text:
Let ${n}$ be a large integer, and let ${M_n}$ be the Gaussian Unitary Ensemble (GUE), i.e. the random Hermitian matrix with probability distribution

$\displaystyle C_n e^{-\hbox{tr}(M_n^2)/2} dM_n$

where ${dM_n}$ is a Haar measure on Hermitian matrices and ${C_n}$ is the normalisation constant required to make the distribution of unit mass. The eigenvalues ${\lambda_1 < \ldots < \lambda_n}$ of this matrix are then a coupled family of ${n}$ real random variables. For any ${1 \leq k \leq n}$, we can define the ${k}$-point correlation function ${\rho_k( x_1,\ldots,x_k )}$ to be the unique symmetric measure on ${{\bf R}^k}$ such that

$\displaystyle \int_{{\bf R}^k} F(x_1,\ldots,x_k) \rho_k(x_1,\ldots,x_k) = {\bf E} \sum_{1 \leq i_1 < \ldots < i_k \leq n} F( \lambda_{i_1}, \ldots, \lambda_{i_k} ).$

A standard computation (given for instance in these lecture notes of mine) gives the Ginebre formula

$\displaystyle \rho_n(x_1,\ldots,x_n) = C'_n (\prod_{1 \leq i < j \leq n} |x_i-x_j|^2) e^{-\sum_{j=1}^n |x_j|^2/2}.$

for the ${n}$-point correlation function, where ${C'_n}$ is another normalisation constant. Using Vandermonde determinants, one can rewrite this expression in determinantal form as

$\displaystyle \rho_n(x_1,\ldots,x_n) = C''_n \det(K_n(x_i,x_j))_{1 \leq i, j \leq n}$

where the kernel ${K_n}$ is given by

$\displaystyle K_n(x,y) := \sum_{k=0}^{n-1} \phi_k(x) \phi_k(y)$

where ${\phi_k(x) := P_k(x) e^{-x^2/4}}$ and ${P_0, P_1, \ldots}$ are the (${L^2}$-normalised) Hermite polynomials (thus the ${\phi_k}$ are an orthonormal family, with each ${P_k}$ being a polynomial of degree ${k}$). Integrating out one or more of the variables, one is led to the Gaudin-Mehta formula

$\displaystyle \rho_k(x_1,\ldots,x_k) = \det(K_n(x_i,x_j))_{1 \leq i, j \leq k}. \ \ \ \ \ (1)$

(In particular, the normalisation constant ${C''_n}$ in the previous formula turns out to simply be equal to ${1}$.) Again, see these lecture notes for details.

The functions ${\phi_k(x)}$ can be viewed as an orthonormal basis of eigenfunctions for the harmonic oscillator operator

$\displaystyle L \phi := (-\frac{d^2}{dx^2} + \frac{x^2}{4})\phi;$

indeed it is a classical fact that

$\displaystyle L \phi_k = (k + \frac{1}{2}) \phi_k.$

As such, the kernel ${K_n}$ can be viewed as the integral kernel of the spectral projection operator ${1_{(-\infty,n+\frac{1}{2}]}(L)}$.

From (1) we see that the fine-scale structure of the eigenvalues of GUE are controlled by the asymptotics of ${K_n}$ as ${n \rightarrow \infty}$. The two main asymptotics of interest are given by the following lemmas:

Lemma 1 (Asymptotics of ${K_n}$ in the bulk) Let ${x_0 \in (-2,2)}$, and let ${\rho_{sc}(x_0) := \frac{1}{2\pi} (4-x_0^2)^{1/2}_+}$ be the semicircular law density at ${x_0}$. Then, we have

$\displaystyle K_n( x_0 \sqrt{n} + \frac{y}{\sqrt{n} \rho_{sc}(x_0)}, x_0 \sqrt{n} + \frac{z}{\sqrt{n} \rho_{sc}(x_0)} )$

$\displaystyle \rightarrow \frac{\sin(\pi(y-z))}{\pi(y-z)} \ \ \ \ \ (2)$

as ${n \rightarrow \infty}$ for any fixed ${y,z \in {\bf R}}$ (removing the singularity at ${y=z}$ in the usual manner).

Lemma 2 (Asymptotics of ${K_n}$ at the edge) We have

$\displaystyle K_n( 2\sqrt{n} + \frac{y}{n^{1/6}}, 2\sqrt{n} + \frac{z}{n^{1/6}} )$

$\displaystyle \rightarrow \frac{Ai(y) Ai'(z) - Ai'(y) Ai(z)}{y-z} \ \ \ \ \ (3)$

as ${n \rightarrow \infty}$ for any fixed ${y,z \in {\bf R}}$, where ${Ai}$ is the Airy function

$\displaystyle Ai(x) := \frac{1}{\pi} \int_0^\infty \cos( \frac{t^3}{3} + tx )\ dt$

and again removing the singularity at ${y=z}$ in the usual manner.

The proof of these asymptotics usually proceeds via computing the asymptotics of Hermite polynomials, together with the Christoffel-Darboux formula; this is for instance the approach taken in the previous notes. However, there is a slightly different approach that is closer in spirit to the methods of semi-classical analysis, which was briefly mentioned in the previous notes but not elaborated upon. For sake of completeness, I am recording some notes on this approach here, although to focus on the main ideas I will not be completely rigorous in the derivation (ignoring issues such as convegence of integrals or of operators, or (removable) singularities in kernels caused by zeroes in the denominator).

