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Papadopoulos, P., Matiadou, N., Pappas, A. (2015). Global existence, stability results and compact invariant sets for a quasilinear nonlocal wave equation on $mathbb{R}^{N}$. International Journal of Nonlinear Analysis and Applications, 6(1), 85-95. doi: 10.22075/ijnaa.2015.220
P. Papadopoulos; N.L. Matiadou; A. Pappas. "Global existence, stability results and compact invariant sets for a quasilinear nonlocal wave equation on $mathbb{R}^{N}$". International Journal of Nonlinear Analysis and Applications, 6, 1, 2015, 85-95. doi: 10.22075/ijnaa.2015.220
Papadopoulos, P., Matiadou, N., Pappas, A. (2015). 'Global existence, stability results and compact invariant sets for a quasilinear nonlocal wave equation on $mathbb{R}^{N}$', International Journal of Nonlinear Analysis and Applications, 6(1), pp. 85-95. doi: 10.22075/ijnaa.2015.220
Papadopoulos, P., Matiadou, N., Pappas, A. Global existence, stability results and compact invariant sets for a quasilinear nonlocal wave equation on $mathbb{R}^{N}$. International Journal of Nonlinear Analysis and Applications, 2015; 6(1): 85-95. doi: 10.22075/ijnaa.2015.220
Global existence, stability results and compact invariant sets for a quasilinear nonlocal wave equation on $mathbb{R}^{N}$
1adepartment of electronics engineering, school of technological applications, technological educational institution (tei) of piraeus, gr 11244, egaleo, athens, greece
2Department of Electronics Engineering, School of Technological Applications, Technological Educational Institution (TEI) of Piraeus, GR 11244, Egaleo, Athens, Greece
3Civil Engineering Department, School of Technological Applications, Technological Educational Institution (TEI) of Piraeus, GR 11244, Egaleo, Athens, Greece.
Abstract
We discuss the asymptotic behaviour of solutions for the nonlocal quasilinear hyperbolic problem of Kirchhoff Type \[ u_{tt}-\phi (x)||\nabla u(t)||^{2}\Delta u+\delta u_{t}=|u|^{a}u,\, x \in \mathbb{R}^{N} ,\,t\geq 0\;,\]
with initial conditions $u(x,0) = u_0 (x)$ and $u_t(x,0) = u_1 (x)$, in the case where $N \geq 3, \; \delta \geq 0$ and $(\phi (x))^{-1} =g (x)$ is a positive function lying in $L^{N/2}(\mathbb{R}^{N})\cap L^{\infty}(\mathbb{R}^{N})$. It is proved that, when the initial energy \ $ E(u_{0},u_{1})$, which corresponds to the problem, is non-negative and small, there exists a unique global solution in time in the space \ ${\cal{X}}_{0}=:D(A) \times {\cal{D}}^{1,2}(\mathbb{R}^{N})$. When the initial energy $E(u_{0},u_{1})$ is negative, the solution blows-up in finite time. For the proofs, a combination of the modified potential well method and the concavity method is used. Also, the existence of an absorbing set in the space ${\cal{X}}_{1}=:{\cal{D}}^{1,2}(\mathbb{R}^{N}) \times L^{2}_{g}(\mathbb{R}^{N})$ is proved and that the dynamical system generated by the problem possess an invariant compact set ${\cal {A}}$ in the same space.
Finally, for the generalized dissipative Kirchhoff's String problem \[ u_{tt}=-||A^{1/2}u||^{2}_{H} Au-\delta Au_{t}+f(u) ,\; \; x \in \mathbb{R}^{N}, \;\; t \geq 0\;,\] with the same hypotheses as above, we study the stability of the trivial solution $u\equiv 0$. It is proved that if $f'(0)>0$, then the solution is unstable for the initial Kirchhoff's system, while if $f'(0)<0$ the solution is asymptotically stable. In the critical case, where $f'(0)=0$, the stability is studied by means of the central manifold theory. To do this study we go through a transformation of variables similar to the one introduced by R. Pego.
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Doping methods available in QuantumATK¶
The effect of electrostatic doping is generally modelled by either increasing (n-doping) or decreasing (p-doping) the number of electrons of the system with respect to the neutral case. QuantumATK offers two general methods to introduce electrostatic doping. Both lead to the same additional electrons (holes) introduced in the system, but differ in the way these extra electrons (holes) are redistributed within it.
General background¶
In QuantumATK the ElectronDensity of the system is expressed as a sum of two contributions:
\(\Delta \rho (\mathbf{r})\) is the ElectronDifferenceDensity, which for a neutral system integrates to zero:
\(\sum_i^{N_{atoms}} \rho_i^{atom} (\mathbf{r})\) is the sum of the densities associated with the individual atoms. For a neutral system, this sum integrates to the total number of electrons \(N\):
It follows that the number of electrons in a generic system with charge \(Q\), which is the integral over the electron density in real space, can be also expressed as:
where \(\tilde{N}\) are the extra electrons leading to a charge \(Q = -e \cdot \tilde{N}\). It follows that
In QuantumATK, the system can be doped in two different ways, either by adding an Explicit charge or by adding Atomic compensation charges. The former modifies the value of \(\tilde{N}\), whereas the latter modifies the value of \(N\). However, both lead to the following result:
Explicit charge¶
An explicit charge can be added by setting the
charge parameter to avalue \(Q\) in the LCAOCalculator. In this case, \(\tilde{N}\)is set to a non-zero value \(\tilde{M} = -Q/e\). The total number ofelectrons in the system is then determined by
Note
There are two important consequences of using this method:
The densities of the individual atoms \(\rho_i^{atom} (\mathbf{r})\) are those associated with the individual neutral species; The extra charge is not bound to any atom. Therefore, it will redistribute in order to minimize the total energy of the system. Atomic compensation charges¶
Using the AtomicCompensationCharge method, \(\tilde{N}\) is set tozero, but the density of the
i-th atom is rescaled by a factor:
where \(n_i\) and \(\tilde{n_i}\) are the number of valence electronsand the extra compensation charge associated with the
i-th isolated atom.The individual values of \(\tilde{n}_i\), and therefore the individualcoefficients \(c_i\), can be set arbitrarily from atom to atom, even withinthe same chemical species. For sake of comparison, we assume here that they areall the same and that they sum up to \(\tilde{M}\):
In this case, the integral of the total density can be expressed as a sum over the individual densities
where
Due to the use of the rescaled atomic densities \(\tilde{\rho}_i^{atom} (\mathbf{r})\), the electronic density of the system will be the same as that obtained using an explicit doping:
Note
Using this method, the extra charge is introduced by modifying the densities of the individual atoms. As a consequence:
The extra electron density will be bound to those atoms for which \(n_i \neq 0\); The electrostatic potential will be locally modified in the vicinity of such charged atoms, becoming more attractive (repulsive) if \(n_i > 0\) (\(n_i < 0\)).
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The population Variance Formula in mathematics is sigma squared equals the sum of x minus the mean squared divided by n.
\[\ \huge \sigma^{2}=\frac{\sum\left(x-\overline{x}\right)^{2}}{n}\]
The variance of a sample is defined by slightly different formula:
\[\ \huge S^{2}=\frac{\sum\left(x-\overline{x}\right)^{2}}{n-1}\]
Where,
\[\ \sigma^{2}$= Variance\] x = Item given in the data \[\ \overline{x} = Mean of the data\] n = Total number of items. \[\ s^{2} = Sample variance\]
Let us understand the concept of population variance in detail below. Take an example, where one teacher needs to find out the average speed for the students taking in reading the comprehension pages. Each student is assigned six pages in the classrooms to read.
A teacher could find the average time here but this information is not just the right technique to analyze the clear picture of her students. Few of them read content very fast while others are just the average. It gives the average taken to read the total content not the actual details about each of the students.
So, you need to find the sample variance of the collected data here. Variance in simple words could be defined as the how far a set of numbers are spread out. This is actually very different from calculating the average or mean of data from a set of number. The variance formula is already given at the top for your reference.
You just need to plugin values in the formula and calculated the needed output. When there is equal difference between first, second, and the third term then find the average or mean here. In case, the difference between terms is larger then you should find the variance for a particular set of data.
The two most common categories of Variance include – Population Variance and the Sample Variance. Population is defined as the total number of members in a particular group while sample is just a part of the population that is used to describe the whole group.
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LHCb Collaboration; Aaij, R; Adeva, B; Adinolfi, M; Anderson, J; Bernet, R; Bowen, E; Bursche, A; Chiapolini, N; Chrzaszcz, M; Dey, B; Elsasser, C; Graverini, E; Lionetto, F; Lowdon, P; Mauri, A; Müller, K; Serra, N; Steinkamp, O; Storaci, B; Straumann, U; Tresch, M; Vollhardt, A; Weiden, A; et al (2015).
Measurement of forward $J/\psi$ production cross-sections in $pp$ collisions at $\sqrt{s}=13$ TeV. Journal of High Energy Physics, 2015(10):172. Abstract
The production of $J/\psi$ mesons in proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=13$ TeV is studied with the LHCb detector. Cross-section measurements are performed as a function of the transverse momentum $p_\mathrm{T}$ and the rapidity $y$ of the $J/\psi$ meson in the region $p_\mathrm{T} < 14 \mathrm{GeV}/c$ and $2.0 < y < 4.5$, for both prompt $J/\psi$ mesons and $J/\psi$ mesons from $b$-hadron decays. The production cross-sections integrated over the kinematic coverage are $15.30\pm 0.03\pm 0.86$ $\mu$b for prompt $J/\psi$ and $2.34\pm 0.01\pm 0.13$ $\mu$b for $J/\psi$ from $b$-hadron decays, assuming zero polarization of the $J/\psi$ meson. The first uncertainties are statistical and the second systematic. The cross-section reported for $J/\psi$ mesons from $b$-hadron decays is used to extrapolate to a total $b\bar{b}$ cross-section. The ratios of the cross-sections with respect to $\sqrt{s}=8$ TeV are also determined.
Abstract
The production of $J/\psi$ mesons in proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=13$ TeV is studied with the LHCb detector. Cross-section measurements are performed as a function of the transverse momentum $p_\mathrm{T}$ and the rapidity $y$ of the $J/\psi$ meson in the region $p_\mathrm{T} < 14 \mathrm{GeV}/c$ and $2.0 < y < 4.5$, for both prompt $J/\psi$ mesons and $J/\psi$ mesons from $b$-hadron decays. The production cross-sections integrated over the kinematic coverage are $15.30\pm 0.03\pm 0.86$ $\mu$b for prompt $J/\psi$ and $2.34\pm 0.01\pm 0.13$ $\mu$b for $J/\psi$ from $b$-hadron decays, assuming zero polarization of the $J/\psi$ meson. The first uncertainties are statistical and the second systematic. The cross-section reported for $J/\psi$ mesons from $b$-hadron decays is used to extrapolate to a total $b\bar{b}$ cross-section. The ratios of the cross-sections with respect to $\sqrt{s}=8$ TeV are also determined.
Additional indexing
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A conformal transformation is one which alters the metric up to a factor, i.e.
$$g_{\mu\nu}(x)\to\Omega^2(x)g_{\mu\nu}(x)$$
A field theory described by a Lagrangian invariant up to a total derivative under a conformal transformation is said to be a
conformal field theory. These transformations include Scaling or dilations $x^\mu \to \lambda x^\mu$ Rotations $x^\mu \to M^\mu_\nu x^\nu$ Translations $x^\mu \to x^\mu + c^\mu$
In addition to these, the conformal group includes a set of special conformal transformations given by,
$$x^\mu \to \frac{x^\mu-b^\mu x^2}{1-2b \cdot x + b^2 x^2}$$
If you compute the generators of the conformal transformations, and the algebra they satisfy, with some manipulation it may be shown there is an isomorphism between the conformal group in $d$ dimensions and the group $SO(d+1,1)$. In two dimensions, the conformal group is rather special; it is simply the group of all analytic maps; this set is infinite-dimensional since one requires an infinite number of parameters to specify all functions analytic in some neighborhood. The global variety of conformal transformations, i.e. those which are not functions of the coordinates but constants, in $d=2$ are equivalent to $SL(2,\mathbb{C})$.
On the other hand, a topological field theory is one which is invariant under all transformations which do not alter the topology of spacetime, e.g. they may not puncture it and increase the genus. The correlation functions do not depend on the metric, and are in fact topological invariants.
Hence, a topological field theory is invariant under conformal transformations by the fact that it does not even depend on the metric. However, not all conformal field theories are topological field theories.This post imported from StackExchange Physics at 2015-03-04 16:11 (UTC), posted by SE-user JamalS
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Consider a $C^k$, $k\ge 2$, Lorentzian manifold $(M,g)$ and let $\Box$ be the usual wave operator $\nabla^a\nabla_a$. Given $p\in M$, $s\in\Bbb R,$ and $v\in T_pM$, can we find a neighborhood $U$ of $p$ and $u\in C^k(U)$ such that $\Box u=0$, $u(p)=s$ and $\mathrm{grad}\, u(p)=v$?
The tog is a measure of thermal resistance of a unit area, also known as thermal insulance. It is commonly used in the textile industry and often seen quoted on, for example, duvets and carpet underlay.The Shirley Institute in Manchester, England developed the tog as an easy-to-follow alternative to the SI unit of m2K/W. The name comes from the informal word "togs" for clothing which itself was probably derived from the word toga, a Roman garment.The basic unit of insulation coefficient is the RSI, (1 m2K/W). 1 tog = 0.1 RSI. There is also a clo clothing unit equivalent to 0.155 RSI or 1.55 tog...
The stone or stone weight (abbreviation: st.) is an English and imperial unit of mass now equal to 14 pounds (6.35029318 kg).England and other Germanic-speaking countries of northern Europe formerly used various standardised "stones" for trade, with their values ranging from about 5 to 40 local pounds (roughly 3 to 15 kg) depending on the location and objects weighed. The United Kingdom's imperial system adopted the wool stone of 14 pounds in 1835. With the advent of metrication, Europe's various "stones" were superseded by or adapted to the kilogram from the mid-19th century on. The stone continues...
Can you tell me why this question deserves to be negative?I tried to find faults and I couldn't: I did some research, I did all the calculations I could, and I think it is clear enough . I had deleted it and was going to abandon the site but then I decided to learn what is wrong and see if I ca...
I am a bit confused in classical physics's angular momentum. For a orbital motion of a point mass: if we pick a new coordinate (that doesn't move w.r.t. the old coordinate), angular momentum should be still conserved, right? (I calculated a quite absurd result - it is no longer conserved (an additional term that varies with time )
in new coordinnate: $\vec {L'}=\vec{r'} \times \vec{p'}$
$=(\vec{R}+\vec{r}) \times \vec{p}$
$=\vec{R} \times \vec{p} + \vec L$
where the 1st term varies with time. (where R is the shift of coordinate, since R is constant, and p sort of rotating.)
would anyone kind enough to shed some light on this for me?
From what we discussed, your literary taste seems to be classical/conventional in nature. That book is inherently unconventional in nature; it's not supposed to be read as a novel, it's supposed to be read as an encyclopedia
@BalarkaSen Dare I say it, my literary taste continues to change as I have kept on reading :-)
One book that I finished reading today, The Sense of An Ending (different from the movie with the same title) is far from anything I would've been able to read, even, two years ago, but I absolutely loved it.
I've just started watching the Fall - it seems good so far (after 1 episode)... I'm with @JohnRennie on the Sherlock Holmes books and would add that the most recent TV episodes were appalling. I've been told to read Agatha Christy but haven't got round to it yet
?Is it possible to make a time machine ever? Please give an easy answer,a simple one A simple answer, but a possibly wrong one, is to say that a time machine is not possible. Currently, we don't have either the technology to build one, nor a definite, proven (or generally accepted) idea of how we could build one. — Countto1047 secs ago
@vzn if it's a romantic novel, which it looks like, it's probably not for me - I'm getting to be more and more fussy about books and have a ridiculously long list to read as it is. I'm going to counter that one by suggesting Ann Leckie's Ancillary Justice series
Although if you like epic fantasy, Malazan book of the Fallen is fantastic
@Mithrandir24601 lol it has some love story but its written by a guy so cant be a romantic novel... besides what decent stories dont involve love interests anyway :P ... was just reading his blog, they are gonna do a movie of one of his books with kate winslet, cant beat that right? :P variety.com/2016/film/news/…
@vzn "he falls in love with Daley Cross, an angelic voice in need of a song." I think that counts :P It's not that I don't like it, it's just that authors very rarely do anywhere near a decent job of it. If it's a major part of the plot, it's often either eyeroll worthy and cringy or boring and predictable with OK writing. A notable exception is Stephen Erikson
@vzn depends exactly what you mean by 'love story component', but often yeah... It's not always so bad in sci-fi and fantasy where it's not in the focus so much and just evolves in a reasonable, if predictable way with the storyline, although it depends on what you read (e.g. Brent Weeks, Brandon Sanderson). Of course Patrick Rothfuss completely inverts this trope :) and Lev Grossman is a study on how to do character development and totally destroys typical romance plots
@Slereah The idea is to pick some spacelike hypersurface $\Sigma$ containing $p$. Now specifying $u(p)$ is trivial because the wave equation is invariant under constant perturbations. So that's whatever. But I can specify $\nabla u(p)|\Sigma$ by specifying $u(\cdot, 0)$ and differentiate along the surface. For the Cauchy theorems I can also specify $u_t(\cdot,0)$.
Now take the neigborhood to be $\approx (-\epsilon,\epsilon)\times\Sigma$ and then split the metric like $-dt^2+h$
Do forwards and backwards Cauchy solutions, then check that the derivatives match on the interface $\{0\}\times\Sigma$
Why is it that you can only cool down a substance so far before the energy goes into changing it's state? I assume it has something to do with the distance between molecules meaning that intermolecular interactions have less energy in them than making the distance between them even smaller, but why does it create these bonds instead of making the distance smaller / just reducing the temperature more?
Thanks @CooperCape but this leads me another question I forgot ages ago
If you have an electron cloud, is the electric field from that electron just some sort of averaged field from some centre of amplitude or is it a superposition of fields each coming from some point in the cloud?
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This problem is taken from Linear Algebra by Fueng Zhang. R is real numbers, C is complex numbers and Q is rational numbers.
Let $V = \{(x,y) \mid x,y \in \mathbb{C}\}$. Under the standard addition and scalar multiplication for ordered pairs of complex numbers, is $V$ a vector space over $\mathbb{C}$? Over $\mathbb{R}$? Over $\mathbb{Q}$ If so, find the dimension of $V$.
The answer key says:
Yes, over $\mathbb{C}, \mathbb{R}$ and $\mathbb{Q}$. The dimensions are $2,4,\infty$, respectively.
I do understand that the dimension over $\mathbb{C}$ is 2 and over $\mathbb{R}$ is 4, but how come the dimension over $\mathbb{Q}$ is infinite?
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A conformal transformation is one which alters the metric up to a factor, i.e.
$$g_{\mu\nu}(x)\to\Omega^2(x)g_{\mu\nu}(x)$$
A field theory described by a Lagrangian invariant up to a total derivative under a conformal transformation is said to be a
conformal field theory. These transformations include Scaling or dilations $x^\mu \to \lambda x^\mu$ Rotations $x^\mu \to M^\mu_\nu x^\nu$ Translations $x^\mu \to x^\mu + c^\mu$
In addition to these, the conformal group includes a set of special conformal transformations given by,
$$x^\mu \to \frac{x^\mu-b^\mu x^2}{1-2b \cdot x + b^2 x^2}$$
If you compute the generators of the conformal transformations, and the algebra they satisfy, with some manipulation it may be shown there is an isomorphism between the conformal group in $d$ dimensions and the group $SO(d+1,1)$. In two dimensions, the conformal group is rather special; it is simply the group of all analytic maps; this set is infinite-dimensional since one requires an infinite number of parameters to specify all functions analytic in some neighborhood. The global variety of conformal transformations, i.e. those which are not functions of the coordinates but constants, in $d=2$ are equivalent to $SL(2,\mathbb{C})$.
On the other hand, a topological field theory is one which is invariant under all transformations which do not alter the topology of spacetime, e.g. they may not puncture it and increase the genus. The correlation functions do not depend on the metric, and are in fact topological invariants.
Hence, a topological field theory is invariant under conformal transformations by the fact that it does not even depend on the metric. However, not all conformal field theories are topological field theories.This post imported from StackExchange Physics at 2015-03-04 16:11 (UTC), posted by SE-user JamalS
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calculateLCAOBasisSpilling¶
calculateLCAOBasisSpilling(
configuration, k_point_sampling=None, basis=None, bands_above_fermi_level=<class 'NL.ComputerScienceUtilities.NLFlag._NLFlag.All'>, processes_per_kpoint=None)¶
Method for calculating the spilling factors per band (and spin) for a given configuration with an attached plane wave calculator and a given LCAO basis.
param configuration: The configuration with an attached plane wave calculator. type configuration:
BulkConfiguration
param k_point_sampling: The k-point sampling to use in the sum over k. By default, the k-points of the ground state configuration are used. The eigensystem is solved again, if different k-points are requested. Default:None. type k_point_sampling: None |
MonkhorstPackGrid
param basis: The LCAO basis to project onto. It must contain basis functions for the elements in the configuration. If it is set to None, the LCAO basis attached to the calculator of the configuration is used. Default:None. type basis: None |
BasisSet
param bands_above_fermi_level: The number of bands above the Fermi level to include in the sum. By default all unoccupied bands present in the ground state calculation are used. The eigensystem is solved again, if more unoccupied bands are requested. Default:
All
type bands_above_fermi_level:
All| int
param processes_per_kpoint: The number of processes used for one k-point. Default:None. type processes_per_kpoint: None | int |
Automatic
returns: The spilling factors per band as a numpy array of floats. In the case of a spin-polarized calculation, the shape is (2, N), otherwise its shape is (1, N). rtype: numpy.ndarray Usage Examples¶
Calculate the average spilling factor for a Medium and High silicon basis set.
# Set up latticelattice = FaceCenteredCubic(5.4306 * Angstrom)# Define elementselements = [Silicon, Silicon]# Define coordinatesfractional_coordinates = [[ 0. , 0. , 0. ], [ 0.25, 0.25, 0.25]]# Set up configurationbulk_configuration = BulkConfiguration( bravais_lattice=lattice, elements=elements, fractional_coordinates=fractional_coordinates, )k_point_sampling = MonkhorstPackGrid( na=7, nb=7, nc=7, )numerical_accuracy_parameters = NumericalAccuracyParameters( k_point_sampling=k_point_sampling, )# PlaneWaveCalculator with a high wave function cutoff to get a complete plane wave basis.calculator = PlaneWaveCalculator( wave_function_cutoff=50*Hartree, numerical_accuracy_parameters=numerical_accuracy_parameters, )bulk_configuration.setCalculator(calculator)bulk_configuration.update()# For each band get the LCAO basis spilling factor for a Medium basis.# Only include the occupied states.spilling_per_band_medium = calculateLCAOBasisSpilling( bulk_configuration, basis=BasisGGAPseudoDojo.Medium, bands_above_fermi_level=0,)# Total averaged spilling.spilling_medium = spilling_per_band_medium.sum() / spilling_per_band_medium.size# For each band get the LCAO basis spilling factor for a High basis.# Only include the occupied states.spilling_per_band_high = calculateLCAOBasisSpilling( bulk_configuration, basis=BasisGGAPseudoDojo.High, bands_above_fermi_level=0,)# Total averaged spilling.spilling_high = spilling_per_band_high.sum() / spilling_per_band_high.sizeprint('Spilling per band, Medium (%) : ', spilling_per_band_medium * 100)print('Spilling per band, High (%) : ', spilling_per_band_high * 100)print('')# Print the average spilling factorsprint('Average spilling, Medium (%) : ', spilling_medium * 100)print('Average spilling, High (%) : ', spilling_high * 100)
The final output of the script is:
Spilling per band, Medium (%) : [[ 0.07159806 0.04691684 0.04034596 0.04223707]]Spilling per band, High (%) : [[ 0.03255014 0.01255693 0.02256928 0.0250443 ]]Average spilling, Medium (%) : 0.0502744832443Average spilling, High (%) : 0.0231801630042
showing that the Medium basis set has a spilling factor of 0.050 % while the High basis has a smaller spilling of 0.023 %. The printed spillings per band show that the lowest band (first index) has the largest spilling factor for both the Medium and High basis sets.
Notes¶
Following [SPAS95], we define the spilling factor for band \(i\) as
\[S_i = \frac{1}{N_k} \sum_k^{N_k} w_k \langle \psi_i(k) | 1 - P(k) |\psi_i(k)\rangle,\]
where \(w_k\) is the weight of k-point \(k\) and the projection operator \(P\) is defined as
\[P(k) = \sum_{\mu\nu} |\phi_\mu(k)\rangle (O(k)^{-1})_{\mu\nu}\langle \phi_\nu(k)|\]
with \(|\phi_\mu(k)\rangle\) being the Fourier transform of the basis orbital \(\mu\) and \(O\) being the basis orbital overlap matrix defined as
\[O_{\mu\nu}(k)=\langle \phi_\mu(k)|\phi_\nu(k)\rangle.\]
The total spilling factor is given by the average over the considered bands, i.e.
\[S = \frac{1}{N_b}\sum_i^{N_b}S_i.\]
The function
calculateLCAOBasisSpilling can be used to calculate the completeness of aa particular basis set. The smaller the spilling factor, the more complete is the basis set.
It also allows for a per band analysis since
calculateLCAOBasisSpilling returns thespilling factor for each band. The number of unoccupied bands included in the calculationcan be set in the function call as:
spilling_per_band_high = calculateLCAOBasisSpilling( configuration, basis=BasisGGAPseudoDojo.Medium, bands_above_fermi_level=5,)
where five unoccupied bands above the Fermi level would be included.
References¶
[SPAS95] D. Sanchez-Portal, E. Artacho, and J. M. Soler. Projection of plane-wave calculations into atomic orbitals. Solid State Communications, 95(10):685 – 690, 1995. URL: http://www.sciencedirect.com/science/article/pii/003810989500341X, doi:https://doi.org/10.1016/0038-1098(95)00341-X.
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Few weeks ago, a Reddit user
xieta posted really neat results of flow past a cylinder to
/r/CFD. These results were computed using the
vorticity – stream function method. My background in CFD is quite limited, and to be honest, this was the first time I’ve heard of this method. Turns out it’s a rather popular method for 2D incompressible flows [1].
In 2014, I tried to write my first CFD solver to model purge effectiveness in a cylindrical cavity. Preliminary results from that effort were presented at the SPIE Optics & Photonics conference. That simulation was done using the projection method, and for whatever reason, I just could not get the outflow boundary to behave correctly. This Reddit post had me research the vorticity method as a possible alternative. I found many articles and example codes online, unfortunately, they all seem to be developed for the Cartesian (x-y) coordinates. I was interested in modeling axisymmetric flow, and thus needed formulation in the cylindrical (r-z) system. This article outlines the method I ended up with. Keep in mind that I am in no way a CFD expert. There may be mistakes, and there are likely better ways to treat external boundaries.
I am partly posting this article to generate feedback and garner community review, so please leave a comment if something is not right. Formulation Velocity Components
To keep the number of equations to minimum, details of the formulation were moved to a separate pdf. But in brief, for two-dimensional incompressible flows it is possible to express velocity components as derivatives of a scalar “stream” function \(\psi\) such the continuity equation is automatically satisfied. Salih [2] does great job deriving the formulation for a Cartesian flow in which \(u=\partial_y\psi\) and \(v=-\partial_x\psi\). For an axisymmetric flow, we instead use the Stokes stream function,
$$u_z=u=\frac{1}{r}\frac{\partial \psi}{\partial r}\\ u_r=v=-\frac{1}{r}\frac{\partial \psi}{\partial z}$$ Vorticity
Vorticity is defined as \(\omega=\nabla \times \vec{v}\). For axisymmetric flow any \(\partial_\theta=0\) and \(u_\theta\), if present, is independent of the other velocity components. Then,
$$\omega=\omega_\theta=\frac{\partial v}{\partial z} – \frac{\partial u}{\partial r}$$ Stream Function Governing Equation
The definition of velocity components in terms of the stream function can then be substituted into the vorticity equation to obtain
$$ \frac{\partial^2\psi}{\partial z^2} + \frac{\partial^2\psi}{\partial r^2} – \frac{1}{r}\frac{\partial \psi}{\partial r} = -\omega r$$ Note that although this looks like the Poisson’s equation, the sign on the last term on the left hand side is different. Vorticity Transport Equation
What ties this method to the Navier Stokes equations is the vorticity transport equation. This equation is derived by taking curl of the momentum equation as demonstrated in [2]. For axisymmetric flow it is given by
$$\frac{\partial \omega}{\partial t} + u\frac{\partial w}{\partial z} + v\frac{\partial w}{\partial r} = \nu\left[ \frac{\partial^2\omega}{\partial z^2} + \frac{\partial^2\omega}{\partial r^2} + \frac{1}{r}\frac{\partial \omega}{\partial r} \right]$$ Finite Difference Form
The stream function and vorticity equations can be solved using the finite difference method. The stream function equation is discretized using the standard central difference, and can be solved using an iterative elliptic solver, such as Jacobi or Gauss-Seidel. The vorticity equation is a PDE that is marched forward in time. The obvious method is using forward time integration and central space differencing (FTCS). The vorticity transport equation is basically advection-diffusion equation. It can be shown (see [1] or other CFD texts) that the FTCS method is unconditionally unstable for
advection equation (inviscid flow). Adding diffusion allows the use of FTCS, but still a restrictive condition on time step remains. I ended up using the 4th order Runge Kutta (RK4) method here. The transport equation is rewritten as \(\partial_t\omega=R(\omega)\) and then three intermediate values of \(\omega\) are computed. Take a look at the pdf for detail. Boundary Conditions
To demonstrate the method, let’s apply it to a flow in cylindrical cavity. The cavity, shown in Figure 1 below, contains inlet on one side. The other side is open to the ambient environment. This setup may look like a simplified version of the SPIE purge problem since that was the intention! We can see that we have five types of boundary conditions to consider. Here I describe them only at top level, again take a look at the pdf for details.
Inlet: here we assume that flow is parallel with the z axis, giving us \(\partial_z\psi=0\) Axis of revolution: there can be now flow across the axis and also \(\partial_r\)=0 so we set \(\psi=0\) Solid walls: no slip condition gives us \(\psi=\psi_{wall}\) Dirichlet boundary on walls, with value set from prescribed inlet flow, since \(Q=2\pi\psi\) Open boundary on zmax: Here \(\psi\) is set from the definition of velocity and \(\partial_z\omega=0\) is used on vorticity Open boundary on rmax: I am not sure what is the appropriate condition here but setting \(u=0\) and \(\partial_r\omega=0\) gives reasonable results Demo Code in Python
I did most of the initial development in Python. The source code is linked at the bottom of the article. Let’s take a look at the main algorithm. As you surely already noticed, the actual program has additional code here for setting simulation parameters and doing screen output.
#generate geometry node_type = makeGeometry() #set streamfunction boundary conditions psi = initPsi() #set initial values to zeros w = np.zeros_like(psi) u = np.zeros_like(psi) v = np.zeros_like(psi) #iterate print ("Starting main loop") for it in range (1001): #solve psi psi = computePsi(w,psi,u,v) #update u and v u,v = computeVel(psi) #advance w w = advanceRK4(w,psi,u,v) if (it%100==0): flux,u_ave = make_plot(it) print ("Done!")
The first function
makeGeometry flags node types as wall, inlet and so on. Then based on these, stream function boundary conditions are set in
initPsi. I am setting initial vorticity to zero since the simulation begins with zero flow. The simulation then dives into the main loop.
computePsi solves the “Poisson-like” stream function equation using the Jacobi method. Velocity components are then computed in the entire domain by differencing
psi. Vorticity transport equation is then marched forward using the RK4 scheme. Finally every 100 iterations, the screen output is updated (this only works if you run the code in Python console).
Here is the RK4 vorticity marcher:
#advances vorticity equation using RK4 def advanceRK4(w,psi,u,v): applyVorticityBoundaries(w,psi,u,v) #compute the four terms of RK4 Rk = R(w) w1 = w + 0.5*dt*Rk R1 = R(w1) w2 = w + 0.5*dt*R1 R2 = R(w2) w3 = w + dt*R2 R3 = R(w3) w_new = w + (dt/6.0)*(Rk + 2*R1 + 2*R2 +R3) #return new value return w_new
First, vorticity is set along all boundaries by calling
applyVorticityBoundaries. The
R function is used to evaluate \(R=\nu\nabla^2\omega – \vec{v}\cdot\nabla\omega\):
#computes RHS for vorticity equation def R(w): dz2 = dz*dz dr2 = dr*dr #make copy so we use consistent data r = np.zeros_like(w) for i in range(1,ni-1): for j in range(1,nj-1): if (node_type[i,j]>0): continue #viscous term, d^2w/dz^2+d^2w/dr^2+(1/r)dw/dr A = nu*( (w[i-1][j]-2*w[i][j]+w[i+1][j])/dz2 + (w[i][j-1]-2*w[i][j]+w[i][j+1])/dr2 + (w[i][j+1]-w[i][j-1])/(2*dr*pos_r[j])) #convective term u*dw/dz B = u[i][j]*(w[i+1][j]-w[i-1][j])/(2*dz) #convective term v*dw/dr C = v[i][j]*(w[i][j+1]-w[i][j-1])/(2*dr) r[i][j] = A - B - C return r
Take a look at the source code for the other functions.
Results from Python
Below you will find results from the example Python code after 200 and 1000 time steps. The top plot is the stream function. The bottom half plots flow speed and velocity vectors.
Results from Java purge code
While Python is great for prototyping, I find it too slow for actual simulations with large number of nodes. As such this vorticity – streamfunction method was implemented in the Java purge code. You can see the results below. The flow inside the cavity looks similar (identical?) to the one obtained previously with the projection method but now the external flow also looks right. Success!
Empty Tube
The two videos below show results for an empty tube. They both show the same result, but top plot is on a log scale from 0.002 to 0.16 m/s, while the second one is on linear scale with minimum speed 0.02 m/s.
Tube with Cup and Baffle
Here is the result with the cup and baffle included. Speed is shown on log scale.
Source Code
You can download the source code here: vorticity-rz.py
References
[2] Tannehill, J., Anderson, D., Pletcher, R., Computational Fluid Mechanics and Heat Transfer, Taylor & Francis, 2nd ed., 1997
[2] Salih, A., Streamfunction-Vorticity Formulation, Department of Aerospace Engineering Indian Institute of Space Science and Technology, March 2013
And that’s it. Again, please leave a comment below. I am very much looking for critiques and review of the boundary conditions. I plan on submitting results from the purge code to a journal soon so this way I hope to catch issues with the CFD solver before the reviewers.
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A
set is one of the simplest structures we can define in math. One may intuitively think of a set as a collection of objects (there are formal definitions that are used to define this more rigorously). With only this structure, we can do some things, but not much. If we have two sets, we may consider maps between them. We are often interested in structure preserving maps. At the level of sets, the only structure we really have is the cardinality of the set (i.e. the number of elements in it). So, the structure preserving map at the level of sets is one that preserves cardinality, or a bijection.
We can, however, add more structure to our sets. Suppose we have a set $X$. One of the next most basic structures we can add on top of $X$ is a
topology. A topology is a collection $\mathcal{T}$ of subsets of $X$ that we are treating as "special" in some sense. Really, they are special just because we say that they are. We further require this collection of subsets to satisfy a few properties. $\emptyset$ and $X$ are in $\mathcal{T}$ If $U, V$ are in $\mathcal{T}$, then so is $U \cup V$. If a finite collection of sets $U_i$ are in $\mathcal{T}$, then so is $\bigcap_i U_i$
That's all that we need to define open sets. So a subset of $\mathbb{R}$ may be open if it is in some $\mathcal{T}$ , as long as the chosen $\mathcal{T}$ satisfies the above three properties. We are still interested in structure preserving maps, but now we have more structure. We may take our old sets and give them each a topology. Then, we may take our old bijection and add a further requirement that if we give the map one of our 'special' (open) sets in the domain, it gives us one of our special sets in the image. These structure preserving functions are called
continuous functions, and we say that continuous functions map open sets to open sets.
You may be familiar with the notion of open sets from analysis. These sets are really only open in the sense that they are in a certain topology (one of many we could define for $\mathbb{R}$). This topology that gives us the familiar notion of open sets from real analysis is called the
standard topology. Call an open interval a set of the form$$(a, b) = \{ x \in \mathbb{R} \mid a < x < b \}$$
Then the standard topology $\mathcal{T}_{st}$ is defined as the collection of open intervals.
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Let $R$ be an integral domain, and $A , B$ be non trivial ideals of $R$. Then prove that $|A \cap B|>1$.
Let $0 \ne a \in A$, $0 \ne b \in B$; we know such $a, b$ exist by virtue of the assumption that $A$ and $B$ are nontrivial. Then since $A$ and $B$ are ideals of the integral domain $R$, we have $0 \ne ab \in A$ and $0 \ne ab \in B$, whence $ab \in A \cap B$. Furthermore, for any $r \in R$, we have $abr \in A$ and $abr \in B$, so that in fact $abr \in A \cap B$, or $(ab) = abR \subset A \cap B$. Now for nonzero $s \in R$ the map $\phi_s: R \to R$ defined by $\phi_s(c) = sc$ is clearly injective, for if $\phi_s(c_1) = \phi_s(c_2)$ then $sc_1 = sc_2$, so by the cancellation property of integral domains we have $c_1 = c_2$. Also, we see that $\phi_{ab}(R) = abR \subset A \cap B$. Thus $A \cap B$ has a subset, $abR$, with $\mid abR \mid = \mid R \mid$. Since nontrivial integral domains $R$ must have $0 \ne 1$, $\mid R \mid \ge 2$, whence $\mid abR \mid \ge 2$. Since $abR \subset A \cap B$, we have $\mid A \cap B \mid \ge 2$; in fact, we have shown that $\mid A \cap B \mid \ge \mid R \mid$; but since $A \cap B \subset R$, we indeed obtain $\mid A \cap B \mid = \mid R \mid$ as well.
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
We know that $0\in A$ and $0\in B$. Let $a\in A$ and $b\in B$. Then $ab\in A\cap B$. Since $R$ is a integral domain, $ab\neq 0$.
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(38.) $\mathbb{Z} \rightarrow S_3$?
Let $φ(n) = \begin{cases} \mathrm{id} \in S_3 &, \text{for all $n$ even,} \\ \mathrm{transposition} (1,2) &, \text{for all $n$ odd integers.} \end{cases}$
Note that (1, 2) is of order 2, isomorphic to Z2.
(41.) $D_4 \rightarrow S_3$?
View D4 as a group of permutations. Same answer as (43.) underneath to $D_4$, just change $S_4$ to $D_4$.
(43.) $S_4 \rightarrow S_3$?
Viewing D4 as a group of permutations, let $φ(p) = \begin{cases} \mathrm{id} \in S_4 &, \text{for all $p$ even permutations,} \\ (1,2) &, \text{for all $p$ odd permutations.} \end{cases}$ Note that (1, 2) is a subgroup of S3 of order 2, isomorphic to Z2.
(1.) I see the image is $S_3$ every time. However I don't understand why the same homomorphism works in all three questions? What's the connection between them? What's the intuition?
(2.) How do you magically envisage and envision this hard piecewise-defined homomorphism?
(3.) What other homomorphisms work for all three? How many are there?
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Now showing items 1-10 of 33
Reaction dynamics of excited states of helium and magneto-optical trapping of helium metastable atoms
(1993)
The temperature dependence of conversion of $\rm He(2\sp3S\sb1)$ metastable atoms to $\rm He\sb2(a\sp3\Sigma\sbsp{u}{+})$ metastable molecules in the three-body reaction $\rm He(2\sp3S\sb1)+ 2He(1\sp1S\sb0)\to He\sb2(a\s ...
FT-ICR studies of giant carbon fullerenes
(1992)
FT-ICR studies of high mass $\rm (C\sb{>150})$ carbon clusters have brought insight to the controversial structures of carbon fullerenes. Laser vaporization followed by supersonic beam technique produced carbon clusters ...
Lifetime of helium metastable spin-states in a helium discharge
(1965)
The lifetime of a helium 23s1 metastable atom electronic spin-state is measured in helium gas using optical pumping techniques. The metastable atoms are created by an RF electrical discharge. The spin-state lifetime is ...
Surface studies using spin-polarized electron energy loss spectroscopy
(1990)
Spin-polarized electron energy loss spectroscopy (SPEELS) has been used to investigate several paramagnetic surfaces. In this technique, a low energy beam of spin-polarized electrons from a GaAs source is directed at the ...
A method of polarization analysis of electrons from optically pumped He_
(1967)
An experiment in progress to extract a polarized beam of electrons from an optically pumped helium source gas is described. Possible ionization mechanisms in helium gas are outlined and the methods of optical pumping are ...
