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A) Let $\tau$ be a Grothendieck pretopology on any category with fiber products. Define a new Grothendieck pretopology $\tau'$ where a cover $\{U_i \to X\}$ is a $\tau'$ cover if and only if there exists a refinement $\{V_{ij} \to X\}$ (i.e. there exists for each $ij$ an $X$-morphism $V_{ij} \to U_i$) such that $\{V_{ij} \to X\}$ is a $\tau$ cover. Then $\tau$ and $\tau'$ generate the same topologies.
B) The operation (lets say $\Phi$) that takes topologies to pretopologies composed with the operation (say $\Psi$) that takes pretopologies to topologies gives back the same topology that you started with (i.e. $\Psi \Phi = id$). So for any pretopology $\tau$, the pretopology $\Phi\Psi \tau$ gives the same topology as $\tau$.
C) Related to (A), on the category of affine schemes, we can define an fppf cover as a family $\{U_i \to X\}$ which is jointly surjective, and such that each morphism is flat, and finitely presented. We could also include the requirement that in addition the morphisms be quasi-finite. Corollary 17.16.2 in EGA IV implies that these two pretopologies gives rise to the same topology.
In response to the first question, there are surely people more expert than me, but it seems to me that Grothendieck topologies in algebraic geometry are almost always defined via Grothendieck pretopologies. I think topologies are the nicer concept, but while the idea is suited to proving very general things about topoi, if you have a specific pretopology such as the étale, Zariski, Nisnevich, flat, cdh, envelopes (also called proper cdh - see 18.3 in Fulton's "Intersection theory" and the Mazza, Voevodsky, Weibel book), etc having actual scheme morphisms in your hands allows you to apply the strong algebr-geometric results which are largely responsible for making topologies such a powerful tool in algebraic geometry. Of course, I'm sure someone in logic would have a very different point of view.
In relation to this line of thought, Voevodsky - while working with the Nisnevich and cdh topologies - found it useful to further simplify the data leading to a topology in the notion of a cd structure (see the papers "Unstable motivic homotopy categories inNisnevich and cdh-topologies" and "Homotopy theory of simplicial sheaves in completely decomposable topologies").
While I'm talking about Voevodsky's work, he also uses from time to time an idea which he calls "covers of normal form" in "Homology of schemes I", but the same idea is used implicitely in the appendix to "Singular homology of abstract algebraic varieties". The way I understand this phenomenon is the following. If you have two pretopologies $\sigma$ and $\rho$ on a category with fiber products, then the covers of the pretopology generated by $\sigma$ and $\rho$ are finite compositions of covers which are either a $\sigma$ cover or a $\rho$ cover. Lets denote the new pretopology by $\langle \sigma, \rho \rangle$. Many pretopologies $\tau$ in common use, are actually generated by two other pretopologies in the sense that $\langle \sigma, \rho \rangle$ and $\tau$ give rise to the same topology.
For example, the cdh pretopology is generated in this way (by definition) by the Nisnevich pretopology and the pretopology of envelopes. Voevodsky shows that the h-pretopology is generated like this Zariski and the pretopology whose covers are jointly surjective families of proper morphisms. The qfh pretopology is generated like this by étale and the pretopology whose covers are jointly surjective families of finite morphisms. |
I have been wondering about some of the different uses of Generalized Complex Geometry (GCG) in Physics. Without going into mathematical detail (see Gualtieri's thesis for reference), a Generalized Complex Geometry attempts to unify symplectic and complex geometry by considering the bundle $TM\oplus T^* M$ with its natural metric $\langle X+\xi, Y+\eta\rangle = \frac{1}{2} \left( \eta(X) + \xi(Y)\right)$ and the Courant Bracket.
The first hints of the necessity of GCGs in Physics came up in a famous paper by Gates, Hull and Roc̆ek, in which they found an 'extra' supersymmetry in the $(2,2)$ supersymmetric model. This extra symmetry turns out to be related to specifying two (integrable) complex structures $J_1, J_2$ which in turn are covariantly constant under
torsionful connections. This means that the manifold need not be Kähler (which is Hermitian and Torsion-free) and led Nigel Hitchin (and his students) to propose more general geometries that could be useful in physics.
More recently, a connection between GCGs and AdS/CFT has been discovered. Recall that in AdS/CFT, we consider a spacetime that is a warped product of $AdS_4$ and a 6-manifold. It turns out that it is natural to consider a 5-manifold $Y^5$ whose cone has some special geometry. If this geometry is Calabi-Yau then such a manifold is known as a
Sasaki-Einstein manifold. As such, we start out with a metric of the form,
$ g_{ij} = g_{AdS_5} + g_{Y^5} = e^{2\Delta + \phi/2}r^2 \left(g_{\mathbb{R}^{1,3}} + r^{-4} g_{C(Y^5)} \right) $
where $g_{C(Y^5)} = dr^2 + r^2 g_{Y^5}$ (the
metric cone of $Y^5$). If we want to obey $\mathcal{N}=1$ supersymmetry, we must enforce on the dilatino and gravitino which eventually leads to a condition on pure spinors. In Generalized Complex Geometry, $TM\oplus T^*M$ naturally acts as a Clifford Algebra on the Clifford Module $\wedge^{\bullet} T^*M$. It turns out that in this situation, we can represent the pure spinors over a Generalized Complex Manifold as the sum of differential forms of different degree (polyforms). As such GCGs can be good candidates for $C(Y^5)$.
Related to this is the result of Graña, et. al which can be poorly paraphrased as:
All $\mathcal{N}=1$ solutions of IIB string theory are described by a pair of pure spinors $\Omega_{\pm}$(up to $B$ transform) that satisfy a pair of differential constaints, $d \Omega_+ = 0$, $d\Omega_- = dA \wedge \Omega_+ + \frac{i}{8}e^{3A}e^{-B}\star (F_1 - F_3 + F_5)$, where $F_k$ is the $k$-form flux and $A = 2\Delta + \phi/2$
I was wondering if there were any other significant uses of GCGs in physics that I have not mentioned. I have seen a variety of papers that do mention GCGs, but outside of these examples, I have not been particularly compelled by their usage.
Thanks! |
Volume 4, Issue 1
March 1983, pages 1-63
pp 1-9 March 1983
The effective temperatures of the classical Cepheids RT Aur and T Vul have been determined by a comparison of their spectral scans with appropriate model atmospheres. The radii of the stars have been determined through the Wesselink method. Using these temperatures and the Wesselink radii, the luminosities of the stars have been determined. These radii estimates, including the radii of SU Cas (Joshi & Rautela 1980) and
ζ Gem (unpublished) fit better in the theoretical period-radius relationship given by Cogan (1978), as compared to earlier determinations of Wesselink radii. The pulsation masses and evolutionary masses of the stars have been calculated. The pulsation to evolutionary mass ratio is derived to be 0.85.
Based on the effective temperatures obtained by us at different phases of the stars a
θ c − ( B-V) 0 relationship is found of the form,$$\begin{gathered} \theta _e = 0.274 (B - V)_0 + 0.637 \hfill \\ \pm 0.011 \pm 0.007 \hfill \\ \end{gathered} $$
pp 11-17 March 1983
We have calculated the effects of irradiation from a point source observed at infinity. Plane-parallel approximation and spherically-symmetric approximations are employed in calculating the self-radiation field for the sake of comparison. It is found that there are considerable changes in the radiation received at infinity between the approximation of plane-parallel stratification and spherical symmetry.
pp 19-21 March 1983
Photoelectric radial-velocity measurements show that HD 107742 is a spectroscopic binary with a period of 875 days and a small (but definitely non-zero) eccentricity. The star is not a member of the Coma Cluster, against which it is seen projected.
pp 23-26 March 1983
Photoelectric radial-velocity measurements show that HD 121844 is a spectroscopic binary with a period of 303 days, a high eccentricity (0.49) and a high negative γ-velocity (- 61 km s
-1).
pp 27-33 March 1983
The study of the variation of equivalent width in a Rayleighscattering planetary atmosphere along the intensity equator and along the mirror meridian on which
μ = μ 0 shows that the equivalent widths decrease monotonically towards the poles, the limb and the terminator with the following characteristics: (i) the weakest lines exhibit the maximum change; (ii) the I e r component shows more change than the I e r component; (iii) the decrease towards the limb or the terminator is not as sharp as that towards the poles; (iv) I e r component shows more decrease towards the limb while I e r component shows more decrease towards the terminator; and (v) the relation W ( μ, φ; μ 0, φ 0) = W ( μ 0, φ 0; μ, φ) holds for the total intensity. These results are qualitatively in agreement with the observations of absorption bands in the spectra of Venus, Jupiter and Saturn.
pp 35-46 March 1983
The variation of the polarization profiles, the Stokes parameters Q and
U, and the angle defining the plane of polarization along the intensity equator and along the mirror meridian, on which μ = μ 0, in a Rayleighscattering atmosphere is studied. It is found that these variations are more complex than thought hitherto, particularly at large phase angles.
pp 47-51 March 1983
We have observed the region of the Coma cluster at 34.5 MHz with a resolution of 26 arcmin × 40 arcmin. A map of the diffuse halo (Coma C) is presented. The size of the halo is found to be 54 arcmin × 30 arcmin. The position angle is 50° ± 10° and the integrated flux is 60 ± 11 Jy.
We have also found an extended source to the south of Coma A. The measured half-power widths of this source are 30 arcmin × 40 arcmin. The position angle is 135° and the integrated flux is ~ 15 Jy at 34.5 MHz. The spectral index in the frequency range 408 to 34.5 MHz is -1.0. It is suggested that this source also belongs to the Coma cluster.
pp 53-58 March 1983
The true potential energy curves for the electronic ground states of astrophysically important AlH and CaH molecules are constructed by the Rydberg-Klein-Rees method. Empirical potential functions, of three-parameters by Lippincott, of five-parameters by Hulburt and Hirsch-felder and, of electronegativity by Szöke and Baitz, are examined for the adequacy to represent the true curve. From the best-fitting function, the dissociation energies
D 0 0 of AlH and CaH molecules are estimated to be 2.99 ± 0.08 and 2.72 ± 0.06 eV respectively. The force constants indicate that these values are of correct magnitude.
pp 59-63 March 1983
IUE observations of HD 62001 do not confirm the ultraviolet variability of 0.15 mag seen from the ANS observations. These intriguing variations in ANS observations seem to be caused by a nearby (∼ 70 arcsec away) B star getting in and out of the field of view of the telescope, which had a jitter of 20 arcsec. There seems to be no evidence for the presence of a hot companion, either from the energy distribution or from UV variability. However, visual light variations are present. This, coupled with the radial velocity variations, indicates a binary nature probably of Algol type.
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Volume 40 | Issue 5 October 2019
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I would like to know what is the induced matrix norm of a matrix $A$, when the domain space is equipped with $l_2$ norm and the range space is equipped with $l_\infty$ norm, i.e., what is $\lvert| A \rvert|_{2 \infty}$? The expression for $\lvert| A \rvert|_{\infty 2}$ is given here.
In general, when $X$ is a normed space, the $X\to \ell_\infty$ norm of a matrix $A$ is computed as
$$ \sup_{\|x\|_X\le 1}\|Ax\|_\infty = \sup_{\|x\|_X\le 1} \max_i \left|\sum_{j} a_{ij}x_j\right| = \max_i \sup_{\|x\|_X\le 1} \left|\sum_{j} a_{ij}x_j\right| = \max_i \|A_{i*}\|_{X^*} $$ where $A_{i*}$ is the $i$th row of $A$, and $X^*$ is the dual of $X$. (Since $A_{i*}$ is a map from $X$ to scalars, its $X^*$ norm makes sense.) Simply put, when the target is $\ell_\infty$, the rows (components of the map) can be considered independently of one another.
In the special case $\ell_2\to\ell_\infty$ we end up with $$ \|A\|_{2\infty} = \max_i \sqrt{\sum_j |a_{ij}|^2} $$ which is the maximal Euclidean length of the rows of $A$. |
Does the sum $$\sum_{n=1}^{\infty}\frac{\tan n}{n^2}$$ converge?
note
$$ \cos(x) = \prod_{n=1}^{\infty}\left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right) $$
then $$ \log(\cos(x)) = \sum_{n=1}^{\infty}\log\left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right) $$ which gives $$ =-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{4^k x^{2k}}{\pi^{2k}{k} \ {(2n-1)}^{2k}} $$ summing over n yields $$ =-\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^k x^{2k}}{\pi^{2k}{k}} $$ where $ \xi(s) = \zeta(s)\left(1 - \frac{1}{2^s}\right)$ and $\zeta(s)$ is the Riemann Zeta Function. Derivation on both sides with respect to $x$ yields $$ -\frac{\sin(x)}{\cos(x)} = -2\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^kx^{2k-1}}{\pi^{2k}} $$ Cancelling the negative, dividing by $x^2$, and summing over $x$ gives $$ \sum_{x=1}^{\infty}\frac{\tan(x)}{x^2} = 2\sum_{x=1}^{\infty}\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^k x^{2k-3}}{\pi^{2k}}=2\sum_{k=1}^{\infty}\frac{{\xi(2k)} \zeta(3-2k)\ 4^{k}}{\pi^{2k}} $$ At $k = 1$ we have $\zeta(3 - 2(1)) = \zeta(1) = \infty \therefore$ the R.H.S must diverge.
It may be useful to compare two other, similar, series. Mathworld gives definitions and analysis of the Cookson Hill and Flint Hill series, defined as the infinite sum of $\frac{\sec^2 n}{n^3}$ and $\frac{\csc^2 n}{n^3}$, respectively. The article for the Flint Hill series links to a paper by M. A. Alekseyev at arxiv.org, who demonstrates the the general convergence of such series can be linked to the irrationality measure of $\pi$. Giving it a quick look, this might be helpful to sketch out a proof- one could bound tan(x) with other trigonometric functions and then determine what implications such a bounding would have on $\mu(\pi)$, and then determine whether or not this satisfies the current best bounding (around 7.5, IIRC), in which case the series converges, or whether it violates it, in which case the problem is undetermined right now. |
Overview
The derivatives of functions are used to determine what changes to input parameters correspond to what desired change in output for any given point in the forward propagation and cost, loss, or error evaluation &mdash whatever it is conceptually the learning process is attempting to minimize. This is the conceptual and algebraic inverse of maximizing valuation, yield, or accuracy.
Back-propagation estimates the next best step toward the objective quantified in the cost function in a search. The result of the search is a set of parameter matrices, each element of which represents what is sometimes called a connection weight. The improvement of the values of the elements in the pursuit of minimal cost is artificial networking's basic approach to learning.
Each step is an estimation because the cost function is a finite difference, where as the partial derivatives express the slope of a hyper-plane normal to surfaces that represent functions that comprise forward propagation. The goal is to set up circumstances so that successive approximations approach the ideal represented by minimization of the cost function.
Back-propagation Theory
Back-propagation is a scheme for distribution of a correction signal arising from cost evaluation after each sample or mini-batch of them. With a form of Einsteinian notation, the current convention for distributive, incremental parameter improvement can be expressed concisely.
$$ \Delta P = \dfrac {c(\vec{o}, \vec{\ell}) \; \alpha} {\big[ \prod^+ \! P \big] \; \big[ \prod^+ \!a'(\vec{s} \, P + \vec{z}) \big] \; \big[ c'(\vec{o}, \vec{\ell}) \big]} $$
The plus sign in $\prod^+\!$ designates that the factors multiplied must be downstream in the forward signal flow from the parameter matrix being updated.
In sentence form, $\Delta P$ at any layer shall be the quotient of cost function $c$ (given label vector $\vec{\ell}$ and network output signal $\vec{o}$), attenuated by learning rate $\alpha$, over the product of all the derivatives leading up to the cost evaluation. The multiplication of these derivatives arise through the recursive application of the chain rule.
It is because the chain rule is a core method for feedback signal evaluation that partial derivatives must be used. All variables must be bound except for one dependent and one independent variable for the chain rule to apply.
The derivatives include three types.
All layer input factors, the weights in the parameter matrix used to attenuate the signal during forward propagation, which are equal to the derivatives of those signal paths All the derivatives of activation functions $a$ evaluated at the sum of the matrix vector product of the parameters and signal at that layer plus the bias vector The derivative of the cost function $c$ evaluated at the current output value $\vec{o}$ with the label $\vec{\ell}$
Answer to the Question
Note that, as a consequence of the above, the derivatives of both the cost (or loss or error) function and any activation functions are necessary.
Redundant Operation Removal for an Efficient Algorithm Design
Actual back propagation algorithms save computing resources and time using three techniques.
Temporary storage of the value used for evaluation of the derivative (since it was already calculated during forward propagation) Temporary storage of products to avoid redundant multiplication operations (a form of reverse mode automatic differentiation) Use of reciprocals because division is more costly than multiplication at a hardware level
In addition to these practical principles of algorithm design, other algorithm features arise from extensions of basic back-propagation. Mini-batch SGD (stochastic gradient descent) applies averaging to improve convergence reliability and accuracy in most cases, provided hyper-parameters and initial parameter states are well chosen. Gradual reduction of learning rates, momentum, and various other techniques are often used to further improve outcomes in deeper artificial networks. |
Research Open Access Published: Global well-posedness for nonlinear fourth-order Schrödinger equations Boundary Value Problems volume 2016, Article number: 25 (2016) Article metrics
981 Accesses
Abstract
This paper studies a class of nonlinear fourth-order Schrödinger equations. By constructing a variational problem and the so-called invariant of some sets, we get global existence and nonexistence of the solutions.
Introduction
This paper concerns the initial value problems for the nonlinear fourth-order Schrödinger equations
where \(0< p<\frac{8}{(n-4)^{+}}\) (we use the convention: \(\frac {8}{(n-4)^{+}}=+\infty\) when \(2\leq n\leq4\); \(\frac{8}{(n-4)^{+}}=\frac {8}{n-4}\) when \(n\geq5\)), \(u(t,x):\mathbb{R}\times\mathbb {R}^{n}\rightarrow\mathbb{C}\), \(n\geq2\), is the unknown function and Δ is the Laplace operator, \(t\in[0,+\infty)\).
where
p is an integer and
Pausader [3] established the global well-posedness for the energy critical fourth-order Schrödinger equation
in the radial case by Strichartz-type estimates, while a specific nonlinear fourth-order Schrödinger equation as above (1.3) had been recently discussed by Fibich
et al. [4]. They described various properties of the equation in the subcritical regime.
Moreover, for \(f(|u|^{2})u=|u|^{p-1}u\) one could also consider the focusing equation
Motivated by the above works, Pausader and Xia [7] proved the scattering theory for the defocusing fourth-order Schrödinger equation
in low spatial dimensions (\(1\leq n \leq4\)) by a virial-type estimate and Morawetz-type estimate.
Recently, Wang [8] proved the small data scattering and large data local well-posedness for the fourth-order nonlinear problem
where \(\mu=\pm1\), in critical \(H^{s_{c}}\) space and in particular, for some \(s_{c} \leq0\) by Fourier restriction theory and Strichartz-type estimates, but the sharp conditions of the global existence and blow up for the problem by potential well theory is still not considered for \(\mu=1\). In this paper we try to solve this problem by a concavity method and potential well theory. Recently, the concavity method and potential well theory were applied by Shen
et al. [9] to study the initial boundary value problem for fourth-order wave equations with nonlinear strain and source terms at high energy level. For other related results, we refer the reader to [10–18].
The plan of this paper is as follows. In the second section, we state some propositions, lemmas, and definitions and prove some invariant sets. In the third section, we state the sharp condition for the global existence and nonexistence of problem (1.1).
Throughout this paper, the \(H^{2}(\mathbb{R}^{n})\)-norm will be designated by \(\|\cdot\|_{H^{2}}\), also, the \(L^{p}(\mathbb{R}^{n})\)-norm will be denoted by \(\|\cdot\|_{L^{p}}\) (if \(p=2\), \(\|\cdot\|_{L^{2}}\) is denoted \(\|\cdot\|\)). For simplicity, hereafter, \(\int_{\mathbb {R}^{n}}\cdot\,dx\) is denoted ∫⋅.
Preliminaries
For problem (1.1), we define the energy space in the course of nature by
Proposition 2.1 Let \(u_{0}\in H\). Then there exists a unique solution u of the Cauchy problem (1.1) in \(C([0, T]; H)\) for some \(T\in(0,\infty]\) ( maximal existence time). Furthermore, we can get alternatives: \(T =\infty\) or \(T<\infty\) and Moreover, u satisfies Lemma 2.2 Suppose \(u_{0}\in H\), \(u\in C([0,T); H)\) be a solution to problem (1.1). Let \(J(t)=\int|x|^{2}|u|^{2}\), then
Furthermore, we consider the following steady-state equation:
For any solution of (2.5), we define the following functionals:
When \(\varphi_{0}\in H\) and
φ are a solution of problem (1.1) in \(C([0,T]; H)\), we have
and we define the set
Now, we study the following constrained variational problem:
By a similar argument to [21], we get the following lemmas.
Lemma 2.3 Solution of (2.5) belongs to M. Proof
Let \(\varphi(x)\) be a solution of steady-state equation (2.5). Then we get
from which
Hence \(\varphi\in M\). □
Lemma 2.4 When \(\varphi(x)\in M\), we have \(d>0\). Proof
Combined with (2.9), we obtain the conclusion. □
Lemma 2.5 Let \(\varphi\in H \), \(\lambda>0\), and \(\varphi_{\lambda}(x)=\lambda \varphi(x)\). Then there exists a unique \(\lambda^{*}>0\) ( depending on φ) such that \(I(\varphi_{\lambda^{*}})=0\) and \(I(\varphi _{\lambda})>0\), for \(\lambda\in(0, \mu)\); \(I(\varphi_{\lambda})<0\), for \(\lambda>\lambda^{*}\). Furthermore, \(P(\varphi_{\lambda^{*}})\geq P(\varphi_{\lambda})\), for any \(\lambda>0\). Proof
and
Furthermore, there exists a unique positive constant \(\lambda^{*}>0\) (depending on
φ) such that \(I(\varphi_{\lambda^{*}}) = 0\) and we can easily see that
and
Combining
with
we obtain
This completes the proof of the lemma. □
Now we discuss the invariant sets of solution for problem (1.1).
Theorem 2.6 Let If \(u_{0}\in V\), then the solution \(u(x,t)\) of problem (1.1) also belongs to V for any t in the interval \([0, T)\). Proof
It means that \(P(u_{0})< d\) is equivalent to \(P(u)< d\) for any \(t\in[0, T)\).
we have \(u(x, t_{1}) \neq0\). Otherwise \(P(u(x, t_{1})) = 0\), which contradicts \(P(u(x, t_{1}))> 0\). From (2.9), it follows that \(P(u(x, t_{1}))\geq d\). This contradicts \(P(u(x, t)) < d\) for any \(t\in[0, T)\), since \(I(u(x, t)) < 0\). In other words, \(u(x, t) \in V\) for any \(t\in[0, T)\). So,
V is an invariant manifold of (1.1). □
By a proof similar to that of Theorem 2.6, we can obtain the following theorem.
Theorem 2.7 Define Then W is an invariant set of problem (1.1). The conditions for global well-posedness Theorem 3.1
(Global existence)
Let \(u_{0}\in W\), then the existence time of solution \(u(x,t)\) for problem (1.1) is infinite. Proof
If \(u_{0}\in W\), from Theorem 2.7, we know that \(u(x, t)\in W\) for \(t\in[0, T)\). For fixed \(t\in[0, T)\), we denote \(u(x, t) = u\). Then we have \(P(u) < d\), \(I(u) > 0\). It follows from (2.6) and (2.7) that
which indicates
Lemma 3.2 Assume that \(\varphi\in H\) and \(\lambda^{*}>0\) satisfy \(I(\varphi _{\lambda^{*}}) = 0\). Suppose \(\lambda^{*}<1\), then it follows that Proof
From the proof of Lemma 2.5 we know
where \(a=\int(|\Delta\varphi|^{2}+|\varphi|^{2})\), \(b=\int|\varphi|^{p+2}\).
Notice that \(I(\varphi_{\lambda^{*}}) = 0\) requires
Observe that \(\varphi= \varphi_{\lambda^{*}=1}\) and \(I(\varphi) = a -b\), using (3.5), we get
Theorem 3.3
(Blow up in finite time)
Let \(p>1\), \(n>\frac{2(p+2)}{p}\), \(u_{0}\in V\), \(E(0)< d\), then any solution \(u(x, t)\) to problem (1.1) blows up in finite time. Proof
Since \(u_{0}\in V\), for \(t\in[0,T)\), from Theorem 2.6 we have \(u(x,t)\in V\),
i.e., \(I(u)<0\). Then we obtain
Since \(u_{0}\in L^{2}(\mathbb{R}^{n})\), \(u\in L^{2}(\mathbb{R}^{n})\), by Lemma 2.2 it follows that
For fixed \(t\in[0,T)\), \(u=u(t)\). Let \(\lambda^{*}>0\) be such that
Since \(I(u)<0\), we know from Lemma 2.5 that \(\lambda^{*}<1\). Because
By Lemma 3.2, we get
where \(c_{0}\) is a positive constant. Furthermore, we can get
Hence there exists a \(t_{0}\geq0\), such that \(J'(t)< J'(0)<0\) for \(t>t_{0}\) and
since \(J(0)>0\) (by \(I(u_{0})<0\)). From (3.11), we know that there exists a \(T_{1}>0\) such that \(J(t)>0\) for \(t\in[0,T_{1})\),
From (3.12), the Hölder inequality, and the Hardy inequality, we have
and it follows that
which contradicts \(T= +\infty\). Finally, we can get
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46, 113-129 (1987) 20.
Cazenave, T: Semilinear Schrödinger Equations, vol. 10. Am. Math. Soc., Providence (2003)
21.
Jiang, XL, Yang, YB, Xu, RZ: Family potential wells and its applications to NLS with harmonic potential. Appl. Math. Inf. Sci.
6, 155-165 (2012) Acknowledgements
This work was supported by the National Natural Science Foundation of China (41306086), the Fundamental Research Funds for the Central Universities. Many thanks go to the reviewers for their revision suggestions, which improved the paper a lot. The authors appreciate Prof. Weike Wang for his valuable suggestions.
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed to the writing of this paper. XP found the motivation of this paper. YN and JL finished the proof of the main theorems and wrote the manuscript. JS and MZ provided many good ideas and assisted with writing this paper. All authors read and approved the final manuscript. |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
I'm trying to understand the definition of tensor product of two vector spaces. So far, I've read the one using free vector spaces and a quotient space (here), and I think I understand it well. However, I want to understand the other definitions I can find, and it seems that a very common way to define it is through the universal property (some category theory included, I suspect). Does anyone here know of a good treatment of this? I have no knowledge of category theory though, but would love to read some about it. I'm a second-year undergrad, so not too much of a high level would be nice.
Just some definitions, in case you're unfamiliar with them: Let $\hat{V}$ denote the vector space of linear functions from a vector space $V$ to the scalar field. Remember, a multilinear map is one of the form $V \times V \times \cdots \times V \to W$ (with $n$ copies of $V$), where $W$ is another vector space, such that if we fix $n-1$ of the arguments, the function becomes a linear function from $V$ to $W$ in the argument not fixed. A multilinear form is one in which $W=K$, the scalar field (you can replace $K$ with $\mathbb{R}$ or $\mathbb{C}$ if you like). For example, the inner product on $\mathbb{R}^n$ is a bilinear form on $\mathbb{R}^n$, since if we fix one argument, it becomes linear in the other. If we view an $n \times n$ matrix as a conglomeration of $n$ columns, then the determinant is an $n$-form.
Then $\hat{V} \otimes \hat{V}$ corresponds to the set of bilinear forms, and in general, a tensor product of multiple copies of $\hat{V}$ corresponds to the set of $n$-linear forms (i.e. multilinear forms with $n$ arguments). That, there is a
concrete description of tensor products of the dual space with itself, and many books which do not wish to develop the notion of tensor product will use this in place of tensor products. That is, all they must do is define a certain kind of map, and then the tensor product is just the set of maps of that kind. Then how do we explain the tensor product $V \otimes V$ (or more generally $V \otimes U$, where $U$ is another vector space)? We could note that $V$ is canonically isomorphic to its double dual, i.e. the dual space of $\hat{V}$, and then view $V \otimes V$ as the set of bilinear forms on $\hat{V}$. But there is a nicer way, and this uses the universal property.
A bilinear map $V \times V \to W$ corresponds to a linear map $V \otimes V \to W$. If $f(-,-)$ denotes the bilinear map, and $x,y \in V$, then our
linear map sends $x \otimes y$ to $f(x,y)$. You could try to think of the tensor product as pairs of vectors, but the tensor product contains elements which are not $x \otimes y$ for some $x,y \in V$. We do have that $x_1 \otimes y_1 + x_2 \otimes y_2$ maps to $f(x_1,y_1)+f(x_2,y_2)$. In more generality, if $W$ and $U$ are two other vector spaces, linear maps $U \otimes V \to W$ correspond to bilinear maps $U \times V \to W$. Then what is an element of $U \otimes V$? It is a thing you stick into a bilinear map. This is the key idea which helped me understand tensor products. I repeat, an element of a tensor product is simply a thing you stick into a bilinear map. In general, elements of some universal construction defined by maps going out of a certain object have some description as "things you stick into some kind of map (or a collection of multiple maps)."
You might like Brian Conrad's handouts for a sophomore differential geometry course. Especially relevant are
Construction of tensor products and the two handouts after that one. They have some nice examples and a heavy emphasis on the universal property.
(I don't think this warranted more than a comment, but I can't post those yet.)
My view of the pedagogy, based on teaching this to second year undergraduates at Cambridge.
The tensor product of vector spaces is defined by generators and relations. Also generators and relations, as a way of defining anything, is a method depending on a universal property (to make much sense).
If you take these two parts one at a time, you have a chance of understanding what is happening. The generators and relations are just bilinearity spelled out. The remark about generators and relations as a mode of defining anything can be learned anywhere you like (e.g. group theory): the reason that there is a universal property is just "stuff", "abstract nonsense", "mathematical maturity" even.
I believe, quite strongly, that the eliding of the punctuation between the two sentences is a negative in teaching this material. (I really do not care if this spoils Mac Lane's or anyone else's view of category theory and its role: "universal property" is only a stepping stone there, not the ultimate goal.)
I'm pretty much in your spot. I think part of the way there is learning to think with universal properties. I recently found a really good book (Algebra: Chapter 0, link below) on 'basic' algebra using category theory to unify things. All the basic stuff like products, disjoint union, surjections and injections are treated rigorously and in great generality through their universal properties. If you already know your group and set theory reading through the first few chapters can be done quickly, and should get you in the right mode of thought. I'm doing this myself right now, and so far I recommend you do the same.
EDIT: A nice application of the tensor product can be found in the first few pages of 'Differential Forms in Topology', that is, if $\Omega^*$ is the algebra generated by the formal symbols $dx_j,j=1,\dots,n$ under the relations $dx^2=0$ and $dx_idx_j=-dx_jdx_i$, then $\Omega^*(U)=C^\infty(U)\otimes\Omega^*$ is the algebra of differential forms on the open set $U$ (under the wedge product). I'm not sure if that's how it's primarily used.
A fully categorical approach that emphasizes the universal properties of the tensor product ,as well as a great deal of multilinear algebra, can be found in T.S.Blyth's
Module Theory:An Approach to Linear Algebra. There's also a discussion in Steven Roman's Advanced Linear Algebra,but the presentation in Blyth's book isn't as dry and formal.
By the way,if anyone has a serious interest in algebra,Blyth's books are some of the great unsung textbooks in the subject. They really should be better known and used in the U.S. then they are.
Thank you all. Your documents has been most helpful. I also saw some papers on the tensor product of modules, especially: http://www.math.ucsb.edu/~mckernan/Teaching/05-06/Winter/220B/l_7.pdf was helpful, and http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html gave some good info too.
Now I'm considering TeXing a file where I try to motivate
why one defines the tensor product in the first place. I think that might help me learn the definition even more. I really like the definition in some strange way, even though I find it kind of hard. I want to learn.
So once again, Thank you. |
Yes there is a (slightly more) rigorous definition:
Given a model with a set of parameters, the model can be said to be overfitting the data if after a certain number of training steps, the training error continues to decrease while the out of sample (test) error starts increasing.
In this example out of sample (test/validation) error first decreases in synch with the train error, then it starts increasing around the 90th epoch, that is when overfitting starts
Another way to look at it is in terms of bias and variance. The out of sample error for a model can be decomposed into two components:
Bias: Error due to the expected value from the estimated model being different from the expected value of the true model. Variance: Error due to the model being sensitive to small fluctuations in the data set.
Overfitting occurs when the bias is low, but the variance is high. For a data set $X$ where the true (unknown) model is:
$ Y = f(X) + \epsilon $ - $\epsilon$ being the irreducible noise in the data set, with $E(\epsilon)=0$ and $Var(\epsilon) = \sigma_{\epsilon}$,
and the estimated model is:
$ \hat{Y} = \hat{f}(X)$,
then the test error (for a test data point $x_t$) can be written as:
$Err(x_t) = \sigma_{\epsilon} + Bias^2 + Variance$
with $Bias^2 = E[f(x_t)- \hat{f}(x_t)]^2$and $Variance = E[\hat{f}(x_t)- E[\hat{f}(x_t)]]^2$
(Strictly speaking this decomposition applies in the regression case, but a similar decomposition works for any loss function, i.e. in the classification case as well).
Both of the above definitions are tied to the model complexity (measured in terms of the numbers of parameters in the model): The higher the complexity of the model the more likely it is for overfitting to occur.
See chapter 7 of Elements of Statistical Learning for a rigorous mathematical treatment of the topic.
Bias-Variance tradeoff and Variance (i.e. overfitting) increasing with model complexity. Taken from ESL Chapter 7 |
Steric hinderance is a major component in determining the feasibility and the rate of a chemical reaction. Wouldn't it be useful to measure it quantitatively then? This would make it easier to compare the property of two molecules. Are there currently ways to measure steric hindrance, or is it not possible for some reason?
There are two ways I know of to measure steric hindrance. Once such quantitative measurement are A values, which give the energetic preference of having the substituent on a substituted cyclohexane ring in the equatorial position versus the axial position. This preference is a result of 1,3-diaxial strain.
Additionally, the Taft equation is a linear free-energy relationship used to measure the steric effects of a substituent on the rate of reaction. The equation is given by:
$${\displaystyle \log \left({\frac {k_{s}}{k_{\ce {CH3}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}}$$
where $\displaystyle \frac {k_{s}}{k_{\ce {CH3}}}$ is the ratio of the substituents rate of reaction relative to a methyl group, $\sigma^*$ is the polar substituent constant that describes the field and inductive effects of the substituent, $E_s$ is the steric substituent constant, $\rho^*$ is the sensitivity factor for the reaction to polar effects, and $\delta$ is the sensitivity factor for the reaction to steric effects.$^{[1]}$
Though A values are useful as a quick reference in deducing the major products of an unfamiliar reaction, the Taft equation would be used preferentially when doing an in depth mechanistic study of a reaction for publication.
$^{[1]}$ Wikipedia, Taft equation There is a full way but it's not easy to calculate.
Steric hindrance is indeed a generic term for a quantifiable phenomenon: electron-electron repulsion, or (much) more broadly, chemical physics.
Electron-electron repulsion can be measured simply/crudely by Coulomb's law: \begin{align} E &= \frac{q_1 q_2}{4\pi \epsilon_0 r}, & F &= \frac{q_1 q_2}{4\pi \epsilon_0 r^2}. \end{align}
Here, e.g. you can approximate atomic charges with some atomic volume assigned and quantify the forces involved between molecules.
But the full dynamics requires quantum mechanical calculations (ignoring things like relativity and gravity).
Coulomb's energy does appear in QM equations but electron-electron interactions are more complicated than simple point-charges so more terms appear for correlation/exchange energies etc. The general principle is the same as for classical physics though: you calculate total kinetic and potential energy of your system.
The quantitative expression would be
the way potential energy changes with intermolecular distance. So you could set up a series of systems with different distances between molecules and see how the energy changes.
One can calculate QM energies with QM software like NWChem, QChem, Orca, Gaussian and others.
While not directly "measuring" steric hindrance but rather "distortion energy" the computational Interaction/Distortion model from K.N. Houk (1, 2) and Bickelhaupt who calls it "Activation Strain" model (3) can give pretty good insight in such things. Especially since steric hindrance is only half (or even less) of the story.
For similar reactions this distortion energy can (!) be related to steric hindrance, but is also inflenced by electronic effects the substituents might have on the reactive center. Over all this model provides pretty good insight if a reaction proceeds fast/slow due to electronic effects or energy needed for distortion.
So how does it work? We pretty much arbitrarily devide our energy of activation into two parts, the interaction energy and the distortion energy. Distortion energy is the energy needed to "twist" the reactants from ground state geometry into geometry at the transition state structure. For this energy is needed. If you then bring this distorted reactants together they will interact and what you get out here is the interaction energy. Sum of all distortion energies and the interaction energy is the energy of activation.
$$\Delta E^{\ddagger}_\mathrm{act}=\Delta E^{\ddagger}_\mathrm{dist} + \Delta E^{\ddagger}_\mathrm{int}$$
Houk himself explains this very well in this talk.
References: |
I found this interesting problem on calculating the limit of $\frac{\sin(xy^2)}{xy}$ on the positive coordinate axes $x$ and $y$. That is, compute the limit on the points $(x_0, 0)$ and $(0,y_0)$ when $x_0 > 0$ and $y_0 > 0$.
My approach was this:
If we first calculate the limit for $x$ axis, the the $x$ is a constant $x=x_0$ and therefore the function is essentially a function of one variable:
$$f(y) = \frac{\sin(x_0y^2)}{x_0y}$$
Using L'Hospital's rule:
$$\lim_{y\to0}f(y)=\frac{\lim_{y\to0}\frac{d\sin(x_0y^2)}{dy}}{\lim_{y\to0}\frac{dx_0y}{dy}}$$
$$=\frac{\lim_{y\to0}2yx_0\cos(x_0y^2)}{\lim_{y\to0}x_0}=0$$
The same idea applies to the other limit.
But in the sheet where this problem was listed, this was listed as a "tricky" limit to calculate. It seemed quite simple, so I would like to hear whether this approach is correct. Thank you! |
I'm going to find an example of
uniform algebra and show that satisfying the definition. Example: Show that The Gelfand transform $\widehat{f}$ is uniform algebra.
We know that:
A uniform algebra is a
closed subalgebra$\mathcal A$ of the complex algebra $C(X)$ that contains the constantsand separates points. Here $X$ is a compact Hausdorff space.
The Gelfand transform of $f$ is the function $\widehat{f}$ defined on $M_{\mathcal A}$ in the following way:
$$\begin{align}\widehat{f}: M_{\mathcal A} &\to \Bbb C\\ \varphi &\mapsto \widehat{f}(\varphi)=\varphi(f), \forall \varphi \in M_{\mathcal A} \end{align}$$
How can we show that $\widehat{A}$ such that $3$ conditions above?
=================================
After a period of time to think about my problem which I posted above, I'll write it here:
Let $\widehat{\mathcal A}=\{\widehat{f}: \ f \in \mathcal A \}$
$1$. $\widehat {\mathcal A} \ $
contains the constants
Because $e \in \mathcal A \implies \widehat{e}(\varphi)=\varphi (e)=1, \ \forall \varphi \in M_{\mathcal A}$.
Therefore, $\widehat{(\lambda e)}(\varphi)=\varphi(\lambda e)=\lambda \varphi(e)=\lambda,\ \forall \lambda \in \Bbb C$.
Hence, $\widehat {\mathcal A} \ $
contains the constants
$2$. $\widehat {\mathcal A} \ $
separates points
We assume that $\varphi_1,\ \varphi_2 \in M_{\mathcal A}$ such that $\widehat{f}(\varphi_1)= \widehat{f}(\varphi_2),\ \forall f \in \mathcal A$.
Whence, $\varphi_1(f)= \varphi_2(f),\ \forall f \in \mathcal A$. So $\varphi_1= \varphi_2$.
Hence, $\widehat {\mathcal A} \ $
separates points
===========================
$3$. Now I have stuck when I try to show $\widehat {\mathcal A}$ is a
closed subalgebra of algebra Banach $C(M_{\mathcal A})$
I think that we have $\widehat{f}$ is continuous, because $\left |\widehat{f}(\varphi ) \right |=\left | \varphi (f) \right |\le \left \| \varphi \right \|\cdot \left \| f \right \|=\left \| f \right \|$
But How can we prove The first condition (i.e $\widehat {\mathcal A}$ is a
closed subalgebra of algebra Banach $C(M_{\mathcal A})$)
I don't remember the definition of
closed subalgebra. Can anyone post it help me!
Any help will be appreciated! Thanks! |
So starting from the time dependent schrodinger equation I perform separation of variables and obtain a time and spatial part. The spatial part is in effect the time independent schrodinger equation.
Since we are dealing with a free particle I can take the time independent equation, set V = 0 and solve.
I can do this successfully to obtain :
$Ae^{+i\sqrt{{2mE}/{\hbar^{2}}}x}+Be^{-i\sqrt{{2mE}/{\hbar^{2}}}x}$
My lecturer has a small section titled :
Solving for the Free Schrodinger Equation
$$V=0$$
$$\frac{\hbar^{2}}{2m}\frac{\partial^2\psi}{\partial x^2}+E\psi=0$$
$$E=\frac{p^2}{2m}$$
$$\psi=Ce^{-{iEt}/{\hbar}+{ipr}/{\hbar}}$$
This is the solution to the free TISE and TDSE.
He seems to be doing the same thing as me initially but he's obtained a different result ?
Also, the first section of his answer :$e^{-{iEt}/{\hbar}}$ is the solution to the time part of the equation (described above). |
I want to show that a rigid body, with two components of its angular velocity vector and one component of its linear velocity vector,
in the absence of external forces and torques, has helical trajectories. This is usually taken for granted in various papers I have read (see for example this p.4 section 4, or this p.206 first paragraph).
I consider a moving rigid body whose centre of mass is located at a point $\boldsymbol{r}_{0}(t)$ in the laboratory frame of reference defined by the basis vectors $\boldsymbol{e}_{x},\boldsymbol{e}_{y},\boldsymbol{e}_{z}$. I attach a moving frame of reference $\left\{ \boldsymbol{r}_{0}(t);\boldsymbol{e}_{x}^{\prime}(t),\boldsymbol{e}_{y}^{\prime}(t),\boldsymbol{e}_{z}^{\prime}(t)\right\} $ to the body; this frame is centred at $\boldsymbol{r}_{0}(t)$ and its axes rotate with the frame of the body. The rotating triad $\left(\boldsymbol{e}_{x}^{\prime},\boldsymbol{e}_{y}^{\prime},\boldsymbol{e}_{z}^{\prime}\right)$ can be characterised by three Euler angles $\theta_{1},\theta_{2},\theta_{3}$ and the following transformation rules:
$$ \mathbf{e}_{i}^{\prime}=\boldsymbol{L}\left(\theta_{1},\theta_{2},\theta_{3}\right)\cdot\mathbf{e}_{i} $$
I adopt the Tait-Bryan angle convention, with an $y-x^{\prime\prime}-z^{\prime}$ intrinsic definition:$\theta_{1}$ is the yaw (anticlockwise around $e_{y}$), $\theta_{2}$ the pitch (anticlockwise around $e_{x}^{\prime}$) and $\theta_{3}$ the bank (roll, anticlockwise around $e_{z}^{\prime}$). For brevity, I define the vector $\boldsymbol{\theta}=\left(\theta_{1},\theta_{2},\theta_{3}\right)$ of the three independent Euler angles. The matrix of the transformation is $$\boldsymbol{L}\left(\boldsymbol{\theta}\right)=\boldsymbol{R}_{Z}\left(\theta_{3}\right)\boldsymbol{R}_{X}\left(\theta_{1}\right)\boldsymbol{R}_{Y}\left(\theta_{2}\right)$$ where the $\bf R_i$ are the rotation matrices in $\mathbb{R}^3$. I assume that the body has linear velocity $\boldsymbol{U}(t)=U\boldsymbol{e}_{z}^{\prime}(t)$ oriented along its anterior-posterior axis $\boldsymbol{e}_{z}^{\prime}(t)$, and rotational velocity vector $\boldsymbol{\omega}(t)$. In the chosen representation, the components of angular velocity in the body's frame of reference are
\begin{align*} \omega_{x} & =\dot{\theta}_{2}\cos\theta_{1}-\dot{\theta}_{3}\sin\theta_{1}\cos\theta_{2}\\ \omega_{y} & =\dot{\theta}_{1}+\dot{\theta}_{3}\sin\theta_{2}\\ \omega_{z} & =\dot{\theta}_{2}\sin\theta_{1}+\dot{\theta}_{3}\cos\theta_{1}\cos\theta_{2}. \end{align*}
These can be found by noticing that $$\boldsymbol{L}^{T}\dot{\boldsymbol{L}}=\left(\begin{array}{ccc} 0 & -\omega_{z} & \omega_{y}\\ \omega_{z} & 0 & -\omega_{x}\\ -\omega_{y} & \omega_{x} & 0 \end{array}\right)$$
The trajectory of the body is therefore given by the curve traced by its centre of mass $\boldsymbol{r}_{0}$ as it moves through space. In the laboratory frame, this reads $$\boldsymbol{r}_{0}(t)=\intop_{0}^{t}\boldsymbol{U}\left(\tau\right)d\tau=U\intop_{0}^{t}\boldsymbol{e}_{z}^{\prime}\left(\tau\right)d\tau$$
Clearly, $\boldsymbol{e}_{z}^{\prime}$ rotates with the body, and therefore is a function of time through the angular velocity components $\omega_{i}$: \begin{align*} \boldsymbol{e}_{z}^{\prime}(t) & =\boldsymbol{L}(t)\cdot\boldsymbol{e}_{z}\\ & =\left(\cos\theta_{3}\sin\theta_{1}+\cos\theta_{1}\sin\theta_{2}\sin\theta_{3}\right)\boldsymbol{e}_{x}+\left(-\cos\theta_{1}\cos\theta_{3}\sin\theta_{2}+\sin\theta_{1}\sin\theta_{3}\right)\boldsymbol{e}_{y}+\left(\cos\theta_{1}\cos\theta_{2}\right)\boldsymbol{e}_{z} \end{align*}
$$ \text{where }\ \theta_{i}=\intop_{0}^{t}\dot{\theta}_{i}d\tau. $$
The problem is complicated, because in order to integrate $\boldsymbol{e}_{z}^{\prime}$I have to invert the relationships between $\omega_{i}$ and $\theta_{j}$. Notice that I am keeping all three components of the angular velocity here, to see at what point and to what extent having only
two rotational degrees of freedom is necessary.