— 1. The bulk asymptotics —

We begin with the bulk asymptotics, Lemma 1. Fix ${x_0}$ in the bulk region ${(-2,2)}$. Applying the change of variables

$\displaystyle x = x_0 \sqrt{n} + \frac{y}{\sqrt{n} \rho_{sc}(x_0)}$

we see that the harmonic oscillator ${L}$ becomes

$\displaystyle - n \rho_{sc}(x_0)^2 \frac{d^2}{dy^2} + \frac{1}{4} (x_0 \sqrt{n} + \frac{y}{\sqrt{n} \rho_{sc}(x_0)})^2$

Since ${K_n}$ is the integral kernel of the spectral projection to the region ${L \leq n+\frac{1}{2}}$, we conclude that the left-hand side of (2) (as a function of ${y,z}$) is the integral kernel of the spectral projection to the region

$\displaystyle - n \rho_{sc}(x_0)^2 \frac{d^2}{dy^2} + \frac{1}{4} (x_0 \sqrt{n} + \frac{y}{\sqrt{n} \rho_{sc}(x_0)})^2 \leq n + \frac{1}{2}.$

Isolating out the top order terms in ${n}$, we can rearrange this as

$\displaystyle -\frac{d^2}{dy^2} \leq \pi^2 + o(1).$

Thus, in the limit ${n \rightarrow \infty}$, we expect (heuristically, at least) that the left-hand side of (2) to converge as ${n \rightarrow \infty}$ to the integral kernel of the spectral projection to the region

$\displaystyle -\frac{d^2}{dy^2} \leq \pi^2.$

Introducing the Fourier dual variable ${\xi}$ to ${y}$, as manifested by the Fourier transform

$\displaystyle \hat f(\xi) = \int_{\bf R} e^{-2\pi i \xi y} f(y)\ dy$

and its inverse

$\displaystyle \check F(y) = \int_{\bf R} e^{2\pi i \xi y} F(\xi)\ d\xi,$

then we (heuristically) have ${\frac{d}{dy} = 2\pi i \xi}$, and so we are now projecting to the region

$\displaystyle |\xi|^2 \leq 1/4, \ \ \ \ \ (4)$

i.e. we are restricting the Fourier variable to the interval ${[-1/2,1/2]}$. Back in physical space, the associated projection ${P}$ thus takes the form

$\displaystyle P f(y) = \int_{[-1/2,1/2]} e^{2\pi i \xi y} \hat f(\xi)\ d\xi$

$\displaystyle = \int_{\bf R} \int_{[-1/2,1/2]} e^{2\pi i \xi y} e^{-2\pi i \xi z}\ d\xi f(z)\ dz$

$\displaystyle = \int_{\bf R} \frac{\sin(\pi(y-z))}{y-z} f(z)\ dz$

and the claim follows.

Remark 1 From a semiclassical perspective, the original spectral projection ${L \leq n+\frac{1}{2}}$ can be expressed in phase space (using the dual frequency variable ${\eta}$ to ${x}$) as the ellipse

$\displaystyle 4 \pi^2 \eta^2 + \frac{x^2}{4} \leq n+\frac{1}{2} \ \ \ \ \ (5)$

which after the indicated change of variables becomes the elongated ellipse

$\displaystyle \xi^2 + \frac{1}{2n \rho_{sc}(x_0)(4-x_0^2)} y + \frac{1}{4n^2 \rho_{sc}(x_0)^2 (4-x_0^2)} y^2$

$\displaystyle \leq \frac{1}{4} + \frac{1}{2n (4-x_0^2)}$

which converges (in some suitably weak sense) to the strip (4) as ${n \rightarrow \infty}$.

— 2. The edge asymptotics —

A similar (heuristic) argument gives the edge asymptotics, Lemma 2. Starting with the change of variables

$\displaystyle x = 2 \sqrt{n} + \frac{y}{n^{1/6}}$

the harmonic oscillator ${L}$ now becomes

$\displaystyle - n^{1/3} \frac{d^2}{dy^2} + \frac{1}{4} (2 \sqrt{n} + \frac{y}{n^{1/6}})^2.$

Thus, the left-hand side of (3) becomes the kernel of the spectral projection to the region

$\displaystyle - n^{1/3} \frac{d^2}{dy^2} + \frac{1}{4} (2 \sqrt{n} + \frac{y}{n^{1/6}})^2 \leq n + \frac{1}{2}.$

Expanding out, computing all terms of size ${n^{1/3}}$ or larger, and rearranging, this (heuristically) becomes

$\displaystyle - \frac{d^2}{dy^2} + y \leq o(1)$

and so, heuristically at least, we expect (3) to converge to the kernel of the projection to the region

$\displaystyle - \frac{d^2}{dy^2} + y \leq 0. \ \ \ \ \ (6)$

To compute this, we again pass to the Fourier variable ${\xi}$, converting the above to

$\displaystyle 4 \pi^2 \xi^2 + \frac{1}{2\pi i} \frac{d}{d\xi} \leq 0$

using the usual Fourier-analytic correspondences between multiplication and differentiation. If we then use the integrating factor transformation

$\displaystyle F(\xi) \mapsto e^{8 \pi^3 i \xi^3 / 3} F(\xi)$

we can convert the above region to

$\displaystyle \frac{1}{2\pi i} \frac{d}{d\xi} \leq 0$

which on undoing the Fourier transformation becomes

$\displaystyle y \leq 0,$

and the spectral projection operation for this is simply the spatial multiplier ${1_{(-\infty,0]}}$. Thus, informally at least, we see that the spectral projection ${P}$ to the region (6) is given by the formula

