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http://mathhelpforum.com/calculus/15334-volume-solid-print.html | # volume of a solid
• May 24th 2007, 04:47 PM
viet
volume of a solid
The region bounded by $y = e^{-x^{2}}, y = 0, x = 1,$ and is revolved about the y-axis. Find the volume of the resulting solid.
im not quite sure how to do this one :confused:
• May 24th 2007, 05:12 PM
Jhevon
Quote:
Originally Posted by viet
The region bounded by $y = e^{-x^{2}}, y = 0, x = 1,$ and is revolved about the y-axis. Find the volume of the resulting solid.
im not quite sure how to do this one :confused:
are you sure those are the right limits? as it is, the area is not bounded. maybe it should be $x=0$ and $x = 1$ and the $x$-axis?
• May 24th 2007, 07:19 PM
ThePerfectHacker
I think you want unbounded area.
Consinder a long interval $[0,N]$.
Then the volume is given by,
$2\pi \int_0^N xe^{-x^2} dx$
To find the unbounded area (which is actually bounded).
Find,
$2\pi \int_0^{\infty} xe^{-x^2} dx$
Hint: Use $t=-x^2$.
• May 24th 2007, 07:28 PM
viet
sorry i seem to forgot something, the correct question is:
The region bounded by $y = e^{-x^{2}}, y = 0, x = 0, x = 1$ and is revolved about the y-axis. Find the volume of the resulting solid.
• May 24th 2007, 07:31 PM
Jhevon
1 Attachment(s)
Quote:
Originally Posted by viet
sorry i seem to forgot something, the correct question is:
The region bounded by $y = e^{-x^{2}}, y = 0, x = 0, x = 1$ and is revolved about the y-axis. Find the volume of the resulting solid.
that's what i thought you meant to say, thought it probably would be cool to do it TPH's way if you didn't make a typo
We proceed by the Method of Cylindrical Shells:
$V = 2 \pi \int_{0}^{1} xe^{-x^2} dx$
Let $u = -x^2$
$\Rightarrow du = -2x dx$
$\Rightarrow - \frac {1}{2} du = x dx$
So our integral becomes:
$V = - 2 \pi \cdot \frac {1}{2} \int_{x=0}^{x=1} e^u du$
$\Rightarrow V = - \pi \left[ e^u \right]_{x=0}^{x=1} = - \pi \left[ e^{-x^2} \right]_{0}^{1}$
$\Rightarrow V = \pi - \frac { \pi}{e}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754257202148438, "perplexity": 699.2911496974438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463108.89/warc/CC-MAIN-20150226074103-00086-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://www.gravityquantum.com/ | Geometría Cuántica del Espaciotiempo.htm Quanten Geometrie der Raumzeit.htm La Géométrie Quantique de l'Espace-temps.htm Квантовая геометрия пространства Время.htm Quantum Geometry of the Spacetime.htm
Quantum Geometry of the Space-time
Model of Unification of General Relativity and Quantum Mechanics
Authors: Rafael Javier Martínez Olmo & Pilar Ávila Barredo (Piska)
E-mail: [email protected]
Vigo
Spain
Vigo, Monday, 2 July 2012
Certainty: “Physically nothing is infinite.
Not even human stupidity is infinite, although it seems so, is not physically possible.”
Physics and the Infinite
Hypothesis.
The purpose of this work is to unify Quantum Mechanics and the General Theory of Relativity, eliminating the randomness in Quantum Mechanics through a model of "hidden variables", geometric of curvature variables "Op", replacing the set of the Real numbers R[1] by the set of the "Real-Natural" numbers R(N) [2] which would help the physical description of nature; under the hypothesis that there is a natural physical unit of Spacetime (VTN), tetra dimensional, elementary, indivisible and the smallest one.
Max Planck devised a dimensional procedure to determine the absolute units of the Nature, since they are obtained from universal constants. As the Universe is four-dimensional we seek the unique combination of constants which gives us a Volume in Time (VT) = G • h • c-2 = cm+3 • sc-1.
The hypothesis is that this volume-time is the basic atom of spacetime.
VTN= G • h • c-2 = 4,9205 ·10E-55 cm+3· sg-1.[3]
[1] The Real number (R) always has an infinite number of decimal numbers.
[2] The Real-Natural number R(N) always has a finite number of decimal numbers.
[3] Fundamental physical constants, (G) of gravity, (h) Planck's constant (c) speed of light.
Introduction.
For centuries the hypothesis that the universe is continuous that has been considered a clear and unquestionable truth. Not only matter and energy have been considered so but also the proper spacetime which contains them.
With the birth of quantum mechanics, we understood that matter is not continuum [represented by the set of real numbers (R)]. For example, an iron bar cannot be divided into smaller pieces and these in turn into smaller ones and so on. If we kept dividing it only an atom of iron would remain and this one cannot be divided in “1/n atoms of iron ".
However, since an atom is not elementary, we can even split an atom of iron. (In the context of this work ‘elementary’ means that it has no parts and consequently is indivisible).
The atom can be divided in protons, neutrons and electrons. There is a strong assumption that protons and neutrons are not elementary. They have an internal structure composed of three quarks. At present, there is no one who has raised the hypothesis that a quark is infinitely divisible.
The electron does not have an internal structure (it has no parts) and is indivisible. We can say it is elementary, in the sense expressed in this work. Quantum Mechanics also ensures that energy is emitted and absorbed in the form of all discreet Quantum [represented only by the set of Natural numbers (N)].
We can affirm that the hypothesis of a continuum universe is neither fulfilled with matter nor with energy. When I use the term “continuous”, “continuum” or “continuity”, I mean that the physical reality of the phenomenon to describe can only be represented by the set of the Real numbers (R). Here a question arises: What would happen if spacetime was not continuum? What will happen if spacetime cannot be represented by (R)?
Geometry of the Spacetime
Foreword.
We know that the two fundamental theories on our understanding of the universe, General Relativity and Quantum Mechanics, are not right, since they describe two antagonistic worlds, therefore one of them or perhaps both must be wrong.
The cause for the infinites to appear when we join General Relativity and Quantum Mechanics without the possibility of being eliminated (General Relativity is not renormalizable), has its origin in the use of the set (R) to describe physical nature.
This is in my opinion a fundamental and previous mathematical reason, which must be solved to achieve the unification of both models, replacing the set of the Real numbers (R) by the set of the Real-Natural numbers R(N) to represent physical reality.
We use the set of Real numbers, for both theories, in fact for all Physics. This is completely wrong, absurd and unnecessary, since it introduces mathematical concepts into Physics. These are not measurable (Physical) such as moment, point, infinitely small, infinite. We should remember that (R) is an “actual infinite” set (Georg Cantor), that is to say that it fulfils the following:
1. Its elements lack ordinal number, that is, an number (N) associated with each element of (R), which would indicate your order number according to structure of construction. This property implies that there is no number R previous or posterior to any other R number.
2. The use of (R) implies that any physical measurement, such as distance, is infinitely divisible. It means that a distance can take any value.
3. In (R) the part is equal to the whole.
All these properties, in my opinion, do not exist in nature. If we use the Real - natural numbers R(N) these three unnatural properties disappear. For this purpose, it is enough if we give up the idea that real numbers have infinite decimal places (that physically we do not need) and we replace it by R(N) which has a sufficiently large number of decimal places.
This choice (R) allows a physical variable to take any value, but it also implies that the value of the variable may be worth infinite (singularity) and be confined to a point that lacks dimensions.
Gravitational and electromagnetic forces are inversely proportional to the square of the distance between the masses and the electrical particles. These forces increase when distance decreases. If distance is zero, the value of the Physical variable (gravitational force or electromagnetic force) is infinite (division by zero).
These are the infinites that prevent the unification, this is the error that we continue to make, an error in the essence of Logic.
This error appears through the mathematical procedure of abstraction, which causes geometric objects of smaller dimensions than those of Physical Nature to appear, that is to say four, three spatial dimensions and a temporary dimension. I mean the mathematical concepts of point, zero dimension, line (dimension one), plane (dimension two)... I also mean the mathematical concepts and physical meaning of zero and infinite.
This Error is in the ambiguity of the mathematical point, when applied to physical concepts as distance, speed, acceleration...
The concept of point is physically absurd, since it has two exclusive properties; it exists or does not exist.
It is absurd to think that physical objects exist and do not exist. However, the mathematical concept of point of dimension = 0, when it is applied to the physics presents this paradox:
The point exists (physically). Because it has a real and exact position in any system of spacetime reference which is determined by its four coordinates in space time. In fact the point is used to indicate a position in spacetime (R4).
The point does not exist (physically). Because it does not have dimensions (dimension = 0). It has no height, no length, no width, and does not exist in time. This incoherence is subtle, but clear after a brief reflection.
Physical Hypotheses and Prediction Model
The universe and all his contents has four dimensions.
Spacetime has four dimensions. Matter has always has three spatial dimensions (volume) and exists in time; energy always occupies a volume of space and exists in time, therefore both of them have four dimensions.
Absolute vacuum lacking both matter and energy does not exist, since there is no way to isolate the gravity of a “spacetime volume”.
Relative vacuum, which at least contains gravity, has therefore, four dimensions.
There is no physical example, (that exists and could be measured), of objects in our universe which has not four dimensions.
Following Euclides, it is not possible to construct, simply by adding (addition, sum), geometric objects of a different dimension from its construction elements, for example, it is not possible to build a line adding a large enough number of points.
If in Nature there are only four dimensions elements, the smallest and indivisible element of our universe must have four dimensions, that is, a volume that changes with time.
There are no objects in Nature of less than four dimensions; they only exist in the Platonic world of ideas as mathematical abstractions of reality.
To calculate the value of the lowest spacetime we use the fundamental constants of nature, c, h, y G, in the same way as Max Karl Ernst Ludwig Planck did a century ago.
We combined them looking in this case for a time volume, for example, a unit of volume during a unit of time, i.e. the volume of a box during an hour. This can be done with the following combination of fundamental constants, which is also unique:
G • h • c-2 = 4,920 551 532 644 910 • 10 -55 cm + 3 • sc -1
Smallest volume-time (without parts), indivisible, elementary.
As we know, the universe is isotropic, does not have favourite spatial directions, it presents the same appearance and properties in any directions. The distinctions we make between length, width and height are merely semantic, since we can share their names and they still represent the same physical reality.
If we associate the idea of isotropy to the elementary and lowest order of the universe, this can only be a sphere.
As a sphere it is easy to calculate the radio: Lmo = 4,8975E-19 centimeters minimum distance and any measurement of distance is equal to the Lmo product by a Natural number (N)
This distance is covered at the speed of light in Tmo = 1,6336E-29 seconds. This is the lowest time interval, and any measurement of time is equal to the product of Tmo by a Natural number (N).
As the distance is very small, there cannot exist a wavelength shorter than Lmo, 4,8975E-19 centimeters. As it is a the smallest wavelength it is the highest value of Emo energy = 253,1776TeV.
Any wavelength is equal to the product of Lmo by a Natural number (N). The value of any measurement of a quantity of energy is achieved dividing Emo by a Natural number (N).
First hundred levels of energy, predicted by the model, in color energy levels detected at the LHC (CERN). Maximum 7 TeV (1 st phase) and 14 TeV (2nd phase).
Table: Quantum´s Elementary Levels of Energy
Energy Level Energy in TeV E. Level Energy in TeV E. Level Energy in TeV E. Level Energy in TeV 1 253,177660585902 26 9,73760233 51 4,964267855 76 3,331285008 2 126,588830292951 27 9,376950392 52 4,868801165 77 3,288021566 3 84,392553528634 28 9,042059307 53 4,776936992 78 3,245867443 4 63,294415146476 29 8,730264158 54 4,688475196 79 3,204780514 5 50,635532117180 30 8,439255353 55 4,603230192 80 3,164720757 6 42,196276764317 31 8,167021309 56 4,521029653 81 3,125650131 7 36,168237226557 32 7,911801893 57 4,441713344 82 3,087532446 8 31,647207573238 33 7,672050321 58 4,365132079 83 3,05033326 9 28,130851176211 34 7,446401782 59 4,29114679 84 3,014019769 10 25,317766058590 35 7,233647445 60 4,219627676 85 2,978560713 11 23,016150962355 36 7,032712794 61 4,150453452 86 2,943926286 12 21,098138382159 37 6,842639475 62 4,083510655 87 2,910088053 13 19,475204660454 38 6,662570015 63 4,018693025 88 2,87701887 14 18,084118613279 39 6,491734887 64 3,955900947 89 2,844692816 15 16,878510705727 40 6,329441515 65 3,895040932 90 2,813085118 16 15,823603786619 41 6,175064892 66 3,83602516 91 2,782172094 17 14,892803563877 42 6,028039538 67 3,778771054 92 2,751931093 18 14,065425588106 43 5,887852572 68 3,723200891 93 2,722340436 19 13,325140030837 44 5,754037741 69 3,669241458 94 2,693379368 20 12,658883029295 45 5,626170235 70 3,616823723 95 2,665028006 21 12,056079075519 46 5,503862187 71 3,565882543 96 2,637267298 22 11,508075481177 47 5,386758736 72 3,516356397 97 2,610078975 23 11,007724373300 48 5,274534596 73 3,468187131 98 2,583445516 24 10,549069191079 49 5,166891032 74 3,421319738 99 2,557350107 25 10,127106423436 50 5,063553212 75 3,375702141 100 2,531776606
Notice how the energy levels are closer to each other as the number of energy level increases. We can see that there are 21 levels of energy in the range of 3 TeV (from level 64 to 84 in blue) and only a level of 13 TeV (19 in red), according to the above, these discreet levels of energy (if we look for them) will become more evident in the second phase (yellow levels). I suggest the search of the level 19 as it is the only level that exists in the status of 13 TeV. These levels are so close to the energy levels in our daily life that they are not detectable and they seem to have the continuity that real numbers idealize. Remember that these energies are of quantum elementary events.
A century ago Max Planck took into account length, mass, time, electric charge and the temperature as fundamental elements to describe The Nature, using Coulomb's and Boltzmann's constants in addition to G, h, c.
These last two constants are not used in this work, since they are considered unnecessary for the description of Quantum Geometry (elementary = not divisible) of the Spacetime.
The difference between Max Planck's dimensional procedure and that established in this work, is in what we consider elementary. For Planck these were length, mass and time.
This can be argued about, against the idea accepted at present, as it is reflected in the following paragraph:
“The system measures several of the fundamental magnitudes of the universe: time, length, mass, electrical charge and temperature. Planck's units are often called (in joke) “God’s units” by physicists. This eliminates any anthropocentric arbitrariness of the system of units”.
Extracted from Wikipedia, Planck's units.
The Hypothesis of work Tetra Dimensional, considers a volume of “spacetime” to be the elementary magnitude (without parts, indivisible).
VT = [G • h • c-2]
In the following table we see the differences between both calculations.
Max Planck Hypothesis Tetra-Dimensional
Dimensional formulae Values(System c,g,s) Dimensional formulae Values(System c,g,s) =5.3912·10-44seconds tt = 1.63·10-29seconds =1.6162·10-33centimeters lt = 4.89·10-19centimeters =2.1764·10-5grams mt = 4.51·10-19 grams
The magnitudes calculated by Max Planck for length and time are extremely small and they involve quantities of inaccessible energy to our technology.
On the contrary the magnitudes calculated in this work are experimentally contrasted at the levels of energy of the current particle accelerators (14 TeV LHC; CERN).
The highest quantity of energy 253,177 TeV is only to two orders of magnitude from the previous particle accelerators, quantum geometry has probably been photographed, but it has remained unnoticed between the trillions of events and data obtained in the experiments of these big accelerators.
For the first time, Quantum Gravitation will be within reach of the experimental physicists, particularly those currently working at the CERN (LHC), which are those that will be able to determine if the prophecies of this work are in accordance with Nature or are wrong
Details
On the words and concepts that I use I would like to avoid any ambiguity, firstly because these terms are easily mistaken due to their colloquial use, secondly the smallest because this ambiguity is also implicit in the physical-mathematical concepts of continuum, infinite and vacuum (zero), which we handle in General Relativity and Quantum Mechanics, when we refer to Space or to Time (Quantum), or to Spacetime (Relativity).
1st This text is essentially mathematical. When I say 'point' I refer to its physical-mathematical concept, geometric object of zero dimension [0,0,0,0]. In this geometrical object it has neither height, width, nor depth and does not exist in time.
2nd When I use the term continuous, continuum or continuity, I mean that the physical reality of the phenomenon to describe, can only be represented by the set of Real numbers (R).
3rd For all matters concerning the concept of infinite, types of infinite sets, sets of Real numbers (R) and Natural numbers (N), I base on the work "The Theory of Transfinite Sets" by Georg Cantor.
Extract of the model.
If we understand the Differential calculus as a Physical Theory (reality), not as a mathematical model, which brings us near to that one infinitely, but without reaching it using the Mythological and physically absurd set of Real numbers (R), we will verify that that is the logical way that the elementary and quantum nature of Spacetime Geometry shows us. Its elementary components (without parts), are volumes in time (VT). They have four dimensions (D4) and are smallest (e > 0) but never zero , of curved topology (p).
They can only be represented by the set of natural numbers R(N).
This route based on the Differential calculus is what I call the Quantum Geometry of Spacetime.
The work tries to lay the foundations and theoretical essentials for this route, which unifies the experimental results of Quantum Mechanics with the theory of the General Relativity, by including the hidden variable Op (Model of hidden variables).
Op = spacetime curvature of an elementary quantum event.
The model is relational and independent from the system of reference (Quantum Relativity), where both the General Theory of Relativity and Quantum Mechanics are modified.
Neither the structure, nor the variables or the results of both models can be expressed as belonging to the set of the Real numbers (R), all these belong to the set of the numbers R(N).
General Relativity can be expressed as an elliptical geometry with spherical topology (p) and radial dynamics. Quantum Mechanics is completed introducing causality, replacing the statistical treatment (chance, randomness) by a geometric treatment (causal, determinist) since we include the variable Op.
This variable Op gives a causal explanation of Quantum Mechanics, since it establishes one to one (Bijection) connection, between the statistical results and the elementary geometries (without parts) of Spacetime that contain the above mentioned quantum events.
Likewise, it also determines the geometry (its form) and the metrics of Nature in its elementary, smallest or indivisible scale. It specifies a model of Quantum Relativity (to give geometrical form to Quantum Mechanics).
The attempts of unification of both theories, have always failed due to the appearance of the infinites. The origin of these indeterminations is in a mistaken reading of the "Calculus" as a consequence of the error of expressing the continuity from points of zero dimension (D = 0; R).
The set of real numbers (R) has allowed us to represent the physical reality to our scale, but it is not right in the description of the elementary or indivisible nature of "Spacetime"; this can only be described in elementary terms, by four dimensional geometric objects, Volume-time (VT) and only represented by the set of the natural numbers (N).
The term continuous is only opposed to discreet if we use the concept of mathematical point to create the continuum (D = 0; R) (infinite divisibility).
There are no physical singularities in Nature; they are the result of using the set of the real numbers out of context (quantum scale). There are therefore mathematical singularities (division by zero) and not physical.
Other Predictions
1. - Quantum Mechanics describes the universe when the radio of volume-times containing the event, stretches to 4,897506921037260 E-19 cm. lowest Lmo distance. At this distance of an elementary mass Mmo, the universe has the highest curvature, or equivalently, is the highest energy of a quantum event.
This wavelength corresponds at an energy level of 253,177660585902000 TeV. This level of energy has an asymptotic nature as unattainable or supreme. The highest energy of a quantum elementary event = 253,1776 TeV
2.-The hidden variables of Quantum Mechanics are in the geometry, a very small scale, of the spacetime that contains the quantum event. Up to date we have thought that the spacetime curvature on this scale was almost flat, therefore with very little influence. On the contrary, the Quantum Geometry places the origin of the curvature and therefore of geometry, in any particle with mass. Consequently, the highest curvature or maximum energy is at a Lmo smallest distance, of any "volume-time" occupied by an elementary mass (Mmo). The highest curvature is: 1/Lmo = 4,8975E+19 cm-1
3.-As we know from Feynman, the quantum electrodynamics (Q.E.D.) was characterised in its beginnings because all its results were infinite. The reason for this is that the sum of stories must bear in mind all the possible ways, and these ways depend on the distance between particles. This distance becomes zero; the zero introduces the infinite in the results.
Feynman determined not to take the calculations up to zero to avoid the indeterminacy, replacing it by a very small number 10-100 centimeters. and stopping the sums on stories on this value. This supposed a solution to indeterminations, but other problems arise. The uniqueness of the probability is lost and infinitesimal terms with negative energy appear.
These problems disappear, if instead of using a very small and arbitrary distance, we use the nature lowest distance. Lmo = 4,8975E-19 cm. to interrupt the sums on stories. Both problems appear when adding probabilities of nonexistent interactions, particularly all those calculated for distances lower than 4,8975E-19 centimeters.
4.-The tetra dimensional geometry at a quantum level behaves as if it were two-dimensional, (only two grades of freedom), since the radio determines three spatial dimensions and time the dynamic evolution. Moreover, the radio has the same ordinal that as time, (the tag of the spherical ring and the time since when it was radiated coincide).
History
Historically this incoherence has appeared before us three times. It is always concealed under the concept of continuity.
The first time that the incoherence showed itself was in the four sophisms by Zeno of Elea, 2.600 years ago. His logic is perfect, closed and conclusive. These sophisms were presented to the local philosophers in Athens, perhaps before the proper Socrates. We know that Aristotle and Plato knew the speech given by Zeno perfectly well.
The second time that the contradiction appeared again, was with the invention of the infinitesimal calculation. At that time many intellectuals were against the logic of such a theory, since there is no way of explaining the movement or its associated variables, speed and acceleration, in term of zero dimensional points.
The mathematical point does not have another point coming after it, neither a previous one. There is no second point neither a fifth one, no previous one to any given point. Because the set of the Real number (R) lacks an ordinal number associated to its elements. In spite of being a completely tidy set, it lacks good order. How can we explain the movement from the point A, to the point B, if we cannot go out of A because the following point does not exist?
This contradiction is implicit in the incoherence of the mathematical point (dimensional = zero) with which the mathematical "continuum" is created (R).
But since the calculation worked and it still works, and the subtle contradiction was not discovered, those who did not agree had to keep silent unwillingly.
Due to this, the movement and its associated magnitudes, speed and acceleration entered the club of the incoherence, since the set of the numbers Real (R) is continuous (by definition) but the infinitesimal Calculation cannot be demonstrated mathematically in terms of points, dimension = zero, it can only be demonstrated mathematically in terms of intervals (e > 0, dimension = 1).
The third time the incoherence appeared was with the birth of Quantum Mechanics (Beginning of Heisenberg uncertainty). The incoherent reality of the point concept went unnoticed again and divided Physics in two, introducing randomness in the quantum world.
Since then the universe possesses simultaneously two mutually exclusive properties: it is random and it is causal, depending on the size of the object of the Nature that we are studying. It is random if this one belongs to the microcosm and determinist if the object belongs to the macrocosm.
This model is finite, quantitative and predictive. It is possible to contrast experimentally at levels of energy between 1 and 14 TeV. The level 19=13,325 140 TeV is the only level of energy that exists between 13 and 14 TeV . The highest level of energy is 253,177 660 585 902 TeV. (Supreme value of energy of an Elementary Quantum event)
QUANTUM GEOMETRY OF THE SPACETIME Hypothesis Tetra-Dimensional of the Geometry of the Space-time (1 tetrahedron and 4 spheres) of a neutron, 1 up quark (blue) and 2 down quarks, (3 spheres with the center in three vertex of the tetrahedron), and his associate lepton, 1 sphere (clearer color) with his center in the quarter vertex. GRAVITY QUANTUM
Autor: Rafael Javier Martínez Olmo Quantum Geometry of the Space-time
Updated on Monday, 13 de Abril de 2015
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https://planetmath.org/ConjugateModule | # conjugate module
If $M$ is a right module over a ring $R$, and $\alpha$ is an endomorphism of $R$, we define the conjugate module $M^{\alpha}$ to be the right $R$-module whose underlying set is $\{m^{\alpha}\mid m\in M\}$, with abelian group structure identical to that of $M$ (i.e. $(m-n)^{\alpha}=m^{\alpha}-n^{\alpha}$), and scalar multiplication given by ${m^{\alpha}}\cdot r=(m\cdot\alpha(r))^{\alpha}$ for all $m$ in $M$ and $r$ in $R$.
In other words, if $\phi:R\to{\rm End}_{\mathbb{Z}}(M)$ is the ring homomorphism that describes the right module action of $R$ upon $M$, then $\phi\alpha$ describes the right module action of $R$ upon $M^{\alpha}$.
If $N$ is a left $R$-module, we define ${{}^{\alpha}N}$ similarly, with $r\cdot{{}^{\alpha}n}={{}^{\alpha}(\alpha(r)\cdot n)}$.
Title conjugate module ConjugateModule 2013-03-22 11:49:47 2013-03-22 11:49:47 antizeus (11) antizeus (11) 9 antizeus (11) Definition msc 16D10 msc 41A45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 24, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9742990136146545, "perplexity": 278.18186648372597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00203.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/a-transverse-harmonic-wave-travels-on-a-rope-according-to-the-following-expression-yxt-013-q3311720 | ## spring equation
A transverse harmonic wave travels on a rope according to the following expression: y(x,t) = 0.13sin(2.3x + 17.2t) The mass density of the rope is ? = 0.138 kg/m. x and y are measured in meters and t in seconds. A= 0.13 F= 2.737 Hz 3) What is the wavelength of the wave? 4) What is the speed of the wave? 5) What is the tension in the rope? 6) At x = 3.8 m and t = 0.48 s, what is the velocity of the rope? (watch your sign) 7) At x = 3.8 m and t = 0.48 s, what is the acceleration of the rope? (watch your sign) 8) What is the average speed of the rope during one complete oscillation of the rope? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9516571760177612, "perplexity": 438.8534871014845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368701638778/warc/CC-MAIN-20130516105358-00053-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathonline.wikidot.com/the-vector-space-of-the-solution-set-to-a-linear-homogeneous | The V.S. of the Sol. Set to a Lin. Homo. System of First Order ODEs
The Vector Space of the Solution Set to a Linear Homogeneous System of First Order ODEs
Recall from the Linear Homogeneous and Linear Nonhomogeneous Systems of First Order ODEs page that if $J = (a, b)$ and $A : J \to \mathbb{R}^{n \times n}$ is a continuous function then a linear homogeneous system of $n$ first order ODEs is of the form:
(1)
\begin{align} \quad \mathbf{x}' = A(t) \mathbf{x} \end{align}
We now prove a very important result which says that all of the solutions to the linear homogeneous system above forms an $n$-dimensional vector space.
Theorem 1: The set of all solutions to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t) \mathbf{x}$ forms an $n$-dimensional vector space with the operations of function addition $+$ and scalar multiplication $\cdot$ over the field $\mathbb{R}$.
• Proof: We break this proof up into three parts. In the first part we show that the set of solutions to $\mathbf{x}' = A(t)\mathbf{x}$ under the operations of function addition $+$ and scalar multiplication $\cdot$ is a vector space over the field $\mathbb{R}$.
• In the second part we show that this vector space has dimension of at least $n$.
• In the third part we show that this vector space has dimension of at most $n$.
• Let $V$ denote the solution set to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$.
• Part 1: Showing that $V$ is a vector space over $\mathbb{R}$. We first note that $V$ is a collection of continuous functions from $J$ to $\mathbb{R}^n$ and $C(J, \mathbb{R}^n)$ is itself a vector space over $\mathbb{R}$. Thus we show that $V$ is a vector subspace over $\mathbb{R}$.
• Clearly $0 \in V$ since $0$ is the trivial solution to $\mathbf{x}' = A(t)\mathbf{x}$.
• Now let $\alpha, \beta \in \mathbb{R}$ and let $\phi^{[1]}, \phi^{[2]} \in V$. Then:
(2)
\begin{align} (\alpha \phi^{[1]} + \beta \phi^{[2]})' &= \alpha \phi^{[1]'} + \beta \phi^{[2]'} \\ &= \alpha A(t) \phi^{[1]} + \beta A(t) \phi^{[2]} \\ &= A(t)[\alpha \phi^{[1]} + \beta \phi^{[2]}] \end{align}
• This show that $(\alpha \phi^{[1]} + \beta \phi^{[2]}) \in V$. So $V$ is closed under function addition $+$ and scalar multiplication $\cdot$. Therefore $V$ is a subspace of $C(J, \mathbb{R})$ and is itself a vector space.
• Part 2: Showing that $\dim V \geq n$.
• Let $\xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \in \mathbb{R}^n$ be such that $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ is a linearly independent set in $\mathbb{R}^n$. (It is possible to choose such vectors in $\mathbb{R}^n$ since $\dim \mathbb{R}^n = n$) Furthermore, $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ must be a basis of $\mathbb{R}^n$.
• Let $\tau \in J$. For each $i \in \{ 1, 2, ..., n \}$, the initial value problem $\mathbf{x}' = A(t)\mathbf{x}$ with $x(\tau) = \xi^{[i]}$ has a solution, call it $\phi^{[i]}$. So $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ is a collection of elements from $V$. We will show this is a linearly independent set of elements from $V$.
• Let $\alpha_1, \alpha_2, ..., \alpha_n \in \mathbb{R}$ and suppose that for all $t \in J$ we have that:
(3)
\begin{align} \quad \alpha_1\phi^{[1]}(t) + \alpha_2\phi^{[2]}(t) + ... + \alpha_n\phi^{[n]}(t) = 0 \end{align}
• Plugging in $t = \tau$ and we get:
(4)
\begin{align} \quad \alpha_1 \phi^{[1]}(\tau) + \alpha_2 \phi^{[2]}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) &= 0 \\ \quad \alpha_1 \xi^{[1]} + \alpha_2 \xi^{[2]} + ... + \alpha_n\xi^{[n]} &= 0 \end{align}
• Since $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ is a linearly independent set of vectors in $\mathbb{R}^n$ the above equation implies that $\alpha_1, \alpha_2, ..., \alpha_n = 0$. Therefore $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ is a linearly independent set in $V$. Therefore:
(5)
\begin{align} \quad \dim V \geq n \end{align}
• Part 3: Showing that $\dim V \leq n$.
• We use the sets $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ and $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ as before.
• Let $\psi \in V$. Then $\psi$ solves the linear homogeneous system $\mathbf{x}' = A(t) \mathbf{x}$. Furthermore, for the element $\tau \in J$ there is an element $\xi \in \mathbb{R}^n$ such that $\psi(\tau) = \xi$.
• Now since $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ is a basis of $\mathbb{R}^n$ there exists $\alpha_1, \alpha_2, ..., \alpha_n \in \mathbb{R}^n$ such that:
(6)
\begin{align} \quad \xi = \alpha_1\xi^{[1]} + \alpha_2\xi^{[2]} + ... + \alpha_n\xi^{[n]} \end{align}
• Let:
(7)
\begin{align} \quad \eta = \alpha_1\phi^{[1]} + \alpha_2\phi^{[2]} + ... + \alpha_n\phi^{[n]} \end{align}
• Note that $\eta \in V$ as it is a linear combination of elements in $V$. So $\eta$ solves the linear homogeneous system $\mathbf{x} = A(t) \mathbf{x}$ and furthermore:
(8)
\begin{align} \quad \eta(\tau) &= \alpha_1\phi^{[1]}(\tau) + \alpha_2\phi^{[2]}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) \\ &=\alpha_1\xi^{[1]} + \alpha_2\xi^{[2]} + ... + \alpha_n\xi^{[n]} \\ &= \xi \end{align}
• Since $\psi$ and $\eta$ are solutions to the IVP $\mathbf{x}' = A(t)\mathbf{x}$ with $x(\tau) = \xi$, we have by the uniqueness theorem that $\psi = \eta$. Therefore:
(9)
\begin{align} \quad \psi = \alpha_1\phi^{[1]} + \alpha_2\phi^{[2]} + ... + \alpha_n\phi^{[n]} \end{align}
• So every element $\psi \in V$ can be expressed as a linear combination of elements from $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$. So $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ spans $V$. Hence:
(10)
\begin{align} \quad \dim V \leq n \end{align}
• Conclusion: So $V$ is a vector space with $\dim V = n$. $\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 10, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999785423278809, "perplexity": 194.4610596447803}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948515165.6/warc/CC-MAIN-20171212041010-20171212061010-00584.warc.gz"} |
https://www.jiskha.com/questions/1429589/the-area-of-a-square-field-is-3-2-hectares-find-the-length-of-its-diagonal-in-metres | # math
The area of a square field is 3/2 hectares find the length of its diagonal in metres please help.
1. 👍
2. 👎
3. 👁
1. 3/2 hectarces = 1500 sq meters
diagnol= side*sqrt2=sqrt(1500) sqrt2
= sqrt(3000) meters= 10sqrt(30) meters
1. 👍
2. 👎
👤
bobpursley
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https://correacamila.com/code/ | Code
• SCOMMAH stands for Scatter in the COncentration- Mass relation and Mass Accretion History. It is a python module that relates the residual in halo concentration, ∆c200 (defined as the difference between halo concentration c200 and the median relation c(M200)), halo formation time, zf (defined as the redshift at which half the present-day halo mass was assembled), last time a halo experienced a major merger event, zlast,major, and the reference density, ρcrit, used in the halo mass definition. The relation is combined with a halo mass history, M(z), model to estimate halo formation times.
`>Scommah will be released soon!`
• COMMAH is a python module that follows the analytic model described in Correa et al. (2015a) to calculate the mass accretion history of a halo of mass $M_{0}$ at $z=0$ in any given redshift interval (e.g. M(z) between z=0 and 10). Also, COMMAH calculates halo concentrations following the semi-analytic model described in Correa et al. (2015c), and outputs the c−M relation at any given redshift. In addition, it also computes the dark matter accretion rate, the rms of the density field, peak height, and the integral of the NFW density profile, suitable for DM annihilation calculations. COMMAH calculates concentration solving a series of equations by performing a Levenberg-Marquardt method. COMMAH is suitable for any cosmology.
`>Learn how to use commah` | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9839568138122559, "perplexity": 3647.0517228632243}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583668324.55/warc/CC-MAIN-20190119135934-20190119161934-00429.warc.gz"} |
https://tcsmath.org/2010/06/19/random-partitions-of-metric-spaces/ | # Random partitions of metric spaces
In addition to the current sequence on Talagrand’s majorizing measures theory, I’m also going to be putting up a series of lecture notes about embeddings of finite metric spaces. The first few will concern random partitions of metric spaces, and their many applications.
Random partitions
Often when one is confronted with the problem of analyzing some problem on a metric space ${(X,d)}$, a natural way to proceed is by divide and conquer: Break the space into small pieces, do something (perhaps recursively) on each piece, and then glue these local solutions together to achieve a global result. This is certainly an important theme in areas like differential geometry, harmonic analysis, and computational geometry.
In the metric setting, “small” often means pieces of small diameter. Of course we could just break the space into singletons ${X = \{x_1\} \cup \{x_2\} \cup \cdots}$, but this ignores the second important aspect in a “divide and conquer” approach: what goes on at the boundary. If we are going to combine our local solutions together effectively, we want the “boundary” between distinct pieces of the partition to be small. Formalizing this idea in various ways leads to a number of interesting ideas which I’ll explore in a sequence of posts.
Example: The unit square
Consider the unit square ${[0,1]^2}$ in the plane, equipped with the ${\ell_1}$ metric. If we break the square into pieces of side length ${\delta/2}$, then every piece has diameter at most ${\delta}$, and a good fraction of the space is far from the boundary of the partition. In fact, it is easy to see that e.g. 60% of the measure is at least distance ${\delta/20}$ from the boundary (the red dotted line).
2
But there is a problem with using this as a motivating example. First, observe that we had both a metric structure (the ${\ell_1}$ distance) and a measure (in this case, the uniform measure on ${[0,1]^2}$). In many potential applications, there is not necessarily a natural measure to work with (e.g. for finite metric spaces, where counting the number of points is often a poor way of measuring the “size” of various pieces).
To escape this apparent conundrum, note that the uniform measure is really just a distraction: The same result holds for any measure on ${[0,1]^2}$. This is a simple application of the probabilistic method: If we uniformly shift the ${\delta/2}$-grid at random, then for any point ${x \in [0,1]^2}$, we have
$\displaystyle {\mathbb P}(x \textrm{ is } \delta/20\textrm{-far from the boundary}) \geq 0.6.$
Thus, by averaging, for any measure ${\mu}$ on ${[0,1]^2}$ there exists a partition where 60% of the ${\mu}$-measure is ${\delta/20}$-far from the boundary. This leads us to the idea that, in many situations, instead of talking about measures, it is better to talk about distributions over partitions. In this case, we want the “boundary” to be small on “average.”
Lipschitz random partitions
We will work primarily with finite metric spaces to avoid having to deal with continuous probability spaces; much of the theory carries over to general metric spaces without significant effort (but the development becomes less clean, as this requires certain measurability assumptions in the theorem statements).
Let ${(X,d)}$ be a finite metric space. If ${P}$ is a partition of ${X}$, and ${x \in X}$, we will write ${P(x)}$ for the unique set in ${P}$ containing ${x}$. We say that ${P}$ is ${\Delta}$-bounded if ${\mathrm{diam}(S) \leq \Delta}$ for all ${S \in P}$. We will also say that a random partition ${\mathcal P}$ is ${{\Delta}}$-bounded if it is supported only on ${\Delta}$-bounded partitions of ${X}$.
Let’s now look at one way to formalize the idea that the “boundary” of a random partition ${\mathcal P}$ is small.
A random partition ${\mathcal P}$ of ${X}$ is ${{L}}$-Lipschitz if, for every ${x,y \in X}$,
$\displaystyle \mathbb P\left(\mathcal P(x) \neq \mathcal P(y)\right) \leq L \cdot d(x,y).$
Intuitively, the boundary is small if nearby points tend to end up in the same piece of the partition. There is a tradeoff between a random partition ${\mathcal P}$ being ${\Delta}$-bounded and ${{L}}$-Lipschitz. As ${\Delta}$ increases, we expect that we can make ${{L}}$, and hence the “boundary effect,” smaller and smaller. The following theorem can be derived from work of Leighton and Rao, or Linial and Saks. The form stated below comes from work of Bartal.
Theorem 1
If ${(X,d)}$ is an ${n}$-point metric space and ${n \geq 2}$, then for every ${\Delta > 0}$, there is an ${\frac{O(\log n)}{\Delta}}$-Lipschitz, ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$.
We will prove a more general fact that will be essential later. For positive numbers ${a,b}$, define ${H(a,b) = \sum_{i=a+1}^b \frac{1}{i}}$ (note that ${H(a,a)=0}$). The next theorem and proof are from Calinescu, Karloff, and Rabani.
Theorem 2
For every ${\Delta > 0}$, there is a ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$ which satisfies, for every ${x \in X}$ and ${0 < r \leq \Delta/8}$,
Observe that Theorem 1 follows from Theorem 2 by setting ${r = d(x,y)}$, and noting that ${\mathcal P(x) \neq \mathcal P(y)}$ implies ${B(x,r) \nsubseteq \mathcal P(x)}$. We also use ${H(1,n) = O(\log n)}$ for ${n \geq 2}$.
Proof:
Suppose that ${|X|=n}$. Let ${\alpha \in [\frac14, \frac12]}$ be chosen uniformly at random, and let ${\pi}$ be a uniformly random bijection ${\pi : \{1,2,\ldots,n\} \rightarrow X}$. We will think of ${\pi}$ as giving a random ordering of ${X}$. We define our random partition ${\mathcal P}$ as follows.
For ${i \geq 1}$, define
$\displaystyle C_i = B(\pi(i), \alpha \Delta) \setminus \bigcup_{j=1}^{i-1} C_j.$
It is straightforward that ${\mathcal P = C_1 \cup C_2 \cup \cdots \cup C_n}$ is a partition of $X$, with perhaps many of the sets ${C_i}$ being empty. Furthermore, by construction, ${\mathcal P}$ is ${\Delta}$-bounded. Thus we are left to verify (1).
To this end, fix ${x \in X}$ and ${r > 0}$, and enumerate the points of ${X = \{y_1, y_2, \ldots, y_n\}}$ so that ${d(x,y_1) \leq d(x,y_2) \leq \cdots \leq d(x,y_n)}$. Let ${B = B(x,r)}$. We will say that a point ${y_i}$ sees ${B}$ if ${d(y_i, x) \leq \alpha \Delta + r}$, and we will say that ${y_i}$ cuts ${B}$ if ${\alpha \Delta - r\leq d(y_i, x) \leq \alpha \Delta + r}$.
(In the picture below, $y_1$ and $y_2$ see $B$, while $y_3$ does not. Only $y_2$ cuts $B$.)
Observe that for any ${y \in X}$,
Using this language, we can now reveal our analysis strategy: Let ${y_{\min}}$ be the minimal element (according to the ordering ${\pi}$) which sees ${B}$. Then ${B(x,r) \nsubseteq \mathcal P(x)}$ can only occur if ${y_{\min}}$ also cuts ${B}$. The point is that the fate of ${B(x,r)}$ is not decided until some point sees ${B}$. Then, if ${y_{\min}}$ does not cut ${B}$, we have ${B(x,r) \subseteq C_{\pi^{-1}(y_{\min})}}$, hence ${B(x,r) \subseteq \mathcal P(x)}$.
Thus we can write,
To analyze the latter sum, first note that if ${y \notin B(x,\Delta/2+r)}$, then ${y}$ can never see ${B}$ since ${\alpha \Delta \leq \Delta/2}$ always. On the other hand, if ${y \in B(x,\Delta/4-r)}$ then ${y}$ always sees ${B}$, but can never cut ${B}$ since ${\alpha \Delta \geq \Delta/4}$ always.
Recalling that ${r \leq \Delta/8}$ by assumption, and setting ${a=|B(x,\Delta/8)|}$ and let ${b=|B(x,\Delta)|}$, we can use (3) to write
Now we come to the heart of the analysis: For any ${i \geq 1}$,
The idea is that if ${y_i}$ cuts ${B}$, then ${\alpha \Delta \geq d(x,y_i)-r}$. Thus if any ${y_j}$ for ${j < i}$ comes before ${y_i}$ in the ${\pi}$-ordering, then ${y_j}$ also sees ${B}$ since ${d(x,y_j) \leq d(x,y_i)}$, hence ${y_i \neq y_{\min}}$. But the probability that ${y_i}$ is the ${\pi}$-minimal element of ${\{y_1, \ldots, y_i\}}$ is precisely ${\frac{1}{i}}$, proving (5).
To finish the proof, we combine (2), (4), and (5), yielding
$\displaystyle \mathbb P\left(B(x,r) \nsubseteq \mathcal P(x)\right) \leq \sum_{i=a+1}^b \mathbb P\left(y_i \textrm{ cuts } B\right) \cdot \mathbb P\left(y_i = y_{\min}\,|\, y_i \textrm{ cuts } B\right) \leq \frac{8r}{\Delta} \sum_{i=a+1}^b \frac{1}{i}\,.$
$\Box$
Tightness for expanders. Before ending this section, let us mention that Theorem 1 is optimal up to a constant factor. Let ${\{G_k\}}$ be a family of degree-3 expander graphs, equipped with their shortest-path metric ${d_k}$. Assume that ${|E(S, \bar S)| \geq c|S|}$ for some ${c > 0}$ and all ${|S| \leq \frac12 |G_k|}$. Let ${\Delta_k = \frac{\log |G_k|}{10}}$, and suppose that ${G_k}$ admits a ${\Delta_k}$-bounded ${{L}}$-Lipschitz random partition. By averaging, this means there is a fixed ${\Delta_k}$-bounded partition ${P}$ of ${G_k}$ which cuts at most an ${{L}}$-fraction of edges. But every ${S \in P}$ has ${\mathrm{diam}(S) \leq \Delta_k}$, which implies ${|S| < n/2}$. Hence the partition must cut an ${\Omega(1)}$ fraction of edges, implying ${L = \Omega(1)}$.
In the next post, we’ll see how these random partitions allow us to reduce may questions on finite metric spaces to questions on trees.
## 4 thoughts on “Random partitions of metric spaces”
1. Nirman says:
Maybe there is a typo : Should the condition of seeing be : $d(y_i,x) \leq \alpha\triangle + r$ and at least from the figure ($\alpha \triangle \geq r$) the condition for cutting appears to be
$\alpha \triangle – r \leq d(y_i,x) \leq \alpha \triangle + r$.
[Yes, thanks! Fixed. -j]
2. Anonymous says:
I wonder if fixing the previous typo might have broken a part of the proof. The proof says:
“if {y \notin B(x,\Delta/2)}, then {y} can never see {B} since {\alpha \Delta \leq \Delta/2} always.”
This seems not quite right. If {y \notin B(x,\Delta/2+r)} then I agree that {y} can never see {B}.
Also, the proof says:
“if {y \in B(x,\Delta/4)} then {y} always sees {B}, but can never cut {B} since {\alpha \Delta \geq \Delta/4} always”
This too seems not quite right. If {y \in B(x,\Delta/4-r)} then I agree that {y} always sees {B} but can never cut {B}.
3. OK, this is finally fixed. I also fixed the tree embedding analysis based on the updated theorem statement.
4. Son P. Nguyen says:
In the example of the unit square $$[0,1]^2$$. Let S be the set of all the points which distance at least $$\frac{\delta}{10}$$ from the boundary. Then S is a square with side $$\frac{\delta}{2}-\frac{\delta}{5}=\frac{3\delta}{10}$$. Thus, $$vol(S)=\frac{9\delta^2}{100}$$ which is only 36% of $$vol([0,1]^2)=\frac{\delta^2}{4}$$. I am sorry but I don’t see why vol(S) is at least 60% of that of [0,1]^2. Thank you.
[Fixed by changing it to $\delta/20$. –j] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 163, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9886021018028259, "perplexity": 223.5388131486141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00122.warc.gz"} |
http://clay6.com/qa/19230/an-object-of-mass-m-and-radius-r-is-performing-pure-rolling-motion-on-smoot | Browse Questions
# An object of mass M and Radius R is performing pure rolling. motion on smooth horizontal surface under the action of a constant force F. The object may be
$\begin{array}{1 1} a) disc \\ b) ring \\ c) solid \;cylinder \\ d) hollow\; sphere \end{array}$
$F_{cr}=ring$
$F= Ma_{cm}$--------(1)
$FR= I \alpha$
=> $F= MR \alpha$----------(2)
Only for a ring $acm= R \alpha$ condition will be satisfied
edited Jun 22, 2014 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8714665770530701, "perplexity": 823.1017665813969}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281151.11/warc/CC-MAIN-20170116095121-00089-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/65737/setting-the-filename-of-a-file-embedded-in-a-pdf-using-the-attachfile-package?answertab=votes | # setting the 'filename' of a file embedded in a PDF using the attachfile package
I'm using the attachfile LaTeX package (in PDFLaTeX) to embed files in a PDF and I've run into a difficulty regarding what I suspect is the 'filename' of the embedded file.
I can attach/embed files in the following manner:
\textattachfile[author=Zapp Brannigan, color=0 0 0]{my_directory/my_file.dat}{\detokenize{my_file.dat}}
The file seems to be embedded successfully in the resulting PDF and is accessible, but the file name suggested by the Evince PDF viewer when selecting to "Save Attachment As..." on the link for the attached file is the following:
my_directory/my_file.dat
So, you can see that it is including the path of the original file in this 'filename'. Obviously, there are difficulties when one tries to save the attached file under this suggested filename.
How might I go about changing this suggested filename to something such as my_file.dat? I've experimented and I've consulted documentation on the attachfile package and have not been able to come up with an answer.
Just as a quick note: I am disinclined to use a package other than attachfile because, in reality, my code is more complicated than that shown; I'm including custom icons for various file types, as permitted by the attachfile package. However, I'll keep an open mind.
Thank you very much for any thoughts you may have on the matter.
-
Moving my_file.dat one directory level upwards, i.e. \textattachfile[author=Zapp Brannigan, color=0 0 0]{my_file.dat}{\detokenize{my_file.dat}}, should solve the problem (if you are in a hurry), but of course this is no real solution. The attachfile2 package "is a further development of Scott Pakin’s package attachfile for pdfTEX. Apart from bug fixes, package attachfile2 adds support for dvips, some new options, gets and writes meta information data about the attached files." – Stephen Aug 3 '12 at 18:10
With Adobe Reader (on Mac OS X) I get my_file.dat without any reference to my_directory. – egreg Aug 3 '12 at 20:13
I haven't tried to get TeX to automatically parse the filename component for you (but if you're interested, I'm sure someone else could help with that), but the following patches the \textattachfile macro to require an extra argument, the name of the file as it should appear in the attachment, which has no relation to the name of the file on the hard disk when the PDF is created. That is, the syntax is now \textattachfile[options]{physicalfilepath}{text}{virtualfilename}.
\documentclass{article}
\usepackage{attachfile}
\makeatletter
\def\atfi@textattachfile@i#1#2#3#4{%
\setkeys{AtFi}{#1}%
\atfi@embedfile{#2}%
\def\atfi@textcolor(##1 ##2 ##3)##4{%
\textcolor[rgb]{##1,##2,##3}{##4}}%
\atfi@set@appearance{%
\expandafter\atfi@textcolor\expandafter
(\atfi@color@rgb){#3\strut}}%
\atfi@flags@to@int
\atfi@insert@file@annot{\detokenize{#4}}%
\endgroup
}
\makeatother
\begin{document}
% now have to pass an extra argument to \textattachfile, the filename you want the viewer to see (note that Adobe Reader appears to 'hide' the path component)
\textattachfile[author=Zapp Brannigan, color=0 0 0]{my_directory/my_file.dat}{\detokenize{my_file.dat}}{filename_seen_by_viewer.txt}
\end{document}
In your case, you would probably want to change filename_seen_by_viewer.txt to my_file.dat, but I left it like this to illustrate more clearly.
(Just grepping the PDF file for my_directory should be sufficient to indicate that it works.)
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.894385576248169, "perplexity": 2089.2132716485794}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007301.29/warc/CC-MAIN-20141125155647-00117-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/6/lesson/6.2.2/problem/6-72 | ### Home > CALC > Chapter 6 > Lesson 6.2.2 > Problem6-72
6-72.
$\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}y^\prime=0$
$\frac{2}{3}x^{-1/3}=-\frac{2}{3}y^{-1/3}y^\prime$
$y^\prime=-\frac{y^{1/3}}{x^{1/3}}$
$\frac{dy}{dx}=-\sqrt[3]{\frac{y}{x}}$
Look at the derivative equation.
Now look at the graph. There are two coordinate points in which y = 0. Does the derivative exist at those points? | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9321834444999695, "perplexity": 1446.04572201589}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00048.warc.gz"} |
http://mathhelpforum.com/trigonometry/183912-finding-all-angles-when-sec-3x-print.html | # Finding all of the angles when sec(3x)
• July 1st 2011, 03:12 PM
Zellator
Finding all of the angles when sec(3x)
Hi Forum
I have some ideas myself on this one, but I'd love to hear how you guys would proceed in this scenario.
Find the sum of all solutions in the interval $[0,2\pi)$ for the equation $sec(3x)=\sqrt2$
First, we can put this in the form $cos(3x)=\frac{1}{\sqrt2}$ for better viewing.
We know that $cos(x)=\frac{1}{\sqrt2}$ is $\frac{\pi}{4}$ and $\frac{7\pi}{4}$.
Then our period will equal $\frac{2\pi}{3}$, so the distance from one $\sqrt2$ to another is $\frac{\pi}{3}$
Since our $\sqrt2$ will appear 3 times faster because of the $3x$, our first $\sqrt2$ is $\frac{\pi}{12}$
Now, can we just go summing it to $\frac{\pi}{3}$ until we get to $2\pi$?
What do you guys think? :)
Thanks!
• July 1st 2011, 03:17 PM
mr fantastic
Re: Finding all of the angles when sec(3x)
Quote:
Originally Posted by Zellator
Hi Forum
I have some ideas myself on this one, but I'd love to hear how you guys would proceed in this scenario.
Find the sum of all solutions in the interval $[0,2\pi)$ for the equation $sec(3x)=\sqrt2$
First, we can put this in the form $cos(3x)=\frac{1}{\sqrt2}$ for better viewing.
We know that $cos(x)=\frac{1}{\sqrt2}$ is $\frac{\pi}{4}$ and $\frac{7\pi}{4}$.
Then our period will equal $\frac{2\pi}{3}$, so the distance from one $\sqrt2$ to another is $\frac{\pi}{3}$
Since our $\sqrt2$ will appear 3 times faster because of the $3x$, our first $\sqrt2$ is $\frac{\pi}{12}$
Now, can we just go summing it to $\frac{\pi}{3}$ until we get to $2\pi$?
What do you guys think? :)
Thanks!
I think you should find all the values of x that lie in the given interval and then add them up.
• July 1st 2011, 03:25 PM
pickslides
Re: Finding all of the angles when sec(3x)
Hi Zellator, you have the right idea
Quote:
Originally Posted by Zellator
First, we can put this in the form $cos(3x)=\frac{1}{\sqrt2}$ for better viewing.
We know that $cos(x)=\frac{1}{\sqrt2}$ is $\frac{\pi}{4}$ and $\frac{7\pi}{4}$.
$\cos(3x)=\frac{1}{\sqrt2}\implies 3x = \frac{\pi}{4},\frac{7\pi}{4}\implies x = \frac{\pi}{12},\frac{7\pi}{12}$
Now add $\frac{2\pi}{3}$ to these solutions, up to $2\pi$
After that add them up as suggested in post #2.
• July 1st 2011, 03:44 PM
Zellator
Re: Finding all of the angles when sec(3x)
Hey mr fantastic and pickslides!
It's great to know that I am on the right track, then.
Thanks for your second opinions, that was really appreciated :) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8217577338218689, "perplexity": 383.178806618914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768378.98/warc/CC-MAIN-20141217075248-00142-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/soapbox-racer-heavier-is-better.322614/ | # Soapbox racer - heavier is better?
1. Jun 30, 2009
### sir_manning
Hello all,
The company that I work for was involved in a soapbox competition over the weekend. During the design of the car, the consensus was that adding more weight inside the car (without affecting the shape) was a good thing. With more mass, the racer would have more momentum, so when the force of air resistance acted against it, the relative change in momentum would be less than for a lighter vehicle.
Anyways, we didn't make it to the finals, so now people are saying we should have made the car lighter, since adding more weight increases the amount of friction (neglecting air resistance). The analogy made was that four wheels roll down a hill are faster than four wheels on a car in neutral rolling down a hill. Personally, I'm not too sure about this one, because the air resistance on a car and the air resistance on four wheels is different. What about four heavier wheels vs. four lighter wheels both of the same dimensions on a windy day?
So what do you guys think? Is heavier better? I'm inclined to think that there is a happy medium - some amount of weight that doesn't increase the rolling friction too much but gives the car enough momentum so that it won't be stopped by a strong gust of wind.
Thanks!
2. Jun 30, 2009
### KLoux
I found this very interesting - I've never done any soapbox racing and haven't previously thought about it. Here's my best guess:
I assumed the soapbox was a block sitting on a slope inclined at some angle $$\theta$$ from horizontal. It starts from rest with only two forces acting on it - gravity and drag (and at the beginning when it's not moving, drag is zero). I assumed the drag force is proportional to the square of the velocity (with constant of proportionality k). If you do the force balance in the direction of the hill, taking downhill as positive, and solve for the acceleration, you get this:
$$a=g sin\left(\theta\right) - \frac{k v^2}{m}$$
So the only place that the mass appears is in the denominator of a negative term - this means that the heavier cars will accelerate down the hill faster than lighter cars.
If you want to find the top speed of the car down the hill, you can set the acceleration to zero (find the speed at which the drag and gravity balance out). You get this relationship:
$$v=\sqrt{\frac{m g}{k} sin\left(\theta\right)}$$
So the heavier cars actually have a higher top speed!
I think the wheels, however, should be as light as possible, or more accurately they should have the least amount of rotational inertia as possible. They will have be accelerated, too, and aren't affected by drag the same way.
Including things like rolling friction and losses in the wheel bearings might change these effects slightly (and might prove that there is actually and optimal weight), but I suspect these effects are very small compared to gravity and drag. Both losses in the bearings and rolling resistance could be modeled as a force that is proportional to the load on the wheel.
Any other thoughts?
-Kerry
3. Jun 30, 2009
### Bob S
This is a good question, and there are several parts to the answer.
The downward (gravitational) accelerating force on a hill of slope angle θ is
Fg = d(mv)/dt = mg sin(θ)
Here, g = 9.81 m/sec2, and m is the total mass, including soapbox, driver, bricks, wheels, etc.
The primary retarding force is air drag. There are several types of air drag. The major one in automobile fuel efficiency is the turbulent air drag, proportional to velocity squared. The other one, at low turbulence is Stokes Law drag, linear in velocity. Both are independent of vehicle mass m, so it represents a relatively larger retarding force for light vehicles. See http://en.wikipedia.org/wiki/Drag_(physics [Broken])
Fd = -(1/2)ρACdv2
Here, ρ is the air density (about 1.2 Kg/m3), Cd is the drag coefficient. A is the frontal area (m2), and v is the velocity (m/sec).
Tire rolling resistance in automobiles is linear in both velocity and vehicle mass, and has a CR (rolling resistance coefficient) of about 0.01. Soapbox Derby hard rubber wheels probably have a lower CR. It is proportional to the force (Newtons) normal to the track. See http://en.wikipedia.org/wiki/Rolling_resistance. For all four wheels, we get
FR = -CRmg cos(θ)
The last force, other than friction, is the effect of the moment of inertia of the wheels. A solid disk wheel of mass mw has a moment of inertia I = (1/2) mwr2, where r is the radius of the wheel. If the wheel is rolling without slipping, r2 becomes v2. If all of the mass is in the rim (rubber), it is mwr2. When all of the mass is in the rim, half the total kinetic energy is in the rotation, and half in the translation. The best way to minimize this is to use light wheels. This creates a slowing down of the soapbox acceleration by an effective force (for a wheel with all the mass in the rubber). For all four wheels we get (recall that mw is included in m).
Fw = -(1/2)mwg sin(θ)
Thus if all of the soapbox mass were in the wheels, the net downward effective translational accelerating force would be +(1/2)mg sin(θ). Putting all these together we get
Fsum = mg sin(θ) - (1/2)ρACdv2 - CRmg cos(θ) - (1/2)mwg sin(θ)
All of these, the accelerating forces and retarding forces, are proportional to the total mass m except the air drag, and the wheel mass. Other than reducing the drag coefficient, the best way to improve acceleration is to load the soapbox up with bricks.
