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I came across the sum: $$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1} $$ where $m>0$.
It's easy to see that this sum is equal to: $$\int_0^1 \frac{1}{1+x^m}dx$$ for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1} = \sum_{k=0}^\infty A_k\ln|1-e^{-(i(2k+1))/m}| $$Where $A_k$ equals
$$A_k= \frac{1}{\prod_{n=0}^k(e^{i(2k+1)}-e^{i\pi(2n+1)/m})\prod_{n=k}^\infty(e^{ik\pi/m}-e^{i(2n+3)}) } $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
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A geometric series is a sires of the form:$$\sum_{n = 0}^{\infty} ar^n$$
And the only time such a series converges is when $|r|<1$. In that case, what it converges to can be found using the following formula:
$$\sum_{n = 0}^{\infty} ar^n=\frac{a}{1-r}$$
To convert the given series into a geometric one (or ones), just start by shifting the index $n$ down by $1$ to make it begin at $0$:
$$\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}=\sum_{n = 0}^{\infty} \frac{6 - 2^{2(n+1) - 1}}{3^n}=\sum_{n = 0}^{\infty} \frac{6 - 2^{2n+1}}{3^n}=\\\sum_{n = 0}^{\infty} \left(\frac{6}{3^n}-\frac{2\cdot2^{2n}}{3^n}\right)=6\sum_{n = 0}^{\infty} \left(\frac{1}{3}\right)^n-2\sum_{n = 0}^{\infty}\left(\frac{4}{3}\right)^n$$
The first of the two geometric series converges because $|1/3|<1$. The second geometric series diverges because $|4/3|$ is not less then $1$. If one series in a sum diverges, the whole thing diverges.
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I have a (simple?) question about Fourier transforms.
Consider a 1D Hamiltonian of the form \begin{equation} H = -JS\sum_{j = 1}^{N-1}a_{j+1}^\dagger a_j + a_j^\dagger a_{j+1} - a^\dagger_{j+1}a_{j+1} - a^\dagger_j a_j \end{equation} where $J$ is a coupling between two nearest neighbours, and $S$ is the spin projection along some z-axis, i.e a standard ferromagnetic chain with $N$ lattice sites and lattice spacing $d$.
To diagonalize this one typically introduces the fourier transformed creation/annhilation operators \begin{equation} a_j = \frac{1}{\sqrt{N}} \sum_k e^{ik jd}a_{k}. \end{equation} This is fine as long as we assume periodic boundary conditions such that $a_{j+N}=a_{j}$.
Now consider the case when we let $N\rightarrow \infty$. In this case, it no longer makes sense to use periodic boundary conditions. How then do we define a Fourier transform in order to diagonalize such a problem?
Is it as simple as just writing \begin{equation} a_j = \int dk e^{ikjd}a_{k} \end{equation}?
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Yes, it has been: , or more stylized , the depression made by the tip of a Babylonian wedge shaped stylus on a clay tablet. When a circular stylus was used (rarely) the symbol was just $\bigcirc$. The earliest positional system was sexagesimal, with base 60, so it had cuneiform symbols for all digits from 1 to 59. Babylonians used it since before 2000 BC for commercial bookkeeping, etc. The absence of zero caused ambiguities, e.g. 1 and 60 had the same symbol. In the medial positions, however, zero was represented by a blank and later by a placeholder symbol . Greek astronomers replaced it with $o$ in the second century BC, that could be used at the end of a number also, removing the ambiguities. They also used their letter $\iota$ for 10 instead of Babylonian cuneiform, even though it was the 9th letter of their alphabet (archaic letter $\digamma$ was used for 6).
So the use of zero as a placeholder much predates its use as a number, and Indians learned about it from Ptolemy's Almagest, if not earlier. Therefore, they would have needed no separate symbol for 10 when they switched to decimal notation. But in any case, zero is already used in 3rd century AD even as a number, while the decimal notation first appears much later, around 458 AD, so the problem never arose.
But 10 was not special among sexagesimal digits, so perhaps more in the spirit of the question is the symbol $\big|$ used in Chinese proto-decimal system before 4th century BC. That system had hieroglyphs for digits from 1 to 9, and for the powers of 10. Although digits were written in order we write them today to form a number, the symbol for a power of ten was placed above or below each, so its value wasn't indicated by position alone. This allowed for non-ambiguous representation without even placeholder 0, and the number could still be recovered if the digits got scrambled.
By the way, hexadecimal system was also used in China, mostly for calculations with weights. This was done on abacus (counting board) starting around 190 AD, and since abacus is not paper what represented 10 was not a symbol, but an arrangement of beads.
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The transport equation describes how a scalar quantity is transported in a space. Usually, it is applied to the transport of a scalar field (e.g. chemical concentration, material properties or temperature) inside an incompressible flow. From the mathematical point of view, the transport equation is also called the convection-diffusion equation, which is a first order PDE (partial differential equation). The convection-diffusion equation is the basis for the most common transportation models.
Mathematical derivation¶
The transport equation (or convection-diffusion equation) can be seen as the generalization of the continuity equation\(^1\). While the continuity equation (extensively described in the article about incompressible flow) usually describes the conservation of mass, the convection-diffusion equation describes the continuity/conservation of any scalar field in any space. Let’s consider an infinitesimal portion of space and its boundaries, as described in figure 1.
The continuity principle states that the rate of change for a scalar quantity in any differential control volume is given by flow and diffusion into and out of that part of the system, along with any generation or consumption inside the control volume. In practice, it means that the variation of concentration of a certain quantity in the volume is given by the balance of this quantity flow across the boundary and the amount of quantity produced or removed in the volume. From the mathematical point of view, this balance is expressed by the following equation:
where \(c\) is the scalar field to be analyzed, \(j\) is the flux of \(c\) through the boundary, and \(S\) is the source/sink term inside \(\Omega\). Equation 1 is nothing more than a balance of a scalar quantity inside the volume: the first term (\(\frac{\partial c}{\partial t}\)) represents the variation of quantity of the scalar field inside the control volume, the second term (\(\nabla\cdot j\)) represents the net balance of the quantity of field which enters and exits the control volume, and the third term (\(S\)) represents the amount of the scalar quantity “created” or “destroyed” inside the volume.
Equation 1 can be further detailed by developing its terms:
The flux \(j\) can be divided into two terms: the convective and the diffusive terms. The convection term is the quantity of the transported field which moves across the boundaries because of the flow; thus it is proportional to the velocity and can be written as \(j_{convection}=cu\), where \(c\) is the transported scalar quantity and \(u\) is the velocity of the means which transports this quantity. The diffusion term is the transportation of the scalar quantity according to its gradient, so \(j_{diffusion}=D\nabla c\), where \(D\) is the diffusivity. The source term can be divided into a pure source term and a reaction term. The pure source term (\(S_S\)) represents the creation/destruction rate of the field inside the volume. The reaction term (\(S_R\)) describes the creation/destruction of the transported quantity as a reaction to this quantity itself; it is, therefore, proportional to the transported field and can be written as \(S_R=f(c)\), where \(f(c)\) is a function of the transported scalar field. \(S_R\) is quite uncommon for engineering applications, so it is often neglected.
Equation 1 can thus be re-written in its fully developed form as:
Heat Transfer¶
During thermal simulations, the temperature field (which is scalar) is transported according to the convection diffusion equation. In this specific case, the following notation is commonly used by the science community:
where:
\(\rho\) is the material density \(c_p\) is the heat capacity \(T\) is the temperature \(k\) is the thermal conductivity \(Q\) is the volumetric heat flux \(\epsilon\) is the emissivity \(\sigma\) is the Stefan-Boltzmann constant \(A\) is the boundary surface in which heat is exchanged by radiation.
Equations 2 and 3 differ only for the notation and for the complexity of the reaction term, coming from the physical modelling of heat transfer phenomena\(^3\).
Equation 3 is often simplified in common engineering applications thanks to the following hypotheses:
Homogeneous material (i.e. constant and uniform material parameters): Incompressible flow (i.e. \(\nabla\cdot u=0\)): Neglection of heat exchange by radiation: Steady state (i.e. \(\frac{\partial T}{\partial t}=0\)): Chemical concentration¶
Transportation models are commonly used to analyze the dispersion of a certain chemical component in a fluid — some pollution particles, for instance. In this case, the transport is defined as “passive transport”, because the presence of the chemical concentration does not affect the fluid flow. Referring to equation 2, the convection term represents the transportation of the chemical component with the fluid, while the diffusive term represents a chemical reaction or molecular diffusion phenomena which could occur in the flow. When a simple recirculation of a certain amount of chemical components is modeled, no explicit boundary conditions are needed for the simulation and only the initial chemical concentration distribution is required (figure 2).
On the other side, explicit boundary conditions are needed when the injection of a chemical component in a region is simulated, e.g. a chimney stack (figure 3); in this case, a boundary condition on the inlet boundary is needed to be imposed (for instance \(c=1\), if pure smoke is emitted into the atmosphere from the chimney or \(c=0.5\) if the chimney output is a mixture of 50% smoke and 50% air).
Level-set method¶
The level set is a particular family of transportation models in which a distance function (named level-set function) is transported \(^{2,3}\). This distance function is computed with respect to an interface (surface for 3D problems, line for 2D problems, and point for 1D problems) and must have 2 characteristics:
it must be an eulerian function, i.e. \(||\nabla\varphi||=1\) where \(\varphi\) is the level-set function it is signed, so it is positive on one side of the interface and negative on the other side.
For the sake of clarity, examples of the 1D and 2D distance function to be transported are shown in figure 4 and figure 5 respectively.
Level-set methods follow the principles of the classic transport equation, but they require a further computation step (called re-initialization) in order to maintain the Eulerian distance function condition. The main advantage of the use of a level-set function is the possibility to solve the transport equation just once, but to obtain many results in the post-processing. For instance, for multiphase flows (see the following section), it is possible to compute the transported viscosity and density fields, the surface tension, and to impose a mixing law across the interface with very little computational effort. In addition, the use of the level-set function allows dealing with fields which show high gradients (e.g. the density field for the simulation of a water droplet falling through the air). The transport of the density field would lead to numerical instabilities across the interface, where the density gradient is very high, while the transport of the level-set is more stable and enables computing the density as post-processing. The transportation of a distance function has several engineering applications; in the following sections, we will report the most common ones.
Image segmentation¶
The purpose of level-set methods in image processing is to identify and export the contour of a certain profile in the image. This process (called image segmentation) is widely used in computer graphics and medical applications\(^4\). It is commonly used to analyze ultrasounds and tomographies’ outputs in the localization of tumors; in figure 6 it is shown the segmentation process for an ultrasound of lungs. By the segmentation process, it is possible to identify objects inside a digital image (i.e. the lungs) through the identification of its boundaries and, eventually, use this information for the analysis of the pathology.
In the case of image segmentation, convection and source terms in equation 3 are not considered; the only means of transport is diffusion according to the gradient of pixels density of the image.
Multiphase flow¶
The level-set method is also commonly used to simulate multiphase flows. In this case, the transport model is defined as “active transport” because convection/diffusion of the interface (through the distance function) affects the flow. The distance from the interface influences the flows because it is used to compute the material parameters in the computational domain; given a flow composed by two fluids characterized by different material properties \(\eta_1\) and \(\eta_2\), the value in any given position (or element, in the case of numerical simulation) is given by:
where \(H(\phi)\) is the Heaviside function defined as
In this way, any complex multiphase flow with non-uniform material properties can be modeled through the simple transport of a scalar field (the level-set function) and the interface can be tracked implicitly as the zero Isocontour of this function. In figure 7, the results of the simulation of a rising bubble are shown; in this case, the interface represents the \(\varphi=0\) isoline, while the region where \(\varphi>0\) is colored in blue and the region where \(\varphi<0\) is colored in red.
Microstructure analysis¶
Similarly to the simulation of multiphase flows, the level-set method is used for the modelling of material crystallization. The transport of the level-set function is the starting point to analyze the phase change and crystals growth within a material. In figure 8, an example of a 2D representation of the crystal structure of a material through level-set method is presented: each microstructural grain is identified through its boundaries (the zero-level of the level-set function), while the color is based on the level-set function in order to distinguish the grain.
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
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 ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
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Extension of block matrix representation of the geometric mean
J. Korean Math. Soc. Published online August 22, 2019
Hana Choi, Hayoung Choi, Sejong Kim, and Hosoo LeeSungkyunkwan University, ShanghaiTech University, Chungbuk National Universit, Jeju National University
Abstract : To extend the well-known extremal characterization of the geometric mean of two $n \times n$ positive definite matrices $A$ and $B$, we solve the following problem.\begin{equation*}\max \Bigg\{ X:X=X^*,~\begin{pmatrix}A & V & X \\V & B & W \\X & W & C\end{pmatrix}\geq 0\Bigg \}\end{equation*}We find an explicit expression of the maximum value with respect to the matrix geometric mean of Schur complements.
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Let $X$ be a smooth complex projective variety of dimension $n$.
Under the duality between $N_1(X)$ and $N^1(X)$ we know that closure of cone of effective curves $\overline{NE}(X)$ is dual to closure of ample cone $\overline{Amp}(X)$.
It was proved in 2004 that the closure of cone of effective divisors $\overline{Eff}(X)$ is dual to the closure of cone of movable curves $\overline{Mov}(X)$. A movable curve by definition is a curve class $C \in N_1(X)$ such that $C=\pi_*(H_1 H_2 \cdots H_{n-1})$, where $\pi: X' \rightarrow X$ is a birational morphism and $H_i$'s are ample classes on $X'$.
My question: Let $Q(X)$ be the cone obtained by curve classes $H_1 H_2 \cdots H_{n-1}$ where $H_i$ are ample divisors on $X$ itself. Is it true/false that $\overline{Q}(X)=\overline{Mov}(X)$? i.e. as long as I am interested only in the closure of these cones; do I really miss some curve class if I only restrict my self to intersection of ample classes on $X$ itself.
Can any body give an example where $\overline{Q}(X) \neq \overline{Mov}(X)$?
Meanwhile, I am only interested in $n=3$ case.
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Discrete Schrödinger equation and ill-posedness for the Euler equation
1.
304 Fine Hall, Princeton, NJ 08544, USA
2.
Kassar House, 151 Thayer street, Providence, RI 02912, USA
DEuler equation with periodic boundary conditions in a family of Banach spaces based on the Fourier coefficients, and show that it is ill-posed in the sense that 'norm inflation' occurs. The proof is based on the observation that the evolution of certain perturbations of the 'Kolmogorov flow' given in velocity by
$U(x,y) = \left( {\begin{array}{*{20}{c}}{\cos \;y}\\0\end{array}} \right)$ Keywords:The Euler equation, ill-posedness, norm inflation, discrete Schrödinger equation, Kolmogorov flow. Mathematics Subject Classification:Primary:35Q31, 76D03;Secondary:37B5. Citation:In-Jee Jeong, Benoit Pausader. Discrete Schrödinger equation and ill-posedness for the Euler equation. Discrete & Continuous Dynamical Systems - A, 2017, 37 (1) : 281-293. doi: 10.3934/dcds.2017012
References:
[1] [2] [3] [4] [5] [6]
A. Grünrock and S. Herr,
Low regularity local well-posedness of the derivative nonlinear Schrödinger equation with periodic initial data,
[7] [8] [9]
J. Mattingly and Ya. Sinai,
An elementary proof of the existence and uniqueness theorem for the Navier-Stokes equations,
show all references
References:
[1] [2] [3] [4] [5] [6]
A. Grünrock and S. Herr,
Low regularity local well-posedness of the derivative nonlinear Schrödinger equation with periodic initial data,
[7] [8] [9]
J. Mattingly and Ya. Sinai,
An elementary proof of the existence and uniqueness theorem for the Navier-Stokes equations,
[1]
Tsukasa Iwabuchi, Kota Uriya.
Ill-posedness for the quadratic nonlinear Schrödinger
equation with nonlinearity $|u|^2$.
[2] [3] [4] [5] [6]
Yonggeun Cho, Gyeongha Hwang, Soonsik Kwon, Sanghyuk Lee.
Well-posedness and ill-posedness for the cubic fractional Schrödinger equations.
[7]
G. Fonseca, G. Rodríguez-Blanco, W. Sandoval.
Well-posedness and ill-posedness results for the regularized Benjamin-Ono equation in weighted Sobolev spaces.
[8] [9] [10]
Marcel Braukhoff.
Semiconductor Boltzmann-Dirac-Benney equation with a BGK-type collision operator: Existence of solutions vs. ill-posedness.
[11]
J. Colliander, M. Keel, G. Staffilani, H. Takaoka, T. Tao.
Polynomial upper bounds for the instability of the nonlinear Schrödinger equation below the energy norm.
[12]
Tetsu Mizumachi, Dmitry Pelinovsky.
On the asymptotic stability of localized modes in the discrete nonlinear Schrödinger equation.
[13]
Yuanyuan Ren, Yongsheng Li, Wei Yan.
Sharp well-posedness of the Cauchy problem for the fourth order nonlinear Schrödinger equation.
[14]
Takafumi Akahori.
Low regularity global well-posedness for the nonlinear Schrödinger equation on closed manifolds.
[15]
Seckin Demirbas.
Local well-posedness for 2-D Schrödinger equation on irrational tori and bounds on Sobolev norms.
[16]
Nobu Kishimoto.
Local well-posedness for the Cauchy problem of the quadratic Schrödinger equation with nonlinearity $\bar u^2$.
[17]
Daniela De Silva, Nataša Pavlović, Gigliola Staffilani, Nikolaos Tzirakis.
Global well-posedness for a periodic nonlinear Schrödinger equation in 1D and 2D.
[18]
Zihua Guo, Yifei Wu.
Global well-posedness for the derivative nonlinear Schrödinger equation in $H^{\frac 12} (\mathbb{R} )$.
[19]
Daniela De Silva, Nataša Pavlović, Gigliola Staffilani, Nikolaos Tzirakis.
Global well-posedness for the $L^2$ critical nonlinear Schrödinger equation in higher dimensions.
[20]
Jun-ichi Segata.
Well-posedness and existence of standing waves
for the fourth order nonlinear Schrödinger type equation.
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Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:299-317, 2019.
Abstract
In the dictionary learning (or sparse coding) problem, we are given a collection of signals (vectors in $\mathbb{R}^d$), and the goal is to find a "basis" in which the signals have a sparse (approximate) representation. The problem has received a lot of attention in signal processing, learning, and theoretical computer science. The problem is formalized as factorizing a matrix $X (d \times n)$ (whose columns are the signals) as $X = AY$, where $A$ has a prescribed number $m$ of columns (typically $m \ll n$), and $Y$ has columns that are $k$-sparse (typically $k \ll d$). Most of the known theoretical results involve assuming that the columns of the unknown $A$ have certain incoherence properties, and that the coefficient matrix $Y$ has random (or partly random) structure. The goal of our work is to understand what can be said in the absence of such assumptions. Can we still find $A$ and $Y$ such that $X \approx AY$? We show that this is possible, if we allow violating the bounds on $m$ and $k$ by appropriate factors that depend on $k$ and the desired approximation. Our results rely on an algorithm for what we call the threshold correlation problem, which turns out to be related to hypercontractive norms of matrices. We also show that our algorithmic ideas apply to a setting in which some of the columns of $X$ are outliers, thus giving similar guarantees even in this challenging setting.
@InProceedings{pmlr-v99-bhaskara19a,title = {{Approximate Guarantees for Dictionary Learning}},author = {Bhaskara, Aditya and Tai, Wai Ming},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {299--317},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/bhaskara19a/bhaskara19a.pdf},url = {http://proceedings.mlr.press/v99/bhaskara19a.html},abstract = {In the dictionary learning (or sparse coding) problem, we are given a collection of signals (vectors in $\mathbb{R}^d$), and the goal is to find a "basis" in which the signals have a sparse (approximate) representation. The problem has received a lot of attention in signal processing, learning, and theoretical computer science. The problem is formalized as factorizing a matrix $X (d \times n)$ (whose columns are the signals) as $X = AY$, where $A$ has a prescribed number $m$ of columns (typically $m \ll n$), and $Y$ has columns that are $k$-sparse (typically $k \ll d$). Most of the known theoretical results involve assuming that the columns of the unknown $A$ have certain incoherence properties, and that the coefficient matrix $Y$ has random (or partly random) structure. The goal of our work is to understand what can be said in the absence of such assumptions. Can we still find $A$ and $Y$ such that $X \approx AY$? We show that this is possible, if we allow violating the bounds on $m$ and $k$ by appropriate factors that depend on $k$ and the desired approximation. Our results rely on an algorithm for what we call the threshold correlation problem, which turns out to be related to hypercontractive norms of matrices. We also show that our algorithmic ideas apply to a setting in which some of the columns of $X$ are outliers, thus giving similar guarantees even in this challenging setting.}}
%0 Conference Paper%T Approximate Guarantees for Dictionary Learning%A Aditya Bhaskara%A Wai Ming Tai%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-bhaskara19a%I PMLR%J Proceedings of Machine Learning Research%P 299--317%U http://proceedings.mlr.press%V 99%W PMLR%X In the dictionary learning (or sparse coding) problem, we are given a collection of signals (vectors in $\mathbb{R}^d$), and the goal is to find a "basis" in which the signals have a sparse (approximate) representation. The problem has received a lot of attention in signal processing, learning, and theoretical computer science. The problem is formalized as factorizing a matrix $X (d \times n)$ (whose columns are the signals) as $X = AY$, where $A$ has a prescribed number $m$ of columns (typically $m \ll n$), and $Y$ has columns that are $k$-sparse (typically $k \ll d$). Most of the known theoretical results involve assuming that the columns of the unknown $A$ have certain incoherence properties, and that the coefficient matrix $Y$ has random (or partly random) structure. The goal of our work is to understand what can be said in the absence of such assumptions. Can we still find $A$ and $Y$ such that $X \approx AY$? We show that this is possible, if we allow violating the bounds on $m$ and $k$ by appropriate factors that depend on $k$ and the desired approximation. Our results rely on an algorithm for what we call the threshold correlation problem, which turns out to be related to hypercontractive norms of matrices. We also show that our algorithmic ideas apply to a setting in which some of the columns of $X$ are outliers, thus giving similar guarantees even in this challenging setting.
Bhaskara, A. & Tai, W.M.. (2019). Approximate Guarantees for Dictionary Learning. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:299-317
This site last compiled Sat, 17 Aug 2019 00:05:37 +0000
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Sine of Half-Integer Multiple of Pi
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Theorem
Let $x \in \R$ be a real number.
Let $\sin x$ be the sine of $x$.
Then: $\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$
or:
\(\displaystyle \forall m \in \Z: \ \ \) \(\displaystyle \map \sin {2 m + \dfrac 1 2} \pi\) \(=\) \(\displaystyle 1\) \(\displaystyle \forall m \in \Z: \ \ \) \(\displaystyle \map \sin {2 m - \dfrac 1 2} \pi\) \(=\) \(\displaystyle -1\) Proof
From the discussion of Sine and Cosine are Periodic on Reals, we have that:
$\map \sin {x + \dfrac \pi 2} = \cos x$
The result then follows directly from the Cosine of Multiple of Pi.
$\blacksquare$
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(EDIT: My original post took $p$ to be the smallest prime that divides $k$. This is unnecessary. You can take $p$ to be
any number other than 1 that divides $k$. Doing so gives you not just one solution but $d(k)-1$ solutions, one for each of the divisors of $k$ other than 1.)
(SECOND EDIT: See the answer to this question for all solutions in $m$ and $n$.)
First,$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$$Then $$s_n = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1}.$$
Thus, for any $m,n \geq 1$, $$s_m - s_n = \frac{1}{n+1} - \frac{1}{m+1} = \frac{m-n}{(m+1)(n+1)}.$$
We want to find $m,n$ such that this last expression is equal to $\frac{1}{k}$. Let $p$ be any positive integer other than 1 that divides $k$. Take $m = (p-1)k-1$. Clearly $m$ is a positive integer.
Then we want to show that the resulting $n$ that solves $$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k}$$ is also an integer.
We have
$$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k} \Rightarrow (m-n)k = (m+1)(n+1) $$$$\Rightarrow ((p-1)k-1 - n)k = (p-1)k(n+1) \Rightarrow (p-1)k-1 - n = (p-1)n + p-1 $$$$\Rightarrow pn = (p-1)k - p \Rightarrow n = \frac{(p-1)k}{p} - 1,$$
which means that $n$ is an integer because $p|k$.
Thus we have a family of solutions $$m = (p-1)k-1, n = \frac{(p-1)k}{p} - 1,$$where $p$ is any positive integer other than 1 that divides $k$.
This works even in the case $k = 2$ because then we just get $m = 1$, $n = 0$.
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I wonder if anybody can help me with this problem.
I'm trying to compute the Mertens function for large $n$. The most obvious algorithm is just to compute all primes up to $\sqrt{n}$ and then to sieve. That takes at least an order of $n\log n$ operations, and really even more.
The most recent article that I could find that discusses methods to compute the function directly is dated 1994, and it proposes to do exactly that.
Are there any known algorithms that let you compute Mertens faster than by sieving? I know that $\pi(n)$ can be computed in $O(n^{2/3})$, I looked into that algorithm but it does not seem to be easily adaptable to my task.
Alternatively, I could use an algorithm to compute $M(n+dn)-M(n)$ for $dn\ll n$ (say $dn\sim \sqrt{n}$ ) in $O(\sqrt{n})$ time or less.
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Asymptotic behaviour for wave equations with memory in a noncylindrical domains
1.
Departamento de Matemática-DMA, Universidade Estadual de Maringá-UEM, Campus Universitário, Av. Colombo, 5790-Zona 7, CEP 87020-900, Maringá-Pr., Brazil
2.
Departamento de Matemática, Universidade Federal do Pará, Campus Universitário do Guamá, Rua Augusto Corrêa 01, Cep 66075-110, Pará, Brazil
$u_{t t}-\Delta u+\int^t_0g(t-s)\Delta u(s)ds + \alpha u_{t}=0$ in $\hat Q$
where $\hat Q$ is a non cylindrical domains of $\mathbb R^{n+1}$ $(n\ge1)$ with the lateral boundary $\hat{\sum}$ and $\alpha$ is a positive constant.
Mathematics Subject Classification:35K55, 35F30, 34B1. Citation:Jorge Ferreira, Mauro De Lima Santos. Asymptotic behaviour for wave equations with memory in a noncylindrical domains. Communications on Pure & Applied Analysis, 2003, 2 (4) : 511-520. doi: 10.3934/cpaa.2003.2.511
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What are the examples of "confined" $Z_N$ gauge theory?
From these two questions:
Phase Structure of (Quantum) Gauge Theory
http://physics.stackexchange.com/questions/102541
http://physics.stackexchange.com/questions/29359/
We learn that there are deconfined $Z_N$ gauge theory such as the $Z_N$-toric code (Kitaev) or $Z_N$-topological order (Wen). There are fractionalized anyons as excitations separated from the ground state by an order O(J) gap where $J$ is the coupling of lattice Hamiltonian. These "deconfined" $Z_N$ gauge theories are beyond Landau-Ginzburg theory, and the "deconfined" $Z_N$ gauge theories cannot be classified by global symmetry-breaking pattern.
However, there are discussions in the posts above concerning
"confined" $Z_N$ gauge theory. What are the examples of "confined" $Z_N$ gauge theory? Are "confined" $Z_N$ gauge theory within Landau-Ginzburg theory, and the "confined" $Z_N$ gauge theories can be classified by global symmetry-breaking pattern?
For example, is there an example of "confined" $Z_N$ gauge theory in 1+1d, 2+1d, 3+1d, etc?
Is the 1+1d example accessible through the transverse magnetic field on Ising Hamiltonian:
$$H(\sigma) = - \sum_{\langle i~j\rangle} J_{ij} \sigma_i \sigma_j -\mu \sum_{j} h_j\sigma_j$$
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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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@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11...
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr...
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One of my discoveries as a physicist was that, despite all attempts at clarity, we still have different meanings for the same words and use different words to refer the the same thing. When Alice says measurement, Bob hears a `quantum to classical channel', but Alice, a hard-core Everettian, does not even believe such channels exist. When Charlie says non-local, he means Bell non-local, but string theorist Dan starts lecturing him about non-local Lagrangian terms and violations of causality. And when I say non-local measurements, you hear ???? ?????. Let me give you a hint, I do not mean 'Bell non-local quantum to classical channels', to be honest, I am not even sure what that means.
So what do I mean when I say measurement? A measurement is a quantum operation that takes a quantum state as its input and spits out a quantum state
and a classical result as an output (no, I am not an Everettian). For simplicity I will concentrate of a special case of this operation, a projective measurement of an observable A.
The classical result of a projective measurement is an eigenvalue of \(A\), but what is the outgoing state?
The Lüders Measurement
Even the term projective measurement can lead to confusion, and indeed in the early days of quantum mechanics it did. When von Neumann wrote down the mathematical formalism for quantum measurements he missed an important detail about degenerate observables (i.e., Hermitian operators with a degenerate eigenvalue spectrum). In the usual projective measurement the state of the system after the measurement is uniquely determined by the classical result (an eigenvalue of the observable). Consequently, if we don't look at the classical result the quantum channel is a standard dephasing channel. In the case of a degenerate observable, the same eigenvalue corresponds to two or more orthogonal eigenstates. Seemingly the state of the system should correspond to one of those eigenstates, and the channel is a standard dephasing channel. But a degenerate spectrum means that the set of orthogonal eigenvectors is not unique, instead each eigenvalue has a corresponding subspace of corresponding eigenvectors. What Lüders suggested is that the dephasing channel does nothing within these subspaces.
Example
Consider the two qubit observable \(A=|00\rangle\langle 00 |\). It has eigenvalues \(1,0,0,0\). A 1 result in this measurement corresponds to "The system is in the state \(|{00}\rangle\)." Following a measurement with outcome \(1\), the outgoing state will be \(|00\rangle\). Similarly, a 0 result corresponds to "The system is not in the state \(|{00}\rangle\)". But here is where the Lüders rule kicks in. Given a generic input state \(\alpha|{00}\rangle+\beta|{01}\rangle+\gamma|{10}\rangle+\delta|{11}\rangle\) and a Lüders measurement of \(A\) with outcome 0, the outgoing state will be \(\frac{1}{\sqrt{|\beta|^2+|\gamma|^2+|\delta|^2}}\left[\beta|{01}\rangle+\gamma|{10}\rangle+\delta|{11}\rangle\right]\).
Non-local measurements
The relation to non-locality may already be apparent from the example, but let me start with some definitions. A system can be called non-local if it has parts in different locations, e.g., one part on Earth and the other on the moon. A measurement is non-local if it reveals something about a non-local system as a whole. In principle these definitions apply to classical and quantum systems. Classically a non-local measurement is trivial, there is no conceptual reason why we can't just measure at each location. For a quantum system the situation is different. Let us use the example above, but now consider the situation where the two qubits are in separate locations. Local measurements of \(\sigma_z\) will produce the desired measurement statistics (after coarse graining) but reveal too much information and dephase the state completely, while a Lüders measurement should not. What is quite neat about this example is that the Lüders measurement of \(|{00}\rangle\) cannot be implemented without entanglement (or quantum communication) resources and two-way classical communication. To prove that entanglement is necessary, it is enough to give an example where entanglement is created during the measurement. To show that communication is necessary, it is enough to show that the measurement (even if the outcome is unknown) can be used to transmit information. The detailed proof is left as an exercise to the reader. The lazy reader can find it here.
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Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:1649-1682, 2019.
Abstract
Robust mean estimation is the problem of estimating the mean $\mu \in \mathbb{R}^d$ of a $d$-dimensional distribution $D$ from a list of independent samples, an $\varepsilon$-fraction of which have been arbitrarily corrupted by a malicious adversary. Recent algorithmic progress has resulted in the first polynomial-time algorithms which achieve \emph{dimension-independent} rates of error: for instance, if $D$ has covariance $I$, in polynomial-time one may find $\hat{\mu}$ with $\|\mu - \hat{\mu}\| \leq O(\sqrt{\varepsilon})$. However, error rates achieved by current polynomial-time algorithms, while dimension-independent, are sub-optimal in many natural settings, such as when $D$ is sub-Gaussian, or has bounded $4$-th moments. In this work we give worst-case complexity-theoretic evidence that improving on the error rates of current polynomial-time algorithms for robust mean estimation may be computationally intractable in natural settings. We show that several natural approaches to improving error rates of current polynomial-time robust mean estimation algorithms would imply efficient algorithms for the small-set expansion problem, refuting Raghavendra and Steurer’s small-set expansion hypothesis (so long as $P \neq NP$). We also give the first direct reduction to the robust mean estimation problem, starting from a plausible but nonstandard variant of the small-set expansion problem.
@InProceedings{pmlr-v99-hopkins19a,title = {How Hard is Robust Mean Estimation?},author = {Hopkins, Samuel B. and Li, Jerry},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {1649--1682},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/hopkins19a/hopkins19a.pdf},url = {http://proceedings.mlr.press/v99/hopkins19a.html},abstract = {Robust mean estimation is the problem of estimating the mean $\mu \in \mathbb{R}^d$ of a $d$-dimensional distribution $D$ from a list of independent samples, an $\varepsilon$-fraction of which have been arbitrarily corrupted by a malicious adversary. Recent algorithmic progress has resulted in the first polynomial-time algorithms which achieve \emph{dimension-independent} rates of error: for instance, if $D$ has covariance $I$, in polynomial-time one may find $\hat{\mu}$ with $\|\mu - \hat{\mu}\| \leq O(\sqrt{\varepsilon})$. However, error rates achieved by current polynomial-time algorithms, while dimension-independent, are sub-optimal in many natural settings, such as when $D$ is sub-Gaussian, or has bounded $4$-th moments. In this work we give worst-case complexity-theoretic evidence that improving on the error rates of current polynomial-time algorithms for robust mean estimation may be computationally intractable in natural settings. We show that several natural approaches to improving error rates of current polynomial-time robust mean estimation algorithms would imply efficient algorithms for the small-set expansion problem, refuting Raghavendra and Steurer’s small-set expansion hypothesis (so long as $P \neq NP$). We also give the first direct reduction to the robust mean estimation problem, starting from a plausible but nonstandard variant of the small-set expansion problem.}}
%0 Conference Paper%T How Hard is Robust Mean Estimation?%A Samuel B. Hopkins%A Jerry Li%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-hopkins19a%I PMLR%J Proceedings of Machine Learning Research%P 1649--1682%U http://proceedings.mlr.press%V 99%W PMLR%X Robust mean estimation is the problem of estimating the mean $\mu \in \mathbb{R}^d$ of a $d$-dimensional distribution $D$ from a list of independent samples, an $\varepsilon$-fraction of which have been arbitrarily corrupted by a malicious adversary. Recent algorithmic progress has resulted in the first polynomial-time algorithms which achieve \emph{dimension-independent} rates of error: for instance, if $D$ has covariance $I$, in polynomial-time one may find $\hat{\mu}$ with $\|\mu - \hat{\mu}\| \leq O(\sqrt{\varepsilon})$. However, error rates achieved by current polynomial-time algorithms, while dimension-independent, are sub-optimal in many natural settings, such as when $D$ is sub-Gaussian, or has bounded $4$-th moments. In this work we give worst-case complexity-theoretic evidence that improving on the error rates of current polynomial-time algorithms for robust mean estimation may be computationally intractable in natural settings. We show that several natural approaches to improving error rates of current polynomial-time robust mean estimation algorithms would imply efficient algorithms for the small-set expansion problem, refuting Raghavendra and Steurer’s small-set expansion hypothesis (so long as $P \neq NP$). We also give the first direct reduction to the robust mean estimation problem, starting from a plausible but nonstandard variant of the small-set expansion problem.
Hopkins, S.B. & Li, J.. (2019). How Hard is Robust Mean Estimation?. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:1649-1682
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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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Semigroup and Blow-Up Dynamics of Attraction Keller-Segel Equations in Scale of Banach Spaces
Semigroup and Blow-Up Dynamics of Attraction Keller-Segel Equations in Scale of Banach Spaces
Abstract
In this paper, we study the asymptotic and blow-up dynamics of the attraction Keller-Segel chemotaxis system of equations in scale of Banach spaces $E_{q}^{\alpha}=H^{2\alpha,q}(\Omega)$, $-1\leq \alpha\leq 1, 10$, if the chemo-attractivity coefficient dominates the Moser-Trudinger threshold value. An analysis of the finite time bounds for blow-up of solutions in norm of $L^{2p}(\Omega), 1\leq p\leq 6$ and $\Omega\subset {\mathbb R}^{N}, N=2,3$, is also furnished.
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AI News, Machine Learning FAQ Machine Learning FAQ
Without going into much depth regarding information gain and impurity measures, we can think of the decision as “is feature x_i >= some_val?” Intuitively, we can see that it really doesn’t matter on which scale this feature is (centimeters, Fahrenheit, a standardized scale – it really doesn’t matter).
Some examples of algorithms where feature scaling matters are: There are many more cases than I can possibly list here … I always recommend you to think about the algorithm and what it’s doing, and then it typically becomes obvious whether we want to scale your features or not.
Machine Learning FAQ
Without going into much depth regarding information gain and impurity measures, we can think of the decision as “is feature x_i >= some_val?” Intuitively, we can see that it really doesn’t matter on which scale this feature is (centimeters, Fahrenheit, a standardized scale – it really doesn’t matter).
Some examples of algorithms where feature scaling matters are: There are many more cases than I can possibly list here … I always recommend you to think about the algorithm and what it’s doing, and then it typically becomes obvious whether we want to scale your features or not.
About Feature Scaling and Normalization
The result of standardization (or Z-score normalization) is that the features will be rescaled so that they’ll have the properties of a standard normal distribution with \mu = 0 and \sigma = 1 where \mu is the mean (average) and \sigma is the standard deviation from the mean;
standard scores (also called z scores) of the samples are calculated as follows: Standardizing the features so that they are centered around 0 with a standard deviation of 1 is not only important if we are comparing measurements that have different units, but it is also a general requirement for many machine learning algorithms.
with features being on different scales, certain weights may update faster than others since the feature values x_j play a role in the weight updates so that w_j := w_j + \Delta w_j, where
Without going into much depth regarding information gain and impurity measures, we can think of the decision as “is feature x_i >= some_val?” Intuitively, we can see that it really doesn’t matter on which scale this feature is (centimeters, Fahrenheit, a standardized scale – it really doesn’t matter).
For example, if we initialize the weights of a small multi-layer perceptron with tanh activation units to 0 or small random values centered around zero, we want to update the model weights “equally.” As a rule of thumb I’d say: When in doubt, just standardize the data, it shouldn’t hurt.
Another prominent example is the Principal Component Analysis, where we usually prefer standardization over Min-Max scaling, since we are interested in the components that maximize the variance (depending on the question and if the PCA computes the components via the correlation matrix instead of the covariance matrix;
In the following section, we will go through the following steps: In this step, we will randomly divide the wine dataset into a training dataset and a test dataset where the training dataset will contain 70% of the samples and the test dataset will contain 30%, respectively.
Let us think about whether it matters or not if the variables are centered for applications such as Principal Component Analysis (PCA) if the PCA is calculated from the covariance matrix (i.e., the k principal components are the eigenvectors of the covariance matrix that correspond to the k largest eigenvalues.
Let’s assume we have the 2 variables \bf{x} and \bf{y} Then the covariance between the attributes is calculated as Let us write the centered variables as The centered covariance would then be calculated as follows: But since after centering, \bar{x}' = 0 and \bar{y}' = 0 we have \sigma_{xy}' = \frac{1}{n-1} \sum_{i}^{n} x_i' y_i' which is our original covariance matrix if we resubstitute back the terms x'
Let c be the scaling factor for \bf{x} Given that the “original” covariance is calculated as the covariance after scaling would be calculated as: \sigma_{xy}' = \frac{1}{n-1} \sum_{i}^{n} (c \cdot x_i - c \cdot \bar{x})(y_i - \bar{y}) =
\sigma_{xy}' = c \cdot \sigma_{xy} Therefore, the covariance after scaling one attribute by the constant c will result in a rescaled covariance c \sigma_{xy} So if we’d scaled \bf{x} from pounds to kilograms, the covariance between \bf{x} and \bf{y} will be 0.453592 times smaller.
In the section on linear classification we computed scores for different visual categories given the image using the formula \( s = W x \), where \(W\) was a matrix and \(x\) was an input column vector containing all pixel data of the image.
In the case of CIFAR-10, \(x\) is a [3072x1] column vector, and \(W\) is a [10x3072] matrix, so that the output scores is a vector of 10 class scores.
There are several choices we could make for the non-linearity (which we’ll study below), but this one is a common choice and simply thresholds all activations that are below zero to zero.
Notice that the non-linearity is critical computationally - if we left it out, the two matrices could be collapsed to a single matrix, and therefore the predicted class scores would again be a linear function of the input.
three-layer neural network could analogously look like \( s = W_3 \max(0, W_2 \max(0, W_1 x)) \), where all of \(W_3, W_2, W_1\) are parameters to be learned.
The area of Neural Networks has originally been primarily inspired by the goal of modeling biological neural systems, but has since diverged and become a matter of engineering and achieving good results in Machine Learning tasks.
Approximately 86 billion neurons can be found in the human nervous system and they are connected with approximately 10^14 - 10^15 synapses.
The idea is that the synaptic strengths (the weights \(w\)) are learnable and control the strength of influence (and its direction: excitory (positive weight) or inhibitory (negative weight)) of one neuron on another.
Based on this rate code interpretation, we model the firing rate of the neuron with an activation function \(f\), which represents the frequency of the spikes along the axon.
Historically, a common choice of activation function is the sigmoid function \(\sigma\), since it takes a real-valued input (the signal strength after the sum) and squashes it to range between 0 and 1.
An example code for forward-propagating a single neuron might look as follows: In other words, each neuron performs a dot product with the input and its weights, adds the bias and applies the non-linearity (or activation function), in this case the sigmoid \(\sigma(x) = 1/(1+e^{-x})\).
As we saw with linear classifiers, a neuron has the capacity to “like” (activation near one) or “dislike” (activation near zero) certain linear regions of its input space.
With this interpretation, we can formulate the cross-entropy loss as we have seen in the Linear Classification section, and optimizing it would lead to a binary Softmax classifier (also known as logistic regression).
The regularization loss in both SVM/Softmax cases could in this biological view be interpreted as gradual forgetting, since it would have the effect of driving all synaptic weights \(w\) towards zero after every parameter update.
The sigmoid non-linearity has the mathematical form \(\sigma(x) = 1 / (1 + e^{-x})\) and is shown in the image above on the left.
The sigmoid function has seen frequent use historically since it has a nice interpretation as the firing rate of a neuron: from not firing at all (0) to fully-saturated firing at an assumed maximum frequency (1).
Also note that the tanh neuron is simply a scaled sigmoid neuron, in particular the following holds: \( \tanh(x) = 2 \sigma(2x) -1 \).
Other types of units have been proposed that do not have the functional form \(f(w^Tx + b)\) where a non-linearity is applied on the dot product between the weights and the data.
TLDR: “What neuron type should I use?” Use the ReLU non-linearity, be careful with your learning rates and possibly monitor the fraction of “dead” units in a network.
For regular neural networks, the most common layer type is the fully-connected layer in which neurons between two adjacent layers are fully pairwise connected, but neurons within a single layer share no connections.
Working with the two example networks in the above picture: To give you some context, modern Convolutional Networks contain on orders of 100 million parameters and are usually made up of approximately 10-20 layers (hence deep learning).
The full forward pass of this 3-layer neural network is then simply three matrix multiplications, interwoven with the application of the activation function: In the above code, W1,W2,W3,b1,b2,b3 are the learnable parameters of the network.
Notice also that instead of having a single input column vector, the variable x could hold an entire batch of training data (where each input example would be a column of x) and then all examples would be efficiently evaluated in parallel.
Neural Networks work well in practice because they compactly express nice, smooth functions that fit well with the statistical properties of data we encounter in practice, and are also easy to learn using our optimization algorithms (e.g.
Similarly, the fact that deeper networks (with multiple hidden layers) can work better than a single-hidden-layer networks is an empirical observation, despite the fact that their representational power is equal.
As an aside, in practice it is often the case that 3-layer neural networks will outperform 2-layer nets, but going even deeper (4,5,6-layer) rarely helps much more.
We could train three separate neural networks, each with one hidden layer of some size and obtain the following classifiers: In the diagram above, we can see that Neural Networks with more neurons can express more complicated functions.
For example, the model with 20 hidden neurons fits all the training data but at the cost of segmenting the space into many disjoint red and green decision regions.
The subtle reason behind this is that smaller networks are harder to train with local methods such as Gradient Descent: It’s clear that their loss functions have relatively few local minima, but it turns out that many of these minima are easier to converge to, and that they are bad (i.e.
Conversely, bigger neural networks contain significantly more local minima, but these minima turn out to be much better in terms of their actual loss.
In practice, what you find is that if you train a small network the final loss can display a good amount of variance - in some cases you get lucky and converge to a good place but in some cases you get trapped in one of the bad minima.
When should I apply feature scaling for my data
In my view the question about scaling/not scaling the features in machine learning is a statement about the measurement units of your features.
If the height is measured in nanometers then it's likely that any k nearest neighbors will merely have similar measures of height.
This again will have an influence on knn and might drastically reduce performance if your data had more noisy constant values compared to the ones that vary.
So this is like with everything else in machine learning - use prior knowledge whenever possible and in the case of black-box features do both and cross-validate.
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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 ...
Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays
(Elsevier, 2014-11)
The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity |y|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ...