$\displaystyle P = M^{-1} 1_{(-\infty,0]} M$

where the Fourier multiplier ${M}$ is given by the formula

$\displaystyle \widehat{Mf}(\xi) := e^{8 \pi^3 i \xi^3 / 3} \hat f(\xi).$

In other words (ignoring issues about convergence of the integrals),

$\displaystyle Mf(y) = \int_{\bf R} (\int_{\bf R} e^{2\pi i y \xi} e^{8 \pi^3 i \xi^3 / 3} e^{-2\pi i z \xi}\ d\xi) f(z)\ dz$

$\displaystyle = 2 \int_{\bf R} (\int_0^\infty \cos( 2\pi (y-z) \xi + 8 \pi^3 \xi^3 / 3 )\ d\xi) f(z)\ dz$

$\displaystyle = \frac{1}{\pi} \int_{\bf R} (\int_0^\infty \cos( t (y-z) + t^3 / 3 )\ dt) f(z)\ dz$

$\displaystyle = \int_{\bf R} Ai(y-z) f(z)\ dz$

and similarly

$\displaystyle M^{-1} f(z) = \int_{\bf R} Ai(y-z) f(y)\ dy$

(this reflects the unitary nature of ${M}$). We thus see (formally, at least) that

$\displaystyle P f(y) = \int_{\bf R} (\int_{(-\infty,0]} Ai(y-w) Ai(z-w)\ dw) f(z)\ dz.$

To simplify this expression we perform some computations closely related to the ones above. From the Fourier representation

$\displaystyle Ai(y) = \frac{1}{\pi} \int_0^\infty \cos(ty + t^3/3)\ dt$

$\displaystyle = \int_{\bf R} e^{2\pi i y \xi} e^{8 \pi i \xi^3/3}\ d\xi$

we see that

$\displaystyle \widehat{Ai}(\xi) = e^{8 \pi^3 i \xi^3/3}$

which means that

$\displaystyle (4 \pi^2 \xi^2 + \frac{1}{2\pi i} \frac{d}{d\xi}) \widehat{Ai}(\xi) = 0$

and thus

$\displaystyle (- \frac{d^2}{dy^2} + y) Ai(y) = 0,$

thus ${Ai}$ obeys the Airy equation

$\displaystyle Ai''(y) = y Ai(y).$

Using this, one soon computes that

$\displaystyle \frac{d}{dw} \frac{Ai(y-w) Ai'(z-w) - Ai'(y-w) Ai(z-w)}{y-z} = Ai(y-w) Ai(z-w).$

Also, stationary phase asymptotics tell us that ${Ai(y)}$ decays exponentially fast as ${y \rightarrow +\infty}$, and hence ${Ai(y-w)}$ decays exponentially fast as ${w \rightarrow -\infty}$ for fixed ${y}$; similarly for ${Ai'(z-w), Ai'(y-w), Ai(z-w)}$. From the fundamental theorem of calculus, we conclude that

$\displaystyle \int_{(-\infty,0]} Ai(y-w) Ai(z-w)\ dw = \frac{Ai(y) Ai'(z) - Ai'(y) Ai(z)}{y-z},$

(this is a continuous analogue of the Christoffel-Darboux formula), and the claim follows.

Remark 2 As in the bulk case, one can take a semi-classical analysis perspective and track what is going on in phase space. With the scaling we have selected, the ellipse (5) has become

$\displaystyle 4 \pi^2 n^{1/3} \xi^2 + \frac{(2\sqrt{n} + y/n^{1/6})^2}{4} \leq n+\frac{1}{2},$

which we can rearrange as the eccentric ellipse

$\displaystyle 4\pi^2 \xi^2 + y \leq \frac{1}{2n^{1/3}} - \frac{y^2}{4 n^{2/3}}$

which is converging as ${n \rightarrow \infty}$ to the parabolic region

$\displaystyle 4\pi^2 \xi^2 + y \leq 0$

which can then be shifted to the half-plane ${y \leq 0}$ by the parabolic shear transformation ${(y,\xi) \mapsto (y+4\pi^2\xi^2,\xi)}$, which is the canonical relation of the Fourier multiplier ${M}$. (The rapid decay of the kernel ${Ai}$ of ${M}$ at ${+\infty}$ is then reflected in the fact that this transformation only shears to the right and not the left.)

Remark 3 Presumably one should also be able to apply the same heuristics to other invariant ensembles, such as those given by probability distributions of the form

$\displaystyle C_n e^{-\hbox{tr}(P(M_n))} dM_n$

for some potential function ${P}$. Certainly one can soon get to an orthogonal polynomial formulation of the determinantal kernel for such ensembles, but I do not know if the projection operators for such kernels can be viewed as spectral projections to a phase space region as was the case for GUE. But if one could do this, this would provide a heuristic explanation as to the universality phenomenon for such ensembles, as Taylor expansion shows that all (reasonably smooth) regions of phase space converge to universal limits (such as a strip or paraboloid) after rescaling around either a non-critical point or a critical point of the region with the appropriate normalisation.