α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
Last edited by a moderator: May 4, 2017
4. Jun 30, 2009
### Bacat
This is a good analysis Kerry, but I think we can't neglect rolling friction. If the wheels are frictionless, then the heavier car will always win, but since this doesn't happen we see that these effects are important. Since we're dealing with an actual vehicle in reality at a competition, we should try to think about this in the correct context.
You can calculate the rolling friction of your wheels by some simple experiments.
http://www.lastufka.net/lab/cars/why/xrollf.htm
But be aware that the friction will be different for different surfaces, so asphalt may produce more higher friction than a board.
Once you can calculate your coefficient of rolling friction, you can find the negative acceleration due to friction on your vehicle as a function of the mass from:
$$F=\mu_{k}\times N = \mu_{k} \times mgCos\theta$$
$$a=\frac{F}{m}=\frac{\mu_{k} \times mgCos\theta}{m}=\mu_{k} \times gCos\theta$$
where $$\mu_{k}$$ is the coefficient of rolling friction, g is gravity, and theta is the incline of the road.
This is a force of friction, so it opposes your downhill acceleration, and we see that the heavier the car is, the more friction it has.
You should be able to combine these to get an expression for total acceleration. I would suggest using calculus to get a look at that function and see how the mass influences the acceleration and the top speed for different time intervals.
5. Jun 30, 2009
### cepheid
Staff Emeritus
I'm a bit confused by how your final expression shows this. It looks like the mass cancels out.
6. Jun 30, 2009
### Bacat
The mass cancels out of the acceleration, but not the friction. Hey wait, that does seem strange.
I think Bob S has the best reply so far. If you want to understand the physics, try to work through his equations.
Bob, is my response correct?
7. Jun 30, 2009
### cepheid
Staff Emeritus
Yeah, this paragraph of Bob's seems to clear things up. Even if you take friction into account, the the heavier car is not a detriment.
8. Jun 30, 2009
### Bob S
Yup. Your expression for the rolling friction force is the same as my tire rolling friction force. They both increase linearly with the vehicle mass, and are independent of velocity.
9. Jun 30, 2009
### RonL
There should be a few bells ringing in the minds of anyone interested in electric cars. A little suction to the frontal area and a positive displacment in the draft area.
10. Jun 30, 2009
### Staff: Mentor
I don't know about soapbox cars, but pinewood derby cars have an advantage if they have their weight in the rear. (More potential energy).
11. Jun 30, 2009
### DaveC426913
Does the length of the track not play in?
Intuitively, I'm thinking a heavier car starts off with a lower acceleration, but ends up with a higher velocity.
(No that can't be right. That implies the acceleration is not constant - which can't be right. If one car ends up farther ahead at the end, it would have show this higher acceleration right from the starting line. And that means the length of the track does not change the outcome of the race.)
12. Jun 30, 2009
### Phrak
Has anyone questioned if the coefficient of friction (both rolling and static) depends upon weight?
13. Jun 30, 2009
### clustro
This is a very interesting piece of physics. I learned about this a few weeks ago when I was watching that old movie Cool Runnings on tv, about the Jamaican bobsled team. In the movie, it is revealed that John Candy's character was banned from bobsledding because he cheated by putting weights in the front of the sled to make it go faster.
This absolutely puzzled me; the force of gravity (so I erroneously thought) was the only significant force at play, and thus adding more weight seemed immaterial, even detrimental. But lo and behold, when drag terms are added in, it does indeed prove that greater acceleration is produce with increasing mass.
Although there were other benefits of greater mass I read. A heavier sled can get greater total momentum behind it during the running start, and a heavier sled is easier to maneuver/handle. Whether that carries over at all into this problem, I don't know.
The coefficient doesn't (its a constant), but the total frictional force does depend on mass.
However, so does gravity. When you divide by m to get the total acceleration, the masses would cancel. The kv^2 damping term would still get divided by m.
14. Jun 30, 2009
### Pinu7
Don't try it.Remember, that there is friction between the kart and the wheels, putting a lot of pressure on it would make it go really really slow(I remember from my mistake with my 8th grade project).Of course, there is air resistance but at low-ish speeds, its not going to matter as much as keeping your wheels free.
Also, you will increase the rolling friction.
15. Jul 1, 2009
### Phrak
The coefficients of friction are not constant with load. Only nominally and ideally are they constant. When a slim margin exists between competitors, small variation could make a difference.
On top of this, no one has brought up unsuspended load as effecting friction, have they?
16. Jul 1, 2009
### mordechai9
I agree it looks like the OP's basic intuition is correct -- more weight in the car means more momentum, so proportionally the decelerating effect of air resistance is reduced. Then again, if you can make the car very streamlined, air resistance might not matter that much anyways for the speeds that you will be considering.
Just wanted to add something else that I didn't see mentioned here. I'm not sure how you get the cart going, but if you start out the race by pushing the car, then consider the following. If the car is heavier, it will take longer to accelerate, because it will require a greater force. This could be crucial.
17. Jul 1, 2009
### mordechai9
Also, I wanted to make a comment about the discussion of "rolling friction". First of all, the friction between the wheels and the ground is essentially a good thing. You need that friction for the wheels to transmit torque and move your vehicle whatsoever. The punchline is that "rolling friction" is essentially negligible compared to sliding friction, which is the more commonly considered type. I just wanted to mention this because it seems like the comments may be exaggerating this effect just slightly.
Insofar as it is a negative effect, it is a fairly complex effect mainly due to elastic deformation. Values for the coefficient of rolling friction according to Wikipedia look to be very small, on the order of .001-.03 for driving surfaces. For sliding friction, coefficients are generally much higher.
Ref:
http://en.wikipedia.org/wiki/Rolling_resistance
http://en.wikipedia.org/wiki/Coefficient_of_friction#Coefficient_of_friction
18. Jul 1, 2009
### Bob S
Actually not. If a mass m is sliding down a slope of angle θ, the downward accelerating force is mg sin(θ). If a wheel of mass m and radius R (with all the mass in the rim) is rolling down the slope, its moment of inertia is mR2, and the downhill accelerating force is only (1/2) mg sin(θ), because the other half of the downward force is accelerating the rotational inertia. So a wheel sliding w/o friction has twice the downward accelerating force of a wheel rotating w/o slipping.
α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
19. Jul 1, 2009
### sir_manning
So assuming that the CR term isn't ridiculous, then the best thing to do is load it up with weight, but you want the wheels as light as possible. However, I don't quite understand how you calculated the effective force due to the angular momentum of the wheels.
For wheels with all the weight in the rim, I=(1/2)mwr2, so L=Iω=(1/2)mwrv and F=d(L)/dt = (1/2)mwrg sin(θ) . Where am I going wrong?
20. Jul 1, 2009
### Naty1
I had the same thought process....
That's the thought I had when reading posts...seems like it should but the net result may be insignificant.... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8606554865837097, "perplexity": 950.9015598054201}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00636.warc.gz"} |
http://nacretv.easylia.com/topic/eigenvalue-calculator-2x2-0b3bc9 | The eigenvector X and the eigenvalue A are then said to belong to each other. so clearly from the top row of the equations we get The calculator will perform symbolic calculations whenever it is possible. Find more Mathematics widgets in Wolfram|Alpha. 1 3 4 5 , l = 1 11. Eigenvalue Calculator Online tool compute the eigenvalue of a matrix with step by step explanations.Start by entering your matrix row number and column number in the input boxes below. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors.It decomposes matrix using LU and Cholesky decomposition. Matrices are the foundation of Linear Algebra; which has gained more and more importance in science, physics and eningineering. So let's do a simple 2 by 2, let's do an R2. Select the size of the matrix and click on the Space Shuttle in order to fly to the solver! This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. the eigenvalue equation is Hx =Ax, with X~ 0. Matrix A: Find. An application A = 10.5 0.51 Given , what happens to as ? Next story Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant; Previous story … Let's say that A is equal to the matrix 1, 2, and 4, 3. Icon 3X3. Display decimals, number of significant digits: Clean. We will see how to find them (if they can be found) soon, but first let us see one in action: Choose your matrix! what it does, what input to enter, what output it gives, and how it is useful). Eigenvalue Calculator. :) https://www.patreon.com/patrickjmt ! Step 4: From the equation thus obtained, calculate all the possible values of λ \lambda λ which are the required eigenvalues of matrix A. Decomposition of Eigenvalues. Calculate. To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the "id" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. 14. Eigenvalue Calculator Online tool compute the eigenvalue of a matrix with step by step explanations.Start by entering your matrix row number and column number in the input boxes below. Eigenvector and Eigenvalue. A simple example is that an eigenvector does not change direction in a transformation:. For background on these concepts, see 7. Learn more Accept. (1.1) (1.2) Ifx is a solution (called an eigenvector), so is any multiple KX, so long as 1<: is not zero. Calculate the eigenvalues and eigenvectors of a 5-by-5 magic square matrix. For 2x2 case we have a simple formula:, where trA is the trace of A (sum of its diagonal elements) and detA is the determinant of A. Clean Cells or Share Insert in. Icon 4X4. Eigenvalue Calculator(2x2) Added Aug 29, 2013 by venkateshb in none Enter a description of your widget (e.g. How to Use the Eigenvalue Calculator? 2 6 1 3 , l =0 12. Male or Female ? That is, For other cases you can use Faddeev–LeVerrier algorithm as it is done in Characteristic polynomial calculator. Let's find the eigenvector, v 1, associated with the eigenvalue, λ 1 =-1, first. Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I'll write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0. Eigenvalue Calculator is a free online tool that displays the eigenvalue of the given matrix. Similarly, we can find eigenvectors associated with the eigenvalue λ = 4 by solving Ax = 4x: 2x 1 +2x 2 5x 1 −x 2 = 4x 1 4x 2 ⇒ 2x 1 +2x 2 = 4x 1 and 5x 1 −x 2 = 4x 2 ⇒ x 1 = x 2. 2 4 4 1 3 1 3 1 2 0 5 3 5, l =3 13. • STEP 2: Find x by Gaussian elimination. Tags: characteristic polynomial complex eigenvalue diagonalization diagonalize a matrix eigenspace eigenvalue eigenvector linear algebra. They have many uses! You da real mvps! BYJU’S online eigenvalue calculator tool makes the calculation faster, and it displays the eigenvalue in a fraction of seconds. Get the free "Eigenvalue and Eigenvector (2x2)" widget for your website, blog, Wordpress, Blogger, or iGoogle. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … An easy and fast tool to find the eigenvalues of a square matrix. For a square matrix A, an Eigenvector and Eigenvalue make this equation true:. 9. !Part 1 http://www.youtube.com/watch?v=G4N8vJpf7hMThis is the second video on Eigenvalues and EigenVectors.Finding Eigenvalues and Eigenvectors : 2 x 2 Matrix Example.In this video I outline the general procedure for finding eigenvalues and eigenvectors for an n x n matrix and work an example using a 2 x 2 matrix. If . Thanks to all of you who support me on Patreon. Calculation precision. Eigenvector calculator Square matrix. Eigenvalues and eigenvectors calculator. 15. then the characteristic equation is . Thus, x really identifies an eigendirection. Once you get the characteric equation in polynomial form, you can solve it for eigenvalues. Works with matrix from 2X2 to 10X10. Thanks to all of you who support me on Patreon. Digits after the decimal point: 2. and the two eigenvalues are . By using this website, you agree to our Cookie Policy. All that's left is to find the two eigenvectors. what it does, what input to enter, what output it gives, and how it is useful). The Mathematics Of It. The l =2 eigenspace for the matrix 2 4 3 4 2 1 6 2 1 4 4 3 5 is two-dimensional. Find the eigenvalues of the matrix 2 2 1 3 and find one eigenvector for each eigenvalue. Icon 2X2. 3 1 2 4 , l =5 10. Click on the Space Shuttle and go to the 3X3 matrix solver! Calculate eigenvalues and eigenvectors. $1 per month helps!! the given eigenvalue. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. That is, convert the augmented matrix A −λI...0 to row echelon form, and solve the resulting linear system by back substitution. eigenvalue of A eigenvectors are the same as those associated with λ for A facts about eigenvaluesIncredible. 2X2 Eigenvalue Calculator. First eigenvalue: Second eigenvalue: Discover the beauty of matrices! Second calculator - the Eigenvalue calculator solves that equation to find eigenvalues (using analytical methods, that's why it works only up to 4th degree), and the calculator below calculates eigenvectors for each eigenvalue found. Matrix Eigenvalue Calculator calculates eigenvalues and eigenvectors for 2x2 symmetric and non-symmetric square matrices. Find a basis for this eigenspace. You da real mvps! Calculate eigenvalues. To improve this 'Eigenvalues and Eigenvectors Calculator', please fill in questionnaire. More: Diagonal matrix Jordan decomposition Matrix exponential. The calculator will find the eigenvalues and eigenvectors (eigenspace) of the given square matrix, with steps shown. A = magic(5) A = 5×5 17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9 [V,D] = eig(A) V = 5×5-0.4472 0.0976 -0.6330 0.6780 -0.2619 … Hence the set of eigenvectors associated with λ = 4 is Some theory can be found below the calculator. and all other eigenvectors corresponding to the eigenvalue (−3) are simply scalar multiples of u 1 — that is, u 1 spans this set of eigenvectors. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. This website uses cookies to ensure you get the best experience. Show Instructions . The computation of eigenvalues and eigenvectors for a square matrix is known as eigenvalue decomposition. We work through two methods of finding the characteristic equation for λ, then use this to find two eigenvalues. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. Enter a description of your widget (e.g. Eigenvalues and Eigenvectors Calculator for a 3 X 3 Real Matrix ... Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue λ.$1 per month helps!! :) https://www.patreon.com/patrickjmt !! Free Matrix Eigenvectors calculator - calculate matrix eigenvectors step-by-step. This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. Eigenvalue and Eigenvector Calculator. λ 1 =-1, λ 2 =-2. Eigenvalues … To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. • STEP 1: For each eigenvalue λ, we have (A −λI)x= 0, where x is the eigenvector associated with eigenvalue λ. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. \$1 per month helps!! The basic representation of the relationship between an eigenvector and its corresponding eigenvalue is given as Av = λv, where A is a matrix of m rows and m columns, λ is a scalar, and v is a vector of m columns. →Below is a calculator to determine matrices for given Eigensystems. Click on the Space Shuttle and go to the 2X2 matrix solver! Thanks to all of you who support me on Patreon. Vectors that map to their scalar multiples, and the associated scalars In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). :) https://www.patreon.com/patrickjmt !! 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https://math.stackexchange.com/questions/1867912/for-pm-sqrt-1-pm-sqrt-2-pm-sqrt-3-pm-cdots-pm-sqrt-2009-show-there-is-a | # For $\pm\sqrt 1\pm\sqrt 2 \pm\sqrt 3 \pm\cdots\pm\sqrt {2009}$, show there is a choice of signs such that it is irrational [closed]
Considering $$\pm\sqrt 1\pm\sqrt 2 \pm\sqrt 3 \pm\cdots\pm\sqrt {2009}$$ where you can replace each $\pm$ with $+$ or $-$. Prove that there is at least one choice of signs such that the number is irrational.
How can I prove that?
## closed as off-topic by Did, heropup, Matthew Conroy, Chill2Macht, Will JagyJul 23 '16 at 1:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, heropup, Matthew Conroy, Chill2Macht
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• Let $a$ be the sum with all positive. Let $b$ be the sum with $+$ for $\sqrt{2}$ and the rest $-$. If both $a$ and $b$ are rational, so is $a+b$. But $a+b=2\sqrt{2}$, irrational. – André Nicolas Jul 22 '16 at 20:28
• As an aside, the proof used in your previous question still applies here. $\sqrt{2003}+(\sum\limits_{i=1}^{2009} a_i \sqrt{i})$ where $a_i\in\Bbb Q$ for all $i$ and $a_{2003}=0$ is guaranteed to be irrational. Your question is simply a special case where all of $a_i$ (except $a_{2003}$ are $+1$ or $-1$) – JMoravitz Jul 22 '16 at 20:28
• The answer here shows that if the choices of signs "respect products" in the sense that the sign of $\sqrt{mn}$ is the product of the signs of $\sqrt m$ and $\sqrt n$ whenever $m$ and $n$ are coprime integers, you do get an irrational number. The non-constructive but very clever two answers posted here show that my answer is too high-tech for the purposes of this question! I only added this comment because you get an explicit choice from there :-) – Jyrki Lahtonen Jul 22 '16 at 20:30
• I am voting for reopening since even in the absence of efforts from the OP, I think the question is interesting and the answers here can be useful to other MSE users. – Jack D'Aurizio Jul 23 '16 at 14:26
Assume there is a choice of signs that makes the expression rational. If not, there are $2^{2009} \ge 1$ choices that make the expression irrational. Let $r$ be the rational. Now if you change the sign on $\sqrt 2$ you have either $r+2\sqrt 2$ or $r- 2 \sqrt 2$. Each of these will be irrational.
Hint: assume the contrary. Then you have a whole bunch of numbers corresponding to different choices of signs ($2^{2009}$ of them), and all these numbers are rational. Then any algebraic combinations of these numbers, such as sums and products, will also be rational. Can you use this to arrive at a contradiction?
You may exploit the fact that $2003$ is a prime number. Take every sign as positive, except the sign of $\sqrt{2003}$, and assume that: $$\sqrt{1}+\sqrt{2}+\ldots+\sqrt{2002}\color{red}{-\sqrt{2003}}+\sqrt{2004}+\ldots+\sqrt{2009} = \frac{p}{q}. \tag{1}$$
Now take a uber-huge prime $P$ such that $P>q$ and every prime in the range $[2,2009]$, with the exception of $2003$, is a quadratic residue $\!\!\pmod{P}$. Such monster exists by Dirichlet's theorem and the quadratic reciprocity theorem. So $\sqrt{1},\sqrt{2},\ldots,\sqrt{2002},\sqrt{2004},\sqrt{2005},\ldots,\sqrt{2009}$, $p$ and $q^{-1}$ can be intepreted as elements of $\mathbb{F}_P$, but $\sqrt{2003}$ does not belong to $\mathbb{F}_P$ by construction, hence $(1)$ is not possible and $$-2\sqrt{2003}+\sum_{k=1}^{2009}\sqrt{k}\not\in\mathbb{Q}.\tag{2}$$ The same argument also works by replacing $2003$ with $1999,1997,1993$ or any prime in the range $[1004,2009]$.
• Just to confirm, this shows in fact that every choice of signs would lead to an irrational number and moreover it would work for every cut-off other than $2009$ (except $1$ of course)? – quid Jul 23 '16 at 14:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9224383234977722, "perplexity": 187.49357781985404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00060.warc.gz"} |
http://openwetware.org/wiki/Physics307L_F07:People/Knockel/Notebook/071107 | # Speed of light!
Experimentalists: us
## Acknowledgments
Thanks Tomas for help with the software!
## Goal
All light travels at the same constant speed, c=2.998x108 m/s. We want to measure this speed by seeing how the time of travel varies when changing the distance traveled. Measuring this time is difficult since light moves 3/10 of a meter every nanosecond (the distances we can create are only a few meters). Luckily! We have a very sensitive time-to-amplitude converter (TAC) that converts a time interval to a voltage (the voltage is proportional to the time interval). This voltage can easily be measured by a quality oscilloscope.
## Equipment
Here is our high-voltage power supply for the PMT (left), our delay box (center), and the TAC (right).
Cool stuff:
• LED that can fire brief pulses of light
• Photomultiplier tube (PMT) to detect the light from the LED
• 2 light polarizers for the PMT and LED
• Delay box (adds a delay on the order of nanoseconds to a signal)
• Time-to-amplitude converter (TAC - pronounced "tack")
• Oscilloscope
• Computer hardware and software that act as a pulse hight analyzer
Boring stuff:
• High-voltage power supply (about 2500 V)
• Low-voltage power supply (about 200 V)
• Many long and short coaxial cables with BNC connectors
• 3 meter tube that LED and PMT can fit inside
## Setup
Connecting stuff:
1. Put on the polarizers on the front of the LED and PMT
2. Put the LED and PMT inside opposite sides of the tube facing each other
3. Connect the LED to the low-voltage power supply (power is turned off)
4. Connect the PMT to the high-voltage power supply (power is turned off)
5. Connect the LED and the PMT to the TAC
6. Connect the TAC to the computer
7. Connect the PMT to the oscilloscope
1. Give the LED about 200 V, but no more than this or it will break
2. Give the PMT 2200 V (no more or it will break) (no less or the TAC will not perceive its signal) (make sure the PMT is in the DARK tube!)
3. Get the oscilloscope to observe the output from the PMT
4. Separate the LED and PMT as much as possible, and then rotate the PMT (thereby rotating the polarizer) so that the PMT signal is the strongest
5. Create 32 ns delay on the delay box (to make sure that the PMT signal always comes AFTER the LED signal)
6. Set the TAC to a range of 100 ns
7. Open the computer software and check to see if the pulses from the TAC are making a Gaussian or Poisson distribution. If not, fiddle with stuff until it does. Make sure the voltage to the PMT is high enough since we had it at only 2000V and we could not get a good signal from the TAC. Keep fiddling and fiddling until you are about to kill Koch for not showing up, and then it will start working. I promise.SJK 00:22, 20 November 2007 (CST)
00:22, 20 November 2007 (CST)
I did it for your own good, to give you that satisfaction of success and not having to kill Koch
## Procedure
We had everything setup as described above, we started by choosing a voltage to keep the PMT's output signal at (the voltage was chosen by rotating the PMT until the polarizers are aligned while the LED and PMT are at a maximum distance). This voltage must be kept constant due to a strange thing called "time walk" that causes weaker signals to have a longer delay. We chose a voltage of -608 units. I think it was measured in volts or maybe millivolts, but I'm not sure, and it doesn't matter because we kept this number constant.
Having chosen a PMT voltage, we needed to calibrate the results from the software. The software measured the average pulse height (H) as well as the full-width-at-half-maximum (FWHM) of the distribution of pulse heights, but it gave these values as dimensionless numbers that were proportional to the voltage of the pulses. Since the TAC made sure the pulses were proportional to the time delay (in fact, given a 100 ns range, 100 ns corresponded to 10 V), the H is proportional to the time delay (t). To find the proportionality constant (k) where H = kt, we used the delay box to create known delays and then measured the response of H while keeping everything else the same. The slope of the line plotted from this is k.
Having found k, we knew from the chain rule that
$c=\frac{dx}{dH}\frac{dH}{dt}$
and from H = kt that
$c=k\frac{dx}{dH}$.
However, since x = mH, where m is the slope of this line,
$c=km\,$.
To find m, we simply varied x and measured the resulting H's. The slope of the line plotted from x vs. H gives m. In doing this, we approximately called the farthest x as x=3 meters, and this is not perfect, but we only care about the slope of the plot of x vs. H, and not the actual values.
For each H measurement, we made sure the PMT voltage was correct and let the software collect about n=1,000,000 pulse height measurements.
Lastly, we compared the H for a certain x with the PMT at the right to voltage to the H at the same x at a very different voltage to get an idea of the amount of error to expect from the "time walk."
## Data, Calculations, and Error
For a Gaussian distribution, FWHM = 2.36σ, where σ is the standard deviation. Using standard error of the mean ($SE=\frac{\sigma}{\sqrt{n}}$) to calculate random uncertainty in H, I got a standard error of about 0.035 for every measurement since my n=1,000,000. Since this value is so small compared to H, I do not provide any random uncertainty error bars in the data below. Also, my random error in the x measurements is negligible since it is only 1 or 2 millimeters.
I WILL analyze random error in my calculations. This error will simply be the standard error of the slope of the linear regression I do in Excel.
### Finding k
t (ns) H 0 8 16 24 32 386.6 475.3 563.8 652.8 743.5
$k=slope=1.114(3)\times10^{10}$ 1/s
### Finding m
x (m) H 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 637.9 649.2 655.2 665.8 674.7 693 697.3 700.3 713.3 718.6 727.7 729.9
$m=slope=2.838(11)\times10^{-2}$ m
### Finding c = km
Simply multiplying k and m and then propagating the error gives...
$c=3.16(13)\times10^{8}$ m/s
## A closer look at error
The presence of this error is explained by the "time walk," which is the largest source of random error. In our procedure, we altered the PMT voltage from -608 to -856 and noticed that H decreased by 106, which is larger than the amount it changed throughout the entire 3-meter range. Clearly, the time walk is a huge problem. Since our oscilloscope's averaging ability was not very good and since the oscilloscope would only display discreet voltages (e.g. -600,-608,-614), we could not always make sure the PMT was perfectly at -608, hence the error.
Random error due to not enough significant digits should also make the fit worse and contributed to the error, but this is negligible.
SJK 00:28, 20 November 2007 (CST)
00:28, 20 November 2007 (CST)
Yeah, Excel uses the exact linear regression formulas we talked about in class last month, so you are correct that 32% of the time the accepted value will be outside your range of uncertainty (if your random errors have gaussian distribution).
The actual value of c is not within my error bars. My uncertainty is 0.13x108 m/s, and the absolute error between the actual and my speed of light is 0.16x108 m/s. This could be due to faulty equipment or some other weird systematic error, but it might not. I am not sure how Excel calculates its standard error of the slope of a linear regression, but perhaps it forms this value so that there is a 68% chance that the correct value is within this distance from the calculated value, so there is a 32% chance that the correct value is not in that interval. Since the actual value of c is only 1.2 times the standard error, I believe there are no large sources of systematic error.
## Conclusion
I have calculated that the speed of light is
$c=3.16(13)\times10^{8}$ m/s.
The actual value of c=3.00x108 m/s is slightly outside of the error interval I have provided, but there is no reason why error intervals should include the actual value even if no significant systematic error exists because the actual value is barely outside this interval.SJK 00:36, 20 November 2007 (CST)
00:36, 20 November 2007 (CST)
I like the way your explained it above better: there is a reason to expect it to be in the interval, but 32% of the time it shouldn't. So this is just one of those times, and like you said, it's very close. Great job with this lab! I wonder if your problem initially was that you had the LED too far towards the end? I think you get better data with it closer to the PMT. Good idea with using the time delay generator to calibrate the unknown proportionality. Great presentation of final result...pretty cool that you can get such good data, even with the time walk problem! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9051066040992737, "perplexity": 915.315145945968}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00567-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/quadratic-drag-a-baseball-is-thrown-upwards.834324/ | # Quadratic drag — a baseball is thrown upwards
Tags:
1. Sep 24, 2015
### jbeatphys
1. The problem statement, all variables and given/known data
A baseball of mass m is thrown straight up with an initial velocity v0. Assuming that the air drag on the ball varies quadratically with speed (f = cv^2), show that the speed varies with height according to the equations.
[Attached]
Where x_{0} is the highest point and k = c/m. Note: x is measured positive upward, and the gravitational force is assumed to be constant.
2. Relevant equations
3. The attempt at a solution
As I see it, F_netup = F(v_intial) - normal force - drag force(-v) & F_netdown = normal force - drag force.
F_up
(1) Integrate to v(t) (from v0 to 0) and then set v(0) and solve for t, which is the time that it takes to get the peak of the throw.
(2) Integrate to x(t) (from 0 to x0) and then sub in t_{peak}, and solve for v^2.
This is what I have in my head. But I when I complete these calculations I get nothing like what I should be getting [according to the equations] — I get a very complicated equation with hyperbolic trig functions.
Thanks for any help that you can provide.
2. Sep 24, 2015
### SteamKing
Staff Emeritus
Unfortunately, we can't peek inside your head. You'll have to show your work.
3. Sep 24, 2015
### jbeatphys
Okay, I can understand that, not a problem (and probably fortunately).
4. Sep 24, 2015
### Staff: Mentor
What are the (two) forces acting on the ball when it is rising? What does F_(v initial) mean? A normal force is usually considered a contact force. Is anything contacting the ball as it rises?
Chet
5. Sep 24, 2015
### jbeatphys
Oops. Yes. There is nothing in contact with the ball (other than the fluid, which is air) and so there is no normal force. I meant to say gravitational force, sorry.
The two forces acting upon the ball whilst it rises are the gravitational and the fluid resistance, which are both opposing the initial velocity. You are right to question what is the F_(v initial) it is definitely the wrong conceptualisation of the problem.
To reiterate:
m dv/dt = -mg-cv^2 (up) &
m dv/dt = mg-cv^2 (down) ... I am pretty sure now.
I am going through my calculations again at the moment.
6. Sep 24, 2015
### jbeatphys
And thanks for correcting my misconceptions.
7. Sep 24, 2015
### Staff: Mentor
Much better. Nice job.
Have something to add?
Draft saved Draft deleted
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https://www.physicsforums.com/threads/satellite-moving-around-a-planet.745163/ | Satellite moving around a planet
1. Mar 25, 2014
utkarshakash
1. The problem statement, all variables and given/known data
A satellite is describing a circular orbit around a massive planet of radius R. The altitude of the satellite above the surface of planet is 3R and its speed is v_0. To place the satellite in an elliptical orbit which will bring it closer to the planet, its velocity is reduced from v_0 to βv_0., when β<1. The smallest permissible value of β if satellite is not to crash on the surface of planet is √(2/K), find K.
3. The attempt at a solution
I think the angular momentum shall be conserved.
$16mR^2 v_o /4R = mR^2 \beta v_0 / R$
But this equation gives incorrect value of β.
I also tried using conservation of energy but the expression for β does not come out to be in the same format as asked in the question.
2. Mar 25, 2014
Rellek
Ok, so what would the second radius of the equation be? Are we supposed to assume the satellite comes right to the surface of the planet without actually touching it?
3. Mar 25, 2014
nasu
Why do you think that the angular momentum is the same on the two trajectories?
The velocity is decreased by a factor of β. Is the distance from the planet increased by the same factor during the deceleration ?
The angular momentum is conserved on each trajectory but not between them. A tangential force (so a torque) was applied to reduce the velocity.
4. Mar 25, 2014
Rellek
R1mV0 + ∫Mdt = R2mβV0 ?
5. Mar 25, 2014
nasu
I would assume that the radius is the same after the velocity is reduced. The equation you wrote is OK in principle. But not useful here.
Focus on conservation laws written for the elliptic trajectory itself.
The circular orbit is just to find a relationship between vo and R.
6. Mar 25, 2014
utkarshakash
Will applying energy conservation help me in this case?
7. Mar 25, 2014
nasu
You are not assuming correctly.
The two positions are on the opposite sides of the major axis. The minor axis is irrelevant here.
Draw a diagram.
8. Mar 25, 2014
utkarshakash
Is the length of major axis 5R?
9. Mar 25, 2014
nasu
Didn't you say that you assume semi-minor axis is R?
10. Mar 25, 2014
utkarshakash
I later realized that it is wrong and that I don't need to know the length of minor axis for this question.
I think that the centre of planet will be one of the focii of the ellipse. If I apply conservation of angular momentum around the focus, I can write
$4Rv_0 = R \beta v_0$
But I don't think that my logic is correct as the above equation still gives me the wrong answer. :(
11. Mar 25, 2014
nasu
Please don't delete parts of your posts after someone else posted. It makes the replies to your post look confusing or silly.
Yes, the planet is in one of the focal points of the ellipse.
The equation you wrote is not correct though.
At one end the satellite has velocity βvo and radius 4R. At the other end the velocity is some value, v2 (unknown yet) and the radius is R.
You need two more equations to solve the problem:
1. Conservation of energy between the two extreme point on the ellipse
2. Newton's second law for the initial circle
12. Mar 25, 2014
utkarshakash
Thank you so much !
Draft saved Draft deleted
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http://libros.duhnnae.com/2017/jul5/150008269654-The-isocohomological-property-higher-Dehn-functions-and-relatively-hyperbolic-groups-Mathematics-Group-Theory.php | # The isocohomological property, higher Dehn functions, and relatively hyperbolic groups - Mathematics > Group Theory
The isocohomological property, higher Dehn functions, and relatively hyperbolic groups - Mathematics > Group Theory - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: The property that the polynomial cohomology with coefficients of a finitelygenerated discrete group is canonically isomorphic to the group cohomology iscalled the weak isocohomological property for the group. In the case when agroup is of type $HF^\infty$, i.e. that has a classifying space with thehomotopy type of a cellular complex with finitely many cells in each dimension,we show that the isocohomological property is equivalent to the universal coverof the classifying space satisfying polynomially bounded higher Dehn functions.If a group is hyperbolic relative to a collection of subgroups, each of whichis polynomially combable respectively $HF^\infty$ and isocohomological, thenwe show that the group itself has these respective properties too. Combiningwith the results of Connes-Moscovici and Dru{\c{t}}u-Sapir we conclude that agroup satisfies the Novikov conjecture if it is relatively hyperbolic tosubgroups that are of property RD, of type $HF^\infty$ and isocohomological.
Autor: R. Ji, B. Ramsey
Fuente: https://arxiv.org/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9244134426116943, "perplexity": 1909.3747608574718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813109.8/warc/CC-MAIN-20180220224819-20180221004819-00759.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/182436-center-endomorphism-algebra.html | # Thread: Center of endomorphism algebra
1. ## Center of endomorphism algebra
If I have an R-algebra A where R is a ring, why the center of the endomorphism algebra $E_{A}(A)$ is the algebra of endomorphism with coefficient in the enveloping algebra $A* \otimes A$:
$Z(E_{A}(A))=E_{A* \otimes A}(A)$ ?
2. Originally Posted by yavanna87
If I have an R-algebra A where R is a ring, why the center of the endomorphism algebra $E_{A}(A)$ is the algebra of endomorphism with coefficient in the enveloping algebra $A* \otimes A$:
$Z(E_{A}(A))=E_{A* \otimes A}(A)$ ?
well, $End_A(A) \cong A$ and so you basically want to prove that $Z(A) \cong End_{A^* \otimes_R A}(A)$. this is a special case of a more general problem. for a solution to the general case see this theorem in my blog. the notation $C_A(B)$ means the centralizer of $B$ in $A$.
3. Thank you very much, i was trying to do that with the left regular representation right now, so I was on the right path! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9680697321891785, "perplexity": 179.78968388224044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320049.84/warc/CC-MAIN-20170623100455-20170623120455-00315.warc.gz"} |
https://community.oracle.com/thread/1520455?tstart=0&messageID=6359780 | # How to compile a single file in a project
I have placed my java connector file called "xxxTargetconnector .class" in the below mentioned path
D:\FSCM9DMO\webserv\FSCMDMO9\applications\peoplesoft\PSIGW\WEB-INF\classes\com\peoplesoft\pt\integrationgateway\targetconnector
I have converted the above class file in to a java file using a decompiler.Now i would like to recompile the java file to get the byte code.This class file imports many class files from jar files which is placed in the below mentioned path.
D:\FSCM9DMO\webserv\Cybersource.
Before compiling i set the path to include the jar files
for ex , set path = D:\FSCM9DMO\webserv\Cybersource
For compilation i went to the below path using command prompt and compiled the particular file.
D:\FSCM9DMO\webserv\FSCMDMO9\applications\peoplesoft\PSIGW\WEB-INF\classes\com\peoplesoft\pt\integrationgateway\targetconnector>javac xxxTargetConnector.java
Unfortunately i got around 99 errors saying
xxxTargetConnector.java:187: cannot find symbol
symbol : variable Logger
location: class com.peoplesoft.pt.integrationgateway.targetconnector.xxxTargetConnector
How to compile a single file in a project ?
• ###### 1. Re: How to compile a single file in a project
According to [http://java.sun.com/javase/6/docs/technotes/tools/windows/javac.html],
``"As a special convenience, a class path element containing a basename of * is considered equivalent to specifying a list of all the files in the directory with the extension .jar or .JAR.``
I have never tried using this feature.
So if all of your dependent classes are in in jar files that are in D:\FSCM9DMO\webserv\Cybersource, a command like the following should work.
javac -cp D:\FSCM9DMO\webserv\Cybersource\* xxxTargetConnector.java
Note that setting your operating system path variable has no impact on where Javac looks for dependent classes. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8328043222427368, "perplexity": 4773.609228644969}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398448506.69/warc/CC-MAIN-20151124205408-00090-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://twistedone151.wordpress.com/2009/08/ | ## Archive for August, 2009
### Monday Math 87
August 31, 2009
For a sequence an, n=1,2,3,…, the series is known as the Dirichlet series generating function (or sometimes simply the Dirichlet series) for the sequence. For example, if we take the trivial sequence 1,1,1,… (an=1 for all n), then the Dirichlet series is the Riemann zeta function
.
For the alternating sequence 1,-1,1,-1,… (an=(-1)n-1), the Dirichlet series is the Dirichlet eta function
.
Now, consider the case where an=f(n), where f(n) is a multiplicative function.
Then we have Dirichlet series generating function:
.
Using the prime factorization of n, , we see:
.
Now, we procede as we did in this post, which is just the above with multiplicative function . The above can be expressed as a product over the primes:
.
When f(n) is a completely multiplicative function, , so then the series within the summation becomes a geometric series:
.
Note that if our function is the constant function f(n)=1 (it is completely multiplicative), then the above gives us the Euler product for the Riemann zeta function
which we found here.
Note that even if the function is just multiplicative, and not completely multiplicative, it may still be possible to find a non-series expression for the series found in the product over primes. One example can be seen this past post.
### Physics Friday 87
August 28, 2009
Continuing from last week with our simple fermion system, we now consider the system enclosed in walls which are impermiable to our fermions, so that the particle number Ñ is now constant. While this is no longer the grand canonical formalism, the fundamental equation is still valid, as it is an attribute of the system independent of its boundary conditions. However, the chemical potential μ, usually called the Fermi level when speaking of a system of indistinguishable fermions, will not be a constant, but will instead vary with temperature so as to keep Ñ constant. Using our work from last time, we have particle number . While this gives Ñ as a one-to-one function of μ (as Ñ(μ) is an increasing function), it is not one that has an analytic inverse, and solving for μ given Ñ and T must generally be done numerically.
We can also consider the low temperature limit kT→0 (β increasing without bound). Recall that the probability of occupation for a given state in our system is , and since we are dealing with fermions, this gives the occupancy (expectation value for the number of particles in the state): . We see that for μ=εn, the occupancy f(μ,T) of our state is exactly 1/2, independent of the temperature. For εn>μ, increases without bound as β does, so that the probability of occupation goes to zero; for εn<μ, as β increases, so that the probability of occupation goes to one. Thus, as the temperature approaches absolute zero, we find the states with energy below the Fermi level fully occupied and those with energy above the Fermi level empty.
This is what we should expect from a system of fermions. As the temperature approaches absolute zero, the particles should go to the lowest energy state allowed; however, being fermions, the Pauli exclusion principle applies, and thus the ground state can hold only a limited number of particles; when it fills, then the particles begin filling the next level. Then it fills, and so on, until we reach the total number of particles; the threshold between filled and empty states is thus our Fermi level. The Fermi level at a temperature of absolute zero is known as the Fermi energy: .
We also see that as the temperature begins increasing away from absolute zero, the transition at the Fermi level becomes less sharp; the occupancy for states of energy just below the Fermi level decreases and that for states just above the Fermi level increases.
Figure 1: Occupancy of a state as a function of state energy at various low temperatures.
In particular, the energy range over which this transfer occurs is on the order of 4kT: expanding in a Taylor series about the point εn=μ, we get .
We should also note that the equation for occupancy as a function of energy possesses inversion symmetry about the point εn=μ, pn,m=1/2.
Now, suppose we have the special case ε1<ε2=ε3, so that ε1 is the ground state, and the excited state is degenerate. Let Ñ=2, so that at absolute zero, the ground state is filled, the excited states are empty, and the Fermi level lies somewhere between
ε1 and ε2. Now, for very low T, we use the approximation for occupancy:
.
In this case, our equation for Ñ becomes:
. Plugging in Ñ=2, we can solve the above for μ, to get
(the ellipsis indicates terms of higher order in kT, which are neglected in our approximation). This illustrates an important general principle: as the temperature increases, the Fermi level is ‘repelled’ by energy regions with higher densities of states.
### Monday Math 86
August 24, 2009
A notable multiplicative function is the Möbius function μ(n) defined as follows:
μ(n)=1 when n is a square-free integer with an even number of distinct prime factors
μ(n)=-1 when n is a square-free integer with an odd number of distinct prime factors
μ(n)=0 when n is not square-free.
Note that μ(1)=1, as 1 has an even number of prime factors; namely, zero.
In terms of the distinct prime factor counting function ω(n) (see here), we can write the Möbius function as:
One can easily confirm that μ(n) is a multiplicative function (but not a completely multiplicative function); if either m or n is not square-free (or both), then mn is not square-free [and μ(mn)=μ(m)μ(n)=0]. If m and n are both square-free and coprime, they have no common prime factors, so mn is square-free and ω(mn)=ω(m)+ω(n) [so ].
In terms of the powers of primes, we see:
.
### Physics Friday 86
August 21, 2009
Let us consider a model quantum system with three permitted spacial ‘orbital’ states; a particle in state n=1,2, or 3 has energies ε1, ε2, and ε3, respectively. Now suppose are particles are spin-½ fermions. Then the spin projection has two possible states, ms=½ and ms=-½; these are designated “up” and “down”. Thus, there are six possible states (n,ms), with n=1,2,3, and ms=±½. Since the particles are fermions, no more than one particle may be found in each of these states at a given time.
Now, let us consider this system in contact with a thermal reservoir with a fixed temperature T, and a reservoir of our fermions with Gibbs potential μ; this is to say, in the grand canonical ensemble. Compute the grand canonical partition sum. To compute it, we note that the grand canonical partition sum factors across non-interacting states, like the canonical partition sum does. Thus,
$\mathcal{Z}=z_{1,-\frac{1}{2}}z_{1,\frac{1}{2}}z_{2,-\frac{1}{2}}z_{2,\frac{1}{2}}z_{3,-\frac{1}{2}}z_{3,\frac{1}{2}}$. Now, each orbital state partition sum has only two terms: one for the empty state, with E=0 and N=0; and one for the occupied state, with E=εn and N=1. Thus, (assuming no magnetic field, so energy is independent of the spin direction):
$z_{n,m_s}=e^{-\beta(0-\mu\cdot0)}+e^{-\beta(\epsilon_n-\mu\cdot1)}=1+e^{-\beta(\epsilon_n-\mu)}$.
Now, we could also pair the states with the same spatial orbit n (but differing spin alignments):
$z_{n}=z_{n,-\frac{1}{2}}z_{n,\frac{1}{2}}=\left(1+e^{-\beta(\epsilon_n-\mu)}\right)^2=1+2e^{-\beta(\epsilon_n-\mu)}+e^{-2\beta(\epsilon_n-\mu)}$.
We can interpret this expanded product in terms of the four possible states of n: the empty state, two singly occupied states (of opposite spin), and one doubly-occupied state.
The probability that the state (n,ms) is empty is thus $\frac{1}{z_{n,m_s}}$, and the probability that it is occupied is $p_{n,m}=\frac{e^{-\beta(\epsilon_n-\mu)}}{z_{n,m_s}}=\frac{1}{e^{\beta(\epsilon_n-\mu)}+1}$.
Now, our grand canonical partition sum is
$\mathcal{Z}=\left[1+e^{-\beta(\epsilon_1-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_2-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_3-\mu)}\right]^2$. Thus, the grand potential is
$\Psi=-\frac{\ln\mathcal{Z}}{\beta}=-\frac{2}{\beta}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]$.
Now, the mean number of particles ⟨N⟩ can be found from Ψ:
$\begin{array}{rcl}\langle{N}\rangle&=&-\frac{\partial\Psi}{\partial\mu}\\&=&\frac{2}{\beta}\frac{\partial}{\partial\mu}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_1-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_2-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\beta{e^{-\beta(\epsilon_1-\mu)}}}{1+e^{-\beta(\epsilon_1-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_2-\mu)}}}{1+e^{-\beta(\epsilon_2-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_3-\mu)}}}{1+e^{-\beta(\epsilon_3-\mu)}}\right]\\&=&\frac{2}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_3-\mu)}+1}\end{array}$
Note that this is the same result we get were we to sum the probability of occupation over all six states, as we should expect.
Similarly, we could find the energy via $U=-\frac{\partial}{\partial\beta}(\ln\mathcal{Z})+\mu\langle{N}\rangle$, or we can use the occupation probabilities to get:
$U=\sum_{n,m}E_{n,m}p_{n,m}=\frac{2\epsilon_1}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2\epsilon_2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2\epsilon_3}{e^{\beta(\epsilon_3-\mu)}+1}$.
We can also find entropy via $S=k\left(\ln\mathcal{Z}+\beta{U}-\beta\mu\langle{N}\rangle\right)$ (see here), getting:
$S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.
### A Question
August 18, 2009
Does anyone know what the current international law and policy says concerning the legal status of, and conflict with, extraterrestrial or other non-human intelligences? Would the Geneva Conventions apply to alien invaders?
Addendum: Seeing the comments, I feel I must clarify. No, I don’t think that UFO sightings are alien visitors, nor do I believe that any contact has been made. I am referring to where international law and treaty currently stands with regards to a purely hypothetical future situation. More precisely, something like Robert J. Freitas Jr.’s analysis, only with regards to international rather than United States law.
### Monday Math 85
August 17, 2009
In number theory, an arithmetic function f(n) is called a multiplicative function if it possesses the property that f(ab)=f(a)f(b) for any positive integers a and b which are coprime (gcd(a,b)=1). Note that since gcd(n,1)=1 for any positive integer n, we see that for any multiplicative function, f(1)=1.
If the property f(ab)=f(a)f(b) holds for all positive integers a and b, coprime or not, then the function is said to be completely multiplicative.
One example of a multiplicative function is Euler’s totient function φ(n) (some others may be found here).
Now, consider the prime factorization of n:
. Plugging this into a multiplicative function, we see:
.
Thus, a multiplicative function is determined completely by its values for the powers of the prime numbers. This simplifies the computation of multiplicative functions. For example, consider φ(n). For a power of a prime pk, k>0, the only positive integers less than or equal to this number which aren’t coprime with it are multiples of p: p, 2p, 3p, … , (pk-1)p=pk; this is pk-1 numbers, so the totient is
. This allows us to compute the totient of any number from it’s prime factorization, e.g.:
φ(720)=φ(24325)
=φ(24)φ(32)φ(5)
=24-1(2-1)·32-1(3-1)·51-1(5-1)
=8·6·4=192
Let f(n) and g(n) be multiplicative functions, and let h(n)= f(n)g(n) be their product. Then for any coprime positive integers a and b,
,
so the product of two multiplicative functions is multiplicative.
### Physics Friday 85
August 14, 2009
One of the simplest problems solved by applying the time-independent Schrödinger equation is the one-dimensional “particle in a box”, where the potential V(x) is zero for 0<x<L, and infinite everywhere else. Then we have ψ(x)=0 for x≤0 and xL, and for 0<x<L, Schrödinger equation becomes
with boundary conditions ψ(0)=ψ(L)=0. The general solution to the differential equation, which can be rewritten as
, is
, where , or solving for energy, . The boundary condition ψ(0)=0 tells us that B=0, and the condition ψ(L)=0 thus tells us that , and thus , n a positive integer. Thus, in terms of that n,
, and , and the energy is quantized. Normalizing via , we get normalized wavefunction .
Now, let us consider the momentum. The one-dimensional quantum momentum operator is , and we see ; our energy eigenstates are obviously not momentum eigenstates. However, note that the eigenfunctions of are of the form (with eigenvalues ). Noting that , we see that our energy eigenstate is the linear combination of a state of momentum and a state of momentum , with equal probabilities for these two possible measurements. Compare this to the classical particle in a one dimensional box, where it has kinetic energy , half the time travelling with momentum of magnitude p in one direction, the other half moving in the opposite direction with momentum of magnitude p.
Now, let us consider a three-dimensional cubic box:
V(x, y, z)=0 when 0<x<L, 0<y<L, and 0<z<L, and is infinite at all other points.
Then we have for the interior of our cube:
,
with boundary conditions .
using the definition of nabla for Cartesian coordinates:
,
and using separation of variables via ,
we get
with . These are just copies of the one-dimensional problem, so we obtain normalized 3d wavefunction
, where nx, ny, and nz are positive integers. The energy is . A given state is defined entirely by the triplet of positive integers nx, ny, nz. Note that except for the ground state nx=ny=nz=1 (), all of the energy levels are degenerate; all states for which have the same value will have the same energy.
[Note also that the proper linear combination of the plane wave with momentum and the seven other plane waves formed by reflections in x, y, z, and combinations thereof, will produce our energy eigenfunction.
Again compare to the possible momentum values of a classical particle in a box making perfectly elastic collisions with the walls.]
### Goo
August 11, 2009
I just got in from watching the G.I. Joe movie. Is it wrong of me that my first thought afterwards was wanting to make some red goo of my own?
### Monday Math 84
August 10, 2009
Last week, I showed that for a fraction that converts to a pure repeating decimal, one with a denominator coprime with 10, the number of digits in the repeat is equal to the smallest whole number n such that 10n-1 is divisible by the fraction denominator d. In terms of modular arithmetic, this means that 10n≡1 (mod d). This is called the multiplicative order of 10 modulo d, and denoted as ordd(10) or Od(10). Let us make a chart of the multiplicative orders of 10 modulo the first few whole numbers coprime with 10:
d ordd(10)
3 1
7 6
9 1
11 2
13 6
17 16
19 18
21 6
You might note that in several cases, all prime, ordd(10) is d-1; however, this is obviously not a general rule. For the other primes, such as 3, though, we have ordd(10) dividing d-1; this does hold for d a prime. Note however, that while ord9(10)=1 divides 9-1=8, ord21(10)=6 does not divide 21-1=20. So what is our general rule for all d coprime with 10?
Consider Euler’s totient function φ(n), and note that for p prime, φ(p)=p-1. So, redoing the chart to add φ(d):
d ordd(10) φ(d)
3 1 2
7 6 6
9 1 6
11 2 10
13 6 12
17 16 16
19 18 18
21 6 12
From this, we see that in every case here, ordd(10) divides
φ(d). This is the general rule. However, proving that this is true requires group theory. This rule does limit the valid possibilities for ordd(10). For example, φ(23)=22, so ord23(10) must be 1, 2, 11, or 22. Now, we can rule out 1 and 2, as 9 and 99 are not divisible by 23. Thus ord23(10) must be 11 or 22 (ord23(10)=22).
.
### Physics Friday 84
August 7, 2009
The canonical ensemble is a powerful tool for statistical mechanics. Let us now illustrate another powerful formalism, an extension known as the “grand canonical” formalism.
In the canonical formalism, our system is in contact with a thermal reservoir with which it may exchange energy; the reservir is large enough that it’s temperature be treated as a constant (thus giving it an entropy which varies linearly with energy). In the grand canonical formalism, we add the ability to exchange particles, as well as energy, with the particle reservoir large enough that the chemical potential is treated as a constant.
If we treat the combination of our system and the reservoir as a closed system with every microstate equally probable, we have for a state where our system has energy Ei and particle number Ni, the fractional occupation is the number of microstates of the reservoir for which its energy is EtEi and it’s particle number is NtNi divided by the total number of microstates for the system and reservoir combination with total energy Et and a total of Nt particles:
However, using the entropy definition , we can rewrite this in terms of the reservoir and total entropies (as functions of energy and particle number):
.
As with our canonical formalism, we can expand the entropy as a Taylor series in energy about EtU and in particle number about the average particle number N
, (using ).
And via additivity of the entropy, .
Thus
,
where is the thermodynamic beta and is the “grand potential”. As with the canonical ensemble, we normalize this: , where
is the “grand canonical partition function”, and the sum is over all states, with state s having energy Es and Ns particles.
We see that just as , we can solve for the grand potential:
.
Let us define . Now, the expectation value for the number of particles is
,
or, using the chain rule,
.
Similarly, one can show that
,
the variance in the particle number.
Now, consider the logarithmic partial derivative of Ƶ with respect to β:
,
and we have our internal energy. We can also show by a second differentiation that
.
From our equation for the grand potential, we see
.
Next, consider the fundamental thermodynamic relation . Soving for the mechanical work term,
. Now, if we continue to hold β (and thus T) and μ constant, the right hand side of the previous relation is just –:
.
From this, one can obtain the equation of state
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https://asmedigitalcollection.asme.org/appliedmechanics/article-abstract/34/4/921/425349/Application-of-Kirchhoff-s-Integral-Equation?redirectedFrom=fulltext | Interaction of a plane compressional step wave with a circular cylindrical obstacle embedded in an elastic medium is studied. The obstacle is rigid, stationary, and of infinite length. The incident wave travels in a direction perpendicular to the axis of the cylinder. Using Kirchhoff’s theorem, surface integral equations are formulated for the displacement potential derivatives in the scattered field and on the cylinder boundary. The wave-front solution obtained for the illuminated zone on the cylinder is identical to that obtained by high-frequency wave-front analysis. Boundary stresses in the shadow zone as well as the initial behavior of the wave-front stresses at the boundary between the illuminated and shadow zones are obtained. The integral equations for both illuminated and shadow-zone boundary stresses are reduced to successive linear matrix equations for numerical analysis. The numerical methods developed in this paper can be applied to interaction problems for obstacles of arbitrary geometrical configuration. They are also readily extended to the case where the medium exhibits bilinear or multilinear stress-strain behavior.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9505454301834106, "perplexity": 520.2681888852336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986658566.9/warc/CC-MAIN-20191015104838-20191015132338-00198.warc.gz"} |
https://rd.springer.com/chapter/10.1007/978-3-642-88208-1_3 | # Cybernetic Queueing and Storage Systems
• A. Ghosal
Part of the Lecture Notes in Operations Research and Mathematical Systems book series (LNE, volume 23)
## Abstract
Any system comprises a set of elements A = (a1,..., an) which are subject to some specified behaviour pattern; for example, a!s may be stochastic following any particular d.f. or deterministic or there may be a set of inter-relationships among a!s etc. All these behaviour patterns can be denoted by the set R-which takes into account any auto-regulatory process, rules, d.f.’s, inter-relations, etc. Thus we denote a system S by {A, R{ (see Klir and Valach, 1965). From this angle let us review a simple queueing or storage system <Z, X,Y,k> or <Z, u, k>; this system is subject to an input process and subject to some rules gives out an output process — in some cases the input and release processes are fed into the system which, subject to the rules of the system, govern the level of the dam or an inventory system {Z{. The simple model (1.1) or (1.2) can be regarded as the regulatory process of the system. Thus if Zt (the level of dam or waiting time) is regarded as an output process, then we can represent the process as follows [A = (X,Y), R = (rules of operation, capacity k, etc.)]:
$$S:\{ A,B\} \to Z$$
(3.1)
Thus we can have a set of systems S, S,..., such that
$${S_{i}}:\{ {A_{i}},{R_{i}}\} \to {Z_{i}}$$
(3.2)
Again Z. may be a vector instead of being a single valued variable, Zi = (Zii, Zi2,..., Zip). Thus in a queueing process we may write Z = (w,Q) where w is the waiting time process and Q is the queue size process. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557868003845215, "perplexity": 1726.9450925044066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594790.48/warc/CC-MAIN-20180723012644-20180723032644-00026.warc.gz"} |
https://access.openupresources.org/curricula/our6-8math/en/grade-6/unit-6/family.html | # Unit 6Big Ideas
## Equations in One Variable
This week your student will be learning to visualize, write, and solve equations. They did this work in previous grades with numbers. In grade 6, we often use a letter called a variable to represent a number whose value is unknown. Diagrams can help us make sense of how quantities are related. Here is an example of such a diagram:
Since 3 pieces are labeled with the same variable , we know that each of the three pieces represent the same number. Some equations that match this diagram are and
A solution to an equation is a number used in place of the variable that makes the equation true. In the previous example, the solution is 5. Think about substituting 5 for in either equation: and are both true. We can tell that, for example, 4 is not a solution, because does not equal 15.
Solving an equation is a process for finding a solution. Your student will learn that an equation like can be solved by dividing each side by 3. Notice that if you divide each side by 3, , you are left with , the solution to the equation.
Draw a diagram to represent each equation. Then, solve each equation.
Solution:
## Equal and Equivalent
This week your student is writing mathematical expressions, especially expressions using the distributive property.
In this diagram, we can say one side length of the large rectangle is 3 units and the other is units. So, the area of the large rectangle is . The large rectangle can be partitioned into two smaller rectangles, A and B, with no overlap. The area of A is 6 and the area of B is . So, the area of the large rectangle can also be written as . In other words, This is an example of the distributive property.
Draw and label a partitioned rectangle to show that each of these equations is always true, no matter the value of the letters.
Solution:
## Expressions with Exponents
This week your student will be working with exponents. When we write an expression like , we call the exponent. In this example, 7 is called the base. The exponent tells you how many factors of the base to multiply. For example, is equal to . In grade 6, students write expressions with whole-number exponents and bases that are
• whole numbers like
• fractions like
• decimals like
• variables like
Remember that a solution to an equation is a number that makes the equation true. For example, a solution to is 2, since . On the other hand, 1 is not a solution, since does not equal . Find the solution to each equation from the list provided.
List:
Solution:
1. 7, because . (Note that -7 is also a solution, but in grade 6 students aren’t expected to know about multiplying negative numbers.)
2. 3, because
3. 1, because
4. , because means
5. 0.008, because means
6. , because
7. Any number! is true no matter what number you use in place of .
8. 5, because this can be rewritten . What would we have to divide by 9 to get 27? 243, because . .
## Relationships Between Quantities
This week your student will study relationships between two quantities. For example, since a quarter is worth 25ȼ, we can represent the relationship between the number of quarters, , and their value in cents like this:
We can also use a table to represent the situation.
1 25
2 50
3 75
Or we can draw a graph to represent the relationship between the two quantities:
A shopper is buying granola bars. The cost of each granola bar is $0.75. 1. Write an equation that shows the cost of the granola bars, , in terms of the number of bars purchased, . 2. Create a graph representing associated values of and 3. What are the coordinates of some points on your graph? What do they represent? Solutions 1. . Every granola bar costs$0.75 and the shopper is buying of them, so the cost is . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8160771727561951, "perplexity": 421.2545145824893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655887319.41/warc/CC-MAIN-20200705090648-20200705120648-00282.warc.gz"} |
https://math.stackexchange.com/questions/1228586/what-does-the-term-undefined-actually-mean | What does the term "undefined" actually mean?
I have read many articles on many sites and in many books to understand what undefined means? On some sites of Maths, I read that it could be any number. and on some sites, I read that it may be some undefined thing; and there are more definitions. But they all have clashes with each other that all are defining the term "undefined" in different ways. So which concept is right for the word "undefined" in criteria of Maths among the following five?