However, even if I assume that the angular velocities are constant $\theta_{i}=\theta_{i}t$, I do not get expressions that contain $\omega_i$ explicitly. Is there another approach that makes use of $\boldsymbol{L}^{T}\dot{\boldsymbol{L}} = \boldsymbol{\omega}\times$? |
Business Insider's (long) article SpaceX's biggest rival has a 'genius' plan to cut its rocket launch costs more than 70% contains the statements sourced from ULA's CEO Tory Bruno:
Vulcan should lift 40 tons (nearly three school buses) into low-Earth orbit. That's less than SpaceX's Falcon Heavy, which can lift more than 70 tons — nearly five school buses — for one-fourth the price. But Bruno said there are big differences between the two systems that will make Vulcan competitive.
The key difference is the rocket's upper stage. Falcon Heavy currently uses a rocket-grade RP-1 kerosene as fuel, but it can freeze in space after a few hours. Vulcan's upper stage will use cryogenic oxygen and hydrogen, which are more resilient to the punishing temperatures of space.
LH2 and LOX have about the same molar density, but the stoichiometry requires twice as many moles of LH2. If the tanks are end-to-end, it would mean the LH2 tank intercepts close to twice as much geometrical exposure to the environment as the LOX tank. However, the Enthalpy of vaporization of LH2 on a
molar basis is only one quarter that of LOX (see LH2 and LOX).
Further, at 1 atmosphere for example, LH2 boils at around 20K while LOX boils at around 90K. That means without active refrigeration, the heat loading would have to be of the order of $(4.5)^4$ times lower if LH2 was used (assuming passive radiation for cooling), which would be a real challenge in sunlight.
There are
two three parts to this question. If I have to split them I will but it's possible an answer can address both at the same time. Is ULA likely to consider putting the 2nd stage LH2 tank insidethe LOX tank (or at least be surrounded by it coaxially)? Unless the second stage is going to Jupiter or beyond, isn't the heating from Sunlight boiling the LH2 a more challenging problem than "freezing" of the RP-1? (See Does the NK-33 engine require subcooled kerosene so cold that it turns to wax? for some density vs temperature plots.) In order to keep the LH2 cold for months, would the 2nd stage end up looking a little bit like the the JWST with those large metallized polymer layers deployed to block the Sun? Temperature of a Spherical Cow in Space:
Spherical cow as illustrated by a 1996 meeting of the American Astronomical Association, in reference to astronomy modeling. From here: "The image was created by Ingrid Kallick for the program cover of the 1996 annual meeting of the American Astronomical Association. An earlier version was created for the National Center for Supercomputing Applications. The artist gave permission for use to the University of Wisconsin Department of Astronomy. The STScI subsequently used the image. http://www.ikallick.com"
Equilibrium temperature happens when
average power in equals average power out, or $\bar{P}_{in} - \bar{P}_{out}$. Averaging should be done over short term variations in attitude relative to the Sun and take into account eclipses for most orbits near the Earth, Moon, or another planet.
$$\bar{P}_{in} = I_{Sun} (1-a) \ \pi R^2$$
$$\bar{P}_{out} = \sigma \epsilon T^4 \ 4 \pi R^2 $$
where $a_{vis}$ is the visible light albedo, $e_{ir}$ is the infrared emissivity (both should really be weighted averages over the appropriate wavelength ranges; @Tristan's and @Puffin's comments explain this better than their associated answers do), $\sigma$ is the Stefan–Boltzmann constant (about 5.67E-08 W m^-2 K^-4), and I is the intensity of sunlight, and for 1AU is the solar constant and about 1360 W/m^2. Solving for the average equilibrium temperature of said cow gives:
$$T \sim \left( \frac{(1-a_{vis})}{e_{ir}} \frac{I_{Sun}}{4 \sigma} \right)^{1/4}$$
For an average
visible-light albedo of 0.95, and an average infrared emissivity of 0.95. this turns out to be about 130 Kelvin at 1 AU, and about 110 Kelvin near Mars, and because of the fourth-root, this varies only slowly with any of the parameters. It seems that space is much more LOX-friendly than LH2-friendly, and only moderate Sun-shielding measures would be necessary to get the LOX down below boiling at 1 atmosphere pressure, like simply having the 2nd stage face the Sun end-on, because cows are not actually spherical. But what about the RP-1?
If the albedo of a hypothetical "LOX compartment" were 0.1 instead of 0.95 (if it were 18 times more absorbing of sunlight), the temperature would rise by the fourth root of 18, or about a factor of two (see my thoughtful (and
unnecessarily down-voted) tutorial on the use of power laws in physics). That would put the RP-1 up near a balmy 273 K or 0C, "sub-cooled" and ready to go! This can be confirmed by the plot of equilibrium temperature of temperature for a spherical blackbody around each of the planets (ignoring eclipses and planetary albedo) found in this answer. |
This is a problem from Schlogl's book in the chapter on the HJM model:
Price option of the RAN instrument with 3 month coupons and maturity 3 years using Monte Carlo(Exercise 4 Range Accrual Note).
Is the code in the Quantlib library? If so can you tell me its location, thank you. If not can you give me some suggestions on how to approach it.
The background
A bond's value and payoff are of the form:
$$V=\sum_{i}^{N} c_{i} B(T_{i-1},T_{i}) \text{ and } [V-K]^{+},$$ where K is the strike, $c_{i}$ are the coupons and $B(T_{i-1},T_{i})$ is the price of a zero bond at time $T_{i-1}$ with maturity $T_{i+1}$.
In the FRN the coupons $c_{i}$ are determined by some asset or floating rate. Specifically in R AN with 3-month coupons (90 days) and interest compounded daily we have:
$$c_{i}=\bar{r}\frac{1}{90}\sum_{k=90\cdot i}^{90\cdot (i+1)}L(t_{k},t_{k+1}),$$
where the Libor rate is $L(t_{k},t_{k+1})=\frac{1}{t_{k+1}-t_{k+1}}(\frac{1}{B(t_{k},t_{k+1})}-1)$ and $\bar{r}$ is a fixed number or the Libor rate plus a spread s.
So we are searching to price the option with payoff $(\sum_{i}^{N} c_{i} B(T_{i-1},T_{i})-K)^{+}$ with the above $c_{i}$.
Glasserman in his Monte Carlo book section 3.7, describes how to go about the Libor rate option. I will try that as well. |
I'm currently reading on self-balancing robots that use an IMU (gyroscopes + accelerometers) to estimate their current tilt angle.
Most documents that I have found say the same things:
You can't just take the arc-tangent of the accelerometers data to find the gravity direction because they are affected by "inertial noises". You can't just integrate the output of the gyroscope over time because it drifts. There are two generally accepted solutions to merge those data: A Kalman filter estimating the current tilt along with the current gyroscope bias. A complimentary filter applying a low-pass filter on the accelerometer data (they can be trusted in the long term), and a high-pass filter on the gyroscope data (it can be trusted in the short term).
All sources that I have found seem to use the raw data from the accelerometers in those filters, disregarding the fact that, in a self-balancing robot, we can have a very good estimate of the "inertial noise" mentioned above.
Here's my though
Let's model our robot with an inverted pendulum with a moving fulcrum and use this poor drawing as a reference.
The inertial forces felt by the accelerometers at C can be derived from (if I didn't make any mistake) $$ \begin{pmatrix} \ddot{c_r} \\ \ddot{c_\Theta} \end{pmatrix} = \begin{pmatrix} -\ddot{x}\sin(\Theta)-R\dot{\Theta}^2 \\ -\ddot{x}\cos(\Theta)+R\ddot{\Theta} \end{pmatrix} $$
Assuming that
Our robot is rolling without slipping We can measure x (either by using stepper motors or DC motors with encoders)
Then we can have a good estimate of all those variables:
$\hat{\ddot{x}}_k$ : Finite differences over our current and previous measures of $x$ $\hat{\dot{\Theta}}_k$ : The current gyroscope reading $\hat{\Theta}_k$ : Previous estimation of $\Theta$ plus the integration of $\hat{\dot{\Theta}}_k$ and $\hat{\dot{\Theta}}_{k-1}$ over one $\Delta t$ $\hat{\ddot{\Theta}}_k$ : Finite differences over $\hat{\dot{\Theta}}_k$ and $\hat{\dot{\Theta}}_{k-1}$ Once we have that, we can negate the effect of the inertial forces in the accelerometers, leaving only a much better measure of the gravity.
It probably still is a good idea to use this as the input of the usual Kalman filter as in 1. above.
Maybe we can even build a Kalman filter that could estimate all those variable at once? I'm going to try that.
What do you think? Am I missing something here?
I think self-balancing-robot could be a good tag, but I can't create it |
A local system of coefficients on a space $X$ is a functor $F\colon \Pi(X)\rightarrow Ab$ from the fundamental groupoid to the category of abelian groups. From this, one can define the homology groups of $X$ with local coefficients $H_*(X,F)$ as done e.g. in chapter VI of Whitehead's book "Elements of homotopy theory". Therein, it is also shown that this construction yields functors $H_i$ from the category of spaces with local coefficients (a morphism between (X,F) and (Y,G) is a continuous map $f\colon X\rightarrow Y$ with a natural transformation $\eta\colon F\rightarrow f^*G$) to the category of abelian groups.
Suppose now that $X$ is path-connected and has a base point $x\in X$. Then the inclusion $\pi_1(X,x)\rightarrow \Pi(X)$ is an equivalence of categories, so we can pick an inverse functor $i\colon \Pi(X)\rightarrow \pi_1(X,x)$. If we have an module $F$ over $\pi_1(X,x)$ (which is the same as a functor $\pi_1(X,x)\rightarrow Ab$), we get a functor $\Pi(X)\rightarrow Ab$ by $F\circ i$.
Hence, we can define the homology groups of $X$ with local coefficients in the $\pi_1(X,x)$-module $F$ as $H_*(X,F):=H_*(X,F\circ i)$.
First question: Why is this definition independent of the chosen inverse $i$?
It seems to be a common thing to identify local coefficient systems over a path-connected space with modules over the fundamental group at a chosen point, so this has to be true.
Second question: Does this definition extend to a functor $H_i$ from the category of pointed path-connected spaces $(X,x)$ with modules over $\pi_1(X,x)$, where morphisms between $(X,x,F)$ and $(Y,y,G)$ are continuous maps $f\colon X\rightarrow Y$ together with an $\pi_1(f)$-equivariant map $F\rightarrow G$.
As the identification mentioned above is abundant in books and papers within algebraic topology, this should be true as well, but I encounter difficulties in proving this. The main problem seems to be that it might not be possible to choose the inverses $i\colon\Pi(X)\rightarrow \pi_1(X,x)$, such that they form a natural transformation. |
Extraordinary claims require extraordinary proofs which really is the reason why this sorts of discussion is important. Similarly, sometimes, you are so blinded to some sorts of a truth and are faced with something so different that you can misread entirely what is being said. If you read this morning's entry, you might get a feel that I am little ambivalent about the true interesting nature of a paper entitled Statistical physics-based reconstruction in compressed sensing by Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová. Let's put this in perspective, our current understanding so far is that the universal phase transition observed by Donoho and Tanner seems to be seen with all the solvers featured here, that there are many ensembles for which it fits (not just Gaussian, I remember my jaw dropping when Jared Tanner showed it worked for the ensembles of Piotr Indyk, Radu Berinde et al) and that the only way to break it is to now consider structured sparsity as shown by Phil Schniter at the beginning of the week. In most people's mind, the L_1 solvers are really a good proxy to the L_0 solvers since even greedy solvers (the closest we can find to L_0 solvers) seem to provide similar results. Then there are results like the ones of Shrinivas Kudekar and Henry Pfister. ( Figure 5 of The Effect of Spatial Coupling on Compressive Sensing) that look like some sort of improvement (but not a large one). In all, a slight improvement over that phase transition could, maybe, be attributed to a slightly different solver, or ensemble (measurement matrices). So this morning I made the point that given what I
understoodabout the graphs displayed in the article, it may be at besta small improvement over the Donoho-Tanner phase transition known to hold for not only Gaussian but other types of matrices and for different kinds of solvers, including greedy algorithms and SL0 (that simulate some sorts of L_0 approach). At bestis really an overstatement but I was intrigued mostly because of the use of an AMP solver, so I fired off an inquisitive e-mail on the subject to the corresponding author: Dear Dr. Krzakala, ... I briefly read your recent paper on arxiv with regards to your statistical physics based reconstruction capability and I am wondering if your current results are within the known boundary of what we know of the phase transition found by Donoho and Tanner or if it is an improvement on it. I provided an explanation of what I meant in today's entry (http://nuit-blanche.blogspot.com/2011/09/this-week-in-compressive-sensing.html). If this is an improvement, I'd love to hear about it. If it is is not an improvement, one wonders if some of the deeper geometrical findings featured by the Donoho-Tanner phase transition have a bearing on phase transition on real physical systems.Best regards,Igor.
The authors responded quickly with:
Dear Igor, Thanks for writing about our work in your blog.Please notice, however, that our axes in the figure you show are not the same as those of Donoho and Tanner. For a signal with N components, we define \rho N as the number of non-zeros in the signal, and \alpha N as the number of measurements. In our notation Donoho and Tanner's parameters are rho_DT = rho/alpha and delta_DT = alpha. We are attaching our figure plotted in Donoho and Tanner's way. Our green line is then exactly the DT's red line (since we do not put any restriction of the signal elements), the rest is how much we can improve on it with our method. Asymptotically (N\to \infty) our method can reconstruct exactly till the red line alpha=rho, which is the absolute limit for exact reconstruction (with exhaustive search algorithms).So we indeed improve a lot over the standard L1 reconstruction!We will be of course very happy to discuss/explain/clarify details if you are interested.With best regardsFlorent, Marc, Francois, Yifan, and Lenka
The reason I messed up reading the variables is because I was probably not expecting something that stunning.
Thank you for Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová for their rapid feedback.
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Can you please help me and tell, how should I move on? Can this be proved by induction?
Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.
Thank you in advance
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Hint $\rm\quad n\ =\ 3\:\bigg(\!\!\dfrac{n-5\:k}{3}\!\bigg)\: +\ 5\:k\ \ $ for $\rm\ \ k\: = \ {-}n\:\ mod\:\ 3,\ \ k \in \{0,1,2\}$ Note $\ $ For the underlying geometric intuition see this post on the Frobenius problem.
Per request in the comments, here are some examples with further detail.
$\qquad\ \ \begin{eqnarray} \rm n =\ 8,\ \ \ k &=& -8&\equiv\ 1,\ &\ \ 8\ &=\ 3\ (\ 8&-&5\cdot 1)/3 + 5\cdot 1\ =\ 3\cdot 1 + 5\cdot 1 \\ \rm n =\ 9,\ \ \ k &=& -9&\equiv\ 0,&\ \ 9 &=\ 3\ (\ 9&-&5\cdot 0)/3 + 5\cdot 0\ =\ 3\cdot 3 + 5\cdot 0 \\ \rm n = 10,\ \ k &=& -10&\equiv\ 2,& 10 &=\ 3\ (10&-&5\cdot 2)/3 + 5\cdot 2\ =\ 3\cdot 0 + 5\cdot 2 \\ \end{eqnarray}$
Thus $\rm\ \ 3\:m + \ \ 8\ =\ 3\:m + 3\cdot 1 + 5\cdot 1\ =\ 3\:(m+1) + 5\cdot 1$
and $\rm\ \ \ \ 3\:m + \ \ 9\ =\ 3\:m + 3\cdot 3 + 5\cdot 0\ =\ 3\:(m+3) + 5\cdot 0$ and $\rm\ \ \ \ 3\:m + 10\ =\ 3\:m + 3\cdot 0 + 5\cdot 2\ =\ 3\:(m+0) + 5\cdot 2$
yields the desired representations for all integers $\ge 8\:.$
Clearly, $8$ can be represented as $3k+5\ell$, by taking $k=1$ and $\ell=1$.
Likewise, $9=3(3) + 5(0)$; and $10=3(0) + 5(2)$. So we can represent $8$, $9$, and $10$.
Now, assume that for $m\gt 10$, and you can represent all numbers strictly smaller than $m$ that are $8$ or larger (the induction hypothesis). To show that you can represent $m$, consider $m-3$ first.
We can avoid an explicit appeal to induction by using the fact that every natural number $n$ has remainder $0$, $1$, or $2$ on division by $3$. Let $n \ge 8$.
If $n$ has remainder $2$ on division by $3$, then $n-8$ is divisible by $3$, say $n-8=3m$. Represent $8$ using $8=3\cdot 1+5\cdot 1$. Then add $m$ $3$'s.
If $n$ has remainder $0$ on division by $3$, then $n-9$ is divisible by $3$, say $n-9=3m$. Represent $9$ using $9=3\cdot 3 +5\cdot 0$. Then add $m$ $3$'s.
If $n$ has remainder $1$ on division by $3$, then $n-10$ is divisible by $3$, say $n-10=3m$. Represent $10$ using $10=3\cdot 0 +5\cdot 2$. Then add $m$ $3$'s.
The argument for remainder $0$ was a little silly, since if $n$ has remainder $0$ on division by $3$, we can clearly use a bunch of $3$'s to represent $n$. But we wanted the solutions for the three cases to use a single template. |
People are always fascinated by intelligent devices, and today they are software “chatbots”, which are becoming more and more human-like and automated. The combination of an immediate response and a permanent connection makes them an attractive way to expand or replace the web applications. The high-level diagram of an architecture for a chat-bot shown in Figure 1. Natural Language Understanding component tries to determine the purpose of the user’s input (intent) and useful data (entities). Also, the chatbot has a Dialog Manager to define the flow of conversation and paths that the system will take. Finally, a Language Generation component generates user output.
The main function of a chatbot us to generate an answer for the user request. It can be a direct answer to the question, requesting missed slots, fallback standard input or anything else. Natural Language Understanding module detects intents (which represent the purpose of a user’s input) and entities (an object that provides specific information for an intent). There are many useful NLU tools, like Dialogflow, wit.ai, LUIS or IBM Watson that can help you classify intents and parse entities using its custom deep learning and rule-based models. They have convenient web interfaces and easily integrate with different messaging platforms
There are three main types of dialog management systems:
Switch statements Finite state machine Frame-based system Partially Observable Markov Decision Processes
Switch statements is the simplest way to implement a dialog manager. Every intent triggers different responses. The initiative of the conversation always comes from the user and it’s easy to design and implement. But it has significant disadvantages: it doesn’t keep a conversation context and can’t take initiative of the dialogue.
Finite state machine helps your bot to navigate through the conversation flow. The nodes of the graph represent system questions, the transitions between the nodes represent answers to questions, and the graph specifies all valid dialogues. FSMs are useful if you have a limited number of conversational scenarios and structured dialogue. Many commercial bots have this model under its hood, but it lacks flexibility and naturalness. So it’s problematic to develop more complex flow, adding new states and transitions between them. Finally, you can get an unmanageable diagram with a bunch of circles and elements.
In a frame (or template) based system, the user is asked questions that enable the system to fill slots in a template in order to perform a task. This model helps to provide more flexibility to the dialog control because it allows a user to input slots in a different order and combination. Google’s Dialogflow chatbot framework uses a frame-based system with the context control mechanism. Contexts represent the current state of a user’s request and allow your agent to carry information from one intent to another. So you create intents with some entities and design context chains. Dialogflow provides a simple and user-friendly interface for creating and managing your bot agents.
Partially Observable Markov Decision Processes (POMDP) helps us to combine the idea of statistical dialogue management approach and handling unexpected results. POMDP characterizes by:
a set of states San agent can be in a set of actions Athe agent can take a set of observations O a matrix of transition probabilities T a reward R(a, s)that the agent receives for taking an action in a state a dialogue state reward J(s) a discount factor \gamma \in \left[0,1\right]
Classical MDP model defines rewards as a function of belief states, but POMDP – as a function of the environment state s , because the performance of the model should be measured by the user really wants. Distribution over possible states called a belief state b_t is maintained where b(s_t) indicates the probability of being in a particular state s_t . The aim of the system is then to optimize the expected dialogue reward using this adjusted reward function. The reward could be obtained using the following formula:
J(s_{i}) = \underset{a}{max}\mathbf{\Bigg(}R(s_{i}, a)+\gamma\sum_{j=1}^{N}p(s_{j}\mid p(s_{i}, a)) \times J(s_{j})\Bigg)
So, first of all, you should understand potential users and nature of the tasks through user surveys, research on similar systems and study related human dialogues. Then you need to build a prototype of your conversation flow and try to figure out which dialogue management system will be most appropriate for this task. |
I want to point out another interesting solution method, which also generalizes the result to the expectation of $X^{-m}$ for integer $m=1,2,3,\dots$. I will use moment generating functions (mgf) and the results from the paper by N Cressie et.al http://amstat.tandfonline.com/doi/abs/10.1080/00031305.1981.10479334?journalCode=utas20 "The Moment-Generating Function and Negative Integer Moments".
They give the result that when $X$ is a
positive random variable with mgf $M_X(t)$ which is defined in an open neighbourhood of the origin, then we have$$\DeclareMathOperator{\E}{\mathbb{E}}\E X^{-m} = \Gamma(m)^{-1} \int_0^\infty t^{m-1} M_X(-t) \; dt$$for positive integers $m$.
It is known that for the beta distribution, the mgf is given by a confluent hypergeometric function as$$ M_X(t) = {}_1F_1(\alpha;\alpha+\beta;t)$$so using the result above gives that$$ \E X^{-m} = \Gamma(m)^{-1} \int_0^\infty t^{m-1} {}_1F_1(\alpha;\alpha+\beta;-t)\; dt$$I evaluated that integral with the help of maple:
assume( a>0, b>0 );assume(m-1,posint)
GAMMA(m)^(-1) * int( t^(m-1)*hypergeom([a],[a+b],-t), t=0..infinity )
GAMMA(a + b) GAMMA(a - m)
-------------------------
GAMMA(a) GAMMA(a + b - m)
so finally we can write the result as $$\E X^{-m} = \frac{\Gamma(\alpha+\beta)\Gamma(\alpha-m)}{\Gamma(\alpha)\Gamma(\alpha+\beta-m)}$$which coinsides with the other answer for $m=1$. Then some human mathematics is needed to conclude that we need the assumption $\alpha > m$ for this to be valid. |
Maass forms of levels 100 to 1000 with $0 \leq R\leq 1$
The horizontal axis is the spectral parameter $R$ with the Laplace eigenvaluesatisying $\lambda=1/4+R^2$. The vertical axis is the level $N$. Each pointcorresponds to a Maass form of weight 0 and trivial character on $\Gamma_0(N)$with the color showing whether the symmetry is
evenor odd. For $N>100$ there are only results for prime level.
Examples of some ranges with complete data:
$1\leq N\leq10, \, 0\leq R\leq 10$ $N$ prime and $100\leq N \leq 250, \, 0\leq R\leq 2$ $N$ prime and $100\leq N \leq 1000, \, 0\leq R\leq 1$ Clicking on a dot takes youto the homepage of the Maass form. |
Explanation of notation:
$H$ is the enthalpy of the system.
$\Delta$ means change of, so $\Delta H$ means change of the enthalpy
This symbol means standard condition, standard condition is defined as a pressure of $10\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\ominus$.
Enthalpy changes:
Enthalpy of dilution – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.
Enthalpy of ($n\text{th}$) electron affinity – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms.$$\ce{Li(g) + e-(g) -> Li-(g) +60\ \text{kJ}}$$$$\ce{F(g) + e-(g) -> F-(g) +328\ \text{kJ}}$$
Enthalpy of ($n\text{th}$) ionization – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.
$$\ce{Li(g) +520\ \text{kJ} -> Li+(g) +e-(g)}$$$$\ce{He(g) +2372\ \text{kJ} -> He+(g) +e-(g)}$$
Enthalpy of lattice dissociation – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions.An example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\ \mathrm{kJ/mol}$$$\ce{NaCl(s) +787\ \text{kJ} -> Na+(g) + Cl-(g)}\qquad\Delta H=787\ \text{kJ/mol}$$
Enthalpy of lattice formation – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.
$$\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\ \text{kJ}\qquad\Delta H=-787\ \text{kJ/mol}$$
Which means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation
Enthalpy of hydration($\Delta_\text{hyd}H^\ominus$) – Enthalpy change when when one mole of ions undergo hydration.
Enthalpy of mixing – The enthalpy change from a substance when mixed.
Enthalpy of neutralisation – The enthalpy change when an acid is completely neutralised by a base.
For a strong acid, like $\ce{HCl}$ and strong base, like $\ce{NaOH}$, they disassociate almost completely $\ce{Cl-}$ and $\ce{Na+}$ are spectator ions so what is actually happening is $$\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\ \text{kJ/mol}}$$
However, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.
For example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\ \text{kJ/mol}$
Enthalpy of precipitation – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.
Enthalpy of solution – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.
Enthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps
Breaking of solute-solute attraction (endothermic) Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF Forming solvent-solute attraction (exothermic)
An example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\ \text{kJ/mol}$
Enthalpy of (Solid$\rightarrow$Liquid: $\Delta_\text{fus}H^\ominus$, Liquid$\rightarrow$Solid:$\Delta_\text{freezing}H^\ominus$, Liquid$\rightarrow$Gas: $\Delta_\text{vap}H^\ominus$, Gas$\rightarrow$Liquid: $\Delta_\text{cond}H^\ominus$, Solid$\rightarrow$Gas: $\Delta_\text{sub}H^\ominus$, Gas$\rightarrow$Solid: $\Delta_\text{deposition}H^\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state.$$\Delta_\text{fus}H^\ominus=-\Delta_\text{freezing}H^\ominus,\Delta_\text{vap}H^\ominus=-\Delta_\text{cond}H^\ominus,\Delta_\text{sub}H^\ominus=-\Delta_\text{deposition}H^\ominus$$ Standard enthalpy of atomization($\Delta_\text{at}H_T^\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\circ K$.$$\ce{S_8 -> 8S}\qquad\Delta_{at}H^\ominus=278.7\ \text{kJ/mol}$$
Standard enthalpy of combustion($\Delta_\text{c}H_T^\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\circ K$ and $10\ \mathrm{kPa}$.$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)}+572\ \text{kJ}\qquad\Delta_cH^\ominus = -286\ \text{kJ/mol}$$
Standard enthalpy of formation($\Delta_\text{f}H_T^\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\ \mathrm{kPa}$ at $T^\circ K$.
It can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\ce{C}$ and $\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\ \text{kJ/mol}$$$\ce{\frac{1}{2}N2(g) +\frac{1}{2}O2(g) -> NO(g)}\qquad\Delta_\text{f}H^\ominus=90.25\ \text{kJ/mol}$$
Standard enthalpy of reaction($\Delta_\text{r}H^\ominus_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\circ K$ and $10\ \mathrm{kPa}$.$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)} +572\ \text{kJ}\qquad\Delta_\text{r}H^\ominus = −572\ \text{kJ/mol}$$$$\Delta_\text{r}H^\ominus=\sum H^\ominus_\text{products}-\sum H^\ominus_\text{reactants}$$$$\Delta_\text{r}H_\text{forward}=-\Delta_\text{r}H_\text{backwards}$$Laws:
Le Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.
Yellow $\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\ce{SCN-}$ to form deep red $\ce{[Fe(SCN)]^2+}$ ions:$$\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$When $\ce{NH4Cl}$ is added, $\ce{Cl-}$ reacts with $\ce{Fe^3+}$ ions to form $\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\ce{Cl-}$ is added into a solution of deep red $\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.
Kirchhoff's Law – Enthalpy of any substance increases with temperature.
Hess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.
Henry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.
Processes:
Constant Entropy – IsoentropicConstant Pressure – Isobaric
Constant Volume – Isovolumetric
Side note: This is probably not a complete list as I may have missed some |
Amplitude of oscillator function
\(A(n)= \psi_n(0)\)
where \(\psi_n(z)=\frac{1}{\sqrt{N_n}}\)HermiteH\(_n(z)\, \exp(-z^2/2)\) is oscillator function, normalised solution of the stationary Schroedinger equation for the Harmonic oscillator. here, HermiteH\(_n\) denotes the \(n\)th Hermite polynomial and \(N_n\) is its norm.
For the asymptotic expansions of various functions with the oscillator function, the asymptotic behaviour of \(A(n)\) at large values of \(n\) is important.
For given number \(n\) of the oscillator function, In fomulas, this function can be denoted with \(A_n\), or with \(A(n)\), or, if letter \(A\) is already occupied for other object, the more long name amos. In general, function Amos=amos can be treated as function of complex argument. For real values of the argument, this function and its asymptotic approximation are shown in figure at right, together with two its asymptotic approximations.
Contents
There is special name for
\(H_n=H_n(0)=\mathrm{HermiteH}[n,0]\).
\(\displaystyle H_n= \frac {2^n \sqrt{\pi}} {\displaystyle \mathrm{Factorial}\left(- \frac{1\!+\!n}{2}\right)}\) \(\displaystyle = \left\{ \begin{array}{ccc} 0 & \mathrm{for ~ odd} & n \\ \displaystyle (-1)^{n/2} \frac{n!}{(n/2)!} &\mathrm{for ~ even} & n \end{array} \right.\)
Then, the amplitude of oscillation of the oscillator function in vicinity of zero, at least for even \(n\), can be expressed through the Hermite number as follows
\(\displaystyle A_n=|F_n(0)|= \frac{|H_n|}{\sqrt{N_n}} \) \(= \displaystyle \frac{\frac{n!}{(n/2)!}}{\sqrt{2^n n! \sqrt{\pi}}} \) \(= \displaystyle \frac{\sqrt{n! / \sqrt{\pi}}} {2^{n/2} (n/2)! } \)
Generalisation
In order to test various representations of the Amplitude of oscillator function \(A_n\), as well as the modifications for similar cases, it worth to treat it as holomorphic suction of complex argument \(n\).
In particular, this provides the amplitude of oscillations of the oscillator function \(F_n(x)\) at large \(n\) and moderate \(x\) for even \(n\). This may be especially important for large \(n\), id huge amount of modes is necessary to take into account in order to describe, for example, the efficiency of channelling of a quantum particle by absorbing walls.
The range of holorpism or the original representation of \(A_n\) through \(\sqrt{n!}\) is limited by the cut line of function \(\sqrt{\,}\).
For this reason, it worth to redefine Amplitude of oscillator function as follows:
\(\displaystyle A_n=A(n)=\mathrm{amos}(n)=\pi^{-1/4} \) \(\displaystyle \exp\left(\frac{1}{2}\mathrm{lof}(n)-\mathrm{lof}(n/2)-\ln(2)\, n/2\big)\right)\)
where lof is holomorphic extension of logarithm of factorial; in vicinity of the positive part of the real axis,
\(\mathrm{lof}(z)= \ln(\mathrm{Factorial}(z)) = \ln(z!)\)
Function amos can be interpreted complex function of complex argument. The map of this function is shown in figure at right with lines \(u\!=\)const\(~\) and lines \(v\!=\)const\(~\) for \(u\!+\!\mathrm i v=\,\)amos\((x\!+\!\mathrm i y)\)
Asymptotics at large values of the argument
For large values of argument, amplitude of oscillator function can be represented with asymptotic series
\(A_n\approx \frac{(2/n)^{1/4}}{\sqrt{\pi}} \) \( \left( 1-\frac{1}{8n}+\frac{1}{128 n^2}+\frac{21}{1024 n^3}-\frac{85}{32768 n^4} -\frac{6511}{262144 n^5}+\frac{13989}{4194304n^6}+\frac{2539085}{33554432 n^7} -\frac{21469709}{2147483648 n^8} -\frac{7385563163 }{17179869184 n^9} +\frac{15306326687 }{274877906944 n^{10}} %+\frac{8625737936387 }{2199023255552 n^{11}} %-\frac{35288423873601 }{70368744177664 n^{12}} %-\frac{29542243391383163 }{562949953421312 n^{13}} %+\frac{59967901184881517 }{9007199254740992 n^{14}} %+\frac{69743016827113388925 }{72057594037927936 n^{15}} %-\frac{1127507014234512448061 }{9223372036854775808 n^{16}} %-\frac{1736837605178928412140179 }{73786976294838206464 n^{17}} %+\frac{3500257705125381294286523 }{1180591620717411303424 n^{18}} %+\frac{6893220525032150345744187527 }{9444732965739290427392 n^{19}} %-\frac{27734856867682323334288371163 }{302231454903657293676544 n^{20}} +O\left(\frac{1}{n^{11}}\right) \right) \)
For large values of number of mode, say \)n>20$, such a representation provides of order of 16 decimal digits, that corresponds to the best precision available with complex double arithmetics. References http://mathworld.wolfram.com/HermiteNumber.html Weisstein, Eric W. "Hermite Number." From MathWorld--A Wolfram Web Resource. |
There is a finite, circular area $A=\pi \times r^2$ with a given radius $r$ and a variable number of points $P_n$ ($n \in \mathbb{N}_{>0}$) that are to be placed inside of this area.
What is the distance $d$ to the closest dots when placing them in a way to maximise this closest distence while also placing them at minimum the same distance $d$ to the edge of the area?
This is distinct from this and similar problems by trying to also maximise the distance between the edge and points and the points at the same time while also using $n$ points instead of 2.
For $n=1$ the obvious solution is $d=r$ while $P_1$ is also the center of the circle.
For $n=2$ I get a distance of $d= r \frac{2}{3} $ from the simple geometry: They have to be on the opposite sides of the center of the circle, so they are both on a line of $ 2 \times r$, which they divide into 3 areas of equal length.
For $n=3$ to $n=6$ I lack a good formula, but I think the solution might be to place them on polygons with a number of corners equal to $n$.
For $n=7$ I found, that one of the points has to end up in the center of the circle, while the other 6 form a hexagon around it, that should have a side length of $d=\frac{r}{2}$. I derived this from experimental drawings, using a set of 7 circles of $R=\frac{r}{2}$ that have their centers always on the edge of 3 others |
I'm reading through Halzen and Martin Chapter 6.3 and I have a few questions about what they're doing. We're calculating the invariant amplitude for $e^-\mu^- \rightarrow e^-\mu^-$ scattering. The text is (up to some rearranging to make the question clearer) as follows:
It is convienent to separate the sums over the electron and muon spins by
$$\overline{|\mathfrak{R}|^2} = \frac{1}{(2s_A+1)(2s_B+1)}\sum_{\substack{all\ spin \\ states}}|\mathfrak{R}|^2 = \frac{e^4}{q^4} L_e^{\mu\nu}L_{\mu\nu}^{muon}$$
where $s_A$, $s_B$ are the spins of the incoming particles, and the tensor associated with the electron vertex is
$$L_e^{\mu\nu} = \frac{1}{2}\sum_{e\ spins}[\bar{u}(k')\gamma^\mu u(k)][\bar{u}(k')\gamma^\nu u(k)]^*$$
and with a similar expression for $L_{\mu\nu}^{muon}$.
It's not clear to me why the second equality in the first line is true. It sounds as if the authors know a priori that we can separate the sums by writing it as a contraction of two tensors. Is this always possible? I'm not sure how to verify this on my own, and the authors don't really do much to explain it.
The book continues:
$$L_e^{\mu\nu} = \frac{1}{2}\sum_{s'}\bar{u}_\alpha^{(s')}(k')\gamma_{\alpha\beta}^{\mu}\sum_{s}\bar{u}_\beta^{(s)}(k)\bar{u}_\gamma^{(s)}(k)\gamma_{\gamma\delta}^{\mu}\bar{u}_\delta^{(s')}(k')$$ $$=\frac{1}{2}\sum_{s'}\bar{u}_\delta^{(s')}(k')\bar{u}_\alpha^{(s')}(k')\gamma_{\alpha\beta}^{\mu}\sum_{s}\bar{u}_\beta^{(s)}(k)\bar{u}_\gamma^{(s)}(k)\gamma_{\gamma\delta}^{\mu}$$
justifying this repositioning of $\bar{u}_\delta$ from the back to the front of the equation by saying
The completeness relations allows the sums over both the initial and the final electron spins to be preformed. The repositioning of $\bar{u}_\delta$ makes this evident; it can be moved as the matrix character is recorded by the component.
I don't understand why we can reposition $\bar{u}_\delta$ like this. Does it commute with all of the other $u$'s and $\gamma$'s? The second sentence makes no sense to me at all. What is the matrix character? |
Assume I have two neural networks, abstracted as two feature maps, parametrized by $\theta_x,\theta_y$ respectively. $\phi_x(x;\theta_x) \in \mathbb{R}^{h_1}$, $\phi_y(x;\theta_y) \in \mathbb{R}^{h_2}$ and we would like to perform canonical correlation analysis on the results of those feature maps, which means we want to maximize:
$$\max_{\theta_x, \theta_y, w_x, w_y} w_x^T C_{xy} w_y + \alpha [\text{min} (0, 1 - w_x^T C_{xx} w_x) + \text{min}(0, 1 - w_y^T C_{yy} w_y)] $$
where $C_{xy} = \mathbb{E}[\phi_x \phi_y^T]$, $C_{xx} = \mathbb{E}[\phi_x\phi_x^T]$, $C_{yy} = \mathbb{E}[\phi_y \phi_y^T]$
I now have to calculate the gradient of this function w.r.t. $\theta_x$ as a function of the Jacobian matrix $\frac{\partial \phi_x}{\partial \theta_x}$
Unfortunately, I don't really now to approach this. For example, what is the gradient of an expectation as a function of the Jacobian? |
We develop all the necessary theory here without recourse to cycles/orbits/etc.
You are about to enter another dimension. A dimension not only of
sight and sound, but of mind. A journey into a wondrous land of
imagination.
Next stop, the Transparent-Transposition-Transformation-Tour!
Recall that the symmetric group $S_n$ has $n!$ elements; we assume here that the transformations in $S_n$ operate on the set $\{1,2,\dots,n\}$. We also have
Proposition 1: There is a canonical injection ${\iota_{n+1}^n}: S_n \to S_{n+1}$ sending the bijections in $S_n$ to the bijections in $S_{n+1}$ that keep $n+1$ fixed.
Let ${S_n}^{'} = {\iota_{n+1}^n}(S_n) \subset S_{n+1}$.
Theorem 2: For every bijection $\sigma \in S_{n+1}$ not in ${S_n}^{'}$ there exists a unique $\sigma^{'} \in {S_n}^{'}$ and transposition of the form $(n+1 \, k) \,\text{with}\, 1 \le k \le n$ such that
$\tag 1 \sigma = \sigma^{'} \circ (n+1 \, k)$Proof
Assume that ${{\sigma}_0}^{'} \circ (n+1 \, k_0) = {{\sigma}_1}^{'} \circ (n+1 \, k_1)$. Now ${{\sigma}_0}^{'} \circ (n+1 \, k_0)$ applied to $k_0$ must be $n + 1$ and ${{\sigma}_1}^{'} \circ (n+1 \, k_1)$ applied to $k_1$ must also be $n + 1$. Since we are dealing with injective mappings, $k_0$ must be equal to $k_1$. But then of course, ${{\sigma}_0}^{'} = {{\sigma}_1}^{'}$.
Since $(n+1)! = (n + 1) n!$, a counting argument shows that these unique representations 'cover' all the permutations in $S_{n+1}$ not in ${S_n}^{'}$. $\quad \blacksquare$
To assist in understanding and increase the 'aesthetics quotient', you can regard the (1) representation as true for any element in $S_{n+1}$, where you allow $k$ to be $0$ and agree that $(n+1 \, 0)$ is the identity transformation.
Corollary 3: Every permutation in $S_{n}$ not equal to the identity transformation $1_{Id}$ can be expressed as the composition of $n-1$ or fewer transpositions.
Proof (Sketch) Use induction and apply theorem 2. $\quad \blacksquare$
Corollary 4: Let $A: \{2,3,\dots,n\} \to \{1, 2,3,\dots,n\}$ be any function satisfying $A(k) \lt k$. Then the $n-1$ transpositions $(k \, A(k))$ generate the symmetric group $S_n$.
Proof: Exercise.
We are now prepared to answer the OP's question.
Lemma 5: Let $\sigma \in S_{n+1}$ be represented as the composition of $M$ transpositions. Then it can also be expressed as the composition of $M$ transpositions,
$\tag 2 \sigma = \alpha_1 \alpha_2 \cdots \alpha_s \beta_1 \beta_2 \cdots \beta_t \text{ with } M = s + t$satisfying
$\qquad \text{Each } \alpha_u \in {S_n}^{'}$
And $\qquad \text{Each transposition } \beta_u \text{ acts on } n + 1$
Proof (Sketch)
Using elementary facts about transpositions (see next section), we can keep 'pushing to the right side' transpositions that act on $n+1$ to new transpositions that still act on $n +1$. $\quad \blacksquare$
The idea now is, see what happens when you try to push $\beta_1$ to the right. The answer is either $M$ remains the same with $s$ increasing by 1 and $t$ decreasing by $1$, or $M$ goes down by $2$ with $t$ decreasing by $2$.
Example 6: Let $n + 1 = 10$ and $\beta_1 = (10 \, 5)$ and $\beta_2 = (10 \, 6)$. Then $\beta_1 \circ \beta_2 = (6 \, 5) \circ (10 \, 5)$.
Example 7: Let $n + 1 = 10$ and $\beta_1 = (10 \, 5)$ and $\beta_2 = (10 \, 5)$. Then $\beta_1 \circ \beta_2$ collapses to the identity transformation.
Lemma 8: Let $\sigma \in S_{n+1}$ be represented as the composition of $M$ transpositions. Then it can also be expressed as the composition of $s + 1$ transpositions,
$\tag 3 \sigma = \alpha_1 \alpha_2 \cdots \alpha_s \beta$satisfying
$\qquad \text{Each } \alpha_u \in {S_n}^{'}$
And $\qquad \text{The transposition } \beta \text{ acts on } n + 1$ (or $\beta = \text{Identity}$) And $\qquad s + 1 \le M \text{ and } M = s + 1 \text{ mod 2}$ when $\beta \ne \text{Identity}$ And $\qquad s \le M \text{ and } M = s \text{ mod 2}$ when $\beta = \text{Identity}$
Proof: Exercise.
Theorem 9: Let a permutation $\sigma$ be expressed as the product of $k$ transpositions and also as the product of $j$ transpositions. Then either $k$ and $j$ are both even or they are both odd.
Proof We shall use induction on the number of elements $\{1,2,\dots,n\}$ that are being permuted.
One can see immediately that the theorem holds for $n = 2$.
Suppose the theorem is true for $n$ and let $\sigma$ be a permutation in $S_{n+1}$ represented as the product of $M$ transpositions. If $\sigma(n+1) = n+1$, then $\sigma \in {S_n}^{'}$ and the argument falls right into a case that can be directly handled by our induction hypothesis. Otherwise, we can apply lemma 8 and put $\sigma$ into the form (3), with $\beta$ not the identity. By our induction hypothesis, the length of the $\alpha\text{-stuff}$ can only be odd or even. Adding $1$ for $\beta$ changes nothing in terms of stating that the parity is well-defined. $\quad \blacksquare$
The OP mentioned these elementary facts about transpositions:
\begin{align}(ab)(ab)&= 1_{Id}\\(ab)(ac)&=(bc)(ab)\\(ab)(cd)&=(cd)(ab)\\(ab)(bc)&=(bc)(ac)\end{align}
With $a = n + 1$, you see how you can 'push to the right' the '$n + 1\text{-transpositions}$', occasionally getting a 'collapse' upon encountering $(ab)(ab)$, since the inverse transformations cancel out.
We prove the following lemma that was analyzed by the OP:
Lemma: If $1_{Id} =\tau_1\tau_2\cdots\tau_M$ where the $\tau$'s are transpositions on $\{1,2,\dots,n+1\}$, then $M$ is even.
Proof The lemma is true for $n = 0$ and also (if you are skeptical) for $n = 1$. Assume that the lemma holds for any $n \gt 1$. By lemma 8, for the identity transformation $1_{Id} \in S_{n+1}$ we have a (2) representation,
$\tag 1 1_{Id} = \alpha_1 \alpha_2 \cdots \alpha_s \beta$
respecting the original parity of $M$. Now $\alpha_1 \alpha_2 \cdots \alpha_s$ is in ${S_n}^{'}$. By theorem 2 there is one and only one such representation, so $\beta$ must be the identity (not really there). So we have
$\tag 2 1_{Id} = \alpha_1 \alpha_2 \cdots \alpha_s$
But each $\alpha_u \in {S_n}^{'}$, so by our inductive hypothesis $s$ is even and so $M$ must also be even. $\quad \blacksquare$ |
Research Open Access Published: On a uniqueness theorem of Sturm–Liouville equations with boundary conditions polynomially dependent on the spectral parameter Boundary Value Problems volume 2018, Article number: 28 (2018) Article metrics
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3 Citations
Abstract
Inverse nodal problems for Sturm–Liouville equations associated with boundary conditions polynomially dependent on the spectral parameter are studied. The authors show that a twin-dense subset \(W_{B}([a,b])\) can uniquely determine the operator up to a constant translation of eigenparameter and potential, where \([a,b]\) is an arbitrary interval which contains the middle point of the domain of the operator and
B is a subset of \(\mathbb {N}\) which satisfies some condition (see Theorem 4.2). Introduction
The inverse problems of the differential operator \(L:=L(q,U_{0},U_{1})\):
are considered, where
λ is called the spectral parameter, q is a real-valued \(L^{2}\)-function on \((0,1)\) and
are arbitrary polynomials of degree \(r_{\xi}\) with real coefficients such that \(R_{\xi1}(\lambda)\) and \(R_{\xi 0}(\lambda)\) have no common zeros for \(\xi=0,1\). The inverse spectral problem for the Sturm–Liouville equation with boundary conditions dependent on the spectral parameter was studied in [1–7] respectively. In particular, Freiling and Yurko [4] studied three inverse spectral problems for
L and showed that this operator L can be recovered either from the Weyl function, or from discrete spectral data, or from two spectra. Recently, the inverse spectral problem for L was studied with mixed spectral data in [7–9]. For the case \(R_{00}(\lambda)=1\), \(R_{01}(\lambda)=-h\) in (1.2) and \(R_{10}(\lambda)=1\), \({R_{11}(\lambda)}=H\) in (1.3), the operator \(U(q,U_{0},U_{1})\) turns to a classical Sturm–Liouville problem \(L(q,h,H)\). Inverse spectral problems and inverse nodal problems of \(L(q,h,H)\) have been well studied, the readers can refer to [2, 10–21] and the references therein.