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This function generates an FLFleetsExt object, given the data inputs as arrays. Supported formats are Excel (xls and xlsx) and R format (RData).

create.fleets.arrays(stk_objs, caa_objs, caa_objs_path, price_objs,
price_objs_path, catch_obj, effort_obj, flt_obj = NULL, stk_nms = NA,
flt_nms, flt_mt_nms, flt_mt_stk_nms, ages = NULL, hist.yrs, sim.yrs,
mean.yrs, new_hist.yrs = hist.yrs, update_catch_effort = TRUE,
update_price = TRUE, update_weight = TRUE,
caa_flt_mt_correspondences = NULL, paa_flt_mt_correspondences = NULL,
caaOpt, priceOpt, excel = TRUE)

Arguments

stk_objs A character vector with the names of the files containing the stocks data. See create.biol.arrays for more detail. Supported format is only Excel (xls and xlsx), each stock can be in different format. A character vector with the names of the files containing the catch at age data (in numbers), both for landings and discards. Supported formats are Excel (xls and xlsx) and R format (RData), each file can be in different format. The number of required files depend of the value of caaOpt argument: caaOpt = 1 or 2 => one per stock and fleet. Named vector stknm_fltnm caaOpt = 3, 4 or 5=> one per stock. Named vector stknm If NULL, the function looks for "caa_stknm_flnm.xlsx" in caaOpt = 1 or 2 and "caa_stknm.xlsx" in caaOpt = 3 ,4 or 5. A character vector with the caa_objs file path. A character vector with the names of the files containing the price at age data. Supported formats are Excel (xls and xlsx) and R format (RData), each prices file can be in different format. The number of required files depend of the value of caaOpt argument: caaOpt = 1 or 2 => one per stock and fleet. Named vector stknm_fltnm. caaOpt = 3, 4 or 5=> one per stock. Named vector stknm. A character vector with the price_objs file path. A character vector with the names of the files containing the catch data in Fcube format. Supported formats are Excel (xls and xlsx) and R format (RData) and required columns are 'year', 'fleet', 'metier', 'stock', 'category' and 'catch'. A character vector with the names of the files containing the effort data in Fcube format. Supported formats are Excel (xls and xlsx) and R format (RData) and required columns are 'year', 'fleet', 'metier' and 'effort'. Both catch_obj and effort_obj must be in the same format. An FLFleets object (optional) with the structure of the fleet and which may contain historical data. If this object is provided, then the arguments stk_nms, flt_nms, flt_mt_nms and ages will not be used and excel argument must be set to FALSE. A character vector (optional) with the name of all the stocks caugth by the different fleets. A character vector with the name of the fleets. A list with one element per fleet. In turn, each element is a character vector with the names of the metiers in the corresponding fleet. A list with one element per fleet and metier. In turn, each element is a character vector with the names of the stocks in the corresponding fleet and metier. A list with one element per stock, with the age classes of the stock. A vector with the historical years. A vector with the simulation years. A vector with the years used to compute the mean to condition the parameters in the projection period. A vector with the years from input files that will be used to condition the parameters in the historic years. If a value is not provided, the it is set equal to hist.yrs. Logical. If TRUE (default), catch and effort must be provided and catch_obj, effort_obj arguments are required. Logical. If TRUE (default), prices must be provided and price_objs and price_objs_path arguments are required. Logical. If TRUE (default), weights at age for landings and discards must be provided, so wl and wd sheets are required in files listed in stk_objs argument. An Excel file name. This file must contain one sheet per stock, with the correspondences between the fleet segments used in caaa_obj data and the fleet metier segmentation used in the analysis. If the file does not exist, it is supposed that the caa data is given by fleet and metier. An Excel file name. This file must contain information on prices correspondences, with same format and requirements as caa_flt_mt_correspondences argument. A code number to determine the way in wich catch at age data are provided. The option to be used depends on the data availabiltiy, from data rich to data-poor and the following codes are available: 1If catch at age data is available at metier level for all the metiers. 2If catch at age data is only available at fleet level. 3If catch at age data is disaggregated but the segments do not correspond exactly with the metiers/fleets considered in the case study. 4If catch at age data is only available at stock level. 5If we want to use the data available previously in the FLCatch objects from flt_obj to derive catch profiles at age and then apply caaOpt==3 using only one fleet segment, fseg, which represents all the fleets and metiers. Note: This approach could lead to a different total catch at age profile derived from the fleets to those in the stocks. A code number to determine the way in wich price at age data are provided. The option to be used depends on the data availabiltiy, from data rich to data-poor and the following codes are available: 1If price data is available at metier level for all the metiers. 2If price data is only available at fleet level. 3If price data is disaggregated but the segments do not correspond exactly with the metiers/fleets considered in the case study. 4If price data is only available at stock level. Logical. If TRUE (default), the data in the Excel file is used to create the stucture of the FLFleets object; whereas if FALSE, the flt_obj object is used instead.

Value

An FLFleetsExt.

FLFleetsExt, create.biol.arrays

format: plain


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zbMATH — the first resource for mathematics

Harmonic maps and representations of non-uniform lattices of $$PU(m,1)$$. (English) Zbl 1147.22009
The authors make use of harmonic maps techniques to study representations of lattices of $$PU(m,1)$$ into $$PU(n,1)$$ in order to provide rigidity results. This goes as follows: first one shows the existence of a harmonic map between the corresponding symmetric spaces. Then one must prove, under certain constraints, that the harmonic map is pluriharmonic, holomorphic, totally geodesic, isometric, depending on the situation. The existence is proved of a harmonic map from $$\mathbb H^m_C$$ to $$\mathbb H^n_C$$ making use of Corlette’s results. Finally the authors study a new invariant associated to representations of fundamental groups of finite topological type and negative Euler characteristic into $$PU(n,1)$$.