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Now showing items 11-20 of 26
Measurement of azimuthal correlations of D mesons and charged particles in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-04)
The azimuthal correlations of D mesons and charged particles were measured with the ALICE detector in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV at the Large Hadron Collider. ...
Production of $\pi^0$ and $\eta$ mesons up to high transverse momentum in pp collisions at 2.76 TeV
(Springer, 2017-05)
The invariant differential cross sections for inclusive $\pi^{0}$ and $\eta$ mesons at midrapidity were measured in pp collisions at $\sqrt{s}=2.76$ TeV for transverse momenta $0.4<p_{\rm T}<40$ GeV/$c$ and $0.6<p_{\rm ...
Measurement of the production of high-$p_{\rm T}$ electrons from heavy-flavour hadron decays in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Elsevier, 2017-08)
Electrons from heavy-flavour hadron decays (charm and beauty) were measured with the ALICE detector in Pb–Pb collisions at a centre-of-mass of energy $\sqrt{s_{\rm NN}}=2.76$ TeV. The transverse momentum ($p_{\rm T}$) ...
Flow dominance and factorization of transverse momentum correlations in Pb-Pb collisions at the LHC
(American Physical Society, 2017-04)
We present the first measurement of the two-particle transverse momentum differential correlation function, $P_2\equiv\langle \Delta p_{\rm T} \Delta p_{\rm T} \rangle /\langle p_{\rm T} \rangle^2$, in Pb-Pb collisions at ...
$\phi$-meson production at forward rapidity in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV and in pp collisions at $\sqrt{s}$ = 2.76 TeV
(Elsevier, 2017-03)
The first measurement of $\phi$-meson production in p-Pb collisions at a nucleon-nucleon centre-of-mass energy $\sqrt{s_{\rm NN}}$ = 5.02 TeV has been performed with the ALICE apparatus at the LHC. $\phi$ mesons have been ...
J/$\psi$ suppression at forward rapidity in Pb–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2017-03)
The inclusive J/ψ production has been studied in Pb–Pb and pp collisions at the centre-of-mass energy per nucleon pair View the MathML source, using the ALICE detector at the CERN LHC. The J/ψ meson is reconstructed, ...
Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Elsevier, 2017-09)
We present the charged-particle pseudorapidity density in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02\,\mathrm{Te\kern-.25exV}$ in centrality classes measured by ALICE. The measurement covers a wide pseudorapidity ...
Azimuthally differential pion femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-06)
We present the first azimuthally differential measurements of the pion source size relative to the second harmonic event plane in Pb-Pb collisions at a center-of-mass energy per nucleon-nucleon pair of $\sqrt{s_{\rm ...
Measurement of electrons from beauty-hadron decays in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV and Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Springer, 2017-07)
The production of beauty hadrons was measured via semi-leptonic decays at mid-rapidity with the ALICE detector at the LHC in the transverse momentum interval $1<p_{\rm T}<8$ GeV/$c$ in minimum-bias p–Pb collisions at ...
Linear and non-linear flow modes in Pb-Pb collisions at $\sqrt{s_{\rm NN}} =$ 2.76 TeV
(Elsevier, 2017-10)
The second and the third order anisotropic flow, $V_{2}$ and $V_3$, are mostly determined by the corresponding initial spatial anisotropy coefficients, $\varepsilon_{2}$ and $\varepsilon_{3}$, in the initial density ...
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If $x,y$ are positive integers satisfying $x^2 - ny^2 = 1$, then $\frac{x}{y}$ is a good rational approximation to $\sqrt{n}$. A
really good rational approximation:
\begin{align}&& x^2 - ny^2 &= 1 \\&\iff& \frac{x^2}{y^2} - n &= \frac{1}{y^2} \\&\iff& \frac{x}{y} - \sqrt{n} &= \frac{1}{y^2\bigl(\sqrt{n} + \frac{x}{y}\bigr)}\end{align}
Since $\sqrt{n} + \frac{x}{y} > 2\sqrt{n} > 2$, it is such a good rational approximation to $\sqrt{n}$ that it must be one of the convergents. (Per a theorem of Legendre, if
$$\biggl\lvert \alpha - \frac{p}{q}\biggr\rvert < \frac{1}{2q^2}$$
then $\frac{p}{q}$ is a convergent of $\alpha$. Legendre's more precise theorem easily implies this corollary. Digging a little deeper, if $x^2 - ny^2 = m$ with $\lvert m\rvert < \sqrt{n}$ for positive integers $x,y$, then $\frac{x}{y}$ must be a convergent of $\sqrt{n}$.
Note: If $m$ is not squarefree, then $\frac{x}{y}$ need not be in reduced form.)
And if we look at the (simple) continued fraction expansion of $\sqrt{n}$, we can write the $k^{\text{th}}$ complete quotient as
$$\xi_k = \frac{\sqrt{n} + b_k}{c_k}$$
with integers $0 \leqslant b_k < \sqrt{n}$ (and $b_k = 0$ only for $k = 0$) and $0 < c_k < 2\sqrt{n}$, where $b_{k+1}, c_{k+1}$ are given by
\begin{align}b_{k+1} &= a_kc_k - b_k, \\c_{k+1} &= \frac{n - b_{k+1}^2}{c_k},\end{align}
as is seen from
$$\xi_{k+1} = \frac{1}{\xi_k - a_k} = \frac{c_k}{\sqrt{n} + b_k - a_kc_k} = \frac{c_k}{\sqrt{n} - b_{k+1}} = \frac{c_k(\sqrt{n} + b_{k+1})}{n - b_{k+1}^2}\,.$$
One inductively verifies that $c_k$ divides $n - b_{k+1}^2 = n - b_k^2 + 2a_kb_kc_k - a_k^2c_k^2$ and that $b_k,c_k$ satisfy the given inequalities. Further one checks that the period of the continued fraction is complete at the first $k > 0$ with $c_k = 1$.
Now it is an interesting fact that if we write the $r^{\text{th}}$ convergent (in reduced form) as
$$\frac{x_r}{y_r} = [a_0, a_1, \dotsc, a_r]$$
then
$$x_r^2 - n y_r^2 = (-1)^{r+1}\cdot c_{r+1}\,.$$
So the continued fraction of $\sqrt{n}$ has period length $k$, then the positive integer solutions to $x^2 - ny^2 = 1$ are given precisely by the numerator and denominator of the convergents with index $ik - 1$, where $i$ is a positive integer which must be even in case $k$ is odd. If $i$ and $k$ are both odd, we get the solutions to $x^2 - ny^2 = -1$. [So in your question you have a small mistake, it must be $ik - 1$ instead of $(k-1)i$.]
Summary: There are no solutions outside that set because being a solution implies giving such a good rational approximation to $\sqrt{n}$ that it must be a convergent, and the convergents with other indices (than one less than a multiple of the period length) have $\lvert x_r^2 - n y_r^2\rvert > 1$.
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AI News, Memristors power quick-learning neural network On Sunday, June 3, 2018 By Read More Memristors power quick-learning neural network
The network, called a reservoir computing system, could predict words before they are said during conversation, and help predict future outcomes based on the present.
The research team that created the reservoir computing system, led by Wei Lu, professor of electrical engineering and computer science at the University of Michigan, recently published their work in Nature Communications.
Reservoir computing systems, which improve on a typical neural network's capacity and reduce the required training time, have been created in the past with larger optical components.
In this process of what's called supervised learning, the connections between nodes are weighted more heavily or lightly to minimize the amount of error in achieving the correct answer.
For example, a system can process a new photo and correctly identify a human face, because it has learned the features of human faces from other photos in its training set.
When a set of data is inputted into the reservoir, the reservoir identifies important time-related features of the data, and hands it off in a simpler format to a second network.
This second network then only needs training like simpler neural networks, changing weights of the features and outputs that the first network passed on until it achieves an acceptable level of error.
Using only 88 memristors as nodes to identify handwritten versions of numerals, compared to a conventional network that would require thousands of nodes for the task, the reservoir achieved 91 percent accuracy.
New quick-learning neural network powered by memristors
To train a neural network for a task, a neural network takes in a large set of questions and the answers to those questions.
In this process of what’s called supervised learning, the connections between nodes are weighted more heavily or lightly to minimize the amount of error in achieving the correct answer.
For example, a system can process a new photo and correctly identify a human face, because it has learned the features of human faces from other photos in its training set.
This second network then only needs training like simpler neural networks, changing weights of the features and outputs that the first network passed on until it achieves an acceptable level of error.
Neural network powered by memristors
University of Michigan researchers created a reservoir computing system that reduces training time and improves capacity of similar neural networks.
This second network then only needs training like simpler neural networks, changing weights of the features and outputs that the first network passed on until it achieves an acceptable level of error.
Using only 88 memristors as nodes to identify handwritten versions of numerals, compared to a conventional network that would require thousands of nodes for the task, the reservoir achieved 91% accuracy.
“We could actually predict what you plan to say next.” In predictive analysis, Lu hopes to use the system to take in signals with noise, like static from far-off radio stations, and produce a cleaner stream of data.
Reservoir computing using dynamic memristors for temporal information processing AbstractReservoir computing systems utilize dynamic reservoirs having short-term memory to project features from the temporal inputs into a high-dimensional feature space.
We show that the internal ionic dynamic processes of memristors allow the memristor-based reservoir to directly process information in the temporal domain, and demonstrate that even a small hardware system with only 88 memristors can already be used for tasks, such as handwritten digit recognition.
We show experimentally that even a small reservoir consisting of 88 memristor devices can be used to process real-world problems such as handwritten digit recognition with performance comparable to those achieved in much larger networks.
A similar-sized network is also used to solve a second-order nonlinear dynamic problem and is able to successfully predict the expected dynamic output without knowing the form of the transfer function.
Indeed, adding vertical scan can improve the classification accuracy to 92.1% as verified through simulation using the device model, although the system also becomes larger and requires 672 inputs.
The computing capacity added by the memristor-based reservoir layer was analyzed by comparing the RC system performance with networks having the same connectivity patterns, by replacing the reservoir layer with a conventional nonlinear downsampling function.
For the second-order dynamic problem that is more naturally suited for the RC system, our analysis shows that the small RC system significantly outperforms a conventional linear network, with orders-of-magnitude improvements in prediction NMSE.
The demonstration of memristor-based RC systems will stimulate continued developments to further optimize the network performance toward broad applications in areas, such as speech analysis, action recognition and prediction.
Future algorithm and experimental advances that can take full advantage of the interconnected nature of the crossbar structures, by utilizing the intrinsic sneak paths and possible loops in the system may further enhance the computing capacity of the system.
Memristors to Power Quick-Learning Neural Networks
Imagine a class full of artificially intelligent machines (let’s say robot doctors) that have attended a lecture, on a new procedure of conducting surgery.
As a key concept in training machines to think like humans -as in, without prior programming, neural networks are the new target for research and improvement.
Now, the team is using memristors chip -which ideally requires minimal space -and can also be integrated straightforwardly and fast into pre-existing silicon-based electronics.
Technically, this contrasts with usual computer systems, whereby processors execute logic separate from memory modules. Wei Lu and team employed a special memristor with abilities to memorize events –especially those in the near future.
For instance, a system can bring up the exact photo when asked to identify a human face, this is because it has learned the distinct features of human faces from the photos providing during the training session.
The interesting part is that reservoir-computing systems that use memristors can skip those expensive training processes and still give the network the capability to remember details with over 98 percent accuracy.
After a set of data is inputted, the reservoir identifies vital time-related features of the data, then hands it off in a new simpler format to the next network in-line.
Now, it is this second network that will require a bit of training to ideally alter the weights of the features and outputs availed on the first network until it attains an acceptable level of error.
Reservoir Computing: Harnessing a Universal Dynamical System
Gauthier There is great current interest in developing artificial intelligence algorithms for processing massive data sets, often for classification tasks such as recognizing a face in a photograph.
dynamical system to predict the dynamics of a desired system is one approach to this problem that is well-suited for a reservoir computer (RC): a recurrent artificial neural network for processing time-dependent information (see Figure 1).
While researchers have studied RCs for well over 20 years [1] and applied them successfully to a variety of tasks [2], there are still many open questions that the dynamical systems community may find interesting and be able to address.
Mathematically, an RC is described by the set of autonomous, time-delay differential equations given by \[\frac{dx_i}{dt} = -\gamma_i x_i + \gamma_i f_i \big[\sum\limits_{j=1}^j W^{in}_{i,j}u_j(t)+ \sum\limits_{n=1}^N W^{res}_{i,n}x_n (t - \tau_{i,n}) + b_i \big], \\y_k(t) = \sum\limits_{m=1}^N W^{out}_{k,m} \mathcal{X}_m,\: \: \: \: \:i = 1, ..., N
\: \: \: \: \: k = 1, ..., K, \tag1 \] with \(J\) inputs \(u_j\), \(N\) reservoir nodes \(x_i\),and \(K\) outputs with values \(y_k\).
Here, \(\gamma_i\) are decay constants, \(W^{in}_{i,j} (W^{res}_{i,n})\)are fixed input (internal) weights, \(\tau_{i,n}\) are link time delays, \(b_i\) are biases, and \(W^{out}_{k,m}\)are the output weights whose values are optimized for a particular task.
We can solve \((2)\) in a least-square sense using pseudo-inverse matrix routines that are often included in a variety of computer languages, some of which can take advantage of the matrices’
We can also find a solution to \((2)\) using gradient descent methods, which are helpful when the matrix dimensions are large, and leverage toolkits from the deep learning community that take advantage of graphical processing units.
Furthermore, we can utilize the predicted time series as an observer in a control system [4] or for data assimilation of large spatiotemporal systems without use of an underlying model [6].
The following is an open question: how can we optimize the parameters in \((1)\) and \((2)\) to obtain the most accurate prediction in either the prediction or classification tasks, while simultaneously allowing the RC to function well on data that is similar to the training data set?
Early studies focused on the so-called echo state property of the network—where the output should eventually forget the input—and the consistency property, where outputs from identical trials should be similar over some period.
However, this scenario ignores the input dynamics and is mostly a statement of the stability of \(\mathbf{X}= 0\).Recent work is beginning to address this shortcoming for the case of a single input channel, demonstrating that there must be a single entire output solution given the input [5].
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This vignette illustrates the use of
mitml for the treatment of missing data at Level 2. Specifically, the vignette addresses the following topics:
Further information can be found in the other vignettes and the package documentation.
For purposes of this vignette, we make use of the
leadership data set, which contains simulated data from 750 employees in 50 groups including ratings on job satisfaction, leadership style, and work load (Level 1) as well as cohesion (Level 2).
The package and the data set can be loaded as follows.
In the
summary of the data, it becomes visible that all variables are affected by missing data.
# GRPID JOBSAT COHES NEGLEAD WLOAD # Min. : 1.0 Min. :-7.32934 Min. :-3.4072 Min. :-3.13213 low :416 # 1st Qu.:13.0 1st Qu.:-1.61932 1st Qu.:-0.4004 1st Qu.:-0.70299 high:248 # Median :25.5 Median :-0.02637 Median : 0.2117 Median : 0.08027 NA's: 86 # Mean :25.5 Mean :-0.03168 Mean : 0.1722 Mean : 0.04024 # 3rd Qu.:38.0 3rd Qu.: 1.64571 3rd Qu.: 1.1497 3rd Qu.: 0.79111 # Max. :50.0 Max. :10.19227 Max. : 2.5794 Max. : 3.16116 # NA's :69 NA's :30 NA's :92
The following data segment illustrates this fact, including cases with missing data at Level 1 (e.g., job satisfaction) and 2 (e.g., cohesion).
# GRPID JOBSAT COHES NEGLEAD WLOAD# 73 5 -1.72143400 0.9023198 0.83025589 high# 74 5 NA 0.9023198 0.15335056 high# 75 5 -0.09541178 0.9023198 0.21886272 low# 76 6 0.68626611 NA -0.38190591 high# 77 6 NA NA NA low# 78 6 -1.86298201 NA -0.05351001 high
In the following, we will employ a two-level model to address missing data at both levels simultaneously.
The specification of the two-level model, involves two components, one pertaining to the variables at each level of the sample (Goldstein, Carpenter, Kenward, & Levin, 2009; for further discussion, see also Enders, Mister, & Keller, 2016; Grund, Lüdtke, & Robitzsch, in press).
Specifically, the imputation model is specified as a list with two components, where the first component denotes the model for the variables at Level 1, and the second component denotes the model for the variables at Level 2.
For example, using the
formula interface, an imputation model targeting all variables in the data set can be written as follows.
The first component of this list includes the three target variables at Level 1 and a fixed (
1) as well as a random intercept (
1|GRPID). The second component includes the target variable at Level 2 with a fixed intercept (
1).
From a statistical point of view, this specification corresponds to the following model \[ \begin{aligned} \mathbf{y}_{1ij} &= \boldsymbol\mu_{1} + \mathbf{u}_{1j} + \mathbf{e}_{ij} \\ \mathbf{y}_{2j} &= \boldsymbol\mu_{2} + \mathbf{u}_{1j} \; , \end{aligned} \] where \(\mathbf{y}_{1ij}\) denotes the target variables at Level 1, \(\mathbf{y}_{2j}\) the target variables at Level 2, and the right-hand side of the model contains the fixed effects, random effects, and residual terms as mentioned above.
Note that, even though the two components of the model appear to be separated, they define a single (joint) model for all target variables at both Level 1 and 2. Specifically, this model employs a two-level covariance structure, which allows for relations between variables at both Level 1 (i.e., correlated residuals at Level 1) and 2 (i.e., correlated random effects residuals at Level 2).
Because the data contain missing values at both levels, imputations will be generated with
jomoImpute (and not
panImpute). Except for the specification of the two-level model, the syntax is the same as in applications with missing data only at Level 1.
Here, we will run 5,000 burn-in iterations and generate 20 imputations, each 250 iterations apart.
By looking at the
summary, we can then review the imputation procedure and verify that the imputation model converged.
# # Call:# # jomoImpute(data = leadership, formula = fml, n.burn = 5000, n.iter = 250, # m = 20)# # Level 1:# # Cluster variable: GRPID # Target variables: JOBSAT NEGLEAD WLOAD # Fixed effect predictors: (Intercept) # Random effect predictors: (Intercept) # # Level 2:# # Target variables: COHES # Fixed effect predictors: (Intercept) # # Performed 5000 burn-in iterations, and generated 20 imputed data sets,# each 250 iterations apart. # # Potential scale reduction (Rhat, imputation phase):# # Min 25% Mean Median 75% Max# Beta: 1.001 1.001 1.001 1.001 1.001 1.001# Beta2: 1.001 1.001 1.001 1.001 1.001 1.001# Psi: 1.000 1.001 1.003 1.001 1.003 1.009# Sigma: 1.000 1.003 1.004 1.004 1.006 1.009# # Largest potential scale reduction:# Beta: [1,3], Beta2: [1,1], Psi: [4,3], Sigma: [3,1]# # Missing data per variable:# GRPID JOBSAT NEGLEAD WLOAD COHES# MD% 0 9.2 12.3 11.5 4.0
Due to the greater complexity of the two-level model, the output includes more information than in applications with missing data only at Level 1. For example, the output features the model specification for variables at both Level 1 and 2. Furthermore, it provides convergence statistics for the additional regression coefficients for the target variables at Level 2 (i.e.,
Beta2).
Finally, it also becomes visible that the two-level model indeed allows for relations between target variables at Level 1 and 2. This can be seen from the fact that the potential scale reduction factor (\(\hat{R}\)) for the covariance matrix at Level 2 (
Psi) was largest for
Psi[4,3], which is the covariance between cohesion and the random intercept of work load.
The completed data sets can then be extracted with
mitmlComplete.
When inspecting the completed data, it is easy to verify that the imputations for variables at Level 2 are constant within groups as intended, thus preserving the two-level structure of the data.
# GRPID JOBSAT NEGLEAD WLOAD COHES# 73 5 -1.72143400 0.83025589 high 0.9023198# 74 5 -2.80749991 0.15335056 high 0.9023198# 75 5 -0.09541178 0.21886272 low 0.9023198# 76 6 0.68626611 -0.38190591 high -1.0275552# 77 6 1.52825873 -1.11035850 low -1.0275552# 78 6 -1.86298201 -0.05351001 high -1.0275552
Enders, C. K., Mistler, S. A., & Keller, B. T. (2016). Multilevel multiple imputation: A review and evaluation of joint modeling and chained equations imputation.
Psychological Methods, 21, 222–240. doi: 10.1037/met0000063 (Link)
Goldstein, H., Carpenter, J. R., Kenward, M. G., & Levin, K. A. (2009). Multilevel models with multivariate mixed response types.
Statistical Modelling, 9, 173–197. doi: 10.1177/1471082X0800900301 (Link)
Grund, S., Lüdtke, O., & Robitzsch, A. (in press). Multiple imputation of missing data for multilevel models: Simulations and recommendations.
Organizational Research Methods. doi: 10.1177/1094428117703686 (Link)
# Author: Simon Grund (grund@ipn.uni-kiel.de)# Date: 2019-01-02
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1. Explain why we can always evaluate the determinant of a square matrix.
2. Examining Cramer’s Rule, explain why there is no unique solution to the system when the determinant of your matrix is 0. For simplicity, use a [latex]2\times 2[/latex] matrix.
3. Explain what it means in terms of an inverse for a matrix to have a 0 determinant.
4. The determinant of [latex]2\times 2[/latex] matrix [latex]A[/latex] is 3. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer.
For the following exercises, find the determinant.
5. [latex]|\begin{array}{cc}1& 2\\ 3& 4\end{array}|[/latex]
6. [latex]|\begin{array}{rr}\hfill -1& \hfill 2\\ \hfill 3& \hfill -4\end{array}|[/latex]
7. [latex]|\begin{array}{rr}\hfill 2& \hfill -5\\ \hfill -1& \hfill 6\end{array}|[/latex]
8. [latex]|\begin{array}{cc}-8& 4\\ -1& 5\end{array}|[/latex]
9. [latex]|\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 3& \hfill -4\end{array}|[/latex]
10. [latex]|\begin{array}{rr}\hfill 10& \hfill 20\\ \hfill 0& \hfill -10\end{array}|[/latex]
11. [latex]|\begin{array}{cc}10& 0.2\\ 5& 0.1\end{array}|[/latex]
12. [latex]|\begin{array}{rr}\hfill 6& \hfill -3\\ \hfill 8& \hfill 4\end{array}|[/latex]
13. [latex]|\begin{array}{rr}\hfill -2& \hfill -3\\ \hfill 3.1& \hfill 4,000\end{array}|[/latex]
14. [latex]|\begin{array}{rr}\hfill -1.1& \hfill 0.6\\ \hfill 7.2& \hfill -0.5\end{array}|[/latex]
15. [latex]|\begin{array}{rrr}\hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill -3\end{array}|[/latex]
16. [latex]|\begin{array}{rrr}\hfill -1& \hfill 4& \hfill 0\\ \hfill 0& \hfill 2& \hfill 3\\ \hfill 0& \hfill 0& \hfill -3\end{array}|[/latex]
17. [latex]|\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}|[/latex]
18. [latex]|\begin{array}{rrr}\hfill 2& \hfill -3& \hfill 1\\ \hfill 3& \hfill -4& \hfill 1\\ \hfill -5& \hfill 6& \hfill 1\end{array}|[/latex]
19. [latex]|\begin{array}{rrr}\hfill -2& \hfill 1& \hfill 4\\ \hfill -4& \hfill 2& \hfill -8\\ \hfill 2& \hfill -8& \hfill -3\end{array}|[/latex]
20. [latex]|\begin{array}{rrr}\hfill 6& \hfill -1& \hfill 2\\ \hfill -4& \hfill -3& \hfill 5\\ \hfill 1& \hfill 9& \hfill -1\end{array}|[/latex]
21. [latex]|\begin{array}{rrr}\hfill 5& \hfill 1& \hfill -1\\ \hfill 2& \hfill 3& \hfill 1\\ \hfill 3& \hfill -6& \hfill -3\end{array}|[/latex]
22. [latex]|\begin{array}{rrr}\hfill 1.1& \hfill 2& \hfill -1\\ \hfill -4& \hfill 0& \hfill 0\\ \hfill 4.1& \hfill -0.4& \hfill 2.5\end{array}|[/latex]
23. [latex]|\begin{array}{rrr}\hfill 2& \hfill -1.6& \hfill 3.1\\ \hfill 1.1& \hfill 3& \hfill -8\\ \hfill -9.3& \hfill 0& \hfill 2\end{array}|[/latex]
24. [latex]|\begin{array}{ccc}-\frac{1}{2}& \frac{1}{3}& \frac{1}{4}\\ \frac{1}{5}& -\frac{1}{6}& \frac{1}{7}\\ 0& 0& \frac{1}{8}\end{array}|[/latex]
For the following exercises, solve the system of linear equations using Cramer’s Rule.
25. [latex]\begin{array}{l}2x - 3y=-1\\ 4x+5y=9\end{array}[/latex]
26. [latex]\begin{array}{r}5x - 4y=2\\ -4x+7y=6\end{array}[/latex]
27. [latex]\begin{array}{l}\text{ }6x - 3y=2\hfill \\ -8x+9y=-1\hfill \end{array}[/latex]
28. [latex]\begin{array}{l}2x+6y=12\\ 5x - 2y=13\end{array}[/latex]
29. [latex]\begin{array}{l}4x+3y=23\hfill \\ \text{ }2x-y=-1\hfill \end{array}[/latex]
30. [latex]\begin{array}{l}10x - 6y=2\hfill \\ -5x+8y=-1\hfill \end{array}[/latex]
31. [latex]\begin{array}{l}4x - 3y=-3\\ 2x+6y=-4\end{array}[/latex]
32. [latex]\begin{array}{r}4x - 5y=7\\ -3x+9y=0\end{array}[/latex]
33. [latex]\begin{array}{l}4x+10y=180\hfill \\ -3x - 5y=-105\hfill \end{array}[/latex]
34. [latex]\begin{array}{l}\text{ }8x - 2y=-3\hfill \\ -4x+6y=4\hfill \end{array}[/latex]
For the following exercises, solve the system of linear equations using Cramer’s Rule.
35. [latex]\begin{array}{l}\text{ }x+2y - 4z=-1\hfill \\ \text{ }7x+3y+5z=26\hfill \\ -2x - 6y+7z=-6\hfill \end{array}[/latex]
36. [latex]\begin{array}{l}-5x+2y - 4z=-47\hfill \\ \text{ }4x - 3y-z=-94\hfill \\ \text{ }3x - 3y+2z=94\hfill \end{array}[/latex]
37. [latex]\begin{array}{l}\text{ }4x+5y-z=-7\hfill \\ -2x - 9y+2z=8\hfill \\ \text{ }5y+7z=21\hfill \end{array}[/latex]
38. [latex]\begin{array}{r}4x - 3y+4z=10\\ 5x - 2z=-2\\ 3x+2y - 5z=-9\end{array}[/latex]
39. [latex]\begin{array}{l}4x - 2y+3z=6\hfill \\ \text{ }-6x+y=-2\hfill \\ 2x+7y+8z=24\hfill \end{array}[/latex]
40. [latex]\begin{array}{r}\hfill 5x+2y-z=1\\ \hfill -7x - 8y+3z=1.5\\ \hfill 6x - 12y+z=7\end{array}[/latex]
41. [latex]\begin{array}{l}\text{ }13x - 17y+16z=73\hfill \\ -11x+15y+17z=61\hfill \\ \text{ }46x+10y - 30z=-18\hfill \end{array}[/latex]
42. [latex]\begin{array}{l}\begin{array}{l}\hfill \\ -4x - 3y - 8z=-7\hfill \end{array}\hfill \\ \text{ }2x - 9y+5z=0.5\hfill \\ \text{ }5x - 6y - 5z=-2\hfill \end{array}[/latex]
43. [latex]\begin{array}{l}\text{ }4x - 6y+8z=10\hfill \\ -2x+3y - 4z=-5\hfill \\ \text{ }x+y+z=1\hfill \end{array}[/latex]
44. [latex]\begin{array}{r}\hfill 4x - 6y+8z=10\\ \hfill -2x+3y - 4z=-5\\ \hfill 12x+18y - 24z=-30\end{array}[/latex]
For the following exercises, use the determinant function on a graphing utility.
45. [latex]|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 8& \hfill 9\\ \hfill 0& \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 3& \hfill 0\\ \hfill 0& \hfill 2& \hfill 4& \hfill 3\end{array}|[/latex]
46. [latex]|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill -9& \hfill 1& \hfill 3\\ \hfill 3& \hfill 0& \hfill -2& \hfill -1\\ \hfill 0& \hfill 1& \hfill 1& \hfill -2\end{array}|[/latex]
47. [latex]|\begin{array}{rrrr}\hfill \frac{1}{2}& \hfill 1& \hfill 7& \hfill 4\\ \hfill 0& \hfill \frac{1}{2}& \hfill 100& \hfill 5\\ \hfill 0& \hfill 0& \hfill 2& \hfill 2,000\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2\end{array}|[/latex]
48. [latex]|\begin{array}{rrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 2& \hfill 3& \hfill 0& \hfill 0\\ \hfill 4& \hfill 5& \hfill 6& \hfill 0\\ \hfill 7& \hfill 8& \hfill 9& \hfill 0\end{array}|[/latex]
For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution.
49. Two numbers add up to 56. One number is 20 less than the other.
50. Two numbers add up to 104. If you add two times the first number plus two times the second number, your total is 208
51. Three numbers add up to 106. The first number is 3 less than the second number. The third number is 4 more than the first number.
52. Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined.
For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule.
53. You invest $10,000 into two accounts, which receive 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account?
54. You invest $80,000 into two accounts, $22,000 in one account, and $58,000 in the other account. At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts?
55. A movie theater needs to know how many adult tickets and children tickets were sold out of the 1,200 total tickets. If children’s tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many children’s tickets and adult tickets were sold?
56. A concert venue sells single tickets for $40 each and couple’s tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many couple’s tickets were sold?
57. You decide to paint your kitchen green. You create the color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix?
58. You sold two types of scarves at a farmers’ market and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold?
59. Your garden produced two types of tomatoes, one green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have?
60. At a market, the three most popular vegetables make up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share?
61. At the same market, the three most popular fruits make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold.
62. Three bands performed at a concert venue. The first band charged $15 per ticket, the second band charged $45 per ticket, and the final band charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band?
63. A movie theatre sold tickets to three movies. The tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie were sold?
64. Men aged 20–29, 30–39, and 40–49 made up 78% of the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20–29 age group increased by 20%, the 30–39 age group increased by 2%, and the 40–49 age group decreased to [latex]\frac{3}{4}[/latex] of their previous population. Originally, the 30–39 age group had 2% more prisoners than the 20–29 age group. Determine the prison population percentage for each age group last year.
65. At a women’s prison down the road, the total number of inmates aged 20–49 totaled 5,525. This year, the 20–29 age group increased by 10%, the 30–39 age group decreased by 20%, and the 40–49 age group doubled. There are now 6,040 prisoners. Originally, there were 500 more in the 30–39 age group than the 20–29 age group. Determine the prison population for each age group last year.
For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate-covered cashews. The nutritional information for these items is shown below.
Fat (g) Protein (g) Carbohydrates (g) Almonds (10) 6 2 3 Cranberries (10) 0.02 0 8 Cashews (10) 7 3.5 5.5
66. For the special “low-carb”trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix?
67. For the “hiking” mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix?
68. For the “energy-booster” mix, there are 1,000 pieces in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix?
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This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization.
A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the momentum operator, they can have further quantum numbers. These states should exist even in an interactiong field theory, describing a single particle away from any interaction.In a local quantum field theory, these states are associated with local field operators: $$| p, \sigma \rangle = \int e^{ipx} \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x$$Where $\psi $ is the field corresponding to the particle and $\sigma$ describes the set of other quantum numbers additional to the momentum.A symmetry generator $Q$ being the integral of a charge density according to the Noether's theorem$$Q = \int j_0(x') d^3x'$$should generate a local field when it acts on a local field:$[Q, \psi_1(x)] = \psi_2(x)$(In the case of internal symmetries $\psi_2$ depends linearly on the components of $\psi_1(x)$, in the case of space time symmetries it depends on the derivatives of the components of $\psi_1(x)$)
Thus in general:
$$[Q, \psi_{\sigma}(x)] = \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x)])$$
Where the dependence of the coefficients $ C_{\sigma\sigma'}$ on the momentum operator $\nabla$ is due to the possibility that $Q$ contains a space-time symmetry.Thus for an operator $Q$ satisfying $Q|0\rangle = 0$, we have$$ Q | p, \sigma \rangle = \int e^{ipx} Q \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x = \int e^{ipx} [Q , \psi_{\sigma}^{\dagger}(x)] |0\rangle d^4x = \int e^{ipx} \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) \int e^{ipx} \psi_{\sigma'}^{\dagger}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) | p, \sigma' \rangle $$Thus the action of the operator $Q$ is a representation in the one particle states. The fact that $Q$ commutes with the Hamiltonian is responsible for the energy degeneracy of its action, i.e., the states $| p, \sigma \rangle$ and $Q| p, \sigma \rangle$ have the same energy.This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user David Bar Moshe
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I know that this function ($g$ means coupling) is non-analytical in $g=0$, so this function is only appreciable under non-perturbative calculations, so is a non-perturbative phenomena.This function is present on many critical/cross temperatures like in Kondo problem and Superconductors. This functions happens in QCD, when we fix physical coupling equal one.Is always that:$$E=E_0\,e^{-\frac{1}{\rho |g|}}$$or, replacing $|g|$, $g^2$ and $\rho$ is some density of state.
When we perceive (physically) that the perturbative series don't converge (like Dyson argument), we treat our series as assymptotic. If the series diverges as $n!$, we can use Borel summation and come up with some integration over a meromorphic function in $(0,\,\infty)$. After some calculation, the poles of this meromorphic function gives contributions like $e^{-\frac{1}{\rho |g|}}$.
From this site, seems to me that only instanton made this contribution (instanton corrections). But renormalons could give the same contribution (no?). Bound states, nearly-bound states and tunneling mechanisms that connect different nearly-bound states seems to me the reason of the appearance of this terms and the divergence of perturbative calculation. But is very interesting that this corrections added in perturbative calculations are very tiny, exponentially tiny,...a far scale,... the typical scale of the bound state or the width of a tunneling barrier that holds nearly-bound state.
In the physical examples that I gave, the Kondo temperature tells us the size of the cloud around the impurity, the QCD energy gives us the size of the proton, Cooper instability gives the size of the electron-electron pair, a QM double well problem gives the distance of the wells,...so on, so on. Always a formation of bound state through scales. Short distance plus small interactions giving long distance bounded states.
I came with this by physical intuition. Can someone can give a mathematical proof of that?This post imported from StackExchange Physics at 2016-05-31 07:24 (UTC), posted by SE-user Nogueira
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But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
I have problem with showing that the limit of the following function$$\frac{\sqrt{\frac{3 \pi}{2n}} -\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$equal to $1$, with $n \to \infty$.
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
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Faddeeva Package From AbInitio
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Line 26: Line 26: :<math>\mathrm{erfi}(x) = -i\mathrm{erf}(ix) = -i[e^{x^2} w(x) - 1]</math> (imaginary error function) :<math>\mathrm{erfi}(x) = -i\mathrm{erf}(ix) = -i[e^{x^2} w(x) - 1]</math> (imaginary error function) :<math>F(x) = \frac{i\sqrt{\pi}}{2} \left[ e^{-z^2} - w(z) \right]</math> ([[w:Dawson function|Dawson function]]) :<math>F(x) = \frac{i\sqrt{\pi}}{2} \left[ e^{-z^2} - w(z) \right]</math> ([[w:Dawson function|Dawson function]]) - :<math>\mathrm{Voigt}(x,y) = \mathrm{Re}[w(x+iy)]</math> (real [[w:Voigt function|Voigt function]], up to scale factor) + :<math>\mathrm{Voigt}(x,y) = \mathrm{Re}[w(x+iy)] \!</math> (real [[w:Voigt function|Voigt function]], up to scale factor) + Note that in the case of erf and erfc, we provide different equations for positive and negative ''x'', in order to avoid numerical problems arising from multiplying exponentially large and small quantities. == Wrappers: Matlab, GNU Octave, and Python == == Wrappers: Matlab, GNU Octave, and Python == Revision as of 22:47, 29 October 2012
Contents Faddeeva / complex error function
Steven G. Johnson has written free/open-source C++ code (with wrappers for other languages) to compute the
scaled complex error function w( z) = e − z2erfc(− iz), also called the Faddeeva function(and also the plasma dispersion function), for arbitrary complex arguments zto a given accuracy. Given the Faddeeva function, one can easily compute Voigt functions, the Dawson function, and similar related functions.
Download the source code from:
http://ab-initio.mit.edu/Faddeeva_w.cc (updated 29 October 2012) Usage
To use the code, add the following declaration to your C++ source (or header file):
#include <complex> extern std::complex<double> Faddeeva_w(std::complex<double> z, double relerr=0);
The function
Faddeeva_w(z, relerr) computes
w( z) to a desired relative error
relerr.
Omitting the
relerr argument, or passing
relerr=0 (or any
relerr less than machine precision ε≈10
−16), corresponds to requesting machine precision, and in practice a relative error < 10 −13 is usually achieved. Specifying a larger value of
relerr may improve performance (at the expense of accuracy).
You should also compile
Faddeeva_w.cc and link it with your program, of course.
In terms of
w( z), some other important functions are: (scaled complementary error function) (complementary error function) (error function) (imaginary error function) (Dawson function) (real Voigt function, up to scale factor)
Note that in the case of erf and erfc, we provide different equations for positive and negative
x, in order to avoid numerical problems arising from multiplying exponentially large and small quantities. Wrappers: Matlab, GNU Octave, and Python
Wrappers are available for this function in other languages.
Matlab (also available here): A function
Faddeeva_w(z, relerr), where the arguments have the same meaning as above (the
relerrargument is optional) can be downloaded from Faddeeva_w_mex.cc (along with the help file Faddeeva_w.m. Compile it into an octave plugin with:
mex -output Faddeeva_w -O Faddeeva_w_mex.cc Faddeeva_w.cc GNU Octave: A function
Faddeeva_w(z, relerr), where the arguments have the same meaning as above (the
relerrargument is optional) can be downloaded from Faddeeva_w_oct.cc. Compile it into a MEX file with:
mkoctfile -DMPICH_SKIP_MPICXX=1 -DOMPI_SKIP_MPICXX=1 -s -o Faddeeva_w.oct Faddeeva_w_oct.cc Faddeeva_w.cc Python: Our code is used to provide
scipy.special.wofzin SciPy starting in version 0.12.0 (see here).
Algorithm
This implementation uses a combination of different algorithms. For sufficiently large |
z|, we use a continued-fraction expansion for w( z) similar to those described in Walter Gautschi, "Efficient computation of the complex error function," SIAM J. Numer. Anal. 7(1), pp. 187–198 (1970). G. P. M. Poppe and C. M. J. Wijers, "More efficient computation of the complex error function," ACM Trans. Math. Soft. 16(1), pp. 38–46 (1990); this is TOMS Algorithm 680.
Unlike those papers, however, we switch to a completely different algorithm for smaller |
z|: Mofreh R. Zaghloul and Ahmed N. Ali, "Algorithm 916: Computing the Faddeyeva and Voigt Functions," ACM Trans. Math. Soft. 38(2), 15 (2011). Preprint available at arXiv:1106.0151.
(I initially used this algorithm for all z, but the continued-fraction expansion turned out to be faster for larger |
z|. On the other hand, Algorithm 916 is competitive or faster for smaller | z|, and appears to be significantly more accurate than the Poppe & Wijers code in some regions, e.g. in the vicinity of | z|=1 [although comparison with other compilers suggests that this may be a problem specific to gfortran]. Algorithm 916 also has better relative accuracy in Re[ z] for some regions near the real- z axis. You can switch back to using Algorithm 916 for all z by changing
USE_CONTINUED_FRACTION to
0 in the code.)
Note that this is SGJ's
independent re-implementation of these algorithms, based on the descriptions in the papers only. In particular, we did not refer to the authors' Fortran or Matlab implementations (respectively), which are under restrictive "semifree" ACM copyright terms and are therefore unusable in free/open-source software.
Algorithm 916 requires an external complementary error function erfc(
x) function for real arguments x to be supplied as a subroutine. More precisely, it requires the scaled function erfcx( x) = e erfc( x2 x). Here, we use an erfcx routine written by SGJ that uses a combination of two algorithms: a continued-fraction expansion for large xand a lookup table of Chebyshev polynomials for small x. (I initially used an erfcx function derived from the DERFC routine in SLATEC, modified by SGJ to compute erfcx instead of erfc, by the new erfcx routine is much faster.) Test program
To test the code, a small test program is included at the end of
Faddeeva_w.cc which tests
w( z) against several known results (from Wolfram Alpha) and prints the relative errors obtained. To compile the test program,
#define FADDEEVA_W_TEST in the file (or compile with
-DFADDEEVA_W_TEST on Unix) and compile
Faddeeva_w.cc. The resulting program prints
SUCCESS at the end of its output if the errors were acceptable.
License
The software is distributed under the "MIT License", a simple permissive free/open-source license:
Copyright © 2012 Massachusetts Institute of Technology Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
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Definition:Indexing Set/Term Definition
Let $I$ and $S$ be sets.
Let $x: I \to S$ be a mapping.
The family of elements $x$ of $S$ indexed by $I$ is often seen with one of the following notations:
$\family {x_i}_{i \mathop \in I}$ $\paren {x_i}_{i \mathop \in I}$ $\set {x_i}_{i \mathop \in I}$ There is little consistency in the literature, but $\paren {x_i}_{i \mathop \in I}$ is perhaps most common.
The preferred notation on $\mathsf{Pr} \infty \mathsf{fWiki}$ is $\family {x_i}_{i \mathop \in I}$.
The subscripted $i \in I$ is often left out, if it is obvious in the particular context.
As $x$ is actually a mapping, one would expect the conventional notation $\map x i$.
However, this is generally not used, and $x_i$ is used instead.
Also known as
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Yu Cheng,Ilias Diakonikolas,Rong Ge,David P. Woodruff;
Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:727-757, 2019.
Abstract
We study the problem of estimating the covariance matrix of a high-dimensional distribution when a small constant fraction of the samples can be arbitrarily corrupted. Recent work gave the first polynomial time algorithms for this problem with near-optimal error guarantees for several natural structured distributions. Our main contribution is to develop faster algorithms for this problem whose running time nearly matches that of computing the empirical covariance. Given $N = \tilde{\Omega}(d^2/\epsilon^2)$ samples from a $d$-dimensional Gaussian distribution, an $\epsilon$-fraction of which may be arbitrarily corrupted, our algorithm runs in time $\tilde{O}(d^{3.26})/\mathrm{poly}(\epsilon)$ and approximates the unknown covariance matrix to optimal error up to a logarithmic factor. Previous robust algorithms with comparable error guarantees all have runtimes $\tilde{\Omega}(d^{2 \omega})$ when $\epsilon = \Omega(1)$, where $\omega$ is the exponent of matrix multiplication. We also provide evidence that improving the running time of our algorithm may require new algorithmic techniques.
@InProceedings{pmlr-v99-cheng19a,title = {Faster Algorithms for High-Dimensional Robust Covariance Estimation},author = {Cheng, Yu and Diakonikolas, Ilias and Ge, Rong and Woodruff, David P.},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {727--757},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/cheng19a/cheng19a.pdf},url = {http://proceedings.mlr.press/v99/cheng19a.html},abstract = { We study the problem of estimating the covariance matrix of a high-dimensional distribution when a small constant fraction of the samples can be arbitrarily corrupted. Recent work gave the first polynomial time algorithms for this problem with near-optimal error guarantees for several natural structured distributions. Our main contribution is to develop faster algorithms for this problem whose running time nearly matches that of computing the empirical covariance. Given $N = \tilde{\Omega}(d^2/\epsilon^2)$ samples from a $d$-dimensional Gaussian distribution, an $\epsilon$-fraction of which may be arbitrarily corrupted, our algorithm runs in time $\tilde{O}(d^{3.26})/\mathrm{poly}(\epsilon)$ and approximates the unknown covariance matrix to optimal error up to a logarithmic factor. Previous robust algorithms with comparable error guarantees all have runtimes $\tilde{\Omega}(d^{2 \omega})$ when $\epsilon = \Omega(1)$, where $\omega$ is the exponent of matrix multiplication. We also provide evidence that improving the running time of our algorithm may require new algorithmic techniques. }}
%0 Conference Paper%T Faster Algorithms for High-Dimensional Robust Covariance Estimation%A Yu Cheng%A Ilias Diakonikolas%A Rong Ge%A David P. Woodruff%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-cheng19a%I PMLR%J Proceedings of Machine Learning Research%P 727--757%U http://proceedings.mlr.press%V 99%W PMLR%X We study the problem of estimating the covariance matrix of a high-dimensional distribution when a small constant fraction of the samples can be arbitrarily corrupted. Recent work gave the first polynomial time algorithms for this problem with near-optimal error guarantees for several natural structured distributions. Our main contribution is to develop faster algorithms for this problem whose running time nearly matches that of computing the empirical covariance. Given $N = \tilde{\Omega}(d^2/\epsilon^2)$ samples from a $d$-dimensional Gaussian distribution, an $\epsilon$-fraction of which may be arbitrarily corrupted, our algorithm runs in time $\tilde{O}(d^{3.26})/\mathrm{poly}(\epsilon)$ and approximates the unknown covariance matrix to optimal error up to a logarithmic factor. Previous robust algorithms with comparable error guarantees all have runtimes $\tilde{\Omega}(d^{2 \omega})$ when $\epsilon = \Omega(1)$, where $\omega$ is the exponent of matrix multiplication. We also provide evidence that improving the running time of our algorithm may require new algorithmic techniques.