The aim of this article is to investigate the inverse spectral and nodal problems for the BVP
L. We show that the result on the Weyl m-function for L also holds by an alternative approach, which is a generalization of the result for the classical Sturm–Liouville operator in [16]. Moreover, the authors show that the operator \(L(q,U_{0},U_{1})\) can be uniquely determined up to constant translation by the twin-dense subset in the interior interval. Preliminaries
Let \(u_{-}(x,\lambda)\) and \(u_{+}(x,\lambda)\) be solutions of equation (1.1) with initial conditions
Denote \(\lambda=\rho^{2}\), \(\tau=|\operatorname{Im} \rho|\), for sufficiently large \(|\lambda|\), we have
Denote
where \([y,z](x):=y(x)z'(x)-y'(x)z(x)\) is the Wronskian of
y and z. Then
Define the Weyl
m-function \(m_{\pm}(x,\lambda)\) by
then
uniformly in \(x\in[\delta,1]\) (resp. \(x\in[0,1-\delta]\)) for \(|\lambda |\rightarrow\infty\) in any sector \(\varepsilon<\arg(\lambda)<\pi-\varepsilon\) for \(\varepsilon>0\), where \(\delta\in(0,1)\).
Denote the spectrum \(\sigma(L):=\{\lambda_{n}\}_{n=0}^{\infty}\) of
L, \(\sigma(L)\) consisting of the zeros (counting with multiplicities) of the entire function \(\Delta(\lambda)\). For n sufficiently large, \(\lambda_{n}\) are real and simple and satisfy the asymptotic formulae (see [4])
where
Inverse spectral problems
For convenience, let \(\widetilde{L}=L(\widetilde{q},\widetilde {U}_{0},\widetilde{U}_{1})\), where \(L(\widetilde{q},\widetilde {U}_{0},\widetilde{U}_{1})\) is the operator of the same form as
L. If a certain symbol γ denotes an object related to L, then the corresponding symbol γ̃ denotes the analogous object related to L̃ and \(\widehat{\gamma}=\gamma-\widetilde {\gamma}\). Theorem 3.1 Let \(m_{-}(a_{0},\lambda)\) be the Weyl m- function of the BVP L. Then \(m_{-}(a_{0},\lambda)\) can uniquely determine functions \(R_{0k}(\lambda)\) for \(k=0 \textit{ and } 1\) as well as q ( a. e.) on the interval \([0,a_{0}]\), \(0< a_{0}\leq1\). Proof
Denote by \(L_{D}\) the boundary value problem (1.1), (1.2) together with \(\Delta_{D}(\lambda):=u_{-}(a_{0},\lambda)=0\) and \(\{\mu_{a_{0},n}\}_{n=1}^{\infty}\), the zeros (counting with multiplicities) of the entire function \(\Delta_{D}(\lambda)\) (see [10]). Then \(\mu_{a_{0},n}\) is real and simple for sufficiently large
n and
where \(\omega_{1}=\frac{1}{2}\int_{0}^{a_{0}}q(x)\,dx-a_{000}\). Thus we have
By virtue of Hadamard’s factorization theorem,
where \(C_{a_{0},0}\) is a constant and \(m_{0}\geq0\). Let \(G_{\delta_{0}}:=\{ \lambda:|\rho-\frac{\pi}{a_{0}} (k-r_{0}+\frac{1}{2} )|>\delta_{0}, k\in\mathbb{Z}\}\), where \(\delta_{0}\) is sufficiently small, then there exists a constant \(C_{a_{0},\delta_{0}}\) (see [10, 11]) such that
Similarly, denote by \(L_{N}\) the boundary value problem (1.1), (1.2) together with \(\Delta_{N}(\lambda):=u_{-}'(a_{0},\lambda)=0\) and \(\lambda_{a_{0},n}\), the zeros (counting with multiplicities) of the entire function \(\Delta_{N}(\lambda)\). Then \(\{\mu_{a_{0},n}\}_{n=1}^{\infty}\) are real and simple for sufficiently large
n and
where \(c_{0}=2r_{0}+\omega_{1}\). Therefore we have
Let \(G_{\delta_{1}}:=\{\lambda:|\rho-\frac{(k-r_{0})\pi}{a_{0}}|>\delta_{1}, k\in\mathbb{Z}\}\), where \(\delta_{1}\) is sufficiently small, then there exists a constant \(C_{a_{0},\delta_{1}}\) such that, for sufficiently large \(|\lambda|\),
Thus we have
where \(m_{1}\geq0\), \(C_{a_{0},1}\) is a constant. Under the assumption \(m_{-}(a_{0},\lambda)=\widetilde{m}_{-}(a_{0},\lambda)\), we obtain
Since \(u_{-}(a_{0},\lambda)\) and \(u_{-}'(a_{0},\lambda)\) (resp. \(\widetilde {u}_{-}(a_{0},\lambda)\) and \(\widetilde{u}_{-}'(a_{0},\lambda)\)) have no common zeros, \(\frac{\widetilde{u}_{-}(a_{0},\lambda)}{u_{-}(a_{0},\lambda)}\) and \(\frac {\widetilde{u}_{-}'(a_{0},\lambda)}{u_{-}'(a_{0},\lambda)}\) are two entire functions in
λ and
Using the maximum modulus principle and Liouville’s theorem, we have
Therefore,
This implies
Thus (3.11) shows
Therefore the proof of Theorem 3.1 is completed. □
Analogously, we prove the following theorem on the Weyl
m-function \(m_{+}(a_{0},\lambda)\). Theorem 3.2 Let \(m_{+}(b_{0},\lambda)\) be the Weyl m- function of the BVP L. Then \(m_{+}(b_{0},\lambda)\) can uniquely determine functions \(R_{1k}(\lambda)\) for \(k=0,1\) as well as q ( a. e.) on the interval \([b_{0},1]\), \(0\leq b_{0}<1\). Inverse nodal problems
By virtue of Lemma 3.1 in [22], we see that, for \(n\gg1\), the eigenfunction \(u_{-}(x,\lambda_{n})\) has exactly \(n-r_{0}-r_{1}\) zeros \(0< x_{n}^{1}< x_{n}^{2}<\cdots<x_{n}^{j}<\cdots<x_{n}^{n-r_{0}-r_{1}}<1\) inside the interval \((0,1)\) and satisfy the following asymptotic formula:
for \(0< j< n-r_{0}-r_{1}\), where
w is as that in (2.10). Denote \(x_{n}^{0}=0\), \(x_{n}^{n-r_{0}-r_{1}+1}=1\). Note that \(\sigma(L)\) might contain non-real eigenvalues, hence we write
where \(\sigma_{R}(L)\) consists of real eigenvalues of
L. Denote by X the collection of all zeros of all eigenfunctions of L. Let \(B=\{n_{k}\}_{k=1}^{\infty}\) be a strictly increasing sequence in \(\mathbb {N}\), where \(\lambda_{n_{k}}\in{\sigma_{R}(L)}\). For \(0\leq a< b\leq 1\), we call the subset \(W_{B}([a,b])\) of \(X\cap[a,b]\) an interior twin-dense nodal subset on the interval \([a,b]\) if the following conditions hold: (1)
For all \(n_{k}\in B\), there exists some \(j_{k}\) such that \(x_{n_{k}}^{j_{k}}\in W_{B}([a,b])\).
(2)
The nodal subset \(W_{B}([a,b])\) is twin on the interval \([a,b]\), i.e., if \(x_{n_{k}}^{j_{k}}\in W_{B}([a,b])\), then \(x_{n_{k}}^{j_{k}+1}\in W_{B}([a,b])\) or \(x_{n_{k}}^{j_{k}-1}\in W_{B}([a,b])\).
(3)
The nodal subset \(W_{B}([a,b])\) is dense on the set \([a,b]\), i.e., \(\overline{W}_{B}([a,b])=[a,b]\), where \(\overline{W}_{B}([a,b])\) denotes the closure of \(\overline{W}_{B}([a,b])\).
The following Lemma 4.1 is necessary for us to prove our main results.
Lemma 4.1
(Theorem 3.2 [8])
If \(W_{B}([a,b])=\widetilde{W}_{\widetilde{B}}([a,b])\), then
Let \(S_{B}=\{\lambda_{n_{k}}:n_{k}\in B\}\). For any sequence \(S=\{x_{n}\} _{n=0}^{\infty}\) of positive real numbers, we define
The following theorem is our main result which concerns the unique determination of the operator from a twin-dense nodal subset and a partial spectrum.
Theorem 4.2
\(0< a<\frac{1}{2}<b<1\).
Suppose \(W_{B}([a,b])=\widetilde{W}_{\widetilde{B}}([a,b])\) and for \(a_{1}=a\) and \(1-b\), where \(k_{0}\) is the number of elements in \(\sigma _{c}(L)\). Then and some constant c. Proof
and
Denote
Then
Moreover, we can choose \(\{x_{n_{k}}^{j_{n_{k}}}\}\in W_{B}([a,b])\) and apply Green’s formula to obtain
i.e.,
Define the functions
and
Then we know \(\Delta_{C}(\lambda)\) is a polynomial of degree \(k_{0}\) and
for some constant
K.
Next, we shall use the technique in Appendix B of [15] to get an estimate of \(|G_{S_{B}}(ix)|\). Without loss of generality, we may assume \(\lambda>1\) for \(\lambda\in\sigma_{R}(L)\) (it can be done by a shift of the parameter
λ in L). Then \(N_{S_{B}(t)}=N_{\sigma_{R}(L)}(t)=0\) for \(t\le1\), and
Hence
By (4.6), we know that there exist a \(t_{0}\) and a positive number
K so that
for \(a_{1}=a\text{ or } 1-b\). This leads to
where \(L=[2(1-a_{1})(r_{0}+r_{1}+\frac{1}{2})+2a_{1}k_{0}-1]\). Hence
i.e.,
and \(a_{1}=a\) and \(1-b\), where
c is a constant. Therefore
is an entire function and
In addition, we easily prove that the following formula
holds for sufficiently large \(R_{k}\), some \(0<\varepsilon_{0}<1\), \(c_{0}\) and \(C_{0}\) are two positive constants. Thus, we have
Therefore, we get
This implies
Similarly, we can define
Then one can repeat the same arguments as above on \(H(b,u_{+},\tilde{u}_{+},\lambda)\) to show
This leads to \(H(b,u_{+},\tilde{u}_{+},\lambda)=0\) and
Hence
This completes the proof of Theorem 4.2. □
Corollary 4.3 Under the assumptions of Theorem 4.2, if \(\lambda_{n_{k}}=\widetilde{\lambda}_{n_{k}}\) for \(n_{k}\gg1 \), then \(q(x)=\widetilde{q}(x)\) and \(R_{ij}(\lambda)=\widetilde{R}_{i,j}(\lambda )\) for \(i,j=0,1\). Remark
The readers might be interested in the inverse nodal problem for a more general equation
where \(\frac{d^{2}}{dx^{2}}+A(\lambda)\) is an operator on \(H^{2}(a,b)\). Some of such problems arise from PDE (please refer to [23, 24] for details). Same arguments for Theorem 4.2 seem to work for (4.28) if \(A(\lambda)\) is an appropriate operator.
Conclusion
In this paper, the authors show that a twin-dense subset \(W_{B}([a,b])\), \(0< a< 1/2<b <1\), can uniquely determine (up to a constant translation on both boundary conditions and potential) the Sturm–Liouville operator associated with boundary conditions polynomially dependent on the spectral parameter. The theorem leads to the same conclusion for classical Sturm–Liouville equation when the coefficient polynomials \(R_{ij}(\lambda)\) are all of degree 0 (refer to [25]), but the translation effect on boundary conditions only appears when one of \(R_{ij}(\lambda)\) is a non-trivial polynomial.
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255, 2002–2017 (2013) Acknowledgements
The authors acknowledge the reviewers for their helpful comments.
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Research Open Access Published: The regularity criterion for weak solutions to the n-dimensional Boussinesq system Boundary Value Problems volume 2017, Article number: 44 (2017) Article metrics
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Abstract
We consider the Boussinesq system in the homogeneous spaces of degree −1. To narrow the gap for the existence of small regular solutions in \(\dot{B}^{-1}_{\infty,\infty}(\mathbb{R}^{n})\), the biggest homogeneous space of degree −1 among those embedded in the space of tempered distributions, we show small solutions in the homogeneous Besov space \(\dot{B}^{-1+\frac{n}{p}}_{p,\infty}(\mathbb{R}^{n})\), with \(n\geq2\), \(n\leq p<\infty\).
Introduction and main results
The Cauchy problem of the Boussinesq system in \(\mathbb{R}^{n}\) (\(n\geq 2\)) reads
where \(u=u(x,t)\) and \(\theta=\theta(x,t)\) denote the unknown velocity field and the scalar temperature in the content of thermal convection, respectively, and \(\pi=\pi(x,t)\) the scalar density of the geophysical fluids,
μ the constant kinematic viscosity, \(\kappa >0\) the thermal diffusivity, and \(e_{n}=(0,0,\ldots,1)^{T}\). While \(u_{0}\), \(\theta_{0}\) are given initial data, with \(\nabla\cdot u_{0}=0\) in the sense of distribution.
The Boussinesq system is extensively used in the atmospheric sciences and oceanographic turbulence (see [1–3] and references therein). The problem of the global regularity of the weak solutions of the 3D Boussinesq equations is a big open problem. It is meaningful to study the regularity of the weak solutions under additional critical growth conditions on the velocity or the pressure. Based on some analysis technique, there are some regularity criteria via the velocity of weak solutions in Besov spaces have been obtained in [4–6]. The pressure criterion is in [7–9]. By the velocity criterion, for the n-dimensional Boussinesq system, Yao
et al. in [10] showed the local well-posedness and blow-up criteria in Besov-Morrey spaces \(N_{p,q,r}^{s}(\mathbb{R}^{n})\) in supercritical case \(s > 1 + \frac{n}{p}\), \(1 < q \leq p < \infty\), \(1 \leq r\leq\infty\), and critical case \(s=1+\frac{n}{p}\), \(1 < q \leq p < \infty\), \(r=1\). Zhang et al. [11] got the existence of the 2-dimensional inviscid Boussinesq equations in critical Besov spaces \(B^{\frac{2}{p}+1}_{p,1}(\mathbb{R}^{2})\) and some blow-up criteria.
The global regularity of smooth solution of the 2D Boussinesq equations with the fractional dissipation has been researched recently in [12–18]. Several Beale-Kato-Majada-type regularity criterion have been obtained in [19–25]. There are also some results for the blow-up criteria for the Boussinesq equations (see [26, 27] and the references therein).
For the Navier-Stokes equations, Xin and Chen in [28] researched the small regular solutions in \(\dot{B}^{-1}_{\infty,\infty}\), which is the biggest homogeneous space of degree −1. The authors studied small solutions in the homogeneous Besov space \(\dot{B}^{-1+\frac {n}{p}}_{p,\infty}(\mathbb{R}^{n})\) and a homogeneous space defined by \(\widehat{M}_{n}(\mathbb{R}^{n})\). Here, motivated by the results in [28], our aim is to do some work addressing the small regular solutions of the n-dimensional Boussinesq system in subcritical spaces \(\dot{B}^{-1+\frac{n}{p}}_{p,\infty}(\mathbb{R}^{n})\). The corresponding content in the space \(\widehat{M}_{n}(\mathbb{R}^{n})\) is of our further interest. This result partially extends the result in [28] in another system. More precisely, we will prove the following.
Theorem 1.1 Suppose \(n\geq2\), \(n\leq p<\infty\), \(2-\frac{n}{p}<\alpha<2\), \(u_{0}\in\dot{B}^{-1+\frac{n}{p}}_{p,\infty}(\mathbb{R}^{n})\), \(\theta _{0}\in\dot{B}^{-1+\frac{n}{p}}_{p,\infty}(\mathbb{R}^{n})\cap\dot {B}^{-3+\frac{n}{p}}_{p,\infty}(\mathbb{R}^{n})\) with \(\operatorname{div} u_{0}=0\) in \(\mathbb{R}^{n}\), and \(\Vert u_{0} \Vert _{\dot{B}^{-1+\frac {n}{p}}_{p,\infty}}+ \Vert \theta_{0} \Vert _{\dot{B}^{-1+\frac {n}{p}}_{p,\infty}}+ \Vert \theta_{0} \Vert _{\dot{B}^{-3+\frac {n}{p}}_{p,\infty}}\leq\epsilon\) for some small constant \(\epsilon =\epsilon(n,p,\alpha)\). Then system (1.1) admits a unique regular solution satisfying Preliminary lemmas
We denote (
cf. [28])
with the heat kernel \(h(x,t)=(4\pi t)^{-n/2}e^{- \vert x \vert ^{2}/(4t)}\). And the Fourier transform
f̂ of \(f\in\mathcal {S}\) is defined by
Here \(\mathcal{S}(\mathbb{R}^{n})\) stands for the Schwartz class of rapidly decreasing smooth functions and \(\mathcal{S}'(\mathbb{R}^{n})\) is the space of tempered distributions. The fractional order of the Laplacian is showed by the Fourier transform. For \(\alpha\in\mathbb{R}\),
Due to the homogeneous counterpart Theorem 2.12.2 and the lifting property Theorem 5.2.3/1 in [29], the homogeneous Besov spaces can be given as follows.
Definition 2.1
For \(1\leq p,q\leq\infty\) and \(-\infty<\alpha<\infty\), the homogeneous Besov spaces are defined
where
An important property of the homogeneous spaces is its invariance under the following space scaling:
And furthermore, if \(s<0\), the homogeneous Besov spaces \(\dot {B}^{s}_{p,q}(\mathbb{R}^{n})\) can be equivalently defined as follows (
cf. [30]). Lemma 2.1 Suppose \(1\leq p,q\leq\infty\), \(s<0\). Then \(f\in\dot {B}^{s}_{p,q}(\mathbb{R}^{n})\) if and only if
In particular, for the degree of −1, we have
It is well known that \(\dot{B}^{-1}_{\infty,\infty}(\mathbb{R}^{n})\) is the biggest critical homogeneous space of degree −1, and as shown by Frazier, Jaweth and Weiss [31], any critical homogeneous space continuously embedded in \(\mathcal{S}'(\mathbb{R}^{n})\) is also continuously embedded into \(\dot{B}^{-1}_{\infty,\infty}(\mathbb{R}^{n})\).
Lemma 2.2 Suppose \(0<\theta<1\), \(1\leq p,q\leq\infty\), \(-\infty<\alpha<\beta <\infty\), and Then where \((\cdot,\cdot)_{\theta,q}\) denotes the real interpolation functor.
We express (1.1)
1,2 in the integral form as
where \(\mathbb{P}\) is the Helmholtz-Weyl projection onto a divergence free vector fields defined by
here \(\delta_{j,k}\) is the Kronecker symbol and \(R_{j}=\partial _{j}(-\Delta)^{-\frac{1}{2}}\) are the Riesz transforms. To prove the existence of the regular solution in \(L^{\infty }((0,\infty),L^{n}(\mathbb{R}^{n}))\), we need the \(L^{p}-L^{q}\) type estimate for \(e^{t\Delta}\) in Lebesgue spaces and Besov spaces. See [30] for the proof of the following lemma.
Lemma 2.3 Suppose \(1\leq p,q\leq\infty\). Thus, the following estimates hold: If \(s_{1},s_{2}<\frac{n}{p}\) and \(s_{1}+s_{2}+n\min\{0,1-\frac{2}{p}\}>0\), then, for a positive constant C, we have
Due to the linearization of the Boussinesq system (1.1), we consider a priori estimates for the Stokes equations. Chen and Xin in [28] gave the estimates in homogeneous spaces. The Stokes equations in \(\mathbb{R}^{n}\), \(n\geq2\) read
We now state the main estimate about the Stokes equations in homogeneous spaces of Chen and Xin in [28], which will be used later.
Lemma 2.4 Let \(0<\alpha<2\), \(2\leq n\leq p\leq\infty\), \(1\leq q\leq\infty\), and Then the solution of equation (2.1) in the following integral formulation: satisfies the estimates provided that the right- hand sides of the above inequalities are finite, respectively.
Similar to Lemma 2.4, we have the analogous results.
Lemma 2.5 Under the same conditions with α, p, q in Lemma 2.4, and \(f\in\dot{B}_{p,\infty}^{\alpha-3+\frac{n}{p}}\), the following integral equation: satisfies the estimation Proof
By Definition 2.1, we estimate the term \(\Vert \int_{0}^{t} e^{(t-s)\Delta}\mathbb{P}f(s)\,ds \Vert _{\dot{B}^{\alpha-1+\frac {n}{p}}_{p,\infty}(\mathbb{R}^{n})}\) as follows:
For the term \({I=\int_{0}^{\frac{t}{2}}(\tau +t-s)^{-2+(\alpha-1+\frac{n}{p})/2}s^{-\frac{\alpha}{2}}\,ds}\), due to \(0< s<\frac{t}{2}\), thus
and the fact that \(-2+(\alpha-1+\frac{n}{p})/2<0\), we have
As regards the term \(\mathit{II}=\int_{\frac{t}{2}}^{t}(\tau +t-s)^{-2+(\alpha-1+\frac{n}{p})/2}s^{-\frac{\alpha}{2}}\,ds\), since \(\frac{t}{2}< s< t\),
therefore
Thereby, the estimate becomes
Due to the definite integral \(\int_{0}^{t} s^{-\frac{\alpha}{2}}\,ds=\frac {2}{2-\alpha}t^{1-\frac{\alpha}{2}}\), we obtain
According to the fact that
thus
Finally, we get
For the estimate of the initial term \(\Vert e^{t\Delta}\mathbb {P}a \Vert _{\dot{B}^{\alpha-1+\frac{n}{p}}_{p,\infty}}\), we have
because of the inequality
then we obtain
On the other hand, by Lemma 2.3, we get
Lemma 2.6 Suppose \(2-\frac{n}{p}<\alpha<2\), \(2\leq n\leq p\leq\infty\), and Thus equation (2.2) satisfies the estimate Proof
the way to deal with the above estimate is similar to the one in Lemma 2.4, we omit it here. Finally, we get
For the initial term, according to Lemma 2.3, we obtain
Therefore, we get the proof. □
The proof of Theorem 1.1
This section is devoted to the proof of Theorem 1.1. For simplicity, without loss of generality, we assume \(\mu=\kappa=1\). In order to deal with the convection terms, we observe the interesting interpolation results
Thus by the embedding and interpolation results, we have
and
And then due to the inequality \(a^{1-\beta}b^{\beta}\leq a+b\), \(0<\beta<1\), \(a,b>0\), we have
By a variable transformation, (3.1) implies
Equation (3.2) means
Thus to prove the two equations in Theorem 1.1, we only need to show that
Set
with
and
We will show
M and M̄ are the contraction operators mapping a ball of U into itself and a ball of Θ into itself, respectively. Now we deal with the operator M̄. Observing that (3.1) and (3.2) are also true for θ, we have
And using Lemma 2.4, we obtain the estimates of
θ. It is more simple for θ here, as the absence of the operator \(\mathbb{P}\), but the estimates are still true, which are
Define two complete metric spaces by
Suppose \(\theta_{1},\theta_{2}\in\Theta\), by (3.5), we have
According to the contraction mapping principle, equation (2.1) admits a unique solution \(\theta\in\Theta_{\epsilon}\), provided that \(C \Vert \theta_{0} \Vert _{\dot{B}^{-1+\frac {n}{p}}_{p,\infty}}\leq\frac{\epsilon}{2}\) and \(\epsilon>0\) is sufficiently small.
Remark 3.1
To show the boundedness of the term \({\sup_{s>0}s^{\frac {\alpha}{2}} \Vert \theta \Vert _{\dot{B}^{\alpha-3+\frac{n}{p}}_{p,\infty }}}\), we choose \(f(s)=u(s)\otimes\theta(s)\) in Lemma 2.6, and by Lemma 2.3, we have
thus we get
that is to say, if \(\Vert u \Vert _{U}\) is small enough, the term \({\sup_{s>0}s^{\frac{\alpha}{2}} \Vert \theta(s) \Vert _{\dot{B}^{\alpha-3+\frac{n}{p}}_{p,\infty}}}\) is bounded.
that is, due to \(\mathbb{P}Mu(t)=Mu(t)\), we obtain \(\nabla\cdot Mu(t)=0\). By the definition of \(U_{\epsilon}\), the previous analysis shows that
Therefore, on the basis of the contraction mapping principle, equation (2.1) admits a unique solution \(u\in U_{\epsilon}\), provided that \(C \Vert u_{0} \Vert _{\dot{B}^{-1+\frac {n}{p}}_{p,\infty}}\in\frac{\epsilon}{3}\), \(C \Vert \theta_{0} \Vert _{\dot{B}^{-3+\frac{n}{p}}_{p,\infty}}\in\frac{\epsilon }{3}\) and \(\epsilon>0\) is sufficiently small.
The proof of Theorem 1.1 is done.
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Acknowledgements
This research is supported by the Foundation for Ph.D. of Henan Normal University of China under Grant 5101019170156.
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Research Open Access Published: Existence and multiplicity of non-trivial solutions for the fractional Schrödinger–Poisson system with superlinear terms Boundary Value Problems volume 2019, Article number: 4 (2019) Article metrics
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Abstract
In this paper, we study the following fractional Schrödinger–Poisson system with superlinear terms
where \(s,t\in (0,1)\), \(4s+2t>3\). Under certain assumptions of external potential \(V(x)\), nonnegative density charge \(K(x)\) and superlinear term \(f(x,u)\), using the symmetric mountain pass theorem, we obtain the existence and multiplicity of non-trivial solutions.
Introduction and main results
In this paper, we are concerned with the fractional Schrödinger–Poisson system
where \((-\Delta )^{s}\) is fractional Laplacian operator, \(s,t\in (0,1)\), \(4s+2t>3\).
On the potential \(V(x)\), we make the following assumptions:
(\(V_{1}\)):
\(V(x)\in C(\mathbb{R}^{3},\mathbb{R})\), \(\inf_{x\in \mathbb{R}^{3}}V(x)>0\).
(\(V_{2}\)):
For any \(b>0\) such that the set \(\{x\in \mathbb{R} ^{3}: V(x)< b\}\) is nonempty and has finite Lebesgue measure. In some previous papers, except for \((V_{1})\)–\((V_{2})\), the following, \((V_{3})\), is needed.
(\(V_{3}\)):
\(\varOmega =\operatorname{int}V^{-1}(0)\) is nonempty and has smooth boundary and \(\overline{\varOmega }=V^{-1}(0)\).
When \(\phi =0\), the system (1.1) reduces to a fractional Schrödinger equation, which is a fundamental equation of fractional quantum mechanics. It was firstly introduced by Laskin [9, 10] as a result of extending the Feynman path integral, from the Brownian-like to the Lévy-like quantum mechanical paths, where the classical Schrödinger equation changes into the fractional Schrödinger equation. Recently, nonlocal fractional problems have attracted much attention, we refer to [12].
When \(s=t=1\), \(K(x)=1\), the system (1.1) reduces to the following Schrödinger–Poisson system (or Schrödinger–Maxwell system):
Due to the real physical meaning, it has been extensively investigated. Benci and Fortunato [4] firstly proposed the system like (1.2) to describe solitary waves for nonlinear Schrödinger type equations and look for the existence of standing waves interacting with unknown electrostatic field. Kristály and Repovš [8] studied a coupled Schrödinger–Maxwell system with the nonlinear term \(f:\mathbb{R}\rightarrow \mathbb{R}\) being superlinear at zero and sublinear at infinity. Under different conditions, they proved a non-existence result and obtained the existence of at least two non-trivial solutions.
There are plenty of results for system (1.2), we refer the interested reader to [3, 5, 13, 17, 19, 25] and the references therein, the main tool is the mountain pass theory [15]. However, to the best of our knowledge, similar results on the fractional Schrödinger–Poisson systems are not so rich as the Schrödinger–Poisson systems (1.2). Zhang, do Ó and Squassina [24] studied the fractional Schrödinger–Poisson system with a general nonlinearity in the subcritical and critical case,
where \(\lambda > 0\), \(s,t \in [0, 1]\), \(4s + 2t \geq 3\). With some hypotheses, a non-trivial positive radial solution is admitted. Very recently, Teng [21] considered the following nonlinear fractional Schrödinger–Poisson system with critical Sobolev exponent:
under some appropriate conditions on \(V(x)\), where \(\mu \in \mathbb{R^{+}}\) is a parameter, \(1< q<2^{\ast }_{s}-1= \frac{3+2s}{3-2s}\), \(s,t\in (0,1)\) and \(2s+2t>3\), the existence of a non-trivial ground state solution of system (1) can be proved. Later, Li [11] studied the nonlinear fractional Schrödinger–Poisson equation
where \(s,t\in (0,1]\), \(4s + 2t >3\). Under some assumptions on
f, the existence of non-trivial solutions for this system is obtained.
Motivated by all the works just described above, we want to find the existence and multiplicity of non-trivial solutions for the fractional Schrödinger–Poisson with superlinear terms, the following assumptions are needed:
( K):
\(K(x)\in L^{\frac{6}{4s+2t-3}}(\mathbb{R}^{3})\bigcup L ^{\infty }(\mathbb{R}^{3})\), \(s, t\in (0,1)\), \(4s+2t>3\), \(K\geq 0\), \(\forall x\in \mathbb{R}^{3}\).
\((f_{1})\) :
\(\lim_{|t|\rightarrow \infty } F(x,t)/{t^{4}}=+ \infty \) a.e. \(x\in \mathbb{R}^{3}\), and there exists \(r_{1}>0\) such that$$ F(x,t)\geq 0, \quad \forall x \in \mathbb{R}^{3}, \vert t \vert \geq r_{1}, $$
where \(F(x,t)=\int ^{x}_{0} f(x,t)\,\mathrm{d}x\).
(\(f_{2}\)):
There exist constant \(a>0\), \(p\in (2,2^{\ast }_{2})\) such that$$\begin{aligned} \bigl\vert f(x,t) \bigr\vert \leq a\bigl(t+{ \vert t \vert }^{p-1}\bigr), \quad \forall (x,t) \in \mathbb{R}^{3} \times \mathbb{R}, \end{aligned}$$
where \(2^{\ast }_{s}=\frac{6}{3-2s}\).
(\(f_{3}\)):
There exists \(L>0\) such that$$\begin{aligned} \frac{1}{4}f(x,t)-F(x,t)\geq 0, \quad \forall x\in \mathbb{R}, \vert t \vert \geq L. \end{aligned}$$
(\(f_{4}\)):
\(f(x,t)=f(x,-t)\), \(\forall (x,t) \in \mathbb{R} ^{3}\times \mathbb{R}\).
Now we are ready to state the main result of this paper as follows.
Theorem 1.1 Variational settings and preliminaries
Let \(L^{r}(\mathbb{R}^{3})(0\leq r<\infty )\) be the usual Lebesgue space with the standard norm \(\|u\|_{r}\) and
û as the Fourier transform of u. Firstly let us introduce some necessary variational settings for system (1.1). A complete introduction to fractional Sobolev spaces can be found in [1]. Recall that the fractional Sobolev spaces \(H^{s}( \mathbb{R}^{3})\) can be described by the Fourier transform, that is,
equipped with the norm
According to Plancherel’s theorem [7], we have \(\|u\|_{2}=\|\hat{u}\|_{2}\), \(\|(-\Delta )^{\frac{s}{2}}u\|_{2}=\|\xi ^{s}\hat{u}\|_{2}\). Thus
Following [14], the fractional Laplacian \((-\Delta )^{s}\) can be viewed as
where \(P.V\). is the principal value and \(C(s)>0\) is a normalization constant.
For \(s\in (0,1) , D^{s,2}(\mathbb{R}^{3})\) is a homogeneous fractional Sobolev space defined as
which is the completion of \(C^{\infty }_{0}(\mathbb{R}^{3})\) with respect to the norm
We use “→” and “⇀” to denote strong and weak convergence in the related function spaces, respectively. The symbol “↪” means that a function space is continuously embedded into another function space.
Let
E is endowed with the following inner product and norm: Lemma 2.1
(Lemma 2.3 in [20])
Suppose that \(V(x)\) satisfies \((V_{1})\) –\((V_{2})\), the Hilbert space E is compactly embedded in \(L^{r}(\mathbb{R}^{3})\) \((2\leq r < 2^{ \ast }_{s})\).
As a consequence of Lemma 2.1, there is constant \(C_{r}>0\) such that
For any \(u\in H^{s}(\mathbb{R}^{3})\), one can use the Lax–Milgram theorem [6] to find that there exists a unique \(\phi ^{t}_{u}\in D^{t,2}(\mathbb{R}^{3})\) such that
In other words, \(\phi ^{t}_{u}\) is the weak solution of the fractional Poisson equation
and the representation formula holds, that is,
which is called the
t-Riesz potential (Chap. 5.1 in [18]), where Lemma 2.2
\(\forall u\in H^{s}(\mathbb{R}^{3})\),
there exists \(C_{0}>0\) such that Proof
In (2.1), let \(v=\phi ^{t}_{u}\), using the Hölder inequality,
The result follows. □
The energy functional associated to problem (1.1) is given by
Moreover, its differential is
It is clear that the pair \((u,\phi ^{t}_{u})\) is a solution to the system (1.1) if and only if
u is a critical point of \(I(u)\). Lemma 2.3
(Symmetric mountain pass theorem)
Let E be a real infinite dimensional Banach space such that \(E=Y\oplus Z\), where Y is finite dimensional subspace. Suppose \(\varPhi \in C^{1}(E,\mathbb{R})\) is an even functional satisfying the Palais–Smale condition, \(\varPhi (0)=0\); if (i) there exist constant ρ, α such that\(\varPhi |_{\partial {B_{\rho }}\bigcap Z \geq \alpha }\), where\(B_{\rho }\) denotes the open ball in E of radius ρ about0 and\(\partial B_{\rho }\) denotes its boundary; (ii) for arbitrary finite dimensional subspace\(\tilde{E} \subset E\), there exists constant\(R=R(\tilde{E})>0\) such that\(\varPhi (u) \leq 0\) if\(u\in \tilde{E}/{B_{R}}\); then the functional Φ possesses an unbounded sequence of critical values. Proof of Theorem 1.1 Lemma 3.1 Under the assumptions \((V_{1})\) –\((V_{2})\) and \((f_{1})\) –\((f_{3})\), \(I(u)\) satisfies the Palais–Smale condition. Proof
Let \(\{u_{n}\}\subset E\) be the Palais–Smale sequence of
I, we assert that \(\{u_{n}\}\) is bounded. Otherwise, there exists a subsequence (for the sake of convenience, we still write it as \(\{u_{n}\}\)) such that \(\|u_{n}\|\rightarrow \infty (n\rightarrow \infty )\). Define \(\omega _{n}:=u_{n}/\|u_{n}\|\), there exists a subsequence such that
Case 1. \(\omega =0\). The proof of this case is almost the same as the one of Lemma 2.3 in [23], so we omit it.
Case 2. \(\omega \neq 0\). We have
then
By \((f_{1})\),
Let \(\varOmega _{n}(a,b)=\{x\in \mathbb{R}^{3}:a\leq |u_{n}(x)|< b, 0 \leq a< b \}\); we have
Take the infimum of the inequality, then
□
If \(\omega \neq 0\), \(|u_{n}(x)|\rightarrow \infty \ (n\rightarrow \infty )\), then, for
n sufficiently large, \(\{x\in \mathbb{R}^{3}:\omega (x)\neq 0\}\subset \varOmega _{n}(r_{1},+\infty )\). By \((f_{1})\) and the Fatou lemma,
a contradiction, so the sequence \({u_{n}}\) is bounded.
Since the sequence \(\{u_{n}\}\) is bounded, there exists a subsequence (we still write it as \(u_{n}\)) such that
To prove \(u_{n}\rightarrow u\) in
E, we need to prove \(\|u_{n}\| \rightarrow \|u\|\) (this is because E is a Hilbert space). By (2.2)
With \((f_{1})\) and the second limit of (3.4),
For \(K\in L^{\infty }\), by \((K)\), we have
For \(K\in L^{\frac{6}{4s+2t-3}}\), we have
Lemma 3.2 Suppose that \((V_{1})\) –\((V_{2})\) and \((f_{1})\) –\((f_{2})\) hold, then, for every finite subspace \(\tilde{E}\subset E\), it follows that Proof
Suppose there exists \(\{u_{n}\}\subset E\) such that \(\|u_{n}\|\rightarrow \infty \) and \(\inf_{n\rightarrow \infty } J(u_{n})>\infty \). Let \(w_{n}:=u_{n}/\|u_{n}\|\), \(\|w_{n}\|=1\) and there exists \(w\in E \backslash \{0\}\) such that
Similarly to the proof of (3.3), we can get a contradiction. □
Corollary 3.3 Suppose that \((V_{1})\) –\((V_{2})\) and \((f_{1})\) –\((f_{2})\) hold, then, for every finite subspace \(\tilde{E}\subset E\), there exists \(R=R(\tilde{E})>0\) such that
Let \(\{e_{i}\}^{\infty }_{i=1}\) be orthogonal bases of space
E, let \(X_{i}=\mathbb{R}e_{i}:=\{\alpha e_{i}:\alpha \in \mathbb{R}\}\), Lemma 3.4
(Lemma 3.8 in [22])
Suppose that \((V_{1})\) –\((V_{2})\) hold, if \(2\leq r<2^{\ast }\), we have Proof
It is obvious that \(0<\beta _{k+1}\leq \beta _{k}\), so that \(\beta _{k} \rightarrow \beta \geq 0\), \(k\rightarrow \infty \). For every \(k\geq 0\), there exists \(u_{k}\in Z_{k}\) such that \(\|u\|=1\) and \(\|u_{k}\|_{r} \geq \beta _{k}/2\). By definition of \(Z_{k}, u_{k} \rightarrow 0\) in \(H^{s}(\mathbb{R}^{3})\). The Sobolev embedding theorem implies that \(u_{k}\rightarrow 0\) in \(L^{r}(\mathbb{R}^{3})\). Thus we have proved that \(\beta =0\). □
Lemma 3.4 shows there exists positive constant \(k_{1},k_{2} \geq 1\) such that
Lemma 3.5 Suppose that \((V_{1})\) –\((V_{2})\) and \((f_{2})\) hold, let \(k_{3}= \max \{k_{1},k_{2}\}\), there exists constant \(\beta ,\alpha >0\) such that \(\varPhi |_{\partial B_{\rho }\bigcap Z_{k_{3}}\geq \alpha }\). Proof
As a result, let \(\rho :=8^{\frac{1}{(2-p)}}\), we can conclude
□
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70, 2150–2164 (2009) Acknowledgements
The authors would like to thank the editor and referees for their valuable suggestions, which improved the structure and the presentation of the paper.
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This work was supported by the National Natural Science Foundation of China (No. 11301148, No. 11671120).
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Rather than going to talk about algorithms and more about SAT, this answer tries to show that there is more than just one kind of logic and that there are also different kinds of approaches.
I won't go into detail much but rather provide a list with some topics and keywords that should form a good basis to do some research:
Higher order logic
There are many ways of extending the propositional logic as you know it, for example (first order logic) with the quantifiers $\forall$ and $\exists$ quantifiers.
Keywords: First/second order logic (with equality), free/bound variables, Skolem normal form Type theory
This is basically just another higher order logic, but it's a very rich topic and in my opinion deserves its own category since it plays a special role in CS.
Keywords: $\Pi$-/$\Sigma$-types, subtypes, quotient types, Curry-Howard correspondence, (un)typed $\lambda$-calculus, System F, Martin-Löf type theory, $\lambda$-cube Proof systems
These can vary from classical logic as you know, but can be more abstract working with other syntactical objects than Boolean values or predicates. Generally they define some set $\mathcal{X}$ of objects (formulas/syntax), a truth-assigning function $t: \mathcal{X} \to \{\texttt{false},\texttt{true}\}$ (defines semantics), some notion of proofs and a way to (efficiently) check whether a proof is valid.
Keywords: Completeness & soundness, Gentzen/Hilbert style proofs, resolution calculus, Sequent calculus, Intuitionistic logic, Formal verification, temporal logic etc.
If this was not specific enough, here is one thing I found fascinating when I first learned about logic:
$$\lnot \exists x . \forall y . \bigl ( F(x,y) \longleftrightarrow \lnot F(y,y) \bigr )$$
The above formula is a tautology (ie. always true) for any domain $\mathcal{U}$ and predicate $F$. Now usually this is not very interesting, but
define $\mathcal{U}$ as the set of all sets, $F(x,y) = y \in x$ and we get Russell's paradox, or define $\mathcal{U}$ some fixed enumeration of $\{0,1\}^\infty$, $F(x,y) = \text{$x$th bit of $y$}$ (now interpreting $\longleftrightarrow$ as equality on $\{0,1\}$) and we get Cantor's diagonalisation argument
which I find quite interesting. |
So Kirchhoff's second law states the sum of the emfs around a circuit is equal to the sum of potential drops. All explanations point to the conservation of energy. Fine. But why do electrons have to lose
all their gained energy from batteries to the components. Conservation of energy would still hold if they lost a bit of the energy to components yet retained the remainder as kinetic energy as it passed back through the battery?
So Kirchhoff's second law states the sum of the emfs around a circuit is equal to the sum of potential drops. All explanations point to the conservation of energy. Fine. But why do electrons have to lose
$\newcommand{\vect}[1]{{\bf #1}}$
More than conservation of energy, this law is related with the fact that electric field is a conservative one, indeed, if $V$ is the electric potential, then in a static situation
$$ \vect{E} = -\nabla V $$
therefore if $a$ and $b$ are two points and $\gamma$ is a path connecting these points, the work to move a charge $q$ from $a$ to $b$ is
$$ \int_\gamma{\rm d}\vect{l}\cdot (q\vect{E}) = -q\int_\gamma{\rm d}\vect{l}\cdot \nabla V = -q\int_a^b{\rm d}V = -q(V_a - V_b) $$
If $\gamma$ is a
closed curve, then this last term vanishes and
$$ \oint_\gamma{\rm d}V = 0 $$
And this is actually the reason behind Kirchhoff's second Law. Imagine that you split the path $\gamma$ into a bunch of segments $p_1\to p_2 \to \cdots p_N \to p_1$. This last equations ensures that after you measure the potential between consecutive poitns $p_i p_{i+1}$ and then add the values up, the result should be exactly zero |
Neural networks (NNs) are used as approximators in reinforcement learning (RL). To update the policy in RL, the actor network's gradients w.r.t its weights are needed. Since NN doesn't have a mathematical expression to work with, how can its derivatives be calculated?
I think what you mean to ask is
how can differentiation occur when there's no obvious neural network function to differentiate?
Don't worry - lots of people get confused about this, because it seems like an obvious hole in the puzzle. As mentioned by @AtillaOzgur, neural networks use partial differentiation through backpropagation.
First, take the output of all the neurons (except the one you're about to differentiate by) as a function:
The above diagram represents the output of one neuron. Do this for every neuron in your network until you have a set. Let's call this set function NN. The output of NN (given all your neuron outputs) is what you'd normally plug into your RL policy.
You then
differentiate NN by a single neuron (n) as shown:
$$\frac{\partial NN}{\partial n} = \lim_{h\to0} \left(\frac{NN(\text{all other neuron outputs}, n + h) - NN(\text{all other neuron outputs}, n)}{h} \right)$$
In reality however, it's the partial derivative of the activation function (A) with respect to the output of a single neuron (n):
$$\frac{\partial A}{\partial n}$$
So, depending on your activation function, you just plug in your neuron output to a certain expression and you've found the value by which to update your neural network.
I hope this helps. Deep learning is definitely a field with a learning curve, but places like StackExchange are great resources to help you out. |
So I'm working a problem that states:
A function $f$ is analytic in an open set $U$. Define $g$ by $g(z)=\overline{f(\overline{z})}$ (just because the notation can be hard to read, this is the the complex conjugate of the function $f$ defined at the complex conjugate of $z$). Show that $g$ is analytic in the oopen set $U^{\star}=\{z:\overline{z}\in{U}\}$ and that $g^{\prime}(z)=\overline{f^{\prime}(\overline{z})}$ (this is the complex conjugate of the derivative of $f$ evaluated at the complex conjugate of $z$) for $z\in{U^{\star}}$
Now, my book gives the definition of an analytic function as one that is a function defined over an open set on which it is differentiable at every point of that set and that every derivative of that function satisfying the preceeding properties has derivatives that are analytic on that domain as well.
So, I know that being differentiable at a point means that a function is continuous at that point, so I see straight away that this means $\forall{\epsilon}>0$ there exists a $\delta>0$ such that if $|z-z_0|<\delta$ then $|f(z)-f(z_0)|<\epsilon$. Since $U$ is an open set, it follows that $U\cap{z_0}$ is non-empty. Because $f$ is analytic in $U$, I can choose two elements $z$ and $z_0$ of $U$ that satisfy $|z-z_0|=|\overline{z}-\overline{z_0}|<\delta$ and $|f(z)-f(z_0)|<\epsilon$. Since $|\overline{z}-\overline{z_0}|<\delta$, this means that $|f(\overline{z})-f(\overline{z_0})|<\epsilon$. Observing that $|f(\overline{z})-f(\overline{z_0})|=|\overline{f(\overline{z})-f(\overline{z_0})}|$, I conclude that $\overline{f(\overline{z})}$ is continuous.
This being the case, then (and I think I've done this part correct) I can compute the derivative by using the limit definition as:
$g'(z)=\lim\limits_{h\rightarrow{0}}\frac{\overline{f(\overline{z}+\overline{h})}-\overline{f(\overline{z})}}{h}=\lim\limits_{\overline{h}\rightarrow{0}}\overline{({f(\overline{z}+\overline{h})-f(\overline{z})})\ \ /\ \ {\overline{h}}}$, as I know that as $h\rightarrow{0}$, so does $\overline{h}\rightarrow{0}$. Since the function $f$ is analytic for all complex numbers in $U$, its derivative there, as given above, is the complex conjugate of the limit which is $\overline{f'(\overline{z})}$.