1-A number divided by zero may be any number (real or imaginary)
2-A number divided by zero may be an entity which is not defined yet.
3-Division of a number by zero does not make sense.
4-If $$x=a/0$$, then no solution exists!
5-It may give a third type of number other than real or imaginary [I have not read this definition in any book or site but it's my thought.]
This query popped up into my mind while my teacher was solving a question from my book, I am showing it to you along with my teacher's work.
Q- Prove that the roots of the following equation are real.
$$x^2-2x(m+\frac{1}{m})+3=0$$ where, $$m$$ is any real number.
Teacher's attempt:
For roots to be real,
$$b^2-4ac>0$$
$$\implies 4(m^2+\frac{1}{m^2}-1)>0$$
$$\implies 4(m^2+\frac{1}{m^2}-2+1)>0$$
$$\implies 4[ (m-\frac{1}{m})^2+1 ]>0$$.
My teacher let us write that inequality is satisfied for all $$m$$ belongs to real number however if $$m=0$$, $$\frac{1}{m}$$ is undefined. So if "undefined" means that "a number divided by zero may be any real or imaginary number" so then I can confess only for real numbers that inequality is satisfied for all $$m$$ belonging to real numbers and not for imaginary numbers since we can't make sense of a statement like this $$i>0$$ but if the term "undefined", in Maths, is defined as "Senseless" or "something else not known" then I strongly apprehend that why my teacher let us write that Inequality is satisfied for all $$m$$ belonging to real numbers?
• if "undefined" means that "a number divided by zero may be any real number" then... It doesn't mean that. Every answer below is telling you that is not what "undefined" means. Before you try to understand what "undefined" means, perhaps you should try to understand what "defined" means because "undefined" means nothing more or less than "not defined." Apr 10, 2015 at 17:56
• Don't make a confusion between $a/0$ where $a$ is a nonzero value (this quotient is not a real number) and the notation $0/0$ which is indeterminate (usually this quotient does not appear as such but as a limit of some function). I have never heard that "a number divided by zero can be any real number" - on the opposite, it is no real number.
– user65203
Apr 10, 2015 at 18:58
• I see this is related to this earlier question of yours. I would have recommended linking to the other question (while explaining the relationship) rather than just cutting and pasting. That question seems to be caused by imprecision in the statement of a fact (which is said to be true for all real $m$ when it is only true for $m \neq 0$). If there is some doubt or confusion that is not based on such an imprecise statement then it would be better to have an example that is not based on such an imprecise statement. Apr 10, 2015 at 22:07
• I believe the recommended way to handle duplication of your own question is to flag one of the questions and request that a moderator merge it with the other. Apr 11, 2015 at 14:55
• Do not make edits unless absolutely necessary. In particular, changing "mathematics" to "maths" is not absolutely necessary!
– Pedro
Apr 13, 2015 at 2:52
Saying that 1 divided by 0 is undefined, does not mean that you can carry out the division and that the result is some strange entity with the property “undefined”, but simply that dividing 1 by 0 has no defined meaning. That is just like when you ask whether the number 1.9 is odd or even: That is not defined. Or when you ask what colour the number 7 has.
• @SufyanSheikh If you want to be precise, you should specify right at the beginning that $m\neq 0$, since $\frac1m$ is not defined when $m=0$. It would be better if the question read something like this "Prove [...] $x^2-2x(m+\frac1m)+3 = 0$, where $m$ is any real number except $0$." Or perhaps "Prove [...], where $m$ is any real number where the expression is defined."
– Eff
Apr 10, 2015 at 14:42
• @SufyanSheikh You can't drastically modify your question and then expect people to make commensurate modifications on their answers. Apr 10, 2015 at 17:51
• @SufyanSheikh, thank you, I think I read the answer before the edit from top to bottom. Apr 10, 2015 at 17:52
• As a programmer, I like to think of the operation as undefined, not the result. The result is not undefined, as there is no result. You can't even get to a result other than to say, "I can't do what you want me to do, so I stopped." 1 / 0? The operation is undefined, so there is no result. In a computer, it generates an exception, since there is no defined algorithm (or set of instructions) for dividing by 0. Such an exception prevents the computer from trying to check the result, which has no representation. Apr 10, 2015 at 21:59
• @goblin, I have treated this as a beginner's question. Apr 10, 2015 at 22:33
To put matters straight: Division is a function $$q:\quad{\mathbb R}\times{\mathbb R}^*, \qquad(a,b)\mapsto q(a,b)=:{a\over b}\ ,$$ whereby $q(a,b)$ is the unique number $x\in{\mathbb R}$ such that $b \>x=a$.
When we say that $\displaystyle{a\over0}$ is undefined then this means no more and no less than that the pair $(a,0)$ is not in the domain of the function $q(\cdot,\cdot)$.
Now to your three ways of understanding "undefined" in the realm of division by $0$:
1. If $\displaystyle{a\over0}$ could be any number, say $=13$, then this would enforce $13\cdot0=a$, which is wrong when $a\ne0$.
2. This is even worse. Why should $\displaystyle{7\over0}$ be the Eiffel tower?
3. There are circumstances where division by zero makes sense, e.g. in connection with maps of the Riemann sphere, or with meromorphic functions. There one has $\infty$ as an additional point in the universe of discourse. But these circumstances require special exception handling measures, and the "usual rules of algebra" are not valid when dealing with $\infty$.
• This is of course a more useful answer than mine. Apr 11, 2015 at 12:19
• Note that in particular $0/0$ is undefined. The literal division of $0$ by $0$ is not a defined number; it is not indeterminate. What is indeterminate, in the sense "it could be anything", is a limit of the form $\lim_{n\to\infty}\frac{a_n}{b_n}$ where $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=0$ - the statement "$0/0$ is indeterminate" is a (slightly misleading) abbreviation of this claim. Apr 11, 2015 at 22:15
• @SufyanSheikh This could be interesting. Apr 12, 2015 at 5:37
What does the term "undefined" actually mean?
In light of the already great answers provided by Carsten and Christian, I thought a more linguistic analysis of "undefined" may be in order. The following two terms are explained in the book Origins of Mathematical Words by Anthony Lo Bello:
indeterminate$\quad$ The Latin noun terminus means the end of something. From it was formed the denominative verb termino, terminare, terminavi, terminatus meaning to set bounds to. The addition of the prefix de- emphasizes that the separation is from something else and produced the compound verb determino, determinare, determinavi, determinatus meaning to fix the limits of. The addition of the negating prefix in- to the past participle of this verb resulted in the Latin adjective indeterminatus meaning undefined, unlimited.
Now to the analysis of the word originally in question:
undefined $\quad$ The Germanic negative prefix un- has been added to the word defined of Latin origin to produce the hybrid undefined. It would have been better to say indefined as we say indefinite, but it is too late now. Defined is from the verb definio, definere, definivi, definitus, which means to set the boundaries. The plural noun finus in Latin means enclosed area, territory. The force of the prefix de- is to add the sense of thoroughness to the action.
Both of these terms, especially the mentioning of "to set the boundaries" in the analysis of the term undefined, should make Christian's answer even more lucid, especially his response to your third way of understanding "undefined."
From the algebraic point of view, the Real numbers form a field under multiplication and addition. If we look at the Reals as a field, there is no separate operation of "division", instead we multiply by the reciprocal. Since $0$ has no reciprocal in $\mathbb{R}$ (in fact the additive identity in a field never does), there exists no element available to multiply by to perform what would be commonly called "division by zero".
We know $0$ has no reciprocal in $\mathbb{R}$ because there is no real number $z$ that makes the following equation true: $$0z=1$$
If there were such a $z$, then we could multiply by that number and obtain an answer for "dividing by zero", but since there is no $z$, we have no way to "divide by zero".
To expand on user38858's answer, it would be like "dividing by orange". "Orange" is not a Real number, so the idea that we could perform any mathmatics with "Orange" in the context of Real numbers makes no sense. There is no real number that represent the reciprocal of $0$, so it also makes no sense to attempt to perform any mathematics on such an object.
From the algebraic standpoint, I would say that "Division by zero does not make sense" is the closest of your three concepts. Maybe I would change it slightly to say "Division by zero is not even possible".
On some site of Mathematics, I read that it could be any number. and on some sites, I read that it may be anything. These are two major meanings of undefined but I think, they have a clash.
I just wanted to address this because it looked like none of the other (very good) answers have.
It looks like you're seeing two different definitions on those various websites:
1. Undefined = not yet defined. As in, this may have a value or meaning or definition, but we haven't gotten there yet. A lot of sites may do this with a variable X, saying it hasn't yet been defined, or we haven't found the value yet, but we can make some statements.
2. Undefined = not defined; we have not given an explanation for what this means. This is how I typically see undefined used. Undefined means there does not exist an answer in the way we defined it.
For example, if we define a function $f$ $|$ $\mathbb R \mapsto \mathbb R$ by $f(x) = 2x$, then $f(x)$ is "defined" for $x$ a real number. $x = 1$, $f(x) = 2$. $f(4) = 8$.
However, $f(orange)$ makes no sense. It's undefined since orange is not an element of $\mathbb R$.
We could start out the problem by saying $f$ $|$ $\mathbb R$ $U$ $(orange) \mapsto \mathbb R$ where $f(x) = 2x$ for $x$ in $\mathbb R$, and $f(orange) = 0$, then $f(orange)$ is not undefined. I've given it an answer.
Similarly, you could start out your problem by giving an output for $\frac{1}{m}$, for $m = 0$, maybe by saying $\frac{1}{m} = 0$. Whether or not you'd find that construction very useful is altogether another matter, but it is possible.
• If I didn't get you wrong, you are saying that the right definition for the term"undefined" is that "A number divided by zero may be any number". Isn't it? Apr 10, 2015 at 17:36
• Sort of, but not really. Sorry this was confusing. In math, for me, there isn't "right" and "wrong" definitions. There's "useful" and "less useful" definitions. "A number divided by zero may be any number" is not a useful definition for me. You can choose to use that definition if you like, but you may find in your career that it is "less useful" than you may have initially thought. I can't come up with a good reason to use such a definition. If you want a concrete way to think about I would consider this more useful: "undefined" = "has no meaning". "orange + 6 = New York" "has no meaning" Apr 10, 2015 at 17:56
• I do know one useful application of defining $1/0$: in doing so, you get a total division function, which is often easier to work with in compound operations, esp. in symbolic manipulation algorithms. A meadow is a particular kind of field where the division operator is total and satisfies $a/0=0$. This has a nice equational theory, and has been studied by J. Bergstra and others. Feb 11, 2016 at 9:41
Something is undefined because none has defined it (yet). It is easy to define addition and multiplication of real numbers.
One can then show the theorem that for arbitrary $a,b\in\mathbb R$, the equation $a+x=b$ has exactly one solution. By virtue of this theorem, one can define $b-a$ as this unique solution.
One can also show that theorem that for $b\in \mathbb R$ and $a\in\mathbb R\setminus\{0\}$, the equation $ax=b$ has a unique solution. By virtue of this theorem, one can define $\frac ba$ as this unique solution under these circumstances. In principle, none could prevent you from defining $\frac ba$ as something completely different, for example $\frac ba:=a^2+b^2-7$. However, this makes little sense as one cannot express nice theorems with this (well, you'd have that this "division" is commutative and there'd we funny equations such as $\frac55=\frac71$). At any rate, such a definition, though feasible, would not be related to multiplication.
So what if we consider only definitions of division $\frac ab$ that coincide with our usual division (unique solution of $bx=a$) as long as we divide by something nonzero? Whatever you define $\frac 10$ to be (say, $\frac 10:=\star$), you'd lose nice properties of division (i.e., this modified notion of division is less helpful to express general statements succinctly). For example, following general rules we should have $\frac10=\frac1{2\cdot 0}=\frac12\cdot \frac10$, so $\star=\frac12\star$. So either we no longer have "$\frac a{2b}=\frac12 \frac ab$ whenever $\frac ab$ is defined"; or we no longer have "the unique solution of $x=\frac12 x$ is $x=0$"; or the symbol $\star$ is actually just $0$ and we no longer have "$\frac ab=c$ implies $a=bc$". Some of these options would be especially awful because the uniqueness of solutions of similar equations was what motivated the introduction of "division" in the first place!
So the more complete answer is: Division by zero is undefined, and is left undefined on purpose because whatever custom extended definition someone might want to introduce, it would not be useful (let alone meaningful) in the sense of an inverse of multiplication.
"a number divided by zero may be any real number" is a wrong statement and is never said.
A number divided by zero is just not a real number.
"zero divided by zero may be any real number" is an informal way to express that certain indeterminate expressions have a limit.
• Hey I have copied it from there. Let me show it to you that from where I got concept of that "dividing by zero may give us any number". Here it is, taken from mathsisfun.com/numbers/dividing-by-zero.html >$0/0$ is like asking "how many $0s$ in $0$?" Are there no zeros in zero at all? Or perhaps there is exactly one zero in zero? Or many zeros? So 0/0 is indeterminate (it could be any value). Apr 11, 2015 at 11:51
• Exactly what I said. You didn't read that post carefully nor my answer. $1/0$ is undefined and $0/0$ is indeterminate.
– user65203
Apr 11, 2015 at 17:31
When referring to the result of a mathematical operation, undefined means there is no meaningful result.
When referring to a mathematical object which satisfies some mathematical relationship, undefined means no object satisfies that relationship.
The operation $f$(🍊) has no meaningful result.
There is no $x$ that satisfies the relationship $x=\frac{a}{0}$; equivalently, there is no $x$ that satisfies the relationship $x \cdot 0 = a$.
• I don't understand your question. Maybe it's better to put it as @phyrfox did: "I like to think of the operation as undefined, not the result. The result is not undefined, as there is no result." Apr 14, 2015 at 0:21
• Your original question may be conflating two questions: "what does 'undefined' mean?" and "why is division by zero undefined?". These are not the same question. I aimed to answer the question in the title. Apr 14, 2015 at 0:27
• The grammar in some of your comments suggests that english is not your first language; it might be easier to understand if you could get someone to explain it in your first language. Apr 14, 2015 at 0:30
• @SufyanSheikh When I explain something to somebody else and they repeat it back to me, I have to listen carefully to find exactly where the disconnect is (if exists). When I'm listening to them speak to me and they can't express all angles of a concept, it's sometimes because they don't know the concept, and other times because they don't know the words. You use phrases which signal ESL, which indicates more strongly a disconnect in words rather than concepts. This isn't meant to be insulting: confronting a disconnect is, in my experience, the swiftest way to learn. Apr 16, 2015 at 15:38
• "English is not my first language but also not the last" I like your attitude. I didn't respond to your last comment on my answer only because I had nothing to say. As for "grumbles about your English"; there are none. Not from me, and presumably not from Polyergic. Carefully reread what I wrote. I wasn't talking about you specifically until I used the word "you". It seems like you're struggling more with the words than the concepts; your very question here is one of language. As such, I thought I would reinforce Polyergic's suggestion to see help from an expert in your first language. Apr 16, 2015 at 17:43
In this case, undefined means "not in the domain." The division operator, $div(numerator, denominator)$ is defined over the domain of real numbers except that y=0 is not in the domain.
Now, that very specific wording aside, I think you might get an intuitive grasp of the meaning if I can draw from a non-mathematics topic, programming languages. Programming language specifications have two terms which are highly related to your query: unspecified behavior and undefined behavior
• Unspecified behavior - the exact result of an operation is not specified, but the general behavior is bounded to some reasonable results. If you're writing software that uses unspecified behavior, you cannot rely on its exact behavior to be the same from compiler to compiler (example: there's a feature in C++ called a "null pointer," which is often represented with the symbol 0. Its actual value is unspecified. On most machines, its value is 0, but on some microcontrollers, a value of 65535 turns out to be surprisingly more useful, so those compilers use that.)
• Undefined behavior - The behavior can be absolutely anything. The compiler is allowed to give you an error. You may get an error at runtime. You may get an error before you even execute the command that has undefined behavior. You may silently get really weird and obtuse behavior, or it may even work exactly like you think it should do. Undefined behavior is truly undefined in every way shape and form. The presence of undefined behavior invalidates the entire program, just like using an undefined value in mathematics invalidates your entire proof.
Now there is some room for undefined behavior to be defined. As a classic example, taking the square root of a negative number is undefined behavior when you're considering the set of real numbers. The square root of -1 simply has no definition if you're only looking at reals. However, if you extend your interest to not only real numbers, but complex numbers, the square root of -1 becomes defined (its value is $i$), but only as long as your concerned with complex numbers.
You could define division by 0 to result in an unspecified value. In fact, there are schools of thought that do. However, it turns out that doing so causes all sorts of problems with the rules of arithmetic that we are so used to. For example math on a Riemann sphere allows the statement $1/0=\infty$. However, they pay a dark price in doing so: the normal rules of arithmetic you are used to do not apply to arithmetic done on Riemann spheres. In particular, you lose a guarantee of a multiplicative inverse for equations that result in $\infty$. $0/0$ and $\infty/\infty$ remain undefined.
Generally speaking though, people do not use such number systems unless they explicitly call them out. For normal every day use, undefined is simply undefined.
The answer by user38858 is very much to the point:
It looks like you're seeing two different definitions on those various websites:
1. Undefined = not yet defined. As in, this may have a value or meaning or definition, but we haven't gotten there yet. A lot of sites may do this with a variable X, saying it hasn't yet been defined, or we haven't found the value yet, but we can make some statements.
2. Undefined = not defined; we have not given an explanation for what this means. This is how I typically see undefined used. Undefined means there does not exist an answer in the way we defined it.
I want to stress here that this is really relevant. After reading the answer by Christian Blatter, I can't help but wonder Why is mathematics fond of infinity, but dismissive towards partially (un)defined operations? Here are contrast answers for Christian Blatter comments on your three ways of understanding "undefined" for division by zero
1-A number divided by zero may be any number (real or imaginary)
In Meadows and the equational specification of division, Jan Bergstra et al rediscover among others the fact that one of the most useful way to define division by zero is that the result is zero. Now that I write this answer, I begin to appreciate why he introduced the new name zero totalized field for a field where division by zero is defined as zero. I'm still not convinced that introducing the name meadows for commutative inverse rings (which are also known as strongly von Neuman regular rings) was a good idea.
2-A number divided by zero may be any thing.
Even so Jan Bergstra should know that a number divided by zero is zero, he introduced common meadows where $0^{-1}=a$.
3-Division of a number by zero does not make sense.
Yes, in the most common context, division of a number by zero simply does not make sense.
Suppose you have a function $f:A \to C$, but you regard $A$ as a subset of some other set $i:A \hookrightarrow B$, and you do all your work from $B$. It is natural to ask if the function $f$ extends to a function $\tilde f:B \to C$ such that $f = \tilde f \circ i$.
Now, when dealing with subsets, we like to abuse notation and talk about $x \in i(A)$ and talk about $f(x)$. But $f$ is not a function on $B$, so in a rigorous sense, this is nonsense. A type error in the first place, and this is where our errors are coming from. This undefined notion comes in when we try to take $x \not \in i(A)$ and ask what $f(x)$ is. We can't pull $x$ back to an element on $x' \in A$ and then compute $f(x')$, so we say that $f(x)$ is undefined.
For example, let $A = \mathbb R^*$, $C = \mathbb R$, and $f: \mathbb R^* \to \mathbb R$ with $f(x) = 1/x$, and $B = \mathbb R$. In this case, we can't extend $f$ to a continuous function $\tilde f$ on $B = \mathbb R$. If $x \ne 0$, then we can pull $x$ back to A and compute $f(x)$, but if $x=0$, then we can't, so we say $f(x)$ is "undefined".
In general, an expression $x$ is undefined iff there is no $y$ such that $y=x$, or
$$x\text{ is undefined}\Longleftrightarrow\neg\exists y:y=x$$
For example, let $f$ be a function with the set of natural numbers as its domain and codomain. $f$ consists of a set of ordered pairs representing the inputs and outputs, as follows:
$$f=\{(1,0),(2,1),(3,2),(4,3),(5,4),(6,5),\ldots\}$$
Hence $f(1)=0$, $f(2)=1$, $f(3)=2$, $f(4)=3$, etc.
You may recognize this as the predecessor function on the natural numbers. The expression is undefined for $0$ in the sense that there is no ordered pair in $f$ such that the first element of the pair is $0$, hence (trivially) there is no $n$ in the codomain of $f$ such that there is an ordered pair in $f$ such that the first element of the pair is $0$ and the second element of the pair is $n$.
One could, of course, naturally extend the definition of $f$ to include $0$ (introducing the negative integers). One could do the same with $\frac{a}{0}$ for nonzero $a$, introducing a point at infinity. In doing so you may lose certain algebraic properties, such as $0\cdot a=a\cdot0=0$ for all $a$.
I think what you need to understand is that not everything that your teacher says should be taken ex cathedra. 0/0 not only can but has been defined in various ways in various contexts. The limit of x/y as x and y go to zero could be defined as infinity if, say, x converges linearly, while y converges exponentially. If the converse is the case, then we might choose (Note that it is a choice. It is not written in stone that 0/0 is undefined!) to define 0/0 as zero. If x and y both converge factorially, we might for some purpose at hand define 0/0 as being equal to 1. Sometimes the limit can even oscillate between two values and in such a case it has been found convenient to define the limit as being half-way between them! It all depends on what works for you in the case at hand.
I'd like to suggest another approach to explaining what an "undefined term" is in the context of term-rewriting systems.
Given an alphabet $A$ (i.e. $A$ is a non-empty, finite set), denote by $A^*$ the set consisting of all finite strings over $A$. Suppose you are given a set $T \subseteq A^*$ of terms over $A$, and a relation $R \subseteq T\times T$, which determines whether a term $s$ can be reduced to a term $t$ in a single step. Denote by $R^*$ the reflexive-transitive closure of $R$, i.e. $s\ R^*\ t$ iff $s = t$, or $s\ R\ t$, or there is a sequence of terms $r_1, \dots, r_n \in T$ ($n \in \{2, 3, \dots\}$) with $r_1 = s$, $r_n = t$, and for every $i \in {1, \dots, n-1}$ $r_i\ R\ r_{i+1}$. Suppose, additionally, you are given a subset of values (or terminals) $V \subseteq T$. Then the set of undefined terms, $U$, is the set of all terms that cannot be reduced to a value, i.e. $U := \{t \in T: \forall v\in V.\neg(t\ R^*\ v)\}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8247082829475403, "perplexity": 306.1535151864531}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00268.warc.gz"} |
https://math.stackexchange.com/questions/502211/is-it-a-unitary-self-adjoint-and-normal-operator/502225 | Is it a unitary, self adjoint and normal operator?
Let $A\colon H\to H$ be a bounded linear operator on a complex Hilbert space such that $\|Ax\|=\|A^*x\|\forall x$, given that there is a nonzero $x$ for which $A^*x=(2+3i)x$. Then I need to know whether it is unitary, self adjoint and normal operator.
How do I proceed?
The norm is at least $|2+3i|$ by definition, so it is not unitary which implies norm $1$. It can't be selfadjoint, because the eigenvalues of selfadjoint operators must be real. It could be normal though, because $A^*$ being the scalar times the identity is suitable.
Note, the definition of a normal operator is precisely the property your operator has, namely: $$\|Nx\|=\|N^*\|\text{ for all }x\in\mathcal{D}(N)=\mathcal{D}(N^*)$$ Moreover, this is equivalent to the usually stated: $$N^*N=NN^*\iff N\text{ normal}$$ Try to adjust the proof from that sketch: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9705458283424377, "perplexity": 157.83698029352158}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00513.warc.gz"} |
http://mathhelpforum.com/discrete-math/104242-help-combinatorics-problem.html | # Thread: Help with combinatorics problem
1. ## Help with combinatorics problem
Hi
I would appreciate if someone could explain how to solve the following problem:
"How many different five letter "words" can you write with the letters BILBANA"
The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:
A+B = 5!
A+A = 5!/2!
B+B = 5!/2!
A+B+B = ?
A+A+B = ?
A+A+B+B = ?
2. Hello Drdumbom
Originally Posted by Drdumbom
Hi
I would appreciate if someone could explain how to solve the following problem:
"How many different five letter "words" can you write with the letters BILBANA"
The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:
A+B = 5!
A+A = 5!/2!
B+B = 5!/2!
A+B+B = ?
A+A+B = ?
A+A+B+B = ?
You're right so far. Let me continue:
With A + B + B there will be two other letters chosen from the remaining 3. So we have two processes:
• Choose 2 out of 3 remaining letters. This can be done in ${^3C_2}=3$ ways.
• Arrange the 5 letters, which include a repeated B. This can be done in $\frac{5!}{2!}= 60$ ways.
So A + B + B can be done in $3\times 60 =180$ ways.
Similarly A + A + B can be done in $180$ ways.
And then in a similar way A + A + B + B + one other letter can be chosen and arranged in ${^3C_1}\times\frac{5!}{2!2!}=90$ ways.
Therefore the total no of 'words' $= 120 + 60 + 60 + 180 + 180 + 90 = 690$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.893997073173523, "perplexity": 398.437528288085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00524.warc.gz"} |
https://brilliant.org/problems/trig-complex/ | # Trig & Complex
Algebra Level pending
Let $$z$$ be a complex number such that $z = 2(\cos 3 ^{\circ} + i \cos 87 ^{\circ}).$ Then $$z^{5}$$ can be expressed as $$r( \sin \alpha^{\circ} + i \cos \alpha^{\circ}) ,$$ where $$r$$ is a real number and $$0 \leq \alpha \leq 90$$. What is the value of $$r + \alpha ?$$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9869328141212463, "perplexity": 95.67790048771086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719045.47/warc/CC-MAIN-20161020183839-00118-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-2x-2-5x-3-by-factoring | Algebra
Topics
# How do you solve 2x^2-5x-3 by factoring?
Apr 3, 2018
$\left(x - 3\right) \left(2 x + 1\right)$
#### Explanation:
$2 x 3 = 6$ so then you must find two numbers (a negative and a positive since there are two negatives in the original problem) that will add up to $5$ and multiply to $6$.
The numbers are then $- 6$ and $+ 1$ which gives you:
$2 {x}^{2} + 1 x - 6 x - 3$
Split down the middle and factor out an $x$ and a $- 3$
$x \left(2 x + 1\right) + - 3 \left(2 x + 1\right)$
Take the leading coefficients and you'll finish with
$\left(x - 3\right) \left(2 x + 1\right)$
Hope this helps!
##### Impact of this question
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http://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-5-logarithmic-exponential-and-other-transcendental-functions-5-3-exercises-page-343/27 | ## Calculus 10th Edition
$f(x)$ has an inverse function.
We find the derivative of $f(x)$: $f'(x) = \frac{1}{x-3} , x \gt; 3$ We see that on the domain of $f(x)$, $(3, \infty)$, $f'(x)$ is always positive so $f(x)$ is strictly monotonic. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9906002283096313, "perplexity": 192.40861355410425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00034.warc.gz"} |
https://www.beatthegmat.com/the-probability-that-event-m-will-not-occur-is-0-8-and-the-probability-that-event-r-will-not-occur-is-t326311.html?sid=5132ccf4284dffb94c3516c3f4f24699 | ## The probability that event $$M$$ will not occur is $$0.8$$ and the probability that event $$R$$ will not occur is
##### This topic has expert replies
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### The probability that event $$M$$ will not occur is $$0.8$$ and the probability that event $$R$$ will not occur is
by M7MBA » Sat Sep 11, 2021 2:33 pm
00:00
A
B
C
D
E
## Global Stats
The probability that event $$M$$ will not occur is $$0.8$$ and the probability that event $$R$$ will not occur is $$0.6.$$ If events $$M$$ and $$R$$ cannot both occur, which of the following is the probability that either event $$M$$ or event $$R$$ will occur?
A) $$\dfrac15$$
B) $$\dfrac25$$
C) $$\dfrac35$$
D) $$\dfrac45$$
E) $$\dfrac{12}{25}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8404634594917297, "perplexity": 364.94734038492567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585199.76/warc/CC-MAIN-20211018062819-20211018092819-00427.warc.gz"} |
https://micronanomanufacturing.asmedigitalcollection.asme.org/IMECE/proceedings-abstract/IMECE2020/84591/V011T11A053/1099528?searchresult=1 | Abstract
Computational fluid dynamics simulations for water desalination using forward osmosis were conducted on a flat membrane module. In the simulations, the effect of the porous support layer is assumed negligible. The simulations were performed with two values of flow rate such that the Reynolds number equals 200 and 800 in each channel. The working temperatures of both the feed and the draw solutions were varied from 20°C to 40°C. The feed solution had a concentration of 0.00355 solute mass fraction while the draw concentration was set to 0.0355 solute mass fraction. In all simulations, the laminar model was utilized. The results of the simulations suggest that the osmotic pressure is not the only factor that affects the water flux in forward osmosis when there is a temperature difference between the two sides of the membrane. The solution properties have a significant effect on the separation process. As the solution temperature increases, the viscosity decreases, which in turn increases the water permeation through the membrane. The feed temperature had a more substantial influence on the water flux compared to the draw temperature. Also, the effect of changing the flow rate did not change the results substantially.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9449036717414856, "perplexity": 502.67884099985207}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00672.warc.gz"} |
https://www.meritnation.com/cbse-class-12-science/math/rd-sharma-xi-2018/values-of-trigonometric-functions-at-sum-or-difference-of-angles/textbook-solutions/46_1_3396_24537_7.19_70574 | RD Sharma XI 2018 Solutions for Class 12 Science Math Chapter 7 Values Of Trigonometric Functions At Sum Or Difference Of Angles are provided here with simple step-by-step explanations. These solutions for Values Of Trigonometric Functions At Sum Or Difference Of Angles are extremely popular among class 12 Science students for Math Values Of Trigonometric Functions At Sum Or Difference Of Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2018 Book of class 12 Science Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2018 Solutions. All RD Sharma XI 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.
Question 1:
If and , where 0 < A, $\mathrm{B}<\frac{\mathrm{\pi }}{2}$, find the values of the following:
(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A − B)
(iv) cos (A − B)
Now,
Question 2:
(a) If , where $\frac{\mathrm{\pi }}{2}$< A < π and 0 < B < $\frac{\mathrm{\pi }}{2}$, find the following:
(i) sin (A + B)
(ii) cos (A + B)
(b) If , where A and B both lie in second quadrant, find the value of sin (A + B).
Question 3:
If , where π < A < $\frac{3\mathrm{\pi }}{2}\mathrm{and}\frac{3\mathrm{\pi }}{2}$< B < 2π, find the following:
(i) sin (A + B)
(ii) cos (A + B)
Question 4:
If , where π < A < $\frac{3\mathrm{\pi }}{2}$and 0 < B < $\frac{\mathrm{\pi }}{2}$, find tan (A + B).
Question 5:
If , where $\frac{\mathrm{\pi }}{2}$< A < π and $\frac{3\mathrm{\pi }}{2}$ < B < 2π, find tan (AB).
Question 6:
If , where $\frac{\mathrm{\pi }}{2}$ < A < π and 0 < B < $\frac{\mathrm{\pi }}{2}$, find the following:
(i) tan (A + B)
(ii) tan (AB)
Question 7:
Evaluate the following:
(i) sin 78° cos 18° − cos 78° sin 18°
(ii) cos 47° cos 13° − sin 47° sin 13°
(iii) sin 36° cos 9° + cos 36° sin 9°
(iv) cos 80° cos 20° + sin 80° sin 20°
Question 8:
If , where A lies in the second quadrant and B in the third quadrant, find the values of the following:
(i) sin (A + B)
(ii) cos (A + B)
(iii) tan (A + B)
Question 9:
Prove that: $\frac{7\pi }{12}+\mathrm{cos}\frac{\pi }{12}=\mathrm{sin}\frac{5\pi }{12}-\mathrm{sin}\frac{\pi }{12}$
LHS = cos105o + cos15o
= cos(90o + 15o) + cos(90o $-$ 75o)
= - sin 15o + sin 75o [As cos(90o+A) = $-$ sin A and cos(90o $-$ B) = sin B]
= sin 75o $-$ sin 15o
= RHS
Hence proved.
Prove that .
Prove that
(i) .
(ii)
(ii)
(i)
Question 12:
Prove that:
(i) $\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{cos}\left(\frac{\pi }{6}+x\right)+\mathrm{cos}\left(\frac{\pi }{3}-x\right)\mathrm{sin}\left(\frac{\pi }{6}+x\right)=1$
(ii) $\mathrm{sin}\left(\frac{4\mathrm{\pi }}{9}+7\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{9}+7\right)-\mathrm{cos}\left(\frac{4\mathrm{\pi }}{9}+7\right)\mathrm{sin}\left(\frac{\mathrm{\pi }}{9}+7\right)=\frac{\sqrt{3}}{2}$
(iii) $\mathrm{sin}\left(\frac{3\mathrm{\pi }}{8}-5\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{8}+5\right)+\mathrm{cos}\left(\frac{3\mathrm{\pi }}{8}-5\right)\mathrm{sin}\left(\frac{\mathrm{\pi }}{8}+5\right)=1$
(i)
(ii)
(iii)
Prove that .
Question 14:
(i) If , prove that $A+B=\frac{\mathrm{\pi }}{4}$.
(ii) If , then prove that $A-B=\frac{\mathrm{\pi }}{4}$.
(i)
(ii)
Question 15:
Prove that:
(i) ${\mathrm{cos}}^{2}45°-{\mathrm{sin}}^{2}15°=\frac{\sqrt{3}}{4}$
(ii) sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
(i)
Hence proved.
(ii)
Question 16:
Prove that:
(i)
(ii)
(iii)
(iv) sin2 B = sin2 A + sin2 (AB) − 2 sin A cos B sin (A B)
(v) cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
(vi)
Question 17:
Prove that:
(i) tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
(ii) $\mathrm{tan}\frac{\pi }{12}+\mathrm{tan}\frac{\pi }{6}+\mathrm{tan}\frac{\pi }{12}\mathrm{tan}\frac{\pi }{6}=1$
(iii) tan 36° + tan 9° + tan 36° tan 9° = 1
(iv) tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
Prove that:
Question 19:
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
Question 20:
If tan A = x tan B, prove that .
Question 21:
If tan (A + B) = x and tan (AB) = y, find the values of tan 2A and tan 2B.
Question 22:
If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.
Question 23:
If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) $\frac{1}{a}-\frac{1}{b}$.
Given:
Question 24:
If x lies in the first quadrant and , then prove that:
Question 25:
If tan x + , then prove that .
Question 26:
If sin (α + β) = 1 and sin (α − β)$=\frac{1}{2}$, where 0 ≤ α, $\mathrm{\beta }\le \frac{\mathrm{\pi }}{2}$, then find the values of tan (α + 2β) and tan (2α + β).
Question 27:
If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).
Question 28:
If sin α + sin β = a and cos α + cos β = b, show that
(i)
(ii)
(i)
Now,
From (1) and (2), we have
(ii)
Prove that:
(i)
(ii)
(iii)
Question 30:
If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.
Given:
Question 31:
If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
Question 32:
If angle $\theta$ is divided into two parts such that the tangents of one part is $\lambda$ times the tangent of other, and $\varphi$ is their difference, then show that $\mathrm{sin}\theta =\frac{\lambda +1}{\lambda -1}\mathrm{sin}\varphi$. [NCERT EXEMPLER]
Let $\alpha$ and $\beta$ be the two parts of angle $\theta$. Then,
$\theta =\alpha +\beta$ and $\varphi =\alpha -\beta$ (Given)
Now,
Applying componendo and dividendo, we get
Question 33:
If $\mathrm{tan}\theta =\frac{\mathrm{sin}\alpha -\mathrm{cos}\alpha }{\mathrm{sin}\alpha +\mathrm{cos}\alpha }$, then show that $\mathrm{sin}\alpha +\mathrm{cos}\alpha =\sqrt{2}\mathrm{cos}\theta$. [NCERT EXEMPLER]
$\mathrm{tan}\theta =\frac{\mathrm{sin}\alpha -\mathrm{cos}\alpha }{\mathrm{sin}\alpha +\mathrm{cos}\alpha }$
Dividing numerator and denominator on the RHS by $\mathrm{cos}\alpha$, we get
Now,
$\mathrm{sin}\alpha +\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}+\theta \right)+\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}+\theta \right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\frac{\mathrm{\pi }}{4}\mathrm{cos}\theta +\mathrm{cos}\frac{\mathrm{\pi }}{4}\mathrm{sin}\theta +\mathrm{cos}\frac{\mathrm{\pi }}{4}\mathrm{cos}\theta -\mathrm{sin}\frac{\mathrm{\pi }}{4}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}\mathrm{cos}\theta +\frac{1}{\sqrt{2}}\mathrm{sin}\theta +\frac{1}{\sqrt{2}}\mathrm{cos}\theta -\frac{1}{\sqrt{2}}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{2}}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\sqrt{2}\mathrm{cos}\theta$
Question 34:
If α and β are two solutions of the equation a tan x + b sec x = c, then find the values of sin (α + β) and cos (α + β).
Question 1:
Find the maximum and minimum values of each of the following trigonometrical expressions:
(i) 12 sin x − 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii)
(iv) sin x − cos x + 1
(i)
(ii)
(iii)
(iv)
Question 2:
Reduce each of the following expressions to the sine and cosine of a single expression:
(i)
(ii) cos x − sin x
(iii) 24 cos x + 7 sin x
Question 3:
Show that sin 100° − sin 10° is positive.
Question 4:
Prove that lies between .
Question 1:
If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ.
Question 2:
If x cos θ = y cos , then write the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
Question 3:
Write the maximum and minimum values of 3 cos x + 4 sin x + 5.
Question 4:
Write the maximum value of 12 sin x − 9 sin2 x.
Question 5:
If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
Question 6:
Write the interval in which the value of 5 cos x + 3 cos $\left(x+\frac{\pi }{3}\right)+3$ lies.
Question 7:
If tan (A + B) = p and tan (AB) = q, then write the value of tan 2B.
Question 8:
If , then write the value of tan x tan y.
Question 9:
If a = b , then write the value of ab + bc + ca.
Question 10:
If A + B = C, then write the value of tan A tan B tan C.
Question 11:
If sin α − sin β = a and cos α + cos β = b, then write the value of cos (α + β).
Question 12:
If tan $\mathrm{\alpha }=\frac{1}{1+{2}^{-x}}$and , then write the value of α + β lying in the interval .
Question 1:
The value of ${\mathrm{sin}}^{2}\frac{5\mathrm{\pi }}{12}-{\mathrm{sin}}^{2}\frac{\mathrm{\pi }}{12}$ is
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) 1
(d) 0
(b) $\frac{\sqrt{3}}{2}$
Question 2:
If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
(a) 0
(b) −1
(c) 1
(d) None of these
(b) −1
π = 180°
$\mathrm{sec}A\left(\mathrm{cos}B\mathrm{cos}C-\mathrm{sin}B\mathrm{sin}C\right)=\frac{\mathrm{cos}B\mathrm{cos}\left(\mathrm{\pi }-\left(A+B\right)\right)-\mathrm{sin}B\mathrm{sin}\left(\mathrm{\pi }-\left(A+B\right)\right)}{\mathrm{cos}A}$
We know that, ,
$\therefore \mathrm{sec}A\left(\mathrm{cos}B\mathrm{cos}C-\mathrm{sin}B\mathrm{sin}C\right)=\frac{\mathrm{cos}B\mathrm{cos}\left(A+B\right)-\mathrm{sin}B\mathrm{sin}\left(A+B\right)}{\mathrm{cos}A}$
Now, using the identities $\mathrm{cos}\left(A+B\right)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B$ and $\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B$, we get
$\mathrm{sec}A\left(\mathrm{cos}B\mathrm{cos}C-\mathrm{sin}B\mathrm{sin}C\right)=\frac{-\mathrm{cos}A\mathrm{cos}{B}^{2}+\mathrm{cos}B\mathrm{sin}A\mathrm{sin}B-\mathrm{sin}B\mathrm{sin}A\mathrm{cos}B-{\mathrm{sin}}^{2}B\mathrm{cos}A}{\mathrm{cos}A}$
$⇒\mathrm{sec}A\left(\mathrm{cos}B\mathrm{cos}C-\mathrm{sin}B\mathrm{sin}C\right)=\frac{-\mathrm{cos}A\left({\mathrm{cos}}^{2}B+{\mathrm{sin}}^{2}B\right)}{\mathrm{cos}A}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A\left(\mathrm{cos}B\mathrm{cos}C-\mathrm{sin}B\mathrm{sin}C\right)=\frac{-\mathrm{cos}A}{\mathrm{cos}A}=-1$
Question 3:
tan 20° + tan 40° + $\sqrt{3}$ tan 20° tan 40° is equal to
(a) $\frac{\sqrt{3}}{4}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\sqrt{3}$
(d) 1
(c) $\sqrt{3}$
Question 4:
If , then the value of A + B is
(a) 0
(b) $\frac{\mathrm{\pi }}{2}$
(c) $\frac{\mathrm{\pi }}{3}$
(d) $\frac{\mathrm{\pi }}{4}$
(d) $\frac{\mathrm{\pi }}{4}$
Question 5:
If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
(a) 0
(b) 5
(c) 1
(d) None of these
(a) 0
Question 6:
If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) None of these
(c) $\frac{1}{6}$
In triangle ABC,
If tan A+tan B+tan C =6,
tan A tan B tan C =6
Question 7:
tan 3A − tan 2A − tan A =
(a) tan 3 A tan 2 A tan A
(b) −tan 3 A tan 2 A tan A
(c) tan A tan 2 A − tan 2 A tan 3 A − tan 3 A tan A
(d) None of these
(a)
Question 8:
If A + B + C = π, then is equal to
(a) tan A tan B tan C
(b) 0
(c) 1
(d) None of these
(c) 1
π = 180°
Using tan(180 – A) = -tan A, we get:
Question 9:
If , where P and Q both are acute angles. Then, the value of PQ is
(a) $\frac{\mathrm{\pi }}{6}$
(b) $\frac{\mathrm{\pi }}{3}$
(c) $\frac{\mathrm{\pi }}{4}$
(d) $\frac{\mathrm{\pi }}{12}$
(b) 60⁰ = $\frac{\mathrm{\pi }}{3}$
$=\frac{1}{7}×\frac{13}{14}+\frac{4\sqrt{3}}{7}×\frac{3\sqrt{3}}{14}\phantom{\rule{0ex}{0ex}}=\frac{13+36}{98}\phantom{\rule{0ex}{0ex}}$
$=\frac{49}{98}$
Hence, the correct answer is option B.
Question 10:
If cot (α + β) = 0, sin (α + 2β) is equal to
(a) sin α
(b) cos 2 β
(c) cos α
(d) sin 2 α
(a)
(a) tan 55°
(b) cot 55°
(c) −tan 35°
(d) −cot 35°
(a)
Question 12:
The value of is
(a)
(b) 0
(c)
(d) $\frac{1}{2}$
(a)
Question 13:
If tan θ1 tan θ2 = k, then
(a) $\frac{1+k}{1-k}$
(b) $\frac{1-k}{1+k}$
(c) $\frac{k+1}{k-1}$
(d) $\frac{k-1}{k+1}$
(a) $\frac{1+k}{1-k}$
$\frac{1+\mathrm{tan}{\theta }_{1}\mathrm{tan}{\theta }_{2}}{1-\mathrm{tan}{\theta }_{1}\mathrm{tan}{\theta }_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1+k}{1-k}$
Question 14:
If sin (π cos x) = cos (π sin x), then sin 2 x =
(a) $±\frac{3}{4}$
(b) $±\frac{4}{3}$
(c) $±\frac{1}{3}$
(d) none of these
Question 15:
If $\mathrm{tan}\theta =\frac{1}{2}$ and $\mathrm{tan}\varphi =\frac{1}{3}$, then the value of $\theta +\varphi$ is
(a) $\frac{\mathrm{\pi }}{6}$ (b) $\mathrm{\pi }$ (c) 0 (d) $\frac{\mathrm{\pi }}{4}$
It is given that $\mathrm{tan}\theta =\frac{1}{2}$ and $\mathrm{tan}\varphi =\frac{1}{3}$.
Now,
Hence, the correct answer is option D.
Question 16:
The value of cos (36° − A) cos (36° + A) + cos (54° + A) cos (54° − A) is
(a) sin 2A
(b) cos 2A
(c) cos 3A
(d) sin 3A
(b) cos 2A
Question 17:
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
(a) a2 + 1
(b) a2 + 2
(c) a2 − 2
(d) None of these
(c) ${a}^{2}-2$
Question 18:
If tan (AB) = 1 and sec (A + B) = $\frac{2}{\sqrt{3}}$, the smallest positive value of B is
(a)
(b)
(c) $\frac{13\mathrm{\pi }}{24}$
(d)
(b)
Question 19:
If AB = π/4, then (1 + tan A) (1 − tan B) is equal to
(a) 2
(b) 1
(c) 0
(d) 3
(a) $2$
Question 20:
The maximum value of ${\mathrm{sin}}^{2}\left(\frac{2\pi }{3}+x\right)+{\mathrm{sin}}^{2}\left(\frac{2\pi }{3}-x\right)$ is
(a) 1/2
(b) 3/2
(c) 1/4
(d) 3/4
(b) $\frac{3}{2}$
$\frac{2\pi }{3}=120°$
Question 21:
If cos (AB)$=\frac{3}{5}$and tan A tan B = 2, then
(a)
(b)
(c)
(d)
(a) $\frac{1}{5}$
Question 22:
If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =
(a) −1
(b) $\frac{1}{2}$
(c) $-\frac{1}{2}$
(d) None of these
(c)$\frac{-1}{2}$
Question 23:
If $\mathrm{tan}\alpha =\frac{x}{x+1}$ and $\mathrm{tan}\beta =\frac{1}{2x+1}$, then $\alpha +\beta$ is equal to
(a) $\frac{\mathrm{\pi }}{2}$ (b) $\frac{\mathrm{\pi }}{3}$ (c) $\frac{\mathrm{\pi }}{6}$ (d) $\frac{\mathrm{\pi }}{4}$
It is given that $\mathrm{tan}\alpha =\frac{x}{x+1}$ and $\mathrm{tan}\beta =\frac{1}{2x+1}$.
$=\frac{2{x}^{2}+2x+1}{2{x}^{2}+2x+1}\phantom{\rule{0ex}{0ex}}=1$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 110, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9881904125213623, "perplexity": 4262.18885797675}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00335.warc.gz"} |
https://www.groundai.com/project/complete-reducibility-of-subgroups-of-reductive-algebraic-groups-over-nonperfect-fields-3/ | Complete reducibility of subgroups of reductive algebraic groups over nonperfect fields \@slowromancapiii@
# Complete reducibility of subgroups of reductive algebraic groups over nonperfect fields 3
## Abstract
Let be a nonperfect separably closed field. Let be a (possibly non-connected) reductive group defined over . We study rationality problems for Serre’s notion of complete reducibility of subgroups of . In our previous work, we constructed examples of subgroups of that are -completely reducible but not -completely reducible over (and vice versa). In this paper, we give a theoretical underpinning of those constructions. To illustrate our result, we present a new such example in a non-connected reductive group of type in characteristic . Then using Geometric Invariant Theory, we generalize the theoretical result above obtaining a new result on the structure of -(and -) orbits in an arbitrary affine -variety. We translate our result into the language of spherical buildings to give a new topological view. A problem on centralizers of completely reducible subgroups and a problem concerning the number of conjugacy classes are also considered.
Keywords: algebraic groups, geometric invariant theory, complete reducibility, rationality, spherical buildings
## 1 Introduction
Let be a field. We write for an algebraic closure of . Let be a (possibly non-connected) affine algebraic -group: we regard as a -defined algebraic group together with a choice of -structure in the sense of Borel [9, AG. 11]. We say that is reductive if the unipotent radical of is trivial. Throughout, is always a (possibly non-connected) reductive -group. In this paper, we continue our study of rationality problems for complete reducibility of subgroups of [37][34]. By a subgroup of we mean a (possibly non--defined) closed subgroup of . If a subgroup of needs to be -defined (or needs to be connected) in some statement, we explicit say so. Recall [26, Sec. 3]
###### Definition 1.1.
A subgroup of is called -completely reducible over (-cr over for short) if whenever is contained in a -defined -parabolic subgroup of , then is contained in a -defined -Levi subgroup of . In particular if is not contained in any proper -defined -parabolic subgroup of , is called -irreducible over (-ir over for short).
We define -parabolic subgroups and -Levi subgroups in the next section (Definition 2.2). These concepts are essential to extend the notion of complete reducibility (initially defined only for subgroups of connected [26, Sec. 3]) to subgroups of non-connected [3][4, Sec. 6]. We defined complete reducibility for a possibly non--defined subgroup of . This is because for a subgroup of , some closely related important subgroups of are not necessarily -defined even if is -defined. For example, centralizers or normalizers of -subgroups of are not necessarily -defined; see [37, Thm. 1.2 and Thm. 1.7] for such examples. If is connected and is a subgroup of , our notion of complete reducibility agrees with the usual one of Serre.
###### Definition 1.2.
Let be an algebraic extension of . We say that a subgroup of is -completely reducible over (-cr over for short) if whenever is contained in a -defined -parabolic subgroup of , is contained in a -defined -Levi subgroup of . In particular, if is not contained in any proper -defined -parabolic subgroup of , is -irreducible over (-ir over for short). We simply say that is -cr (-ir for short) if is -cr over (-ir over ).
So far, most studies on complete reducibility is for complete reducibility over only; see [19][29][30] for example. Not much is known on complete reducibility over (especially for nonperfect ) except a few theoretical results and important examples in [4, Sec. 5][1][37][34]. In particular, in [36, Thm. 1.10] [35, Thm. 1.8][37, Thm. 1.2], we found several examples of -subgroups of that are -cr over but not -cr (and vice versa). The main result of this paper is to give a theoretical underpinning for our (possibly somewhat mysterious) construction of those examples. For an algebraic extension of and an affine group , we denote the set of -points of by . We write for where .
###### Theorem 1.3.
Let be a nonperfect separably closed field. Suppose that a subgroup of is -cr but not -cr over . Let be a minimal -defined -parabolic subgroup of containing . Then there exists a unipotent element such that is -cr over .
###### Theorem 1.4.
Let be a nonperfect separably closed field. Suppose that a subgroup of is -cr over but not -cr. Let be a minimal -defined -parabolic subgroup of containing . Then
1. is -ir over for some -defined -Levi subgroup of .
2. Moreover, there exists an element such that is not -cr over .
To illustrate our theoretical results (Theorems 1.31.4) and ideas in the proofs, we present a new example of a -subgroup of that is -cr over but not -cr (and vice versa).
###### Theorem 1.5.
Let be a nonperfect separably closed field of characteristic . Let be a simple -group of type . Let be a non-trivial element in the graph automorphism of . Let . Then there exists a -subgroup of that is -cr over but not -cr (and vice versa).
A few comments are in order. First, the non-perfectness of is (almost) essential in Theorems 1.31.4, and 1.5 in view of the following [5, Thm. 1.1]:
###### Proposition 1.6.
Let be connected. Let be a subgroup of . Then is -cr over if and only if is -cr over .
So in particular if is perfect and is connected, a subgroup of is -cr over if and only if it is -cr. The forward direction of Proposition 1.6 holds for non-connected . The reverse direction depends on the recently proved center conjecture of Tits [26][31][21] in spherical buildings, but this method does not work for non-connected ; the set of -parabolic subgroups does not form a simplicial complex in the usual sense of Tits [32] as we have shown in [34, Thm. 1.12]. In the following we assume that is separably closed. So every maximal -torus of splits over , thus is -split. This simplifies arguments in many places. For the theory of complete reducibility over arbitrary , see [1][2].
Second, note that the -definedness of in Theorem 1.5 is important. Actually it is not difficult to find a -subgroup with the desired property. For our construction to work, it is essential for to be nonseparable in . We write or for the Lie algebra of . Recall [7, Def. 1.1]
###### Definition 1.7.
A subgroup of is nonseparable if the dimension of is strictly smaller than the dimension of (where acts on via the adjoint action). In other words, the scheme-theoretic centralizer of in (in the sense of [13, Def. A.1.9]) is not smooth.
We exhibit the importance of nonseparability of in the proof of Theorem 1.5. Nonseparable -subgroups of are hard to find, and only a handful examples are known [7, Sec. 7][36, Thm. 1.10] [35, Thm. 1.8][37, Thm. 1.2]. Note that if characteristic of is very good for connected , every subgroup of is separable [7, Thm. 1.2]. Thus, to find a nonseparable subgroup we are forced to work in small (at least for connected ). See [7][16] for more on separability.
Our second main result is a generalization of Theorems 1.3 and 1.4 using the language of Geometric Invariant Theory (GIT for short) [22]. Let be a (possibly non-connected) affine -variety. When acts on -morphically, we say that is a -variety. One of the main themes of GIT is to study the structure of -orbits (and -orbits) in [18][8][1]. Recently studies on completely reducibility (over ) via GIT have been very fruitful; GIT gives a very short and uniform proof for many results [4][8][1]. This makes a striking contrast to traditional representation theoretic methods (which depend on long case-by-case analyses) [19][29][30].
We recall the following algebro-geometric characterization for complete reducibility (over ) via GIT ([4, Prop. 2.16, Thm. 3.1] and [1, Thm. 9.3]). This turns problems on complete reducibility into problems on the structure of -(or -) orbits. Let be a subgroup of such that for some and . Suppose that (and ) acts on via simultaneous conjugation.
###### Proposition 1.8.
is -cr if and only if is Zariski closed in . Moreover, is -cr over if and only if is cocharacter closed over .
The definition of a cocharacter closed orbit is given in the next section (Definition 2.7). Using Proposition 1.8 and various techniques from GIT we can sometimes generalize results on complete reducibility (over ) to obtain new results on GIT where (or ) acts on an arbitrary affine -variety rather than on some tuple of ; see [8][1] for example. We follow the same philosophy here and generalize Theorems 1.3 and 1.4.
###### Theorem 1.9.
Let be nonperfect. Suppose that there exists such that is Zariski closed but is not cocharacter closed over . Let be the set of -cocharacters of destabilizing over . Pick such that is minimal among -parabolic subgroups for . Then there exists a unipotent element such that is cocharacter closed over .
###### Theorem 1.10.
Let be nonperfect. Suppose that there exists such that is cocharacter closed over but is not Zariski closed. Let be the set of -cocharacters of destabilizing over . Pick such that is minimal among -parabolic subgroups for . Then
1. There exists such that and fixes .
2. Any -defined cocharacter of destabilizing over is central in .
3. Moreover, there exists an element such that is not cocharacter closed over .
Roughly speaking, we say that is destabilized over by a -cocharacter of if is taken outside of by taking a limit of along in the sense of GIT [18][22]; see Definition 2.8 for the precise definition. Note that if is perfect, Theorems 1.9 and 1.10 have no content: in that case is cocharacter closed if and only if is Zariski closed [1, Cor. 7.2, Prop. 7.4].
To complement the paper we also investigate the structure of centralizers of completely reducible subgroups of . In particular we ask [34, Open Problem 1.4]:
###### Open Problem 1.11.
Suppose that a -subgroup of is -cr over . Is -cr over ?
We have some partial answer [34, Thm. 1.5]:
###### Proposition 1.12.
Let be connected. Suppose that a -subgroup of is -cr over . If is reductive, then it is -cr over .
We need the connectedness assumption in Proposition 1.12 since it depends on the center conjecture of Tits. If (or more generally if is perfect), the answer to Open Problem 1.11 is “yes” by [4, Cor. 3.17] (and Proposition 1.6). A trouble arises for nonperfect since is not necessarily reductive even if is -cr over [37, Rem. 3.11]. This does not happen if by [4, Prop. 3.12] that depends on a deep result of Richardson [24, Thm. A]. The reductivity of was crucial in the proof of [4, Cor. 3.17] to apply a tool from GIT.
In general, if is nonperfect, even if a -subgroup of is -cr over , it is not necessarily reductive [37, Prop. 1.10]. This pathology happens because the classical construction of Borel-Tits [11, Prop. 3.1] fails over nonperfect ; see [37, Sec. 3.2]. This does not happen if ; a -cr subgroup is always reductive [26, Prop. 4.1].
Here is our third main result in this paper. Let be connected. Fix a maximal -torus of . We write for an automorphism of which normalizes and induces on (the set of positive roots of ). It is known that for of not type , , or , and for of type , , or where is the longest element of the Weyl group of and is a suitable graph automorphism of (cf. [19, Proof of Thm. 4.1]).
###### Theorem 1.13.
Let be connected. Suppose that a semisimple -subgroup of is -cr over . Let be a minimal -parabolic subgroup containing , and be a -Levi subgroup of . If the automorphism of extends to an automorphism of (in particular if is not of type , , or ), then is -cr over .
Here is the structure of the paper. In Section 2, we set out the notation and show some preliminary results. Then in Section 3, we prove our first main result (Theorems 1.3 and 1.4). In Section 4, we present the example (Theorem 1.5). In Section 5, we generalize Theorems 1.3 and 1.4 and prove our second main result (Theorems 1.9 and 1.10). In Section 6, we translate Theorems 1.3 and 1.4 into the language of spherical buildings, and prove Theorems 6.4 and 6.5. This gives a new topological perspective for the rationality problems for complete reducibility and GIT. Then in Section 7, we attack Open Problem 1.11, and prove Theorem 1.13 in a purely combinatorial way. In Section 8, we consider a problem on the number of conjugacy classes and prove Theorem 8.1. We note that nonseparability comes into play in a crucial way in the proof of Theorem 8.1.
## 2 Preliminaries
Throughout, we denote by a separably closed field. Our references for algebraic groups are [9][10][13][17], and [28].
Let be a (possibly non-connected) affine algebraic group. We write for the identity component of . It is clear that if is -defined, is -defined. We write for the derived group of . A reductive group is called simple as an algebraic group if is connected and all proper normal subgroups of are finite. We write and ( and ) for the set of -characters and -cocharacters (-characters and -cocharacters) of respectively. For -characters and -cocharacters we simply say characters and cocharacters of .
Fix a maximal -torus of (such a exists by [9, Cor. 18.8]). Then splits over since is separably closed. Let denote the set of roots of with respect to . We sometimes write for . Let . We write for the corresponding root subgroup of . We define . Let . Let be the coroot corresponding to . Then is a -homomorphism such that for some . Let denote the reflection corresponding to in the Weyl group of . Each acts on the set of roots by the following formula [28, Lem. 7.1.8]: By [12, Prop. 6.4.2, Lem. 7.2.1] we can choose -homomorphisms so that
We recall the notions of -parabolic subgroups and -Levi subgroups from [25, Sec. 2.1–2.3]. These notions are essential to define -complete reducibility for subgroups of non-connected and also to translate results on complete reducibility into the language of GIT; see [3] and [4, Sec. 6].