Probing depths of low energy electrons in metals
(1992)
Spin-polarized electron energy-loss spectroscopy has been used to investigate the probing depth of low energy ($\sim$30 eV) electrons in metals. A beam of spin-polarized electrons is directed at the surface of the sample ...
Studies of electron exchange collisions and polarized electron production in a flowing helium afterglow
(1989)
A flowing helium afterglow apparatus has been used to study thermal-energy electron exchange collisions between spin-polarized electrons and O$\sb2$ or NO molecules. Penning ionization of CO$\sb2$ by spin-polarized ...
Studies of paramagnetic metal surfaces and thin films using spin-sensitive electron spectroscopies
(1991)
Spin-sensitive electron spectroscopies were used to study paramagnetic metal surfaces and thin films. In these experiments, a beam of low-energy spin-polarized electrons from a GaAs source is directed at the target surface, ...
Study of low energy electron inelastic scattering mechanisms using spin sensitive techniques
(1995)
Spin sensitive electron spectroscopies were used to study low energy electron inelastic scattering from metal surfaces and thin films. In these experiments, a beam of spin polarized electrons from a GaAs source is directed ...
Spin polarized metastable deexcitation spectroscopy as a probe of gases absorbed on metal surface
(1992)
Spin polarized metastable deexcitation spectroscopy provides an important surface probe, in which a beam of thermal energy metastable noble gas atoms is deexcited at the target surface under study, releasing its energy ...
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Degenerate elliptic equation
$ \newcommand{\abs}[1]{\left|#1\right|} $
A partial differential equation \begin{equation} \label{eq1} F(x,Du) = 0 \end{equation} where the real-valued function $F(x,q)$ satisfies the condition \begin{equation} \label{eq2} \sum_{\abs{\alpha} = m} \frac{\partial F(x,Du)}{\partial q_\alpha} \xi^\alpha \geq 0 \end{equation} for all real $\xi$, and there exists a $\xi \neq 0$ for which \ref{eq2} becomes an equality. Here, $x$ is an $n$-dimensional vector $(x_1,\ldots,x_n)$; $u$ is the unknown function; $\alpha$ is a multi-index $(\alpha_1,\ldots,\alpha_n)$; $Du$ is a vector with components $$ D^\alpha u = \frac{\partial^{\abs{\alpha}}u}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n}}; $$ the derivatives in equation \ref{eq1} are of an order not exceeding $m$; the $q_\alpha$ are the components of a vector $q$; $\xi$ is an $n$-dimensional vector $(\xi_1,\ldots,\xi_n)$; and $\xi^\alpha = \xi_1^{\alpha_1} \cdots \xi_n^{\alpha_n} $. If strict inequality in equation \ref{eq2} holds for all $x$ and $Du$ and for all real $\xi \neq 0$, equation \ref{eq1} is elliptic at $(x,Du)$. Equation \ref{eq1} degenerates at the points $(x,Du)$ at which inequality \ref{eq2} becomes an equality for any real $\xi \neq 0$. If equality holds only on the boundary of the domain under consideration, the equation is called degenerate on the boundary of the domain. The most thoroughly studied equations are second-order degenerate elliptic equations $$ \sum a^{ik}(x) u_{x_i x_k} + \sum b^i(x) u_{x_i} + c(x)u = f(x), $$ where the matrix $\left[ a^{jk}(x) \right]$ is non-negative definite for all $x$-values under consideration.
See also Degenerate partial differential equation and the references given there.
How to Cite This Entry:
Degenerate elliptic equation.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Degenerate_elliptic_equation&oldid=25860
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Consider a $C^k$, $k\ge 2$, Lorentzian manifold $(M,g)$ and let $\Box$ be the usual wave operator $\nabla^a\nabla_a$. Given $p\in M$, $s\in\Bbb R,$ and $v\in T_pM$, can we find a neighborhood $U$ of $p$ and $u\in C^k(U)$ such that $\Box u=0$, $u(p)=s$ and $\mathrm{grad}\, u(p)=v$?
The tog is a measure of thermal resistance of a unit area, also known as thermal insulance. It is commonly used in the textile industry and often seen quoted on, for example, duvets and carpet underlay.The Shirley Institute in Manchester, England developed the tog as an easy-to-follow alternative to the SI unit of m2K/W. The name comes from the informal word "togs" for clothing which itself was probably derived from the word toga, a Roman garment.The basic unit of insulation coefficient is the RSI, (1 m2K/W). 1 tog = 0.1 RSI. There is also a clo clothing unit equivalent to 0.155 RSI or 1.55 tog...
The stone or stone weight (abbreviation: st.) is an English and imperial unit of mass now equal to 14 pounds (6.35029318 kg).England and other Germanic-speaking countries of northern Europe formerly used various standardised "stones" for trade, with their values ranging from about 5 to 40 local pounds (roughly 3 to 15 kg) depending on the location and objects weighed. The United Kingdom's imperial system adopted the wool stone of 14 pounds in 1835. With the advent of metrication, Europe's various "stones" were superseded by or adapted to the kilogram from the mid-19th century on. The stone continues...
Can you tell me why this question deserves to be negative?I tried to find faults and I couldn't: I did some research, I did all the calculations I could, and I think it is clear enough . I had deleted it and was going to abandon the site but then I decided to learn what is wrong and see if I ca...
I am a bit confused in classical physics's angular momentum. For a orbital motion of a point mass: if we pick a new coordinate (that doesn't move w.r.t. the old coordinate), angular momentum should be still conserved, right? (I calculated a quite absurd result - it is no longer conserved (an additional term that varies with time )
in new coordinnate: $\vec {L'}=\vec{r'} \times \vec{p'}$
$=(\vec{R}+\vec{r}) \times \vec{p}$
$=\vec{R} \times \vec{p} + \vec L$
where the 1st term varies with time. (where R is the shift of coordinate, since R is constant, and p sort of rotating.)
would anyone kind enough to shed some light on this for me?
From what we discussed, your literary taste seems to be classical/conventional in nature. That book is inherently unconventional in nature; it's not supposed to be read as a novel, it's supposed to be read as an encyclopedia
@BalarkaSen Dare I say it, my literary taste continues to change as I have kept on reading :-)
One book that I finished reading today, The Sense of An Ending (different from the movie with the same title) is far from anything I would've been able to read, even, two years ago, but I absolutely loved it.
I've just started watching the Fall - it seems good so far (after 1 episode)... I'm with @JohnRennie on the Sherlock Holmes books and would add that the most recent TV episodes were appalling. I've been told to read Agatha Christy but haven't got round to it yet
?Is it possible to make a time machine ever? Please give an easy answer,a simple one A simple answer, but a possibly wrong one, is to say that a time machine is not possible. Currently, we don't have either the technology to build one, nor a definite, proven (or generally accepted) idea of how we could build one. — Countto1047 secs ago
@vzn if it's a romantic novel, which it looks like, it's probably not for me - I'm getting to be more and more fussy about books and have a ridiculously long list to read as it is. I'm going to counter that one by suggesting Ann Leckie's Ancillary Justice series
Although if you like epic fantasy, Malazan book of the Fallen is fantastic
@Mithrandir24601 lol it has some love story but its written by a guy so cant be a romantic novel... besides what decent stories dont involve love interests anyway :P ... was just reading his blog, they are gonna do a movie of one of his books with kate winslet, cant beat that right? :P variety.com/2016/film/news/…
@vzn "he falls in love with Daley Cross, an angelic voice in need of a song." I think that counts :P It's not that I don't like it, it's just that authors very rarely do anywhere near a decent job of it. If it's a major part of the plot, it's often either eyeroll worthy and cringy or boring and predictable with OK writing. A notable exception is Stephen Erikson
@vzn depends exactly what you mean by 'love story component', but often yeah... It's not always so bad in sci-fi and fantasy where it's not in the focus so much and just evolves in a reasonable, if predictable way with the storyline, although it depends on what you read (e.g. Brent Weeks, Brandon Sanderson). Of course Patrick Rothfuss completely inverts this trope :) and Lev Grossman is a study on how to do character development and totally destroys typical romance plots
@Slereah The idea is to pick some spacelike hypersurface $\Sigma$ containing $p$. Now specifying $u(p)$ is trivial because the wave equation is invariant under constant perturbations. So that's whatever. But I can specify $\nabla u(p)|\Sigma$ by specifying $u(\cdot, 0)$ and differentiate along the surface. For the Cauchy theorems I can also specify $u_t(\cdot,0)$.
Now take the neigborhood to be $\approx (-\epsilon,\epsilon)\times\Sigma$ and then split the metric like $-dt^2+h$
Do forwards and backwards Cauchy solutions, then check that the derivatives match on the interface $\{0\}\times\Sigma$
Why is it that you can only cool down a substance so far before the energy goes into changing it's state? I assume it has something to do with the distance between molecules meaning that intermolecular interactions have less energy in them than making the distance between them even smaller, but why does it create these bonds instead of making the distance smaller / just reducing the temperature more?
Thanks @CooperCape but this leads me another question I forgot ages ago
If you have an electron cloud, is the electric field from that electron just some sort of averaged field from some centre of amplitude or is it a superposition of fields each coming from some point in the cloud?
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Integration or Integral Calculus is usually passed as the Integral calculus and defined as the reverse operation of the differential calculus. It is true in certain cases but does not capture the true essence of it. Integration is an important part of mathematics that was introduced earlier to differentiation.
\[\large \int_{a}^{\infty}f(x)dx=\lim_{b\rightarrow \infty}\left [ \int_{a}^{b}f(x)dx\right ]\]
\[\large \int_{a}^{b}f(x)dx=F(b)-F(a)\]
a and ∞, b are the lower and upper limits, F(a) is the lower limit value of the integral, F(b) is the upper limit value of the integral.
\[\large \int_{a}^{\infty }\frac{dx}{x^{2}+a^{2}}=\frac{\pi }{2a}\]
\[\large \int_{a}^{\infty }\frac{x^{m}dx}{x^{n}+a^{n}}=\frac{\pi a^{m-n+1}}{n\sin \left ( \frac{(m+1)\pi }{n} \right )},0< m+1< n\]
\[\large \int_{a}^{\infty }\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )},0< p< 1\]
\[\large \int_{a}^{\infty }\frac{x^{m}dx}{1+2x\cos \beta +x^{2}}=\frac{\pi \sin (m\beta )}{\sin (m\pi )\sin \beta }\]
\[\large \int_{a}^{\infty }\frac{dx}{\sqrt{a^{2}-x^{2}}}=\frac{\pi }{2}\]
\[\large \int_{a}^{\infty }\sqrt{a^{2}-x^{2}}dx=\frac{\pi a^{2}}{4}\]
\[\large \int_{0}^{\pi }\sin(mx)\sin (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers\]
\[\large \int_{0}^{\pi }\cos (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers\]
\[\large \int_{0}^{\pi }\sin (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m+n\;even\\ \frac{2m}{m^{2}-n^{2}} & if\;m+n\;odd \end{matrix}\right.\;m,n\;integers\]
The concept was proposed in 5
thcentury BC where the area of a shape was given by imagining small polygonstogether. This is a topic of debate usually who discovered Integrationactually. Still, the name of Isaac Newton is taken often because he was theperson behind the advancements in Integration. In simple words, Integralcalculus is the term that is used to calculate the area under a curve.
Moving ahead, Fourier was the person who used the limits to the top or bottom of integral symbol or to mark the start or end point of the integration. This is termed as the definite integral or more applied form of the integration. The concept of definite integrals is frequently used for the real-world problems because it helps to measure or calculate the finite area in a plenty of cases. It is also asked frequently in competitive exams too like JEE or AIEEE etc.
You should be completely focused to solve a problem related to the definite integrals. The only way to succeed in this domain is continuous practice and efforts. Most of the formulas in definite and indefinite integrals are the same but you should know the difference among too when they need to be used. Once you are sure on definite integral properties then solving problems would be easier for you.
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Triangular summation method
$$A=\| a_{nk}\|,\quad n,k=1,2,\ldots,$$
that is, by a matrix for which $a_{nk}=0$ for $k>n$. A triangulation summation method is a special case of a row-finite summation method. A triangular matrix $A$ is called normal if $a_{nn}\neq0$ for all $n$. The transformation
$$\sigma_n=\sum_{k=1}^na_{nk}s_k$$
realized by a normal triangular matrix $A$ has an inverse:
$$s_n=\sum_{k=1}^na_{nk}^{-1}\sigma_k,$$
where $A^{-1}=\| a_{nk}^{-1}\|$ is the inverse of $A$. This fact simplifies the proof of a number of theorems for matrix summation methods determined by normal triangular matrices. Related to the triangular summation methods are, e.g., the Cesàro summation methods and the Voronoi summation method.
References
[Ba] S.A. Baron, "Introduction to the theory of summability of series", Tartu (1966) (In Russian) [Co] R.G. Cooke, "Infinite matrices and sequence spaces", Macmillan (1950) [Ha] G.H. Hardy, "Divergent series", Clarendon Press (1949) How to Cite This Entry:
Triangular summation method.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Triangular_summation_method&oldid=25646
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Here is a well-known interview/code golf question: a knight is placed on a chess board. The knight chooses from its 8 possible moves uniformly at random. When it steps off the board it doesn’t move anymore. What is the probability that the knight is still on the board after \( n \) steps?
We could calculate this directly but it’s more interesting to frame it as a Markov chain.
Calculation using the transition matrix
Model the chess board as the tuples \( \{ (r, c) \mid 0 \leq i, j \leq 7 \} \).
Here are the valid moves and a helper function to check if a move
\( (r,c) \rightarrow (u,v) \) is valid and if a cell is on the usual \( 8 \times 8 \) chessboard: moves = [(-2, 1), (-1, 2), (1, 2), (2, 1), (2,-1), (1,-2), (-1,-2), (-2,-1)] def is_move(r, c, u, v): for m in moves: if (u, v) == (r + m[0], c + m[1]): return True return False def on_board(x): return 0 <= x[0] < 8 and 0 <= x[1] < 8
The valid states are all the on-board positions plus the immediate off-board positions:
states = [(r, c) for r in range(-2, 8+2) for c in range(-2, 8+2)]
Now we can set up the transition matrix.
def make_matrix(states): """ Create the transition matrix for a knight on a chess board with all moves chosen uniformly at random. When the knight moves off-board, no more moves are made. """ # Handy mapping from (row, col) -> index into 'states' to_idx = dict([(s, i) for (i, s) in enumerate(states)]) P = np.array([[0.0 for _ in range(len(states))] for _ in range(len(states))], dtype='float64') assert P.shape == (len(states), len(states)) for (i, (r, c)) in enumerate(states): for (j, (u, v)) in enumerate(states): # On board, equal probability to each destination, even if goes off board. if on_board((r, c)): if is_move(r, c, u, v): P[i][j] = 1.0/len(moves) # Off board, no more moves. else: if (r, c) == (u, v): # terminal state P[i][j] = 1.0 else: P[i][j] = 0.0 return to_idx, P
We can visualise the transition graph using graphviz (full code here):
Oops! The corners aren’t connected to anything so we have 5 communicating classes (the 4 corners plus the rest). We never reach these nodes from any of the starting positions so we can get rid of them:
corners = [(-2,9), (9,9), (-2,-2), (9,-2)] states = [(r, c) for r in range(-2, 8+2) for c in range(-2, 8+2) if (r,c) not in corners]
Here’s the new transition graph:
Intuitively, the knights problem is symmetric, and this graph is symmetric, so it’s likely that we’ve set things up correctly.
Let \( X_0 \), \( X_1 \), \( \ldots \), \( X_n \) be the positions of the knight. Then then probability of the knight moving from state \( i \) to \( j \) in \( n \) steps is
\[
P(X_n = j \mid X_0 = i) = (P^n)_{i,j} \]
So the probability of being on the board after \( n \) steps, starting from \(i\), will be
\[
\sum_{k \in \mathcal{B}} (P^n)_{i,k} \]
where \( \mathcal{B} \) is the set of on-board states. This is easy to calculate using Numpy:
start = (3, 3) n = 5 idx = to_idx[start] Pn = matrix_power(P, n) pr = sum([Pn[idx][r] for (r, s) in enumerate(states) if on_board(s)])
For this case we get probability \( 0.35565185546875 \).
Here are a few more calculations:
start: (0, 0) n: 0 Pr(on board): 1.0 start: (3, 3) n: 1 Pr(on board): 1.0 start: (0, 0) n: 1 Pr(on board): 0.25 start: (3, 3) n: 4 Pr(on board): 0.48291015625 start: (3, 3) n: 5 Pr(on board): 0.35565185546875 start: (3, 3) n: 100 Pr(on board): 5.730392258771815e-13
It’s always good to do a quick Monte Carlo simulation to sanity check our results:
def do_n_steps(start, n): current = start for _ in range(n): move = random.choice(moves) new = (current[0] + move[0], current[1] + move[1]) if not on_board(new): break current = new return on_board(new) N_sims = 10000000 n = 5 nr_on_board = 0 for _ in range(N_sims): if do_n_steps((3,3), n): nr_on_board += 1 print('pr on board from (3,3) after 5 steps:', nr_on_board/N_sims)
The estimate is fairly close to the value we got from taking power of the transition matrix:
pr on board from (3,3) after 5 steps: 0.3554605 Absorbing states
An
absorbing state of a Markov chain is a state that, once entered, cannot be left. In our problem the absorbing states are precisely the off-board states.
A natural question is: given a starting location, how many steps (on average) will it take the knight to step off the board?
With a bit of matrix algebra we can get this from the transition matrix \( \boldsymbol{P} \). Partition \( \boldsymbol{P} \) by the state type: let \( \boldsymbol{Q} \) be the transitions of transient states (here, these are the on-board states to other on-board states); let \( \boldsymbol{R} \) be transitions from transient states to absorbing states (on-board to off-board); and let \( \boldsymbol{I} \) be the identity matrix (transitions of the absorbing states). Then \( \boldsymbol{P} \) can be written in block-matrix form:
\[
\boldsymbol{P}= \left( \begin{array}{c|c} \boldsymbol{Q} & \boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \]
We can calculate powers of \( \boldsymbol{P} \):
\[
\boldsymbol{P}^2= \left( \begin{array}{c|c} \boldsymbol{Q} & \boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \left( \begin{array}{c|c} \boldsymbol{Q} & \boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) = \left( \begin{array}{c|c} \boldsymbol{Q}^2 & (\boldsymbol{I} + \boldsymbol{Q})\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \]
\[
\boldsymbol{P}^3= \left( \begin{array}{c|c} \boldsymbol{Q}^3 & (\boldsymbol{I} + \boldsymbol{Q} + \boldsymbol{Q}^2)\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \]
In general:
\[
\boldsymbol{P}^n= \left( \begin{array}{c|c} \boldsymbol{Q}^n & (\boldsymbol{I} + \boldsymbol{Q} + \cdots + \boldsymbol{Q}^{n-1})\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \]
We want to calculate \( \lim_{n \rightarrow \infty} \boldsymbol{P}^n \) since this will tell us the long-term probability of moving from one state to another. In particular, the top-right block will tell us the long-term probability of moving from a transient state to an absorbing state.
Here is a handy result from matrix algebra:
Lemma. Let \( \boldsymbol{A} \) be a square matrix with the property that \( \boldsymbol{A}^n \rightarrow \mathbf{0} \) as \( n \rightarrow \infty \). Then \[ \sum_{n=0}^\infty = (\boldsymbol{I} – \boldsymbol{A})^{-1}. \]
Applying this to the block form gives:
\[
\begin{align*} \lim_{n \rightarrow \infty} \boldsymbol{P}^n &= \left( \begin{array}{c|c} \boldsymbol{Q}^n & (\boldsymbol{I} + \boldsymbol{Q} + \cdots + \boldsymbol{Q}^{n-1})\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \\ &= \left( \begin{array}{c|c} \lim_{n \rightarrow \infty} \boldsymbol{Q}^n & \lim_{n \rightarrow \infty} (\boldsymbol{I} + \boldsymbol{Q} + \cdots + \boldsymbol{Q}^{n-1})\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \\ &= \left( \begin{array}{c|c} \mathbf{0} & (\boldsymbol{I} – \boldsymbol{Q})^{-1}\boldsymbol{R} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right) \end{align*} \]
where \( \lim_{n \rightarrow \infty} \boldsymbol{Q}^n = 0\) since all of the entries in \( \boldsymbol{Q} \) are transient.
The top-right corner also contains the
fundamental matrix as defined in the following theorem: Theorem Consider an absorbing Markov chain with \( t \) transient states. Let \( \boldsymbol{F} \) be a \( t \times t \) matrix indexed by the transient states, where \( \boldsymbol{F}_{i,j} \) is the expected number of visits to \( j \) given that the chain starts in \( i \). Then
\[
\boldsymbol{F} = (\boldsymbol{I} – \boldsymbol{Q})^{-1}. \]
Taking the row sums of \( \boldsymbol{F} \) gives the expected number of steps \( a_i \) starting from state \( i \) until absorption (i.e. we count the number of visits to each transient state before eventual absorption):
\[
a_i = \sum_{k} \boldsymbol{F}_{i,k} \]
Back in our Python code, we can rearrange the states vector so that the transition matrix is appropriately partitioned. Taking the \( \boldsymbol{Q} \) matrix is very quick using Numpy’s slicing notation:
states = [s for s in states if on_board(s)] + [s for s in states if not on_board(s)] (to_idx, P) = make_matrix(states) # k states k = len(states) # t transient states t = len([s for s in states if on_board(s)]) Q = P[:t, :t] assert Q.shape == (t, t) assert Q.shape == (64, 64) F = linalg.inv(np.eye(*Q.shape) - Q) # example calculation for a_(3,3): state = (3, 3) print(F[to_idx[state], :].sum())
Again, compare to a Monte Carlo simulation to verify that the numbers are correct:
start: (0, 0) Avg nr steps to absorb (MC): 1.9527606 start: (0, 0) Avg nr steps (F matrix): 1.9525249995183136 start: (3, 3) Avg nr steps to absorb (MC): 5.4187947 start: (3, 3) Avg nr steps (F matrix): 5.417750460813215
So, on average, if we start in the corner \( (0,0) \) we will step off the board after \( 1.95 \) steps; if we start in the centre at \( (3,3) \) we will step off the board after \( 5.41 \) steps.
Further reading
The theoretical parts of this blog post follow the presentation in chapter 3 of Introduction to Stochastic Processes with R (Dobrow).
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Let $A$ and $B$ be two real $n\times n$ matrices, and let $C(x)=B(I-xB)^{-1}$, where $I$ is the identity matrix of order $n$, for any real scalar $x$ such that $I-xB$ is invertible. Denote by $\mathrm{tr}$ the trace operator.
Is it possible to get a closed form solution for
$$\int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx},$$
at least when $B$ is diagonalizable?
What I've done so far:
Assuming $B$ is diagonalizable, $C$ is too, and hence it admits the spectral decomposition
$$ C(x)=\sum_{\lambda\in\mathrm{Sp}(B)}\frac{\lambda}{1-x\lambda}Q_{\lambda}, $$
where $\mathrm{Sp}(B)$ denotes the set of distinct eigenvalues of $B$, and $Q_{\lambda}$ is the projector onto $\mathrm{null}(B-\lambda I)$ along $\mathrm{col}(B-\lambda I)$ ($\mathrm{null}$ and $\mathrm{col}$ stand for the null and the column spaces). Hence
$$ \int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}=\sum_{\lambda\in \mathrm{Sp}(B)}\mathrm{tr}(AQ_{\lambda})\int\frac{\lambda}{1-x\lambda}\,dx $$
If all the eigenvalues of $B$ are real we obtain
$$ \int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}=\sum_{\lambda\in \mathrm{Sp}(B)}\mathrm{tr}(AQ_{\lambda})\int\frac{\lambda}{1-x\lambda }\,dx=\sum_{\lambda\in\mathrm{Sp}(B)}\mathrm{ln}(\left\vert 1-x\lambda \right\vert )\mathrm{tr}(AQ_{\lambda}) $$
But what about the case in which not all eigenvalues of $B$ are real? Do we get any simplification from the fact that the eigenvalues of real matrices come in complex conjugate pairs, and the eigenvectors (and hence the projectors $Q_{\lambda}$) associated to complex conjugate eigenvalues are complex conjugates? Can we still use the formula in the last display above for $\int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}$?
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1. Precise determination of the mass of the Higgs boson and tests of compatibility of its couplings with the standard model predictions using proton collisions at 7 and 8 TeV
European Physical Journal C, ISSN 1434-6044, 05/2015, Volume 75, Issue 5, p. 1
Properties of the Higgs boson with mass near 125 GeV are measured in proton-proton collisions with the CMS experiment at the LHC. Comprehensive sets of...
TRANSVERSE-MOMENTUM | TOP-PAIR | RATIOS | NLO | RESUMMATION | ELECTROWEAK CORRECTIONS | BROKEN SYMMETRIES | HADRON COLLIDERS | LHC | QCD CORRECTIONS | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment
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Journal Article
2. Measurements of properties of the Higgs boson decaying to a W boson pair in pp collisions at s=13TeV
Physics Letters B, ISSN 0370-2693, 04/2019, Volume 791, pp. 96 - 129
Measurements of the production of the standard model Higgs boson decaying to a W boson pair are reported. The candidates are selected in events with an...
Higgs
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3. Searches for a heavy scalar boson H decaying to a pair of 125 GeV Higgs bosons hh or for a heavy pseudoscalar boson A decaying to Zh, in the final states with h -> tau tau
PHYSICS LETTERS B, ISSN 0370-2693, 04/2016, Volume 755, pp. 217 - 244
A search for a heavy scalar boson H decaying into a pair of lighter standard-model-like 125 GeV Higgs bosons hh and a search for a heavy pseudoscalar boson A...
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4. Searches for a heavy scalar boson H decaying to a pair of 125 GeV Higgs bosons hh or for a heavy pseudoscalar boson A decaying to Zh, in the final states with $h o au au
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Nuclear and Particle Physics Proceedings, ISSN 2405-6014, 04/2016, Volume 273-275, pp. 2451 - 2453
We present a phenomenological evaluation of the ratio , from the LHC combined value, we get This value is close to one with a precision of the order ∼ 1%....
Higgs
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7. Precise determination of the mass of the Higgs boson and tests of compatibility of its couplings with the standard model predictions using proton collisions at 7 and 8TeV
ISSN 1434-6044, 2015
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Journal of High Energy Physics, ISSN 1126-6708, 6/2015, Volume 2015, Issue 6, pp. 1 - 36
The importance of off-shell contributions is discussed for H → V V (∗) with V ∈ {Z,W} for large invariant masses m VV involving a standard model (SM)-like...
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9. The particle at the end of the universe
: how the hunt for the Higgs boson leads us to the edge of a new world
2013, ISBN 9780142180303, 353
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10. Search for a CP-odd Higgs boson decaying to Zh in pp collisions at root s=8 TeV with the ATLAS detector
PHYSICS LETTERS B, ISSN 0370-2693, 05/2015, Volume 744, pp. 163 - 183
A search for a heavy, CP-odd Higgs boson, A, decaying into a Z boson and a 125 GeV Higgs boson, h, with the ATLAS detector at the LHC is presented. The search...
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11. Search for H -> gamma gamma produced in association with top quarks and constraints on the Yukawa coupling between the top quark and the Higgs boson using data taken at 7 TeV and 8 TeV with the ATLAS detector
PHYSICS LETTERS B, ISSN 0370-2693, 01/2015, Volume 740, pp. 222 - 242
A search is performed for Higgs bosons produced in association with top quarks using the diphoton decay mode of the Higgs boson. Selection requirements are...
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12. Search for a standard model-like Higgs boson in the mu(+)mu(-) and e(+)e(-) decay channels at the LHC
PHYSICS LETTERS B, ISSN 0370-2693, 05/2015, Volume 744, pp. 184 - 207
A search is presented for a standard model-like Higgs boson decaying to the mu(+)mu(-) or e(+)e(-) final states based on proton-proton collisions recorded by...
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JOURNAL OF HIGH ENERGY PHYSICS, ISSN 1029-8479, 11/2015, Issue 11
A search for a charged Higgs boson is performed with a data sample corresponding to an integrated luminosity of 19.7 +/- 0.5 fb(-1) collected with the CMS...
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2013, ISBN 1616148012, 325
"Two leading physicists discuss the importance of the Higgs Boson, the future of particle physics, and the mysteries of the universe yet to be unraveled. On...
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2012, 1st ed., ISBN 9780199603497, xxi, 277
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16. Search for lepton flavour violating decays of the Higgs boson to $\mu\tau$ and e$\tau$ in proton-proton collisions at $\sqrt{s}=$ 13 TeV
ISSN 1029-8479, 2018
coupling: Yukawa | experimental results | CMS | Higgs particle: decay modes | CERN LHC Coll | Higgs particle: branching ratio: upper limit | Higgs particle: leptonic decay | lepton: flavor: violation | Higgs particle --> electron tau | 4/3 | Higgs particle: hadroproduction | p p: colliding beams | p p: scattering | 13000 GeV-cms | coupling constant: upper limit | Higgs particle --> muon tau
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17. Constraints on off-shell Higgs boson production and the Higgs boson total width in $ZZ\to4\ell$ and $ZZ\to2\ell2\nu$ final states with the ATLAS detector
2018
A measurement of off-shell Higgs boson production in the $ZZ\to4\ell$ and $ZZ\to2\ell2\nu$ decay channels, where $ℓ$ stands for either an electron or a muon,...
muon: pair production | gluon gluon: fusion | experimental results | Z0: leptonic decay | width: upper limit | neutrino: pair production | Higgs particle: coupling | CERN LHC Coll | electron: pair production | Higgs particle: decay | Z0: pair production | ATLAS | Higgs particle: off-shell | Higgs particle: width | p p: colliding beams | Higgs particle: hadroproduction | p p: scattering | 13000 GeV-cms | mass spectrum: (4lepton) | dilepton: mass spectrum
muon: pair production | gluon gluon: fusion | experimental results | Z0: leptonic decay | width: upper limit | neutrino: pair production | Higgs particle: coupling | CERN LHC Coll | electron: pair production | Higgs particle: decay | Z0: pair production | ATLAS | Higgs particle: off-shell | Higgs particle: width | p p: colliding beams | Higgs particle: hadroproduction | p p: scattering | 13000 GeV-cms | mass spectrum: (4lepton) | dilepton: mass spectrum
Paper
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There are 3 units that are used to express charge.
This is most commonly used in macroscopic situations.
1 coulomb of charge is that which, if placed at a separation of 1 metre from another charge of magnitude 1 coulomb, experiences a force of $8.99\times 10^9$N along the line joining the charges.
This definition results from the observation-based Coulomb's Law of Electrostatic Forces:$F_e=k_e\frac{q_1 q_2}{r^2}$
This is a unit of charge most commonly used in microscopic situations due to its convenience.
It is the magnitude of charge present on 1 electron, or 1 proton - it is equal in magnitude.
$e\approx 1.602\times 10^{-19}$
Electrostatic unit of charge (esu)
The least used of the three, it is the CGS unit of charge. It is also known as franklin (Fr), or statcoulomb (statC).
1 esu of charge is that which, if placed at a separation of 1 centimetre from another charge of magnitude 1 esu, experiences a force of magnitude 1 dyne along the line joining the two charges.
Again, from Coulomb's Law of Electrostatic Forces, we can determine that
$1 esu\approx 3.335\times 10^{-10} C$
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Let's say you have a number $x$, and a priori, you know that $x \in [0, 1)$ (each value from 0 to 1 is equally likely.) Then a wizard comes and tells you that $x \in [a, b) \subseteq [0, 1)$. How much information does this give you?
It would seem to be $-\log_2(b-a)$ bits, but I don't know how to prove this, since both a priori and a posteriori have an infinite amount of entropy.
The reason I think $-\log_2(b-a)$ bits seems reasonably is it seems to agree with examples.
$-\log_2(1-0) = 0$, which is true, since no information is conveyed.
$-\log_2(\frac12 - 0) = 1$, which seems reasonably, since it would give you the first binary digit of $x$
In general, $-\log_2(b-a=\frac1{2^n})=n$, seems reasonable, as it gives you about $n$ binary digits.
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Can you express the fraction $\frac{1949}{1999}$ in the form $\frac 1a+\frac 1b$? Give reasons supporting your answer.
I think the only way to do this is keep trying numbers but then I will never get the answer. I cry every time.
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Can you express the fraction $\frac{1949}{1999}$ in the form $\frac 1a+\frac 1b$? Give reasons supporting your answer.
I think the only way to do this is keep trying numbers but then I will never get the answer. I cry every time.
$$\frac 1a+\frac 1b = \frac{a+b}{ab},$$ so you would need $ab$ to divide $1999$. But…
Given $$\frac 1a+\frac 1b=\frac {1949}{1999}$$ Combining the fraction gives $$\frac {a+b}{ab}=\frac {1949}{1999}$$ Setting terms gives the system $$\begin{cases}a+b=1949\\ab=1999\end{cases}$$ With $a,b$ can be solved by a quadratic. Namely $$b^2-1949b+1999$$ Where $a=1949-b$.
Since $1999$ cannot be factored, the roots are really ugly looking numbers. Namely, the two possible values of $b$ are $$b_1=\frac {1949+\sqrt{3790605}}2\\b_2=\frac {1949-\sqrt{3790605}}2$$and with the $a$ values as $$a_1=\frac {1949-\sqrt{3790605}}{2}\\a_2=\frac {1949+\sqrt{3790605}}2$$
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Learning Objectives
Create and interpret graphs of potential energy Explain the connection between stability and potential energy
Often, you can get a good deal of useful information about the dynamical behavior of a mechanical system just by interpreting a graph of its potential energy as a function of position, called a
potential energy diagram. This is most easily accomplished for a one-dimensional system, whose potential energy can be plotted in one two-dimensional graph—for example, U(x) versus x—on a piece of paper or a computer program. For systems whose motion is in more than one dimension, the motion needs to be studied in three-dimensional space. We will simplify our procedure for one-dimensional motion only.
First, let’s look at an object, freely falling vertically, near the surface of Earth, in the absence of air resistance. The mechanical energy of the object is conserved, E = K + U, and the potential energy, with respect to zero at ground level, is U(y) = mgy, which is a straight line through the origin with slope mg . In the graph shown in Figure 8.11, the x-axis is the height above the ground y and the y-axis is the object’s energy.
The line at energy E represents the constant mechanical energy of the object, whereas the kinetic and potential energies, K
A and U A, are indicated at a particular height y A. You can see how the total energy is divided between kinetic and potential energy as the object’s height changes. Since kinetic energy can never be negative, there is a maximum potential energy and a maximum height, which an object with the given total energy cannot exceed:
$$K = E - U \geq 0,$$
$$U \leq E \ldotp$$
If we use the gravitational potential energy reference point of zero at y
0, we can rewrite the gravitational potential energy U as mgy. Solving for y results in
$$y \leq \frac{E}{mg} = y_{max} \ldotp$$
We note in this expression that the quantity of the total energy divided by the weight (mg) is located at the maximum height of the particle, or y
max. At the maximum height, the kinetic energy and the speed are zero, so if the object were initially traveling upward, its velocity would go through zero there, and y maxwould be a turning point in the motion. At ground level, y 0 = 0, the potential energy is zero, and the kinetic energy and the speed are maximum:
$$U_{0} = 0 = E - K_{0},$$
$$E = K_{0} = \frac{1}{2} mv_{0}^{2},$$
$$v_{0} = \pm \sqrt{\frac{2E}{m}} \ldotp$$
The maximum speed ±v
0 gives the initial velocity necessary to reach y max, the maximum height, and −v 0 represents the final velocity, after falling from y max. You can read all this information, and more, from the potential energy diagram we have shown.
Consider a mass-spring system on a frictionless, stationary, horizontal surface, so that gravity and the normal contact force do no work and can be ignored (Figure 8.12). This is like a one-dimensional system, whose mechanical energy E is a constant and whose potential energy, with respect to zero energy at zero displacement from the spring’s unstretched length, x = 0, is U(x) = \(\frac{1}{2}\)kx
2.
You can read off the same type of information from the potential energy diagram in this case, as in the case for the body in vertical free fall, but since the spring potential energy describes a variable force, you can learn more from this graph. As for the object in vertical free fall, you can deduce the physically allowable range of motion and the maximum values of distance and speed, from the limits on the kinetic energy, 0 ≤ K ≤ E. Therefore, K = 0 and U = E at a
turning point, of which there are two for the elastic spring potential energy,
$$x_{max} = \pm \sqrt{\frac{2E}{k}} \ldotp$$
The glider’s motion is confined to the region between the turning points, −x
max ≤ x ≤ x max. This is true for any (positive) value of E because the potential energy is unbounded with respect to x. For this reason, as well as the shape of the potential energy curve, U(x) is called an infinite potential well. At the bottom of the potential well, x = 0, U = 0 and the kinetic energy is a maximum, K = E, so v max = ± \(\sqrt{\frac{2E}{m}}\).
However, from the slope of this potential energy curve, you can also deduce information about the force on the glider and its acceleration. We saw earlier that the negative of the slope of the potential energy is the spring force, which in this case is also the net force, and thus is proportional to the acceleration. When x = 0, the slope, the force, and the acceleration are all zero, so this is an
equilibrium point. The negative of the slope, on either side of the equilibrium point, gives a force pointing back to the equilibrium point, F = ±kx, so the equilibrium is termed stable and the force is called a restoring force. This implies that U(x) has a relative minimum there. If the force on either side of an equilibrium point has a direction opposite from that direction of position change, the equilibrium is termed unstable, and this implies that U(x) has a relative maximum there.
Example 8.10
Quartic and Quadratic Potential Energy Diagram
The potential energy for a particle undergoing one-dimensional motion along the x-axis is U(x) = 2(x
4 − x 2), where U is in joules and x is in meters. The particle is not subject to any non-conservative forces and its mechanical energy is constant at E = −0.25 J. (a) Is the motion of the particle confined to any regions on the x-axis, and if so, what are they? (b) Are there any equilibrium points, and if so, where are they and are they stable or unstable? Strategy
First, we need to graph the potential energy as a function of x. The function is zero at the origin, becomes negative as x increases in the positive or negative directions (x
2 is larger than x 4 for x < 1), and then becomes positive at sufficiently large |x|. Your graph should look like a double potential well, with the zeros determined by solving the equation U(x) = 0, and the extremes determined by examining the first and second derivatives of U(x), as shown in Figure 8.13.
You can find the values of (a) the allowed regions along the x-axis, for the given value of the mechanical energy, from the condition that the kinetic energy can’t be negative, and (b) the equilibrium points and their stability from the properties of the force (stable for a relative minimum and unstable for a relative maximum of potential energy). You can just eyeball the graph to reach qualitative answers to the questions in this example. That, after all, is the value of potential energy diagrams.
You can see that there are two allowed regions for the motion (E > U) and three equilibrium points (slope \(\frac{dU}{dx}\) = 0), of which the central one is unstable \(\left( \dfrac{d^{2}U}{dx^{2}} < 0 \right)\), and the other two are stable \(\left(\dfrac{d^{2}U}{dx^{2}} > 0 \right)\).
Solution To find the allowed regions for x, we use the condition $$K = E - U = - \frac{1}{4} - 2(x^{4} - x^{2}) \geq 0 \ldotp$$If we complete the square in x 2 , this condition simplifies to \(2 \left(x^{2} − \dfrac{1}{2} \right)^{2} \leq \frac{1}{4}\), which we can solve to obtain $$\frac{1}{2} - \sqrt{\frac{1}{8}} \leq x^{2} \leq \frac{1}{2} + \sqrt{\frac{1}{8}} \ldotp$$This represents two allowed regions, x p≤ x ≤ x Rand −x R≤ x ≤ − x p, where x p= 0.38 and x R= 0.92 (in meters). To find the equilibrium points, we solve the equation $$\frac{dU}{dx} = 8x^{3} - 4x = 0$$and find x = 0 and x = ±x Q, where x Q= \(\frac{1}{\sqrt{2}}\) = 0.707 (meters). The second derivative $$\frac{d^{2}U}{dx^{2}} = 24x^{2} - 4$$is negative at x = 0, so that position is a relative maximum and the equilibrium there is unstable. The second derivative is positive at x = ±x Q, so these positions are relative minima and represent stable equilibria. Significance
The particle in this example can oscillate in the allowed region about either of the two stable equilibrium points we found, but it does not have enough energy to escape from whichever potential well it happens to initially be in. The conservation of mechanical energy and the relations between kinetic energy and speed, and potential energy and force, enable you to deduce much information about the qualitative behavior of the motion of a particle, as well as some quantitative information, from a graph of its potential energy.
Exercise 8.10
Repeat Example 8.10 when the particle’s mechanical energy is +0.25 J.
Before ending this section, let’s practice applying the method based on the potential energy of a particle to find its position as a function of time, for the one-dimensional, mass-spring system considered earlier in this section.