That's what I'm concerned about, however, because I haven't actually shown yet that $\overline{f(\overline{z})}$ is, in fact, analytic on its domain, all I know is that it is continuous on its domain. My thought is that it might only be analytic for the set $U$ if $U$ is a subset of the real numbers, since that is the only place that $\overline{z}$ is analytic and thus the composition $(a\circ{b})(z)$ for $b(z)=f(\overline{z})$ (analytic for $\overline{z}\in{U}$) and $a(z)=\overline{z}$, so the composition would be analytic on the set given by the intersection of the real numbers with $U$ but I don't believe that is right?
I've tried considering an equivalent case given by the conjugation $\overline{g(z)}=\overline{\overline{f(\overline{z})}}=f(\overline{z})$ to no avail.
Essentially, I'm asking to see if anyone can point me in the right direction and tell me if what I've done so far is on the right track.
I should add that these problems are from a section in the book preceeding the discussion about the Cauchy-Riemann Equations, so while I'm not sure if they will help, I don't want to use them.
I'm tagging this as homework although it's not 'homework' in the sense that I'm taking a class, but the problem is from a book, so I feel it is appropriate to tag it as such. I invite any admins to remove the tag if they feel it is unnecessary. |
What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?
As everyone above has pointed out, the expected value is $0$.
I expect that the original poster might have wanted to know about how big the determinant is. A good way to approach this is to compute $\sqrt{E((\det A)^2)}$, so there will be no cancellation.
Now, $(\det A)^2$ is the sum over all pairs $v$ and $w$ of permutations in $S_n$ of $$(-1)^{\ell(v) + \ell(w)} (1/2)^{2n-\# \{ i : v(i) = w(i) \}}$$
Group together pairs $(v,w)$ according to $u := w^{-1} v$. We want to compute $$(n!) \sum_{u \in S_n} (-1)^{\ell(u)} (1/2)^{2n-\# (\mbox{Fixed points of }u)}$$
This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in $$e^{2x-x^2/2+x^3/3 - x^4/4 + \cdots} = e^x (1+x).$$
So $\sqrt{E((\det A)^2)}$ is $$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n-1)! \right)} = \sqrt{(n+1)!}/ 2^n$$
If $N \ge 2$, then the expected value is $0$ since interchanging two rows preserves the distribution but negates the determinant.
It is a little more convenient to work with random (-1,+1) matrices. A little bit of Gaussian elimination shows that the determinant of a random n x n (-1,+1) matrix is $2^{n-1}$ times the determinant of a random n-1 x n-1 (0,1) matrix. (Note, for instance, that Turan's calculation of the second moment ${\bf E} \det(A_n)^2$ is simpler for (-1,+1) matrices than for (0,1) matrices, it's just n!. It is also clearer why the determinant is distributed symmetrically around the origin.)
The log $\log |\det(A_n)|$ of a (-1,+1) matrix is known to asymptotically be $\log \sqrt{n!} + O( \sqrt{n \log n} )$ with probability $1-o(1)$; see this paper of Vu and myself. A more precise result should be that the logarithm is asymptotically normally distributed with mean $\log \sqrt{(n-1)!}$ and variance $2 \log n$. This result was claimed by Girko; the proof is unfortunately not quite complete, but the result is still likely to be true.
For some further results of this nature, see Exercise 5.64 of
Enumerative Combinatorics, vol. 2. This exercise deals with the uniform distribution on (0,1)-matrices or $(-1,1)$-matrices, but the arguments can be carried over to other distributions where the matrix entries are i.i.d. The proofs are similar to the argument in David Speyer's comment.
5.64. a.[2+] Let $\mathcal D_n$ be the set of all $n\times n$ matrices of $+1$'s and $-1$'s. For $k\in\mathbb P$ let \begin{align*} f_k(n)&= 2^{-n^2} \sum_{M\in\mathcal D_n} (\det M)^k \\ g_k(n)&= 2^{-n^2} \sum_{M\in\mathcal D_n} (\operatorname{per} M)^k, \end{align*} where $\operatorname{per}$ denotes the permanent function defined by $$\operatorname{per}(m_{ij})= \sum_{n\in\mathfrak{S}_n} m_{1,\pi(1)} m_{2,\pi(2)} \dots m_{n,\pi(n)}.$$ Find $f_k(n)$ and $g_k(n)$ explicitly when $k$ is odd or $k=2$. b.[3-] Show that $f_4(n)=g_4(n)$, and show that $$\sum_{n\ge 0} f_4(n) \frac{x^n}{n!} = (1-x)^{-3} e^{-2x}. \tag{5.120}$$ HINT. We have $$\sum_m (\det M)^4 = \sum_M \left(\sum_{\pi\in\mathfrak S_n} \pm m_{1,\pi(1)}\dots m_{n,\pi(n)}\right)^4.$$ Interchange the order of summation and use Exercise 5.63. c.[2+] Show that $f_{2k}(n)<g_{2k}(n)$ if $k\ge 3$ and $n\ge 3$. d.[3-] Let $\mathcal D'_n$ be the set of all $n\times n$ 0-1 matrices. Let $f'_k(n)$ and $g'_k(n)$ be defined analogously to $f_k(n)$ and $g_k(n)$. Show that $f'_k(n)=2^{-kn} f_k(n+1)$. Show also that \begin{align*} g'_1(n) &= 2^{-n} n!\\ g'_2(n) &= 4^n n!^2 \left(1+\frac1{1!}+\frac1{2!}+\dots+\frac1{n!}\right) \end{align*}
Unless I'm missing something, this also follows immediately from linearity and multiplicativity of expectation, treating each entry as independently $0-1$ with probability $1/2$. Every permutation yields the same expected value in the sum, $\pm (1/2)^n$ depending on sign, and the number of even and odd permutations is identical (for $n \ge 2$, as noted above).
It's probably worth mentioning that an old result of Komlos shows that despite this, the probability the determinant is actually 0 is $o(1)$.
Is it not zero whenever $n \geq 2$? Let $A$ be a $n \times n$ permutation matrix with determinant $-1$ (which requires $n \geq 2$). Then the uniform distribution of a random $n \times n$ $(0,1)$-matrix $X$ is the same as the distribution of $AX$. The determinant is multiplicative, hence Det$(AX)=$Det$(A)$Det$(X)=-$Det$(X)$. Hence the probability of Det$(X)=x$ is the same as the probability of Det$(X)=-x$.
Miodrag Zivkovic has actually done a classification on small orders of 0-1 matrices by rank and absolute determinant value. You may be interested in the tables in his Arxiv paper http://arxiv.org/abs/math.CO/0511636 .
Gerhard "Ask Me About System Design" Paseman, 2010.01.26 |
Exercise:Suppose that $a_k \geq 0$ for $k$ large and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges.
Prove that $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = 0$$
Attempt in proof:
Suppose $a_k \geq 0$ for any large $k$ and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges. . Then give $\epsilon >0$ there is $N \in N$ such that $\left|\sum_{k = n}^{\infty} \frac{a_k}{k}\right| < \epsilon$.
Let $s_n = \frac{a_n}{j+k}$ denote the partial sums.
Then $\sum_{k = 1}^{\infty} \frac{a_k}{j+k} $ will converge to zero if and only if its partial sum converges to zero as n approaches infinity.
Taking the limit of $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = \lim_{j \to \infty}\frac{a_1}{j + k} + \cdots+ \frac{a_n}{j + k} + \cdots = \lim_{j \to \infty}\frac{a_1/j}{1 + k/j} + \cdots+ \frac{a_n/j}{1 + k/j} +\cdots $$
Can someone please help me finish? I don't know if this a right way. Any help/hint/suggestion will be really appreciate it. Thank you in advance. |
Published inAlgebras and Representation Theory
Let J(C) be the poset of order ideals of a cominuscule poset C where C comes from two of the three infinite families of cominuscule posets or the exceptional cases. We show that the Auslander-Reiten translation τ on the Grothendieck group of the bounded derived category for the incidence algebra of the poset J(C), which is called the Coxeter transf...
Published inAlgebras and Representation Theory
In this note, we investigate the representation type of the cambrian lattices and some other related lattices. The result is expressed as a very simple trichotomy. When the rank of the underlined Coxeter group is at most 2, the lattices are of finite representation type. When the Coxeter group is a reducible group of type A13\documentclass[12pt]{mi...
The exchange graph of a 2-acyclic quiver is the graph of mutation-equivalent quivers whose edges correspond to mutations. When the quiver admits a nondegenerate Jacobi-finite potential, the exchange graph admits a natural acyclic orientation called the oriented exchange graph, as shown by Br\"ustle and Yang. The oriented exchange graph is isomorphi...
Given partitions $\alpha$, $\beta$, $\gamma$, the short exact sequences $0\to N_\alpha \to N_\beta \to N_\gamma \to 0$ of nilpotent linear operators of Jordan types $\alpha$, $\beta$, $\gamma$, respectively, define a constructible subset $\mathbb V_{\alpha,\gamma}^\beta$ of an affine variety. Geometrically, the varieties $\mathbb V_{\alpha,\gamma}^...
Published inAlgebras and Representation Theory
We introduce and study the higher tetrahedral algebras, an exotic family of finite-dimensional tame symmetric algebras over an algebraically closed field. The Gabriel quiver of such an algebra is the triangulation quiver associated to the coherent orientation of the tetrahedron. Surprisingly, these algebras occurred in the classification of all alg...
Published inAlgebras and Representation Theory
Formulas for the dimension vectors of all objects M in the category S(6~)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathcal {S}(\tilde {6})$\end{document} of nilpo...
Using only the combinatorics of its defining ribbon graph, we classify the two-term tilting complexes, as well as their indecomposable summands, of a Brauer graph algebra. As an application, we determine precisely the class of Brauer graph algebras which are tilting-discrete.
Published inForum Mathematicum
When Γ is a row-finite digraph, we classify all finite-dimensional modules of the Leavitt path algebra L ( Γ ) {L(\Gamma)} via an explicit Morita equivalence given by an effective combinatorial (reduction) algorithm on the digraph Γ. The category of (unital) L ( Γ ) {L(\Gamma)} -modules is equivalent to a full subcategory of quiver representati...
Published inArchiv der Mathematik
We characterize the indecomposable transjective modules over an arbitrary cluster-tilted algebra that do not lie on a local slice, and we provide a sharp upper bound for the number of (isoclasses of) these modules.
Published inEuropean Journal of Mathematics
We prove that every toric quiver flag variety Y is isomorphic to a fine moduli space of cyclic modules over the algebra End(T)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{docum... |
Angular velocity fixed at birth
Because of the Born rigidity of black holes, it follows directly from a consideration of the Ehrenfest paradox[5], that the angular velocity of a black hole can never be changed - it is fixed at birth. To explain this a little more fully, imagine a disk of radius \(R\) rotating with constant angular velocity \(\omega\).
Now imagine a reference frame fixed at the centre of the disk, The relative velocity of any point on the disk is given by \(\omega R\). So the circumference will undergo Lorentz contraction by a factor of
\[\sqrt{(1-(\omega R)^2/c^2)}\]
whereas the diameter does not.
So we have
\[\frac{circumference}{diameter}=\pi\sqrt{(1-(\omega R)^2/c^2)}\]
This is paradoxical for a truly rigid body and so in the general case, the body must deform. In the special case of a black hole, because of Born rigidity, deformation is not possible, and hence, the only alternative is that \(\omega\) is fixed. (Thanks to Wikipedia for this explanation)
When black holes increase in mass, they must also increase in angular momentum in order to keep the angular velocity constant as it grows.
Agree or disagree, or have any questions or observations about this, and I would love to hear from you, so please This email address is being protected from spambots. You need JavaScript enabled to view it., or leave a comment. Your views are always most welcome. |
Abstract
We prove that when subjected to periodic forcing of the form $$p-{\mu ,\rho ,\omega } (t) = \mu (\rho h(x,y) + \sin (\omega t)),$$ certain two-dimensional vector fields with dissipative homoclinic loops generate strange attractors with Sinai-Ruelle-Bowen measures for a set of forcing parameters (μ, ρ, ω) of positive Lebesgue measure. The proof extends ideas of Afraimovich and Shilnikov and applies the recent theory of rank 1 maps developed by Wang and Young. We prove a general theorem and then apply this theorem to an explicit model: a forced Duffing equation of the form $${{d^2 q} \over {dt^2 }} + (\lambda - \gamma q^2){{dq} \over {dt}} - q + q^3 = \mu \sin (\omega t).$$
Fingerprint ASJC Scopus subject areas Mathematics(all) Applied Mathematics Cite this
Research output: Contribution to journal › Article
Communications on Pure and Applied Mathematics, vol. 64, no. 11, pp. 1439-1496. https://doi.org/10.1002/cpa.20379
}
TY - JOUR
T1 - Dissipative homoclinic loops of two-dimensional maps and strange attractors with one direction of instability
AU - Wang, Qiu-Dong
AU - Ott, William
PY - 2011/11
Y1 - 2011/11
N2 - We prove that when subjected to periodic forcing of the form $$p-{\mu ,\rho ,\omega } (t) = \mu (\rho h(x,y) + \sin (\omega t)),$$ certain two-dimensional vector fields with dissipative homoclinic loops generate strange attractors with Sinai-Ruelle-Bowen measures for a set of forcing parameters (μ, ρ, ω) of positive Lebesgue measure. The proof extends ideas of Afraimovich and Shilnikov and applies the recent theory of rank 1 maps developed by Wang and Young. We prove a general theorem and then apply this theorem to an explicit model: a forced Duffing equation of the form $${{d^2 q} \over {dt^2 }} + (\lambda - \gamma q^2){{dq} \over {dt}} - q + q^3 = \mu \sin (\omega t).$$
AB - We prove that when subjected to periodic forcing of the form $$p-{\mu ,\rho ,\omega } (t) = \mu (\rho h(x,y) + \sin (\omega t)),$$ certain two-dimensional vector fields with dissipative homoclinic loops generate strange attractors with Sinai-Ruelle-Bowen measures for a set of forcing parameters (μ, ρ, ω) of positive Lebesgue measure. The proof extends ideas of Afraimovich and Shilnikov and applies the recent theory of rank 1 maps developed by Wang and Young. We prove a general theorem and then apply this theorem to an explicit model: a forced Duffing equation of the form $${{d^2 q} \over {dt^2 }} + (\lambda - \gamma q^2){{dq} \over {dt}} - q + q^3 = \mu \sin (\omega t).$$
UR - http://www.scopus.com/inward/record.url?scp=80052023566&partnerID=8YFLogxK
UR - http://www.scopus.com/inward/citedby.url?scp=80052023566&partnerID=8YFLogxK
U2 - 10.1002/cpa.20379
DO - 10.1002/cpa.20379
M3 - Article
VL - 64
SP - 1439
EP - 1496
JO - Communications on Pure and Applied Mathematics
JF - Communications on Pure and Applied Mathematics
SN - 0010-3640
IS - 11
ER - |
Areas Related to Circles Class 10 Notes i.e. for chapter 12 provided here are extremely useful for the class 10 students to prepare and revise this chapter in a more efficient way. The most important topics from this chapter include the following topics which are explained in detail below.
Area of a circle Perimeter or circumference of a circle Area of a sector of a circle Area of a segment of a circle Area and Circumference of a Circle
For a circle, its area can be calculated by measuring the region that is enclosed by its boundary. The perimeter of a circle or its circumference is defined as the length of the circle’s boundary. The formula for area and perimeter of a circle is given as-
Area of a Circle \(\pi r^{2}\) Circumference of a Circle \(2\pi r\) Area of a Sector of a Circle
A sector of a circle can be defined as the part of that particular circle which is in between any two chosen radii.
In the above diagram, the shaded region is the sector of a triangle and its area and the length of an arc of a sector is given by-
\(Area\, of\, the\, sector\, of\, angle\, \theta\) \(\frac{\theta}{360}\times \pi\times r^{2}\) \(Length\, of\, an\, arc\, of\, a\, sector\, of\, angle\, \theta\) \(\frac{\theta}{360}\times 2\pi r\) Area of a Segment of a Circle
Consider the same figure as given above-
In this, AB is a chord which divides the circle into two parts i.e. a major part and a minor part. Now, the Area of the segment APB can be found by-
Area of the segment APB = Area of the sector OAPB – Area of ∆ OAB
Or,\(Area\, of\, the\, segment\, APB = \frac{\theta}{360}\times \pi r^{2} – area\, of\, triangle\, \bigtriangleup OAPB\)
Example Questions
Also Access NCERT Solutions for class 10 Maths Chapter 12 NCERT Exemplar for class 10 Maths Chapter 12 Practice Questions Calculate the radius of a circle if its circumference is 528 cm. What is the area of the sector of angle 60 degrees if the radius of the circle is 10 cm? Find the area of the following figure: (Note: ABCD is a square of side 14 cm and APD and BPC are semicircles).
Access CBSE Class 10 Maths Sample Papers Here.
Access NCERT Class 10 Maths Book Here.
Also Check:
Keep visiting BYJU’S to get more such maths notes for class 10 along with other preparation materials and prepare for the class 10 board exam in a more effective way. |
Quadratic Formula Complete the Square Factor
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
\[ x = \frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \]
Back to Top
Move c to the right side of the equation. \[ax^2+bx = -c \]
Factor a out of both terms on the left side. \[a(x^2+\frac{b}{a}x) = -c \] Divide both sides by a. \[x^2+\frac{b}{a}x = \frac{-c}{a} \] Divide the coefficient of the x term by 2. Square that and add to both sides. \[x^2+\frac{b}{a}x + (\frac{b}{2a})^2 = \frac{-c}{a} + (\frac{b}{2a})^2 \] Factor the left side. Combine the terms on the right side. \[(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2} \] Square root both sides. \[x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a} \] Solve for x. \[x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \] Back to Top
Rewrite the middle term using two factors whose product equals a*c and sum equals b. Group the terms in pairs. \[ (ax^2+sx)(+tx+c) = 0 \]
Factor each pair. \[mx(jx+k)+n(jx+k) \] Factor the shared binomial. \[ (jx+k)(mx+n) = 0 \] Set each binomial equal to 0. Solve for x. \[jx+k=0\]\[mx+n=0\] Back to Top |
Intuitive argument: by symmetry we should have $P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}$. However, $P(f(1, W_1) \ge 0 \mid W_1 = 0)$ is either $0$ or $1$ depending on whether $f(1,0) \ge 0$.
Of course this is not really a proof because I conditioned on an event of probability 0. So let's try to use the same idea in an actual proof.
Consider the conditional probability $Y = P(X_1 \ge 0 \mid W_1)$.
Suppose $X_1 = f(1, W_1)$ where $f$ is Borel; then we have $Y = 1_B(W_1)$ a.s., where $B = \{x : f(1,x) \ge 0\}$. In particular we have $$P(Y \in \{0,1\}) = 1. \tag{*}$$
But since $X_1, W_1$ are both linear functionals of the Gaussian process $\{W_t\}$, they are jointly Gaussian with mean 0. Of course $W_1$ has variance 1, and as shown in your link, $X_1$ has variance $1/3$. Also, using Fubini's theorem we have$$\begin{align*}E[W_1 X_1] &= E\left[ W_1 \int_0^1 W_t\,dt\right] \\ &= E\left[\int_0^1 W_1 W_t\,dt\right] \\ &= \int_0^1 E[W_1 W_t]\,dt \\ &= \int_0^1 t\,dt = \frac{1}{2}.\end{align*}$$So if we set $Z = W_1 - 2 X_1$, then $Z$ is a normal random variable with mean 0 and variance $1/3$ which is independent of $W_1$. We can write the event $\{X_1 \ge 0\}$ as $\{Z \le W_1\}$. So by independence we have $$Y = P(X_1 \ge 0 \mid W_1) = P(Z \le W_1 \mid W_1) = \Phi(\sqrt{3}W_1) \quad \text{a.s.}$$where $\Phi$ is the standard normal cdf. In particular $0 < Y < 1$ almost surely, contradicting (*). |
Belyi's theorem states that the following properties of a nonsingular projective algebraic curve $X$ are equivalent:
1) $X$ is defined over $\overline{\mathbb{Q}};$
2) There exists a meromorphic function $\phi: X\to\mathbb{P}^1\mathbb{C} $ ramified at most at $0,1,$ and $\infty$;
3) $X$ is isomorphic to $\Gamma \backslash \mathbb{H}$ (compactified at cusps) for a finite index subgroup $[\mathrm{PSL}_2(\mathbb{Z}), \Gamma]<\infty.$
The remaining question is: $$\boxed{\text{ Is there a way to treat singularities in this or a similar framework? }}$$
The following of my original questions have been answered:
Can this be generalized to arbitrary projective nonsingular varities of higher dimensions? (I discussed this with one professor here in Goettingen. That seems to be ongoing research. Please see also comment of David Roberts.)
What compactification do they mean here? $\Gamma \backslash \mathbb{H} \cup \mathbb{Q}$! (see the answer of Robin Chapman)
What is a nice reference for the proof of Belyi's theorem? (see answer of YBL and Koeck + the comments of Emerton)
How does the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ enter the picture? (see comment of Ariyan and answer of YBL)
Where can I find nice examples where these computations have been done explicitly? (see answer of Andy Putnam and JSE) |
What can we say about the quantum state from the number of zero and non-zero eigenvalues of the corresponding density matrix?
The number of zero eigenvalues has no significance, and is not really well defined anyway.
If the number of non-zero eigenvalues is not one, then there are many different ways to write the density matrix $\rho$ as a coherent decompositions of the form $\rho = \sum_k p_k|\psi_k\rangle\langle\psi_k|$ with $\langle\psi_k|\psi_k\rangle=1$ and $p_i\geq p_j \geq 0$ for $i \leq j$. Iff $\langle\psi_i|\psi_j\rangle=\delta_{ij}$, then this decomposition is an eigendecomposition. Because $\rho$ is Hermitian and positive, an eigendecomposition is also a singular value decomposition, and hence describes all optimal low rank approximations (with respect to the Euclidean norm) in a succinct form. Hence this decomposition is somtimes called optimal coherent decomposition by some communities.
More pragmatically, I recently explained this as follows:
For practical computations, one can just decompose the density matrix into a sum of pure states. The optimal way to do this (i.e. that you get the least error for the number of pure states that you use) is the optimal coherent decomposition, where you compute the eigenvalue decomposition of the density matrix. The dynamics of Schrödinger equations is such that any such decomposition stays valid (and optimal) during time propagation, i.e. you can just propagate each individual pure state.
The last sentence of this pragmatic explanation assumes that $\langle\psi_i(t)|\psi_j(t)\rangle=\langle\psi_i(t_0)|\psi_j(t_0)\rangle$ is preserved during time propagation, which is valid for "closed" systems.
Anything related to entanglement or any other properties? Does they vary with the nature of states such as it is pure or mixed?
As others pointed out, an entangled state is also a pure state. If you compute a partial trace over an entangled state, you get a mixed state, but this is not really related to the eigendecomposition. But this is an interesting observation nevertheless, because the optimal coherent decomposition for the corresponding subsystem won't be preserved in general during time propagation, and hence there can be some sort of quantum leap from the perspective of the subsystem in terms of the optimal coherent decomposition. But the optimal coherent decomposition is only unique if $p_i> p_j \geq 0$ for $i < j$ anyway. |
Indeed, in one very specific sense we
don't have more freedom one way versus the other. So the basic rule of modeling a dynamical system says, "any Lagrangian or Hamiltonian or force law is allowed, as long as we recover the same equations of motion each way." And that freedom is the same for both the Lagrangian and Hamiltonian approaches.
But the Lagrangian path is really straightforward, it says "choose some generalized coordinates that embody your constraints," for example if you're dealing with a gyroscope you probably want its axial tilt in spherical coordinates $\theta,\phi$ but also you need to express its spin about that axis as some angle $\lambda.$ Then it says "express the normal Newtonian kinetic energy $K$ and potential energy $U$ in terms of those generalized coordinates and their time derivatives; just divide the object up into particles and express $v_i$ in your coordinates and write down e.g. $K = \frac12 \sum_i m_i v_i^2$. Then a no-nonsense Lagrangian is $L = K - U.$"
When you're done with this you've got a Lagrangian and Then the equations of motion for a generalized coordinate $q$ are,$$\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot q}.$$ In the most simple cases we can see that the left-hand-side is basically a net force and the right hand side is basically $dp/dt$, for example the 1D harmonic oscillator has $L = \frac12 m \dot x^2 - \frac12 k x^2$ and we see that the left hand side is $kx$ while the right hand side is $\frac{d}{dt} m \dot x.$
So we define the canonical momentum conjugate to $q$ as $\partial L/\partial \dot q,$ and this simple approach has defined our generalized momentum without us having much say in it. In
that respect the choice of momentum is "forced" in the Lagrangian picture, whereas it is not forced in the Hamiltonian picture.
Now there is a recipe for taking a Lagrangian theory and preparing it into a Hamiltonian theory, though I'm not sure there is a reverse! It is a somewhat complicated recipe. You have an expression for these variables $q_i$ and their momenta $p_i = \frac{\partial L}{\partial \dot q_i},$ so in general $p_i = p_i(q_i, \dot q_i).$ What you're looking for is a sort of one-sided inverse of this function, let's call it $v_i(q, p),$ such that $p_i(q, v_i(q,p)) = p.$
With these sorts of complicated expressions I like to remember that in the mathematically pure realm we have partial derivatives of functions with respect to their first or second arguments, not with respect to other symbols. Say $f^{[i]}(x_1, x_2, \dots, x_n) = \partial f/\partial x_i$ for short. Then total derivatives of the above expression with respect to $q$ and $p$ say $p_i^{[1]} + p_1^{[2]} v_i^{[1]} = 0$ and $p_i^{[2]} v_i^{[2]} = 1,$ where the arguments to the functions are the obvious ones (each $p_i$ derivative gets $q$ and $v_i$, each $v_i$ derivative gets $q$ and $p$).
So then the claim is that the canonical Hamiltonian is:$$H = - L\big(q_1, v_1(q_1, p_1), q_2, v_2(q_2, p_2), \dots\big) + \sum_i v_i(q_i, p_i) ~p_i.$$I've written this out for a lot of variables but I hope you can see that it is just repeating the same process for each one individually, so we can just focus on one variable at a time, so I will drop the subscripts $i$ but keep the notation $v(q, p)$ for $v_i(q_i, p_i).$ The equations of motions come from: $$\left(\frac{\partial H}{\partial q}\right)_{p} = -\frac{\partial L}{\partial q} - \frac{\partial L}{\partial \dot q} ~ \frac{\partial v}{\partial q} + \frac{\partial v}{\partial q}~p = -\frac{\partial L}{\partial q} = -\frac{dp}{dt},$$ with that last step coming from the Euler-Lagrange equations. Meanwhile, $$\left(\frac{\partial H}{\partial p}\right)_{q} = - \frac{\partial L}{\partial \dot q} ~ \frac{\partial v}{\partial p} + \frac{\partial v}{\partial p}~p + v = v = \frac{dq}{dt},$$with that last step essentially being a sort of fundamental postulate of Hamiltonian mechanics (that this function $v_i(q_i, p_i)$ we derived should be identified with $dq_i/dt$ since it takes the role of $\dot q_i$ from the Lagrangian theory).
So the Lagrangian picture picks out a very special "canonical momentum" which you must use to derive the right Hamiltonian picture, but there is
no general requirement that the Hamiltonian picture has that momentum precisely. Any momentum and position coordinates which generate the same evolution equations is the same. You technically had this freedom in the Lagrangian picture but it was so easy to choose your Lagrangian as $K - U$ and simply know that it was correct, that you didn't think twice about it. |
Research Open Access Published: Blow-up criteria for Boussinesq system and MHD system and Landau-Lifshitz equations in a bounded domain Boundary Value Problems volume 2016, Article number: 90 (2016) Article metrics
856 Accesses
1 Citations
Abstract
In this paper, we prove some blow-up criteria for the 3D Boussinesq system with zero heat conductivity and MHD system and Landau-Lifshitz equations in a bounded domain.
Introduction
Let Ω be a bounded, simply connected domain in \(\mathbb{R}^{3}\) with smooth boundary
∂Ω, and ν be the unit outward normal vector to ∂Ω. First, we consider the regularity criterion of the Boussinesq system with zero heat conductivity:
where
u, π, and θ denote the unknown velocity vector field, pressure scalar, and temperature scalar of the fluid, respectively. \(\omega:=\operatorname {curl}u\) is the vorticity, and \(e_{3}:=(0,0,1)^{t}\). Theorem 1.1 Let \(u_{0}\in H^{3}\) and \(\theta_{0}\in W^{1,p}\) with \(3< p\leq6\) and \(\operatorname {div}u_{0}=0\) in Ω and \(u_{0}\cdot\nu=0\), \(\operatorname {curl}u_{0}\times\nu=0\) on ∂Ω. Let \((u,\theta)\) be a strong solution of problem (1.1)-(1.5). If u satisfies with \(0< T<\infty\), then the solution \((u,\theta)\) can be extended beyond \(T>0\). Here BMO denotes the space of bounded mean oscillation.
Secondly, we consider the blow-up criterion for the 3D MHD system
Here
b is the magnetic field of the fluid.
It is well known that problem (1.7)-(1.11) has a unique local strong solution [4]. But whether this local solution can exist globally is an outstanding problem. Kang and Kim [3] proved some Serrin-type regularity criteria.
Theorem 1.2 Let \(u_{0},b_{0}\in H^{3}\) with \(\operatorname {div}u_{0}=\operatorname {div}b_{0}=0\) in Ω and \(u_{0}\cdot\nu=b_{0}\cdot\nu=0\), \(\operatorname {curl}u_{0}\times\nu=\operatorname {curl}b_{0}\times\nu =0\) on ∂Ω. Let \((u,b)\) be a strong solution to problem (1.7)-(1.11). If (1.6) holds, then the solution \((u,b)\) can be extended beyond \(T>0\). Remark 1.1
When \(\Omega:=\mathbb{R}^{3}\), our result gives the well-known regularity criterion
Next, we consider the following 3D density-dependent MHD equations:
For this problem, Wu [8] proved that if the initial data \(\rho_{0}\), \(u_{0}\), and \(b_{0}\) satisfy
When \(b=0\), Kim [2] proved the following regularity criterion:
Here \(L_{w}^{s}\) denotes the weak-\(L^{s}\) space, and \(L_{w}^{\infty}=L^{\infty}\).
The aim of this paper is to refine (1.20) as follows.
Theorem 1.3 Let \(\rho_{0}\), \(u_{0}\), and \(b_{0}\) satisfy (1.18). Let \((\rho, u, b)\) be a strong solution of problem (1.12)-(1.17) in the class (1.19). Suppose that u satisfies one of the following two conditions: with \(0< T<\infty\). Then the solution \((\rho,u,b)\) can be extended beyond \(T>0\).
Finally, we consider the 3D Landau-Lifshitz system:
Carbou and Fabrie [9] showed the existence and uniqueness of local smooth solutions. When \(\Omega:=\mathbb{R}^{n}\) (\(n=2,3,4\)), Fan and Ozawa [10] proved some regularity criteria. The aim of this paper is to prove a logarithmic blow-up criterion for problem (1.23)-(1.25) when Ω is a bounded domain. We will prove the following:
Theorem 1.4 and \(0< T<\infty\), then the solution can be extended beyond \(T>0\).
In Section 2, we give some preliminary lemmas, which will be used in the following sections. The proof of Theorem 1.1 for problem (1.1)-(1.5) will be given in Section 3. The new regularly criterion of Theorem 1.2 for the 3D MHD problem (1.7)-(1.11) will be proved in Section 4. In Section 5, we prove Theorem 1.3, and in Section 6, we give the main proof of final Theorem 1.4.
Preliminary lemmas
In the following proofs, we will use the logarithmic Sobolev inequality [11]
and the following three lemmas.
Lemma 2.1
([12])
Let \(\Omega\subseteq\mathbb{R}^{3}\) be a smooth bounded domain, let \(b:\Omega\rightarrow\mathbb{R}^{3}\) be a smooth vector field, and let \(1< p<\infty\). Then Lemma 2.2 Let Ω be a smooth and bounded open set, and let \(1< p<\infty\). Then we have the estimate for all \(b\in W^{1,p}(\Omega)\). Lemma 2.3 We have for all \(f\in W_{0}^{1,4}(\Omega)\). Proof
Then we have [13], p.71,
and it is obvious that
Thus, (2.4) is proved. □
Finally, when
b satisfies \(b\cdot\nu=0\) on ∂Ω, we will also use the identity
for any sufficiently smooth vector field
b. Proof of Theorem 1.1
which gives
Applying curl to (1.2) and setting \(\omega:=\operatorname {curl}u\), we find that
which implies
and therefore
provided that
and \(y(t):=\sup_{[t_{0},t]}\Vert u\Vert _{H^{3}}\) for any \(0< t_{0}\leq t\leq T\), and \(C_{0}\) is an absolute constant.
Applying \(\partial_{t}\) to (1.2), we deduce that
which yields
which implies
which leads to
which leads to
which gives
and
This completes the proof of Theorem 1.1.
Proof of Theorem 1.2
We only need to prove a priori estimates.
which implies
with the same
y and ϵ as in (3.5).
which implies
Applying \(\partial_{t}\) to (1.8), we have
for any \(\delta\in(0,1)\).
Applying \(\partial_{t}\) to (1.9), we have
for any \(\delta\in(0,1)\).
Similarly, testing (4.11) by \(-\Delta b_{t}\), we infer that
which yields
This completes the proof of Theorem 1.2.
Proof of Theorem 1.3
We only need to establish a priori estimates.
(I) Let (1.21) hold.
and the generalized Hölder inequality [7]
with \(\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}\) and \(\frac{1}{q}=\frac {1}{q_{1}}+\frac{1}{q_{2}}\), we derive
which yields
from which it follows that
with
for any \(0< t_{0}\leq t\leq T\), where \(C_{0}\) is an absolute constant, provided that
We first compute \(I_{2}\):
for any \(0<\delta<1\).
for any \(0<\delta<1\).
which gives
for any \(0<\delta<1\).
It is easy to compute that
for any \(0<\delta<1\).
Similarly to (5.12), we deduce that
which leads to
Similarly, we have
which implies
and thus
Now it is standard to prove that
(II) Let (1.22) hold.
We still have (5.9), (5.10), (5.11) with \(s=\infty \), (5.12) with \(s=\infty\), (5.13) with \(s=\infty\), and (5.14), (5.15) with \(s=\infty\), and then using (5.27) and (2.4), we arrive at (5.16) and (5.17). Then by the same calculations as those in (5.18)-(5.26), we conclude that (5.18)-(5.26) hold.
This completes the proof of Theorem 1.3.
Proof of Theorem 1.4
We only need to establish a priori estimates.
First, using the formula \(a\times(b\times c)=(a\cdot c)b-(a\cdot b)c\) and the fact that \(\vert d\vert =1\) implies \(d\Delta d=-\vert \nabla d\vert ^{2}\), we have the following equivalent equation:
Testing (6.1) by \(d_{t}\) and using \((a\times b)\cdot b=0\) and \(d\cdot d_{t}=0\), we get
Testing (1.23) by \(-\Delta d_{t}\) and using \(\vert d\vert =1\), we find that
for any \(0<\delta<1\). Here we have used the Gagliardo-Nirenberg inequalities
Applying \(\partial_{i}\) to (1.23), we get
which yields
which implies
provided that
with \(y(t):=\sup_{[t_{0},t]}\Vert d\Vert _{H^{3}}\) for any \(0< t_{0}\leq t\leq T\), where \(C_{0}\) is an absolute constant.
which implies
which leads to
This completes the proof of Theorem 1.4.
References 1.
Giga, Y: Solutions for semilinear parabolic equations in \(L^{p}\) and regularity of weak solutions of the Navier-Stokes system. J. Differ. Equ.
62, 186-212 (1986) 2.
Kim, H: A blow-up criterion for the nonhomogeneous incompressible Navier-Stokes equations. SIAM J. Math. Anal.
37, 1417-1434 (2006) 3.
Kang, K, Kim, J: Regularity criteria of the magnetohydrodynamic equations in bounded domains or a half space. J. Differ. Equ.
253(2), 764-794 (2012) 4.
Sermange, M, Temam, R: Some mathematical questions related to the MHD equations. Commun. Pure Appl. Math.
36(5), 635-664 (1983) 5.
Chen, Q, Miao, C, Zhang, Z: The Beale-Kato-Majda criterion to the 3D magneto-hydrodynamics equations. Commun. Math. Phys.
275, 861-872 (2007) 6.
Kozono, H, Ogawa, T, Taniuchi, Y: The critical Sobolev inequalities in Besov spaces and regularity criterion to some semilinear evolution equations. Math. Z.
242, 251-278 (2002) 7.
Triebel, H: Theory of Function Spaces. Birkhäuser, Basel (1983)
8.
Wu, H: Strong solution to the incompressible MHD equations with vacuum. Comput. Math. Appl.
61, 2742-2753 (2011) 9.
Carbou, G, Fabrie, P: Regular solutions for Landau-Lifshitz equation in a bounded domain. Differ. Integral Equ.
14, 213-229 (2001) 10.
Fan, J, Ozawa, T: Logarithmically improved regularity criteria for Navier-Stokes and related equations. Math. Methods Appl. Sci.
32, 2309-2318 (2009) 11.
Ogawa, T, Taniuchi, Y: A note on blow-up criterion to the 3D Euler equations in a bounded domain. J. Differ. Equ.
190, 39-63 (2003) 12.
Beirão da Veiga, H, Cripo, F: Sharp inviscid limit results under Navier type boundary conditions. An \(L^{p}\) theory. J. Math. Fluid Mech.
12, 397-411 (2010) 13.
Adams, RA, Fournier, JF: Sobolev Spaces, 2nd edn. Pure and Appl. Math. (Amsterdam), vol. 140. Elsevier, Amsterdam (2003)
14.
Lunardi, A: Interpolation Theory, 2nd edn. Lecture Notes. Scuola Normale Superiore di Pisa (New Series). Edizioni della Normale, Pisa (2009)
15.
Ogawa, T: Sharp Sobolev inequality of logarithmic type and the limiting regularity condition to the harmonic heat flow. SIAM J. Math. Anal.
34, 1318-1330 (2003) Acknowledgements
J. Fan is partially supported by NSFC (No. 11171154), Junpin Yin is supported by the NSFC (Grant No. U1430103) and Beijing Center for Mathematics and Information Interdisciplinary Sciences (BCMIIS). The authors would like to thank the referee for reading the paper carefully and for the valuable comments, which improved the presentation of the paper.
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript. |
GL(2)
Curves
group.alternating
The alternating group, denoted $A_n$ is the subgroup of the symmetric group of even permutations. It is of order $\frac{n!}{2}$, and is index $2$ in $S_n$. The group is solvable for $n\leq 4$ and simple for $n\geq 5$.
Not referenced anywhere at the moment. |
When there are more than one type of parentheses (say
[] and
()), the problem is NP-complete. The proof is the same as the proof for the palindrome problem except that in the reduction, 0s and 1s in $w$ are changed to
[s and
(s respectively, 0s in $v_i$ are changed to
]s and 1s in $r_k$ are changed to
)s.
When there are only one type of parentheses (say
()), the problem can be solved in polynomial time.
For a character $c\in\{$
($,$
)$\}$, denote $$t(c)=\begin{cases}1 & \text{if }c\text{ is '('},\\-1 & \text{if }c\text{ is ')'}.\end{cases}$$For a string $s=s_1s_2\ldots$ where $s_i\in\{$
($,$
)$\}$, denote $f(s)=\sum_{i=1}^{|s|}t(s_i)$ and $g(s)=\max_j\sum_{i=j}^{|s|}t(s_i)$. We have$$\begin{align}f(s_1s_2)&=f(s_1)+f(s_2),\\g(s_1s_2)&=\max\{g(s_1)+f(s_2), g(s_2)\}.\end{align}$$
We say a string $s$ is
potentially accepted if for any prefix $p$ of $s$, $f(p)\ge 0$. Note $s$ is potentially accepted if and only if $f(s)\ge g(s)$. Also note $s\in L$ if and only if $f(s)=0$ and $s$ is potentially accepted.
Note if $\sigma$ is a valid permutation, and $f(x_{\sigma(i)})<0$ and $f(x_{\sigma(i+1)})\ge0$ for some $i$, we can swap the two elements to get another valid permutation. This suggests that we can arrange all $x_i$'s satisfying $f(x_i)\ge 0$ (say $x_{r_1},\ldots,x_{r_k}$) on the first $k$ positions, then arrange other $x_i$'s on the rest positions.
For the first $k$ positions, we process greedily. That is, say $\sigma(1),\ldots,\sigma(j)$ is determined, if there is some $i$ such that $x_{\sigma(1)}x_{\sigma(2)}\ldots x_{\sigma(j)}x_{r_i}$ is potentially accepted, we arrange $x_{r_i}$ on position $j+1$, i.e. $\sigma(j+1)=r_i$. Otherwise, we cannot arrange these $x_{r_i}$'s on positions $1,\ldots,j$ too, so we assert there is no valid permutation.
For the rest positions, we arrange them according to $g$ from big to small. That is, $g(x_{\sigma(k+1)})\ge g(x_{\sigma(k+2)})\ge\cdots\ge g(x_{\sigma(n)})$. If this arrangement is invalid, i.e. the final string does not belong to $L$, we assert there is no valid permutation. The correctness of this process comes from the following lemma:
If for some valid $\sigma$ such that $f(x_{\sigma(k+1)}),\ldots,f(x_{\sigma(n)})<0$ and for some $i$, $g(x_{\sigma(k+i)})<g(x_{\sigma(k+i+1)})$, we can swap $x_{\sigma(k+i)}$ and $x_{\sigma(k+i+1)}$ to get another valid permutation.
Proof of the lemma. For convenience, denote $s_0=x_{\sigma(1)}\ldots x_{\sigma(k+i-1)}$ and $s=x_{\sigma(1)}\ldots x_{\sigma(n)}$. Note after swapping, $f(p)$ is not changed for prefix $p$ of $s$ such that $|p|\le |s_0|$ or $|p|\ge |s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)}|$, we need only to show $s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}$ is potentially accepted. Note$s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)}$ is potentially accepted, we have$$\begin{align}g(s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)})\le f(s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)})\end{align}$$or$$\begin{align}g(s_0)+f(x_{\sigma(k+i)})+f(x_{\sigma(k+i+1)})&\le f(s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)}),\\g(x_{\sigma(k+i)})+f(x_{\sigma(k+i+1)})&\le f(s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)}),\\g(x_{\sigma(k+i+1)})&\le f(s_0x_{\sigma(k+i)}x_{\sigma(k+i+1)}),\end{align}$$or$$\begin{align}g(s_0)+f(x_{\sigma(k+i+1)})+f(x_{\sigma(k+i)})&\le f(s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}),\\g(x_{\sigma(k+i+1)})+f(x_{\sigma(k+i)})<g(x_{\sigma(k+i+1)}) &\le f(s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}),\\g(x_{\sigma(k+i)})\le g(x_{\sigma(k+i+1)})&\le f(s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}),\end{align}$$which means$$\begin{align}g(s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)})\le f(s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}).\end{align}$$Hence $s_0x_{\sigma(k+i+1)}x_{\sigma(k+i)}$ is potentially accepted. |
Edit: I'm leaving the old post below, but before I want to write the proof as suggested by Bruce from his book, which uses the ideas in a more efficient way.
Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, $$\|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1.$$So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^*x=qpq$, so $qx^*x=x^*xq$, and then $qz=zq$. Then$$pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq.$$So $q=u^*pu$.
=============================================
(the old post starts here)
(A good friend pointed me to the ideas in this answer, so I'm sharing them here)
The result holds in any unital C$^*$-algebra. So assume that $\|p-q\|<1$, with $p,q$ in a unital C$^*$-algebra $A\subset B(H)$.
Claim 1: There is a continuous path of projections joining $p$ and $q$.
Proof. Let $\delta\in(0,1)$ with $\|p-q\|<\delta$. For each $t\in[0,1]$, let $x_t=tp+(1-t)q$. Then$$\|x_t-p\|=\|(1-t)(p-q)\|<\delta(1-t),$$$$\|x_t-q\|=\|t(p-q)\|<\delta t.$$This, together with the fact that $x_t$ is selfadjoint, implies that $\sigma(x_t)\subset K=[-\delta/2,\delta/2]\cup[1-\delta/2,1+\delta/2]$ (since $\min\{t,1-t\}\leq1/2$). Now let $f$ be the continuous function on $K$ defined as $0$ on $[-\delta/2,\delta/2]$ and $1$ on $[1-\delta/2,1+\delta/2]$. Then, for all $t\in[0,1]$, $f(x_t)\in A$ is a projection. And$$t\to x_t\to f(x_t)$$is continuous, completing the proof of the claim. Edit: years later, I posted this answer to a question on MSE that proves the continuity.
Claim 2: We may assume without loss of generality that $\|p-q\|<1/2$.
This is simply a compacity argument, using that each projection in the path $f(x_t)$ is very near another projection in the path. The compacity allows us to make the number of steps finite, and so if we find projectons $p=p_0,p_1,\ldots,p_n=q$ and unitaries with $u_kp_ku_k^*=p_{k+1}$, we can multiply the unitaries to get the unitary that achieves $q=upu^*$.
Claim 3: If $\|p-q\|<1/2$, there exists a unitary $u\in A$ with $q=upu^*$.
Let $x=pq+(1-p)(1-q)$. Then $$\|x-1\|=\|2pq-p-q\|=\|p(q-p)+(p-q)q\|\leq2\|p-q\|<1,$$so $x$ is invertible. Let $x=uz$ be the polar decomposition. Then $u$ is a unitary. Note that$$qx^*x=q(qpq+(1-q)(1-p)(1-q))=qpq,$$so $q$ commutes with $x^*x$ and then with $z=(x^*x)^{1/2}$. Note also that $px=xq$, so $puz=uzq=uqz$. As $z$ is invertible, $pu=uq$, i.e.$$q=u^*pu.$$Note that $u=xz^{-1}\in A$. |
To give you a bit of background on this, what you're trying to do is essentially to solve Richardsons first weather prediction problem (which he did by hand on a 3x3 grid laid over central Europe). For more on this you can see wikipedia.