Cheng, Y., Diakonikolas, I., Ge, R. & Woodruff, D.P.. (2019). Faster Algorithms for High-Dimensional Robust Covariance Estimation. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:727-757
This site last compiled Sat, 17 Aug 2019 00:05:37 +0000
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Let $A = Z[\sqrt{-2}]$ be the ring with elements of the form $\{ a +b \epsilon\mid a,b \in\mathbb Z \}$ and where $\epsilon^2 = -2$.
Let $(1- \epsilon)$ denote the ideal generated by $1- \epsilon$.
I am tasked to show that $A / (1 - \epsilon) \simeq \mathbb Z_3$.
My attempt:
First I see that all elements in the ideal $(1- \epsilon)$ are given by $\{ (a+b\epsilon) (1- \epsilon) | a,b \in Z \}$ so a generic element has the form $a+2b + \epsilon(b -a )$.
Now I want to show that there are 3 equivalence classes in $A / (1 - \epsilon)$ where two elements $x,y$ are equivalent $\iff (y_1 +y_2\epsilon) -(x_1+x_2 \epsilon) \in (1 - \epsilon) $.
So I obtain the condition that $y_1 - x_1$ must be of the form $a+2b$ and $y_2 -x_2$ must be of the form $b-a$ but I can't seem to find a way to utilize this information.
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Now showing items 1-1 of 1
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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Definition:Imaginary Number Informal Definition
In particular, this applies to the equation $x^2 + 1 = 0$.
In order to be able to allow such equations to have solutions, the concept $i = \sqrt {-1}$ is introduced.
$i$ does not exist on the real number line, but is a completely separate concept.
It can be treated as a number, and combined with real numbers in algebraic expressions.
When $a, b$ are real numbers, we have: $a i = i a$ $a + i = i + a$ $i a + i b = i \left({a + b}\right) = \left({a + b}\right) i = a i + b i$
etc.
In engineering applications, $j$ is usually used instead. Numbers of the form $a i$ (or $i a$), where $a \in \R$, are known as imaginary numbers.
Numbers of the form $a + b i$ are known as complex numbers.
When considering the roots of $x^2 + 40 = 10 x$, and determining that they are $5 \pm \sqrt {-15}$, he concluded:
These quantities are "truly sophisticated" and that to continue working with them would be "as subtle as it would be useless". These Imaginary Quantities (as they are commonly called) arising from the Supposed Root of a Negative Square (when they happen) are reputed to imply that the Case proposed is Impossible. $e^{i \pi} + 1 = 0$
and introduced the letter $i$ to mean $\sqrt {-1}$.
Sources 1972: Frank Ayres, Jr.: Theory and Problems of Differential and Integral Calculus(SI ed.) ... (previous) ... (next): Chapter $1$: Variables and Functions: The Set of Real Numbers
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It’s been a while, so let’s include a recap : a (transitive) permutation representation of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is determined by the conjugacy class of a cofinite subgroup $\Lambda \subset \Gamma $, or equivalently, to a dessin d’enfant. We have introduced a
quiver (aka an oriented graph) which comes from a triangulation of the compactification of $\mathbb{H} / \Lambda $ where $\mathbb{H} $ is the hyperbolic upper half-plane. This quiver is independent of the chosen embedding of the dessin in the Dedeking tessellation. (For more on these terms and constructions, please consult the series Modular subgroups and Dessins d’enfants).
Why are quivers useful? To start, any quiver $Q $ defines a noncommutative algebra, the
path algebra $\mathbb{C} Q $, which has as a $\mathbb{C} $-basis all oriented paths in the quiver and multiplication is induced by concatenation of paths (when possible, or zero otherwise). Usually, it is quite hard to make actual computations in noncommutative algebras, but in the case of path algebras you can just see what happens.
Moreover, we can also
see the finite dimensional representations of this algebra $\mathbb{C} Q $. Up to isomorphism they are all of the following form : at each vertex $v_i $ of the quiver one places a finite dimensional vectorspace $\mathbb{C}^{d_i} $ and any arrow in the quiver [tex]\xymatrix{\vtx{v_i} \ar[r]^a & \vtx{v_j}}[/tex] determines a linear map between these vertex spaces, that is, to $a $ corresponds a matrix in $M_{d_j \times d_i}(\mathbb{C}) $. These matrices determine how the paths of length one act on the representation, longer paths act via multiplcation of matrices along the oriented path.
A
necklace in the quiver is a closed oriented path in the quiver up to cyclic permutation of the arrows making up the cycle. That is, we are free to choose the start (and end) point of the cycle. For example, in the one-cycle quiver
[tex]\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar[ld]^b \\ & \vtx{} \ar[lu]^c &}[/tex]
the basic necklace can be represented as $abc $ or $bca $ or $cab $. How does a necklace act on a representation? Well, the matrix-multiplication of the matrices corresponding to the arrows gives a square matrix in each of the vertices in the cycle. Though the dimensions of this matrix may vary from vertex to vertex, what does not change (and hence is a property of the necklace rather than of the particular choice of cycle) is the
trace of this matrix. That is, necklaces give complex-valued functions on representations of $\mathbb{C} Q $ and by a result of Artin and Procesi there are enough of them to distinguish isoclasses of (semi)simple representations! That is, linear combinations a necklaces (aka super-potentials) can be viewed, after taking traces, as complex-valued functions on all representations (similar to character-functions).
In physics, one views these functions as potentials and it then interested in the points (representations) where this function is extremal (minimal) : the
vacua. Clearly, this does not make much sense in the complex-case but is relevant when we look at the real-case (where we look at skew-Hermitian matrices rather than all matrices). A motivating example (the Yang-Mills potential) is given in Example 2.3.2 of Victor Ginzburg’s paper Calabi-Yau algebras.
Let $\Phi $ be a super-potential (again, a linear combination of necklaces) then our commutative intuition tells us that extrema correspond to zeroes of all partial differentials $\frac{\partial \Phi}{\partial a} $ where $a $ runs over all coordinates (in our case, the arrows of the quiver). One can make sense of differentials of necklaces (and super-potentials) as follows : the partial differential with respect to an arrow $a $ occurring in a term of $\Phi $ is defined to be the
path in the quiver one obtains by removing all 1-occurrences of $a $ in the necklaces (defining $\Phi $) and rearranging terms to get a maximal broken necklace (using the cyclic property of necklaces). An example, for the cyclic quiver above let us take as super-potential $abcabc $ (2 cyclic turns), then for example
$\frac{\partial \Phi}{\partial b} = cabca+cabca = 2 cabca $
(the first term corresponds to the first occurrence of $b $, the second to the second). Okay, but then the vacua-representations will be the representations of the quotient-algebra (which I like to call the
vacualgebra)
$\mathcal{U}(Q,\Phi) = \frac{\mathbb{C} Q}{(\partial \Phi/\partial a, \forall a)} $
which in ‘physical relevant settings’ (whatever that means…) turn out to be
Calabi-Yau algebras.
But, let us return to the case of subgroups of the modular group and their quivers. Do we have a natural
super-potential in this case? Well yes, the quiver encoded a triangulation of the compactification of $\mathbb{H}/\Lambda $ and if we choose an orientation it turns out that all ‘black’ triangles (with respect to the Dedekind tessellation) have their arrow-sides defining a necklace, whereas for the ‘white’ triangles the reverse orientation makes the arrow-sides into a necklace. Hence, it makes sense to look at the cubic superpotential $\Phi $ being the sum over all triangle-sides-necklaces with a +1-coefficient for the black triangles and a -1-coefficient for the white ones. Let’s consider an index three example from a previous post [tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\ i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\ & & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]
In this case the super-potential coming from the triangulation is
$\Phi = -aid+agd-cge+che-bhf+bif $
and therefore we have a noncommutative algebra $\mathcal{U}(Q,\Phi) $ associated to this index 3 subgroup. Contrary to what I believed at the start of this series, the algebras one obtains in this way from dessins d’enfants are
far from being Calabi-Yau (in whatever definition). For example, using a GAP-program written by Raf Bocklandt Ive checked that the growth rate of the above algebra is similar to that of $\mathbb{C}[x] $, so in this case $\mathcal{U}(Q,\Phi) $ can be viewed as a noncommutative curve (with singularities).
However, this is not the case for all such algebras. For example, the vacualgebra associated to the second index three subgroup (whose fundamental domain and quiver were depicted at the end of this post) has growth rate similar to that of $\mathbb{C} \langle x,y \rangle $…
I have an outlandish conjecture about the growth-behavior of all algebras $\mathcal{U}(Q,\Phi) $ coming from dessins d’enfants :
the algebra sees what the monodromy representation of the dessin sees of the modular group (or of the third braid group). I can make this more precise, but perhaps it is wiser to calculate one or two further examples…
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Let ${\mathbb F}:({\mathcal A},{\mathcal E}_{\mathcal A})\to({\mathcal B},{\mathcal E}_{\mathcal B})$ be an exact functor between exact categories, and suppose ${\mathbb F}$ has both a left adjoint ${\mathbb F}_\lambda$ and a right adjoint ${\mathbb F}_\rho$. Then the class
${\mathcal E} := \{X\to Y\to Z\in{\mathcal E}_{\mathcal A}\ |\ {\mathbb F}X\to{\mathbb F}Y\to{\mathbb F}Z\in{\mathcal E}_{\mathcal B}\}$
defines another exact structure on ${\mathcal A}$.
It is interesting to ask for criteria to decide when this exact structure is Frobenius. One such criterion is the following:
Suppose ${\mathcal E}_{\mathcal B}$ is the split exact structure, and that for each $X\in{\mathcal A}$ the unit $\eta_X: X\to{\mathbb F}_\rho{\mathbb F}X$ is an ${\mathcal E}_{\mathcal A}$-monomorphism, while the counit ${\mathbb F}_\lambda{\mathbb F}X\to X$ is an ${\mathcal E}_{\mathcal A}$-epimorphism. Then $({\mathcal A},{\mathcal E})$ has enough projectives and injectives, and the classes ${\mathcal P}$/${\mathcal I}$ of projective/injective objects in $({\mathcal A},{\mathcal E})$ are given by
${\mathcal P} = \{X\in{\mathcal A}\ |\ X\text{ is a summand of some }{\mathbb F}_\lambda , Y\in{\mathcal B}\}$ and
${\mathcal I} = \{X\in{\mathcal A}\ |\ X\text{ is a summand of some }{\mathbb F}_\rho Y, Y\in{\mathcal B}\}$,
respectively. Consequently, if ${\mathbb F}_\lambda$ and ${\mathbb F}_\rho$ have the same image, then $({\mathcal A},{\mathcal E})$ is Frobenius.
This seems very restrictive, but in fact there are at least two cases I know where it can be applied:
(1) If ${\mathcal A}$ is a dg-category, then the forgetful functor ${\mathbb F}: \text{dg-mod}({\mathcal A})\to\text{gr-mod}({\mathcal A})$ fulfills the requirements of the criterion above and thus can be used to construct a Frobenius structure on $\text{dg-mod}({\mathcal A})$ (for pretriangulated dg-categories ${\mathcal A}$, this structure can in turn be restricted to ${\mathcal A}$ itself). (2) If $G$ is a finite group, $H$ is a subgroup, then the fortgetful functor $G\text{-mod}\to H\text{-mod}$ has left adjoint $\text{Ind}^G_H$ and right adjoint $\text{Coind}^G_H$, and these two functors coincide for $(G:H)<\infty$. In this case, the above criterion therefore applies to provide $G\text{-mod}$ with a Frobenius structure "relative to $H$". Question
Do you know more criteria for constructing Frobenius structures and situations where they can be applied?
For example, I would be interested in a criterion which can be applied to show that the category of maximal Cohen-Macaulay modules over a Gorenstein ring is Frobenius.
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On a second order boundary value problem with singular nonlinearity
DOI: http://dx.doi.org/10.12775/TMNA.2006.001
Abstract
In this paper we investigate in a variational setting, the elliptic
boundary value problem $-\Delta u={\textt{sign}u}/{|u|^{\alpha+1}}$ in $\Omega$, $u=0$ on $\partial\Omega$, where $\Omega$ is an open connected bounded subset of ${\mathbb R}^N$, and $\alpha> 0$. For the positive solution, which is checked as a minimum point of the formally associated functional $$ E(u)=\frac 12\int_\Omega|\nabla u|^2+\frac{1}{\alpha} \int_\Omega \frac1{|u|^\alpha}, $$ we prove dependence on the domain $\Omega$. Moreover, an approximative functional $E_\varepsilon$ is introduced, and an upper bound for the sequence of mountain pass points $u_\varepsilon$ of $E_\varepsilon$, as $\varepsilon\to 0$, is given. For the onedimensional case, all sign-changing solutions of $-u''={\text{sign}u}/{|u|^{\alpha+1}}$ are characterized by their nodal set as the mountain pass point and $n$-saddle points ($n> 1$) of the functional $E$.
boundary value problem $-\Delta u={\textt{sign}u}/{|u|^{\alpha+1}}$ in
$\Omega$, $u=0$ on $\partial\Omega$, where $\Omega$ is an open connected
bounded subset of ${\mathbb R}^N$, and $\alpha> 0$. For the positive solution,
which is checked as a minimum point of the formally associated
functional
$$
E(u)=\frac 12\int_\Omega|\nabla u|^2+\frac{1}{\alpha}
\int_\Omega \frac1{|u|^\alpha},
$$
we prove dependence
on the domain $\Omega$. Moreover, an approximative functional $E_\varepsilon$ is
introduced, and an upper bound for the sequence of mountain pass points
$u_\varepsilon$ of $E_\varepsilon$, as $\varepsilon\to 0$, is given. For the onedimensional
case, all sign-changing solutions of
$-u''={\text{sign}u}/{|u|^{\alpha+1}}$ are characterized by their
nodal set as the mountain pass point and $n$-saddle points ($n> 1$) of
the functional $E$.
Keywords
Variational methods; elliptic problems; singular nonlinearity
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On the monotonicity of the period function of a quadratic system
1.
Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275
$ \dot x=- y + x y,\quad \dot y=x + 2 y^2-c x^2, \quad -\infty < c < +\infty.$
We show that this system has two isochronous centers for $c=1/2$, and its period function has only one critical point for $c\in(7/5, 2)$. For all other cases, the period function is monotone. This improves the results in [1].
Mathematics Subject Classification:34C07, 34C08, 37G1. Citation:Yulin Zhao. On the monotonicity of the period function of a quadratic system. Discrete & Continuous Dynamical Systems - A, 2005, 13 (3) : 795-810. doi: 10.3934/dcds.2005.13.795
[1]
Yu Chen, Yanheng Ding, Suhong Li.
Existence and concentration for Kirchhoff type equations around topologically critical points of the potential.
[2] [3] [4]
Tsung-Fang Wu.
On semilinear elliptic equations involving critical Sobolev exponents and sign-changing weight function.
[5]
Gioconda Moscariello, Antonia Passarelli di Napoli, Carlo Sbordone.
Planar ACL-homeomorphisms : Critical points of their components.
[6]
P. Candito, S. A. Marano, D. Motreanu.
Critical points for a class of nondifferentiable functions and applications.
[7] [8] [9]
Cristian Bereanu, Petru Jebelean.
Multiple critical points for a class of periodic lower semicontinuous functionals.
[10] [11] [12] [13] [14]
Zhiyou Wu, Fusheng Bai, Guoquan Li, Yongjian Yang.
A new auxiliary function method for systems of nonlinear equations.
[15]
Alexander Blokh, Michał Misiurewicz.
Dense set of negative Schwarzian maps whose critical points have minimal limit sets.
[16]
Jaume Llibre, Jesús S. Pérez del Río, J. Angel Rodríguez.
Structural stability of planar semi-homogeneous polynomial vector fields applications to critical points and to infinity.
[17]
Fengjuan Meng, Chengkui Zhong.
Multiple equilibrium points in global attractor for the weakly damped wave equation with critical exponent.
[18]
M. L. Miotto.
Multiple solutions for elliptic problem in $\mathbb{R}^N$
with critical Sobolev exponent and weight function.
[19]
Iuliana Oprea, Gerhard Dangelmayr.
A period doubling route to spatiotemporal chaos in a system of Ginzburg-Landau equations for nematic electroconvection.
[20]
Primitivo B. Acosta-Humánez, J. Tomás Lázaro, Juan J. Morales-Ruiz, Chara Pantazi.
On the integrability of polynomial vector fields in the plane by means of Picard-Vessiot theory.
2018 Impact Factor: 1.143
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4 Methods to Account for Radiation in Participating Media
Radiative heat transfer in semitransparent media is described by the radiative transfer equation (RTE). Solving this equation is challenging in terms of computational costs. However, depending on a medium’s radiation properties, simplifications exist that allow the solving of such models in a fraction of the time. This blog post gives an overview of the available methods and when they can be applied.
Defining the Radiative Transfer Equation
An incident beam that travels in direction \Omega through a participating medium interacts with the medium. Part of its intensity, I(\Omega), is absorbed by the fraction \kappa I(\Omega), where \kappa (m^{-1}) is the absorption coefficient. Another fraction is scattered in another direction, \sigma_s I(\Omega), where \sigma_s(m^{-1}) is the scattering coefficient. The intensity in a given direction is attenuated by scattering in a different direction and augmented by radiation coming from a different direction. This is described by:
(1)
where \phi(\Omega^{\prime},\Omega) is the scattering phase function that describes the probability of a ray from direction \Omega^{\prime} being scattered into direction \Omega. The medium itself can emit radiation in all directions by the factor \kappa I_b, where I_b is the intensity of a blackbody.
Radiation interacting with a semitransparent medium.
All of these effects are fully described by an integro-differential equation called RTE:
(2)
The key to solving this equation lies in the approximation of the scattering integral. In combination with heat transfer, the incident radiation
(3)
and the radiative heat flux
(4)
are important quantities.
Before we discuss the different methods to solve this equation, we introduce another quantity that describes the participating medium — the optical thickness or optical depth:
(5)
It describes how transparent the medium is to radiation. If \tau\ll 1, the medium is called “optically thin”, and if \tau\gg1, “optically thick”.
Methods for Solving the RTE
The following sections give an overview of the four methods available with the COMSOL Multiphysics® software to solve the RTE:
Discrete ordinates method (DOM) P1 approximation Rosseland approximation Beer–Lambert law
Except for the last one, we owe all of these methods to astrophysics and its analysis of stellar atmospheres. A comprehensive book about this complex topic is
Radiative Heat Transfer by M.F. Modest (Ref. 1). It contains detailed explanations and derivations of the solution methods that go beyond the scope of this blog post. Method 1: The Discrete Ordinates Method
The most general method of solving the RTE is the DOM. Its name indicates the idea behind this method. The integral over the angular space is divided into discrete directions. Thus, one partial differential equation (PDE) per discrete ordinate remains to be solved for the intensity, I:
(6)
s
where \mathbf{S}_i is the i
th discrete ordinate and w_j, the quadrature weights.
The details of how to divide the angular space into discrete ordinates are described in this previous blog post. The default
S N method uses a symmetric quadrature set of order Nand divides the 3D angular space in N(N+2) directions. The figure below illustrates the discrete ordinates for the symmetric even quadrature set and different order, N. Discrete ordinates for the level symmetric even quadrature set from S2 up to S12 (8–168 directions).
The default S4 method is sufficient for many applications but already introduces 24 dependent variables for the intensities. The major advantage of the DOM over the other methods is the high accuracy in arbitrary configurations because of its discretization of the angular space. Additionally, the method can handle various forms of the scattering phase functions: isotropic, linear or polynomial anisotropic, and Henyey–Greenstein.
Because this method is computationally expensive and the required memory easily exceeds the available memory on common workstations for complex 3D geometries, we want to talk briefly about a simple way to tweak the solver — the performance index, which is available at the interface level.
Let’s say we need to be very accurate and use the S8 method, which adds 80 additional equations to the model. The solver splits these variables into segregated groups, and each group is computed in a single iterative step before the solver moves on to the next group. The performance index controls the number of groups that are created. For a performance index of 0 (minimum value), 10 groups are created, where each group contains 8 intensity variables. If the performance index is set to 1 (maximum value), each intensity variable goes into a separate, segregated group and the required memory remains low. This approach also works for larger models, but the computational time increases.
Performance index to control the solver. Method 2: The P1 Approximation
Instead of using discrete ordinates, the P1 method is based on spherical harmonics to discretize the angular space. They are eigenfunctions of the Laplace operator in spherical coordinates. The P1 approximation uses linear terms only and from this, it emerges that solving the following equation for Eq. (3) is equivalent to:
(7)
D_\textrm{P1} is the P1 diffusion coefficient, defined as:
(8)
with the linear Legendre coefficient, a_1, for the scattering phase function.
Hence, with the P1 method, isotropic and linear anisotropic scattering can be considered. The second term on the left-hand side corresponds to the radiative heat source, Q_\textrm{r}. Thus, only one additional equation is needed to take radiation transport into account.
Method 3: The Rosseland Approximation
First, let’s recall the stationary heat equation for a medium with density \rho(kg/m^3), heat capacity C_p(J/(kg\cdot K)), and thermal conductivity k(W/(m\cdot K):
(9)
The first term on the left-hand side is the convective term, and on the right side is the heat source term.
Let’s take a closer look at the conductive term where the heat flux, \mathbf{q}, follows Fourier’s law of heat conduction:
(10)
Getting back to radiation in participating media, light propagation behaves similarly to heat conduction under the assumption of large optical depths (\tau\gg 1), and we can rewrite Eq. (10) as follows:
(11)
with the highly nonlinear “radiative conductivity”, k_\textrm{R}:
(12)
with \beta_\textrm{R}=\kappa+\sigma_\textrm{s} being the Rosseland mean extinction coefficient and \sigma(W/(m^2K^4)), the Stefan–Boltzmann constant.
From a computational point of view, no additional equation is required to account for radiation in participating media. Just a highly nonlinear conductivity term appears. However, the number of problems for which this approximation is valid is limited: mainly, radiation that depends on the temperature and its gradient only and not on its direction or distance to the source, which is valid at very large optical depths. This applies to stellar atmospheres, for which the method was first developed by Rosseland. It is also a common method in the glass industry for a wide range of uses.
Because the Rosseland approximation inserts an additional term to Eq. (9), it is available as an extension of the
Solid feature and is called Optically Thick Participating Medium. Subfeature to consider radiation at large optical depths. Method 4: Beer–Lambert Law
This law is a great simplification of the RTE but still provides an accurate solution if the following conditions are fulfilled:
The radiation source is described by collimated, almost monochromatic beams Refraction, reflection, or scattering in the medium can be neglected There is no emission in the wavelength range of the incident beam
This is the case for photometry and the analysis of chemical compositions. The RTE then simplifies to:
(13)
with the beam’s orientation \mathbf{e}_i.
As the beam travels through the medium, energy is absorbed and the radiative heat source term is defined by:
(14)
Hence, the Beer–Lambert law describes the attenuation of the radiation intensity by absorption as the beam travels through the medium.
Verification Examples: Accounting for Radiation in Participating Media
This section shows examples for the different methods we discussed earlier and compares the results for various properties of the participating medium.
Cooling Glass Melt
To compare the DOM, P1 approximation, and Rosseland approximation for a typical scenario, let’s take a look at a glass melt that is cooled down from 600°C to 20°C. Due to the high temperatures, this cooling occurs mainly via radiation. The resulting temperature distribution after 10 seconds is compared for low and high absorption coefficients.
Temperature profile at the centerline and in the glass plate for \kappa=5\ m^{-1}. The Rosseland approximation is not sufficient for a low optical thickness. The P1 approximation provides a very accurate solution. Temperature profile at the centerline and in the glass plate for \kappa=120\ m^{-1}. The Rosseland approximation provides a reasonable result despite its simplicity. The P1 approximation is still very accurate but not as accurate as it is for a low \kappa.
This example shows that the P1 approximation can be very accurate over a large range of \tau and gives good results for optically thin media as well. It also shows the weakness of the Rosseland approximation for small optical depths that are present at the walls and for a low \kappa. The computational costs for the DOM are significantly higher, taking about 10 times longer to solve than the other methods.
Scattering in a Cylinder
To investigate the accuracy of the P1 approximation and DOM for different scattering effects and wall properties, a verification model is investigated for three different cases:
Constant surface emissivity, \epsilon_r=0.5, with isotropic scattering Radially varying emissivity, \epsilon_r=0.5(1-y/R), with isotropic scattering Radially varying emissivity, \epsilon_r=0.5(1-y/R), with linear anisotropic scattering
The scattering albedo \omega=\sigma / (\sigma+\kappa) is used to parameterize the model.
Case 1: Incident radiation for a varying isotropic scattering albedo in the radial direction. Case 2: Incident radiation for a varying isotropic scattering albedo in the azimuthal direction. Case 3: Incident radiation for a varying linear anisotropic scattering along the normalized optical thickness.
This example shows that the P1 approximation approaches the accurate DOM solution for larger optical thicknesses for \omega \rightarrow 1. The error for small optical thickness increases. In particular, for linear anisotropic scattering (case 3), the method reproduces the results only roughly. Nevertheless, the P1 method is still a good approximation, especially if you consider the lower computational effort.
Concluding Thoughts on the Methods in COMSOL Multiphysics®
The methods we discussed cover the entire range of methods for computing radiation in participating media under different assumptions. To conclude this blog post, we summarize our findings for each method:
DOM Is the most versatile method, solving the full RTE in discrete directions (up to 512) and providing high accuracy Can include complex forms of anisotropic scattering Computational costs increase as the number of discrete ordinates increase P1 method Is reasonably accurate for many configurations where the directional aspect of the radiation propagation does not dominate Can only include isotropic and linear anisotropic scattering Is computationally inexpensive by adding only one additional scalar equation to the system Rosseland approximation Provides good results for very large optical depths Cannot consider scattering Is computationally inexpensive by using a radiative conductivity in the heat transfer equation Beer-Lambert law Provides good results for applications that fulfill the assumptions for the underlying theory Has a narrow range of applications Is computationally inexpensive Additional Resources
For more information about the functionality available for heat transfer modeling, click the button below to go to the Heat Transfer Module product page.
Learn more about modeling radiation by checking out the following tutorials:
Reference M.F. Modest, Radiative Heat Transfer, Academic Press, 2003. Comments (0) 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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In a continuance of this series today, I wanted to talk about the one-sided limits of calculus. Limits and where they lead are one of the fundamental concepts of calculus. Since limits show what happens near any particular point on a graph we need to take a look at them. Read on with me and I will explain their behavior so you can easily solve them.
One-Sided Limits
Our current idea of the limit definition goes back around 200 years ago by Bernard Bolzano. He was instrumental in several mathematical ideas. One of which was the definition of one-sided limits which I will be talking about today. Now on to some problems!
Problem 1.
Use the graph to see if the following statements are true or false. Look at the graph carefully because it can be tricky to see what is really going on sometimes.
We can see there is a function \(y=f(x)\). There is also a small portion of the positive x-axis defined from [3 to 6].
A. True or False: \(lim_{x \to -3+}(f(x)=9\)
F(x) is your [y] value. X is of course your [x] value. Does [y] or [f(x)] get closer to 9 as [x] gets closer to 3 when coming from larger numbers than +3 ?
We can see that it sure does. The answer is [true].
B. True or False: \(lim_{x \to 0-}(f(x)=3\)
F(x) is your [y] value. X is your [x] value. The question is asking if [y=3] as [x] approaches 0 from the negative side of the [x] axis ?
It appears that [y=0] when [x=0] so the answer is false for this question.
C. True or False: \(lim_{x \to 0-}(f(x)\) = \(lim_{x \to 0+}(f(x)\) ?
This question is asking if the left limit is the same as the right limit. What does [y] equal when [x] is coming from the negative side? It equals 0. Now what does [y] equal as [x] is coming from the positive side? It equals 0 also. So both one-sides limits equal each other and the answer to the question is [true].
D. True or False: \(lim_{x \to 0}(f(x)\) exists?
We have basically answered this question already. When the limit is listed without directional signs(- or +) it is asking if the limit exists and is equal from both sides of the graph. It is equivalent to asking if the limit from the left side equal the limit from the right side? As mentioned before we have answered this. We did see that the limit from the left side equaled the limit from the right side. So the answer to this question is [true].
E. True or False: \(lim_{x \to 0}(f(x)=0\)
This is another version of the question above. Since there are no directional signs(- or +) it is implying the full limit. The full limit is each one-sided limit being equal to each other. The left hand limit=0. The right hand limit=0. Therefore the limits equal each other and both equal 0. So the answer is also true for this question.
F. True or False: \(lim_{x \to 3}(f(x)=9\)
F(x) is your [y] value. X is your [x] value. The question is asking if [y=9] whether [x] approaches 3 from the negative or positive side?
When [x] is coming from the positive side of the [x] axis things are pretty clear and defined. However, there is a hole in the graph when [x] approaches 3 from the negative side. Therefore our limits are not equal and the answer is [false].
G. True or False: \(lim_{x \to 6-}(f(x)=6\)
F(x) is your [y] value. X is your [x] value. This is asking if [y=6] when [x] approaches 6 from the negative side? We can plainly see that it does not. When [y=6] it looks like [x] is somewhere between 2 and 3. So the answer to this question is [false].
Problem 2.
F(x) = 6-x when x < 2 and (x/2) +1 when x > 2.
Find \(lim_{x \to 2+}(f(x)\) and \(lim_{x \to 2-}(f(x)\).
This is what I call a conditional function. Certain sections of the function give different values depending on the condition.
We will start by looking at the one-sided limits. First lets look at the limit as [x] approaches 2 from the positive side of the x-axis. The expresison (x/2) + 1 is used to evaluate when x > 2.
The \(lim_{x \to 2+}(f(x)=((x/2)+1)\) = (2/2) + 1 = 2.
The expression 6-x is used to evaluate the expression when x < 2.
The \(lim_{x \to 2-}(f(x)=6-x\) = 6-2 = 4.
Does the limit exist? For a full limit to exist its one-sided limits have to exist and be equal. The two limits above that we evaluated are not equal. Therefore the limit does not exist.
Problem 3.
Use the relation \(lim_{\theta \to 0}{\frac{\sin\theta}{\theta}}=1\) to find the limit of this function.
\(\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}\)
As \(\theta\) gets closer to 0 the square root of \(\theta\) also gets closer to 0.
We now have : \(lim_{\sqrt{2}\theta \to 0}{\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}}\)
So \(x = \sqrt{2} \theta \). Now substitute in the values.
\(lim_{x \to 0} \frac{\sin x}{x} =1 \). That is our answer using the relation.
Problem 4.
Find the limit of this expression. \(lim_{y \to 0}{\frac{\sin 3y}{7y}}\)
Again we will use the theorem \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \)
To get the form that we want so we can solve easily we need to multiply the numerator and denominator by \(\frac{3}{7}\).
Once you see how everything cancels out you will see that your answer is also \(\frac{3}{7}\).
Problem 5.
Use the relation \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1 \) for the following one-sided limits problem.
\( f(x) = \frac{2x + 2x\cos(2x)}{\sin(2x)\cos(2x)} \) as x approaches 0.
We should write this expression as the sum of two terms.
\(\frac{2x}{\sin(2x)\cos(2x)} + {2x}{\sin(2x)}\)
Now rewrite the first fraction as the product of two fractions and then factor the result.
\(\frac{2x}{\sin(2x)} * \frac{1}{\cos(2x)} + \frac{2x}{\sin(2x)}\)
\(= \frac{2x}{\sin(2x)} \left(\frac{1}{\cos(2x)} + 1\right) \)
Multiply this out and then simplify. I did not show all the Algebra steps there because it is a pain but if you are in calculus this should be trivial for you.
You will this expression: \(\frac{1}{1} \left(\frac{1}{1} +1\right)\)
\(=2\)
Problem 6.
Use the relation \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1\) to find out the limit of the function below.
\(f(\theta)=\frac{\sin \theta}{\sin (2\theta)}\)
Multiply \(\frac{\sin \theta}{\sin (2\theta)} by \frac{2\theta}{2\theta} \).
This equals : \(\frac{1}{2} * \frac{\sin \theta}{\theta} * \frac{2\theta}{\sin (2\theta)}\).
We now have: \(\frac{1}{2} * 1 * \frac{1}{1} \).
So you should get \(\frac{1}{2}\) as your answer.
Conclusion
After doing a few problems you should be able to see how useful one-sided limits can be to you. This will be one of your premier tools later on to help you solve problems. Limits often exhibit some wild behavior. That is why we must look at both sides of any particular limit. Sometimes they may be exactly equal on both sides but many times they will not be. Graphing will help when it gets more complicated so I will be using more graphs as we progress.
I hope you enjoyed this article and that it will help your understanding and appreciation of Calculus.
If you are already not a member of my site then please subscribe. I would appreciate it very much!
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Definition:Isolated Point (Topology)/Subset
Jump to navigation Jump to search
$x \in H$ is an
Definition
$x \in H$ is an
isolated point of $H$ if and only if: $\exists U \in \tau: U \cap H = \left\{{x}\right\}$ Also see Results about isolated pointscan be found here.
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We can then define the parallel transporter for matter fields to be a linear map \({\parallel_{C}\colon V_{p}\rightarrow V_{q}}\), where \({C}\) is a curve in \({M}\) from \({p}\) to \({q}\). Choosing a gauge, the parallel transporter can be viewed as a (gauge-dependent) map \({\parallel^{\beta}{}_{\alpha}\colon\left\{ C\right\} \rightarrow GL(n,\mathbb{C})}\). This determines the (gauge-dependent) matter field connection 1-form \({\Gamma^{\beta}{}_{\alpha}\left(v\right)\colon T_{x}M\rightarrow gl(n,\mathbb{C})}\), which can also be written when acting on a \({\mathbb{C}^{n}}\)-valued 0-form as \({\check{\Gamma}\left(v\right)\vec{\Phi}}\). The values of the parallel transporter are again viewed as a rep of the gauge group \({G}\), so that the values of the connection are a rep of the Lie algebra \({\mathfrak{g}}\), and if \({G}\) is compact we can choose a unitary gauge so that \({\mathfrak{g}}\) is represented by anti-hermitian matrices. We then define the
gauge potential (AKA gauge field, vector potential) \({\check{A}}\) by
\(\displaystyle \check{\Gamma}\equiv-iq\check{A}, \)
where \({q}\) is called the
coupling constant (AKA charge, interaction constant, gauge coupling parameter). Note that \({A^{\beta}{}_{\alpha}}\) are then hermitian matrices in a unitary gauge. The gauge covariant derivative is then
\(\displaystyle \nabla_{v}\vec{\Phi}=\mathrm{d}\vec{\Phi}\left(v\right)-iq\check{A}\left(v\right)\vec{\Phi}, \)
which can be generalized to \({\mathbb{C}^{n}}\)-valued \({k}\)-forms in terms of the exterior covariant derivative as
\(\displaystyle \mathrm{D}\vec{\Phi}=\mathrm{d}\vec{\Phi}-iq\check{A}\wedge\vec{\Phi}. \)
For a matter field (0-form), this is often written after being applied to \({e_{\mu}}\) as
\(\displaystyle \mathrm{D}_{\mu}\vec{\Phi}=\partial_{\mu}\vec{\Phi}-iq\check{A}{}_{\mu}\vec{\Phi}, \)
where \({\mu}\) is then a spacetime index and \({\check{A}_{\mu}\equiv\check{A}(e_{\mu})}\) are \({gl(n,\mathbb{C})}\)-valued components. This expression is not coordinate-dependent, and we may therefore also write it using an abstract index.
This connection defines a curvature \({\check{R}\equiv\mathrm{d}\check{\Gamma}+\check{\Gamma}\wedge\check{\Gamma}}\), which lets us define the
field strength (AKA gauge field) \({\check{F}}\) by
\(\displaystyle \begin{aligned}\check{R} & \equiv-iq\check{F}\\ \Rightarrow\check{F} & =\mathrm{d}\check{A}-iq\check{A}\land\check{A}. \end{aligned} \)
Δ The definition \({\check{\Gamma}\equiv-iq\check{A}}\) is the convention with a mostly pluses metric signature; with a mostly minuses signature the sign is reversed. However, one also finds this definition in terms of an elementary charge \({e\equiv\pm q}\), which may be positive or negative depending on convention, again reversing the sign.
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In what particular situation would the following be true?
$AB = B^TA$
where $A$ is symmetrical, $B$ is not.
I also know that $BB = B$.
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We know that $$(AB)^T=B^TA^T=B^TA=AB$$therefore the symmetry of $AB$ is an equivalent condition. No more general condition can be implied.
Let us do it in dimension $2\times 2$.
$A=\pmatrix{a &b\\ b&c}$ and $B=\pmatrix{s&p\\q&r}$.
The conditions on $B$ are:
It's not symmetric
It's a projection
$AB$ is symmetric
First deduction: $r+s= 1$. Indeed, if not then $p=q=0$ but we know that $p\neq q$.
Second deduction: $r$ and $s$ are the two solutions to the equation $x^2-x+pq=0$. That is $$r,s=\frac{1\pm \sqrt{1-4pq}}{2}$$ and in particular, we must have $1-4pq\geq 0$.
Third deduction: $$ap+b\sqrt{1-4pq}=qc$$
From this we can derive a method to produce all possible matrices $B$:
Example: pick $q=0$. Assuming $a\neq 0$, the relevant solution to the quadratic equation is $p=-\frac ba$. Assuming $b\neq 0$ we have $p\neq q$. Here, $1-4pq\geq 0$ comes for free. $r=1$ and $s=0$. This gives $$B=\pmatrix{0 & -\frac ba\\ 0 & 1}$$ You can check that $AB$ is symmetric as required: $$\pmatrix{a &b\\ b&c}\pmatrix{0 & -\frac ba\\ 0 & 1}=\pmatrix{0 &0\\ 0&c-\frac{b^2}{a}}$$ and that the other requirements are met as well.
Conclusion: In general, there are a lot of such matrices (potentially several $1$-parameter families, which can be described explicitly at least in dimension $2$).
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I think you mean "perpendicular to", not "parallel to", in the question.
It is clear from the formula for $p(x)$ in terms of $x$, $a$, and the inner product, namely$$
p(x) = \frac{\langle x, a\rangle}{\langle a, a\rangle} a,
$$that $x - p(x)$ is perpendicular to $a$. Just do a short calculation with the formula:$$
\begin{align}
\langle x - p(x), a\rangle & = \langle x, a\rangle - \langle p(x), a\rangle \\
& = \langle x ,a\rangle - \left\langle \frac{\langle x, a\rangle}{\langle a, a\rangle} a, a \right\rangle \\
& = \langle x,a\rangle - \frac{\langle x, a\rangle}{\langle a, a\rangle} \langle a, a\rangle \\
& = \langle x, a\rangle - \langle x, a\rangle \\
& = 0.
\end{align}
$$So perhaps your question is really
why $p(x)$ is given by that formula. To answer that we'd need to know whatever definition you had of $p(x)$. In any case, for what you want, it's enough to just take the above formula as the definition of $p(x)$. To give the argument:
Fix any $y$ in $\{a\}^{\perp}$. The above calculation shows that $x - p(x)$ is perpendicular to $a$, and so is $y$, so the difference $x - p(x) - y$ is also perpendicular to $a$, and hence to $p(x)$ (as $p(x)$ is a scalar multiple of $a$ by definition). So by the Pythagorean theorem$$
\begin{align}
d(x,y)^2 & = \|x - y\|^2 \\
& = \|x - p(x) - y + p(x)\|^2\\
& = \|x - p(x) - y\|^2 + \|p(x)\|^2 \\
& \geq \|p(x)\|^2
\end{align}
$$and taking square roots one deduces$$
d(x,y) \geq \|p(x)\| = \left\|\frac{\langle x,a\rangle}{\|a\|^2} a\right \| = \frac{|\langle x,a\rangle|}{\|a\|^2} \|a\| = \frac{|\langle x,a\rangle|}{\|a\|}.
$$The argument given for this inequality also shows that equality holds if and only if $\|x - p(x) - y\|^2 = 0$, ie, if and only if $y = x - p(x)$ (which we know is possible, as we proved $x - p(x)$ was indeed in $\{a\}^{\perp}$ ). This establishes$$
d(x,\{a\}^{\perp}) = \frac{|\langle x,a\rangle|}{\|a\|}.
$$
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The structure of a compound has a big effect on its properties. But how do we know what that structure is? The most useful methods of determining molecular structure involve the interaction of electromagnetic radiation, or light, with matter. Visible light, ultraviolet and infrared radiation, and even microwaves and radio waves interact with matter. They can each tell us different kinds of information about the materials they interact with.
How does light interact with matter?
Light has wave properties. much like the waves you could see at the ocean shore. The different colors of light that we see have different wavelengths; blue light has a shorter wavelength than red light, for example. These different wavelengths of light have different amounts of energy. This idea is described in the Planck-Einstein relation:
\[ E = h \nu \]
(where E = energy, h = Planck's constant, n = frequency)
or
\[ E = \dfrac{h c}{ \lambda}\]
where
\(c\) is the speed of light \(\lambda\) is the wavelength
This equation means:
Higher frequencies of light are more energetic than lower frequency ones (when the number n gets bigger, the number E also gets bigger). Higher frequencies correspond to shorter wavelengths (so when the length l gets longer, E gets smaller).
There are a couple of important and surprising points about the interaction of photons with matter:
Light is quantized; it travels in packages, called photons, and different photons have specific amounts of energy. Absorption of light by matter is also quantized; only specific packages or "quanta" can be absorbed by a specific material. Consequently, specific compounds absorb specific frequencies of light and don't absorb others.
When ultraviolet and visible light are absorbed, the energy from the light is transferred to an electron. The electron is excited to a higher energy level. Only certain energy levels are available in a material, and so the material can only absorb certain photons. That means:
A photon with not enough energy to reach another energy level is not absorbed. A photon with too much energy to reach another energy level is not absorbed, either; the electron cannot absorb some of the energy from a photon and have a little left over for later. The wavelength or frequency of a photon that is absorbed by the electron corresponds to the amount of energy needed to reach another energy level.
The same sort of event can happen "backwards": an electron can lose energy by falling to a lower energy level. The lost energy can be given up by the electron as a photon of light. The wavelength or frequency of the photon corresponds to the difference between electron energy levels. This phenomenon, in which light is absorbed by a material and then given off again, is called "fluorescence".
There are many kinds of electromagnetic radiation that can provide different kinds of information about structure. For example:
UV-Visible spectroscopy tells us something about the electronic levels in a material. In another chapter on molecular orbital theory, we will see more about this type of spectroscopy. X-rays can be used to construct an exact three-dimensional map of where the atoms lie in a crystalline material based upon how the x-rays scatter as they pass through the crystal. X-ray crystallography is a little bit too complicated for us, however. Radio waves interact with nuclear particles in a way that is similar to the absorption of UV light by electrons. However, this phenomenon only occurs in a strong magnetic field. The absorption of radio waves by the hydrogen nuclei in water molecules in human tissues is referred to as magnetic resonance imaging (MRI). The observation of nuclei in small molecules by a similar technique is referred to as nuclear magnetic resonance (NMR). NMR spectroscopy will be the subject of another chapter. Infrared light is absorbed by different bonds in a molecule. Infrared spectroscopy is the subject of another chapter.
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Simple Linear Regression
Let’s continue with the cars data, but this time let’s formally recognize that a car’s weight might have some effect on city MPG. Assume
We again focus on the expected value, not $\sigma$, but we will start dropping the $\mathbb{E}(Y)$ notation because it quickly becomes cumbersome and needlessly repetitive. Notice that this model implicitly states that the expected city MPG depends linearly on a car’s weight.
The following code reads in the dataset, plots the $\texttt{mpgCity}$ variable against the $\texttt{weight}$ data, and calculates an estimate of the population mean dependent on the sampled cars’ weights. Here the estimated expected city MPG is formed by a linear combination of $\hat{\beta}_0, \hat{\beta}_1$ and $\texttt{weight}$.
import numpy as npimport pandas as pdimport bplot as bpfrom scipy.optimize import minimizefrom scipy.stats import norm as normalimport patsybp.LaTeX()bp.dpi(300)cars = pd.read_csv("https://raw.githubusercontent.com/roualdes/data/master/cars.csv")
bp.scatter(cars['weight'], cars['mpgCity'])
<matplotlib.collections.PathCollection at 0x124a35b70>
def ll(beta, yX): y = yX[:, 0] X = yX[:, -2:] yhat = np.full(y.shape, np.nan) for r in range(X.shape[0]): yhat[r] = np.sum(beta * X[r,:]) d = y - yhat return np.sum(d * d)pX = patsy.dmatrix("~ weight", data=cars)yX = np.c_[cars['mpgCity'].values, np.asarray(pX)]beta_hat = minimize(ll, normal.rvs(size=2), args=(yX))['x']
We write the estimated linear model as
What’s nice about this model is that we can easily interpret the parameters $\hat{\beta}_0 = 50.14$ and $\hat{\beta}_1 = -0.01$. For instance, $\hat{\beta}_0$, better known as an intercept, is an estimate of (the population) city MPG for a car that weighs $0$ pounds. While this doesn’t make sense conceptually, this is the literal interpretation of the estimated intercept in the context of these data. Notice that there’s not data near $\text{weight} = 0$. When this is the case, you can generally expect the intercept to not make much sense.