Example 8.11
Sinusoidal Oscillations
Find x(t) for a particle moving with a constant mechanical energy E > 0 and a potential energy U(x) = \(\frac{1}{2}\)kx
2, when the particle starts from rest at time t = 0. Strategy
We follow the same steps as we did in Example 8.9. Substitute the potential energy U into Equation 8.14 and factor out the constants, like m or k. Integrate the function and solve the resulting expression for position, which is now a function of time.
Solution
Substitute the potential energy in Equation 8.14 and integrate using an integral solver found on a web search:
$$t = \int_{x_{0}}^{x} \frac{dx}{\sqrt{\left(\dfrac{k}{m}\right) \Big[ \left(\dfrac{2E}{k}\right) - x^{2} \Big]}} = \sqrt{\frac{m}{k}} \Bigg[ \sin^{-1} \left( \dfrac{x}{\sqrt{\frac{2E}{k}}}\right) - \sin^{-1} \left(\frac{x_{0}}{\sqrt{\frac{2E}{k}}}\right) \Bigg] \ldotp$$From the initial conditions at t = 0, the initial kinetic energy is zero and the initial potential energy is \(\frac{1}{2}\)kx
0 2 = E, from which you can see that \(\frac{x_{0}}{\sqrt{\left(\dfrac{2E}{k}\right)}}\) = ±1 and sin −1 (±) = ±90°. Now you can solve for x:
$$x(t) = \sqrt{\left(\dfrac{2E}{k}\right)} \sin \Big[\left(\sqrt{\dfrac{k}{m}}\right)t \pm 90^{o} \Big] = \pm \sqrt{\left(\dfrac{2E}{k}\right)} \cos \Big[ \left(\sqrt{\dfrac{k}{m}}\right)t \Big] \ldotp$$
Significance
A few paragraphs earlier, we referred to this mass-spring system as an example of a harmonic oscillator. Here, we anticipate that a harmonic oscillator executes sinusoidal oscillations with a maximum displacement of \(\sqrt{\left(\dfrac{2E}{k}\right)}\) (called the amplitude) and a rate of oscillation of \(\left(\dfrac{1}{2 \pi}\right) \sqrt{\frac{k}{m}}\) (called the frequency). Further discussions about oscillations can be found in Oscillations.
Exercise 8.11
Find x(t) for the mass-spring system in Example 8.11 if the particle starts from x
0 = 0 at t = 0. What is the particle’s initial velocity? Contributors
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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This is a theoretical computer science question, regarding the proof of whether or not context-free languages are closed under an operation. This means basically that any context-free language which undergoes this operation would still be context-free.
For a language $A$ which is a subset $\Sigma^*$, define the language $A_+$ as
$\qquad\displaystyle A_+ = \{xyz | y\in \Sigma \wedge xz \in A\}$.
Prove that the set of context-free languages is closed under the $+$ operator.
So $A_+$ contains all strings that can be obtained by inserting one symbol into a string in $A$. Show that the class of context-free languages is closed under the operation $+$ (i.e., show that if $A$ is context free, then $A_+$ is also context free).
Note that this is not a homework question. Examples of closure proofs are sparse online and in my textbook.
First, let's think about the different ways to prove closure of CFLs:
Constructing a "template" Push Down Automata that can incorporate any other Push Down Automata and add on a part to the beginning, middle, or end of the PDA and still be able to accept the language accepted by the original CFL, with the new operation on it. Basically, if for any given PDA P, if a PDA P' can be created which accepts the language of P with the new operation (in our case, the "+" operator) performed on it, then that operation must be under closure.
Solving the problem in this manner is quite simple to think about. Imagine a PDA P which accepts strings from the CFL L. In application to our problem, this would be a PDA which can successfully read in the string 'xz', where x and z are simply any string conforming to our alphabet. The PDA P' would similarly have the ability to read xz, but each state of the PDA could have an additional self loop which reads the character in the string y.
I have selected the answer which I find to be most appropriate for this question. It simply involves using the "tempate PDA" strategy which I outline above; however, my construction did not achieve the goals of the new language (think about why before looking at the answer below).
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I'm trying understand where electricians came to agreement that Ie = Ib + Ic. I know that the base of a NPN transistor is very thin, allowing electrons to flow through to the collector which explains why $I_{c} = \beta I_{b}$. Mathematically, I can also see that when we have $\alpha = \frac{I_{c}}{I_{e}} = \frac{\beta}{1+\beta}$ But I want to understand the actual physics behind this phenomenon, so that I could remember it for my E5 advancement exam.
Charge conservation.
Emitter
emits electrons and collector collect the electrons. When emitted electron goes into base region which is p type material (for npn transistor) this electron have a nonzero probability to recombine. However, after recombination charge neutral base becomes negative charged and pull hole from the base. This mechanism is one of the component of base current.
Think like you have
5 electron in emitter and you emit through them to collector. In the base this electrons have %20 recombination probability. You will lose one electron at the base and pull one hole to base to preserve charge neutrality. At the end you have 4 electron at collector side.
$$I_E = I_B+I_C$$
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If Einstein traveled 99.99999999% the speed of light, and shined a flashlight backwards, he would see photons moving at the speed of light. A stationary observer would also see photons moving at the speed of light. This is one of the fundamental principles of Special Relativity:
The speed of light is the same in all reference frames.
This principle isn't just plucked out of thin air. It's supported by various experiments done at the end of the 19th century, chief among them being the Michelson-Morley experiment, that failed to show any variation in the speed of light with direction, even though we should by all accounts be moving at fantastic speeds relative to, for example, the center of the galaxy.
In order for this principle to make any sense, we have to relax the "common sense" notion that time and space are absolute. In reality, the speed of time's passage and the length of objects are both dependent on the speed of the observer relative to the object. In particular, the
Lorentz factor $\gamma$ defines the magnitude of this shift:
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
For the speed you quoted (.9999999999c), the Lorentz factor is $\gamma=70710$. This has four implications:
Einstein will observe that clocks in the rest of the world tick 70710 times slower than the watch in his pocket;
Einstein will observe that pens in the rest of the world are 70710 times shorter (in his direction of motion) than the pen in his pocket;
The stationary observer will observe that Einstein's watch ticks 70710 times slower than his watch; and
The stationary observer will observe that Einstein's pen is 70710 times shorter (in the direction of motion) than his pen.
If the first two seem to contradict the last two ("Who's right?" you may ask), remember that
you can't be both Einstein and the stationary observer at the same time, just like you can't travel at two speeds at once. You have to pick a speed to travel at, and that fixes your notion of space and time in a way that is consistent with the laws of physics in your frame of reference.
So, what will the stationary observer think of the light beam that propagates backwards from Einstein's flashlight? He will observe that time is moving much, much more slowly, which, in isolation, might imply that the light beam would move slower.
he also observes that lengths are much, much shorter, which, in isolation, would mean that the light beam can cover a much greater distance in each timestep (which would make it move faster). It turns out that, for objects traveling at the speed of light, these two effects exactly cancel out! This means that, even though time is slowed down, the distance is also contracted in just such a way that light travels the same distance in the same amount of time. Hence, light has the same speed for both observers. BUT
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Motivation:
It is a well known fact that the gravitational field (in General Relativity and direct generalizations of it) has no local energy-momentum density.
Usually there are two reasons stated, one is heuristic, one is mathematical.
Heuristic reasoning:The gravitational field can be "turned off" at any one $x\in M$ by a proper choice of reference frame, which implies there are problems defining a local density. Mathematical reasoning:The (Einstein-Hilbert) stress-energy (SEM) tensor of a matter field $\psi$ is defined as $$ T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}} $$ where $S_m$ is the matter action. It is, in some sense, the response of the matter field action to changes in the metric. On the other hand, the response of the gravitational actionto a change in the metric is $$ \frac{1}{16\pi G}G_{\mu\nu}=\frac{\delta S_{EH}}{\delta g^{\mu\nu}}, $$ which is basically the vacuum EoMs of GR, clearly this is not to be interpreted as a SEM tensor. So the gravitational field has no SEM tensor.
Moreover, the covariant conservation law $$ \nabla_\mu T^{\mu\nu}=0 $$ is usually interpreted not as a genuine conservation law, but rather the exchange of energy-momentum between the matter field and the gravitational field.
Background:
Let $\psi$ be a matter field, whose target space carries a representation of a Lie group $G$, with action $$ S_m [\psi]=\int d^4x\ \mathcal L_m(\psi,\partial\psi)$$ such that the Lie group $G$ is a group of symmetries of $S_m$. From Noether's theorem, one may obtain $k$ (we have $h=\dim G$) conserved currents satisfying $$ \partial_\mu\mathcal J^\mu _a=0. $$
Let us turn the global symmetry into a gauge symmetry, introduce the gauge connection/gauge field $\mathcal A_\mu=A^a_\mu T_a$ ($T_a$ are a set of generators for $\mathfrak g$) through the covariant derivative $$ D_\mu\psi=\partial_\mu \psi + \mathcal A_\mu\psi, $$ and introduce a self-action $$ S_{GF}[A]=\int d^4 x\ \mathcal L_{GF}(A,F) $$ for the gauge field.
One may show the following:
The current $$ j^\mu_a=\frac{\delta S_m}{\delta A^a_\mu} $$ is an equivalent current to the Noether current $\mathcal J^\mu _a$
provided that $A\rightarrow 0$. This includes the conservation law $\partial_\mu j^\mu_a=0$.
With $A_\mu^a\neq 0$, the current satisfies the identity $$ D_\mu j^\mu_a=\partial_\mu j^\mu_a-A_\mu^cC_{c\ a}^{b}j^\mu_b=0, $$ where $ D_\mu $ is the covariant derivative in the coadjoint bundle.
Question:
The situation here is clearly analogous to the situation with GR but with $T^{\mu\nu}$ replaced with $j^\mu_a$. In particular:
The conservation law $D_\mu j^\mu_a=0$ cannot be interpreted as a real conservation law, it expresses the fact that the matter field can exchange charge with the gauge field.
It does not make sense to define a gauge current for the gauge field $A$, only for the matter field $\psi$.
Does that meanthat in a nonabelian gauge theory, the local charge current density of the gauge field is ill-defined?
If so, I would find it odd, since it is not something that's usually stated, plus unlike gravity, where you can "turn off" the "field strength" $\Gamma^\rho_{\mu\nu}$ by a choice of reference frame at a point, the gauge curvature $\mathcal F=d\mathcal A+\frac{1}{2}[\mathcal A,\mathcal A]$ cannot be eliminated at points via gauge transformations.
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Let's say I have a gas of Li atoms in the $1s^2 2s^1$ ground state configuration and I want to excite the $2s \to 3s$ transition of the outermost electron with a laser pulse on resonance. (That is the mean photon energy is equal to the energy difference of the two levels.)
There are some things I don't understand about this basic situation, and I'd be happy if somebody could explain me that.
The wavefunction of this interacting system is \begin{equation} \Psi=\Psi(\textbf{r}_1,\sigma_1,\textbf{r}_2,\sigma_2,\textbf{r}_3,\sigma_3) \neq \Psi_{100}(\textbf{r}_1,\sigma_1) \Psi_{100}(\textbf{r}_2,\sigma_2)\Psi_{200}(\textbf{r}_3,\sigma_3) \end{equation} that is it isn't separable into products of wavefunctions containing the spin and spatial coordinates of the individual electrons. Because of this configuration seems to have no physical or any kind of meaning to me, because it seems to assume this separability. What is the point of using it then?
What does it mean to excite the outermost electron? There is only an entangled system of 3 electrons in which all of them are indistinguishable, thus it seems to me that there is no meaning in talking about "inner" or "outer" electrons or their transitions. The only meaningful thing is to talk about the 3-electron system as a whole and about the excitations of this whole system. Why is this language of "exciting the 2s electron to the 3s state" is used then? What is the meaning of it?
If the usage of the language "exciting the outermost 2s electron to the 3s state" is correct, what physical measurement can I use to tell that this excitation has happened? Maybe somehow I could measure the energies of each of the electrons before and after the excitations and then I'd find that two energy values out of the 3 are the same before and after the laser pulse, but one energy I get has increased by the photon energy. But this method doesn't seem to help much, first because in an entangled system the individual particles are in mixed states and their energy measurement yield a statistical mixture of values (so at best all I could say is that the mean of one of the distributions has increased), and second, because I cannot label the electrons and associate the energy-value distributions to them (because they are indistinguishable) so again I can't say that there was an outermost electron in a certain state and now it is in another state. All I can say that the energy measurements on individual electrons previously yielded a distribution of values which was the sum of three probability distributions with certain means (e.g. $\overline{E}_1=\overline{E}_2<\overline{E}_3$) and deviations and now it's the sum of some other 3 distributions from which one maybe has a higher mean than before. [E.g. $\overline{E}'_1=\overline{E}'_2<\overline{E}'_3(>\overline{E}_3)$] But nothing says that the distribution with the higher mean comes from the electron which gave the highest mean before.
In summary, please help me clear up the following. Taking into account what I've said previously,
What is the physical meaning of the electron configuration of an N-electron system?
What is the meaning of exciting the outermost electron from one state to the other?
What measurement tells me that the outermost electron has been excited? (Actually even better, if somebody can tell me how to measure if the outermost electron is in a mixed state or in a pure superposition state, the superposition being made up from the eigenstates of the 1-electron energy operator. The operator that corresponds to observing the 1-electron energy.)
This became a bit long, if something is not clear, please indicate, and I'll try to clear up. Thank you in advance.
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Revised Answer
(I have again misinterpreted your question, which is simpler than I had anticipated.)
This question is Problem 3.44 in Griffiths' "Introduction to Electrodynamics". A solution is given as Exercise 2 in Physics Pages using
Green's Reciprocity Theorem, but it is not easy to understand.
In the simplified problem suggested by your author, we have 3 parallel conducting plates A, B, C. The middle plate B carries total charge $+Q$. The other two plates are either neutral or grounded.
Suppose the area of each face of each conductor is $A$ and the surface charges on the left and right faces of plate B are $+\sigma_1$ and $+\sigma_2$ where $\sigma_1+\sigma_2=\frac{Q}{A}$. The surface charges on the adjacent faces of the outer plates A, C must be $-\sigma_1, -\sigma_2$ respectively. This is because there is no electric field inside any of the conductors, so all electric field lines starting on one face of B must end on the adjacent face of A or C.
The electric field due to a plane face with surface charge density $\sigma$ is $E=\frac{\sigma}{2\epsilon_0}$. So the field between B and A is $E_1=\frac{\sigma_1}{\epsilon}$ and that between B and C is $E_2=\frac{\sigma_2}{\epsilon}$.
Assuming plates A and C are either grounded or at the same potential, then the potential differences with plate B are equal : $V_{BA}=V_{BC}$ which means that $$E_1 x=E_2 (\ell-x)$$ where $\ell$ is the distance between A and C.
Using the above expressions for surface charge we get $$\sigma_1 x=\sigma_2 (\ell-x)$$ $$Q_1 x=Q_2 (\ell-x)$$ Now $Q_1+Q_2=Q$. Therefore the charges induced on A and C are $$Q_1=-\frac{\ell-x}{\ell}Q$$ $$Q_2=-\frac{x}{\ell}Q$$
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I want to evaluate the integral:
$$I=\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2 \ \Theta(x_1-x_2) \ e^{i(ax_1+bx_2)}$$ where $\Theta(x)$ is the Heaviside function.
What I was doing now was taking the relation for $\Theta$: $\Theta (x)=-\frac{1}{2\pi i}\int_{-\infty}^{\infty}d\tau \frac{1}{\tau + i\epsilon} e^{-ix\tau}$ and I got: $$I= -\frac{1}{2\pi i}\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}d\tau \ \frac{1}{\tau + i\epsilon}e^{-i(\tau-a)x_1}e^{-i(\tau+b)x_2}\\=2\pi i\int_{-\infty}^{\infty}d\tau\ \frac{1}{\tau + i\epsilon}\delta(\tau-a)\delta(\tau+b) =2\pi i\frac{\delta(a+b)}{a + i\epsilon}$$ I didn't know if the integral was convergent and I could simply interchange the integrals, so I tried it in a different form with $X=x_1+x_2$ and $x=x_1-x_2$ : $$I=\frac{1}{2}\int_{-\infty}^{\infty}dX \int_{-\infty}^{\infty}dx \ \Theta(x) \ e^{ia\frac{x+X}{2}+ib\frac{X-x}{2}} \\ =\pi \int_{-\infty}^{\infty}dx \ \Theta(x) e^{-2\pi i\frac{b-a}{4\pi}}\delta(\frac{b+a}{2})$$ With the Fourier Transform of the Heaviside function $\int_{-\infty}^{\infty}dk\ \Theta(k)e^{-2\pi i kx}=\frac{1}{2}(\delta(x)-\frac{i}{\pi k}) $ I get $$I=\pi \left(2\pi\delta(b-a)-\frac{4 i}{b-a}\right)\delta(a+b)=2\pi^2\delta(a)\delta(b)+2\pi i\frac{\delta(a+b)}{a}$$ I don't know yet where the $\delta(a)\delta(b)$ should come from in the first method. When I want to check that now and integrate $I$ over $a$ and $b$ I get from the first line: $$\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}db \ I = \int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2 \Theta(x_1-x_2) \delta(x_1)\delta(x_2) \\ = \Theta(0)=\frac{1}{2}$$ and from the second result: $$\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}db \ I=4\pi^2-\int_{-\infty}^{\infty}db \frac{1}{b}=-\infty$$ Where did it go wrong? Is the integral correct?
Thank you in advance.
EDIT: corrected mistake in derivation because of comment.
This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user gaugi
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The electrostatic force between the electron and the proton (in classical terms) varies as $1/r^2$ so when the electron and proton are separated by a large distance the force goes to 0 therefore at large distance the electron and proton become free particles. Note that when the electron and proton are very close the force between them increases to infinity. Also note that for the electrostatic force there are two charges: $+$ and $-$.
The strong force is called a color force because there are 3 different kinds of color "charges" in Quantum Chromodynamics (QCD), unlike the case of electrostatics which only has 2 charges. (There are no "real" colors, or course, physicists just use the term color since there are 3 primary colors which matches up with the names needed for the 3 different "charges" of quarks.) One of the reasons why the strong color force between two or three quarks is different than the electrostatic force between an electron and proton is that the force carriers of the strong force (the massless gluons) are also colored therefore the gluons are also strongly attracted to each other. Whereas in the electrostatic case, the force carriers (virtual photons) are uncharged so two virtual photons do not attract each other.
The color force between two (or three) quarks is quite different than the electrostatic force between two charges. In a very simplified model you can think about the force between two quarks as varying like $r$ or $r^2$. First of all note that when $r\rightarrow 0$ the force goes to 0. This is the asymptotic freedom of the color force which was discovered in 1973 and for which Gross, Wilczek and Politzer were awarded the Nobel Prize in Physics in 2004. This means that at very high energies (and short distances) the quarks act like free particles and the color force is small.
However when $r\rightarrow \infty$ the force goes to $\infty$. This model of a force that increases with distance is another statement of the principle of color confinement in QCD. The fact that gluons will interact with other gluons with the same strong color force that attracts quarks is thought to be the reason for color confinement. So, if you start with the three different colored quarks bound into a colorless proton and if you try to pull one of the quarks out of the proton, it will take more and more force and thus more and more energy as you pull the quark out. Thus, as you try to separate the quark out of the proton, at some point when enough energy has been added to the system it becomes energetically favorable to create a new pair of quarks ($q\bar{q}$) in the region between the quark and the residual "proton". Now the the newly created $\bar{q}$ will be attracted to the quark that is being pulled out of the proton whereas the other newly created $q$ will be pulled back into the proton which will then constitute a normal proton again with 3 quarks. Meanwhile the $q$ that is being pulled out and the newly created $\bar{q}$ will become bound together as a meson - therefore the attempt to pull a quark out of a proton will result in a final state that has a meson and a proton. This is called color confinement - because you can never separate a single colored quark (or gluon) out of a proton or other hadron - all composite particles must be colorless - either a $q\bar{q}$ that is colorless (a meson) or three differently colored $q$'s that create a colorless proton or hadron. This strong color force is responsible for binding 3 quarks into hadrons (like protons or neutrons) or a $q\bar{q}$ into mesons.
Now when protons and neutrons are bound together in a nucleus, even though the proton and neutron as a whole is colorless, when they are close to each other a residual part of the color force will attract the proton and neutron together. This can be modeled as the exchange of $\pi$ mesons between the nucleons and since the pion has a mass, this will result in a short range force that will vary as:
$F(r)=\tfrac{\pm K}{r^2}e^{-r/m} \ \ \ $ where $m$ is the mass of the pion.
This residual color force is responsible for nuclear binding.
Now the weak interactions are mediated by the $W$ and $Z$ mesons which are much heavier than the pion by a factor of about 600 ($m_\pi \approx 130-135 MeV$ but $m_W \approx 91 GeV$ and $m_Z \approx 80 GeV$). Thus the weak force will also be of the form:
$F(r)=\tfrac{\pm K}{r^2}e^{-r/m} \ \ \ $ where $m$ is the mass of the $W$ or $Z$.
Now the coupling constant $K$ is about the same as the electromagnetic coupling constant, but since the range of the force is so small, it is a very weak force. In fact there are no known bound states that are held together by the weak force. The weak force mostly changes one type of particle into another type of particle. For example an electron can be converted into a neutrino ($\nu$) by a $W$ meson and one type of quark can change into another type of quark via a $W$ meson. This, for example, is how a free neutron decays into a proton plus an electron and a neutrino:
It is this ability to change the types of particles through the weak interactions that is most significant for "force", the fact that the range of the force is so small is one of the reasons why the weak interaction force is so weak and the actual "force" part of the weak force is largely insignificant since it does not result in any bound states.This post imported from StackExchange Physics at 2015-03-23 09:23 (UTC), posted by SE-user FrankH
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I've solved this circuit but i got some issues about the alternatives.
English question at this picture:
Switch \$S_1\$ had been closed for long time and after that \$S_2\$ was opened. Therefore, at \$t=0\$, the switch \$S_2\$ has closed.
The current \$i_1(t)\$ is equal for \$t\geq 0\$
My attempt:
\$L\frac{di_1(t)}{dt} + i_1(t).1 = 2\Rightarrow\$ \$i(t) = C.e^{-t} + 2\$
\$i_1(t=0^-) = i_1(t=0^+) = 2V/2\Omega = 1A\$
\$C = -1\$
\$i_1(t) = -e^{-t} + 2\Rightarrow \boxed{i_1(t) = 2\left(1-\frac{e^{-t}}{2}\right)}\$
\$\tau = 1s\$
Is this alternatives about that question all wrong respecting time constant?
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The Eulerian Number $A(n,k)$ is the number of permutations of $1$ to $n$ with exactly $k$ rises (or ascents, i.e., positions $i$ such that $x_i<x_{i+1}$).
The article The Boundary of the Eulerian Number Triangle by Alexander Gnedin and Grigori Olshanski, presents the following asymptotic relation (Remark 12):
$$\lim_{n\rightarrow\infty} A(n,k)\sim (k+1)^n\qquad \text{for fixed $k=0,1,\ldots$}$$
According to them:
The relation can be readily checked directly. For instance, it follows from formula $$A(n,k)=\sum_{j=0}^k(-1)^j\binom{n+1}{j}(k+1-j)^n$$
I'm not an expert on asymptotic theory so I don't know if this is straightforward or not. Should I really compute the ratio $$\frac{A(n,k)}{(k+1)^n}=\frac{\sum_{j=0}^k(-1)^j\binom{n+1}{j}(k+1-j)^n}{(k+1)^n}$$ and manually prove that it converges to $1$? Or are there techniques to approach these kind of problems? I see some properties regarding products, quotients and powers... but the Eulerian Number formula is a sum, and sum does not behave well with asymptotics
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My textbook asked me to evaluate the integral $\int_1^5\left(\frac{x}{\sqrt{2x-1}}\right)$ using u-substitution. I rewrote the integrand as $x*(2x-1)^{-\frac{1}{2}}$, but I soon became stuck because I couldn't settle on what was $g'(x), g(x),$ or $F(x)$ (I didn't know how to define u or du). I checked the book (this was an example problem) and it said to set u = $\sqrt{2x-1}$. How did it come to this conclusion? Thanks for your help.
They are trying to get you to use the trick that setting $$u=\sqrt{2x-1}\implies\frac{u^2+1}{2}=x\implies u \space du=dx$$ Then $$I=\int_{x=1}^{x=5}\frac{x}{\sqrt{2x-1}}dx=\int_{u=1}^{u=3}\frac{\left(\frac{u^2+1}{2}\right)}{u}\cdot u\space du=\frac{1}{2}\int_1^3(u^2+1)\space du$$
In general, it can be difficult to identify what substitutions to perform, and even harder in most cases to identify them in terms of a sort of function composition. For some cases, composition is simple to notice: $$\int 2x e^{x^2}dx=\int e^udu=e^{x^2}+C$$ But in more complicated examples, $u$-substitutions can be difficult to immediately see. It is often helpful, instead of considering function composition, to see what parts of the integrand may be simplified or cancelled through a particular substitution. In your integral, substituting $u=\sqrt{2x-1}\implies du=\frac{dx}{\sqrt{2x-1}}=\frac{dx}{u}$ is helpful - the substitution transforms the $x$ in the numerator into a quadratic in $u$ (since $u=\sqrt{2x-1}\implies x=\frac{u^2+1}{2}$), the denominator simply becomes $u$, and we multiply by $u$ when we change from $dx$ to $u\,du$. The integral becomes: $$\int_1^5\frac{x}{\sqrt{2x-1}}\,dx=\frac{1}{2}\int_1^3\frac{u^2+1}{u}\,u\,du=\frac{1}{2}\int_1^3\left(u^2+1\right)du\text,$$ which we can integrate very easily. In this case, it may not have been obvious that this integral was an example of function composition, but it was much easier to play around with different substitutions and notice that this particular one led to some very convenient simplifications and cancellations.
Alternatively: $$\int_1^5\left(\frac{x}{\sqrt{2x-1}}\right) \ dx=\frac12\int_1^5\left(\frac{2x-1+1}{\sqrt{2x-1}}\right) \ dx=\\ \frac12\int_1^5\left[\sqrt{2x-1}+ \frac1{\sqrt{2x-1}} \right] \ dx=\\ \frac12\left[\frac13(2x-1)^{3/2}+(2x-1)^{1/2}\right]\bigg{|}_1^5=\\ \frac12\left[(9+3)-(\frac13+1)\right]=\frac{16}{3}.$$
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Maybe a little bit obvious but what I would do is the following:
$\textrm{Let:}$
$$\omega + \phi + \psi = \pi$$
Then take the tangent function to both sides.
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Since $\tan \pi = 0$, then:
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Now group:
$$\tan \left ( \left ( \omega + \phi + \right) + \psi \right) = \tan \left ( \pi \right)$$
Resolving we have:
$$\frac{\tan \left (\omega + \phi \right) + \tan \left (\psi \right) }{1-\tan \left( \omega + \phi \right) \tan \left (\psi \right)} = 0$$
In order to make the whole equation to zero, the numerator has to be zero as well, therefore just replace:
$$ \tan \left (\omega + \phi \right) + \tan \left (\psi \right) = 0$$
By expanding it:
$$\tan \left (\omega + \phi \right) = - \tan \left (\psi \right )$$
$$\frac{\tan \omega + \tan \phi}{1- \tan \omega \tan \phi} = - \tan \psi$$
$$\tan \omega + \tan \phi = \left( 1- \tan \omega \tan \phi \right) \left ( - \tan \psi \right )$$
Don't worry, we're almost there:
$$\tan \omega + \tan \phi = - \tan \psi + \tan \omega \tan \phi \tan \psi$$
$$\tan \omega + \tan \phi + \tan \psi = \tan \omega \tan \phi \tan \psi$$
Therefore we have proved the identity!.
By the way I used the angles $\omega$, $\phi$ and $\psi$ as I feel more comfortable working with them but in your case you may want them to be replaced by the letters $\textrm{A, B and C}$.
I hope this have helped you.
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78Skills to Develop
Calculate position vectors in a multidimensional displacement problem. Solve for the displacement in two or three dimensions. Calculate the velocity vector given the position vector as a function of time. Calculate the average velocity in multiple dimensions.
Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions. However, now they are vector quantities, so calculations with them have to follow the rules of vector algebra, not scalar algebra.
Displacement Vector
To describe motion in two and three dimensions, we must first establish a coordinate system and a convention for the axes. We generally use the coordinates x, y, and z to locate a particle at point P(x, y, z) in three dimensions. If the particle is moving, the variables x, y, and z are functions of time (t):
$$x = x(t) \quad y = y(t) \quad z = z(t) \ldotp \label{4.1}$$
The position vector from the origin of the coordinate system to point P is \(\vec{r}\)(t). In unit vector notation, introduced in Coordinate Systems and Components of a Vector, \(\vec{r}\)(t) is
$$\vec{r} (t) = x(t)\; \hat{i} + y(t)\; \hat{j} + z(t)\; \hat{k} \ldotp \label{4.2}$$
Figure 4.2 shows the coordinate system and the vector to point P, where a particle could be located at a particular time t. Note the orientation of the x, y, and z axes. This orientation is called a right-handed coordinate system (Coordinate Systems and Components of a Vector) and it is used throughout the chapter.
With our definition of the position of a particle in three-dimensional space, we can formulate the three-dimensional displacement. Figure 4.3 shows a particle at time t
1 located at P 1 with position vector \(\vec{r}\)(t 1). At a later time t 2, the particle is located at P 2 with position vector \(\vec{r}\)(t 2). The displacement vector \(\Delta \vec{r}\) is found by subtracting \(\vec{r}\)(t 1) from \(\vec{r}\)(t 2):
$$\Delta \vec{r} = \vec{r} (t_{2}) - \vec{r} (t_{1}) \ldotp \label{4.3}$$
Vector addition is discussed in Vectors. Note that this is the same operation we did in one dimension, but now the vectors are in three-dimensional space.
The following examples illustrate the concept of displacement in multiple dimensions
Example 4.1
Polar Orbiting Satellite
A satellite is in a circular polar orbit around Earth at an altitude of 400 km—meaning, it passes directly overhead at the North and South Poles. What is the magnitude and direction of the displacement vector from when it is directly over the North Pole to when it is at −45° latitude?
Strategy
We make a picture of the problem to visualize the solution graphically. This will aid in our understanding of the displacement. We then use unit vectors to solve for the displacement.
Solution
Figure 4.4 shows the surface of Earth and a circle that represents the orbit of the satellite. Although satellites are moving in three-dimensional space, they follow trajectories of ellipses, which can be graphed in two dimensions. The position vectors are drawn from the center of Earth, which we take to be the origin of the coordinate system, with the y-axis as north and the x-axis as east. The vector between them is the displacement of the satellite. We take the radius of Earth as 6370 km, so the length of each position vector is 6770 km.
In unit vector notation, the position vectors are
$$\vec{r}(t_{1}) = 6770 \ldotp \; km\; \hat{j}$$
$$\vec{r}(t_{2}) = 6770 \ldotp \; km (\cos 45°)\; \hat{i} + 6770 \ldotp \; km (\sin(−45°))\; \hat{j} \ldotp$$
Evaluating the sine and cosine, we have
$$\vec{r}(t_{1}) = 6770 \ldotp \hat{j}$$
$$\vec{r}(t_{2}) = 4787\; \hat{i} − 4787\; \hat{j} \ldotp$$
Now we can find \(\Delta \vec{r}\), the displacement of the satellite:
$$\Delta \vec{r} = \vec{r} (t_{2}) - \vec{r} (t_{1}) = 4787\; \hat{i} - 11,557\; \hat{j} \ldotp$$
The magnitude of the displacement is \(|\Delta \vec{r}| = \sqrt{(4787)^{2} + (-11,557)^{2}} = 12,509\; km\). The angle the displacement makes with the x-axis is \(\theta = \tan^{-1} \left(\dfrac{-11,557}{4787}\right) = -67.5^{o}\).
Significance
Plotting the displacement gives information and meaning to the unit vector solution to the problem. When plotting the displacement, we need to include its components as well as its magnitude and the angle it makes with a chosen axis—in this case, the x-axis (Figure 4.5).
Note that the satellite took a curved path along its circular orbit to get from its initial position to its final position in this example. It also could have traveled 4787 km east, then 11,557 km south to arrive at the same location. Both of these paths are longer than the length of the displacement vector. In fact, the displacement vector gives the shortest path between two points in one, two, or three dimensions.
Many applications in physics can have a series of displacements, as discussed in the previous chapter. The total displacement is the sum of the individual displacements, only this time, we need to be careful, because we are adding vectors. We illustrate this concept with an example of Brownian motion.
Example 4.2
Brownian Motion
Brownian motion is a chaotic random motion of particles suspended in a fluid, resulting from collisions with the molecules of the fluid. This motion is three-dimensional. The displacements in numerical order of a particle undergoing Brownian motion could look like the following, in micrometers (Figure 4.6):
$$\Delta \vec{r}_{1} = 2.0\; \hat{i} + \hat{j} + 3.0 \hat{k}$$
$$\Delta \vec{r}_{2} = - \hat{i} + 3.0\; \hat{k}$$
$$\Delta \vec{r}_{3} = 4.0\; \hat{i} -2.0\; \hat{j} + \hat{k}$$
$$\Delta \vec{r}_{4} = -3.0\; \hat{i} + \hat{j} + 3.0\; \hat{k} \ldotp$$
What is the total displacement of the particle from the origin?
Solution
We form the sum of the displacements and add them as vectors:
$$\begin{split} \Delta \vec{r}_{Total} & = \sum \Delta \vec{r}_{i} = \Delta \vec{r}_{1} + \Delta \vec{r}_{2} + \Delta \vec{r}_{3} + \Delta \vec{r}_{4} \\ & = (2.0 - 1.0 + 4.0 - 3.0)\; \hat{i} + (1.0 + 0 - 2.0 + 1.0)\; \hat{j} + (3.0 +3.0 + 1.0 + 2.0)\; \hat{k} \\ & = 2.0\; \hat{i} + 0\; \hat{j} + 9.0\; \hat{k}\; \mu m \ldotp \end{split}$$
To complete the solution, we express the displacement as a magnitude and direction,
$$| \Delta \vec{r}_{Total}| = \sqrt{2.0^{2} + 0^{2} + 9.0^{2}} = 9.2 \mu m, \quad \theta = \tan^{-1} \left(\dfrac{9}{2}\right) = 77^{o},$$
with respect to the x-axis in the xz-plane.
Significance
From the figure we can see the magnitude of the total displacement is less than the sum of the magnitudes of the individual displacements.
Velocity Vector
In the previous chapter we found the instantaneous velocity by calculating the derivative of the position function with respect to time. We can do the same operation in two and three dimensions, but we use vectors. The instantaneous
velocity vector is now
$$\vec{v} (t) = \lim_{\Delta t \rightarrow 0} \frac{\vec{r} (t + \Delta t) - \vec{r} (t)}{\Delta t} = \frac{d \vec{r}}{dt} \ldotp \label{4.4}$$
Let’s look at the relative orientation of the position vector and velocity vector graphically. In Figure 4.7 we show the vectors \(\vec{r}\)(t) and \(\vec{r}\)(t + \(\Delta\)t), which give the position of a particle moving along a path represented by the gray line. As \(\Delta\)t goes to zero, the velocity vector, given by Equation 4.4, becomes tangent to the path of the particle at time t.
Equation 4.4 can also be written in terms of the components of \(\vec{v}\)(t). Since
$$\vec{r} (t) = x(t)\; \hat{i} + y(t)\; \hat{j} + z(t)\; \hat{k},$$
we can write
$$\vec{v} (t) = v_{x} (t)\; \hat{i} + v_{y} (t)\; \hat{j} + v_{z} (t)\; \hat{k} \label{4.5}$$
where
$$v_{x} (t) = \frac{dx(t)}{dt}, \quad v_{y} (t) = \frac{dy(t)}{dt}, \quad v_{z} (t) = \frac{dz(t)}{dt} \ldotp \label{4.6}$$
If only the average velocity is of concern, we have the vector equivalent of the one-dimensional average velocity for two and three dimensions:
$$\vec{v}_{avg} = \frac{\vec{r} (t_{2}) - \vec{r} (t_{1})}{t_{2} - t_{1}} \ldotp \label{4.7}$$
Example 4.3
Calculating the Velocity Vector
The position function of a particle is \(\vec{r}\)(t) = 2.0t
2 \(\hat{i}\) + (2.0 + 3.0t) \(\hat{j}\)+ 5.0t \(\hat{k}\) m. (a) What is the instantaneous velocity and speed at t = 2.0 s? (b) What is the average velocity between 1.0 s and 3.0 s? Solution
Using Equation 4.5 and Equation 4.6, and taking the derivative of the position function with respect to time, we find
$$v(t) = \frac{d\vec{r} (t)}{dt} = 4.0t\; \hat{i} + 3.0\; \hat{j} + 5.0\; \hat{k}\; m/s$$$$\vec{v} (2.0\; s) = 8.0\; \hat{i} + 3.0\; \hat{j} + 5.0\; \hat{k}\; m/s$$$$Speed\; |\vec{v} (2.0\; s)| = \sqrt{8^{2} + 3^{2} + 5^{2}} = 9.9\; m/s \ldotp$$ From Equation 4.7, $$\begin{split} \vec{v}_{avg} & = \frac{\vec{r} (t_{2}) - \vec{r} (t_{1})}{t_{2} - t_{1}} = \frac{\vec{r} (3.0\;s ) - \vec{r} (1.0\; s)}{3.0\; s - 1.0\; s} = \frac{(18\; \hat{i} + 11\; \hat{j} + 15\; \hat{k}) m - (2\; \hat{i} + 5\; \hat{j} + 5\; \hat{k}) m}{2.0\; s} \\ & = \frac{(16\; \hat{i} + 6\; \hat{j} + 10\; \hat{k}) m}{2.0\; s} = 8.0\; \hat{i} + 3.0\; \hat{j} + 5.0\; \hat{k}\; m/s \ldotp \end{split}$$ Significance
We see the average velocity is the same as the instantaneous velocity at t = 2.0 s, as a result of the velocity function being linear. This need not be the case in general. In fact, most of the time, instantaneous and average velocities are not the same.
Exercise 4.1
The position function of a particle is \(\vec{r}\)(t) = 3.0t
3 \(\hat{i}\) + 4.0 \(\hat{j}\). (a) What is the instantaneous velocity at t = 3 s? (b) Is the average velocity between 2 s and 4 s equal to the instantaneous velocity at t = 3 s? The Independence of Perpendicular Motions
When we look at the three-dimensional equations for position and velocity written in unit vector notation, Equation 4.2 and Equation 4.5, we see the components of these equations are separate and unique functions of time that do not depend on one another. Motion along the x direction has no part of its motion along the y and z directions, and similarly for the other two coordinate axes. Thus, the motion of an object in two or three dimensions can be divided into separate, independent motions along the perpendicular axes of the coordinate system in which the motion takes place.
To illustrate this concept with respect to displacement, consider a woman walking from point A to point B in a city with square blocks. The woman taking the path from A to B may walk east for so many blocks and then north (two perpendicular directions) for another set of blocks to arrive at B. How far she walks east is affected only by her motion eastward. Similarly, how far she walks north is affected only by her motion northward.
Independence of Motion
In the kinematic description of motion, we are able to treat the horizontal and vertical components of motion separately. In many cases, motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.
An example illustrating the independence of vertical and horizontal motions is given by two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and it follows a curved path. A stroboscope captures the positions of the balls at fixed time intervals as they fall (Figure 4.8).
It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies vertical motion is independent of whether the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, not by any horizontal forces.) Careful examination of the ball thrown horizontally shows it travels the same horizontal distance between flashes. This is because there are no additional forces on the ball in the horizontal direction after it is thrown. This result means horizontal velocity is constant and is affected neither by vertical motion nor by gravity (which is vertical). Note this case is true for ideal conditions only. In the real world, air resistance affects the speed of the balls in both directions.
The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called
projectile motion, is to resolve it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. Contributors
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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In ASK, a carrier is multiplied by a set of discrete amplitudes, depending on the information bits. In practice, binary ASP (BASK) is often used, where one of the amplitudes is zero, i.e. for a digital \$0\$, the modulated signal is zero, and for a digital \$1\$ the modulated signal is the carrier multiplied with some fixed amplitude. This important special case of ASK is called on-off keying (OOK).