The first weather models used the primitive equations (which are still used for teaching purposes and other simple problems today). We might simplify those further by assuming your setup is 2D, dropping coriolis force and advective terms, so that the differential equations you want to solve are
The continuity equation $$\frac{\partial \rho}{\partial t} + \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} = 0 $$ The Euler equations $$\frac{\partial v_x}{\partial t} = - \frac{1}{\rho} \frac{\partial P}{\partial x}$$$$\frac{\partial v_y}{\partial t} = - \frac{1}{\rho} \frac{\partial P}{\partial y}$$and the Equation of state $P=c_s^2 \rho$, where $c_s$ is the constant sound speed.
As next step you need to discretize the differential terms, so that they become solvable on a grid. Without going into details, let us denote any quantity $Q(t,x,y)$ at timestep n and at coordinates $i,j$, where $n,i,j$ are integer values, as $Q^n_{i,j}$.
Then a time difference is approximated in the finite difference framework as $$\frac{\partial Q(t,x,y)}{\partial t} \approx \frac{Q^{n+1}_{i,j}-Q^n_{i,j}}{\Delta t}$$ where $Q^{n+1}_{i,j}$ indicates the Quantity at time n+1, after performing the time integration with timestep $\Delta t$.Similarly for spatial derivatives we can take $$\frac{\partial Q(t,x,y)}{\partial x} \approx \frac{Q^{n}_{i+1,j}-Q^n_{i,j}}{\Delta x}$$ if $v_x >0$ and $$\frac{\partial Q(t,x,y)}{\partial x} \approx \frac{Q^{n}_{i,j}-Q^n_{i-1,j}}{\Delta x}$$ if $v_x < 0$ (note the change in indices $i,j$). We introduced $\Delta x$ here, which is your cell size in x-direction. Similar for $\Delta y$.Same for the y-direction with the j-index. If now the evolution equation for Q is $$\frac{Q^{n+1}_{i,j}-Q^n_{i,j}}{\Delta t} = R,$$ where R is some arbitrary right-hand side, then the computation you have to do is $$Q^{n+1}_{i,j} = Q^n_{i,j} + \Delta t \cdot R,$$for all quantities Q (in this case $\rho, v_x, v_y$), for all cells (one loop for each direction).
If you want to evolve this system for longer time than only one timestep, then you have to observe that $\Delta t$ underlies the so-called CFL condition.Also observe that no matter how many timesteps you integrate your problem for, you need to consider what your simulation boundaries should do, and what the initial conditions look like.
For the initial conditions I would recommend you to solve the system of discretized equations in such a way, that you have $\partial Q/\partial t=0$ everywhere (so that you start in a steady state), and then you can perturb this steady state by making one $v_{x,ij}$, or $v_{y,ij}$ too large, which seems what you wanted to inspect initially. In order to see that the boundaries are a problem, consider the following: How do you compute $\partial v_x/\partial x$ at the leftmost cell value? There is no cell value left of it, but you need one, if $v_x < 0$ there. So then you might want to look into periodic boundaries or reflective boundaries.
Summary
Yeah, this is the absolute minimum you need to do to solve your problem. You might do faster by hand. |
This is from Nielsen and Chuang. If we have an ensemble of pure states that obey the following relationship for all $i, j$
$\vert \psi_i \rangle = \sum_{j} u_{ij}\vert \phi_j \rangle$
where $u_{ij}$ corresponds to the entries of a unitary matrix, we will obtain the same density matrix for both. That is, we will have
$\rho = \sum_{i}\vert \psi_i \rangle\langle \psi_i \vert = \sum_{i}\vert \phi_i \rangle\langle \phi_i \vert$
The proof is straightforward and given in the book but I have no idea what the correct way to interpret this result is. Any help is appreciated. |
Get your free trial content now! Video Transcript Transcript Combining Rational Number Like Terms
We're in Cambodia, located in Southeast Asia, to check in with our friend Laura. Laura’s an archaeologist and she's busy excavating a site that hasn't been touched in hundreds of years. While digging, Laura's discovered lots of interesting artifacts. Unfortunately they are all broken, so she has to put the pieces back together correctly. Just like putting together pieces of broken artifacts, combining rational number like terms works similarly by piecing parts of an expression together. Here's the first expression that we need to simplify: First, let's find the like terms and use the Commutative Property of Addition to reorder the expression writing the x’s together and the y’s together. Next, combine the like terms giving us 15x plus 2y. Let’s try another problem with rational numbers. Because the coefficients for this problem are fractions, we may have a few more steps but the strategy to simplify the expression is the same. First, use the Commutative Property of Addition to reorder the terms. Next, we want to combine the like terms but we can't add the coefficients of 'h' just yet. First, we'll have to find a common denominator and modify the expression. Nine works. Now, use the Multiplicative Identity Property of 1 to modify the expression so both coefficients have 9 as the denominator. To do this, we multiply three thirds by two thirds giving us six ninths. Now that both coefficients of 'h' have the same denominator we can combine the like terms and simplify the fractions.
Notice, the expression contains terms that are not multiplied with a variable.These terms are called constants.
Constants are numbers that are multiplied by a variable raised to the zero power which is always equal to 1. Even though the variable raised to the zero power is not written we know it's there. In other words, a constant is simply a number by itself. And just in time to try another problem. To make the problem easier to solve first change the mixed number into an improper fraction. Next, use the Distributive Property to get rid of the parentheses. Now, reorder the terms into Standard Form using the Commutative Property. If we want to combine like terms, we have to first find a common denominator for the constant terms. In this case, it's 18.
Now, we can combine the like terms. The last step, as always, make sure all the fractions are written in lowest terms and that the expression is written in standard form. Looks like we're good to go! So, how's Laura doing putting together all of the broken pieces? How about that? She's put together enough pieces to host a tea party with her new pal.
Combining Rational Number Like Terms Übung Du möchtest dein gelerntes Wissen anwenden? Mit den Aufgaben zum Video Combining Rational Number Like Terms kannst du es wiederholen und üben. Describe how to combine like terms, and simplify the expression $\frac{2}{3}h+\frac{1}{4}-\frac{1}{9}h+\frac{1}{4}$. Tipps
The commutative property of addition states that $a+b=b+a$.
The commutative property of multiplication states that $a\times b=b\times a$.
Reducing a fraction means dividing the numerator as well as the denominator by the same number.
You can add terms such as $3h$ and $4h$ as follows: $3h+4h=(3+4)h=7h$.
Lösung
To simplify $\frac23h+\frac14-\frac19h+\frac14$, you first reorder the terms using the commutative property of addition:
$\frac23h-\frac19h+\frac14+\frac14$.
Next, combine the like terms. For this you have to multiply $\frac23$ by $\frac33$. So both coefficients of $h$ have the same denominator:
$\frac33\cdot\frac23h-\frac19h+\frac14+\frac14=\frac69h-\frac19h+\frac14+\frac14$.
Now you are able to combine like terms:
$\frac69h-\frac19h+\frac14+\frac14=\frac59h+\frac24=\frac59h+\frac12$.
The second fraction $\frac24$ is reduced by dividing both the numerator and the denominator by $2$:
$\frac24=\frac{2\div2}{4\div 2}=\frac12$.
State how to combine like terms with integers. Tipps
The commutative property of addition is given by $a+b=b+a$.
Just look at the following examples for like terms:
$4x$ and $-3x$ are like terms. $4x$ and $4y$ aren't like terms at all. $3y$ and $-4y$ are like terms. Lösung
To simplify this expression we first find the like terms and reorder the expression using the commutative property.
So the $x$-terms as well as the $y$-terms are grouped together: $2x+13x-7y+9y$.
Now we can combine the like terms:
$2x+13x=15x$ $-7y+9y=2y$ Explain the mathematical terms. Tipps
The commutative property for multiplication states that $a\times b=b\times a$.
Expanding a fraction means multiplying the numerator as well as the denominator by the same number. This doesn't change the value of the fraction, as what you are multiplying by is equal to $1$.
In this expression $3$ is the coefficient of $x^2$, $4$ is the coefficient of $x$, and $-4$ is a constant.
Lösung
To combine rational number like terms we use some mathematical terms:
A constant is a number that is multiplied by a variable raised to the zeroth power. A coefficient is a constant by which a variable is multiplied. We use the commutative property of addition, $a+b=b+a$, to change the order of terms. If we want to factor any term, like $a\times(b+c)$, we use the distributive property, which tells us that $a\times(b+c)=a\times b+a\times c$. The multiplicative identity property of $1$ is that multiplying by it doesn't change the value. Combine and simplify the given rational expressions as much as possible. Tipps
Use the commutative property, $a+b=b+a$, as well as the distributive property, $a\times(b+c)=a\times b+a\times c$.
Here you see an example for changing mixed terms into fractions.
Lösung $\mathbf{\frac23\left(\frac12x-\frac14+x\right)}$
$\frac23\left(\frac12x-\frac14+x\right)=\frac26x-\frac2{12}+\frac23x$.
We can now reduce the fractions and reorder the expression to get
$\frac13x+\frac23x-\frac16=x-\frac16$.
$~$
$\mathbf{\frac45x-\frac13y+\frac23y-y+\frac15x}$
$\frac45x+\frac15x-\frac13y+\frac23y-y$.
Now we can combine the like terms to get
$\frac55x+\frac13y-y=x+\frac13y-\frac33y=x-\frac23y$.
$~$
$\mathbf{3\left(\frac25y-\frac23x+2x\right)-\frac15y}$
$\frac65y-\frac63x+6x-\frac15y$.
Now we reduce the fractions if possible, reorder the expression, and combine the like terms:
$-2x+6x+\frac65y-\frac15y=4x+\frac55y=4x-y$.
$~$
$\mathbf{\frac38y+\frac25x+\frac56y-\frac4{15}x}$
$\begin{array}{rcl} \frac38y+\frac25x+\frac56y-\frac4{15}x&=&\frac25x-\frac4{15}x+\frac38y+\frac56y\\ &=&\frac6{15}x-\frac4{15}x+\frac9{24}y+\frac{20}{24}y\\ &=&\frac2{15}x+\frac{29}{24}y \end{array}$
$~$
$\mathbf{2\frac12\left(\frac15x-\frac35y+\frac32x+y\right)}$
$\frac15x-\frac35y+\frac32x+y=\frac2{10}x+\frac{15}{10}x-\frac35y+\frac55y=\frac{17}{10}x+\frac25y$.
The mixed number $2\frac12$ can be written as a fraction, $\frac52$. Last we use the distributive property:
$\frac52\left(\frac{17}{10}x+\frac25y\right)=\frac{85}{20}x+\frac{10}{10}y=\frac{17}4x+y$.
$~$
$\mathbf{3\frac13x+2\frac14y-2\frac12x-1\frac15y}$
$\frac{10}3x+\frac94y-\frac52x-\frac65y$.
Now we reorder the expression to
$\frac{10}3x-\frac52x+\frac94y-\frac65y$.
Next we expand each fraction to get the common denominator:
$\frac{20}{6}x-\frac{15}6x+\frac{45}{20}y-\frac{24}{20}y=\frac56x+\frac{21}{20}y$.
Give the steps for combing like terms. Tipps
The distributive property states that $a\times(b+c)=a\times b+a\times c$.
The commutative property of addition states that a+b=b+a.
For the standard form you reorder the terms from terms with the largest exponent to terms with the smallest exponent. For instance, $2x^2+2x-3y+4$ is in standard form.
Lösung
First we rewrite the mixed fraction:
$\frac{1}{3}+\frac{1}{2}\left(1\frac{1}{3}x-\frac{1}{4}\right)=\frac{1}{3}+\frac{1}{2}\left(\frac{4}{3}x-\frac{1}{4}\right)$.
Next we use the distributive property to get rid of the parentheses:
$\frac{1}{3}+\frac{1}{2}\left(\frac{4}{3}x-\frac{1}{4}\right)=\frac{1}{3}+\frac{4}{6}x-\frac{1}{8}$.
Reordering the expression depending on the like terms leads to
$\frac{1}{3}+\frac{4}{6}x-\frac{1}{8}=\frac{1}{3}-\frac18+\frac{4}{6}x$.
So, now we are in the position to combine the like terms. For this we have to find the common denominator of the first two fractions, it's $24$, and expand the fraction as follows:
$\begin{array}{rcl} \frac{1}{3}-\frac18+\frac{4}{6}x&=&\frac{1\cdot 8}{3\cdot 8}-\frac{1\cdot 3}{8\cdot 3}+\frac{4}{6}x\\ &=&\frac{8}{24}-\frac{3}{24}+\frac{4}{6}x\\ &=&\frac{5}{24}+\frac{4}{6}x \end{array}$
Last we can reduce the second fraction and rewrite the expression in standard form:
$\frac{5}{24}+\frac{4}{6}x=\frac{5}{24}+\frac23x=\frac23x+\frac{5}{24}$.
Calculate the total amount of ingredients by combining like terms. Tipps
Let $e$ be the number of eggs, $f$ the number of cups of flour, and $s$ the number of cups of sugar.
For the muffins, Laura needs $2\frac13f+\frac12e$.
For the cookies, Laura needs $\frac12\left(1\frac13s+\frac34f\right)+\frac14e$.
Just add like terms and combine.
Lösung
First we establish both expressions describing the ingredients for the muffins as well as for the cookies:
muffins: $2\frac13f+\frac12e$ cookies: $\frac12\left(1\frac13s+\frac34f\right)+\frac14e$
We add both expressions together to get,
$2\frac13f+\frac12e+\frac12\left(1\frac13s+\frac34f\right)+\frac14e$.
Rewrite the mixed terms as fractions and use the distributive property:
$\frac73f+\frac12e+\frac12\left(\frac43s+\frac34f\right)+\frac14e=\frac73f+\frac12e+\frac46s+\frac38f+\frac14e$.
Now we reorder the expression to get,
$\frac73f+\frac38f+\frac12e+\frac14e+\frac46s$.
Almost done! We still have to find the common denominators to combine like terms:
$\begin{array}{rcl}\frac73f+\frac38f+\frac12e+\frac14e+\frac46s&=&\frac{7\cdot 8}{3\cdot 8}f+\frac{3\cdot 3}{8\cdot 3}f+\frac{1\cdot2}{2\cdot 2}e+\frac14e+\frac46s\\ &=&\frac{56}{24}f+\frac{9}{24}f+\frac{2}{4}e+\frac14e+\frac46s\\ &=&\frac{65}{24}f+\frac34e+\frac23s \end{array}$
So in total Laura needs:
$\frac34$ eggs $\frac{65}{24}$ cups of flour $\frac23$ cups of sugar |
February 15th, 2019, 01:32 PM
# 1
Banned Camp
Joined: Feb 2019
From: Casablanca
Posts: 23
Thanks: 2
weird series
Hello everyone
Here is a $Z_n$ Suite
$\displaystyle Z_3 = 1/2$
$\displaystyle Z_n = Z_ {n + 1} / \cos ({\pi} / {n})$
It's a bizarre suite decreasing to 0 without ever going.
I will define the n with which I work in my suite so as not to fall into absurdities like 1/0 which gives false calculations.
find that
$\displaystyle n = \pi / \arccos (Z_{n + 1} / Z_n) $
Note that $Z_n \ne 0$ and $(Z_{n + 1} / Z_n) \ne 1$,
so the $Z_n$ limit is non-zero.
And since I have a decreasing sequence, $\displaystyle Z_{n + 1} / Z_n < 1$, so the series $Z_n$ is convergent according to the d'Alembert criterion.
Is there an error in this reasoning?
$Z_n$ is a positive term and is decreasing and minus 0 tends to 0 minus bound 0, which is impossible because $Z_n \ne 0$, no?
Last edited by skipjack; February 15th, 2019 at 01:48 PM.
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I'm not getting what acceleration concept is and how it relates to motion and how motion and acceleration can be in different direction? And what's behind the concept of negative and positive acceleration?
Let's say that we move along the straight line. Acceleration shows how fast velocity changes, it doesn't matter how fast you move:
If velocity increases, acceleration is positive If velocity doesn't change, acceleration is 0 If velocity decreases (slows down), acceleration is negative
So when you're in the car and you step on brakes, you keep moving forward for some time, but acceleration is negative (points backwards) - it opposes the forward motion.
If you want an image in you head - instead of brakes think of Hulk stopping the car - he pushes it into opposite direction, the car keeps moving but slows down.
How motion and acceleration can be in different direction?
This isn't surprising. Hitting the brakes on your car is not the same as putting it in reverse.
Think of a satellite in orbit. At any point in time it is moving "horizontally" (tangential to the earth). However, its acceleration is always directly towards the centre of the earth, in other words, at $90^{\circ}$ to its direction of motion.
"Motion" is how the object is currently moving. Acceleration can be in any direction; it depends on the direction of the force. For a satellite, the only force is the earth's gravity, the direction of which is towards the centre of the earth.
Let's start with Newton's 2nd Law:
$$ \vec{F} = m\,\vec{a}$$
Here Symbols have usual meaning. Now consider the left hand side of equation. It gives Force acting on particle. It can be Gravitational or Electromagnetic Forces (or Nuclear Forces too but we keep ourselves restricted to the former two.
Now right hand side gives the acceleration of particle. Acceleration is defined as:
$${a }= \frac{d^2 x}{dt^2}$$
Now let's take mass to be positive.
Then the first equation says that Force acting on body is in the direction of acceleration of the body on which force is applied. Further notice that mass is scalar quantity and the Force and acceleration are vector quantities. Then Force formula simply relates the force in each direction to acceleration of body in that direction.
Now consider the spring force acting on body (for brevity consider S.H.M motion) Then Hooke's law applied to mass (on which spring force is applied) gives:
$$ \vec{F} = -k\,\vec x$$
Now, the body is moving forward but force is acting backward on the object. That is object is retarding. Therefore we have:
$$-k\,\vec{x} = m\,\vec{a}$$
Which says force is in direction opposite to retardation or equivalentaly since retardation is negative of acceleration, force is in the direction of acceleration. If the particle is moving away from origin, then the above equation says that particle is pulled in the direction toward origin and hence its acceleration is also towards origin.That is the body is slowing down.
Hope this helps |
In this example we use the package to infer the bias and coefficients in a logistic regression model using stochastic gradient Langevin Dynamics with control variates. We assume we have data \(\mathbf x_1, \dots, \mathbf x_N\) and response variables \(y_1, \dots, y_N\) with likelihood \[ p(\mathbf X, \mathbf y | \beta, \beta_0 ) = \prod_{i=1}^N \left[ \frac{1}{1+e^{-\beta_0 + \mathbf x_i \beta}} \right]^{y_i} \left[ 1 - \frac{1}{1+e^{-\beta_0 + \mathbf x_i \beta}} \right]^{1-y_i} \]
First let’s load in the data, we will use the cover type dataset commonly used to benchmark classification models. We use the dataset from LIBSVM, which transforms the problem from multiclass to binary. The covertype dataset can be downloaded using the
sgmcmc function
getDataset as follows:
library(sgmcmc)# Download and load covertype datasetcovertype = getDataset("covertype")
First we’ll remove about 10000 observations from the original dataset to form a test set, this will be used to check the validity of the algorithm. Then we’ll separate out the response variable
y and the explanatory variables
X. The response variable is the first column in the dataset.
set.seed(13)testObservations = sample(nrow(covertype), 10^4)testSet = covertype[testObservations,]X = covertype[-c(testObservations),2:ncol(covertype)]y = covertype[-c(testObservations),1]dataset = list( "X" = X, "y" = y )
In the last line we defined the dataset as it will be input to the relevant
sgmcmc function. A lot of the inputs to functions in
sgmcmc are defined as lists. This improves flexibility by enabling models to be specified with multiple parameters, datasets and allows separate tuning constants to be set for each parameter. We assume that observations are always accessed on the first dimension of each object, i.e. the point \(x_i\) is located at
X[i,] rather than
X[,i]. Similarly the observation \(i\) from a 3d object
Y would be located at
Y[i,,].
Now we want to set the starting values and shapes for our parameters. We can see from the likelihood equation we have two parameters, the bias \(\beta_0\) and the coefficients \(\beta\). We’ll just set these to start from zero. Similar to the data, these are just a list with the relevant names.
# Get the dimension of X, needed to set shape of params$betad = ncol(dataset$X)params = list( "bias" = 0, "beta" = matrix( rep( 0, d ), nrow = d ) )
Now we’ll define the functions
logLik and
logPrior. It should now become clear why the list names come in handy. The function
logLik should take two parameters as input:
params and
dataset. These parameters will be lists with the same names as those you defined for
params and
dataset earlier. There is one difference though, the objects in the lists will have automatically been converted to
TensorFlow objects for you. The
params list will contain
TensorFlow tensor variables; the
dataset list will contain
TensorFlow placeholders. The
logLik function should take these lists as input and return the value of the log likelihood as a tensor at point
params given data
dataset. The function should do this using
TensorFlow operations, as this allows the gradient to be automatically calculated; it also allows the wide range of distribution objects as well as matrix operations that
TensorFlow provides to be taken advantage of. A tutorial of
TensorFlow for
R is beyond the scope of this article, for more details we refer the reader to the website of TensorFlow for R.
Specifying the
logLik and
logPrior functions regularly requires specifying specific distributions.
TensorFlow already has a number of distributions implemented in the
TensorFlow Probability package. All of the distributions implemented in TensorFlow Probability are located in
tf$distributions, a list is given on the TensorFlow Probability website. More complex distributions can be specified by coding up the
logLik and
logPrior functions by hand, examples of this, as well as using various distribution functions, are given in the other tutorials. With this in place we can define the
logLik function as follows
logLik = function(params, dataset) { yEstimated = 1 / (1 + tf$exp( - tf$squeeze(params$bias + tf$matmul(dataset$X, params$beta)))) logLik = tf$reduce_sum(dataset$y * tf$log(yEstimated) + (1 - dataset$y) * tf$log(1 - yEstimated)) return(logLik)}
Next we want to define our log-prior density, we assume each \(\beta_i\) has an independent Laplace prior distribution, with location 0 and scale 1, so that \(\log p( \beta ) = - \sum_{i=0}^d | \beta_i|\). Similar to the log-likelihood function, the log-prior density is defined as a function with input
params. In our case the definition is
logPrior = function(params) { logPrior = - (tf$reduce_sum(tf$abs(params$beta)) + tf$reduce_sum(tf$abs(params$bias))) return(logPrior)}
Finally, we’ll set the stepsize parameters for the algorithm, along with the minibatch size.
sgldcv relies on two stepsize parameters, one for the optimization step and one for the MCMC step. To allow stepsizes to be set for different parameters, the form of the stepsizes for the MCMC will be lists with names corresponding to each of the names in
params. The optimization step will just be one value as the stepsize is automatically tuned
stepsizesMCMC = list("beta" = 5e-6, "bias" = 5e-6)stepsizesOptimization = 5e-6
Alternatively, we can simply use the shortcut
stepsizesMCMC = 2e-5 which would set the stepsizes for each parameter to
2e-5. The optimization step is performed using the
TensorFlow
AdamOptimizer.
Now we can run our SGLD-CV algorithm using the function
sgldcv from the
sgmcmc package, which returns a list of Markov chains for each parameter as output. Use the argument
verbose = FALSE to hide the output of the function. To make the results reproducible we’ll set the seed to 13. As the dataset size is quite large, we’ll change the
minibatchSize from its default
0.01 * N to
500. To allow a small 1000 iteration burn-in we’ll set the number of iterations to be 11000
output = sgldcv(logLik, dataset, params, stepsizesMCMC, stepsizesOptimization, logPrior = logPrior, minibatchSize = 500, nIters = 11000, verbose = FALSE, seed = 13 )
A common performance measure for a classifier is the log loss. To check the algorithm converged, we’ll plot the log loss of the data from our test set every 10 iterations. Let \[\hat \pi_i^{(j)} := \frac{1}{1 + \exp\left[-\beta_0^{(j)} - \mathbf x_i \beta^{(j)}\right]},\] here \(\hat \pi_i^{(j)}\) denotes the probability that the \(j^{\text{th}}\) iteration of our MCMC chain assigned to observation \(i\) is in our test set. Define our test set by \(T\), the log loss is given by \[A := \frac{1}{|T|} \sum_{y_i \in T} \left[ y_i \log \hat \pi_i^{(j)} + (1 - y_i) \log(1 - \hat \pi_i^{(j)}) \right]\]
To check convergence, we’ll plot the log loss every 10 iterations as follows
yTest = testSet[,1]XTest = testSet[,2:ncol(testSet)]# Remove burn-inoutput$bias = output$bias[-c(1:1000)]output$beta = output$beta[-c(1:1000),,]iterations = seq(from = 1, to = 10^4, by = 10)logLoss = rep(0, length(iterations))# Calculate log loss every 10 iterationsfor ( iter in 1:length(iterations) ) { j = iterations[iter] # Get parameters at iteration j beta0_j = output$bias[j] beta_j = output$beta[j,] for ( i in 1:length(yTest) ) { pihat_ij = 1 / (1 + exp(- beta0_j - sum(XTest[i,] * beta_j))) y_i = yTest[i] # Calculate log loss at current test set point LogPred_curr = - (y_i * log(pihat_ij) + (1 - y_i) * log(1 - pihat_ij)) logLoss[iter] = logLoss[iter] + 1 / length(yTest) * LogPred_curr }}library(ggplot2)plotFrame = data.frame("iteration" = iterations, "logLoss" = logLoss)ggplot(plotFrame, aes(x = iteration, y = logLoss)) + geom_line() + ylab("Log loss of test set") |
Get your free trial content now! Video Transcript Transcript Long Division of Polynomials “The World --- will end on ---th 30--! No more --- for ---- of the year! ---”Scientists have intercepted a cryptic radio message and rush to save mankind.Luckily, the scientists’ robot was able to detect a code embedded in the message. The message is coded in math and tells them the exact day, month and year when the world will end. To solve the riddle and save the world, the scientists must use their knowledge of long division of polynomials. The long division of polynomials
These are the
equations the Robot overheard. The result of the first equation will tell us the month and date while the combined result of the next two equations will give us the year. Let's get started.
The first part of the message is the month and date:
Long division with polynomials is no different than long division with numbers. Let's look at the easy example 428 divided by 6. In this example we first look at how many times 6 goes into 4. Since 6 does not go into 4 we have to look at the next digit. How many times does 6 go into 42. 6 goes into 42 7 times. Now multiply 7 times 6 which gives us 42. We subtract 42 from 42 and get zero. In the same way, we should ask ourselves, "How many times does 'x' go into x squared?" x' times 'x' gives us 'x' squared, so we write 'x'. Next, we multiply 'x' times 'x' plus 3. We subtract this result from the polynomial under the long division symbol. 'x' squared minus 'x' squared gives us zero. Negative 17x minus 3x gives us negative 20x.
Now, just like with regular long division, we bring the next number down and repeat the process of asking the question - How many times does the
divisor go into the dividend? - until there are no numbers left. In this example, six goes into eight exactly one time. Then, we multiply 1 times 6 and get 6. This 6 is subtracted from the 8 and we're left with 2.
In the same way, we bring the negative 51.2 down to get the
expression -20x - 51.2. Now we have to see how many times x goes into -20x. It goes in -20 times, which we write here. Then, we multiply -20 by 'x' plus 3 to get -20x minus 60.This is subtracted from the binomial negative 20x minus 51.2. This gives us 8.8.
What do we do with this? We know it's a remainder, but how do we know how many times 'x' goes into it?!? We don't even know what 'x' is! Don't worry, just like long
division with numbers, you write the remaining numbers over the divisor. Normally, you have to simplify your fraction, if you can. But when your divisor has a variable, you can't simplify any further, so you're done! Remember, this remainder 8.8 tells us the month and date of the End of the world.
The second part of the message reveals the first two numbers of the year. Just like before, we ask, "how many times does
'x' go into 'x' squared?"We multiply our answer by our dividend, subtract and bring down the next term.We repeat the process. And finally, write the leftovers over our dividend, giving us the remainder. This remainder is the first 2 digits of the year the world will end. The scientists are almost done! Once they find out the last two digits of the year, they can save the world!
The third and final part of the message reveals the last two numbers of the year. Even though the scientists have to hurry, they stick to the plan and go step by step. * bring the next term down * multiply * subtract * and write the dividend over the divisor, and there you have it: the remainder!
So 1 and 6 are the last two digits of the year. With the code cracked, the scientists rush to enter the date into their time machine “8.8.3016”. That's strange, it sure doesn't look like the world's about to end. “The World Series will end on August 8th, 3016! No more games for the remainder of the year! Get your tickets now!.” Oh! Game 7 of the World Series!!!
Long Division of Polynomials Übung Du möchtest dein gelerntes Wissen anwenden? Mit den Aufgaben zum Video Long Division of Polynomials kannst du es wiederholen und üben. Explain how to solve the equation using long division of polynomials. Tipps
Long division with polynomials is a lot like regular long division.
Pay close attention to signs.
$x$ goes into $x^2$ $x$ times.
Lösung
Long division with polynomials is a lot like regular long division.
You start by seeing how many times the first term of the divisor goes into the first term of the dividend. In this case, you find out how many times $x$ goes into $x^2$.
Next, you multiply the result by the divisor. You have that $x$ times $x+3$ gives us $x^2-3x$.
Then you subtract this result from the polynomial, as you can see in the image.
Then just like in regular long division, you bring down the next term in from the dividend. So here you bring $-51.2$ down.
The process above gets repeated, dividing the divisor into the result. So in this case you find out how many times $x$ goes into $-20x$. Then you multiply $-20$ by $x+3$. You subtract the result from the binomial $-20x -51.2$, giving you the remainder.
You divide the remainder $8.8$ by $x+3$ and add the result to binomial on top of the division sign.
Define the term "polynomial." Tipps
$5x^2 + 3x -7$ is a polynomial.
Polynomials can have an infinite number of terms.
Although some terms in a polynomial are negative, they are all still being added together.
Lösung
A polynomial is a type of algebraic expression. It must have two ore more terms. A binomial is a special kind of polynomial that has two terms.
The terms of a polynomial are added together and can be positive or negative. For example, you can think of the polynomial $5x^2 + 3x -7$ as "seven $x$ squared plus three $x$ plus negative seven."
The terms of a polynomial can contain both variables and constants. Using the above example, the term $5x^2$ contains both the variable $x$ and the constant $5$.
Polynomials can contain many variables. So the statement "Polynomials can only contain one variable" is false.
Any number of terms in a polynomial can be positive or negative. So the statement "More terms in a polynomial must be positive than negative" is false.
Polynomials can contain any number of terms. So the statement "Polynomials can contain up to five terms" is false.
Calculate the remainder using long division of polynomials. Tipps
Long division with polynomials is a lot like regular long division.
Pay close attention to signs.
$x$ goes into $x^2$ $x$ times.
Lösung
Long division with polynomials is a lot like regular long division.
You start by seeing how many times the first term of the divisor goes into the first term of the dividend. In this case, you find out how many times $x$ goes into $x^2$.
Next, you multiply the result by the divisor. $x$ times $x-11$ gives us $x^2-11x$.
Then you subtract this result from the polynomial, as you can see in the image.
Then just like in regular long division, you bring down the next term in from the dividend. So here you bring $115$ down.
The process above gets repeated, dividing the divisor into the result. So in this case you find out how many times $x$ goes into $-9x$. Then you multiply $-9$ by $x-11$. You subtract the result from the binomial $-9x +115$, giving you the remainder.
You divide the remainder $16$ by $x-11$ and add the result to binomial on top of the division sign.
Determine the result. Tipps
Try doing the polynomial long division on paper.
The first step is to see how many times the first term of the divisor $x$ goes into the first term of the dividend $x^4$.
The second step is to subtract $x^3(x+2)$ from the first two terms of the dividend, and bring down the next term in the dividend ($-15x^2$) next to the result.
The middle steps of polynomial long division get repeated, just like in regular long division.
Lösung
The first step is to see how many times the first term of the divisor $x$ goes into the first term of the dividend $x^4$. The result is $x^3$.
The second step is to subtract $x^3(x+2)$ from the first two terms of the dividend, and bring down the next term in the dividend, $-15x^2$, next to the result.
The third step is to see how many times the first term of the divisor $x$ goes into the first term of the result from the second step, $-16x^3-15x^2$. The result is $-16x^2$.
The second and third steps get repeated over and over, until the last term in the dividend is reached. At this point, the remainder, $50$, is found, and divided by the divisor to get the last term in the solution.
Determine the result of each division operation. Tipps
Try dividing the divisor into the dividend and seeing what result you get. You can do these calculations using long division with polynomials.
It may help to try doing the math without looking at the possible results.
Lösung
You can divide the divisor into the dividend and see what result you get. This will show you which result matches each dividend.
You can do this division using long division with polynomials.
Let's use the example of $\frac{x+5}{4x+16}$.
First we see how many times $x$ goes into $4x$. The result is $4$.
Then we multiply this result by the entire divisor. $4 \times (x+5)$ gives us $4x+20$. We subtract this result from the entire dividend, which gives us $-4$. Then $-4$ is the remainder.
We divide the remainder by the divisor, and add it to our solution. This gives us the overall result of $4-\frac{4}{x+5}$ from the long division.
So we can pair the dividend $4x+16$ with the result $4-\frac{4}{x+5}$.
Similarly,
The dividend $2x-12$ gives the result $2-\frac{22}{x+5}$.
The dividend $7x+35$ gives the result $7$.
The dividend $5x-19$ gives the result $5-\dfrac{44}{x+5}$.
Correct the calculations. Tipps
There are no mistakes in the divisor or the dividend.
There is one mistake in the final result.
There are two mistakes in the calculations that lead to the answer.
Lösung
Start by seeing how many times the first term of the divisor goes into the first term of the dividend. In this case, you find out how many times $x$ goes into $x^2$. The result is $x$.
Next, you multiply the result by the divisor. $x$ times $x+3$ gives us $x^2+3x$.
Then you subtract this result from the polynomial; as you can see in the image, this gives you $0+4x$.
Then just like in regular long division, you bring down the next term from the dividend. So here you bring down $-8$.
The process above gets repeated, dividing the divisor into the result. So in this case you find out how many times $x$ goes into $4x$. Then you multiply $4$ by $x+3$. You subtract the result from the binomial $4x-8$, giving you the remainder.
You divide the remainder $20$ by $x+3$ and add the result to the binomial on top of the division sign. |
Research Open Access Published: Schwarz boundary value problem for the Cauchy-Riemann equation in a rectangle Boundary Value Problems volume 2016, Article number: 7 (2016) Article metrics
912 Accesses
3 Citations
Abstract
In this paper, Schwarz problem for the inhomogeneous Cauchy-Riemann equation in a rectangle is investigated explicitly. By the parqueting reflection principle and the Cauchy-Pompeiu formula, a modified Schwarz-Poisson representation formula in a rectangle is constructed. In particular, the boundary behaviors for the related Schwarz-type and Pompeiu-type operators at the corner points are discussed in detail.
Introduction
There are many investigations of boundary value problems for its theoretical significance and extensive applications in mathematical physics, such as elasticity theory, potential theory, medical imaging. Numerous results have been achieved for boundary value problems in different particular domains; see [1–10]. The basic boundary value problems are the Schwarz, Dirichlet, Robin and Neumann problems. In general, the well-known conformal invariance of the reflection across circles and lines can be used to find a Schwarz operator for simple domains like circles, half circles, rings, half planes,
etc. But the explicit Schwarz-Poisson formula in a rectangle cannot be simply obtained from the classical Schwarz-Poisson formula on the unit disc or the half-plane by conformal mapping. In [11, 12], the authors introduced a method of plane parqueting, which can be used to determine a Schwarz operator, the harmonic Green and Neumann functions, and requires that reflecting the domain at all of these circular or straight curves produces a parqueting of whole complex plane with possible exception of singular points. In [7, 10, 13], the authors discussed Schwarz and Dirichlet problems in an isosceles orthogonal triangle and an equilateral triangle via different series. Also the Green and Neumann functions for a strip and a rectangle were established in detail in [6]. In the present paper, we construct a modified Schwarz-Poisson representation formula by the parqueting reflection principle and Cauchy-Pompeiu formula in a rectangle, discussing the related Schwarz problem explicitly.
Let Ω be a domain in the complex plane \(\mathbb{C}\) defined by
where 0,
a, \(\mu=a+ib\), \(\nu=ib\) are four corner points of the domain Ω. The boundary \(\partial\Omega=[0,a]\cup[a,\mu]\cup[\mu,\nu]\cup[\nu,0]\) is oriented counter-clockwise. That is, the oriented line \([0,a]\) is parameterized by \(t\mapsto t\), \(t\in[0,a]\), the segment \([a,\mu]\) is parameterized by \(t\mapsto a+it\), \(t\in[0,b]\), the oriented segment \([\mu,\nu]\) is parameterized by \(t\mapsto t+ib\), \(t\in[a,0]\), and the oriented segment \([\nu,0]\) is parameterized by \(t\mapsto it\), \(t\in[b,0]\). Lemma 1.1 If \(w\in C^{1}(\Omega;\mathbb{C} )\cap C(\overline{\Omega};\mathbb{C})\), then and where Ω is the rectangle defined by (1) and \(\zeta=\xi +i\eta\), \(\xi,\eta\in\mathbb{R}\).
Reflecting the point \(z\in\Omega\) at \([a,\mu]\), the symmetric point is \(z_{1}=2a-\overline{z}\) and the domain Ω is bijectively mapped onto a rectangle \(\Omega_{1}=\{z=x+iy: a\leq x\leq2a, 0\leq y\leq b\}\). Furthermore, the symmetric point of \(z_{1}\) at \([\mu,\nu]\) is \(z_{2}=-z+2a+2bi\) and the domain \(\Omega_{1}\) is bijectively mapped onto a rectangle \(\Omega_{2}=\{z=x+iy: a\leq x\leq2a, b\leq y\leq2b\}\). Continuing the reflection at \([a,\mu]\), the symmetric point of \(z_{2}\) is \(z_{3}=\overline {z}+2bi\) and the domain \(\Omega_{2}\) is bijectively mapped onto a rectangle \(\Omega_{3}=\{z=x+iy: 0\leq x\leq a, b\leq y\leq2b\}\). Finally, reflecting \(z_{3}\) at \([\mu,\nu]\), the symmetric point of \(z_{3}\) is just point
z. Therefore, let \(\Omega_{0,0}=\Omega\cup\Omega_{1}\cup\Omega_{2}\cup\Omega _{3}\), we know that \(\Omega_{0,0}\) can be a basic rectangle and the reflection along the horizontal and vertical direction is equivalent to extending the basic rectangle \(\Omega_{0,0}\) by double periods \(2bi\), 2 a, respectively (see Figure 1).
Denoting \(\omega_{mn}=2(ma+nbi)\), \(m,n\in\mathbb{Z}\), and
then \(\mathbb{C}=\bigcup^{+\infty}_{m,n=-\infty}\overline{\Omega} _{m,n}\). By the technique of plane parqueting, the complex plane is divided into infinitely rectangles, which are congruent to the domain Ω.
Schwarz-Poisson formula
Suppose \(R_{M,N}=\{(k,j): |k|\leq M, |j|\leq N, k,j\in\mathbb{Z}\}\) for \(M,N\in\mathbb{N}\) be a finite set of double series. If the limit \(\lim_{(M,N)\rightarrow(\infty,\infty)}\sum_{(m,n)\in R_{M,N}}f_{m,n}(\zeta,z)\) exists, then the double series \(\sum_{(m,n)\in\mathbb{Z}\times\mathbb{Z}}f_{m,n}(\zeta ,z)\), simply \(\sum_{m,n}f_{m,n}(\zeta,z)\) is convergent along the rectangle, that is,
Lemma 2.1 Suppose that \(S,E,W\subset\mathbb{C}\) be three bounded sets. If \(S\cap E_{m,n}=\emptyset\), \(S\cap W_{m,n}=\emptyset\) for all \((m,n)\in\mathbb{Z}\times\mathbb{Z}\), then the double series is uniformly convergent for \((\zeta,z,w)\in S\times E\times W\) with \(E_{m,n}=\{z+\omega_{mn}: z\in E\}\) and \(W_{m,n}=\{z+\omega_{mn}: z\in W\}\).
Let
then we have the following result.
Theorem 2.1 Any \(w\in C^{1}(\Omega;\mathbb{C} )\cap C(\overline{\Omega};\mathbb{C})\) can be represented as and, for \(z\in\Omega\), where α is a fixed point in Ω and Proof
For \(z\in\Omega\), we know \(\pm\overline{z}+\omega_{mn}\), \(-z+\omega_{mn}\notin\Omega\) for \((m,n)\in\mathbb{Z}\times\mathbb {Z}\), and \(z+\omega_{mn}\notin\Omega\) for \((m,n)\neq(0,0)\). Then by Lemma 1.1, one has, for \(z\in\Omega\),
which implies that
where
Let
then from the relationship
we obtain
where
Obviously,
with
and
and
Schwarz problem
The classical Schwarz kernel for the upper half-plane \(\mathbb{C}^{+}\) is
and satisfies (see [14])
with \(\rho\in C([c,d],\mathbb{C})\), \(c< d\).
Lemma 3.1 For \(\rho\in C(\partial\Omega,\mathbb{C})\), \(z\in\Omega\), we obtain Proof
When \(\xi\in[a,\mu]\) and letting \(\xi=a+iy\), then \(y\in[0,b]\) and
For \(z\in\Omega\) and \(z\rightarrow t\in(a,\mu)\), we obtain \((a-z)i\in \mathbb{C}^{+}\) and \(\rightarrow(a-t)i\in(0,b)\). Thus, from the classical Schwarz kernel theory (23), the first limit in Lemma 3.1 equals \(\rho(t)\).
For \(\xi\in[\mu,\nu]\), we take \(\xi=y+ib\), then \(y\in[a,0]\) and the second equation is
When \(z\in\Omega\) and \(z\rightarrow t\in(\mu,\nu)\), we obtain \(\overline{z}+i b\in\mathbb{C}^{+}\) and \(\rightarrow\overline{t}+i b\in(0,a)\). Hence, from (23), the second equation is true. In the same way, when \(\xi\in[\nu,0]\), suppose \(\xi=iy\), then \(y\in[b,0]\) and the third limit equals
The proof is completed. □
Define the following Pompeiu-type operator:
for \(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\) with \(q_{m,n}\) given by (4) and \(\alpha\in\Omega\).
Lemma 3.2 If \(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\), then \(A_{\alpha}[f](z)\in C(\overline{\Omega},\mathbb{C})\) and \(\frac{\partial A_{\alpha}[f](z)}{\partial\overline{z}}=f(z)\) for \(z\in\Omega\). Proof
From the classical Pompeiu-type operator in [14],
and \(\frac{\partial T[f](z)}{\partial\overline{z}}=f(z)\) for \(z\in\Omega\). When \(z\in\Omega\), then \(z+\omega_{mn}\notin\Omega\) for \((m,n)\neq(0,0)\) and \(\pm\overline{z}+\omega_{mn}, -z+\omega_{mn}\notin\Omega\). Thus by (4) and (24), we have \(A_{\alpha}[f](z)\in C(\overline {\Omega},\mathbb{C})\) and the integral in (24) is analytic for \(z\in\Omega\) except for one term \(T[f](z)\), therefore,
□
Lemma 3.3 For \(f\in L_{p}(\Omega,\mathbb{C})\), \(p>2\), Proof
By (24), we obtain
with
and
Obviously, we know \(\Phi(\zeta,\alpha)=\Phi(\overline{\zeta},\alpha)=0\) for \(\zeta\in\Omega\). Therefore,
Furthermore, when \(z\in\partial\Omega\), it satisfies equation (17) by replacing
ζ with z. For \(\zeta\in\Omega\), \(z\in[\mu ,\nu]\), we have \(\overline{z}=z-2bi\), then
Similarly,
Also, for \(\zeta\in\Omega\) and \(z\in\partial\Omega\setminus[\mu,\nu]\),
thus the proof is completed. □
Consider the Schwarz-type operator
where \(q_{m,n}\) is given by (4) and \(\gamma\in C(\partial\Omega ,\mathbb{R})\). Then we have
By (18),
and
therefore,
Lemma 3.4 For \(\gamma\in C(\partial\Omega,\mathbb{R})\), \(S_{\alpha}[\gamma](z)\) is analytic in Ω, i. e., Proof
From (25), the sum in integrand can be rewritten as
which is convergent for \(\zeta\in\partial\Omega\), \(z\in\Omega\) by Lemma 2.1. Obviously, from the expression of \(q_{m,n}\) in (4), the integrand in \(S_{\alpha}[\gamma](z)\) is analytic for \(z\in\Omega\), hence the proof is completed. □
Lemma 3.5 For \(z\in\Omega\), \(\gamma\in C(\partial\Omega,\mathbb{R})\), where \(t\in L^{0}\cup\{t_{0}\}\), \(L^{0}\) is L except for two endpoints and Furthermore, for \(t\in\partial\Omega\setminus L^{0}\), Proof
When \(L=[\mu,\nu]\), \(t_{0}=\nu\) and \(t\in L^{0}\cup\{t_{0}\}=(\mu,\nu]\),
where
By \(\Gamma\in C([\mu,\nu]\cup[\nu,-a+i b],\mathbb{R})\) and from the second equation in Lemma 3.1, the above limit is \(\Gamma(t)=\gamma (t)-\gamma(\nu)\). That is,
Similarly,
Furthermore, by (17), we obtain for \(z\in[0,a]\cup[a,\mu)\cup (\nu,0]\) and \(\zeta\in[\mu,\nu]\), \(\sum_{m,n} [q_{m,n}(\zeta, z)-q_{m,n}(\zeta,\overline{z}) ]=0\), therefore, for \(t\in[0,a]\cup[a,\mu)\cup(\nu,0]\),
Combining the result (27), we obtain (29) is also true for corner points \(t=\mu,\nu\), hence (28) holds for \(L=[\mu ,\nu]\) and \(t\in\partial\Omega\setminus(\mu,\nu)\). Similarly, (28) is also true for
L, t in the other cases. The proof is completed. □ Lemma 3.6 For \(\gamma\in C(\partial\Omega,\mathbb{R})\) and \(t\in\partial\Omega\), Proof
From (26), we only need to prove
and
In particular, for \(\alpha\in\Omega\), \(\pm\overline{\alpha}+\omega _{mn}\) do not belong to Ω, so we have
Combining the above three equations and taking \((M,N)\rightarrow(\infty ,\infty)\), thus
Then from (26),
hence,
When \(t\in(\mu,\nu)\), we rewrite \(\operatorname{Re}S_{\alpha}[\gamma](z)\) as
thus from the results for \(L=[\mu,\nu]\), \(t_{0}=\mu\) in (27) and for \(L=[0,a]\), \([a,\mu]\), and \([\nu,0]\) in (28), we get \(\lim_{ z\in\Omega, z\rightarrow t}\operatorname{Re}S_{\alpha}[\gamma ](z)=\gamma(t)\), \(t\in(\mu,\nu)\). Furthermore, by (32) and Lemma 3.5, (30) obviously holds for \(t=\mu,\nu\). Similarly, when \(t\in \partial\Omega\setminus[\mu,\nu]\), the result is also true. Then the proof is completed. □
Theorem 3.1 The Schwarz problem for \(f\in L_{p}(\Omega;\mathbb{C})\), \(p>2\), \(\gamma\in C(\partial\Omega;\mathbb{R}) \), and α is a fixed point in Ω, is uniquely solvable by Proof
Then from Theorem 2.1, we know \(\phi(z)=i\operatorname{Im}w(\alpha)=ic\), thus, the proof is completed. □
References 1.