###### Definition 2.1.
Let be a affine -variety. Let be a -morphism of affine -varieties. We say that exists if there exists a -morphism (necessarily unique) whose restriction to is . If this limit exists, we set .
###### Definition 2.2.
Let . Define
We call an -parabolic subgroup of , an -Levi subgroup of . Note that the unipotent radical of . If is -defined, , , and are -defined [25, Sec. 2.1-2.3]. Any -defined parabolic subgroups and -defined Levi subgroups of arise in this way since is separably closed. It is well known that . Note that -defined -Levi subgroups of a -defined -parabolic subgroup of are -conjugate [8, Lem. 2.5(iii)]. Let be a reductive -subgroup of . Then, there is a natural inclusion of -cocharacter groups. Let . We write or just for the -parabolic subgroup of corresponding to , and for the -parabolic subgroup of corresponding to . It is clear that and . If is connected, -parabolic subgroups and -Levi subgroups are parabolic subgroups and Levi subgroups in the usual sense [28, Prop. 8.4.5].
The next result is used repeatedly to reduce problems on -complete reducibility to those on -complete reducibility where is an -Levi subgroup of .
###### Proposition 2.3.
Suppose that a subgroup of is contained in a -defined -Levi subgroup of . Then is -cr over if and only if it is -cr over .
###### Proof.
This follows from Proposition 1.8 and [1, Thm. 5.4(ii)]. ∎
The next result shows how complete reducibility behaves under central isogenies.
###### Definition 2.4.
Let and be reductive -groups. A -isogeny is central if is central in where is the differential of at the identity of and is the Lie algebra of .
###### Proposition 2.5.
Let and be reductive -groups. Let and be subgroups of and be subgroups of and respectively. Let be a central -isogeny.
1. Suppose that and (in particular if is connected). If is -cr over , then is -cr over .
2. Suppose that and . If is -cr over , then is -cr over .
###### Proof.
Proposition 1.8 and [1, Cor. 5.3] show that a subgroup of a reductive is -cr over if and only if it is -cr over . Now the result follows from the connected case [37, Prop. 1.12]. ∎
###### Remark 2.6.
In Proposition 2.5 if we know that a -defined -parabolic subgroup of always arises as the inverse image of a -defined -parabolic subgroup of , then a similar argument as in the proof of [37, Prop. 1.12] goes through and we can omit the assumptions “ and ” in Part 1 and “ and ” in Part 2. We do not know this is the case or not.
Now we recall some terminology form GIT [1, Def. 1.1, Sec. 2.4]. Let be a -variety. Let .
###### Definition 2.7.
We say that is cocharacter closed over if for every such that exists, is -conjugate to . Moreover, we say that is cocharacter closed if for every cocharacter of such that exists, is -conjugate to .
Note that by the Hilbert-Mumford theorem [18], is cocharacter closed if and only if it is Zariski closed.
###### Definition 2.8.
Let . We say that destabilize over if exists. Moreover if exists and is not -conjugate to , we say that properly destabilizes over . Similarly, for , if exists, we say that destabilize . If exists for and is not -conjugate to , we say that properly destabilizes .
We use the following very useful results from GIT [8, Thm. 3.3] and [1, Cor. 5.1].
###### Proposition 2.9.
Let be perfect. Suppose that exists for and is -conjugate to . Then is -conjugate to .
For nonperfect , we do not know whether Proposition 2.9 still holds [1, Question 7.8]. It is known that if the centralizer of in is separable, it holds for nonperfect [1, Thm. 7.1]. See [1] and [2] for details.
###### Proposition 2.10.
Suppose that exists for and is -conjugate to . If is cocharacter closed over , then is -conjugate to .
## 3 G-cr over k vs G-cr
We prove theorems 1.3 and 1.4. Our proof works for both connected and non-connected in a uniform way.
###### Proof of Theorem 1.3.
Since is not -cr over , there exists a proper -defined -parabolic subgroup of containing . Let be a minimal such -defined -parabolic subgroup where . Since is -cr and -Levi subgroups of are -conjugate by [4, Cor. 6.7], there exists such that is contained in . Then is contained in . Suppose that is not -cr over . Then it is not -cr over by Proposition 2.3. So there exists a proper -defined -parabolic subgroup of containing . Thus is contained in a -defined -parabolic subgroup of . Then is contained in . Note that . Thus and we have . It is clear that is strictly contained in . This contradicts the minimality of . So we conclude that is -cr over . ∎
###### Proof of Theorem 1.4.
We start with Part 1. Let be a minimal -defined -parabolic subgroup of containing . Since is -cr over , there exists a -defined -Levi subgroup of containing . Since -defined -Levi subgroups of are -conjugate, there exists such that . Set . Suppose that is not -ir over . So there exists a -defined proper -parabolic subgroup of containing . Then we have . Since is a -defined -parabolic subgroup of strictly contained in , this contradicts the minimality of .
For part 2, let be a minimal -defined -Levi subgroup containing . Since is not -cr, there exists a proper -parabolic subgroup of containing . Let be a minimal such -parabolic subgroup of . Since an -parabolic subgroup of is -conjugate to a -defined -parabolic subgroup of , there exists such that is a -defined -parabolic subgroup of . Then . Suppose that is -cr over . Then is -cr over by Proposition 2.3, so there exist a -defined -Levi subgroup of containing . Note that is not -cr since is not -cr. Then there exists a proper -parabolic subgroup of containing . Thus is an -parabolic subgroup of containing . Then is an -parabolic subgroup of containing . It is clear that is a proper subgroup of . This contradicts the minimality of . Thus is not -cr over . ∎
###### Remark 3.1.
Although Theorems 1.3 and 1.4 (and ideas in the proofs) explain necessary conditions to have examples of a subgroup of that is -cr over but not -cr (or vice versa), it is still a difficult problem to find concrete such examples with a -defined . In the next section, we use the converse of Theorems 1.3 and 1.4: we start with some subgroup of and conjugate it by (or ) as in the proof of Theorems 1.3 and 1.4 to obtain a subgroup with the desired property. For our construction to work, (or ) needs to be chosen very carefully and the choice is closely related to the nonseparability of . We show all details in the next section. The same idea was used in [7][37][35], and [36].
## 4 The D4 example
In this section we prove Theorem 1.5. We use the triality of in an essential way.
Let be a simple algebraic group of type defined over a nonperfect field of characteristic . Fix a maximal -torus of and a -defined Borel subgroup of . let be the set of roots corresponding to , and be the set of positive roots of corresponding to and . The following Dynkin diagram defines the set of simple roots of .
Let where is the non-trivial element of the graph automorphism group of (normalizing and ) as the diagram defines; we have , and is fixed by . We label in the following. The corresponding negative roos are defined accordingly. Note that Roots 1, 2, 3, 4 correspond to , , , respectively.
Define . Then
Pλ =⟨T,σ,Uζ∣ζ∈Ψ(~G)+∪{−1,−2,−3}⟩, Lλ =⟨T,σ,Uζ∣ζ∈{±1,±2,±3}⟩, Ru(Pλ) =⟨Uζ∣ζ∈Ψ(~G)+∖{1,2,3}⟩.
Let . Let . Define
H:=v(√a)⋅⟨(nασ),(α+γ)∨(¯¯¯k∗)⟩.
Here is our first main result in this section.
###### Proposition 4.1.
is -defined. Moreover, is -cr but not -cr over .
###### Proof.
First, we have Using this and the commutation relations [17, Lem. 32.5 and Prop. 33.3], we obtain
v(√a)⋅(nασ)=(nασ)ϵ12(a).
An easy computation shows that commutes with . Now it is clear that is -defined.
Now we show that is -cr. It is sufficient to show that is -cr since it is -conjugate to . Since is contained in , by Proposition 2.3 it is enough to show that is -cr. We actually show that is -ir. Note that . We have
(nασ)⋅(α+γ)∨(¯¯¯k∗)=(γ+δ)∨(¯¯¯k∗),(nασ)3=nαnγnδ.
Thus contains , , and . Now it is clear that is -ir.
Next, we show that is not -cr over . Suppose the contrary. Clearly is contained in a -defined -parabolic subgroup . Then there exists a -defined -Levi subgroup of containing . Then by [8, Lem. 2.5(iii)] there exists such that is contained in . Thus . So . By [28, Prop. 8.2.1], we set
u:=∏ζ∈Ψ(Ru(Pλ))ϵζ(xζ).
We compute how acts . Using the labelling of the positive roots above, we have . We compute how acts on :
nασ=(45811107)(69)(12). (4.1)
Using this and the commutation relations,
u−1⋅(nασϵ12(a))= nασϵ7(x4+x7)ϵ10(x7+x10)ϵ9(x6+x9)ϵ11(x10+x11) ϵ6(x6+x9)ϵ8(x8+x11)ϵ4(x4+x5)ϵ5(x5+x8) ϵ12(x5x10+x5x11+x7x8+x7x11+x8x10+x92+a).
Thus if we must have
x4=x5=x7=x8=x10=x11,x6=x9, x5x10+x5x11+x7x8+x7x11+x8x10+x92+a=0.
Set . Then we have . Thus . This is impossible since and . We are done. ∎
###### Remark 4.2.
From the computations above we see that the curve is not contained in , but the corresponding element in , that is, is contained in . Then the argument in the proof of [36, Prop. 3.3] shows that is strictly smaller than . So is non-separable in .
Now we move on to the second main result in this section. We use the same , , and as above. We also use the same labelling of the roots of . Let . Let
K:=v(√a)⋅⟨nασ,(α+γ)∨(¯k∗)⟩=⟨nασϵ−12(a),(α+γ)∨(¯¯¯k∗)⟩.
Define
H:=⟨K,ϵ11(1)⟩.
###### Proposition 4.3.
is -defined. Moreover, is -ir over but not -cr.
###### Proof.
is clearly -defined. First, we show that is -ir over . Note that
v(√a)−1⋅H=⟨nασ,(α+γ)∨(¯¯¯k∗),ϵ11(1)ϵ2(√a)⟩.
Thus we see that is contained in . So is contained in .
###### Lemma 4.4.
is the unique proper -parabolic subgroup of containing .
###### Proof.
Suppose that is a proper -parabolic subgroup containing . In the proof of Proposition 4.1 we have shown that is -cr. Then there exists a -Levi subgroup of containing since is contained in . Since -Levi subgroups of are -conjugate by [8, Lem. 2.5(iii)], without loss, we set . Then , so centralizes . Recall that by [28, Thm. 13.4.2], is an open set of where is the opposite of containing .
.
###### Proof.
First of all, from Equation (4.1) we see that is contained in . Since , is also contained in . So is contained in . Set for some . Using Equation 4.1 and the commutation relations, we obtain
(nασ)⋅u =ϵ4(x7)ϵ5(x4)ϵ6(x9)ϵ7(x10)ϵ8(x5)ϵ9(x6)ϵ10(x11)ϵ11(x8)ϵ12(x5x10+x6x9+x12).
So, if we must have . But , so for . Then
(nασ)⋅u=ϵ6(x6)ϵ9(x6)ϵ12(x26+x12).
So we must have if | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9826029539108276, "perplexity": 1119.3908352063124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578675477.84/warc/CC-MAIN-20190424234327-20190425020327-00417.warc.gz"} |
https://www.physicsforums.com/threads/hamiltonian-mechanics-electromagnetic-field.568986/ | # Hamiltonian mechanics electromagnetic field
1. Jan 19, 2012
### Liquidxlax
1. The problem statement, all variables and given/known data
Let (V(x,t) , A(x,t)) be a 4-vector potential that constructs the electromagnetic field (in gaussian Units) by
E(x,t) = -∇V(x,t) - (1/c)δtA(x,t) , B = ∇xA , (x,t) elements of R3xRt
Consider the lagrangian
L=.5mv2 - eV(x,t) + (ev/c)(dot)A(x,t)
a) compute and interpret the Euler-lagrange equatinons of motion for this system
b) determine the hamiltonian
c) determine hamilton's equations of motion. Are they gauge invariant?
2. Relevant equations
3. The attempt at a solution
when i apply the euler-lagrange equations do i take the curl of A or the gradient? if gradient what does that mean to take the ∇A?
is there only one equation of motion because i fail to see the other possibility if there is one.
Otherwise i can solve the rest
thanks
Can you help with the solution or looking for help too?
Similar Discussions: Hamiltonian mechanics electromagnetic field | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.912251889705658, "perplexity": 2074.2136615837976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00024-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/orbital-velocities-in-the-schwartzschild-geometry.409241/ | # Orbital velocities in the Schwartzschild geometry
1. Jun 10, 2010
### espen180
I'm trying to use the tensor formulation of GR to calculate the velocity of a particle in a circular orbit around a black hole.
Here is the work I have done so far.
What concerns me is that I end up getting zero velocity when applying the metric to the differential equations I get from the geodesic equation. I wonder if I have made a miscalculation, but I am unable to find any, so maybe there is a misunderstanding on my part.
Any help is appreciated.
2. Jun 10, 2010
### George Jones
Staff Emeritus
3. Jun 10, 2010
### bcrowell
Staff Emeritus
There seems to be a problem with signs in (13), because both terms are negative-definite.
Using (12), and choosing t=0 to coincide with $\tau=0$, you can set $t=\beta \tau$, where $\beta$ is a constant. Let's also set $\omega=d\phi/der t$. Then (13) gives $\omega=\pm i\sqrt{2m/r^3}$, where $m=r_s/2$. This seems sort of right, since it coincides with Kepler's law of periods. However, it's imaginary due to the sign issue. It would also surprise me if Kepler's law of periods was relativistically exact when expressed in terms of the Schwarzschild coordinates, but maybe that's the case.
If you can fix the sign problem, then you seem to have the right result in the nonrelativistic limit of large r. You might then want to check the result in the case of r=3m, where I believe you should obtain lightlike circular orbits.
4. Jun 10, 2010
### starthaus
You are right, his equation (13) is in the "not-even-wrong" category. It is easu to see that since the correct Lagrangian, for the simplified case he's considering is:
$$L=(1-r_s/r)\frac{dt^2}{ds^2}-r^2\frac{d\phi^2}{ds^2}$$
5. Jun 10, 2010
### espen180
@George Jones , bcrowell
Thanks for pointing that out, and thank you for the reference. I traced the sign error back to a differentiation error when calculating the Christoffel symbol.
As for writing $t=\beta\tau$ I don't see how that will help since t does not appear in the other equations. I must be missing something.
I will try to arrive at a result and do the "reality checks" you mentioned.
@Starthaus
Thanks for your input. I am afraid I don't know how to arrive at or what to do with the Lagrangian. From it's appearance it looks just like the metric, so L=1 here, I imagine?
EDIT:
I arrived at $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}$$. The units match, but I doubt this is correct, since letting r=3m gives $$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{\sqrt{2m}}$$ while my intuition tells me it should be c.
Last edited: Jun 10, 2010
6. Jun 10, 2010
### starthaus
No, the correct equations are:
$$\frac{d^2t}{ds^2}=0$$
and
$$\frac{d^2\phi}{ds^2}=0$$
Hint: in your writeup you made $$dr=d\theta=0$$, remember? You need to think what that means.
7. Jun 10, 2010
### espen180
Yes, I arrived at these as well when I made the assumptions you mentioned. As for their meaning, I interpret it as circular (constant radial coordinate) motion around the equator ($\theta=\pi/2$) with constant velocity. In addition, there was a third equation I arrived at,
$$\frac{c^2r_s}{r^2}\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-r\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0$$
which, when I used the substitution
$$\left(\frac{\text{d}t}{d\tau}\right)^2=1+\frac{r^2}{c^2}\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2$$
which I got from the metric, gave me
$$\frac{c^2r_s}{r^2}+(r_s-r)\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0$$
which solves to
$$\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}$$
8. Jun 10, 2010
### starthaus
There are only two independent equations, the ones I mentioned to you.
The third Lagrange equation, exists only if r is variable and its correct form would have been:
$$\frac{r_s}{r^2}\frac{dt^2}{ds^2}-2r\frac{d\phi^2}{ds^2}=0$$
But you made $$dr=0$$ (this is why I gave you the hint), so the third equation does not exist. This is the root of your errors.
Last edited: Jun 10, 2010
9. Jun 11, 2010
### espen180
I don't understand why it shouldn't exist. I derived the general case, then assumed dr=0 and substituted that into the equations. dr doesn't even appear in that equation anymore, do I don't see how it has any influence.
EDIT: Where does the factor of 2 come from there? I somehow didn't appear in my calculations.
For the velocity, I obtained
$$v=\frac{\text{d}\phi}{\text{d}\tau}r=c\sqrt{\frac{r_s}{r-r_s}}=\sqrt{\frac{2GM}{r-\frac{2GM}{c^2}}}$$
My intuition says that this is wrong by a factor of $\sqrt{2}[/tex], since then it would give v=c at [itex]r=\frac{3GM}{c^2}$, but I don't see how that factor dissapeared.
Last edited: Jun 11, 2010
10. Jun 11, 2010
### starthaus
You started with the metric that has $$dr=0$$. therefore all your attempts to differentiate wrt $$r$$ should result in null terms. Yet, you clearly differentaite wrt $$r$$ in your derivation and this renders your derivation wrong.
From the Euler-Lagrange equations.
Yes, it is very wrong.
From the correct equation $$\frac{d^2\phi}{ds^2}=0$$ you should obtain (no surprise):
$$\frac{d\phi}{ds}=constant=\omega$$
The trajectory is completed by the other obvious equation
$$r=R=constant$$
You get one more interesting equation, that gives u the time dilation. Start with:
$$ds^2=(1-r_s/R)dt^2-(Rd\phi)^2$$ and you get:
$$\frac{dt}{ds}=\sqrt{\frac{1+(R\omega)^2}{1-r_s/R}}$$
or:
$$\frac{ds}{dt}=\sqrt{1-r_s/R}\sqrt{1-\frac{(R\omega)^2}{1-r_s/R}$$
The last equation gives you the hint that:
$$v=\frac{R\omega}{\sqrt{1-r_s/R}}$$
The last expression is what you were looking for.
11. Jun 12, 2010
### espen180
The only thing I said is that I have constant r. If you have a function, say f(x)=x2, you can say that x is fixed at three, but you can still get the slope at x=3 by differentiating wrt x.
As for the last equation you posted. I don't doubt its validity, but I don't see how it will get me anywhere, since neither v nor $$\omega$$ are known. In fact, since $$v=\omega R$$, v cancels on both sides.
12. Jun 12, 2010
### starthaus
Umm, no. If you did things correctly, then you'd have realised that $$dr=d\theta=0$$ reduces the metric to :
$$ds^2=(1-r_s/R)dt^2-R^2d\phi^2$$
So, your Christoffel symbols need to reflect that. They don't.
Yet, the result is extremely important since it tells you that the orbiting object has constant angukar speed and the trajectory is a circle.
No, $$v$$ is not $$\omega R$$.
Last edited: Jun 12, 2010
13. Jun 12, 2010
### espen180
How do you define v?
Of course the trajectory is a circle. I imposed that restriction by setting r=constant after deriving the general case geodesic equations. The equations saying d2t/ds2=0 and d2φ/ds2=0 are neccesary consequences.
What I am seeking is an expression which gives the orbital velocity as a function of r. I define $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$, and except for the factor $$\sqrt{2}$$ my result reduces to the Newtonian formula at large r, which makes me beleive my derivation is valid, the erronous factor $$\sqrt{2}$$ notwithstanding.
Last edited: Jun 12, 2010
14. Jun 12, 2010
### starthaus
You don't get to "define", you need to "derive" :
$$\frac{ds}{dt}=\sqrt{1-r_s/R}\sqrt{1-\frac{(R\omega)^2}{1-r_s/R}$$
The last equation gives you the hint that:
$$v=\frac{R\omega}{\sqrt{1-r_s/R}}$$
since, in GR:
$$\frac{ds}{dt}=\sqrt{1-r_s/r}\sqrt{1-v^2}$$
15. Jun 12, 2010
### espen180
And is v, in your case, measured by an observer from infinity? You have to give a definition. $$v=\frac{\text{d}\phi}{\text{d}\tau}r$$ and $$\frac{\text{d}\phi}{\text{d}t}r$$, for example, aren't the same, so you have to specify.
Aside from that, how can your equation be used to calculate the orbital period as a function of r only?
16. Jun 12, 2010
### starthaus
You can do it all by yourself by remembering that $$\frac{d\phi}{d\tau}=\omega$$ (see the derivation from the Euler-Lagrange equation)
$$\phi=\omega \tau$$. Make $$\phi=2\pi$$
The orbital period is not a function of r.
17. Jun 12, 2010
### espen180
For circular motion, it is easy to obtain the relationship $$v=\omega r$$. You get it directly from the definition of the radian. Do you claim $$v=\frac{\text{d}\phi}{\text{d}\tau} r$$ is not a valid definition of v? If so, please explain.
Please link to the Euler-Lagrange derivation, and I'll do my best to understand it.
How do you arrive at that conclusion? It is obvious that the orbital period is a function of r. That's why Mercury's orbital period is shorter that Earth's.
18. Jun 12, 2010
### starthaus
Not in GR. You are fixated on galilean physics. I am sorry, untill you get off your fixations, I can't help you.
19. Jun 12, 2010
### espen180
Then please explain the situation in GR.
20. Jun 12, 2010
### starthaus
What do u think I've been doing for you starting with post 4? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9490458965301514, "perplexity": 787.9871488316203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00501.warc.gz"} |
http://www.pveducation.org/pvcdrom/solar-cell-operation/fill-factor | # Fill Factor
The short-circuit current and the open-circuit voltage are the maximum current and voltage respectively from a solar cell. However, at both of these operating points, the power from the solar cell is zero. The "fill factor", more commonly known by its abbreviation "FF", is a parameter which, in conjunction with Voc and Isc, determines the maximum power from a solar cell. The FF is defined as the ratio of the maximum power from the solar cell to the product of Voc and Isc. Graphically, the FF is a measure of the "squareness" of the solar cell and is also the area of the largest rectangle which will fit in the IV curve. The FF is illustrated below.
Graph of cell output current (red line) and power (blue line) as function of voltage. Also shown are the cell short-circuit current (Isc) and open-circuit voltage (VOC) points, as well as the maximum power point (Vmp, Imp). Click on the graph to see how the curve changes for a cell with low FF.
As FF is a measure of the "squareness" of the IV curve, a solar cell with a higher voltage has a larger possible FF since the "rounded" portion of the IV curve takes up less area. The maximum theoretical FF from a solar cell can be determined by differentiating the power from a solar cell with respect to voltage and finding where this is equal to zero. Hence:
giving:
${V}_{MP}={V}_{OC}-\frac{nkT}{q}\mathrm{ln}\left(\frac{{V}_{MP}}{nkT/q}+1\right)$
The equation above does not yield a simple or closed form solution and requires iteration to calculate VMP. The equation above only relates Voc to VMP and extra equations are needed to find IMP and FF. A more commonly used expression for the FF can be determined empirically as:1
## Fill Factor - Empirical
$FF=\frac{{v}_{oc}-\mathrm{ln}\left({v}_{oc}+0.72\right)}{{v}_{oc}+1}$
where voc is defined as a "normalized Voc":
## Voc normalized
${v}_{oc}=\frac{q}{nkT}{V}_{oc}$
Fill Factor Calculator 1
###### Results
The above equations show that a higher voltage will have a higher possible FF. However, large variations in open-circuit voltage within a given material system are relatively uncommon. For example, at one sun, the difference between the maximum open-circuit voltage measured for a silicon laboratory device and a typical commercial solar cell is about 120 mV, giving maximum FF's respectively of 0.85 and 0.83. However, the variation in maximum FF can be significant for solar cells made from different materials. For example, a GaAs solar cell may have a FF approaching 0.89.
The above equation also demonstrates the importance of the ideality factor, also known as the "n-factor" of a solar cell. The ideality factor is a measure of the junction quality and the type of recombination in a solar cell. For the simple recombination mechanisms discussed in Types of Recombination, the n-factor has a value of 1. However, some recombination mechanisms, particularly if they are large, may introduce recombination mechanisms of 2. A high n-value not only degrades the FF, but since it will also usually signal high recombination, it gives low open-circuit voltages.
A key limitation in the equations described above is that they represent a maximum possible FF, although in practice the FF will be lower due to the presence of parasitic resistive losses, which are discussed in Effects of Parasitic Resistances. Therefore, the FF is most commonly determined from measurement of the IV curve and is defined as the maximum power divided by the product of Isc*Voc, i.e.:
## Fill Factor
$FF=\frac{{V}_{MP}{I}_{MP}}{{V}_{OC}{I}_{SC}}$
Fill Factor Calculator 2 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9219250082969666, "perplexity": 951.7571421605024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946453.89/warc/CC-MAIN-20180424022317-20180424042317-00151.warc.gz"} |
https://en.wikipedia.org/wiki/Dimensionless_physical_constant | # Dimensionless physical constant
In physics, a dimensionless physical constant, sometimes called a fundamental physical constant, is a physical constant that is dimensionless. It has no units attached and has a numerical value that is independent of the system of units used. Perhaps the best-known example is the fine-structure constant, α, which has approximate value of 1137.036.
The term fundamental physical constant is also used to refer to universal but dimensioned physical constants such as the speed of light c, vacuum permittivity ε0, Planck constant h, and the gravitational constant G.[1] Increasingly,[year needed] physicists reserve the use of the term fundamental physical constant for dimensionless physical constants that cannot be derived from any other source.[citation needed]
## Characteristics
There is no exhaustive list of such constants. But it is meaningful to ask about the minimal number of fundamental constants necessary to determine a given physical theory. Thus, the Standard Model requires 25 physical constants, about half of them the masses of fundamental particles (which become "dimensionless" when expressed relative to the Planck mass or, alternatively, relative to the electron mass along with the gravitational coupling constant).
Fundamental physical constants cannot be derived but have to be measured. Development in physics may lead to either a reduction or an extension of their number: discovery of new particles, or new relationships between physical phenomena, would introduce new constants, while on the other hand, the development of a more fundamental theory might allow the derivation of several constants from a more fundamental constant.
A long-sought goal of theoretical physics is to find first principles ("Theory of Everything") from which all of the fundamental dimensionless constants can be calculated and compared to the measured values.
The large number of fundamental constants required in the Standard Model has been regarded as unsatisfactory since the theory's formulation in the 1970s. The desire for a theory that would allow the calculation of particle masses is a core motivation for the search for "Physics beyond the Standard Model".
The mathematician Simon Plouffe has made an extensive search of computer databases of mathematical formulae, seeking formulae for the mass ratios of the fundamental particles.
Arthur Eddington set out alleged mathematical reasons why the reciprocal of the fine structure constant had to be exactly 136.[year needed] When its value was discovered to be closer to 137, he changed his argument to match that value.[citation needed] Experiments have since shown that Eddington was wrong; to six significant digits, the reciprocal of the fine-structure constant is 137.036.
An empirical relation between the masses of the electron, muon and tau has been discovered by physicist Yoshio Koide, but this formula remains unexplained.
## Examples
Dimensionless fundamental physical constants include:
### Fine structure constant
One of the dimensionless fundamental constants is the fine structure constant:
${\displaystyle \alpha ={\frac {e^{2}}{\hbar c\ 4\pi \varepsilon _{0}}}\approx {\frac {1}{137.03599908}},}$
where e is the elementary charge, ħ is the reduced Planck's constant, c is the speed of light in a vacuum, and ε0 is the permittivity of free space. The fine structure constant is fixed to the strength of the electromagnetic force. At low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127. There is no accepted theory explaining the value of α; Richard Feynman elaborates:
There is a most profound and beautiful question associated with the observed coupling constant, e – the amplitude for a real electron to emit or absorb a real photon. It is a simple number that has been experimentally determined to be close to 0.08542455. (My physicist friends won't recognize this number, because they like to remember it as the inverse of its square: about 137.03597 with about an uncertainty of about 2 in the last decimal place. It has been a mystery ever since it was discovered more than fifty years ago, and all good theoretical physicists put this number up on their wall and worry about it.) Immediately you would like to know where this number for a coupling comes from: is it related to pi or perhaps to the base of natural logarithms? Nobody knows. It's one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the "hand of God" wrote that number, and "we don't know how He pushed his pencil." We know what kind of a dance to do experimentally to measure this number very accurately, but we don't know what kind of dance to do on the computer to make this number come out, without putting it in secretly!
The analog of the fine structure constant for gravitation is the gravitational coupling constant. This constant requires the arbitrary choice of a pair of objects having mass. The electron and proton are natural choices because they are stable, and their properties are well measured and well understood. If αG is calculated from the masses of two protons, its value is ≈10−38.
### Standard model
The original standard model of particle physics from the 1970s contained 19 fundamental dimensionless constants describing the masses of the particles and the strengths of the electroweak and strong forces. In the 1990s, neutrinos were discovered to have nonzero mass, and a quantity called the vacuum angle was found to be indistinguishable from zero.
The complete standard model requires 25 fundamental dimensionless constants (Baez, 2011). At present, their numerical values are not understood in terms of any widely accepted theory and are determined only from measurement. These 25 constants are:
### Cosmological constant
One constant is required for cosmology:
### Barrow and Tipler
Barrow and Tipler (1986) anchor their broad-ranging discussion of astrophysics, cosmology, quantum physics, teleology, and the anthropic principle in the fine structure constant, the proton-to-electron mass ratio (which they, along with Barrow (2002), call β), and the coupling constants for the strong force and gravitation.
### Martin Rees's Six Numbers
Martin Rees, in his book Just Six Numbers, mulls over the following six dimensionless constants, whose values he deems fundamental to present-day physical theory and the known structure of the universe:
N and ε govern the fundamental interactions of physics. The other constants (D excepted) govern the size, age, and expansion of the universe. These five constants must be estimated empirically. D, on the other hand, is necessarily a nonzero natural number and cannot be measured. Hence most physicists would not deem it a dimensionless physical constant of the sort discussed in this entry.
Any plausible fundamental physical theory must be consistent with these six constants, and must either derive their values from the mathematics of the theory, or accept their values as empirical.
## References
1. ^
2. ^ a b Rees, M. (2000), p. .
3. ^ Rees, M. (2000), p. 53.
4. ^ Rees, M. (2000), p. 110.
5. ^ Rees, M. (2000), p. 118.
## External articles
General
Articles on variance of the fundamental constants | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9639179110527039, "perplexity": 636.6353150002956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00292-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://crypto.stackexchange.com/questions/29698/x509-certificates-two-algorithm-identifiers | # X509 certificates: Two Algorithm Identifiers?
I'm wondering why there are two Algorithm Identifiers in a X509 certificate. One is in the "to be signed" part and the other one is at the end right before the actual signature.
Are they used for the same purpose? Do they have to be the same? Can implementations choose which of these two fields they evaluate?
• signature
• signatureAlgorithm
following the names defined in RFC 5280 section 4.1. And section 4.1.1.2 then goes on to state (for signatureAlgorithm): | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557644486427307, "perplexity": 1579.028168469124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00560.warc.gz"} |
http://mathhelpforum.com/new-users/216613-hi-everyone-mhf.html | ## Hi Everyone on MHF
New to the forum and very rusty with my math skills. It's been a long time since college.
I'm in the Analytical Chemistry field but haven't done much in the theorectical / math area in a long while.
Thanks in advance and I hope I may be able to help someone else along the way.
Enzo | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8336984515190125, "perplexity": 616.5269431036028}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096686.2/warc/CC-MAIN-20150627031816-00041-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-applied-math/4938-effect-moon.html | Math Help - Effect of moon..
1. I have a test next week...
The moon pulls on the earth with one-sith the force that the earth exerts on the moon.
a) true
b) false
2. Originally Posted by babygirl
The moon pulls on the earth with one-sith the force that the earth exerts on the moon.
a) true
b) false
b) False.
Consider Newton's third law. If the Earth is pulling on the Moon with a force F, the Moon has to pull on the Earth with an equal, but oppositely directed force -F.
-Dan | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8710898756980896, "perplexity": 1058.2402373951454}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701159031.19/warc/CC-MAIN-20160205193919-00117-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-can-i-calculate-the-amount-of-matter-in-a-given-volume#107665 | Chemistry
Topics
# How can I calculate the amount of matter in a given volume?
Aug 9, 2014
Use the density equation: density = mass/volume.
#### Explanation:
The amount of matter in a substance is its mass. In order to calculate the amount of matter in a given volume of a sample, you will use the density equation: density = mass/volume.
To find mass using the density equation, you must know the volume of your sample and its density. In this example, I'm going to use the element copper. The density of the elements are known and can be looked up in charts.
To find mass from density and volume, manipulate the equation so that mass = (density)(volume).
Example: What is the mass of a sample of copper that has a volume of $8.00 c {m}^{3}$?
Known or Given:
Cu volume = $8.00 c {m}^{3}$
density of copper at ${20}^{\text{o}}$C = $8.96$g/$c {m}^{\text{3}}$
http://www.chemicalelements.com/elements/cu.html
Equation:
density = mass/volume
Solution:
mass = (density)(volume) = ($8.96$g/$c {m}^{\text{3}}$)($8.00 c {m}^{3}$) = 1.12g Cu
##### Impact of this question
10217 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9285739660263062, "perplexity": 709.234717432357}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00313.warc.gz"} |
https://arxiv.org/abs/math/9903129 | math
(what is this?)
# Title: Partial Representations and Partial Group Algebras
Abstract: The partial group algebra of a group G over a field K, denoted by K_{par}(G), is the algebra whose representations correspond to the partial representations of G over K-vector spaces. In this paper we study the structure of the partial group algebra K_{par}(G), where G is a finite group. In particular, given two finite abelian groups G_1 and G_2, we prove that if the characteristic of K is zero, then K_{par}(G_1) is isomorphic to K_{par}(G_2) if and only if G_1 is isomorphic to G_2.
Comments: LaTeX, 25 pages, no figures Subjects: Group Theory (math.GR) MSC classes: 20C99 Cite as: arXiv:math/9903129 [math.GR] (or arXiv:math/9903129v1 [math.GR] for this version)
## Submission history
From: Ruy Exel [view email]
[v1] Mon, 22 Mar 1999 19:29:27 GMT (22kb) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.855278730392456, "perplexity": 1999.6642308577732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738662400.75/warc/CC-MAIN-20160924173742-00172-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://link.springer.com/chapter/10.1007%2F978-3-642-23881-9_79 | Chapter
Artificial Intelligence and Computational Intelligence
Volume 7002 of the series Lecture Notes in Computer Science pp 618-627
# Fuzzy Soft Matrices and their Applications
• Yong YangAffiliated withCarnegie Mellon UniversityCollege of Mathematics and Information Science, Northwest Normal University
• , Chenli JiAffiliated withCarnegie Mellon UniversityCollege of Mathematics and Information Science, Northwest Normal University
* Final gross prices may vary according to local VAT.
## Abstract
In this paper, we define fuzzy soft matrices and study their basic properties. We then define products of fuzzy soft matrices that satisfy commutative law and present a decision making method. This method can solve decision making problems which consider many observers’ views. We finally offer some examples to show that the presented method is more reasonable and reliable in solving practical problems.
### Keywords
Soft sets Fuzzy soft sets Fuzzy soft matrices Products of fuzzy soft matrices Decision making | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8060449361801147, "perplexity": 2350.600343654973}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737940789.96/warc/CC-MAIN-20151001221900-00208-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://dsp.stackexchange.com/questions/19609/convolution-matrix-for-image-scaling | # convolution matrix for image scaling
Is there any way how to compute convolution matrix for Nearest Neighbor (bicubic, bilinear) image scaling (upscale/downscale)?
• Convolution cannot change image size (except adding borders). Dec 14 '14 at 7:37
This can't be done as a convolution because some of the operations you mentioned are not linear. Additionally, convolution by definition (multiplication by a toplitz matrix) shouldn't ever change the input and output size.
For bicubic:
http://en.wikipedia.org/wiki/Bicubic_interpolation#Bicubic_convolution_algorithm
Notice the high order terms.
As others have mentioned, convolution operation cannot change the size of the image. However, interpolation is a convolution operation. You just need to upsample by zero padding the image before performing the convolution (filtering) operation.
For example, say you want to interpolate a 1D signal times 2. The original signal is given by $a_0, a_1, .. , a_n$. You first zero pad it to get $a_0, 0, a_1, 0, ..$, then you filter the result. If the filter taps are 0.5, 1, 0.5, you will get simple linear interpolation. Similar scheme can be used in 2D images for biliniear interpolation, spline interpolation, etc.
• I'm not sure bilinear interpolation can be done in the Fourier domain. For example, there is a constant term and a production of the two directions and is hence nonlinear. Most interesting, useful, things aren't linear :-) Jan 6 '15 at 3:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8710080981254578, "perplexity": 772.1715548638358}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00658.warc.gz"} |
https://kluedo.ub.uni-kl.de/frontdoor/index/index/docId/1079 | ## On the critical behaviour of hermitean f-matrix models in the double scaling limit with f >= 3
• An algorithm for the isolation of any singularity of f-matrix models in the double scaling limit is presented. In particular it is proved by construction that only those universality classes exist that are known from 2-matrix models.
$Rev: 13581$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9371795058250427, "perplexity": 864.3361628498196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647299.37/warc/CC-MAIN-20180320052712-20180320072712-00012.warc.gz"} |
https://www.lmfdb.org/EllipticCurve/2.2.17.1/676.1/ | ## Results (displaying all 47 matches)
Label Base field Conductor norm Conductor label Isogeny class Weierstrass coefficients
676.1-a1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-a $$\bigl[1$$ , $$-1$$ , $$1$$ , $$-213$$ , $$-1257\bigr]$$
676.1-a2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-a $$\bigl[1$$ , $$-1$$ , $$1$$ , $$-3$$ , $$3\bigr]$$
676.1-b1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$a + 1$$ , $$a + 1$$ , $$343 a + 530$$ , $$-24564 a - 38363\bigr]$$
676.1-b2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$a - 1$$ , $$a + 1$$ , $$7 a - 17$$ , $$-21 a + 53\bigr]$$
676.1-b3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$a + 1$$ , $$a + 1$$ , $$-77 a - 120$$ , $$326 a + 509\bigr]$$
676.1-b4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$a + 1$$ , $$a + 1$$ , $$-397 a - 620$$ , $$-6334 a - 9891\bigr]$$
676.1-b5 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$-a$$ , $$a$$ , $$-8 a - 9$$ , $$20 a + 33\bigr]$$
676.1-b6 $$\Q(\sqrt{17})$$ 676 676.1 676.1-b $$\bigl[1$$ , $$-a - 1$$ , $$a + 1$$ , $$-342 a + 872$$ , $$24905 a - 63799\bigr]$$
676.1-c1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$-a + 1$$ , $$a$$ , $$152 a - 394$$ , $$-1348 a + 3456\bigr]$$
676.1-c2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$-a + 1$$ , $$a$$ , $$5372 a - 13884$$ , $$315198 a - 808120\bigr]$$
676.1-c3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$-a + 1$$ , $$a$$ , $$172 a - 429$$ , $$-1001 a + 2573\bigr]$$
676.1-c4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$a$$ , $$a + 1$$ , $$-173 a - 257$$ , $$1000 a + 1572\bigr]$$
676.1-c5 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$a$$ , $$a + 1$$ , $$-5373 a - 8512$$ , $$-315199 a - 492922\bigr]$$
676.1-c6 $$\Q(\sqrt{17})$$ 676 676.1 676.1-c $$\bigl[1$$ , $$a$$ , $$a + 1$$ , $$-153 a - 242$$ , $$1347 a + 2108\bigr]$$
676.1-d1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-d $$\bigl[1$$ , $$a + 1$$ , $$a$$ , $$-130 a + 337$$ , $$-3387 a + 8677\bigr]$$
676.1-d2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-d $$\bigl[1$$ , $$-a - 1$$ , $$a$$ , $$-9 a - 28$$ , $$26 a + 56\bigr]$$
676.1-d3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-d $$\bigl[1$$ , $$-a - 1$$ , $$a$$ , $$-49 a - 308$$ , $$-1038 a - 320\bigr]$$
676.1-d4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-d $$\bigl[1$$ , $$a - 1$$ , $$a$$ , $$136 a - 353$$ , $$991 a - 2539\bigr]$$
676.1-e1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-e $$\bigl[1$$ , $$-a - 1$$ , $$a$$ , $$131 a + 206$$ , $$3256 a + 5084\bigr]$$
676.1-e2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-e $$\bigl[1$$ , $$-a$$ , $$a + 1$$ , $$-137 a - 217$$ , $$-992 a - 1548\bigr]$$
676.1-e3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-e $$\bigl[1$$ , $$a + 1$$ , $$a$$ , $$10 a - 37$$ , $$-17 a + 45\bigr]$$
676.1-e4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-e $$\bigl[1$$ , $$a + 1$$ , $$a$$ , $$50 a - 357$$ , $$1087 a - 1715\bigr]$$
676.1-f1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-f $$\bigl[1$$ , $$a - 1$$ , $$1$$ , $$807 a + 1259$$ , $$-9405 a - 14687\bigr]$$
676.1-g1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$-a + 1$$ , $$0$$ , $$16 a - 43$$ , $$68 a + 269\bigr]$$
676.1-g2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$-a + 1$$ , $$0$$ , $$71 a - 158$$ , $$-2475 a - 5608\bigr]$$
676.1-g3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$a$$ , $$1$$ , $$-4361 a - 6815$$ , $$-110961 a - 173274\bigr]$$
676.1-g4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$1$$ , $$a$$ , $$71 a - 563$$ , $$2278 a - 4487\bigr]$$
676.1-g5 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$a$$ , $$1$$ , $$-18766 a - 29330$$ , $$-1992071 a - 3110764\bigr]$$
676.1-g6 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$0$$ , $$a + 1$$ , $$588 a - 1510$$ , $$-3188 a + 8164\bigr]$$
676.1-g7 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$a$$ , $$1$$ , $$-298776 a - 466970$$ , $$-127981903 a - 199853580\bigr]$$
676.1-g8 $$\Q(\sqrt{17})$$ 676 676.1 676.1-g $$\bigl[1$$ , $$1$$ , $$a$$ , $$-94 a - 88$$ , $$503 a + 860\bigr]$$
676.1-h1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-h $$\bigl[1$$ , $$-a$$ , $$1$$ , $$-807 a + 2066$$ , $$9405 a - 24092\bigr]$$
676.1-i1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-i $$\bigl[1$$ , $$a - 1$$ , $$1$$ , $$28 a - 71$$ , $$-137 a + 351\bigr]$$
676.1-i2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-i $$\bigl[1$$ , $$-a - 1$$ , $$1$$ , $$-a - 2$$ , $$3 a + 5\bigr]$$
676.1-j1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-j $$\bigl[1$$ , $$0$$ , $$1$$ , $$-460$$ , $$-3830\bigr]$$
676.1-j2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-j $$\bigl[1$$ , $$0$$ , $$1$$ , $$-5$$ , $$-8\bigr]$$
676.1-j3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-j $$\bigl[1$$ , $$0$$ , $$1$$ , $$0$$ , $$0\bigr]$$
676.1-k1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$1$$ , $$a + 1$$ , $$-72 a - 492$$ , $$-2279 a - 2209\bigr]$$
676.1-k2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$1$$ , $$a + 1$$ , $$93 a - 182$$ , $$-504 a + 1363\bigr]$$
676.1-k3 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$0$$ , $$a$$ , $$-589 a - 921$$ , $$3187 a + 4977\bigr]$$
676.1-k4 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$a$$ , $$0$$ , $$-71 a - 87$$ , $$2475 a - 8083\bigr]$$
676.1-k5 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$-a + 1$$ , $$1$$ , $$4361 a - 11176$$ , $$110961 a - 284235\bigr]$$
676.1-k6 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$a$$ , $$0$$ , $$-16 a - 27$$ , $$-68 a + 337\bigr]$$
676.1-k7 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$-a + 1$$ , $$1$$ , $$18766 a - 48096$$ , $$1992071 a - 5102835\bigr]$$
676.1-k8 $$\Q(\sqrt{17})$$ 676 676.1 676.1-k $$\bigl[1$$ , $$-a + 1$$ , $$1$$ , $$298776 a - 765746$$ , $$127981903 a - 327835483\bigr]$$
676.1-l1 $$\Q(\sqrt{17})$$ 676 676.1 676.1-l $$\bigl[1$$ , $$a + 1$$ , $$0$$ , $$3 a - 3$$ , $$-a + 5\bigr]$$
676.1-l2 $$\Q(\sqrt{17})$$ 676 676.1 676.1-l $$\bigl[1$$ , $$-a$$ , $$1$$ , $$-28 a - 43$$ , $$137 a + 214\bigr]$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000001192092896, "perplexity": 643.1098585548787}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00342.warc.gz"} |
https://learn.careers360.com/ncert/question-without-using-the-pythagoras-theorem-show-that-the-points-4-4-3-5-and-minus-1-minus-1-are-the-vertices-of-a-right-angled-triangle/ | Q
# Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Q ; 6 Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1),$ are the vertices of a right angled triangle.
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It is given that point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
Now,
We know that the distance between two points is given by
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
Length of AB $= |\sqrt{(4-3)^2+(4-5)^2}|= |\sqrt{1+1}|= \sqrt2$
Length of BC $= |\sqrt{(3+1)^2+(5+1)^2}|= |\sqrt{16+36}|= \sqrt{52}$
Length of AC $= |\sqrt{(4+1)^2+(4+1)^2}|= |\sqrt{25+25}|= \sqrt{50}$
Now, we know that Pythagoras theorem is
$H^2= B^2+L^2$
Is clear that
$(\sqrt{52})^2=(\sqrt{50})^2+(\sqrt 2)^2\\ 52 = 52\\ i.e\\ BC^2= AB^2+AC^2$
Hence proved
Exams
Articles
Questions | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8959934115409851, "perplexity": 1407.3875004818294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00391.warc.gz"} |
http://www.math.stonybrook.edu/solenoid/par-abs.html | Andrzej Bis.
Pseudo-orbits, pseudoleaves and geometric entropy of foliations.
We show that the entropy of a finitely generated pseudogroup (resp., of a foliation on a compact Riemannian manifold) can be calculated by suitable counting separated pseudo-orbits (resp., pseudoleaves).
Since several years, pseudo-orbits play an important role in the theory of classical dynamical systems. In particular, it is shown ([Mi] and [BS]) that pseudo-orbits can be used to calculate the topological entropy of transformations. More recently, the similar result was obtained [Hur] for the inverse-image entropy introduced earlier by Remi Langevin and the second author [LW1].
On the other hand, geometric entropy $h(\Cal F)$ of a foliation $\Cal F$ of a compact Riemannian manifold has been introduced [GLW] and shown to be a handful tool to study topology and dynamics of foliated manifolds ( [Hu1 - 3], [ GLW], [ LW2], [Eg], [IT], etc.). The entropy $h(\Cal F)$ can be calculated either by counting the number of points separated along the leaves or by counting the number of separated orbits of holonomy pseudogroups generated by nice coverings by charts distinguished by $\Cal F$. Also, pseudoleaves of foliations have been defined by Takashi Inaba [In] who has shown (among the other results) that expansive C$^1$-foliations (in the sense of [IT]) of codimension-one which have the pseudoleaf tracing property are topologically stable.
In this article, we show that the geometric entropy $h(\Cal F)$ of any foliation $\Cal F$ of a compact Riemannian manifold $M$ coincides with that calculated by suitable counting separated pseudoleaves (Theorem 2 in Section 4). To this end, we study the entropy of finitely generated pseudogroups of local transformations of local transformations of compact metric spaces and we show that it can be calculated by counting (again, in a suitable way) separated pseudo-orbits (Theorem 1 in Section 2). In Section 5, we sketch an easy proof of Theorem 2 for foliated bundles. Finally, in Section 6 we provide an example of a group acting on $S^1$ for which the "usual" formula for the entropy in terms of pseudo-orbits does not work: It gives the number strictly bigger than the entropy calculated in terms of separated orbits.
Some significance of these results could be observed if one would try to calculate (or, to estimate) entropies of pseudogroups or foliations with the aid of computers.
Phil Boyland.
New dynamical invariants on hyperbolic manifolds.
The rotation measure is an asymptotic dynamical invariant assigned to a typical point of a flow in a fiber bundle over a hyperbolic manifold. The total mass of the rotation measure is the average speed of the orbit and its direction'' is the ergodic invariant probability measure of the hyperbolic geodesic flow which best captures the asymptotic dynamics of the given point. The rotation measure exists almost everywhere and is constant for an ergodic measure of the given flow and so it may be viewed as assigning an ergodic measure of the geodesic flow to one of the given flow. It generalizes the usual notion of homology rotation vector by encoding homotopy information. Using suspension flows, applications are given to diffeomorphisms on surfaces that are isotopic to the identity or to pseudoAnosov maps.
Use of meanders and train tracks for description of statics and dynamics of liquid crystals and 2+1 gravity.
In this talk the qualitative analysis of statics and dynamics of textures in liquid crystals is performed with help of meanders(V.Arnold) and train tracks (W.Thurston).It is argued ,that similar analysis can be applied to 2+1 gravity.The train tracks alone are sufficient for the description of time evolution of liquid crystal textures and gravity and the master equation which describes such evolution could be used in principle instead of more familiar Wheeler-DeWitt equation for gravity .The solution of the master equation is possible but requires the large scale numerical work. To by-pass this difficulty the approximation of train tracks by the meandritic labyrinths is used which allows to study possible phases of such systems using Peierls-like arguments.
Fransois Labourie.
The space of all convex surfaces as an hyperbolic lamination.
For every negatively curved 3-manifold, I constructed a canonical compact space laminated by riemann surfaces havng the following properties (i) for every g, the set of all compact leaves of genus greater than g is dense (ii) a generic leaf is dense, (iii) the space is stable in the following sense : if we deform the metric on the mainfold there is an homeomorphism between the two laminated spaces which preserve the leaves. Of course, these properties are reminiscence of that of the geodesic flow, who lives, so to say, on the boundary of this space. The general construction is part of a technique to construct laminations associated to what I call Monge-Ampere problems and which uses holomorphic curves rechniques. The fact that properties (i) (ii) (iii) are true, heavily depends on the negative curvature assumption.
Bernhard Leeb
On the rigidity of nonpositively curved spaces of higher rank.
The main result is a characterisation of irreducible Riemannian symmetric spaces and Euclidean buildings of rank $\geq2$ among singular spaces of nonpositive curvature by their asymptotic geometry: Namely, these are precisely the geodesically complete Hadamard spaces whose Tits boundary is a connected irreducible spherical Tits building.
Dynamical laminations of polynomials.
We discuss generalizations to higher degrees of the following fundamental fact due to W. Thurston: any gap of a quadratic lamination is either pre-periodic, or pre-critical.
Under the assumption that the corresponding quotient space (Julia set) is locally connected, we prove the same statement for higher degrees with one multiple ctitical point. The method is purely topological and transparent. It seems to allow a generalization for any number of the critical points. Corollaries are given.
Viorel Nitica (with A. Torok)
An open dense set of stably ergodic diffeomorphisms around a non-ergodic one.
Let A be a C1 Anosov diffeomorphism of a compact manifold M, and f a C1 circle diffeomorphism such that the product map A x f is a partially hyperbolic diffeomorphism. Then there is a C1 neighborhood of Axf that contains an open and dense set of partially hyperbolic diffeomorphisms with the accesibility property. If, in addition, A and f are C^2, A has a smooth invariant measure, and f has a smooth invariant measure, then there is a C2 neighborhood of A x f that contains an open and dense set of stably ergodic diffeomorphisms. This partially answers a question posed by Pugh and Shub.
Leonid Potyagailo
Foliated complexes and hierarchical accessibility of finitely presented groups.
In this joint work with Thomas Delzant we define a class $\cal C$ of subgroups of a finitely presented group $G$ which we call elementary. In, particular this can be elementary subgroups (virtually abelian or finite) of a discrete group of isometries of the hyperbolic space or elementary (virtually cyclic or finite) subgroups of a word (Gromov) hyperbolic group etc. To the group $G$ we associate an invariant $c(G)$ called complexity which is strongly decreasing with respect to splittings of $G$ as amalgamated free product or $HNN$-extension over elementary subgroups. This implies that the group $G$ admits a finite hierarchy of graph of groups decompositions over its elementary subgroups. This is in a sense similar to well-known hierarchies of $3$-dimensional Haken manifolds along a system of incompressible surfaces.
Yo'av Rieck (with Eric Sedgewick)
Finiteness of Heegaard Surfaces for Most 3-Manifolds (Preliminary Report).
We show that most" non-Haken 3-manifolds contain only finitely many isotopy classes of Heegaard surfaces of any bounded genus.
Useing Dehn surgery techniques, we show that when filling a 3-manifold containing no closed essential surface but a torus boundary component, for all but finitely many fillings the resulting manifold contains only finitely many isotopy classes of Heegaard surfaces that are not surfaces for $X$. In 1990 Klaus Johannson had shown finiteness for Haken manifolds (including manifolds with boundary) of isotopy classes of surfaces of bounded genus. Combining his solution (for $X$) and our work provides a proof for the above statement.
This is a part of a broader project, still in progress, where the bounded genus assumption will be omitted.
This proves, for most 3-manifolds, the following conjecture:
Conjecture (Waldhausen '78): any closed 3-manifold contains only finitely many isotopy classes of minimal genus Heegaard surfaces.
Ernest Rosales-Gonzales
Om countable number of limit cycles for polynomial equations in C^2.
In this talk I consider polynomial differential equations in $\C^2$ of the form $dw \over dz}={P_n/ Q_n$, where $P_n$, $Q_n$ are polynomials with complex coefficients of degree at most $n$. These equations can be identified with the space ${\bf A}_n$ of coefficients of the polynomials $P_n$ and $Q_n$. Let $\varphi$ be a solution of equation~(\ref{eq0.1}). A cycle on $\varphi$ is a non trivial element of the fundamental group $\pi_1(\varphi)$ of $\varphi$. Then, generic equations of the class ${\bf A}_n, \ n \geq 3$ have a countable set of homologically independent limit cycles. The exceptional set (the subset of ${\bf A}_n$ excluded by the genericity assumptions) is given by a finite number of analytical and real algebraic conditions of real codimension at least two.
Pawel Walczak.
Virtual leaves
Foliations, finitely generated groups and pseudogroups of transformations of manifolds (or, topological spaces) carry some dynamics. In particular, geometric (or, topological) entropy of such systems can be defined [GLW]. For instance, if $G$ is a group of homeomorphisms of a compact metric space $(X, d)$ and $G_1$ is a finite symmetric generating set, then points $x$ and $y$ of $X$ are said to be $(n, \epsilon )$-separated when $d(gx, gy)\ge\epsilon$ for some $g = g_1\cdot\dots\cdot g_n$, where $g_i\in G_1$. If $N(n, \epsilon )$ denotes the maximal cardinality of a subset of $X$ consisting of pairwise $(n, \epsilon )$-separated points, then the {\it entropy} $h(G, G_1)$ w.r.t. $G_1$ can be defined as $$h(G, G_1) = \lim_{\epsilon\to 0}\limsup_{n\to\infty}\frac{1}{n}\log N(n, \epsilon ).$$ The similar definition can be written for pseudogroups and, when applied to holonomy pseudogroups, for foliations.
Attie and Hurder [AH] applied entropy to produce an example of a complete Riemannian manifold of bounded geometry which cannot be quasi-isometric to a leaf of a foliation of a compact manifold (see also [Hu]). A similar but simpler example of this sort has been obtained by Zeghib [Ze]. Following this line, we constructed [Wa] an example of a {\it virtual leaf}, i.e. a complete surface $N$ of bounded geometry which is not quasi-isometric to a leaf as above but which, for any $\delta > 0$, is quasi-isometric to a $\delta$-pseudoleaf on a fixed compact foliated manifold $(M, \Cal F)$. (Recall that a $\delta$-{\it pseudoleaf} is an immersed submanifold which cuts the leaves at an angle less than $\delta$.) The key observation leading to this construction is that the dynamics of pseudo-orbits of a finitely generated group can be more chaotic than the dynamics of true orbits.
In this lecture, we intend to decsribe our example of a virtual leaf and to give a brief review of those results concerning the entropy of foliations which are necessary to understand the reasoning.
[AH] O. Attie and S. Hurder, Manifolds which cannot be leaves of foliation \jour Topology \vol 35 \yr 1996\pages 335 -- 353\endref
[GLW] E. Ghys, R. Langevin and P. Walczak, Entropie g\'eom\'etrique des feuilletages\jour Acta Math. \vol 160 \yr 1988\pages 105 -- 142\endref
[Hu] S. Hurder, Coarse geometry of foliations \inbook Geometric Study of Foliations, Proc. Tokyo 1993\publ World Sci.\publaddr Singapore\yr 1994\pages 35 -- 96\endref
[Wa] P. Walczak, A virtual leaf info preprint \yr 1997\endref
[Ze] A. Zeghib, An example of a 2-dimensional no leaf \inbook Geometric Study of Foliations, Proc. Tokyo 1993\publ World Sci. \publaddr Singapore\yr 1994\pages475 -- 477\endref
Anton Zorich.
Topological dynamics of orientable measured foliations and Teichmuller geodesic flow.
Consider generic orientable measured foliation on a closed orientable surface of genus $g$. We study how do the leaves of the foliation wind around the surface.
It follows from the results of H.Masur and W.Veech that generically all the leaves asymptotically follow one and the same asymptotic cycle, determined by the foliation. We give the precise bound for the deviation from the asymptotic cycle; it is expressed in terms of the Lyapunov exponents of the Teichmuller geodesic flow.
The similar results are valid for orientable measured foliations having prescribed types of saddles. In collaboration with M.Kontsevich we classified the corresponding connected components of the strata of the moduli spaces of holomorphic and quadratic differentials. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9172949194908142, "perplexity": 518.8417486428976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00074.warc.gz"} |
https://eng.libretexts.org/Bookshelves/Materials_Science/TLP_Library_I/09%3A_Brittle_Fracture/9.04%3A_Section_4- | # 9.4: Inglis and the crack tip stress idea
In 1913, Inglis calculated what the stresses and strains were in an elastic plate containing an elliptical crack, with semi-axes b and c, and under an applied stress σ - applied vertically in this case.