MSC:
 22E40 Discrete subgroups of Lie groups 32Q05 Negative curvature complex manifolds 32Q20 Kähler-Einstein manifolds 53C24 Rigidity results 53C35 Differential geometry of symmetric spaces 53C43 Differential geometric aspects of harmonic maps
Full Text:
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text:
# zbMATH — the first resource for mathematics

Continuity of separately continuous mappings. (English) Zbl 1349.54034
Summary: By means of a topological game, a class of topological spaces which contains compact spaces, $$q$$-spaces and $$W$$-spaces was defined in A. Bouziad [Topology Appl. 50, No. 1, 73–80 (1993; Zbl 0827.54018)]. We will show that if $$Y$$ belongs to this class, every separately continuous function $$f\:X\times Y\to Z$$ is jointly continuous on a dense subset of $$X \times Y$$ provided that $$X$$ is $$\sigma$$-$$\beta$$-unfavorable and $$Z$$ is a regular weakly developable space.

##### MSC:
 54C05 Continuous maps 54E30 Moore spaces 54E52 Baire category, Baire spaces 54C99 Maps and general types of topological spaces defined by maps
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## Abstract

Let $\phi$ be a compactly supported symmetric refinable function in $L_2(\RR)$ with a finitely supported symmetric mask on $\ZZ$. Under the assumption that the shifts of $\phi$ are stable, in this paper we prove that one can always construct three wavelet functions $\psi^1$, $\psi^2$ and $\psi^3$ such that \begin{enumerate} \item[{(i)}] All the wavelet functions $\psi^1$, $\psi^2$ and $\psi^3$ are compactly supported and are finite linear combinations of the functions $\phi(2\cdot-k), k\in \ZZ$; \item[{(ii)}] Each of the wavelet functions $\psi^1$, $\psi^2$ and $\psi^3$ is either symmetric or antisymmetric; \item[{(iii)}] $\{\psi^1, \psi^2, \psi^3\}$ generates a tight wavelet frame in $L_2(\RR)$, that is, $$\| f\|^2 =\sum_{\ell=1}^3 \sum_{j\in \ZZ} \sum_{k\in \ZZ} |\la f, \psi^\ell_{j,k}\ra|^2 \qquad \forall\; f\in L_2(\RR),$$ where $\psi^\ell_{j,k}:=2^{j/2}\psi^\ell(2^j\cdot-k)$, $\ell=1, 2, 3$ and $j, k\in \ZZ$; \item[{(iv)}] Each of the wavelet functions $\psi^1$, $\psi^2$ and $\psi^3$ has the highest possible order of vanishing moments, that is, its order of vanishing moments matches the order of the approximation order provided by the refinable function $\phi$. \end{enumerate} %Our result in this paper generalizes several results in the %literature on symmetric tight wavelet frames that are derived from %refinable functions via a multiresolution analysis. We shall give an example to demonstrate that the assumption on stability of the refinable function $\phi$ cannot be dropped. Some examples of symmetric tight wavelet frames with three generators will be given to illustrate the results and construction in this paper.

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## Precalculus (6th Edition)

RECALL: In a monomial $Cxy$ where $C$ is a real number, C is called the numerical coefficient. Thus, the missing expression in the given statement is: numerical coefficient.


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## LU Decomposition Calculator

Given an $m\times n$ real or complex matrix $A$, this application calculates an LU decomposition of $A$ in the form $A = PLU,$ where $P$ is a permutation matrix, $L$ is a lower triangular matrix with unit diagonal entries, and $U$ is an upper triangular matrix.

See Input Data for the description of how to enter matrix or just click Example for a simple example.


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## Sequence Fractals Part IV #5

$z^2+s_i*c$
si = -1.27i, 1, (repeating)
Center:1.0+0.0i , Zoom = 0.5

I have to confess that I am undecided on where to go next. The ‘*si‘ and ‘+si‘ cases are different. These deserve the full treatment as the +si case in Part I and Part II. That includes looking at the Julia sets. Finding Cycles and Misiurewicz points. Exploring how the C and M points move around as the sequence changes. Look at pictures of the other critical points. (These all start at 0 which is always a critical point, the other critical points depend on c and move around. That means extra programming for me.)

On the other hand, both of these classes, ‘*si‘ and ‘+si‘ are part of the universe of slices of the connectivity locus of quartic polynomials. In some sense these two classes are two parts of some much larger whole. I want to avoid “been there, done that”.

To be clear, there are an infinite number of genuinely unique ways to parameterize the space of quartic polynomials. And in each of them, there are an infinite number of genuinely unique 2D cross sections, or slices. And any one of those infinite times infinite number of slices can be selected to explore with pan and zoom.

Contrast this the quadratic case. There are only a small number (afaik) of useful parameterizations, z2+c being ubiquitous. There, the whole universe exists in a single 2 dimensional slice of the parameter space, which we explore with pan and zoom.

To be continued.

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DRAFT

# Dynamic balances

TODO

## Key formulas

##### Characteristic function
$$\ldots$$

##### Indifference curve
$$\ldots$$

##### Price response function
$$\ldots$$

##### AMM portfolio value
$$\ldots$$

##### Divergence loss
$$\ldots$$

Whilst reasonable care has been taken to verify the above formulas they may still contain errors. Please do not use them without independent verification.


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# Does the Dirac ground state of the H atom have an electron current?