The slope, $\hat{\beta}_1$, describes the linear relationship between a car’s weight and city MPG. This estimate too is relatively easy to interpret in the context of these data. For every 1 unit increase in a car’s weight (a 1 pound increase), we estimate a $0.01$ decrease in the expected city MPG.
Be careful to not overly interpret these estimates as describing a causal relationship. Determining causal relationships from non-experimental data is no easy task, and we won’t even try to broach this topic in this course. Peter Norvig, Director of Research at Google, wrote an essay describing many of the difficulties of applied statistics practice surrounding observational data. His essay Warning Signs in Experimental Design and Interpretation enumerates common warning signs for when a practioner of applied statistics might be misinterpreting their data. For a more theoretical approach to determining causal relationships from data, see Judea Pearl’s book Causality.
Just like before, the estimates $\hat{\beta}_0, \hat{\beta}_1$ are simply one set of estimates based on one random sample. The values we produced could be due to pure random chance. To better understand our uncertainty in these estimates, we will calculate confidence intervals. We will let the function $\texttt{boot::boot}$ do the random sampling for us, and for that we need write a function that accepts our data and a vector of indices.
N = cars['mpgCity'].sizeR = 999betas = np.full((R, 2), np.nan)for r in range(R): idx = np.random.choice(N, N) betas[r, :] = minimize(ll, normal.rvs(size=2), args=(yX[idx, :]))['x']
beta_p = np.percentile(betas, [10, 90], axis=0)
axs = bp.subplots(1, 2)bp.current_axis(axs[0])bp.density(betas[:, 0])bp.percentile_h(betas[:, 0], y=0)bp.rug(beta_p[:, 0])bp.labels(x='Intercept', y='Density')bp.current_axis(axs[1])bp.density(betas[:, 1])bp.percentile_h(betas[:, 1], y=0)bp.rug(beta_p[:, 1])bp.labels(x='Slope', y='Density')bp.tight_layout()
<matplotlib.axes._subplots.AxesSubplot at 0x124fb9710>
The $90\%$ confidence intervals for $\beta_0$ and $\beta_1$ are
np.round(beta_p, 3)
array([[ 4.6007e+01, -1.0000e-02], [ 5.3851e+01, -8.0000e-03]])
Some classes will make a big deal about the slope estimate not including $0$. There’s no doubt such conclusions have some appeal. However, this too often encourages binary thinking such as, “is the true population slope equal to zero or isn’t it?” When a statistic, known as p-value, is smaller than $0.05$ or a confidence interval excludes zero, the common phrase is, statistically significantly different from zero, as if zero or not are the only options.
Increasingly, statisticians are warning against such binary decision making; e.g. “It’s time to talk about ditching statistical significance”, “Moving to a World Beyond ‘p < 0.05’”, or “Scientists rise up against statistical significance”.
In this class, we’ll focus on predictions, understanding uncertainty in our predictions, and making decisions in the face of this uncertainty.
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Accelerated Stochastic Power Iteration by Christopher De Sa, Bryan He, Ioannis Mitliagkas, Christopher Ré, and Peng Xu and referencing work by many other members of Hazy Research
and referencing work by many other members of Hazy Research
Surprisingly, standard acceleration doesn’t always work for stochastic PCA. We provide a very simple stochastic PCA algorithm, based on adding a momentum term to the power iteration, that achieves the optimal sample complexity andan accelerated iteration complexity in terms of the eigengap. Importantly, it is embarrassingly parallel, allowing accelerated convergence in terms of wall-clock time. Our results hinge on a tight variance analysis of a stochastic two-term matrix recurrence, which implies acceleration for a wider class of non-convex problems.
Principal component analysis (PCA) is one of the most powerful tools in machine learning. It is used in practically all data processing pipelines to allow data to be compressed and visualized. When Gene Golub and William Kahan presented a practical SVD algorithm for in 1964 they were set for life.
The current simplicity vs speed trade-off Currently, no simple methods achieve an accelerated convergence rate in the stochastic setting.
Classic methods, like the extremely simple power iteration and the Lanczos algorithm, achieve linear rates with respect to the target accuracy, ε.Lanczos is
fast, achieving an accelerated rate: the number of iterations (full passes over the data) only depends on the square root of the eigengap, Δ, between the top two eigenvalues. However, doing full passes over the data is not practical for large-scale applications. State-of-the-art stochastic PCA algorithms only iterate on a handful of samples at a time, which is much more practical and makes large-scale PCA possible.This is extremely simple and effective, resulting in popular industry use.
QUESTION: Do we really need these complex methods to get acceleration? Let's just apply some good old momentum and go home. The optimization zeitgeist says we can accelerate anything!!
Our results
Turns out it's not that simple, but it is possible! We wrote a paper about it. In this post, we summarize the following results:
Applying acceleration techniques from optimization to the (full-pass) power method achieves quadratic acceleration. Specifically, we get this improvement by adding a momentum term. Naively adding momentum in the stochastic setting does notalways yield acceleration! However, we can design stochastic versions of the full-pass accelerated method that are guaranteed to be accelerated. Our novel variance analysis, based on orthogonal polynomials, proves that we can achieve acceleration for small, but non-zero, variance in the iterates. We describe two algorithms that ensure this condition holds. The first one uses a mini-batch approach, and the second one uses SVRG-style variance reduction. In the latter case the batch-size is independent of the target accuracy, ε. This means that simple methods are enough to get the optimal sample complexity and an accelerated iteration complexity.
Unlike previous methods, each iteration of our algorithms is parallelizable.Deployed in parallel, our methods achieve true acceleration in wall clock time!
The notebooks reproducing our experiments can be found here. These theoretical insights are part of a broader body of work. In YellowFin, we use this theoretical understanding of momentum to provide an automatic tuner for SGD that is competitive with state-of-the-art adaptive methods on training ResNets and LSTMs. Accelerating the Power Method
The goal of PCA is to find the top eigenvector of a symmetric positive semidefinite matrix \(A\in \mathbb R^{d\times d}\), often known as the sample covariance matrix. A popular way to find this is the power method, which iteratively runs the update \(\mathbf{w}_{t+1} = A \mathbf{w}_t\) and converges to the top eigenvector in \(\tilde{\mathcal O}(1/\Delta)\) steps, where \(\Delta\) is the eigen-gap between the top two eigenvalues of \(A\). As we mentioned earlier, this convergence is really slow if the matrix is poorly conditioned.
Adding momentum to the power method achieves quadratic acceleration.
Motivated by momentum methods used to accelerate convex optimization, we tried to apply momentum to the power method. In particular, we tried replacing the power method update with $$\mathbf{w}_{t+1} = A\mathbf{w}_t - \beta \mathbf{w}_{t-1},$$ where the extra \(\beta \mathbf{w}_{t-1}\) is the analogue of the momentum term from convex optimization. Given its similarity to known acceleration schemes, perhaps it is not surprising that with appropriate settings of \(\beta\), the momentum update converges in \(\tilde{\mathcal O}(1/\sqrt\Delta)\) steps.
The magic behind the scenes is the Chebyshev polynomials. The entire dynamics of the momentum update can be captured by the two-term recurrence of the (scaled) Chebyshev polynomials $$p_{t+1}(x) = x p_t(x) - \beta p_{t-1}(x).$$ In particular, the Chebyshev polynomial evaluated at \(\lambda_i\) gives the decay of the eigen-direction corresponding to eigenvalue \(\lambda_i\). To help illustrate the effect of momentum on eigenvectors with different eigenvalues, we show the (scaled) Chebyshev polynomials at \(t=100\) for several different momentum parameters. This lets us visually compare how quickly different eigenvectors will decay.
From this, we can see that the (scaled) Chebyshev polynomials have two regions:
a bounded region and an exponential region.In the bounded region, the Chebyshev polynomials are small, and in the exponential region, the Chebyshev polynomial grows rapidly.
For power iteration, where \(\beta = 0\), the corresponding polynomial is \(p_t(\lambda) = \lambda^t\). The recurrence reduces mass on small eigenvalues quickly. However, eigenvalues near the largest eigenvalue decay relatively slowly, yielding slow convergence.
As \(\beta\) is increased, an elbow appears in \(p_t(\lambda)\). For values of \(\lambda\) smaller than the knee, \(p_t(\lambda)\) remains small, which implies that these eigenvalues decay quickly. For values of \(\lambda\) greater than the knee, \(p_t(\lambda)\) grows rapidly, which means that these eigenvalues will remain. By selecting a \(\beta\) value that puts the elbow close to \(\lambda_2\) so that \(\lambda_1\) can stay as far as possible away from the calm region, our recurrence quickly eliminates mass on all but the largest eigenvector.
Generalizing to the Stochastic Setting
As we mentioned before, we would like our method to work for the stochastic setting, which is the streaming PCA setting. Given the simplicity of the deterministic method we just discussed, one natural option is to add momentum to the stochastic power method (also known as Oja’s algorithm), rather than the deterministic power method, leading to the recurrence To test this approach, we run this update on a randomly generated matrix.
However, this is what we get from the experiment: even after searching for a good learning rate \(\eta\), we did
NOT obtain acceleration by adding momentum! Naively adding momentum in the stochastic setting does notalways work!
What's wrong? From the experiment, we see that adding momentum does accelerate convergence to the noise ball (For a constant step size scheme, the stochastic algorithm won't converge to the true solution. Instead, it converges to a level where it fluctuates, and we refer to the fluctuation level as the noise ball). However, it also increases the size of the noise ball. We have to decrease the step size to remedy this increase in the noise ball, but this roughly cancels out the acceleration from momentum. This phenomenon has independently been observed for stochastic optimization. Now, we need to figure out what exactly is happening in the stochastic case, since we know we can obtain quadratic acceleration in the deterministic case.
A simple statistic to quantify the stochasticity of the problem is the variance of the iterates. Can we characterize the connection between the variance and momentum parameter? Yes we can! And once again, the magic bullet is the Chebyshev polynomials, which has a closed-form expression to help with the variance analysis. Specifically, the stochastic update can be reduced to analyzing the stochastic matrix sequence $$F_{t+1} = A_{t+1} F_t - \beta F_{t-1},$$ where \(A_{t}\) is an i.i.d random matrix with \(\mathbb{E}[A_t] = A\).
You have to have low variance for momentum to work in the stochastic setting.
It turns out that we can obtain an almost exact expression for the variance of each iterate \(F_t\), which we show in our work. There are two technical tricks that help the analysis:
Kronecker productto characterize the covariance of the random matrix. Since this is a linear recurrence, the second order (and in fact, any order) moment also follows a linear recurrence. closed-form expression of Chebyshev polynomials.
With these two points, we can build up the variance bound, which can then be used for any further analysis for momentum methods in the stochastic setting. Using this, we are able to show that under the condition of low variance, this momentum scheme is able to achieve acceleration.
How to Control the Variance
The variance condition for obtaining acceleration with momentum over the stochastic power method is pretty strict. However, there are several ways that the variance of the iterates can be controlled to obtain acceleration more generally.
You can get acceleration in the stochastic setting by controlling the variance.
We consider two common ways to lower the variance:
mini-batching and variance reduction.Mini-batching is a popular way to speed up computation in stochastic optimization and is embarrassingly parallelizable.By using large enough mini-batch sizes, we can guarantee acceleration from momentum.However, to achieve very accurate solutions, the required mini-batch size grows in size.
The second technique we consider is variance reduction, as in SVRG. Variance reduction makes it possible to use a batch size that does not grow when we increase accuracy. By using a small number of full passes (possible when the data is not sampled at random but available in a fixed dataset), variance reduction significantly reduces the required mini-batch size while still achieving an accelerated linear convergence.
Next Steps
Our variance analysis generalizes as a tool for other problems. In particular, these techniques should work for problems such as randomized iterative solvers for linear systems and convex quadratic optimization problems.
Our work points out two interesting directions:
Can we extend our convergence analysis to the stochastic convex optimization setting? For general convex objectives, the difficulty comes from the fact that the gradient is no longer a linear or affine function of the iteration. Can we characterize the behavior of momentum with respect to the noise of the system like in the streaming PCA setting? PCA is intrinsically a non-convex optimization problem, so can we extend the momentum scheme further to other non-convex problems where we can obtain acceleration under certain conditions?
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The alternating group $A_5 $ has two conjugacy classes of order 5 elements, both consisting of exactly 12 elements. Fix one of these conjugacy classes, say $C $ and construct a graph with vertices the 12 elements of $C $ and an edge between two $u,v \in C $ if and only if the group-product $u.v \in C $ still belongs to the same conjugacy class.
Observe that this relation is symmetric as from $u.v = w \in C $ it follows that $v.u=u^{-1}.u.v.u = u^{-1}.w.u \in C $. The graph obtained is the icosahedron, depicted on the right with vertices written as words in two adjacent elements u and v from $C $, as indicated.
Kostant writes : “Normally it is not a common practice in group theory to consider whether or not the product of two elements in a conjugacy class is again an element in that conjugacy class. However such a consideration here turns out to be quite productive.”
Still, similar constructions have been used in other groups as well, in particular in the study of the largest sporadic group, the monster group $\mathbb{M} $.
There is one important catch. Whereas it is quite trivial to multiply two permutations and verify whether the result is among 12 given ones, for most of us mortals it is impossible to do actual calculations in the monster. So, we’d better have an alternative way to get at the icosahedral graph using only $A_5 $-data that is also available for the monster group, such as its character table.
Let $G $ be any finite group and consider three of its conjugacy classes $C(i),C(j) $ and $C(k) $. For any element $w \in C(k) $ we can compute from the character table of $G $ the number of different products $u.v = w $ such that $u \in C(i) $ and $v \in C(j) $. This number is given by the formula
$\frac{|G|}{|C_G(g_i)||C_G(g_j)|} \sum_{\chi} \frac{\chi(g_i) \chi(g_j) \overline{\chi(g_k)}}{\chi(1)} $
where the sum is taken over all irreducible characters $\chi $ and where $g_i \in C(i),g_j \in C(j) $ and $g_k \in C(k) $. Note also that $|C_G(g)| $ is the number of $G $-elements commuting with $g $ and that this number is the order of $G $ divided by the number of elements in the conjugacy class of $g $.
The character table of $A_5 $ is given on the left : the five columns correspond to the different conjugacy classes of elements of order resp. 1,2,3,5 and 5 and the rows are the character functions of the 5 irreducible representations of dimensions 1,3,3,4 and 5.
Let us fix the 4th conjugacy class, that is 5a, as our class $C $. By the general formula, for a fixed $w \in C $ the number of different products $u.v=w $ with $u,v \in C $ is equal to
$\frac{60}{25}(\frac{1}{1} + \frac{(\frac{1+\sqrt{5}}{2})^3}{3} + \frac{(\frac{1-\sqrt{5}}{2})^3}{3} – \frac{1}{4} + \frac{0}{5}) = \frac{60}{25}(1 + \frac{4}{3} – \frac{1}{4}) = 5 $
Because for each $x \in C $ also its inverse $x^{-1} \in C $, this can be rephrased by saying that there are exactly 5 different products $w^{-1}.u \in C $, or equivalently, that the valency of every vertex $w^{-1} \in C $ in the graph is exactly 5.
That is, our graph has 12 vertices, each with exactly 5 neighbors, and with a bit of extra work one can show it to be the icosahedral graph.
For the monster group, the Atlas tells us that it has exactly 194 irreducible representations (and hence also 194 conjugacy classes). Of these conjugacy classes, the involutions (that is the elements of order 2) are of particular importance.
There are exactly 2 conjugacy classes of involutions, usually denoted 2A and 2B. Involutions in class 2A are called “Fischer-involutions”, after Bernd Fischer, because their centralizer subgroup is an extension of Fischer’s baby Monster sporadic group.
Likewise, involutions in class 2B are usually called “Conway-involutions” because their centralizer subgroup is an extension of the largest Conway sporadic group.
Let us define the
monster graph to be the graph having as its vertices the Fischer-involutions and with an edge between two of them $u,v \in 2A $ if and only if their product $u.v $ is again a Fischer-involution.
Because the centralizer subgroup is $2.\mathbb{B} $, the number of vertices is equal to $97239461142009186000 = 2^4 * 3^7 * 5^3 * 7^4 * 11 * 13^2 * 29 * 41 * 59 * 71 $.
From the general result recalled before we have that the valency in all vertices is equal and to determine it we have to use the character table of the monster and the formula. Fortunately GAP provides the function ClassMultiplicationCoefficient to do this without making errors.
gap> table:=CharacterTable("M"); CharacterTable( "M" ) gap> ClassMultiplicationCoefficient(table,2,2,2); 27143910000
Perhaps noticeable is the fact that the prime decomposition of the valency $27143910000 = 2^4 * 3^4 * 5^4 * 23 * 31 * 47 $ is symmetric in the three smallest and three largest prime factors of the baby monster order.
Robert Griess proved that one can recover the monster group $\mathbb{M} $ from the monster graph as its automorphism group!
As in the case of the icosahedral graph, the number of vertices and their common valency does not determine the monster graph uniquely. To gain more insight, we would like to know more about the sizes of minimal circuits in the graph, the number of such minimal circuits going through a fixed vertex, and so on.
Such an investigation quickly leads to a careful analysis which other elements can be obtained from products $u.v $ of two Fischer involutions $u,v \in 2A $. We are in for a major surprise, first observed by John McKay:
Printing out the number of products of two Fischer-involutions giving an element in the i-th conjugacy class of the monster,
where i runs over all 194 possible classes, we get the following string of numbers : 97239461142009186000, 27143910000, 196560, 920808, 0, 3, 1104, 4, 0, 0, 5, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
That is, the elements of only 9 conjugacy classes can be written as products of two Fischer-involutions! These classes are :
1A = { 1 } written in 97239461142009186000 different ways (after all involutions have order two) 2A, each element of which can be written in exactly 27143910000 different ways (the valency) 2B, each element of which can be written in exactly 196560 different ways. Observe that this is the kissing number of the Leech lattice leading to a permutation representation of $2.Co_1 $. 3A, each element of which can be written in exactly 920808 ways. Note that this number gives a permutation representation of the maximal monster subgroup $3.Fi_{24}’ $. 3C, each element of which can be written in exactly 3 ways. 4A, each element of which can be written in exactly 1104 ways. 4B, each element of which can be written in exactly 4 ways. 5A, each element of which can be written in exactly 5 ways. 6A, each element of which can be written in exactly 6 ways.
Let us forget about the actual numbers for the moment and concentrate on the orders of these 9 conjugacy classes : 1,2,2,3,3,4,4,5,6. These are precisely the components of the fundamental root of the extended Dynkin diagram $\tilde{E_8} $!
This is the content of
John McKay’s E(8)-observation : there should be a precise relation between the nodes of the extended Dynkin diagram and these 9 conjugacy classes in such a way that the order of the class corresponds to the component of the fundamental root. More precisely, one conjectures the following correspondence
This is similar to the classical McKay correspondence between finite subgroups of $SU(2) $ and extended Dynkin diagrams (the binary icosahedral group corresponding to extended E(8)). In that correspondence, the nodes of the Dynkin diagram correspond to irreducible representations of the group and the edges are determined by the decompositions of tensor-products with the fundamental 2-dimensional representation.
Here, however, the nodes have to correspond to conjugacy classes (rather than representations) and we have to look for another procedure to arrive at the required edges! An exciting proposal has been put forward recently by John Duncan in his paper Arithmetic groups and the affine E8 Dynkin diagram.
It will take us a couple of posts to get there, but for now, let’s give the gist of it : monstrous moonshine gives a correspondence between conjugacy classes of the monster and certain arithmetic subgroups of $PSL_2(\mathbb{R}) $ commensurable with the modular group $\Gamma = PSL_2(\mathbb{Z}) $. The edges of the extended Dynkin E(8) diagram are then given by the configuration of the arithmetic groups corresponding to the indicated 9 conjugacy classes! (to be continued…)7 Comments
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Background
Let $\tau$ be the tension and $\mu$ be a linear mass density (i.e., mass per unit length), then the wave equation for a string is given by:$$\partial_{tt} \psi \left(x,t\right) - \frac{ \tau }{ \mu } \partial_{xx} \psi \left(x,t\right) = 0 \tag{0}$$where $\partial_{jj} \equiv \partial^{2}/\partial j^{2}$ and $\psi \left(x,t\right)$ is a general solution to this equation, called the wave equation.
This has a simple solution of the form:$$\psi \left(x,t\right) = A \ e^{i \left( \pm \mathbf{k} \cdot \mathbf{x} \pm \omega t \right)} \tag{1}$$where $A$ is some amplitude and the phase speed of the wave is given by:$$\frac{\omega}{k} = \sqrt{\frac{ \tau }{ \mu }} \equiv C \tag{2}$$
We want to find solutions of the form $f\left( x - C \ t \right)$, but this only works for non-dispersive waves and does not work for nonlinear waves. In other words, the solution applies when the wave's phase speed is $C$ = constant.
Reflection and Transmission
First, assume $\tau$ is uniform throughout the string to avoid any unwanted acceleration. Next, let us define a general form:$$\psi_{j} \left(x,t\right) = f_{j} \left(x - v_{j} t\right) = f_{j} \left(t - \frac{x}{v_{j}} \right) \tag{3}$$where the subcript $j$ = $i$ for incident, $r$ for reflected, and $t$ for transmitted waves. Now let us assume there is some boundary at $x$ = 0 and that our string has different mass densities on either side. Let's define $\mu_{1}$ for Region 1 (-$\infty < x < 0$) and $\mu_{2}$ for Region 2 ($0 < x < \infty$). Then we have:$$\begin{align} v_{1} & = \sqrt{\frac{ \tau }{ \mu_{1} }} \tag{4a} \\ v_{2} & = \sqrt{\frac{ \tau }{ \mu_{2} }} \tag{4b}\end{align}$$
Note that the reflected wave, $\psi_{r} \left(x,t\right)$, will have a negative $v_{r}$ and thus a positive sign in the expression for $f$. Since the waves are linear, we can just write them a linear superposition of two waves for Region 1. Then we have:$$\begin{align}\psi_{1} \left(x,t\right) = \psi_{i} \left(x,t\right) + \psi_{r} \left(x,t\right) = f_{i} \left(t - \frac{x}{v_{1}} \right) + f_{r} \left(t + \frac{x}{v_{1}} \right) \tag{5a} \\\psi_{2} \left(x,t\right) = \psi_{t} \left(x,t\right) = f_{t} \left(t - \frac{x}{v_{2}} \right) \tag{5b}\end{align}$$
Boundary Conditions
There are two boundary conditions (BCs) that must be met:
The string is continuous The slope of the string is continuous
These can be written mathematically as:$$\begin{align}\psi_{1} \left(0,t\right) & = \psi_{2} \left(0,t\right) \tag{6a} \\\partial_{x} \psi_{1} \left(x,t\right) \vert_{x = 0} & = \partial_{x} \psi_{2} \left(x,t\right) \vert_{x = 0} \tag{6b}\end{align}$$where these equations can be rewritten in terms of $f_{j}$ (and integrating the second) to find:$$\begin{align}f_{i} \left(t - \frac{x}{v_{1}} \right) + f_{r} \left(t + \frac{x}{v_{1}} \right) & = f_{t} \left(t - \frac{x}{v_{2}} \right) \tag{7a} \\v_{2} \left[ \ f_{i} \left(t\right) - f_{r} \left(t\right) \right] & = v_{1} f_{t} \left(t\right) \tag{7b}\end{align}$$We can solve these two equations for $f_{r}$ and $f_{t}$ in terms of $f_{i}$ to find:$$\begin{align}f_{r} & = \left( \frac{ v_{2} - v_{1} }{ v_{1} + v_{2} } \right) \ f_{i} \tag{8a} \\f_{t} & = \left( \frac{ 2 \ v_{2} }{ v_{1} + v_{2} } \right) \ f_{i} \tag{8b}\end{align}$$
Coefficients/Amplitudes
We can see from the last two equations that the amplitudes of the reflected ($R$) and transmitted ($T$) wave are given by:$$\begin{align}R & = \left( \frac{ v_{2} - v_{1} }{ v_{1} + v_{2} } \right) = \left( \frac{ \sqrt{ \mu_{1} } - \sqrt{ \mu_{2} } }{ \sqrt{ \mu_{1} } + \sqrt{ \mu_{2} } } \right) \tag{9a} \\T & = \left( \frac{ 2 \ v_{2} }{ v_{1} + v_{2} } \right) = \left( \frac{ 2 \ \sqrt{ \mu_{1} } }{ \sqrt{ \mu_{1} } + \sqrt{ \mu_{2} } } \right) \tag{9b}\end{align}$$
Massless Ring
A massless ring
1 at one end of a string 2 is treated as a form of impedence. Because the ring is massless, we require that the net transverse (i.e., orthogonal to the direction of wave propagation, say, along the x/horizontal direction) force be zero. A finite transverse force would result in an infinite acceleration. The only difference in this case is that we need to use a non-uniform tension. So we just follow the same steps as above but use $\tau_{j}$ for Region $j$ and so we have:$$\begin{align}\tau_{1} \ \sin{\theta_{1}} & = \tau_{2} \ \sin{\theta_{2}} \tag{10a} \\\tau_{1} \ \partial_{x} \psi_{1} \left(x,t\right) \vert_{x = 0} & = \tau_{2} \ \partial_{x} \psi_{2} \left(x,t\right) \vert_{x = 0} \tag{10b}\end{align}$$where the angles, $\theta_{j}$, are relative to the x/horizontal direction. We can define the impedence as $Z_{j} = \tau_{j}/v_{j} = \sqrt{ \tau_{j} \ \mu_{j} }$, which allows us to redefine the reflection ($R$) and transmission ($T$) coefficients as:$$\begin{align}R & = \left( \frac{ Z_{1} - Z_{2} }{ Z_{1} + Z_{2} } \right) \tag{11a} \\T & = \left( \frac{ 2 \ Z_{1} }{ Z_{1} + Z_{2} } \right) \tag{11b}\end{align}$$ Massive Ring
In contrast to a massless ring, a massive ring requires an alteration of the BCs since we now need to include Newton's laws. We can assume the string applies a force and the massive ring undergoes an acceleration, allowing us to write:$$\begin{align}F & = \tau \partial_{x} \psi \left(x,t\right) \tag{12a} \\m \ a & = m \ \partial_{tt} \psi \left(x,t\right) \tag{12b}\end{align}$$Note that $F$ in Equation 12a is the vertical force on the ring due to the tension in the string, $m$ in Equation 12b is the mass of the ring, and $a$ in Equation 12b is the acceleration of the ring
3.
We can see that in the limit as $m \rightarrow 0$ we have $\partial_{x} \psi \rightarrow 0$, thus the force is null as is necessary for a massless system. We also see that as $m \rightarrow \infty$ we have $\partial_{tt} \psi \rightarrow 0$, which implies a constant velocity for the massive ring (i.e., it would be zero here as the initial condition is that it starts from rest).
Boundary Examples
Now we can provide a few useful examples:
Uniform String: $\mu_{1} = \mu_{2}$ or $v_{1} = v_{2}$ Solid(inifinite?) Wall at $x = 0$: $\mu_{2} \rightarrow \infty$ or $v_{2} = 0$ Zero mass string for $x > 0$: $\mu_{2} \rightarrow 0$ or $v_{2} = \infty$ Massless ring on vertical, frictionless pole at $x = 0$: $\tau_{2} = 0$ $\rightarrow$ $Z_{2} = 0$ Massive ring on vertical, frictionless pole at $x = 0$: $\lim_{m \rightarrow 0} \ \partial_{x} \psi = 0$ $\Rightarrow$ $R$ = 1 and $T$ = 2 $\lim_{m \rightarrow \infty} \ \partial_{tt} \psi = 0$ $\Rightarrow$ $R$ = -1 and $T$ = 0 Footnotes The ring must be massless to maintain the boundary conditions without requiring an infinite force to do so. This results because we require continuity in slope and tension at the junction. A finite mass would also result in $Z_{2} \neq 0$, as it would act like an second tension. Assume the ring is on a frictionless vertical rod. As an aside, one should note that BCs and differential equations are the primary constituents of a problem. This is relevant because the superposition rule was not used in Equations 12a and 12b in contrast to the approach used in earlier sections. The use of a superposition is just one of many possible methods one can use to solve the differential equations but is not required and may not apply in some circumstances. That is, the BCs and differential equations exist independent of whether one can apply the superposition rule. References French, A.P. (1971), Vibrations and Waves, New York, NY: W. W. Norton & Company, Inc.; ISBN:0-393-09936-9. Whitham, G.B. (1999), Linear and Nonlinear Waves, New York, NY: John Wiley & Sons, Inc.; ISBN:0-471-35942-4.
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Differentiating
I want to integrate -ln|cos(x)|+k, I and I should end with tan(x).
However, I'm stuck. I don't know what I need to do after $\displaystyle -x^{-1}*|cos(x)|-ln(x)*sin(x)$
This is how far I've gotten so far: http://img504.imageshack.us/img504/6739/tanjf9.gif Hello
Something I do not understand : the title of your post is "differentiating" and you want to integrate -ln|cos(x)|+k ?
I think that you want to differentiate -ln|cos(x)|+k
Differentiating ln(u(x)) gives u'(x) / u(x)
Here it gives sin(x)/cos(x) = tan(x) Hello, I think you have a little problem with the chain rule of differentiation.
Quote:
Originally Posted by
No Logic Sense
It is stated as below :
$\displaystyle [f(g(x))]'=g'(x)f'(g(x))$
To let you visualize it, if you let $\displaystyle t=g(x)$, it'll be :
$\displaystyle [f(t)]'=g'(x)f'(t)$
So here, what do you have ?
$\displaystyle \ln |\cos(x)|$
so $\displaystyle f(t)=\ln |\cos(x)|$
so f is the logarithm function.
and $\displaystyle t=g(x)=\cos(x)$
by using the formula, you'll have the derivative of your function :
$\displaystyle =\underbrace{(-\sin(x))}_{g'(x)} \cdot \overbrace{\frac{1}{t}}^{f'(t)}=-\sin(x) \frac{1}{\cos(x)}$
it's not formal, but it helps you understand how to use the formula.
Note that you just have to multiply the final result by -1 to get the derivative you're looking for Visual AND formal... http://www.ballooncalculus.org/mhf24.gif
As usual, straight continuous lines differentiate with respect to x and the straight dashed line with respect to the dashed balloon expression, so that the triangular network satisfies the chain rule.
Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
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AliPhysics 608b256 (608b256)
#include <AliFMDCorrNoiseGain.h>
AliFMDCorrNoiseGain () AliFMDCorrNoiseGain (const AliFMDFloatMap &map) Float_t Get (UShort_t d, Char_t r, UShort_t s, UShort_t t) const void Set (UShort_t d, Char_t r, UShort_t s, UShort_t t, Float_t x) const AliFMDFloatMap & Values ()
AliFMDFloatMap fValues
Get the noise calibration. That is, the ratio
\[ \frac{\sigma_{i}}{g_{i}k} \]
where \( k\) is a constant determined by the electronics of units DAC/MIP, and \( \sigma_i, g_i\) are the noise and gain of the \( i \) strip respectively.
This correction is needed because some of the reconstructed data (what which have an AliESDFMD class version less than or equal to 3) used the wrong zero-suppression factor. The zero suppression factor used by the on-line electronics was 4, but due to a coding error in the AliFMDRawReader a zero suppression factor of 1 was assumed during the reconstruction. This shifts the zero of the energy loss distribution artificially towards the left (lover valued signals).
So let's assume the real zero-suppression factor is \( f\) while the zero suppression factor \( f'\) assumed in the reconstruction was (wrongly) lower. The number of ADC counts \( c_i'\) used in the reconstruction can be calculated from the reconstructed signal \( m_i'\) by
\[ c_i' = m_i \times g_i \times k / \cos\theta_i \]
where \(\theta_i\) is the incident angle of the \( i\) strip.
This number of counts used the wrong noise factor \( f'\) so to correct to the on-line value, we need to do
\[ c_i = c_i' - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor \]
which gives the correct number of ADC counts over the pedestal. To convert back to the scaled energy loss signal we then need to calculate (noting that \( f,f'\) are integers)
\begin{eqnarray} m_i &=& \frac{c_i \times \cos\theta_i}{g_i \times k}\\ &=& \left(c_i' - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor\right)\frac{\cos\theta}{g_i \times k}\\ &=& \left(\frac{m_i'\times g_i\times k}{\cos\theta} - \lfloor f'\times n_i\rfloor + \lfloor f\times n_i\rfloor\right) \frac{\cos\theta}{g_i \times k}\\ &=& m_i' + \frac{1}{g_i \times k} \left(\lfloor f\times n_i\rfloor- \lfloor f'\times n_i\rfloor\right)\cos\theta\\ &=& m_i' + \frac{\lfloor n_i\rfloor}{g_i \times k} \left(f-f'\right)\cos\theta \end{eqnarray}
inline
inline
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protected
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In 1934 Witt became Helmut Hasse‘s assistent in Gottingen, where he qualified as a university lecturer in 1936. By 1938 he has made enough of a name for himself to be offered a lecturer position in Hamburg and soon became an associate professor, the down-graded position held by Emil Artin (2) until he was forced to emigrate in 1937.
A former fellow student of him in Gottingen, Erna Bannow (4), had gone earlier to Hamburg to work with Artin. She continued her studies with Witt and finished her Ph.D. in 1939. In 1940 Erna Bannow and Witt married.
So, life was smiling on Ernst Witt that sunday january 28th 1940, both professionally and personally. There was just one cloud on the horizon, and a rather menacing one. He was called up by the Wehrmacht and knew he had to enter service in february. For all he knew, he was spending the last week-end with his future wife… (later in february 1940, Blaschke helped him to defer his military service by one year).
Still, he desperately wanted to finish his paper before entering the army, so he spend most of that week-end going through the final version and submitted it on monday, as the published paper shows.
In the 70ties, Witt suddenly claimed he did discover the Leech lattice $ {\Lambda} $ that sunday. Last time we have seen that the only written evidence for Witt’s claim is one sentence in his 1941-paper Eine Identität zwischen Modulformen zweiten Grades. “Bei dem Versuch, eine Form aus einer solchen Klassen wirklich anzugeben, fand ich mehr als 10 verschiedene Klassen in $ {\Gamma_{24}} $.”
But then, why didn’t Witt include more details of this sensational lattice in his paper?
Ina Kersten recalls on page 328 of Witt’s collected papers : “In his colloquium talk “Gitter und Mathieu-Gruppen” in Hamburg on January 27, 1970, Witt said that in 1938, he had found nine lattices in $ {\Gamma_{24}} $ and that later on January 28, 1940, while studying the Steiner system $ {S(5,8,24)} $, he had found two additional lattices $ {M} $ and $ {\Lambda} $ in $ {\Gamma_{24}} $. He continued saying that he had then given up the tedious investigation of $ {\Gamma_{24}} $ because of the surprisingly low contribution
$ \displaystyle | Aut(\Lambda) |^{-1} < 10^{-18} $
to the Minkowski density and that he had consented himself with a short note on page 324 in his 1941 paper.”
In the last sentence he refers to the fact that the sum of the inverse orders of the automorphism groups of all even unimodular lattices of a given dimension is a fixed rational number, the Minkowski-Siegel mass constant. In dimension 24 this constant is
$ \displaystyle \sum_{L} \frac{1}{| Aut(L) |} = \frac {1027637932586061520960267}{129477933340026851560636148613120000000} \approx 7.937 \times 10^{-15} $
That is, Witt was disappointed by the low contribution of the Leech lattice to the total constant and concluded that there might be thousands of new even 24-dimensional unimodular lattices out there, and dropped the problem.
If true, the story gets even better : not only claims Witt to have found the lattices $ {A_1^{24}=M} $ and $ {\Lambda} $, but also enough information on the Leech lattice in order to compute the order of its automorphism group $ {Aut(\Lambda)} $, aka the Conway group $ {Co_0 = .0} $ the dotto-group!
Is this possible? Well fortunately, the difficulties one encounters when trying to compute the order of the automorphism group of the Leech lattice from scratch, is one of the better documented mathematical stories around.
The books From Error-Correcting Codes through Sphere Packings to Simple Groups by Thomas Thompson, Symmetry and the monster by Mark Ronan, and Finding moonshine by Marcus du Sautoy tell the story in minute detail.
It took John Conway 12 hours on a 1968 saturday in Cambridge to compute the order of the dotto group, using the knowledge of Leech and McKay on the properties of the Leech lattice and with considerable help offered by John Thompson via telephone.
But then, John Conway is one of the fastest mathematicians the world has known. The prologue of his book On numbers and games begins with : “Just over a quarter of a century ago, for seven consecutive days I sat down and typed from 8:30 am until midnight, with just an hour for lunch, and ever since have described this book as “having been written in a week”.”
Conway may have written a book in one week, Ernst Witt did complete his entire Ph.D. in just one week! In a letter of August 1933, his sister told her parents : “He did not have a thesis topic until July 1, and the thesis was to be submitted by July 7. He did not want to have a topic assigned to him, and when he finally had the idea, he started working day and night, and eventually managed to finish in time.”
So, if someone might have beaten John Conway in fast-computing the dottos order, it may very well have been Witt. Sadly enough, there is a lot of circumstantial evidence to make Witt’s claim highly unlikely.
For starters, psychology. Would you spend your last week-end together with your wife to be before going to war performing an horrendous calculation?
Secondly, mathematical breakthroughs often arise from newly found insight. At that time, Witt was also working on his paper on root lattices “Spiegelungsgrupen and Aufzähling halbeinfacher Liescher Ringe” which he eventually submitted in january 1941. Contained in that paper is what we know as Witt’s lemma which tells us that for any integral lattice the sublattice generated by vectors of norms 1 and 2 is a direct sum of root lattices.
This leads to the trick of trying to construct unimodular lattices by starting with a direct sum of root lattices and ‘adding glue’. Although this gluing-method was introduced by Kneser as late as 1967, Witt must have been aware of it as his 16-dimensional lattice $ {D_{16}^+} $ is constructed this way.
If Witt wanted to construct new 24-dimensional even unimodular lattices in 1940, it would be natural for him to start off with direct sums of root lattices and trying to add vectors to them until he got what he was after. Now, all of the Niemeier-lattices are constructed this way, except for the Leech lattice!
I’m far from an expert on the Niemeier lattices but I would say that Witt definitely knew of the existence of $ {D_{24}^+} $, $ {E_8^3} $ and $ {A_{24}^+} $ and that it is quite likely he also constructed $ {(D_{16}E_8)^+, (D_{12}^2)^+, (A_{12}^2)^+, (D_8^3)^+} $ and possibly $ {(A_{17}E_7)^+} $ and $ {(A_{15}D_9)^+} $. I’d rate it far more likely Witt constructed another two such lattices on sunday january 28th 1940, rather than discovering the Leech lattice.
Finally, wouldn’t it be natural for him to include a remark, in his 1941 paper on root lattices, that not every even unimodular lattices can be obtained from sums of root lattices by adding glue, the Leech lattice being the minimal counter-example?
If it is true he was playing around with the Steiner systems that sunday, it would still be a pretty good story he discovered the lattices $ {(A_2^{12})^+} $ and $ {(A_1^{24})^+} $, for this would mean he discovered the Golay codes in the process!
Which brings us to our next question : who discovered the Golay code?
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When one starts learning about physics, vectors are presented as mathematical quantities in space which have a direction and a magnitude. This geometric point of view has encoded in it the idea that under a change of basis the components of the vector must change contravariantly such that the magnitude and direction remain constant. This restricts what physical ideas may be the components of a vector (something much better explained in Feynman's Lectures), so that three arbitrary functions de not form an honest vector $\vec{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$ in some basis. So, in relativity a vector is defined "geometrically" as directional derivative operators on functions on the manifold $M$ and this implies, if $A^{\mu}$ are the components of a vector in the coordinate system $x^\mu$, then the components of the vector in the coordinate system $x^{\mu'}$ are$$A^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\mu}A^\mu$$(this all comes from the fact that the operators $\frac{\partial}{\partial x^\mu}=\partial_\mu$ form a basis for the directional derivative operators, see Sean Carrol's Spacetime and Geometry)
My problem is the fact that too many people use the coordinates $x^\mu$ as an example of a vector, when, on an arbitrary transformation, $$x^{\mu'}\neq\frac{\partial x^{\mu'}}{\partial x^\mu}x^\mu$$ I understand that this equation is true if the transformation beween the two coordinates is linear (as is the case of a lorentz transformation between cartesian coordinate systems) but I think it can´t be true in general. Am I correct in that the position does not form a four-vector? If not, can you tell me why my reasoning is flawed?
When one starts learning about physics, vectors are presented as mathematical quantities in space which have a direction and a magnitude. This geometric point of view has encoded in it the idea that under a change of basis the components of the vector must change contravariantly such that the magnitude and direction remain constant. This restricts what physical ideas may be the components of a vector (something much better explained in Feynman's Lectures), so that three arbitrary functions de not form an honest vector $\vec{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$ in some basis. So, in relativity a vector is defined "geometrically" as directional derivative operators on functions on the manifold $M$ and this implies, if $A^{\mu}$ are the components of a vector in the coordinate system $x^\mu$, then the components of the vector in the coordinate system $x^{\mu'}$ are$$A^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\mu}A^\mu$$(this all comes from the fact that the operators $\frac{\partial}{\partial x^\mu}=\partial_\mu$ form a basis for the directional derivative operators, see Sean Carrol's Spacetime and Geometry)
You are correct.
Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to $\mathbb{R}^n$, and nonlinear maps do not preserve the linear structure.
On a manifold, there is no sense in attempting to "vectorize" points. A point is a point, an element of the manifold, a vector is a vector, element of a tangent space at a point. Of course you can map points into $n$-tuples, that is part of the definition of a topological manifold, but there is no reason why the inverse of this map should carry the linear structure over to the manifold.
And now, for a purely personal opinion: While Carroll's book is really good, the physicist's way of attempting to categorize
everything by "transformation properties" is extremely counterproductive, and leads to such misunderstandings as you have overcome here. If one learns proper manifold theory, this is clear from the start...
Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place.
Co-ordinates are absolutely not vectors, they are
labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as wrong is this: in flat space ( i.e. Euclidean, Minkowski or generally signatured spaces), affine co-ordinates for positions can have two roles: they are both labels and (once one has chosen an origin) superposition weights that combine linear basis tangents to the Euclidean (Minkowski ...) manifold linearly to yield a general tangent to the manifold. If you think about it, what I have just said is a slightly different take on Uldreth's second paragraph that begins "Position is a vector ...".
It's worth saying that I definitely recall the following learning sequence as a teenager. When beginning high school at about age 11, I was first shown co-ordinates (Cartesian of course) as
labels. I suspect that this is how they are introduced to all children. I distinctly recall the idea that only two years later was the notion (that only works for Cartesian and generally affine co-ordinates) of a point's co-ordinates as a position vector introduced. Before that I had a very clear idea of a vector as a displacement or link between two points, an idea that, through the appropriate limit, leads to the tangent idea in a general manifold. On reading your question, I laugh when I recall the teacher's implying that the second role of co-ordinates as position vectors was a "new and advanced" way to look at vectors, whereas on the contrary it is a way of thinking that you correctly understand to be very limited and only workable in the affine case.
Here is a bare bones easy way to see that coordinate tuples are not 4-vectors.
Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation:
$x' = x + A $ $y' = y$ $z' = z$ $t' = t$
Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors don't change at all in this case, while the coordinate tuples do.
Right -- vectors in general relativity live in some tangent space. This is the point of differential geometry, and of calculus in general -- you approximate non-linear things, which are
not vector spaces (like curvy manifolds) with linear things (like their tangent spaces), which are vector spaces. This is exactly the motivation for defining the basis vectors as $\partial_\mu$, as you describe.
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Faddeeva Package From AbInitio
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Line 26: Line 26: :<math>\mathrm{erfi}(x) = -i\mathrm{erf}(ix) = -i[e^{x^2} w(x) - 1]</math> (imaginary error function) :<math>\mathrm{erfi}(x) = -i\mathrm{erf}(ix) = -i[e^{x^2} w(x) - 1]</math> (imaginary error function) :<math>F(x) = \frac{i\sqrt{\pi}}{2} \left[ e^{-x^2} - w(x) \right]</math> ([[w:Dawson function|Dawson function]]) :<math>F(x) = \frac{i\sqrt{\pi}}{2} \left[ e^{-x^2} - w(x) \right]</math> ([[w:Dawson function|Dawson function]]) - :<math>\mathrm{Voigt}(x,y) = \mathrm{Re}[w(x+iy)] \!</math> (real [[w:Voigt function|Voigt function]], up to a scale factor) Note that in the case of erf and erfc, we provide different equations for positive and negative ''x'', in order to avoid numerical problems arising from multiplying exponentially large and small quantities. Note that in the case of erf and erfc, we provide different equations for positive and negative ''x'', in order to avoid numerical problems arising from multiplying exponentially large and small quantities. Revision as of 22:54, 29 October 2012
Contents Faddeeva / complex error function
Steven G. Johnson has written free/open-source C++ code (with wrappers for other languages) to compute the
scaled complex error function w( z) = e − z2erfc(− iz), also called the Faddeeva function(and also the plasma dispersion function), for arbitrary complex arguments zto a given accuracy. Given the Faddeeva function, one can easily compute Voigt functions, the Dawson function, and similar related functions.