He found that the stress at the crack tip, σt , was given by
$\sigma_{\mathrm{t}}=\sigma\left(1+2 \frac{c}{b}\right)$
For a sharp crack, i.e. c >> b, the stress would be much greater at the crack tip So failure could occur by cracking because it’s only at a crack tip that the stress required to break a bond, or simultaneously break a given number of bonds in a unit area of material, is reached.
So far, so good, but σt depends on the SHAPE of the crack, i.e. c / b – and we know that the length is important.
By looking at crack tip stress fields using photoelasticity, we can see that stresses are indeed concentrated at crack tips. Inglis was correct about this.
It's the idea that a critical stress to break a bond is needed that is wrong. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8393785357475281, "perplexity": 1166.186207851533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00185.warc.gz"} |
https://www.cqc2t.org/quantum-metrology-kerr-metric/ | # Publication
#### Quantum metrology in the Kerr metric
11/06/2019
SP Kish, TC Ralph
Physical Review D, 99, 124015 (2019)
A surprising feature of the Kerr metric is the anisotropy of the speed of light. The angular momentum of a rotating massive object causes co- and counterpropagating light paths to move at faster and slower velocities, respectively, as determined by a far-away clock. Based on this effect we derive the ultimate quantum limits for the measurement of the Kerr rotation parameter a using an interferometric setup. As a possible implementation, we propose a Mach-Zehnder interferometer to measure the “one-way height differential” time effect. We isolate the effect by calibrating to a dark port and rotating the interferometer such that only the direction-dependent Kerr-metric-induced phase term remains. We transform to the zero angular momentum observer (ZAMO) flat metric where the observer sees c=1. We use this metric and the Lorentz transformations to calculate the same Kerr phase shift. We then consider nonstationary observers moving with the planet’s rotation, and we find a method for canceling the additional phase from the classical relative motion, thus leaving only the curvature-induced phase.
University: The University of Queensland
Authors Centre Participants: Mr. Sebastian Kish, Prof. Tim C Ralph
Source: Physical Review D
Publication Type: Refereed Journal article | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9305770993232727, "perplexity": 1817.712075565718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146341.16/warc/CC-MAIN-20200226084902-20200226114902-00390.warc.gz"} |
https://cstheory.stackexchange.com/questions/25367/how-to-translate-the-axiom-schema-of-induction-by-curry-howard/25373#25373 | # How to translate the axiom schema of induction by Curry-Howard?
I'm trying to understand the Curry-Howard correspondence. I am comfortable with it for propositional logic, but get confused when $\forall, \exists$ quantifiers come in the picture.
The axiom schema of induction (in second-order logic) is $\forall P. \left[(P(0) \implies \forall k\in \mathbb N. (P(k)\implies P(k+1)))\implies \forall n\in \mathbb N. P(n)\right]$.
To my understanding, via Curry-Howard every statement gets translated to a type. Proving a statement means showing that the type is inhabited. For example, +, = are dependent types taking two parameters of type nat; every $\forall k\in A$ gets translated into a dependent product $\prod_{k::A}$. So the axiom schema should become something like $\left[\prod_{P::?} \left((P\, 0) \rightarrow \left(\prod_{k::\mathbb N} (P \, k)\rightarrow P \, (+\, k\, 1) \right)\right)\right]\rightarrow \prod_{n::\mathbb N}(P \,n)$
A difficulty arises when translating $\forall P$: what is the type of $P$? The problem is that, in order for the above to make sense, $(P\, k)$ is supposed to be a type for all $k::\mathbb N$ (so we are allowed to say $(P\, 0) \rightarrow \cdots$, for instance). So $P$ seems to me like a (dependent) type constructor taking a nat as parameter. But then we are taking a product over $P$, so we must be able to treat $P$ as an element of a type! The problem seems to boil down to, are type constructors themselves of some type? Can we write this statement without having to refer to such a type? Perhaps higher-order logic translates to higher-order type theory, but I do not know the latter. References would also be appreciated.
• I am not sure this is research level, but I am equally unsure how this would be answered at cs.stackexchange, so I gave an answer below (actually, mostly references). Jul 26, 2014 at 17:20
• It would be answered on cs.stackexchange, and it is not research-level. Jul 28, 2014 at 10:09
First a correction. The statement $$\forall P \,.\, (P(0) \Rightarrow (\forall n . P(n) \Rightarrow P(S n)) \Rightarrow \forall m . P(m)),$$ is not the schema of induction, but rather induction written in second-order arithmetic. To get the schema you need to avoid $\forall P$. This is done by saying that there are infinitely many axioms of the form $$P(0) \Rightarrow (\forall n . P(n) \Rightarrow P(S n)) \Rightarrow \forall m . P(m),$$ one for each formula $P(x)$. That is, whereas in your axiom $P$ refers to arbitrary subsets of $\mathbb{N}$, in the schema it refers to sytnactic entities, namely formulas with parameter $x$. It is called a "schema" because it prescribes a shape from which infinitely many axioms may be generated by plugging in all possible syntactic entities $P$.
To answer your question, we just have to look at the Curry-Howard correspondence to see what a predicate $P(x)$ correspond to. The answer is type family or dependent type. Because you wrote your axiom in second-order arithmetic, it became unclear how to quantify over dependent types. This is possible by using universes, for instance Daimano in his answer used $\mathsf{Prop}$, the universe of propositions. However, under the pure Curry-Howard there is no $\mathsf{Prop}$, there are only types.
We have two options. If we translate the schema, then it is this: for every dependent type $$x : \mathbb{N} \vdash P(x) \, \mathsf{type}$$ we get the type $$\textstyle P(0) \to (\prod_{n:\mathbb{N}} P(n) \to P(S n)) \to \prod_{m:\mathbb{N}} P(m)$$ If we translate the second-order formulation of induction, we need to use a type universe $\mathsf{Type}$, and then we get $$\textstyle \prod_{P : \mathbb{N} \to \mathsf{Type}} P(0) \to (\prod_{n:\mathbb{N}} P(n) \to P(S n)) \to \prod_{m:\mathbb{N}} P(m)$$
What are these types inhabited by? In the case of schema, they are inhabited by primitive recursors $\mathsf{rec}_P$ which satisfy the equations \begin{align*} \mathsf{rec}_P \; z \; f \; 0 &= z \\ \mathsf{rec}_P \; z \; f \; (S n) &= f \; n \; (\mathsf{rec}_P \; z \; f \; n) \end{align*} Depending on the details of your formalism we may be able to drop the subscript $P$ on $\mathsf{rec}_P$, and just write $\mathsf{rec}$ -- but secretly there is a different $\mathsf{rec}$ for each $P$.
In the case of the second-order induction we would get a single primitive recursor $\mathsf{rec}$ satisfying \begin{align*} \mathsf{rec} \; P \; z \; f \; 0 &= z \\ \mathsf{rec} \; P \; z \; f \; (S n) &= f \; n \; (\mathsf{rec} \;P \; z \; f \; n) \end{align*}
If I spoke to a Haskell programmer I would say that in the case of schema each $\mathsf{rec}$ was obtained from $P$ using a "deriving" clause, whereas in the second-order case there is a single polymorphic $\mathsf{rec}$.
Yes, in type theory second order variables do have a type, it may be called $Prop$ (the type of propositions) or $Set$ (the type of sets) depending on the context. In your case, I would write $P:\mathbb N\rightarrow Prop$. The nLab entry on type theory and the Wikipedia entry on the calculus of constructions contain a host of references which you may find interesting. Also, the lecture notes by Sørensen and Urzyczyn are an excellent starting place for learning about the Curry-Howard isomorphism, independently of type theory.
An additional note: if you are interested in second order Peano arithmetic and Curry-Howard, you may also want to look into System F before type theory. System F, which corresponds to propositional second order, has an enormous expressive power (precisely related to second order Peano arithmetic) and data types representing inductive structures may be obtained by the following "trick": take the second order formula describing the desired induction principle and "erase" all first order information. In particular, from usual induction (the formula you consider), you get $\forall P.(P\Rightarrow (P\Rightarrow P)\Rightarrow P)$, whose normal forms are exactly the Church integers. See Girard, Lafont and Taylor's book Proofs and Types and Krivine's book Lambda-calculus, Types and Models for a formal justification of the above "trick".
• Thanks! I corrected the question - I am learning Haskell so that was on my mind when I was thinking about this... Jul 26, 2014 at 18:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9825645089149475, "perplexity": 333.4761162533376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00604.warc.gz"} |
https://www.physicsforums.com/threads/gamma-function-help.379235/ | # Gamma Function help!
1. Feb 17, 2010
### Integral8850
1. The problem statement, all variables and given/known data
I (2n,m) = Integral cos^(2n)O sin^(m)O cosO dO
limits are 0 to 2pi and O = theta
0.4/3 = 0.1333
show that I 2n,m = 2n/ m+1 (I2n-2, m+2)
2. Relevant equations
I really have no idea how to work with this problem. It is under Gamma function of the instructors notes. I have searched for an hour for an example similar to this problem. If anyone could point me in the right direction or a direction I would be greatful!
Thanks
3. The attempt at a solution
2. Feb 17, 2010
### vela
Staff Emeritus
Usually to establish this kind of relationship, you can use integration by parts. Have you tried that?
Similar Discussions: Gamma Function help! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9546219110488892, "perplexity": 1458.5692805237468}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812855.47/warc/CC-MAIN-20180219231024-20180220011024-00418.warc.gz"} |
http://www.intechopen.com/books/matlab-a-fundamental-tool-for-scientific-computing-and-engineering-applications-volume-1/matlab-based-algorithm-for-real-time-analysis-of-multiexponential-transient-signals | InTech uses cookies to offer you the best online experience. By continuing to use our site, you agree to our Privacy Policy.
Computer and Information Science » Numerical Analysis and Scientific Computing » "MATLAB - A Fundamental Tool for Scientific Computing and Engineering Applications - Volume 1", book edited by Vasilios N. Katsikis, ISBN 978-953-51-0750-7, Published: September 26, 2012 under CC BY 3.0 license. © The Author(s).
# Matlab-Based Algorithm for Real Time Analysis of Multiexponential Transient Signals
By Momoh-Jimoh E. Salami, Ismaila B. Tijani, Abdussamad U. Jibia and Za'im Bin Ismail
DOI: 10.5772/50006
Article top
## Overview
Figure 1. Overview of Techniques for Multiexponential signal analysis
Figure 2. Flowchart of the MATLAB-based algorithm for Multicomponent transient signal analysis
Figure 3. Block diagram of the multicomponent signal analyzer prototype.
Figure 4. NI sbRIO-9642 for real-time hardware target
Figure 5. Block Diagram of the real-time set-up
Figure 6. Lab view Block Diagram
Figure 7. Embedded MATLAB set-up for algorithm simulation
Figure 8. NI Labview MATLAB script nodes
Figure 9. Schematic diagram of the SPECTRAMAX Gemini Spectrofluorometer operation (SPECTRAmax®)
Figure 10. Sample simulation results with embedded MATLAB function
Figure 11. Power distributions for Quinine in water
Figure 12. Power distribution Quinine plus Arcridine Sodium in water
Figure 13. Power distribution for Acridine Orange + Fluorescein + Sodium and quinine in water
# Matlab-Based Algorithm for Real Time Analysis of Multiexponential Transient Signals
Momoh-Jimoh E. Salami1, Ismaila B. Tijani1, Za’im Bin Ismail1 and Abdussamad U. Jibia2
## 1. Introduction
Multiexponential transient signals are particularly important due to their occurrences in many natural phenomena and human applications. For instance, it is important in the study of nuclear magnetic resonance (NMR) in medical diagnosis (Cohn-Sfetcu et al., 1975)), relaxation kinetics of cooperative conformational changes in biopolymers (Provencher, 1976), solving system identification problems in control and communication engineering (Prost and Guotte, 1982), fluorescence decay of proteins (Karrakchou et al., 1992), fluorescence decay analysis (Lakowicz, 1999). Several research work have been reported on the analysis of multicomponent transient signals following the pioneer work of Prony in 1795 (Prony, 1975) and Gardner et al. in 1959 (Gardner, 1979). Detailed review of several techniques for multicomponent transient signals’ analysis was recently reported in (Jibia, 2010).
Generally, a multiexponential transient signal is represented by a linear combination of exponentials of the form
S(τ)=∑kMAkexp(−λkτ)+n(τ) (1)
where M is the number of components, Ak and λk respectively represent the amplitude and real-valued decay rate constants of the kth component and n(τ) is the additive white Gaussian noise with variance σn2. The exponentials in equation (1) are assumed to be separable and unrelated. That is, none of the components is produced from the decay of another component. Therefore, in determination of the signal parameters, M, Ak and λk from equation (1), it is not sufficient that equation (1) approximates data accurately; it is also important that these parameters are accurately estimated.
There are many problems associated with the analysis of transient signals of the form given in equation (1) due to the nonorthogonal nature of the exponential function. These problems include incorrect detection of the peaks, poor resolution of the estimated decay and inaccurate results for contaminated or closely-related decay rate data as reported in (Salami et.al., 1985). These problems become increasingly difficult when the level of noise is high. Although Gardner transform eliminated the nonorthogonality problem, it introduced error ripples due to short data record and nonstationarity of the preprocessed data. Apart from these problems, analysis of multiexponential signal is computationally intensive and requires efficient tools for its development and implementation in real-time.
To overcome these problems, modification of Gardner transform has been proposed recently with Multiple Signal Classification (MUSIC) superposition modeling technique (Jibia et al., 2008); with minimum norm modeling technique (Jibia and Salami, 2007), with homomorphic deconvolution, with eigenvalues decomposition techniques, and the Singular Value Decomposition (SVD) based-Autoregressive Moving Average (ARMA) modeling techniques (Salami and Sidek, 2000; Jibia, 2009). As reported in (Jibia, 2009), performance comparison of these four modeling techniques has been investigated. Though, all the four techniques were able to provide satisfactory performances at medium and high signal-to-noise ratio (SNR), the SVD-ARMA was reported to have the highest resolution, especially at low SNR.
Hence, the development of SVD-ARMA based algorithm for multiexponential signal analysis using MATLAB software package is examined in this chapter.
MATLAB provides computational efficient platform for the analysis and simulation of complex models and algorithms. In addition, with the aid of inbuilt embedded MATLAB Simulink block, it offers a tool for the integration of developed algorithm/model in an embedded application with little programming efforts as compared to the use of other programming languages (Mathworks, 2008). This functionality is explored in integrating the developed MATLAB-based algorithm into National Instrument (NI) Labview embedded programming tool. Hence, an integrated MATLAB-Labview software interface is proposed for real-time deployment of the algorithm. To this end, the analytical strength of MATLAB together with simplicity and user-friendly benefits of the National Instrument (NI), Labview design platforms are explored in developing an efficient, user-friendly algorithm for the real-time analysis of multiexponential transient signal.
The rest of the chapter is organized as follows. Section 2 provides brief review of techniques for multiexponential signal analysis. The MATLAB algorithm development for the signal analysis is presented in section 3. The development of an integrated MATLAB-Labview real-time software interface is then examined in section 4. Section 5 presents sample real-time data collection together with results and analysis. The chapter is concluded in section 6 with recommendation for future study.
## 2. Techniques of multiexponential transient signal analysis
Several techniques have been reported for the analysis of transient multiexponential signal. They are classified as: (i) time domain or frequency domain, and (ii) parametric or nonparametric techniques. The main objective of these techniques in analyzing the multiexponentially decaying signals is to estimate the signal parameters as accurately as possible and to get better display of the signal spectra. Time-domain techniques constitute the oldest methods of multiexponential signal analysis prior to the advent of Gardner transformation technique (Gardner et al, 1959; Salami, 1985). Gardner transformation is one of the most important methods of the transient signal analysis based on spectral analysis. Generally, spectral analysis involves transformation of a time-domain signal to a frequency domain so that certain features of the signals that characterized them are easily discerned such as its decay constant, λk and amplitude, Ak. In other words, it is the process of obtaining the frequency content (spectrum) of a signal (Proakis and Manolakis 1996). The spectral analysis approach is further categorized into nonparametric and parametric techniques. Nonparametric technique is a frequency-domain technique that obtains the signal spectra directly from the deconvolved data, while the parametric technique obtains the signal spectra indirectly by determining a finite set of parameters that defines a closed form mathematical model for the deconvolved data. Therefore the techniques of multiexponential transient signal analysis are sub-divided into time-domain, nonparametric frequency domain and parametric frequency domain techniques as shown in Figure 1 with their associated methods.
Among the earliest time-domain technique is the peeling technique. However, this technique produces poor results when S(τ)contains more than two components. Other time-domain techniques such as Prony’s method and its variants produce better performance than the peeling technique, however they are very sensitive to noise ((Smyth,2002; Salami, 1995). The nonlinear least squares technique (Smyth,2002) is computationally inefficient and the solution sometimes fails to converge, which means the estimate of the signal parameters cannot be accurately obtained.
The nonparametric techniques of spectral analysis are introduced to overcome some limitations of the time-domain techniques. The Gardner transformation technique produces error ripples which obscure the real peaks of the spectrum due to the cutoff points. This technique is good in analyzing signal with high SNR. The fast Fourier transform (FFT) technique, which is an improvement over the original Gardner transformation produces improved resolution. However, the problem of error ripples still exists. The extension of this technique involving the use of digital signal processing and Gaussian filtering is sensitive to noise and its data range has to be limited to get accurate estimates of the signal parameters. Whilst the modified FFT technique is better than the previous methods, it often fails to estimate λk correctly especially when the peaks are closely related. The differential technique (Swingler, 1977) provides some improvements over the existing techniques such as better resolution display but it is not suitable for analyzing noisy signal. Furthermore, modifying the FFT technique by incorporating integration procedure (Balcou,1981) does not produce better results as compared to the previous digital technique and the modified FFT technique. Moreover, this technique is still affected by error ripples.
#### Figure 1.
Overview of Techniques for Multiexponential signal analysis
Parametric techniques such as Autoregressive (AR), Moving Average (MA) and ARMA models are introduced to the analysis of multiexponential signals to alleviate the drawbacks of the nonparametric techniques. The AR modeling technique requires less computation than the ARMA modeling technique as its model parameters are relatively easy to estimate. However, it is sensitive to noise due to the assumed all-pole model. On the other hand, MA parameters are difficult to estimate and the resultant spectral estimates have poor resolution. Although, the ARMA modeling technique is much better in estimating noisy signal than the AR modeling technique, it requires a lot of computation. A detailed review of these techniques can be obtained in (Jibia, 2009; Salam and Sidek, 2000).
## 3. MATLAB-based Algorithm development
The systematic process involved in the development of the MATLAB-based SVD-ARMA algorithm for multiexponential transient signal analysis suitable for real-time application is discussed in this section. Apart from performance evaluation and signal generation, the algorithm consists of five major steps: obtaining convolution integral of the exponential signal using modified Gardner transformation; signal interpolation using spline technique; generation of deconvolved data; SVD-ARMA modeling of the deconvolved data; and power spectrum computation to finally estimate the transient signal parameters. A brief summary of the steps involved are as shown in Figure 2, and briefly highlighted as follows:
Step 1. Signal generation
Generate the required signal, S(τ) from MATLAB or from the fluorescence substances. For simulation data, MATLAB inbuilt function is used to generate the white Gaussian noise and the DC offset.
#### Figure 2.
Flowchart of the MATLAB-based algorithm for Multicomponent transient signal analysis
Step 2. Signal preprocessing via Gardner transformation:
This involves the conversion of the measured or generated signal S(τ) and p(τ) to y(t) and h(t) respectively using modified Gardner transformation (Salami and Sidek, 2003). This yields a convolution integral as subsequently described.
In general, equation (1) is expressed as
S(τ)=∑k=1MAkp(λkτ)+n(τ), (2)
where the basis function, p(τ) = exp(). This equation can also be expressed as
S(τ)=∫0∞g(λ)p(λτ)dλ+n(τ), (3)
where
g(λ)=∑k=1MAkδ(λ−λk). (4)
Both sides of equation (3) are multiplied by ταin the modified Gardner transformation instead of only τ in the original Gardner transformation. The value of the modifying parameter,α is carefully chosen based on the criteria given by Salami (1995) to avoid poor estimation of the signal parameters because α modifies the amplitude of the signal. Salami (1995) suggested to use 0<α1 in the analysis of multiexponential signals. According to Salami (1985), the choice of α can lead to improved signal parameters estimation as it produces noise reduction effect. The nonlinear transformation λ = e-r and τ = et are applied to equation (3), resulting in a convolution integral of the form
y(t)=∫−∞∞x(λ)h(t−λ)dλ+v(t), (5)
where the output function, y(t) = exp(αt) S{exp(t)}, the input function, x(t) = exp{(α-1)t}g(e-t), the impulse response function of the system, h(t) = exp(αt)p(et) and the additive noise, v(t) = exp(αt)n(et).
The discrete impulse response function, h[n] is obtained by sampling ταp(λkτ) at 1/Δt Hz. Later, H(k) is obtained from the discrete Fourier transform (DFT) of h[n]. Next, equation (4) is converted into a discrete-time convolution. This is done by sampling y(t) at a rate of 1/Δt Hz to obtain
y[n]=∑m=−nminnmaxx[m]h[n−m]+v[n], (6)
where the total number of samples, N equals nmax -nmin+1, both nmax and nmin represent respectively the upper and lower data cut-off points. The criteria for the selection of these sampling conditions have been thoroughly discussed by Salami (1987) and (Sen 1995) and these are not discussed further here. Equation (5) forms the basis of estimating the signal parameters since ideally x(n) can be recovered from the observed data by deconvolution. It is necessary to interpolate the discrete time convolution of y[n] since the log samples, τn = exp(n∆t) are not equally spaced.
Step 3. Cubic spline Interpolation
The discrete-time signal y[n] obtained in (5) consists of non-equally spaced samples which would be difficult to digitally process. A cubic interpolation is therefore applied to y[n] using MATLAB function ‘spline’ to obtain equally spaced samples of y[n].
Step 4. Generation of deconvolved data
In this stage, deconvolved data is generated from y[n] using optimally compensated inverse filtering due to its ability to handle noisy signal when compared to conventional inverse filtering (Salami,1995).
Conventionally, this is done by taking the DFT of equation (5) to produce:
Y(k)=X(k)H(k)+V(k) (7)
The deconvolved data, X(k)can be obtained by computing Y(k)/H(k), that is
X⌢(k)=Y(k)H(k)=X(k)+V(k)H(k)for0≤k≤N−1 (8)
where Y(k), X(k), H(k) and V(k) represent the discrete Fourier transform of y(n), x(n), h(n) and v(n) respectively. This inverse filtering operation is called the conventional inverse filtering. It yields deconvolved data with decreasing SNR for increasing values of k. Therefore, the accuracy of X(k) deteriorates when the noise variance level is high.
To overcome this problem, an optimally compensated inverse filtering is introduced. In this approach, H(k) is modified by adding an optimally selected value, μ into it. This procedure is done according to Riad and Stafford (1980) by initially designing a transfer function, HT (k) that yields a betterX (k) in equation (7), where HT (k) is given by:
HT(k)=H∗(k)[|H(k)|2+μ], (9)
where denotes the complex conjugate. Substituting equation (8) into equation (7) yields
X∧(k)=Y(k)H∗(k)[|H(k)|2+μ], (10)
which is referred to as the optimally compensated inverse filtering. It is noted that a small value of μ has a very little effect in the range of frequency when |H(k)|2 is significantly larger thanμ. However, if|H(k)|2 is very small, the effect of μ on the deconvolved data is quite substantial, that is μ tends to make X(k) less noisy. Therefore, μ puts limit on the noise amplification because the denominator becomes lower bounded according to Dabóczi and Kollár (1996). The parameter μ is carefully selected according to the SNR of the data to obtain good results. The choice of the optimum value of μ according to Salami and Sidek (2001) is best determined by experimental testing.
Equation (9) shows one-parameter compensation procedure, however, multi-parameter compensation is considered in this study. Thus, a regularization operator L(k) is introduced into (8), that is
F(k)=H∗(k)[|H(k)|2+α|L(k)|2] (11)
where α is the controlling parameter and the regularization operator, L(k) is the discrete Fourier transform of the second order backward difference sequence. |L(k)|2 is given as
|L(k)|2=16sin4(πkN), (12)
where N is the number of samples. Using both the second and fourth order backward difference operations in equation (10) yield
F(k)=H∗(k)[|H(k)|2+μ+α|L(k)|2+β|L(k)|4] (13)
where |L(k)|4denotes the fourth order backward difference operator and μ, α and β are the varied compensation parameters to improve the SNR of the deconvolved data.
Unwanted high frequency noise can still be introduced by this optimally compensated inverse filtering which can make some portion of X(k)unusable. Therefore, a good portion of X^(k)denoted as f (k), is given as
f(k)=∑i=1MBiexp(j2πkΔflnλi)+V(k), (14)
where 1 ≤ k ≤ 2Nd+1, Ak2, Nd(N/2)1, Nd is the number of useful deconvolved data points, N is the number of data samples and V(k) is the noise samples of the deconvolved data. Equation (13) is interpreted as a sum of complex exponential signals. The number of deconvolved data points, Nd is carefully selected to produce good results from f (k).
Step 5. Signal parameters estimation using SVD-ARMA modeling
SVD-ARMA algorithm is applied to f(k) to estimate the signal parameters M and λk. as it provides consistent and accurate estimates of AR parameters with minimal numerical problem which is necessary for real-time application. In addition, it is a powerful computational procedure for matrix analysis especially for solving over determined system of equations. The detailed mathematical analysis of this algorithm is reported in (Salami, 1985).
Generally, the ARMA model assumes that f (k) satisfies the linear difference equation (Salami 1985)
f(k)=−∑i=1pa(i)f(k−i)+∑i=1qb(i)V(k−i) (15)
where V(k) is the input driving sequence, f(k) is the output sequence, a[i] and b[i] are the model coefficients with AR and MA model order of p and q respectively. Usually, the white Gaussian noise becomes the input driving sequence in the analysis of exponentially decaying transient signals.
One of the most effective procedures for estimating these model parameters is by solving a modified Yule-Walker equation (Kay and Marple 1981). This procedure is subsequently discussed.
Equation (14) is multiplied by f *(k-m) and taking the expectation yields:
Rff(k)=−∑n=1pa[n]Rff(k−n)+∑n=0qb[n]h(k−m), (16)
where Rff (k) is the autocorrelation function of f(k) and h(k) is the impulse response function of the ARMA model. Next, considering the AR portion of equation (15) leads to the modified Yule-Walker equation
Rff(k)+∑n=1pa[n]Rff(k−n)=0;k≥q+1. (17)
Equation (16) may not hold exactly in practice because both p and q are unknown prior to analysis and Rff (k) has to be estimated from noisy data. This problem is solved by using an SVD algorithm. This algorithm is used by first expressing equation (16) in matrix form as Ra = e with R having elements r(i,l)=Rff(qe+1+il), where 1 ≤ ir ; 1 ≤ lpe + 1. Note that both pe and qe are the guess values of the AR and MA model order respectively, a is a pe × 1 and e is a r × 1 error vector with r > pe. The SVD algorithm is applied to R to produce
R=USVT =∑n=1peσnunvnH (18)
where the r × (pe+1) unitary matrix U = [u1 u2upe+1], (pe+1) × (pe+1) unitary matrix V = [v1 v2vpe+1] and Σ is a diagonal matrix with diagonal elements (σ 1, σ 2σpe+1). These diagonal elements are called singular values and are arranged so that σ 1 > σ 2 > … >σpe+1> 0. Only the first M singular values will be nonzero so that σ M+1 = σ M+2 = … =σpe+1= 0. However, σ M+1σ M+2 ≠ … ≠σpe+1≠ 0 due to noise contamination. The problem is solved by constructing a lower rank matrix RL from R using the first M singular values, that is
RL=UMSMVMH=∑n=1MσnunvnH, (19)
where UM, ΣM and VM are the truncated version of U, Σ and V respectively. The AR coefficients are then estimated from the relation a = -RL#r, where r corresponds to the first column of RL and RL# is given as
RL#=∑n=1Mσn−1uvnHn. (20)
The estimated AR coefficients are then used to generate the residual error sequences:
β(k)=∑l=0pe∑m=0pea[l]a*[m]Rff(k+m−l) (21)
from which the actual MA parameters are obtained directly from equation (20). However, MA spectra can be obtained from the DFT of the error sequences, β(k). An exponential window is applied to β(k) to ensure that the MA spectra derived from the error sequences are positive definite. Next, the ARMA spectrum is computed from
Sf(z)=∑k=−pepeβ(k)z−k|A(z)|2 (22)
and the desired power distribution of x(t), denoted as Px(t) is obtained by evaluating Sf (z) on the unit circlez=exp{2πjtNΔt}, that is:
Px(t)=Sf(z)|z=exp{j2πtNΔt}=∑k=1MBk2δ(t−lnλk). (23)
Eventually, M and ln λk are obtained from Px(t).
Step 6. Graphical presentation of output
Power distribution graph has been used to display the results of multiexponential signal analysis. This is computed from the power spectrum of the resulting output signal from SVD-ARMA modeling method as shown in equation (21).
Step 7. Performance evaluation
The efficiency of the algorithm with SVD-ARMA modeling technique in estimating λk correctly is determined by the Cramer-Rao lower bound (CRLB) expressed as:
CRLB(λk)=6(1+0.7(λkN)3/2)2N3SNR, (24)
where N is the number of data points, the CRB(λk) is the CRLB for estimating λk and SNR equals to Ak2 divided by the variance of the white Gaussian noise.
The Cramer-Rao lower bound will determine whether the estimator is efficient by comparing the variance of the estimator, var(λk) with the Cramer-Rao lower bound. Variance that approaches the CRLB is said to be optimal according to Kay and Marple (1981) and Sha’ameri (2000). Consequently, the closer the variance of the estimator is to the CRLB, the better is the estimator.
A MATLAB-mfile code has been written to implement the steps 1 to 6 described above, details of this have been thoroughly discussed in (Jibia, 2009).
## 4. Integrated MATLAB-labview for real-time implementation
This section discusses the development of proposed integrated Labview-MATLAB software interface for real-time (RT) implementation of the algorithm for multicomponent signal analysis as described in section 3. Real-time signal analysis is required for most practical applications of multicomponent signal analysis. In this study, reference is made to the application to fluorescence signal analysis. A typical multicomponent signal analyzer comprises optical sensor which is part of the spectrofluorometer system, signal conditioning/data acquisition system, embedded processor that runs the algorithm in real-time, and display/storage devices as shown in Figure 3. National Instrument (NI) real-time hardware and software are considered for the development of this system due to its ease of implementation.
#### Figure 3.
Block diagram of the multicomponent signal analyzer prototype.
### 4.1. Labview real-time module and target
The Labview real-time module (RT-software) together with NI sbRIO-9642 (RT-hardware) has been adopted in this study. The Labview Real-Time software module allows for the creation of reliable, real-time applications which are easily downloaded onto the target hardware from Labview GUI programming tool.
The NI sbRIO-9642 is identical in architecture to CompactRIO system, only in a single circuit board. Single-Board RIO hardware features a real-time processor and programmable FPGA just as with CompactRIO, and has several inputs and outputs (I/O) modules as shown in Figure 4.
System development involves graphical programming with Labview on the host Window computer, which is then downloaded and run on real-time hardware target. Since the algorithm has been developed with MATLAB scripts, an integrated approach was adopted in the programme development as subsequently described.
### 4.2. MATLAB-labview software integration
The developed algorithm with MATLAB scripts was integrated inside Labview for real-time embedded set-up as shown in Figure 5. Labview front-panel and block diagrams were developed with inbuilt Labview math Scripts to run the MATLAB algorithm for the analysis of multicomponent signals. The use of Labview allows for ease of programming, and real-time deployment using the Labview real-time module described in section 4.1. Also, it provides a user friendly software interface for real-time processing of the fluorescence signals.
As shown in Figure 5, the sampled data produced by the spectrofluorometer system are pre-processed by the NI-DAQ Cards. These signals are then read by the embedded real-time software, analyze the signals and display the results in a user friendly manner. The user is prompted to enter the number of samples to be analyzed from the front-panel using the developed SVD-ARMA algorithm.
### Figure 4.
NI sbRIO-9642 for real-time hardware target
### Figure 5.
Block Diagram of the real-time set-up
Generally, the developed algorithm with MATLAB scripts was integrated inside Labview for real-time embedded set-up as follows:
1. Pre-simulation of the MATLAB algorithm inside embedded MATLAB Simulink block: this requires re-structuring of the codes to be compatible for embedded Simulink implementation, and hence deployment inside Labview MathScript. Figure 7 shows the MATLAB Simulink blocks configuration with embedded MATLAB function together with cross-section of the algorithm. The DATA with time vector is prepared inside the workspace and linked to the model input. Figure 10 (a-f) shows the simulation results with experimental data described in section 5.
### Figure 6.
Lab view Block Diagram
### Figure 7.
Embedded MATLAB set-up for algorithm simulation
1. Labview programming: Development of Labview front-panel and block diagram is as shown in Figure 4. In the block diagram programming, the Labview MathScript node is employed to integrate the MATLAB codes in the overall Labview programme. The script (Figure 8) invokes the MATLAB software script server to execute scripts written in the MATLAB language syntax
### Figure 8.
NI Labview MATLAB script nodes
1. The integrated software interface is evaluated with the real-time fluorescence data collected from a spectrofluorometer.
## 5. Sample collection, results and discussion
Due to unavailability of spectrofluorometer system that can be directly linked with the set-up, sampled data collected from fluorescence decay experiment conducted using Spectramax Germini XS system were used to test the performance of the integrated system. The schematic diagram of the spectrofluorometer operation is shown in Figure 9, and itemized as follows:
Step 1. The excitation light source is the xenon flash lamp.
Step 2. The light passes through the excitation cutoff wheel. This wheel reduces the amount of stray light into the movable grating.
Step 3. The movable grating selects the desired excitation wavelength. Then, this excitation light enters a 1.0 mm diameter fiber.
Step 4. This 1.0 mm diameter fiber focuses the excitation light before entering the sample in the micro-plate well. This focusing prevents part of the light from striking adjacent wells.
Step 5. The light enters the wheel and if fluorescent molecules are present, the two mirrors focus the light from the well into a 4.0 mm optical bundles.
Step 6. The movable, focusing grating allows light of chosen emission wavelength to enter the emission cutoff wheel.
Step 7. This emission cutoff wheel will further filter the light before the light enters the photomultiplier tube.
Step 8. The photomultiplier tube detects the emitted light and passes a signal to the instrument’s electronics which then send the signal to the data acquisition system inside the spectrofluorometer.
Three intrinsic fluorophores (Acridine Orange; Fluorescein Sodium and Quinine) were used in the experiment. The details of the substances are given in Table 1.
Acridine Orange Fluorescein Sodium Quinine Molecular formulaManufacturerMolecular weightPurity by HPLCFormColourSolubility in waterSolutility in ethanol C17H12ClN3Merck301.8g/mol99.1%SolidOrange red28g/lSoluble C20H10Na2O5Merck376.28Extra pureSolidReddish brown500g/l140g/l C20H24N2O2Merck324.43Extra purePowder,White0.5g/l1200g/l
#### Table 1.
Characteristics of the fluorophores samples
The simulation results for Acridine orange, Fluorescein Sodium, Quinine, Quinine plus Arcridine, Fluorescein Sodium plus Acridine orange, and Fluorescein Sodium plus Acridine orange plus Quinin in water are shown in Figure 10 (a-f) respectively.
Figure 11 to Figure 13 show the sample of results obtained from the integrated MATLAB-Labview real-time software which has been developed. Both results of simulation and real-time software interface yield accurate estimates of the fluorescence data as shown in Figure 10-Figure 13, and presented in Table 2. The singular values for each of the samples combination are given in Table 3. The results indicate accurate determination of the constituent samples
#### Figure 9.
Schematic diagram of the SPECTRAMAX Gemini Spectrofluorometer operation (SPECTRAmax®)
#### Figure 10.
Sample simulation results with embedded MATLAB function
## 6. Conclusion/Future study
The development of MATLAB-based algorithm for real-time analysis of multicomponent transient signal analysis based on SVD-ARMA modeling technique has been presented in this chapter. To enhance real-time interface and rapid prototyping on target hardware, complementary benefits of MATLAB and Labview were explored to develop real-time software interface downloadable into single board computer by NI Labview. In the absence of the spectrofluorometer system, the developed user friendly software for real-time deployment was validated with the collected real-time data. The obtained results indicate the effectiveness of the proposed integrated software for practical application of the proposed algorithm.
Future direction of this research will be directed towards development of customized spectrofluorometer sub-system that can be directly integrated to the overall system. This will eventually facilitate direct application of the developed algorithm in practical applications involving transient signal analysis. Also, other algorithms based on homomorphic and eigenvalues decomposition developed by the authors in similar study are to be made available as option on the user interface.
Mixture Expected value SVD-ARMA Percentage error Acridine orange 0.5978 0.625 4.55 Fluorescein Sodium -1.4584 -1.438 1.40 Quinine -0.6419 -0.625 2.63 AcridineOrange+Fluorescein Sodium 0.6539 0.6563 0.37 -1.4584 -1.438 1.40 AcridineOrange+Quinine 0.7750 0.761 1.81 -0.5539 -0.5313 4.08 Acridine Orange + Fluorescein Sodium+Quinine 0.5105 0.5325 4.31 -0.6152 -0.5938 3.48 -1.5260 -1.533 0.46
#### Table 2.
Estimated Log of decay rates and percentage error from fluorescence decay experiment (Inλ)
#### Table 3.
Singular values for SVD-ARMA using experimental data
#### Figure 11.
Power distributions for Quinine in water
#### Figure 12.
Power distribution Quinine plus Arcridine Sodium in water
#### Figure 13.
Power distribution for Acridine Orange + Fluorescein + Sodium and quinine in water
## References
1 - AbdulsammadJibia,2010Real Time Analysis of Multicomponent Transient Signal using a Combination of Parametric and Deconvolution Techniques”, A PhD thesis submitted to Department of ECE, IIUM,Malaysia.
2 - V. Arunachalam, 1980Multicomponent Signal Analysis. PhD dissertation, Dept. of Electrical Engineering, The University of Calgary, Alberta, Canada.
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7 - R. Gutierrez-Osuna, A. Gutierrez-Galvez, N. Powar, 2003Transient response analysis for temperature-modulated hemoresistors. Sensors and Actuators Journal B (Chemical), 93 (1-3), 57-66.
8 - F. B. Hildebrad, 1956Introduction to numerical analysis. New York: McGraw-Hill.
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11 - A. U. Jibia, M. J. E. Salami, O. O. Khalifa, 2009Multiexponential Data Analysis Using Gardner Transform. (a draft ISI review journal paper)
12 - A. U. Jibia, M. J. E. Salami, 2007Performance Evaluation of MUSIC and Minimum Norm Eigenvector Algorithms in Resolving Noisy Multiexponential Signals. Proceedings of World Academy of Science, Eng. and Tech 26Bangok, Thailand. 24-28.
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14 - S. M. . Kay, S. L. Marple, 1981Spectrum analysis- A modern perspective. Labview usr manual,2010. www.ni.com/pdf/manuals.accessed Jan., 2011
15 - J. R. Lakowicz, 1999Principles of fluorescence spectroscopy. 2nd Ed. New York: Kluwer Academic/Plenum Publishers.
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17 - S. W. Provencher, 1976A Fourier method for the analysis of exponential decay curves. Biophysical journal. 162739
18 - R. Prost, R. Goutte, 1982Kernel splitting method in support constrained deconvolution for super-resolution. Institut National des Science, France.
19 - R. Prony, (1795, Essai Experimentals et Analytique. J. ecole Polytech., Paris, 12476
20 - J. G. Proakis, D. G. Manolakis, 2007Digital signal processing: Principles, algorithms, and applications. New Jersey: Prentice-Hall.
21 - R. Prony, (1795, Essai Experimentals et Analytique. J. ecole Polytech., Paris, 12476
22 - S. W. Provencher, 1976A Fourier method for the analysis of exponential decay curves. Biophysical journal, 162739
23 - M. J. E. Salami, (1985, 1985Application of ARMA models in multicomponent signal analysis. Ph.D. Dissertation, Dept. of Electrical Engineering, University of Calgary, Calgary, Canada
24 - M. J. E. Salami, 1995Performance evaluation of the data preprocessing techniques used in multiexponential signal analysis. AMSE Press. 33 (2), 23-38.
25 - M. J. E. Salami, 1999Analysis of multicomponent transient signal using eigen- decomposition based methods.Proc. Intern. Wireless and Telecom. Symp.,3 (5), 214-217.
26 - M. J. E. Salami, Y. I. Bulale, 2000Analysis of multicomponent transient signals: A MATLAB approach. 4th International Wireless and Telecommunications Symposium/Exhibition, 253256Proceedings of the IEEE, 69 (11), 1380-1419.
27 - M. J. E. Salami, Z. Ismail, 2003Analysis of multiexponential transient signals using interpolation-based deconvolution and parametric modeling techniques. IEEE International conference on Industrial Technology, 1271276
28 - Salami M.J.E. and Sidek S.N.2000Performance evaluation of the deconvolution techniques used in analyzing multicomponent transient signals. Proc. IEEE Region 10 International Conference on Intelligent System and Technologies for the New Millennium (TENCON 2000), 1, 487-492.
29 - T. K. Sarkar, D. D. Weiner, V. K. Jain, 1981Some mathematical considerations in dealing with the inverse problem. IEEE transactions on antennas and propagation, 29 (2), 373-379.
30 - J. Schlesinger, 1973Fit to experimental data with exponential functions using the fast Fourier transform. Nuclear instruments and methods. 106503508
31 - R. Sen, 1995On the identification of exponentially decaying signals. IEEE Transaction Signal Processing. 43819361945
32 - G. K. Smyth, 2002Nonlinear regression. Encyclopedia of environmetrics. 314051411
33 - SPECTRAmax® GEMINI XS product literature. http://www.biocompare.com/Product-Reviews/41142 -Molecular-Devices-8217-SPECTRAmax-GEMINI-XS-Microplate-Spectrofluorometer/ accessed onDecember, 2008
34 - D. N. Swingler, 1999Approximations to the Cramer-Rao lower bound for a single damped exponential signal.. Signal Processing, 75 (2), 197-200.
35 - W. F. Trench, 1964An algorithm for the inversion of finite Toeplitz matrices. J. SIAM, 12 (3), 515-522.
36 - T. J. Ulrych, 1971Application of homomorphic deconvolution to seismology. Geophysics, 4 (36), 650-660. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8536948561668396, "perplexity": 2811.874110246672}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399428.8/warc/CC-MAIN-20160624154959-00136-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/28362-solved-limit-we-ve-seen-number-times.html | # Math Help - [SOLVED] Limit we've seen a number of times
1. ## [SOLVED] Limit we've seen a number of times
I should just look this up, but I can't figure out how to do a search for it. It's cropped up a couple of times, but I don't recall the answer, much less how to find it.
Anyway, here it is
$\lim_{n \to \infty} \sqrt[n]{n}$
(I'm remembering something about taking the log of the limit perhaps?)
-Dan
2. ## The answer is...
1
I will try to remember how to get this!
A little later...
Rewrite this as:
e^(ln(n)/n)
Then use L'Hostpital on ln(n)/n.
Lim e^(1/n) = 1
n→∞
3. Originally Posted by spammanon
1
I will try to remember how to get this!
Rewrite this as:
e^(ln(n)/n)
Then use L'Hostpital on ln(n)/n.
Lim e^(1/n)
n→∞
Aaah! I'd forgotten about that little trick. Thank you. (Yay! I remembered the answer correctly!)
-Dan
4. Do it without L'Hopital.
$\ln n = \int_1^n \frac{dx}{x}$.
But, $0\leq \frac{1}{x} \leq \frac{1}{\sqrt{x}}$ for $x\in [1,n]$ for all $n\in \mathbb{Z}^+$.
Thus, $0\leq \int_1^n \frac{dx}{x} \leq \int_1^n \frac{dx}{\sqrt{x}} = 2\sqrt{n} - 2 \leq 2\sqrt{n}$.
Thus, $0 \leq \frac{\ln n}{n} \leq \frac{2\sqrt{n}}{n} = \frac{2}{\sqrt{n}}$.
Now use the Squeeze Theorem. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8807447552680969, "perplexity": 1670.8785259186577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246646036.55/warc/CC-MAIN-20150417045726-00044-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/why-amperes-law-cannot-be-applied-to-certain-situations.163967/ | # Why ampere's law cannot be applied to certain situations
• #1
Ja4Coltrane
225
0
I know this question is asked a lot, but I am confused as to why ampere's law cannot be applied to certain situations. In particular: why can't it be used to calculate the field from a wire of finite length?
• #2
Ja4Coltrane
225
0
Is it perhaps because as the current moves through the wire, there is changing electric flux due to the fact that the charge on one side is going away (even at the instant when the charge is still uniform)?
• #3
Crosson
1,256
4
No, its because Ampere's law (in the integral form i.e. amperian loop that you are familiar with) involves solving for an unknown function B(x,y,z) which is inside a path integral. I don't know of any numerical method for doing this, and it only works symbollically in the toroid, solenoid, slab of current, and infinite wire cases.
The equation is pretty to look at, but in almost all cases e.g. (finite wire) no one has steps for trying to find a solution or an approximation to one...
...because we just transform it into a different equation which we can solve beatuifully: we define the MAGNETIC VECTOR POTENTIAL $$A$$ so that its CURL is the magnetic field $$B$$, and then $$A$$ satisfies LAPLACES EQUATION, which is a standard (easy) PARTIAL DIFFERENTIAL EQUATION.
If numerical integration of laplaces equation cannot proceed for some reason then you can always integrate the field of current elements according to the biot-savaart law.
• #4
gabee
175
0
I was thinking about this too since we just learned Ampere's law...I had the idea that it didn't work perhaps because some "information" about the B-field configuration is not included in the calculation for the path integral. In other words, it's hard to "undo" the integral to get the proper B.
There's a good explanation by nrqed in this thread:
https://www.physicsforums.com/archive/index.php/t-118298.html
nrqed said:
Ampere's law is valid (well, if you include the term added by Maxwell), no matter what. *BUT* the integral over \vec B \cdot \dl is in general difficult to do. It is only in a simple case like an (idealized) infinite wire that you can say that (with a circular closed loop centered around the wire) that vec B \cdot \dl = B dl and that, moreover, the magnitude of the B field is a constant which may be taken out of the integral.
It's the same as for Gauss' law for the electric field. It is *always* valid but the integral is easy to do only in specific cases with a lot of symmetry (spherical, cylindrical or planar symmetry). When a system is more complex and there is no obviosu symmetry, it is not that the law fails, it is rather than it is not very useful because it involves an integral very difficult to do. The reason books look at those special cases (infinte planes, infinite wires, infinite cylinders, etc) is that these are the only cases where the integrals involved in Gauss' and Ampere's laws are easy to carry out. It does not mean that the laws are not valid al the time, simply that they are not terribly useful.
• #5
lpfr
388
3
What is the shape of your finite length wire? It cannot be just a straight segment because current must come from somewhere and go elsewhere (This is not a problem if the wire is infinite, infinity is so far!). You must add other bits of wire to carry the coming and the going current. But, unless the new wires are aligned, you break the symmetry and you now know that B will no more be the same at constant distance and you do not know any more the direction of B.
In fact, out of some elementary and high symmetric problems, you must use Biot-Savart.
• #6
Ja4Coltrane
225
0
Thanks guys, but I figured it out.
I FEEL SO GOOD!!
I USED THE AMPERE MAXWELL EQUATION TO FIND THE B FIELD FROM A FINITE WIRE!!!
• #7
gabee
175
0
Care to post your solution? I'm interested.
• #8
Ja4Coltrane
225
0
rather than posting the full solution, I'll just explain the concept of each step.
First, we must consider that the meaning of a current in a finite wire is the same as that of a staticaly charged rod moving with uniform velocity. With this in mind, one can determine the rate of change of electric flux through some surface by first finding the electric flux from a point charge a distance x from the surface (a circle). Then one can integrate over the length of the rod to find an expression from the flux of a full rod of length "L" and with distance "a" from the cross sectional area of the circular path. After doing this, one can find the derivative of this with respect to time (noting that this is an implicit differentiation). Now you have the rate of change of electric flux with respect to time. Plugg it into the Ampere-Maxwell law, and you will find that you can use it to determine the magnetic field.
• #9
lpfr
388
3
There is a small problem: the electric field created by a static charge is not the same when the charge is moving.
This is the electric field created by a charge q in movement:
$$\vec E={-q\over 4\pi \epsz}\left[{\vec e_{r'}\over r'^2}+ {r'\over c}{d\ \over dt}\left({\vec e_{r'}\over r'^2}\right) + {1\over c^2}{d^2\ \over dt^2}\left(\vec e_{r'}\right)\right]$$
This formula and its user manual can be found in tome II chap. 21-1 of the Feynman Lectures on Physics.
You are ignoring the second term (the wake field). The third is zero if speed is constant.
• #10
Ja4Coltrane
225
0
uh oh...............
I'm sort of in high school...get back to me in a year or two...lol
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3K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9317560195922852, "perplexity": 371.7253473457523}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00681.warc.gz"} |
https://depositonce.tu-berlin.de/items/6652de6f-176e-4410-b456-c5d928a7170f | # Approximability of 3- and 4-hop bounded disjoint paths problems
## Inst. Mathematik
A path is said to be l-bounded if it contains at most l edges. We consider two types of l-bounded disjoint paths problems. In the maximum edge- or node-disjoint path problems MEDP(l) and MNDP(l), the task is to find the maximum number of edge- or node-disjoint l-bounded (s,t)-paths in a given graph G with source s and sink t, respectively. In the weighted edge- or node-disjoint path problems WEDP(l) and WNDP(l), we are also given an integer k and non-negative edge weights, and seek for a minimum weight subgraph of G that contains k edge- or node-disjoint l-bounded (s,t)-paths. Both problems are of great practical relevance in the planning of fault-tolerant communication networks, for example. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8733446598052979, "perplexity": 1718.2816133742704}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00194.warc.gz"} |
https://brilliant.org/practice/applying-the-arithmetic-mean-geometric-mean/?subtopic=classical-inequalities&chapter=classical-inequality-statements | Algebra
# Applying the Arithmetic Mean Geometric Mean Inequality
What is the smallest possible value of the expression $\frac{x^2+361}{x}$ for positive real $x$?
What is the minimum value of $y=\frac{x^2}{x-9}$ when $x>9$?
How many ordered pairs of real numbers satisfy
$36 ^ { x^2 + y} + 36 ^{ y^2 + x } = \frac{2} {\sqrt{6} } ?$
Let $y_1, y_2, y_3, \ldots, y_{8}$ be a permutation of the numbers $1, 2, 3, \ldots, 8$. What is the minimum value of $\displaystyle \sum_{i=1}^8 (y_i + i )^2$?
If $a$ and $b$ are positive numbers such that $a\ne 1, b\ne 1$ and $a+b \neq 1,$ what is the order relation of the following $3$ expressions $A, B$ and $C?$ \begin{aligned} A &= \frac{1}{\log_a 2}+\frac{1}{\log_b 2}, \\ B &= 2\left(\frac{1}{\log_{a+b} 2}-1\right), \\ C &= 2\left(1+\frac{1}{\log_{ab} 2}-\frac{1}{\log_{a+b} 2}\right) \end{aligned}
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000087022781372, "perplexity": 223.81098987207264}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00158.warc.gz"} |
https://www.physicsforums.com/threads/what-exactly-makes-mesons-unstable.716039/ | # What exactly makes Mesons unstable?
1. Oct 12, 2013
### 4nn4
As the title says- what is it that makes all mesons unstable? I know some quarks are less stable than others obviously, but why is there no stable meson? What is it about any combination of quark and anti-quark that makes a particle decay nearly immediately?
In a related topic- what exactly is it about uud that makes a proton stable?
Thank you in advance to any respondents.
2. Oct 12, 2013
### fzero
In general an initial state $|i\rangle$ will be unstable whenever there exists a final state $|f\rangle$, such that:
1. $|i\rangle$ and $|f\rangle$ are connected by some interaction term $\Gamma$. Namely, the matrix element $\langle f |\Gamma| i \rangle$ is nonzero.
2. The potential energy of the final state is less than the energy of the initial state. Here we really mean that the mass of the particles in the final state are less than the initial state particle. Any excess energy will contribute to the kinetic energy of the final state particles.
For the mesons, the electroweak interactions always exist between quarks and antiquarks of the same or different type. For the same type of quark, we have electromagnetic interactions $\bar{q} A_\mu \gamma^\mu q$, while for different quarks, there are weak interactions $\bar{q} W_\mu \gamma^\mu q'$. These interactions can switch a quark $q$ to the same quark $q$ plus a photon or to a different type quark $q'$ plus a W boson. They can also destroy a quark-antiquark pair $\bar{q}q$ and create a photon, or destroy a $\bar{q}q'$ to create a W. The final state photon can create an electron-positron pair in the final state, while the W bosons will themselves decay to electrons and neutrinos in the final state, both of which are much, much lighter than the observed masses of mesons.
What is not explained by the above is why the mesons are heavier than electrons and neutrinos. It turns out that this is a consequence of the strength of the strong interactions, but there is no fundamental reason why. It is just an experimental fact.
The proton is stable because it is the lowest energy (lightest) configuration of 3 quarks. Because of charge conservation and color confinement, any electroweak interaction term like the ones above applied to the proton must lead to another final state composed of 3 quarks. Remember, the electroweak interactions either switch a quark for a quark + a gauge boson or replace a quark-antiquark pair for a gauge boson. So acting on a baryon $q_1q_2q_3$, since we don't have an antiquark to destroy, we obtain another state
$$q_1q_2q_3' + W,Z, \gamma.$$
There is no state $q_1q_2q_3'$ with mass less than the proton, so a decay of this type is forbidden by energy conservation.
The strong interactions are a slightly different story. These have a form similiar to the EM interaction, between same type quarks: $\bar{q} G_\mu \gamma^\mu q$. The difference is the color indices on the quarks, which I haven't drawn in. The difference between this operator and the electroweak ones is that it operates on gluons. Since the proton is bound by the strong interactions, there are lots of virtual gluons present at any given time. So we can have strong interactions where we convert an internal gluon to a quark-antiquark pair. We can represent this as an interaction
$$q_1q_2q_3g \rightarrow q_1q_2q_3 + \bar{q}_4 q_4.$$
So the strong interactions can generate an additional meson in the final state. However, since we still have the $q_1q_2q_3$ particle in the final state, energy conservation still prevents the photon from "decaying" in this manner.
3. Oct 12, 2013
### 4nn4
Thank you for taking the time to write such a detailed and clearly explained response.
Thanks again
4. Oct 13, 2013
### kloptok
I find this statement interesting as I've always been told that it is baryon number conservation which leads to the proton being stable. Something I've never been entirely satisfied with as the baryon number conservation is in a sense not as fundamental as e.g. charge conservation.
Could you explain further why charge conservation and confinement leads to the proton being stable (and perhaps, if it is too much asked, comment on why this is different in GUT theories, where the proton is unstable? I would be grateful for any answer!
5. Oct 13, 2013
### dauto
fzero did not mention it, but Baryon number conservation is essential to explain proton stability. Baryon number is not conserved in GUT theories, hence the prediction that protons do decay if GUT turns out to be correct.
6. Oct 13, 2013
### Staff: Mentor
It is expected that baryon number conservation breaks down at high energies - there has to be some reason why we are surrounded by baryons without significant amounts of antibaryons. If there would be no such process, it would be as fundamental as charge conservation.
7. Oct 13, 2013
### fzero
Baryon number conservation is the explanation for why there are no interaction terms in the standard model that turn a quark into something else, such as $\bar{e} V_\mu \gamma^\mu q$ or the like. In the SM, baryon number is assigned to quarks directly and so is part of "charge conservation," though I didn't elaborate since my post was already very long and liable to be confusing. It's clear that confinement is necessary, else one could imagine a baryon splitting up into free quarks.
Before the quark model, baryon number conservation was much more directly connected with the fact that, in the decay chain of any heavy baryon, the end state always contains a proton. In the quark model, the end result is the same.
8. Oct 13, 2013
### arivero
Incidentally, if the mass of charged pion were equal to the mass of muon and the electron were massless, it would be stable at first order, because of "helicity suppression". Actually I am not sure if it is unstable beyond first order.
9. Oct 14, 2013
### Staff: Mentor
How does a massless, charged particle look like?
If electron and muon both had at least the mass of a pion, the charged pion would be the lightest charged particle, and electrons and muons would decay to pions (+neutrino).
10. Oct 17, 2013
### Doofy
Tonight I've been unsuccessfully googling around for a document that walks you through those nucleon decay predictions without requiring too much background knowledge, do you know of any good ones?
11. Oct 18, 2013
### dauto
I'm not aware of any good ones. The basic concept is quite simple though. In grande unified theories there are multiplets that include leptons and quarks in them. That means that quarks can turn into leptons and vice-verse and would emit a virtual boson in the process. That's analog to the way that an electron can turn in a neutrino and vice-verse while emitting a W boson in the process. The W boson is very massive suppressing decays that depend on the weak force (That is why it is weak). The Gut bosons are expected to be about a million billion times more massive than the W boson suppressing decays that depend on them (such as the proton decay) by a much larger factor. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9660069346427917, "perplexity": 477.3827798436691}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00365.warc.gz"} |
http://math.stackexchange.com/questions/253902/sum-m-0n-m-np2-n-choose-m-pm-qn-m-npq | $\sum_{m=0}^n (m-np)^2 {n \choose m} p^m q^{n-m} = npq$
How to show that:
$\sum_{m=0}^n (m-np)^2 {n \choose m} p^m q^{n-m} = npq$
-
do you really want to show it this way or can you use the binomial random variable? – Jean-Sébastien Dec 8 '12 at 18:11
Here is a way to see it without having to expand anything in the sum. Consider a random variable $X$ following a binomial distribution with parameters $n,p$. By definition, it is the sum of $n$ i.i.d. Bernoulli$(p)$ random variable, who have expectation $p$ and variance $pq$. So since $$X=X_1+X_2+\cdots X_n,$$ it's expectation is the sum of the expectation of all $X_m$, so $E[X]=\sum_{m=0}^nE[X_m]=np$. Since the $X_m's$ are independant, the variance of their sum is the sum of their variance so $$Var[X]=npq,$$ $(q=1-p)$. By definition of the variance, you also have $$Var[X]=\sum_{k=0}^{n}(m-np)^2\binom{n}{m}p^{m}q^{n-m}.$$ thus, both are equal.