1. Jan 5, 2014

I have confounded myself with the following observation. Take the standard expression for the lowest energy Dirac 4-spinor solution of the Dirac equation with a Coulomb potential (the H atom ground state). Plug this into the standard expression for the 4-vector current and use values for the gamma matrices in spherical polar coordinates. In the r and θ directions the current is zero. No surprise there. But the $\phi$ component of current is non-zero (unless I've screwed up). Moreover, this current results in a magnetic dipole moment which can be calculated by multiplying by r.sin($\theta$) and integrating over all space. The result is a magnetic moment of exactly one Bohr magneton. The reason why this result confuses me is that the ground state has (roughly speaking) an orbital angular momentum quantum number of zero* and so one would not expect there to be any orbital current or a magnetic moment associated with an orbital current. However, one Bohr magneton is the correct magnetic moment for the ground state - but it arises entirely from the electron spin. So I have tentatively concluded that the $\phi$ current is a manifestation of the spin distribution. Is this right? And is this current 'real'? Can a current truly arise from spin only when the electron 'cloud' is stationary? The total current is of prodigious magnitude, ~1 milliamp for a single electron!

*To be more precise, the ground state does not have a well defined value for the operator L^2 since it does not commute with the Dirac Hamiltonian. The upper 2-spinor has L = 0 and whilst the lower 2-spinor has L = 1 the lower2-spinor is very small in magnitude so the expectation value of the orbital angular momentum is tiny, about 10^-5 in h-bar units. This would align with a similarly tiny magnetic moment, which also confirms that the above magnetic moment is not due to orbital motion.

2. Jan 6, 2014

### edguy99

I believe your calculations are correct and I am also struck by the implications. I have not seen much discussion on the issue so it is hard to evaluate its meaning. Certainly the "Dirac Electron" is one example of discussing this, where one equation for a massive particle can be rewritten as two equations for two interacting massless particles, where the coupling constant of the interaction is the mass of the electron.

The visual image you have constructed matches a trapped current (in blue rotating horizontally, with the axis shown in blue), trapping an orthogonal magnetic field (in red rotating vertically).

3. Jan 10, 2014

### tom.stoer

I guess this can be resolved by treating the hydrogen atom as a two-particle system.

Unfortunately this is often ignored, but in non.-rel. QM there's a nice procedure via unitary transformations to map a system S with canonical variables r1, r2, p1, p2 and H[p1,p2,r1-r2] to a new system S' with, r,R,p,P and H'[p,P,r]. Note that H' does not depend on R, therefore [H',P] = 0. The system S' still lives in a two-particle Hilbert space, but usually (in many textbooks) this information gets lost. Therefore they only deal with one-particle states, i.e. |nlm>, neglecting the important information that there is another degree of freedom, namely the canonical pair R,P. The complete quantum state including the total momentum reads |nlm,K> and b/c [H',P] = 0 we have conservation of P which means plane waves exp(iKX).

For the binding energy only |nlm> is relevant, but the total spectrum is Enlm + K2/2M.

Now I would guess that something similar might by relevant for other operators. So it could very well be that there are additional contributions to certain operators sensitive to the two-particle nature of the problem. Therefore I would suggest to check what happens to the two-particle operators for spin, current etc. when one applies the above mentioned transformation.


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# Calling NAG Library Routines from Octave

NAG Technical Report TR 4/09

Anna Kwiczala
Numerical Algorithms Group, May 2009

## Abstract

This report gives detailed instructions on how to call routines in the NAG C and Fortran Libraries from the Octave programming environment.

## Contents

1. Introduction

Octave [1] is a freely redistributable programming language for numerical computations. It is mostly compatible with MATLAB. The NAG Libraries [2] contain a large selection of numerical and statistical routines, which can be accessed from different languages [3]. This article gives a brief introduction to one of the methods and examples of calling NAG routines from within Octave.

Octave can be extended by dynamically linked functions [4], which are used in the same way as Octave functions. Octave has been written in C++, so it is not surprising that the easiest way of extending it is by writing the linking code in that language. It can be called by Octave directly through its native oct-file interface. Due to the fact that C functions can easily be called from C++, our best choice is to use C++ code together with the NAG C Library.

2. C++ code and Oct-Files

Octave comes with a ready-made script mkoctfile, which can be used to compile C++ code into oct-files. All we have to do is to write a C++ function calling one of the NAG routines and then use the script. To define the entry point into the dynamically linked function, we need to use the Octave DEFUN_DLD macro in our C++ code. Therefore, our function will look like this:

  #include <octave/oct.h>

DEFUN_DLD (function_name, args, nargout,
"Function description")
{
octave_value_list retval;

// retrieve input arguments from args
// call NAG routine
// assign output arguments to retval

return retval;
}

where:

• oct.h contains most of the necessary definitions for an oct-file in Octave;
• function_name will be our function's name seen in Octave and it must match the file's name, but be different from any NAG routine names;
• args is the list of arguments to the function of type octave_value_list;
• nargout is the number of output arguments;
• "Function description" is the string that will be seen as the function help text;
• the return type is always octave_value_list.

3. Examples

The rest of the code required will be easier to demonstrate with the use of examples. The code for this article was tested on a Linux machine running 64-bit Fedora 8 (Werewolf) with Octave 3.0.3, plus c++ (GCC) 4.3.3, gcc (GCC) 4.3.3, GNU Fortran (GCC) 4.3.3, Mark 8 of the NAG C Library and Mark 22 of the NAG FORTRAN Library.

3.1 Example 1  Log gamma function ln G(x)

 The simplest example: we call a function with only one return value - the ln G (x) logarithm gamma function routine nag_log_gamma (s14abc) from the NAG C Library.