Digital PAM is a digital modulation format where a pulse is multiplied by the current data symbol. In this sense, ASK can be seen as a special case of digital PAM, where the pulse is a sinusoid with the carrier frequency over one symbol interval. In general, the digital PAM signal can be written as
$$s(t)=\sum_{k=0}^{\infty}A_kp(t-kT)\tag{1}$$
where \$A_k\$ are the discrete symbols, \$p(t)\$ is the pulse function, and \$T\$ is the symbol interval. Note that \$A_k\$ can be any set of discrete symbols. Usually the number of symbols is a power of \$2\$. E.g., if the number of symbols is \$2^m\$, each symbol carries \$m\$ bits. Note that the signal \$s(t)\$ in (1) can be modulated by a carrier. In the general case, the symbols \$A_k\$ can be complex-valued, and a passband signal is generated by
$$\tilde{s}(t)=\Re\{s(t)e^{j\omega_c t}\}\tag{2}$$
where \$\omega_c\$ is the carrier frequency in radians. Equations (1) and (2) are the general representation of digital passband PAM. Quadrature amplitude modulation (QAM) and phase shift keying (PSK) are special cases.
Note that the type of PAM explained in nidhin's answer is
analog PAM.
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Now showing items 11-20 of 51
Multiplicity dependence of jet-like two-particle correlations in pp collisions at $\sqrt s$ =7 and 13 TeV with ALICE
(Elsevier, 2017-11)
Two-particle correlations in relative azimuthal angle (Δ ϕ ) and pseudorapidity (Δ η ) have been used to study heavy-ion collision dynamics, including medium-induced jet modification. Further investigations also showed the ...
Electroweak boson production in p–Pb and Pb–Pb collisions at $\sqrt{s_\mathrm{NN}}=5.02$ TeV with ALICE
(Elsevier, 2017-11)
W and Z bosons are massive weakly-interacting particles, insensitive to the strong interaction. They provide therefore a medium-blind probe of the initial state of the heavy-ion collisions. The final results for the W and ...
Investigating the Role of Coherence Effects on Jet Quenching in Pb-Pb Collisions at $\sqrt{s_{NN}} =2.76$ TeV using Jet Substructure
(Elsevier, 2017-11)
We report measurements of two jet shapes, the ratio of 2-Subjettiness to 1-Subjettiness ($\it{\tau_{2}}/\it{\tau_{1}}$) and the opening angle between the two axes of the 2-Subjettiness jet shape, which is obtained by ...
Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
Multiplicity dependence of identified particle production in proton-proton collisions with ALICE
(Elsevier, 2017-11)
The study of identified particle production as a function of transverse momentum ($p_{\text{T}}$) and event multiplicity in proton-proton (pp) collisions at different center-of-mass energies ($\sqrt{s}$) is a key tool for ...
Probing non-linearity of higher order anisotropic flow in Pb-Pb collisions
(Elsevier, 2017-11)
The second and the third order anisotropic flow, $V_{2}$ and $V_3$, are determined by the corresponding initial spatial anisotropy coefficients, $\varepsilon_{2}$ and $\varepsilon_{3}$, in the initial density distribution. ...
The new Inner Tracking System of the ALICE experiment
(Elsevier, 2017-11)
The ALICE experiment will undergo a major upgrade during the next LHC Long Shutdown scheduled in 2019–20 that will enable a detailed study of the properties of the QGP, exploiting the increased Pb-Pb luminosity ...
Neutral meson production and correlation with charged hadrons in pp and Pb-Pb collisions with the ALICE experiment at the LHC
(Elsevier, 2017-11)
Among the probes used to investigate the properties of the Quark-Gluon Plasma, the measurement of the energy loss of high-energy partons can be used to put constraints on energy-loss models and to ultimately access medium ...
Azimuthally differential pion femtoscopy relative to the second and thrid harmonic in Pb-Pb 2.76 TeV collisions from ALICE
(Elsevier, 2017-11)
Azimuthally differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze-out, provide very important information on the ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
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Consider the functional $$J(y)=y^2(1)+\int_0^1y'^2(x)\,dx$$ with $y(0)=1$ , where $y\in C^2[0,1]$. If $y$ extremizes $J$ then find the value of $y(x)$.
I tried through Bolza problem. Firstly Euler-Lagrange equation gives , $y''=0$. So general solution is $y(x)=C_1x+C_2$. Now , $y(0)=1$ gives $C_2=1$. Then $y(x)=C_1x+1$.
Now Transversality condition gives , $\left[2y'(x)-2y(1)\right]_{x=1}=0\implies C_2=0.$ which is NOT possible, also I can not find $C_1$.
So how can I solve the problem ?
According to 'daw's' answer I am confused about the sign..
From Euler-Lagrange equation the solution is $y(x)=C_1x+1$. Putting this in the given $J(y)$ then for extreme value putting $\frac{dJ}{dC_1}=0$ find $C_1$. But I want to solve by Bolza-method..
Please help anyone.....
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One of the first articles on this blog was a sample Particle In Cell code on structured Cartesian mesh. In this article, I will try to outline the steps for developing a PIC code that runs on an unstructured mesh. Although I don’t purposely plan to glance over important details, there is much more math going into developing an unstructured code. In fact, the example code was originally developed for the Advanced PIC course, and we spent four lectures and some 200 slides covering the details. Feel free to leave a comment if something is not clear. Or even better,
register for the next Advanced PIC course (registration for 2016 course will go live soon)! Pros and Cons
Before getting into the implementation details, I wanted to say few things about pros and cons of unstructured PIC codes. Their primary benefit is the ability to resolve complex geometries. On a Cartesian mesh, we either let the smooth geometry degenerated into a sugar cube representation as shown in Figure 1, or we may develop techniques such as cut-cells. I tried this latter approach in my thesis with mixed results, at best. Unstructured meshes additionally allow us to naturally vary the cell size based on the expected plasma density. Accomplishing something similar with a structured mesh would require adaptive mesh refinement (AMR), leading to a complex mesh description.
Now to the cons. Firstly, particle pushing is much simpler on a structured mesh. This is due to the fact that with a simple (non-AMR) Cartesian mesh, there exists a direct analytical relationship between the particle position and the cell index. Thus, having the particle position, you can easily compute the cell index using relationship such as \(i=(x-x_0)/\Delta x\). No such relationship exists for unstructured meshes. Now each cell has a single index, and the cell index doesn’t have any meaning in regards to its position. In general, we need to find the particle cell by looping through all mesh elements and determining whether a point (the particle) lies within the cell volume. Obviously this is a slow process, given a typical mesh may have tens of thousands of elements. Below I describe a faster algorithm based on neighbor search.
The other drawback is that in the Finite Element Method, the derivative of the solution is available at one less
order than the order used in computing the solution. The example posted here uses linear (or first-order) elements. The electric field, the derivative of the potential, is defined only by zeroth-order functions. Zeroth order is a fancy word for constant. Thus, the electric field will be constant in each volume element. This means that as a particle moves from one side of a cell to the order, it does NOT experience a gradual variation in electric field, as is the case with the Finite Difference Method. Mesh Generation
One other difference between structured and unstructured codes is that in the latter case, you also need some way to generate the mesh. In the course, we used Salome, an open source mesher / library platform. The example models flow of ions past a charged sphere. We used Salome’s geometry module to first draw a 3D brick to represent the bounds of the simulation domain. We then used Boolean subtraction to cut a spherical hole in the brick. Next we switched to the meshing module and converted the geometry object into a tetrahedral mesh. This also resulted in creation of triangles on the outer surface. We sorted the triangles into two groups:
inlet on the \(z=0\) face, and sphere for triangles on the sphere. This allows us to later set appropriate Dirichlet boundary conditions. Salome supports export in several formats. We used the “dat” format, which is quite simple, but doesn’t support multiple groups in one file. As such, the volume and the two surface mesh groups were exported into three files:
mesh.dat,
inlet.dat, and
sphere.dat.
Mesh Loading
The listing below shows piece of
main function of the program. It starts with loading of the volume and surface meshes, exported from Salome.
/**************** MAIN **************************/ int main() { /*instantiate volume*/ Volume volume; if (!LoadVolumeMesh("mesh.dat",volume) || !LoadSurfaceMesh("inlet.dat",volume,INLET) || !LoadSurfaceMesh("sphere.dat",volume,SPHERE)) return -1; ... }
For example, the surface mesh loader is given below. The surface and volume mesh share same nodes so this code simply loads the node ids and sets the corresponding node to the prescribed type.
/*loads nodes from a surface mesh file and sets them to the specified node type*/ bool LoadSurfaceMesh(const string file_name, Volume &volume, NodeType node_type) { /*open file*/ ifstream in(file_name); if (!in.is_open()) {cerr<<"Failed to open "<<file_name<<endl; return false;} /*read number of nodes and elements*/ int n_nodes, n_elements; in>>n_nodes>>n_elements; cout<<"Mesh contains "<<n_nodes<<" nodes and "<<n_elements<<" elements"<<endl; int nn = volume.nodes.size(); /*read the nodes*/ for (int n=0;n<n_nodes;n++) { int index; double x, y, z; in >> index >> x >> y >> z; if (index<1 || index>nn) {cerr<<"Incorrect node number "<<index<<endl;continue;} volume.nodes[index-1].type=node_type; } cout<<" Done loading "<<file_name<<endl; return true; } Particle Pusher
The PIC code needs to know the cell containing each particle so that we can scatter particle densities and also gather electric fields. How do we find the containing cell on a tetrahedral mesh? Let’s consider a 2D analogy: triangular mesh. Imagine you have a triangle 123 as shown in Figure 3 below. Also imagine a particle located at point
p. We can subdivide the main triangle into 3 sub-triangles, in which one vertex is replaced by the point p. If p is inside the triangle, \(A_{p23}+A_{p31}+A_{p12}\approx A_{123}\). The approximation is for numerical errors associated with computer arithmetic. Alternatively, we can write \(L_1+L_2+L_3\approx 1\) where \(L_1 \equiv \dfrac{A_{p23}}{A_{123}}\) and so on.
We can do something similar for a tetrahedron. The difference is that since we are now dealing with 3D elements, we need to consider element volumes, not areas. Volume of a tetrahedron is given by a 4×4 determinant. A 4×4 determinant can be rewritten as a linear combination of 3×3 determinants as shown in the second slide in Figure 3. Following thesis of Santi, let point “1” correspond to our particle position and points 2,3,4 be the remaining volume nodes. This then allows us to pre-compute the four 3×3 determinants as they are functions of mesh-geometry only.
Finding the matching cell using a brute force search where we loop over all cells is obviously not computationally feasible. Instead, we can utilize a certain property of the volumes. The sub-volume such as \(V_{p234}\) will be negative if the point is completely behind the face, such 234. We can thus implement the following recursive neighbor search algorithm:
if inserting a particle for the first time, perform brute force search otherwise, check if particle still located in last known cell if not, find most negative basis check location in the neighbor cell corresponding to this basis
This neighbor search resulted in 43x speed up compared to the brute force for this example mesh. Listing for the XtoL function is given below.
The four basis functions
part.lc are used later to scatter particle weights to the mesh.
/*converts physical coordinate to logical, returns true if particle matched to a tet*/ bool XtoLtet(Particle &part, Volume &volume, bool search) { /*first try the current tetrahedron*/ Tetra &tet = volume.elements[part.cell_index]; bool inside = true; /*loop over vertices*/ for (int i=0;i<4;i++) { part.lc[i] = (1.0/6.0)*(tet.alpha[i] - part.pos[0]*tet.beta[i] + part.pos[1]*tet.gamma[i] - part.pos[2]*tet.delta[i])/tet.volume; if (part.lc[i]<0 || part.lc[i]>1.0) inside=false; } if (inside) return true; if (!search) return false; /*we are outside the last known tet, find most negative weight*/ int min_i=0; double min_lc=part.lc[0]; for (int i=1;i<4;i++) if (part.lc[i]<min_lc) {min_lc=part.lc[i];min_i=i;} /*is there a neighbor in this direction?*/ if (tet.cell_con[min_i]>=0) { part.cell_index = tet.cell_con[min_i]; return XtoLtet(part,volume); } return false; } Finite Element Field Solver
The particle pusher is an important part of the FEM-PIC code, but not the only one. The other important piece is the potential solver. Previously, I described the Finite Volume Method which can be used on non-Cartesian meshes. FVM Poisson solver requires secondary control volumes centered on face centroids, which may be difficult to compute for a completely unstructured mesh. For this reason, the Finite Element Method (FEM) is commonly used on unstructured meshes.
The FEM method is quite math heavy, and this section does not come even close to presenting a coherent description. I merely copied several of the 100+ slides used in the course to describe the method. The formulation is based on the Finite Element Method book by Hughes.
The first slide in Figure 4 shows what’s known as “Strong” form of a boundary value problem. This is the typical formulation, in which the differential equation is given along with a set of Dirichlet and Neumann boundary conditions. The objective is to find the function \(u\). The Finite Element Method is on the other hand based on Weak, or variational, formulation of the problem. This form, defines the problem using several integrals containing test functions (or variations). This weak form can be discretized using element shape functions for the variations, giving us the Galerkin form. This can be further written in a matrix form, giving us the Matrix form, shown in the last slide of Figure 4.
The Matrix form shows that the entire problem reduces to a matrix equation \(Kd=F\), where \(K\) is the global
stiffness matrix, \(F\) is the force vector, and \(d\) is the solution vector on the unknown nodes. It is possible to generate the global stiffness matrix by assembling in smaller stiffness matrixes defined for each element. The idea behind this assembly is summarized in Figure 5. The bulk of any FEM code is spent in building the element stiffness matrixes and force vectors, and then assembling them into the right place in the global matrix.
There are many additional details that are still missing. One such a detail is computing integrals from numerical data. The trapezoid rule could be used, but another rule known as Gaussian quadrature is generally superior. It is summarized in the first part of Figure 6. Computing the element stiffness matrix also requires computing derivatives of the shape functions. Part of this process is presented in the second image of Figure 6.
Implementation
As an example, the listing below shows the code used to compute the global K matrix and also pieces of the force vector F that are not dependent on the ion density. That piece is added later, in the Newton-Raphson solver.
/*preassembles the K matrix and "h" and "g" parts of the force vector*/ void FESolver::preAssembly() { /*loop over elements*/ for (int e=0;e<n_elements;e++) { Tetra &tet = volume.elements[e]; double ke[4][4]; for (int a=0;a<4;a++) for (int b=0;b<4;b++) { ke[a][b] = 0; /*reset*/ /*perform quadrature*/ for (int k=0;k<n_int;k++) for (int j=0;j<n_int;j++) for (int i=0;i<n_int;i++) { double nax[3],nbx[3]; double xi = 0.5*(l[i]+1); /*not used!*/ double eta = 0.5*(l[j]+1); double zeta = 0.5*(l[k]+1); getNax(nax,e,a,xi,eta,zeta); getNax(nbx,e,b,xi,eta,zeta); /*dot product*/ double dot=0; for (int d=0;d<3;d++) dot+=nax[d]*nbx[d]; ke[a][b] += dot*detJ[e]*W[i]*W[j]*W[k]; } } /*we now have the ke matrix*/ addKe(e,ke); /*force vector*/ double fe[4]; for (int a=0;a<4;a++) { /*second term int(na*h), always zero since support only h=0*/ double fh=0; /*third term, -sum(kab*qb)*/ double fg = 0; for (int b=0;b<4;b++) { int n = tet.con[b]; double gb = g[n]; fg-=ke[a][b]*gb; } /*combine*/ fe[a] = fh + fg; } addFe(F0, e,fe); } /*end of element*/ }
The above code is called from
main prior to commencing the main loop,
/**************** MAIN **************************/ int main() { /*instantiate volume*/ Volume volume; if (!LoadVolumeMesh("mesh.dat",volume) || !LoadSurfaceMesh("inlet.dat",volume,INLET) || !LoadSurfaceMesh("sphere.dat",volume,SPHERE)) return -1; /*instantiate solver*/ FESolver solver(volume); /*set reference paramaters*/ solver.phi0 = 0; solver.n0 = PLASMA_DEN; solver.kTe = 2; int n_elements = volume.elements.size(); int n_nodes = volume.nodes.size(); /*initialize solver "g" array*/ for (int n=0;n<n_nodes;n++) { if (volume.nodes[n].type==INLET) solver.g[n]=0; /*phi_inlet*/ else if (volume.nodes[n].type==SPHERE) solver.g[n]=-100; /*phi_sphere*/ else solver.g[n]=0; /*default*/ } /*sample assembly code*/ solver.startAssembly(); solver.preAssembly(); /*this will form K and F0*/ double dt = 1e-7; /*ions species*/ Species ions(n_nodes); ions.charge=1*QE; ions.mass = 16*AMU; ions.spwt = 2e2; /*main loop*/ int ts; for (ts=0;ts<200;ts++) { /*sample new particles*/ InjectIons(ions, volume, solver, dt); /*update velocity and move particles*/ MoveParticles(ions, volume, solver, dt); /*check values*/ double max_den=0; for (int n=0;n<n_nodes;n++) if (ions.den[n]>max_den) max_den=ions.den[n]; /*call potential solver*/ solver.computePhi(ions.den); solver.updateEf(); if ((ts+1)%10==0) OutputMesh(ts,volume, solver.uh, solver.ef, ions.den); cout<<"ts: "<<ts<<"t np:"<<ions.particles.size()<<"t max den:"<<max_den<<endl; } /*output mesh*/ OutputMesh(ts,volume, solver.uh, solver.ef, ions.den); /*output particles*/ OutputParticles(ions.particles); return 0; } Results
And putting it all together, we end up with a FEM PIC code. Results from the example code are shown below in Figure 7. The first image shows plasma potential and ion density on a center cut. The second image shows an animation of ion density versus time. I have yet to compare these results against the structured PIC code (this same problem is studied in PIC Fundamentals using a Cartesian code), so user beware.
Source Code
I have yet to do a comparison against the Cartesian code so “use at your risk.” Let me know if you find any bugs.
The code requires a subdirectory called
results for storing the VTK files. You can visualize these with Paraview. Use the following to compile and run:
mkdir results g++ -std=c++11 -O2 fem-pic.cpp -o fem-pic ./fem-pic
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OpenCV 4.1.2-pre
Open Source Computer Vision
Here, the matter is straight forward. For every pixel, the same threshold value is applied. If the pixel value is smaller than the threshold, it is set to 0, otherwise it is set to a maximum value. The function
cv.threshold is used to apply the thresholding. The first argument is the source image, which should be a grayscale image. The second argument is the threshold value which is used to classify the pixel values. The third argument is the maximum value which is assigned to pixel values exceeding the threshold. OpenCV provides different types of thresholding which is given by the fourth parameter of the function. Basic thresholding as described above is done by using the type cv.THRESH_BINARY. All simple thresholding types are:
See the documentation of the types for the differences.
The method returns two outputs. The first is the threshold that was used and the second output is the
thresholded image.
This code compares the different simple thresholding types:
The code yields this result:
In the previous section, we used one global value as a threshold. But this might not be good in all cases, e.g. if an image has different lighting conditions in different areas. In that case, adaptive thresholding thresholding can help. Here, the algorithm determines the threshold for a pixel based on a small region around it. So we get different thresholds for different regions of the same image which gives better results for images with varying illumination.
Additionally to the parameters described above, the method cv.adaptiveThreshold three input parameters:
The
adaptiveMethod decides how the threshold value is calculated:
The
blockSize determines the size of the neighbourhood area and C is a constant that is subtracted from the mean or weighted sum of the neighbourhood pixels.
The code below compares global thresholding and adaptive thresholding for an image with varying illumination:
Result:
In global thresholding, we used an arbitrary chosen value as a threshold. In contrast, Otsu's method avoids having to choose a value and determines it automatically.
Consider an image with only two distinct image values (
bimodal image), where the histogram would only consist of two peaks. A good threshold would be in the middle of those two values. Similarly, Otsu's method determines an optimal global threshold value from the image histogram.
In order to do so, the cv.threshold() function is used, where cv.THRESH_OTSU is passed as an extra flag. The threshold value can be chosen arbitrary. The algorithm then finds the optimal threshold value which is returned as the first output.
Check out the example below. The input image is a noisy image. In the first case, global thresholding with a value of 127 is applied. In the second case, Otsu's thresholding is applied directly. In the third case, the image is first filtered with a 5x5 gaussian kernel to remove the noise, then Otsu thresholding is applied. See how noise filtering improves the result.
Result:
This section demonstrates a Python implementation of Otsu's binarization to show how it works actually. If you are not interested, you can skip this.
Since we are working with bimodal images, Otsu's algorithm tries to find a threshold value (t) which minimizes the
weighted within-class variance given by the relation:
\[\sigma_w^2(t) = q_1(t)\sigma_1^2(t)+q_2(t)\sigma_2^2(t)\]
where
\[q_1(t) = \sum_{i=1}^{t} P(i) \quad \& \quad q_2(t) = \sum_{i=t+1}^{I} P(i)\]
\[\mu_1(t) = \sum_{i=1}^{t} \frac{iP(i)}{q_1(t)} \quad \& \quad \mu_2(t) = \sum_{i=t+1}^{I} \frac{iP(i)}{q_2(t)}\]
\[\sigma_1^2(t) = \sum_{i=1}^{t} [i-\mu_1(t)]^2 \frac{P(i)}{q_1(t)} \quad \& \quad \sigma_2^2(t) = \sum_{i=t+1}^{I} [i-\mu_2(t)]^2 \frac{P(i)}{q_2(t)}\]
It actually finds a value of t which lies in between two peaks such that variances to both classes are minimal. It can be simply implemented in Python as follows:
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$$ \lim_{x \to 0} \frac{\lvert2x-1\rvert - \lvert2x+1\rvert}{x} $$
Defining the function piecewise reveals the limit is in fact, continuous about 0
However when I go to solve it in a normal algebraic manner, the $2x$ terms are canceling, leaving me with an undefined output - 0 in the denominator.
Any hints would be fantastic, I've solved this every way I can think of and I keep getting different answers, none of which are the correct answer.
I did graph this, and it does show a continuous function around $0$ where $f(x) = -4$
My problem here is that I'm being a huge dunce about absolute values.
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(originally a comment, now extended and probably worth being an answer)
The lower right Dini derivate (what you want) would be $-2$ if the function was equal to $2x\sin(1/x)$ for $x>0,$ with $f(0)$ still equal to $0$ (and it doesn't matter what $f(x)$ is for $x < 0).$ See Calculating Dini derivatives for $f(x)=\begin{cases}x\,\sin{\left(\frac{1}{x}\right)} & x\neq 0\\ 0 & x=0\end{cases}$ for how the Dini derivates of the modified function can be found. Perhaps the place where you found this had incorrectly copied from the last page of this document or a similar one?
More generally, if
$$ g(x) = \begin{cases} ax \cdot \sin\left(\frac{1}{x}\right), & x < 0 \\ 0, & x=0 \\ bx \cdot \sin\left(\frac{1}{x}\right), & x > 0 \end{cases} $$
then the Dini derivates $D_{-},$ $D^{-},$ $D_{+},$ $D^{+}$ at $x=0$ are equal to $-|a|,$ $+|a|,$ $-|b|,$ $+|b|$, respectively.
And if $\alpha$ and $\beta$ are real numbers each greater than $1,$ and
$$ h(x) = \begin{cases} a|x|^{\alpha} \cdot \sin\left(\frac{1}{x}\right), & x < 0 \\ 0, & x=0 \\ bx^{\beta} \cdot \sin\left(\frac{1}{x}\right), & x > 0 \end{cases} $$
then the Dini derivates $D_{-},$ $D^{-},$ $D_{+},$ $D^{+}$ at $x=0$ are all equal to $0,$ and thus the two-sided derivative of $h(x)$ exists and equals $0$ at $x=0.$ (The reason for the absolute values when $x<0$ is to avoid problems with negative numbers raised to irrational powers.)
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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May 19th, 2016, 03:28 AM
# 1
Newbie
Joined: Jun 2014
From: UK
Posts: 20
Thanks: 0
Acceleration components of gravity vector
Hello,
for "fun", I thought I'd have a go at modelling the orbit of planets around the sun in Excel. I'm trying to complete this worksheet I found online:
Building Planetary Orbits
I've stumbled at the first hurdle, though; the solution requires me to devise equations for the $\displaystyle x$ and $\displaystyle y$ components of acceleration of a planet orbiting the Sun, from the equation for vector gravitational force:
$\displaystyle \vec{F_g}=-G\frac{M.m}{r^2}\hat{r}$
...where $\displaystyle G$ is a gravitational constant, $\displaystyle M$ is the mass of the Sun, $\displaystyle m$ is the mass of the planet, $\displaystyle \hat{r}$ is the unit vector between the Sun and the planet and $\displaystyle r$ is the radius of the planets orbit around the Sun.
I've come up with a solution (because I had another model that did something similar, so I just copied that) but I don't understand why it works:
$\displaystyle a_x=\vec{F_g}\frac{x}{r}$
$\displaystyle a_y=\vec{F_g}\frac{y}{r}$
...where $\displaystyle x$ and $\displaystyle y$ are the planets position in 2D space - I have a known starting position.
So...why do these two equations give me acceleration? I appeciate that this might be a very elementary maths question and, before anybody suggests it, I have an introductory book on orbital mechanics on order from Amazon
Any help would be appreciated, though,
Thanks
May 19th, 2016, 06:18 AM
# 2
Math Team
Joined: Jan 2015
From: Alabama
Posts: 3,264
Thanks: 902
What you have, $\displaystyle a_x= \vec{F_g}\frac{x}{r}$ and $\displaystyle a_y= \vec{F_g}\frac{y}{r}$, are clearly
incorrect for two different reasons. I presume you know that "force equals mass times acceleration" so that "acceleration equals force divided by mass". Since force is proportional to mass, acceleration due to gravity does not depend upon mass. But M is still part of $\displaystyle \vec{F_g}$ so that your components of acceleration are proportional to mass. Second, since both $\displaystyle a_x$ and $\displaystyle a_y$ are numbers times $\displaystyle \vec{F_g}$ you have both components pointing parallel to the force vector not parallel to the x and y axes as they should be.
May 19th, 2016, 06:33 AM
# 3
Newbie
Joined: Jun 2014
From: UK
Posts: 20
Thanks: 0
Well, that's a very definite $\displaystyle WRONG!$ then - thanks for clearing that up
Can you explain why they work? I use them to model the path of a planet around the Sun and the orbit is as expected...
More helpful, though, would be a suggestion as to what the formulae for $\displaystyle a_x$ and $\displaystyle a_y$ should be...
Thanks for your thoughts on this
May 19th, 2016, 08:07 AM
# 4
Newbie
Joined: Jun 2014
From: UK
Posts: 20
Thanks: 0
OK...
I've had a sit and a think about this (plus I read all of this - page 19 is very helpful!).
I've revised my calculations as follows:
Because $\displaystyle r=\sqrt{r_x^2+r_y^2}$ and $\displaystyle \hat{r}=\frac{\vec{r}}{r}$, we can derive $\displaystyle \hat{r_x}=\frac{r_x}{r}$ and $\displaystyle \hat{r_x}=\frac{r_x}{r}$
So $\displaystyle \vec{F_g}=\frac{G.M.m}{r^2}$ can be expanded into its x/y components as follows:
$\displaystyle F_g(x)=\frac{-r_x}{r}.F_g$
and
$\displaystyle F_g(y)=\frac{-r_y}{r}.F_g$
where $\displaystyle r_x$ and $\displaystyle r_y$ are the $\displaystyle x,y$ co-ordinates of the planet.
Thus, as we know $\displaystyle \vec{a}=\frac{\vec{F_g}}{M}$;
$\displaystyle a_x=\frac{F_g(x)}{M}$
and
$\displaystyle a_y=\frac{F_g(y)}{M}$
I've substituted these equations into my model and they also return the orbits expected - hooray!
The reason my model worked was because the mass $\displaystyle M$ I'm using is set to 1 for simplicity (i.e. the Sun=1 Solar Mass), so $\displaystyle \frac{F_g}{1}=F_g$ in this case...
I'm off to read the rest of this Newtonian Mechanics paper, anyway...
Thanks
Last edited by eigenvexed; May 19th, 2016 at 08:10 AM.
Reason: A sort of sudden epiphany...
May 19th, 2016, 09:40 AM
# 5
Math Team
Joined: Jul 2011
From: Texas
Posts: 3,018
Thanks: 1603
sorry, but $a \ne \dfrac{F_g}{M}$
$a = \dfrac{F_g}{m}$, where $m$ represents the planetary mass.
$F_g = -\dfrac{GMm}{r^2} \hat{r}$
$ma_r = -\dfrac{GMm}{r^2}$
$a_r = -\dfrac{GM}{r^2}$
$a_x = a_r \cos{\theta} = -\dfrac{GM \cos{\theta}}{r^2}$
$a_y = a_r \sin{\theta} = -\dfrac{GM \sin{\theta}}{r^2}$
where $\theta$ is the angle between positive x-axis and the orbital radius
May 19th, 2016, 10:51 PM
# 6
Newbie
Joined: Jun 2014
From: UK
Posts: 20
Thanks: 0
Ah, Ok - to be honest, I found the use of $\displaystyle M$ and $\displaystyle m$ a bit ambiguous in what I was reading - thanks for clearing that up.
The original equation for the gravitational force vector is:
$\displaystyle \vec{F_g}=G\frac{M.m}{r^2}$
I've read that, where $\displaystyle M$ is significantly greater than $\displaystyle m$, a gravitational field around the larger mass can be defined as $\displaystyle g=-\frac{GM}{r^2}\hat{r}$ which is where the equations I'm using then derive from - perhaps this is why they work, despite my transposing $\displaystyle M$ and $\displaystyle m$...?
The equations...
$\displaystyle a_x=a_rcos\theta$
and
$\displaystyle a_y=a_rsin\theta$
...were my first thought on how to approach this, too - the problem I found was that there was no easy way of calculating $\displaystyle \theta$...how do you do it?
Thanks for your help
May 20th, 2016, 01:26 AM
# 7
Math Team
Joined: Jul 2011
From: Texas
Posts: 3,018
Thanks: 1603
you don't need to calculate $\theta$ ...
$a_x=a_r\cos{\theta}=a_r \cdot \dfrac{x}{r} = -\dfrac{GM}{r^2} \cdot \dfrac{x}{r} = -\dfrac{GMx}{r^3}$
similarly, $a_y=-\dfrac{GMy}{r^3}$
Tags acceleration, components, gravity, vector
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The niche I have chosen is very competitive…
I can count the keywords on just one hand…
There are tons of irrelevant low comp keywords…
ABC Manuals, ABC Apps, etc…
however, I do not have apps or manuals on my site, so if
i were to target these keywords, then I feel they would not be buyer intent; they are simply looking up a manual, app, etc…
In other words, I would feel that it would be misleading, because if
they are looking for app or manual stuff, then land on unrelated…
Should I go for Irrelevant keywords?
I have a web form that is split in to several pages, it is an adaptation from a hard copy paper form, and the requirements are that the data captured should be the same.
For my web form I have asked for the users salutation/firstname/surname, I must also capture if they have a previous name.
The next page asks for their address, and also a previous address.
One thought that has been put forward is that most users won’t have a previous name, so this field is largely redundant and therefore wouldn’t need to be shown by default.
However, in my mind, if you are asking the user to action something “if” they have a previous name, then you are increasing the cognitive load, when it would actually just be easier to ignore the part of the form that isn’t relevant?
I will try and show you what I mean…
Or, Perhaps it would be better to do some user testing to find out? Would a simple user test be the best way to find out what works best?
Reviews and post from customers for a juice center are getting posted in the Google maps which aren’t relevant to my business profile. The juice center is below my office apartment. Please assist how to fix
Is there any way to format numbers in 123/Excel-compatible spreadsheet applications, Google Spreadsheets in particular, in a way that hides the fractional part
and the preceding decimal separator (i.e. dot or comma depending on locale) if it is zero (i.e. the number is an integer)? 123
0.0 → 123.0
0.# → 123.
this is the closest I get
0.? → 123.␠
0_.0 → 123 0
0_.# → 123
0_.? → 123 ␠
1.23
0.0% → 123.0%
0.#% → 123.%
0.?% → 123. %
0_.0% → 123 0%
0_.#% → 123 %
0_.?% → 123 %
I know that one can format positive, negative and zero-value numbers separately with semicolons:
+0.0%;−0.0%;±0_?%
I am currently trying to run a Java Project in Eclipse called Person (2 classes: Person and Greeter). When I run the code it is trying to run a project that is not even open. I went into that project, edited the code so that there no errors, and went back into my current project called Person to run it and the same issue arose. Something is sticking me to this old code project and I can’t run anything new. Even if I go into another project it still gives me errors on this other coding project, HelloGlobe, and I can’t go on!!!
Eclipse
Suppose I have two independent Brownian motions $ B^1_t, B^2_t$ and $ \mathbb F_t$ be the natural filtration generated by them. Let $ T > 0$ be a fixed finite number. Let $ q_t$ be a $ [-1,1]$ valued $ \mathbb{F}_t$ martingale that the analyst controls.
Let $ \mathcal Q$ be the set of $ [-1,1]$ valued $ \mathbb F_t$ martingales. The control problem is: $ $ \sup_{q \in \mathcal Q} E_{0,q_0} f(q_T, B^1_T)$ $
where $ q_0$ is the value of $ q$ at time $ 0$ . $ f(\cdot, \cdot)$ is linear in each argument. As we can see, the objective function does not depend on $ B^2_t$ .
I would conjecture that, given the linearity of $ f(\cdot, \cdot)$ in each of its argument and no dependence on $ B^2_t$ , that we can restrict attention to $ [-1,1]$ valued $ \sigma(B^1_t)$ martingales where $ \sigma(B^1_t)$ is a coarser filtration generated only by $ B^1_t$ .
Does that sound like a reasonable conjecture? And how can I argue this if true?
I have a problem concerning the “coverage” on Google Search Console. The website on the company I’m working for have weird errors on this page. Google is telling that it had issues referencing pages with an odd URL like you can see above :
http://www.rbs-france.fr/fr/assembleuse.php?ACF-510SOLO_+A+STAR+WARS+STORY+Extended+TV+Spot+Trailer+2018+Donald+Glover+Sci-Fi+MoEge+Kalbimdeki+A http://www.rbs-france.fr/fr/agrafeuse_manuelle.php?B40/4.Grapadora+Stanley+FMHT6-70868+TR75+Grapadora+6+en+1https://www.amazon.es/Stanley-FMHT6-70868-TR75-Grapadora/dp/B00975VA12/ref=sr_1_10?s=tools&ie=UTF8&qid=1480319901&sr=1-10&keywords=grapadora+cables.21+Ways+to+Hide+and+Organize+Things+in+your+House+ http://www.rbs-france.fr/fr/Thermorelieur_Fastback_Planax.php?FB-9-E-La+replica+della+Seredova+su+Twitter+e+fbLa http://www.rbs-france.fr/fr/destructeur_document_personnel.php?Securio-C18-1.9x15Dragon+Ball+Fighterz+VEGETA+62+HIT+COMBO+AND+C18+73+HIT+COMBOHope After looking a these URLs, I noticed that, the concerned pages are products currently online on the company’s website and an irrelevant suffix. I tried to reproduce this error but I couldn’t. For most of the errors, I’ve seen that the suffix concerns videos. I don’t know why these URLs have a such irrelevant suffix and how it happened. I tried to search on the Internet but I couldn’t see anyone with an issue like mine.
If someone here knows a way to understand or to solve my problem, I’ll be very thankfull !
I get too many irrelevant Facebook notifications from friends of the activities which do not involve me at all. This include status updates, posting pictures, sometime comment replies and what not. I am not sure if I get these from all friends or a chosen few.
This is what my notifications look like
These are my notification settings
I tried contacting Facebook support so many times. Never received any response.
“Get Notifications” option is already unchecked for these friends. I even tried enabling “Get Notifications” and then disabled it just to make Facebook aware of it but it didn’t stop notifications.
It looks a like a way to keep me active but instead it has made me leave Facebook for a very long time. What settings should I do to disable these notifications?
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Probability -- A Measure Theoretic Approach Probability using Measure Theory
A mathematically rigorous definition of probability, and some examples therein.
The Traditional Definition:
Consider a set \( \Omega \) (called the
sample space), and a function \( X:\Omega\rightarrow\mathbb{R} \) (called a random variable.
If \( \Omega \) is countable (or finite), a function \( \mathbb{P}:\Omega\rightarrow\mathbb{R} \) is called a
probability distribution if it satisfies the following 2 conditions:
For each \( x \in \Omega \), \( \mathbb{P}(x) \geq 0 \)
If \( A_i\cap A_j = \emptyset \), then \( \mathbb{P}(\bigcup\limits_0^\infty A_i) = \sum\limits_0^\infty\mathbb{P}(A_i) \)
And if \( \Omega \) is uncountable, a function \( F:\mathbb{R}\rightarrow\mathbb{R} \) is called a
probability distribution or a cumulative distribution function if it satisfies the following 3 conditions:
For each \( a,b\in\mathbb{R} \), \( a < b \rightarrow F(a)\leq F(b) \)
\( \lim\limits_{x\to -\infty}F(x) = 0 \)
\( \lim\limits_{x\to\infty}F(x) = 1 \)
The Intuition:
What idea are we even trying to capture with these seemingly disparate definitions for the same thing? Well, with the two cases taken separately it's somewhat obvious, but they don't seem to marry very well. The discrete case is giving us a pointwise estimation of something akin to the proportion of observations that should correspond to a value (in a perfect world). The continuous case is the same thing, but instead of corresponding to that particular value (which doesn't really even make sense in this case), the proportion corresponds to the point in question and everything less than it. The shaded region in the top picture below and the curve in the picture directly below it denote the cumulative density function of a standard normal distribution (don't worry too much about what that means for this post, but if you're doing anything with statistics, you should probably know a bit about that).
Another way to define a continuous probability distribution is through something called a probability density function, which is closer to the discrete case definition of a probability distribution (or
probability mass function). A probability density function is a function \( f:\mathbb{R}\rightarrow\mathbb{R} + \) such that \( \int{-\infty}^xf(t)dt = F(x) \). In other words, \( \frac{dF}{dX} = f \). This new function has some properties of our discrete case probability function, but lacks some others. On the one hand, they’re both defined pointwise, but on the other, this one can be greater than one in some places – meaning the value of the probability density function isn’t really the probability of an event, but rather (as the name “suggests”) the density therein. Does it measure up?
Now let’s check out the measure theoretic approach…
Let \( \Omega \) be our sample space, \( S \) be the \( \sigma \)-algebra on \( \Omega \) (so \( S \) is the collection of measurable subsets of \( \Omega \)), and \( \mu:S\to\mathbb{R} \) a measure on that measure space. Let \( X:\Omega\rightarrow\mathbb{E} \) be a random variable (\( \mathbb{E} \) is generally taken to be \( \mathbb{R} \) or \( \mathbb{R}^n \)). We define the function \( \mathbb{P}:\mathcal{P}(\mathbb{E})\rightarrow\mathbb{R} \) (where \( \mathcal{P}(\mathbb{E}) \) is the powerset of \( \mathbb{E} \) – the set of all subsets) such that if \( A\subseteq\mathbb{E} \), we have \( \mathbb{P}(A)=\mu(X^{-1}(A)) \). We call \( \mathbb{P} \) a
probability distribution if the following conditions hold:
\( \mu(\Omega) = 1 \)
for each \( A\subseteq\mathbb{E} \) we have \( X^{-1}(A)\in S \).
Why do this?
Well, right off the bat we have a serious benefit: we no longer have two disparate definitions of our probability distributions. Furthermore, there is the added benefit of having a natural separation of concerns: the measure \( \mu \) determines the what we might intuitively consider to be the probability distribution while the random variable is used to encode the aspects of the events that we care about.
To further illustrate this
The Examples A fair die All even
Let’s consider a fair die. Our sample space will be \( {1,2,3,4,5,6} \). Since our die is fair, we’ll define our measure fairly: for any \( x \) in our sample space, \( \mu({x}) = \frac{1}{6} \). If we want to know, for instance, what the probability of getting each number is, we could use a very intuitive random variable \( X(a) = a \) (so \( X(1)=1 \), etc.). Then we see that \( \mathbb{P}({1}) = \mu({1}) = \frac{1}{6} \), and the rest are found similarly.
Odds and Evens?
What if we want to consider the fair die of yester-paragraph, but we only care if the face of the die shows an odd or an even number? Well, since the actual distribution of the die hasn’t changed, we won’t have to change our measure. Instead we’ll change our random variable to capture just those aspects we care about. In particular, \( X(a) = 0 \) if \( a \) is even, and \( X(a) = 1 \) if \( a \) is odd. We then see \( \mathbb{P}(1) = \mu({1,3,5}) = \frac{1}{2} \) and \( \mathbb{P}(0) = \mu({2,4,6}) = \frac{1}{2} \)
Getting loaded All even
Now let’s consider the same scenario of wanting to know the probability of getting each number, but now our die is loaded. Being as how we’re changing the distribution itself and not just the aspects we’re choosing to care about, we’re going to want to change the measure this time. For simplicity, let’s consider a kind of degenerate case scenario. Let our measure be: \( \mu(A) = 1 \) if \( 1\in A \) and \( \mu(A)=0 \) if \( 1\notin A \). Basically, we’re defining our probability to be such that the only possible outcome is a roll of 1. So since we are concerned with the same things we were concerned with last time, we can take that same random variable. We note \( \mathbb{P}(1) = 1 \) and \( \mathbb{P}(a) = 0 \) for any \( a \neq 1 \).