Download the source code from:
http://ab-initio.mit.edu/Faddeeva_w.cc (updated 29 October 2012) Usage
To use the code, add the following declaration to your C++ source (or header file):
#include <complex> extern std::complex<double> Faddeeva_w(std::complex<double> z, double relerr=0);
The function
Faddeeva_w(z, relerr) computes
w( z) to a desired relative error
relerr.
Omitting the
relerr argument, or passing
relerr=0 (or any
relerr less than machine precision ε≈10
−16), corresponds to requesting machine precision, and in practice a relative error < 10 −13 is usually achieved. Specifying a larger value of
relerr may improve performance (at the expense of accuracy).
You should also compile
Faddeeva_w.cc and link it with your program, of course.
In terms of
w( z), some other important functions are: (scaled complementary error function) (complementary error function) (error function) (imaginary error function) (Dawson function)
Note that in the case of erf and erfc, we provide different equations for positive and negative
x, in order to avoid numerical problems arising from multiplying exponentially large and small quantities. Wrappers: Matlab, GNU Octave, and Python
Wrappers are available for this function in other languages.
Matlab (also available here): A function
Faddeeva_w(z, relerr), where the arguments have the same meaning as above (the
relerrargument is optional) can be downloaded from Faddeeva_w_mex.cc (along with the help file Faddeeva_w.m. Compile it into an octave plugin with:
mex -output Faddeeva_w -O Faddeeva_w_mex.cc Faddeeva_w.cc GNU Octave: A function
Faddeeva_w(z, relerr), where the arguments have the same meaning as above (the
relerrargument is optional) can be downloaded from Faddeeva_w_oct.cc. Compile it into a MEX file with:
mkoctfile -DMPICH_SKIP_MPICXX=1 -DOMPI_SKIP_MPICXX=1 -s -o Faddeeva_w.oct Faddeeva_w_oct.cc Faddeeva_w.cc Python: Our code is used to provide
scipy.special.wofzin SciPy starting in version 0.12.0 (see here).
Algorithm
This implementation uses a combination of different algorithms. For sufficiently large |
z|, we use a continued-fraction expansion for w( z) similar to those described in Walter Gautschi, "Efficient computation of the complex error function," SIAM J. Numer. Anal. 7(1), pp. 187–198 (1970). G. P. M. Poppe and C. M. J. Wijers, "More efficient computation of the complex error function," ACM Trans. Math. Soft. 16(1), pp. 38–46 (1990); this is TOMS Algorithm 680.
Unlike those papers, however, we switch to a completely different algorithm for smaller |
z|: Mofreh R. Zaghloul and Ahmed N. Ali, "Algorithm 916: Computing the Faddeyeva and Voigt Functions," ACM Trans. Math. Soft. 38(2), 15 (2011). Preprint available at arXiv:1106.0151.
(I initially used this algorithm for all z, but the continued-fraction expansion turned out to be faster for larger |
z|. On the other hand, Algorithm 916 is competitive or faster for smaller | z|, and appears to be significantly more accurate than the Poppe & Wijers code in some regions, e.g. in the vicinity of | z|=1 [although comparison with other compilers suggests that this may be a problem specific to gfortran]. Algorithm 916 also has better relative accuracy in Re[ z] for some regions near the real- z axis. You can switch back to using Algorithm 916 for all z by changing
USE_CONTINUED_FRACTION to
0 in the code.)
Note that this is SGJ's
independent re-implementation of these algorithms, based on the descriptions in the papers only. In particular, we did not refer to the authors' Fortran or Matlab implementations (respectively), which are under restrictive "semifree" ACM copyright terms and are therefore unusable in free/open-source software.
Algorithm 916 requires an external complementary error function erfc(
x) function for real arguments x to be supplied as a subroutine. More precisely, it requires the scaled function erfcx( x) = e erfc( x2 x). Here, we use an erfcx routine written by SGJ that uses a combination of two algorithms: a continued-fraction expansion for large xand a lookup table of Chebyshev polynomials for small x. (I initially used an erfcx function derived from the DERFC routine in SLATEC, modified by SGJ to compute erfcx instead of erfc, by the new erfcx routine is much faster.) Test program
To test the code, a small test program is included at the end of
Faddeeva_w.cc which tests
w( z) against several known results (from Wolfram Alpha) and prints the relative errors obtained. To compile the test program,
#define FADDEEVA_W_TEST in the file (or compile with
-DFADDEEVA_W_TEST on Unix) and compile
Faddeeva_w.cc. The resulting program prints
SUCCESS at the end of its output if the errors were acceptable.
License
The software is distributed under the "MIT License", a simple permissive free/open-source license:
Copyright © 2012 Massachusetts Institute of Technology Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
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It’s been a while, so let’s include a recap : a (transitive) permutation representation of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is determined by the conjugacy class of a cofinite subgroup $\Lambda \subset \Gamma $, or equivalently, to a dessin d’enfant. We have introduced a
quiver (aka an oriented graph) which comes from a triangulation of the compactification of $\mathbb{H} / \Lambda $ where $\mathbb{H} $ is the hyperbolic upper half-plane. This quiver is independent of the chosen embedding of the dessin in the Dedeking tessellation. (For more on these terms and constructions, please consult the series Modular subgroups and Dessins d’enfants).
Why are quivers useful? To start, any quiver $Q $ defines a noncommutative algebra, the
path algebra $\mathbb{C} Q $, which has as a $\mathbb{C} $-basis all oriented paths in the quiver and multiplication is induced by concatenation of paths (when possible, or zero otherwise). Usually, it is quite hard to make actual computations in noncommutative algebras, but in the case of path algebras you can just see what happens.
Moreover, we can also
see the finite dimensional representations of this algebra $\mathbb{C} Q $. Up to isomorphism they are all of the following form : at each vertex $v_i $ of the quiver one places a finite dimensional vectorspace $\mathbb{C}^{d_i} $ and any arrow in the quiver [tex]\xymatrix{\vtx{v_i} \ar[r]^a & \vtx{v_j}}[/tex] determines a linear map between these vertex spaces, that is, to $a $ corresponds a matrix in $M_{d_j \times d_i}(\mathbb{C}) $. These matrices determine how the paths of length one act on the representation, longer paths act via multiplcation of matrices along the oriented path.
A
necklace in the quiver is a closed oriented path in the quiver up to cyclic permutation of the arrows making up the cycle. That is, we are free to choose the start (and end) point of the cycle. For example, in the one-cycle quiver
[tex]\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar[ld]^b \\ & \vtx{} \ar[lu]^c &}[/tex]
the basic necklace can be represented as $abc $ or $bca $ or $cab $. How does a necklace act on a representation? Well, the matrix-multiplication of the matrices corresponding to the arrows gives a square matrix in each of the vertices in the cycle. Though the dimensions of this matrix may vary from vertex to vertex, what does not change (and hence is a property of the necklace rather than of the particular choice of cycle) is the
trace of this matrix. That is, necklaces give complex-valued functions on representations of $\mathbb{C} Q $ and by a result of Artin and Procesi there are enough of them to distinguish isoclasses of (semi)simple representations! That is, linear combinations a necklaces (aka super-potentials) can be viewed, after taking traces, as complex-valued functions on all representations (similar to character-functions).
In physics, one views these functions as potentials and it then interested in the points (representations) where this function is extremal (minimal) : the
vacua. Clearly, this does not make much sense in the complex-case but is relevant when we look at the real-case (where we look at skew-Hermitian matrices rather than all matrices). A motivating example (the Yang-Mills potential) is given in Example 2.3.2 of Victor Ginzburg’s paper Calabi-Yau algebras.
Let $\Phi $ be a super-potential (again, a linear combination of necklaces) then our commutative intuition tells us that extrema correspond to zeroes of all partial differentials $\frac{\partial \Phi}{\partial a} $ where $a $ runs over all coordinates (in our case, the arrows of the quiver). One can make sense of differentials of necklaces (and super-potentials) as follows : the partial differential with respect to an arrow $a $ occurring in a term of $\Phi $ is defined to be the
path in the quiver one obtains by removing all 1-occurrences of $a $ in the necklaces (defining $\Phi $) and rearranging terms to get a maximal broken necklace (using the cyclic property of necklaces). An example, for the cyclic quiver above let us take as super-potential $abcabc $ (2 cyclic turns), then for example
$\frac{\partial \Phi}{\partial b} = cabca+cabca = 2 cabca $
(the first term corresponds to the first occurrence of $b $, the second to the second). Okay, but then the vacua-representations will be the representations of the quotient-algebra (which I like to call the
vacualgebra)
$\mathcal{U}(Q,\Phi) = \frac{\mathbb{C} Q}{(\partial \Phi/\partial a, \forall a)} $
which in ‘physical relevant settings’ (whatever that means…) turn out to be
Calabi-Yau algebras.
But, let us return to the case of subgroups of the modular group and their quivers. Do we have a natural
super-potential in this case? Well yes, the quiver encoded a triangulation of the compactification of $\mathbb{H}/\Lambda $ and if we choose an orientation it turns out that all ‘black’ triangles (with respect to the Dedekind tessellation) have their arrow-sides defining a necklace, whereas for the ‘white’ triangles the reverse orientation makes the arrow-sides into a necklace. Hence, it makes sense to look at the cubic superpotential $\Phi $ being the sum over all triangle-sides-necklaces with a +1-coefficient for the black triangles and a -1-coefficient for the white ones. Let’s consider an index three example from a previous post [tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\ i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\ & & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]
In this case the super-potential coming from the triangulation is
$\Phi = -aid+agd-cge+che-bhf+bif $
and therefore we have a noncommutative algebra $\mathcal{U}(Q,\Phi) $ associated to this index 3 subgroup. Contrary to what I believed at the start of this series, the algebras one obtains in this way from dessins d’enfants are
far from being Calabi-Yau (in whatever definition). For example, using a GAP-program written by Raf Bocklandt Ive checked that the growth rate of the above algebra is similar to that of $\mathbb{C}[x] $, so in this case $\mathcal{U}(Q,\Phi) $ can be viewed as a noncommutative curve (with singularities).
However, this is not the case for all such algebras. For example, the vacualgebra associated to the second index three subgroup (whose fundamental domain and quiver were depicted at the end of this post) has growth rate similar to that of $\mathbb{C} \langle x,y \rangle $…
I have an outlandish conjecture about the growth-behavior of all algebras $\mathcal{U}(Q,\Phi) $ coming from dessins d’enfants :
the algebra sees what the monodromy representation of the dessin sees of the modular group (or of the third braid group). I can make this more precise, but perhaps it is wiser to calculate one or two further examples…
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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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I have to design a feedback linearization for the model of a car with a single trailer. This picture shows the system:
And this is the kinematic model I got, where
q_dot = [x_dot; y_dot; thA_dot; thP_dot; thR_dot].
I chose
y1 = x and
y2 = y as outputs. I proved that the system is controllable and observable. To complete the exercise, then, I calculated the time derivatives of y1 and y2. The classic iteration - if I'm not wrong - says to continue the derivatives until a dependance from both inputs comes out. So I calculated the second derivatives with respect to time, the third ones and the fourth ones, until I got two functions (the fourth derivatives of y1 and y2) depending on
u1,
du1/dt,
d²u1/dt²,
d³u1/dt³ and
u2 (the second input comes out only at this step). With these results, I thought to make a system extension:
ξ1 = u1,
ξ2 = du1/dt and
ξ3 = d²u1/dt². The new inputs are
w1 = d³u1/dt³ and
w2 = u2. Finally I can write:
where
v is a new control variable.This was to explain the problem and how I got to this point. Now, my question is: how do I choose
v? It should depend on 8 variables, I suppose: q1, q2, q3, q4, q5, ξ1, ξ2, ξ3. But I don't know how.
I have to design a feedback linearization for the model of a car with a single trailer. This picture shows the system:
You used feedback linearization, so you could use techniques for linear (time invariant) systems, such as pole placement or LQR. Namely you have to have LTI state space model which is two quadruple integrators
$$ \dot{x} = \begin{bmatrix} 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & 0 & 0 & I \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 0 \\ I \\ \end{bmatrix}v, $$
with $x = \begin{bmatrix}y_1 & y_2 & \dot{y}_1 & \dot{y}_2 & \ddot{y}_1 & \ddot{y}_2 & \dddot{y}_1 & \dddot{y}_2\end{bmatrix}^\top$. You could try to find an expression for the derivatives of $y_1$ and $y_2$ as a function of $q$. If this does not have an explicit solution you could also use an observer, since if $y_1$ and $y_2$ are known the system is observable.
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I want to take the weak gradient operator
$$ \begin{aligned} \nabla: L^2(\Omega) &\to W^{-1,2}(\Omega,\mathbb{R}^d) \\ \langle \nabla u, \phi \rangle_{W^{-1,2},W_0^{1,2}}&:=(u,\text{div}\phi)_{L^2}=\int_\Omega u(x) \,\text{div}\phi(x) ~\text{d}x \end{aligned} $$ for all $\phi \in W_0^{1,2}(\Omega,\mathbb{R}^d)$.
Assume $\nabla u=0$. Show that $u$ is constant.
I know that if I interpret $u$ as a tempered distribution and take the gradient as a distributional derivative it should be possible to conclude $u$ is constant. But I'd like to take the definition of the weak gradient operator above.
We have $$ \int_\Omega u(x) \, \text{div}\phi(x) ~dx=0 \text{ for all } \phi \in W_0^{1,2}(\Omega,\mathbb{R}^d).$$
If $d=1$ I could conclude $u=\text{const}$ by the fundamental lemma of calculus of variations. But in general $$ \int_\Omega u(x) \, (\partial_{1}\phi_1(x)+...+\partial_d \phi_d(x)) ~d(x_1,...,x_d)=0 \text{ for all } \phi \in W_0^{1,2}(\Omega,\mathbb{R}^d)$$ and I don't know if there is a similar theorem.
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If all numbers are real, it is decidable. This follows, for example from the general result in the paper
Vorobʹev, N. N., Jr. Deciding the consistency of a system of inequalities...
https://link.springer.com/chapter/10.1007%2F978-1-4612-0441-1_33.
If complex numbers are allowed it is not clear to me what the answer is.Chebotarev has a generalization of Sturm's theorem to functionsof the form $P(x,\cos x,\sin x)$ where $P$ is a polynomial, but I do notknow a reference for a general result with complex $\lambda_k$. I suspect it might be wrong: with complex $\lambda$ the question might be undecidable.
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In Statistics, the researcher checks the significance of the observed result, which is known as
test static. For this test, a hypothesis test is also utilized. The P-value or probability value concept is used everywhere in the statistical analysis. It determines the statistical significance and the measure of significance testing. In this article, let us discuss its definition, formula, table, interpretation and how to use P-value to find the significance level etc. in detail. P-value Definition
The P-value is known as the probability value. It is defined as the probability of getting a result that is either the same or more extreme than the actual observations. The P-value is known as the level of marginal significance within the hypothesis testing that represents the probability of occurrence of the given event. The P-value is used as an alternative to the rejection point to provide the least significance at which the null hypothesis would be rejected. If the P-value is small, then there is stronger evidence in favour of the alternative hypothesis.
P-value Table
The P-value table shows the hypothesis interpretations:
P-value Description Hypothesis Interpretation P-value ≤ 0.05 It indicates the null hypothesis is very unlikely. Rejected P-value > 0.05 It indicates the null hypothesis is very likely. Accepted or it “fails to reject”. P-value > 0.05 The P-value is near the cut-off. It is considered as marginal The hypothesis needs more attention. P-value Formula
We Know that P-value is a statistical measure, that helps to determine whether the hypothesis is correct or not. P-value is a number that lies between 0 and 1. The level of significance(α) is a predefined threshold that should be set by the researcher. It is generally used as 0.05. The formula for the calculation for P-value is
\(z = \frac{\hat{p}-p0}{\sqrt{\frac{po(1-p0)}{n}}}\)
Step 1: Find out the test static Z is
Where,\(\hat{p}\) = Sample Proportion
P0 = assumed population proportion in the null hypothesis
N = sample size
Step 2: Look at the Z-table to find the corresponding level of P from the z value obtained. P-Value Example
An example to find the P-value is given here.
Question: A statistician wants to test the hypothesis H 0: μ = 120 using the alternative hypothesis Hα: μ > 120 and assuming that α = 0.05. For that, he took the sample values as
n =40, σ = 32.17 and x̄ = 105.37. Determine the conclusion for this hypothesis?
Solution:
We know that,\(\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}\)
Now substitute the given values\(\sigma _{\bar{x}}=\frac{32.17 }{\sqrt{40}}\) = 5.0865
Now, using the test static formula, we get
t = (105.37 – 120) / 5.0865
Therefore, t = -2.8762
Using the Z-Score table, we can find the value of P(t>-2.8762)
From the table, we get
P (t<-2.8762) = P(t>2.8762) = 0.003
Therefore,
If P(t>-2.8762) =1- 0.003 =0.997
P- value =0.997 > 0.05
Therefore, from the conclusion, if p>0.05, the null hypothesis is accepted or fails to reject.
Hence, the conclusion is “fails to reject H
0.”
Stay tuned with BYJU’S – The Learning App for related concepts on P-value and examples and explore more videos.
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The quantum mechanical treatment of the hydrogen atom can be extended easily to other one-electron systems such as \(He^+\), \(Li^{2+}\), etc. The Hamiltonian changes in two places. Most importantly, the potential energy term is changed to account for the charge of the nucleus, which is the atomic number of the atom or ion, \(Z\), times the fundamental unit of charge, \(e\). As shown in Equation \ref{8.6.1}, the energy of attraction between the electron and the nucleus increases (i.e., \(V\) gets more negative) as the nuclear charge increases.
\[ \hat {V} (r) = - \dfrac {Z e^2}{4 \pi \epsilon _0 r} \label {8.6.1}\]
The other effect is a very slight change in the reduced mass included in the kinetic energy operator. In fact, the larger the nucleus, the better the approximation that the reduced mass is given by the mass of the electron.
Exercise \(\PageIndex{1}\)
Compare the reduced mass of the \(Li^{+2}\) ion to that of the hydrogen atom.
The effects of the change in V show up in the wavefunctions and the energy eigenvalues. The expression for the energy becomes
\[ E_n = - \frac {Z^2 \mu e^4}{8 \epsilon ^2_0 h^2 n^2} = Z^2 E_{n, H} \label {8.6.2}\]
where \(E_{n, H}\) is the energy of the hydrogen atom. The forms of the wavefunctions are identical to those of the hydrogen atom, except for the fact that Z in the radial functions is no longer equal to 1. The selection rules are unchanged, and the Zeeman effect still occurs.
Exercise \(\PageIndex{2}\)
Use the orbital energy level expression in Equation \ref{8.6.2} to predict quantitatively the relative energies (in \(cm^{-1}\)) of the spectral lines for H and \(Li^{2+}\).
As the plots in Figure \(\PageIndex{1}\) reveal, the increased charge on the nucleus creates a stronger attraction for the electron and thus the electron charge density distributions shift to smaller values of r. These other systems look a lot like compressed hydrogen atoms.
Figure \(\PageIndex{1}\): Radial distribution functions plotted for the 2s orbitals of H (blue), \(He^+\) (red) and \(Li^{2+}\) (black) on the same axis, demonstrating compression of the orbital as Z is increased from 1 to 3.
Exercise \(\PageIndex{3}\)
Determine whether or not the angular momentum values, the spherical harmonic functions, and the spectroscopic selection rules that describe the electron in H are the same or are different for \(Li^{2+}\). Write a paragraph to justify your answer.
Contributors
Adapted from "Quantum States of Atoms and Molecules" by David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski
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While the verdict on a neolithic Scottish icosahedron is still open, let us recall Kostant’s group-theoretic construction of the icosahedron from its rotation-symmetry group $A_5 $.
The alternating group $A_5 $ has two conjugacy classes of order 5 elements, both consisting of exactly 12 elements. Fix one of these conjugacy classes, say $C $ and construct a graph with vertices the 12 elements of $C $ and an edge between two $u,v \in C $ if and only if the group-product $u.v \in C $ still belongs to the same conjugacy class.
Observe that this relation is symmetric as from $u.v = w \in C $ it follows that $v.u=u^{-1}.u.v.u = u^{-1}.w.u \in C $. The graph obtained is the icosahedron, depicted on the right with vertices written as words in two adjacent elements u and v from $C $, as indicated.
Kostant writes : “Normally it is not a common practice in group theory to consider whether or not the product of two elements in a conjugacy class is again an element in that conjugacy class. However such a consideration here turns out to be quite productive.”
Still, similar constructions have been used in other groups as well, in particular in the study of the largest sporadic group, the monster group $\mathbb{M} $.
There is one important catch. Whereas it is quite trivial to multiply two permutations and verify whether the result is among 12 given ones, for most of us mortals it is impossible to do actual calculations in the monster. So, we’d better have an alternative way to get at the icosahedral graph using only $A_5 $-data that is also available for the monster group, such as its character table.
Let $G $ be any finite group and consider three of its conjugacy classes $C(i),C(j) $ and $C(k) $. For any element $w \in C(k) $ we can compute from the character table of $G $ the number of different products $u.v = w $ such that $u \in C(i) $ and $v \in C(j) $. This number is given by the formula
$\frac{|G|}{|C_G(g_i)||C_G(g_j)|} \sum_{\chi} \frac{\chi(g_i) \chi(g_j) \overline{\chi(g_k)}}{\chi(1)} $
where the sum is taken over all irreducible characters $\chi $ and where $g_i \in C(i),g_j \in C(j) $ and $g_k \in C(k) $. Note also that $|C_G(g)| $ is the number of $G $-elements commuting with $g $ and that this number is the order of $G $ divided by the number of elements in the conjugacy class of $g $.
The character table of $A_5 $ is given on the left : the five columns correspond to the different conjugacy classes of elements of order resp. 1,2,3,5 and 5 and the rows are the character functions of the 5 irreducible representations of dimensions 1,3,3,4 and 5.
Let us fix the 4th conjugacy class, that is 5a, as our class $C $. By the general formula, for a fixed $w \in C $ the number of different products $u.v=w $ with $u,v \in C $ is equal to
$\frac{60}{25}(\frac{1}{1} + \frac{(\frac{1+\sqrt{5}}{2})^3}{3} + \frac{(\frac{1-\sqrt{5}}{2})^3}{3} – \frac{1}{4} + \frac{0}{5}) = \frac{60}{25}(1 + \frac{4}{3} – \frac{1}{4}) = 5 $
Because for each $x \in C $ also its inverse $x^{-1} \in C $, this can be rephrased by saying that there are exactly 5 different products $w^{-1}.u \in C $, or equivalently, that the valency of every vertex $w^{-1} \in C $ in the graph is exactly 5.
That is, our graph has 12 vertices, each with exactly 5 neighbors, and with a bit of extra work one can show it to be the icosahedral graph.
For the monster group, the Atlas tells us that it has exactly 194 irreducible representations (and hence also 194 conjugacy classes). Of these conjugacy classes, the involutions (that is the elements of order 2) are of particular importance.
There are exactly 2 conjugacy classes of involutions, usually denoted 2A and 2B. Involutions in class 2A are called “Fischer-involutions”, after Bernd Fischer, because their centralizer subgroup is an extension of Fischer’s baby Monster sporadic group.
Likewise, involutions in class 2B are usually called “Conway-involutions” because their centralizer subgroup is an extension of the largest Conway sporadic group.
Let us define the
monster graph to be the graph having as its vertices the Fischer-involutions and with an edge between two of them $u,v \in 2A $ if and only if their product $u.v $ is again a Fischer-involution.
Because the centralizer subgroup is $2.\mathbb{B} $, the number of vertices is equal to $97239461142009186000 = 2^4 * 3^7 * 5^3 * 7^4 * 11 * 13^2 * 29 * 41 * 59 * 71 $.
From the general result recalled before we have that the valency in all vertices is equal and to determine it we have to use the character table of the monster and the formula. Fortunately GAP provides the function ClassMultiplicationCoefficient to do this without making errors.
gap> table:=CharacterTable("M"); CharacterTable( "M" ) gap> ClassMultiplicationCoefficient(table,2,2,2); 27143910000
Perhaps noticeable is the fact that the prime decomposition of the valency $27143910000 = 2^4 * 3^4 * 5^4 * 23 * 31 * 47 $ is symmetric in the three smallest and three largest prime factors of the baby monster order.
Robert Griess proved that one can recover the monster group $\mathbb{M} $ from the monster graph as its automorphism group!
As in the case of the icosahedral graph, the number of vertices and their common valency does not determine the monster graph uniquely. To gain more insight, we would like to know more about the sizes of minimal circuits in the graph, the number of such minimal circuits going through a fixed vertex, and so on.
Such an investigation quickly leads to a careful analysis which other elements can be obtained from products $u.v $ of two Fischer involutions $u,v \in 2A $. We are in for a major surprise, first observed by John McKay:
Printing out the number of products of two Fischer-involutions giving an element in the i-th conjugacy class of the monster,
where i runs over all 194 possible classes, we get the following string of numbers :
97239461142009186000, 27143910000, 196560, 920808, 0, 3, 1104, 4, 0, 0, 5, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
That is, the elements of only 9 conjugacy classes can be written as products of two Fischer-involutions! These classes are :
1A = { 1 } written in 97239461142009186000 different ways (after all involutions have order two) 2A, each element of which can be written in exactly 27143910000 different ways (the valency) 2B, each element of which can be written in exactly 196560 different ways. Observe that this is the kissing number of the Leech lattice leading to a permutation representation of $2.Co_1 $. 3A, each element of which can be written in exactly 920808 ways. Note that this number gives a permutation representation of the maximal monster subgroup $3.Fi_{24}’ $. 3C, each element of which can be written in exactly 3 ways. 4A, each element of which can be written in exactly 1104 ways. 4B, each element of which can be written in exactly 4 ways. 5A, each element of which can be written in exactly 5 ways. 6A, each element of which can be written in exactly 6 ways.
Let us forget about the actual numbers for the moment and concentrate on the orders of these 9 conjugacy classes : 1,2,2,3,3,4,4,5,6. These are precisely the components of the fundamental root of the extended Dynkin diagram $\tilde{E_8} $!
This is the content of
John McKay’s E(8)-observation : there should be a precise relation between the nodes of the extended Dynkin diagram and these 9 conjugacy classes in such a way that the order of the class corresponds to the component of the fundamental root. More precisely, one conjectures the following correspondence
This is similar to the classical McKay correspondence between finite subgroups of $SU(2) $ and extended Dynkin diagrams (the binary icosahedral group corresponding to extended E(8)). In that correspondence, the nodes of the Dynkin diagram correspond to irreducible representations of the group and the edges are determined by the decompositions of tensor-products with the fundamental 2-dimensional representation.
Here, however, the nodes have to correspond to conjugacy classes (rather than representations) and we have to look for another procedure to arrive at the required edges! An exciting proposal has been put forward recently by John Duncan in his paper Arithmetic groups and the affine E8 Dynkin diagram.
It will take us a couple of posts to get there, but for now, let’s give the gist of it : monstrous moonshine gives a correspondence between conjugacy classes of the monster and certain arithmetic subgroups of $PSL_2(\mathbb{R}) $ commensurable with the modular group $\Gamma = PSL_2(\mathbb{Z}) $. The edges of the extended Dynkin E(8) diagram are then given by the configuration of the arithmetic groups corresponding to the indicated 9 conjugacy classes! (to be continued…)
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In Part II, we saw how Alice can blindly evaluate the hiding :math:`E(P(s))` of her polynomial :math:`P` of degree :math:`d`, at a point :math:`s` belonging to Bob. We called this “blind” evaluation, because Alice did not learn :math:`s` in the process.
However, there was something missing in that protocol – the fact that Alice is
able to compute :math:`E(P(s))` does not guarantee she will indeed send :math:`E(P(s))` to Bob, rather than some completely unrelated value.
Thus, we need a way to “force” Alice to follow the protocol correctly. We will explain in part IV precisely how we achieve this. In this post, we focus on explaining the basic tool needed for that – which we call here the
Knowledge of Coefficient (KC) Test.
As before, we denote by :math:`g` a generator of a group :math:`G` of order :math:`|G|=p` where the discrete log is hard. It will be convenient from this post onwards to write our group additively rather than multiplicatively. That is, for :math:`\alpha\in\mathbb{F}_p`, :math:`\alpha\cdot g` denotes the result of summing :math:`\alpha` copies of :math:`g`.
The KC Test
For :math:`\alpha\in\mathbb{F}_p^*` [1], let us call a pair of elements :math:`(a,b)` in :math:`G` an :math:`\alpha`-pair if :math:`a,b \neq 0` and :math:`b=\alpha\cdot a.`
The KC Test proceeds as follows.
Bob chooses random :math:`\alpha\in\mathbb{F}_p^*` and :math:`a\in G.` He computes :math:`b=\alpha\cdot a.` He sends to Alice the “challenge” pair :math:`(a,b).` Note that :math:`(a,b)` is an :math:`\alpha`-pair. Alice must now respond with a differentpair :math:`(a’,b’)` that is also an :math:`\alpha`-pair. Bob accepts Alice’s response only if :math:`(a’,b’)` is indeed an :math:`\alpha`-pair. (As he knows :math:`\alpha` he can check if :math:`b’=\alpha\cdot a’.)`
Now, let’s think how Alice could successfully respond to the challenge. Let’s assume for a second that she
knew :math:`\alpha.` In that case, she could simply choose any :math:`a’` in :math:`G,` and compute :math:`b’=\alpha\cdot a’;` and return :math:`(a’,b’)` as her new :math:`\alpha`-pair.
However, as the only information about :math:`\alpha` she has is :math:`\alpha\cdot a` and :math:`G` has a hard discrete log problem, we expect that Alice cannot find :math:`\alpha.`
So how can she successfully respond to the challenge without knowing :math:`\alpha?`
Here’s the natural way to do it: Alice simply chooses some :math:`\gamma\in\mathbb{F}_p^*,` and responds with :math:`(a’,b’)=(\gamma\cdot a,\gamma\cdot b).`
In this case, we have:
:math:`b’=\gamma \cdot b = \gamma \alpha \cdot a = \alpha (\gamma\cdot a) =\alpha \cdot a’,`
so indeed :math:`(a’,b’)` is an :math:`\alpha`-pair as required.
Note that if Alice responds using this strategy, she
knows the ratio between :math:`a` and :math:`a’`. That is, she knows the coefficient :math:`\gamma` such that :math:`a’=\gamma\cdot a.`
The Knowledge of Coefficient Assumption [2] (KCA) states that
this is always the case, namely: KCA: If Alice returns a valid response :math:`(a’,b’)` to Bob’s challenge :math:`(a,b)` with non-negligible probability over Bob’s choices of :math:`a,\alpha`, then she knows :math:`\gamma` such that :math:`a’=\gamma\cdot a.`
The KC Test and Assumption will be important tools in Part IV.
What does “Alice knows” mean exactly
You may wonder how we can phrase the KCA in precise mathematical terms; specifically, how do we formalize the notion that “Alice knows :math:`\gamma`” in a mathematical definition?
This is done roughly as follows: We say that, in addition to Alice, we have another party which we call
Alice’s Extractor. Alice’s Extractor has access to Alice’s inner state.
We then formulate the KCA as saying that: whenever Alice successfully responds with an :math:`\alpha`-pair :math:`(a’,b’),` Alice’s Extractor outputs :math:`\gamma` such that :math:`a’=\gamma\cdot a.` [3]
[1] :math:`\mathbb{F}_p^*` denotes the non-zero elements of :math:`\mathbb{F}_p`. It is the same as :math:`\mathbb{Z}_p^*` described in Part I.
[2] This is typically called the Knowledge of Exponent Assumption in the literature, as traditionally it was used for groups written multiplicatively.
[3] The fully formal definition needs to give the Extractor “a little slack” and states instead that the probability that Alice responds successfully but the Extractor does not output such :math:`\gamma` is negligible.
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Fierz-Pauli action can be written as
$$ \tag{1} S = \int d^Dx \; \frac{1}{2} h_{\mu \nu} \mathcal{O}^{\mu \nu, \alpha \beta}h_{\alpha \beta} $$
where the operator
$$ \tag{2} \mathcal{O}^{\mu \nu} {}_{\alpha \beta} = \left( \eta^{( \mu}{}_{\alpha}\eta^{\nu )}{}_{\beta} - \eta^{\mu \nu} \eta_{\alpha \beta}\right) \left(\Box - m^2 \right) - 2\partial^{(\mu}\partial_{(\alpha}\eta^{\nu )}{}_{\beta )} + \partial^{\mu}\partial^{\nu}\eta_{\alpha \beta} + \partial_{\alpha}\partial_{\beta}\eta^{\mu \nu} $$
satisfies the symmetries
$$ \tag{3} \mathcal{O}^{\mu \nu , \alpha \beta} = \mathcal{O}^{\nu \mu , \alpha \beta} = \mathcal{O}^{\mu \nu , \beta \alpha} = \mathcal{O}^{\alpha \beta , \mu \nu}. $$
Now, to find the propagator one rewrites this operator in momentum space as
$$ \tag{4} \mathcal{O}^{\mu \nu}{}_{\alpha \beta}(\partial \to ip) = -\left( \eta^{( \mu}{}_{\alpha}\eta^{\nu )}{}_{\beta} - \eta^{\mu \nu} \eta_{\alpha \beta}\right) \left(p^2 + m^2 \right) \\ + 2p^{(\mu}p_{(\alpha}\eta^{\nu )}{}_{\beta )} - p^{\mu}p^{\nu}\eta_{\alpha \beta} - p_{\alpha}p_{\beta}\eta^{\mu \nu}. $$
I know that propagator $ \mathcal{D}_{\alpha \beta, \sigma \lambda}$ satisfies the same symmetries above so it will have the same structure as $\mathcal{O}^{\mu \nu, \alpha \beta}$. I also know that their multiplication should give me the identity operator, namely
$$ \tag{5} \mathcal{O}^{\mu \nu, \alpha \beta}\mathcal{D}_{\alpha \beta, \sigma \lambda} = \frac{i}{2}\left(\delta^{\mu}_{\sigma} \delta^{\nu}_{\lambda} + \delta^{\nu}_{\sigma}\delta^{\mu}_{\lambda}\right). $$
To find the propagator, I should assume a form for it, put that form into the above equation and find the coefficients. My question is,
what should be the form of the propagator in that case?
I suppose, after determining the form of the propagator, calculations will be lengthy. So, before trying something wrong and wasting my time, I want to learn the correct form (the correct assumption) for this propagator. I'll be glad if anybody can help.
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Một số bài tập hình học vi phân
Đã gửi 04-11-2006 - 09:11
of latitude on http://dientuvietnam...imetex.cgi?S^2.
Proof:
No loss of generality, we can consider [the latitude tex]c(t)$ to be the
curve http://dientuvietnam...mimetex.cgi?c(t) is the intersection of http://dientuvietnam...mimetex.cgi?S^2
with the hyperplane z=r).
Let http://dientuvietnam...mimetex.cgi?V(t)=(a(t),b(t),c(t)) to
be the parallel transport of http://dientuvietnam...mimetex.cgi?V_0 on http://dientuvietnam...mimetex.cgi?S^2 with http://dientuvietnam...mimetex.cgi?V(0)=V_0. We know
that the tangent vector of http://dientuvietnam...mimetex.cgi?c(t) is http://dientuvietnam...mimetex.cgi?S^2 at the
point http://dientuvietnam...mimetex.cgi?c(t) is http://dientuvietnam.net/cgi-bin/mimetex.cgi?n(t) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?c'(t), we can choose a basis for http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_{c(t)}S^2 as http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(t)=a(t)u(t)+b(t)w(t). By Problem 4a) we have http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dV}{dt}
must have the same direction as n(t) so we have the equation
V'(t)=K(t)n(t) or equivalently
http://dientuvietnam.net/cgi-bin/mimetex.cgi?-b'(t)\sqrt{1-r^2}=K(t)r. Comparing the
x and y coordinates we get
http://dientuvietnam.net/cgi-bin/mimetex.cgi?rb'(t)-a(t)=K(t)\sqrt{1-r^2},a'(t)+rb(t)=0.
Substituting http://dientuvietnam.net/cgi-bin/mimetex.cgi?K(t)=-\dfrac{\sqrt{1-r^2}}{r}b'(t) we get a linear
system of ODEs:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?a'(t)=-rb(t),b'(t)=ra(t).
Solving this system we get http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(0)=V_0 we get
http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(0,1,0)-A(r,0,-\sqrt{1-r^2})=V_0.
The geometric meaning of the "từ cấm" is clear from the above
decompose of http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(t) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?V_0 in to the directions of u(t) and
w(t).
The Buddha
Đã gửi 10-11-2006 - 08:46
do Carmo.)
Problem 1/Chapter 3: We have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?X the first column of http://dientuvietnam...ex.cgi?f'(v)\not=0 and http://dientuvietnam...-f^2=|f'|^2 which
is a contradiction to http://dientuvietnam.../mimetex.cgi?U. If cos u=0 then
http://dientuvietnam.net/cgi-bin/mimetex.cgi?u"=-2\dfrac{ff'}{f^2}u'v' and
http://dientuvietnam.net/cgi-bin/mimetex.cgi?v"(|f'|^2+|g'|^2)=ff'|u'|^2-(f'f"+g'g")|v'|^2 we can see that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?r=|f(v)| hence $r\cos (\beta (t))=\dfrac{u'(t)f^2}{|\gamma
'(t)|}$. Hence since we showed that http://dientuvietnam...9;f^2=constant. But we have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{d}{dt}(u'f^2(v(t)))=u"f^2+2u'v'ff'=0.
d) From the proof of c) we have in this case
http://dientuvietnam.net/cgi-bin/mimetex.cgi?u'(t)>0.
Remark: Here do Carmo require for a geosedic with time t in http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{C_1^2}{v^2}+v'^2(1+4v^2)=C_2,u'=\dfrac{C_1}{v^2}.
From the first equation we see that except the case where v(t) is
a constant there exists K>0 such that for |t|>K we have v'(t)
does not change its sign. In particular we may assume that v(t) is
strictly increasing for |t|>K (in fact it can be shown from the equation that
). We can show also along with |t|. Henc the two branches for t>0 and t<0 cut infinitely many of times.
The Buddha
Đã gửi 10-11-2006 - 13:48
Đã gửi 13-11-2006 - 07:44
d) Since http://dientuvietnam.net/cgi-bin/mimetex.cgi?\rightarrow) We just choose in the condition above t=0. Then
since http://dientuvietnam.net/cgi-bin/mimetex.cgi?\leftarrow) If http://dientuvietnam...a_{ij} a_{ji}=0 for every i,j iff A is
anti-symmetric.
c) Since f is an isometry, there is a 1-1 correspondence between
vector fields in M and vector fields in N. Moreover, we have
(see proof of Problem 3/Chapter 2)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?[X_n,X_i]=[X_n,X_j]=0, above equation is equivalent to
, and we are
done.
b) By definition (see page 70 of do Carmo), showing that X is
tangent to the geodesic spheres centered at q is equivalent to
showing that \gamma (t))>=0" [/tex] for any geodesics
with . We have
\gamma (t))>_{\gamma
(t)}=<\dfrac{D}{dt}\gamma '(t),\gamma (t))>_{\gamma (t)}+<\gamma
'(t),\dfrac{D}{dt}\gamma (t))>_{\gamma (t)}=0.
" [/tex]
This is because since is a geodesics, and d).
Hence . At t=0 we have
The Buddha
Đã gửi 13-11-2006 - 23:30
Đã gửi 19-11-2006 - 02:49
Problem 6/Chapter 3: If M is connected, http://dientuvietnam...n/mimetex.cgi?X is a Killing field, and X(q)= 0, http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{d}{ds}[A.B(s)]=A.\dfrac{d}{ds}B(s). This
thing shows that http://dientuvietnam...mimetex.cgi?p_n to show that http://dientuvietnam...tex.cgi?X|_V=0. In particular,
Now A is not empty and both open and closed in a connected space
M, hence A=M.
Bài viết đã được chỉnh sửa nội dung bởi
toilachinhtoi: 19-11-2006 - 02:53
The Buddha
Đã gửi 21-11-2006 - 11:48
Problem: Let M be a compact manifold of dimension n. Let N_1 and N_2 be two closed totally geodesic submanifolds of M, with dimensions n_1 and n_2 satisfying n_1+n_2>n. Prove that N_1 and N_2 has a non-empty intersection.
The Buddha
Đã gửi 02-01-2007 - 03:19
$d(ins(X)\nu )(p)=(div~X)\nu (p)$. For this end we choose a geodesic
frame $E_1,...,E_n$ at p. We also arrange that $E_1,...,E_n$ is
the (local) orientation for M . Then
$\nu (E_1,...,E_n)\equiv 1.$
Now we can write $X(q)=\sum _{i=1}^nf_i(q)E_i(q)$. Then for
$Z_i=(E_1,\ldots ,E_{i-1},\widehat{E_i},E_{i+1},\ldots E_n)$ we have
$ins(X)\nu (Z_i)(q)=\sum _{j=1}^nf_j\nu (E_j,Z)(q)=\sum
_{j=1}^{n}(-1)^{j-1}f_j\nu (E_1,\ldots ,E_{i-1},E_j,E_{i+1},\ldots
,E_n)(q)=(-1)^{i-1}f_i(q).$
The formula above shows that $ins(X)\nu =\sum
_{i=1}^n(-1)^{i+1}f_i\theta _i$ as hinted by do Carmo (we use
$(-1)^{i-1}=(-1)^{i+1}$). Then
$d(ins (X)\nu )&=&\sum _{i=1}^n(-1)^{i+1}df_i\wedge \theta _i+\sum
_{i=1}^n(-1)^{i+1}f_i\wedge d\theta _i=\sum _{i=1}^n(E_if_i)\nu +\sum _{i=1}^n(-1)^{i+1}f_i d\theta _i.
$
Now we show that $d\theta _i(p)=0$ for any i. We need to show this
only for $\theta _n$. We have
$d\theta _n=\sum _{j=1}^{n}(-1)^{j-1}\omega _1\wedge \ldots \wedge
d\omega _i\wedge \ldots \wedge \omega _{n-1}.$
Hence to show that $d\theta _n(p)=0$ it suffices to show that
$d\omega _i(p)=0$ for any i.
Claim: If $\omega$ is a 1-form then $[d\omega ](X,Y)(q)=X[\omega
(Y)]-Y[\omega (X)]-\omega ([X,Y])$. Proof of Claim: We write locally
$\omega =\sum _{i=1}^nf_idu _i$ where $du _1,...,du _n$ is the dual
basis for $\dfrac{\partial}{\partial u_1},...,\dfrac{\partial
}{\partial u_n}$. Then $d\omega =\sum _{i=1}^n\sum
_{j=1}^n\dfrac{\partial f_i}{\partial x_j}du _j\wedge du _i$. Now we
need only to check the Claim for the case
$X=g\dfrac{\partial}{\partial u_i},Y=h\dfrac{\partial}{\partial u_j}$.
For this case we have
$[d\omega ](g\dfrac{\partial}{\partial u_i},h\dfrac{\partial}{\partial
u_j})&=&gh[\dfrac{\partial f_j}{\partial x_i}-\dfrac{\partial
f_i}{\partial x_j}],g\dfrac{\partial}{\partial u_i}[\omega (h\dfrac{\partial}{\partial
u_j})]&=&g\dfrac{\partial}{\partial u_i}[hf_j]=gh\dfrac{\partial
f_j}{\partial u_i}-gf_j\dfrac{\partial h}{\partial u_i},
h\dfrac{\partial}{\partial u_j}[\omega (g\dfrac{\partial}{\partial
u_i})]&=&h\dfrac{\partial}{\partial u_j}[gf_i]=gh\dfrac{\partial
f_i}{\partial u_j}-hf_i\dfrac{\partial g}{\partial u_j},
\omega ([g\dfrac{\partial}{\partial u_i},h\dfrac{\partial}{\partial
u_j}])&=&\omega (g\dfrac{\partial h}{\partial
u_i}\dfrac{\partial}{\partial u_j}-h\dfrac{\partial g}{\partial
u_j}\dfrac{\partial}{\partial u_i})=gf_j\dfrac{\partial h}{\partial
u_i}-hf_i\dfrac{\partial g}{\partial u_j}.
$
These equations complete the proof of the Claim.
Now we continue the proof of Problem 11. By the Claim, the
properties of Riemannian connection and geodesic frame at p we
have
$[d\omega _i](E_j,E_k)(p)&=&E_j[\omega _i(E_k)]-E_k[\omega
_i(E_j)]-\omega [E_j,E_k](p)=E_j[\delta _{ik}]-E_k[\delta
_{ij}]-\omega (\nabla _{E_j}E_k(p)
&&-\nabla _{E_k}E_j(p))=0-0-\omega (0)=0.
$
Hence by Problem 8a), at p
$d(ins(X)\nu )(p)=(\sum _{i=1}E_i(f_i)(p))\nu =div~X(p)\nu .
$
Problem 12/Chapter 3 (Maximum principle): If $f:M\rightarrow R $ is subharmonic, where M is connected, compact without boundary then f is a constant.
By Stokes theorem, since M has no boundary, we have
$\int _{M}d\omega =\int _{\partial M}\omega =\int _{\emptyset}\omega
=0$
for any $(n-1)$-form.
For each function g if we define $Y=grad~g$ then by definition and
Problem 11 we have $(\Delta g)\nu =(div ~Y)\nu =d(ins(Y)\nu )$.
Hence by above observation we have
$\int _M(\Delta g)\nu =\int _Md(ins(Y)\nu )=0$
for any function g.