3.2 Example 2  Scaled complex complement of error function, exp(-z2)erfc(-iz)

 We call a function with complex input and output values - nag_complex_erfc (s15ddc), which computes values of the function w(z)=exp(-z2)erfc(-iz), for complex z. Complex numbers are stored differently in Octave and NAG C Library, so extra code is required to copy input data from Octave complex type to NAG complex type and the result back from NAG complex type to Octave complex type.

3.3 Example 3  Approximate solution of complex simultaneous linear equations AX = B

 We call a function which takes and returns complex arrays - nag_complex_lu_solve_mult_rhs (f04akc). The function calculates the approximate solution of a set of complex linear equations with multiple right-hand sides AX=B, where A has been factorized by nag_complex_lu (f03ahc). Again, some copying between Octave and NAG complex types is required.

3.4 Example 4  Solving a nonlinear minimization problem

 In this example we call the nonlinear minimization routine nag_opt_nlp (e04ucc). The routine takes as input two user-supplied functions. We show two ways of supplying the functions - either as C code or Octave code. We also demonstrate how to pass Octave arrays and matrices into C.

3.5 Example 5  Univariate time series, seasonal and non-seasonal differencing

 We call a function from NAG Fortran Library - G13AAF. The function carries out non-seasonal and seasonal differencing on a time series. It returns the differenced values and data which allow the original series to be reconstituted from the differenced series. We use NAG header files [5] for the NAG Fortran Library.

5. Summary

Octave contains many tools that make extending it fairly easy. The method used in this article is just one of a few possible approaches which we believe is the easiest one. If you want to try different methods, consult Dynamically Linked Functions section on Octave website [4].

References


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# American Institute of Mathematical Sciences

May  2008, 9(3&4, May): 493-516. doi: 10.3934/dcdsb.2008.9.493

## Minimal subsets of projective flows

 1 Department of Mathematics and Statistics, Queen's University, Kingston, ON Canada K7L 3N6, Canada 2 Dipartimento di Sistemi e Informatica, Università di Firenze, Via di S. Marta 3, 50139 Firenze

Received  January 2007 Revised  June 2007 Published  February 2008

We study the minimal subsets of the projective flow defined by a two-dimensional linear differential system with almost periodic coefficients. We show that such a minimal set may exhibit Li-Yorke chaos and discuss specific examples in which this phenomenon is present. We then give a classification of these minimal sets, and use it to discuss the bounded mean motion property relative to the projective flow.
Citation: Kristian Bjerklöv, Russell Johnson. Minimal subsets of projective flows. Discrete & Continuous Dynamical Systems - B, 2008, 9 (3&4, May) : 493-516. doi: 10.3934/dcdsb.2008.9.493
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# 6.5 Physical applications  (Page 5/11)

 Page 5 / 11

A tank is in the shape of an inverted cone, with height $10$ ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?

Approximately $43,255.2$ ft-lb

## Hydrostatic force and pressure

In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called pascals .

Let’s begin with the simple case of a plate of area $A$ submerged horizontally in water at a depth s ( [link] ). Then, the force exerted on the plate is simply the weight of the water above it, which is given by $F=\rho As,$ where $\rho$ is the weight density of water (weight per unit volume). To find the hydrostatic pressure    —that is, the pressure exerted by water on a submerged object—we divide the force by the area. So the pressure is $p=F\text{/}A=\rho s.$

By Pascal’s principle , the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal’s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula $F=\rho As$ directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.

Suppose a thin plate is submerged in water. We choose our frame of reference such that the x -axis is oriented vertically, with the downward direction being positive, and point $x=0$ corresponding to a logical reference point. Let $s\left(x\right)$ denote the depth at point x . Note we often let $x=0$ correspond to the surface of the water. In this case, depth at any point is simply given by $s\left(x\right)=x.$ However, in some cases we may want to select a different reference point for $x=0,$ so we proceed with the development in the more general case. Last, let $w\left(x\right)$ denote the width of the plate at the point $x.$

Assume the top edge of the plate is at point $x=a$ and the bottom edge of the plate is at point $x=b.$ Then, for $i=0,1,2\text{,…},n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of the interval $\left[a,b\right],$ and for $i=1,2\text{,…},n,$ choose an arbitrary point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ The partition divides the plate into several thin, rectangular strips (see the following figure).

Let’s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth, $s\left({x}_{i}^{*}\right).$ We then have

${F}_{i}=\rho As=\rho \left[w\left({x}_{i}^{*}\right)\text{Δ}x\right]s\left({x}_{i}^{*}\right).$

Adding the forces, we get an estimate for the force on the plate:

$F\approx \sum _{i=1}^{n}{F}_{i}=\sum _{i=1}^{n}\rho \left[w\left({x}_{i}^{*}\right)\text{Δ}x\right]s\left({x}_{i}^{*}\right).$

A Function F(X)=Sinx+cosx is odd or even?
neither
David
Neither
Lovuyiso
f(x)=1/1+x^2 |=[-3,1]
apa itu?
fauzi
determine the area of the region enclosed by x²+y=1,2x-y+4=0
Hi
MP
Hi too
Vic
hello please anyone with calculus PDF should share
Which kind of pdf do you want bro?
Aftab
hi
Abdul
can I get calculus in pdf
Abdul
How to use it to slove fraction
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how can we scatch a parabola graph
Ok
Endalkachew
how can I solve differentiation?
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
what is the procedures in solving number 1?
review of funtion role?
for the function f(x)={x^2-7x+104 x<=7 7x+55 x>7' does limx7 f(x) exist?
find dy÷dx (y^2+2 sec)^2=4(x+1)^2
Integral of e^x/(1+e^2x)tan^-1 (e^x)
why might we use the shell method instead of slicing
fg[[(45)]]²+45⅓x²=100
find the values of c such that the graph of f(x)=x^4+2x^3+cx^2+2x+2
anyone to explain some basic in calculus
I can
Debdoot