Odds or evens
Try to do this one yourself. I’m going to go get some sleep now. Please feel free to contact me with any questions. I love doing this stuff, so don’t be shy!
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Calmet, Xavier, Fragkakis, Dionysios and Gausmann, Nina (2011)
The flavor of quantum gravity. European Physical Journal C, C71 (11). pp. 1781-1784. ISSN 1434-6044 Abstract
We develop an effective field theory to describe the coupling of non-thermal quantum black holes to particles such as those of the Standard Model. The effective Lagrangian is determined by imposing that the production cross section of a non-thermal quantum black hole be given by the usual geometrical cross section. Having determined the effective Lagrangian, we estimate the contribution of a virtual hole to the anomalous magnetic moment of the muon, $\mu \to e \gamma$ transition and to the electric dipole moment of the neutron. We obtain surprisingly weak bounds on the Planck mass due to a chiral suppression factor in the calculated low energy observables. The tightest bounds come from $\mu \to e \gamma$ and the limit on the neutron electric dipole moment. These bounds are in the few TeV region. However, the bound obtained from proton decay is much more severe and of the order of $1 \times 10^6$ GeV.
Item Type: Article Schools and Departments: School of Mathematical and Physical Sciences > Physics and Astronomy Subjects: Q Science Depositing User: Xavier Calmet Date Deposited: 17 Oct 2012 15:10 Last Modified: 29 Jul 2019 10:57 URI: http://sro.sussex.ac.uk/id/eprint/41179 📧Request an update
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Note this version of the R package and tutorial addresses two errors in the original paper: 1) The harmonic mean
p-value for set \(\mathcal{R}\) should be compared against a threshold \(\alpha_L\), not \(\alpha_{|\mathcal{R}|}\). 2) The mechanism for calculating asymptotically exact harmonic mean p-values was wrongly stated. The harmonicmeanp package version 2.0 only corrected error 1. Here both errors are addressed, and many of the results have changed quantitatively. In particular, correcting the errors makes the test less powerful than previously claimed for narrowing down regions of significance. See the updated correction for further information: http://blog.danielwilson.me.uk/2019/07/correction-harmonic-mean-p-value-for.html
The harmonic mean
p-value (HMP) is a method for performing a combined test of the null hypothesis that no p-value is significant (Wilson, 2019). It is more robust to dependence between p-values than Fisher’s (1934) method, making it more broadly applicable. Like Bonferroni correction, the HMP procedure controls the strong-sense family-wise error rate (ssFWER). It is more powerful than common alternatives including Bonferroni and Simes procedures when combining large proportions of all the p-values, at the cost of slightly lower power when combining small proportions of all the p-values. It rivals the power of the BH procedure (Benjamini and Hochberg, 1995) which controls both the weak-sense family-wise error rate (wsFWER) and the false discovery rate (FDR), in the sense that the HMP is expected to find one or more p-values or groups of p-values significant more often than the BH procedure finds one or more p-values significant.
Method Robust to dependence Indicative power 1 Controls Significance very rare Significance uncommon Significance common FDR wsFWER ssFWER Fisher \(\times\) \(\bullet\) \(\bullet\,\bullet\,\bullet\,\bullet\) \(\bullet\,\bullet\,\bullet\,\bullet\,\bullet\) \(\checkmark\) \(\checkmark\) \(\checkmark\) HMP \(\checkmark\) \(\bullet\,\bullet\,\circ\) \(\bullet\,\bullet\,\bullet\) \(\bullet\,\bullet\,\bullet\,\bullet\) \(\checkmark\) \(\checkmark\) \(\checkmark\) BH \(\checkmark\checkmark\) \(\bullet\,\bullet\,\circ\) \(\bullet\,\bullet\,\circ\) \(\bullet\,\bullet\,\circ\) \(\checkmark\) \(\checkmark\) \(\times\) Bonferroni \(\checkmark\checkmark\checkmark\) \(\bullet\,\bullet\,\circ\) \(\bullet\,\bullet\) \(\bullet\,\bullet\) \(\checkmark\) \(\checkmark\) \(\checkmark\)
There are two approaches to applying the HMP:
One compares \(\overset{\circ}{p}_{\mathcal{R}}\), the HMP of a set of
p-values \(\mathcal{R}\), to significance threshold \(\alpha_L\,w_\mathcal{R}\).
Equivalently, one compares \(p_{\overset{\circ}{p}_{\mathcal{R}}}\), an asymptotically exact HMP, to the significance threshold \(\alpha\,w_\mathcal{R}\).
In the above,
\(\overset{\circ}{p}_{\mathcal{R}}=\left(\sum_{i\in\mathcal{R}}w_{i}\right)/\left(\sum_{i\in\mathcal{R}}w_{i}/p_{i}\right)\) defines the HMP for the set of
p-values \(\mathcal{R}\). It is computed in R by \(\texttt{hmp.stat}\).
\(\alpha_L\) is a significance threshold that depends on the desired family-wise error rate \(\alpha\) and the total number of individual
p-values \(L\). It is computed in R by \(\texttt{qharmonicmeanp}\).
\(w_1 \dots w_L\) are weights for the individual
p-values that must sum to one, i.e. \(\sum_{i=1}^{L}w_{i}=1\), and \(w_\mathcal{R} = \sum_{i\in\mathcal{R}} w_i\) equals the sum of weights of the individual p-values in set \(\mathcal{R}\). The HMP is robust to the choice of weights, so it is reasonable to start with equal weights (\(w_{i}=1/L\)).
The asymptotically exact
p-value is computed by R using the \(\texttt{p.hmp}\) command, which takes into account the total number of individual p-values, \(L\).
In this tutorial, I will focus on the second approach because one usually wishes to quote in reports a
p-value that can be directly compared to the usual significance threshold, e.g. \(\alpha=0.05\). When the subscript \(\mathcal{R}\) is dropped from \(\overset{\circ}{p}_{\mathcal{R}}\) and \(p_{\overset{\circ}{p}_{\mathcal{R}}}\) it means \(\mathcal{R}\) includes all the p-values. This is the “headline” HMP.
The method may be used as follows:
The headline HMP is deemed significant when \(\overset{\circ}{p}\leq\alpha_L\) or, equivalently, \(p_{\overset{\circ}{p}}\leq\alpha\). Here significant means that we reject the null hypothesis that none of the
p-values are significant.
If the headline HMP is not significant, neither is the HMP for any subset. If the headline HMP is significant, subsets may also be significant. The significance thresholds are all pre-determined so the number of subsets that are tested does not affect them.
The HMP for a subset of
p-values is deemed significant when \(\overset{\circ}{p}_{\mathcal{R}}\leq\alpha_L\,w_{\mathcal{R}}\) or, equivalently, \(p_{\overset{\circ}{p}_{\mathcal{R}}}\leq\alpha\,w_{\mathcal{R}}\). Here significant means that we reject the null hypothesis that none of the p-values in subset \(\mathcal{R}\) are significant.
To use this tutorial, copy and paste the R code from your web browser to the R console. In the HTML version, you can select and copy R code simply by clicking within the code snippet (as long as JQuery is enabled in your web browser and the page was compiled with Rmarkdown v2). To ensure the tutorial runs correctly, execute each code snippet once, in order. This tutorial was compiled using Rmarkdown (Allaire et al., 2018), with equations rendered by Mathjax.
If you haven’t already installed the package, type at the R console:
install.packages("harmonicmeanp")
Once you have installed the package, load it in the usual way:
library(harmonicmeanp)
## Loading required package: FMStable
To check you have version 3.0 installed, type
stopifnot(packageVersion("harmonicmeanp")>=3.0)
Download the 312457
p-values from chromosome 12 of the genome-wide association study (GWAS) for neuroticism (Okbay et al., 2016). This file is an excerpt of http://ssgac.org/documents/Neuroticism_Full.txt.gz. For usage conditions see http://ssgac.org/documents/ReadMe_genetic_variants_associated_with_swb.txt. It took me a few seconds to download the data excerpt. The 8 megabyte file contains rs identifiers and SNP positions as per human genome build GRCh37/hg19 as well as the p-values.
system.time((gwas = read.delim("http://www.danielwilson.me.uk/files/Neuroticism_ch12.txt", header=TRUE,as.is=TRUE)))
## user system elapsed ## 0.730 0.024 0.881
head(gwas)
## rs pos p## 1 rs7959779 149478 0.3034## 2 rs4980821 149884 0.5905## 3 rs192950336 150256 0.1125## 4 rs61907205 151213 0.4896## 5 rs2368809 151236 0.7066## 6 rs4018398 151469 0.9420
The harmonic mean
p-value (HMP) is a statistic with which one can perform a combined test of the null hypothesis that none of the p-values is significant even when the p-values are dependent. In GWAS, p-values will often be dependent because of genetic linkage. The HMP can be used to test the null hypothesis that no SNPs on chromosome 12 are significant. Let’s do it manually by first calculating the HMP, assuming equal weights. Note that a total of L=6524432 tests were performed genome-wide, so this number must be used to determine the weights if we are to control the genome-wide ssFWER, even though we are only analysing the 312457 SNPs on chromosome 12 in this example.
L = 6524432gwas$w = 1/LR = 1:nrow(gwas)(HMP.R = sum(gwas$w[R])/sum(gwas$w[R]/gwas$p[R]))
## [1] 0.0008734522
One of the remarkable properties of the HMP is that for small values (e.g. below 0.05), the HMP can be directly interpreted as a
p-value after adjusting for multiple comparisons. That the HMP equals \(\overset{\circ}{p}_{\mathcal{R}}=0.0008734522\) suggests it is strongly significant before multiple testing correction. To test this formally, first the HMP significance threshold is computed. For that I will assume a false positive rate of \(\alpha=0.05\), i.e. 5%.
# Specify the false positive ratealpha = 0.05# Compute the HMP significance threshold(alpha.L = qharmonicmeanp(alpha, L))
## [1] 0.02593083
The multiple testing-adjusted threshold against which to evaluate the significance of the combined test is determined by the sum of the weights for the
p-values being combined. The HMP for subset \(\mathcal{R}\) is significant when \(\overset{\circ}{p}_\mathcal{R}\leq \alpha_L w_\mathcal{R}\).
# Test whether the HMP for subset R is significancew.R = sum(gwas$w[R])alpha.L * w.R
## [1] 0.001241835
Therefore
after adjusting for multiple comparison we can reject the null hypothesis of no association on chromosome 12 at level \(\alpha=0.05\) because 0.0008734522 is below 0.001241835.
An equivalent approach is to calculate an asymptotically exact
p-value based on the HMP.
# Use p.hmp instead to compute the HMP test statistic and# calculate its asymptotically exact p-value in one step# Note this line has changed because of a previous error.w.R*pharmonicmeanp(HMP.R/w.R, L=L, lower.tail=TRUE)
## [1] 0.001343897
# Compare it to the multiple testing thresholdw.R*alpha
## [1] 0.002394515
The asymptotically exact
p-value of \(p_{\overset{\circ}{p}_{\mathcal{R}}}=0.001343897\) is close to the HMP of \(\overset{\circ}{p}_{\mathcal{R}}=0.0008734522\) and also significant because it is below \(0.002394515\). Note however that direct interpretation of the HMP is anti-conservative compared to the asymptotically exact test, which is why the HMP had to be compared directly to the more stringent threshold \(\alpha_L=0.02593083\). The asymptotically exact p-value can be computed in one step:
# Note that the p.hmp function has been redefined to take argument L. Omitting L will issue a warning.R = 1:nrow(gwas)p.hmp(gwas$p[R],gwas$w[R],L)
## p.hmp ## 0.001343897
The combined
p-value for chromosome 12 is useful because if the combined p-value is not significant, neither is any constituent p-value, after multiple testing correction, as always. Conversely, if the combined p-value is significant, there may be one or more subsets of constituent p-values that are also significant. These subsets can be hunted down because another useful property of the HMP is that the significance thresholds of these further tests are the same no matter how many combinations of subsets of the constituent p-values are tested. Specifically, for any subset \(\mathcal{R}\) of the L p-values, the HMP is compared against a threshold \(\alpha_L\,w_{\mathcal{R}}\) (equivalently, the asymptotically exact HMP is compared against a threshold \(\alpha\,w_{\mathcal{R}}\)), where \(w_{\mathcal{\mathcal{R}}}=\sum_{i\in\mathcal{R}}w_{i}\) and the \(w_{i}\)s are the weights of the individual p-values, constrained to sum to one. Assuming equal weights, \(w_{i}=1/L\), meaning that \(w_{\mathcal{R}}=\left|\mathcal{R}\right|/L\) equals the fraction of all tests being combined. In what follows I will mainly use the asymptotically exact p-values, rather than directly interpreting the HMP.
For example, separately test the
p-values occurring at even and odd positions on chromosome 12:
R = which(gwas$pos%%2==0)p.hmp(gwas$p[R],gwas$w[R],L)
## p.hmp ## 0.002658581
w.R = sum(gwas$w[R])alpha*w.R
## [1] 0.001200587
R = which(gwas$pos%%2==1)p.hmp(gwas$p[R],gwas$w[R],L)
## p.hmp ## 0.00230653
w.R = sum(gwas$w[R])alpha*w.R
## [1] 0.001193928
Neither of the two tests is significant individually: for even positions, the combined
p-value was \(p_{\overset{\circ}{p}_{\mathcal{R}}}=0.002658581\) which was above the significance threshold of \(\alpha\,w_{\mathcal{R}}=0.001200587\) and for odd positions, the combined p-value was \(p_{\overset{\circ}{p}_{\mathcal{R}}}=0.00230653\) which was above the significance threshold of \(\alpha\,w_{\mathcal{R}}=0.001193928\).
Comparing
p-values with different significance thresholds can be confusing. Instead, it is useful to calculate adjusted p-values, which are compared directly to \(\alpha\), the intended strong-sense familywise error rate. An adjusted p-value is simply divided by its weight w. For example:
R = which(gwas$pos%%2==0)p.R = p.hmp(gwas$p[R],gwas$w[R],L)w.R = sum(gwas$w[R])(p.R.adjust = p.R/w.R)
## p.hmp ## 0.11072
R = which(gwas$pos%%2==1)p.R = p.hmp(gwas$p[R],gwas$w[R],L)w.R = sum(gwas$w[R])(p.R.adjust = p.R/w.R)
## p.hmp ## 0.09659422
Now it is easy to see that both tests are non-significant, assuming \(\alpha=0.05\).
Of course it makes little sense to combine
p-values according to whether their position is an even or odd number. Instead we might wish to test the first 156229 SNPs on chromosome 12 separately from the second 156228 SNPs to begin to narrow down regions of significance.
R = 1:156229p.R = p.hmp(gwas$p[R],gwas$w[R],L)w.R = sum(gwas$w[R])(p.R.adjust = p.R/w.R)
## p.hmp ## 1
R = 156230:312457p.R = p.hmp(gwas$p[R],gwas$w[R],L)w.R = sum(gwas$w[R])(p.R.adjust = p.R/w.R)
## p.hmp ## 0.02842931
This is much clearer: only in the second half of the chromosome can we reject the null hypothesis of no significant
p-values at the \(\alpha=0.05\) level. For the first half of the chromosome, the adjusted p-value was \(p_{\overset{\circ}{p}_{\mathcal{R}}}/w_{\mathcal{R}}=1\). By the corrected definition of asymptotically exact HMPs, the adjusted p-value will not exceed 1, although in general while p-values must be 1 or below, adjusted p-values need not be. For the second half of the chromosome, the adjusted p-value was \(p_{\overset{\circ}{p}_{\mathcal{R}}}/w_{\mathcal{R}}=0.02842931\) which is below the standard significance threshold of \(\alpha=0.05\).
Note that it was completely irrelevant that we had already performed tests of even- and odd-positioned SNPs: as mentioned above, the significance thresholds are pre-determined by the \(w_{\mathcal{R}}\)’s no matter how many subsets of
p-values are tested and no matter in what combinations. We can test any subset of the p-values without incurring further multiple testing penalties. For example, let’s test 50 megabase windows overlaping at 10 megabase intervals. Testing overlapping versus non-overlapping windows has no effect on the significance thresholds, but of course it has an effect on the resolution of our conclusions and on the computational time.
# Define overlapping sliding windows of 50 megabase at 10 megabase intervalswin.50M.beg = outer(0:floor(max(gwas$pos/50e6-1)),(0:4)/5,"+")*50e6win.50M.beg = win.50M.beg[win.50M.beg+50e6<=max(gwas$pos)]# Calculate the combined p-values for each windowsystem.time({ p.50M = sapply(win.50M.beg,function(beg) { R = which(gwas$pos>=beg & gwas$pos<(beg+50e6)) p.hmp(gwas$p[R],gwas$w[R],L) })})
## user system elapsed ## 0.054 0.010 0.064
# Calculate sums of weights for each combined testsystem.time({ w.50M = sapply(win.50M.beg,function(beg) { R = which(gwas$pos>=beg & gwas$pos<(beg+50e6)) sum(gwas$w[R]) })})
## user system elapsed ## 0.042 0.006 0.049
# Calculate adjusted p-value for each windowp.50M.adj = p.50M/w.50M
Now plot them
# Took a few seconds, plotting over 312k pointsgwas$p.adj = gwas$p/gwas$wplot(gwas$pos/1e6,-log10(gwas$p.adj),pch=".",xlab="Position on chromosome 12 (megabases)", ylab="Adjusted significance (-log10 adjusted p-value)", ylim=sort(-log10(range(gwas$p.adj,p.50M.adj,na.rm=TRUE)))) arrows(win.50M.beg/1e6,-log10(p.50M.adj),(win.50M.beg+50e6)/1e6,len=0,col="#D3C991",lwd=2)# Superimpose the significance threshold, alpha, e.g. alpha=0.05abline(h=-log10(0.05),col="black",lty=2)# When using the HMP to evaluate individual p-values, the HMP threshold must be used,# which is slightly more stringent than Bonferroni for individual testsabline(h=-log10(qharmonicmeanp(0.05,L)),col="grey",lty=2)# For comparison, plot the conventional GWAS threshold of 5e-8. Need to convert# this into the adjusted p-value scale. Instead of comparing each raw p-value# against a Bonferonni threshold of alpha/L=0.05/6524432, we would be comparing each# against 5e-8. So the adjusted p-values p/w=p*L would be compared against# 5e-8*L = 5e-8 * 6524432 = 0.3262216abline(h=-log10(0.3262216),col="grey",lty=3)
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For a differential equation like this with real coefficients: $$\frac{d^2y}{dx^2}+p \frac{dy}{dx}+q y = 0$$ By solving $$\lambda ^2+p\lambda +q = 0$$ we obtain two eigenvalue $\lambda_1 = a+bi$ and $\lambda_2 = a-bi$, if the discriminant $\Delta$ is smaller than 0. then if we want to obtain two real function solutions linearly independent,
$e^{ax}\cos(bx), e^{ax}\sin(bx)$
Then question is for a differential equation with solution space of dimension $d$, is there any guarantee that the real sub-solution space is also a space of dimension $d$. Is it always possible to find $d$ linearly independent real solutions?
I have to add that the above is just an example. What I'm asking is the general nth order ODE with n linearly independent complex solutions. Is the dimension of its real solution space the same with the complex one?
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Banach Journal of Mathematical Analysis Banach J. Math. Anal. Volume 3, Number 1 (2009), 131-142. A generalization of the weak amenability of Banach algebras Abstract
Let $A$ be a Banach algebra and let $\varphi$ and $\psi$ be continuous homomorphisms on $A$. We consider the following module actions on $A$, $$a\cdot x=\varphi(a)x , \hspace{0.7cm} x\cdot a=x\psi(a) \hspace{1.5cm} (a,x\in A).$$ We denote by $A_{(\varphi,\psi)}$ the above $A$-module. We call the Banach algebra $A$, $(\varphi,\psi)$-weakly amenable if every derivation from $A$ into $(A_{(\varphi,\psi)})^*$ is inner. In this paper among many other things we investigate the relations between weak amenability and $(\varphi,\psi)$-weak amenability of $A$. Some conditions can be imposed on $A$ such that the $(\varphi'',\psi'')$-weak amenability of $A^{**}$ implies the $(\varphi,\psi)$-weak amenability of $A$.
Article information Source Banach J. Math. Anal., Volume 3, Number 1 (2009), 131-142. Dates First available in Project Euclid: 21 April 2009 Permanent link to this document https://projecteuclid.org/euclid.bjma/1240336430 Digital Object Identifier doi:10.15352/bjma/1240336430 Mathematical Reviews number (MathSciNet) MR2461753 Zentralblatt MATH identifier 1163.46034 Citation
Bodaghi, A.; Eshaghi Gordji, M.; Medghalchi, A. R. A generalization of the weak amenability of Banach algebras. Banach J. Math. Anal. 3 (2009), no. 1, 131--142. doi:10.15352/bjma/1240336430. https://projecteuclid.org/euclid.bjma/1240336430
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Minimum Time to Move Between Fixed Points Through a Point Moving Along x Axis
\[x\]axis between points A and B.Line segments joint to point P to points R and S as shown, and the lines PR and PS make angles
\[\theta\]and
\[\phi\]with the positive and negative
\[y\]axes respectively. How can you find the minimum time to move from R to Q?
\[tan \theta =\frac{AB-x}{b} \rightarrow sec^2 \theta \frac{d \theta}{dx}=- \frac{1}{b} \rightarrow \frac{d \theta}{dx}=- \frac{cos^2 \theta}{b}\]
\[tan \phi =\frac{x}{a} \rightarrow sec^2 \phi \frac{d \phi}{dx}= \frac{1}{a} \rightarrow \frac{d \phi}{dx}= \frac{cos^2 \phi}{a}\]
Then
\[\frac{d \theta}{d \phi}=\frac{\frac{d \theta}{dx}}{\frac{d \phi}{dx}}=\frac{- \frac{cos^2 \theta}{b}}{ \frac{cos^2 \phi}{a}}=- \frac{a cos^2 \theta}{b cos^2 \phi}\]
Let a particle move from R to P with speed
\[v_1\]and from P to Q with speed
\[v_2\].
The time taken to move from R to P is
\[t_1=\frac{b}{v_1 cos \theta}\].
The time taken to move from P to Q is
\[t_2=\frac{a}{v_2 cos \phi}\].
The time taken to move from R to Q is then
\[T=\frac{b}{v_1 cos \theta}+ \frac{a}{v_2 cos \phi}\].
\[\begin{equation} \begin{aligned} \frac{dT}{d \theta}&= \frac{b sec \theta tan \theta}{v_1 } + \frac{a}{v_2 sec \phi tan \phi} \frac{d \phi}{d \theta} \\ &= \frac{b sec \theta tan \theta}{v_1 } + \frac{a}{v_2 sec \phi tan \phi} (- \frac{b cos^2 \phi}{a cos^2 \theta}) \\ &= b sec^2 \theta (\frac{sin \theta}{v_1}-\frac{sin \phi}{v_2}) =0 \end{aligned} \end{equation}\].
Hence
\[\frac{sin \theta}{v_1}-\frac{sin \phi}{v_2} =0 \rightarrow \frac{sin \theta}{sin \phi}=\frac{v_1}{v_2}\]is a condition for what is obviously a minimum time.
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The Torque Construct: The Rotational Analog to Force
Consider a force
F exerted tangentially on the rim of a wheel or disk. The rim is at a distance r from the axis of rotation. We can formally define torque, represented by the Greek letter \(\tau\) , in terms of the force F and the distance r:
\[ \tau = r_{\perp}F = rF_{tangential}\]
where r is sometimes referred to as the
moment arm of this applied force—the further away from the axis a particular force is applied, the more torque is exerted, producing more change in rotational motion. Torque can be thought of as the “turning effectiveness of a force” or “rotational force.”
Figure 7.5.1
What happens if the applied force is not purely tangential, as in the second circle in figure 7.5.1? This force can be broken down into its tangential and radial components, \(F_{tangential}\) and \(F_{radial}\) . Note that the radial component \(F_{radial}\) of this force has no effect on the rotational motion of this disk! So, for any general force exerted a distance r from a rotation axis, it is only the tangential component of this force \((F_{tangential})\) that will affect rotational motion. The tangential component of the force can always be found with the appropriate trig function. If \(\theta\) is the angle between the applied force,
F, and r, the tangential component is \(F \sin\theta\) .
Figure 7.5.2
Torque, along with other angular variables, has vector properties. If we imagine the torque causing the object to rotate about an axis perpendicular to the plane defined by the force and the moment arm, \(r\) , we can use the same right-hand-rule introduced for finding the direction of \(\theta\) and \(\omega\) to find the direction of the torque \(\tau\) . If you curl the fingers of your right hand in the direction of rotation that the torque would cause, then your thumb points in the direction of the torque.
The Angular Momentum Construct:The rotational analog to momentum
It is useful to consider the angular momentum of both a point object as well as the angular momentum of extended objects. In either case, we need to be clear about the axis (or point) about which we are calculating the angular momentum
A particle with momentum p, located at position r from some point in space has
angular momentum L about that point with a magnitude given by
\[L=r_{\perp}p=rp_{tangential}\]
Note that the angular momentum is related to the linear momentum the same way as torque is related to force. Both
L and \(\tau\) depend on the choice of the point in space to which they are referenced. Like torque, angular momentum is a vector. Its direction is perpendicular to both r and p and is given by the RHR.
Figure 7.5.3
If a system has many parts, its total angular momentum is the vector sum of the angular momenta of all the parts:
\[ L = L_1 + L_2 + L_3 \dots = \sum L_i \]
A rigid object with rotational inertia I about some particular axis has an angular momentum about the same axis given by
\[ L = I\omega \]
The direction of \(L\) is parallel to the direction of \(\omega\). These directions are shown in the figure 7.5.4.
Figure 7.5.4 Rotational Inertia: The Rotational Analog to Mass
Recall that for translational motion an object with a large amount of inertia has a greater momentum than an object with a small amount of inertia, both moving at the same speed. Mass,
m, is the measure of inertia in translational motion. The rotational motion analogy to inertia is rotational inertia (or rotational mass), or in very technical language, moment of inertia. With a given net torque, \(\Sigma\tau\), different objects will experience different rotational accelerations.
The rotational inertia of an object does not depend solely on the amount of mass in the object, but on
how this mass is distributed about the axis of rotation.
For the simplest case of a point mass
m moving in a circle of radius r, its rotational inertia is given by:
\[I = mr^2 \]
This definition allows us to calculate the rotational inertia of
any object, provided we know the position r of every portion of its mass as measured perpendicularly with respect to the axis of rotation:
\[ I = m_1r_1^2 +m_2r_2^2 +m_3r_3^2 + \dots = \sum m_ir_i^2 \]
This looks a lot like calculus (which it is in the limit of infinitesimally small mass increments.) The table below gives the rotational inertia of several simple geometric shapes, as calculated in the limit of infinitesimal increments of mass using this equation.
Object Rotational Inertia Illustration Point mass m moving in radius r I=mr 2 Thin ring of mass m, radius r rotating about center I=mr 2 Thin rod of mass m, length L rotating about one end perpendicular to the rod I=\(\frac{1}{3}\)mL 2 Thin rod of mass m, length L rotating about one e perpendicular to the rod I=\(\frac{1}{12}\)mL 2 Disk of mass s, radius r, rotating about an axis perpendicular to disk though the center I=\(\frac{1}{2}\)mr 2 Sphere of mass m, radius r, rotating about an axis thorough the center I=\(\frac{2}{5}\)mr 2 Thin hollow spherical shell of mass m, radius r, about an axis through the center I=\(\frac{2}{3}\)mr 2
As seen from the formulas in the table, objects with the same mass can have very different rotational inertias, depending on how the mass is distributed with respect to the axis of rotation.
Also, it is possible for an object to
change its rotational inertia (e.g., a gymnast tucking in or extending arms and legs), which can lead to dramatic results as net torques are applied. The rotational inertia of a composite object is the sum of the rotational inertias of each component, all calculated about the same axis.
\[ I_{total} = I_1 + I_2 + I_3 + \dots \]
So for a ring and a disk stacked upon each other and rotating about the symmetry axis of both, the rotational inertia is :
\[ I_{total} = I_{ring} + I_{disk} \]
The SI units of rotational inertia are \( kg \cdot{m^2}\)
Angular Impulse: The Rotational Analog to Impulse
The angular analog to impulse, is
angular impulse:
\[ AngJ_{ext} = \int \tau_{ext}(t)dt \]
or, if the torque is constant with time, or we define an average torque \( \tau_{avg}\)
\[ AngJ_{ext} = \tau_{avg}\Delta t \]
A Statement of Angular Momentum Conservation:
\[ AngJ_{ext} = \int\tau_{ext}(t)dt = \Delta L_{system}\]
or
\[ AngJ_{ext} = \tau_{ext}\Delta t= \Delta L_{system}\]
Note
If the net external angular impulse acting on a system is zero, then there is no change in the total angular momentum of that system; otherwise, the change in angular momentum is equal to the net external angular impulse.
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In Bohmian mechanics, the initial configurations and the pilot wave determine the future of the system. Given a set of initial positions for particles and a set of arbitrary trajectories, can we define a pilot wave that guides the particles on that trajectories?
A funny question which has not really much meaning for physics, but serves more as a mathematical puzzle, I guess.
Let's look at the guiding equation for the configuration $Q \in \mathbb{R}^{3N}$ (without spin for simplicity, $\hbar=2m=1$):
$$ \frac{dQ(t)}{dt} = \Im \frac{\psi^* \nabla \psi}{|\psi|^2}(Q(t)) $$
Now, given an arbitrary differentiable (this at least should hold!) trajectory $Q(t)$, the mathematical question you ask is if there exists a function $\psi: \mathbb{R}^{3N} \to \mathbb{C}$ such that this equation is satisfied. I can't provide a proof here, but you can see that the left hand side is given and the right hand side is thus determined only at the trajectories, so there is a large freedom and there should be a large number of possible choices for $\psi$ that work.
It may be that there are possible integral curves obtainable from the guiding wave equation showing arbitrary trajectories to allow for quantum randomness, but this would be purely mathematical. Even in standard quantum theory, we can obtain solutions to the Schrödinger equation, would be wave functions, that are physically impossible and are discarded on that basis.
This is more pronounced in Bohmian Mechanics which is a realistic theory where the mathematics describes physical processes; including that the wave function describes the pilot wave. For a system the wave function can be known precisely and does not change state, (unless a measurement is made). This must apply to the pilot wave also. Each particle in a system is represented by the same wave function and thus guided by an identical pilot wave.
With the wave function known precisely the possible particle positions will show a probabilistic distribution according to the wave function squared. But we have no way of knowing a particle’s position within that distribution. If we take a measurement to find out, the wave function collapses to a different state and so we would now have a different pilot wave but for a different system, not the one under consideration. So while the initial particle positions are random the pilot wave is not.
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When I was first learning abstract algebra, the professor gave the usual sequence of results for polynomials over a field: the Division Algorithm, the Remainder Theorem, and the Factor Theorem, followed by the Corollary that if $D$ is an integral domain, and $E$ is any integral domain that contains $D$, then a polynomial of degree $n$ with coefficients in $D$ has at most $n$ distinct roots in $E$.
He then challenged us, as a homework, to go over the proof of the Factor Theorem and to point out exactly which, where, and how the axioms of a field used in the proof.
Every single one of us missed the fact that commutativity is used.
Here's the issue: the division algorithm (on either side),
does hold in $\mathbb{H}[x]$ (in fact, over any ring, commutative or not, in which the leading coefficient of the divisor is a unit). So given a polynomial $p(x)$ with coefficients in $\mathbb{H}$, and a nonzero $a(x)\in\mathbb{H}[x]$, there exist unique $q(x)$ and $r(x)$ in $\mathbb{H}[x]$ such that $p(x) = q(x)a(x) + r(x)$, and $r(x)=0$ or $\deg(r)\lt\deg(a)$. (There also exist unique $q'(x)$ and $s(x)$ such that $p(x) = a(x)q'(x) + s(x)$ and $s(x)=0$ or $\deg(s)\lt\deg(a)$.
The usual argument runs as follows: given $a\in\mathbb{H}$ and $p(x)$, divide $p(x)$ by $x-a$ to get $p(x) = q(x)(x-a) + r$, with $r$ constant.
Evaluating at $a$ we get $p(a) = q(a)(a-a)+r = r$, so $r=p(a)$. Hence $a$ is a root if and only if $(x-a)$ divides $p(x)$.
If $b$ is a root of $p(x)$, $b\neq a$, then
evaluating at $b$ we have $0=p(b) = q(b)(b-a)$; since $b-a\neq 0$, then $q(b)=0$, so $b$ must be a root of $q$; since $\deg(q)=\deg(p)-1$, an inductive hypothesis tells us that $q(x)$ has at most $\deg(p)-1$ distinct roots, so $p$ has at most $\deg(p)$ roots.
And
that is where we are using commutativity: to go from $p(x) = q(x)(x-a)$ to $p(b) = q(b)(b-a)$.
Let $R$ be a ring, and let $a\in R$. Then $a$ induces a set-theoretic map from $R[x]$ to $R$, "evaluation at $a$", $\varepsilon_a\colon R[x]\to R$ by evaluation:$$\varepsilon_a(b_0+b_1x+\cdots + b_nx^n) = b_0 + b_1a + \cdots + b_na^n.$$This map is a group homomorphism, and
if $a$ is central, also a ring homomorphism; if $a$ is not central, then it is not a ring homomorphism: given $b\in R$ such that $ab\neq ba$, then we have $bx = xb$ in $R[x]$, but $\varepsilon_a(x)\varepsilon_a(b) = ab\neq ba = \varepsilon_a(xb)$.
The "evaluation" map also induces a set theoretic map from $R[x]$ to $R^R$, the ring of all $R$-valued functions in $R$, with the pointwise addition and multiplication ($(f+g)(a) = f(a)+g(a)$, $(fg)(a) = f(a)g(a)$); the map sends $p(x)$ to the function $\mathfrak{p}\colon R\to R$ given by $\mathfrak{p}(a) = \varepsilon_a(p(x))$. This map is a group homomorphism, but it is
not a ring homomorphism unless $R$ is commutative.
This means that from $p(x) = q(x)(x-a) + r(x)$ we
cannot in general conclude that $p(c) = q(c)(c-a) +r(c)$ unless $c$ commutes in $R$ with $a$. So the Remainder Theorem may fail to hold (if the coefficients involved do not commute with $a$ in $R$), which in turn means that the Factor Theorem may fail to hold So one has to be careful in the statements (see Marc van Leeuwen's answer). And even when both of them hold for the particular $a$ in question, the inductive argument will fail if $b$ does not commute with $a$, because we cannot go from $p(x) = q(x)(x-a)$ to $p(b)=q(b)(b-a)$.
This is exactly what happens with, say, $p(x) = x^2+1$ in $\mathbb{H}[x]$. We are fine as far as showing that, say, $x-i$ is a factor of $p(x)$, because it so happens that when we divide by $x-i$, all coefficients involved centralize $i$ (we just get $(x+i)(x-i)$). But when we try to argue that any root different from $i$ must be a root of $x+i$, we run into the problem that we cannot guarantee that $b^2+1$ equals $(b+i)(b-i)$
unless we know that $b$ centralizes $i$. As it happens, the centralizer of $i$ in $\mathbb{H}$ is $\mathbb{R}[i]$, so we only conclude that the only other complex root is $-i$. But this leaves the possibility open that there may be some roots of $x^2+1$ that do not centralize $i$, and that is exactly what occurs: $j$, and $k$, and all numbers of the form $ai+bj+ck$ with $a^2+b^2+c^2=1$ are roots, and if either $b$ or $c$ are nonzero, then they don't centralize $i$, so we cannot go from $x^2+1 = (x+i)(x-i)$ to "$(ai+bj+ck)^2+1 = (ai+bj+ck+i)(ai+bj+ck-i)$".
And
that is what goes wrong, and there is where commutativity is hiding.
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Consider the following system consisting of a box sliding down a plane. The coefficient of friction between the plane and the box is $\mu$. A pendulum is attached to the top of the box as shown.
The acceleration of the box+pendulum is $a=g(\sin\theta-\mu\cos\theta)$ I believe.
In a non-inertial frame attached to the box, the free-body diagram for the pendulum is
My goal is to find the angle $\phi$ from the equilibrium of these 3 forces. I have to pick x and y axes to decompose these forces. If I pick the x axis along the fictitious force and the y perpendicular to it I get (I think)
$$m_P a+T\sin\phi=m_P g\sin\theta$$
and
$$T\cos\phi=m_{P}g\cos\theta$$
which I can then solve for $\phi$ :
$$\tan\phi=\frac{g\sin\theta-a}{g\cos\theta}=\mu$$
Is this right?
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An important piece of information is:
Theorem: $f$ is not continuous.
Proof: Observe that $f$ is invertible, because
$$f(f(f(f(x)))) = f(f(-x)) = x$$
and so $f \circ f \circ f = f^{-1}$. Any
continuous invertible function on $\mathbb{R}$ is either strictly increasing or strictly decreasing.
If $f$ is strictly increasing, then:
$1 < 2$ $f(1) < f(2)$ $f(f(1)) < f(f(2))$ $-1 < -2$
contradiction! Similarly, if $f$ is strictly decreasing, then:
$1 < 2$ $f(1) > f(2)$ $f(f(1)) < f(f(2))$ $-1 < -2$
contradiction! Therefore, we conclude $f$ is not continuous. $\square$
For the sake of completeness, the entire solution space for $f$ consists of functions defined as follows:
Partition the set of all positive real numbers into ordered pairs $(a,b)$ Define $f$ by, whenever $(a,b)$ is one of our chosen pairs, $f(0) = 0$ $f(a) = b$ $f(b) = -a$ $f(-a) = -b$ $f(-b) = a$
To see that every solution is of this form, let $f$ be a solution. Then we must have $f(0) = 0$ because:
Let $f(0) = a$. Then $f(a) = f(f(0)) = 0$ but $-a = f(f(a)) = f(0) = a$, and so $f(0) = 0$
If $a \neq 0$, then let $f(a) = b$. We have:
$f(b) = f(f(a)) = -a$ $f(-a) = f(f(b)) = -b$ $f(-b) = f(f(-a)) = a$
From here it's easy to see the set $\{ (a,f(a)) \mid a>0, f(a)>0 \}$ partitions the positive real numbers and so is of the form I describe above.
One particular solution is
$$ f(x) = \begin{cases} 0 & x = 0\\ x+1 & x > 0 \wedge \lceil x \rceil \text{ is odd}\\ 1-x & x > 0 \wedge \lceil x \rceil \text{ is even}\\ x-1 & x < 0 \wedge \lfloor x \rfloor \text{ is odd}\\ -1-x & x < 0 \wedge \lfloor x \rfloor \text{ is even}\end{cases}$$
e.g. $f(1/2) = 3/2$, $f(3/2) = -1/2$, $f(-1/2) = -3/2$, and $f(-3/2) = 1/2$.
(This works out to be Jyrki Lahtonen's example)
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Everyone keeps claiming that integer factoring is in $NP$ but I just don't get it... Even with the simplest algorithm (division with all integers up to $\sqrt{n}$) the complexity should be $\sqrt{n}\log(n)$... How is that not in $P$? Is there something I'm missing?
One of the things to remember when dealing with natural numbers (and others, but naturals are the central things here) is the encoding, and that the definitions of $P$ and $NP$ reference the length of the encoding of the input on a Turing Machine (or something closely equivalent).
So the input to integer factoring, as a decision problem, is typically two numbers $n$ and $k$ in $\mathbb{N}$, and the question is whether $n$ has a factor $d \leq k$.
So the magnitude of $n$ is $n$, but the
size of its encoding may be only $O(\log n)$ (for example, in binary). This is exponentially smaller than $n$ (i.e. if we take $n' = \log_{2} n$, then $n = 2^{n'}$).
So then the $\sqrt{n}\log n$ "obvious" algorithm runs in time $2^{\frac{n'}{2}}\cdot n'$, which is exponential in the input size.
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where $A,B\in\mathbb{R}^{n\times n}$.
My current solution is that this will only work iff $A$ and $B$ commute. Since:
$(AB)^\top = B^\top A^\top = B A$ $\ $ ($=AB$. iff $A$ and $B$ commute.)
I tried to ...
come up with a counterexample of a product of two symmetric matrices, that does not commute. However, my examples always commuted. prove that the product of two symmetric matrices does commute, but i didn't succeed there either: $(AB)_{ij}=\sum_{k=1}^n a_{ik}b_{kj} = \sum_{k=1}^n a_{ki}b_{jk} = \sum_{k=1}^n b_{jk}a_{ki} = (BA)_{ji}$
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The Lagrangian of chiral perturbation theory (with two quark flavors) is written using the following matrix $U$
$$U=e^{i\sigma^i\phi_i/f}$$ where $\sigma^i$ are the Pauli matrices, $\phi_i$ are three scalar fields and $f$ is a constant with mass dimension. $U$ is unitary, which makes the $\phi_i$ fields real.