$(m-np)^2\binom n m=\{m(m-1)+(1-2np)m+n^2p^2\}\binom n m=n(n-1)\binom{n-2}{m-2}+(1-2np)n\binom {n-1}{m-1}+n^2p^2\binom n m$
as $m\binom n m=m\frac{n!}{m!(n-m)!}=mn\frac{(n-1)!}{m\cdot(m-1)!\{(n-1)-(m-1)\}!}=n\binom{n-1}{m-1}$ for $m\ge 1$
and $m(m-1)\binom n m=m(m-1)\frac{n!}{m!(n-m)!}$ $=m(m-1)n(n-1)\frac{(n-1)!}{m(m-1)\cdot(m-2)!\{(n-2)-(m-2)\}!}=n(n-1)\binom{n-2}{m-2}$ for $m\ge 2$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9866496920585632, "perplexity": 108.1640869370218}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500832032.48/warc/CC-MAIN-20140820021352-00358-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://eccc.weizmann.ac.il/author/219/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > AUTHORS > AARON POTECHIN:
All reports by Author Aaron Potechin:
TR21-091 | 29th June 2021
Gil Cohen, Dor Minzer, Shir Peleg, Aaron Potechin, Amnon Ta-Shma
#### Expander Random Walks: The General Case and Limitations
Cohen, Peri and Ta-Shma (STOC'21) considered the following question: Assume the vertices of an expander graph are labelled by $\pm 1$. What "test" functions $f : \{\pm 1\}^t \to \{\pm1 \}$ can or cannot distinguish $t$ independent samples from those obtained by a random walk? [CPTS'21] considered only balanced labelling, ... more >>>
TR16-058 | 12th April 2016
Boaz Barak, Samuel Hopkins, Jonathan Kelner, Pravesh Kothari, Ankur Moitra, Aaron Potechin
#### A Nearly Tight Sum-of-Squares Lower Bound for the Planted Clique Problem
We prove that with high probability over the choice of a random graph $G$ from the Erd\H{o}s-R\'enyi distribution $G(n,1/2)$, the $n^{O(d)}$-time degree $d$ Sum-of-Squares semidefinite programming relaxation for the clique problem will give a value of at least $n^{1/2-c(d/\log n)^{1/2}}$ for some constant $c>0$.
This yields a nearly tight ... more >>>
TR12-185 | 29th December 2012
Siu Man Chan, Aaron Potechin
#### Tight Bounds for Monotone Switching Networks via Fourier Analysis
We prove tight size bounds on monotone switching networks for the NP-complete problem of
$k$-clique, and for an explicit monotone problem by analyzing a pyramid structure of height $h$ for
the P-complete problem of generation. This gives alternative proofs of the separations of m-NC
from m-P and of m-NC$^i$ from ... more >>>
TR09-142 | 17th December 2009
Aaron Potechin
#### Bounds on Monotone Switching Networks for Directed Connectivity
Revisions: 1
We prove that any monotone switching network solving directed connectivity on $N$ vertices must have size $N^{\Omega(\log N)}$
more >>>
ISSN 1433-8092 | Imprint | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9269170761108398, "perplexity": 3783.735476555696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00622.warc.gz"} |
http://tex.stackexchange.com/questions/26217/which-ide-editor-and-compiler-do-you-use-for-tex-latex-how-do-macros-work/26266 | # Which IDE/Editor and “compiler” do you use for TeX/LaTeX? - How do macros work? [closed]
I've been using LaTeX for a few weeks now, with TeXworks, and have found it quite frustrating.
I'm running MiKTeX 2.8 on Windows 7 x64 with TeXworks 0.3.
Reasons why I find TeX frustrating:
• Very difficult to install packages, and get them running in my code (sometimes impossible, i.e. with xy-pic)
• When I write code, it takes a while (dollar signs everywhere)
• I also end up using \\[2mm] all over the place
Quick related question, how do I create macros? — Basically I want to replace $\rightarrow$ with \im (implies) etc. It would also be helpful if I could define keyboard shortcuts for this type of thing (Alt+A = $\forall%, Alt+E = $\exists$, Alt+. = $\rightarrow$, Alt+, = $\leftrightarrow$ etc.). So which TeX, LaTeX or other package would you recommend I use for typesetting, with which IDE/text-editor? Purpose: For typing up Discrete Mathematics Lectures (Graph Theory, Logic Theory etc.) - ## closed as not a real question by Joseph Wright♦Mar 27 '12 at 18:44 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. You can tell MiKTeX to automatically install packages. Simply go to Start -> MiKTeX 2.8 -> Maintenance (Admin) -> Settings (Admin). Under "General" you find "package installation", you can set it to install on the fly. This of course requires an online repository, if you haven't already go to the package manager (-> Maintenance (Admin) -> Package Manager), and select "Repository" "Change Repository" "Internet" and choose a server near you – Tom Bombadil Aug 21 '11 at 10:28 and for the implies: write \newcommand{\imp}{$\rightarrow$} in the preamble. Shortcuts are either customizable in your editor, so have a look at the TexWorks manual, or consider using Autohotkey, it can assign anything to any key combo – Tom Bombadil Aug 21 '11 at 10:36 Based on your remark about Alt+A = $\forall$, etc., I'll ask: are you putting a dollar sign around each character in mathematics? You ought to be doing something more like $\forall\epsilon\exists\delta:\epsilon < \delta, or the like. If you have a long list of equations, or a chained equation, you might want to be using displayed math environments instead, e.g. the align or gather environments provided by amsmath.sty. (If these are not your problems, my apologies.) – Niel de Beaudrap Aug 21 '11 at 10:54 If this is a question about IDEs, then it's a duplicate of LaTEX editors/IDEs. If it's a question about installing packages on MikTeX Then it's a duplicate of this question. If it's a question about TeXworks shortcuts then BAM. In any of these cases, this should be closed. – Seamus Aug 22 '11 at 10:42 This is far too many questions (and complaints) in one to fit our format: 'not a real question' I think. – Joseph Wright Mar 27 '12 at 18:44 ## 3 Answers Regarding your reasons: • Package installation mainly depends on the TeX distribution and on the operation system. • TeX Live has been providing a package manager since 2008, it's called TeX Live Manager aka tlmgr, running on Windows, Linux, Mac and further Unixes. • MiKTeX also provides a package manager, mpm, even for a longer time • Manual installation should always be possible, there are explanations and howtos, also on this site, see for example: How can I manually install a package on MiKTeX (Windows) • Writing code with many signs is usual - you could use $$ and $$ instead though I guess you don't like that either
• Writing \\[2mm] is a sign that your layout requires improvement. Spacing should be adjusted in the preamble, for example using the setspace package or by customizing environments.
Create your own macros for example by \newcommand or \newenvironment. Not only for having shortcuts, but also for a consistent document which can easily be changed later. You can use \ensuremath in your macro definitions, so your macros will work both in math and in text mode, saving you typing \$.
Generally, I recommend using TeX Live. The version 2011 has just been released. for the editor, I recommend TeXworks as easy and quick editor and TeXnicCenter as complex and very capable editor. TeXnicCenter provides auto-completion of source code which can be useful in your case, and of course shortcuts you asked for. You can even extend the auto-completion yourself, see How to add new auto-complete options in TeXnicCenter.
You might have a further look at:
-
\\[2mm] should be replaced by setting \parskip, it isn't the line spacing. – Leo Liu Aug 21 '11 at 11:17
@Leo Liu: Possibly, if the OP uses it between paragraphs, he didn't write it. So, good addition. In general \\ is for breaking lines, not paragraphs, in text as well as in tables and multiline math where \parskip would not be so meaningful. – Stefan Kottwitz Aug 21 '11 at 11:26
Thank you very much. Quick question before I move to TeXLive, how does the compilation speed compare? Is there a x64 version? – A T Aug 28 '11 at 7:27
Here are some thoughts to the points you made in your post:
packages: As Tom Bombadil said in his comment, MiKTeX usually installs packages on the fly, if you set it up correctly. Installing fonts is a bit more tricky sometimes. I don't know anything about specific pitfalls of xy-pic. If you can't get it running for some reason, check if there's a question about it on tex.sx, and if not, you're welcome to ask a question about it.
dollar-signs: I don't use LaTeX for math, but my understanding is that math typesetting is actually one of the main strengths of it. I'd be surprised if there were any easier and similarly comprehensive way of math typesetting.
macros: Defining shortcuts
keyboard shortcuts in TeXworks: Can I configure shortcut keys in TeXworks? might or might not help.
editors: LaTeX Editors/IDEs lists a lot of editors, TeX Community Polls has a poll regarding the popularity of editors in this community.
General Thoughts: We strive to have only one question per post. If you have several questions and they all haven't been asked on here, it's absolutely no problem to post several questions. This way, other users can benefit maximally from the problems and solutions other users have come up with. Hence, I think this question here should be closed. This doesn't mean we don't want to answer your questions, this question 1) simply doesn't fit the question format and 2) contains questions that have already been asked and thus partially is a duplicate. If, after reading the questions I pointed you at and searching for other questions on tex.sx regarding your problems, you still can't figure something out, just go ahead and ask one or several questions.
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I agree with your last comment. People on TeX.SE are a lot more friendly to simple questions than over on stackoverflow.com – benregn Aug 21 '11 at 19:54
With regards to "which editor to use" and "how to configure keyboard shortcuts"... I came to LaTeX only recently, but I think I can recommend Texmaker.
Texmaker is a cross-platform editor that integrates nicely with a MiKTeX installation on Windows. Make miktex\bin show up in PATH, and Texmaker has everything it needs. If a package is missing, MiKTeX's excellent auto-installation feature will come up and install the package for you... I actually use this setup under WINE in Linux, because IMHO it beats any other combination of editor and LaTeX distribution hands-down.
Texmaker has a good auto-completion feature, and allows free definition of keyboard shortcuts ("Options" / "Configure" / "Editor") as well as user macros and commands (menu "User"). Together with some smart \newcommand LaTeX macros, it has saved me much typing this way. The integrated PDF viewer is nice too, opening the PDF at the current source position and allowing to jump from any place of the PDF to the corresponding LaTeX source file / line.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053963422775269, "perplexity": 1788.7166009247978}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768982.112/warc/CC-MAIN-20141217075248-00091-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/how-many-spacetime-quanta-are-there-depends-how-you-count.748316/ | # How many spacetime quanta are there? (depends how you count.)
Tags:
1. Apr 11, 2014
### marcus
http://arxiv.org/abs/1404.1750
How many quanta are there in a quantum spacetime?
Seramika Ariwahjoedi, Jusak Sali Kosasih, Carlo Rovelli, Freddy P. Zen
(Submitted on 7 Apr 2014)
Following earlier insights by Livine and Terno, we develop a technique for describing quantum states of the gravitational field in terms of coarse grained spin networks. We show that the number of nodes and links and the values of the spin depend on the observables chosen for the description of the state. Hence the question in the title of this paper is ill posed, unless further information about what is being measured is given.
16 pages, 9 figures
==quote page 13, section VII==
VII. HOW MANY QUANTA OF SPACE ARE THERE IN QUANTUM SPACETIME?
Armed with the observations of Section II and with the technology developed in Sections III to VI, let us return to the question of the number of quanta in a given classical geometry. The second example of Section II (the oscillators) shows that the number of quanta is not an absolute property of a quantum state: it depends on the basis on which the state is expanded. In turn, this depends on the way we are interacting with the system.
... For instance, an antenna tuned into a wavelength λ is insensitive to the high-frequency components of the field, if any of these is excited. We cannot treat the gravitational field in the same manner at all scales, because Fourier analysis requires a background geometry, which is, in general, not available in gravity.
Sections III to VI, however, provide a viable alternative: areas and volumes of big links and big nodes, AL and VN , capture large scale features of the field, and are insensitive to higher frequency components of the field, in a way similar to the long wavelength Fourier modes. In fact, notice that this is what we mean when we refer to macroscopic areas and volume.
The area of a table is not the sum of individual areas of all microscopic elements of its boundary; it is the area of a coarse-grained description of the table where the surface is assumed to be flat, even if in reality it is not, at small scales. When we measure the gravitational field, that is, geometrical quantities, we routinely refer to its long wavelength modes. For instance, we can measure the Earth-Moon distance with a laser. What we are measuring is a non-local, integrated value of the gravitational field, in the same manner in which an antenna measures a single wavelength of the electromagnetic field.
The quanta of the gravitational field we interact with, are those described by the quantum numbers of coarse-grained operators like AL and VN , not the maximally fine-grained ones.
Therefore we can begin to answer the question of the title. The number of quanta we see in a system depends on the way we interact with it. When interacting with a gravitational field at large scales we are probing coarse grained features of space, which can be described by the quantum numbers JL,VN of a coarse-grained graph γ0. Probing the field as shorter scales tests higher modes, which can be described by more fine grained subset graphs γ1,...,γm.
The relevance of this construction for understanding the scaling the dynamics, cosmology and black holes will be studied elsewhere.
==endquote==
Last edited: Apr 11, 2014
2. Apr 11, 2014
### marcus
==quote page 2, end of section II==
Quantum discreteness is not the existence of elementary “bricks” of nature. It is the appearance of discreteness in the way a system interacts. An interaction depends on a variable of the system and this variables my have discrete spectrum. If a different variables interact, different kinds of discreteness (classically incompatible with one another) show up...
==endquote==
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http://www.chegg.com/homework-help/questions-and-answers/current-state-economy-bull-inflation-rate-inflation-worry-period-stable-prices-bull-unempl-q1312873 | State of Economy
What is the current state of the economy?
• What is the inflation rate? Is inflation a worry, or are we in a period of stable prices?
• What is the unemployment rate? Will a high unemployment rate cause deflation? What is the current structure of the labor market? How does the current structure affect the threat of cost push inflation or deflation?
• What is the growth rate of GDP? Can you find predictions on the future growth rate?
• What is the income distribution? How has this changed over recent history? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8337730169296265, "perplexity": 2133.7085540168728}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700212265/warc/CC-MAIN-20130516103012-00040-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/53912/scalar-field-lagrangian-and-potential/53915 | # Scalar field lagrangian and potential
This question is a continuation of this Phys.SE post.
Scalar field theory does not have gauge symmetry, and in particular, $\phi\to\phi−1$ is not a gauge transformation. but why? and
I want see the mathematics that will represent the interaction potential from the Lagrangian for scalar field.
Details in the paper: see check the equation (3) please.
-
## 1 Answer
Because there is no physical redundancy in the description. A gauge theory (and related gauge transformations) only occurs if there are different field configurations corresponding to a certain physical configuration. Such redundant configurations can be related by local gauge transformations (see for example the vector potential $A_\mu$ in electrodynamics). In your case (which the other question referred to), the Lagrangian is not invariant under your transformation, since the quantity $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$$ changes under $\phi\rightarrow\phi-1$ to
$$U(\phi-1)= \frac{1}{8} (\phi-1)^2 (\phi -3)^2.$$
-
Where from the equation $U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$ arises or more precisely why did we write in this form? – Unlimited Dreamer Feb 14 '13 at 11:51
That's the interaction potential you wrote down in your other question: physics.stackexchange.com/questions/52590/… – Frederic Brünner Feb 14 '13 at 11:51
My curiosity is why did he write the equation like this? – Unlimited Dreamer Feb 14 '13 at 11:52
Such a term (i.e. one of a higher order than 2) is necessary in order to get interactions in your theory. – Frederic Brünner Feb 14 '13 at 12:01
arxiv.org/abs/0802.3525 in this article, check the equation(3) please. – Unlimited Dreamer Feb 14 '13 at 12:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9572129845619202, "perplexity": 797.7415138034143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096301.47/warc/CC-MAIN-20150627031816-00307-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/non-measurable-set.70810/ | # Non measurable set
1. Apr 10, 2005
### Hurkyl
Staff Emeritus
Is the statement "All subsets of R are measurable" consistent with the axioms of ZF? (Specifically, of course, the omission of the axiom of choice)
I've never really gotten a straight answer to this question, and I'd be very happy to hear it.
2. Apr 10, 2005
### matt grime
I cannot provide any definite answer I'm afraid, but I suspect the answer is yes it is consistent, as is its negation.
Since R may not be a set in some model of ZF, I don't know how this works. But, I seem to remember reading somewhere that in order to construct a non-measurable set you need to use the axiom of choice.
3. Apr 10, 2005
### fourier jr
yes it's impossible to construct a nonmeasurable set without the axiom of choice. solovay proved it in 1970
4. Apr 10, 2005
### mathwonk
so, to repeat the obvious: it seems that the existence of a non measurable set is an axiom equivalent to the axiom of choice?
now that i like. the axiom of choice is a stupidly obvious, highly useful, property I can barely live without. (e.g every surjective function has a right inverse: duhhh.)
measurability and non measurable sets are interesting and not obvious at all though!
5. Apr 10, 2005
### Hurkyl
Staff Emeritus
It would surprise me if it was equivalent -- I can't even begin to imagine how I could get from the existance of a single nonmeasurable set to being able to, say, construct a well-ordering on any (arbitrarily large!) set.
6. Apr 10, 2005
### mathman
From the above discussion, it seems that without the axiom of choice, existence of non-measurable sets cannot be proven. However, it still doesn't follow that you can prove all subsets of R are measurable.
7. Apr 11, 2005
### Palindrom
For the ignorant (me)- what is ZF?
8. Apr 11, 2005
### fourier jr
9. Apr 11, 2005
### Jimmy Snyder
Jerry Bona once said,
The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma? ;)
10. Apr 11, 2005
### Palindrom
11. Apr 11, 2005
### Hurkyl
Staff Emeritus
I don't know what people have against the well-ordering principle. I wield transfinite induction far more easily than I wield Zorn's lemma for proving things!
The funny thing is that sometimes I don't even notice that the problem is practically already in the form needed for Zorn's lemma until I'm done with the transfinite induction proof.
12. Apr 11, 2005
### Jimmy Snyder
They're equivalent. ;)
13. Apr 11, 2005
### mathwonk
i thought it was obvious that was the content of post 3, but i could be oversimplifying.
14. Jan 12, 2010
### futurebird
I was thinking about this same question today. Though, to be very honest I'm yet to find a construction of a non-measurable set that I "get" -- I can follow the steps of the proof but I don't see the big picture. You know?
It is very interesting to me that something sensible like the axiom of choice is linked to something annoying and not intuitive like non-measurable sets.
15. Jan 12, 2010
### wofsy
I don't know anything about the axiom of choice but it seems that whenever you are dealing with infinite sets it would have to come up. Measurability is just a question about an infinite set - is it not?
Also measurable or non-measurable seems to mean Lebesque measurable on R^n - is that what we are talking about here?
The Continuum Hypothesis implies the existence of a non-lebesque measurable subset of the plane. Is the Continuum Hypothesis equivalent to the Axiom of Choice?
A friend of mine once tried to resurrect St. Anselm's proof of the existence of God using the Hausdorff Maximal Principle. The set of virtues are partially ordered by inclusion. The union of any set of virtues is a set of virtues so any chain in the partial ordering has a maximal element. Therefore there is a set of virtues greater than which none can be conceived. So the existence of the idea of God as consumately virtuous depends upon the Axiom of Choice
Last edited: Jan 12, 2010
16. Jan 12, 2010
### wofsy
For the real line a non-measurable set is a basis for the reals over the rationals. The existence of this basis is an application of the Axiom of Choice. But is there a proof that every non-measurable subset of the line is obtained this way?
Last edited: Jan 12, 2010
17. Jan 12, 2010
### mathman
No - they are independent of each other.
Could you back up your first statement?
18. Jan 12, 2010
### wofsy
Sure.
Assume the Continuum Hypothesis and map the real numbers from 0 to 1 bijectively onto the first uncountable ordinal.
In the unit square, define the function f(x,y) = 1 if x<y in the induced ordering, 0 otherwise.
f is the characteristic function of the set (x,y) with x<y so this set is measurable if f is a measurable function. If f is measurable then we can integrate it and find the measure of the set. By Fubini's theorem we can calculate the integral as a double integral and can integrate along either the x or y axis first and get the same answer.
Intgerating along the x axis first:
For each y, there are only countable many x < y since the ordering is onto the first uncountable ordinal. Thus the integral of f over the x axis is zero since any countable set has measure zero. The double integral is thus zero.
Integrating along the y axis first:
For each x, there are only countable many y less that it so the measure of the set of y greater than x is 1. Thus the integral along the y axis is 1 for each fixed x. Thus the double integral equals 1.
This contradicts Fubini's Theorem so f is not measurable.
Last edited: Jan 12, 2010
19. Jan 12, 2010
### Hurkyl
Staff Emeritus
That is a well-ordering of R. Why does it exist?
20. Jan 12, 2010
### wofsy
That is a good question. I am not sure of the answer. I think though that the first uncountable ordinal can be constructed inductively and does not require the Well Ordering Hypothesis. It inherits a well ordering from its construction. The bijection of the reals onto it - i.e. the assumption of the Continuum Hypothesis - automatically transfers this ordering to the reals.I think the well ordering Hypothesis really says that any set can be mapped bijectively onto one of the ordinals - countable or uncountable. My reading of Cantor's construction of the ordinals is that the well ordering is an automatic by product of the way he constructs them. For instance, the inductive construction of the integers automatically gives you a well ordering of them. But... I am not sure.
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# The Stiffness Of Springs
Extracts from this document...
Introduction
The Stiffness Of Springs
The spring constant is a measure of the stiffness of an elastic system.
How is the stiffness of a single spring related to the stiffness of springs in series and parallel?
Plan an experiment that will enable you to make a comparison of the stiffness for identical springs in series and parallel from your results.
Plan
The task of this experiment is to determine the relationship between the stiffness of springs in series an in parallel.
The stiffness of a spring can be shown as:
## F = kx
Where F is the force on a spring, x is the extension of the spring and k is the spring constant or the spring’s stiffness. This means that the force on a spring is proportional to the extension with k being the constant. Therefore as more force is put onto a spring the more it will extend. By using this simple formula we can find the spring constant.
## k = F/x
By dividing the force by the spring extension we can find k. Both the force and the spring extension are easily measurable. We can show the relationship between the force and the extension in a graph.
x
. F
The graph shows that as the force gets bigger the extension does. The gradient of the line is the spring constant. It is a straight line, as the spring constant does not change up to a certain point. There is a point when a certain force will create a very large extension. This point is called the elastic limit. Once a spring reaches this point it becomes permanently deformed. This means it does not return to its original shape or that its spring constant becomes altered.
The point at which the gradient of the line changes in the elastic limit.
Middle
321
6
355
Springs in Series
Mass No. Spring I + II (mm) 1 239 2 307 3 376 4 445 5 513 6 580
Mass No. Spring III + IV (mm) 1 249 2 330 3 395 4 462 5 520 6 601
Mass No. Spring I + III (mm) 1 249 2 315 3 382 4 448 5 514 6 582
Mass No. Spring I + II + III (mm) 1 317 2 420 3 523 4 622 5 724 6 824
Mass No. Spring II + III + IV (mm) 1 329 2 433 3 535 4 637 5 737 6 840
Mass No. Spring I + II + III + IV (mm) 1 404 2 539 3 674 4 811 5 946 6 1080
Springs in Parallel
Mass No. Spring I + II (mm) 1 158 2 173 3 192 4 209 5 226 6 243
Mass No. Spring II + III (mm) 1 162 2 177 3 196 4 213 5 229 6 246
Mass No. Spring III + IV (mm) 1 167 2 184 3 200 4 218 5 235 6 252
Mass No. Spring I + II + III (mm) 1 175 2 183 3 193 4 204 5 215 6 226
Mass No. Spring II +III + IV (mm) 1 176 2 186 3 198 4 209 5 220 6 231
Mass No. Spring I + II + III + IV (mm) 1 158 2 163 3 170 4 177 5 186 6 195
On the following pages are graphical interpretations of these results.
Analysis
These are the spring constants. These have been worked out by first drawing the force/extension on a graph. Then by drawing a line of best fit through these points. Then the inverse of the gradient of the line gives you the spring constant. This is because the gradient of a line is y/x. This would mean that the gradient showed that spring constant was extension/force when it should be force/extension. To solve this I have to use the inverse function of the gradient as my spring constant.
The lines on these graphs do not pass trough the origin. This is due to the fact that I have not taken away the original length of the springs from the totals. This gives every result an offset which will vary from spring systems. However I found that this is irrelevant, as this would not affect the gradient from which I will obtain the spring constant.
Spring constant should be measured in Nm however as I have measured all my extensions in millimetres I will have to change everything into metres, as it is a standard unit of measurement.
Conclusion
If I had used identical springs instead of similar springs then I would have had more conclusive results. Also if the springs became slightly deformed during the experiment then my results could be affected.
Where I took the results there might have been a breeze which might have fractionally affected my results which could lead to more errors. If I did this experiment in an environment where air pressure was normal and there was no breeze then I could achieve more accurate results.
There are several anomalous results in my experiment. These have either been marked with ** or with a circle. These results do not fit the pattern which the other results do. This could be due to many different errors. If I had repeated these results I might have a got a results that did fit the pattern.
On my graphs I have quantified some of the errors by drawing error boxes around the points I have plotted. This is because the masses were ± 3 of 100g. Also I left some space for error for the extension. This was ± 1mm. This accounted for if I had incorrectly read any results or if there was a breeze or friction.
To improve my results I also could have repeated the experiment and taken averages for each value or taken more readings. However I did not have sufficient time for this.
The formulas that I had found out might not be correct as none of the results actually correlated with the formula exactly. However this is probably due to the many errors that could affect my results. I based my formulas on rough patterns which seemed to fit with my results.
Another avenue to explore in this experiment would be mixed systems with both series and parallel systems within it.
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# Related AS and A Level Waves & Cosmology essays
1. ## Black Holes Research and Report
4 star(s)
Finding the Radius of the Event Horizon To find the radius of the event horizon of a black hole, physicists first need use equations of potential gravitational field strength and kinetic energy...
2. ## Investigate the effect of mass on the extension of a spring.
3 star(s)
I must also make sure I measure the length only when it has stretched as much as it can. Variable I will also endeavour to keep the characteristics of the spring the same, such as the spring's original length, the spring's material as well as its thickness.
1. ## Investigating the relationship between the mass and time period in a spring-mass system
When I was conducting my experiment I quickly realised how easily an error in timing may have been introduced. First of all, when I started with my lightest weight (which mass, with the weight hanger, was about 150g) I was unable to count the number of oscillations properly, because they were too quick.
2. ## The aim of this investigation is to examine the effect on the spring constant ...
During this time, the spring obeys Hooke's Law, meaning that the extensions is equal to the stretch force applied divided by the spring's 'Spring Constant' (which will be discussed later). a - b: After point a there is a small section of the graph where the spring is still within
1. ## Measuring spring constant using oscilations of a mass.
the following reasons among with others: * I couldn't measure the time properly for the period meaning that I took more time to react after I saw the pendulum go across the fiducial marker. * It could also be because my height of small h could have been measured wrongly
2. ## Determine the value of 'g', where 'g' is the acceleration due to gravity.
By substituting for m in the equation above: =() = = = + By measuring the extension (e) and the corresponding time period (T) using several different masses in turn, a graph of e against T2 can be drawn as shown below: However the value of k can be calculated by plotting a graph of T2 against m.
1. ## In this experiment, I am going to find out the relationship between Force and ...
Record my results on a table and then plot my graph of extension against force and to see how this material behaves before performing my experiment. EXPERIMENT DIAGRAM METHOD I will set my apparatus as shown in the diagram above.
2. ## Investigation on how putting springs in series and parallel affects their extension.
How I will conduct my experiment I will set up a clamp stand with my springs on it, attach weights of 1N,2N,3N,4N,5N,6N and 7N to a single spring and then to two springs in parallel. When I attach weights to the two springs in series however I shall only go
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https://charmestrength.com/what-is-another-word-for-torque/ | # What Is Another Word For Torque?
What is another word for torque? What is another word for torque?
revolution spin
turn twist
force rotation
turning twisting
rotational force rotating force
## What is the best definition of torque?
1. 1. Torque is a twisting force, generally that causes something to rotate around an axis or other point. The force that causes a wheel to rotate around an axle is an example of torque. noun.
## What is a torque in English?
1 : a force that produces or tends to produce rotation or torsion an automobile engine delivers torque to the drive shaft also : a measure of the effectiveness of such a force that consists of the product of the force and the perpendicular distance from the line of action of the force to the axis of rotation.
## What is the other name of torque in physics?
torque, also called moment of a force, in physics, the tendency of a force to rotate the body to which it is applied.
## What is torque simple words?
Torque is a twisting force that speaks to the engine's rotational force and measures how much of that twisting force is available when an engine exerts itself. This applies torque, or a twisting force, to the bolt. While horsepower is simply measured in horsepower, torque is typically measured in pounds feet (lb. -ft).
## Related guide for What Is Another Word For Torque?
### What are some examples of torque?
Torque is the application of force where there is rotational motion. The most obvious example of torque in action is the operation of a crescent wrench loosening a lug nut, and a close second is a playground seesaw.
### What does torque mean in slang?
(chiefly US, idiomatic, slang) Annoyed, upset, angry.
### What is the symbol of torque called?
τ
symbol quantity symbol
τ, τ torque N m
I moment of inertia kg m2
L, L angular momentum kg m2/s
H, H angular impulse N m s
### What is a torque in physics?
Torque is a measure of how much a force acting on an object causes that object to rotate. In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F.
### What is HP and torque?
Torque, simply, is the ability of a vehicle to perform work — specifically, the twisting force applied by the crankshaft. Horsepower is how rapidly the vehicle can perform that work. For instance, a lightweight sports car that operates at high rpm may have high horsepower, but low torque.
### Is torque the same as moment?
Difference Between Torque and the Moment. The turning or the twisting effect of a body is called the torque. Torque is considered the force that rotates the body about the axis. A moment is the force that causes the body to move (not rotate).
### What is another word for horsepower?
What is another word for horsepower?
power force
vigorUS vigourUK
potence firepower
sinew forcefulness
stamina brawn
### What is torque in human body?
Torque is the driving force for human movement. Being able to manipulate the target muscle torque will allow for a more specific intervention. Moment Arm of a force system is the perpendicular distance from an axis to the line of action of a force. Torque is the ability of a force to cause rotation on a lever.
### What is another word for powerful?
What is another word for powerful?
forceful violent
strong vigorous
almighty hard
mighty ferocious
explosive fierce
### What is torque in layman's terms?
In simple terms, the definition of torque is the engine's rotational force. It differs from horsepower as it refers to the amount of work an engine can exert, while horsepower defines how quickly that work can be delivered. It's why torque is often referred to in layman's terms as 'pulling power', 'oomph', or 'grunt'.
### What is torque in real life?
Torque is the measure of the force that can cause an object to rotate about an axis. In physics, torque is simply the tendency of a force to turn or twist. Different terminologies such as moment or moment of force are interchangeably used to describe torque.
### Why is torque important in life?
Torque is a crucial part of generating power from a car's engine, as it represents the load an engine can handle to generate a certain amount of power to rotate the engine on its axis. The force is measured in pounds (lb) per foot (ft) of rotation around one point. Multiply this torque force (in lb-ft.)
### What is torquing a bolt?
Torque tightening is the accurate application of torque to a nut so that a bolt can hold its load securely without breaking. Too much torque can break the bolt. When you apply the right amount of torque, the bolt is properly stretched so that it can act as a solid spring to clamp the two materials together.
### What does Tworking mean?
New Word Suggestion. A type of dance or exercise that emphasizes a female' buttocks.
### Is torque a scalar or vector quantity?
Torque is a vector quantity. The direction of the torque vector depends on the direction of the force on the axis.
### What Greek letter is used for torque?
or τ, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.
### What is the Greek letter for torque?
Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically \boldsymbol \tau or τ, the lowercase Greek letter tau.
### Why does torque exist?
conservation of energy. As a lever pivots, one end moves further than the other, so in order for conservation of energy to be satisfied, the force has to be lower on that side. Torque is defined by cross product of radial distance and force r×F, since F exists torque exists.
### Why is torque measured in NM?
Torque tells you how strong an engine is. Torque is measured in Newton metres (Nm) or you might see the imperial measurement of lb-ft (pounds-feet). If you want to calculate the conversion for yourself, 1 Nm is equivalent to 0.738 lb/ft.
### What's the difference between work and torque?
Torque – Torque is the vector quantity. It is the rotational force which causes an object to rotate about an axis. Work is said to be done when the object moves the certain distance when the force is applied.
### Is torque better than BHP?
How BHP And Torque Affect Car Performance. To keep it short and sweet, BHP affects the top speed and acceleration of a car while torque affects the amount of load you can carry without performance degradation.
### What Does BHP stand for horsepower?
What is Brake Horsepower (bhp) However, Brake Horsepower (bhp) is often used as a more realistic measurement of power. This is because bhp considers the power left after other car parts are working such as the gearbox, alternator, and water pump as well as any loss of power due to friction.
### What is difference between torque and torsion?
Main Difference – Torque vs. Torsion. The main difference between torque and torsion is that torque describes something that is capable of producing an angular acceleration, whereas torsion describes the twist formed in a body due to a torque.
### What is the difference between stress and pressure?
The important difference between stress and pressure. The internal resistive force to the deformation per unit area is termed as stress. The amount of force applied per unit area is termed as pressure. The magnitude of the pressure at a point in all the directions remains the same.
### What is difference between couple and torque?
A couple is the moment that is the resultant of two forces of the same magnitude, acting in opposite direction at the same distance from the reaction point. Torque is a moment that is applied in such a way that it tends to rotate a body around its axis.
### What is the symbol for horsepower?
hp
Horsepower
One mechanical horsepower lifts 550 pounds (250 kg) by 1 foot high in 1 second.
General information
Unit of power
Symbol hp | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8490394949913025, "perplexity": 814.7047683476894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00255.warc.gz"} |
http://radfordmathematics.com/calculus/integration/power-rule-integration/power-rule-integration.html | ☰ Related
# Power Rule for Integration
The power rule for integration provides us with a formula that allows us to integrate any function that can be written as a power of $$x$$.
By the end of this section we'll know how to evaluate integrals like: $\int 4x^3 dx$ $\int \frac{3}{x^2}dx$ $\int \begin{pmatrix} 2x + 3 \sqrt{x} \end{pmatrix} dx$ We start by learning the power rule for integration formula, before watching a tutorial and working through some exercises.
We also treat each of the "special cases" such as negatitive and fractional exponents to integrate functions involving roots and reciprocal powers of $$x$$.
The power rule for integration is an essential step in learning integration, make sure to work through all of the exercises and to watch all of the tutorials.
## Power Rule for Integration
Given a function, which can be written as a power of $$x$$, we can integrate it using the power rule for integration: $\text{if} \quad f(x) = a.x^n$ \begin{aligned} \text{then} \quad F(x) &= \int a.x^n dx \\ F(x) &= \frac{a}{n+1}x^{n+1} + c \end{aligned}
## Exercise 1
Integrate each of the following
1. $$\int 5x^3 dx$$
2. $$\int 3x dx$$
3. $$\int 7x^2 dx$$
4. $$\int -4x^5 dx$$
5. $$\int 12 x^7 dx$$
6. $$\int -2x^5 dx$$
7. $$\int 2 dx$$
8. $$\int 10 x^4 dx$$
## Solution
1. $$\int 5x^3 dx = \frac{5}{4}x^4 + c$$
2. $$\int 3x dx = \frac{3}{2}x^2 + c$$
3. $$\int 7x^2 dx = \frac{7}{3}x^3 + c$$
4. $$\int -4x^5 dx = -\frac{2}{3}x^6 + c$$
5. $$\int 12x^7 dx = \frac{3}{2}x^8 + c$$
6. $$\int -2x^5 dx = -\frac{1}{3}x^6 + c$$
7. $$\int 2 dx = 2x + c$$
8. $$\int 10x^4 dx = 2x^5 + c$$
## Solution
1. We integrate $$5x^3$$ as follows: \begin{aligned} \int 5x^3 dx &= \frac{5}{3+1}x^{3+1}dx \\ & = \frac{5}{4}x^4 + c \end{aligned}
2. We integrate $$3x$$ using the power rule for integration and the fact that $$x = x^1$$: \begin{aligned} \int 3x dx & = \int 3x^1 dx \\ & = \frac{3}{1+1}x^{1+1} + c \\ & = \frac{3}{2}x^2 + c \end{aligned}
3. We integrate $$7x^2$$ as follows: \begin{aligned} \int 7x^2 dx & = \frac{7}{2+1}x^{2+1} + c \\ & = \frac{7}{3}x^3 + c \end{aligned}
4. We integrate $$-4x^5$$ as follows: \begin{aligned} \int - 4x^5 dx & = -\frac{4}{5+1}x^{5+1} +c \\ & = - \frac{4}{6}x^6 + c \\ & = - \frac{2}{3}x^6 + c \end{aligned}
5. We integrate $$12x^7$$ as follows: \begin{aligned} \int 12x^7 dx & = \frac{12}{7+1}x^{7+1} + c \\ & = \frac{12}{8}x^8 + c \\ & = \frac{3}{2}x^8 + c \end{aligned}
6. We integrate $$-2x^5$$ as follows: \begin{aligned} \int -2x^5 dx & = - \frac{2}{5+1}x^{5+1} + c \\ & = - \frac{2}{6}x^{6} + c \\ & = - \frac{1}{3} x^6 + c \end{aligned}
7. We integrate $$2$$ using the fact that $$2 = 2x^0$$: \begin{aligned} \int 2 dx & = \int 2x^0 dx \\ & = \frac{2}{0+1}x^{0+1} + c \\ & = \frac{2}{1}x^1+c \\ & = 2x+c \end{aligned}
8. We integrate $$10x^4$$ as follows: \begin{aligned} \int 10x^4 dx & = \frac{10}{4+1}x^{4+1} + c \\ & = \frac{10}{5}x^5+c\\ & = 2x^5+c \end{aligned}
## Negative Exponents
Any function looking like $$f(x) = \frac{a}{x^n}$$ can be written using a negative exponent: $f(x) = a.x^{-n}$ Using this fact we can integrate any function written as: $f(x) = \frac{a}{x^n}$ Except for $$\frac{a}{x}$$! Indeed, we'll soon learn about that special case.
This formula is illustrated wih some worked examples in Tutorial 2.
Now that we've seen that we can integrate functions looking like $$f(x)=\frac{a}{x^n}$$ using negative powers of $$x$$, let's work through the exercise below.
Useful Trick: it's often useful to use the fact that $$\int ax^n dx = a \int x^n dx$$, particularly when $$a$$ is a fraction like in question 5. in the following exercise. It's useful to write: $$\int \frac{5}{2x^3}dx = \frac{5}{2}\int \frac{1}{x^3}dx$$ to not let the fraction $$\frac{5}{2}$$ lead to a error in arithmetic.
## Exercise 2
1. $$\int \frac{2}{x^3}dx$$
2. $$\int \frac{3}{x^5}dx$$
3. $$\int -\frac{1}{x^2} dx$$
4. $$\int \frac{6}{x^5}dx$$
5. $$\int \frac{5}{2x^3} dx$$
## Solution Without Working
1. $$\int \frac{2}{x^3} dx = - \frac{1}{x^2} + c$$
2. $$\int \frac{3}{x^5} dx = - \frac{3}{4x^4} + c$$
3. $$\int - \frac{1}{x^2} dx = \frac{1}{x} + c$$
4. $$\int \frac{6}{x^5} dx = - \frac{3}{2x^4} + c$$
5. $$\int \frac{5}{2x^3} dx = \frac{2}{7x^2} + c$$
## Solutions With Working
1. To integrate $$\frac{2}{x^3}$$ we use the fact that $$\frac{2}{x^3} = 2x^{-3}$$: \begin{aligned} \int \frac{2}{x^3} dx &= \int 2x^{-3} dx \\ & = \frac{2}{-3+1}x^{-3+1}+c \\ & = \frac{2}{-2}x^{-2}+c \\ & = -x^{-2}+c \\ & = - \frac{1}{x^2} + c \end{aligned}
2. To integrate $$\frac{3}{x^5}$$ we use the fact that $$\frac{3}{x^5} = 3x^{-5}$$: \begin{aligned} \int \frac{3}{x^5} dx & = \int 3x^{-5} dx \\ & = \frac{3}{-5+1}x^{-5+1} + c \\ & = \frac{3}{-4}x^{-4} + c \\ & = - \frac{3}{4} x^{-4} + c \\ & = - \frac{3}{4}.\frac{1}{x^4} + c \\ & = - \frac{3}{4x^4} + c \end{aligned}
3. To integrate $$-\frac{1}{x^2}$$ we use the fact that $$-\frac{1}{x^2} = -x^{-2}$$: \begin{aligned} \int - \frac{1}{x^2} dx & = \int -x^{-2}dx \\ & = \frac{-1}{-2+1}x^{-2+1} + c \\ & = \frac{-1}{-1}x^{-1} +c \\ & = x^{-1} + c \\ & = \frac{1}{x} + c \end{aligned}
4. To integrate $$\frac{6}{x^5}$$ we use the fact that $$\frac{6}{x^5} = 6x^{-5}$$: \begin{aligned} \int \frac{6}{x^5} dx & = \int 6x^{-5} dx \\ & = \frac{6}{-5+1}x^{-5+1} + c \\ & = \frac{6}{-4}x^{-4} + c \\ & = - \frac{3}{2}x^{-4}+ c \\ & = - \frac{3}{2}.\frac{1}{x^4} + c \\ & = - \frac{3}{2x^4} + c \end{aligned}
5. To integrate $$\frac{5}{2x^3}$$, we use the fact that $$\frac{5}{2x^3} = \frac{5}{2}x^{-3}$$: \begin{aligned} \int \frac{5}{2x^3} dx & = \int \frac{5}{2}x^{-3} dx \\ & = \frac{5}{2}\int x^{-3}dx \\ & = \frac{5}{2} \times \frac{1}{-3+1}x^{-3+1}+c \\ & = \frac{5}{2} \times \frac{1}{-2}x^{-2}+c \\ & = \frac{5}{-4}.x^{-2}+c \\ & = - \frac{5}{4}.x^{-2}+c\\ & = - \frac{5}{4}.\frac{1}{x^2}+c \\ \int \frac{5}{2x^3} dx & = - \frac{5}{4x^2}+c \end{aligned}
## Fractional Exponents
Functions looking like $$f(x) = a.\sqrt[n]{x^m}$$ can be written as powers of $$x$$ using fractional exponents: $f(x) = a.x^{\frac{m}{n}}$ we can therefore use the power rule for integration to integrate any function looking like $$f(x)=a.\sqrt[n]{x^m}$$.
This formula is illustrated wih some worked examples in Tutorial 3.
Now that we've seen how to integrate roots using fractional powers of $$x$$, let's work through a few more questions.
## Exercise 3
Integrate each of the following:
1. $$\int \sqrt{x} dx$$
2. $$\int 2.\sqrt[3]{x} dx$$
3. $$\int 4. \sqrt[5]{x^4}dx$$
4. $$\int 6 .\sqrt{x}dx$$
5. $$\int \frac{\sqrt[3]{x}}{4} dx$$
## Solutions Without Working
1. The answer can be written in two ways:
$$\int \sqrt{x} dx = \frac{2}{3} \sqrt{x^3}+c$$ or $$\int \sqrt{x} dx = \frac{2}{3}.x^{\frac{3}{2}}+c$$
2. The answer can be written in two ways:
$$\int 2. \sqrt[3]{x} dx = \frac{3}{2} \sqrt[3]{x^4}+c$$ or $$\int 2. \sqrt[3]{x} dx = \frac{3}{2}.x^{\frac{3}{4}}+c$$
3. The answer can be written in two ways:
$$\int 4.\sqrt[5]{x^4}dx = \frac{20}{9}.\sqrt[5]{x^9}+c$$ or $$\int 4.\sqrt[5]{x^4}dx = \frac{20}{9}.x^{\frac{9}{5}}+c$$
4. The answer can be written in two ways:
$$\int 6 \sqrt{x}dx = 4.\sqrt{x^3}+c$$ or $$\int 6 \sqrt{x}dx = 4.x^{\frac{3}{2}}+c$$
5. The answer can be written in two ways:
$$\int \frac{\sqrt[3]{x}}{4}dx = \frac{3}{16}.\sqrt[3]{x^4}+c$$ or $$\int \frac{\sqrt[3]{x}}{4}dx = \frac{3}{16}.x^{\frac{4}{3}}+c$$
## Solutions With Working
1. We integrate $$\int \sqrt{x} dx$$ as follows: \begin{aligned} \int \sqrt{x} dx & = \int x^{\frac{1}{2}} dx \\ & = \frac{1}{\frac{1}{2} + 1}.x^{\frac{1}{2}+1} + c \\ & = \frac{1}{\frac{3}{2}}.x^{\frac{3}{2}}+c \\ & = \frac{2}{3}.x^{\frac{3}{2}}+c \\ \int \sqrt{x} dx & = \frac{2}{3}.\sqrt{x^3}+c \end{aligned}
2. We integrate $$\int 2. \sqrt[3]{x} dx$$ as follows: \begin{aligned} \int 2. \sqrt[3]{x} dx & = \int 2.x^{\frac{1}{3}} dx \\ & = \frac{2}{\frac{1}{3}+1}.x^{\frac{1}{3}+1}+c \\ & = \frac{2}{\frac{4}{3}}x^{\frac{4}{3}}+c \\ & = 2 \times \frac{3}{4}.x^{\frac{4}{3}} + c \\ & = \frac{3}{2}.x^{\frac{4}{3}}+c \\ \int 2. \sqrt[3]{x} dx & = \frac{3}{2}.\sqrt[3]{x^4}+c \end{aligned}
3. We integrate $$\int 4 \sqrt[5]{x^4} dx$$ as follows: \begin{aligned} \int 4 \sqrt[5]{x^4} dx & = \int 4.x^{\frac{4}{5}} dx \\ & = \frac{4}{\frac{4}{5}+1}.x^{\frac{4}{5}+1}+c \\ & = \frac{4}{\frac{9}{5}}.x^{\frac{9}{5}}+c \\ & = 4 \times \frac{5}{9}.x^{\frac{9}{5}}+c \\ & = \frac{20}{9}.x^{\frac{9}{5}}+C \\ \int 4 \sqrt[5]{x^4} dx & = \frac{20}{9}.\sqrt[5]{x^9} + c \end{aligned}
4. We integrate $$\int 6.\sqrt{x} dx$$ as follows: \begin{aligned} \int 6.\sqrt{x} dx & = \int 6.x^{\frac{1}{2}}+c \\ & = 6\times \frac{1}{\frac{1}{2}+1}.x^{\frac{1}{2}+1}+c \\ & = 6 \times \frac{1}{\frac{3}{2}}.x^{\frac{3}{2}}+c \\ & = 6 \times \frac{2}{3}.x^{\frac{3}{2}}+c \\ & = 4.x^{\frac{3}{2}}+c \\ \int 6.\sqrt{x} dx & = 4.\sqrt{x^3}+c \end{aligned}
5. We integrate $$\int \frac{\sqrt[3]{x}}{4} dx$$ as follows: \begin{aligned} \int \frac{\sqrt[3]{x}}{4} dx & = \frac{1}{4} \int \sqrt[3]{x} dx \\ & = \frac{1}{4} \int x^{\frac{1}{3}} dx \\ & = \frac{1}{4} \times \frac{1}{\frac{1}{3}+1}.x^{\frac{1}{3}+1}+c \\ & = \frac{1}{4} \times \frac{1}{\frac{4}{3}}x^{\frac{4}{3}}+c \\ & = \frac{1}{4}\times \frac{3}{4}.x^{\frac{4}{3}}+c \\ & = \frac{3}{16}.x^{\frac{4}{3}}+c\\ \int \frac{\sqrt[3]{x}}{4} dx & = \frac{3}{16}.\sqrt[3]{x^4}+c \end{aligned}
### Integrands with more than one term
We now look at integrals in which the integrand has more than one term.
For instance: $$\int \begin{pmatrix} x^2 + x^3 \end{pmatrix}dx$$.
To evaluate such integrals, we integrate each term as though it was on its own: $\int \begin{pmatrix} x^2 + x^3 \end{pmatrix}dx = \int x^2 dx + \int x^3 dx$ Keeping that in mind, let's work through the following exercise.
## Exercise 4
Integrate each of the following:
1. $$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx$$
2. $$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix}dx$$
3. $$\int \frac{4+2x}{x^3} dx$$
4. $$\int \frac{3x^3 - 1}{x^2} dx$$
5. $$\int \frac{dx}{4\sqrt{x}}$$
## Solution Without Working
1. We can write the solution in two ways:
$$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx = 4x + \frac{1}{x} + c$$, or
$$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx = 4x + x^{-1}+c$$
2. We can write the solution in two ways:
$$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix} dx = \frac{x^3}{3} - \frac{3}{2x^2}+c$$, or
$$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix} dx = \frac{1}{3}x^3 - \frac{3}{2}.x^{-2}+c$$
3. We can write the solution in two ways:
$$\int \frac{4+2x}{x^3} dx = - \frac{2}{x^2} - \frac{2}{x} + c$$, or
$$\int \frac{4+2x}{x^3} dx = -2.x^{-2}-2.x^{-1}+c$$
4. We can write the solution in two ways:
$$\int \frac{3x^3 - 1}{x^2} dx = \frac{3}{2}x^2 + \frac{1}{x}+c$$, or
$$\int \frac{3x^3 - 1}{x^2} dx = \frac{3}{2}x^2 + x^{-1}+c$$
5. We can write the solution in two ways:
$$\int \frac{dx}{4\sqrt{x}} = \frac{1}{2}\sqrt{x} + c$$, or
$$\int \frac{dx}{4\sqrt{x}} = \frac{1}{2}x^{\frac{1}{2}}+c$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000100135803223, "perplexity": 685.2578853365935}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742906.49/warc/CC-MAIN-20181115182450-20181115204450-00245.warc.gz"} |
https://www.physicsforums.com/threads/earth-and-moon-where-does-gravity-cancel-out.646564/ | Earth and moon where does gravity cancel out
1. Oct 23, 2012
lozzajp
1. The problem statement, all variables and given/known data
where does the gravity cancel out between the earth and moon?
2. Relevant equations
s=3.84405x10^8m
earth m = 5.98x10^24kg
moon m =7.35x10^22kg
G = 6.67x10^-11N m^2/kg^2
mass of object between = 1kg for simplicity
F = G x m / r^2
r = ?
F earth = F moon
G x earth / r^2 = G x moon / (s-r)^2
3. The attempt at a solution
I get so far with that and ive ended up with:
earth = (moon x r^2) / (s-r)^2
not sure if its right and how do i isolate r?
Last edited: Oct 23, 2012
2. Oct 23, 2012
aftershock
Let's call the distance away from Earth r, and the total distance between the Earth and moon R.
A mass in between the Earth and moon will experience a gravitational force from both.
Fearth = GMem/r2
Fmoon = GMmm/(R-r)^2
For the net force to be zero the two above forces must equal each other.
3. Oct 23, 2012
lozzajp
i understand that but i dont know how to solve for r (R = s in my part)
4. Oct 23, 2012
Mesmerr
You need to use simple algebra to isolate r.
start with multiplying both sides by the rights side denominator. then you need to multiply out Earth and divide everything by r2. By doing this you should be able to get a single r2 term. then just isolate r.
5. Oct 23, 2012
aftershock
I worked it out, and it's definitely doable but the algebra is a little more intense than it probably needs to be. I thought of an easier way to do it. Let R be the total distance between the moon and the earth.
The distance away you are from the earth must be some percentage of total R. So if you're halfway of the total distance then .5R , if you're a quarter then .25R, etc. So let's say you're distance from earth is aR where a is some number <1
The distance from the moon must then be 1-a
So we can say this.
GM/aR = Gm/(1-a)R here I called M the mass of earth and m the mass of the moon.
Now you can just solve for a, that is significantly easier. Multiply a by the total distance and that's how far from earth you need to be (well the center of the earth)
6. Oct 23, 2012
lozzajp
thats a good way to do it! thanks for that.
do post the intense algebra? :)
7. Oct 23, 2012
aftershock
You cancel G and cross multiply to get
(R-r)2/r2 = m/M
Foil out R-r
(R2-2Rr+r2)/r2 = m/M
You cross multiply the m/M to bring it to the left and r2 to bring it to the right.
Distribute the new M/m term on the left. Now subtract r2 so it's back on the left. Give r2 a common denominator wit the rest of the terms. Now you can factor out r2
The end result will be a constant plus r times some constant plus r2 times some constant = 0
So you can plug that into the quadratic formula.
8. Oct 23, 2012
haruspex
Oops.
Last edited: Oct 23, 2012
9. Oct 23, 2012
aftershock
I'm not sure what you mean? What does being inside the earth have to do with this? You're in space located in between the moon and the earth.
10. Oct 23, 2012
haruspex
Oops, was thinking of a different question entirely. Sorry for the noise.
11. Oct 23, 2012
lozzajp
So it is impossible without quadratic formula.. guess I have some reading to do thanks for all your help!
Similar Discussions: Earth and moon where does gravity cancel out | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8660283088684082, "perplexity": 984.7429558386442}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00519.warc.gz"} |
http://physics.stackexchange.com/questions/31897/whats-next-after-higgs-boson-discovery | # What's next after Higgs Boson discovery? [duplicate]
Possible Duplicate:
Practical matter of the Higgs-Mechanism
As everybody knows that the Higgs Boson was discovered on July 4th,2012, I am so curious about it.
1. What are the possible outcome applications of Higgs Boson?
2. How Higgs Boson is related to Quantum Mechanics?
3. Does Higgs Boson fits in General Theory of Relativity?
4. Does Higgs Boson fits in String Theory?
Please share the knowledge. I am very curious about it.
-
## marked as duplicate by dmckee♦Jul 12 '12 at 17:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Please do not ask lists of questions in a single "question" like this, and please do search before you post, especially on something as topical as this. – dmckee Jul 12 '12 at 17:52
The title is misleading, I thought you were asking what next is expected to show up at the LHC or something like that. – Dilaton Jul 12 '12 at 22:08
## 1 Answer
1. There are no known applications and no imaginable applications of the Higgs boson which is not surprising given the fact that its lifetime is a zeptosecond.
2. The Higgs boson is a particular particle – state in the Hilbert space – in a correct quantum mechanical theory describing Nature. I mean the Standard Model or its extensions. Any discussion of the Higgs boson would be totally impossible without quantum mechanics. All properties of the Higgs boson crucially depend on principles and special effects of quantum mechanics.
3. The Higgs boson is a particle associated with the Higgs field. To see the emergence of particles from fields, one has to discuss physics at the level of quantum mechanics; see the previous point. However, even in classical physics, one may add the Higgs field to the general theory of relativity, much like the electromagnetic fields. The Higgs field is a source of gravity and other things. But it's just "another added player"; the main field in the general theory of relativity is the metric tensor, i.e. the spacetime geometry, not the Higgs field.
4. All realistic models of string theory that are or were trying to describe the Universe around us contain a Higgs field and a Higgs boson; string theory seems incompatible with all the alternatives. In this sense, the discovery of the Higgs is a victory for string theory and a confirmation of one of its predictions (a relatively uncontroversial prediction). Scalar fields – a broader family of fields similar to the Higgs field (fields without spin) – are also generic and important in string theory, see
http://motls.blogspot.cz/2012/06/higgs-boson-scalar-fields-and-victory.html
I think that none of the points you asked answers the question in the title, "what's next". We don't know what will be the next discovery. Physics researcher is not Stalin's five-year plan. If we knew what the next breakthrough would be, we would already to it tonight. One may only discuss what ideas people are studying in their effort to make the big breakthrough. When it comes to physics that is as analogous to the Higgs boson as possible, supersymmetry would probably be the top pick.
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https://www.physicsforums.com/threads/derivatives-of-log-and-exponential-functions.58011/ | # Derivatives of Log and Exponential Functions
1. Dec 28, 2004
### majinknight
Hi i was wondering if someone could check my work for these couple questions:
Find dy/dx (do not simplify)
a)y=e^sin3x
dy/dx= e^sin3x (cos3x)(3)
=3cos3xe^sin3x
b)y=5^(square rootx) x^2
dy/dx=x5^(square rootx) (xln5+2)
c)y=ln(x^2 / (2x+5)^3 )
y=lnx^2 - ln(2x+5)^3
dy/dx= 2x/x^2 - 6(2x+5)^2 / (2x=5)^3
=2/x - 6/(2x+5)
d)y=4log base 2 (square rootx+1)
y=4logbase2 (x+1)^-1
dy/dx= 4/lnbase2(square rootx+1)
e)y=ln[x^2 - e^x / x^2 +e^x]
y= ln(x^2 -e^x) - ln(x^2 +e^x)
dy/dx= [2x -e^x / x^2 - e^x] - [2x +e^x / x^2+e^x]
f)e^x^2 multiplied by y^3=x (isolate dy/dx)
I am unsure of how to do this one, i have never seen one like this before, could someone show what i would do.
g)Use logarthimic differentiation to find dy/dx if
y=[e^x cosx / (square root x)]^5
lny=5xlne +5lncosx -5/2lnx
dy/dx= y[5x-5tanx-5/2x]
dy/dx= =[e^x cosx / (square root x)]^5 [5x-5tanx-5/2x]
h)A radioactive substance decays in such a way that the amount in grams present at time t years is given by A(t)=100e^-0.2t
i)What is the initial amount of radioactive material, A(0)?
A(0)= 100
ii)Find the rate of decay function, A'(t).
A'(t)=-20e^-0.2t
iii)How mucgh radio active material is present when t-50 years? How fast is the material decaying at this time?
A(50)=100e^-10
=0.0045399
A'(50)=-20e^-10
=-0.0009079
iv)At what time t is one half of the original substance remaining? What is the decay rate at this time?
I am not sure what to do here also if someone could show me how id really appreciate it.
*For this question and parts of the question i am not sure if some of my equations are correct as the numbers i am getting do not seem like it should be what they are.
j)solve the logarithmic equation logbase3(x-3) + logbase3(x) = logbase3(4)
I am also not sure what to do for this question.
Thanks in advanced for the help! If someone could show me how to do those few questions and check my work on those ones cuz my text book does not show what the answers are or how to do some of them.
2. Dec 28, 2004
### dextercioby
This is gonna be a hell of a long post. :tongue2: This part is correct.
$$\ln y=\sqrt{x}\ln5 +2\ln x;\frac{d\ln y}{dx}=\frac{1}{y}\frac{dy}{dx}$$
So:$$\frac{dy}{dx}=y\frac{d\ln y}{dx}=(5^{\sqrt{x}}x^{2})(\frac{\ln 5}{2\sqrt{x}}+\frac{2}{x})=...$$
Okay...