Now if f is a smooth function such that $\Delta f\geq 0$, then
since
$\int _M(\Delta f)\nu =0$
by above observation, we have that $\Delta f\equiv 0$. Substituting
this into formula of Problem 9b) we have $\Delta (f^2)=2f\Delta
f+2|grad f|^2= 2|grad f|^2$. Hence
$2\int _M|grad f|^2\nu =\int _M\Delta f^2\nu =0,$
which implies that $|grad f|\equiv 0$. Since M is connected, we
have that f must be a constant.
The Buddha
Đã gửi 06-01-2007 - 08:57
Proof: Fixed $p\in M$. We will prove that $divG(p)=0.$
First we construct the normal coordinates as hinted by do Carmo. We
choose a normal neighborhood $U=exp_p(B_{\epsilon }(0))$ of p.
Then since we have the exponential map $exp_{p}:~B_{\epsilon
}(0)\subset T_pM\rightarrow U$ is a diffeomorphism, we can use
$(u_1,...,u_n)$ to be the coordinates for a point
$q=exp_p(u_1e_1+...+u_ne_n)$ where $e_1,...e_n$ is a normal basis of
$T_pM.$ We compute the vector fields $\dfrac{\partial }{\partial
u_i}.$ To this end, we choose a function $f:~U\rightarrow \RR.$ Then
we have, where $q=exp _p(v),$
$\dfrac{\partial }{\partial u_i}=\dfrac{\partial }{\partial u_i}(f\circ
exp _{p}(v+te_i))|_{t=0}=Tf\circ (Texp _p)_{v}\circ e_i.$
This shows that $\dfrac{\partial }{\partial u_i}|_q=(Texp
_p)_{v}\circ e_i.$
Now we can see that $\Gamma _{ij}^k(p)=0$ for this coordinate
because any geodesic starting from p will be written by a linear
system of ODEs (the geodesic will be a straight line in this
coordinate)
$\dfrac{d^2x_k}{dt^2}=0.$
Comparing this with the equation (1) page 62 at the start point p
we see that
$\sum _{i,j}\Gamma _{ij}^k(p)v_iv_j=0$
for any initial velocity $(v_1,...,v_n)=\dfrac{dx}{dt}(0).$ Then we
must have $\Gamma _{ij}^k(p)=0$ for any i,j,k.
$\Gamma _{ij}^k(p)=0$ (If $\Gamma _{ij}^k(p)$ were not 0 then the
solution to this system $(1)$ will not be a straight line, for we
can choose an initial $dx(0)=v$ such that ).
We denote by $X_i=\dfrac{\partial }{\partial u_i},Y_j=\dfrac{\partial
}{\partial v_j}$ as tangent vectors of TM (We abuse the same
$X_1,...,X_n$ for either vectors in $T_pM$ and tangent vectors of
$TM$). Consider a point $(q,w)\in T_qM$ with $w=w_1X_1+...+w_nX_n.$
If we take the curve $\alpha (t)=(q(t),w(t))$ with $q(0)=q,
q'(0)=X_i,w(0)=v,w'(0)=0$ then $\alpha '(0)=X_i.$ Moreover we have
$\dfrac{Dw}{dt}(0)=\nabla _{X_i}v.$
If we take the curve $\alpha (t)=(q,w(t))$ with $w(0)=v,w'(0)=X_j$
then $\alpha '(0)=Y_j.$ Moreover we have $\dfrac{Dw}{dt}(0)=X_j.$
Hence for the natural metric we have
$<X_i,X_j>^{S}_{q,v}=<X_i,X_j>_q+<\nabla _{X_i}v,\nabla _{X_j}v>_q,<X_i,Y_j>^{S}_{q,v}=<\nabla _{X_i}v,X_j>_q,
<Y_i,Y_j>^{S}_{q,v}=<X_i,X_j>_q.$
For the product metric we have
$<X_i,X_j>^{P}_{q,v}=<Y_i,Y_j>_{q,v}=<X_i,X_j>_q,<X_i,Y_j>^{P}_{q,v}=0.$
If we define $\tilde{X_i}=(X_i,\nabla _{X_i}v),\tilde{Y_j}=(0,X_j)$
then we have
$<X_i,X_j>^{S}_{q,v}=<\tilde{X_i},\tilde{X_j}>^{P}_{q,v},<X_i,Y_j>^{S}_{q,v}=<\tilde{X_i},\tilde{Y_j}>^{P}_{q,v},
<Y_i,Y_j>^{S}_{q,v}&=&<\tilde{Y_i},\tilde{Y_j}>^{P}_{q,v}.$
(Here we use super-script S and P for natural and product
metric). We can find some $n\times n$ matrix J (depending on
q,v) such that $\tilde{X_i}=A.X_i,\tilde{Y_i}=A.Y_i$ where A is
the linear map represented by the matrix $(2n)\times (2n)$ with the
form
$A=\left (\begin{array}{ll}I_n&J\&I_n\end{array}\right ),$
with $I_n$ the $n\times n$ identity matrix. We see that $det (A)=1.$
This show that in the product metric
$vol(X_1,...,X_n,Y_1,...,Y_n)=vol(\tilde{X_i},...,\tilde{X_n},\tilde{Y_1},...,\tilde{Y_n}).$
Thus we have the volume form in natural metric and in product metric
is the same.
Now we can proceed as in do Carmo. We compute divG in the product
metric. Using formula (1') page 62 in the text book, we see that
$G(u_i)=v_i,~G(v_j)=-\sum _{ik}\Gamma _{ik}^jv_iv_k.$
Since in product metric, $\nabla _{X_i}Y_j\equiv 0,$ and at the
point p we has shown that $\nabla _{X_i}X_j(p)=\nabla
_{Y_iY_j}=0(p),$ we have $\Gamma _{ij}^k(p)=0$ in the product
metric. Hence, since $\Gamma _{ij}^k$ independent of $v_1,...,v_n$
we have as computed in do Carmo that $divG=0.$
Now fix p as above and let $f (t,u,v)$ be the geodesic flow. If we
write $f (t,u,v)=(f_1(t,u,v),...,f_{2n}(t,u,v),)$ then we have
$0=div G(t,u,v)=\sum _{i=1}^{2n}\dfrac{\partial\dfrac{df_i}{dt}}{\partial x_i}=\dfrac{d}{dt}[div ~f].$
Hence $div ~f$ is a constant with respect to t. At $t=0$ we have
f is the identity map, thus $div ~f(0,u,v)=2n$ for all u,v.
Hence $div~f(t,u,v)=2n=\mbox{dimension of }TM$ for all $t,u,v.$
The Buddha
Đã gửi 13-01-2007 - 10:20
a) Let $\varphi (s,p)$ be the flow of X, that is $\dfrac{d\varphi }{ds}=X\circ \varphi$. From definition, since X is a Killing field
we have that $\varphi (s,.)$ is an isomorphism for every s. Now we consider $f(t,s)=\varphi (s,\gamma (t))$. Since $\gamma$ is a
geodesic, for any s we have $f(s,.)$ is a geodesic. Hence
$X\circ \gamma (t)=X\circ \varphi (0,\gamma (t))=\dfrac{\partial f}{\partial s}(0,t)$
is a Jacobian field.
b) Since M is connected, to show that X=0 we need only to show that for every $p\in M$, X=0 in a neighborhood of p. Fixed $p\in M$.
Choose U to be a totally normal neighborhood of p. To show
that X=0 in U, we need to show only that X=0 on any geodesic
emanating from p. Define $J(t)=X\circ \gamma (t)$. Then J is a
Jacobian field with $J(0)=X\circ \gamma (0)=X(p)=0$, and
$J'(0)=\nabla _{\gamma '(0)}X(p)=0$. Hence by the uniqueness of
Jacobian fields we have $J\equiv 0$.
The Buddha
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— Written by Triangles on April 01, 2017 • ID 52 —
A collection of practical tips and tricks to improve the gradient descent process and make it easier to understand.
Introduction to machine learning — What machine learning is about, types of learning and classification algorithms, introductory examples.
Linear regression with one variable — Finding the best-fitting straight line through points of a data set.
The gradient descent function — How to find the minimum of a function using an iterative algorithm.
The gradient descent in action — It's time to put together the gradient descent with the cost function, in order to churn out the final algorithm for linear regression.
Multivariate linear regression — How to upgrade a linear regression algorithm from one to many input variables.
Introduction to classification and logistic regression — Get your feet wet with another fundamental machine learning algorithm for binary classification.
The cost function in logistic regression — Preparing the logistic regression algorithm for the actual implementation.
The problem of overfitting in machine learning algorithms — Overfitting makes linear regression and logistic regression perform poorly. A technique called "regularization" aims to fix the problem for good.
Real-world data can come up in different orders of magnitude. For example, your age ranges from 0 to 100 years, while your yearly income from €10,000 to €10,000,000 (and more). Using such unprocessed data as input features for a linear regression system might slow down the gradient descent algorithm to a crawl.
It happens because — as we will see shortly — such not normalized data warps the cost function the gradient descent has to process, making the minimum point really difficult to reach.
Because of that, an important trick in machine learning and in linear regression is to make sure that all the input features are
on a similar scale. This is a preparatory step you do in order to optimize the input data, known as feature scaling.
In feature scaling you basically normalize your input values. For example, say you have two features:
Below you will find a contour plot for the cost function [texi]J(\theta_1, \theta_2)[texi] as if we were using the raw, unprocessed values. As you may see the result is a very thin and stretched version of it. The gradient descent algorithm would oscillate a lot back and forth, taking a long time before finding its way to the minimum point.
With feature scaling we will bring back the original bowl-shaped figure in order to let the gradient descent algorithm do its job efficiently. You have to options here:
min-max scaling or standardization.
The idea is to get every input feature into approximately a [texi][-1,1][texi] range. The name comes from the use of [texi]\min[texi] and [texi]\max[texi] functions, namely the smallest and greatest values in your dataset. It requires dividing the input values by the range (i.e. the maximum value minus the minimum value) of the input variable:
[tex] x_i^{\prime} = \frac{x_i - \min(x_i)}{\max(x_i) - \min(x_i)} [tex]
where [texi]x_i[texi] is the original [texi]i[texi]-th input value, [texi]x_i^{\prime}[texi] is the normalized version.
For example, say I'm dealing with the yearly income [texi]x_1[texi] and in particular I want to normalize the value of $30,000:
[tex] x_1^{\prime} = \frac{30,000 - 10,000}{10,000,000 - 10,000} \approx 0.002 [tex]
Just rinse and repeat such normalization for every value in your dataset. Of course if you are in a multivariate scenario remember to skip feature [texi]x_0[texi], since [texi]x_0 = 1[texi] as seen in the previous episode.
This technique goes also under the name of
z-score normalization and many other confusing aliases I wish I could forget. In brief, you transform your data set so that the values follow the property of a normal distribution, namely with mean 0 ([texi]\mu = 0[texi]) and standard deviation 1 ([texi]\sigma = 1[texi]). Unlike min-max scaling, with standardization you are thinking in terms of how many standard deviations a value is far from the mean of the entire data set.
The general formula for standardization:
[tex] x_i = \frac{x_i - \mu_i}{\sigma_i} [tex]
Following the links to my previous articles above I'm able to compute the mean and the standard deviation on my data set. I'll show you an example with the yearly income [texi]x_1[texi]. Suppose I have collected five samples: $10,000, $30,000, $32,000, $35,000, $150,000. The mean and the standard deviation are:
[tex] \begin{equation} \mu_1 = 51,400 \qquad \sigma_1 \approx 50,078 \end{equation} [tex]
Now, let's apply the standardization to the value of $30,000 as I did before with the min-max scaling:
[tex] x_1 = \frac{30,000 - 51,400}{50,078} \approx -0.4 [tex]
You can read it as [texi]-0.4[texi] standard deviations ([texi]-0.4\text{STD}[texi]) from the mean.
Rinse and repeat the procedure for every value in your dataset as for the min-max scaling, and remember to skip [texi]x_0[texi] in multivariate problems.
Using standardization is important when you are comparing measurements that have different units, like years and dollars. It is also a general requirement for many machine learning algorithms besides linear regression. As a rule of thumb I'd say: when in doubt, just standardize the data, it shouldn't hurt.
You want to know if the gradient descent is working correctly. Since the job of the gradient descent is to find the value of [texi]\theta[texi]s that minimize the cost function, you could plot the cost function itself (i.e. its output) and see how it behaves as the algorithm runs.
The image below shows what I mean. The number of iterations on the horizontal axis, the cost function output on the vertical one. On each iteration the gradient descent churns out new [texi]\theta[texi]s values: you take those values and evaluate the cost function [texi]J(\theta)[texi]. You should see a descending curve if the algorithm behaves well: it means that it's minimizing the value of [texi]\theta[texi]s correctly.
More generally, the gradient descent works properly when [texi]J(\theta)[texi] decreases after every iteration.
Plotting [texi]J(\theta)[texi] also tells you whether or not the gradient descent has converged. Different problems require different number of iterations until convergence, so in general you can assume that the algorithm has found a minimum when [texi]J(\theta)[texi] decreases less than some small value [texi]\epsilon[texi] in one iteration.
Choosing a proper value [texi]\epsilon[texi] is not an easy task. Some people set it to value [texi]10^-3[texi] and also automatize the task in what is called
automatic convergence test: their algorithm stops when [texi]J(\theta)[texi] has decreased less than [texi]\epsilon[texi] in one iteration.
If your [texi]J(\theta)[texi] plot seen before starts to look weird — upward curves, dramatically slow decreasing, ... — the gradient descent is not working properly: it is time to fix [texi]\alpha[texi], by using a
smaller value.
It has been proved mathematically that for sufficiently small [texi]\alpha[texi], [texi]J(\theta)[texi] decreases on every iteration. On the other hand if [texi]\alpha[texi] is too small the gradient descent can be slow to converge.
The rule of thumb here is to try a range of [texi]\alpha[texi] values. Start with [texi]\alpha = 0,001[texi] and look at the [texi]J(\theta)[texi] plot. Does it decrease properly and rapidly? You are done with it. Otherwise, switch to [texi]\alpha = 0,01[texi] ([texi]\times 10[texi] scale), rinse and repeat until the algorithm works fine.
Machine Learning @ Coursera -
Gradient Descent in Practice I - Feature Scaling (link) Machine Learning @ Coursera - Gradient Descent in Practice II - Learning Rate (link) Wikipedia - Feature Scaling (link) Sebastianraschka.com - About Feature Scaling and Normalization (link)
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In Wikipedia about difference equations, there is some description about correspondence between ODE and difference equation:
If you consider the Taylor series of the solution to a linear differential equation:$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$ you see that the coefficients of the series are given by the nth derivative of $f(x)$ evaluated at the point $a$. The differential equation provides a linear difference equation relating these coefficients.
The rule of thumb (for equations in which the polynomial multiplying the first term is non-zero at zero) is that:$$ y^{[k]} \to f[n+k]$$and more generally $$x^m*y^{[k]} \to n(n-1)(n-m+1)f[n+k-m]$$
Example: The recurrence relationship for the Taylor series coefficients of the equation: $$(x^2 + 3x -4)y^{[3]} -(3x+1)y^{[2]} + 2y = 0\,$$ is given by$$n(n-1)f[n+1] + 3nf[n+2] -4f[n+3] -3nf[n+1] -f[n+2]+ 2f[n] = 0\, $$ or $$-4f[n+3] +2nf[n+2] + n(n-4)f[n+1] +2f[n] = 0.\,$$
My questions are:
I was wondering what the rationale behind this conversion from an ODE to a difference equation is? Although having tried to read it several times, I was not able to understand it. In reverse direction, can a difference equation be converted to an ODE using this correspondence? How to? Is the conversion of an ODE into a difference equation in numerical methods for solving an ODE related to the correspondence between the two mentioned above?
This equivalence can be used to quickly solve for the recurrence relationship for the coefficients in the power series solution of a linear differential equation.
Problems generally solved using the power series solution method taught in normal differential equation classes can be solved in a much easier way.
I was wondering how exactly the correspondence can make solving an ODE or a difference equation easier?
Thanks and regards!
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I suppose the right way to do C (charge), T (time reversal), P(parity) transformation on the state $\hat{O}| v \rangle$ with operators $\hat{O}$ is that:
$$C(\hat{O}| v \rangle)=(C\hat{O}C^{-1})(C| v \rangle)\\P(\hat{O}| v \rangle)=(P\hat{O}P^{-1})(P| v \rangle)\\T(\hat{O}| v \rangle)=(T\hat{O}T^{-1})(T| v \rangle)$$
Thus to understand how an operator $\hat{O}$ transforms under C,P,T, we care about the following form$$\hat{O} \to (C\hat{O}C^{-1})\\\hat{O} \to (P\hat{O}P^{-1})\\\hat{O} \to (T\hat{O}T^{-1})$$
Here $\hat{O}=\hat{O}(\hat{\Phi},\hat{\Psi},a,a^\dagger)$ contains possible field operators ($\hat{\Phi},\hat{\Psi}$), or $a,a^\dagger$ etc.
To understand how a state $|v \rangle$ transforms, we care about$$| v \rangle\to C| v \rangle\\| v \rangle \to P| v \rangle\\| v \rangle\to T| v \rangle$$
However, in Peskin and Schroeder QFT book, throughout Chap 3, the transformation is done on the fermion field $\hat{\Psi}$(operator in the QFT) :$$\hat{\Psi} \to (C\hat{\Psi}C)? (Eq.3.145)\\\hat{\Psi} \to (P\hat{\Psi}P)? (Eq.3.128)\\\hat{\Psi} \to (T\hat{\Psi}T)? (Eq.3.139)$$
I suppose one should take one side as inverse operator ($(C\hat{\Psi}C^{-1}),(P\hat{\Psi}P^{-1}),(T\hat{\Psi}T^{-1})$). What have been written there in Peskin and Schroeder QFT Chap 3 is incorrect, especailly because $T \neq T^{-1}$, and $T^2 \neq 1$ in general. ($T^2=-1$ for spin-1/2 fermion)
Am I right?(P&S incorrect here) Or am I wrong on this point? (Why is that correct? I suppose S. Weinberg and M. Srednicki and A Zee use the way I described.) This post imported from StackExchange Physics at 2014-06-04 11:39 (UCT), posted by SE-user Idear
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Normal Distribution Introduction
The first formal model in these notes happened so fast, you might have missed it. By assuming $X_n \sim_{iid} \text{Bernoulli}(p)$, we created a single model. This one statistical model assumed the Bernoulli distribution. Our data consisted of multiple independent observations from the identical distribution (iid), a Bernoulli distribution with unknown population parameter $p$.
In this section, we change the assumed distribution to the Normal distribution. Because the support for the Normal distribution is all real numbers, this distribution applies to data that could potentially take on any value in the real line. We complete the section by rehearsing our use of the likelihood function as it applies to Normal data. When, out in the real world on your own, if don’t know what model to apply, assume normality.
Warm Up numeric: a type of variable that represents numbers, meaningfully mathable values. For example, adult heights. continuous random variable: a random variable that can take on any value within a given range. For example, a random varible that measures time. density plot: a visualization of an approximation of the probability density function. standardize: a mathematical trick to make unitless a dataset. standard normal distribution: a instance of the normal distribution where the mean is $0$ and the standard deviationis $1$. sampling distribution: the name of the distribution of a statistic. Central Limit Theorem: a theorem that says the statistic the mean looks normal if the sample size is big enough. Density Plot
It’s common to assume any single numerical variable data follows a Normal distribution, $ Y_n \sim \text{Normal}(\mu, \sigma)$ for $n = 1, 2, \ldots, N$. Note that we will call this a formal model although I understand that it is difficult to separate the ideas of distribution and model at this point. For now, let’s push forward and begin as we ought to begin any analysis, plot the data.
import numpy as npimport pandas as pdfrom scipy.stats import norm as normalimport bplot as bpbp.LaTeX()bp.dpi(300)
Let’s return to our dataset about the Order Carnivora. Consider the variable
BW, which records birth weight in grams. These are numeric data and it’s reasonable to assume the data came from a continuous random variable, because grams can theoretically take on any positive value in a reasonable range for birth weights from the Order Carnivora.
Since our dataset has missing values, recorded as
NaNs, let’s first drop them. Next we’ll approximate the density function from which these data may have come from. A density plot is an approximation, from the available data, of the probability density function itself. A common alternative plot is a histogram.
carnivora = pd.read_csv("https://raw.githubusercontent.com/roualdes/data/master/carnivora.csv")bw = carnivora['BW'].dropna()bp.density(bw)bp.histogram(bw)bp.labels(x='Birth weight (g)', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x11cd38ac8>
The above plot tells us that the majority of our data consist of observations below $500$ grams with a few observations showing up sporadically above $500$. How many observations are above $500$ is not immediately clear, so let’s see if we can modify the plot above to help us visualize all of the data.
bw = carnivora['BW'].dropna()bp.density(bw)bp.rug(bw)bp.labels(x='Birth weight (g)', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x11cde0128>
Using vectorization, try to produce code that counts the number of observations greater than $500$. And then turn this into a proportion. What does this proportion represent, in terms of a Bernoulli distribution?
Properties of the Normal Distribution
If $Y \sim \text{Normal}(\mu, \sigma)$, then one can standardize the random variable $Y$ by subtracting off the mean $\mu$ and scaling by the standard deviation $\sigma$,
This linear transformation of a random variable that follows a normal distribution is so common, that the random variable $Z$ has a special name. A random variable that follows the $\text{Normal}(0, 1)$ distribution is called standard Normal; $Z$ follows a standard Normal distribution with mean $\mu = 0$ and variance $\sigma = 1$.
The probability density function of the $\text{Normal}(\mu, \sigma)$ distribution is
A plot of the standard Normal probability density function is displayed below. Try to change the code to help you better understand that the $\text{Normal}(\mu, \sigma)$ distribution indexes an uncountable number of distributions via $\mu$ and $\sigma$. For each specific choice of $(\mu, \sigma)$, think of it as an instantiation of a new random variable.
x = np.linspace(-4, 4, 101)mu = 0; sigma = 1;fx = normal.pdf(x, loc=mu, scale=sigma)bp.curve(x, fx)bp.labels(x='$x$', y='$f(x)$', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x11cde09b0>
Likelihood Example
To estimate the population mean birth weight of animals from the Order Carnivora $\mu$, we’ll assume a Normal distribution, $Y_n \sim_{iid} \text{Normal}(\mu, \sigma)$. Find the MLE of $\mu$ and then $\sigma$ by hand.
Example
Find the MLE of $(\mu, \sigma)$ using a computer.
Assumed Normality
Notice from the plot above of birth weights from the Order Carnivora, the data don’t obviously come from a Normal distribution, and yet we modeled these data with a Normal distribution. Such an assumption doesn’t offend nearly any statistician, and yet it’s almost offensive that no statistician is bothered by this. In this subsection, we’ll explore why statisticians are often happy to assume normality.
Statisticians are not often bothered by assuming Normal data, because they are trained to not think about data statically. The randomness of data, despite the data appearing to be fixed quantities, comes from imagining that the process that produced these data could be repeated (even if it can’t). In the case of a random sample of animals from the Order Carnivora, this would mean that you could (but wouldn’t) randomly sample a new set of data from the same population of animals from the Order Carnivora.
Recall that earlier in these notes, we already saw this idea. Our operational definition of probability is the limitting relative frequency of repeating the process an infinite number of times. We are now expanding on this idea to imagine that an entire new dataset comes from each iteration.
Statisticians recognize that collecting new data is unlikely to happen, but based on our operational definition of probability, this theoretical resampling is just the natural, logical extension. Since each new dataset comes from the same population, we’d assume the Normal distribution for each new dataset. Based on the model $Y_n \sim \text{Normal}(\mu, \sigma)$, the likelihood dictates that we use the sample mean to estimate $\mu$. The punchline to all of this is that statisticians can prove mathematically the shape of the multiple estimates of $\mu$ that would come about based on this infinite resampling.
Just like we can use a computer to approximate the probability a coin flip turning up heads to be $1/2$, we can simulate the shape of multiple esimates of $\mu$ from a population of animals from the Order Carnivora. Let’s look at the code and a plot, and then I’ll explain what’s going on.
N = bw.sizeR = 1001mus = np.full((1001, ), np.nan)seed = np.random.seed(1234)for r in range(R): idx = np.random.choice(bw._index, N) mus[r] = bw[idx].mean() bp.density(mus)bp.rug(mus)x = np.linspace(np.min(mus), np.max(mus), 101)fx = normal.pdf(x, loc=bw.mean(), scale=bw.std()/np.sqrt(N))bp.curve(x, fx, color='tab:orange')bp.labels(x='Mean birth weight (g)', y='Density', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x11d8976a0>
Let’s explain the code above in English, before we dive into the details of what just happened statistcally. We are interested in birth weights from animals of the Order Carnivora, but we are not interested in missing values, encoded as
NaNs in Python. We use the library
pandas to select only specific variables we are currently interested in, namely
BW, and then drop the
NaNs.
The next paragraph (if you will) of code, stores the sample size and the number of resamples to take. The number of resamples is analogous to how many coin flips you want to simulate. In this case,
R is specifically how many new data sets you want to simulate. Next, we preallocate a chunk of memory to hold our
R estimates of $\mu$. Preallocation should be necessary in statically typed languages and remains an incredibly good idea in any language that allows it.
The for loop iterates
R times. In each iteration, we randomly sample integers that correspond the index of specific observations in our vector of data, recognizing that the variable
bw came from the dataframe
carnivora. Sampling by index seems like a pain right now, but it is more memory efficient and we’ll learn in the future that is the more robust solution. Each loop creates a new vector of indices
idx, that we use to index our original vector of data and then calculate the sample mean. By the end of the for loop, we have
R estimates of $\mu$.
The plot displays a density plot of the multiple estimates of $\mu$. This density plot (in blue) graphically represents an estimate of the sampling distribution of the sample mean. The sampling distribution of an estimator is the theoretical distribution for the collection of statistics one would obtain if they infinitely resampled from the population of interest. In the scenario above, we resampled and calculated the statistic the sample mean. Because we resampled only
R times, this is an approximation of the sampling distribution for the sample mean.
Pay particular attention to the fact that our original data is not Normal, but the collection of multiple estimates of $\mu$ are in fact nearly Normal. This phenomenon, named the Central Limit Theorem is a classic result of mathematical statistics. For an arbitrary population distribution with finite variance, the sampling distribution for the sample mean will be approximately $\text{Normal}(\mu, \sigma^2/N)$. Notice the square root of the sample size in the denominator of the standard deviation. This says that as the sample size tends to infinity, the standard deviation of the sampling distribution will collapse on the true population mean $\mu$; when you collect all the individuals from the population mean, you will know the population mean, no more estimating. The plot above also contains the Central Limit Theorem approximation (in orange) to the sampling distribution of the sample mean.
Sampling Distributions
Recall that a statistic is any function calculated from a set of random variables (data). By applying the logic above, all statistics have sampling distributions. These sampling distributions come about by imagining infinitely resampling from the same population and calculating the same statistic on each new sample.
It takes about a semester to accept this fact. You need to keep in mind that each statistic, calculated from what you once thought was a static dataset, is now to be thought of as a single random variable. Each statistic is a function applied to random variables. Because the arguments to the function are random variables, the output of the function, the statistic, is a random variable. Since random variables follow distributions, the statistic follows a distribution. The name of this distribution is the sampling distribution.
Even after you accept that every statistic should be thought of as a random variable, it still doesn’t quite help you imagine the sampling distribution for that statistic. It’s hard to imagine sampling distributions for statistics, because we don’t know their shape. We don’t know their shape and yet we know they exist, so they are given a general name, the sampling distribution. The key points that you should keep in mind are that sampling distributions
exist, are different for each statistic, and are easiest to remember when you’ve grasped how they come about; data are not static. Central Limit Theorem
A more formal definition of the Central Limit Theorem goes like this. Assume $X_n \sim_{iid} F(\theta)$ for $n = 1, \ldots, N$ where $\mathbb{E}(X) = \mu$ and $\mathbb{V}(X) = \sigma^2 < \infty$. Let $\hat{\mu}$ denote the sample mean. Then
In English, we’d read the sampling distribution of the sample mean approaches a normal distribution with mean $\mu$ and standard deviation $\mathbb{D}(X)/\sqrt{N}$ as the sample size increases, so long as the population from which the independent data were sampled has finite variance.
There are a few important facts about the Central Limit Theorem above. As stated above, the Central Limit Theorem
depends on unknown population parameters, $\mu$ and $\sigma$, is an approximation of the sampling distribution of the sample mean that depends on the sample size $N$, and doesn’t tell us about other statistics than the sample mean.
Notice that I didn’t claim that the finiteness of the variance of the population as an important fact. This assumption is generally a reasonable assumption that most applied statisticians are willing to accept.
In most introductory statistics courses, the majority of the course material is based around the Central Limit Theorem. The Central Limit Theorem is a mathematical theorem, with it’s own assumptions, but by focusing on it, a course is making assumptions about the student’s future use of statistics. Using only statistics that follow the CLT limits students application of statistics. On the other hand, statisticians maintain their jobs by showing that statistics other than the sample mean follow the Central Limit Theorem, and there’s been no shortage of papers on this topic.
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The classical series $e = \lim_{n \to \infty} X_n$ where $X_n = \sum_{k=0}^n 1/k!$ is incredibly efficient. But is it known to be the most efficient series in terms of denominators for using fractions in the sum? In other words, is it known whether there is another series of fractions $e = \lim_{n \to \infty} Y_n$ where $Y_n = \sum_{k=0}^n a_k/b_k$ ($b_k > 0$, $a_k,b_k$ integers) where $\lim_{k \to \infty} b_k/k! \leq 1$ and $\lim_{n \to \infty} |(Y_n - e)/(X_n -e)| < 1$?
The classical formula $X_n = \sum_{k=0}^{k=n} \frac{1}{k!}$ was used by A. Yee, in 2010, to calculate the first 500 billion digits of e. Thus I guess using $X_n$ is still the state of art method to calculate $e$.
I have a vague reason for $X_n$ being a really good choice to calculate $e$. Observe the following , $\frac{1}{10!} = 2.7557319e^{-7}$, $\quad$ $\frac{1}{11!} = 2.5052108e^{-8}$, $\quad$$\frac{1}{12!} = 2.0876756e^{-9}$
As $k$ increases, the largest decimal value which $1/k!$ can effect increases. Therefore, one can calculate the first $l$ decimal places accurately by computing $X_n$ for finite $n$. Also, $n$ scales linearly with respect to $l$
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I came across John Duffield Quantum Computing SE via this hot question. I was curious to see an account with 1 reputation and a question with hundreds of upvotes.It turned out that the reason why he has so little reputation despite a massively popular question is that he was suspended.May I ...
@Nelimee Do we need to merge? Currently, there's just one question with "phase-estimation" and another question with "quantum-phase-estimation". Might we as well use just one tag? (say just "phase-estimation")
@Blue 'merging', if I'm getting the terms right, is a specific single action that does exactly that and is generally preferable to editing tags on questions. Having said that, if it's just one question, it doesn't really matter although performing a proper merge is still probably preferable
Merging is taking all the questions with a specific tag and replacing that tag with a different one, on all those questions, on a tag level, without permanently changing anything about the underlying tags
@Blue yeah, you could do that. It generally requires votes, so it's probably not worth bothering when only one question has that tag
@glS "Every hermitian matrix satisfy this property: more specifically, all and only Hermitian matrices have this property" ha? I though it was only a subset of the set of valid matrices ^^ Thanks for the precision :)
@Nelimee if you think about it it's quite easy to see. Unitary matrices are the ones with phases as eigenvalues, while Hermitians have real eigenvalues. Therefore, if a matrix is not Hermitian (does not have real eigenvalues), then its exponential will not have eigenvalues of the form $e^{i\phi}$ with $\phi\in\mathbb R$. Although I'm not sure whether there could be exceptions for non diagonalizable matrices (if $A$ is not diagonalizable, then the above argument doesn't work)
This is an elementary question, but a little subtle so I hope it is suitable for MO.Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$.The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form:$$ J = \begin...
@Nelimee no! unitarily diagonalizable matrices are all and only the normal ones (satisfying $AA^\dagger =A^\dagger A$). For general diagonalizability if I'm not mistaken onecharacterization is that the sum of the dimensions of the eigenspaces has to match the total dimension
@Blue I actually agree with Nelimee here that it's not that easy. You get $UU^\dagger = e^{iA} e^{-iA^\dagger}$, but if $A$ and $A^\dagger$ do not commute it's not straightforward that this doesn't give you an identity
I'm getting confused. I remember there being some theorem about one-to-one mappings between unitaries and hermitians provided by the exponential, but it was some time ago and may be confusing things in my head
@Nelimee if there is a $0$ there then it becomes the normality condition. Otherwise it means that the matrix is not normal, therefore not unitarily diagonalizable, but still the product of exponentials is relatively easy to write
@Blue you are right indeed. If $U$ is unitary then for sure you can write it as exponential of an Hermitian (time $i$). This is easily proven because $U$ is ensured to be unitarily diagonalizable, so you can simply compute it's logarithm through the eigenvalues. However, logarithms are tricky and multivalued, and there may be logarithms which are not diagonalizable at all.
I've actually recently asked some questions on math.SE on related topics
@Mithrandir24601 indeed, that was also what @Nelimee showed with an example above. I believe my argument holds for unitarily diagonalizable matrices. If a matrix is only generally diagonalizable (so it's not normal) then it's not true
also probably even more generally without $i$ factors
so, in conclusion, it does indeed seem that $e^{iA}$ unitary implies $A$ Hermitian. It therefore also seems that $e^{iA}$ unitary implies $A$ normal, so that also my argument passing through the spectra works (though one has to show that $A$ is ensured to be normal)
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
@Blue fair enough - as with @Semiclassical I was thinking about it with the t parameter, as that's what we care about in physics :P I can possibly come up with a number of non-Hermitian matrices that gives unitary evolution for a specific t
Or rather, the exponential of which is unitary for $t+n\tau$, although I'd need to check
If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.Then$$\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.$$Now observe that $e^U$ is upper ...
There's 15 hours left on a bountied question, but the person who offered the bounty is suspended and his suspension doesn't expire until about 2 days, meaning he may not be able to award the bounty himself?That's not fair: It's a 300 point bounty. The largest bounty ever offered on QCSE. Let h...
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Re: Use [tex] and [/tex] tags NOT [math] and [/math] tags for latex.
$\displaystyle \[\sqrt {{b^2} - 4ac} \]$
Re: Use [tex] and [/tex] tags NOT [math] and [/math] tags for latex.
am quite new to this forum and i have little idea on how to use the editor but please this is my question
Please am stuck with the integration I kindly need assistance in solving this equation, I will really appreciate if anyone could help with the solution or a guide on what i should do
[tex]C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz[\tex]
Re: Use [tex] and [/tex] tags NOT [math] and [/math] tags for latex.
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]
Re: Use [tex] and [/tex] tags NOT [tex] and [/tex] tags for latex.
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:
Quote:
Originally Posted by
Loser66
$\displaystyle \int_1^e \dfrac{1}{x}dx=lne-ln1=1$
Re: Use [tex] and [/tex] tags NOT [tex] and [/tex] tags for latex.
Re: Use [tex] and [/tex] tags NOT [tex] and [/tex] tags for latex.
\$\int_1^e \dfrac{1}{x}dx=lne-ln1=1\$ using the dollar sign gives a better rendering.
Quote:
Originally Posted by
Mr.MathType
$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
Re: Use [tex] and [/tex] tags NOT [tex] and [/tex] tags for latex.
Actually I just noticed something else that would make it look better: Code the log as \ln, not simply ln. So now we have either $\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$ or
$\displaystyle \int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$,
depending on whether it's inline or not.
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Answer
$A=\frac{1}{2} \times a \times b$ square inch.
Work Step by Step
By using the formula $A=\frac{1}{2} \times a \times b \times sin\gamma$ to calculate the area of the triangle, we can see that we have to maximize the $sin\gamma$. The maximum is 1 where $\gamma = 90^{\circ}$ So the maximum area of the triangle is $A=\frac{1}{2} \times a \times b \times 1$.
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L Hospital rule is a general method of evaluating indeterminate forms such as 0/0 or ∞/∞. To evaluate the limits of indeterminate forms for the derivatives in calculus, L Hospital’s rule is used. L’ hospital’s rule states that
If \(\lim_{x\rightarrow c}f(x)= \lim_{x\rightarrow c}g(x)=0 \ or\pm \infty\) and \(\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\), then \(\lim_{x\rightarrow c}\frac{f(x)}{g(x)}= \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\)
Note: The limit of the quotient of function is equivalent to the limit of the quotient of their derivatives, given that the provided conditions are satisfied. L Hospital Rule proof
By the use of Extended Mean Value Theorem or Cauchy’s Mean Value Theorem, the L Hospital rule can be proved.
If f and g are two continuous functions on the interval [a, b] and differentiable on the interval (a, b), the
f’(c)/g’(c) = [f(b)-f(a)]/[g(b)-g(a)] such that c belong to (a, b).
Assume that the two functions f and g are defined on the interval (c, b) in such a way that f(x)→0 and g(x)→0, as x→c
+.
But we have f’(c) / g’(c) tends to finite limits. The functions f and g are differentiable, and f’(x) and g’(x) exists on the set [ c, c+k], and also f’ and g’ are continuous on the interval [c, c+k] provided with the conditions f(c)= g(c) = 0 and g’(c) ≠ 0 on the interval [c, c+k].
By Cauchy Mean Value Theorem states that there exists c
k∈ (c, c+k), such that
f’(c
k)/g’(c k) = [f(c+k)-f(c)]/[g(c+k)-g(c)] = f(c+k)/g(c+k)
Now, k→0
+ ,
While, \(\lim_{k\rightarrow 0^{+}}\frac{f(c+k)}{g(c+k)}=\lim_{x\rightarrow 0^{+}}\frac{f(x)}{g (x)}\)
So, we have \(\lim_{x\rightarrow c^{+}}\frac{f(x)}{g(x)}= \lim_{x\rightarrow c^{+}}\frac{f'(x)}{g'(x)}\)
Hence, L Hospital rule for x→c
+ is proved. Similarly, we can prove it for x→c – . Indeterminate Forms
Using L Hospital rule, we can solve the problem in 0/0, ∞/∞, ∞ – ∞, 0 x ∞, 1∞, ∞
0, or 0 0 forms. These forms are known as indeterminate forms. To remove the indeterminate forms in the problem, we can use L Hospital rule. L Hospital Rule Examples
Some examples of L Hospital rule are given below:
Example 1:
Evaluate \(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x)
Solution:
Given:\(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x)
Differentiate the above form, we get
= \(\lim_{x\rightarrow 0}\) (2 cos x – 2 cos 2x) / (1 – cos x)
= \(\lim_{x\rightarrow 0}\) (-2 sin x + 4 sin 2x) / (sin x)
= \(\lim_{x\rightarrow 0}\) (-2 cos x + 8 cos 2x) / (cosx)
Now substitute the limit,
= -2+8/1 = 6/1 = 6
Therefore, \(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x) = 6.
Example 2:
Evaluate \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x
Solution:
Given: \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x
= \(\lim_{x\rightarrow 0}\) 3cos 3x /4 cos 4x
Now substitute the limit,
= 3 cos 0/ 4 cos 0
= ¾
Therefore, \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x = ¾.
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Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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It is messy because you have misunderstood the problem. While $q(\underline{v})$ is induced by the bilinear form $f(\underline{u}, \underline{v})=\underline{v}^TA\underline{u}$, where $A$ is your $3\times 3$ coefficient matrix, $q$ is quadratic, not bilinear, also not a linear transformation. So, what you are asked to do is to find a decomposition of the form $A = P^TDP$ (where $P$ is invertible and the diagonal of $D$ does not necessarily contain any eigenvalue of $A$), but you have confused this with an eigenvalue decomposition $A = P^{-1}DP$. Surely, as your matrix $A$ is real symmetric, you can do
both by performing an orthogonal decomposition $A=Q^TDQ$ where $QQ^T=I$ and $D$ contains the eigenvalues of $A$, but this is simply not required.
In general, you can find a decomposition $A = P^TDP$ by using elementary row/column operations. This is somewhat akin to finding a row-reduced echelon form of a matrix, but here we need to perform both an elementary row operation and a corresponding elementary column operation at each step. In other words, if, in a certain step, you multiply $A$ by an elementary matrix $E$ on the left, you should also mutiply $A$ by $E^T$ on the right.
For the problem you describe, however, simple inspection plus some completing-square trick is enough. Note that$$
\begin{eqnarray}
&&x_1^2 + x_2^2 + 9x_3^2 + 2x_1x_2 - 6x_1x_3 - 5x_2x_3\\
&=&(x_1 + x_2 - 3x_3)^2 + x_2x_3\\
&=&(x_1 + x_2 - 3x_3)^2 + \frac14[(x_2 + x_3)^2 - (x_2 - x_3)^2].
\end{eqnarray}
$$So you may take $B=\{(x_1 + x_2 - 3x_3),\ (x_2 + x_3),\ (x_2 - x_3)\}$. You may verify that $A = P^TDP$ where$$
P=\begin{pmatrix}
1&1&-3\\0&1&1\\0&1&-1
\end{pmatrix},
\ D=\begin{pmatrix}
1\\&\frac14\\&&-\frac14
\end{pmatrix}.
$$
|
We obtain a partial solution of the problem on the growth of the norms of exponential functions with a continuous phase in the Wiener algebra. The problem was posed by J.-P. Kahane at the International Congress of Mathematicians in Stockholm in 1962. He conjectured that (for a nonlinear phase) one can not achieve the growth slower than the logarithm of the frequency. Though the conjecture is still not confirmed, the author obtained first nontrivial results.
Let Γ be an arithmetic group of affine automorphisms of the n-dimensional future tube
T. It is proved that the quotient space T/Γ is smooth at infinity if and only if the group Γ is generated by reflections and the fundamental polyhedral cone (“Weyl chamber”) of the group dΓ in the future cone is a simplicial cone (which is possible only for n ≤ 10). As a consequence of this result, a smoothness criterion for the Satake–Baily–Borel compactification of an arithmetic quotient of a symmetric domain of type IV is obtained.
To an arbitrary involutive stereotype algebra
A the continuous envelopeoperation assigns its nearest, in some sense, involutive stereotype algebra Env C A so that homomorphisms to various C*-algebras separate the elements of Env C A but do not distinguish between the properties of A and those of Env C A.
If A is an involutive stereotype subalgebra in the algebra
C( M) of continuous functions on a paracompact locally compact topological space M, then, for C( M) to be a continuous envelope of A, i.e., Env C A = C( M), it is necessary but not sufficient that A be dense in C( M). In this note we announce a necessary and sufficient condition for this: the involutive spectrum of A must coincide with M up to a weakening of the topology such that the system of compact subsets in M and the topology on each compact subset remains the same.
It is well--known that certain properties of continuous functions on the circle T, related to the Fourier expansion, can be improved by a change of variable, i.e., by a homeomorphism of the circle onto itself. One of the results in this area is the Jurkat--Waterman theorem on conjugate functions, which improves the classical Bohr--P\'al theorem. In the present work we propose a short and technically very simple proof of the Jurkat--Waterman theorem. Our approach yields a stronger result.
Asymptotic properties of products of random matrices ξ k = X k …X 1 as k → ∞ are analyzed. All product terms X i are independent and identically distributed on a finite set of nonnegative matrices A = {A 1, …, A m }. We prove that if A is irreducible, then all nonzero entries of the matrix ξ k almost surely have the same asymptotic growth exponent as k→∞, which is equal to the largest Lyapunov exponent λ(A). This generalizes previously known results on products of nonnegative random matrices. In particular, this removes all additional “nonsparsity” assumptions on matrices imposed in the literature.We also extend this result to reducible families. As a corollary, we prove that Cohen’s conjecture (on the asymptotics of the spectral radius of products of random matrices) is true in case of nonnegative matrices.
This work is motivated by the observation that the character of an irreducible gl
n-module (a Schur polynomial), being the sum of exponentials of integer points in a Gelfand–Tsetlin polytope, can be expressed by using Brion’s theorem. The main result is that, in the case of a regular highest weight, the contributions of all nonsimplicial vertices vanish, while the number of simplicial vertices is n! and the contributions of these vertices are precisely the summands in Weyl’s character formula.
We give an alternative proof of the main result of [1]; the proof relies on Brion’s theorem about convex polyhedra. The result itself can be viewed as a formula for the character of the Feigin-Stoyanovsky subspace of an integrable irreducible representation of the affine Lie algebra widehatsln(C). Our approach is to assign integer points of a certain polytope to vectors comprising a monomial basis of the subspace and then compute the character by using (a variation of) Brion’s theorem.
The degenerate Lie group is a semidirect product of the Borel subgroup with the normal abelian unipotent subgroup. We introduce a class of the highest weight representations of the degenerate group of type A, generalizing the PBW-graded representations of the classical group. Following the classical construction of the flag varieties, we consider the closures of the orbits of the abelian unipotent subgroup in the projectivizations of the representations. We show that the degenerate flag varieties $\Fl^a_n$ and their desingularizations $R_n$ can be obtained via this construction. We prove that the coordinate ring of $R_n$ is isomorphic to the direct sum of duals of the highest weight representations of the degenerate group. In the end, we state several conjectures on the structure of the highest weight representations.
The Thoma cone is a certain infinite-dimensional space that arises in the representation theory of the infinite symmetric group. The present note is a continuation of a paper by A. M. Borodin and the author (Electr. J. Probab. 18 (2013), no. 75), where a 2-parameter family of continuous-time Markov processes on the Thoma cone was constructed. The purpose of the note is to show that these processes are diffusions.