The Lagrangian at $O(p^2)$ order is
$$\mathcal{L}_2=\frac{f^2}{4}tr(D_{\mu}U^{\dagger}D^{\mu}U)$$ the Lagrangian at next to leading order $O(p^4)$ is $$\mathcal{L}_4=\frac{l_1}{4}tr{}(D_{\mu}U^{\dagger}D^{\mu}U)tr(D_{\nu}U^{\dagger}D^{\nu}U)+\frac{l_2}{4}tr(D_{\mu}U^{\dagger}D_{\nu}U)tr(D^{\mu}U^{\dagger}D^{\nu}U)\\ +\frac{l_3+l_4}{16}\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\frac{l_4}{8}tr(D_{\mu}UD^{\mu}U^{\dagger})tr(\chi{}U^{\dagger}+U\chi^{\dagger})\\ +l_5tr(U^{\dagger}F_R^{\mu\nu}UF_{L\mu\nu})+i\frac{l_6}{2}tr(F_R^{\mu\nu}D_{\mu}UD_{\nu}U^{\dagger}+F_L^{\mu\nu}D_{\mu}U^{\dagger}D_{\nu}U)\\ -\frac{l_7}{16}\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2+\frac{h_1+h_3-l_4}{4}tr(\chi^{\dagger}\chi)\\ +\frac{h_1-h_3-l_4}{16}\bigg(\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2\\ -2tr(\chi{}U^{\dagger}\chi{}U^{\dagger}+U\chi^{\dagger}U\chi^{\dagger})\bigg)-\frac{4h_2+l_5}{2}tr(F_{R\mu\nu}F_R^{\mu\nu}+F_{L\mu\nu}F_L^{\mu\nu})$$
at next to leading order the coefficients $l_i$ and renormalize with loops with vertices coming only from the $O(p^2)$ Lagrangian like
$$l_i=l_i^r(\mu)+\frac{\gamma_i}{32\pi^2}\big(-\frac{1}{\epsilon}-\ln4\pi+\gamma_e-1\big)$$
where the gammas are
$$\gamma_1=1/3\qquad\gamma_2=2/3\qquad\gamma_3=-1/2\qquad\gamma_4=2\qquad\gamma_5=-1/6\qquad\gamma_6=-1/3\qquad\gamma_7=0$$
Now comes the question. Let's consider the same theory with the same Lagrangian but allowing the $\phi$ fields in $U$ to be complex.
My question is, would the $\gamma_i-s$ change?
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Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!!
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Is it possible to solve the linear PDE analytically \begin{equation} \frac{\partial u}{\partial z} + a \frac{\partial u}{\partial t} + \int_{0}^{t} e^{-\beta (t-t')} u(z,t') dt'=f(z,t), \end{equation} subject to the conditions \begin{equation} u(z,0) = 0; \quad u(L,t) = u_0(t), \end{equation} by applying the Laplace transform in time to reduce this to a regular inhomogeneous 1st order ODE in space?
Applying the Laplace transform we get
$$ U_z(z,s) +a(s U(z,s)-u(z,0))+\frac{1}{s+\beta}U(z,s) = F(z,s) $$
or
$$ U_z(z,s)+\left(a s+\frac{1}{s+\beta}\right)U(z,s) = F(z,s) $$
with the condition $U(L,s) = U_0(s)$
or
$$ U(z,s) = e^{-\frac{z (s (\beta +s)+1)}{\beta +s}} \left(\int_1^z e^{\frac{\zeta (s (\beta +s)+1)}{\beta +s}} F(\zeta ,s) \, d\zeta -\int_1^L e^{\frac{\zeta (s (\beta +s)+1)}{\beta +s}} F(\zeta ,s) \, d\zeta +U_0(s) e^{L \left(\frac{1}{\beta +s}+s\right)}\right) $$
After that you can find the anti-transform using residue theory.
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Take a branch of the function $f(t) = \sqrt{1-t^2}$ on the closed upper-half plane ($\bar{H}$) so that $f(i) = \sqrt{2} > 0$. Then, we can define
$\phi(z) = \int_0^{z} \frac{dt}{f(t)}$, where the path is taken in $\bar{H}$.
This function maps $\bar{H}$ to a half-infinite rectangular strip $[-\pi/2,\pi/2] \times [0,\infty)$.
I don't completely understand why this is the case.
Wouldn't
$\psi(z) = \int_0^{z} f(t) dt$ map $\bar{H}$ onto $[-\pi/4,\pi/4] \times (-\infty,0]$?
The argument for this is the same for that of $\phi$: The function $\psi$ maps $[-1,1]$ to $[-\pi/4,\pi/4]$. Every time $z$ passes a branch point along the real axis, the argument changes by $\pm\pi$, so this results in a $\pm\frac{\pi}{2}$ turn.
So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$? (More generally, why does the Scwarz Christoffel mapping need all those things appear in the denominator rather than in the numerator?)
p.s. I don't really understand why the the turns happen in the directions that they do rather than making like a stair-case like shape in the case of $\phi$ or $\psi$.
EDIT: Here are some references that helped me clarify this:
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You're doing a titration of the carbonic acid $\ce{H_2CO_3}$ which is the product of the dissolution of $\ce{CO_2}$ in water by a base here the sodium hydroxide. According to wikipedia for carbonic acid, $\ce{pKa_1=3.6}$, $\ce{pKa_2=6.3}$ and $\ce{pKa_3=10.32}$.
When the carbon dioxyde is in water, the solution is acid then your phenolphtalein is colorless. During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around $\ce{pH=8.2}$ (Ref). Then when you finish you're titration the color of the solution will change.
You need to make some prediction to find the theorical pH at the equivalence. In this case at the equivalence $\ce{[H^+]=\sqrt{Ka_1Ka_2}}=1.26 \times 10^{-10}$ then $$\ce{pH_{eq}=-\log_{10}(1.26 \times 10^{-10})=9.9}$$
So at this point the phenolphtalein must be pink and then no more acid (so carbone dioxyde) will be in your solution. You'll have only carbonate ion which will not react with sodium hydroxyde.
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The below-given problem is in black box setting means input is given by set of generators.
Given an abelian $p$-group $A$ and two matrices $U_1$ and $U_2$ in $R(A)$ such that the order of $U_1$ and $U_2$ are coprime with $p$, output an element $U \in R(A)$ such that $UU_1=U_2U$ if such an element exists.
$R(A)$ denotes the set of all automorphisms of $A$.
The brute force way seems to give a $\Theta(|A|^{\log |A|})$ bound.
Question: Is the above problem solvable in the polynomial time?
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I came across this problem
Find the number of solutions of $|\cos(x)|=\cot(x)+\frac{1}{\sin(x)}$ that lie in $[-4\pi,4\pi]$
in a PreCalculus book and I am looking for a simple solution.
What I can say: let $\displaystyle f(x)=|\cos(x)|$ and $\displaystyle g(x)=\cot(x)+\frac{1}{\sin(x)}$
The functions $f$ and $g$ are $2\pi$-periodic so it is enough to study the problem on $[0,2\pi]$. The function $g$ is undefined when $x$ is a multiple of $\pi$. $g(x)=\frac{1+\cos(x)}{\sin(x)}$, so $g$ has the same sign than $\sin$. Therefore, there is no solution on $(\pi,2\pi)$ and e can assume that $x\in(0,\pi)$. Let $c=\cos(x)$. Then $c$ is a root of $P(c)=c^4+2c+1$ and $c\in[-1,1]$. By Descartes' Rule of Signe, $P$ has no positive root and $0$ or $2$ negative roots (counted with multiplicity). $c=-1$ is one of them and Synthetic Division gives $P(c)=(c+1)(c^3-c^2+c+1)$. Let $Q(c)=c^3-c^2+c+1$. $Q(-1)\neq 0$, so $P$ has another negative root. By the Lower Bound Rule, $c=-1$ is a lower bound for the roots of $P$ and the other root, say $a$ is in $(-1,0)$. This means that the original equation has at most one solution ($x=\arccos(a)$) in $(\frac{\pi}{2},\pi)$. On the other hand, $f$ is decreasing $(\frac{\pi}{2},\pi)$, $g(x)=\cot(\frac{x}{2})$, so $g$ is decreasing on $(\frac{\pi}{2},\pi)$. Since $f(\frac{\pi}{2})=0<1=g(\frac{\pi}{2})$ and $f(\pi)=1>0=\cot(\frac{\pi}{2})$, we obtain that $f(x)=g(x)$ for exactly one value in $(\frac{\pi}{2},\pi)$.
Finally, there is exactly one solution in $[0,2\pi]$, so there are $4$ solutions in the given interval.
That's quite convoluted, so I was hoping for a way to shortcut the solution. Any idea?
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2019-07-18 17:03
Precision measurement of the $\Lambda_c^+$, $\Xi_c^+$ and $\Xi_c^0$ baryon lifetimes / LHCb Collaboration We report measurements of the lifetimes of the $\Lambda_c^+$, $\Xi_c^+$ and $\Xi_c^0$ charm baryons using proton-proton collision data at center-of-mass energies of 7 and 8 TeV, corresponding to an integrated luminosity of 3.0 fb$^{-1}$, collected by the LHCb experiment. The charm baryons are reconstructed through the decays $\Lambda_c^+\to pK^-\pi^+$, $\Xi_c^+\to pK^-\pi^+$ and $\Xi_c^0\to pK^-K^-\pi^+$, and originate from semimuonic decays of beauty baryons. [...] arXiv:1906.08350; LHCb-PAPER-2019-008; CERN-EP-2019-122; LHCB-PAPER-2019-008.- 2019-08-02 - 12 p. - Published in : Phys. Rev. D 100 (2019) 032001 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-07-02 10:45
Observation of the $\Lambda_b^0\rightarrow \chi_{c1}(3872)pK^-$ decay / LHCb Collaboration Using proton-proton collision data, collected with the LHCb detector and corresponding to 1.0, 2.0 and 1.9 fb$^{-1}$ of integrated luminosity at the centre-of-mass energies of 7, 8, and 13 TeV, respectively, the decay $\Lambda_b^0\to \chi_{c1}(3872)pK^-$ with $\chi_{c1}\to J/\psi\pi^+\pi^-$ is observed for the first time. The significance of the observed signal is in excess of seven standard deviations. [...] arXiv:1907.00954; CERN-EP-2019-131; LHCb-PAPER-2019-023; LHCB-PAPER-2019-023.- 2019-09-03 - 21 p. - Published in : JHEP 1909 (2019) 028 Article from SCOAP3: PDF; Fulltext: LHCb-PAPER-2019-023 - PDF; 1907.00954 - PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-06-21 17:31
Updated measurement of time-dependent CP-violating observables in $B^0_s \to J/\psi K^+K^-$ decays / LHCb Collaboration The decay-time-dependent {\it CP} asymmetry in $B^{0}_{s}\to J/\psi K^{+} K^{-}$ decays is measured using proton-proton collision data, corresponding to an integrated luminosity of $1.9\,\mathrm{fb^{-1}}$, collected with the LHCb detector at a centre-of-mass energy of $13\,\mathrm{TeV}$ in 2015 and 2016. Using a sample of approximately 117\,000 signal decays with an invariant $K^{+} K^{-}$ mass in the vicinity of the $\phi(1020)$ resonance, the {\it CP}-violating phase $\phi_s$ is measured, along with the difference in decay widths of the light and heavy mass eigenstates of the $B^{0}_{s}$-$\overline{B}^{0}_{s}$ system, $\Delta\Gamma_s$. [...] arXiv:1906.08356; LHCb-PAPER-2019-013; CERN-EP-2019-108; LHCB-PAPER-2019-013.- Geneva : CERN, 2019-08-22 - 42 p. - Published in : Eur. Phys. J. C 79 (2019) 706 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-06-21 17:07
Measurement of $C\!P$ observables in the process $B^0 \to DK^{*0}$ with two- and four-body $D$ decays / LHCb Collaboration Measurements of $C\!P$ observables in $B^0 \to DK^{*0}$ decays are presented, where $D$ represents a superposition of $D^0$ and $\bar{D}^0$ states. The $D$ meson is reconstructed in the two-body final states $K^+\pi^-$, $\pi^+ K^-$, $K^+K^-$ and $\pi^+\pi^-$, and, for the first time, in the four-body final states $K^+\pi^-\pi^+\pi^-$, $\pi^+ K^-\pi^+\pi^-$ and $\pi^+\pi^-\pi^+\pi^-$. [...] arXiv:1906.08297; LHCb-PAPER-2019-021; CERN-EP-2019-111.- Geneva : CERN, 2019-08-07 - 30 p. - Published in : JHEP 1908 (2019) 041 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-05-16 14:53 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-05-16 14:31
Measurement of $CP$-violating and mixing-induced observables in $B_s^0 \to \phi\gamma$ decays / LHCb Collaboration A time-dependent analysis of the $B_s^0 \to \phi\gamma$ decay rate is performed to determine the $CP$-violating observables $S_{\phi\gamma}$ and $C_{\phi\gamma}$, and the mixing-induced observable $\mathcal{A}^{\Delta}_{\phi\gamma}$. The measurement is based on a sample of $pp$ collision data recorded with the LHCb detector, corresponding to an integrated luminosity of 3 fb$^{-1}$ at center-of-mass energies of 7 and 8 TeV. [...] arXiv:1905.06284; LHCb-PAPER-2019-015; CERN-EP-2019-077; LHCb-PAPER-2019-015; CERN-EP-2019-077; LHCB-PAPER-2019-015.- 2019-08-28 - 10 p. - Published in : Phys. Rev. Lett. 123 (2019) 081802 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-04-10 11:16
Observation of a narrow pentaquark state, $P_c(4312)^+$, and of two-peak structure of the $P_c(4450)^+$ / LHCb Collaboration A narrow pentaquark state, $P_c(4312)^+$, decaying to $J/\psi p$ is discovered with a statistical significance of $7.3\sigma$ in a data sample of $\Lambda_b^0\to J/\psi p K^-$ decays which is an order of magnitude larger than that previously analyzed by the LHCb collaboration. The $P_c(4450)^+$ pentaquark structure formerly reported by LHCb is confirmed and observed to consist of two narrow overlapping peaks, $P_c(4440)^+$ and $P_c(4457)^+$, where the statistical significance of this two-peak interpretation is $5.4\sigma$. [...] arXiv:1904.03947; LHCb-PAPER-2019-014 CERN-EP-2019-058; LHCB-PAPER-2019-014.- Geneva : CERN, 2019-06-06 - 11 p. - Published in : Phys. Rev. Lett. 122 (2019) 222001 Article from SCOAP3: PDF; Fulltext: PDF; Fulltext from Publisher: PDF; Related data file(s): ZIP; Supplementary information: ZIP; External link: SYMMETRY დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-04-01 11:42
Observation of an excited $B_c^+$ state / LHCb Collaboration Using $pp$ collision data corresponding to an integrated luminosity of $8.5\,\mathrm{fb}^{-1}$ recorded by the LHCb experiment at centre-of-mass energies of $\sqrt{s} = 7$, $8$ and $13\mathrm{\,Te\kern -0.1em V}$, the observation of an excited $B_c^+$ state in the $B_c^+\pi^+\pi^-$ invariant-mass spectrum is reported. The state has a mass of $6841.2 \pm 0.6 {\,\rm (stat)\,} \pm 0.1 {\,\rm (syst)\,} \pm 0.8\,(B_c^+) \mathrm{\,MeV}/c^2$, where the last uncertainty is due to the limited knowledge of the $B_c^+$ mass. [...] arXiv:1904.00081; CERN-EP-2019-050; LHCb-PAPER-2019-007.- Geneva : CERN, 2019-06-11 - 10 p. - Published in : Phys. Rev. Lett. 122 (2019) 232001 Article from SCOAP3: PDF; Fulltext: PDF; Fulltext from Publisher: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-04-01 09:46
Near-threshold $D\bar{D}$ spectroscopy and observation of a new charmonium state / LHCb Collaboration Using proton-proton collisiondata, corresponding to an integrated luminosity of 9 fb$^{-1}$, collected with the~LHCb detector between 2011 and 2018, a new narrow charmonium state, the $X(3842)$ resonance, is observed in the decay modes $X(3842)\rightarrow D^0\bar{D}^0$ and $X(3842)\rightarrow D^+D^-$. The mass and the natural width of this state are measured to be \begin{eqnarray*} m_{X(3842)} & = & 3842.71 \pm 0.16 \pm 0.12~ \text {MeV}/c^2\,, \\ \Gamma_{X(3842)} & = & 2.79 \pm 0.51 \pm 0.35 ~ \text {MeV}\,, \end{eqnarray*} where the first uncertainty is statistical and the second is systematic. [...] arXiv:1903.12240; CERN-EP-2019-047; LHCb-PAPER-2019-005; LHCB-PAPER-2019-005.- Geneva : CERN, 2019-07-08 - 23 p. - Published in : JHEP 1907 (2019) 035 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2019-03-22 09:21
Search for lepton-universality violation in $B^+\to K^+\ell^+\ell^-$ decays / LHCb Collaboration A measurement of the ratio of branching fractions of the decays $B^+\to K^+\mu^+\mu^-$ and $B^+\to K^+e^+e^-$ is presented. The proton-proton collision data used correspond to an integrated luminosity of $5.0$ fb$^{-1}$ recorded with the LHCb experiment at centre-of-mass energies of $7$, $8$ and $13$ TeV. [...] arXiv:1903.09252; LHCb-PAPER-2019-009; CERN-EP-2019-043; LHCb-PAPER-2019-009 CERN-EP-2019-043.- Geneva : CERN, 2019-05-14 - 13 p. - Published in : Phys. Rev. Lett. 122 (2019) 191801 Article from SCOAP3: PDF; Fulltext: 1903.09252 - PDF; LHCb-PAPER-2019-009 - PDF; Fulltext from Publisher: PDF; Related data file(s): ZIP; Supplementary information: ZIP; დეტალური ჩანაწერი - მსგავსი ჩანაწერები
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I'm teaching an applied math class this summer and I want to take a short detour into finance (not my specialty at all); specifically the Black-Scholes model of stock movements. I want my students to be able to simulate stock movements and create some fun graphs. But I still have a couple basic questions about it myself.
If I have this right, the change $\Delta S$ in a stock price over a small time interval $\Delta t$ is posited to behave as
$\Delta S = \mu S \Delta t + \sigma \sqrt{\Delta t} \varepsilon S$
where $\mu = \text{drift rate}$, $\sigma = \text{volatility}$ (constant), and $\varepsilon$ is a fair coin flip resulting in $1$ and $-1$ (I prefer this incremental equation to a stochastic one, I'm not up on Ito's lemma and all that). $S_T$, the stock price at time $T$, is then (for fixed $\Delta t$) the random variable
$S_T = S_0 \left(1+\mu \Delta t + \sigma \sqrt{\Delta t} \right)^X \left(1+\mu \Delta t - \sigma \sqrt{\Delta t}\right)^{N-X}$
where $X$ is a binomial R.V. counting the number of $1$'s from the coin flips and $N = T/\Delta t$. Using the normal approximation for $X$ and letting $\Delta t \rightarrow 0$ gets us
$S_T = S_0 e^{(\mu-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$
where $Z$ is standard normal (or we could replace $\sqrt{T} Z$ with brownian motion $W$ for a dynamic model). From this we can compute the following expected values:
$\ln(E[S_T]) = \ln(S_0) + \mu T$
$E[\ln(S_T)] = \ln(S_0) + (\mu - \sigma^2/2) T$
First, a dumb question: if volatility isn't a concern, it's the second of these quantities that an investor is concerned with when deciding to purchase a stock, yes? i.e., the expected value of logarithmic returns is the correct measure of the performance of a stock, not log of the expected value?
My second question maybe isn't as dumb. Assuming what I just said is correct, does this imply that, in a world where this model held perfectly and that $\mu$ and $\sigma$ were known for all stocks and to all investors, that $\mu - \sigma^2/2 = r$, the risk-free rate? It would seem to, since a riskless bond with $\mu = r$ and $\sigma = 0$ is always available to investors. I'm asking because I'm curious and I'd like to say something intelligent to my students about the relationship between $\mu$ and $\sigma$ in this model, e.g higher $\sigma$ means higher $\mu$.
Another question. I'm told there's a magic wand called risk-neutral valuation which allows me instead to write
$\Delta S = r S \Delta t + \sigma \sqrt{\Delta t} \varepsilon^\star S$
where we've replaced $\mu$ with the risk-free rate and $\varepsilon^\star$ is a different random variable derived from the risk-neutral probability measure. I'll buy that for the moment. What confuses me is how, when deriving the Black-Scholes formula for European options, one arrives at the correct formula, even though we've replaced $\mu$ with $r$ but
not replaced $\varepsilon$ with $\varepsilon^\star$.
What I mean is, suppose you write
$S_T = S_0 e^{(r-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$
That is, replacing $\mu$ with $r$ but
not changing $\sigma$ to $0$ or changing anything in the second exponential (i.e. changing it to a different version of browning motion) If you use this expression to compute the discounted expected value of the payout of a European call option at strike price $K$
$e^{-rT}E[\text{max}(S_T - K,0)]$
one arrives at the correct formula for $C$, the Black-Scholes price of a European call option. But why should this be? Why should the fair value of such an option be arrived at by assuming $\mu = r$, but
still assuming $\sigma \neq 0$? I realize that letting $\sigma = 0$ makes an option pointless to begin with, but I really don't get why we are justified in letting $\mu = r$, instead of what $\mu$ actually is (whatever it may be).
Finally, anything intelligent you can tell me about how actual investors react to (their estimation of) $\mu$ and $\sigma$, within the context of this model, would be helpful.
Sorry for my naivety, the closest I've been to a stock market is flipping past CNBC on my couch. Thanks for any help.
UPDATE:
What you've both said makes good sense to me. A quick aside: is the Riesz Representation theorem the essential ingredient one uses to prove the existence of risk-neutral measure?
I'm still fuzzy on one thing though. I've not been through the Black-Scholes PDE/dynamic hedging argument in detail but I get the gist; setting up a self-financing risk-less portfolio by trading back and forth the derivative and the stock. And I'm sure that's the most conceptually sound and insightful way to derive the Black-Scholes formula for European options. But I'm not going to have time to go into this in class, so let's suppose instead we didn't know any of this Black-Scholes PDE stuff, nor the Feynman-Kac formula. Again assuming the model
$S_T = S_0 e^{(\mu - \sigma^2/2)T} e^{\sigma \sqrt{T} Z}$
is there a way to argue from simpler principles that the computation
$e^{-rt} E[\text{max}(S_T-K,0)]$
is a valid pricing for a European option, after replacing $\mu$ with $r$? Because honestly, if I'm actually out there selling these options and need to price them, and I have no education about any of this other than this model, and I know I can sell enough of them for the law of averages to win out, I'm doing this exact computation but leaving $\mu$ right where it is (this was my original guess as to how to price an option by the way, before learning the actual formula or the hedging argument). In fact, it seems to me that, whatever probabilistic model $S_T$ you believe in for a particular stock, this computation should lead you to your best guess as to the value of the option,
without any alterations such as letting $\mu = r$. Where am I going wrong here?
Finally, could you recommend some realistic values of $\mu$ and $\sigma$ for me to play around with my students? Do practical traders actually bother trying to estimate $\mu$ and $\sigma$?
Thanks for both of your help.
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The main problem lies in the "large logarithms". Indeed, suppose you want to calculate some quantity in Quantum Field Theory, for instance a Green Function. In perturbation theory this is something like:
$$\tilde{G}(p_1,...,p_n)=\sum_k g^k F_k(p_1,...,p_n)$$
for some generic functions $F$ and $g$ is the coupling constant. It's not enough to require a small $g$. You need small $g$ AND small $F$, for every value of the momenta $p$ (so for every value of the energy scale of your system).
A nice little calculation to understand this point. It's obvious that:
$$\int_0^\infty \frac{dx}{x+a}=[log(x+a)]_0^\infty=\infty$$
Let's use a cutoff:$$\int_0^\Lambda \frac{dx}{x+a}=log\frac{(\Lambda+a)}{a}$$
This is still infinite if the (unphysical) cutoff is removed. The whole point of renormalization is to show that a finite limit exist (this is "Fourier-dual" to send the discretization interval of the theory to zero). This quantity is finite:
$$\int_0^\Lambda \frac{dx}{x+a}-\int_0^\Lambda \frac{dx}{x+b} \rightarrow log\frac{b}{a}$$
But if $a \rightarrow \infty $ the infinite strikes back!So for a generic quantity F(p) regularized to F(p)-F(0) we want at least two things: that the coupling is small at that momentum $p$ and that $p$ is not far away from zero. But zero is arbitrary, we can choose an arbitrary (subtraction) scale. So we can vary this arbitrary scale $\mu$ in such a way that it is always near the energy scale we are probing.
Is convenient to take this scale $\mu$ at the same value of the renormalization scale. This is the energy at which you take some finiteness conditions (usually two conditions on the two point Green function and one condition on the 4 point one). The finiteness conditions are real physical measures at an arbitrary energy scale, so they fix the universe in which you live. If you change $\mu$ and you don't change mass, charge, ecc. you are changing universe. The meaning of renormalization group equations is to span the different subtraction points of the theory, remaining in your universe. And of course every physical quantity is independent of these arbitrary scale.
EDIT:Some extra motivations for the running couplings and renormalization group equations, directly for Schwartz:
The continuum RG is an extremely practical tool for getting partial results for high- order loops from low-order loops. [...]
Recall [...] that the difference between the momentum-space Coulomb potential V (t) at two scales, t1 and t2 , was proportional to [...]
ln t1 for t1 ≪ t2. The RG is able to reproduce this logarithm, and similar logarithms of physical quantities. Moreover, the solution to the RG equation is equivalent to summing series of logarithms to all orders in perturbation theory. With these all-orders results, qualitatively important aspects of field theory can be understood quantitatively. Two of the most important examples are the asymptotic behavior of gauge theories, and critical exponents near second-order phase transitions.
[...]
$$e^2_{eff}(p^2)=\frac{e^2_R}{1-\frac{e^2_R}{12 \pi^2}ln\frac{p^2}{\mu^2}}$$
$$e_R=e_{eff}(\mu)$$
 This is the effective coupling including the 1-loop 1PI graphs, This is called leading- logarithmic resummation.
Once all of these 1PI 1-loop contributions are included, the next terms we are missing should be subleading in some expansion. [...] However, it is not obvious at this point that there cannot be a contribution of the form $ln^2\frac{p^2}{\mu^2}$ from a 2-loop 1PI graph. To check, we would need to perform the full zero order calculation, including graphs with loops and counterterms. As you might imagine, trying to resum large logarithms beyond the leading- logarithmic level diagrammatically is extremely impractical. The RG provides a shortcut to systematic resummation beyond the leading-logarithmic level.
Another example: In supersymmetry you usually have nice (theoretically predicted) renormalization conditions at very high energy for your couplings (this is because you expect some ordering principle from the underlying fundamental theory, string theory for instance). To get predictions for the couplings you must RG evolve all the couplings down to electroweak scale or scales where human perform experiments. Using RG equations ensures that the loop expansions for calculations of observables will not suffer from very large logarithms.
A suggested reference: Schwartz, Quantum Field Theory and the Standard model. See for instance pag. 422 and pag.313.This post imported from StackExchange Physics at 2015-11-01 18:07 (UTC), posted by SE-user Rexcirus
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Learning Objectives
Solve for the centripetal acceleration of an object moving on a circular path. Use the equations of circular motion to find the position, velocity, and acceleration of a particle executing circular motion. Explain the differences between centripetal acceleration and tangential acceleration resulting from nonuniform circular motion. Evaluate centripetal and tangential acceleration in nonuniform circular motion, and find the total acceleration vector.
Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. For example, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the second, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating, although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors.
Centripetal Acceleration
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or \(\frac{d\vec{v}}{dt}\) ≠ 0. This is shown in Figure 4.18. As the particle moves counterclockwise in time \(\Delta\)t on the circular path, its position vector moves from \(\vec{r}\)(t) to \(\vec{r}\)(t + \(\Delta\)t). The velocity vector has constant magnitude and is tangent to the path as it changes from \(\vec{v}\)(t) to \(\vec{v}\)(t + \(\Delta\)t), changing its direction only. Since the velocity vector \(\vec{v}\)(t) is perpendicular to the position vector \(\vec{r}\)(t), the triangles formed by the position vectors and \(\Delta \vec{r}\), and the velocity vectors and \(\Delta \vec{v}\) are similar. Furthermore, since |\(\vec{r}\)(t) | = |\(\vec{r}\)(t + \(\Delta\)t)| and |\(\vec{v}\)(t)| = |\(\vec{v}\)(t + \(\Delta\)t)|, the two triangles are isosceles. From these facts we can make the assertion \(\frac{\Delta v}{v}\) = \(\frac{\Delta r}{r}\) or \(\Delta\)v = \(\frac{v}{r}\) \(\Delta\)r.
We can find the magnitude of the acceleration from
$$a = \lim_{\Delta t \rightarrow 0} \left(\dfrac{\Delta v}{\Delta t}\right) = \frac{v}{r} \left(\lim_{\Delta t \rightarrow 0} \dfrac{\Delta r}{\Delta t}\right) = \frac{v^{2}}{r} \ldotp$$
The direction of the acceleration can also be found by noting that as \(\Delta\)t and therefore \(\Delta \theta\) approach zero, the vector \(\Delta \vec{v}\)approaches a direction perpendicular to \(\vec{v}\). In the limit \(\Delta\)t → 0, \(\Delta \vec{v}\) is perpendicular to \(\vec{v}\). Since \(\vec{v}\) is tangent to the circle, the acceleration \(\frac{d \vec{v}}{dt}\) points toward the center of the circle. Summarizing, a particle moving in a circle at a constant speed has an acceleration with magnitude
$$a_{C} = \frac{v^{2}}{r} \ldotp \label{4.27}$$
The direction of the acceleration vector is toward the center of the circle (Figure 4.19). This is a radial acceleration and is called the
centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latin words centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”
Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration.
Example 4.10
Creating an Acceleration of 1 g
A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. What does the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward the center of the circular trajectory?
Strategy
Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetal acceleration.
Solution
Set the centripetal acceleration equal to the acceleration of gravity: 9.8 m/s
2 = \(\frac{v^{2}}{r}\).
Solving for the radius, we find
$$r = \frac{(134.1\; m/s)^{2}}{9.8\; m/s^{2}} = 1835\; m = 1.835\; km \ldotp$$
Significance
To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circular trajectory or increase its speed on its existing trajectory or both.
Exercise 4.5
A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge of the flywheel if it experiences a centripetal acceleration of 900.0 cm/s
2?
Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circular path. Typical centripetal accelerations are given in the following table.
Table 4.1 - Typical Centripetal Accelerations
Object Centripetal Acceleration (m/s 2 or factors of g) Earth around the Sun 5.93 x 10 -3 Moon around the Earth 2.73 x 10 -3 Satellite in geosynchronous orbit 0.233 Outer edge of a CD when playing 5.75 Jet in a barrel roll (2-3 g) Roller coaster (5 g) Electron orbiting a proton in a simple Bohr model of the atom 9.0 x 10 22 Equations of Motion for Uniform Circular Motion
A particle executing circular motion can be described by its position vector \(\vec{r}\)(t). Figure 4.20 shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle \(\theta\) with the x-axis. Vector \(\vec{r}\)(t) making an angle \(\theta\) with the x-axis is shown with its components along the x- and y-axes. The magnitude of the position vector is A = |\(\vec{r}\)(t)| and is also the radius of the circle, so that in terms of its components,
$$\vec{r} (t) = A \cos \omega \hat{i} + A \sin \omega t \hat{j} \ldotp \label{4.28}$$
Here, \(\omega\) is a constant called the
angular frequency of the particle. The angular frequency has units of radians (rad) per second and is simply the number of radians of angular measure through which the particle passes per second. The angle θ that the position vector has at any particular time is \(\omega\)t.
If T is the period of motion, or the time to complete one revolution (2\(\pi\) rad), then
Velocity and acceleration can be obtained from the position function by differentiation:
$$\vec{v} (t) = \frac{d \vec{r} (t)}{dt} = -A \omega \sin \omega t \hat{i} + A \omega \cos \omega t \hat{j} \ldotp \label{4.29}$$
It can be shown from Figure 4.20 that the velocity vector is tangential to the circle at the location of the particle, with magnitude A\(\omega\). Similarly, the acceleration vector is found by differentiating the velocity:
$$\vec{a} (t) = \frac{d \vec{v} (t)}{dt} = -A \omega^{2} \cos \omega t \hat{i} - A \omega^{2} \sin \omega t \hat{j} \ldotp \label{4.30}$$
From this equation we see that the acceleration vector has magnitude A\(\omega^{2}\) and is directed opposite the position vector, toward the origin, because \(\vec{a}\)(t) = −\(\omega^{2} \vec{r}\)(t).
Example 4.11
Circular Motion of a Proton
A proton has speed 5 x 10
6 m/s and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time t = 2.0 x 10 −7 s = 200 ns? At t = 0, the position of the proton is 0.175 m \(\hat{i}\) and it circles counterclockwise. Sketch the trajectory. Solution
From the given data, the proton has period and angular frequency:
$$T = \frac{2 \pi r}{v} = \frac{2 \pi (0.175\; m)}{5.0 \times 10^{6}\; m/s} = 2.20 \times 10^{-7}\; s$$
$$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2.20 \times 10^{-7}\; s} = 2.856 \times 10^{7}\; rad/s \ldotp$$
The position of the particle at t = 2.0 x 10
−7 s with A = 0.175 m is
$$\begin{split} \vec{r} (2.0 \times 10^{-7}\; s) & = A \cos \omega (2.0 \times 10^{-7}\; s) \hat{i} + A \sin \omega (2.0 \times 10^{-7}\; s) \hat{j}\; m \\ & = 0.175 \cos (2.856 \times 10^{7}\; rad/s) (2.0 \times 10^{-7}\; s) \hat{i} + 0.175 \sin (2.856 \times 10^{7}\; rad/s) (2.0 \times 10^{-7}\; s) \hat{j}\; m \\ & = 0.175 \cos (5.712\; rad) \hat{i} + 0.175 \sin (5.172\; rad) \hat{j}\; m \\ & = 0.147 \hat{i} - 0.095 \hat{j}\; m \ldotp \end{split}$$
From this result we see that the proton is located slightly below the x-axis. This is shown in Figure 4.21.
Significance
We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a different starting position were given, we would have a different final position at t = 200 ns.
Nonuniform Circular Motion
Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showing an acceleration in the direction of the motion.
In uniform circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. If the speed of the particle is changing as well, then we introduce an additional acceleration in the direction tangential to the circle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocity vector. If the speed of the particle is changing, then it has a
tangential acceleration that is the time rate of change of the magnitude of the velocity:
$$a_{T} = \frac{d |\vec{v}|}{dt} \ldotp \label{4.31}$$
The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially inward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a
total acceleration that is the vector sum of the centripetal and tangential accelerations:
$$\vec{a} = \vec{a}_{C} + \vec{a}_{T} \ldotp \label{4.32}$$
The acceleration vectors are shown in Figure 4.22. Note that the two acceleration vectors \(\vec{a}_{C}\) and \(\vec{a}_{T}\) are perpendicular to each other, with \(\vec{a}_{C}\) in the radial direction and \(\vec{a}_{T}\) in the tangential direction. The total acceleration \(\vec{a}\) points at an angle between \(\vec{a}_{C}\) and \(\vec{a}_{T}\).
Example 4.12
Total Acceleration during Circular Motion
A particle moves in a circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varies with time according to
$$v(t) = c_{1} - \frac{c_{2}}{t^{2}}, c_{1} = 4.0\; m/s, c_{2} = 6.0\; m \cdotp s \ldotp$$
What is the total acceleration of the particle at t = 2.0 s?
Strategy
We are given the speed of the particle and the radius of the circle, so we can calculate centripetal acceleration easily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of the tangential acceleration by taking the derivative with respect to time of |v(t)| using Equation 4.31 and evaluating it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration.
Solution
Centripetal acceleration is
$$v(2.0\; s) = \left(4.0 - \dfrac{6.0}{(2.0)^{2}}\right) m/s = 2.5\; m/s$$
$$a_{C} = \frac{v^{2}}{r} = \frac{(2.5\; m/s)^{2}}{2.0\; m} = 3.1\; m/s^{2}$$
directed toward the center of the circle. Tangential acceleration is
$$a_{T} = \Big| \frac{d \vec{v}}{dt} \Big| = \frac{2 c_{2}}{t^{3}} = \frac{12.0}{(2.0)^{3}} m/s^{2} = 1.5\; m/s^{2} \ldotp$$
Total acceleration is
$$|\vec{a}| = \sqrt{3.1^{2} + 1.5^{2}} m/s^{2} = 3.44\; m/s^{2}$$
and \(\theta\) = tan
−1 \(\left(\dfrac{3.1}{1.5}\right)\) = 64° from the tangent to the circle. See Figure 4.23. Significance
The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polar coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used for motion along curved paths, is discussed in detail later in the book.
Contributors
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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Effective band structure of random alloy InGaAs¶ Version: 2015.1
Random alloys are becoming technologically important materials. For example, SiGe is one of the best known thermoelectric materials, and InGaAs is a prime candidate to substitute or complement silicon in future CMOS devices. Simulations of both thermoelectric elements as well as transistor devices often refer to band structure parameters such as band gaps and effective masses.
In this tutorial, you will lean how to setup the structure of a random alloy andcalculate the so-called effective bandstructure. The tutorial will focus onIn
0.53Ga 0.47As and calculations will be performed with DFT using the TB09-MGGAexchange-correlation functional, in order to obtain accurate band gaps.
Note
In a normal band structure, each band is a line with zero width/broadening inboth the E- and
k-directions: Bloch’s theorem applies and the wavenumber k is a good quantum number. In a disordered alloy, Bloch’s theorem strictlydoes not apply, and one cannot a priory know if k is still a good quantumnumber or if the whole concept of a band structure is well defined.
In QuantumATK 2015 it is possible to calculate “effective band structures” of random alloys, following the method described in Refs. [PZ10][PZ12][HES11]. The purpose is to investigate whether or not the random alloy band structure is well defined, and if it is reasonable to extract band parameters.
Methodology¶
In the calculation of effective band structures, one considers a large super-cellcontaining many atoms. Here we consider a 3x3x3 repetition of the primitive FCCunit cell with 64 atoms. We model In
0.53Ga 0.47As as InAs with some of the indium atomsreplaced at random with gallium atoms. The In/Ga ratio is kept fixed at 0.53/0.47in all calculations.
From the band structure of the super-cell containing more than 500 electronic bands,one can “unfold” the band structure such that it only contains bands correspondingto the primitive cell. In the unfolding procedure, one calculates a spectral weight\(\mid \langle e^{ i \bf{k} \cdot \bf{r}} \mid \psi_{j,K} \rangle \mid\), where \(\mid \psi_{j,K} \rangle\) is an eigenstate of the super-cell at k-point \(K\).If the super-cell is simply a copy of a simple unit cell (e.g. a super-cell of InAs),the spectral weights will either be 0 or 1, and the band unfolding can be doneexplicitly. However, in the case of a disordered alloy, such as In
0.53Ga 0.47As,the band weights can be any real number between 0 and 1. Band structures of InAs¶
You will first calculate the normal band structure of a InAs, using the primitive (1x1x1) unit cell. Then a 3x3x3 InAs configuration is created, and you will show that the effective band structure of this repeated super-cell reproduces that of the primitive cell.
1x1x1 InAs¶ Select the ATK-DFTcalculator. Use MGGAfor exchange-correlation potential. Set up a 9x9x9 grid for k-point sampling. Use a SingleZetaPolarizedbasis set to speed up calculations. Select the Next, add an block. Open it and set the Brillouin zone route to “L, G, X”. Change the Default output file name to
InAs_1x1x1.hdf5.
Transfer the script to the Editorin order to specify the TB09-MGGA c-parameter. Locate the line
exchange_correlation = MGGA.TB09LDA
and change it to
exchange_correlation = MGGA.TB09LDA(c=0.91)
Run the calculation using the Job Manageror in a Terminal. It takes about eight minutes to complete the simulation.
Note
You can learn more about the TB09-MGGA exchange-correlation potential and the TB09-MGGA c-parameter from the tutorial Meta-GGA and 2D confined InAs.
3x3x3 InAs¶ The k-point sampling in the calculator should be 3x3x3. The default output file name should be
InAs_3x3x3.hdf5.
Results¶
The band structures for the 1x1x1 (left) and 3x3x3 (right) InAs configurations are shown below. The 3x3x3 band structure is folded at the zone boundaries, which makes it hard to locate the valleys corresponding to the X- and L-valley minima of the 1x1x1 configuration.