I don't understand where is the square root and what's that argument for the $\log_{2}$
I don't follow you.If u have:
$$y=\ln(x^{2}-\frac{e^{x}}{x^{2}}+e^{x})$$,then the decomposition u found is wrong.It's correct iff u have
$$y=\ln(\frac{x^{2}-e^{x}}{x^{2}+e^{x}})$$
Then your result would be correct.However,the fact that you're not writing it in 'tex' misleads me.
Here u got me completely lost.Is it
$$e^{x^{2}}y^{3}=x$$????????If so,then take natural log.out of both sides and get:
$$x^{2}+3\ln y=\ln x$$.Differentiate wrt to 'x' to find
$$2x+\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}$$.From there extract the derivative and substitute 'y' by the expression foud from the first equation.
Okay.
So far so good.
You found that,initially,A(0)=100.The question is:find 't' from the eq.
$$A(t)=100e^{-0.2t}=50$$
Raise both sides of the eq. to the power 3.Use the property
$$3^{x+y}=3^{x}3^{y}$$
to find the solution.
Daniel.
3. Dec 28, 2004
### majinknight
Ya some of it is confusing as i dont know how to use that tex thing. For question d) the root is over x+1. And for question e) the equation is the second one you posted which as you said my answer is correct. f) is the equation you said so then i did what you said and this is the answer i get.
dy/dx= (cube root of x-e^x^2)(1/x -2x)
Is that correct?
And for h)iv)
50=100e^-0.2t
ln0.5=-0.2t
t=ln0.5/-0.2
so then:
A'(ln0.5/-0.2)=-20e^ln0.5
Would i leave my answer at that, and is that correct?
For question j) i still don't understand what you mean, could you please show me what the first line of the solution would be please. Thanks so much for the help!
4. Dec 28, 2004
### nolachrymose
For (j), he means this:
$$\log_b{xy} = \log_b{x}+\log_b{y}$$
5. Dec 28, 2004
### dextercioby
d)$$y=4\log_{2}\sqrt{x+1} \Rightarrow 2^{y}=2^{4\log_{2}\sqrt{x+1}}=(2^{\log_{2}\sqrt{x+1}})^{4}=(\sqrt{x+1})^{4}=(x+1)^{2}$$
Take '\ln' from both sides of the eq.to find:
$$y=\frac{2}{\ln 2}\ln(x+1)\Rightarrow \frac{dy}{dx}=\frac{2}{(\ln 2)(x+1)}$$
h)iv) $$t=5\ln 2 \Rightarrow A'(5\ln 2)=20e^{-0.2\cdot 5\cdot\ln 2}=\frac{20}{2}=10$$
It's not correct what u've written.Mass cannot be negative.
j)$$x(x-3)=4 \Rightarrow x=4$$
The solution x=-1 implies logarithm in the base "3" from a negative number...
Daniel.
Last edited: Dec 28, 2004
6. Dec 28, 2004
### dextercioby
For f):
$$\frac{dy}{dx}=y(\frac{1}{x}-2x)=(\sqrt[3]{x e^{-x^{2}}})(\frac{1}{x}-2x)$$
Daniel.
7. Dec 28, 2004
### majinknight
Alright thanks i understand now! Thanks so much. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.924839973449707, "perplexity": 2387.986396728527}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281424.85/warc/CC-MAIN-20170116095121-00395-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.nag.com/numeric/nl/nagdoc_26/nagdoc_fl26/html/f07/f07cvf.html | # NAG Library Routine Document
## 1Purpose
f07cvf (zgtrfs) computes error bounds and refines the solution to a complex system of linear equations $AX=B$ or ${A}^{\mathrm{T}}X=B$ or ${A}^{\mathrm{H}}X=B$, where $A$ is an $n$ by $n$ tridiagonal matrix and $X$ and $B$ are $n$ by $r$ matrices, using the $LU$ factorization returned by f07crf (zgttrf) and an initial solution returned by f07csf (zgttrs). Iterative refinement is used to reduce the backward error as much as possible.
## 2Specification
Fortran Interface
Subroutine f07cvf ( n, nrhs, dl, d, du, dlf, df, duf, du2, ipiv, b, ldb, x, ldx, ferr, berr, work, info)
Integer, Intent (In) :: n, nrhs, ipiv(*), ldb, ldx Integer, Intent (Out) :: info Real (Kind=nag_wp), Intent (Out) :: ferr(nrhs), berr(nrhs), rwork(n) Complex (Kind=nag_wp), Intent (In) :: dl(*), d(*), du(*), dlf(*), df(*), duf(*), du2(*), b(ldb,*) Complex (Kind=nag_wp), Intent (Inout) :: x(ldx,*) Complex (Kind=nag_wp), Intent (Out) :: work(2*n) Character (1), Intent (In) :: trans
#include nagmk26.h
void f07cvf_ ( const char *trans, const Integer *n, const Integer *nrhs, const Complex dl[], const Complex d[], const Complex du[], const Complex dlf[], const Complex df[], const Complex duf[], const Complex du2[], const Integer ipiv[], const Complex b[], const Integer *ldb, Complex x[], const Integer *ldx, double ferr[], double berr[], Complex work[], double rwork[], Integer *info, const Charlen length_trans)
The routine may be called by its LAPACK name zgtrfs.
## 3Description
f07cvf (zgtrfs) should normally be preceded by calls to f07crf (zgttrf) and f07csf (zgttrs). f07crf (zgttrf) uses Gaussian elimination with partial pivoting and row interchanges to factorize the matrix $A$ as
$A=PLU ,$
where $P$ is a permutation matrix, $L$ is unit lower triangular with at most one nonzero subdiagonal element in each column, and $U$ is an upper triangular band matrix, with two superdiagonals. f07csf (zgttrs) then utilizes the factorization to compute a solution, $\stackrel{^}{X}$, to the required equations. Letting $\stackrel{^}{x}$ denote a column of $\stackrel{^}{X}$, f07cvf (zgtrfs) computes a component-wise backward error, $\beta$, the smallest relative perturbation in each element of $A$ and $b$ such that $\stackrel{^}{x}$ is the exact solution of a perturbed system
$A+E x^=b+f , with eij ≤β aij , and fj ≤β bj .$
The routine also estimates a bound for the component-wise forward error in the computed solution defined by $\mathrm{max}\left|{x}_{i}-\stackrel{^}{{x}_{i}}\right|/\mathrm{max}\left|\stackrel{^}{{x}_{i}}\right|$, where $x$ is the corresponding column of the exact solution, $X$.
## 4References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia http://www.netlib.org/lapack/lug
## 5Arguments
1: $\mathbf{trans}$ – Character(1)Input
On entry: specifies the equations to be solved as follows:
${\mathbf{trans}}=\text{'N'}$
Solve $AX=B$ for $X$.
${\mathbf{trans}}=\text{'T'}$
Solve ${A}^{\mathrm{T}}X=B$ for $X$.
${\mathbf{trans}}=\text{'C'}$
Solve ${A}^{\mathrm{H}}X=B$ for $X$.
Constraint: ${\mathbf{trans}}=\text{'N'}$, $\text{'T'}$ or $\text{'C'}$.
2: $\mathbf{n}$ – IntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
3: $\mathbf{nrhs}$ – IntegerInput
On entry: $r$, the number of right-hand sides, i.e., the number of columns of the matrix $B$.
Constraint: ${\mathbf{nrhs}}\ge 0$.
4: $\mathbf{dl}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array dl must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
On entry: must contain the $\left(n-1\right)$ subdiagonal elements of the matrix $A$.
5: $\mathbf{d}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array d must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: must contain the $n$ diagonal elements of the matrix $A$.
6: $\mathbf{du}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array du must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
On entry: must contain the $\left(n-1\right)$ superdiagonal elements of the matrix $A$.
7: $\mathbf{dlf}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array dlf must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
On entry: must contain the $\left(n-1\right)$ multipliers that define the matrix $L$ of the $LU$ factorization of $A$.
8: $\mathbf{df}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array df must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: must contain the $n$ diagonal elements of the upper triangular matrix $U$ from the $LU$ factorization of $A$.
9: $\mathbf{duf}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array duf must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
On entry: must contain the $\left(n-1\right)$ elements of the first superdiagonal of $U$.
10: $\mathbf{du2}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array du2 must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-2\right)$.
On entry: must contain the $\left(n-2\right)$ elements of the second superdiagonal of $U$.
11: $\mathbf{ipiv}\left(*\right)$ – Integer arrayInput
Note: the dimension of the array ipiv must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: must contain the $n$ pivot indices that define the permutation matrix $P$. At the $i$th step, row $i$ of the matrix was interchanged with row ${\mathbf{ipiv}}\left(i\right)$, and ${\mathbf{ipiv}}\left(i\right)$ must always be either $i$ or $\left(i+1\right)$, ${\mathbf{ipiv}}\left(i\right)=i$ indicating that a row interchange was not performed.
12: $\mathbf{b}\left({\mathbf{ldb}},*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the second dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$.
On entry: the $n$ by $r$ matrix of right-hand sides $B$.
13: $\mathbf{ldb}$ – IntegerInput
On entry: the first dimension of the array b as declared in the (sub)program from which f07cvf (zgtrfs) is called.
Constraint: ${\mathbf{ldb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
14: $\mathbf{x}\left({\mathbf{ldx}},*\right)$ – Complex (Kind=nag_wp) arrayInput/Output
Note: the second dimension of the array x must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$.
On entry: the $n$ by $r$ initial solution matrix $X$.
On exit: the $n$ by $r$ refined solution matrix $X$.
15: $\mathbf{ldx}$ – IntegerInput
On entry: the first dimension of the array x as declared in the (sub)program from which f07cvf (zgtrfs) is called.
Constraint: ${\mathbf{ldx}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
16: $\mathbf{ferr}\left({\mathbf{nrhs}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: estimate of the forward error bound for each computed solution vector, such that ${‖{\stackrel{^}{x}}_{j}-{x}_{j}‖}_{\infty }/{‖{\stackrel{^}{x}}_{j}‖}_{\infty }\le {\mathbf{ferr}}\left(j\right)$, where ${\stackrel{^}{x}}_{j}$ is the $j$th column of the computed solution returned in the array x and ${x}_{j}$ is the corresponding column of the exact solution $X$. The estimate is almost always a slight overestimate of the true error.
17: $\mathbf{berr}\left({\mathbf{nrhs}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: estimate of the component-wise relative backward error of each computed solution vector ${\stackrel{^}{x}}_{j}$ (i.e., the smallest relative change in any element of $A$ or $B$ that makes ${\stackrel{^}{x}}_{j}$ an exact solution).
18: $\mathbf{work}\left(2×{\mathbf{n}}\right)$ – Complex (Kind=nag_wp) arrayWorkspace
19: $\mathbf{rwork}\left({\mathbf{n}}\right)$ – Real (Kind=nag_wp) arrayWorkspace
20: $\mathbf{info}$ – IntegerOutput
On exit: ${\mathbf{info}}=0$ unless the routine detects an error (see Section 6).
## 6Error Indicators and Warnings
${\mathbf{info}}<0$
If ${\mathbf{info}}=-i$, argument $i$ had an illegal value. An explanatory message is output, and execution of the program is terminated.
## 7Accuracy
The computed solution for a single right-hand side, $\stackrel{^}{x}$, satisfies an equation of the form
$A+E x^=b ,$
where
$E∞=OεA∞$
and $\epsilon$ is the machine precision. An approximate error bound for the computed solution is given by
$x^-x ∞ x∞ ≤ κA E∞ A∞ ,$
where $\kappa \left(A\right)={‖{A}^{-1}‖}_{\infty }{‖A‖}_{\infty }$, the condition number of $A$ with respect to the solution of the linear equations. See Section 4.4 of Anderson et al. (1999) for further details.
Routine f07cuf (zgtcon) can be used to estimate the condition number of $A$.
## 8Parallelism and Performance
f07cvf (zgtrfs) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
f07cvf (zgtrfs) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The total number of floating-point operations required to solve the equations $AX=B$ or ${A}^{\mathrm{T}}X=B$ or ${A}^{\mathrm{H}}X=B$ is proportional to $nr$. At most five steps of iterative refinement are performed, but usually only one or two steps are required.
The real analogue of this routine is f07chf (dgtrfs).
## 10Example
This example solves the equations
$AX=B ,$
where $A$ is the tridiagonal matrix
$A = -1.3+1.3i 2.0-1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0-2.0i -1.3+1.3i 2.0+1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0+1.0i -1.3+3.3i -1.0+1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 2.0-3.0i -0.3+4.3i 1.0-1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0+1.0i -3.3+1.3i$
and
$B = 2.4-05.0i 2.7+06.9i 3.4+18.2i -6.9-05.3i -14.7+09.7i -6.0-00.6i 31.9-07.7i -3.9+09.3i -1.0+01.6i -3.0+12.2i .$
Estimates for the backward errors and forward errors are also output.
### 10.1Program Text
Program Text (f07cvfe.f90)
### 10.2Program Data
Program Data (f07cvfe.d)
### 10.3Program Results
Program Results (f07cvfe.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 142, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9967120289802551, "perplexity": 4398.444075728406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00583.warc.gz"} |
https://proofwiki.org/wiki/Definition:Uniform_Distribution | Definition:Uniform Distribution
Definition
Discrete Uniform Distribution
Let $X$ be a discrete random variable on a probability space.
Then $X$ has a discrete uniform distribution with parameter $n$ if and only if:
$\Img X = \set {1, 2, \ldots, n}$
$\map \Pr {X = k} = \dfrac 1 n$
That is, there is a number of outcomes with an equal probability of occurrence.
This is written:
$X \sim \DiscreteUniform n$
Continuous Uniform Distribution
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $a, b \in \R$ such that $a < b$.
$X$ is said to be uniformly distributed on the closed real interval $\closedint a b$ if and only if it has probability density function:
$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$
This is written:
$X \sim \ContinuousUniform a b$
Also see
• Results about uniform distributions can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.904338002204895, "perplexity": 133.34678871555718}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00140.warc.gz"} |
https://www.geteasysolution.com/14.444444444_as_a_fraction | # 14.444444444 as a fraction
## 14.444444444 as a fraction - solution and the full explanation with calculations.
Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process.
If it's not what You are looking for, type in into the box below your number and see the solution.
## What is 14.444444444 as a fraction?
To write 14.444444444 as a fraction you have to write 14.444444444 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
14.444444444 = 14.444444444/1 = 144.44444444/10 = 1444.4444444/100 = 14444.444444/1000 = 144444.44444/10000 = 1444444.4444/100000 = 14444444.444/1000000 = 144444444.44/10000000 = 1444444444.4/100000000 = 14444444444/1000000000
And finally we have:
14.444444444 as a fraction equals 14444444444/1000000000 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989253640174866, "perplexity": 1152.204194521795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00233.warc.gz"} |
https://docs.quantumatk.com/manual/Types/PartialElectronDensity/PartialElectronDensity.html | # PartialElectronDensity¶
class PartialElectronDensity(configuration, kpoints=None, band_indices=None, energy_range=None, spectrum_method=None)
Class for calculating the partial electron density.
Parameters: configuration (BulkConfiguration | MoleculeConfiguration) – The configuration with an attached calculator for which to calculate the partial electron density. kpoints – The k-points for which to calculate the eigenvalues. This can either be a MonkhorstPackGrid, RegularKpointGrid, or a list of 3-dimensional fractional k-points e.g. [[0.0, 0.0, 0.0], [0.0, 0.0, 0.1], ...]. Default: MonkhorstPackGrid(nx,ny,nz), where nx, ny, nz is the sampling used for the self-consistent calculation. band_indices (list of int) – The band indices of the Bloch states to include in the partial density. Default: All occupied bands energy_range (list of two PhysicalQuantity of type energy.) – Specifies the band energy range, relative to the Fermi level, of the bands to include in the partial density. Needs to be an interval or None. Default: All energies below the Fermi level. spectrum_method (GaussianBroadening) – The method to use for calculating the partial electron density. Default: GaussianBroadening(0.1*eV)
axisProjection(projection_type='sum', axis='c', spin=None, projection_point=None, coordinate_type=<class 'NL.ComputerScienceUtilities.NLFlag._NLFlag.Fractional'>)
Get the values projected on one of the Cartesian axes.
Parameters: projection_type (str) – The type of projection to perform. Should be either ‘sum’ for the sum over the plane spanned by the two other axes. ‘average’ or ‘avg’ for the average value over the plane spanned by the two other axes. ‘line’ for the value along a line parallel to the axis and through a point specified by the projection_point parameter. Default: ‘sum’ axis (str) – The axis to project the data onto. Should be either ‘a’, ‘b’ or ‘c’. Default: ‘c’ spin (Spin.Sum | Spin.Z | Spin.X | Spin.Y | Spin.Up | Spin.Down | Spin.RealUpDown | Spin.ImagUpDown) – Which spin component to project on. Default: Spin.All projection_point (sequence, PhysicalQuantity) – Axis coordinates of the point through which to take a line if projection_type is ‘projection_point’. Must be given as a sequence of three coordinates [a, b, c]. It the numbers have units of length, they are first divided by the length of the respective primitive vectors [A, B, C], and then interpreted as fractional coordinates. Unitless coordinates are immidiately interpreted as fractional. coordinate_type (Fractional | Cartesian) – Flag to toggle if the returned axis values should be given in units of Angstrom (NLFlag.Cartesian) or in units of the norm of the axis primitive vector (NLFlag.Fractional). Default: Fractional A 2-tuple of 1D numpy.arrays containing the axis values and the projected data. tuple.
bandIndices()
Returns: The band indicies to include in the partial electron density. list of int.
derivatives(x, y, z, spin=None)
Calculate the derivative in the point (x, y, z).
Parameters: x (PhysicalQuantity with type length) – The Cartesian x coordinate. y (PhysicalQuantity with type length) – The Cartesian y coordinate. z (PhysicalQuantity with type length) – The Cartesian z coordinate. spin (Spin.All | Spin.Sum | Spin.Up | Spin.Down | Spin.X | Spin.Y | Spin.Z) – The spin component to project on. Default: Spin.All The gradient at the specified point for the given spin. For Spin.All, a tuple with (Spin.Sum, Spin.X, Spin.Y, Spin.Z) components is returned. PhysicalQuantity of type length-4
energyRange()
Returns: The band energy range, relative to the Fermi level, of the bands to include in the partial density. list of two PhysicalQuantity of type energy.
evaluate(x, y, z, spin=None)
Evaluate in the point (x, y, z).
Parameters: x (PhysicalQuantity with type length) – The Cartesian x coordinate. y (PhysicalQuantity with type length) – The Cartesian y coordinate. z (PhysicalQuantity with type length) – The Cartesian z coordinate. spin (Spin.All | Spin.Sum | Spin.Up | Spin.Down | Spin.X | Spin.Y | Spin.Z) – The spin component to project on. Default: Spin.All The value at the specified point for the given spin. For Spin.All, a tuple with (Spin.Sum, Spin.X, Spin.Y, Spin.Z) components is returned. PhysicalQuantity of type length-3
gridCoordinate(i, j, k)
Return the coordinate for a given grid index.
Parameters: i (int) – The grid index in the A direction. j (int) – The grid index in the B direction. k (int) – The grid index in the C direction. The Cartesian coordinate of the given grid index. PhysicalQuantity of type length.
kpoints()
Returns: The kpoints given as input. MonkhorstPackGrid | RegularKpointGrid | numpy.ndarray
metatext()
Returns: The metatext of the object or None if no metatext is present. str | unicode | None
nlprint(stream=<_io.TextIOWrapper name='<stdout>' mode='w' encoding='UTF-8'>)
Print a string containing an ASCII table useful for plotting the partial electron density.
Parameters: stream (file-like) – The stream to write to. This should be an object that supports strings being written to it using a write method. Default: sys.stdout
primitiveVectors()
Returns: The primitive vectors of the grid. PhysicalQuantity of type length.
scale(scale)
Scale the field with a float.
Parameters: scale (float) – The parameter to scale with.
setMetatext(metatext)
Set a given metatext string on the object.
Parameters: metatext (str | unicode | None) – The metatext string that should be set. A value of “None” can be given to remove the current metatext.
shape()
Returns: The number of grid points in each direction. tuple of three int.
spinProjection(spin=None)
Construct a new GridValues object with the values of this object projected on a given spin component.
Parameters: spin (Spin.All | Spin.Sum | Spin.X | Spin.Y | Spin.Z) – The spin component to project on. Default: Spin.All A new GridValues object for the specified spin. GridValues
toArray()
Returns: The values of the grid as a numpy array slicing off any units. numpy.array
unit()
Returns: The unit of the data in the grid. A physical unit.
unitCell()
Returns: The unit cell of the grid. PhysicalQuantity of type length.
volumeElement()
Returns: The volume element of the grid represented by three vectors. PhysicalQuantity of type length.
## Usage Examples¶
Analyze the partial electron density of a silver (111) surface. This structure is characterized by a surface state around the $$\Gamma$$-point. In the example, the kpoint sampling is specified with a RegularKpointGrid in a kpoint region around the $$\Gamma$$-point.
# -------------------------------------------------------------
# Bulk Configuration
# -------------------------------------------------------------
# Set up lattice
vector_a = [2.88903, 0.0, 0.0]*Angstrom
vector_b = [-1.44451, 2.50197, 0.0]*Angstrom
vector_c = [0.0, 0.0, 40.0]*Angstrom
lattice = UnitCell(vector_a, vector_b, vector_c)
# Define elements
elements = [Silver, Silver, Silver, Silver, Silver, Silver, Silver, Silver,
Silver, Silver, Silver, Silver]
# Define coordinates
fractional_coordinates = [[ 0.749999669084, 0.5 , 0.175654000711],
[ 0.749999669084, 0.5 , 0.352570000323],
[ 0.749999669084, 0.5 , 0.529485999935],
[ 0.749999669084, 0.5 , 0.706401999547],
[ 0.416666424719, 0.833333341397, 0.234626000582],
[ 0.416666424719, 0.833333341397, 0.411542000194],
[ 0.416666424719, 0.833333341397, 0.588457999806],
[ 0.416666424719, 0.833333341397, 0.765373999418],
[ 0.083334237113, 0.166666658603, 0.293598000453],
[ 0.083334237113, 0.166666658603, 0.470514000065],
[ 0.083334237113, 0.166666658603, 0.647429999677],
[ 0.083334237113, 0.166666658603, 0.824345999289]]
# Set up configuration
bulk_configuration = BulkConfiguration(
bravais_lattice=lattice,
elements=elements,
fractional_coordinates=fractional_coordinates
)
# -------------------------------------------------------------
# Calculator
# -------------------------------------------------------------
#----------------------------------------
# Basis Set
#----------------------------------------
basis_set = [
LDABasis.Silver_SingleZetaPolarized,
]
#----------------------------------------
# Exchange-Correlation
#----------------------------------------
exchange_correlation = LDA.PZ
k_point_sampling = MonkhorstPackGrid(
na=11,
nb=11,
)
numerical_accuracy_parameters = NumericalAccuracyParameters(
density_mesh_cutoff=45.0*Hartree,
k_point_sampling=k_point_sampling,
)
calculator = LCAOCalculator(
basis_set=basis_set,
exchange_correlation=exchange_correlation,
numerical_accuracy_parameters=numerical_accuracy_parameters,
)
bulk_configuration.setCalculator(calculator)
nlprint(bulk_configuration)
bulk_configuration.update()
# -------------------------------------------------------------
# Partial Electron Density
# -------------------------------------------------------------
kpoints = RegularKpointGrid(
ka_range=[-0.05, 0.05],
kb_range=[-0.05, 0.05],
kc_range=[0.0, 0.0],
na=5,
nb=5,
nc=1,
)
partial_electron_density = PartialElectronDensity(
configuration=bulk_configuration,
kpoints=kpoints,
energy_range=[-0.5, 0.5]*eV,
band_indices=All,
)
partial_electron_density.py
## Notes¶
The PartialElectronDensity can be calculated by specifying the
1. energy_range. Only electronic states with energies in the specified energy range $$[E_{min}; E_{max}]$$ will contribute to the partial electron density. The Fermi occupation factor used to construct the full electron density is not taken into account for the partial electron density. If $$E_{max}>0$$ unoccupied states will therefore contribute to the partial density.
2. band_indices. Only states from the specified band indices will contribute to the partial electron density. By default, all bands are taken into account and the selection is determined by the energy range.
3. kpoints. The partial electron density is calculated from the eigenstates at the specified k-points.
4. spectrum_method. The method used to integrate the density. Currently only GaussianBroadening is supported. Each electronic state is assigned a weight which is computed by integrating a Gaussian centered at the state’s energy over the specified energy range.
The partial electron density is defined as:
$n(\mathbf{r}) = \sum_{i}\sum_\mathbf{k} w_{\mathbf{k}} |\psi_{i,\mathbf{k}}(\mathbf{r})|^2 w(\epsilon_{i,\mathbf{k}}),$
where the sum over band indices $$i$$ includes only the specified bands (by default all bands). The weights $$w_{\mathbf{k}}$$ and $$w(\epsilon_{i,\mathbf{k}})$$ are the k-point weights and the Gaussian weights, respectively. The energy dependent Gaussian weight is computed as the integral of a normalized Gaussian centered at $$\epsilon_{i,\mathbf{k}}$$ with standard deviation $$\sigma$$ over the specified energy range, i.e.
\begin{split}\begin{align*} w(\epsilon_{i,\mathbf{k}}) &= \int_{E_{min}}^{E_{max}} \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(\frac{\left(E - \epsilon_{i,\mathbf{k}}\right)^2}{2 \sigma^2}\right)\,dE \\ &= {\frac {1}{2}}\left[\mathrm{erf}\left(\frac{E_{max}-\epsilon_{i,\mathbf{k}}}{\sqrt{2} \sigma}\right)-\mathrm{erf}\left(\frac{E_{min}-\epsilon_{i,\mathbf{k}}}{\sqrt{2} \sigma}\right)\right], \end{align*}\end{split}
with $$\operatorname {erf}$$ being the error function.
## Tersoff-Hamann approximation of simulated STM¶
The PartialElectronDensity can be used to simulate scanning tunneling microscopy (STM) images within the Tersoff-Hamann approximation [TH85] assuming that the tip wave function has s-orbital character. When simulating an STM image, the energy_range can be interpreted as the bias difference applied between the tip and the surface.
Notice that the Tersoff-Hamann approximation is the simplest approach to simulated STM images and it is not always applicable. Also bare in mind that far away from a surface the partial density will be zero for an LCAO calculation since the basis orbitals have a finite range.
[TH85] J. Tersoff and D. R. Hamann. Theory of the scanning tunneling microscopy. Phys. Rev. B, 31:805, 1985. doi:10.1103/PhysRevB.31.805. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8126785159111023, "perplexity": 4762.485326951934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514577363.98/warc/CC-MAIN-20190923150847-20190923172847-00413.warc.gz"} |
https://www.physicsforums.com/threads/peaking-factor-and-power-profile.761131/ | # Peaking factor and power profile
1. Jul 9, 2014
### Vnt666Skr
2. Jul 10, 2014
### Staff: Mentor
Peaking factors are developed from normalized axial and radial/lateral power profiles.
One is interested in how the local power relates to the core average power, as well as the absolute magnitude of the power. Local power restricted by some margin to some absolute limit in order to ensure that under certain anticipated anomalies the fuel is not damaged, or in the event of a postulated accident, the fuel damage is limited and not underestimated.
From a fuel performance perspective, one wishes to 'flatten' the radial and axial profiles such that one minimizes corrosion and other irradiation-dependent behavior/consequences.
3. Jul 10, 2014
### Vnt666Skr
Thanks Astronuc.
Is it defined at each axial/radial position? Suppose I have a power profile of a single pin. How do I find out the peaking factors at various locations in the axial and radial direction?
4. Jul 10, 2014
### QuantumPion
FdH is ratio of the total pin power to the total core power divided by number of pins. This is a 2-D (radial) value and each pin as one value for FdH.
Fq(z) is the ratio of power density of the pin divided by the power density of the core. This is a 3-D (axial) value. Each pin has a Fq(z) as a function of height, and a peak Fq.
If your pin power profile is normalized, you need to first multiply by the assembly's relative power density. Fz is the maximum normalized power for the core, assembly, or pin.
Last edited: Jul 10, 2014
5. Jul 10, 2014
### Staff: Mentor
A peaking factor would be determined from the local power density (or linear power) divided by the core average power density (or linear power). The average power density is found from the thermal rating of the reactor core divided by the total length of active fuel. The local power density is calculated with a core simulation code (e.g., SIMULATE or other proprietary code) which solves a multi-group neutron diffusion or transport problem. The codes calculate the neutron flux and local enrichment, which includes effects of depletion and transmutation, and from these determine the fission density, from power density is calculated.
An example of core average power. Given a 3700 MWt core, with 193 assemblies, 264 fuel rods per assembly, and an active fuel length of 12 ft (including blankets), the core average linear power in kW/ft is given by
3700000 kW / (193 * 264 * 12 ft) = 6.05 kW/ft or 19.85 kW/m
Similar Discussions: Peaking factor and power profile | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8973985910415649, "perplexity": 2316.4915341658957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00299.warc.gz"} |
http://mathhelpforum.com/algebra/138574-factor-equation-print.html | # factor the equation
• April 11th 2010, 06:31 PM
kenzie103109
factor the equation
3q^2 + 16q + 5
• April 11th 2010, 06:38 PM
I look for a solution of the form (3q-a)(q-b). Expanding $3q^2 - (3b+a)q +ab$. Equating the coefficients of like terms gives ab=5 and 3b+a=-16. Thinking for a moment will convince you that b=-5 and a=-1 satisfy these equations. So $3q^2 + 16q + 5= (3q+1)(q+5)$.
• April 11th 2010, 06:48 PM
mag6
3q^2 + 16q +5
3q^2 +15q +q +5
3q( q+5) +1 (q+5)
(3q+1) (q+5)
• April 11th 2010, 06:48 PM
CaptainBlack
Quote:
Originally Posted by kenzie103109
3q^2 + 16q + 5
$P(q)=3q^2 + 16q + 5$
The rational roots theorem suggests that $\pm 5$, $\pm 1/3$ and $\pm 5/3$ are the only possible rational roots for this quadratic in $q$. Descartes rule of signs tells you it has no positive roots, so that reduces the list to $-5, -1/3$ and $-5/3$. Now trying these out shows that $P(-5)=0$ and $P(-1/3)=0$, so we have:
$P(q)=k(q+1/3)(q+5)$
and to match the leading coefficient of $P(q)$ we must have $k=3$ so:
$P(q)=(3q+1)(q+5)$
CB | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9592505097389221, "perplexity": 1348.0381971413287}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928965.67/warc/CC-MAIN-20150521113208-00051-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/magnet-back-and-forth.739868/ | # Magnet Back and Forth.
1. Feb 23, 2014
### sahil_time
If i move a magnet back and forth in free space, will it radiate?
If it does not then why?
2. Feb 23, 2014
Staff Emeritus
Yes.
3. Feb 23, 2014
### sahil_time
Why is it not used for practical purposes? Is it because it cannot produce high energy radiation?
4. Feb 23, 2014
### davenn
there is a magnetic field out around a magnet regardless of if it is moving or not
its used for many many different things ... electric motors just to name one
what hi energy radiation ? ... a magnet produces a magnetic field
cheers
Dave
5. Feb 23, 2014
### sahil_time
But that is due to change in flux linkage, imagine a magnet in absolute free space. If i were to
then move it back and forth, will it emit radiation just like an antenna?
6. Feb 23, 2014
### IPNightly
Move a magnet past a coil of wire (or move a coil past a magnet - the choice is yours) and a current flows in the wire. This is the basis of electricity production... essentially all thermal power stations and wind farms generate electricity using this process (solar PV panels don't).
Energy taken from a moving magnet is used for practical purposes all over the world (but in case you're thinking you'd like to get some of that free energy, remember you're only taking out at an absolute maximum the same amount of energy you put in to move the magnet in the first place.)
Far from this not being used for practical purposes, I'm honestly having trouble thinking of a phenomenon in the modern world put to more frequent practical use.
7. Feb 23, 2014
### IPNightly
But now I don't get it - you're thinking of something like, say, the earth - a big magnet moving in space? It has a magnetic field around it that moves with the earth. It doesn't "emit radiation" due to its magnetic core any more than it emits gravitational "radiation". Maybe on a quantum level you're thinking of force carrying particles that we haven't yet identified? Can you explain what kind of practical use you're talking about... propelling a spaceship by jiggling a magnet about or something?
8. Feb 23, 2014
### sahil_time
Okay, i think the question has been a little vague i guess. Let me rephrase.
--An electron, if moved back and forth, emits radiation. This is the basis of antenna theory.
--In contrast, any neutral metal will not radiate if moved back and forth because it is neutral and
hence does not produce a magnetic field.
--Now, if a magnet, which although has neutral charge on the whole, but has a magnetic field,
is moved back and forth, will it radiate?
9. Feb 23, 2014
### IPNightly
I'm not really in my field (no pun intended) but I'd answer that antenna theory starts with an isolated negative charge (if we're using an electron) with a net negative charge spreading out into the universe. A magnetic pole is never isolated - you only ever (so far) see dipole magnets. By analogy you couldn't create an antenna by shaking a piece of net-charge-neutral metal. Happy to be out-argued.
10. Feb 23, 2014
### The_Duck
The frequency of the radiation will be equal to the frequency of the oscillation of the magnet, of order 1 Hz. I don't think there is any practical use for such low-frequency radiation. Furthermore the total radiated power will be extremely low.
11. Feb 24, 2014
### sahil_time
Thank you everyone :)
12. Feb 24, 2014
### davenn
Electrons oscillate back and forward in a conductor when an AC current is applied
the acceleration of that charge generates an electromagnetic field which will be radiated
again, the only thing being radiated is a magnetic field
there's only a magnetic field, not an electromagnetic field because there is no AC current flowing to generate one
Dave
Last edited: Feb 24, 2014
13. Feb 24, 2014
### sophiecentaur
But the fact that there is a varying magnetic field must imply a varying induced electric field, surely.
∇E = -dB/dt
any change that is 'forced' onto one field (as is waggling a magnet around) will induce a change in the other.
It would be interesting to know whether anyone has gone to the trouble of spinning one of these new fangled super magnets at very high speed (100s of kHz would not be impossible) to see if the resulting RF field can be detected at a distance.
14. Feb 24, 2014
Staff Emeritus
There is a changing magnetic dipole, so there is RF (assuming you call a few Hz RF) radiation. As sophiecentaur points out, the changing magnetic field induces an electric field.
$$<P> = \frac{\mu_0 d_0^2 \omega^4}{12 \pi c^3}$$
Where the magnetic dipole is $d = d_0 \cos(\omega t)$.
Note that for a bar magnet being shaken this is very small. Everything in the numerator is small, and everything in the denominator is big.
15. Feb 24, 2014
Staff Emeritus
Plugging in numbers, I get something like 10-37 watts radiated by shaking a bar magnet back and forth.
16. Feb 24, 2014
### davenn
OK
I see where you are coming from
So you are saying that the field outside the magnet is then cutting back through the magnet as the magnet moves within the field and thus generating a current which in turn is generating an EM emission
and OK so its freakin' tiny, nothing like what the OP was indicating or hoping for
Dave
17. Feb 25, 2014
### sahil_time
No i guess, the magnet's movement induces closed electric loops in space everywhere, that is
what Maxwell's equation suggests ∇XE = -∂B/∂t. The fact that if a magnet moves back and forth
, there will be flux change everywhere, and hence an electric field given by ∇XE = -∂B/∂t.
Also consequently, there will be a magnetic field due to the induced electric field, it will be
∇XH = -∂D/∂t . Where J = 0. So everything self propels to create RF.
Am i right? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8131442070007324, "perplexity": 1008.2554760225773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00632.warc.gz"} |
https://www.physicsforums.com/threads/classical-mechanics-problem.212761/ | # Homework Help: Classical Mechanics problem
1. Feb 2, 2008
### neelakash
1. The problem statement, all variables and given/known data
Problem: Derive eigenvalue equation of motion of a system undergoing free small oscillation in three dimensional space.
2. Relevant equations
3. The attempt at a solution
I want to know if I have correctly written the values of T and V---
$$\ T=$$$$\frac{1}{2}$$$$\ m$$ [$$\dot{x}^2 +\dot{y}^2+ \dot{z}^2$$]
And $$\ V=$$$$\frac{1}{2}$$$$\ k$$[$$\ x^2+\ y^2+\ z^2$$]
I m not sure if I have written the write thing...what should be the sign of k.Should I write -ve?And is it OK to assume that the force constants are equal?What would be if they were not equal?Would they simply add like (1/2)(k1+k2) x^2?
If I can form T and V,I am confident that I will be able to do the rest of calculations.
Last edited: Feb 2, 2008
2. Feb 2, 2008
### neelakash
It appears to me that the potential energy would be
$$\ V=$$$$\frac{1}{2}$$[$$\ k_1+k_2$$]$$\ x^2$$+...
From the elimentary knowledge, we know when a spring of natural length is compressed or elongated, workdone on the particle is -(1/2)kx^2. So,...(1/2)kx^2 amount of energy is stored within the spring as potential energy.So, V would be +ve.
Right?
So, we are to omit the factor of half in V in the first post. Please tell me if I am correct.
Last edited: Feb 2, 2008 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9337542653083801, "perplexity": 1026.6444120084718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945232.48/warc/CC-MAIN-20180421145218-20180421165218-00515.warc.gz"} |
https://en.wikipedia.org/wiki/Iverson_bracket | # Iverson bracket
In mathematics, the Iverson bracket, named after Kenneth E. Iverson, is a notation that denotes a number that is 1 if the condition in square brackets is satisfied, and 0 otherwise. More exactly,
$[P] = \begin{cases} 1 & \text{if } P \text{ is true;} \\ 0 & \text{otherwise.} \end{cases}$
where P is a statement that can be true or false. This notation was introduced by Kenneth E. Iverson in his programming language APL,[1][2] while the specific restriction to square brackets was advocated by Donald Knuth to avoid ambiguity in parenthesized logical expressions.[3]
## Uses
The Iverson bracket converts a Boolean value to an integer value through the natural map $\textbf{false}\mapsto 0; \textbf{true}\mapsto1$, which allows counting to be represented as summation. For instance, the Euler phi function that counts the number of positive integers up to n which are coprime to n can be expressed by
$\phi(n)=\sum_{i=1}^{n}[\gcd(i,n)=1],\qquad\text{for }n\in\mathbb N^+.$
More generally the notation allows moving boundary conditions of summations (or integrals) as a separate factor into the summand, freeing up space around the summation operator, but more importantly allowing it to be manipulated algebraically. For example,
$\sum_{1\le i \le 10} i^2 = \sum_{i} i^2[1 \le i \le 10].$
In the first sum, the index $i$ is limited to be in the range 1 to 10. The second sum is allowed to range over all integers, but where i is strictly less than 1 or strictly greater than 10, the summand is 0, contributing nothing to the sum. Such use of the Iverson bracket can permit easier manipulation of these expressions.
Another use of the Iverson bracket is to simplify equations with special cases. For example, the formula
$\sum_{1\le k\le n \atop \gcd(k,n)=1}\!\!k = \frac{1}{2}n\varphi(n)$
which is valid for n > 1 but which is off by 1/2 for n = 1. To get an identity valid for all positive integers n (i.e., all values for which $\phi(n)$ is defined), a correction term involving the Iverson bracket may be added:
$\sum_{1\le k\le n \atop \gcd(k,n)=1}\!\!k = \frac{1}{2}n(\varphi(n)+[n=1])$
## Special cases
The Kronecker delta notation is a specific case of Iverson notation when the condition is equality. That is,
$\delta_{ij} = [i=j].$
The indicator function, often denoted $\mathbf{1}_A(x)$, $\mathbf{I}_A(x)$ or $\chi_A(x)$, is an Iverson bracket with set membership as its condition:
$\mathbf{I}_A(x) := [x\in A]$.
The sign function and Heaviside step function are also easily expressed in this notation:
$\sgn(x) = [x > 0] - [x < 0]$
$H(x) = [x > 0].$
The comparison functions max, min, abs, can be expressed respectively as
$\max(x,y) = x[x>y]+y[x\leq y],$
$\min(x,y) = x[x\leq y]+y[x> y],$
and
$|x| = x[x\geq 0]-x[x<0].$
The floor and ceiling functions can be expressed as
$\lfloor x \rfloor = \sum_{n=-\infty}^{\infty}n[n \le x < n+1]$
and
$\lceil x \rceil = \sum_{n=-\infty}^{\infty}n[n-1 < x \le n].$
The Macaulay brackets can be expressed
$\{x\} = x\cdot[x\geq 0].$
And the trichotomy of the reals is equivalent to the following identity:
$[a < b] + [a = b] + [a > b] = 1.$
## References
1. ^ Kenneth E. Iverson, A Programming Language, New York: Wiley, p. 11, 1962.
2. ^ Ronald Graham, Donald Knuth, and Oren Patashnik. Concrete Mathematics, Section 2.2: Sums and Recurrences.
3. ^ Donald Knuth, "Two Notes on Notation", American Mathematical Monthly, Volume 99, Number 5, May 1992, pp. 403–422. (TeX, arXiv:math/9205211). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9356558918952942, "perplexity": 444.1913494958674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736677497.37/warc/CC-MAIN-20151001215757-00205-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://math.stackexchange.com/tags/cohomology/hot | Tag Info
7
Here is a counterexample: let $$C_i = \begin{cases} \bigoplus_{n \in \mathbb{N}} \mathbb{Z} & i = 0 \\ 0 & i \neq 0 \end{cases}$$ with the zero differential. Then $H_0(C)=\bigoplus_{n \in \mathbb{N}} \mathbb{Z}$ has countably infinite rank; $H^0(C) = \prod_{n \in \mathbb{N}} \mathbb{Z}$ has a rank $2^{\aleph_0}$ by a theorem of Nöbeling. It's ...
4
The simplicial complex you describe is called the Tits building for $GL_n(k)$. When $k=\mathbb{F}_p$ it is homotopy equivalent to a wedge of $p^{n(n-1)/2}$ spheres of dimension $n-2$. I think that theorem is due to Quillen, and am pretty sure that volume II of Benson's Representations and Cohomology contains a proof. I don't know what happens for other ...
3
Related: curved $A_\infty$-algebras $\mathcal A:=(A_\bullet, d)$ are endowed with a pseudo differential $d_1:A_n\rightarrow A_{n+1}$ s.t. $$d_1\circ d_1(a)\pm d_2(a,d_0(1))\pm d_2(d_0(1),a)=0$$ for all $a\in \mathcal A$, denoting by $d_0(1)\in A_2$ the curvature of the algebra ( $1$ is the unit) and with $d_2$ the binary product in $\mathcal A$. Clearly, ...
2
Unless I'm being silly, the torsion subgroup is contained in the kernel of the map $H^{j+1}(X,\Bbb Z)\to H^{j+1}(X,\Bbb C)$, so you're done by exactness of the long exact sequence. (For reasonable spaces, e.g., manifolds or simplicial complexes, the sheaf cohomology and singular cohomology agree.)
2
Well, if those sets are pairwais disjoint, then everything splits as a direct sum: the $k$-forms $$\Omega^k(\cup_i U_i)\simeq\oplus_i\Omega^k(U_i)$$ is given by $\omega\mapsto (\omega|_{U_i})_i$ and the De Rham differential $d$ respects this splitting. So, both the kernel and image of $d$ respect this splitting as well.. Note, however, that for an infinite ...
2
As it came out in the comments, your doubt about cellular cohomology of $\mathbb{RP}^n$ not being isomorphic to singular cohomology was because you switched the even and odd cases in $\partial$, as seen on Hatcher (current online edition) p.144. In fact, cellular (co)homology is always isomorphic to singular (co)homology for CW complexes: Hatcher is again a ...
2
Use the universal coefficient theorem ($\mathbb{Z}_2$ is a PID): $$0 \to \operatorname{Ext}^1_{\mathbb{Z}_2}(H_{i-1}(M; \mathbb{Z}_2), \mathbb{Z}_2) \to H^i(M; \mathbb{Z}_2) \to \hom_{\mathbb{Z}_2}(H_i(M; \mathbb{Z}_2), \mathbb{Z}_2) \to 0$$ Since $\mathbb{Z}_2$ is a field, the $\operatorname{Ext}^1$ term is zero, and since $M$ is a closed (thanks Jack ...
2
Construct $K(G, n)$ by starting with $n$-cells and then attaching cells of higher dimension to kill the homotopy groups above dimension $n$; do the same for $K(G, m)$. The smash product will then have no cells below dimension $n + m$.
2
By the universal coefficient theorem $H^1(X)\cong\operatorname{Hom}(H_1(X),\Bbb Z)$ as $H_0$ is always free. Morphism groups to torsion-free groups are always torsion-free.
2
Are you familiar with the inflation restriction exact sequence? If $N \unlhd G$ and $G$ acts on module $A$, then $$0 → H^1(G/N, A^N) → H^1(G, A) → H^1(N, A)^{G/N} → H^2(G/N, A^N) →H^2(G, A)$$ is exact. Taking $N=H$ in your example, we have $A^N=0$, and $|N|$ and $|A|$ are coprime (I am assuming that $Z_{p^k}$ denotes a cyclic group of order $p^k$), so ...
1
Here are some books I've found helpful: Cohomology of Groups - Kenneth Brown. Very thorough, almost to a fault. It can take a while to actually get through because of how many details are included. If you really want to see the inner-workings of everything, work through this book and its exercises, although you may not develop any intuition in doing so. ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9588720202445984, "perplexity": 253.44658244718556}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500834883.60/warc/CC-MAIN-20140820021354-00372-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://proteinsandwavefunctions.blogspot.com/2017/12/predicting-pka-values-using-pm3.html | ## Thursday, December 28, 2017
### Predicting pKa values using PM3 - conformer search part 2
Disclaimer: These are preliminary results and may contain errors
In a previous post a looked at the effect of conformational search on the accuracy of pKa predictions. This is a follow up from a slightly different angle using data obtained by Mads.
The plot shows
$$\Delta \Delta G(n) = G_{0}(n) - G_{+}(n) - (G_{0}(\min) - G_{+}(\min))$$
where $G_{0}(n)$ is the minimum free energy value found among $n$ conformations of neutral acebutolol generated by RDKit and optimized using either PM3/COSMO (blue) or PM3/SMD (red). $G_{+}(n)$ is the corresponding value for protonated acebutolol. $G_{0}(n)$ (and $G_{+}(n)$) are found using $n = 11, 21, 31, ..., 201$ starting conformations and $G_{0}(\min)$ is the lowest value of found for $G_{0}(n)$ and similarly for $G_{+}(\min)$.
So $G_{0}(\min) - G_{+}(\min)$ is our best estimate of the correct $\Delta G$ value and $\Delta \Delta G(n)$ is the deviation from the best estimate for a give value of $n$.
Keeping in mind that a 1.4 kcal/mol error corresponds to a 1 pH unit error in pKa, we are clearly no where near convergence. The fact that we observe this using two different programs (MOPAC and GAMESS) indicates that the problem is probably not the optimizer.
The next plot shows $\Delta G_X(n) =G_X(n)-G_X(\min)$. All errors are above 1 kcal/mol for $n<50$. The error for COSMO neutral does not dip below 0.5 kcal/mol until $n =$ 181, compared to $n=$ 101 for SMD neutral.
The RDKit starting geometries are not pre-minimized using MMFF. The above plot shows the corresponding plot for MMFF optimized in the gas phase. I think the much better convergence is due to the optimization being done in the gas phase. It's interesting that, again, the neutral is slower to converge, and only does so for $n >$ 140.
I think the next step is do redo this analysis with focus on finding the lowest energy conformer rather than the accuracy of the pKa values. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8695963621139526, "perplexity": 1251.8296672992326}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203123.91/warc/CC-MAIN-20190324002035-20190324024035-00277.warc.gz"} |
https://e2e.ti.com/blogs_/b/powerhouse/posts/power-tips-setting-up-bode-plotter-properly | # Power Tips: How to set up a frequency response analyzer for Bode plot
Loop gain is an important parameter for characterizing a switch-mode power supply. Using a frequency analyzer to measure loop gain gives you a way to stabilize the power supply and optimize transient response.
Before measuring the Bode plot, you need to first break the loop and insert a small resistor at the breaking point, as shown in Figure 1. The frequency analyzer has a source that injects an AC disturbance, ṽds, across this small resistor.
Figure 1: Typical Bode plot measurement setup
As a result, AC fluctuation occurs at the two nodes, A and B, across the breaking point. The frequency analyzer has two receivers to measure signals at nodes A and B, ṽA and ṽB. You can calculate the system loop gain, TV, with Equation 1:
Equation 1
To measure TV accurately, the analyzer must measure ṽA and ṽB accurately. A frequency analyzer receiver has limited signal measurement resolution. In this post, I will use the AP300 from AP Instruments, a widely used frequency response analyzer, as a setup example. Figure 2 shows the AP300’s receiver specifications and Figure 3 shows the source specifications.
Figure 2: AP300 frequency response analyzer receiver specifications
Image source: AP Instruments
Figure 3 AP300 frequency response analyzer source specifications
Image source: AP Instruments
Disturbance injection signal amplitude
According to the receiver specs, the measurable signal should be greater than 5µV. To measure voltages, ṽA and ṽB, accurately, the two signals should be of amplitude greater than that measurable by the frequency response analyzer.
Voltages ṽA and ṽB are related to both the disturbance injection signal and the loop gain itself (Equation 2):
Equation 2
Solving Equation 1 and Equation 2 results in Equation 3 and Equation 4:
Equation 3
Equation 4
At frequencies lower than the crossover, the magnitude of loop gain, |TV|, is much greater than 1. Signal ṽB is approximated by ṽds/|TV|. To guarantee that signal ṽB is greater than the measurable amplitude of 5µV, the disturbance signal ṽds should be greater than 5µV × |TV|. A power converter with tight regulation usually has DC gain greater than 60dB. As a rule of thumb, ṽds starts from 50mV at 100Hz.
Another important specification is the output impedance of the source. The AP300 has 50Ω output impedance. To ensure the delivery of sufficient energy, it’s best to insert a 50Ω matching resistor at the breaking point. Using a smaller resistor is acceptable if you adjust the injection signal amplitude to compensate for the loss of signal strength, but don’t choose too small a resistor. I recommend using a resistor of value greater than one-fifth the output impedance of the frequency response analyzer source output port.
If you inserted a small resistor, use Equation 5 to adjust the disturbance signal amplitude. For example, for a 20Ω resistor, ṽds should start from 88mV at 100Hz.
Equation 5
It is not a good idea to keep large constant disturbance signal amplitude over the whole frequency range. As frequency increases, the magnitude of |TV| decreases, which then makes signal ṽB increase. For some applications, a large disturbance at crossover might saturate the error amplifier or the duty cycle. To keep the signal as small as possible, the disturbance signal should decrease with frequency.
Figure 4 shows the AP300 interface, which provides a programmable source. The green trace in the graph shows the disturbance signal amplitude over frequency.
Figure 4: The AP300 Bode plot graphical user interface, GUI
Figure 5 shows a Bode plot measured with a constant disturbance signal of 25mV. The measured Bode plot shows a gain of only 50dB at 100Hz, while I estimated over 70dB gain at 100Hz from the high-performance controller, TPS53661. DC gain is an important indicator for regulator output DC regulation.
Figure 5: Bode plot of a step-down converter with the TPS53661 controller, with a constant disturbance signal of 25mV
I adjusted the disturbance signal accordingly and measured the Bode plot again. The measured Bode plot shows a much higher gain at 100Hz, as shown in Figure 6.
Figure 6: Bode plot of a step-down converter with the TPS53661 controller, with a programmable disturbance signal
Measurement IF-bandwidth selection
The intermediate frequency, IF-bandwidth, reducing the IF receiver bandwidth reduce the effect of random noise on the measurement. It takes the frequency analyzer longer to complete the measurement.
Figure 6 shows the difference between measurements with different signal bandwidths. The Bode plot measured with a 10Hz bandwidth is clean and smooth. The Bode plot measured with a 100Hz bandwidth shows a lot of glitches for frequencies lower than 1 KHz. For applications with crossover below 10 KHz, I recommend use IF bandwidth less than 10Hz for a clean bode plot.
Figure 7: Bode plots with different measurement bandwidths
Setting disturbance signal at the correct amplitude is important for accurate Bode plot measurement. This post provides equations for engineer to estimate the proper disturbance signal amplitude. For applications with low crossover, the IF bandwidth should be small accordingly to provide a clean bode plot and accurate phase margin.
• Thank you for your interests in TI products and my article.
The answers to your questions can be found in a my 2018 power supply design seminar topic, “ considerations for loop gain measurement in power supplies”. Please feel free to download the slides from TI website. training.ti.com/power-supply-design-seminar-2018-training-series
Thank you
• I have three questions:
1. To keep the signal as small as possible, the disturbance signal should decrease with frequency. Does it have equation?
2. Where we can set IF-bandwidth? in frequency analyzer?
3. The disturbance signal ṽds should be greater than 5µV × |TV|. A power converter with tight regulation usually has DC gain greater than 60dB. As a rule of thumb, ṽds starts from 50mV at 100Hz. How to get 50mV?
thanks. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8868815898895264, "perplexity": 2351.3440056091113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00222.warc.gz"} |
https://infoscience.epfl.ch/record/199926 | Infoscience
Journal article
# Pattern-based sensing of aminoglycosides with fluorescent amphiphiles
A mixture of two amphiphiles with fluorescent head groups can be used as a sensing ensemble for the pattern-based analysis of aminoglycoside antibiotics. In buffered aqueous solution, the amphiphiles form a dynamic mixture of micellar aggregates. In the presence of aminoglycosides, the relative amount and the composition of the micelles is modified. The re-equilibration of the system is analyte-specific, and characteristic fluorescence spectra are obtained for different aminoglycosides. Accurate differentiation in the low micromolar concentration range can be achieved by a principal component analysis of the spectral data. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8278289437294006, "perplexity": 4416.623928370394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718278.43/warc/CC-MAIN-20161020183838-00287-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/208615/distribution-of-mordell-weil-ranks-of-higher-genus-curves | # Distribution of Mordell–Weil ranks of higher genus curves
By "nice curve", I mean a smooth, projective, geometrically integral curve over $\newcommand{\Q}{\mathbb{Q}}\newcommand{\Jac}{\operatorname{Jac}}\Q$ with at least one $\Q$-rational point. The Mordell–Weil rank $r(C)$ of a nice curve $C$ is the rank (as an abelian group) of $\Jac(C)(\mathbb{Q})$, the group of $\Q$-rational points of the Jacobian of $C$.
Let $g$ be a positive integer, and consider the distribution of $r(C)$ among nice curves $C$ of genus $g$. For $g = 1$, the well-known "minimalist conjecture" says that $r(C) \leq 1$ for $100\%$ of elliptic curves, with $0$ and $1$ each having $50\%$ probability.
For fixed $g \geq 2$, what conjectures are there about the distribution of $r(C)$ among nice curves of genus $g$? What heuristics or evidence support such conjectures? (And, if it's not too speculative, what if we replace $\Q$ with some other number field?)
• For the elliptic curve case it is conjectured (a comment by Henri Darmon at this conference: crm.umontreal.ca/Counting14/index.php) that the average rank over a number field should increase with the degree of the number field uniformly over all fields of a fixed degree. Thus, one should expect that higher rank curves become more ubiquitous as the fields get larger. – Stanley Yao Xiao Jun 7 '15 at 2:12
• It strikes me that this and the question it links could be relevant: mathoverflow.net/questions/35060/… although the dimension of the space of $g$-fold products of elliptic curves is small compared to the dimension of moduli of $g$-dimensional PPAV, and still small compared to the dimension of moduli of genus $g$ curves. Still perhaps enough to get a heuristic from. – Joe Berner Jun 8 '15 at 13:01
## 1 Answer
General Katz-Sarnak heuristics suggest that the analogue of the minimalist conjecture should still be true. Let me sketch the reason why from two perspectives - the function field model, where we can establish a version of the conjecture, and the Sarnak-Shin-Templier conjectures on families of automorphic forms.
First note that it is not obvious what you mean by a random nice curve of genus $g$, because there is not always a clear height function on the moduli space of smooth curves of genus $g$. Moreover it is not always rational, so we do not always know how to count curves of bounded height!
It's safer to choose some parameterized family of curves, like hyperelliptic curves or plane curves, and ask for the distribution in that. However, my answers will not depend too much on the family.
In both cases, we will assume the truth of BSD and show that 50% of abelian varieties have analytic rank 0 and 50% have analytic rank 1 (under some heuristic.)
In a function field, an abelian variety $A$ over $\mathbb F_q(T)$ has analytic rank equal to the dimension of the Frobenius-invariant subspace of $H^1(\mathbb P^1_{\overline{\mathbb F_q}}, T_\ell(A))$ where $T_\ell(A)$ is the $\ell$-adic Tate module of $A$. Our family of curves will be parameterized by a space like $Maps(\mathbb P^1, \mathbb P^n)$ where $\mathbb P^n$ parameterizes our family of curves over $\mathbb F_q$. This cohomology group $H^1(\mathbb P^1_{\overline{\mathbb F_q}}, T_\ell(A))$ forms a sheaf on this mapping space and we may compute its monodromy group.
If the monodromy group on a component is the full orthogonal group, the minimalist conjecture is true in the $q \to \infty$ limit for that component, because the distribution of Frobenius is as a random element in the monodromy group, and a random orthogonal matrix has a $0$-dimensional invariant subspace with probability $1/2$ and a $1$-dimensional invariant subspace with probability $1/2$. This argument is explained in Katz's book Twisted L-Functions and Monodromy.
For an explicit family it's not too hard to show that the monodromy group is orthogonal using the moment method. To show the monodromy group contains the special orthogonal group, it's sufficient to show that the first few moments agree with the moments of the orthogonal one. This can be done for sufficiently large degree maps from $\mathbb P^1$ to $\mathbb P^n$ using an independence argument, as long as $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishes, which can be checked for the universal families of hyperelliptic and complete intersection curves by another independence argument. This sort of argument is explained in Katz's book Moments, Monodromy and Perversity.
