? And is it possible to use computers to create a universal library of everything ever written or every photo ever taken? These are just some of the intriguing questions that best-selling popular math writer Paul Nahin tackles in Number-Crunching. Through brilliant math ideas and entertaining stories, Nahin demonstrates how odd and unusual math problems can be solved by bringing together basic physics ideas and today's powerful computers. Some of the outcomes discussed are so counterintuitive they will leave readers astonished.
The Man of Numbers: Fibonacci's Arithmetic Revolution is Devlin's latest book.
"The Man of Numbers," at 156 pages (plus notes, bibliography, and index) and ten chapters is a fairly quick read. Leonardo of Pisa, also known as Fibonacci, is mostly only known for the Fibonacci sequence. Devlin shows us that there was much more to Fibonacci's life and that, in fact, Fibonacci played a very key role in the marketing of arithmetic in 1202 to the world of commerce in Western Europe through his book, Liber Abacci (The book of calculation.)
I'm not going to review the book chapter by chapter as you can find that kind of information on the web. NPR has a nice review, an excerpt from the book, and an audio interview with Mr. Devlin. ScienceNews.org has a review and Amazon.com has several reviews. But, I will point out some items of particular interest.
Welcome to the September 2, 2011 edition of carnival of mathematics.
This is the 81st edition. In the tradition of the Carnival of Mathematics, we provide trivia on the number of the edition.
- 81 is 3^4 and also 9^2.
- The awesome card get, Set, contains 81 cards.
- 81 is the square of the sum of its digits. Thanks to this site.
- 81 is a heptagonal number and a 28-gonal number.
- There are 81 stable chemical elements.
Some more trivia about the number 81 appears here.
That concludes this month's Carnival of Mathematics. Oops, we've not mentioned our submissions. There are lots this month. Here they are ...
Mike Croucher, owner of the Carnival of Mathematics, presents A retrospective of 4 years of mathematical articles at WalkingRandomly. Happy Birthday, WalkingRandomly! I'm a big fan of Mike's blog and I discovered a bunch of neat articles among his most popular.
Mike Croucher also nominated these two articles:
Martin Cohen is blogless yet he wants to share his paper, Exceedingly Elementary Proofs That a^(1/n) -> 1, n^(1/n) -> 1, and (1+1/n)^n -> e. I offered a home for the paper. I created a PDF and have placed it here.
Martin would love your feedback on the paper. Please leave comments here or contact him at the email address in the paper.
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