Effective band structure of 3x3x3 InAs¶ Add the Analysis from File block, open it,and select the existing HDF5 file
InAs_3x3x3.hdf5.
Add also the block. It will be crossed over, indicating that the block needs further editing.
Open the EffectiveBandstructure block and start editing it.
Select a 1x1x1 InAs configuration, either from the
Builderor from the LabFloor, and drag it to the black area named “Primitive configuration”.
Apply the following settings:
Points pr. segment: 101 Brillouin zone route: L, G, X E0: -2.0 eV E1: 2.75 eV Points: 501 The script is now done, so save it and run the calculation. It takes about three minutes to complete on a laptop.
Note
When running the calculation the folllowing warning will appear in the log file:
Warning: The calculation did not converge to the requested tolerance!
This is perfectly OK. A single SCF iteration is executed to obtain the basis functions in the primitive cell, which are used in the effective band structure calculation.
The EffectiveBandstructure object will now have appeared on the
LabFloor. It looks like this:
Select the object and use the
Effective Bandstructure Analyzerplugin to visualize it. Compare the figure (shown below) with the normal band structure plot of the 1x1x1 InAs shown above.
The analyzer shows a band density map of the
k,E( k) plane. Each band willadd a weight between 0 and 1 to the band density. At the \(\Gamma\)-point thereare three degenerate valence bands, and the band density becomes 3 (bright yellowin the figure above). The conduction band has a weight of 1.
In order to see all the bands with a weight close to 0, you can plot the band densityon a logarithmic scale. Right-mouse click on the plot and select
Log scale in thebottom of the menu, and the plot is shown as the left figure below. A lot of bandsappear in red color. By inspecting the colorbar you notice that these bands haveweights around 10 -9, which is essentially just numerical noise.Try to adjust the “Data range” (upper right corner) and set the minimum value to -6.The “noise bands” now disappear, as shown at the right figure below.
Important
So far, you have seen that the effective bandstructure works, and give what we expect: The effective band structure of the 3x3x3 repeated InAs is the same as the normal bandstructure of the simple 1x1x1 InAs.
In the next section you will learn how to setup a random InGaAs alloy – a system wherethe normal bandstructure cannot be calculated for a simple 1x1x1 configuration, and whereit is not
a priori obvious if the whole concept of a band structure is well-defined or not. In 0.53Ga 0.47As random alloy¶
You will now set up a In
0.53Ga 0.47As random alloy structure and calculate the effectivebandstructure. We shall also consider the sample averaged effective band structure,and analyze the consequences of finite band broadening. Random alloy¶
Add a new InAs configuration, and change the lattice constant to 5.8687 Å, which is the experimental value for In
0.53Ga 0.47As.
Right-click the new Stash item and copy it (you will need the copy later on for calculating the effective bandstructure).
Repeat the InAs configuration 3 times in each direction.
Next, select all the indium atoms by using thetool, and open the plugin. Apply the following settings:
Algorithm: Fixed fraction New element: Ga Percentage: 47% Algorithm:
Then click
Create.
In the new configuration, approximately 47% of the indium atoms have beensubstituted with gallium atoms in a random way. Transfer this In
0.53Ga 0.47Asrandom alloy configuration to the Script Generator . Effective band structure calculation¶ MGGA exchange correlation. 3x3x3 k-point sampling. SingleZetaPolarized basis set.
Drag-and-drop the copied 1x1x1 InAs configuration with the 5.8687 Å lattice constant from the
Builderonto the black area.
Apply these settings:
Points pr. segment: 101 Brillouin zone route: L, G, X E0: -2 eV E1: 2.75 eV Points: 501
Transfer the script to the
Editorand add the TB09-MGGA c-parameter. For In 0.53Ga 0.47As with a SingleZetaPolarized basis set, you should use c=1.01 to get approximately the experimental band gap of 0.74 eV. Results¶
Investigate the calculated effective band structure using the
Effective Bandstructure Analyzer.It should be similar to the figure on the left below. The bands are no longer weighted ateither 0 or 1, but with a value in between. In particular around the L- and X-point conductionband minima, there are clear deviations from the simple band structure seen for InAs. Try toadjust the data range and set the maximum value to 0.5. Then some of the bands become morevisible, as shown below on the right. Near the \(\Gamma\)-point, both the valence band and conduction bandsare well-defined, and the effect of the random disorder seems to be rather limited.
Note
The resulting effective bandstructure is not necessarily exactly the same as in the figure above due to the random alloy generation.
Sample averaged effective band structure¶
A quantitatively more reliable effective band structure should be obtained by repeating the above procedure a number of times to create different random configurations. Some python scripting can help to finish the task elegantly and efficiently.
Download the Python script
ingaas_ebs.py, which creates a number of random In
0.53Ga 0.47As alloysand performs the effective band structure calculation for each of them. You can thenaverage over the results. The default number of generated alloys (
number_of_runs) is 10.You will get even better results by increasing that number.
Run the script, and observe the EffectiveBandstructure objects appearin the file
InGaAs_3x3x3.hdf5 on the
LabFloor.Mark all the generated data objects and click the Effective Bandstructure Analyzer.The analyzer will now show the average of all the effective band structure densities.Try to adjust the colours by changing the data range and maximum value:
It is possible to trace the most significant bands with the heighest weight:Draw a box with the mouse like shown in the figure above. A progress bar will appearin the bottom of the window signalling that the effective band structure datais being analyzed for band tracing. After a little while a new window will appear,like the one shown below. The figure shows the traced bands plotted with a finite with.The center of the line is identified as a band. It can be exported to a text fileby right-clicking on the figure and choosing
Export data.
Tip
The band is calculated in the following way:
For each k, the band density is first smoothed by performing a convolution with a Lorentzian function (the broadening in energy is 0.2 eV). The peaks of this smoothed band density is found. A band is formed by selecting the peak energies closest to the peak energy found at the previous k-point (or, for the first k-point, closest to the center of the box you draw). Proceed to the next k-point and repeat the process.
The width is calculated as the second moment of the raw data (i.e. not smoothed) in an interval of -0.15–0.15 eV around the peak energy. The maximum width is thus 0.3 eV. Note that if two or more bands are close to each other, the calculated width will be slightly wrong, since the density from the other bands will contribute to the width. Try to draw a box around the highest valence band. The Band Trace figure will be quickly updated to appear like the figure below. Close to the \(\Gamma\)-point the senond-highest valence band erroneously contribute to the width of the first one.
Finite broadening¶
Contrary to a normal band structure, the effective bands will have a finite broadening.If the broadening is small, the band structure is well-defined; if the broadening is large,the concept of bands is no longer valid. The average effective band structure of anIn
0.53Ga 0.47As random alloy, as shown in the figure above, is well-defined close to the\(\Gamma\)-point. However, the conduction band is not well-defined around the L- and X-valley minima,where the bands have a finite broadening of \(w \approx\) 0.06 eV.
The well-defined effective band structure around the \(\Gamma\)-point can be understood in terms of the long wavelength present there. The local random variations are not “seen” by a long wavelength electron. On the contrary, at the X and L points, the wave functions vary on the atomic scale, and are affected by the disorder.
Analysis and consequences of finite band broadening¶
One can interpret the finite band broadening in terms of a life-time \(\tau = \hbar / w =\) 11 fs. Together with an approximate longitudinal effective mass \(m_l\)=1.5 \(m_e\) and transverse effective mass \(m_t\)=0.2 \(m_e\), both obtained from virtual crystal approximation (VCA) calculations, we can estimate the effective mean free path (MFP) as:
where the band velocity \(v_k = 1/\hbar \cdot \partial E(k)/\partial k\). Assuming a parabolic band \(E(k) = \hbar^2 k^2/2m\) around the valley minima, the mean free path as a function of energy, relative to the X- or L-valley minimum, is then:
At the valley minima, the mean free path vanishes, and at a constant energy surfaceof 0.01 eV above the X- or L-valley minima we obtain estimated mean free pathsof 1.5 nm and 0.5 nm in the transverse and longitudinal directions, respectively.Clearly, the random disorder will be a limiting factor for transport in these valleys.The MFP vs. energy can be plotted with the python script
mean_free_path.py.
Final comments¶
For scientific use we recommend to use even larger super-cells, e.g. 4x4x4 or even larger, which will be more like a real random alloy. The relatively small 3x3x3 cell used in this tutorial was chosen to make the calculations relatively fast. The small cell size might affect the results in a quantitive manner, but we do expect the main conclusions to remain unchanged.
Since the effective band structure close to the \(\Gamma\)-point has a narrow broadening, it is reasonable to assume that Bloch’s theorem applies here and that the random disorder has a small effect on the transport properties. In this case, more efficient studies of the band structure of random alloys can be performed with the virtual crystal approximation as explained in the tutorial vca_for_ingaas.
References¶
[HES11] M. W. Haverkort, I. S. Elfimov, and G. A. Sawatzky. Electronic structure and self energies of randomly substituted solids using density functional theory and model calculations. arXiv, pages 1109.4036, 2011. URL: http://arxiv.org/abs/1109.4036.
[PZ10] V. Popescu and A. Zunger. Effective band structure of random alloys. Phys. Rev. Lett., 104:236403, Jun 2010. doi:10.1103/PhysRevLett.104.236403.
[PZ12] V. Popescu and A. Zunger. Extracting e versus p k effective band structure from supercell calculations on alloys and impurities. Phys. Rev. B, 85:085201, Feb 2012. doi:10.1103/PhysRevB.85.085201.
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Let's consider the very basic of a Mean-Variance Portfolio:
$$ \text{max}_{x} (1-\lambda)\sum_i^n\mu_ix_i-\lambda\sum_i^n\sum_j^n x_i Q_{ij}x_j $$ $$\text{ s.t. }\sum_i^nx_i=1 \text{ , } x_i \geq 0 \text{ (No shorting) }$$ where $\lambda, \mu, Q$ is the risk averse parameter, expected return vector and variance-covariance matrix for all the assets.
I
$\sum_i^n\sum_j^n x_i Q_{ij}x_j$ as the variance of my protfolio, am I correct? Interpret
Let's assume
I invest 10\$ and my first constraint is normalizing the 10\$ to 1 so I have to multiply my results by 10. Let's assume I use software to find the optimal portfolio with the weights $x_1...x_n$ so the value of asset $i$ is $x_i*100\%$ of my total investment.
My expected return is then: $10\$\sum_i^n\mu_ix_i$
QUESTION: What is the variance (risk) of the portfolio then? Becuase of the non-linearity I cant simply multiply by 10.
My first thought is of course to multiply my weights $x_i$ with 10 and simply compute the variance by $\sum_i^n\sum_j^n x_i Q_{ij}x_j$:
EDIT But consider this case: I have a low $\lambda$ (I am "risk loving") so my solution to the problem is to invest all my money in asset $c$ where the expected return is $\mu_c=23\%=0.23$ and variance of the return is $Q_{c,c}=0.9$. This suggests that my variance is $10^2*0.9=90$ while expected to have a profit of 2.3\$. Am I correct? My concern is then that my variance or risk is so much higher initial investment so it makes me question if my math is correct.
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When a ray of light is projected, (say) from the surface of Earth to outside in space. The condition is that, there is no obstruction to it till
infinity(it travels only in vaccum). My question is that how far can that ray of light go?
Also, instead of a
ray of light,if I consider a beam of laserwith same conditions, then how far can a beam of laser go?
Compare both the situations.
And does the light(ray of light and beam of laser)
stopsafter traveling some distance orit has no end?
Theoretically, the photon (or the beam of photons, there really isn't a difference) can go an infinite distance, traveling all the while at a speed $c$.
Since photons contain energy, $E=h\nu$, then energy conservation requires the photon to only be destroyed via interaction (e.g., absorption in an atom). There is nothing that could make the photon simply stop after some distance, it can only be stopped via an interaction of some sort.
Note that some of the light we are seeing from very distant galaxies are some billions of years old and traveled many yottameters to get here. Had they not been absorbed by Hubble space telescope, for instance, they would have continued on their way through our galaxy (until something else stopped it).
A photon will travel "at the speed of light" until obstructed. From the speed, and elapsed time, you can calculate how far the light will travel.
Laser light consists of more than one photon "in phase", which has exactly the same property in this respect, as a solitary photon.
Note that it is correct that a photon can travel an infinite distance in an infinite time, but it can
not reach any desired point in the universe.
This is caused by the expansion of the universe, which also leads to the fact that we can not receive information outside of the observable universe.
One small addition to the other answers: While it is indeed true that the light will never stop if it doesn't hit anything, it
will however get red shifted, and thus become less energetic, due to the expansion of the universe. For example, the cosmic microwave background consists of photons which were emitted back when the atoms formed. However, back then the temperature of the universe was about $3000\,\rm K$ (about the melting point of iron) while today the cosmic microwave background has a temperature of merely $2.7\,\rm K$. So the photons we see in the CMB have travelled for more than 13 billion years without vanishing, however they have shifted in frequency from visible light down to microwaves.
Provided that there is nothing for the photon to interact with (i.e. we look at it in vacuum), the mean free path will be infinite; that is, it will continue travelling forever in a given direction. There's nothing which will stop the photon's path. Hence, it will go arbitrarily far. Whether you have a single photon or a laser, the answer won't change.
The fact that photon lines will never end manifests itself in another relevant fact. If you look at the intensity $I$ of light on a sphere of radius $r$ away from a point source, the intensity drops off as $1/r^2$. More specifically, if $P$ is the power of that source, then $I(r) = \frac{P}{4 \pi r^2}.$ The $4 \pi r^2$ in the denominator is just the surface area of the sphere.
You may think this is relatively trivial, but in fact, it's actually a pretty deep fact. We know from 20th century work that there are particles similar to photons but with some differences. One of these is the $Z$-boson. Unlike the massless photon, the $Z$ boson is massive. Its mass is around $91 GeV/c^2$, which is about 97 times as massive as a proton. If you did the corresponding analysis for $Z$-bosons, you'd find that they decay, and the decay length is on the order of $10^{-18}m$. A $Z$ boson will on average only travel about that far in vacuum. This leads to a different functional form for the above intensity, which will have an exponential dampening. In fact, this mass is essentially equivalent to studying photons in a medium which provides dissipation (e.g. inside a superconductor).
The fact that the photon doesn't suffer this same fate is really a consequence of its masslessness. There are many possible bounds on the photon mass. Of course, just the fact that we see photons from very long distances away provides a (rather strong) upper bound on the photon mass, though it is perhaps a bit deceptive as there are certain unusual models which avoid this strong bound. The most robust, model-independent bounds we have to date are about $10^{-14} eV/c^2$, that is, a factor of about $10^{23}$ lower than the proton mass.
A ray of light or a laser beam will not stop until it reaches an obstruction.
If there is no obstruction, light will NEVER stop.
It has no end.
Whether it be a beam or ray of light, photons will keep traveling until they are absorbed. Photons can't stop because they travel at a constant velocity, the speed of light, i.e., they can't accelerate or decelerate. However, their wavelengths change over time due do the expansion of the universe, i.e, their wavelengths get larger and loose energy as such because $E_{\gamma}$ and $\lambda$ are inversely proportional,
$E_{\gamma} = \frac{hc}{\lambda}$.
A "ray of light" must be respelled as "photon" because here we are talking physics.
Between a single photon and a laser beam, in this case, there is no difference. Every photon will continue his travel until stopped, every single photon is "indistinguible" from others (in the sense that they are no different intrinsecally). The photons of a laser beam are only at the same energy level and travel in the same direction (assuming a perfect laser) but this is of no importance for the question.
A photon can be stopped only by interacting with it with enough energy. If the interaction is of lower energy or is a gravitational field the photon will be deviated but continue "moving".
And does the light(ray of light and beam of laser) stops after traveling some distance or it has no end?
I think that you want to know if a photon can travel outside the Universe. If a photon reach the limit of the Universe it will continue his travel, extending the Universe itself!
Newton's
first law states a particle will have constant velocity unless an external force acts upon it. The photon has no mass, but nonetheless the first law still holds true in the case of light. When a ray of light is projected, (say) from the surface of Earth to outside in space. The condition is that, there is no obstruction to it till infinity(it travels only in vaccum). My question is that how far can that ray of light go?
$$x=vt$$
In this case $c=v$ where $c$ is the speed of light travelling in a vacuum (a constant) and $t$ seems to $\rightarrow \infty$ seconds based on the information given in your question.
The distance the light travels depends on the time it travels for because $c$ is constant in a vacuum which implies:
$$x\rightarrow\infty$$
Also, instead of a ray of light,if I consider a beam of laserwith same conditions, then how far can a beam of laser go?
Same as with 1.
Compare both the situations.
One is a ray of light travelling infinitely in a vacuum and the other is several rays of coherent light travelling infinitely in a vacuum.
The distance that a particle can travel is partly set by its mass.
If the particle has a mass less than something like 7 eV, then it could cross the universe without attenuation.
protected by Qmechanic♦ Dec 15 '14 at 15:21
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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I performed an experiment about the RL circuit, with following circuit:
(I used the PASCO RLC circuit lab.)
By Kirchhoff's voltage law, (denoting the emf of the signal generator $\mathscr E$, inductance of the inductor $L$, the resistance of the resistor $R$, and the current of the circuit $i$) $$\mathscr E - iR - L \frac{\mathrm d i}{\mathrm d t}=0$$ with initial current $i_0 = -\frac{\mathscr E}{R}$ (almost), hence $$i = \frac {\mathscr E}{R} - 2 \frac{\mathscr E}{R} e^{-Rt/L}$$ and then $$\Delta v_L = -L \frac{\mathrm d i}{\mathrm dt} = 2 \mathscr E e^{-Rt/L}.$$ So the natural logarithm of the potential of the inductor must be linear. However, the data of the experiment was strange:
(The blue graph: the emf $\mathscr E$ of the signal generator; the orange graph: $\ln (-\Delta v_L / \mathscr E)$.)
Is this just an error of the experiment? Or is there another reason for this graph?
+) The aim of the experiment is finding the time constant of the circuit, with the gradient of the (linear) graph. The inductance of the coil is $$L = \mu N^2 \frac{A}{\ell} \propto \mu$$ where $\mu$ is the permeability of the space in the coil, $N$ is the number of turns, $A$ is the cross-sectional area and $\ell$ is the length of the coil. The time constant of the other experiments, which has the linear log graphs, are as follows:
$R = 10.0 \; \Omega$, $\mu = \mu_\text{air}$: $\tau=0.400 \; \mathrm{ms}$, $L=4.00 \; \mathrm{mH}$
$R = 10.0 \; \Omega$, with iron core: $\tau=1.06 \; \mathrm{ms}$, $L=10.6 \; \mathrm{mH}$
$R = 300. \; \Omega$, $\mu = \mu_\text{air}$: $\tau=0.0149 \; \mathrm{ms}$, $L=4.47 \; \mathrm{mH}$
If we take the slope of the tangent (that is, the average slope with very short time interval) as of the graph, then we have $L_\text{iron core} = 9.91 \;\text{mH},$ which is similar to the case of $R = 10.0 \; \Omega.$ However if we take the average slope, we have $L_\text{iron core} = 3.11 \;\text{mH}.$
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Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world — including you! You’ll find this week’s puzzle below.
Mull it over on your commute, dissect it on your lunch break and argue about it with your friends and lovers. When you’re ready,
submit your answer using the link below. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!
Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏
Rasmus Ibsen-Jensen 👏 of Vienna, Austria, our big winner. You can find a solution to the previous Riddler at the bottom of this post.
Now here’s this week’s Riddler, a Pokémon Go puzzle that comes to us from
Po-Shen Loh, a math professor at Carnegie Mellon University, the coach of U.S. International Math Olympiad team and the founder of expii.com.
Your neighborhood park is full of Pokéstops — places where you can restock on Pokéballs to, yes, catch more Pokémon! You are at one of them right now and want to visit them all. The Pokéstops are located at points whose (x, y) coordinates are integers on a fixed coordinate system in the park.
For any given pair of Pokéstops in your park, it is possible to walk from one to the other along a path that always goes from one Pokéstop to another Pokéstop adjacent to it. (Two Pokéstops are considered adjacent if they are at points that are exactly 1 unit apart. For example, Pokéstops at (3, 4) and (4, 4) would be considered adjacent.)
You’re an ambitious and efficient Pokémon trainer, who is also a bit of a homebody: You wish to visit each Pokéstop and return to where you started, while traveling the shortest possible total distance. In this open park, it is possible to walk in a straight line from any point to any other point — you’re not confined to the coordinate system’s grid. It turns out that this is a really hard problem, so you seek an approximate solution.
If there are N Pokéstops in total, find the upper and lower bounds on the total length of the optimal walk. (Your objective is to find bounds whose ratio is as close to 1 as possible.)
Advanced extra credit: For solvers who prefer a numerical question with this theme, suppose that the Pokéstops are located at every point with coordinates (x, y), where x and y are relatively prime positive integers less than or equal to 1,000. Find upper and lower bounds for the length of the optimal walk, again seeking bounds whose ratio is as close to 1 as possible.
Submit your answer
Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me. I’m especially on the hunt for Riddler Jr. problems — puzzles that can stoke the curiosity and critical thinking of Riddler Nation’s younger compatriots.
And here’s the solution to last week’s Riddler, concerning a hungry, but persnickety, grizzly bear. The bear wants to maximize its intake of salmon, but is only willing to eat fish that are at least as big as all the fish it’s eaten already. If the bear will see two or three salmon during its fishing expedition, which weigh something uniformly random between 0 and 1 kilogram, it should
always eat every fish it can.
To see why, let’s start with the two-fish case. If the bear eats every fish it can — the “greedy” strategy — its expected fish intake is the expected weight of the first fish plus the expected weight of the second fish
given that it’s willing to eat it. Say the first fish has a weight \(x_1\) and the second fish a weight \(x_2\). The expectation under the greedy strategy is
$$\int_0^1 \left(x_1 +\int_{x_1}^1 x_2 dx_2\right)dx_1=\frac{5}{6}\approx 0.833$$
But if he forgoes the first fish, he’ll eat only the second fish, which is only half a kilogram on average. Therefore, always eat the first fish!
A similar argument holds for the three-fish case. We now know that if the bear forgoes the first fish, we’re back to the two-fish case, where it can expect 5/6 kilograms of salmon total. In fact, a nice pattern emerges. If the bear eats every fish it can on an expedition N fishes long, its expected intake is the first N terms of the sum 1/2 + 1/3 + 1/4 + 1/5 + 1/6 … So, for a three-fish expedition, eating the first fish yields 1/2 + 1/3 + 1/4 = 13/12 kilograms on average. Again, always eat the first fish!
But things change if the fishing expedition gets any longer.
Laurent Lessard compared optimal bear behavior to greedy bear behavior. For three fish or fewer, these strategies are identical. But for more fish, the bear does well to let certain, heavier fish go early on in the expedition, in order to maximize consumption in the future.
The 🏆 Coolest Riddler Extension Award 🏆 this week goes to
Kris Mycroft. Kris transported the problem to the arctic, and looked at the problem faced by polar bears in a network of streams.
Looks like it pays to be the alpha bear at the head of the river.
And, lest you think The Riddler is a mere diversion, comfortably abstracted from the struggles of daily life, the Alaska Salmon Program illustrated the puzzle with real-world bear footage in this delightful Twitter thread:
Elsewhere in the puzzling world:
Some puzzles on Olympic strategy. [The New York Times] Puzzles about a summer birthday party. [The Wall Street Journal] A few Olympics problems, right on time. [Expii] And, appropriately, a R-I-O puzzle. [NPR] A puzzle book of a very different kind. [Wired]
Have a wonderful weekend!
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2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Подробная запись - Подобные записи 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Подробная запись - Подобные записи 2018-08-23 11:31 Подробная запись - Подобные записи 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Подробная запись - Подобные записи 2018-08-23 11:31 Подробная запись - Подобные записи 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Подробная запись - Подобные записи 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Подробная запись - Подобные записи 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Подробная запись - Подобные записи 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Подробная запись - Подобные записи 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Подробная запись - Подобные записи
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Given a set of $N$ assets, the amount of strategies proposed in literature to diversify the investors wealth in order to find the 'optimal' portfolio is overwhelming. However, for example DeMiguel et al point out that estimation error is large in general, leading to bad out-of-sample returns for many sophisticated portfolio strategies.
In my mind it should be a straightforward idea to combine several promising allocation candidates such that the estimation risk is 'diversified' as well. In other words, given a set of portfolio weights proposed by $k$ distinct strategies $\{\omega_1,\ldots,\omega_k\}$, it could be beneficial to compute a vector $c\in\mathbb{R}^k, \left( \sum\limits_{i=1}^{k} c_i =1\right)$ based on some criteria and then investing into $\omega^*:=c_1\omega_1+\ldots +c_k\omega_k$.
I am aware of some ideas combining two portfolios via shrinkage approaches (see for example Posterior Inference for Portfolio Weights and the appendix of Optimal Versus Naive Diversification: How Inefficient is the 1/N Portfolio Strategy?)
but I am would like to learn more about those ideas. Any reference or further comment directing in this direction is welcome to answer my question: What are the benefits and gains from combining several allocation strategies?
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Keeping Track of Element Order in Multiphysics Models
Whenever you are building a finite element model in COMSOL Multiphysics, you should be aware of the element order that is being used. This is particularly important for multiphysics models as there are some distinct benefits to using different element orders for different physics. Today, we will review the key concepts behind element order and discuss how it applies to some common multiphysics models.
What Is Element Order?
Whenever we solve a finite element problem, we are approximating the true solution field to a partial differential equation (PDE) over a domain. The finite element method starts by subdividing the modeling domain up into smaller, simpler domains called
elements. These elements are defined by a set of points, traditionally called nodes, and each node has a set of shape functions or basis functions. Every shape function is associated with some degrees of freedom. The set of all of these discrete degrees of freedom is traditionally referred to as the solution vector.
Note: You can read more about the process of going from the governing PDE to the solution vector in our previous blog posts “A Brief Introduction to the Weak Form” and “Discretizing the Weak Form Equations“.
Once the solution vector is computed, the finite element approximation to the solution field is constructed by interpolation using the solution vector and the set of all of the basis functions in all of the elements. The
element order refers to the type of basis functions that are used.
Let’s now visualize some of the basis functions for one of the more commonly used elements in COMSOL Multiphysics: the two-dimensional Lagrange element. We will look at a square domain meshed with a single quadrilateral (four-sided) element that has a node at each corner. If we are computing a scalar field, then the Lagrange element has a single degree of freedom at each node. You can visualize the shape functions for a first-order Lagrange element in the image below.
The shape functions for a first-order square quadrilateral Lagrange element.
The first-order shape functions are each unity at one node and zero at all of the others. The complete finite element solution over this element is the sum of each shape function times its associated degree of freedom. We’ll now compare our first-order shape functions with our second-order shape functions.
The shape functions for a single second-order square quadrilateral Lagrange element.
Observe that the second-order quadrilateral Lagrange element has node points at the midpoints of the sides as well as in the element center. It has a total of nine shape functions and, again, each shape function is unity at one node and zero elsewhere.
Let’s now look at what happens when our single quadrilateral element represents a domain that is not a perfect square but rather a domain with some curved sides. In such cases, it is common to use a so-called
isoparametric element, meaning that the geometry is approximated with the same shape functions as the one used for the solution. This geometric approximation is shown below for the first- and second-order cases. A domain with curved sides. Single first- and second-order quadrilateral elements are applied.
As we can see in the image above, the first-order element simply approximates the curved sides as straight sides. The second-order element much more accurately approximates these curved boundaries. This difference, known as a geometric discretization error, is discussed in greater detail in an earlier blog post. The shape functions for the isoparametric first- and second-order Lagrange elements are shown below.
The shape functions of a single first-order isoparametric Lagrange element for the domain with curved sides. The shape functions of a single second-order isoparametric Lagrange element for a domain with curved sides.
We can observe from the above two images that the first-order element approximates all sides of the domain as straight lines, while the second-order element approximates the curved shapes much more accurately. Thus, if we are modeling a domain with curved sides, we need to use several linear elements along any curved domain boundaries just so that we can accurately represent the domain itself.
For any real-world finite element model, there will of course always be more than one element describing the geometry. Additionally, keep in mind that regardless of the element order, you will want to perform a mesh refinement study, also called a mesh convergence study. That is, you will use finer and finer meshes (smaller and smaller elements) to solve the same problem and see how the solution converges. You terminate this mesh refinement procedure after achieving your desired accuracy. A good example of a mesh refinement study is presented in the application example of a Stress Analysis of an Elliptic Membrane.
All well-posed, single-physics finite element problems will converge toward the same answer, regardless of the element order. However, different element orders will converge at different rates and therefore require various computational resources. Let’s explore why different PDEs have different element orders.
Element Order in Single-Physics Models
For the purposes of this discussion, let’s consider just the set of PDEs governing common single-physics problems that exhibit no variation in time. We can put all of these PDEs into one of two broad categories:
Poisson-type: Poisson-type PDEs are used to describe heat transfer in solids, solid mechanics, electric currents, electrostatics and magnetostatics, thin-film flow, and flow in porous media governed by Darcy’s law or the Richards’ equation. Such governing PDEs are all of the form:\nabla \cdot (- D \nabla u ) = f
Note that this is a second-order PDE, thus second-order (quadratic) elements are the default choice within COMSOL Multiphysics for all of these types of equations.
Transport-type: Transport-type PDEs are used to describe chemical species transport as well as heat transfer in fluids and porous media. The governing equations here are quite similar to Poisson’s equation, with one extra term — a velocity vector:\nabla \cdot ( -D \nabla u + \mathbf{v} u ) = f
The extra velocity term results in a governing equation that is closer to a first-order PDE. The velocity field is usually computed by solving the Navier-Stokes equation, which is itself a type of transport equation that describes fluid flow. It is often the case that, for such problems, there is a high Péclet number or Reynolds number. This is one of the reasons why the default choice is to use first-order (linear) elements for these PDEs.
Note that for fluid flow problems where the Reynolds number is low, the default is to use the so-called
P2 + P1 elementsthat solve for the fluid velocity via second-order discretization and solve for the pressure via first-order discretization. The P2 + P1 elements are the default for the Creeping Flow, Brinkman Equationsand Free and Porous Media Flowinterfaces. This is also the case for the Two-Phase Flow, Level Setand Two-Phase Flow, Phase Fieldinterfaces. Further, any type of transport or fluid flow interface uses stabilization to solve the problem more quickly and robustly. For an overview of stabilization methods, check out our earlier blog post “Understanding Stabilization Methods“.
So how can we check the default settings for the element order used by a particular physics interface? Within the Model Builder, we first need to go to the
Show menu and toggle on Discretization. After doing so, you will see a Discretization section within the physics interface settings, as shown in the screenshot below. Screenshot showing how to view the element order of a physics interface.
Keep in mind that as long as you’re working with only single physics, it typically does not matter too much which element order you use as long as you remember to perform a mesh convergence study. The solutions with a different element order may require quite varying amounts of memory and time to solve, but they will all converge toward the same solution with sufficient mesh refinement. However, when we start dealing with multiphysics problems, things become a little bit more complicated. Next, we’ll look at two special cases of multiphysics modeling where you should be aware of element order.
Conjugate Heat Transfer: Heat Transfer in Solids with Heat Transfer in Fluids
COMSOL Multiphysics includes a predefined multiphysics coupling between heat transfer and fluid flow that is meant for simulating the temperature of objects that are cooled or heated by a surrounding fluid. The
Conjugate Heat Transfer interface (and the functionally equivalent Non-Isothermal Flow interface) is available with the Heat Transfer Module and the CFD Module for both laminar and turbulent fluid flow.
The
Conjugate Heat Transfer interface is composed of two physics interfaces: the Heat Transfer interface and the Fluid Flow interface. The Fluid Flow interface (whether laminar or turbulent) uses linear element order to solve for the fluid velocity and pressure fields. The Heat Transfer interface solves for the temperature field in the fluid as well as the temperature field in the solid. The same linear element discretization is used throughout the temperature field in both the solid and fluid domains.
Now, if you are setting up a conjugate heat transfer problem by manually adding the various physics interfaces, you do need to be careful. If you start with the
Heat Transfer in Solids interface and add a Heat Transfer in Fluids domain feature to the interface, a second-order discretization will be used for the temperature field by default. This is not generally advised, as it will require more memory than a first-order temperature discretization. The default first-order discretization of the fluid flow field justifies using first-order elements throughout the model.
It is also worth mentioning a related multiphysics coupling: the
Local Thermal Non-Equilibrium interface available with the Heat Transfer Module. This interface is designed to solve for the temperature field of a fluid flowing through a porous matrix medium as well as the temperature of the matrix through which the fluid flows. That is, there are two different temperatures, the fluid and the solid matrix temperature, at each point in space. The interface also uses first-order discretization for both of the temperatures. Thermal Stress: Heat Transfer in Solids with Solid Mechanics
The other common case where a multiphysics coupling uses different element orders from a single-physics problem is when computing thermal stresses. For the
Thermal Stress multiphysics coupling, the default is to use linear discretization for the temperature and quadratic discretization for the structural displacements. To understand why this is so, we can look at the governing Poisson-type PDE for linear elasticity:
where \mathbf{C} is the stiffness tensor and \mathbf{\epsilon} is the strain tensor.
For a problem where temperature variation affects stresses, the strain tensor is:
where \mathbf{\alpha} is a tensor containing the coefficients of thermal expansion, T is the temperature, T_0 is the strain-free reference temperature, and \mathbf{u} is the structural displacement field.
By default, we solve for the structural displacements using quadratic discretization, but we can see from the equation above that the strains are computed by taking the gradients of the displacement fields. This lowers the discretization order of the strains to a linear order. Hence, the temperature field discretization should also be lowered to a linear order.
Closing Remarks
We have discussed the meaning of discretization order in COMSOL Multiphysics and why it is relevant for two different multiphysics cases that frequently arise. If you are putting together your own multiphysics models, you’ll want to keep element order in mind.
Additionally, it is good to address what can happen if you build a multiphysics model with element orders that disagree with what we’ve outlined here. As it turns out, in many cases, the worst thing that will happen is that your model will simply require more memory and converge to a solution more slowly. In the limit of mesh refinement, any combination of element orders in different physics will give the same results, but the convergence may well be very slow and oscillatory. If you do observe any spatial oscillations to the solution (for example, a stress field that looks rippled or wavy), then check the element orders.
Today’s blog post is designed as a practical guideline for element selection in multiphysics problems within COMSOL Multiphysics. A more in-depth discussion of stability criterion for mixed (hybrid) finite element methods can be found in many texts, such as
Concepts and Applications of Finite Element Analysis by Robert D. Cook, David S. Malkus, Michael E. Plesha, and Robert J. Witt. Comments (4) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science TAGS CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
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Kasiki's test and the index of coincidence are used to attack a
Vigenère cipher (or other polyalphabetic ciphers with small alphabet and small key size) - they both try to get the length of the keyword.
Kasiki's test gets probable prime factors of the keyword length, while the coincidence index test gets us an estimation of the absolute length of the keyword.
The original Kasiki test
This is what was done with
CHR in your example: We look at repeated sequences of three or more characters, and at which distances they occur. We collect all these distances, and look at the prime factors of these.
The idea is that probably such a repeated sequence comes from the same plain text sequence, which then randomly hit the same keyword position. They will only hit the same position if their distance is a multiple of the keyword length.
Of course, we also can get random hits here (where different plaintext sequences got encrypted by different key sequences to get the same ciphertext sequence) - so be careful which prime factors to select. Either combine this method with Friedman's test mentioned below, or try different keyword length's for further analysis.
The coincidence index
The coincidence index of a text is defined as
Here
n is the size of the alphabet, m the number of occurrences for the character l l, k the total size of the text. ( δ(x if the characters at position i, x i) = 1 i and j are equals, else 0. But for calculation the second sum is more convenient.)
The coincidence index of a totally random text would be $1/k$ (and this is also the total minimum), while for natural language texts it is higher (0.067 for english, a bit higher for German). For a ciphertext encrypted by a monoalphabetic cipher it is still the same as for the original plaintext, for polyalphabetic ciphers (like Vigenère) it is between those.
(Actually, there are different definitions of coincidence index, but the values you have seem to use this one.)
Column-wise coincidence index to get the keyword length
You now take guesses at the key length
h, and writes the cipher text in h columns. Then (if we guessed right) effectively the letters in each column are encrypted with the same key letter, e.g. monoalphabetically. We now calculate the coincidence index of . If we had a good guess at the key length, the coincidence index is near the coincidence index of the original plaintext column (which, if long enough, is a good approximation for the coincidence index of the used language). each column individually
In your example, for the lengths 1,2,3,4 the coincidence indices are only around 0.045, which is not much higher than 0.385, the expected value for a random 26-letter distribution. For length 5, you see a sudden jump: the indices are around the value for English (0.067). For 6, it will likely get lower again.
This will get you quite similar results as the Kasiki test, but is easier automated.
Friedman's method
Actually, Friedman derived from this a formula which spares us the calculation of each column's coincidence index - we just need the global coincidence index of the ciphertext (and the one of the natural language) to get an estimation of the keyword length.
While the English Wikipedia page does not mention this formula, the german language Wikipedia shows this:
Here $E_S$ is the expected coincidence index for our natural language, $\phi(T)$ the total coincidence index of our text, n the alphabet size, k the text length.
Your example
Using your value for m = 1 (which is the global coincidence index) and 26 for
n, we get approximately
$$h ≈ \frac{(0.067 - 0.0385)\cdot k }{ (k-1)\cdot 0.045 - k\cdot 0.0385 + 0.067} = \frac{0.0285·k}{0.0065·k +0.022}$$
For example, for k = 1000 we would get h ≈ 4.36. This is (together with the previous check from the column-wise coincidence index) a good hint at a keyword size of
5.
When we have the keyword length, we can use other methods, like frequency analysis on each "column", to crack the whole cipher.
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I am trying to solve the coupled differential equation numerically with Mathematica. But the range of values are large so mathematica cannot give correct plot. Here is the code:
a = 4.75388*10^26; b = 5.424*10^-3; d = 4.75388*10^20; {X, Y} = {x, y} /. NDSolve[{x'[ z] == -((a/z) (x[z] - b*z^(3/2) E^(-z)) (BesselK[1, z]/BesselK[2, z])), y'[z] == ((d/z) (x[z] - b *z^(3/2) E^(-z)) (BesselK[1, z]/ BesselK[2, z]) - (a *z/4) (BesselK[1, z]) y[z]), x[0.1] == 1.552*10^-4, y[0.1] == 10^(-9)}, {x, y}, z] // FullSimplify // First LogLogPlot[{X[z], Y[z]}, {z, 0.1, 100}, PlotRange -> All]
I was wondering how to write a function $ F (r, q, n, f) $ in Mathematica, defined in this way:
$ $ F(r,q,n,f):=\sum_{i_0=1}^q f(i_0) \Biggl(\sum_{i_1=i_0+1}^{q+1} f(i_1)\biggl(\sum_{i_2=i_1+1}^{q+2} f(i_2)\Bigl(\ldots(\sum_{i_n=i_{n-1}+1}^{q+n} f(i_n))\ldots \Bigl) \biggl) \Biggl)$ $ es. $ $ \sum_{i_0=1}^2 f(i_0) \Biggl(\sum_{i_1=i_0+1}^{3} f(i_1)\biggl(\sum_{i_2=i_1+1}^{4} f(i_2) \biggl) \Biggl)=f(1)f(2)f(3)+f(1)f(2)f(4)+f(1)f(3)f(4)+ +f(2)f(3)f(4)$ $
does an operator already exist that can be used in this way?
trying to write this function on mathematica I realized that the “recursion” is variable and I don’t know how to program in this case.
thank you
$ \ $
$ \ $
further example
$ $ \sum_{i_0=1}^1 f(i_0) \Biggl(\sum_{i_1=i_0+1}^{2} f(i_1)\biggl(\sum_{i_2=i_1+1}^{3} f(i_2)(\sum_{i_3=i_2+1}^{4} f(i_2)) \biggl) \Biggl)=f(1)f(2)f(3)f(4)$ $
Same as here. Everything very blurry. May be there is any new solution for Mathematica 12 under Windows 10?
I want to solve below equation in Mathematica. I know a little bit about Mathematica. Can somebody help me?
\begin{array}{l} \frac{1}{{{R^2}}}\frac{\partial }{{\partial \theta }}\left( {{P_0}\frac{{\partial \left( {{P_X}} \right)}}{{\partial \theta }} – \Lambda {R^2}{P_X}} \right) + \frac{1}{R}\frac{\partial }{{\partial R}}\left( {R{P_0}\frac{{\partial \left( {{P_X}} \right)}}{{\partial R}} + R{P_X}\frac{{\partial \left( {{P_0}} \right)}}{{\partial R}}} \right) – i2\Lambda \Gamma \left( {{P_X}} \right)\ + \underbrace {\frac{1}{R}\frac{\partial }{{\partial R}}\left( { – 3R{P_0}\frac{{\partial {P_0}}}{{\partial R}}} \right) + i2\Lambda\left( {{P_0}} \right)}_f = 0\ {P_0}\left( R \right) = {\rm{available}}\ \Lambda = 1 \ {P_X}\left( {{R_1},\theta } \right) = {P_X}\left( {{R_1},\theta } \right) = 0\ {P_X}\left( {R,0} \right) = {P_X}\left( {R,2\pi } \right) \end{array}
I realise this is possibly a VERY open ended question, and potentially has answers dotted around e.g. here .