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3, 22-35 (2005) 2.
Akel, M, Hussein, H: Two basic boundary-value problems for the inhomogeneous Cauchy-Riemann equation in an infinite sector. Adv. Pure Appl. Math.
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Begehr, H, Schmersau, D: The Schwarz problem for polyanalytic functions. Z. Anal. Anwend.
24, 341-351 (2005) 4.
Begehr, H, Gaertner, EA: Dirichlet problem for the inhomogeneous polyharmonic equation in the upper half plane. Georgian Math. J.
14(1), 33-52 (2007) 5.
Begehr, H, Vaitekhovich, T: Schwarz problem in lens and lune. Complex Var. Elliptic Equ.
59(1), 76-84 (2014) 6.
Begehr, H: Green function for a hyperbolic strip and a class of related plane domains. Appl. Anal.
93(11), 2370-2385 (2014) 7.
Shupeyeva, B: Some basic boundary value problems for complex partial differential equations in quarter ring and half hexagon. Dissertation, FU, Berlin (2013). http://www.diss.fu-berlin.de/diss/receive/FUDISS-thesis-000000094596
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Vaitekhovich, T: Boundary value problems for complex partial differential equations in a ring domain. Dissertation, FU, Berlin (2008). http://www.diss.fu-berlin.de/diss/receive/FUDISS-thesis-000000003859
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Wang, Y, Wang, Y: Two boundary-value problems for the Cauchy-Riemann equation in a sector. Complex Anal. Oper. Theory
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Wang, Y, Wang, Y: Schwarz-type problem of nonhomogeneous Cauchy-Riemann equation on a triangle. J. Math. Anal. Appl.
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Begehr, H: Complex Analytic Methods for Partial Differential Equations: An Introductory Text. World Scientific, Singapore (1994)
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Begehr, H: Boundary value problems in complex analysis I. Bol. Asoc. Mat. Venez.
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Stein, EM, Shakarchi, R: Complex Analysis. Princeton University Press, Princeton (2003)
Acknowledgements
This paper is supported by the Fundamental Research Funds for the Central Universities. The authors would like to thank the referees for their valuable suggestion and comments.
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the manuscript and typed, read, and approved the final manuscript. |
The symbol $B$ is explained in greater detail here: Letter codes in molecular term symbols. In this context, it indicates that it is the second excited state of the same multiplicity as the ground state. The ground state, labelled with $X$, is a doublet; the first doublet excited state is labelled with $A$, and the second doublet excited state is labelled with $B$.
The superscript $2$ denotes the multiplicity, i.e. this is a doublet state, with total spin $S = 1/2$ (technically the spin
quantum number) such that $2S + 1 = 2$. The dinitrogen cation $\ce{N2+}$ will have one unpaired electron in its low-lying electronic states, so it's not too surprising that the total spin is $1/2$.
Finally, there is a bit more to the symbol $\Sigma$ than just being a substitute or a generalisation of $\mathrm{A}$. In the linear point groups, this denotes the total orbital angular momentum about the internuclear axis. In this state, there is no orbital angular momentum about this axis, and the relevant quantum number is $\Lambda = 0$, which is denoted by the letter $\Sigma$. ($\Lambda = 1$ is a $\Pi$ state, $\Lambda = 2$ a $\Delta$ state, and so on.)
If you think about the ground state electronic configuration of $\ce{N2+}$, it is essentially the same as $\ce{N2}$, but with one unpaired electron in the $\sigma_\mathrm{g}$ orbital. In order to generate any kind of orbital angular momentum, you need to have an asymmetric population of $\pi$-type MOs. Since this isn't the case, the ground state of $\ce{N2+}$ has no orbital angular momentum, and is consequently a $\Sigma$ state. A similar argument can be applied to the second excited state, which seems to arise from a $\sigma\to\sigma$ excitation. |
Symmetries and Dualities in Name-Passing Process Calculi Abstract
We study symmetries and duality between input and output in the \(\pi \)-calculus. We show that in dualisable versions of \(\pi \), including \(\pi \) and fusions, duality breaks with the addition of ordinary input/output types. We illustrate two proposals of calculi that overcome these problems. One approach is based on a modification of fusion calculi in which the name equivalences produced by fusions are replaced by name preorders, and with a distinction between positive and negative occurrences of names. The resulting calculus allows us to import subtype systems, and related results, from the pi-calculus. The second approach consists in taking the minimal symmetrical conservative extension of \(\pi \) with input/output types.
KeywordsType System Operational Semantic Parallel Composition Typing Rule Mobile Process Preview
Unable to display preview. Download preview PDF.
References 1. 2.Fu, Y.: The \(\chi \)-calculus. In: Proc. APDC, pp. 74–81. IEEE Computer Society Press (1997)Google Scholar 3. 4. 5.Hirschkoff, D., Madiot, J.M., Sangiorgi, D.: Name-Passing Calculi: From Fusions to Preorders and Types. long version of the paper presented at LICS’13, in preparation (2014)Google Scholar 6.Hirschkoff, D., Madiot, J.M., Xu, X.: A behavioural theory for a \(\pi \)-calculus with preorders. submitted (2014)Google Scholar 7. 8. 9. 10.Parrow, J., Victor, B.: The fusion calculus: expressiveness and symmetry in mobile processes. In: Proc. of LICS, pp. 176–185. IEEE (1998)Google Scholar 11. 12. 13.Sangiorgi, D.: \(\pi \)-calculus, internal mobility, and agent-passing calculi. In: Selected papers from TAPSOFT ’95, pp. 235–274. Elsevier (1996)Google Scholar 14.Sangiorgi, D.: Lazy functions and mobile processes. In: Proof, Language, and Interaction, pp. 691–720. The MIT Press (2000)Google Scholar 15.Sangiorgi, D., Walker, D.: The Pi-Calculus: a theory of mobile processes. Cambridge University Press (2001)Google Scholar 16.van Bakel, S., Vigliotti, M.G.: An Implicative Logic based encoding of the \(\lambda \)-calculus into the \(\pi \)-calculus (2014). http://www.doc.ic.ac.uk/~svb/ |
Sampling from Phase Space Distributions in 3D Charged Particle Beams
In the previous installment of this series, we explained two concepts needed to model the release and propagation of real-world charged particle beams. We first introduced probability distribution functions in a purely mathematical sense and then discussed a specific type of distribution — the transverse phase space distribution of a charged particle beam in 2D. Now, let’s combine what we’ve learned and find out how to sample the initial positions and velocities of 3D beam particles from this distribution.
Reviewing 2D Phase Space Distributions and Ellipses
To start, let’s briefly review phase space distributions and ellipses in 2D, both of which are fully explained in the previous post in the Phase Space Distributions in Beam Physics series. The particles in real-world nonlaminar charged particle beams occupy a region in phase space that is often elliptical in shape. The equation for this phase space ellipse in 2D depends on the beam emittance ε and Twiss parameters,
(1)
where
x and x’ are the transverse position and inclination angle of the particle, respectively. The Twiss parameters are further related by the Courant-Snyder condition,
(2)
The actual positions of particles in the ellipse can vary. Two of the most common distributions of phase space density are a uniform density within the ellipse and a Gaussian distribution with a maximum at the ellipse’s center, both of which are illustrated below. The blue curve in each case is the phase space ellipse described in Eq.(1), where ε is the
4-rms transverse emittance. For the Gaussian distribution, note that some particles still lie outside the ellipse. Since the Gaussian distribution decreases gradually without reaching exactly zero, there is always a chance that a few particles will lie outside the ellipse, no matter how large it is drawn. When using the 4-rms emittance to define the ellipse in Eq.(1), about 86% of the particles lie inside the ellipse. Comparison of a uniform and Gaussian distribution.
Let’s consider a simpler case in which the probability of finding a particle at any point in phase space is constant inside the ellipse and zero outside of it. For this problem, substituting Eq.(2) into Eq.(1) and solving for
x’ yields
(3)
The probability distribution function is then
(4)
\begin{array}{cc}
C & -\frac{\alpha x}{\beta} -\frac{\sqrt{\varepsilon \beta -x^2}}{\beta} \textless x' \textless -\frac{\alpha x}{\beta} + \frac{\sqrt{\varepsilon \beta -x^2}}{\beta}\\
0 & \textrm{otherwise}
\end{array}\right.
where the constant
C depends on the size of the ellipse. The probability g(x) of the particle having a given x-coordinate is
Considering the locations where Eq.(3) can take on real values, we get
\begin{array}{cc}
2C \frac{\sqrt{\varepsilon \beta -x^2}}{\beta} & -\sqrt{\varepsilon \beta} \textless x \textless \sqrt{\varepsilon \beta}\\
0 & \textrm{otherwise}
\end{array}\right.
Or, more simply,
(5)
Suppose we have a population of model particles that we want to sample using the probability distribution function given by Eq.(4). More specifically, we’d like to first sample the initial transverse positions of the particles according to Eq.(5) and then assign appropriate inclination angles so that the particles lie within the phase space ellipse. One way to accomplish this is to compute a cumulative distribution function starting from Eq.(5) and then use the method of inverse random sampling. Another possible method is using Eq.(5) to define the density of particles, which we can enter directly into the
Inlet and Release features in the particle tracing interfaces. In this case, the normalization is done automatically. Screenshot showing how to input the particle density in the Inlet feature.
Still, the most convenient approach is using the
Particle Beam feature available in the Charged Particle Tracing physics interface. The Particle Beam feature automatically distributes the particles in phase space, allowing you to specify the location of the beam center, emittance, and Twiss parameters. Screenshot showing how to input the particle density in the Particle Beam feature. Simulating Charged Particle Beams in 3D
So far, we’ve only considered charged particle beams as idealized sheet beams where the out-of-plane (
y) component of the transverse position and velocity can be ignored. However, real beams propagate in 3D space and only extend a finite distance in both transverse directions. Thus, in order to get a complete picture of a beam, we must consider two orthogonal transverse directions x and y as well as the inclination angles x' = v_x/ v_z and y' = v_y/v_z. Particle beam propagating in 3D space.
The reason why simulating the release of particle beams in 3D is more complicated than in 2D is that the degrees of freedom for the two transverse directions are often coupled in real-world beams. For example, suppose two particles are released at the same transverse position i.e., the same
x– and y-coordinates. Let’s say that one of these particles has a very large inclination angle in the x direction ( x’) and the other particle has a very small inclination angle in the x direction. The particle with the large inclination angle in the x direction is more likely to have a small inclination angle in the y direction and vice versa. Hence, we can’t just sample from two different distributions for x’ and y’ because the value of each one affects the probability distribution of the other.
To phrase this problem in a more abstract sense: Instead of considering the two transverse directions as separate 2D phase space ellipses, we actually need to think about the transverse particle motion using distributions of phase space in four dimensions! Since we’re used to seeing objects only in 2D or 3D, distributions with more than three space dimensions are much harder to visualize.
This is where the Particle Beam feature is most useful. It includes settings for sampling the initial particle positions and inclination angles from a variety of built-in 4D transverse phase space distributions. Some common distributions are the Kapchinskij-Vladimirskij (KV) distribution, waterbag distribution, parabolic distribution, and Gaussian distribution. First, let’s consider the simplest distribution, the KV distribution, and then visualize the other distributions in this group.
Mathematically, the KV distribution considers the beam particles to be uniformly distributed on an infinitesimally thin, 4D hyperellipsoid in phase space. It’s written as
+\left(\frac{r_x x' -r'_x x}{\varepsilon_x} \right)^2
+\left(\frac{y}{r_y} \right)^2
+\left(\frac{r_y y' -r'_y y}{\varepsilon_y} \right)^2 = 1
where
r x and rare the maximum extents of the beam in the y xand ydirections, εand x εare the beam emittances associated with the two transverse directions, and y r’and x r’are the inclination angles at the edge of the beam envelope. y
Because it is more difficult to visualize 4D probability distribution functions than functions of lower dimensions, it is often convenient to visualize the distribution indirectly by plotting its projection onto lower dimensions. An interesting property of the KV distribution is that its projection onto any 2D plane is an ellipse of uniform density. The projections onto six such planes are shown below. The projections of the 4D hyperellipsoid onto the
x-x’ and y-y’ planes are tilted because nonzero values have been specified for the Twiss parameter α in each transverse direction. The KV distribution projected onto six 2D planes.
Compare the distributions shown above to the following alternatives.
The waterbag, parabolic, and Gaussian distributions projected onto six 2D planes.
We see that the projection onto any 2D plane forms an ellipse-shaped distribution in all cases, but the ellipses are only uniformly filled in the KV distribution.
Concluding Thoughts on Modeling Charged Particle Beams
Even as this blog series on modeling charged particle beams comes to a close, we have only scratched the surface of the intricate and highly technical field of beam physics. While we’ve discussed transverse phase space distributions in 3D, we haven’t examined longitudinal emittance or the related phenomenon of bunching. We also haven’t categorized the phenomena that causes emittance to increase, decrease, or remain constant as the beam propagates.
This series is meant to be an introduction to the way in which random or pseudorandom sampling from probability distribution functions plays an important role in capturing the real-world physics of high-energy ion and electron beams. For a more comprehensive overview of beam physics, references 1-3 provide an excellent starting point. More technical details about each of the 4D transverse phase space distributions described above, including algorithms for sampling pseudorandom numbers from these distributions, can be found in references 4-7. To learn more about how these concepts apply in the COMSOL Multiphysics® software, browse the resources featured below or contact us for guidance.
Other Posts in This Series Sampling Random Numbers from Probability Distribution Functions Phase Space Distributions and Emittance in 2D Charged Particle Beams References Humphries, Stanley. Principles of charged particle acceleration. Courier Corporation, 2013. Humphries, Stanley. Charged particle beams. Courier Corporation, 2013. Davidson, Ronald C., and Hong Qin. Physics of intense charged particle beams in high energy accelerators. Imperial college press, 2001. Lund, Steven M., Takashi Kikuchi, and Ronald C. Davidson. “Generation of initial Vlasov distributions for simulation of charged particle beams with high space-charge intensity.” Physical Review Special Topics — Accelerators and Beams, vol. 12, N/A, November 19, 2009, pp. 114801 12, no. UCRL-JRNL-229998 (2007). Lund, Steven M., Takashi Kikuchi, and Ronald C. Davidson. “Generation of initial kinetic distributions for simulation of long-pulse charged particle beams with high space-charge intensity.” Physical Review Special Topics — Accelerators and Beams, 12, no. 11 (2009): 114801. Batygin, Y. K. “Particle distribution generator in 4D phase space.” Computational Accelerator Physics, vol. 297, no. 1, pp. 419-426. AIP Publishing, 1993. Batygin, Y. K. “Particle-in-cell code BEAMPATH for beam dynamics simulations in linear accelerators and beamlines.” Nuclear Instruments and Methods in Physics Research. Section A: Accelerators, Spectrometers, Detectors and Associated Equipment 539, no. 3 (2005): 455-489. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
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1. Measurement of the ratio of the production cross sections times branching fractions of B c ± → J/ψπ ± and B± → J/ψK ± and ℬ B c ± → J / ψ π ± π ± π ∓ / ℬ B c ± → J / ψ π ± $$ \mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm }{\pi}^{\pm }{\pi}^{\mp}\right)/\mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm}\right) $$ in pp collisions at s = 7 $$ \sqrt{s}=7 $$ TeV
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Get your free trial content now! Video Transcript Transcript Compound Inequalities
Jerry and Larry want to ride the rollercoaster, but they have a "little big" problem. What does this have to do with
compound inequalities? Let me tell you.
Let’s look at the problem first: You are only allowed to ride the rollercoaster if you are at least 4 feet 7 inches in height or shorter than 6 foot 3. Jerry is shorter than 4 foot 7 while Larry is taller than 6 foot 3.
Compound Inequalities on a Number Line: Conjunction and Disjunction
Let’s have a look at the
number line to see how this problem can be represented graphically. It's not that you have to be an exact height to ride a rollercoaster, but you do have to be taller than a certain height as well as shorter than a certain height.
So you can say the height is greater than or equal to 4 foot 7, which is represented by a
filled circle. Also, it has to be less than 6 foot 3, which is represented by an empty circle.
We'll use x to represent height. You can write this as a compound inequality with
AND. x is less than 4 foot AND x is also less than 6 foot 3.
If you don’t look at the people who are allowed to ride the rollercoaster but instead at the people who aren’t allowed, like Jerry and Larry, you have a
compound inequality with OR; either x is more than 4 foot 7 or x is greater than or equal to 6 foot 3.
This is represented graphically by these parts. Be careful: If you flip the wording in the problem, you also have to switch the circles! This means an empty circle at 4 foot 7 and an filled circle at 6 food 3.
How to solve Compound Inequalities: Conjunction
Let's take a look at an example to figure out
how to calculate compound inequalities: 4 < 2x < 10.
This is a
compound inequality with AND. So you can split this into two inequalities and put the word AND in between, ending up with one inequality, 4 < 2x and the other inequality is 2x < 10.
Now you calculate each inequality separately. Since we have 2x in both equations, we should divide both sides of both inequalities by two; the signs stay the same.
You can bring the two solutions, 2 < x and x < 5, back into one compound ineqality by reversing the process earlier. Remove the AND and now line up the Xs to get: 2 < x < 5.
How to solve Compound Inequalities: Disjunction
For compound inequalities with OR, you already have two seperate inequalities to begin with. So you simply calculate each individually.
In order to isolate the Xs, you have to
eliminate 1 in the equation on the left, and 2 in the equation on the right. So you add 1 and 2 to the left and right equations, respectively.
Now you have to divide the equations by the
coefficients, or the numbers in front of the x, which is 3 in the first equation and 4 in the second.
Your solutions, x < 5 and x > 12, can be brought together in the end with the word OR.
Let’s get back to Jerry and Larry. They really want to ride the rollercoaster. Jerry has an idea. Jerry decided to give his shoes to Larry. Now their heights are just right.
Compound Inequalities Übung Du möchtest dein gelerntes Wissen anwenden? Mit den Aufgaben zum Video Compound Inequalities kannst du es wiederholen und üben. Write an inequality to describe the situation. Tipps
The phrases
less than, at most, between, more than, or at leasthave special meanings in math.
For example,
more thanmeans $>$. The equal sign is not included. At most$10$ means $\le 10$ and $10$ is included. Lösung
Jerry and Larry want to ride the rollercoaster. They have a problem with the minimum and maximum height allowed to ride the roller coaster.
At least$4$ feet $7$ inches can be expressed mathematically using $\ge$. So $x\ge 4$ feet $7$ inches or $4$ feet $7$ inches $\le x$. Shorter than$6$ feet $3$ inches can be expressed as $x<6$ feet $3$ inches. Explain what the inequality means. Tipps
For example,
more thanmeans $>$.
$\ge x$ means
at least, $x$ is included
For example,
less thanmeans $<$.
$\le x$ means
at most, $x$ is included
$2\le x\le4$ means that $x$ is $2$ or more as well as $4$ or less.
Lösung
First let's have a look at the signs and their meanings:
$<x$ less than, $x$ is not included $\le x$ at most, $x$ is included $>x$ greater than, $x$ is not included $\ge x$ at least, $x$ is included Simplify the following inequalities. Tipps
To solve an inequality treat it like an equation and isolate the variable $x$ using opposite operations.
Remember, when multiplying with or dividing by a negative value, flip the sign!
Use
Opposite Operations: Multiplication ($\times~\longleftrightarrow~\div$) Division ($\div~\longleftrightarrow~\times$) Addition ($+~\longleftrightarrow~-$) Subtraction ($-~\longleftrightarrow~+$) Lösung
Let's have a look at solving compound inequalities.
We start with a compound inequality: $4<2x<10$.
We can treat this inequality just like an equation by using opposite operations to isolate the variable $x$.
REMEMBER!If we multiply or divide both sides of the equation by a negative value, we have to flip the sign.
The left side divided by $2$ gives us $2<x$, and the right side becomes $x<5$. Because these inequalities are connected by the word
and, we get the following compound inequality: $2<x<5$.
Next we look at $3x-1<14$ and $4x-2>46$.
First we transform the first inequality:
Adding $1$ on both sides gives us $3x<15$ Dividing by $3$ leaves us with $x<5$. Notice that the sign is still pointing in the same direction. Add $2$ to both sides to get $4x>48$ Dividing by $4$ on both sides gives us our solution, $x>12$.
Let's have a look at the difference between
andand or. Andmeans that both inequalities must be fulfilled. This can be seen as the intersection of two sets. Oris a disjunction. This means that at least one of the inequalities has to be fulfilled. Determine the opposite inequality. Tipps
The opposite of inclusion (and) is exclusion (or).
The opposite of
$<$ is $\ge$ $\le$ is $>$ Lösung
The opposite of inclusion (and) is exclusion (or).
If we want to examine all values for $x$ that don't satisfy the compound inequality $12\le3\times(x+2)<24$, we first have to isolate the variable $x$:
$\begin{array}{rcrcl} ~~\\ 12&\le& 3\times(x+2)&<&24&&|\text{ Distributive Property}\\ 12&\le& 3x+6&<&24\\ \color{#669900}{-6}&&\color{#669900}{-6}&&\color{#669900}{-6}&&|\text{ ($+~\leftrightarrow~-$) Opposite Operations}\\ 6&\le& 3x&<&18\\ \color{#669900}{\div3}&&\color{#669900}{\div3}&&\color{#669900}{\div3}&&|\text{ ($\times~\leftrightarrow~\div$) Opposite Operations}\\ 2&\le& x&<&6\\ ~~\\ \end{array}$
But what does this inequality mean? Right, both inequalities, $2\le x$ and $x<6$ have to be satisfied.
The opposite of the first inequality is $2>x$ or $x<2$. The opposite of the second inequality is given by $6\le x$.
The opposite of
andis or. So the resulting compound inequality is
$x<2$ or $x\ge 6$.
Decide which number line belongs to which compound inequality. Tipps
First, isolate the $x$.
$<x$ less than, $x$ is not included $\le x$ at most, $x$ is included $>x$ greater than, $x$ is not included $\ge x$ at least, $x$ is included Lösung
We can represent compound inequalities with number lines. We can indicate included numbers with filled circles and excluded numbers with empty circles.
First, we have to isolate the variable $x$ in each inequality.
$\begin{array}{rclcrccl} 3&<&3x&<&12&&|\text{ ($\times~\leftrightarrow~\div$) Opposite Operations}\\ 1&<&x&<&4 \end{array}$
This is the blue number line with two empty circles $4$ and $14$.
$\begin{array}{rclcrccl} 4&<&3x+1&\le&13&&|\text{ ($+~\leftrightarrow~-$) Opposite Operations}\\ 3&<&3x&\le&12&&|\text{ ($\times~\leftrightarrow~\div$) Opposite Operations}\\ 1&<&~~x&\le&4 \end{array}$
This is the green line number with an empty circle at $1$ and $4$.
$\begin{array}{rclcrccl} 3&\le&2x-5&\le& 9&&|\text{ ($-~\leftrightarrow~+$) Opposite Operations}\\ 8&\le&2x&\le&14&&|\text{ ($\times~\leftrightarrow~\div$) Opposite Operations}\\ 4&\le&~~x&\le &7 \end{array}$
This is the yellow line number.
$\begin{array}{rclcrccl} 5&<&3x-4&\le& 11&&|\text{ ($-~\leftrightarrow~+$) Opposite Operations}\\ 9&<&3x&\le&15&&|\text{ ($\times~\leftrightarrow~\div$) Opposite Operations}\\ 3&<&~~x&\le&5 \end{array}$
This is the red line number.
Solve the inequality. Tipps
To solve an inequality treat it like an equation, but don't forget to consider the sign.
Remember, if you multiply or divide by a negative value, you have to flip the sign!
Or more explicitly:
$<$ changes to $<$, and vice versa $\le$ changes to $\ge$, and vice versa Lösung
We can solve inequalities by following these simple steps:
Use opposite operations If you multiply or divide by a negative number, flip the sign Isolate the variable $x$
This compound inequality can also be written as follows:
$1<x\le4$. |
October 28th, 2018, 09:58 PM
# 1
Newbie
Joined: Jul 2018
From: morocco
Posts: 26
Thanks: 0
Math Focus: algebraic number theory
divisiblity of a binomial coefficient
Hello, how to show the following?? :
$C_{2^{m-1}}^{k}\times 2^k$ is divisible by $2^{m}$.
with for any $k$ such that $2^{m-1}\geq k \geq 1$.
October 29th, 2018, 03:34 AM
# 2
Math Team
Joined: Dec 2013
From: Colombia
Posts: 7,689
Thanks: 2669
Math Focus: Mainly analysis and algebra
I would suggest induction.
I realise now that your $${k \choose m-1}$$ looks strange. $k$ is normally no smaller than $(m-1}$. Do you have an error?
Last edited by v8archie; October 29th, 2018 at 03:41 AM.
October 30th, 2018, 03:45 PM
# 4
Senior Member
Joined: Aug 2017
From: United Kingdom
Posts: 313
Thanks: 112
Math Focus: Number Theory, Algebraic Geometry
I assume you mean: $C_k^{2^{m-1}} \times 2^k$ is divisible by $2^m$.
Let $\nu$ denote the 2-adic valuation. The above statement is then equivalent to $\nu(C_k^{2^{m-1}}) \geq m-k$. See if you can prove $\nu(C_k^{2^{m-1}}) = m - 1 - \nu(k)$, or more generally that $\nu_p(C_s^{p^r}) = r - \nu_p(s)$, where $p$ is a prime and $\nu_p$ is the $p$-adic valuation. The result is immediate from this.
Tags binomial, coefficient, divisiblity
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Effects of localized spatial variations on the uniform persistence and spreading speeds of time periodic two species competition systems New general decay results in a finite-memory bresse system
Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, P.O. Box 546, Dhahran 31261, Saudi Arabia
$ \begin{array}{ll} \rho_1\varphi_{tt}-k_1(\varphi_x+\psi+lw)_x-lk_3(w_x-l\varphi) = 0,\\ \rho_2\psi_{tt}-k_2\psi_{xx}+k_1(\varphi_x+\psi+lw)+ \int_0^tg(t-s)\psi_{xx}(\cdot,s)ds = 0,\\ \rho_1w_{tt}-k_3(w_x-l\varphi)_x+lk_1(\varphi_x+\psi+lw) = 0, \end{array} $
$ (x,t) \in (0,L) \times (0, \infty) $
$ g $
$ \xi $
$ H $
$ g'(t)\leq-\xi(t)H(g(t)),\qquad\forall t\geq0. $
$ \xi $
$ H $ Keywords:Bresse system, viscoelastic, general and ptimal decay, equal and non-equal speeds of wave propagation. Mathematics Subject Classification:Primary: 35B35, 35L05; Secondary: 35B40, 35L20. Citation:Salim A. Messaoudi, Jamilu Hashim Hassan. New general decay results in a finite-memory bresse system. Communications on Pure & Applied Analysis, 2019, 18 (4) : 1637-1662. doi: 10.3934/cpaa.2019078
References:
[1]
M. O. Alves, L. H. Fatori, M. A. Jorge Silva and R. N. Monteriro,
Stability and optimality of decay rate for a weakly dissipative Bresse system,
[2] [3] [4] [5] [6] [7]
A. Guesmia and S. A. Messaoudi,
On the stabilization of Timoshenko systems with memory and different speeds of wave propagation,
[8] [9] [10] [11]
A. Soufyane and B. Said-Houari,
The effect of the wave speeds and the frictional damping terms on the decay rate of the bresse system,
[12]
A. Wehbe and W. Youssef,
Exponential and polynomial stability of an elastic Bresse system with two locally distributed feedbacks,
show all references
References:
[1]
M. O. Alves, L. H. Fatori, M. A. Jorge Silva and R. N. Monteriro,
Stability and optimality of decay rate for a weakly dissipative Bresse system,
[2] [3] [4] [5] [6] [7]
A. Guesmia and S. A. Messaoudi,
On the stabilization of Timoshenko systems with memory and different speeds of wave propagation,
[8] [9] [10] [11]
A. Soufyane and B. Said-Houari,
The effect of the wave speeds and the frictional damping terms on the decay rate of the bresse system,
[12]
A. Wehbe and W. Youssef,
Exponential and polynomial stability of an elastic Bresse system with two locally distributed feedbacks,
[1]
Abdelaziz Soufyane, Belkacem Said-Houari.
The effect of the wave speeds and the frictional damping terms on the decay rate of the Bresse system.
[2]
Ammar Khemmoudj, Taklit Hamadouche.
General decay of solutions of a Bresse system with viscoelastic boundary conditions.
[3]
Belkacem Said-Houari, Salim A. Messaoudi.
General decay estimates for a Cauchy viscoelastic wave problem.
[4]
Ammar Khemmoudj, Yacine Mokhtari.
General decay of the solution to a nonlinear viscoelastic modified von-Kármán system with delay.
[5]
Jing Zhang.
The analyticity and exponential decay of a Stokes-wave coupling system with viscoelastic damping in the variational framework.
[6]
Nguyen Thanh Long, Hoang Hai Ha, Le Thi Phuong Ngoc, Nguyen Anh Triet.
Existence, blow-up and exponential decay estimates for a system of nonlinear viscoelastic wave equations with nonlinear boundary conditions.
[7]
Dongbing Zha, Yi Zhou.
The lifespan for quasilinear wave equations with multiple propagation speeds in four space dimensions.
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Kunimochi Sakamoto.
Destabilization threshold curves for diffusion systems with equal diffusivity under non-diagonal flux boundary conditions.
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Tae Gab Ha.
Global existence and general decay estimates for the viscoelastic equation with acoustic boundary conditions.
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Mohammad M. Al-Gharabli, Aissa Guesmia, Salim A. Messaoudi.
Existence and a general decay results for a viscoelastic plate equation with a logarithmic nonlinearity.
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Well-posedeness and energy decay of solutions to a bresse system with a boundary dissipation of fractional derivative type.
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Cyclic asymptotic behaviour of a population reproducing by fission into two equal parts.
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What is the exact shape of the universe? I know of the balloon analogy, and the bread with raisins in it. These clarify some points, like how the universe can have no centre, and how it can expand equally everywhere in all directions.
But they also raise some questions, like if you are on the surface of a balloon and travel in 1 direction, you will eventually return to your starting point, is it possible our universe have this feature?
If it has, or had, would this be a symmetry of sorts ($\psi(x)=\psi(x+R)$), and as such have a conserved quantity associated with it (by Noether)?
Assuming "small curled up dimensions" wouldn't these dimensions have this type of symmetry, what are the associated conserved quantities?
Is it known exactly what the geometrical shape of the universe is? (on a large scale) (I am not talking about only the observable universe).
How does one define the "size" of a dimension, is this scale only applicable to curled up ones?
Is it possible to describe to a layman the shape of the universe without resorting to inept analogies?
There are a bunch of questions here. Let me try to take them in order:
Is it possible that our Universe has the feature that if you travel far enough you return to where you started?
Yes. The standard Big-Bang cosmological model is based on the idea that the Universe is homogeneous and isotropic. One sort of homogeneous spacetime has the geometry of a 3-sphere (like a regular sphere, but with one more dimension). In these cosmological models, if you travel far enough you get back to where you started.
However, the best available data seem to indicate that the Universe is very nearly spatially flat. This means that, if we do live in a 3-sphere Universe, the radius of the sphere is very large, and the distance you'd have to travel is much larger than the size of the observable Universe. Even if that weren't true, the fact that the Universe is expanding would make it hard or impossible to circumnavigate the Universe in practice: no matter how fast you went (short of the speed of light), you might never make it all the way around. Nonetheless, 3-sphere Universes, with the geometrical property you describe, are definitely viable cosmological models.
Does this give rise to a symmetry by Noether's theorem?
Not really. Noether's theorem is generally applied to continuous symmetries (i.e., ones that can be applied infinitesimally), not discrete symmetries like this. The fact that space is homogeneous gives rise to a symmetry, namely momentum conservation, whether or not space has the 3-sphere geometry, but the symmetry you're talking about here doesn't give rise to anything extra.
Would small curled up dimensions have the same sort of symmetry?
I'll leave this for someone else, I think. Not my thing.
Is it known exactly what the geomtrical shape of the universe is?
No, and don't let anyone tell you otherwise! Sometimes, especially in pop-science writing, people imply that we know a lot more about the global properties of the Universe than we do. We often assume things like homogeneity to make our lives simpler, but in fact we have precisely no idea what things are like outside of our horizon volume.
How to describe the "size" of a dimension?
If the Universe's geometry has enough symmetries, it makes sense to define an overall time coordinate everywhere. Then it makes sense to imagine a "slice" through spacetime that represents the Universe at an instant of time. If some of those slices have the geometrical property you're talking about, that traveling a distance R in a certain direction gets you back to your starting point, then it makes sense to call R the "size" of the corresponding dimension. If you can travel forever, then we say the size in that dimension is infinite.
Is it possible to describe to a layman the shape of the universe without resorting to inept analogies?
All analogies are imperfect. I think the best you can do is use a bunch of them and try to convey the limitations of each.
the answer to the Magellan question for our Universe is actually "No". If the Universe were a static 3-sphere, as Ted Bunn suggests, then you could "swim" around the Universe just like Megallan around the Earth.
But an important fact about our Universe is that its size is changing: the size can't stay constant just like the apple can't sit in the middle of a bedroom (in the air) - that's another thing that gravity according to general relativity guarantees. Already today, it resembles an empty de Sitter space (because the dark energy or cosmological constant is the majority of the energy in the Universe) that is exponentially expanding. The so-called Penrose causal diagram of the de Sitter space
shows that there are no time-like (mostly vertical) trajectories in it that would lead you to the same place, given by the $x$ coordinate. That's because the height of the diagram (square) is so limited.
So it will never be possible to travel along a straight line, even if you approached the speed of light, and return to the same place where you started (and meet your civilization that didn't travel anywhere). The time-dependence of the size of the Universe is important for this conclusion.
The spatial section of our Universe is nearly flat - that's because the total density (including dark energy and dark matter) is very close to the critical density, a fact guaranteed by the cosmic inflation. It may be slightly positively curved, like a 3-sphere (the surface of a 4-dimensional ball), or slightly negative curved (the surface of a hyperboloids resembling a saddle, but 3-dimensional one).
The whole spacetime geometry is close to the 4-dimensional de Sitter space that may be written as the following hyperboloid in 4+1 dimensions: $$ -A^2+B^2+C^2+D^2+E^2=R^2 $$ where $R$ is the radius of the curvature.
Concerning another question you asked: if the Universe is compact, like a sphere, then the wave function may be simply thought of as a wave function of the sphere. Of course, if the sphere is parameterized by some coordinates such as the latitude and longitude, it is a periodic function at least of the latter.
A sphere has a full rotational symmetry, $SU(2)$. Extra dimensions in string/M-theory can't have any continuous symmetries. They're believed to be Calabi-Yau manifolds or very similar manifolds in which all the dimensions are knitted in a nontrivial way.
If you have a compact dimension of space and you write a wave function so that it satisfies $\psi(x) = \psi(x+2\pi R)$, then indeed, it is possible to interpret this periodicity as a symmetry with respect to the (discrete) group of translations by multiples of $2\pi R$. While Noether's theorem is normally designed for continuous symmetries, you may say what the symmetry implies for such discrete symmetries as well.
The multiplicative generator of the symmetry is the translation by $2\pi R$ and it commutes with the Hamiltonian - because it is the indentity, after all. The generator may be written as $\exp(2\pi i R p)$ where $p$ is the momentum in the circular direction. And because it is equal to one, it means that the momentum is quantized in units of $1/R$ which is probably not too surprising. A problem here is that the translation by $2\pi R$ is not really a "nontrivial symmetry" in this case: it is an operation that is doing nothing at all. It keeps all objects invariant, so there is no nontrivial Noether's conserved quantity associated with it.
Since only a fraction of the universe is actually observable, I am not sure if it possible to answer this question.
From the NASA site: "Recent measurements (c. 2001) by a number of ground-based and balloon-based experiments, including MAT/TOCO, Boomerang, Maxima, and DASI, have shown that the brightest spots are about 1 degree across. Thus the universe was known to be flat to within about 15% accuracy prior to the WMAP results. WMAP has confirmed this result with very high accuracy and precision. We now know that the universe is flat with only a 2% margin of error." The WMAP results are given here :http://lambda.gsfc.nasa.gov/product/map/current/
protected by Kyle Kanos Jun 16 '15 at 21:20
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The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
Morias
\(\mathrm{mori}(z)=\displaystyle \frac{ J_0 (L_1 z)}{1-z^2}\)
where \(L_1\approx 2.4\) is first zero of the Bessel function, \(J_0(L_1)\!=\!0~\).
\(u\!+\!\mathrm i v=\mathrm{mori}(x\!+\!\mathrm i y)\)
For integer \(m>0\) and \(|z|\gg 1\), function mori\((z)\) can be approximated with
\(\displaystyle \mathrm{mori}(z) \approx \mathrm{morias}_m(z) \) \(\displaystyle= - \sqrt{\frac{2}{\pi L_1}} z^{-5/2} \, G_m(z^2)\, \cos\left( - \frac{\pi}{4} + L_1 z\, F_m(z^2) \right) \)
where
\(G_m(x) = \sum_{n=0}^{m} g_n x^{-n}\) \(F_m(x) = \sum_{n=0}^{m} f_n x^{-n}\)
The coefficients \(f\) and \(g\) can be calculated from the asymptotic expansion of the Hankel function;
\(\begin{array}{l} f_0=1\\ f_1=- 0.021614383628830615865\\ f_2=0.0019465899152260872595\\ f_3=-0.0010834984344719114778\\ f_4=0.001464410164512283719\\ f_5=-0.003628876399615993660\\ f_6=0.01431760830195380729\\ f_7=-0.0824438982874790057\\ f_8=0.652747801052423657\\ f_9=-6.80376838070624330\\ f_{10}=90.322658904953727\\ f_{11}=-1487.942880868968812\\ f_{12}=29785.50901270392374\\ \end{array}\) \(~\) \(\begin{array}{l} g_0=1 \\ g_1=0.989192808185584692068 \\ g_2=0.99228788615079417081 \\ g_3= 0.989481317221334367489\\ g_4= 0.994709980602617872387\\ g_5= 0.97818700495778240956\\ g_6=1.0575251177784290263\\ g_7= 0.5188843197279991625\\ g_8=5.432808917007474985\\ g_9=-52.5640507009104629\\ g_{10}=807.429675670594971\\ g_{11}= -14844.4023379433794\\ g_{12}=328204.367306340176 \end{array}\)
For \(m\!=\!11\), the complex map of function \(\mathrm{morias}_m\) is shown in figure at right. At this value of \(m\), the last line in the table above is not used, as the following increase of number of terms, taken into account, does not improve the precision of approximation with the complex double arithmetics in the examples and numerical tests.
Overlapping of the first two maps is shown in the third image at right.
Function \(\mathrm{morias}_{11}(z)\) approximates \(\mathrm{mori}(z)\) at \(|z|\!>\!2\), and, for \(|z|\!>\!4\), it provides the precise approximation; the error of this approximation is small compared to the rounding error for the complex double variables.
Application
Function morias seems to be useful for the efficient evaluation of function naga that refers of the loss of the pinhole waveguide. However, for the contour integral, morias of the complex argument should be implemented.
For the numerical testing of the asymptotic formulas of the complex argument, it is neserably to have function implemented in a "dingle piece". Then, for the commercial implementation as a built-in function of some language or software, the appropriate combination of morias and expansion of function mori for small and moderate values of the argument can be arranged.
References |
You are correct on both questions. 1 you answered yourself. It is the correct rate to close out the trade. 2 you use a dollar discount rate because you are discounting dollars. The (1-K/X) term represents one dollar from the first trade and K/X dollars from the close out trade.
I performed spectral analysis on the stock market for disaggregated returns. If $\mu$ is the center of location and anything away from $\mu$ is an "error", then the stock market is in equilibrium once every 20-21 years as an aggregate whole. But, like a musical instrument, the periods of the individual firms could be relatively small. Still, that would ...
That formula is algebraically equivalent to saying different, stochastic assets can have different expected returns.$$ \mathbb{E} \left[ R_i \right] = r_f + \gamma_i $$Some simple algebraLet $X_i$ be a random variable denoting a risky cash flow, $p_i$ be today's price of that risky cash flow, $r_f$ be the risk free rate, and $\gamma_i$ be some risk ...
Carry is typically only associated with known cashflows - its closely related cousin, roll, is typically associated with unknown cashflows, assuming the state of the world is unchanged.Given this, carry is typically only analyzed for the current period of the swap or bond. If we assume your swaps are fixed Semi vs a 6m Ibor index, then the natural period ...
As long as your market is complete and $\tau$ is measurable w.r.t. the filtration generated by the market the continuous cash flow paid until $\tau$ is a hedgeable contingent claim and you have to work under the risk neutral measure.
I think you have a little misunderstanding about treasury futures. I would get this book: http://www.amazon.com/Treasury-Bond-Basis-Depth-Arbitrageurs/dp/0071456104?ie=UTF8&psc=1&redirect=true&ref_=oh_aui_search_detailpage It is the absolute best guide to this product.A few important things to understand:Every treasury future has ...
If the cash flows depend on the (random) interest rates, then the $C_i$ are random variables and so would be the sum $\sum\limits_{i=1}^n C_id(i)$. However, initial market prices need to be constants, they cannot be random (because you need to know how much this claim is worth right now). The $d(i)$ are real numbers though since they are discount factors (...
I have laid out below one way of solving this kind of problem. You have your timeline right and I have reproduced it with the correct amounts. The way to discount your 30Ks is the same as discounting 1,500K if you do it this way. Basically, you need to compute a discount factor. To calculate this discount factor, you need to de-annualize your interest rate ...
We are comparing two situations: (1) an all equity firm, versus (2) the same firm which has decided to do a predefined amount of borrowing. Because the tax shields arise as a result of the borrowing decision (i.e. they are a potential advantage of borrowing) they are discounted at the borrowing rate. It is treated almost as a separate project: to borrow ...
By no arbitrage, market participants need to agree on the values of the discount factor, even if they are using different conventions (day count, compounding period) to convert the discount factor into a rate.For example, consider two discount factors computed using continuous compounding, where one is computed using the 30/360 day count (year fraction $t_{...
Pricing always takes place under the risk neutral probability measure. In fact, this would make the price more conservative (i.e. lower) with respect to risk; if you priced it under the true measure you would be putting a smaller hazard rate for this random time.Completeness make the risk neutral probability measure unique. In your case you might have ...
The conversion factor associated with each bond the futures' delivery basket is constructed such that the invoice prices of the bonds are identical under the assumption that the yield curve is flat at the level of the futures' notional coupon. Therefore, the bond with the highest duration will be the CTD when yields are above the notional coupon and the bond ...
First, I am not sure which exact statement was made. Also, you cannot just say "without CF" because you are essentially creating an artificial market with messed-up utility. In summary the cheapest-to-deliver bond is:The bond that results in the smallest loss or greatest profit for the futures seller.Futures sellers have to buy the bonds they are going ...
Let's consider a single cash flow CF$PV = (\frac{1}{1+i})^n CF$As you wrote $v = -\frac{1}{PV} \frac{d PV}{di}$Taking the derivative of PV with respect to i and plugging it in:$v= - \frac{(1+i)^n}{CF} n \frac{1}{(1+i)^{n-1}}\frac{-1}{(1+i)^2}CF$after simplifying we get$v = \frac{1}{1+i}n$(which is easy to remember, no need to derive it every ...
Another way to write:IRR(A) = x and IRR(0,0,A) = xis:PV(A;x)=0 and PV(0,0,A;x)=0where PV=present value, and x is the discount rate.Since we are using the same discount rate x, we can just add these up:PV(A,0,A;x)=0which means thatIRR(A,0,A) = xIs it clear?
I do not follow your analysis. In the case of either type of annuity the FV is equal to the PV times $(1+r)^n$. This factor is simply the factor which translates any amount in period 0 into an equivalent amount in period n.For an ordinary annuity:$$PVA=PMT \frac{1}{r}[1-\frac{1}{(1+r)^n}]$$When this value is "transferred" to period $n$ by multiplying ...
“You can't compensate for risk by using a high discount rate." - Warren Buffett at the 1998 Berkshire Hathaway Shareholder MeetingThe simple answer to your question is, “yes, many implementations of discounted cash flow analyses which adjust the discount rate for risk are double counting”. This practice is pervasive in academia, but has no basis in the ...
In Wolfram Alpha language ...2.7∗10^6*(1+x)^12 -75000*(1+x)^8 +50,000 -3.1∗10^6 = 0... gives x≈0.0124671 per month, which is 16.03% per annum.I.e. All incomings and outgoings must add up to zero, after adjusting for the monthly interest rate "x" over the number of months invested.2.7m remains in the fund for the full 13-1 = 12 months, and would ...
Your thought process on valuing the tax savings to the whole enterprise at the cost of capital makes perfect sense to me. While I get the thought process on tax savings being funded by debt, the cash flows due to tax savings do not flow back to debt, but rather to the enterprise whether or not the firm is highly leveraged.I think a simple scenario ...