The paper contains an explicit construction of Strebel differentials on one-parameter families of hyperelliptic curves of even genus. Descriptions of the corresponding separatrices are presented.
The boundary of the Gelfand–Tsetlin graph is an infinite-dimensional locally compact space whose points parameterize the extreme characters of the infinite-dimensional group
U(∞). The problem of harmonic analysis on the group U(∞) leads to a continuous family of probability measures on the boundary—the so-called zw-measures. Recently Vadim Gorin and the author have begun to study a q-analogue of the zw-measures. It turned out that constructing them requires introducing a novel combinatorial object, the extended Gelfand–Tsetlin graph. In the present paper it is proved that the Markov kernels connected with the extended Gelfand–Tsetlin graph and its q-boundary possess the Feller property. This property is needed for constructing a Markov dynamics on the q-boundary. A connection with the B-splines and their q-analogues is also discussed.
Let $G$ be a connected reductive algebraic group over $\mathbb{C}$. Let $\Lambda^{+}_{G}$ be the monoid of dominant weights of $G$. We construct the integrable crystals $\mathbf{B}^{G}(\lambda),\ \lambda\in\Lambda^{+}_{G}$, using the geometry of generalized transversal slices in the affine Grassmannian of the Langlands dual group. We construct the tensor product maps $\mathbf{p}_{\lambda_{1},\lambda_{2}}\colon \mathbf{B}^{G}(\lambda_{1}) \otimes \mathbf{B}^{G}(\lambda_{2}) \rightarrow \mathbf{B}^{G}(\lambda_{1}+\lambda_{2})\cup\{0\}$ in terms of multiplication of generalized transversal slices. Let $L \subset G$ be a Levi subgroup of $G$. We describe the restriction to Levi $\operatorname{Res}^G_L\colon\operatorname{Rep}(G)\rightarrow\operatorname{Rep}(L)$ in terms of the hyperbolic localization functors for the generalized transversal slices.
We consider mixed problems for strongly elliptic second-order systems in a bounded domain with Lipschitz boundary in the space Rn. For such problems, equivalent equations on the boundary in the simplest L2-spaces Hs of Sobolev type are derived, which permits one to represent the solutions via surface potentials. We prove a result on the regularity of solutions in the slightly more general spaces Hsp of Bessel potentials and Besov spaces Bsp. Problems with spectral parameter in the system or in the condition on a part of the boundary are considered, and the spectral properties of the corresponding operators, including the eigenvalue asymptotics, are discussed.
In this paper we study attractors of skew products, for which the following dichotomy is ascertained. These attractors either are not asymptotically stable or possess the following two surprising properties. The intersection of the attractor with some invariant submanifold does not coincide with the attractor of the restriction of the skew product to this submanifold but contains this restriction as a proper subset. Moreover, this intersection is thick on the submanifold, that is, both the intersection and its complement have positive relative measure. Such an intersection is called a bone, and the attractor itself is said to be bony. These attractors are studied in the space of skew products. They have the important property that, on some open subset of the space of skew products, the set of maps with such attractors is, in a certain sense, prevalent, i. e., "big." It seems plausible that attractors with such properties also form a prevalent subset in an open subset of the space of diffeomorphisms.
|
We try to analyze the average correlation of a portfolio as it can be found here in section 2 b), the same formula which is also used by the CBOE to calculate implied correlations:
$$ \rho_{av(2)} = \frac{\sigma^2 - \sum_{i=1}^N w_i^2\sigma_i^2}{2 \sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j} $$
EDIT:Assuming that $\sigma^2 = \sum_{i=1}^N \sum_{j=1}^N w_i w_j \sigma_i \sigma_j \rho_{i,j}$, where $\rho_{i,i}=1$, for $i=1,\ldots,N$, the above expression can be written as $$ \rho_{av(2)} = \frac{\sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j \rho_{i,j}}{\sum_{i=1}^N \sum_{j>i}^N w_i w_j \sigma_i \sigma_j}. $$
The following questions arise.
Assuming that $w_i \in \mathbb{R}$, i.e. long/short leverage is allowed, is it possible that $|\rho_{av(2)}|>1 $ ? Note that we don't assume $\sum w_i=1$. Does there already exist the notion of contribution to average correlation? Meaning that e.g. in a long/short portfolio, where average correlation should be close to zero, I can identify positions that drive the average correlation up (in absolute value).
|
Second-order linear structure-preserving modified finite volume schemes for the regularized long wave equation
1.
Graduate School of China Academy of Engineering Physics, Beijing 100088, China
2.
School of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing 210044, China
3.
College of Science, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China
In this paper, we develop four energy-preserving algorithms for the regularized long wave (RLW) equation. On the one hand, we combine the discrete variational derivative method (DVDM) in time and the modified finite volume method (mFVM) in space to derive a fully implicit energy-preserving scheme and a linear-implicit conservative scheme. On the other hand, based on the (invariant) energy quadratization technique, we first reformulate the RLW equation to an equivalent form with a quadratic energy functional. Then we discretize the reformulated system by the mFVM in space and the linear-implicit Crank-Nicolson method and the leap-frog method in time, respectively, to arrive at two new linear structure-preserving schemes. All proposed fully discrete schemes are proved to preserve the corresponding discrete energy conservation law. The proposed linear energy-preserving schemes not only possess excellent nonlinear stability, but also are very cheap because only one linear equation system needs to be solved at each time step. Numerical experiments are presented to show the energy conservative property and efficiency of the proposed methods.
Keywords:Modified finite volume method, discrete variational derivative method, energy quadratization, linear conservative scheme, regularized long-wave equation. Mathematics Subject Classification:Primary: 35Q53, 65M08, 65M06. Citation:Qi Hong, Jialing Wang, Yuezheng Gong. Second-order linear structure-preserving modified finite volume schemes for the regularized long wave equation. Discrete & Continuous Dynamical Systems - B, 2019, 24 (12) : 6445-6464. doi: 10.3934/dcdsb.2019146
References:
[1] [2] [3] [4] [5]
M. Dehghan and R. Salehi,
The solitary wave solution of the two-dimensional regularized long-wave equation in fluids and plasmas,
[6] [7] [8] [9] [10] [11] [12]
L. Mei and Y. Chen,
Numerical solutions of RLW equation using Galerkin method with extrapolation techniques,
[13] [14] [15]
E. Hairer, C. Lubich and G. Wanner,
[16]
C. Bubb and M. Piggot, Geometric integration and its application,
[17] [18] [19] [20] [21]
J. Cai, C. Bai and H. Zhang,
Decoupled local/global energy-preserving schemes for the
[22]
Y. Gong, J. Cai and Y. Wang,
Some new structure-preserving algorithms for general multi-symplectic formulations of Hamiltonian PDEs,
[23] [24] [25]
Q. Hong, Y. Wang and J. Wang,
Optimal error estimate of a linear Fourier pseudo-spectral scheme for the two dimensional Klein-Gordon-Schrödinger equations,
[26]
L. Kong, J. Hong, L. Ji and P. Zhu,
Compact and efficient conservative schemes for coupled nonlinear Schrödinger equations,
[27]
Z. Sun and D. Zhao,
On the
[28]
T. Wang, B. Guo and Q. Xu,
Fourth-order compact and energy conservative difference schemes for the nonlinear Schrödinger equation in two dimensions,
[29] [30]
J. Wang and Y. Wang,
Numerical analysis of a new conservative scheme for the coupled nonlinear Schrödinger equations,
[31]
D. Furihata,
Finite difference schemes for $\frac{\partial u}{\partial t} = (\frac{\partial}{\partial x})^{\alpha}\frac{\delta G}{\delta u}$ that inherit energy conservation or dissipation property,
[32]
D. Furihata,
Dissipative or conservative finite-difference schemes for complex-valued nonlinear partial differential equations,
[33]
J. Cai, Y. Gong and H. Liang,
Novel implicit/explicit local conservative scheme for the regularized long-wave equation and convergence analysis,
[34] [35] [36] [37]
S. Badia, F. Guillen-Gonzalez and J. Gutierrez-Santacreu,
Finite element approximation of nematic liquid crystal flows using a saddle-point structure,
[38]
F. Guillen and G. Tierra,
Second order schemes and time-step adaptivity for Allen-Cahn and Cahn-Hilliard models,
[39]
Y. Gong, J. Zhao and Q. Wang,
Linear second order in time energy stable schemes for hydrodynamic models of binary mixtures based on a spatially pseudospectral approximation,
[40]
X. Yang and D. Han,
Linearly first-and second-order, unconditionally energy stable schemes for the phase field crystal equation,
[41]
J. Zhao, X. Yang, Y. Gong and Q. Wang,
A novel linear second order unconditionally energy stable scheme for a hydrodynamic Q-tensor model of liquid crystals,
[42]
Y. Gong, Y. Wang and Q. Wang, Linear-implicit conservative schemes based on energy quadratization for Hamiltonian PDEs, submitted.Google Scholar
[43]
Y. Gong, Q. Wang, Y. Wang and J. Cai,
A conservative Fourier pseudospectral method for the nonlinear Schrodinger equation,
[44]
J. Cai,
Some linearly and nonlinearly implicit schemes for the numerical solutions of the regularized long-wave equation,
show all references
References:
[1] [2] [3] [4] [5]
M. Dehghan and R. Salehi,
The solitary wave solution of the two-dimensional regularized long-wave equation in fluids and plasmas,
[6] [7] [8] [9] [10] [11] [12]
L. Mei and Y. Chen,
Numerical solutions of RLW equation using Galerkin method with extrapolation techniques,
[13] [14] [15]
E. Hairer, C. Lubich and G. Wanner,
[16]
C. Bubb and M. Piggot, Geometric integration and its application,
[17] [18] [19] [20] [21]
J. Cai, C. Bai and H. Zhang,
Decoupled local/global energy-preserving schemes for the
[22]
Y. Gong, J. Cai and Y. Wang,
Some new structure-preserving algorithms for general multi-symplectic formulations of Hamiltonian PDEs,
[23] [24] [25]
Q. Hong, Y. Wang and J. Wang,
Optimal error estimate of a linear Fourier pseudo-spectral scheme for the two dimensional Klein-Gordon-Schrödinger equations,
[26]
L. Kong, J. Hong, L. Ji and P. Zhu,
Compact and efficient conservative schemes for coupled nonlinear Schrödinger equations,
[27]
Z. Sun and D. Zhao,
On the
[28]
T. Wang, B. Guo and Q. Xu,
Fourth-order compact and energy conservative difference schemes for the nonlinear Schrödinger equation in two dimensions,
[29] [30]
J. Wang and Y. Wang,
Numerical analysis of a new conservative scheme for the coupled nonlinear Schrödinger equations,
[31]
D. Furihata,
Finite difference schemes for $\frac{\partial u}{\partial t} = (\frac{\partial}{\partial x})^{\alpha}\frac{\delta G}{\delta u}$ that inherit energy conservation or dissipation property,
[32]
D. Furihata,
Dissipative or conservative finite-difference schemes for complex-valued nonlinear partial differential equations,
[33]
J. Cai, Y. Gong and H. Liang,
Novel implicit/explicit local conservative scheme for the regularized long-wave equation and convergence analysis,
[34] [35] [36] [37]
S. Badia, F. Guillen-Gonzalez and J. Gutierrez-Santacreu,
Finite element approximation of nematic liquid crystal flows using a saddle-point structure,
[38]
F. Guillen and G. Tierra,
Second order schemes and time-step adaptivity for Allen-Cahn and Cahn-Hilliard models,
[39]
Y. Gong, J. Zhao and Q. Wang,
Linear second order in time energy stable schemes for hydrodynamic models of binary mixtures based on a spatially pseudospectral approximation,
[40]
X. Yang and D. Han,
Linearly first-and second-order, unconditionally energy stable schemes for the phase field crystal equation,
[41]
J. Zhao, X. Yang, Y. Gong and Q. Wang,
A novel linear second order unconditionally energy stable scheme for a hydrodynamic Q-tensor model of liquid crystals,
[42]
Y. Gong, Y. Wang and Q. Wang, Linear-implicit conservative schemes based on energy quadratization for Hamiltonian PDEs, submitted.Google Scholar
[43]
Y. Gong, Q. Wang, Y. Wang and J. Cai,
A conservative Fourier pseudospectral method for the nonlinear Schrodinger equation,
[44]
J. Cai,
Some linearly and nonlinearly implicit schemes for the numerical solutions of the regularized long-wave equation,
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 8.291e-5 3.357e-5 8 3.97993 0.42983 1.633e-4 6.721e-5 12 3.97993 0.42983 2.404e-4 9.791e-5 16 3.97993 0.42983 3.138e-4 1.255e-4
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 8.291e-5 3.357e-5 8 3.97993 0.42983 1.633e-4 6.721e-5 12 3.97993 0.42983 2.404e-4 9.791e-5 16 3.97993 0.42983 3.138e-4 1.255e-4
Time Analytical / / 0 3.97993 0.42979 / / 4 3.97993 0.42979 4.020e-5 1.455e-5 8 3.97993 0.42979 8.265e-5 3.124e-5 12 3.97993 0.42979 1.224e-4 4.673e-5 16 3.97993 0.42979 1.614e-4 6.131e-5
Time Analytical / / 0 3.97993 0.42979 / / 4 3.97993 0.42979 4.020e-5 1.455e-5 8 3.97993 0.42979 8.265e-5 3.124e-5 12 3.97993 0.42979 1.224e-4 4.673e-5 16 3.97993 0.42979 1.614e-4 6.131e-5
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 6.485e-5 2.651e-5 8 3.97993 0.42983 1.270e-4 5.174e-5 12 3.97993 0.42983 1.854e-4 7.413e-5 16 3.97993 0.42983 2.416e-4 9.502e-5
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 6.485e-5 2.651e-5 8 3.97993 0.42983 1.270e-4 5.174e-5 12 3.97993 0.42983 1.854e-4 7.413e-5 16 3.97993 0.42983 2.416e-4 9.502e-5
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 1.998e-4 7.882e-5 8 3.97993 0.42983 3.936e-4 1.574e-4 12 3.97993 0.42983 5.842e-4 2.332e-4 16 3.97993 0.42983 7.671e-4 3.021e-4
Time Analytical / / 0 3.97993 0.42983 / / 4 3.97993 0.42983 1.998e-4 7.882e-5 8 3.97993 0.42983 3.936e-4 1.574e-4 12 3.97993 0.42983 5.842e-4 2.332e-4 16 3.97993 0.42983 7.671e-4 3.021e-4
Method CPU(s) CPU(s) FIEP 2.023e-4 8.298e-5 1.398 1.363e-4 5.520e-5 1.796 LIEP 1.035e-4 3.946e-5 0.993 1.687e-4 6.716e-5 1.268 LICN 1.566e-4 6.316e-5 1.073 9.172e-5 3.545e-5 1.076 LILF 4.896e-4 1.962e-4 1.006 4.241e-4 1.684e-4 1.128 NC-II [34] 2.088e-4 7.532e-5 1.755 1.234e-4 4.236e-5 2.137 AMC-CN [44] 3.944e-4 1.581e-4 1.506 1.828e-4 7.307e-5 1.988 Linear-CN [6] 2.648e-4 1.088e-4 1.046 7.945e-4 2.957e-4 1.108
Method CPU(s) CPU(s) FIEP 2.023e-4 8.298e-5 1.398 1.363e-4 5.520e-5 1.796 LIEP 1.035e-4 3.946e-5 0.993 1.687e-4 6.716e-5 1.268 LICN 1.566e-4 6.316e-5 1.073 9.172e-5 3.545e-5 1.076 LILF 4.896e-4 1.962e-4 1.006 4.241e-4 1.684e-4 1.128 NC-II [34] 2.088e-4 7.532e-5 1.755 1.234e-4 4.236e-5 2.137 AMC-CN [44] 3.944e-4 1.581e-4 1.506 1.828e-4 7.307e-5 1.988 Linear-CN [6] 2.648e-4 1.088e-4 1.046 7.945e-4 2.957e-4 1.108
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Pricing options on investment project contraction and ownership transfer using a finite volume scheme and an interior penalty method.
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We liberate millions of students and teachers from exam-oriented education by reducing attention and time dissipated on training, with the solution entirely directed by mathematical models.
– Hisen Zhang A lot of wastes for printed past paper booklet Great difficulty to make up a paper for internal assessment No focus for students while practicing Reduced wastes An easier life for teachers and students. Digitize quiz bank Automate paper make up process Introduce AI Director
An AI directed question paper assembler. P.A.E. stands for “Paper Assembly Engine”.
Students may use it for exam-training and teachers are allowed to learn about their student’s progress.
The entire process is directed by Artificial Intelligence.
Although many digital quiz banks seem to be handy, a traditional quiz bank requires user to select questions manually. Unlike other quiz banks (such as the one offered by THE INTERNATIONAL BACCALAUREATE®), our system enables 100% automated process, from selection to assembly to response collection. This feature is achieved by the
Data Flow Cycle, a fundamental idea introduced later in this chapter.
To understand the complete system, it would be better to first get familiar with the data flow loop between three main subsystems.
The entire system consists of three subsystems: AI Director, Assembly Engine, and User Interface.
The data work cycle works like this:
The
AI Directorgenerates weights depending on feedback.
The
Assembly Engineassembles question papers referring to the weights.
The
User Interfacedisplays papers assembled and collects feedback.
Note: Chapter 2 describes the three subsystems in details.
Beside the subsystems, other components are labeled in the diagram above. The input of the system is the past paper published on the Internet. The output has two parts: the generated question paper and the AI reports.
For this program, the A-Level series question paper published by Cambridge International Examination is applied. They are available to qualified centers and teachers.
The
User Interface sends AJAX requests to the API. The return is in JSON. Then the Javascript code in user’s browser refreshes the web page partially.
The question paper is presented in HTML.
AI report is both available to students and teacher. The contents are generated for two groups of users respectively. For students, this report details their strength and weakness; for teachers, this shows the overall statistics in the system, and a specific profile for each student as well.
In this project, the quiz bank is implemented with
sqlite3. The database is light enough to operate and stable enough to keep the questions.
In most cases, the database is read-only. When multi-user feature works, another database for storing account data should be deployed separately.
Both
Assembly Engine or
AI Director may access to the quiz bank.
Only the
User Interface is visible to the users. The rest parts are running in the background. Communication across the line of visibility is implemented with HTTP restful API.
The assembly engine takes weights as input. The questions selected are based on weights on different topics. The output is transmitted in JSON via the API.
There are different modes for assembly engine. The mode is specified implicitly (or explicitly in some occasions) with the parameters passed to the API.
The interfaces and usage are listed below:
Interface Method Parameter Comment /paper/ [GET] topic_keyword, similar_to If topic_keyword is specified, fetch the questions with the keyword; else select the questions most similar to the value.
The user interface takes the JSON from paper information to render them in the web page. When users type command into the search bar, the Javascript code loaded into user’s web browser will send AJAX requests to the back end. The API server at the back end calls the assembly engine and response with the information of question in JSON.
Sample Response
1 {
Explanation to some important fields:
Field Type Comment citation int How many times the question is referred id int The unique id of the question in the quiz bank q_ans string The answer to the question q_path string Add this to the images’
src attribute
quality int The mark for quality. Currently reserved.
The main role played by AI Director is to decide which question to be selected from the quiz bank.
Since the quiz bank is designed to be updated once a season, it does not make sense to run AI codes over the quiz bank over and over. The idea here is to pre-process through all records in the database and store this information (mainly matrices) on the disk. This method also accelerates the process for question selecting since the decision is made by referring to the matrices with calculated values.
There will be more introduction to this component in later chapters.
The cutter is responsible for cutting the images of questions and fetch their answers from the input, which is the past paper in PDF format, and append these data into the database.
The cutter should be an important component (or, subsystem). However, since this part of the program is highly curriculum related, and only little usage of this function (the update interval is supposed to be 4 months or so), I decided to classify this function as an accessory.
By the way, this part takes a great portion of time over the development.
Most algorithms applied in this program are included in the pre-processor.
This method lowers the weights for stopwords (i.e. “and”,”the”) and therefore it highlights the featured words. Sorting these words gives a keyword list used to generate word vectors.
The cosine distance applies to vectors in spaces with more than three dimensions as well. This made it possible to calculate the similarity between two vectors. The result of cosine distance gets closer to zero if two vectors are more alike. The similarity is defined by $1-\cos\theta$.
Therefore the difficulty is to generate a vector for each question: and such a vector must be able to display the feature of that question. An assumption is made: the words appeared in the text do have a strong connection to the topic behind. For instance, the word “spring” will come along with the topic “elastic deformation”.
K-Means clustering works based on Euclidean distance. With the given number of K, the output of this algorithm is the K cluster, Such a method is very straightforward.
However, this is also the problem. Although only once hyperparameter needs to be specified, The value of K is difficult to decide. In practical, the K is often decided arbitrarily.
As an improvement to the K-Means algorithm, Iterative Self-Organizing Data Analysis Technique will refer to the K value given, but modify the value during the process as well. This is achieved by splitting and merging the clusters once according to the following conditions:
If the standard deviation within a cluster exceeds $\theta_{split}$ If the distance between two centroids are closer than $\theta_{merge}$ etc.
While programming this system, some functions already integrated into standard or third-party modules do not fit my case perfectly. For instance, I first planned to use CSV (comma separated values) to store the matrix of cosine similarity. However, storing in text not only occupies large storage space but also has a loss in data precision. Therefore I defined a type of file to store this matrix efficiently.
As suggested in earlier chapter, the matrix of cosine similarity looks like this:
1 [[0.2432432],
The calculated cosine similarity matrix is symmetrical. This means it is possible to cut the size in half when it comes to storage, whether in the memory or on the disk. To be precise, the actual elements required to restore a complete n by n matrix is:
$\frac {1}{2}n (n-1)$
In this case, the values are in the type of float. Each float element has the size of 4 bytes. Therefore the total size of the file is $2n(n-1)$ bytes.
The built-in function in Python repr() convert the objects in strings, and they can be restored to object with eval(). However, float data here must experience loss in precision because the data is rounded to some digits for saving storage space.
The comparison below may show how the solution makes a difference. Rounded to 5 digits, a “float” stored in string occupies 8 bytes on average (with “0.” and a comma). n is the dimension of the matrix.
n CSM Round(5) Round(8) 100 19.8 KB 39.6 KB 54.5 KB 1000 2 MB 4 MB 5.5 MB 10000 200 MB 400 MB 550 MB
Some advantages of applying CSM format:
No loss in data precision
More compact storage
In reality, the first 12 bytes are reserved for storing the dimension of the matrix and the values on the symmetrical axis. The following table shows the header of
smc files.
Field Type Size Comment SIZE unsigned long 8 Dimension SYM_VAL float 4 Values on the symmetrical axis DATA float 4 Elements in the matrix
Some projects like PyRadar do integrate the function of ISODATA clustering for image processing. However, there is not a ISODATA library for general purposes. Therefore I started to implement the code from scratch. For making the life easier for other programmers, I decide to make it publicly available.
This module is for general purposes by allowing vectors as input and the vectors in clusters as output.
I have a dream.
I have a dream, that students may have some spare time, doing some coolest things with coolest people. This project helps by raising the efficiency of training.
I have a dream, that teachers don’t have to study for the exams, but to study in the fields they are talented. P.A.E. automate this process with machine learning.
I have a dream, that students may join pure project-based study, instructed by their teachers, instead of being prisoned by examinations…
I have a dream.
This system is designed to be valid based on these following ‘believed-to-be-true’ statements:
“Practice makes perfect.” The keywords show some features of the text.
So far these postulates are widely validated. However, no solid shreds of evidence suggest they are truth. In other words, if these statements are proved to be wrong, the system may not be effective as it was proposed.
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Update: The MathJax Plugin for TiddlyWiki has a new home: https://github.com/guyru/tiddlywiki-mathjax Some time ago I came across MathJax, a nifty, Javascript based engine for displaying TeX and LaTeX equations. It works by “translating” the equation to MathML or HTML+CSS, so it works on all modern browsers. The result isn’t a raster image, like in most LaTeX solutions (e.g. MediaWiki), so it’s scales with the text around it. Furthermore, it’s quite easy to integrate as it doesn’t require any real installation, and you could always use MathJax’s own CDN, which makes things even simpler.
I quickly realized MathJax will be a perfect fit for TiddlyWiki which is also based on pure Javascript. It will allow me to enter complex formulas in tiddlers and still be able to carry my wiki anywhere with me, independent of a real TeX installation. I searched the web for an existing MathJaX plugin for TiddlyWiki but I came up empty handed (I did find some links, but they referenced pages that no longer exist). So I regarded it as a nice opportunity to begin writing some plugins for TiddlyWiki and created the MathJaxPlugin which integrates MathJax with TiddlyWiki.
As I don’t have an online TiddlyWiki, you’ll won’t be able to import the plugin, instead you’ll have to install it manually (which is pretty simple).
Start by creating a new tiddler named
MathJaxPlugin, and tag with
systemConfig (this tag will tell TiddlyWiki to execute the JS code in the tiddler, thus making it a plugin. Now copy the following code to the tiddler content:
/***
|''Name:''|MathJaxPlugin|
|''Description:''|Enable LaTeX formulas for TiddlyWiki|
|''Version:''|1.0.1|
|''Date:''|Feb 11, 2012|
|''Source:''|http://www.guyrutenberg.com/2011/06/25/latex-for-tiddlywiki-a-mathjax-plugin|
|''Author:''|Guy Rutenberg|
|''License:''|[[BSD open source license]]|
|''~CoreVersion:''|2.5.0|
!! Changelog
!!! 1.0.1 Feb 11, 2012
* Fixed interoperability with TiddlerBarPlugin
!! How to Use
Currently the plugin supports the following delemiters:
* """\(""".."""\)""" - Inline equations
* """$$""".."""$$""" - Displayed equations
* """\[""".."""\]""" - Displayed equations
!! Demo
This is an inline equation \(P(E) = {n \choose k} p^k (1-p)^{ n-k}\) and this is a displayed equation:
\[J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha}\]
This is another displayed equation $$e=mc^2$$
!! Code
***/
//{{{
config.extensions.MathJax = {
mathJaxScript : "http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML",
// uncomment the following line if you want to access MathJax using SSL
// mathJaxScript : "https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML",
displayTiddler: function(TiddlerName) {
config.extensions.MathJax.displayTiddler_old.apply(this, arguments);
MathJax.Hub.Queue(["Typeset", MathJax.Hub]);
}
};
jQuery.getScript(config.extensions.MathJax.mathJaxScript, function(){
MathJax.Hub.Config({
extensions: ["tex2jax.js"],
"HTML-CSS": { scale: 100 }
});
MathJax.Hub.Startup.onload();
config.extensions.MathJax.displayTiddler_old = story.displayTiddler;
story.displayTiddler = config.extensions.MathJax.displayTiddler;
});
config.formatters.push({
name: "mathJaxFormula",
match: "\\\\\\[|\\$\\$|\\\\\\(",
//lookaheadRegExp: /(?:\\\[|\$\$)((?:.|\n)*?)(?:\\\]|$$)/mg,
handler: function(w)
{
switch(w.matchText) {
case "\\[": // displayed equations
this.lookaheadRegExp = /\\\[((?:.|\n)*?)(\\\])/mg;
break;
case "$$": // inline equations
this.lookaheadRegExp = /\$\$((?:.|\n)*?)(\$\$)/mg;
break;
case "\\(": // inline equations
this.lookaheadRegExp = /\\\(((?:.|\n)*?)(\\\))/mg;
break;
default:
break;
}
this.lookaheadRegExp.lastIndex = w.matchStart;
var lookaheadMatch = this.lookaheadRegExp.exec(w.source);
if(lookaheadMatch && lookaheadMatch.index == w.matchStart) {
createTiddlyElement(w.output,"span",null,null,lookaheadMatch[0]);
w.nextMatch = this.lookaheadRegExp.lastIndex;
}
}
});
//}}}
After saving the tiddler, reload the wiki and the MathJaxPlugin should be active. You can test it by creating a new tiddler with some equations in it:
This is an inline equation $$P(E) = {n \choose k} p^k (1-p)^{ n-k}$$ and this is a displayed equation:
\[J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha}\]
Which should result in the tiddler that appears in the above image.
Update 2011-08-19: Removed debugging code from the plugin.
Changelog 1.0.1 (Feb 11, 2012 Applied Winter Young’s fix for interoperability with other plugins (mainly TiddlerBarPlugin
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Definition:Euler-Mascheroni Constant Contents Definition
The
Euler-Mascheroni Constant $\gamma$ is the real number that is defined as:
\(\displaystyle \gamma\) \(:=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {\sum_{k \mathop = 1}^n \frac 1 k - \int_1^n \frac 1 x \rd x}\) \(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {H_n - \ln n}\) $\gamma \approx 0 \cdotp 57721 \, 56649 \, 01532 \, 86060 \, 65120 \, 90082 \, 40243 \, 1 \ldots$ Also known as
This constant is otherwise known as
Euler's constant but must not be confused with Euler's number.
It is also sometimes known as
Mascheroni's constant. Source of Name
He calculated it to $6$ decimal places, and published it in $1738$ as $0 \cdotp 577218$ (although only the first $5$ were correct, as Euler himself surmised).
Mascheroni published a calculation to $32$ places of the value of this constant.
Only the first $19$ places were accurate. The remaining ones were corrected in $1809$ by Johann von Soldner.
There exists disagreement over the question of who was first to name it $\gamma$ ( gamma).
Some sources claim it was Euler who named it $\gamma$, in $1781$, while others suggest its first appearance of that symbol for it was in Lorenzo Mascheroni's $1790$ work
Adnotationes ad calculum integrale Euleri.
An early appearance of the symbol $\gamma$ was by Carl Anton Bretschneider in his $1837$ paper
Theoriae logarithmi integralis lineamenta nova ( J. reine angew. Math. Vol. 17: 257 – 285), and this may indeed be the first. Also see Existence of Euler-Mascheroni Constant where its existence is demonstrated. Sources 1738: Leonhard Paul Euler: De Progressionibus Harmonicis Obseruationes( Commentarii Acad. Sci. Imp. Pet. Vol. 7: 150 – 161) 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous) ... (next): $\S 1$: Special Constants: $1.20$ 1972: George F. Simmons: Differential Equations... (previous) ... (next): $\S 3$: Appendix $\text A$: Euler 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach... (previous) ... (next): $\S 14.3 \ (5)$ 1986: David Wells: Curious and Interesting Numbers... (previous) ... (next): $0 \cdotp 577 \, 215 \, 664 \, 901 \, 532 \, 860 \, 606 \, 512 \, 090 \, 082 \, 402 \, 431 \ldots$ 1992: George F. Simmons: Calculus Gems... (previous) ... (next): Chapter $\text {A}.21$: Euler ($1707$ – $1783$) 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms(3rd ed.) ... (previous) ... (next): $\S 1.2.7$: Harmonic Numbers 1997: David Wells: Curious and Interesting Numbers(2nd ed.) ... (previous) ... (next): $0 \cdotp 57721 \, 56649 \, 01532 \, 86060 \, 65120 \, 90082 \, 40243 \, 1 \ldots$ 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics(5th ed.) ... (previous) ... (next): Entry: Euler's constant
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After the answers by joshphysics and user37496, it seems to me that a last remark remains.
The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates
Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.
In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:
(1) They are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)
and
(2) they enjoy the
commutation relations of some Lie algebra $\ell$:$$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$on a common invariant domain ${\cal D}\subset {\cal H}$.
As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique
simply connected Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the universal covering of all the other Lie groups whose Lie algebra is $\ell$ itself.
All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element.
If (1) and (2) hold, the natural question is:
Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group $\cal G$ just admitting $\ell$ as its Lie algebra, such that $$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$
Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.
If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint opertors $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are
conserved quantities.Otherwise the situation is a bit more complicated, nevertheless one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generator when $\cal G$ is $SL(2,\mathbb C)$).
Well, the fundamental theorem by Nelson has the following statement.
THEOREM (Nelson)
Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator$$\Delta := \sum_{i=1}^N A_i^2$$is essentially self-adjoint on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then:
(a) Every $A_i$ is essentially self-adjoint on $\cal D$,
and
(b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements:$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$ where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$.
Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.
The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own Lie algebra.
As another application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected) Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming $X$ and $P$ into the standard operators. Following this way one builds up an alternate proof of Stone-von Neumann's theorem.
As a last comment, I stress that usually ${\cal G}_\ell$ is This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
not the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is the appearance of annoying phases as explained by Josh, and overall phases do not affect states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume quantum states that are coherent superpositions of both integer and semi-integer spin. Otherwise an internal phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize superselection rules.
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Now showing items 1-1 of 1
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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Search
Now showing items 1-10 of 28
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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Q1 T T F, the overlap integral is zero. \(\beta\) pertains to energy, not overlap. T F, \(\beta\) is experimentally found to be negative. F, the resonance integral pertains to energy, so it is \(\beta\) . F, non-adjacent resonance integrals are zero. T Q2
In the ethylene system, there are two p orbitals to consider, \( |1\rangle\) and \(|p\rangle\), each with energy E.
The Hamiltonian, \(\hat{H}\), is given by:
\( \begin{bmatrix}
\alpha& \beta\\ \beta& \alpha\end{bmatrix} \)
We know that \(\hat{H} |\psi\rangle = E|\psi\rangle\) . Factoring, we have \( \hat{H} -EI|\psi\rangle = | 0 \rangle\), where
I is the identity matrix.
This implies that \( |\psi\rangle = (\hat{H} -EI)^{-1} |0\rangle\). For the left side to be non-zero, the inverse matrix that acts on the zero vector must be infinite. From linear algebra, the inverse of a matrix is inversely proportionate to its determinant:
\((\hat{H} -EI)^{-1} = \frac{C}{det(\hat{H} -EI)}\). For the inverse to be infinite, we require \(det(\hat{H} -EI) = 0\).
Q3
a). \( \hat{H}\) for a 5 carbon system by the Huckel Theory, is given by:
\[\begin{bmatrix}\alpha&\beta&0&0&0\\\beta&\alpha&\beta&0&0\\0&\beta&\alpha&\beta&0\\0&0&\beta&\alpha&0\\0&0&0&\beta&\alpha\end{bmatrix} \]
b). Given the following wavefunctions, choose two and show they are normalized:
\( | \pi_1 \rangle = +0.45 | 1 \rangle +0.45 | 2 \rangle + +0.45 | 3 \rangle +0.45 | 4 \rangle +0.45 |5 \rangle \) \( | \pi_2 \rangle = -0.51 | 1 \rangle + 0.63 | 2 \rangle -0.51 | 3 \rangle + 0.20 | 4 \rangle + 0.20 | 5 \rangle \) \( | \pi_3 \rangle = -0.37 | 1 \rangle + 0.37 | 3 \rangle + 0.60 | 4 \rangle + 0.60 | 5 \rangle \) \( | \pi_4 \rangle = -0.21 | 1 \rangle - 0.63 | 2 \rangle -0.20 | 3 \rangle + 0.51 | 4 \rangle + 0.51 | 5 \rangle \) \( | \pi_5 \rangle = +60 | 1 \rangle -0.60 | 3 \rangle + 0.37 | 4 \rangle + 0.37 | 5 \rangle \)
The dot products to show normalization are:
\(\langle1|1\rangle\) is: 1.012500e+00 \(\langle2|2\rangle\) is: 9.971000e-01 \(\langle3|3\rangle\) is: 9.938000e-01 \(\langle4|4\rangle\) is: 1.001200e+00 \(\langle5|5\rangle\) is: 9.938000e-01
c). Show that two are orthogonal to each other:
Actually calculating the dot products we see they are not perfectly orthogonal, some with great intersection.
\(\langle1|2\rangle\) is: 4.635000e-01 \(\langle1|3\rangle\) is: 5.400000e-01 \(\langle1|4\rangle\) is: 1.710000e-01 \(\langle1|5\rangle\) is: 3.330000e-01 \(\langle2|3\rangle\) is: 6.174000e-01 \(\langle2|4\rangle\) is: 1.620000e-02 \(\langle2|5\rangle\) is: -4.640000e-01 \(\langle3|4\rangle\)is: 7.637000e-01 \(\langle3|5\rangle\) is: 0 \(\langle4|5\rangle\) is: 1.314000e-01
d). Use the Hamiltonian matrix to calculate the distinct energies using these eigenfunctions.
We can find the energies of this system by setting \(\hat{H} - EI = 0\).
\[\begin{vmatrix}\alpha-E&\beta&0&0&0\\\beta&\alpha-E&\beta&0&0\\0&\beta&\alpha-E&\beta&0\\0&0&\beta&\alpha-E&0\\0&0&0&\beta&\alpha-E\end{vmatrix} = 0\]
Dividing the matrix by \(\beta\) and using the variable \( x = \frac{\alpha - E}{\beta}\), we can solve a determinant of the form:
\[\begin{vmatrix}x&1&0&0&0\\1&x&1&0&0\\0&1&x&1&0\\0&0&1&x&0\\0&0&0&1&x\end{vmatrix} = 0\]
Continuing to solve for x by breaking down the determinant into smaller determinants, (also known as its "minors" or Laplace's formula) you will arrive after careful factoring (see https://people.richland.edu/james/le...terminant.html):
\[ x(x^2-3)(x^2-1) = 0\]
The energies relate to the other variables via: \(E = \alpha - x\beta\)
The roots are \(x = 0, \pm 1, \pm \sqrt{3}\), so the energies are \( \alpha, \alpha \pm \sqrt{3}\beta, \alpha \pm \beta\)
Q4
The allyl cation can be described by the Hamtilonian:
\[ \hat{H} = \begin{bmatrix}\alpha&\beta&0\\\beta&\alpha&\beta\\0&\beta&\alpha\end{bmatrix}\]
Similar to the previous problem, we must again solve for \(det(\hat{H}-EI) = 0 \), which results in:
\[\begin{vmatrix}x&1&0\\1&x&1\\0&1&x\end{vmatrix}= x(x^2-2)= 0\]
The roots are \(x = 0,\pm\sqrt{2}\), which correspond to energies: \(\alpha + \sqrt{2}\beta, \alpha, \alpha - \sqrt{2}\beta\) which correspond to bonding, non-bonding and anti-bonding orbitals. The deolocalization of the electrons lowers the energy of the system to 0.82 \(\beta\) when compared to ethylene.
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Overview of Integration Methods in Space and Time
Integration is one of the most important mathematical tools, especially for numerical simulations. Partial Differential Equations (PDEs) are usually derived from integral balance equations, for example. Once a PDE needs to be solved numerically, integration most often plays an important role, too. This blog post gives an overview of the integration methods available in the COMSOL software and shows you how you can use them.
The Importance of Integrals
COMSOL uses the finite element method, which transforms the governing PDE into an integral equation — the weak form, in other words. Having a closer look at the COMSOL simulation software, you may realize that many boundary conditions are formulated in terms of integrals. A couple of examples of these are
Total heat flux or floating potential. Integration also plays a key role in postprocessing, as COMSOL provides many derived values based on integration, like electric energy, flow rate, or total heat flux. Of course, our users can also use integration in COMSOL for their own means, and here you will learn how. Integration by Means of Derived Values
A general integral has the form
where [t_0,t_1] is a time interval, \Omega is a spatial domain, and F(u) is an arbitrary expression in the dependent variable u. The expression can include derivatives with respect to space and time or any other derived value.
The most convenient way to obtain integrals is to use the “Derived Values” in the Results section of the new ribbon (or the Model Builder if you’re not running Windows®).
How to add volume, surface, or line integrals as Derived Values.
You can refer to any available solution by choosing the corresponding data set. The Expression field is the integrand and allows for dependent or derived variables. For transient simulations, the spatial integral is evaluated at each time step. Alternatively, the settings window offers Data Series Operations, where
Integration can be selected for the time domain. This results in space-time integration. Example of Surface Integration Settings with additional time integration via the Data Series Operation.
The
Average is another Derived Value related to integration. It equals an integral, which is divided by the volume, area, or length of the considered domain. The Average Data Series Operation additionally divides by the time horizon. Derived Values are very useful, but because they are only available for postprocessing, they cannot handle every type of integration. That is why COMSOL provides more powerful and flexible integration tools. We demonstrate these methods with an example model below. Spatial and Temporal Integration for a Heat Transfer Example Model
We introduce a simple heat transfer model, a 2D aluminum unit square in the (
x, y)-plane. The upper and right sides are fixed at room temperature (293.15 K) and on the left and lower boundary, a General inward heat flux of 5000W/m^2 is prescribed. A stationary solution and a time-dependent solution after 100 seconds are shown in the following figures. Spatial Integration by Means of Component Coupling Operators
Component Coupling Operators are, for example, needed when several integrals are combined in one expression, when integrals are requested during calculation, or in cases where a set of path integrals are required. Component Coupling Operators are defined in the Definitions section of the respective component. At that stage, the operator is not evaluated yet. Only its name and domain selection are fixed.
How to add Component Coupling Operators for later use.
For our example, we first want to calculate the spatial integral over the stationary temperature, which is given by
In the COMSOL software, we use an integration operator, which is named
intop1 by default. Settings window of the integration operator. How to evaluate the Integration operator.
In the next step, we demonstrate how an Integration operator can also be used within the model. We could, for example, ask what heating power we need to apply to obtain an average temperature of 303.15 K, which equals an average temperature increase of 10 K compared to room temperature. First, we need to compute the difference between the desired and the actual average temperature. The average is calculated by the integral over T, divided by the integral over the constant function 1, which gives the area of the domain. Fortunately, this type of calculation can easily be done with an
Average operator in COMSOL. By default, such an operator is named aveop1. (Note that the average over the domain is the same as the integral for our example. That is because the domain has unit area.) The corresponding difference is given by
Next, we need to find the General heat flux on the left and lower boundary, so that the desired average temperature is satisfied. To this end, we introduce an additional degree of freedom named
q_hot and an additional constraint as a global equation. The General inward heat flux is replaced by q_hot. How to add an additional degree of freedom and a global equation, which forces the average temperature to 303.15 K.
Solving this coupled system with a stationary study results in q_{hot}=5881.30 W/m^2. This value has to be prescribed as a General inward heat flux boundary condition to achieve an average temperature of 303.15 K in the whole domain.
Computing the Antiderivative by Means of Integration Coupling
A frequently asked question we receive in Support is: How can one obtain the spatial antiderivative? The following application of integration coupling answers this question. The antiderivative is the counterpart of the derivative, and geometrically, it enables the calculation of arbitrary areas bounded by function graphs. One important application is the calculation of probabilities in statistical analyses. To demonstrate this, we fix y=0 in our example and denote the antiderivative of T(x,0) by u(x). This means that \frac{\partial u}{\partial x}=T(x,0). A representation of the antiderivative is the following integral
where we use \bar x in order to distinguish the integration and the output variable. In contrast to the integrals above, we here have a function as a result, rather than a scalar quantity. We need to include the information that for each \bar x\in[0,1] the corresponding value of u(\bar x) requires an integral to be solved. Fortunately, this is easy to set up in the COMSOL environment and requires only three ingredients, so to speak. First, a logical expression can be used to reformulate the integral as
Second, we need an integration operator that acts on the lower boundary of our example domain. Let’s denote it by
intop2. Third, we need to include the distinction of integration and output variable. The notation for this situation is source and destination for x and \bar x, respectively. When using an integration coupling operator, the built-in operator dest is available, which indicates that the corresponding expression does not belong to the integration variable. More precisely, it means \bar x=dest(x) in COMSOL. Putting the logical expression and the dest operator together, results in the expression T*(x<=dest(x)), which is exactly the input expression that we need for intop2. Altogether, we can calculate the antiderivative by intop2(T*(x<=dest(x))), resulting in the following plot in our example: How to plot the antiderivative by Integration coupling, the dest operator, and a logical expression.
COMSOL provides two other integration coupling operators, namely
general projection and linear projection. These can be used to obtain a set of path integrals in any direction of the domain. In other words, integration is performed only with respect to one dimension. The result is a function of one dimension less than the domain. For a 2D example the result is a 1D function, which can be evaluated on any boundary. Some more details on how to use these operators are subject to a forthcoming blog post on component couplings. Spatial Integration by Means of an Additional Physics Interface
The most flexible way of spatial integration is to add an additional PDE interface. Let’s remember the example of the antiderivative and assume that we want to calculate the antiderivative not only for y=0. The task can be formulated in terms of the PDE
with Dirichlet boundary condition u=0 on the left boundary. The easiest interface to implement this equation is the
Coefficient Form PDE interface, which only needs the following few settings: How to use an additional physics interface for spatial integration.
The dependent variable u represents the antiderivative with respect to x and is available during calculation and postprocessing. Besides flexibility, a further advantage of this method is accuracy, because the integral is not obtained as a derived value, but is part of the calculation and internal error estimation.
Temporal Integration by Means of Built-In Operators
We have already mentioned the Data Series Operations, which can be used for time integration. Another very useful method for time integration is provided by the built-in operators
timeint and timeavg for time integration or time average, respectively. They are readily available in postprocessing and are used to integrate any time-dependent expression over a specified time interval. In our example we may be interested in the temperature average between 90 seconds and 100 seconds, i.e.:
The following surface plot shows the resulting integral, which is a spatial function in
(x,y): How to use the built-in time integration operator timeavg .