What are the most robust methods of testing the result of
NonlinearModelFit[...] in Mathematica?
I know already that there are some methods available in Mathematica such as
AdjustedRSquared,
RSquared,
AIC,
AICc, and
BIC.
The most familiar of these to me personally are
AdjustedRSquared and
RSquared, but I know many methods of testing exist.
A sub-question to this is, when are these tests (and others, whatever they may be), most appropriate? Does this depend on the model fitted, the number of parameters, the amount of data, and so on?
I strongly appreciate that some of these questions can only be answered subjectively, and I even anticipate some potential off-topic or close requests; after all if there was a catch-all test — I wouldn’t need to ask this question! But I feel having a question where goodness of fit specifically in Mathematica is discussed (especially where some of the more experienced members of the community can contribute) might be beneficial.
So I have
f[x_] := g[x]/g'[x] FullSimplify[Integrate[f[x], x]]
and I’m trying to get F(x) in terms of g(x) but it gives the random expression below as output
1/4 (x^2 + 2 Log[x])
How would I get F(x) based on g(x) to be returned?
When saving my notebook as PDF Mathematica 12.0 does not print special caracters (<, >, :, ;, and so on) and numbers correctly. My example:
It never happened in previous versions. Do you know how to fix this?
I have a new approach to SGD optimizer, and thought to try to test it in Mathematica. It uses gradients to simultaneously maintain online parabola model for smarter choice of step size – I only need to
ask for gradients and be able to manually update parameters.
Is it doable within Mathematica neural network framework? I see NetPortGradient, but how to modify parameters?
As it seems a common research direction, maybe there is some simple example?
I asked this question about the analytical solution of the PDE:
$ $ \partial_t c=\partial_x((c-a)(c-b)\partial_xc) $ $
And I was given the answer that Maple gives the following implicit form as a solution:
$ $ \eqalign{&{k_{{1}}}^{2}{k_{{2}}}^{2}{c}^{2} + \left( 2\,{k_{{1}}}^{4}k_{{2}}k_{{3 }}-2\,{k_{{1}}}^{2}{k_{{2}}}^{2}a-2\,{k_{{1}}}^{2}{k_{{2}}}^{2}b \right) c\cr &+ \left( 2\,{k_{{1}}}^{6}{k_{{3}}}^{2}-2\,a{k_{{1}}}^{4}k_{{ 2}}k_{{3}}-2\,b{k_{{1}}}^{4}k_{{2}}k_{{3}}+2\,ab{k_{{1}}}^{2}{k_{{2}}} ^{2} \right) \ln \left( -{k_{{1}}}^{2}k_{{3}}+ck_{{2}} \right)\cr & -2\,{k _{{2}}}^{4}t-2\,k_{{1}}{k_{{2}}}^{3}x-2\,{k_{{2}}}^{3}k_{{3}}-2\,k_{{4 }}{k_{{2}}}^{3} =0} $ $ But when I’m doing the
DSolve on Mathematica (I don’t have Maple) and I get no solution!
So my question is: can Mathematica give me that implicit solution? Can Maple do things that Mathematica cannot?
I am writing a code which is supposed to simulate the heat transfer. All the time, when I run the code Mathematica starts computations, and after some time beeps and clears out all variables. If I for example add
Print["A"] somewhere it does the same, but in a different moment of computations than previously. What can cause the problem? Shouldn’t Mathematica print some error first? Is this normal behaviour of Mathematica? I’ve been using it for 5 years, but this is the first kind of big program.
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I'm following a very good IPython notebook (the whole list can be found here) in which some sampling techniques are explained. However, I don't understand the use of a categorical variable in a change point model.
The posterior distribution for the model is the following:
\begin{aligned} P( \lambda_1, \lambda_2, \tau | \mathbf{y} ) &\propto \left[\prod_{t=1851}^{\tau} \text{Poi}(y_t|\lambda_1) \prod_{t=\tau+1}^{1962} \text{Poi}(y_t|\lambda_2) \right] \text{Gamma}(\lambda_1|\alpha,\beta) \text{Gamma}(\lambda_2|\alpha, \beta) \\ &\propto \left[\prod_{t=1851}^{\tau} e^{-\lambda_1}\lambda_1^{y_t} \prod_{t=\tau+1}^{1962} e^{-\lambda_2} \lambda_2^{y_t} \right] \lambda_1^{\alpha-1} e^{-\beta\lambda_1} \lambda_2^{\alpha-1} e^{-\beta\lambda_2} \\ &\propto \lambda_1^{\sum_{t=1851}^{\tau} y_t +\alpha-1} e^{-(\beta+\tau)\lambda_1} \lambda_2^{\sum_{t=\tau+1}^{1962} y_i + \alpha-1} e^{-\beta\lambda_2} \end{aligned}
After this, the parameters are given by:
$$\lambda_1 \sim \text{Gamma}(\sum_{t=1851}^{\tau} y_t +\alpha, \tau+\beta)$$ $$\lambda_2 \sim \text{Gamma}(\sum_{t=\tau+1}^{1962} y_i + \alpha, 1962-\tau+\beta)$$ $$\tau \sim \text{Categorical}\left( \frac{\lambda_1^{\sum_{t=1851}^{\tau} y_t +\alpha-1} e^{-(\beta+\tau)\lambda_1} \lambda_2^{\sum_{t=\tau+1}^{1962} y_i + \alpha-1} e^{-\beta\lambda_2}}{\sum_{k=1851}^{1962} \lambda_1^{\sum_{t=1851}^{\tau} y_t +\alpha-1} e^{-(\beta+\tau)\lambda_1} \lambda_2^{\sum_{t=\tau+1}^{1962} y_i + \alpha-1} e^{-\beta\lambda_2}} \right)$$
How can we obtain the distributions for these parameters? By marginalizing the posterior distribution? In the case of $\lambda_{1}$ and $\lambda_{2}$ each parameter is described by a Gamma distribution multiplied by some constant, so maybe that is the idea. In the case of $\tau$ I can't see why we obtain a categorical distribution with that argument.
Thanks a lot
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There is a one-dimensional diffusion process in which particles start running at $t = 0$ and from $x_o > 0$.
When particles reach x = 0 they are removed from the system, thus the total concentration is not conserved anymore.
I have to solve the diffusion equation, which is the following partial differential equation:
$$\frac{\partial P (R, t)}{\partial t} = D\triangledown^2P(R,t) $$
Where $P(R, t)$ is the probability that the particles arrive at R at time t.
We have the initial conditions:
$$c(x,0) = N\delta(x - x_o)$$
$$c(0, t) = 0$$
I have been doing some research in how to do so and I came across with a method which is based on a particular Gaussian function:
$$G (R, t) = (\frac{1}{4\pi Dt})^{\frac{d}{2}} e^{\frac{(R-R_0)^2}{4Dt}}$$
Where d is the dimensionality of the system.
But the issue here is that we are working with an 'absorbing boundary' that makes the condition $c(0, t) = 0$ useless because we work from $x_o > 0$.
Then how could I solve the probability (i.e this differential equation)? We have been suggested the method of images, but not sure how it works with differential equations.
NOTE: I know this is more a mathematical issue (solving a differential EQ.) but thought that as it is a mechanics problem it could be useful asking here.
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The space between the plates of a parallel plates capacitor is filled consecutively with two dielectrics layers $1$ and $2$ having the thickness $d_1$ and $d_2$ respectively and permittivities $\epsilon_1$ and $\epsilon_2$. The area of each plate is equal to $S$ find:
1. The capacitance of the capacitor
2. The density $\sigma'$ of the bound charges on the boundary plane if the voltage across the capacitor equals $V$ and the electric field is directed from layer $1$ to layer $2$.
I managed to find $C_{eq}$ easily, for the second part I don't understand that word "
boundary plane", I only know $\sigma'=\sigma\bigg(1-\dfrac{1}{\epsilon}\bigg)$ holds when there would have been just one plate, how to work on two such consecutive plates. Please help.
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I am trying to follow the Optimal Kelly derivation on Wikipedia for two continuous assets: one risky and one risk-free.
The derivation begins by assuming that the risky assets follows a GBM (a special type of exponential semi-martingale). It is assumed that we already know the solution for the expected value of $S_t$ (convexity adjustment and all).
The derivation then says that our expected rate of return from investing $f$ in the risky asset and $(1-f)$ in the risk-free asset is:
(1) $${\displaystyle G(f)=f\mu -{\frac {(f\sigma )^{2}}{2}}+(1-f)\ r}$$
How did $G(f)$ become a quadratic function of $f$? Intuitively, the quadratic form makes more sense because otherwise optimizing for a linear $f^*$ would prescribe maximum leverage and thereby assure Gambler's ruin. My sense is that it follows from Ito's lemma (or some analogue of it).
I follow along with the remainder of the derivation in terms of finding $f^*$.
(Correct answer awarded for proof by Ito’s Lemma)
extra credit
But why is $G(f)$ not the following?
(2) $$G(f)= \ln\left(f\,e^{(\mu -{\frac {\sigma^{2}}{2}})}\ + (1-f)\,e^r\right) $$
Because: $${\mathbb{E}[S_t]} = S_0\,e^{(\mu -{\frac {\sigma^{2}}{2}})t}$$
I.e., how the fractional weights for the risky and riskless asset make their ways into the exponent?
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Wooclap allows you to insert mathematical formulas in a question and its possible answers by using the AsciiMath and LaTeX formats.
AsciiMath
Simply surround formulas by backticks ` or by the elements <math> and </math>.
Examples:
<math>1/(x+1)</math>
or `1/(x+1)`
<math>text(N)_2 + 3 text(H)_2 -> 2 text(NH)_3</math>
or `text(N)_2 + 3 text(H)_2 -> 2 text(NH)_3`
<math>sum_(i=1)^n i^3=((n(n+1))/2)^2</math>
or `sum_(i=1)^n i^3=((n(n+1))/2)^2`
You will find a summary of the available symbols here.
LaTeX
Examples:
$\frac{1}{x+1}$
$e^{i \pi} + 1 = 0$
$\oiint_{\partial \Omega} \mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{1}{\varepsilon_0} \iiint_\Omega \rho \,\mathrm{d}V$
The formulas are rendered using the KaTeX engine. The list of supported functions is available here: https://katex.org/docs/supported.html.
N.B. If you wish to use the dollar sign in your question, here is how you can obtain it, without messing up your LaTeX format.
$$ 500 or \$ 500
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Does there exists an uncountable family $\mathcal{A}$ consisting of uncountable compact subsets of $\mathbb{R}$ and pairwise disjoint?
Let $K$ be the set of reals in $[0,1]$ with only $0$s and $1$s in their decimal expansions. For each $x \in K \setminus\{1\}$ the set $2x + K$ is compact, and this family is pairwise disjoint (since $x$ is recoverable from any element of $2x + K$).
Edit to make the last claim more explicit: suppose $y \in (2x_0 + K) \cap (2x_1 + K)$. The $n$th digit of the decimal expansion of $y$ is at least $2$ iff the $n$th digit of the decimal expansion of $x_0$ is $1$ iff the $n$th digit of the decimal expansion of $x_1$ is $1$. Since $x_0$ and $x_1$ have only $0$s and $1$s in their expansions, this forces them to be equal.
Let $C$ be any Cantor set in the line. $C \times C$ is again a Cantor set, so let $h:C \times C \cong C$ be a homeomorphism. For each $x \in C$ let $C_x = \{x\} \times C$; clearly each $C_x$ is yet another Cantor set, and $\{h[C_x]:x \in C\}$ is a partition of $C$ into $2^\omega$ Cantor sets.
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Hybrid Functionals¶
Hybrid functionals are exchange-correlation functionals that take into account exact Fock exchange[Bec93]. There are different approaches as to how to mix in the exact exchange. Themethod we implemented in our
PlaneWaveCalculator code is the HSE functional described in[HSE03].
Background¶
The HSE functional is an adaptation of the PBE0 functional [PEB96], where the exchange-correlation energy is defined as:
The exact-exchange energy \(E^\text{HF}_x\) is defined in terms of the Kohn-Sham orbitals as:
and is very expensive to evaluate.
For periodic systems, the exact-exchange energy converges very slowly with the distance. For this reason Heyd et al. [HSE03] suggested splitting up the exchange in a long- and short-range part (defined by a screening length parameter \(\mu\)). In which only for the short range part exact exchange is mixed:
The most general version of this functional depends on two parameters, the screening length \(\mu\), and the mixing fraction \(\alpha\), which was \(\frac{1}{4}\) in the PBE0 functional:
ACE Implementation¶
To include the exact exchange energy in the total energy functional in a DFT calculation one needs to add a non-local orbital dependent exact exchange potential, \(V^\text{HF}_x [\psi_{\mathbf{k}n}(\mathbf{r})] = \frac{\delta E^\text{HF}_x}{\delta \psi_{\mathbf{k} n}^*(\mathbf{r})}\) to the Hamiltonian when solving the Kohn-Sham equations. Even though only action of the exchange potential operator is needed when iteratively solving for eigensolutions, the evaluation of the operator is still very expensive compared to the other parts of a DFT calculation, and it has to be carried out several times in the iterative process.
However, using the Adaptively Compressed Exchange operator method [Lin16] we avoid the expensive re-evaluations of the exchange operator in the subsequent steps in the iterative eigensolver. In this approach the exact-exchange operator \(\hat{V}^{\text{HF}}_x\) is applied once to the current best estimate for the ground state wave-functions, \(|\psi_i\rangle\),
From these \(W\) states, a set of so-called ACE projectors \(|p_i\rangle\) are constructed. The projection operator:
has the same action as \(\hat{V}^\text{HF}_x\) on states in the subspace spanned by the wave functions and can be proven to a good approximation for states that are close to the subspace. The advantage is that \(\hat{V}^\text{ACE}_x\) is much cheaper to apply (same computational complexity as the pseudopotential) in the following uses of the Hamiltonian. Note, however, that to construct the ACE projectors, one needs to fully evaluate the action of the full exact exchange operator, \(\hat{V}^\text{HF}_x\), a process which is still very expensive and scales as
where \(N_\tilde{\mathbf{k}}\) is the number of symmetry-reduced \(\mathbf{k}\)-points, \(N_\mathbf{k}\) is total number of unreduced \(\mathbf{k}\)- points, \(N_\text{band}\) is the number of occupied bands and \(N_\mathbf{G}\) is the number of plane wave basis functions.
Instead of only using the ACE operator for a single SCF iteration it has proven more efficient to keep the operator fixed and do a full self-consistent cycle before updating the ACE projectors from the most recent Kohn-Sham wave functions.
In QuantumATK ACE hybrid functional calculations therefore employs a double self-consistent loopalgorithm, see Fig. 118. Starting from a reasonable initial guess for the wavefunctions \(|\psi_i\rangle\), we construct the ACE projectors and perform a regular DFT SCFloop in which the ACE operators are added to the Hamiltonian and are kept constant. The resultingwave functions are then used to update the ACE projectors and a new SCF loop is started. In thisway the ACE operator adaptively improves every outer iteration and the process is continued untilthe exact exchange energy of the system is converged, in which case the ACE operator will beidentical to the real exact exchange operator [LL17]. The tolerance on theexchange energy and the number of outer scf loops taken can be set on
IterationControlParameters.
Usage Examples¶
Running a calculation with HSE is as simple as setting the correct
exchange_correlationon the
PlaneWaveCalculator. By choosing:
exchange_correlation = HybridGGA.HSE06calculator = PlaneWaveCalculator(exchange_correlation=exchange_correlation)
the standard
HSE06 parameters are chosen, i.e.
screening_length=0.11*Bohr**-1and
exx_fraction=0.25.
One can set custom values for these parameters by choosing:
exchange_correlation = HybridGGA.HSECustom(screening_length=0.14*Bohr**-1, exx_fraction=0.20)calculator = PlaneWaveCalculator(exchange_correlation=exchange_correlation)
By default the first SCF loop will have ACE projectors that are set to zero. In a lot of cases it makes sense to start from a PBE calculation, this can be done as follows:
calculator = PlaneWaveCalculator(exchange_correlation=GGA.PBE)configuration.setCalculator(calculator)configuration.update()new_calculator = calculator()(exchange_correlation=HybridGGA.HSE06)configuration.setCalculator(new_calculator, initial_state=configuration)configuration.update()
When restarting, the wave functions of the old configuration will be used to construct ACE projectors, and an exact exchange contribution will already be present in the first SCF loop.
References¶
[Bec93] Axel D. Becke. A new mixing of hartree–fock and local density‐functional theories. The Journal of Chemical Physics, 98(2):1372–1377, 1993. URL: https://doi.org/10.1063/1.464304, doi:10.1063/1.464304.
[HSE03] (1, 2) Jochen Heyd, Gustavo E. Scuseria, and Matthias Ernzerhof. Hybrid functionals based on a screened coulomb potential. The Journal of Chemical Physics, 118(18):8207–8215, 2003. URL: https://doi.org/10.1063/1.1564060, doi:10.1063/1.1564060.
[Lin16] Lin Lin. Adaptively compressed exchange operator. Journal of Chemical Theory and Computation, 12(5):2242–2249, 2016. PMID: 27045571. URL: http://dx.doi.org/10.1021/acs.jctc.6b00092, doi:10.1021/acs.jctc.6b00092.
[LL17] Lin Lin and Michael Lindsey. Convergence of adaptive compression methods for Hartree-Fock-like equations. arXiv:1703.05441 [math], March 2017. arXiv: 1703.05441. URL: http://arxiv.org/abs/1703.05441.
[PEB96] John P. Perdew, Matthias Ernzerhof, and Kieron Burke. Rationale for mixing exact exchange with density functional approximations. The Journal of Chemical Physics, 105(22):9982–9985, 1996. URL: https://doi.org/10.1063/1.472933, doi:10.1063/1.472933.
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The least count of the watch used for the measurement of time period is $0.01$ s
This information is just telling you to round off to the second decimal place, as you correctly did.
The sample mean is $\mu = 0.56$ and the sample standard deviation is $\sigma = 0.02$. The answer the text is referring to is
$$\frac \sigma \mu = 0.0357 = 3.57 \%$$
But I would say that this is not entirely correct. The standard error is not $\sigma$, but
$$\frac \sigma {\sqrt N}$$
Where $N$ is the number of measurements. In our case,
$$\frac \sigma {\sqrt N}=0.009$$
So the real percentage error should be
$$\frac{0.009}{0.56} = 0.0161 = 1.61 \%$$
Update: a more careful discussion
As requested, I will try to explain more why we don't need to explicitly include the resolution of the instrument ($0.01/2$) in our calculation.
In my previous discussion I explained why the solution reported in your text was $35.7 \%$, but actually that reasoning is not really correct.
The sample mean of your data set is not really $\mu=0.56$, but $\mu=0.0556$, as you correctly wrote. But since they (incorrectly) used the standard deviation, $0.02$, as standard error, we have to round off the mean and write our result as
$$0.56 \pm 0.02$$
Because it would clearly be silly to write
$$0.556 \pm 0.02$$
because if we are not sure of the second decimal place we bother writing the third?
But if the correct standard error is used, we get
$$0.556 \pm 0.009$$
You may notice a strange thing:
the number of significative digits has increased, even if our instrument had a resolution of only $0.01/2=0.005$. This is a property of the mean and it is why we use the mean in the first place: via the mean operation, we can increase the number of significative digits and circumvent the limitations of our instrument.
Take for example the case in which we have two measurements: $2$ and $7$, with resolution of $0.5$ clearly. The mean is $9/2=4.5$, so we have gained one significative place.
You can then see that with an infinite number of measurement our result becomes exact,
regardless of the resolution of the instrument, because of the $\sqrt N$ term in the denominator of the standard error.
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Global existence, boundedness and stabilization in a high-dimensional chemotaxis system with consumption
1.
Institut für Mathematik, Universität Paderborn, Warburger Str. 100, 33098 Paderborn, Germany
2.
School of Science, Xihua University, Chengdu 610039, China
$\left\{ \begin{align} & {{u}_{t}}=\Delta u-\chi \nabla \cdot \left( u\nabla v \right)+\kappa u-\mu {{u}^{2}},\ \ \ \ \ \ \ x\in \mathit{\Omega },t>0, \\ & {{v}_{t}}=\Delta v-uv,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in \mathit{\Omega },t>0, \\ \end{align} \right.$
$N$
$μ>0$
$κ>0$
$(\frac{κ}{μ },0)$ Keywords:Chemotaxis, logistic source, global existence, boundedness, asymptotic stability, weak solution. Mathematics Subject Classification:35Q92, 35K55, 35A01, 35B40, 35D30, 92C17. Citation:Johannes Lankeit, Yulan Wang. Global existence, boundedness and stabilization in a high-dimensional chemotaxis system with consumption. Discrete & Continuous Dynamical Systems - A, 2017, 37 (12) : 6099-6121. doi: 10.3934/dcds.2017262
References:
[1]
N. Bellomo, A. Bellouquid, Y. Tao and M. Winkler,
Toward a mathematical theory of Keller-Segel models of pattern formation in biological tissues,
[2]
X. Cao and J. Lankeit, Global classical small-data solutions for a three-dimensional chemotaxis Navier-Stokes system involving matrix-valued sensitivities,
[3]
X. He and S. Zheng,
Convergence rate estimates of solutions in a higher dimensional chemotaxis system with logistic source,
[4] [5] [6] [7]
S. Ishida, K. Seki and T. Yokota,
Boundedness in quasilinear Keller-Segel systems of parabolic-parabolic type on non-convex bounded domains,
[8] [9]
O. A. Ladyženskaja, V. A. Solonnikov and N. N. Ural'ceva,
[10]
J. Lankeit,
Eventual smoothness and asymptotics in a three-dimensional chemotaxis system with logistic source,
[11] [12]
X. Li, Global existence and uniform boundedness of smooth solutions to a parabolic-parabolic chemotaxis system with nonlinear diffusion,
[13] [14] [15]
N. Mizoguchi and M. Winkler, Blow-up in the two-dimensional parabolic Keller-Segel system, Preprint.Google Scholar
[16] [17]
K. Osaki, T. Tsujikawa, A. Yagi and M. Mimura,
Exponential attractor for a chemotaxis-growth system of equations,
[18]
M. M. Porzio and V. Vespri,
Hölder estimates for local solutions of some doubly nonlinear degenerate parabolic equations,
[19] [20] [21]
Y. Tao and M. Winkler,
Eventual smoothness and stabilization of large-data solutions in a three-dimensional chemotaxis system with consumption of chemoattractant,
[22] [23]
L. Wang, S. U. -D. Khan and S. U. -D. Khan, Boundedness in a chemotaxis system with consumption of chemoattractant and logistic source,
[24]
M. Winkler, A three-dimensional Keller-Segel-Navier-Stokes system with logistic source: Global weak solutions and asymptotic stabilization, Preprint.Google Scholar
[25] [26]
M. Winkler,
Boundedness in the higher-dimensional parabolic-parabolic chemotaxis system with logistic source,
[27]
M. Winkler,
Global large-data solutions in a chemotaxis-(Navier-)Stokes system modeling cellular swimming in fluid drops,
[28] [29]
M. Winkler,
Global asymptotic stability of constant equilibria in a fully parabolic chemotaxis system with strong logistic dampening,
[30] [31]
show all references
References:
[1]
N. Bellomo, A. Bellouquid, Y. Tao and M. Winkler,
Toward a mathematical theory of Keller-Segel models of pattern formation in biological tissues,
[2]
X. Cao and J. Lankeit, Global classical small-data solutions for a three-dimensional chemotaxis Navier-Stokes system involving matrix-valued sensitivities,
[3]
X. He and S. Zheng,
Convergence rate estimates of solutions in a higher dimensional chemotaxis system with logistic source,
[4] [5] [6] [7]
S. Ishida, K. Seki and T. Yokota,
Boundedness in quasilinear Keller-Segel systems of parabolic-parabolic type on non-convex bounded domains,
[8] [9]
O. A. Ladyženskaja, V. A. Solonnikov and N. N. Ural'ceva,
[10]
J. Lankeit,
Eventual smoothness and asymptotics in a three-dimensional chemotaxis system with logistic source,
[11] [12]
X. Li, Global existence and uniform boundedness of smooth solutions to a parabolic-parabolic chemotaxis system with nonlinear diffusion,
[13] [14] [15]
N. Mizoguchi and M. Winkler, Blow-up in the two-dimensional parabolic Keller-Segel system, Preprint.Google Scholar
[16] [17]
K. Osaki, T. Tsujikawa, A. Yagi and M. Mimura,
Exponential attractor for a chemotaxis-growth system of equations,
[18]
M. M. Porzio and V. Vespri,
Hölder estimates for local solutions of some doubly nonlinear degenerate parabolic equations,
[19] [20] [21]
Y. Tao and M. Winkler,
Eventual smoothness and stabilization of large-data solutions in a three-dimensional chemotaxis system with consumption of chemoattractant,
[22] [23]
L. Wang, S. U. -D. Khan and S. U. -D. Khan, Boundedness in a chemotaxis system with consumption of chemoattractant and logistic source,
[24]
M. Winkler, A three-dimensional Keller-Segel-Navier-Stokes system with logistic source: Global weak solutions and asymptotic stabilization, Preprint.Google Scholar
[25] [26]
M. Winkler,
Boundedness in the higher-dimensional parabolic-parabolic chemotaxis system with logistic source,
[27]
M. Winkler,
Global large-data solutions in a chemotaxis-(Navier-)Stokes system modeling cellular swimming in fluid drops,
[28] [29]
M. Winkler,
Global asymptotic stability of constant equilibria in a fully parabolic chemotaxis system with strong logistic dampening,
[30] [31]
[1]
Ling Liu, Jiashan Zheng.
Global existence and boundedness of solution of a parabolic-parabolic-ODE chemotaxis-haptotaxis model with (generalized) logistic source.
[2]
Chunhua Jin.
Global classical solution and stability to a coupled chemotaxis-fluid model with logistic source.
[3]
Abelardo Duarte-Rodríguez, Lucas C. F. Ferreira, Élder J. Villamizar-Roa.
Global existence for an attraction-repulsion chemotaxis fluid model with logistic source.
[4]
Hua Zhong, Chunlai Mu, Ke Lin.
Global weak solution and boundedness in a three-dimensional competing chemotaxis.
[5] [6]
Liangchen Wang, Yuhuan Li, Chunlai Mu.
Boundedness in a parabolic-parabolic quasilinear chemotaxis system with logistic source.
[7] [8]
Pan Zheng, Chunlai Mu, Xuegang Hu.
Boundedness and blow-up for a chemotaxis system with generalized volume-filling effect and logistic source.
[9]
Shijie Shi, Zhengrong Liu, Hai-Yang Jin.
Boundedness and large time behavior of an attraction-repulsion chemotaxis model with logistic source.
[10]
Feng Li, Yuxiang Li.
Global existence of weak solution in a chemotaxis-fluid system with nonlinear diffusion and rotational flux.
[11]
Tobias Black.
Global existence and asymptotic stability in a competitive two-species chemotaxis system with two signals.
[12]
Rachidi B. Salako, Wenxian Shen.
Existence of traveling wave solutions to parabolic-elliptic-elliptic chemotaxis systems with logistic source.
[13]
Tong Li, Anthony Suen.
Existence of intermediate weak solution to the equations of multi-dimensional chemotaxis systems.
[14]
Ke Lin, Chunlai Mu.
Convergence of global and bounded solutions of a two-species chemotaxis model with a logistic source.
[15]
Xie Li, Zhaoyin Xiang.
Boundedness in quasilinear Keller-Segel equations with nonlinear sensitivity and logistic source.
[16]
Masaaki Mizukami.
Boundedness and asymptotic stability in a two-species chemotaxis-competition model with signal-dependent sensitivity.
[17] [18] [19]
Masaki Kurokiba, Toshitaka Nagai, T. Ogawa.
The uniform boundedness and threshold for the global existence of the radial solution to a drift-diffusion system.
[20]
Sachiko Ishida.
Global existence and boundedness for chemotaxis-Navier-Stokes systems
with position-dependent sensitivity in 2D bounded domains.
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(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
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In this paper:
in eqns 3,5 the state eqn has the mean removed.
$(z_t-\mu)=A(z_{t-1}-\mu) + \epsilon_t$
$y_t=C z_t + \delta_t$
However I have looked at several implementations of Kalman Filters for state space models I haven't seen this "de-meaned" version any where?
Moreover if you look at implementations such as this by Kevin Murphy he doesn't de-mean the state process either:
Thus Murphy (and most other authors) take the state space model to be i.e the state equation is NOT demeaned:
$z_t=Az_{t-1} + \epsilon_t$
$y_t=Cz_t + \delta_t$
However this model also includes an initial observation x0 which is distributed as $N(\xi, \Lambda)$. (Note that in the paper the choice over what x0 to use is not covered.)
Q1.) From a theoretical point of view are the two representations equivalent? Clearly they are if $\mu=0$ but more generally for non zero values of $\mu$?Does the mean of $x_0=\xi$ offset the process so that the rest of the observations cab be treated as mean zero? If so can someone show me how this makes the processes equivalent algebraically!
When I test the model I use data generated using the non demeaned version of the state equation, with the appropriate x_0, and use the toolbox to estimate it (which also does not demean the state equation) and I get good estimates. This is to be expected as $\mu$ in this case is zero by design. However when I work with empirical data I can't be sure that the empirical state process will be mean zero.
Q.2) Therefore should I always de-mean the state process when working with empirical data.
Kind Regards
Baz
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Let $X$ be a random varaible from a distribution with pdf $$ f(x) = \theta x^{\theta-1}, \quad 0< x < 1. $$
a) Name the distribution of $U=-\ln(X)$ by first finding its density
b) Let $X_1, X_2, \ldots,X_n$ be independent and identically distributed random variables with pdf given by earlier with $\theta$= 3. Using the result from a) and by the central limit theorem (CLT)
i) find an approximation to $P(X_1 \cdot X_2 \cdot \ldots\cdot X_{30} \leq 1.85 \cdot 10^{-5})$
i.e. for the probability of the product of the r.v's.
My attempt
I found part a) by doing the transformation, and got an exponential with parameter $\theta$ where my pdf is: $$ f(x)=\theta e^{-u\theta} . $$ Now where I am struggling is with part i) where I am supposed to find an approximation using the CLT and together with the mean equal to $1/3$ and variance equal to $1/9$. I am able to show that with 30 observation we just take the product of every mean and variance from each individual observation giving us:
mean = $(\frac{1}{3})^{30}$
variance = $(\frac{1}{9})^{30}$
and then we can just substitute these into the z score by using central limit theorem $$ P(X_1 \cdot X_2 \cdot \ldots \cdot X_{30} \leq 1.85 \cdot 10^{-5}) = P\left(\frac{X -\mu }{\sigma}\leq \frac {1.85 \cdot 10^{-5} - (\frac{1}{3})^{30}}{(\frac{1}{9})^{15}} \right) $$ Hence giving me an answer of $ P(Z \leq 3,808,985,943)$ which definitely cannot be correct. Would appreciate it if somebody could point out my mistake.
Reattempt:
Using the hint i got i managed to deduce the following
$P(X_1 \cdot X_2 \cdot \ldots \cdot X_{30} \leq 1.85 \cdot 10^{-5})$
$P(log X_1 \cdot log X_2 \cdot \ldots \cdot log X_{30} \leq log 1.85 \cdot 10^{-5})$
P($\sum_{k=1}^{30} log X_i \leq log 1.85 \cdot 10^{-5})$
And since X random variable can be normally distributed X~ N( $\frac{1}{3}$ , $\frac{1}{9}$)
Substitute into the Z score
$ P(\frac{ log X -\mu }{\frac{\sigma}{\sqrt(n)}})\leq \frac { \frac{log1.85 \cdot 10^{-5}}{30} - (\frac{1}{3})}{(\frac{\frac{1}{3}}{\sqrt(n)})} $
$P(Z \leq 11.45)$
Am I on the right track? Is it correct to use $\mu$ = $\frac{1}{3}$ and $\sigma$ = $\frac{1}{3}$ in the z score?
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Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section. Moon Site Admin Posts: 751 Joined: Tue Nov 02, 2010 7:52 pm Location: Dhaka, Bangladesh Contact:
Problem 10: $ABCD$ is a rectangle where $AB = 3$, $BC = 6$ and $CD=CE$. The area of the part of the quadrangle $CDGF$ that lies outside the circular arc can be expressed as $a-b \sqrt{3}-\frac{c\cdot \pi}{4}$ where $a$, $b$, $c$ are integers. Find $a+b+c$. Attachments 2011_11.png (11.96 KiB) Viewed 1253 times Tahmid Hasan Posts: 665 Joined: Thu Dec 09, 2010 5:34 pm Location: Khulna,Bangladesh.
where is the point $E$ situated?
বড় ভালবাসি তোমায়,মা
Moon Site Admin Posts: 751 Joined: Tue Nov 02, 2010 7:52 pm Location: Dhaka, Bangladesh Contact:
Diagram uploaded.
Mehfuj Zahir
Posts: 78 Joined: Thu Jan 20, 2011 10:46 am
Connect the point C,G.Find the area of triangle CGF=9.SO,BG=6.Then using pythagoreous find BF&get GF.Now the area of triangle CGF&CDG is equal.Find area of quadrangle DGFC.The angleCDF is 30.Then subtract area inscribed by the arch DF from quadrangle DGFC.Ans should be a+b+C=30
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The textbook Elements of Information Theory gives us an example:
For example, if we knew the true distribution p of the random
variable, we could construct a code with average description length
H(p). If, instead, we used the code for a distribution q, we would
need H(p) + D(p||q) bits on the average to describe the random
variable.
To paraphrase the above statement, we can say that if we change the information distribution(from q to p) we need D(p||q) extra bits on average to code the new distribution.
An illustration
Let me illustrate this using one application of it in natural language processing.
Consider that a large group of people, labelled B, are mediators and each of them is assigned a task to choose a noun from
turkey,
animal and
book and transmit it to C. There is a guy name A who may send each of them an email to give them some hints. If no one in the group received the email they may raise their eyebrows and hesitate for a while considering what C needs. And the probability of each option being chosen is 1/3. Toally uniform distribution(if not, it may relate to their own preference and we just ignore such cases).
But if they are given a verb, like
baste, 3/4 of them may choose
turkey and 3/16 choose
animal and 1/16 choose
book. Then how much information in bits each of the mediators on average has obtained once they know the verb? It is:
\begin{align*}D(p(nouns|baste)||p(nouns)) &= \sum_{x\in\{turkey, animal, book\}} p(x|baste) \log_2 \frac{p(x|baste)}{p(x)} \\&= \frac{3}{4} * \log_2 \frac{\frac{3}{4}}{\frac{1}{3}} + \frac{3}{16} * \log_2\frac{\frac{3}{16}}{\frac{1}{3}} + \frac{1}{16} * \log_2\frac{\frac{1}{16}}{\frac{1}{3}}\\&= 0.5709 \space \space bits\\\end{align*}
But what if the verb given is
read? We may imagine that all of them would choose
book with no hesitatation, then the average information gain for each mediator from the verb
read is:
\begin{align*}D(p(nouns|read)||p(nouns)) &= \sum_{x\in\{book\}} p(x|read) \log_2 \frac{p(x|read)}{p(x)} \\&= 1 * \log_2 \frac{1}{\frac{1}{3}} \\& =1.5849 \space \space bits \\\end{align*}We can see that the verb
read can give the mediators more information. And that's what relative entropy can measure.
Let's continue our story. If C suspects that the noun may be wrong because A told him that he might have made a mistake by sending the wrong verb to the mediators. Then how much information in bits can such a piece of bad news give C?
1) if the verb given by A was
baste:
\begin{align*}D(p(nouns)||p(nouns|baste)) &= \sum_{x\in\{turkey, animal, book\}} p(x) \log_2 \frac{p(x)}{p(x|baste)} \\&= \frac{1}{3} * \log_2 \frac{\frac{1}{3}}{\frac{3}{4}} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{\frac{3}{16}} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{\frac{1}{16}}\\&= 0.69172 \space \space bits\\\end{align*}
2) but what if the verb was
read?\begin{align*}D(p(nouns)||p(nouns|baste)) &= \sum_{x\in\{book, *, *\}} p(x) \log_2 \frac{p(x)}{p(x|baste)} \\&= \frac{1}{3} * \log_2 \frac{\frac{1}{3}}{1} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{0} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{0}\\&= \infty \space \space bits\\\end{align*}
Since C never know what would the other two nouns be and any word in the vocabulary would be possible.
We can see that the KL divergence is asymmetric.
I hope I am right, and if not please comment and help correct me. Thanks in advance.
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If a linear operator between two Banach spaces is surjective and bounded, can we get any information about a right inverse? For example, is it bounded?
Thanks, trying to understand trace operator stuff on the Sobolev spaces.
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If a linear operator between two Banach spaces is surjective and bounded, can we get any information about a right inverse? For example, is it bounded?
Thanks, trying to understand trace operator stuff on the Sobolev spaces.
If $T: X \to Y$ is bounded and surjective, the Open Mapping Theorem says there is an isomorphism $S:\; X/\ker(T) \to Y$ such that $T = S \circ \pi$, where $\pi: X \to X/\ker(T)$ is the quotient map. The trouble is, a quotient of $X$ might not be isomorphic to a closed subspace of $X$, so there might not be a bounded right inverse. For example, every separable Banach space is isomorphic to a quotient of $\ell^1$, but not every Banach space is isomorphic to a subspace of $\ell^1$ (e.g. $\ell^2$ is not).
Let $T : X \to Y$ be a surjection between Banach spaces $X$, $Y$.
You can get a linear right inverse, cf. Robert Isreal's comment.
You can get a bounded right inverse: The open mapping theorem yields $r > 0$ with $B_r^Y \supset T \, B_1^X$. Hence, for $y \in Y$ you find $x \in X$ with $T \, x = y$ and $\|x\| \le \|y\|/r$.
You cannot always get the existence of a linear
and bounded right inverse. In Existence of right inverse. it is mentioned that this is true iff the kernel of $T$ is complemented in $X$.
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Solve the initial value problem: $y'+2y=e^{-x}$ with $y(0)=4$
Our teacher gave me a mistake / unclear mark at the part which I will mark red here. But I don't understand why and I no longer have the opportunity to ask him.
This is an extract of my solution (it leads to the correct solution):
$$e^{2x} \cdot \frac{dy}{dx} + e^{2x} \cdot 2y = e^{2x} \cdot e^{-x} \Leftrightarrow$$
$$\Leftrightarrow \color{red}{\frac{d}{dx} \cdot e^{2x} \cdot y} = e^{x} \Leftrightarrow$$
$$\int{\frac{d}{dx} \cdot e^{2x} \cdot y} \text{ } dx = \int{e^x} \text{ } dx$$
$$...$$
I don't understand, is an explanation here really required? So I took that $y$ from $\frac{dy}{dx}$ and wrote it at the end of the term and now we got $\frac{d}{dx}$ at the front. $\frac{d}{dx}$ means "the derivative of.." (whatever comes after). And taking that derivative, we will get what we had before (the line before the red marked line), so they are equal to each other.
But how would you explain that (on paper) in a reasonable way? I don't know if my current explanation is fine.
Edit: When I handed this in, I used brackets too. The reason I got point deduction is "how do you get from here to there?" (first line to red line).
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Let $k$ be a field and $A=k\times k\times ...$ the product of denumerably many many copies of $k$.
Let $I\subset A$ be the ideal of eventually zero sequences and $\mathfrak m\supset I$ a maximal ideal containing it. Since in $A$ every element $a$ is multiple of $a^2$, we certainly have $\mathfrak m=\mathfrak m^2$ but $\mathfrak m$ is not finitely generated: else it would be generated by an idempotent ( by Nakayama ).
Edit Since the OP has edited his question, requesting an example with a local ring, here is such an example.
Consider the domain $A=\mathbb Q[X^{1/n}|\; n=1,2,\cdots]$ consisting of "polynomials" over a field $k$ with positive rational exponents, and its maximal ideal $M=\langle X^{1/n}|n=1,2,\cdots\rangle\subset A$.
Obviously $M=M^2$. If we now localize at $M$ we get the required local ring $R=A_M$, with maximal ideal $\mathfrak m=MA_M$.
Indeed, $\mathfrak m=\mathfrak m^2$ is clear and that ideal is not finitely generated: the simplest argument is again that if it were, it would be generated by a single idempotent element (Nakayama).
But this is impossible, because $R$ is a domain and thus has only $1$ and $0$ as idempotents.
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