I know this question has been answered but I am giving a complete derivation to assist others as well as serve as my notes.We will take a firm with a free cash flow $\mathcal F$ every year perpetually. This is a simplification that can easily be generalized to variable payments.Unlevered FirmThe earnings by the stockholders are reduced by taxes. Thus ...
So, you have this:$$ \sum_{k=1}^{k=15} 0.8 \cdot 1.06^{2k} = 5.3456\ldots $$And you want to know if there's a formula, or closed form. Yes there is.$$ \sum_{k=1}^{k=n} x^{k} = x \frac{x^{n+1}-1}{x-1} $$Where, we're going to set $x=1.06^{-2}$, and sum from 1 to 15 (since you're not including the first period in the valuations).$$ 0.8 \sum_{k=1}^{k=...
In basic instruments, one can ignore the past fixings and price purely on the future cashflows. Hence the term Net Present Value.With more exotic stuff such as range accruals, the past fixings are used to calculate the future payoff. In this case, to find the NPV at an intermediate date between the start date and expiry, you typically need to enter the ...
The idea of considering past cash flows into an NPV calculation is rather adventurous, in my opinion.If an investor wants to invest money into a financial instrument that has already generated positive cash flows before making his investment (e.g. investing into a bond when some interest payment dates have already passed), the investor would not consider ...
A simple query on google could have given you the answer...Let's definelumpsum qperiodic contribution ay periodsa periodic rate i$$q*(1+i)^y + a( ((1+i)^y-1) / i ) - a = f$$Suppose we do not want an initial investment $q=0$.2040 - 2016 = 24 years. As you want to know the monthly contribution, everything needs to be converted to months. Thus 24 ...
The annuity expression $a_{4}^{(12)}$is written as:$$a_{4}^{(12)}= \frac{1-(1+i)^{-4}}{i^{(12)}} = \frac{i}{i^{(12)}} a_4$$where, $i$ is the effective annual rate of interest and $i^{(12)}$ is nominal rate of interest convertible monthly, which is equal to$$i^{(12)}=12((1+i)^{1/12}-1)$$There is no closed formula to get the interest rate, you have to ...
The annuity method is the correct method. I am not familiar with wolframalpha but I assume it is correct.Look at it this way: in the second case (take out 481,000, repay 500,000 after 100 days) you have full use of the borrowed 481,000 for 100 days.In the first case (takeout 481,000, repay with an annuity of 10 payments over 100 day) you effectively ...
In a case like this, where the settlement date is in the middle of the coupon period, it is not right to use PV = -110 (minus the purchase price) in Step 3.Instead you should increase the purchase price by the accrued interest, which is a fraction of the coupon based on how far the settlement date is within the current coupon period. (So for ex if you are ... |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
In Hagan's paper on valuing CMS swaps (Convexity Conundrums: Pricing CMS Swaps, Caps, and Floors), there is:
So the swap rate must also be a Martingale, and
$$E \big[ R_s(\tau) \big| \mathcal{F}_0 \big]=R_s(0) = R_s^0$$
To complete the pricing, one now has to invoke a mathematical model (Black’s model, Heston’s model, the SABR model, . . . ) for how $R_s(\tau)$ is distributed around its mean value $R_s^0$. In Black’s model, for example, the swap rate is distributed according to
$$R_s(\tau) = R_s(0)e^{\sigma x\sqrt{\tau}-\frac{1}{2}\sigma^2\tau}$$
where x is a normal variable with mean zero and unit variance. One completes the pricing by integrating to calculate the expected value.
And he stopped here. So I am trying to "complete the pricing by integrating", but I am not sure of what he actually meant. I think he just wants to get the swap rate $R_s(0)$, but I am not sure.
This integration would lead to:
$$ R_s(0) = E \big[ R_s(\tau) \big| \mathcal{F}_0 \big] = \int_{-\infty}^{+\infty} R_s(\tau) \frac{-\frac{x^2}{2}}{\sqrt{2\pi}}dx = \int_{-\infty}^{+\infty} R_s(0)e^{\sigma\sqrt{\tau}x-\frac{1}{2}\sigma^2\tau} \frac{-\frac{x^2}{2}}{\sqrt{2\pi}}dx $$
This does not make sens, since $R_s(0)$ is on both side of the equation. Maybe I am not looking at it the right way, and he meant something else. Or maybe my integration is wrong.
I am a bit lost here. Any help would be appreciated to understand how to "complete this pricing". |
Advanced Studies in Pure Mathematics Adv. Stud. Pure Math. Galois–Teichmüller Theory and Arithmetic Geometry, H. Nakamura, F. Pop, L. Schneps and A. Tamagawa, eds. (Tokyo: Mathematical Society of Japan, 2012), 579 - 600 $n$-nilpotent obstructions to $\pi_1$ sections of $\mathbb{P}^1 - \{0,1,\infty\}$ and Massey products Abstract
Let $\pi$ be a pro-$\ell$ completion of a free group, and let $\pi = [\pi]_1 \supset [\pi]_2 \supset [\pi]_3 \supset \ldots$ denote the lower central series of $\pi$. Let $G$ be a profinite group acting continuously on $\pi$. First suppose that the action is given by a character. Then the boundary maps $\delta_n: H^1 (G,\pi/[\pi]_n)\to H^2 (G, [\pi]_n/[\pi]_{n+1})$ are Massey products. When the action is more general, we partially compute these boundary maps. Via obstructions of Jordan Ellenberg, this implies that $\pi_1$ sections of $\mathbb{P}^1_k - \{0,1,\infty\}$ satisfy the condition that associated $n^{th}$ order Massey products in Galois cohomology vanish. For the $\pi_1$ sections coming from rational points, these conditions imply that $$\langle x^{-1},\ldots,x^{-1},(1-x)^{-1},x^{-1},\ldots,x^{-1}\rangle = 0$$ where $x$ in $H^1 (\mathrm{Gal}(\overline{k}/k), \mathbb{Z}_\ell(\chi))$ is the image of an element of $k^*$ under the Kummer map. For the $\pi_1$ sections coming from rational tangent vectors at infinity, these conditions imply that $$\langle x^{-1},\ldots,x^{-1},(-x)^{-1},x^{-1},\ldots,x^{-1}\rangle = 0.$$
Article information Dates Received: 31 March 2011 Revised: 3 February 2012 First available in Project Euclid: 24 October 2018 Permanent link to this document https://projecteuclid.org/ euclid.aspm/1540417831 Digital Object Identifier doi:10.2969/aspm/06310579 Mathematical Reviews number (MathSciNet) MR3051256 Zentralblatt MATH identifier 1321.11116 Citation
Wickelgren, Kirsten. $n$-nilpotent obstructions to $\pi_1$ sections of $\mathbb{P}^1 - \{0,1,\infty\}$ and Massey products. Galois–Teichmüller Theory and Arithmetic Geometry, 579--600, Mathematical Society of Japan, Tokyo, Japan, 2012. doi:10.2969/aspm/06310579. https://projecteuclid.org/euclid.aspm/1540417831 |
Linear quadratic mean-field-game of backward stochastic differential systems
1.
School of Mathematics, Shandong University, Jinan 250100, China
2.
Zhongtai Securities Institute for Financial Study, Shandong University, Jinan 250100, China
3.
Department of Applied Mathematics, The Hong Kong Polytechnic University, Hong Kong, China
This paper is concerned with a dynamic game of
N weakly-coupled linear backward stochastic differential equation (BSDE) systems involving mean-field interactions. The backward mean-field game (MFG) is introduced to establish the backward decentralized strategies. To this end, we introduce the notations of Hamiltonian-type consistency condition (HCC) and Riccati-type consistency condition (RCC) in BSDE setup. Then, the backward MFG strategies are derived based on HCC and RCC respectively. Under mild conditions, these two MFG solutions are shown to be equivalent. Next, the approximate Nash equilibrium of derived MFG strategies are also proved. In addition, the scalar-valued case of backward MFG is solved explicitly. As an illustration, one example from quadratic hedging with relative performance is further studied. Keywords:Backward mean-field game (BMFG), $\epsilon$-Nash equilibrium, Hamiltonian-type consistency condition (HCC), Riccati-type consistency condition (RCC), backward stochastic differential equation (BSDE). Mathematics Subject Classification:Primary: 93E20, 91A23; Secondary: 93E03, 91A10, 93E14. Citation:Kai Du, Jianhui Huang, Zhen Wu. Linear quadratic mean-field-game of backward stochastic differential systems. Mathematical Control & Related Fields, 2018, 8 (3&4) : 653-678. doi: 10.3934/mcrf.2018028
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R. Buckdahn, B. Djehiche and J. Li,
A general stochastic maximum principle for SDEs of mean-field type,
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D. Duffie,
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G. Guan and Z. Liang,
Optimal management of DC pension plan under loss aversion and value-at-risk constraints,
[17]
J. Huang, S. Wang and Z. Wu,
Backward mean-field linear-quadratic-gaussian (LQG) games: Full and partial information,
[18]
M. Huang,
Large-population LQG games involving a major player: The Nash certainty equivalence principle,
[19]
M. Huang, P. Caines and R. Malhamé,
Large-population cost-coupled LQG problems with non-uniform agents: Individual-mass behavior and decentralized $\varepsilon$-Nash equilibria,
[20]
M. Huang, R. Malhamé and P. Caines,
Large population stochastic dynamic games: Closed-loop McKean-Vlasov systems and the Nash certainty equivalence principle,
[21]
M. Kohlmann and X. Y. Zhou,
Relationship between backward stochastic differential equations and stochsdtic controls: A linear-quadratic approach,
[22] [23]
T. Li and J. Zhang,
Asymptotically optimal decentralized control for large population stochastic multiagent systems,
[24]
A. E. Lim,
Quadratic hedging and mean-variance portfolio selection with random parameters in an incomplete market,
[25]
A. E. Lim and X. Y. Zhou,
Linear-quadratic control of backward stochastic differential equations,
[26]
M. Loève,
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J. Ma and J. Yong,
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S. Nguyen and M. Huang,
Linear-quadratic-Gaussian mixed games with continuum-parametrized minor players,
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$\epsilon$-Nash mean field game theory for nonlinear stochastic dynamical systems with major and minor agents,
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Fully coupled forward-backward stochastic differential equations and applications to optimal control,
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D. Pirjol and L. Zhu,
Discrete sums of geometric Brownian motions, annuities and asian options,
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B. Wang and J. Zhang,
Mean field games for large-population multiagent systems with markov jump parameters,
[38]
Z. Wu,
Adapted solution of generalized forward-backward stochastic differential equations and its dependence on parameters,
[39]
J. Yong,
A linear-quadratic optimal control problem for mean-field stochastic differential equations,
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D. T. Zhang,
Forward-backward stochastic differential equations and backward linear quadratic stochastic optimal control problem,
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I had this question for math $$ \lim_{x\to 0}\frac{\sin{ax^2}}{\sin{bx^2}} $$
So I used squeeze theorem and got $-1<1/(\sin{bx^2})<1$ and I multiplied by $\sin{ax^2}$ and got $-\sin{ax^2}< \sin{ax^2}/\sin{bx^2}< \sin{ax^2}$
Which then equals $0 < \lim_{x\to 0}\sin{ax^2}/\sin{bx^2}<0$ and the answer I got is $0$, but the answer is supposed to be $a/b$
Where did I go wrong and how can you solve this without using l'Hôpital's rule? |
Suppose $X_1,X_2,\ldots$ is a sequence of Cauchy random variables with density $$f(x)=\frac{1}{\pi(1+x^2)}, \hspace{3mm}x\in \mathbb{R}$$ and let $S_n=X_1+\ldots+X_n$.
It's easy to show that $\frac{S_n}{n}$ converges in distribution using characteristic functions (in fact, $S_n/n$ has the same distribution as $X$ for every $n$).
On the other hand, $S_n/n$ does
not converge in probability. Does this follow from the fact that these random variables are not integrable (if they were, we could apply WLLN)? Or do we need to use the definition of convergence in probability directly? |
Let $A$ be a nilpotent matrix. Prove that $\det(I+A)=1$
Could someone at least give me a clue ?
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Since $A$ is nilpotent, we have
$A^m = 0 \tag{1}$
for some positive interger $m$. This implies every eigenvalue of $A$ vanishes, since the equation
$Av = \lambda v \tag{2}$
for non-zero $v$ (recall eigenvectors are required to be non-zero) implies
$0 = A^mv = \lambda^m v, \tag{3}$
whence
$\lambda^m = 0, \tag{4}$
since $v \ne 0$. (4) forces
$\lambda = 0 \tag{5}$
Now use the fact that for any scalars $\lambda$ and $a$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda + a$ is an eigenvalue of $A + aI$; indeed we have, from (2),
$(A + aI)v = Av + av = (\lambda + a)v. \tag{6}$
(6) allows us to conclude that every eigenvalue of $A + I$ is $1$; hence $\det (A+I)$, being the product of its eigenvalues, satisfies
$\det(A+I) = 1. \tag{7}$
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
If your matrices are over algebraically closed field, consider Jordan canonical form $J$ of $A$. It will have only zeros on the diagonal, since nilpotent matrix has only zeroes as eigenvalues. Thus, $A = S J S^{-1}$ where $J$ is upper triangular and has zeroes on the diagonal. Now, $I + A = S I S^{-1} + S J S^{-1} = S(I + J)S^{-1}$, and so $det(I+A) = det(S)det(I+J)det(S^{-1}) = det(I+J)$, but $I+J$ is upper triangular and has only $1$-s on the diagonal, so $det(I+J) = 1$.
If our base ring is an arbitrary domain (for instance a non algebraically closed field), we can embed this ring into an algebraically closed field, and repeat the argument above.
If our ring is not necessarily domain, I don't know how to prove it, or whether it's true at all.
We have that: $A^m =0$. Then: \begin{equation} (I+A)A^{m-1}=A^{m-1}+A^{m}\\ \therefore (I+A)A^{m-1}=A^{m-1} \end{equation} Now, comptuting: \begin{equation} det(I+A)det (A^{m-1})=det(A^{m-1}); \quad det(A^{m-1})\neq 0\\ \therefore det(I+A)=1 \end{equation} |
The field you are describing is one possible solution to Maxwell's equations - a plane wave. It's a highly useful solution because all nonevanescent (propagating) fields in freespace / homogeneous mediums can be built from a superposition of plane waves.
But real solutions don't have to (and indeed never do) look like this particular solution. Real waves are superpositions of different plane waves; in particular, a superposition involving a spread of directions of constituent plane waves decreases swiftly with transverse distance from the center of the disturbance. Indeed, once we include a spread of directions
and frequencies in the plane wave superposition, the disturbance at any time can have a truly compact support i.e. is precisely zero at any point that cannot be reached from the source travelling at less than light speed for the time since the disturbance began, and thus comply with special relativity.
Another, much more realistic propagating solution to Maxwell's equations is a spherical wave.
$$\mathbf{E}\left(r,\,\theta,\,\phi,\,t\right)=A\frac{\sin\theta}{r}\left[\cos\left(kr-\omega t\right)-\frac{\sin\left(kr-\omega t\right)}{kr}\right]\hat{\boldsymbol{\phi}}$$$$$$$$\mathbf{B}\left(r,\,\theta,\,\phi,\,t\right)=\frac{2A\cos\theta}{r^{2}\omega}\left[\sin\left(kr-\omega t\right)+\frac{\cos\left(kr-\omega t\right)}{kr}\right]\hat{\mathbf{r}}+\frac{A\sin\theta}{r^{3}\omega}\left[\left(\frac{1}{k}-kr^{2}\right)\cos\left(kr-\omega t\right)+r\sin\left(kr-\omega t\right)\right]\hat{\boldsymbol{\theta}} $$
where $(r,\,\theta,\,\phi)$ are the spherical polar co-ordinates.
As $r\to\infty$ such a wave looks exactly like a plane wave with amplitude varying like $1/r$ over regions subtending angles at the source. |
Let $K \subseteq \mathbb{R}^n$ be a "fat" convex body, i.e. one that contains a ball of radius 1. I'm interested in the following question about points $y \in K$: If you take a normally distributed $e \sim \mathcal{N}(0,\sigma^2)^n$ and add $e$ to $y$, what is the probability that $y + e \in K$? This question can be rephrased to: what is the noncentral gaussian measure \begin{equation}\mu_{y,\sigma^2}(K) = \frac{1}{(\sqrt{2 \pi \sigma^2})^n} \int_K e^{\frac{\| x - y \|^2}{2 \sigma^2}} \mathrm{d} \lambda^n(x)\end{equation}of $K$? $\mu_{y,\sigma^2}(K)$ can be interpreted as a measure of how much of the neighborhood of $y$ is in $K$. Now, it's easy to see that if $y$ is an extreme point in $K$, then $\mu_{y,\sigma^2}(K)$ is negligibly small. Take for instance the cube $[0,1]^n$, if $y$ is one of the corners of $[0,1]^n$, then $\mu_{y,\sigma^2}(K) \leq 2^{-n}$ (regardless of $\sigma^2$). However, it seems like $\mu_{y,\sigma^2}(K)$ can be lower-bounded by some non-neglible term for "typical" $y$ if $\sigma = O(1/n)$.
I'm looking for some sort of mass-inequality for $\mu_{y,\sigma^2}(K)$, something like\begin{equation}\mu( \mu_{y,\sigma^2}(K) < \gamma ) < \epsilon\end{equation}where $y$ follows the uniform law on $K$. I need $\gamma$ to be non-negligible and $\epsilon$ to be negligible. I wonder if there has been a treatment of this or a closely related question about convex bodies. I've solved this problem for spheres (trivially) and for cubes and suspect that it is some general property of fat convex bodies or at least symmetric convex bodies. Moreover, I can show that for any convex body this is true for average $y$, i.e. something like $\mathbb{E}[\mu_{y,\sigma^2}(K)] \geq \alpha$ for any constant $\alpha < 1$ if $\sigma = O(1/n)$. One more observation is that that $\mu_{y,\sigma^2}(K)$ does not concentrate around its average, i.e. $\mu_{y,\sigma^2}(K)$ easily varies by a factor of $1/2$ or more. This can be seen in the cube example: If $y$ is uniformly distributed on $[0,1]$, then both $|y - 1/2| < \sigma/2$ and $y < \sigma / 2$ have significant probability. If $y \approx 1/2$, then the probability that $y + e \in [0,1]$ is close to 1. If $y \approx 0$, then the probability of $y + e \in [0,1]$ is close to $1/2$.
I suspect this problem to be somehow related to the study of isoperimetry of convex bodies, as a typical $y$ in $K$ is very close to the boundary of $K$. However, I didn't find any result that would help me with my problem.
This problem originates from an application in lattice-based cryptography. I would be very grateful for any pointers! |
Research Open Access Published: The truncation regularization method for identifying the initial value of heat equation on a spherical symmetric domain Boundary Value Problems volume 2018, Article number: 13 (2018) Article metrics
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2 Citations
Abstract
In this paper, identifying the initial value for high dimension heat equation with inhomogeneous source on a spherical symmetric domain is investigated. The truncation regularization method is a powerful technique for solving this inverse problem. We prove the convergence estimates between the regularization solution and the exact solution under the prior and the posterior regularization parameter choice rulers. A numerical example is presented to validate the effectiveness of this method.
Introduction
The initial value problem is one of the backward heat conduction problems (BHCPs). These problems have been studied over several decades due to their significance in many engineering problems and practical application problems, such as in welding of iron and steel, quenching of solids in liquids and testing of new thermal protective material.
In this paper, we consider an inhomogeneous heat equation on a symmetric domain as follows:
where \(r_{0}\) is the radius, \(\varphi(r)\) is the initial value. We use the additional condition \(u(r,T)=g(r)\) and \(f(r,t)\) to determine the initial value \(\varphi(r)\). The measured data of \(g(r)\) and \(f(r,t)\) are \(g^{\delta }(r)\) and \(f^{\delta}(r,t)\), which satisfy
The initial value problem is one of the backward heat conduction problems (BHCPs). A BHCP is severely ill-posed problem [1]. To overcome this difficulty, many scholars proposed some regularization techniques for the BHCP, such as the kernel-based method [2], the mollification method [3], the Fourier regularization method [4], optimal filtering method [5], the iterative method [6], the quasi-reversibility method [7–9], the central difference method [10], the filter regularization method [11], the method of fundamental solutions [12, 13], the boundary element method [14, 15], the group preserving scheme [16], modified Tikhonov regularization method [17], Quasi-boundary value method [18] and so on. But these references about BHCP, there are some drawbacks as follows: firstly, the regularization parameter is a prior choice rule, according to this choice rule, the parameter depends on the prior bound of the exact solution. But in practice we cannot obtain the exact solution, and the inaccurate prior bound may lead to the bad regularized solution. Secondly, they only considered the one dimensional BHCP; however, about high dimensional BHCP, there is little research results. In [19–21], the authors ever considered the high dimensional BHCP, but the regularization parameter is a prior choice. Thirdly, the equation is homogeneous and the measurement data is only one.
The truncation regularization method has been used to solve several inverse problems. In [22, 23], the authors used the truncation method to solve BHCP. In [24–26], the authors used the truncation method to solve a cauchy problem for the Helmholtz equation and the modified Helmholtz equation. In [27–30], the authors used the truncation method to identify the unknown source. In this paper, we mainly use the truncation regularization method to identify the initial value under two parameter choice rules. Moreover, we give an example to show the effectiveness of this method. We also compare the effectiveness between the posterior choice rule and the prior choice rule.
Using the separation of variables, we obtain the solution of the problem (1.1) as follows:
where
is the orthonormal eigenfunction system with weight \(r^{2}\) on \([0,r_{0}]\). It is also a complete system in \(L^{2}[0,r_{0};r^{2}]\). Now let \(\varphi_{n}=(\varphi(r), \omega_{n}(r)), f_{n}(\tau)=(f(r,t), \omega_{n}(r))\) and \(g_{n}=(g(r),\omega_{n}(r))\), \(h_{n}=\varphi_{n}+\int_{0}^{T}e^{(\frac{n\pi}{r_{0}})^{2}\tau }f_{n}(\tau)\,d\tau\). Using \(u(r,T)=g(r)\), we have
Define operator \(K: h(r)\rightarrow g(r)\), then
The operator
K is a linear self-adjoint compact operator, and
So
We give a prior bound on the initial value,
i.e.,
where \(E>0\) is a constant and \(\Vert \cdot \Vert _{p}\) denotes the norm in Sobolev space which is defined as follows:
This article is organized as follows. Section 2 presents some preliminaries results. Section 3 presents the convergence estimates under two parameter choice rules. In Section 4, a numerical example is proposed to show the effectiveness of this method. In Section 5, a brief conclusion is given.
Some auxiliary results
Throughout this paper, \(L^{2}[0,r_{0};r^{2}]\) denotes the Hilbert space of Lebesgue measurable function
φ with weight \(r^{2}\) on \([0,r_{0}]\). \((\cdot,\cdot)\) and \(\Vert \cdot,\cdot \Vert \) denote the inner and norm on \(L^{2}[0,r_{0};r^{2}]\), respectively, with the norm Lemma 2.1 For any \(n\geq1\), we have where \(C_{1}, C_{2}\) are constants. Lemma 2.2 Suppose \(f\in L^{\infty }(0,T;L^{2}[0,r_{0};r^{2}])\), then there exists a positive M such that where \(M:=\frac{T}{2}(\frac{r_{0}}{\pi})^{2}(1-e^{(\frac{\pi }{r_{0}})^{2}})\). Proof
For \(t\in[0,T]\),
Thus
where \(M:=T\sup_{n\in\mathbb{N}}(\int_{0}^{T}e^{-2(\frac{n\pi }{r_{0}})^{2}(T-\tau)}\,d\tau)=\frac{T}{2}(\frac{r_{0}}{\pi })^{2}(1-e^{(\frac{\pi}{r_{0}})^{2}})\). □
Regularization method and convergence estimate
It is obvious that the instability arises in the components of large
n in the solution. It is natural to imagine that we should replace \(e^{(\frac{n\pi}{r_{0}})^{2}T}\) by a bounded approximation or eliminate the noise in the input data. In this paper, we eliminate all the components of large n from the solution and define the truncation regularized solution as follows: Error estimate under a prior parameter choice rule Theorem 3.1 Let \(\varphi(r)\) given by (1.10) be the exact solution of problem (1.1). Let \(\varphi^{N,\delta}(r)\) given by (3.1) be the regularization solution. Choosing the regularization parameter \(N=[\gamma]\), where \(\gamma=(\frac {E}{\delta})^{\frac{1}{p+1}}\), then we obtain the following estimate: where \([\gamma]\) denotes the largest integer less than or equal to γ. Proof
By the triangle inequality, we have
Thus
Hence
Choosing the regularization parameter \(N=[(\frac{E}{\delta })^{\frac{1}{p+1}}]\), we obtain
Theorem 3.1 is proved. □
Error estimate under a posterior parameter choice rule
Choose \(\vert g^{\delta}(r)-\int_{0}^{T}e^{-(\frac{n\pi}{r_{0}})^{2}(T-\tau )}f^{\delta}(r,\tau)\,d\tau \vert >\tau\delta\). Let \(\psi(r):=g^{\delta}(r)-\int_{0}^{T}e^{-(\frac{n\pi}{r_{0}})^{2}(T-\tau )}f^{\delta}(r,\tau)\,d\tau\). Applying a discrepancy principle, we choose a posterior regularization parameter
N that satisfies
where \(P_{N}:L^{2}[0,r_{0};r^{2}]\rightarrow \operatorname{span}\omega_{n}\vert _{n\leq N}\) is an orthogonal projective operator.
I is an identity operator. Lemma 3.1 Let \(d(N)= \Vert (I-P_{N})(g^{\delta }(r)-\int_{0}^{T}e^{-(\frac{n\pi}{r_{0}})^{2}(T-\tau)}f^{\delta}(r,\tau )\,d\tau) \Vert \), then we have the following conclusions: (a)
\(d(N)\)
is a continuous function; (b)
\(\lim_{N\rightarrow\infty}\,d(N)=0\);
(c)
\(\lim_{N\rightarrow0}\,d(N)= \Vert g^{\delta}(r)-\int _{0}^{T}e^{-(\frac{n\pi}{r_{0}})^{2} (T-\tau)}f^{\delta}(r,\tau)\,d\tau \Vert \);
(d)
\(d(N)\)
is a strictly decreasing function, for any\(N\in [1,\infty)\). Lemma 3.2 Proof
Due to (3.5), we obtain
Thus we obtain
On the other hand,
i.e.,
So
□
Theorem 3.2 Let \(\varphi(r)\) given by (1.10) be the exact solution of problem (1.1). Let \(\varphi^{N,\delta}(r)\) given by (3.1) be the regularization solution. The regularization parameter N is chosen in (3.7). Then we obtain where \(C_{3}:=(\frac{\pi(\tau+\sqrt{2(M+1)})}{C_{1}})^{\frac{p}{p+1}}\), \(C_{4}:=\frac{\sqrt{2(M+1)}}{C_{1}}(\frac{C_{2}}{ (\tau-\sqrt{2(M+1)})})^{\frac{1}{p+1}} \pi^{\frac{p}{p+1}}\). Proof
According to (3.3), we have
Thus we obtain
Using (3.8), we have
where \(C_{3}:=(\frac{\pi(\tau+\sqrt{2(M+1)})}{C_{1}})^{\frac{p}{p+1}}\), \(C_{4}:=\frac{\sqrt{2(M+1)}}{C_{1}}(\frac{C_{2}}{ (\tau-\sqrt{2(M+1)})})^{\frac{1}{p+1}} \pi^{\frac{p}{p+1}}\).
Theorem 3.2 is proved. □
Numerical implementation and numerical example
In this section, we present numerical experiment for above regularization method. The exact solution of problem (3.1) is difficult to obtain. So we use a given \(\varphi(r)\) to solve the positive problem. We have
Let \(T=1\), \(r_{0}=\pi\). Using (1.5), we have
Using (3.1), we have
Time and space of grid step size are \(\Delta t=\frac{1}{P}\) and \(\Delta r=\frac{\pi}{M}\). A grid point on the time interval \([0,T]\) is \(t_{q}=\frac{q-1}{P}\) \((q=1,2,\ldots,P+1)\). \(r_{i}=\frac{i-1}{M}\pi\) \((i=1,2,\ldots,M+1)\) is a grid point on the space interval. We have
where
Noise data is generated by adding a random perturbation, that is,
where
ε is relative error level. In the computational procedure, we take the source function \(f(r,t)=t\sin(r)\). Example 1
Take initial function \(\varphi(r)=\sin(r)\).
Figure 1 shows the comparisons of the numerical effects between the exact solution and its regularization solution for the prior and posterior regularization parameter choice rules. We can find that the smaller
ε, the better the computed approximation is. Moreover, we can also easily find that the posterior parameter choice rule works better than the prior parameter choice rule. This is consistent with our theoretical analysis. Conclusion
We consider an inverse problem to determine an initial date for heat equation with inhomogeneous source on a columnar symmetric domain. Using the truncation method, we construct the regularization solution. Moreover, we obtain the Hölder type error estimate under prior and posterior parameter choice rules. Finally, an example is given to show the effectiveness of the truncation method.
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37, 8518-8532 (2013) Acknowledgements
The work is supported by the National Natural Science Foundation of China (11561045, 11501272) and the Doctor Fund of Lan Zhou University of Technology.
Ethics declarations Competing interests
The authors declare that they have no competing interests.
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This question already has an answer here:
How do you in general derive a formula for summation of n-squared, n-cubed, etc...? Clear explanation with reference would be great.
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This question already has an answer here:
How do you in general derive a formula for summation of n-squared, n-cubed, etc...? Clear explanation with reference would be great.
In general, if you want to derive a formula for the sum of $1^k+\cdots+n^k,$ let $$f(x)=\sum_{j=0}^{k+1}a_jx^j$$ and solve $$f(x)-f(x-1)=x^k$$ for the coefficients $a_j$. The polynomial that results will have the property that $f(n)=1^k+\cdots+ n^k$ for all positive integers $n$.
The Wikipedia article on Faulhaber's Formula will be very useful.
Equivalents of formulas for sums of squares and sums of cubes were already known in classical times. Formulas for the next few powers were developed by Islamic mathematicians. The full story was known by the time of Bernoulli, in the early $18$-th century.
The motivation was usually to develop a formula for
quadrature (integration) of integer powers of $x$.
Start with: $$ \sum_{0 \le k \le n} z^k = \frac{1 - z^{n + 1}}{1 - z} $$ If you do: $$ \left( z \frac{d}{d z} \right)^m \sum_{0 \le k \le n} z^k = \sum_{0 \le k \le n} k^m z^k $$ Compute the derivative, evaluate at $z = 1$ (you will need to apply l'Hôpital for the derivative at the right hand side as a limit) and you have the sum of the first $m$-th powers.
Another way is to apply the Euler-Maclaurin formula to the sum of $n^m$, which gives: $$ \sum_{0 \le k \le n} n^m = \frac{1}{m + 1} \left( \sum_{0 \le k \le m} \binom{m + 1}{k} \, B_k n^{m + 1 - k} \right) $$ Here the $B_k$ are the Bernoulli numbers.
Let $\displaystyle S(n;m) = \sum_{k=1}^n k^m$. Now consider $$(k+1)^{m+1} - k^{m+1} = \sum_{l=0}^m \dbinom{m+1}{l} k^l$$ Now telescopic summation gives $$(n+1)^{m+1} - 1 = \sum_{l=0}^m \dbinom{m+1}{l} \left(\sum_{k=1}^nk^l \right) = \sum_{l=0}^m \dbinom{m+1}{l} S(n;l)$$ $(1)$ Take $m=0$ and get $S(n;0)$.
$(2)$ Take $m=1$, use $S(n;0)$ and get $S(n;1)$.
$(3)$ Take $m=2$, use $S(n;0),S(n;1)$ and get $S(n;2)$.
$(4)$ Take $m=3$, use $S(n;0),S(n;1),S(n;2)$ and get $S(n;3)$.
$(5)$ And so on... |
Find distance between intersections of circle and line?
02-07-2019, 03:48 PM (This post was last modified: 02-07-2019 04:56 PM by kevin3g.)
Post: #1
Find distance between intersections of circle and line?
The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way?
02-07-2019, 06:05 PM (This post was last modified: 02-08-2019 01:08 PM by Albert Chan.)
Post: #2
RE: Find distance between intersections of circle and line?
Midpoint of intersecting points is closest to the circle center, (-1,0), a line with slope -1/2
y = 2x + 3
y = -1/2*x -1/2 ; line passes circle center and midpoint
-> midpoint = (-1.4, 0.2)
Distance from midpoint to circle center = √(0.4² +0.2²) = √(20)/10 ~ 0.447
Distance from midpoint to intersecting points= √(55 - 20/100) = √(274/5) ~ 7.403
Distance between intersecting points ~ 2 * 7.403 ~ 14.8
Intersecting points = (-1.4 ± t, 0.2 ± 2t) for some t
t² + (2t)² = 5 t² = 274/5; matching distance square
t = ±√(274)/5
-> Intersecting points ~ (-1.4 ± 3.31, 0.2 ± 6.62) = (-4.71, -6.42) and (1.91, 6.82)
02-07-2019, 06:13 PM (This post was last modified: 02-07-2019 06:15 PM by kevin3g.)
Post: #3
RE: Find distance between intersections of circle and line?
Is there a way I can subtract/use the answers I get from the solve function easily without having to retype them?
{-4.7106, 1.9106} {-6.4212, 6.8212}
02-07-2019, 06:43 PM (This post was last modified: 02-07-2019 07:06 PM by ijabbott.)
Post: #4
RE: Find distance between intersections of circle and line?
(02-07-2019 03:48 PM)kevin3g Wrote: The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way?
I don't know if it is faster, but:
\[
y=2x+3, (x+1)^2+y^2=55 \\
(x+1)^2+(2x+3)^2=55 \\
x^2+2x+1+4x^2+12x+9=55 \\
5x^2+14x-45=0
\]
If the quadratic equation \(ax^2+bx+c=0\) has real roots, the roots are separated by \(\frac{\sqrt{b^2-4ac}}{a}\), since \(\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a} = \frac{2\sqrt{b^2-4ac}}{2a} = \frac{\sqrt{b^2-4ac}}{a}\). This gives the \(x\) separation of the intersections, \(\Delta x = \frac{\sqrt{b^2-4ac}}{a}\).
The \(y\) separation, \(\Delta y\) will be twice that since \(y=2x+3\), so \(\Delta y = 2\Delta x\) (and the \(3\)s cancel). The distance between the intersection points \(D = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\Delta x)^2 + (2\Delta x)^2} = \sqrt{5(\Delta x)^2} = \sqrt{5\left(\frac{\sqrt{b^2-4ac}}{a}\right)^2} =\sqrt{5\left(\frac{b^2-4ac}{a^2}\right)} = \frac{\sqrt{5(b^2-4ac)}}{a}\).
Plugging in the coefficients of the quadratic equation, \(D = \frac{\sqrt{5(14^2- (4 \cdot 5 \cdot (-45)))}}{5} = \frac{\sqrt{5(196+900)}}{5} = \frac{\sqrt{5480}}{5} \approx 14.81\).
EDIT: Corrected mistake in \(c\) spotted by Albert Chan. \(c\) should be \(-45\), not \(-41\).
— Ian Abbott
02-08-2019, 12:33 AM (This post was last modified: 02-08-2019 02:31 PM by Albert Chan.)
Post: #5
RE: Find distance between intersections of circle and line?
Another way is to find angles of intersection.
Move the circle center to (0,0), line -> y' = 2(x' - 1) + 3 = 2x' + 1
Scale down to create a unit circle, line -> y'' = 2 x'' + 1√55
-> sin(z) = 2 cos(z) + 1√55
Use half angle formulas, t=tan(z/2), and let k=1√55:
2t/(1+t²) = 2 * (1-t²)/(1+t²) + k
2t = 2 - 2t² + k + kt²
(k-2) t² - 2t + (k+2) = 0
-> t = 0.6605, -1.7328
-> z = 66.89°, 239.98°
Distance between intersecting point = 2 r sin(Δz/2) = 2 √(55) sin(173.09°/2) ~ 14.8
Intersecting hi point = (√(55) cos(66.89°) - 1 , √(55) sin(66.89°)) ~ (1.91, 6.82)
Intersecting lo point = (√(55)cos(239.98°) - 1, √(55)sin(239.98°)) ~ (-4.71, -6.42)
Edit: this may be more accurate: |sin(Δz/2)| = |Δt| / √((1+t1²)(1+t2²))
02-08-2019, 11:59 AM
Post: #6
RE: Find distance between intersections of circle and line?
You can see many examples, from ... http://www.hp-prime.de/files/composite_f...ages_m.pdf
02-08-2019, 12:08 PM
Post: #7
RE: Find distance between intersections of circle and line?
(02-07-2019 03:48 PM)kevin3g Wrote: The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way?
Hey @kevin3g,
you can solve the system between the two equations and once you found the intersection points coordinates you apply the usual formula (c=sqrt(((x1-x2)^2)+((y1-y2)^2))) to find their distance
Honestly I dont know if there is a faster way than that given.
Best,
Aries
02-08-2019, 12:24 PM
Post: #8
RE: Find distance between intersections of circle and line?
More easy is ... https://www.wolframalpha.com/input/?i=circle+and+line
02-08-2019, 12:45 PM (This post was last modified: 02-08-2019 01:44 PM by Albert Chan.)
Post: #9
RE: Find distance between intersections of circle and line?
Distance from point (x0, y0) to line Ax + By + C = 0 is abs(Ax0 + By0 + C)/√(A² + B²)
-> Distance from circle center (-1,0) to y=2x+3 is abs(-2 + 0 + 3)/√5 = 1/√5
-> Distance between intersecting points = 2 * √(55 - 1/5) = 2 √(274/5) ~ 14.8
02-08-2019, 01:05 PM
Post: #10
RE: Find distance between intersections of circle and line?
Continuing my previous post ...
02-08-2019, 04:47 PM
Post: #11
RE: Find distance between intersections of circle and line?
(02-08-2019 12:45 PM)Albert Chan Wrote: Distance from point (x0, y0) to line Ax + By + C = 0 is abs(Ax0 + By0 + C)/√(A² + B²)
That's an elegant way to do it!
— Ian Abbott
09-04-2019, 10:16 PM (This post was last modified: 09-10-2019 12:50 AM by Albert Chan.)
Post: #12
RE: Find distance between intersections of circle and line?
I just thought of a way, even if we do not know the distance for point to line formula.
At the end, I accidentally proved the "official" distance formula
First, shift the coordinate so the point (in this case, center of circle), is the origin.
y = 2x + 3 = 2(x+1) + 1 = 2x' + 1
With this shifted coordinates, find where the line hit the axis.
In other words, find points on line, P=(0, y0), Q=(x0, 0).
Let h = distance of origin to the line.
Area ΔOPQ = ½ |x0 y0| = ½ |PQ| h
\(\large \mathbf{h = {|x_0 y_0| \over \sqrt{x_0^2 + y_0^2 }}}\)
But, we can do better !
Assume the line has form y = m x + c, we have (x0, y0) = (-c/m, c)
h = | -c²/m | / √(c²/m² + c²)
\(\large \mathbf{h = {|c| \over \sqrt{m² + 1}}}\)
For this example, h = |1| / √(2² + 1) = 1/√5
Distance of chord = \(\large \mathbf{2 \sqrt{r^2 - \frac{c^2}{m^2+1}}}\) = 2 √(55 - 1/5) ≈ 14.8
For prove of official distance formula, rewrite Ax + By + C = 0
A(x-x0) + B(y-y0) + (Ax0 + By0 + C) = 0
(y-y0) = (-A/B) (x-x0) - (Ax0 + By0 + C)/B
Matching pattern y' = mx' + c, then use h formula → \(\large \mathbf{ h= { |A x_0 + B y_0 + C| \over \sqrt{A^2 + B^2}}} \)
09-06-2019, 05:16 AM (This post was last modified: 09-06-2019 05:21 AM by teerasak.)
Post: #13
RE: Find distance between intersections of circle and line?
In CAS mode, use command
solve({y=2*x+3,(x+1)^2+y^2=55},{x,y})
then the calculator will return you 2 intersection points. Put that in a variable e.g m1
then use command
distance(m1(1),m1(2))
you will get the distance between two points
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Law of Sines Law of Cosines
Used when you know two angles and the included side (ASA), two angles and the non-included side (AAS) or two sides and the non-included angle (SSA).\[ \frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c} \]
Solve by cross multiplying.\[ b\sin{A} = a\sin{B} \]
Solve for a side:\[ b = \frac{a\sin{B}}{\sin{A}} \]
Or solve for an angle:\[ \sin{A} = \frac{a\sin{B}}{b} \]
The angles of a triangle must add to $180^{\circ}{\rm.}$ When you know two, find the third by subtracting from 180.\[ A + B + C = 180^{\circ} \]
Used when you know three sides (SSS) or two sides and the included angle (SAS).\[ a^2 = b^2 + c^2 - 2bc\cos{A} \] \[ b^2 = a^2 + c^2 - 2ac\cos{B} \] \[ c^2 = a^2 + b^2 - 2ab\cos{C} \]
When solving for missing angles, the equations can be rearranged:\[ \cos{A} = \frac{b^2 + c^2 - a^2}{2bc} \] \[ \cos{B} = \frac{a^2 + c^2 - b^2}{2ac} \] \[ \cos{C} = \frac{a^2 + b^2 -c^2}{2ab} \]
The angles of a triangle must add to $180^{\circ}{\rm.}$ When you know two, find the third by subtracting from 180.\[ A + B + C = 180^{\circ} \]
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Let $G=\pi(X,x)$ be the fundamental group of a compact orientable surface of genus $g\ge 2$. It is well known that a presentation of $G$ is $$G=\langle x_1,y_1,\dots,x_g,y_g \ | \ [x_1,y_1]\cdots [x_g,y_g]=1\rangle$$ (where $[x,y]=xyx^{-1}y^{-1}$ is the commutator).
Denote by $F$ be the free group with $2g$ generators $x_1,y_1,\dots,x_g,y_g$ and by $R$ be the normal closure of the relation $r=[x_1,y_1]\cdots [x_g,y_g]$, so $G=F/R$.
It is clear that $r\in [F,F]$.
Question: Is there an elementary proof that $r=[x_1,y_1]\cdots [x_g,y_g]\not\in [F,[F,F]]$?
This result appears when one considers the Stallings exact sequence associated to $$1\to [G,G]\to G\to G^{ab}\to 1$$ to get $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to H_1(G,\mathbb{Z})\to H_1(G^{ab},\mathbb{Z}) \to 0$$
Since $H_1(G,\mathbb{Z})\cong H_1(G^{ab},\mathbb{Z})\cong G^{ab}$ we obtain a short exact sequence $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to 0$$ which should be injective at the left (see at the end some argument why).
Now, Hopf's formula gives $$H_2(F/R,\mathbb{Z})=(R\cap [F,F])/[R,F]=R/[R,F]$$ since $R\subset [F,F]$, hence $H_2(F/R,\mathbb{Z})$ is cyclic and the generator is given by (the class of) $r$. So the map $\psi:H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})$ is either injective or zero. But $$H_2(G^{ab},\mathbb{Z})\cong [F,F]/[F,[F,F]]$$ since $G^{ab}\cong F/[F,F]$, and so the map $\psi$ is given by the natural map $$\psi:R/[R,F]\to [F,F]/[F,[F,F]]$$ coming from the inclusion $R\hookrightarrow [F,F]$, hence $\psi$ is injective if and only if $r\not\in [F,[F,F]]$.
One possibility is to use another description of the map $\psi$ as $$H_2(G,\mathbb{Z})\to \bigwedge H_1(G^{ab},\mathbb{Z})$$ that should correspond to the dual of the cup product in cohomology via Poincaré duality (i.e. dual universal coefficient theorem) (but I am not sure if to consider this approach as really elementary). |
The existing literature on generalized trigonometric functions is scarce and it seems there isn't a comprehensive account of generalized trigonometric functions anywhere. Also there isn't a unified accepted notation and different authors use different notations. The generalized sine and cosine functions $\sin_{pr}x$, $\ \cos_{pr}x$ are defined by the formulas$$x=\int_0^{\sin_{pr}x}\frac{dt}{\sqrt[p]{1-t^r}},\qquad \cos_{pr}x=\sqrt[r]{1-(\sin_{pr}x)^r}.$$For generic values of parameters $p,r$ it appears that these functions are not very interesting, but when these parameters have special values, then $\sin_{pr}x$ can be expressed algebraically through Jacobi elliptic functions, and as a consequence one can establish addition theorems analogous to addition theorems for elliptic functions.
Below I give a unified relatively simple discussion of several such cases. The first case $r=2,\ p=2$ is trivial. The second case $r=3,\ p=\frac{3}{2}$ is due to Cayley and Dixon. The third case $r=4,\ p=\frac{4}{3}$ has been considered relatively recently by Edmunds, Gurka, and Lang (Properties of generalized trigonometric functions, J. Approx. Theory 164 (2012), no. 1, 47-56.) I couldn't find the discussion of the fourth case $r=6,\ p=\frac{6}{5}$ in the literature, but it easily follows from the same considerations as in the first three cases. The parameter $r$ is denoted in analogy with Ramanujan's theory of elliptic functions to alternative bases (see "Ramanujan's notebooks, vol. 5" by Bruce Berndt). I don't know whether it is just a coincidence or there is a deeper connection, but $r=2,3,4,6$ are the four signatures for which such alternative theories of elliptic functions have been developed.