Similar operators are available for integration on spherical objects, namely
ballint, circint, diskint, and sphint. Temporal Integration by Means of Additional Physics Interfaces
If temporal integrals have to be available in the model, you need to define them as additional dependent variables. Similar to the
Coefficient Form PDE example shown above, this can be done by adding an ODE interface of the Mathematics branch. Suppose, for example, that at each time step, the model requests the time integral from start until now over the total heat flux magnitude, which measures the accumulated energy. The variable for the total heat flux is automatically calculated by COMSOL and is named ht.tfluxMag. The integral can be calculated as an additional dependent variable with a Distributed ODE, which is a subnode of the Domain ODEs and DAEs interface. The source term of this domain ODE is the integrand, as shown in the following figure. How to use an additional physics interface for temporal integration.
What is the benefit of such a calculation? The integral can be reused in another physics interface, which may be influenced by the accumulated energy in the system. Moreover, it is now available for all kinds of postprocessing, which is more convenient and faster than built-in operators. For an example, check out the Carbon Deposition in Hetereogeneous Catalysis model, where a domain ODE is used to calculate the porosity of a catalyst as a time-dependent field variable in the presence of chemical reactions.
Integration of Analytic Functions and Expressions
So far, we have shown how to integrate solution variables during calculation or in postprocessing. We have not yet covered integrals of analytic functions or expressions. To this end, COMSOL provides the built-in operator
integrate( expression, integration variable, lower bound, upper bound).
The expression might be any 1D function, such as
sin(x). It is also possible to include additional variables, such as sin(x*y). The second argument specifies over which variable the integral is calculated. For example integrate(sin(x*y),y,0,1) yields a function in x, because integration only eliminates the integration variable y. Note that the operator can also handle analytic functions, which need to be defined in the Definitions node of the current component. How to add an analytic function. How to integrate over an analytic function. Further Reading Model downloads Using Integration coupling operators: Acoustics of a Muffler Using global equations for time integration: Process Control Using a PID Controller Using global equations to satisfy constraints: Using Global Equations to Satisfy Constraints Using domain ODEs for time integration: Capacity Fade of a Li-Ion Battery and Carbon Deposition in Heterogeneous Catalysis Knowledge Base entry: Computing Time and Space Integrals 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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In this tutorial, we write an Alphabets program, starting from a mathematical equation for LU decomposition. Then we will generate code to execute the alphabets program, and test the generated code for correctness.
The equation for LU Decomposition, derived from first principles using simple algebra in Foundations (pg.3), is as follows: $$ U_{i,j}=\begin{cases} 1=i\le j & A_{i,j}\\ 1<i\le j & A_{i,j}-\sum_{k=1}^{i-1}L_{i,k}U_{k,j} \end{cases}\\ L_{i,j}=\begin{cases} 1 = i\le j & \frac{A_{i,j}}{U_{j,j}}\\ 1< i\le j & \frac{1}{U_{j,j}}(A_{i,j}-\sum_{k=1}^{j-1}L_{i,k}U_{k,j}) \end{cases} $$
[Temp note due to : in the last case of L, the condition is “1 < j ⇐ i”]
Let's start from an empty alphabets file, with LUD as the name of the system, and a positive integer N as its parameter. A system (Affine System) takes its name from system of affine recurrence equations, and represents a block of computation. An Alphabets program may contain multiple systems.
Caveat: Remember the phrase, “It's not a bug, it's a feature”? Well, in a tutorial, a feature is called a “learning opportunity.”
Parameters are runtime constants represented with some symbol in the code. In this example, parameter N will be used to define the size of the matrices, which is not known until runtime.
affine LUD {N|N>0} .
In most cases, a computation uses some inputs and produces outputs. Such variables must be declared with a name, a data type, and a shape/size. In Alphabets, the shape/size is represented with polyhedral domains.For this example, the
A matrix is given, and we are computing two triangular matrices
L and
U.
A is an NxN square matrix. The declaration for
A looks as follows:
float A {i,j|1<=(i,j)<=N}; //starting from 1 to be consistent with the equation in the notesSimilarly,
L is a lower triangular matrix of size N (with unit diagonals, implicit) and
U is an upper triangular matrix of size N. The declarations should look like the following:
// The convention is that i is the vertical axis going down, and j is the horizontal axis float L {i,j|1<i<=N && 1<=j<i}; // Note that the diagonal elements of L are not explicitly declared float U {i,j|1<=j<=N && 1<=i<=j};Now these variable declarations need to be placed at appropriate places to specify whether they are input/output/local.
input/
given is the keyword for input,
output/
returns is the keyword for output, and
local/
using is the keyword for local variables.
affine LUD {N|N>0} input float A {i,j|1<=(i,j)<=N}; output float L {i,j|1<=j<i<=N}; float U {i,j|1<=i<=j<=N}; .
Polyhedral domains are represented as { “index names” | “affine constraints using indices and parameters” }, where constraints can be intersected with
“&&”. Sometimes constraints can be expressed with short-hand notation like
“a<b<c” or
“(b,c)<0”.
Unions of such domains can be expressed as “{ a,b | constraints on a and b } || { c,d | constraints on c and d }”. One important point about Alphabets domains is that the names given to indices are only for textual representation. Internally, all analysis/transformation/code generation tools only care about which dimension the constraint applies to.For example, a domain { i,j | 0⇐i<j<N } is equivalent to { x,y | 0⇐x<y<N }, because
i and
x are both names given to the first dimension, and
j and
y are names given to the second dimension.
Now the only remaining step before a complete Alphabets program is writing the equations. After a little experience, the connection from mathematical equations (of a certain form) to Alphabets equations should become increasingly clear. There are two slightly different syntactic conventions for writing equations, one is called the “Show syntax” and the other is called “AShow syntax”. Show syntax is closer to the internal representation of Alphabets programs, and is more expressive when writing complex programs. AShow syntax uses “array notation” so that it is easier for people used to imperative programs.
We will first write the equation for
U in AShow syntax, and then move on to Show as we write the equation for
L.
In this equation,
U is on the left hand side, and the right hand side should define
U for each point in the declared domain of the
U variable.In AShow syntax, the names for indices used appear on the LHS of the equations.
For this example, the following LHS for
U gives
i, j as the names for the first and second dimensions to be used when writing the expressions in the RHS.These names do not have to match the names used in variable declaration. You could use
x,y instead of
i,j if desired.
U[i,j] = RHSexpr;
The first thing you notice in the definition of
U in the mathematical equation is the branch based on values of
i and
j. This branching is expressed with CaseExpression in Alphabets.A CaseExpression starts with the keyword “
case”, ends with keyword “
esac”, and has list of “
;”-delimited expressions, called “clauses” as its children.Often (but not always), each child of a case is a RestrictExpression (whose syntax is “domain : expr”), which restricts the domain to the specified domain.
Using the above expressions, the branching of the definition of
U is as follows :
U[i,j] = case {|1==i} : expr1; {|1<i} : expr2; esac;Note that because index names are already declared in the context (equation LHS), there is nothing to the left of the
| in the AShow syntax.
Moving on to the definitions in each case, the first case is . This is written as
A[i,j] in AShow syntax, similar to accessing an array. A variable without a square bracket, is treated either as a scalar variable (as in,
X[i,j] = 0) or as an access with the identity dependence function, (i.e.,
X[i] = A[i] would be the same as
X[i] = A).
The last piece missing before completing the definition of
U is the summation in the second branch.Mathematically, a reduction is an operation that applies an associative-comutative operator (in general, the operator may only be associative, but In Alphabets, we have only associative-comutative operators) to a set of values, such as summation (sum over a set of numbers).
Reductions are expressed with the following syntax :
reduce(operator, projection, expr);operator: operator to be applied (+, *, max, min, and, or)
In the mathematical equation, summation with one new index
k is used. For each value of
k, the expression
L[i,k]*U[k,j] is computed and added up to produce the result
U[i,j]. Thus, the projection function is
(i,j,k → i,j). (from the three dimensional space indexed by
i,j,k, all values computed at
[i,j,k] are used to compute
U[i,j] in the two dimensional space indexed by
i,j – i.e., the
k is 'projected out')
When the projection function is canonic (e.g.,
(i,j,k→i,j)), then the projection function can be replaced with a simpler syntax (AShow syntax for reductions) that specifies the names of new dimensions surrounded by square brackets.For example, the projection
(i,j,x,y→i,j) can be expressed as
[x,y].
Using the above, summation in the original equation can be written as the following Alphabets fragment.
reduce(+, [k], L[i,k]*U[k,j]);
Putting all this together, the final equation for
U is:
U[i,j] = case {|1==i} : A[i,j]; {|1<i} : A[i,j] - reduce(+, [k], L[i,k]*U[k,j]); esac;
This is exactly like the original equation Caveat
Now we will write the equation for
L, but this time in Show syntax. Unlike the AShow syntax, Show syntax does not rely on the context for naming of indices. Index names can be different in every (sub)expression if it makes sense to do so.Because of this, the LHS does not have square brackets, all we need is the variable name.
L = RHSexpr; //Show syntax
CaseExpression and RestrictExpression are same as AShow syntax. However, since index names are no longer deduced from the context where they occur, they must be explicitly named everywhere. While this may seem cumbersome, it allows expressions to have compositional semantics. In our example, the index names used in the domain of RestrictExpression have to be made explicit.The branch in the definition of
L becomes the following Alphabets :
L = case {i,j|1==j} : expr1; {i,j|1<i} : expr2; esac;
In the array notation in AShow syntax, a DependenceExpression was implicit: just add expressions within square brackets to access variables). In the Show syntax
DependenceExpression is used to explicitly specify which value of a variable is required for a computation. The syntax of
DependenceExpression is “(affine_function)@expr”, where
affine_function is of the form
(list_of_indices → list_of_affine_expressions). For example, the dependence
(i,j→i-1,i+j)@A means that at index point
(i,j) this computation evaluates to the value of
A at index point
(i-1,i+j).
The child of
DependenceExpression can be any Alphabets expression, possibly another DependenceExpression. For example,
(i,j→i,j,i+j,0)@(a,b,c,d→a,c-a)@A is a perfectly legal Alphabets expression.
Reductions in Show syntax are exactly like in the AShow syntax, except that the projection function is specified in the dependence syntax. This is all you need in order to write the rest in of the equation in Show syntax.
L = case {i,j|1==j} : (A / (i,j->j,j)@U); {i,j|1<i} : (A - reduce(+, (i,j,k->i,j), (i,j,k->i,k)@L*(i,j,k->k,j)@U))/(i,j->j,j)@U; esac;
Combine all of the above, and you will get the Alphabets program for LU decomposition. Don't forget the keyword
let/
through before equations the period at the end (since our example has no local variables). Notice how we can mix and match Show and AShow syntax within the program, but each equation must obviously, be consistent.
affine LUD {N|N>0} input float A {i,j|1<=(i,j)<=N}; output float L {i,j|1<i<=N && 1<=j<i}; float U {i,j|1<=j<=N && 1<=i<=j}; let U[i,j] = case {|1==i} : A[i,j]; {|1<i} : A[i,j] - reduce(+, [k], L[i,k]*U[k,j]); esac; L = case {i,j|1==j} : A / (i,j->j,j)@U; {i,j|1<j} : (A - reduce(+, (i,j,k->i,j), (i,j,k->i,k)@L*(i,j,k->k,j)@U))/(i,j->j,j)@U; esac; .
Analyses, transformations, and code generation of Alphabets programs are performed using the AlphaZ system. The normal interface for using AlphaZ is the scripting interface called compiler scripts. Given below is an example script for that does several things using the LUD program we wrote above.
# read program and store the internal representation in variable prog prog = ReadAlphabets("./LUD.ab"); # store string (corresponding to system name) to variable system system = "LUD"; # store output directory name to variable outDir outDir = "./test-out/"+system; # print out the program using Show syntax Show(prog); # print out the program using AShow syntax AShow(prog); # prints out the AST of the program (commented out) #PrintAST(prog); # generate codes (this is demand-driven, memoized code) generateWriteC(prog, system, outDir); generateWrapper(prog, system, outDir); generateMakefile(prog, system, outDir);Save this script with .cs extension, place the alphabets file in the same directory as the script, and then right click on the editor and select
“Run As → Compiler Script” to run the script.
If you get some error message, try looking at the first line of the error messages to find out what it is about. Common problems are:
FileNotFoundException in this case)
xxx does not exist)
In this tutorial, we use two basic code generators, without going into too much detail. The two types of codes generated are
WriteC and
Wrapper.
WriteC code may not be efficient, but it can be generated without any additional specification beyond the program.
Wrapper code is a wrapper around other generated codes that allocates/frees memory for input and output variables, and it also have different options for testing.
Note:Current implementation of the Wrapper prints out the bounding box of the domain of the output variable.
generateMakefile produces a Makefile that should compile the generated codes. You can make with different options.
Congratulations!! You are nearly at the end. Now, you will actually make and execute the code (in a separate terminal window).
compiles the code and produces an executable
xxx (where
xxx is the system name) that executes the program with default input that is 1 everywhere. Compiling with this option does not test very much, but it will test if it compiles and runs and produces no errors.
Compiles the code and produces an executable
xxx.check (
xxx is the system name) that prompts the user for all values of input variables.After executing, it prints out all values of the output variables.This option should be used for testing small to mid-sized input data.
Compiles the code with another code named
xxx_verify.c that defines a function
xxx_verify (
xxx is the system name).Users can provide different program as
xxx_verify to compare outputs.
Same as verify, except the inputs are generated randomly.
You will see that when you execute the code,
. You may be able to easily fix the error in your Alpha program and regenerate correctly executing C code, or you may want a bit of help. In either case, we would like to know. Please email Sanjay.Rajopadhye@colostate.edu with the error message that is produced. it will produce an error
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Perhaps the most useful feature of thermochemical equations is that they can be combined to determine Δ
H m values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is reaction of 1 mol C( s) and 0.5 mol O 2( g) to form 1 mol CO( g):
(Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) In step 2 the mole of CO reacts with an additional 0.5 mol O
2yielding 1 mol CO 2: \[ \text{C} \text{O} (g) + \frac{1}{2} \text{O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) ~~~~~~~~~~~~~~~~~~~~~~~~~ \Delta H_{m} = –283.0 \text{ kJ} = \Delta H_{2} \] The net result of this two-step process is production of 1 mol CO 2 from the original 1 mol C and 1 mol O 2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2.
Using the molecular representations of each reaction pictured above, confirm this conclusion, using the molecules to visually represent what is occurring.
On paper this net result can be obtained by
adding the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction
\[\text{C} (s) + \small\frac{1}{2}\normalsize\text{O}_2 (g) \qquad\quad\quad \space\space\space\rightarrow \cancel{\text{CO} (g)} \qquad \qquad \]
\[ \underline{\qquad\quad\space \small\frac{1}{2} \normalsize\text{O}_2 (g) + \cancel{\text{CO} (g)} \rightarrow \qquad \qquad \space\space \text{CO}_2 (g)} \] \[ \text{C}(s) + \space\space\space \text{O}_2 (g) + \cancel{\text{CO} (g)} \rightarrow \cancel{\text{CO} (g)} + \text{CO}_2 (g) \]
Experimentally it is found that the enthalpy change for the net reaction is the
sumof the enthalpy changes for steps 1 and 2: \[ \Delta H_{net} = –110.5 \text{ kJ} + (–283.0 \text{ kJ} ) = –393.5 \text{ kJ} = \Delta H_{1} + \Delta H_{2} \] That is, the thermochemical equation \[ \text{C} (s) + \text{ O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) \\ \Delta H_{m} = –393.5 \text{ kJ} \]
Notice how the equation above represents the reaction symbolically, while the 3D molecules show the microscopic view, and the final images show this process as we see it, on the macroscopic level.is the correct one for the overall reaction.
In the general case it is always true that
whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as
Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain Δ H m values for reactions which cannot be carried out experimentally, as the next example shows.
Example \(\PageIndex{1}\) : Enthalpy Change
Acetylene (C
2H 2) cannot be prepared directly from its elements according to the equation Hfor this reaction from the following thermochemical equations, all of which can be determined experimentally: m \[ \begin{align} \text{C} (s) &+ \space\space\space\text{O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) &\Delta H_{m} &= –393.5 \text{kJ} \tag{2a} \\ \text{H}_{2} (g) &+ \tfrac{1}{2} \text{O}_{2} (g) \rightarrow \text{H}_{2} \text{O} (l) &\Delta H_{m} &= –285.8 \text{kJ} \tag{2b} \\ \text{C}_{2} \text{H}_{2} (g) &+ \tfrac{\text{5}}{\text{2}} \text{O}_{2} (g) \rightarrow 2 \text{C} \text{O}_{2} (g) + \text{H}_{2} \text{O} (l) &\Delta H_{m} &= –1299.8 \text{kJ} \tag{2c} \end{align}\]
Solution: We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1): a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2 a) by 2. b) Since Eq. (1) has 1 mol H 2 on the left, we leave Eq. (2b) unchanged.
We then have
c) Since Eq. (1) has 1 mol C 2H 2 on the right, whereas there is 1 mol C 2H 2 on the left of Eq. (2c) we write Eq. (2c) in reverse. $$\begin{align} \text{2 C}(s) &+ 2\space\ce{O2}(g) \space\space ~\to~ \ce{2CO2}(g) &\Delta H_{m}&=\text{ 2 (}-\text{393.5) kJ} \\ \ce{H2}(g) &+ \tfrac{1}{2}\space\ce{O2}(g) \space ~\to~ \ce{H2O}(l) &\Delta H_{m} &= -\text{285.8 kJ} \\ \underline{ \ce{2 CO2}(g)} &\underline{\text{ +} \space\space\space\ce{H2O}(l) ~\to~ \ce{C2H2}(g) + \tfrac{5}{2}\ce{O2}(g)} &\Delta H_{m}&=-\text{(}-\text{1299.8 kJ)} \\ \ce{2 C}(s) + \ce{H2}(g) &+ 2\tfrac{1}{2}\ce{O2}(g) ~\to~ \ce{C2H2}(g) +\tfrac{5}{2}\ce{O2}(g) \end{align}$$ \[ \begin{align} & \Delta H_{m} = (-787.0 -285.8 + 1299.8) \text{ kJ} \\ & = 227.0 \text{ kJ}\end{align}\]
\[ 2 \text{C} (s) + \text{H}_{2} (g) \rightarrow \text{C}_{2} \text{H}_{2} (g) ~~~~~~~~~~~~~~~~~~~~~ \Delta H_{m} = 227.0 \text{ kJ}\]
Cancelling 5/2 O 2 on each side, the desired result is
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The Heat of Reaction (also known and Enthalpy of Reaction) is the change in the enthalpy of a chemical reaction that occurs at a constant pressure. It is a thermodynamic unit of measurement useful for calculating the amount of energy per mole either released or produced in a reaction. Since enthalpy is derived from pressure, volume, and internal energy, all of which are state functions, enthalpy is also a state function.
Introduction
ΔH, or the change in enthalpy arose as a unit of measurement meant to calculate the change in energy of a system when it became too difficult to find the ΔU, or change in the internal energy of a system, by simultaneously measure the amount of heat and work exchanged. Given a
constant pressure, the change in enthalpy can be measured as ΔH=q (see enthalpy for a more detailed explanation).
The notation ΔHº or ΔHº
rxn then arises to explain the precise temperature and pressure of the heat of reaction ΔH. The standard enthalpy of reaction is symbolized by ΔHº or ΔHº rxn and can take on both positive and negative values. The units for ΔHº are kiloJoules per mole, or kj/mol. ΔH and ΔHºrxn Δ = represents the change in the enthalpy; (ΔH products-ΔH reactants) a positive value indicates the products have greater enthalpy, or that it is an endothermic reaction (heat is required) a negative value indicates the reactants have greater enthalpy, or that it is an exothermic reaction (heat is produced) º = signifies that the reaction is a standard enthalpy change, and occurs at a preset pressure/temperature rxn= denotes that this change is the enthalpy of reaction The Standard State: The standard state of a solid or liquid is the pure substance at a pressure of 1 bar ( 10 5 Pa) and at a relevant temperature.
The ΔHº
rxn is the standard heat of reaction or standard enthalpy of a reaction, and like ΔH also measures the enthalpy of a reaction. However, ΔHº rxn takes place under "standard" conditions, meaning that the reaction takes place at 25º C and 1 atm. The benefit of a measuring ΔH under standard conditions lies in the ability to relate one value of ΔHº to another, since they occur under the same conditions. How to Calculate ΔH Experimentally
Enthalpy can be measured experimentally through the use of a calorimeter. A calorimeter is an isolated system which has a constant pressure, so ΔH=q=cp
sp x m x (ΔT)
How to calculate ΔH Numerically
To calculate the standard enthalpy of reaction the standard enthalpy of formation must be utilized. Another, more detailed, form of the standard enthalpy of reaction includes the use of the standard enthalpy of formation ΔH
º f:
\[ ΔH^\ominus = \sum \Delta v_p \Delta H^\ominus_f\;(products) - \sum \Delta v_r \Delta H^\ominus_f\; (reactants)\]
with
v p= stoichiometric coefficient of the product from the balanced reaction v r= stoichiometric coefficient of the reactants from the balanced reaction ΔH º f= standard enthalpy of formation for the reactants or the products
Since enthalpy is a state function, the heat of reaction depends only on the final and initial states, not on the path that the reaction takes. For example, the reaction \( A \rightarrow B\) goes through intermediate steps (i.e. \(C \rightarrow D\)), but A and B remain intact.
Therefore, one can measure the enthalpy of reaction as the sum of the ΔH of the three reactions by applying Hess' Law.
Additional Notes
Since the ΔHº represents the total energy exchange in the reaction this value can be either positive or negative.
A positive ΔHº value represents an additionof energy from the reaction (and fromthe surroundings), resulting in an endothermic reaction. A negative value for ΔHº represents a removalof energy from the reaction (and intothe surroundings) and so the reaction is exothermic.
Example \(\PageIndex{1}\): the combustion of acetylene
Calculate the enthalpy change for the combustion of acetylene (\(\ce{C2H2}\))
Solution
1) The first step is to make sure that the equation is balanced and correct. Remember, the combustion of a hydrocarbon requires oxygen and results in the production of carbon dioxide and water.
\[\ce{2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(g)}\]
2) Next, locate a table of Standard Enthalpies of Formation to look up the values for the components of the reaction (Table 7.2, Petrucci Text)
3) First find the enthalpies of the products:
ΔHº
f CO 2 = -393.5 kJ/mole
Multiply this value by the stoichiometric coefficient, which in this case is equal to 4 mole.
v
pΔH º f CO 2 = 4 mol (-393.5 kJ/mole)
= -1574 kJ
ΔH
º f H 2O = -241.8 kJ/mole
The stoichiometric coefficient of this compound is equal to 2 mole. So,
v
pΔH º f H 2O = 2 mol ( -241.8 kJ/mole)
= -483.6 kJ
Now add these two values in order to get the sum of the products
Sum of products (Σ v
pΔHº f(products)) = (-1574 kJ) + (-483.6 kJ) = -2057.6 kJ Now, find the enthalpies of the reactants:
ΔHº
f C 2H 2 = +227 kJ/mole
Multiply this value by the stoichiometric coefficient, which in this case is equal to 2 mole.
v
pΔHº f C 2H 2 = 2 mol (+227 kJ/mole)
= +454 kJ
ΔHº
f O 2 = 0.00 kJ/mole
The stoichiometric coefficient of this compound is equal to 5 mole. So,
v
pΔHº f O 2 = 5 mol ( 0.00 kJ/mole)
= 0.00 kJ
Add these two values in order to get the sum of the reactants
Sum of reactants (Δ v
rΔHº f(reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ The sum of the reactants and products can now be inserted into the formula:
ΔHº = Δ v
pΔHº f(products) - ? v rΔHº f(reactants)
= -2057.6 kJ - +454 kJ
= -2511.6 kJ
Example \(\PageIndex{1}\):
Add text here. For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Practice Problems Calculate ΔH if a piece of metal with a specific heat of .98 kJ·kg−1·K−1 and a mass of 2 kg is heated from 22 oC to 28 oC. If a calorimeter's ΔH is +2001 Joules, how much heat did the substance inside the cup lose? Calculate the ΔH of the following reaction: CO 2 (g)+ H 2O (g) -->H 2CO 3 (g)if the standard values of ΔH fare as follows: CO 2 (g): -393.509 KJ /mol, H 2O (g): -241.83 KJ/mol, and H 2CO 3 (g): -275.2 KJ/mol. Calculate ΔH if a piece of aluminum with a specific heat of .9 kJ·kg−1·K−1 and a mass of 1.6 kg is heated from 286 oK to 299 oK. If the calculated value of ΔH is positive, does that correspond to an endothermic reaction or an exothermic reaction? Solutions ΔH=q=cp spx m x (ΔT) = (.98) x (2) x (+6 o) = 11.76 kJ Since the heat gained by the calorimeter is equal to the heat lost by the system, then the substance inside must have lost the negative of +2001 J, which is -2001 J. ΔH º= ∑Δv pΔH º f(products) - ∑Δ v rΔH º f(reactants) so this means that you add up the sum of the ΔH's of the products and subtract away the ΔH of the products: (-275.2kJ) - (-393.509kJ + -241.83kJ) = (-275.2) - (-635.339) = +360.139 kJ. ΔH=q=cp spx m x (ΔT) = (.9) x (1.6) x (13) = 18.72 kJ Endothermic, since a positive value indicates that the system GAINED heat. References Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007. Zumdahl, Steven S., and Susan A. Zumdahl. Chemistry. Boston: Houghton Mifflin Company, 2007. Contributors Rachel Martin (UCD), Eleanor Yu (UCD)
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I want to run some simulation studies of (linear) factor models and for that reasons I am wondering about the features such a simulation should contain - every suggestion is welcome, I'll do my best to update my code accordingly to provide some general usable framework at the end of the day. Citations are welcome, too!
The question is related to this question, however, I think due to the presence of factors and predictors the setup is sufficiently different to ask a separate question.
This is my approach:
Underlying assumptions and statistical model for the returns
I want to work with linear factor models of the form \begin{align}E[r_t] =& \beta(z_t) E[f_t] \\ \beta(z_t) =& \beta_0 + \beta_1z_t \end{align} which essentially implies a model for the returns which are driven by the factor returns (I'll go for market excess returns, so it is a CAPM model) and the factor-loadings are time-varying, where the dynamics come from some macro-risk variable $z_t$ which may predict business cycles (I'll go for the aggregated dividend yield). Substituting yields $$E[r_t] = \beta_0 E[f_t] + \beta_1(z_t \times f_t) $$ Based on this framework I first estimate appropriate parameters $\beta$ (and prevailing mispricing $\alpha$) using some test asset. In my case I used some B/M and Size portfolio sorts from Kenneth Frenchs Homepage to get estimates (based on simple OLS) $\hat{\alpha}$ and $\hat{\beta}$ which I'll assume to be fixed (probably one could base the simulation on sampled values of these parameters as well). One could also adjust for time-series characteristics by estimating $\hat{\alpha}$ and $\hat{\beta}$ based on some sub-samples of the data, employing rolling windows, state-space models, etc.
The following code is running the regressions. I gathered the FF Market Data and Portfolio-Sort returns (from Kenneth Frenchs Homepage), the dividend yield time-series (from Goyal Welchs Homepage) and combined everything in one data.frame available on a gist:
library(RCurl)library(tidyverse)# Read-in data (contains asset returns, market (factor) returns ---------# and dividend yield (predictive variable)) -----------------------------x <- getURL("https://gist.githubusercontent.com/voigtstefan/ be71f96eea328dc32786a56409362b40/raw /3b8bd42fdc3d327767941e44de6321a80a76a247/ data.csv")data <- read_csv(x)
The linear model is fitted as follows to get estimates of $\hat \alpha$ and $\hat\beta$:
# Fit linear model on predictive variables and market factor -------------fit <- lm(Returns~dy+Market,data)
Simulate (noisy) factor returns
In a next step I simulate factor returns which are distributed similar to the initial market returns. I only matched the first two moment of this distribution, but probably one could either adjust for the presence of fat tails (estimate a student-t), adjust for time-series aspects (estimate time-varying volatilities).
I want to see how factor models react to the inclusion of
new factors which do not have explanatory power. In other words, I'll feed the model with additional simulated factors but the simulated returns itself are only driven by, say, the market data.
# Simulate factors ------------------------------seed.init <- 1 # For reproducabilitysim.factor <- matrix(rnorm(nrow(data), mean(data$Market), sd(data$Market)),nrow=nrow(data),ncol=1)
Simulate returns
Now that I have simulated factors, I can simulate returns by employing the statistical model explained above: I use the predictors (which I do not simulate) and the simulated market returns to sample returns from the regression equation. The factor loadings are replaced with the estimated coefficients.
I gathered (factor) market-returns and
we simulate return time-series by running a time-series regression of the portfolio sorts based on B/M and Size on two explanatory variables: market returns and the dividend yield as predictive variable. In a second step I use the estimated parameters to simulate time-series of 150 years of monthly data. I compute the HJ-measure based on the usual moment conditions (with and without conditioning information) for models which always include the market factor but in addition by adding randomly generated \textit{factors} which are independent from the returns and the market factor. In addition I compute the corresponding model likelihoods of the unrestricted version of our Bayesian model.
# Simulate returns ------------------------------sim.returns <- cbind(1,data$dy,sim.factor)%*%coef(fit) + rnorm(n=nrow(data),sd=sd(fit$residuals))sim.data <- data.frame(returns = sim.returns, factor = sim.factor, dy = data$dy)
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Now showing items 1-10 of 55
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Multiplicity dependence of the average transverse momentum in pp, p-Pb, and Pb-Pb collisions at the LHC
(Elsevier, 2013-12)
The average transverse momentum <$p_T$> versus the charged-particle multiplicity $N_{ch}$ was measured in p-Pb collisions at a collision energy per nucleon-nucleon pair $\sqrt{s_{NN}}$ = 5.02 TeV and in pp collisions at ...
Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Directed flow of charged particles at mid-rapidity relative to the spectator plane in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(American Physical Society, 2013-12)
The directed flow of charged particles at midrapidity is measured in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV relative to the collision plane defined by the spectator nucleons. Both, the rapidity odd ($v_1^{odd}$) and ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
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 ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
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For $k=0,1,...$, let $a_k$ be the rational number defined by the serie expansion
$\frac{1}{e^x -1} = \frac{1}{x}+ \sum_{k=0}^\infty (-1)^k \frac{a_k}{\Gamma(k+1)} x^k$
The computation of the question simply shows that if there would be a "good value" to attach to $\sum_{n=1}^{+ \infty} n^k$, it should be $a_k$. The real question is why this "good value" is the same as the one obtained via the zeta function: $\zeta (-k)$. The answer is that the analytic continuation of the zeta function is defined by essentially the same computation.
The zeta function is defined by the formula $\zeta(s) = \sum_{n \geq 1} n^{-s}$ for $s \in \mathbb{C}$, $Re(s)>1$. To show that this function admits a meromorphic continuation to the all complex plane, a way to do is to use the trick to write
$n^{-s} = \frac{1}{\Gamma(s)} \int_{0}^{+\infty} e^{-nt} t^{s-1} dt$.
(Remark not useful for what follows: this trick is used very often in quantum field theory, where $t$ is called in this context a "Schwinger parameter" or the "Schwinger proper time").
Using this and summing the geometric serie, one finds for $Re(s)>1$.
$ \zeta(s)= \frac{1}{\Gamma(s)} \int_{0}^{+\infty} \frac{t}{e^t - 1} t^{s-2} dt$.
It is possible to do an integration by part to obtain:
$ - \frac{1}{(s-1) \Gamma(s)} \int_{0}^{+\infty} \frac{d}{dt}(\frac{t}{e^t - 1}) t^{s-1} dt$
The point is that this expression makes sense for $Re(s)>0$ so this formula defines a (and so the) analytic continuation of the zeta function to $Re(s)>0$. Doing successive integrations by part, one obtains the "full" zeta function and it is immediate from this that $\zeta(-k)=a_k$.
This computation defining the analytic continuation of the zeta function is really the same as (precisely the Mellin transform of) the OP's computation. It is one of the standard way to guess what $\sum_{n=1}^{+\infty} n^k$ "should be".
Remark: essentially the same computation appears in other ways to "regularize" $\sum_{n=1}^{+\infty} n^k$, for example the exponential regularization $\sum_{n=1}^{+\infty} n^k e^{-n \epsilon}$
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I would like to have the unit circle, centered at the origin, on the Cartesian plane drawn. The origin is to be marked with a dot and labeled "O" and five dots are to be drawn on the circle, one on the x-axis, and the others at
k(2\pi/5) radians from the positive x-axis for each integer
1 \leq k < 5, and they are to be labeled "
w_{k}". Arrows are to be drawn from the origin to each
w_{k}, too. One angle is to be drawn and labeled - the one from the positive x-axis to the ray through
w_{1}.
I have included some of the code. It starts with the following commands.
\tikzset{mydot/.style={fill,circle,inner sep=1.5pt}}
Is there a manual that explains each of these commands? I guess that this is instructing LaTeX how to make the dots indicating the coordinates each time "
\mydot" is in the code.
\documentclass{amsart}\usepackage{amsmath}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsthm}\usepackage{newlfont}\usepackage{mathtools}\usepackage{tikz}\tikzset{mydot/.style={ fill, circle, inner sep=1.5pt }}\begin{document}\begin{tikzpicture}[>=latex]% the coordinates of the vertices\coordinate (O) at (0,0);\coordinate (w_{1}) at (\cos(2\pi/5), \sin(2\pi/5));% the axes\draw[help lines,->] (-1.5,0) -- (1.5,0);\draw[help lines,->] (0,-1.5) -- (0,1.5);% labelling the vertices\node[mydot,label={below:$O$}] at (O) {};\node[mydot,label={right:$w_{1} = \bigl(\cos(2\pi/5), \sin(2\pi/5)\bigr)$}] at (w_{1}) {};% the arcs for the angle\begin{scope}[gray]\draw[->] (1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=2\pi/5] node[midway,right] {$2\pi/5$};\end{scope}\end{tikzpicture}\end{document}
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Methods for Enforcing Inequality Constraints
How do you find the shortest overland distance between two points across a lake? Such obstacles and bounds on solutions are often called inequality constraints. Requirements for nonnegativity of gaps between objects in contact mechanics, species concentrations in chemistry, and population in ecology are some examples of inequality constraints. Previously in this series, we discussed equality constraints on variational problems. Today, we will show you how to implement inequality constraints when using equation-based modeling in COMSOL Multiphysics®.
A Penalty Method for Path Planning
Assume you want to go from a point (0,0.2) to (1,1), but there is a circular obstacle centered at (0.5,0.5) and radius 0.2. The curve (x,u(x)) that minimizes the distance between two end points, (a,u(a)) and (b,u(b)), should minimize the functional
In the Euclidean space, the shortest distance between two points is a straight line. As such, solving for this equation using what we have discussed so far should give a straight line. However, in our case, that straight line goes through the obstacle. This is not feasible. We have to respect the constraint that our path cannot go into the obstacle; that is, the distance from a point on the path to the center of the circular obstacle should be greater than or equal to the radius of the circle.
A basic schematic of the inequality constrained problem.
We want to add a term to our functional to penalize constraint violation. For an equality constraint g=0, we penalize both negative and positive values of g. For an inequality constraint g \leq 0, we have to penalize only positive values of g, while negative values are acceptable.
In the obstacle problem, this means we have to penalize only when the path tries to penetrate the obstacle. It is feasible, although maybe not optimal, to stray far from the shores of our circular lake, thus our penalty-regularized objective function is
where \mu is the penalty parameter and we used the ramp function
x& x\ge 0 \\
0 & x \le 0.
\end{cases}
For a general functional
to be minimized subject to the inequality constraint
the penalized functional becomes
The final step is to take the first variational derivative of the above functional and set it to zero. Here, we need to remember that the derivative of a ramp function is the Heaviside step function H.
In the above derivation, we used the fact that \big \langle x \big \rangle H(x) = xH(x). Note that we are discussing nonpositivity constraints here. Nonnegativity constraints can be dealt with by taking the negative of the constraint equation and solving for the equivalent nonpositivity constraint.
The first term in the above variational equation is the now-familiar contribution from the unconstrained problem. Let’s see how to add the penalty term using the
Weak Contribution node. We will use the simpler Dirichlet Boundary Condition node to fix the ends. Implementing an inequality constraint using the penalty method.
Note that in the above weak contribution, we did not include \frac{\partial g}{\partial u^{\prime}}, since the constraint in this problem depends only on the solution and not on its spatial derivative.
Solving this problem for a sequence of increasing penalty parameters, while reusing the previous solution as an initial estimate, we get the result shown below.
Finding the shortest path around an obstacle using the penalty method. A zero penalty returns the unconstrained solution.
In our discussion of the physical interpretation of the penalty method for equality constraints, we have said that the penalty term introduces a reaction proportional to the constraint violation. That interpretation applies, but for inequality constraints, the reaction is one-sided.
Imagine forcing a bead to stay in place using compressive springs. The springs are not hooked to the bead; they just touch. To keep the bead at x=0, we have to use two springs, one on each side of the origin. If it is acceptable for the bead to go to the left of the origin, our constraint is x \leq 0 and we have to place the compressive springs only on the right side. That is why we penalize \big \langle g \big \rangle instead of |g|.
A spring analogy of penalty enforcement of equality (multilateral) and inequality (unilateral) constraints. The Lagrange Multiplier Method
When we discussed the numerical properties of different constraint enforcement strategies, we said that while the Lagrange multiplier method enforces constraints exactly, it has certain undesirable properties in numerical solutions. Namely, it is sensitive to initial estimates of the solution and can require a direct linear solver. These downsides are still there, but in inequality constraints, there is an additional challenge. Specifically, the constraint may not always be active.
Consider our path planning problem. In parts of the path not touching the obstacle, the constraint is not active. However, we do not know beforehand where the constraints will be active and not active. There are several strategies to deal with this, but we would like to point out two commonly used methods.
Active Set Strategy
Assume some constraints are active and some are not. In a distributed constraint, this means picking points where we expect g=0 and assuming the other points are inside the feasible region with g <0. At the inactive points, the Lagrange multiplier should be zero. On the active set, the Lagrange multiplier should not be negative. These are the so-called Karush-Kuhn-Tucker (KKT) conditions. If, after computation, the active set changes or the KKT conditions are violated, we have to appropriately update the active set and recompute. The following flowchart summarizes this procedure:
An active set strategy for the Lagrange multiplier enforcement of inequality constraints.
Through this iterative process, each iteration solves an equality constrained problem. Lagrange multiplier implementations of equality constraints have been discussed in the previous blog post in this series.
Slack Variables
Another strategy is introducing the so-called
slack variables to change the inequality into equality. The constraint g \leq 0 is equivalent to g + s^2=0. Now we have an equality constraint involving a new slack variable. If the inequality is a distributed (pointwise) constraint, the slack variable as well as the Lagrange multiplier will be functions.
On one hand, the slack variable strategy introduces yet another unknown to be solved for, but solves the problem in one go. On the other hand, the active set strategy does not need this variable but has to solve multiple problems until the active set stops changing.
The Augmented Lagrangian Method
Like we did for equality constraints, let’s use a calculus problem to discuss the basic ideas of this method. Augmenting the unconstrained objective function with a Lagrange multiplier term with an estimated multiplier and a penalty term, we have
The first-order optimality condition for this constraint is
This suggests the multiplier update
If we start with a zero initial estimate for the Lagrange multiplier, the above equation updates the Lagrange multiplier only when g is positive. Thus, the Lagrange multiplier is never negative and is positive only on points that tend to violate the constraint.
The augmented Lagrangian method, being an approximate constraint enforcement strategy, will allow a small violation of the constraint. In fact, if there is no violation of the constraint, the Lagrange multiplier stays zero. As such, this strategy satisfies the dual feasibility part (\lambda \ge 0) of the KKT conditions exactly but satisfies the complementary slackness part (\lambda g = 0) only approximately. Both conditions are exactly satisfied with the Lagrange multiplier method.
Obstacles with Irregular Shapes
In today’s example, we used a very idealized lake whose boundaries could be described by a simple function. Or perhaps it was a pool. Sometimes, inequality constraints have such simple analytical forms. Nonnegativity constraints in chemical reactions or population dynamics are examples of this. However, other constraints do not have such simple forms.
For example, in contact mechanics, we want to keep the gap between contacting objects nonnegative, but the boundaries of the objects are rarely so simple that we can define them by smooth analytical functions. On top of that, for large deformation problems, we want to enforce the contact constraints on the deforming domains, but we definitely don’t have an analytical description of the deforming domains. As such, the constraints have to be satisfied on the discrete version of the objects obtained from the mesh. Complex search operations have to be used to find out, as the objects move and deform, what points come into and out of contact with other points. There are built-in features for this and other contact simulation procedures in the COMSOL Multiphysics® software.
For some simple geometries and deformations, we can conceivably use the General Extrusion operator to compute the distance between deforming objects without using the contact mechanics functionality. More complicated geometries would need the latter.
Up Next…
So far in this series, we have shown how to solve constrained and unconstrained variational problems using COMSOL Multiphysics and discussed the pros and cons of different numerical strategies for constraint enforcement. That said, we limited ourselves to 1D single-physics problems and variational problems with, at most, first-order derivatives in the functional.
With the basics of this subject covered, we hope we are ready to tackle higher dimensions, higher-order derivatives, and multiple fields. This will come in the next and final blog post in this series. Stay tuned!
View More Blog Posts in the Variational Problems and Constraints Series Part 1: Introduction to Modeling Soap Films and Other Variational Problems Part 2: Specifying Boundary Conditions and Constraints in Variational Problems Part 3: Methods for Dealing with Numerical Issues in Constraint Enforcement Part 5: Image Denoising and Other Multidimensional Variational Problems Comments (1) 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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The luminescence spectrum of the hydrogen atom reveals light being emitted at discrete frequencies. These spectral features appear so sharp that they are called lines. These lines, occurring in groups, are found in different regions of the spectrum; some are in the visible, some in the infrared, and some in the vacuum ultraviolet. The occurrence of these lines was very puzzling in the late 1800’s. Spectroscopists approach this type of problem by looking for some regularity or pattern in the observations. Johannes Rydberg recognized a pattern and expressed it in terms of the following formula,
\[ \bar {\nu} = R_H \left ( \frac {1}{f^2} - \frac {1}{i^2} \right ) \label {2-8}\]
Here \(\bar {\nu}\) is the “frequency” of the line in wavenumber units
\[\bar {\nu} = \dfrac {\nu}{c} \label {2-9}\]
\(R_H\) is a constant equal to 109,677.581 cm
-1, now called the Rydberg constant, and \(f\) and \(i\) are positive integers with \(i > f\). Different groups of lines, called Rydberg series, are obtained for different values of f. The lines in each series arise from a range of values for \(i\). This analysis by Rydberg was pretty amazing. It pictured the hydrogen atom as some sort of counting machine that utilized integer numbers for some unknown reason.
Example \(\PageIndex{1}\)
Calculate the wavelength of a line in the hydrogen atom luminescence spectrum corresponding to f = 7 and i = 8. In which region of the electromagnetic spectrum will this line appear?
Since the Rydberg equation was derived empirically (i.e., invented to describe experimental data), the next question was, “Can the Rydberg equation and the origin of the integer values for \(f\) and \(i\) be obtained from theoretical considerations?” This question was enormously difficult for scientists at the time because the nature of the spectrum, discrete lines rather than a continuous frequency distribution, and the very existence of atoms, was not consistent with existing physical theories.
Example
Explain what it means to say a constant or an equation is empirical. Give an example of a value that is determined empirically.
Contributors Adapted from "Quantum States of Atoms and Molecules" by David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski
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Stability in representation theory of the symmetric groups
In the finite-dimensional representation theory of the symmetric groups
$$S_n$$ over the base field $$\mathbb{C}$$, there is an an interesting phenomena of "stabilization" as $$n \to \infty$$: some representations of $$S_n$$ appear in sequences $$(V_n)_{n \geq 0}$$, where each $$V_n$$ is a finite-dimensional representation of $$S_n$$, where $$V_n$$ become "the same" in a certain sense for $$n >> 0$$. One manifestation of this phenomena are sequences $$(V_n)_{n \geq 0}$$ such that the characters of $$S_n$$ on $$V_n$$ are "polynomial in $n$". More precisely, these sequences satisfy the condition: for $$n>>0$$, the trace (character) of the automorphism $$\sigma \in S_n$$ of $$V_n$$ is given by a polynomial in the variables $$x_i$$, where $$x_i(\sigma)$$ is the number of cycles of length $$i$$ in the permutation $$\sigma$$. In particular, such sequences $$(V_n)_{n \geq 0}$$ satisfy the agreeable property that $$\dim(V_n)$$ is polynomial in $$n$$. Such "polynomial sequences" are encountered in many contexts: cohomologies of configuration spaces of $$n$$ distinct ordered points on a connected oriented manifold, spaces of polynomials on rank varieties of $$n \times n$$ matrices, and more. These sequences are called $$FI$$-modules, and have been studied extensively by Church, Ellenberg, Farb and others, yielding many interesting results on polynomiality in $$n$$ of dimensions of these spaces. A stronger version of the stability phenomena is described by the following two settings:
- The algebraic representations of the infinite symmetric group
$$S_{\infty} = \bigcup_{n} S_n,$$ where each representation of $$S_{\infty}$$ corresponds to a ``polynomial sequence'' $$(V_n)_{n \geq 0}$$.
- The "polynomial" family of Deligne categories $$Rep(S_t), ~t \in
\mathbb{C}$$, where the objects of the category $$Rep(S_t)$$ can be thought of as "continuations of sequences $$(V_n)_{n \geq 0}$$" to complex values of $$t=n$$. I will describe both settings, show that they are connected, and explain some applications in the representation theory of the symmetric groups.
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