In the note "Families of L-Functions and their Symmetry", Sarnak, Shin, and Templier conjecture equidistribution results for the L-functions of certain families of automorphic forms. Given any family of curves parameterized by an open subset of $\mathbb A^n$ (e.g. hyperelliptic, plane curves, complete intersection) the (conjectural) automorphic forms associated to their first etale cohomology groups are clearly a "geometric family" by the definition of the authors. In any of these families the local factors will be equidistributed in $SP_{2g}$, because a hyperelliptic/plane/complete intersection curve over a finite field has characteristic polynomial of Frobenius equidistributed in $SP_{2g}$.
Furthermore in their notation the rank of the family will be $0$ - this is actually the same $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishing condition as before and can be checked in the same way.
Also, the $\epsilon$ factors should be equidistributed between $+1$ and $-1$ - I don't immediately see a reason for this but I don't see a reason why not.
Hence assuming Conjecture 2, the distribution of the low-lying zeroes follow the $SO_{even}(\infty)$ law half the time and the $SO_{odd}(\infty)$ law half the time. The first one has average analytic rank $0$ and the second has average analytic rank $1$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8154116272926331, "perplexity": 170.64265557172212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203021.14/warc/CC-MAIN-20190323201804-20190323223804-00413.warc.gz"} |
http://mathoverflow.net/questions/55341/well-ordering-of-countably-branching-well-founded-trees | # Well ordering of countably branching well founded trees
Hello everybody,
I would like to say, before stating the question, that I'm not a mathematician, and therefore i apologize in advance if the question is trivial, or well-known. I'll try to state it as precisely as I can.
Let us define a well founded countably branching tree, as a set
$T\subset {\mathbb{N}}^{*}$, i.e. a subset of finite sequences of naturals, such that
1. $T$ is prefix-closed: if $\vec{n}.m\in T$ then $\vec{n}\in T$, where $\vec{n}\in {\mathbb{N}}^{*}$
2. Writing $\vec{n}< \vec{m}$ for $\vec{n}$ is a prefix of $\vec{m}$, there are no infinite ${<}$-increasing sequences.
So more informally $T$ is a tree whose nodes are labeled by naturals, each node can have at finitely or at most $\omega$ children, and there are no infinite branches.
Let me also define a well founded finite branching tree as above, but with the constraint that each node can have at most finitely many children.
Now let us fix an ordinal $\alpha$. We say that $(T,F)$ is a $\alpha$-labeled tree, if
1. $T$ is a countably branching w.f. tree, and
2. $F: T \rightarrow \alpha$
In other words, $F$ labels the nodes of $T$ with ordinals $\beta<\alpha$.
For the sake of simplicity let us consider from now on $\omega$-labeled trees, but my question extends to the general case.
Now I would like to define a well-ordering on the set of $\omega$-labeled trees. Here I'll be a bit informal, because I don't know how to state exactly my intuition, but the ordering should be something like the lexicographical ordering. So when judging if a tree $T$ (with root labeled with $n$, say) is smaller than $U$ (with root labeled also with $n$, say) I would just check if the leftmost branch is - recursively - smaller, if not, i'll move to the second leftmost branch etc. I'm not sure exactly how to judge coherently two trees having different labels for the root, but of course the trivial tree with just root=leaf labeled with $5$ should be smaller than the some tree where the root is labeled with $6$.
QUESTION:
Let say that we define somehow this ordering, by defining recursively a map $o: \mathcal{T}\rightarrow \gamma$, where $\mathcal{T}$ denotes the set of countably branching $\omega$-labeled trees and $\gamma$ is some ordinal.
What is the smallest ordinal $\gamma$ such that a well ordering $o$ of $\mathcal{T}$ exists? I know that $\gamma\leq\omega_{1}$. However it is perhaps the case that $\gamma<\omega_{1}$.
In particular if we consider $\omega$-labeled finite branching w.f. trees, I know that $\gamma$ would be $\epsilon_{0}$, defined as here. So I wonder if the least $\gamma$ that can well order countably branching $\omega$-labeled trees, is something smaller than $\omega_{1}$, perhaps $\epsilon_{1}$, or something.
As I said, the question generalizes to "what is the least ordinal sufficient for well ordering $\alpha$-labeled countably branching well founded trees?
SIDE QUESTION:
Can you suggest an easy reading about $\epsilon_{0}$, $\epsilon_{0}$-induction and in general the usefulness of $\epsilon$ numbers? By easy I mean accessible to a non expert.
bye
Matteo
-
I think it should be $\epsilon_1$. However I'm not that familiar with proof theoretic ordinals. – Michael Blackmon Feb 13 '11 at 19:34
Thanks Michael, do you have any intuitions supporting your claim? I'm not that familiar either :P – Matteo Mio Feb 13 '11 at 19:36
well, consider what you are defining. The definition of $\epsilon_0$ is a result of coding proofs as trees of finite height, so the next level up would be proofs of countable length, and it seems like this should be the next level up in terms of proof theoretic strength. But I'm probably missing some fine detail here. – Michael Blackmon Feb 13 '11 at 20:32
Yes actually I posted the question out of a similar guess, but that's indeed just a guess. – Matteo Mio Feb 13 '11 at 22:00
I have several observations:
• There is a very commonly used ranking function for well-founded trees, defined so that the rank $\rho(x)$ of a node $x$ in the tree is the supremum of $\rho(y)+1$ for all children $y$ of $x$. This is well-defined exactly because the tree order is well-founded. The rank of the tree itself is $\rho(\langle\rangle)+1$, where $\langle\rangle$ is the root node (the empty string).
• This ranking function has the property that whenever $y$ is a descendent of $x$, then $\rho(y)\lt\rho(x)$. That is, the ranking respects the tree order. This is somewhat different than your weaker notion of labeling.
• In particular, the tree $T_s$ of all strings $t$ such that $st\in T$, which can be thought of as the part of $T$ below node $s$, has a strictly lower rank than $T$, if $s$ is nontrivial.
• The rankings of well-founded countable trees is always a countable ordinal, but it is not bounded by $\epsilon_0$ or indeed, by any countable ordinal. The reason is that if we have a tree of rank $\alpha$, we can build a tree of rank $\alpha+1$ simply by adding a new root (which corresponds to prepending a symbol to every string in the tree); and if we have trees $T_n$ of rank $\alpha_n$, then we can make a tree of rank larger than $\sup_n\alpha_n$ by making a tree $T$ with a root node, having children nodes, whose subsequent descendents make a copy of $T_n$. Thus, by induction, the countable trees have ranks unbounded in the countable ordinals.
• Every well-founded finitely branching tree is actually finite. This is an immediate consequence of Konig's lemma, since every infinite finitely branching tree has an infinite branch, and so cannot be well-founded. The ordinal $\epsilon_0$ arises with such trees, when $\omega$-labeled, by thinking of them as representing countable ordinals according to the Cantor normal form (for example, as used in Goodstein's theorem).
• You didn't specify your intuitive order on trees precisely, but I believe that the natural interpretations of it (on the countably branching trees) will not be a well-order, as you desired. First, one common understanding of the lexical order has longer strings being lower in the order, and so we easily get descending sequences that way. Apart from this, consider the trees $T_n$ consisting of the initial segments of the string $000\cdots1$, with $n$ many $0$s. The left-most branches of these trees are exactly those terminal strings, and these are descending in the lexical order. So this is probably not what you want. Perhaps you might explain your intuitive idea a bit more fully? (In particular, why do you expect a well-order?)
• You ask about the least ordinal sufficient for well-ordering a given set. If you don't insist that the order respect any structure (which on my reading of your question you don't), then the answer is simply the cardinality of the set. For example, since there are continuum many well-founded $\omega$-labeled trees, the smallest well-ordering of them has order type $\beth_1$. Since there are only countably many finitely branching well-founded trees, the smallest ordering of these has order type $\omega$.
• Finally, perhaps the best answer to your question is to point you towards the Kleene-Brouwer ordering on trees. If $T$ is a tree of finite sequences of ordinals, then $s\lt_{KB} t$ if and only if either $t\subset s$ or for the least $i$ such that $s(i)\neq t(i)$, we have $s(i)\lt t(i)$. This turns the tree order into a linear order, using the idea of lexical order, and any well-founded tree order turns into a well-order under the Kleene-Brouwer ordering. One can now compare trees by their Kleene-Brouwer order-types, and this might be the order which you are seeking.
-
Thank you Joel for the excellent answer! In particular points 4 and 8, helped me a lot! – Matteo Mio Feb 14 '11 at 11:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9344068169593811, "perplexity": 274.37285885981595}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011162707/warc/CC-MAIN-20140305091922-00021-ip-10-183-142-35.ec2.internal.warc.gz"} |
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https://mathhelpboards.com/threads/heights-of-the-employees-of-a-company.26945/ | # [SOLVED]Heights of the employees of a company
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Hey!!
In the following table there are the heights of the employees of a company:
1. Calculate the mean value, the variance and the standard deviation of the heights of the employees.
2. Determine the distribution of sampling with replacement of average number of children of each employee for sample size $2$.
3. Which is the mean value and the variance of the sampling average?
4. Which is the mean value of the sampling variances?
5. Which is the mean value of the sampling standard deviations?
I have done the following:
1. The mean value is equal to $\frac{1436}{8}=179.5$.
The variance is equal to $\displaystyle{s^2=\frac{\sum_{i=1}^n\left (x_i-\overline{x}\right )^2}{n}}$.
The standard deviation is equal to $\displaystyle{s=\sqrt{108.5}=10.42}$.
2. I don't really understand this question. How is this related to the given data? Is there maybe a typo and instead of number of children it should be the height? But even in this case, I don't understand what we have to do here. Do you have an idea?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
In the following table there are the heights of the employees of a company:
1. Calculate the mean value, the variance and the standard deviation of the heights of the employees.
2. Determine the distribution of sampling with replacement of average number of children of each employee for sample size $2$.
3. Which is the mean value and the variance of the sampling average?
4. Which is the mean value of the sampling variances?
5. Which is the mean value of the sampling standard deviations?
I have done the following:
1. The mean value is equal to $\frac{1436}{8}=179.5$.
The variance is equal to $\displaystyle{s^2=\frac{\sum_{i=1}^n\left (x_i-\overline{x}\right )^2}{n}}$.
The standard deviation is equal to $\displaystyle{s=\sqrt{108.5}=10.42}$.
2. I don't really understand this question. How is this related to the given data? Is there maybe a typo and instead of number of children it should be the height? But even in this case, I don't understand what we have to do here. Do you have an idea?
Hey mathmari !!
Yes. It looks like a copy-paste mistake. It should be height instead of number of children.
It looks like an exercise about the difference between populations, samples, and the probability distribution when repeatedly drawing a sample (with replacement) from a population.
In this case the complete population exists of 8 employees.
It means that in (1) we are not calculating a sample mean and sample variance, but a population mean and population variance. Conventionally greek symbols are used for those instead of latin symbols.
So we have $\mu = \frac{\sum x_i}{N}=179.5$ and $\sigma^2=\frac{\sum (x_i - \mu)^2}{N}=108.5$. I'm using $N=8$ here to distinguish the total number of people in the population from the $n$ people that we have in a sample.
In (2) and following we are repeatedly drawing samples of size 2 out of the complete population of size 8.
And we look at the sample means of those samples.
The question is then about what we can say about the probability distribution of the sample means.
As it is, the Central Limit Theorem says that it will tend to a normal distribution.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
In (2) and following we are repeatedly drawing samples of size 2 out of the complete population of size 8.
And we look at the sample means of those samples.
The question is then about what we can say about the probability distribution of the sample means.
As it is, the Central Limit Theorem says that it will tend to a normal distribution.
But the Central Limit Theorem assumes that $n$ tends to infinity, or not? But we have $n=2$.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
But the Central Limit Theorem assumes that $n$ tends to infinity, or not? But we have $n=2$.
We are creating a new type of super-sample by repeatedly drawing a sample.
The result is a sample of sample-means with an as yet unspecified size.
Note that we have 3 different $n$'s that we must distinguish.
• The population size $n_\text{population}=8$.
• The single shot sample size $n_x=2$ from which we calculate a sample-mean.
• The size of the sample of sample-means $n_{\bar x}$ that is as yet unspecified.
It is $n_{\bar x}$ that tends to infinity so that we can apply the Central Limit Theorem.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
We are creating a new type of super-sample by repeatedly drawing a sample.
The result is a sample of sample-means with an as yet unspecified size.
Note that we have 3 different $n$'s that we must distinguish.
• The population size $n_\text{population}=8$.
• The single shot sample size $n_x=2$ from which we calculate a sample-mean.
• The size of the sample of sample-means $n_{\bar x}$ that is as yet unspecified.
It is $n_{\bar x}$ that tends to infinity so that we can apply the Central Limit Theorem.
Ahh I see!!
So from the Central Limit Theoremwe get the following:
From the population that has some distribution with mean value $\mu$ and variance $\sigma^2$, we choose random samples of size $n_{\bar x}$ and we calculate the mean, then for big $n_{\bar x}$ (theoretically $n_{\bar x}\rightarrow \infty$) the distribution of these means (sample-means) is approximately normal with mean value $\mu$ and variance $\frac{\sigma^2}{n_{\bar x}}$.
Is this correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
Ahh I see!!
So from the Central Limit Theoremwe get the following:
From the population that has some distribution with mean value $\mu$ and variance $\sigma^2$, we choose random samples of size $n_{\bar x}$ and we calculate the mean, then for big $n_{\bar x}$ (theoretically $n_{\bar x}\rightarrow \infty$) the distribution of these means (sample-means) is approximately normal with mean value $\mu$ and variance $\frac{\sigma^2}{n_{\bar x}}$.
Is this correct?
That doesn't look correct. Or at least not what is intended for the problem.
I'm afraid I've misinterpreted the question after all. Sorry for that.
Let's go back to square one and forget about the Central Limit Theorem for now.
The distribution of the employees looks like this:
This is a actually a histogram with each height that is in range as a separate bin.
We are drawing samples of size 2 with replacement.
So the possible samples are (employee 1, employee 1), (employee 1, employee 2), ..., (employee 8, employee 8).
Each sample has its own average height, sample variance, and sample standard deviation.
The distribution of the average height of those samples is then the list of all possible average heights combined with the frequency that they occur.
We might make a table as follows:
\begin{array}{|c|c|c|c|c|}\hline
\text{Empl1} & \text{Empl2} & \text{AvHeight} & \text{Var} & \text{Stdev} \\ \hline
1 & 1 & 169 \\
1 & 2 & 167 \\
\vdots & \vdots & \vdots \\
8 & 8 & 185 \\ \hline
\end{array}
If we make a histogram with each height that is in range as a separate bin, we get the distribution of sampling with replacement of average height of each employee for sample size 2. That is what the question asks for.
We should see that the resulting histogram looks a little more like a normal distribution than the original histogram.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
We might make a table as follows:
\begin{array}{|c|c|c|c|c|}\hline
\text{Empl1} & \text{Empl2} & \text{AvHeight} & \text{Var} & \text{Stdev} \\ \hline
1 & 1 & 169 \\
1 & 2 & 167 \\
\vdots & \vdots & \vdots \\
8 & 8 & 185 \\ \hline
\end{array}
If we make a histogram with each height that is in range as a separate bin, we get the distribution of sampling with replacement of average height of each employee for sample size 2. That is what the question asks for.
We should see that the resulting histogram looks a little more like a normal distribution than the original histogram.
At the table we have Empl1 1 and Empl2 2. Do we consider also the case Empl1 2 and Empl2 1, or do we consider these cases as the same?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
At the table we have Empl1 1 and Empl2 2. Do we consider also the case Empl1 2 and Empl2 1, or do we consider these cases as the same?
These are different samples. If we want to list all samples we have to distinguish them.
We might compress the list though, but then we need to keep track that this combination occurs twice, and weigh the average height accordingly.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
These are different samples. If we want to list all samples we have to distinguish them.
We might compress the list though, but then we need to keep track that this combination occurs twice, and weigh the average height accordingly.
If I did everything correct, we get the following frequencies for the average heights:
And the corresponding histogram is:
It looks more like a normal distribution than the original histogram.
So, is everything correct so far?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
If I did everything correct, we get the following frequencies for the average heights:
And the corresponding histogram is:
It looks more like a normal distribution than the original histogram.
So, is everything correct so far?
Looks good to me.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Looks good to me.
Great!!
For the question 3, do we calculate the mean value and the variance from the table of #9 ?
Or do we use the table of all possible samples:
And we add all the average heights and divide by the number of them?
I got stuck right now.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
For the question 3, do we calculate the mean value and the variance from the table of #9 ?
Or do we use the table of all possible samples:
And we add all the average heights and divide by the number of them?
I got stuck right now.
We can do both.
However, we get the most accurate results from the table of all possible samples.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
We can do both.
However, we get the most accurate results from the table of all possible samples.
Ok!! We have the following:
Therefore the mean value of the sampling average is equal to $$\frac{\text{Sum of Average Heights}}{\text{Sum of Frequencies}}=\frac{6462}{64}=100.97$$ right?
As for the variance it is $\displaystyle{s^2=\frac{\sum_{i=1}^{64}\left (x_i-100.97\right )^2}{64}}$ or is there also an other way to calculate this?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
Therefore the mean value of the sampling average is equal to $$\frac{\text{Sum of Average Heights}}{\text{Sum of Frequencies}}=\frac{6462}{64}=100.97$$ right?
Not quite.
It should be the same as the average height of the population, which is 179.5.
Did you perhaps forget to weigh the average heights with their frequencies? That would account for a factor of slightly less than 2 as is the case.
As for the variance it is $\displaystyle{s^2=\frac{\sum_{i=1}^{64}\left (x_i-100.97\right )^2}{64}}$ or is there also an other way to calculate this?
That is not the formula of the sample variance.
The correct formula is:
$$s^2=\frac{\sum_{i=1}^n (x_k-\bar x)^2}{n-1}$$
Note in particular the $n-1$ in the denominator.
This assumes we have to estimate the population mean from only this sample, which is $\bar x$.
Next, we need the sample variance of every sample, which uses the average height of every sample.
Suppose we have a sample with employee heights $(x_1, x_2)$.
Then the sample variance is:
$$s^2=\frac{\sum_{i=1}^n (x_k-\bar x)^2}{n-1} = \frac{(x_1-\bar x)^2 + (x_2-\bar x)^2}{2-1}$$
where $\bar x = \frac{x_1+x_2}{2}$, which is in the $\text{Av.Height}$ column that you already have.
When we have the variances of all the samples, we can calculate their mean by summing them and dividing by the total number. Oh, and don't forget to 'weigh' them by their frequency.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Not quite.
It should be the same as the average height of the population, which is 179.5.
Did you perhaps forget to weigh the average heights with their frequencies? That would account for a factor of slightly less than 2 as is the case.
Yes, that was the mistake! Thanks for the hint!
Next, we need the sample variance of every sample, which uses the average height of every sample.
Suppose we have a sample with employee heights $(x_1, x_2)$.
Then the sample variance is:
$$s^2=\frac{\sum_{i=1}^n (x_k-\bar x)^2}{n-1} = \frac{(x_1-\bar x)^2 + (x_2-\bar x)^2}{2-1}$$
where $\bar x = \frac{x_1+x_2}{2}$, which is in the $\text{Av.Height}$ column that you already have.
When we have the variances of all the samples, we can calculate their mean by summing them and dividing by the total number. Oh, and don't forget to 'weigh' them by their frequency.
So we have the following:
From that we get that the mean value of the sampling average is equal to $$\frac{\text{Sum of Average Heights * Frequency}}{\text{Sum of Frequencies}}=\frac{11488}{64}=179.5$$
And the variance of the sampling average is equal to $$\frac{\text{Sum of Variance * Frequency}}{\text{Sum of Frequencies}}=\frac{6944}{64}=108.5$$ which is again the same as variance of population. So these two values have to be always the same, right?
At question 4, the mean value of the sampling variances is asked. Do we have to find the mean value of the $6$th column of the above table?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
From that we get that the mean value of the sampling average is equal to $$\frac{\text{Sum of Average Heights * Frequency}}{\text{Sum of Frequencies}}=\frac{11488}{64}=179.5$$
And the variance of the sampling average is equal to $$\frac{\text{Sum of Variance * Frequency}}{\text{Sum of Frequencies}}=\frac{6944}{64}=108.5$$ which is again the same as variance of population. So these two values have to be always the same, right?
At question 4, the mean value of the sampling variances is asked. Do we have to find the mean value of the $6$th column of the above table?
You just did that. So you have just answered question 4.
Rereading question 3, I see that your original formula was more or less correct after all. It is the variance of the entire population (of sample-means).
That is, it should be:
$$\sigma_{\bar x}^2 = \frac{\sum_{i=1}^{64} f_i\cdot(\bar x_i - \mu_{\bar x})^2}{\sum_{i=1}^{64} f_i}$$
where $\mu_{\bar x}=179.5$, which is the mean value of the sampling average that you just calculated, where $\bar x_i$ are the sampling averages, and where $f_i$ are the frequencies of those sampling averages.
Note the use of Greek letters since we are talking about the entire population of possible samples.
Can we predict what it will be? That is, how it is related to the variance of the original population?
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
You just did that. So you have just answered question 4.
I got stuck right now. Wasn't this the answer of question 3, i.e. the variance of the sampling average?
Rereading question 3, I see that your original formula was more or less correct after all. It is the variance of the entire population (of sample-means).
That is, it should be:
$$\sigma_{\bar x}^2 = \frac{\sum_{i=1}^{64} f_i\cdot(\bar x_i - \mu_{\bar x})^2}{\sum_{i=1}^{64} f_i}$$
where $\mu_{\bar x}=179.5$, which is the mean value of the sampling average that you just calculated, where $\bar x_i$ are the sampling averages, and where $f_i$ are the frequencies of those sampling averages.
Note the use of Greek letters since we are talking about the entire population of possible samples.
Can we predict what it will be? That is, how it is related to the variance of the original population?
It will be the same, or not? But why?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
I got stuck right now. Wasn't this the answer of question 3, i.e. the variance of the sampling average?
Question 3 asks for the variance of the sampling averages.
Question 4 asks for the average of the sampling variances.
For question 3 we have:
$$\mu_{\bar x} = \text{Mean value of sampling averages} = \frac{\sum f_i \cdot \bar x_i}{\sum f_i}$$
and:
$$\sigma_{\bar x}^2 = \text{Variance of sampling averages} = \frac{\sum f_i \cdot (\bar x_i - \mu_{\bar x})^2}{\sum f_i}$$
For question 4 we have:
$$s_i^2 = \text{Sampling variance of sample }i = \frac{\sum_j (x_{i,j} - {\bar x_i})^2}{n_x-1}$$
and:
$$\mu_{s^2} = \text{Mean value of sampling variances} = \text{Average of sampling variances} = \frac{\sum f_i \cdot s_i^2}{\sum f_i}$$
See the difference?
It will be the same, or not? But why?
I don't think that they will be the same. Perhaps calculate it and see?
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Question 3 asks for the variance of the sampling averages.
Question 4 asks for the average of the sampling variances.
For question 3 we have:
$$\mu_{\bar x} = \text{Mean value of sampling averages} = \frac{\sum f_i \cdot \bar x_i}{\sum f_i}$$
and:
$$\sigma_{\bar x}^2 = \text{Variance of sampling averages} = \frac{\sum f_i \cdot (\bar x_i - \mu_{\bar x})^2}{\sum f_i}$$
For question 4 we have:
$$s_i^2 = \text{Sampling variance of sample }i = \frac{\sum_j (x_{i,j} - {\bar x_i})^2}{n_x-1}$$
and:
$$\mu_{s^2} = \text{Mean value of sampling variances} = \text{Average of sampling variances} = \frac{\sum f_i \cdot s_i^2}{\sum f_i}$$
See the difference?
We have this table.
From this we get the following values:
For question 3 :
$$\mu_{\bar x} = \text{Mean value of sampling averages} = \frac{11488}{64}=179.5$$
and:
$$\sigma_{\bar x}^2 = \text{Variance of sampling averages} = \frac{3472}{64}=54.25$$
For question 4 :
$$\mu_{s^2} = \text{Mean value of sampling variances} = \text{Average of sampling variances} = \frac{6944}{64}=108.5$$
For question 5 :
$$\mu_{s} = \text{Mean value of sampling standard deviations} = \text{Average of sampling standard deviations} = \frac{526.09}{64}=8.22$$
Is everything correct?
I don't think that they will be the same. Perhaps calculate it and see?
The variance of the population is $108.5$ which is equal to the mean value of sampling variances. The variance of sampling averages is different.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
We have this table.
From this we get the following values:
For question 3 :
$$\mu_{\bar x} = \text{Mean value of sampling averages} = \frac{11488}{64}=179.5$$
and:
$$\sigma_{\bar x}^2 = \text{Variance of sampling averages} = \frac{3472}{64}=54.25$$
For question 4 :
$$\mu_{s^2} = \text{Mean value of sampling variances} = \text{Average of sampling variances} = \frac{6944}{64}=108.5$$
For question 5 :
$$\mu_{s} = \text{Mean value of sampling standard deviations} = \text{Average of sampling standard deviations} = \frac{526.09}{64}=8.22$$
Is everything correct?
Yep. All correct.
The variance of the population is $108.5$ which is equal to the mean value of sampling variances. The variance of sampling averages is different.
The variance of sampling averages is $54.25$.
Isn't that half of $108.5$? And didn't we have $n_x=2$?
We have:
$$\sigma_{\bar x}^2 = \frac{\sigma^2}{n_x}$$
and:
$$\sigma_{\bar x} = SE = \text{Standard Error} = \frac{\sigma}{\sqrt{n_x}}$$
We see an $SE$ in most statistical tests.
The $SE$ we have just found, is the one that applies to the z-test.
That is, if we want to know if 2 groups have a different population mean, we typically test if their sampling averages differ by at least $1.96 \cdot SE$. If they do, the groups are significantly different.
(The factor $1.96$ is the critical z-value for a 2-sided z-test with significance level $\alpha=5\%$.)
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Yep. All correct.
The mean value of sampling variances is the same as the variance of population. But the mean value of sampling standard deviations is not equal to the standard deviation of the population.
Why is it like that? Or do both have to be equal and I have done something wrong?
The variance of sampling averages is $54.25$.
Isn't that half of $108.5$? And didn't we have $n_x=2$?
We have:
$$\sigma_{\bar x}^2 = \frac{\sigma^2}{n_x}$$
and:
$$\sigma_{\bar x} = SE = \text{Standard Error} = \frac{\sigma}{\sqrt{n_x}}$$
We see an $SE$ in most statistical tests.
The $SE$ we have just found, is the one that applies to the z-test.
That is, if we want to know if 2 groups have a different population mean, we typically test if their sampling averages differ by at least $1.96 \cdot SE$. If they do, the groups are significantly different.
(The factor $1.96$ is the critical z-value for a 2-sided z-test with significance level $\alpha=5\%$.)
I see!!
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
The mean value of sampling variances is the same as the variance of population.
Basically that is why we have $n-1$ in the denominator for the sampling variance.
It is a correction so that we get the correct variance.
But the mean value of sampling standard deviations is not equal to the standard deviation of the population.
Why is it like that? Or do both have to be equal and I have done something wrong?
When adding independent variables, the result has a variance that is the sum of the variances.
And if we divide an independent variable by a constant, the result has a variance that is divided by that constant.
So we can expect that variances behave predictably and consistently. "They add up" so to speak.
The standard deviation is the square root of the variance.
Consequently summing independent variables, or taking their mean, gives results that appear to be unpredictable.
That is, we can expect such standard deviations to be different from the population.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Basically that is why we have $n-1$ in the denominator for the sampling variance.
It is a correction so that we get the correct variance.
When adding independent variables, the result has a variance that is the sum of the variances.
And if we divide an independent variable by a constant, the result has a variance that is divided by that constant.
So we can expect that variances behave predictably and consistently. "They add up" so to speak.
The standard deviation is the square root of the variance.
Consequently summing independent variables, or taking their mean, gives results that appear to be unpredictable.
That is, we can expect such standard deviations to be different from the population.
Ahh ok!!
There is also an other question:
If we have a sample of size $30$ (with replacement), which is the probability that the sample average is over $181$ cm?
I have done the following:
We have that the sample mean value is equal to the population mean $\mu_X=\mu=179.5$.
The sample standard deviation is equal to $\sigma_X=\frac{\sigma}{\sqrt{n}}=\frac{10.42}{\sqrt{30}}=1.90$.
The probability that the sample mean height is over $181$ is equal to \begin{align*}P(Χ > 181) &=1-P(X\leq 181)=1-P\left (Z\leq \frac{181-\mu_X}{\sigma_X}\right )=1-P\left (Z\leq \frac{181-179.5}{1.90}\right ) \\ & =1-P\left (Z\leq \frac{1.5}{1.90}\right )=1-P\left (Z\leq 0.79\right )=1-0.78524 \\ & =0.21476\end{align*} So the possibility that the sample mean height is over $181$ is equal to $21,476\%$.
Is everything correct?
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mar 5, 2012
8,684
There is also an other question:
If we have a sample of size $30$ (with replacement), which is the probability that the sample average is over $181$ cm?
I have done the following:
We have that the sample mean value is equal to the population mean $\mu_X=\mu=179.5$.
Let me nitpick a bit...
The sample mean value is the average height of the sample of size $30$ yes?
Isn't it unlikely that it will be exactly the same as the population mean?
The sample standard deviation is equal to $\sigma_X=\frac{\sigma}{\sqrt{n}}=\frac{10.42}{\sqrt{30}}=1.90$.
Isn't the sample standard deviation equal to $s = \sqrt{\frac{\sum (x_i - \bar x)^2}{n-1}}$?
Did you perhaps mean the standard deviation of the sample-means? Also known as the Standard Error (SE)?
The probability that the sample mean height is over $181$ is equal to \begin{align*}P(Χ > 181) &=1-P(X\leq 181)=1-P\left (Z\leq \frac{181-\mu_X}{\sigma_X}\right )=1-P\left (Z\leq \frac{181-179.5}{1.90}\right ) \\ & =1-P\left (Z\leq \frac{1.5}{1.90}\right )=1-P\left (Z\leq 0.79\right )=1-0.78524 \\ & =0.21476\end{align*} So the possibility that the sample mean height is over $181$ is equal to $21,476\%$.
You wrote 'possibility'. Did you perhaps mean 'probability'?
Otherwise this looks correct.
#### mathmari
##### Well-known member
MHB Site Helper
Apr 14, 2013
4,003
You wrote 'possibility'. Did you perhaps mean 'probability'?
Oh yes, I meant probability.
The sample mean value is the average height of the sample of size $30$ yes?
Isn't it unlikely that it will be exactly the same as the population mean?
Isn't the sample standard deviation equal to $s = \sqrt{\frac{\sum (x_i - \bar x)^2}{n-1}}$?
Did you perhaps mean the standard deviation of the sample-means? Also known as the Standard Error (SE)?
Otherwise this looks correct.
So, did I used at the calculation of the probability the correct values for mean value and standard deviation just the wrong names? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9901114702224731, "perplexity": 573.9819959187097}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655883961.50/warc/CC-MAIN-20200704011041-20200704041041-00327.warc.gz"} |
https://www.physicsforums.com/threads/physics-true-false-questions-v-isolated-parallel-plates.65962/ | # Physics true/false questions v. isolated parallel plates
1. Mar 4, 2005
### Gonger
Which of the statements below are true for two oppositely charged, isolated parallel plates? (C is the capacitance, U is the stored energy, +Q and -Q are the charges on the plates.) Note: isolated plates can not lose their charge. (Enter ALL correct statements, e.g., BCD)
A) When the distance is doubled, U increases.
True. Im thinking that if the distance is doubled there is more to store energy in.
B) Inserting a dielectric increases C.
I have no idea on this one.
C) When the distance is halved, Q stays the same.
True. I dont think distance will affect the charge.
D) Increasing the distance increases the electric field.
True.
E) When the distance is doubled, C increases.
False. It doesnt have any affect on the capacitance
F) Inserting a dielectric decreases U.
Again no idea.
G) Inserting a dielectric increases Q.
I just have no clue with dielectrics.
Thanks for any help in advance.
2. Mar 4, 2005
### Corneo
Some forumlas you need to know is
$$C = \frac {Q}{V} = \epsilon_0 \frac {A}{d}, \ \ U = \frac {Q^2}{2C}$$
So when d increases, C decreases, then U increase.
When ever you are dealing with a dielectric, the capacitance is given by
$$C_d = \kappa \epsilon_0 \frac {A}{d}$$
Where $\kappa$ is a constant called the dielectric constant. This is always greater than 1. This says that $C_d > C$. Which is exactly why dielectrics are inserted, to increase the capacitance.
That should get you started.
3. Mar 4, 2005
### Gonger
cool. I got it thanks.
4. Mar 5, 2005
### Andrew Mason
True. $C \propto \epsilon > \epsilon_0$
False. The field is uniform between the plates and is independent of the separation where the plate size is large compared to separation. The potential depends upon distance (V = Ed)
True. C is proportional to d.
True. The dielectric reduces the field between the plates. So it reduces the potential (V=Ed),
False. Charge is constant.
So: ABCEF
AM
5. Mar 5, 2005
### Corneo
I think $C \propto \frac {1}{d}$.
6. Mar 5, 2005
### Andrew Mason
Right you are. So ABCF.
AM
Similar Discussions: Physics true/false questions v. isolated parallel plates | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9109942317008972, "perplexity": 2609.2441141601516}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00686.warc.gz"} |
https://testbook.com/question-answer/which-of-the-following-of-newtons-laws-expl--608a80918817d2b1860191e1 | # Which of the following of Newton’s laws explains the recoiling of a gun when a bullet is fired?
This question was previously asked in
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1. Law of gravitation
2. Third law of motion
3. First law of motion
4. Second law of motion
Option 2 : Third law of motion
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## Detailed Solution
The correct answer is Third law of motion.
Key Points
• Newton's Third Law of Motion states that 'To every action, there is an equal and opposite reaction'.
• When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is force is called the action force.
• The bullet also exerts an equal and opposite force on the gun in the backward direction.
• Therefore a gun recoils when a bullet is fired from it.
• Newton's Second Law of motion
• It states that the rate of change of momentum of an object is proportional to the applied force in the direction of the force.
• i.e. F=ma.
• Where F is the force applied, m is the mass of the body, and a is the acceleration produced.
• A fielder pulling his hand back while catching a ball is an application of Newton's second law of motion.
• Newton's First law
• It states that Every object remains in uniform motion in a straight line unless compelled to change its state by the action of an external force.
• When a bus starts suddenly, the passengers receive a backward jerk an application of Newton's first law of motion.
• Law of Universal Gravitation:
• It states that all objects attract each other with a force that is proportional to the masses of two objects and inversely proportional to the square of the distance that separates their centers.
• It is given mathematically as follows:
• $$F = \frac{Gm_1m_2}{R^2}$$
• Where m1 and m2 are the mass of two objects, G is the gravitational constant and R is the distance between their centers.
• The gravitational constant G establishes a relationship between gravitational force, mass, and distance.
• The value of G is 6.67 × 10-11 N kg-2 m2. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9487212300300598, "perplexity": 745.9982159963653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337595.1/warc/CC-MAIN-20221005073953-20221005103953-00548.warc.gz"} |
http://www.zora.uzh.ch/id/eprint/108345/ | # Search for $WW$ $\gamma$ and $WZ$ $\gamma$ production and constraints on anomalous quartic gauge couplings in $pp$ collisions at $\sqrt{s}=8\text{ }\text{ }\mathrm{TeV}$
CMS Collaboration; Canelli, M F; Chiochia, V; Kilminster, B; Robmann, P; et al (2014). Search for $WW$ $\gamma$ and $WZ$ $\gamma$ production and constraints on anomalous quartic gauge couplings in $pp$ collisions at $\sqrt{s}=8\text{ }\text{ }\mathrm{TeV}$. Physical Review D (Particles, Fields, Gravitation and Cosmology), 90(032008):online.
## Abstract
A search for $WV$ $\gamma$ triple vector boson production is presented based on events containing a $W$ boson decaying to a muon or an electron and a neutrino, a second $V$ ($W$ or $Z$) boson, and a photon. The data correspond to an integrated luminosity of $19.3\text{ }\text{ }{\mathrm{fb}}^{$-${}1}$ collected in 2012 with the CMS detector at the LHC in $pp$ collisions at $\sqrt{s}=8\text{ }\text{ }\mathrm{TeV}$. An upper limit of 311 fb on the cross section for the $WV$ $\gamma$ production process is obtained at 95% confidence level for photons with a transverse energy above 30 GeV and with an absolute value of pseudorapidity of less than 1.44. This limit is approximately a factor of 3.4 larger than the standard model predictions that are based on next-to-leading order QCD calculations. Since no evidence of anomalous $WW$$\gamma or WWZ \gamma quartic gauge boson couplings is found, this paper presents the first experimental limits on the dimension-eight parameter {f}_{T,0} and the CP-conserving WWZ \gamma parameters {\kappa{}}_{0}^{W} and {\kappa{}}_{C}^{W}. Limits are also obtained for the WW \gamma$$\gamma$ parameters ${a}_{0}^{W}$ and ${a}_{C}^{W}$.
## Abstract
A search for $WV$ $\gamma$ triple vector boson production is presented based on events containing a $W$ boson decaying to a muon or an electron and a neutrino, a second $V$ ($W$ or $Z$) boson, and a photon. The data correspond to an integrated luminosity of $19.3\text{ }\text{ }{\mathrm{fb}}^{$-${}1}$ collected in 2012 with the CMS detector at the LHC in $pp$ collisions at $\sqrt{s}=8\text{ }\text{ }\mathrm{TeV}$. An upper limit of 311 fb on the cross section for the $WV$ $\gamma$ production process is obtained at 95% confidence level for photons with a transverse energy above 30 GeV and with an absolute value of pseudorapidity of less than 1.44. This limit is approximately a factor of 3.4 larger than the standard model predictions that are based on next-to-leading order QCD calculations. Since no evidence of anomalous $WW$$\gamma or WWZ \gamma quartic gauge boson couplings is found, this paper presents the first experimental limits on the dimension-eight parameter {f}_{T,0} and the CP-conserving WWZ \gamma parameters {\kappa{}}_{0}^{W} and {\kappa{}}_{C}^{W}. Limits are also obtained for the WW \gamma$$\gamma$ parameters ${a}_{0}^{W}$ and ${a}_{C}^{W}$.
## Statistics
### Citations
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10 citations in Web of Science®
23 citations in Scopus® | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9955543875694275, "perplexity": 1075.8991065250063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590199.42/warc/CC-MAIN-20180718135047-20180718155047-00420.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/85685-width-confidence-interval.html | # Thread: Width of a Confidence Interval
1. ## Width of a Confidence Interval
I am OK with part a, but I am having some troubles with part b.
Attempt for part b:
note: P(T>t_(n1+n2-2),alpha/2)=alpha/2 where T~t distribution with n1+n2-2 degrees of freedom.
Now, as n1 increases and n2 increases,
(i) t_(n1+n2-2),alpha/2 gets smaller
(ii) denominator gets larger
(iii) the ∑ terms gets larger because the upper indices of summation are n1 and n2, respectively
(i) and (ii) push towards a narrower confidence interval, but (iii) pushes towards a wider confidence interval. How can we determine the ultimate result?
Thank you for any help!
[note: also under discussion in sos math cyberboard]
2. I agree that the "intuitive" answer is that the confidence interval will be "narrower" as the sample sizes increase, but how can we prove this more rigorously? My method above doesn't quite seem to work...
The trouble is that as n1 and n2 increase, the numerator ∑(Xi-Xbar)^2 + ∑(Yi-Ybar)^2 also increases (because now we are summing over more and more non-negative terms as n1 and n2, the upper indices of summation, increase), and this tends to make the CI wider. How can we deal with this matter? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8942172527313232, "perplexity": 1257.050807925146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281450.93/warc/CC-MAIN-20170116095121-00097-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Boundary_conditions_in_fluid_dynamics | # Boundary conditions in fluid dynamics
Boundary conditions in fluid dynamics are the set of constraints to boundary value problems in computational fluid dynamics. These boundary conditions include inlet boundary conditions, outlet boundary conditions, wall boundary conditions, constant pressure boundary conditions, axisymmetric boundary conditions, symmetric boundary conditions, and periodic or cyclic boundary conditions.
Transient problems require one more thing i.e., initial conditions where initial values of flow variables are specified at nodes in the flow domain.[1] Various types of boundary conditions are used in CFD for different conditions and purposes and are discussed as follows.
## Inlet boundary conditions
Showing inlet flow velocity in a pipe
In inlet boundary conditions, the distribution of all flow variables needs to be specified at inlet boundaries mainly flow velocity.[1] This type of boundary conditions are common and specified mostly where inlet flow velocity is known.
## Outlet boundary condition
Showing outlet flow velocity in a pipe
In outlet boundary conditions, the distribution of all flow variables needs to be specified, mainly flow velocity. This can be thought as a conjunction to inlet boundary condition. This type of boundary conditions is common and specified mostly where outlet velocity is known.[1] The flow attains a fully developed state where no change occurs in the flow direction when the outlet is selected far away from the geometrical disturbances. In such region, an outlet could be outlined and the gradient of all variables could be equated to zero in the flow direction except pressure.
## No-slip boundary condition
Showing wall boundary condition
The most common boundary that comes upon in confined fluid flow problems is the wall of the conduit. The appropriate requirement is called the no-slip boundary condition, wherein the normal component of velocity is fixed at zero, and the tangential component is set equal to the velocity of the wall.[1] It may run counter to intuition, but the no-slip condition has been firmly established in both experiment and theory, though only after decades of controversy and debate.[2]
${\displaystyle V_{\text{normal}}=0}$
${\displaystyle V_{\text{tangential}}=V_{\text{wall}}}$
Heat transfer through the wall can be specified or if the walls are considered adiabatic, then heat transfer across the wall is set to zero.
${\displaystyle Q_{\text{Adiabatic Walls}}=0}$
## Constant pressure boundary conditions
Showing constant pressure boundary condition
This type of boundary condition is used where boundary values of pressure are known and the exact details of the flow distribution are unknown. This includes pressure inlet and outlet conditions mainly. Typical examples that utilize this boundary condition include buoyancy driven flows, internal flows with multiple outlets, free surface flows and external flows around objects.[1] An example is flow outlet into atmosphere where pressure is atmospheric.
## Axisymmetric boundary conditions
Showing axisymmetric boundary condition
In this boundary condition, the model is axisymmetric with respect to the main axis such that at a particular r = R, all θs and each z = Z-slice, each flow variable has the same value.[3] A good example is the flow in a circular pipe where the flow and pipe axes coincide.
${\displaystyle V_{r}(R,\theta ,Z)=Constant}$ ${\displaystyle (r=R,\theta ,Z)}$
## Symmetric boundary condition
Showing symmetric boundary condition
In this boundary condition, it is assumed that on the two sides of the boundary, same physical processes exist.[4] All the variables have same value and gradients at the same distance from the boundary. It acts as a mirror that reflects all the flow distribution to the other side.[5] The conditions at symmetric boundary are no flow across boundary and no scalar flux across boundary.
A good example is of a pipe flow with a symmetric obstacle in the flow. The obstacle divides the upper flow and lower flow as mirrored flow.
## Periodic or cyclic boundary condition
A quarter showing cyclic boundary condition
A periodic or cyclic boundary condition arises from a different type of symmetry in a problem. If a component has a repeated pattern in flow distribution more than twice, thus violating the mirror image requirements required for symmetric boundary condition. A good example would be swept vane pump (Fig.),[6] where the marked area is repeated four times in r-theta coordinates. The cyclic-symmetric areas should have the same flow variables and distribution and should satisfy that in every Z-slice.[1] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8611029386520386, "perplexity": 914.0170675533423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335303.67/warc/CC-MAIN-20220929003121-20220929033121-00457.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/12/lesson/12.2.1/problem/12-67 | ### Home > APCALC > Chapter 12 > Lesson 12.2.1 > Problem12-67
12-67.
A function $f$ has derivatives of all orders for all $x$. Some values of $f^{ }′$ are given in the table below. It is known that $f(2) = -5$.
$x$ $f ′(x)$ $2$ $2.2$ $2.4$ $2.6$ $2.8$ $4$ $12$ $23$ $39$ $59$
1. Use a trapezoidal sum with four subintervals to approximate $\int _ { 2 } ^ { 2.8 } f ^ { \prime } ( x ) d x$.
$0.2 ( \frac { 1 } { 2 } ) ( 4 + 2 ( 12 ) + 2 ( 23 ) + 2 ( 39 ) + 59$
2. Use your approximation from part (a) to estimate the value of $f(2.8)$. Justify your estimate with your work.
$f ( 2.8 ) = f ( 2 ) + \int _ { 2 } ^ { 2.8 } f ^ { \prime } ( x ) d x$
3. Use Euler’s Method, starting with $x = 2$, with four steps of equal size, to approximate $f(2.8)$. Show your work.
At $x=2$, the slope is $4$. Since the step size is $dx=.02$, then $\frac { d y } { d x } = \frac { 4 } { 1 } = \frac { 4 ( 0.2 ) } { 0.2 }$
Starting at $(2,-5)$, the next point is: $(2 + 0.2 , - 5 + 4 ( 0.2 ))$
4. If $f^{\prime\prime}(2) = 32$ and $f^{\prime\prime\prime}(2) = 72$, write a third-degree Taylor polynomial about $x = 2$ and use it to approximate $f(2.8)$.
$p _ { 3 } ( x ) = f ( 2 ) + f ^ { \prime } ( 2 ) ( x - 2 ) + \frac { f ^ { \prime \prime } ( 2 ) } { 2 ! } ( x - 2 ) ^ { 2 } + \frac { f ^ { \prime \prime \prime } ( 2 ) } { 3 ! } ( x - 2 ) ^ { 3 }$
5. How close were your estimates of $f(2.8)$ in parts (b) through (d)? Explain why this happened. | {"extraction_info": {"found_math": true, "script_math_tex": 34, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9737648963928223, "perplexity": 333.0534454911777}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00433.warc.gz"} |
https://www.vernier.com/experiment/psv-10_energy-content-of-foods/ | ### Introduction
Energy content is an important property of food. The energy your body needs for running, talking, and thinking comes from the food you eat. Energy content is the amount of heat produced by the burning of 1 gram of a substance, and is measured in joules per gram (J/g).
You can determine energy content by burning a portion of food and capturing the heat released to a known mass of water in a calorimeter. If you measure the initial and final temperatures, the energy released can be calculated using the equation
$H = \vartriangle t \cdot m \cdot C_p$
where H = heat energy absorbed (in J), Δt = change in temperature (in °C), m = mass (in g), and Cp = specific heat capacity (4.18 J/g°C for water). Dividing the resulting energy value by grams of food burned gives the energy content (in J/g).
### Objectives
In this experiment, you will
• Measure temperature.
• Analyze data.
• Use a balance.
• Determine energy content.
• Compare the energy content of different foods. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8956363201141357, "perplexity": 1473.5225511989202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00483.warc.gz"} |
http://mathhelpforum.com/differential-equations/171179-3-dimensional-wave-equation-cauchy-data.html | ## 3-dimensional Wave equation with Cauchy data
Let $u(x,y,t)$ be the solution of the Cauchy problem
$u_{tt}-c^2 u_{xx}-c^2 u_{yy}=0$,
$u(x,y,0)=f(x,y)$,
$u_{t}(x,y,0)=g(x,y)$,
where $c>0$ is a constant and $f(x,y)$ and $g(x,y)$ vanish for $x^2 +y^2 >r^2$ for some $r>0$.
Show that the solution $u(x,y,t)$ vanishes if $x^2+y^2-r^2>c^2 t^2$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9935829639434814, "perplexity": 149.97022587335212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657129229.10/warc/CC-MAIN-20140914011209-00150-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.physicsoverflow.org/25256/interpreting-the-cs-wzw-correspondence | Interpreting the CS/WZW correspondence
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It is understood that there is a correspondence between the 3d Chern-Simons topological quantum field theory (TQFT) and the 2d Wess-Zumino-Witten conformal quantum field theory (CQFT). A good summary is given on this nLab page:
This is often claimed (e.g. in that nLab page) to be an instance of the holographic principle. To me, this carries the implication that the two theories are strictly equivalent---that is, any calculation of a physical quantity in one theory could be carried out just as well in the other theory. This aspect of the holographic principle is frequently emphasized in general accounts of the holographic principle.
However, I do not see that this is delivered by the precise nature of the CS/WZW correspondence. In particular, the states of the bulk CS theory correspond only to the conformal blocks of the WZW theory, which are essentially the space of solutions to the Ward equations. While this is interesting, it is not as strong as one might expect: in particular, I don't see that it gives a way to translate some calculation in the CQFT---for example, the numerical value of a correlator for a given conformal surface with labelled marked points---into a corresponding calculation in the TQFT.
My question is the following: If the CS/WZW correspondence is a holographic duality, why does it not seem possible to replicate every calculation from one theory in the other?
Of course, a good answer to this question might deny the premise, or show some way in which one can indeed replicate calculations from one theory in the other theory.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Jamie Vicary
edited Dec 6, 2014
I am asking me the a similar question, but I am in a hurry at the moment. The point is CS can be constructed only out of the modular tensor category, while WZW also needs the VOA.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Marcel Bischoff
Indeed, that is a closely-related fact that also implies that the CQFT is richer than the TQFT.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Jamie Vicary
physics.stackexchange.com/questions/75260/…
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Carlo Beenakker
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Actually I think the idea of the holographic principle is that, as in a holograph, all the information in the 'bulk' is already present at the 'boundary'. So, it claims that any calculation involving bulk observables can be expressed in terms of boundary observables. It may not claim the reverse, though that could often be taken for granted!
In discussions with Urs Schreiber and Jamie Vicary we seem to have settled on the following formulation. A modular tensor category gives rise to a once extended 3d TQFT and can also be reconstructed from this once extended 3d TQFT. By work of Fuchs, Runkel and Schweigert, a modular tensor category equipped with an equivalence to a category of representations of a vertex operator algebra and equipped with a symmetric Frobenius object gives rise to a rational CFT.
The italicized phrases would then be ways that the rational CFT has more information than the once extended 3d TQFT. The first one gives the 'local information' needed to construct a CFT locally starting from the extended TQFT. The second one gives the 'global information' or 'sewing information' needed to finish the job.
Apparently there may be one, many or no rational CFTs corresponding to a given once extended 3d TQFT.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user John Baez
answered Mar 16, 2014 by (365 points)
In AdS/CFT, at least in the best-understood variants, my understanding is that the theories on the bulk and boundary are exactly dual, and the word `duality' is often used in this context. So in this case, at least, it would be expected that each theory can be interpreted in the other. So are you suggesting that the CS/WZW correspondence is of a lesser nature, and not a duality in the same sense?
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Jamie Vicary
I don't understand the AdS/CFT stuff very well, so for all I know it works the same as CS/WZW, without the precise theorems. Note you only detected the need for a bit of extra data in the CS/WZW correspondence by having a theorem relating a large class of 3d topological field theories to a large class of 2d conformal field theories. I've never heard of theorems or even precise claims like this in the higher-dimensional cases, so it could be that people are 'smuggling in' extra data on one side of the 'duality' to make the correspondence work, without knowing or caring very much.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user John Baez
By the way, I dislike the trend in theoretical physics toward using 'duality' to mean a wide variety of isomorphisms, correspondences, and other relationships. To me a duality happens when we have two categories containing a dualizing object or, as a special case, a closed category containing a dualizing object - or various categorified versions of this theme. The dualities in linear and projective geometry are the prototypes.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user John Baez
It's confusing that the situation in classical field theory is the opposite: a classical WZW model is a bundle gerbe with connection over the group $G$, while a classical Chern-Simons theory is a multiplicative bundle gerbe with connection over $G$, which is additional structure. In particular, there can be different multiplicative structures on the same gerbe, i.e. different TFTs corresponding to one CFT (for that, $G$ has to be non-simple or so).
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Konrad Waldorf
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One of the issues here is that if we define CS by its physics definition (action, fields) and similarly for WZW then it does seem like there is an equivalence. For example, on page 30 of
we apparently see how to go from a state in CS theory to a correlator in WZW. From a mathematical point of view, this uses extra information about the states in CS theory (i.e. they are not just abstract vectors in a vector space, they are functionals on flat connections). In other words, we need information about CS theory which isn't contained merely in its associated bordism representation (= mathematical notion of TQFT) $Z : Bord \rightarrow Vect$.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Bruce Bartlett
answered Mar 17, 2014 by (40 points)
+ 1 like - 0 dislike
The CS/WZW correspondence is of a slightly different nature than the usual AdS/CFT correspondences. First, calling the first one CS/WZW correspondence is misleading, because the "CFT"-dual is not really the WZW model, but as you pointed out, only its "chiral" part, i.e. its chiral conformal blocks. Both are instances of holography, but the real interest of the usual AdS/CFT correspondences is that the theory dual to the CFT contains gravity. Equivalently, the CFT is an honest CFT including a stress-energy tensor (which sources the graviton in the bulk), unlike the CS/WZW correspondence.
There is one setting in which they can be compared instructively. This is the case when the gauge group of the Chern-Simons theory is $SL(2,R) \times SL(2,R)$. In this case, modulo subtleties, the Chern-Simons theory is equivalent to 3d general relativity (see for instance this paper by Witten). It turns out that if you want to obtain an asymptotically AdS spacetime, you have to impose stronger boundary conditions that you would normally impose on a Chern-Simons theory. This is explained for instance clearly in Section 4 of this paper. The effects of these boundary conditions on the asymptotic symmetry algebra is to reduce the $sl(2,R) \oplus sl(2,R)$ Kac-Moody symmetry to the Virasoro symmetry (some kind of Drinfeld-Sokolov reduction).
The CFT dual to pure 3d gravity is not known (see for instance this paper and follow ups), but there have been recent conjectures (and extensive checks of these) about the AdS duals of rather simple and exactly solvable CFTs, including minimal models.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Samuel Monnier
answered Mar 26, 2014 by (60 points)
The word "CS/WZW correspondence" seems quite appropriate given that locally the field/correlator correspondence is that of holography and globally the non-chiral WZW part is precisely encoded by the CS-theory, as for instance manifestly so in the FRS theorem (ncatlab.org/nlab/show/FRS-theorem+on+rational+2d+CFT). (Of course with two-sided boundary.) If anything deserves the word "correspondence", then this should be an example. The word "duality" on the other hand, this seems to better be avoided, both for CS/WZW as well as, I would think, for genuine AdS/CFT.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Urs Schreiber
The relation between 3d CS and 3d gravity is at least subtle. Notice that Witten essentially retracted his 1988 claim to have quantized 3d gravity in 2007 in "Three-Dimensional Gravity Revisited" (arxiv.org/abs/0706.3359). ncatlab.org/nlab/show/…
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Urs Schreiber
I agree with your point about the FRS theorem. But we're now drifting even further from the genuine AdS/CFT story. Concerning the use of the word "duality", it is used in physics when two theories with different formulations turn out to be identical (in the physical sense: there is a bijection of observables, etc...). I feel it is appropriate for genuine AdS/CFT. Concerning your second comment, the discussion about the change in boundary conditions required to obtain AdS in 3d gravity requires only the classical theory, which is still subtle, but understood.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Samuel Monnier
Actually, you should maybe post your comment about the FRS theorem as an answer, it looks like it fully addresses the concerns of the OP about the CS/WZW correspondence.
This post imported from StackExchange MathOverflow at 2014-12-06 09:45 (UTC), posted by SE-user Samuel Monnier
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http://library.kiwix.org/ell.stackexchange.com_en_all_2020-11/A/question/34233.html | ## different usages of et al?
0
History's London is closely bound up with England's history.
History's London is closely bound up with that of England.
History's London is closely bound up with the history of England's.
Our findings are different from the findings of John et al's (2014).
Our findings are different from John et al's findings (2014).
Our findings are different from those of John et al's (2014).
Our findings are different from those of John et als' (2014)
Would you please tell me if these are correct? if not, could you correct me? and which one do you use in academic style?
Many thanks
1John et al. means John and others so it is already plural - there is no need to add any further pluralising characters. Notice the punctuation of et al. the period is after al not between at and al. – Frank – 2014-09-25T06:00:02.497
It is also common to have a space between et and al. – painfulenglish – 2014-09-25T07:27:09.600
so, merely is the last one incorrect? and what about my original question? – nima – 2014-09-25T08:22:30.847
I think you mean London's history, not history's London. – snailplane – 2014-09-25T12:31:14.743
There is no dot between "et" and "al". There is a dot following "al" because it is an abbreviation. (Of "alias" in its declined form "alii".) Also, parentheses snug up against their contents and always have a leading and trailing spaces between them and the rest of their containing sentences. – Codeswitcher – 2015-02-27T05:05:31.510
1
By "history's London" do you mean the history of London? If so, the correct expression is "The history of London" or "London's history".
Correct forms would be:
London's history is closely bound up with England's history.
London's history is closely bound up with that of England.
London's history is closely bound up with the history of England.
"Of" makes it possessive, so you don't also add "'s" to "England".
Our findings are different from the findings of John et al (2014).
or
Our findings are different from the findings of John et al. (2014).
"et al" is short for the Latin "et alii". So "et" of itself is not an abbreviation and does not call for a period. thefreedictionary.com says to put a period after "al". I think I usually see it with no periods at all.
Our findings are different from John et al.'s (2014) findings.
Typically we put the publication date immediately after the title or author, not at the end of the sentence. I suppose this could depend on the style guide that you are using.
Our findings are different from those of John et al. (2014).
Like an earlier one, the "of" indicates possession, so you do not also use an "'s". Or if you were thinking you need an "s" to make it plural, "et al" means "and others" and so is already plural. While most English plurals end with an "s", you shouldn't add one when they don't.
Our findings are different from those of John et al. (2014).
Ends up the same as the previous.
So let's see, what general rules came out of that?
The history of London is "London's history", not "history's London".
To make a possessive, you can use "of" or you can use "'s". Don't use both. "England's history" or "history of England", not "history of England's". To be technical, perhaps I should add that that last could be correct if you were using "history of England" as the "owner" of something. Like, "The history of England's importance can be seen by ..." Here "England" modifies "history", but then "history of England" as a unit modifies "importance". We could also write that, "The importance of the history of England can be seen by ..."
Technicality: It's "et al.", not "et.al". "Al" is the abbreviation; "et" is a complete word.
Because et alii and et al. are in a foreign language, it is best to italicize them (if possible) or underline them (if italicizing is not possible). Et alii can be translated as "among others". – Jasper – 2014-09-25T18:41:32.943 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8608943819999695, "perplexity": 2729.5112939294027}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351374.10/warc/CC-MAIN-20210225153633-20210225183633-00304.warc.gz"} |
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