Let's consider the case when $\frac{1}{p}=1-\frac{1}{r}$. Then by a series of simple changes of variables one obtains\begin{align}\int_0^{u}\frac{dt}{(1-t^r)^{1-1/r}}=\frac{1}{r}\int_0^{u^r}\frac{dt}{(1-t)^{1-1/r}t^{1-1/r}}=\tag{1}\\\frac{1}{r}\int_0^{u^r}\frac{dt}{(t-t^2)^{1-1/r}}=\frac{1}{r}\int_0^{u^r}\frac{dt}{\left(\frac{1}{4}-\left(\frac{1}{2}-t\right)^2\right)^{1-1/r}}=\\\frac{2^{2-2/r}}{r}\int\limits_{1-2u^r}^{1}\frac{dt}{(1-t^2)^{1-1/r}}=\frac{2^{1-2/r}}{r}\int\limits_{(1-2u^r)^2}^{1}\frac{dt}{\sqrt{t}(1-t)^{1-1/r}}=\\\frac{2^{1-2/r}}{r}\int\limits_0^{1-(1-2u^r)^2}\frac{t^{1/r-1}dt}{\sqrt{1-t}}=2^{1-2/r}\cdot\int\limits_0^\sqrt[r]{1-(1-2u^r)^2}\frac{dt}{\sqrt{1-t^r}}\end{align}When $r=2,3,4$ the last integral can be inverted in terms of elliptic functions. But the case $r=6$ also can be inverted, since$$\int\frac{dt}{\sqrt{1-t^6}}=-\frac{1}{2}\int\frac{d\left(\frac{1}{t^2}\right)}{\sqrt{\frac{1}{t^6}-1}}.$$
As an illustration, according to Edmunds,Gurka, and Lang ($r=4$) one has$$\sqrt{1-\sqrt{\frac{1-\left(\sin_{\frac{4}{3},4}x\right)^2}{1+\left(\sin_{\frac{4}{3},4}x\right)^2}}}=\text{sn}\left(x,\frac{1}{\sqrt{2}}\right).$$
Representation in terms of elliptic functions is possible also when $\frac{1}{p}\neq 1-\frac{1}{r}$. This can be seen by substitution $t\to \frac{at+b}{ct+d}$ in the integral (second integral in eq.$(1)$)$$\int\frac{dt}{(1-t)^{1-1/r}t^{1-1/r}}, \quad (r=2,3,4,6),$$and specifying parameters $a,b,c,d$ such that the resulting integral again has he form of an incomplete beta function. |
In Hutchings and Taubes lecture note on Seiberg-Witten equation HERE, above equation (4.20) the authors claim that there is a version of Weitzenbock formula reads (where $\beta \in \Omega^{0,2}(M, E)$, M is a symplectic manifold with compatible $J$, $E$ is a line bundle with $U(1)$ connection $a$)
\begin{equation} \int {{{\left| {\bar \partial _a^*\beta } \right|}^2}} = \frac{1}{2}\int {\left( {{{\left| {\nabla _a^*\beta } \right|}^2} - i\left\langle {\omega ,{F_a}} \right\rangle {{\left| \beta \right|}^2}} \right)} \end{equation}
I try to prove it, by first guessing that it might come from something like a Kahler identity \begin{equation} {{\bar \partial }_a}\bar \partial _a^*\beta = \frac{1}{2}{\nabla _a}\nabla _a^*\beta + \left( {...} \right) \end{equation} and try to figure out if the ... matches the formula in the note. But I could not reproduce the result: the ... I got is of the form $\Lambda \left( {F_a^{2,0} \wedge \beta } \right)$, where $\Lambda = (\omega\wedge)^*$, and a (2,0)-piece of $F$ rather than a (1,1) piece shows up.
So I wonder how to prove the Weitzenbock formula above? Any reference is welcome; I google for a while but may be I am not looking at the write place. |
Research Open Access Published: Quasilinear elliptic equations with Hardy terms and Hardy-Sobolev critical exponents: nontrivial solutions Boundary Value Problems volume 2015, Article number: 171 (2015) Article metrics
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2 Citations
Abstract
In this paper, we obtain one positive solution and two nontrivial solutions of a quasilinear elliptic equation with
p-Laplacian, Hardy term and Hardy-Sobolev critical exponent by using variational methods and some analysis techniques. In particular, our results extend some existing ones. Introduction and main results
We shall study the following quasilinear elliptic equation:
where \(\triangle_{p}u=\operatorname{div}(|\nabla u|^{p-2}\nabla u)\) denotes the
p-Laplacian differential operator, Ω is an open bounded domain in \(\mathbb{R}^{N}\) (\(N\geq3\)) with smooth boundary ∂Ω and \(0\in\Omega\), \(0\leq\mu<\mu_{1}:= (\frac{N-p}{p} )^{p}\), \(0 \leq s< p\), \(1< p< N\), \(p^{\ast}(s)=\frac{p(N-s)}{N-p}\) is the Hardy-Sobolev critical exponent and \(p^{\ast}=p^{\ast}(0)=\frac{Np}{N-p}\) is the Sobolev critical exponent. The conditions of f will be given later.
On the Sobolev space \(W^{1,p}_{0}(\Omega)\), we set
which is well defined on \(W^{1,p}_{0}(\Omega)\) by the Hardy inequality [1]. It is comparable with the standard Sobolev norm of \(W^{1,p}_{0}(\Omega)\), but it is not a norm since the triangle inequality or subadditivity may fail, which has been clarified in [2]. The following minimization problem will be useful in what follows:
which is the best Hardy-Sobolev constant.
Ding and Tang [13] obtained the existence and multiplicity of solutions for (1.4) if \(0\leq\mu< (\frac{N-2}{2} )^{2}\), \(0\leq s<2\) and
f satisfies some suitable conditions. Kang [14] considered another special case of (1.1) with \(f(x,u)=\lambda\frac{|u|^{q-2}u}{|x|^{t}}\); for details, we refer the readers to see Remark 1.1.
Let \(F(x,u):=\int^{u}_{0}f(x,s)\, ds\), \(x\in\Omega\), \(u\in\mathbb{R}\). In order to state our results, we make the following assumptions:
(A 1):
\(f\in C(\overline{\Omega} \times\mathbb{R}^{+},\mathbb{R})\), \(\lim_{u\rightarrow 0^{+}}\frac{f(x,u)}{u^{p-1}}=0\) and \(\lim_{u\rightarrow \infty}\frac{f(x,u)}{u^{p^{\ast}-1}}=0\) uniformly for \(x\in\overline {\Omega}\).
(A 2):
There exists a constant \(\rho_{0}\) with \(\rho_{0}>p\) such that$$0< \rho_{0} F(x,u)\leq f(x,u)u,\quad \forall x\in\overline{\Omega}, \forall u\in \mathbb{R}^{+}\setminus\{0\}. $$
(A 3):
\(f\in C(\overline{\Omega} \times\mathbb{R},\mathbb{R})\), \(\lim_{u\rightarrow 0}\frac{f(x,u)}{|u|^{p-1}}=0\) and \(\lim_{|u|\rightarrow \infty}\frac{f(x,u)}{|u|^{p^{\ast}-1}}=0\) uniformly for \(x\in\overline {\Omega}\).
(A 4):
There exists a constant \(\rho_{0}\) with \(\rho_{0}>p\) such that$$0< \rho_{0} F(x,u)\leq f(x,u)u,\quad \forall x\in\overline{\Omega}, \forall u\in \mathbb{R}\setminus\{0\}. $$
Let \(b(\mu)\) be one of zeroes of the function \(g(t)=(p-1)t^{p}-(N-p)t^{p-1}+\mu\), where \(t\geq0\) and \(0\leq\mu<\mu _{1}\). Now, our main results read as follows.
Theorem 1.1 Suppose that \(N\geq3\), \(0\leq\mu<\mu_{1}\), \(0\leq s< p\), \(1< p< N\). If (A 1), (A 2) and hold, then problem (1.1) has at least one positive solution. Theorem 1.2 Remark 1.1
We extend the special case \(p=2\) in [13] to a more general situation \(1< p< N\). The author [14] obtained one positive solution for a special case of (1.1) with \(f(x,u)=\lambda\frac{|u|^{q-2}u}{|x|^{t}}\), where \(\lambda>0\), \(0\leq t< p\), \(\overline{q}< q< p^{\ast}(t)\) and
Remark 1.2
We prove Theorems 1.1 and 1.2 by critical point theory. Due to the lack of compactness of the embeddings in \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p^{\ast}}(\Omega)\), \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p}(\Omega,|x|^{-p}\,dx)\) and \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p^{\ast}(s)}(\Omega,|x|^{-s}\,dx)\), we cannot use the standard variational argument directly. The corresponding energy functional fails to satisfy the classical Palais-Smale ((PS) for short) condition in \(W^{1,p}_{0}(\Omega)\). However, a local (PS) condition can be established in a suitable range. Then the existence result is obtained via constructing a minimax level within this range and the mountain pass lemma due to Ambrosetti and Rabinowitz (see also [15]).
Notations
For the functional \(I: X\to\mathbb{R}\) (
X is a Banach space), we say that I satisfies the classical Palais-Smale ((PS) for short) condition if every sequence \(\{u_{n}\}\) in X such that \(I(u_{n})\) is bounded in X and \(I'(u_{n})\to0\) as \(n\to\infty\) contains a convergent subsequence. We say that I satisfies (PS) c
The rest of this paper is organized as follows. In Section 2, we establish some preliminary lemmas, which are useful in the proofs of our main results. In Section 3, we give detailed proofs of our main results.
Preliminaries
In what follows, we let \(\|\cdot\|_{p}\) denote the norm in \(L^{p}(\Omega)\). It is obvious that the values of \(f(x,u)\) for \(u<0\) are irrelevant in Theorem 1.1, and we may define
We shall firstly consider the existence of nontrivial solutions for the following problem:
The energy functional corresponding to (2.1) is given by
By the Hardy and Hardy-Sobolev inequalities (see [1, 16]) and (A
1), \(I\in C^{1}(W^{1,p}_{0}(\Omega),\mathbb{R})\). Now it is well known that there exists a one-to-one correspondence between the weak solutions of problem (2.1) and the critical points of I on \(W^{1,p}_{0}(\Omega)\). More precisely, we say that \(u\in W^{1,p}_{0}(\Omega)\) is a weak solution of problem (2.1) if for any \(v\in W^{1,p}_{0}(\Omega)\), there holds
Next, we shall give some lemmas which are needed in proving our main results.
Lemma 2.1
([17])
If \(f_{n}\rightarrow f\) a. e. in Ω and \(\|f_{n}\|_{p}\leq C<\infty\) for all n and some \(0< p<\infty\), then Lemma 2.2
([14])
Suppose \(1< p< N\), \(0\leq s< p\) and \(0\leq\mu<\mu_{1}\). Then the limiting problem has radially symmetric ground states and satisfies where \(\widetilde{U}_{p,\mu}(x)=\widetilde{U}_{p,\mu}(|x|)\) is the unique radial solution of (2.2) satisfying Moreover, \(\widetilde{U}_{p,\mu}\) has the following properties: where \(c_{1}\) and \(c_{2}\) are positive constants depending on p and N; \(a(\mu)\) and \(b(\mu)\) are zeroes of the function \(g(t)=(p-1)t^{p}-(N-p)t^{p-1}+\mu\) (\(t\geq0\), \(0\leq\mu<\mu_{1}\)) satisfying \(0\leq a(\mu)<\frac{N-p}{p}<b(\mu)<\frac{N-p}{p-1}\). Lemma 2.3 Assume that (A 1) and (A 2) hold. If \(c\in (0,\frac{p-s}{p(N-s)}A^{\frac{N-s}{p-s}}_{\mu,s} )\), then I satisfies (PS) c condition. Proof
Suppose that \(\{{u_{n}}\}\) is a (PS)
c I in \(W^{1,p}_{0}(\Omega)\), that is,
By (A
2), we have
where \(\theta=\min\{\rho_{0},p^{\ast}(s)\}\). Hence we conclude that \(\{u_{n}\}\) is a bounded sequence in \(W^{1,p}_{0}(\Omega)\). So there exists \(u\in W^{1,p}_{0}(\Omega)\) such that (going if necessary to a subsequence)
By the continuity of embedding, we have \(\|u_{n}\|_{p^{\ast}}^{p^{\ast}}\leq C_{1}<\infty\). Going if necessary to a subsequence, from [1] one can get that
as \(n\rightarrow\infty\). By (A
1), for any \(\varepsilon>0 \), there exists \(a(\varepsilon)>0\) such that
Set \(\delta:=\frac{\varepsilon}{2a(\varepsilon)}>0\). Let \(E\subset\Omega \) with \(\operatorname{meas}(E)<\delta=\frac{\varepsilon}{2a(\varepsilon)}\), it follows from the fact \(\|u_{n}\|_{p^{\ast}}^{p^{\ast}}\leq C_{1}\) that
It follows from the fact that \(f(x,u_{n})u_{n} \rightarrow f(x,u)u\) as \(n\to\infty\) a.e. in Ω and Vitali’s theorem that
Similarly, we can also get
Let \(v_{n}=u_{n}-u\). By the definition of \(\|\cdot\|\), we get
It follows from \(I'(u_{n})\rightarrow0\), \(\int_{\Omega }f(x,u_{n})u_{n}\,dx\rightarrow\int_{\Omega}f(x,u)u\, dx\) and the Brezis-Lieb lemma [17] that
From (2.3) and \(I'(u_{n})\rightarrow0\), we get
By \(I(u_{n})\rightarrow c\), \(\int_{\Omega}F(x,u_{n})\,dx\rightarrow\int_{\Omega}F(x,u)\,dx\), \(\lim_{n\rightarrow\infty}(\|u_{n}\|^{p}- \|v_{n}\|^{p})=\|u\|^{p}\) and the Brezis-Lieb lemma, we have
That is,
We claim that \(\|v_{n}\|^{p}\rightarrow0\) as \(n\rightarrow\infty\). Otherwise, there exists a subsequence (still denoted by \(v_{n}\)) such that
By the definition of \(A _{\mu,s}\) in (1.3), we have
It follows from (2.7) that \(k\geq A _{\mu,s}k^{\frac{p}{p^{\ast}(s)}}\), so we have \(k\geq A _{\mu,s}^{\frac{N-s}{p-s}}\), which together with (2.6) and \(c<\frac{p-s}{p(N-s)}A^{\frac{N-s}{p-s}}_{\mu,s}\) (see the assumption in Lemma 2.3) implies that
However, (2.5) implies that
it follows from the definition of
I, (1.2) and (A 2) that
So we get a contradiction. Therefore, we can obtain
From the discussion above,
I satisfies (PS) c
In the following, we shall give some estimates for the extremal functions. Define a function \(\varphi\in C_{0}^{\infty}(\Omega) \) such that \(\varphi(x)=1\) for \(|x|\leq R\), \(\varphi(x)=0\) for \(|x|\geq2R\), \(0\leq\varphi(x)\leq1\), where \(B_{2R}(0)\subset\Omega\). Set \(v_{\varepsilon}(x)=\varphi(x)\widetilde{V}_{\varepsilon }(x)\), \(\varepsilon>0\), where \(\widetilde{V}_{\varepsilon}(x)\) see the definition in Lemma 2.2. Then we can get the following results by the method used in [18]:
and
Lemma 2.4 Suppose that \(0\leq s< p\) and \(0\leq\mu<\mu_{1}\). If (A 1), (A 2) and (1.5) hold, then there exists \(u_{0}\in W^{1,p}_{0}(\Omega)\) with \(u_{0}\not\equiv0\) such that Proof
We consider the functions
Since \(\lim_{t\rightarrow\infty}g(t)=-\infty\), \(g(0)=0\) and \(g(t)>0\) for
t small enough, \(\sup_{t\geq0}g(t) \) is attained for some \(t_{\varepsilon}>0\). Therefore, we have
hence
By (A
1), we can easily get
Hence, together with \(t_{\varepsilon}\leq C\), we can get
On the other hand, the function \(\overline{g}(t)\) attains its maximum at \(t_{\varepsilon}^{0}\) and is increasing in the interval \([0,t_{\varepsilon}^{0}]\), together with (2.8), (2.9) and (2.11) and \(F(x,t)\geq C_{5}|t|^{\rho_{0}}\) which is directly got from (A
2), we deduce
Note that \(b(\mu)>\frac{N-p}{p}\) implies
By (1.5), we have \(\rho_{0}>\frac{p [2N-p-b(\mu)p ]}{N-p}\), which implies
and
Therefore, by choosing
ε small enough, we have
Hence the proof of this lemma is then completed by taking \(u_{0}=v_{\varepsilon}\). □
Proofs of our main results Proof of Theorem 1.1
From the Sobolev and Hardy-Sobolev inequalities, we can easily get
The condition (A
1) implies that for any \(\varepsilon>0\) there exist \(\delta_{2}>\delta_{1}>0\) such that
and
Therefore, there exists a constant \(C_{\varepsilon}>0\) such that
Then one gets
for \(\varepsilon>0\) small enough. So there exists \(\beta>0\) such that
By Lemma 2.4, there exists \(u_{0}\in W^{1,p}_{0}(\Omega)\) with \(u_{0}\not\equiv0\) such that
It follows from the nonnegativity of \(F(x,t)\) that
therefore, \(\lim_{t\rightarrow+\infty}I(tu_{0})\rightarrow-\infty\). Hence, we can choose \(t_{0}>0\) such that
By virtue of the mountain pass lemma in [19], there is a sequence \(\{u_{n}\}\subset W^{1,p}_{0}(\Omega)\) satisfying
where
and
Note that
By Lemma 2.3, we can assume that \(u_{n}\rightarrow u\) in \(W^{1,p}_{0}(\Omega)\). From the continuity of \(I'\), we know that
u is a weak solution of problem (2.1). Then \(\langle I'(u),u^{-}\rangle=0\), where \(u^{-}=\min\{u,0\}\). Thus \(u\geq0\). Therefore u is a nonnegative solution of (1.1). By the strong maximum principle, u is a positive solution of problem (1.1), so Theorem 1.1 holds. □ Proof of Theorem 1.2
has at least one positive solution
v. Let \(u_{2}=-v\), then \(u_{2}\) is a solution of References 1.
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Kang, DS: On the quasilinear elliptic problems with critical Sobolev-Hardy exponents and Hardy terms. Nonlinear Anal.
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Acknowledgements
The author thanks the referees and the editors for their helpful comments and suggestions. Research was supported by the National Natural Science Foundation of China (No. 11401011).
Additional information Competing interests
The author declares that he has no competing interests. |
Let $X$ be an affine, complex variety, $A$ be a $\mathbb{C}$-algebra (not necessarily noetherian) and $F_A$ is a coherent sheaf over $X \times \mbox{Spec}(A)$, flat over $\mbox{Spec}(A)$. Denote by $Y \subset X \times \mbox{Spec}(A)$ the scheme-theoretic support of $F_A$. Then, does there exist a non-empty open subset $U \subset \mbox{Spec}(A)$ such that $Y \cap (X \times U)$ is flat over $U$? This is true if $A$ does not have any nilpotent elements.
Let $A$ be a non-reduced, Artinian ring with maximal ideal $\mathfrak{m}$. The underlying topological space of $\text{Spec}\ A$ is a one-point space. Thus, the unique nonempty open is the entire space $\text{Spec}\ A$. Thus, the problem asks, for every $A$-flat coherent sheaf on a finite type $A$-scheme, whether the scheme-theoretic support of the coherent sheaf is also $A$-flat. That is
not true.
Consider the $A$-algebra $\widetilde{B}:= A\epsilon \times A\eta $ that is free of rank $2$ as an $A$-module, i.e., $$\widetilde{B}=A[\epsilon,\eta]/\langle\ \epsilon+\eta-1,\ \epsilon\eta\ \rangle= A\epsilon \oplus A\eta.$$ For every ideal $H\subset A$, consider the $A$-subalgebra of $\widetilde{B}$, $$B_H =\{\ a\epsilon + b\eta\in \widetilde{B} \ |\ a-b\in H\ \}.$$ Finally, consider the following $B_H$-module, $$M := B_H/\left( \langle \epsilon \rangle \cap B_H \right) \oplus B_H/\left( \langle \epsilon \rangle \cap B_H \right).$$
Proposition. 1. For every nonzero ideal $H\subset \mathfrak{m}$, the $A$-module $B_H$ is not flat. 2. The $A$-module $M$ is flat of rank $2$. 3. Also, the $B_H$-annihilator of $M$ equals the zero ideal. Thus, on the scheme $Y=\text{Spec}\ B_H$, the scheme-theoretic support of the $A$-flat coherent sheaf $\widetilde{M}$ equals $Y$, and $Y$ is not $A$-flat. Proof. The $A$-module $B_H$ equals $A\cdot 1 \oplus H\cdot \epsilon$. Since $H$ is not $A$-flat, also $B_H$ is not $A$-flat.
As an $A$-algebra, each of the following quotient $\widetilde{B}$-algebras is isomorphic to $A$ itself, $$\widetilde{B}/\langle \epsilon \rangle, \ \ \widetilde{B}/\langle \eta \rangle.$$ Since the $A$-subalgebra $A\cdot 1$ is contained in $B_H$, also the quotient $B_H$-algebras also equal $A$, $$B_H/\langle \epsilon \rangle \cap B_H = A = B_H/\langle \eta \rangle \cap B_H.$$ Thus, the $A$-module $M$ is free of rank $2$.
Finally, the common intersection $\langle \epsilon \rangle \cap \langle \eta \rangle$ in $B$ is the zero ideal. Thus, the common intersection in $B_H$ is also the zero ideal. Therefore the annihilator ideal of $M$ is the zero ideal. Thus, the scheme-theoretic support of $M$ equals all of $\text{Spec}\ B_H$.
QED |
Abstract
Lingua originale English pagine (da-a) 397-410 Numero di pagine 14 Rivista Dynamic Systems and Applications Volume 22 Stato di pubblicazione Published - 2013 Fingerprint All Science Journal Classification (ASJC) codes Mathematics(all) Cita questo Dynamic Systems and Applications, 22, 397-410.
In: Dynamic Systems and Applications, Vol. 22, 2013, pag. 397-410.
Variational versus pseudomonotone operator approach in parameter-dependent nonlinear elliptic problems. / Livrea, Roberto; Candito, Pasquale; Carl; Livrea, Roberto.
Risultato della ricerca: Article
Dynamic Systems and Applications, vol. 22, pagg. 397-410.
}
TY - JOUR
T1 - Variational versus pseudomonotone operator approach in parameter-dependent nonlinear elliptic problems
AU - Livrea, Roberto
AU - Candito, Pasquale
AU - Carl, null
AU - Livrea, Roberto
PY - 2013
Y1 - 2013
N2 - We study the existence of nontrivial solutions of parameter-dependent quasilinear elliptic Dirichlet problems of the form $-\Delta u = \lambda f(u)$ in $\Omega$, $u = 0$ on $\partial\Omega$, in a bounded domain $\Omega$ with sufficiently smooth boundary, where $\lambda$ is a real parameter and $\Delta_p$ denotes the p-Laplacian. Recently the authors obtained multiplicity results by employing an abstract localization principle of critical points of functional of the form $\Phi-\lambda\Psi$ on open subleveis of $\Phi$, i.e., of sets of the form $\Phi^{-1}(-\infty,r)$, combined with differential inequality techniques and topological arguments. Unlike in those recent papers by the authors, the approach in this paper is based on pseudomonotone operator theory and fixed point techniques. The obtained results are compared with those obtained via the abstract variational principle. Moreover, by applying truncation techniques and regularity results we are able to deal with elliptic problems that involve discontinuous nonlinearities without making use of nonsmooth analysis methods. ©Dynamic Publishers, Inc.
AB - We study the existence of nontrivial solutions of parameter-dependent quasilinear elliptic Dirichlet problems of the form $-\Delta u = \lambda f(u)$ in $\Omega$, $u = 0$ on $\partial\Omega$, in a bounded domain $\Omega$ with sufficiently smooth boundary, where $\lambda$ is a real parameter and $\Delta_p$ denotes the p-Laplacian. Recently the authors obtained multiplicity results by employing an abstract localization principle of critical points of functional of the form $\Phi-\lambda\Psi$ on open subleveis of $\Phi$, i.e., of sets of the form $\Phi^{-1}(-\infty,r)$, combined with differential inequality techniques and topological arguments. Unlike in those recent papers by the authors, the approach in this paper is based on pseudomonotone operator theory and fixed point techniques. The obtained results are compared with those obtained via the abstract variational principle. Moreover, by applying truncation techniques and regularity results we are able to deal with elliptic problems that involve discontinuous nonlinearities without making use of nonsmooth analysis methods. ©Dynamic Publishers, Inc.
KW - p-Laplacian, Dirichlet problem
UR - http://hdl.handle.net/10447/258581
M3 - Article
VL - 22
SP - 397
EP - 410
JO - Dynamic Systems and Applications
JF - Dynamic Systems and Applications
SN - 1056-2176
ER - |
Does this definite integral have a closed-form expression?
\begin{align*} I &= \int_0^\infty \sqrt{ \frac{1}{2} \frac{1}{x} \left( \frac{1}{(1+x)^2} + \frac{z}{(1+xz)^2} \right) } \, dx \\ &= \frac{1}{\sqrt{2}} \int_0^\infty \frac{1}{x} \sqrt{ \frac{x}{1+x} \left( 1-\frac{x}{1+x} \right) + \frac{xz}{1+xz} \left( 1-\frac{xz}{1+xz} \right) } \, dx \end{align*}
where $x>=0$, and $z>=0$ is constant.
I have tried the following substitution:
$$ u = \frac{xz}{1+xz}, \quad x = \frac{1}{z} \frac{u}{(1-u)}, \quad dx = \frac{1}{z} \frac{1}{(1-u)^2} du $$
$$ I = \frac{1}{\sqrt{2}} \int_0^1 \frac{1}{\sqrt{u(1-u)}} \sqrt{ 1 + \frac{z}{((1-u)z+u)^2} } \, du $$
...but I can't seem to make any more progress. Using $u=\frac{x}{1+x}$ yields a similar expression which is also difficult to simplify.
Can you find a way to simplify this further?
Update (Nov 4, 2013): The integral can be rearranged to the following form, which might make it easier to match something from a table of known integrals:
$$ I = \frac{1}{\sqrt{2}} \sqrt{\frac{1{+}z}{z}} \int_0^1 u^{-\frac{1}{2}} (1{-}u)^{-\frac{1}{2}} \left( 1+\frac{1{-}z}{z} u \right)^{-1} \left( 1 + 2\frac{1{-}z}{1{+}z}u + \frac{(1{-}z)^2}{z(1{+}z)}u^2 \right)^{\frac{1}{2}} \, du $$ |
Cleve Moler is the author of the first MATLAB, one of the founders of MathWorks, and is currently Chief Mathematician at the company. He is the author of two books about MATLAB that are available online. He writes here about MATLAB, scientific computing and interesting mathematics.
Denormal floating point numbers and gradual underflow are an underappreciated feature of the IEEE floating point standard. Double precision denormals are so tiny that they are rarely numerically significant, but single precision denormals can be in the range where they affect some otherwise unremarkable computations. Historically, gradual underflow proved to be very controversial during the committee deliberations that developed the standard.
My previous post was mostly about normalized floating point numbers. Recall that normalized numbers can be expressed as$$ x = \pm (1 + f) \cdot 2^e $$The fraction or mantissa $f$ satisfies$$ 0 \leq f < 1 $$$f$ must be representable in binary using at most 52 bits for double precision and 23 bits for single precision.The exponent $e$ is an integer in the interval$$ -e_{max} < e \leq e_{max} $$where $e_{max} = 1023$ for double precision and $e_{max} = 127$ for single precision.The finiteness of $f$ is a limitation on precision. The finiteness of $e$ is a limitation on range. Any numbers that don't meet these limitations must be approximated by ones that do.
Floating Point Format
Double precision floating point numbers are stored in a 64-bit word, with 52 bits for $f$, 11 bits for $e$, and 1 bit for the sign of the number. The sign of $e$ is accommodated by storing $e+e_{max}$, which is between $1$ and $2^{11}-2$.Single precision floating point numbers are stored in a 32-bit word, with 23 bits for $f$, 8 bits for $e$, and 1 bit for the sign of the number. The sign of $e$ is accommodated by storing $e+e_{max}$, which is between $1$ and $2^{8}-2$.The two extreme values of the exponent field, all zeroes and all ones, are special cases. All zeroes signifies a denormal floating point number, the subject of today's post. All ones, together with a zero fraction, denotes infinity, or Inf. And all ones, together with a a nonzero fraction, denotes Not-A-Number, or NaN.
floatgui
My program floatgui, available here, shows the distribution of the positive numbers in a model floating point system with variable parameters. The parameter $t$ specifies the number of bits used to store $f$, so $2^t f$ is an integer. The parameters $e_{min}$ and $e_{max}$ specify the range of the exponent.
Gap Around Zero
If you look carefully at the output from floatgui shown in the previous post you will see a gap around zero. This is especially apparent in the logarithmic plot, because the logarithmic distribution can never reach zero.Here is output for slightly different parameters, $t = 3$, $e_{min} = -5$, and $e_{max} = 2$. Howrever, the gap around zero has been filled in with a spot of green. Those are the denormals.
Zoom In
Zoom in on the portion of these toy floating point numbers less than one-half. Now you can see the individual green denormals -- there are eight of them in this case.
Denormal Floating Point Numbers
Denormal floating point numbers are essentially roundoff errors in normalized numbers near the underflow limit, realmin, which is $2^{-e_{max}+1}$. They are equally spaced, with a spacing of eps*realmin. Zero is naturally included as the smallest denormal.Suppose that x and y are two distinct floating point numbers near to, but larger than, realmin. It may well be that their difference, x - y, is smaller than realmin. For example, in the small floatgui system pictured above, eps = 1/8 and realmin = 1/32. The quantities x = 6/128 and y = 5/128 are between 1/32 and 1/16, so they are both above underflow. But x - y = 1/128 underflows to produce one of the green denormals.Before the IEEE standard, and even on today's systems that do not comply with the standard, underflows would simply be set to zero. So it would be possible to have the MATLAB expression
x == y
be false, while the expression
x - y == 0
be true.On machines where underflow flushes to zero and division by zero is fatal, this code fragment can produce a division by zero and crash.
if x ~= yz = 1/(x-y);end
Of course, denormals can also be produced by multiplications and divisions that produce a result between eps*realmin and realmin. In decimal these ranges are
format compactformat shorte[eps*realmin realmin][eps('single')*realmin('single') realmin('single')]
ans =4.9407e-324 2.2251e-308ans =1.4013e-45 1.1755e-38
Denormal Format
Denormal floating point numbers are stored without the implicit leading one bit,$$ x = \pm f \cdot 2^{-emax+1}$$The fraction $f$ satisfies$$ 0 \leq f < 1 $$And $f$ is represented in binary using 52 bits for double precision and 23 bits for single recision. Note that zero naturally occurs as a denormal.When you look at a double precision denormal with format hex the situation is fairly clear. The rightmost 13 hex characters are the 52 bits of the fraction. The leading bit is the sign. The other 12 bits of the first three hex characters are all zero because they represent the biased exponent, which is zero because $emax$ and the exponent bias were chosen to complement each other. Here are the two largest and two smallest nonzero double precision denormals.
format hex[(1-eps); (1-2*eps); 2*eps; eps; 0]*realminformat longeans
ans =000fffffffffffff000ffffffffffffe000000000000000200000000000000010000000000000000ans =2.225073858507201e-3082.225073858507200e-3089.881312916824931e-3244.940656458412465e-3240
The situation is slightly more complicated with single precision because 23 is not a multiple of four. The fraction and exponent fields of a single precision floating point number -- normal or denormal -- share the bits in the third character of the hex display; the biased exponent gets one bit and the first three bits of the 23 bit fraction get the other three. Here are the two largest and two smallest nonzero single precision denormals.
format hexe = eps('single');r = realmin('single');[(1-e); (1-2*e); 2*e; e; 0]*rformat shorteans
ans =007fffff007ffffe000000020000000100000000ans =1.1755e-381.1755e-382.8026e-451.4013e-450
Think of the situation this way. The normalized floating point numbers immediately to the right of realmin, between realmin and 2*realmin, are equally spaced with a spacing of eps*realmin. If just as many numbers, with just the same spacing, are placed to the left of realmin, they fill in the gap between there and zero. These are the denormals. They require a slightly different format to represent and slightly different hardware to process.
IEEE Floating Point Committee
The IEEE Floating Point Committee was formed in 1977 in Silicon Valley. The participants included representatives of semiconductor manufacturers who were developing the chips that were to become the basis for the personal computers that are so familiar today. As I said in my previous post, the committee was a remarkable case of cooperation among competitors.Velvel Kahan was the most prominent figure in the committee meetings. He was not only a professor of math and computer science from UC Berkeley, he was also a consultant to Intel and involved in the design of their math coprocessor, the 8087. Some of Velvel's students, not only from the campus, but also who had graduated and were now working for some of the participating companies, were involved.A proposed standard, written by Kahan, one of his students at Berkeley, Jerome Coonen, and a visiting professor at Berkeley, Harold Stone, known as the KCS draft, reflected the Intel design and was the basis for much of the committee's work.The committee met frequently, usually in the evening in conference rooms of companies on the San Francisco peninsula. There were also meetings in Austin, Texas, and on the East Coast. The meetings often lasted until well after midnight.Membership was based on regular attendance. I was personally involved only when I was visiting Stanford, so I was not an official member. But I do remember telling a colleague from Sweden who was coming to the western United States for the first time that there were three sites that he had to be sure to see: the Grand Canyon, Las Vegas, and the IEEE Floating Point Committee.
Controversy
The denormals in the KCS draft were something new to most of the committee. Velvel said he had experimented with them at the University of Toronto, but that was all. Standards efforts are intended to regularize existing practice, not introduce new designs. Besides, implementing denormals would reguire additional hardware, and additional transistors were a valuable resource in emerging designs. Some experts claimed that including denormals would slow down all floating point arithmetic.A mathematician from DEC, Mary Payne, led the opposition to KCS. DEC wanted a looser standard that would embrace the floating point that was already available on their VAX. The VAX format was similar to, but not the same as, the KCS proposal. And it did not include denormals.Discussion went on for a couple of years. Letters of support for KCS from Don Knuth and Jim Wilkinson did not settle the matter. Finally, DEC engaged G. W. (Pete) Stewart, from the University of Maryland. In what must have been a surprise to DEC, Pete also said he thought that the KCS proposal was a good idea. Eventually the entire committee voted to accept a revised version.
Denormals Today
Denormal floating point numbers are still the unwanted step children in today's floating point family. I think it is fair to say that the numerical analysis community has failed to make a strong argument for their importance. It is true that they do make some floating point error analyses more elegant. But with magnitudes around $10^{-308}$ the double precision denormals are hardly ever numerically significant in actual computation. Only the single precision denormals around $10^{-38}$ are potentially important.Outside of MATLAB itself, we encounter processors that have IEEE format for the floating point numbers, but do not conform to the 754 standard when it comes to processing. These processors usually flush underflow to zero and so we can expect different numerical results for any calculations that might ordinarily produce denormals.We still see processors today that handle denormals with microcode or software. Execution time of MATLAB programs that encounter denormals can degrade significantly on such processors.The Wikipedia page on denormals has the macros for setting trap handlers to flush underflows to zero in C or Java programs. I hate to think what might happen to MATLAB Mex files with such macros. Kids, don't try this at home. |
Suppose that $p$ is a prime with $p \equiv 7 \pmod 8$. If $t = \frac{p - 1}{2}$ , prove that $$2^t \equiv 1 \pmod p.$$
Any hints will be appreciated. Thanks so much.
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Obviously $p$ is odd.
It is well known that $$\left(\dfrac 2p\right)=(-1)^{\frac{p^2-1}{8}}$$ where $\left(\dfrac ap\right)$ is the Legendre's symbol.
On the other hand $$p=8m+7\Rightarrow p^2-1=64m^2+112m+48\Rightarrow\frac{p^2-1}{8}=8m^2+14m+6\in 2\Bbb Z$$
It follows (
because $\frac{p^2-1}{8}$ is even) $$\left(\dfrac 2p\right)= 1$$ which means that $2$ is a square modulo $p$.
Thus $$\left(2\right)^{\frac{p-1}{2}}=\left(x^2\right)^{\frac{p-1}{2}}=x^{p-1}\equiv 1\pmod p$$
Let $\alpha$ denote a primitive $8$th root of unity in an algebraic closure of $\mathbb F_p$. The element $y=\alpha+\alpha^{-1}$ verifies $y^2=2$ for $y^4=-1$ hence $\alpha^2+\alpha^{-2}=0$. Consequently, $2^{(p-1)/2}=y^{p-1}$. Note also that $y^p=\alpha^p+\alpha^{-p}$. If $p\equiv -1\pmod 8$ then $y^p=y$, hence $2^{(p-1)/2}=y^{p-1}=1$. |
$\newcommand{\HH}{\mathbb{H}}$Here is an expansion of what Anton is saying.
Suppose that $M$ is a closed hyperbolic three-manifold. It follows that the universal cover of $M$ is $\HH^3$: hyperbolic space. The covering map of $M$ comes with a deck group - namely there is an action of $\pi_1(M)$ on $\HH^3$ so that the quotient $\pi_1(M) \backslash \HH^3$ is homeomorphic to $M$.
Now fix a set $S$ of generators for $\pi_1(M)$. Let $\Gamma = \Gamma(M, S)$ be the Cayley graph for $\pi_1(M)$ relative to $S$. Also, fix a point $x$ of $\HH^3$. Let $\rho_x{:}\,\Gamma \to \HH^3$ be the orbit map: namely $\rho_x(g) = g \cdot x$ for vertices of $\Gamma$, and all edges of $\Gamma$ are sent to geodesics. The Švarc-Milnor lemma says that $\rho_x$ is a quasi-isometry.
It follows that the Cayley graphs for any pair of closed hyperbolic three-manifolds are quasi-isometric. Thus homeomorphism types cannot be distinguished this way. |
It's not just about integrating over wavelengths, it's also about integrating over space.
There are quite a few steps to get from the $L$ you have to the $R$ and I'm stuck on one integral's derivation but here goes:
The radiance shown here is the $\textit{spectral radiance}$ for a $\textit{perfect}$ blackbody ($\epsilon=1$) so let's start with that.
$L(\lambda)=\frac{2hc^2}{\lambda^5}\frac{1}{\exp({\frac{hc}{\lambda k T}})-1}$
Using the definition of spectral radiance, what the equation above gives you is the amount of radiation emitted by a surface per unit wavelength, per unit projected area and per unit solid angle.
To get the 'complete' radiance, i.e. radiance across all wavelengths you need to take the integral
$L=\int_{0}^{+\infty}L(\lambda)d\lambda$
With a change of variable $x=\frac{hc}{\lambda kT}$ you get to:
$L=\frac{2k^4T^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}dx$
Noting that both $x$ and the integral are dimensionless, we can do a dimension check:
$[L]=\frac{J^4K^{-4}K^4}{J^3s^3m^2s^{-2}}=Jm^{-2}s^{-1}=Wm^{-2}$
(note that steradian is dimensionless so this makes sense).
In general, radiance can be angle dependent but a blackbody is a Lambertian surface, which means by definition that radiance is direction independent.The radiance itself is direction-independent. However, to relate to exitance you need to take into account the fact that radiation transfer between two surfaces is very much angle dependent.
The radiance $L$ is defined as:
$L\triangleq \frac{d^2\Phi}{\cos\theta dA d\Omega}$
Where $d \Omega$ is the solid angle subtended by the infinitesimal 'receiving' object.$dA$ is the area of the infinitesimal surface emitting the flux$\theta$ is the angle made by the normal to the emitting surface and the direction of propagation.
Note that $d\Omega$ can also be written$d\Omega=\frac{\cos(\theta_s)dA_s}{r^2}$
where $\theta_s$ is the angle made by the normal to the receiving surface and the direction of propagation.
In other words, the flux exchanged from the infinitesimal emitting surface of area $dA$ and a receiving infinitesimal surface of area $dA_s$
$d^2\phi=L\frac{\cos\theta \cos\theta_s dA dA_s}{r^2}$
This equation is symetric in angles and areas which makes sense: If what we called the receiving surface has a radiance $L_s=L$ then we should expect just as much flux going in one direction and in the other direction.
Exitance is by definition the total radiant flux coming out of the surface. To relate radiance to exitance you need to calculate how much radiant flux is transmitted from an infinitesimal surface into a hemisphere 'above' it.
Many people think this means $R=2\pi L$ since the solid angle of a hemisphere is $2\pi$. This is however incorrect because it fails to take into account the direction of propagation i.e. $\theta$.
The radiation exchange is more efficient right above the emitting surface (where $\cos\theta=1$) than it is 'on the extreme side' (where $\cos\theta =0$).
$R\triangleq\iint_{\text{hemisphere}}\frac{L}{r^2}\cos\theta dA_s$
Parametrize the infinitesimal area $dA_s$ using spherical coordinates:$dA_s=r^2\sin\theta d\theta d\phi$
$R=\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{2\pi}L\cos\theta sin\theta d\theta d\phi$
Using
$\sin\theta\cos\theta=\frac{\sin2\theta}{2}$,
you get to
$R=L\pi$
and finally
$R=\frac{2\pi k^4T^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}dx$
Writing
$\sigma=\frac{2\pi k^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}$
You get for a perfect blackbody$R=\sigma T^4$
I'm not good enough at integration to solve the integral above but evidently the limit is $\frac{\pi^4}{15}$ (you can check that with Wolfram Alpha or Mathematica) and you end up with
$\sigma=\frac{2\pi^5 k^4}{15h^3c^2}$ which is indeed $5.67*10^{-8}Wm^{-2}K^{-4}$
Edit, bonus: you can remember the SF constant with '5-6-7-8'.. 'amaze your friends'.
Now if you have a gray body of emittance $\epsilon$ the radiance changes to:
$L(\lambda)=\epsilon \frac{2hc^2}{\lambda^5}\frac{1}{\exp({\frac{hc}{\lambda k T}})-1}$
Using the same derivation, you get:
$R=\sigma \epsilon T^4$
Note that an implicit assumption in the equations you gave is that
$\epsilon(\lambda)=\epsilon \forall \lambda$ i.e. spectral emissivity is independent of wavelength. Not anyways true.. |
If the natural density of $A = \{a_i\}$ exists, then we can show that it must be zero.
Let $\displaystyle S_{n} = \frac{|A \cup [1,n]|}{n}$
Now $\displaystyle \{\frac{n}{a_n}\}$ is a subsequence of $S_{n}$ and so if the limit is $\displaystyle 2\delta > 0 $ then we have that for all $\displaystyle n > N_0$, $\displaystyle \frac{n}{a_n} > \delta$ and so $\displaystyle \frac{1}{a_n} > \frac{\delta}{n}$ for all $\displaystyle n > N_0$ and so $\displaystyle \sum \frac{1}{a_n}$ diverges.
The main problem is actually showing that the limit exists.
It is easy to show that $\liminf$ is zero: If the limit was $\displaystyle 2\delta > 0$ then for all $n > N_{0}$, $S_{n} > \delta$ and an argument similar to above works.
Now suppose $\displaystyle \limsup S_n = 2\delta > 0$. Then there is a subsequence $\displaystyle S_{N_1}, S_{N_2}, ..., S_{N_k}, \dots $ which converges to $\displaystyle 2\delta$.
Now we can choose the subsequence so that $\displaystyle S_{N_i} > \delta$ and $\displaystyle N_{k+1} > \frac{2N_{k}}{\delta}$
Now the number of elements of $\displaystyle A$ in the interval $\displaystyle (N_{k}, N_{k+1}]$ is atleast $\displaystyle \delta N_{k+1} - N_k \ge \delta N_{k+1} - \frac{\delta N_{k+1}}{2} \ge \frac{\delta N_{k+1}}{2}$ and so the sum of reciprocals in that interval is atleast $\displaystyle \frac{\delta N_{k+1}}{2} \frac{1}{N_{k+1}} = \displaystyle \frac{\delta}{2}$
And so the sum of reciprocals must diverge.
Hence $\displaystyle \limsup S_n = 0 = \liminf S_n$ and thus $\displaystyle \lim S_n = 0$ and thus the natural density is zero. |
In general, one extracts a manifold invariant from a TQFT by interpreting the closed manifold as a bordism from the empty set to the empty set. The TQFT sends this bordism to a homomorphism of the ground field, which is a number. Such invariants are always multiplicative under disjoint union, this is a consequence of the TQFT being a
monoidal functor:$$\mathcal{Z}(M_1 \sqcup M_2) = \mathcal{Z}(M_1) \otimes \mathcal{Z}(M_2) = \mathcal{Z}(M_1) \cdot \mathcal{Z}(M_2)$$
Some TQFTs, like the Crane-Yetter invariant (but not, say, the Turaev-Viro model) give manifold invariants that are multiplicative under connected sum $\#$.
One way to see this is to notice that they can be defined (for connected manifolds) with Kirby calculus: Given a manifold, choose a handle decomposition and consider its link diagram. The diagram is then labelled with morphisms from a ribbon fusion category and the whole diagram is evaluated as a morphism from the monoidal identity to itself, again a number. Now the evaluation of the disjoint union of two link diagrams must then give the product of the evaluations of the respective diagrams, since a ribbon fusion category is monoidal. But the disjoint union of two link diagrams of manifolds $M_1$ and $M_2$ is the link diagram of the connected sum $M_1 \# M_2$!
Which leads me to believe that this multiplicativity secretly comes from monoidality of some functor again. Is there a category of bordisms where the monoidal product of morphisms (=bordisms) is the connected sum, and not the disjoint union? Are such TQFTs actually monoidal functors from this bordism category to $\mathrm{Vect}$?
A related, noncategorical question was asked here: Monoid structure of oriented manifolds with connect sum |
I have been considering a generalisation of the cosmological process involved in computing the temperature of the cosmic neutrino background.
It is well-known that (simplistically) this temperature $T_{\nu}$ differs from the temperature of the CMB $T_{\gamma}$ by a factor of $(4/11)^{1/3}$ due to the fact that shortly after neutrino-decoupling at 0.8 MeV, electrons and positrons annihilate and
dump energy to the photon plasma which is still at equilibrium, while leaving the temperature of the decoupled neutrinos constant, as they no longer interact with the plasma.
I was wondering what would happen if the neutrinos decoupled
before any particles became non-relativistic ($T > 100$ GeV) such that they had to go through all of the 'annihilation' processes as follows: top quarks, Higgs + W + Z, bottom quarks, charm quarks, tau leptons, (QCD phase transition), pions, muons, and electrons. Specifically, I am wondering what happens for the annihilation processes which are not of the form
\begin{equation} X^+ + X^- \rightarrow \gamma \gamma, \end{equation}
such as the annihilation of Higgs, W bosons and Z bosons, and what happens at the QCD phase transition (between tau annihilation and pion annihilation). These processes
do change the number of relativistic degrees of freedom in entropy, so if we just used the conservation of entropy to compute the temperature we would treat them the same way as the other processes. However do these processes actually dump energy to the plasma? I can't see why they would. |
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