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
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Pseudo-slant submanifolds of a Sasakian manifold. (English) Zbl 1117.53043
Let $\overline{M}$ be a Riemannian manifold equipped with an almost contact metric structure $(\phi,\xi,\eta,g)$. A submanifold $M$ of $\overline{M}$ is said to be pseudo-slant if the structure vector field $\xi$ is tangent to $M$ everywhere, and if there exist two subbundles $D_1$ and $D_2$ of the tangent bundle $TM$ of $M$ such that $TM$ decomposes orthogonally into $TM = D_1 \oplus D_2 \oplus {\Bbb R}\xi$, $\phi D_1$ is a subbundle of the normal bundle of $M$, and there exists a real number $0 \leq \theta < \pi/2$ such that for each nonzero vector $X \in D_2$ the angle between $\phi X$ and $D_2$ is equal to $\theta$. The authors derive some equations for certain tensor fields and investigate the integrability of some distributions on pseudo-slant submanifolds for the special case that the almost contact metric structure is Sasakian.

##### MSC:
 53C40 Global submanifolds (differential geometry) 53C25 Special Riemannian manifolds (Einstein, Sasakian, etc.)
##### Keywords:
Sasakian manifolds; pseudo-slant submanifolds


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A particle is attached to one end of string the other end of which is fixed. The particle moves in vertical plane. The velocity at the lowest point is such that the particle can just complete the circle. The ratio of tension in string at A&B is

one year ago

Md Mushahid
21 Points
							when an object of mass m which is fixed to an end of an inextensible and light rodand revolving around a circular path about the other end,weight of the object and tension in the rod are getting adjusted to provide centripetal force as shown in figure, Let us consider when the object is at B, as shown in figure.Centripetal force $\frac{m\times v^{2}}{I} =T - mg\cos \theta$where T is tension in the rod, m is mass of object and θ is the angular dispalcementwhen the vertical position is taken as reference. When the object is at A, θ = 0. hence centripetal force $\frac{m\times v^{2}}{I} =T - mg$-------------2When the object is at the top most position C, θ = π, hence centripetal force $\frac{m\times v^{2}}{I} =T + mg$............(3) When the object is at top most position C, minimum required velocity vmin is obtained,by considering tension in the rod T is zero.when we apply this condition in eqn.(3), we get vmin = $\sqrt{gl}$

one year ago
Vikas TU
13786 Points
							The Expression for Velocity of Body Moving in a Vertical Circle: Consider a small body of mass 'm' attached to one end of a string and whirled in a vertical circle of radius 'r'. In this case, the acceleration of the body increases as it goes down the vertical circle and decreases when goes up the vertical circle.The ratio of tension would be = 2mg/8mg = 0.25

one year ago
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text:
In quantum optics, a superradiant phase transition is a phase transition that occurs in a collection of fluorescent emitters (such as atoms), between an state containing few electromagnetic excitations (as in the electromagnetic vacuum), and a superradiant state with many electromagnetic excitations trapped inside the emitters. The superradiant state is made thermodynamically favourable by having strong, coherent interactions between the emitters.

The superradiant phase transition was originally predicted in so called Dicke model of superradiance when it is assumed that atoms have only two energetic levels and they are interacting only with one mode of the electromagnetic field [1] .[2] The phase transition occurs when the strength of the interaction between the atoms and the field is larger than the energy of the non-interacting part of the system which similarly to the case of superconductivity and ferromagnetism leads to the effective dynamical interactions between atoms of the ferromagnetic type and the spontaneous ordering of excitations below the critical temperature. It means that the collective Lamb shift in the system of atoms interacting with the vacuum fluctuations becomes comparable with the energies of atoms alone and the vacuum fluctuations cause the spontaneous self-excitation of matter.

The transition can be readily understood with the use of Holstein-Primakoff transformation[3] applied to two level atom. As the result of this transformation the atoms become the Lorentz harmonic oscillators with the frequency equal to the difference between the energy levels and the whole system becomes the system of the interacting harmonic oscillators of atoms and the field known as Hopfield dielectric which further predicts in the normal state polarons for photons or polaritons. If now the interaction with the field is so strong that the system collapses in the harmonic approximation and complex polariton frequencies (soft modes) appear then the physical system with nonlinear terms of the higher order becomes the system with the Mexican hat-like potential and will undergo ferroelectric-like phase transition.[4] In this model the system is mathematically equivalent for one mode of excitation to the Trojan wave packet when the circularly polarized field intensity corresponds to the electromagnetic coupling constant and above the critical value it changes to the unstable motion of the ionization.

The superradiant phase transition was the subject of a wide discussion if it is only a result of the simplified model of the matter-field interaction and if it can occur for the real physical parameters of physical systems (no-go theorem) [5] .[6] However both the original derivation and the later corrections leading to nonexistence of the transition due to Thomas–Reiche–Kuhn sum rule canceling for the harmonic oscillator the needed inequality to impossible negativity of the interaction were based on the assumption that the quantum field operators are commuting numbers and the atoms do not interact with the static Coulomb forces which generally is not true. It currently can be observed in model systems like Bose-Einstein condensates and artificial atoms [7] .[8]

## Theory

### Criticality of linearized Jaynes-Cummings model

Superradiant phase transition is formally predicted already by the critical behavior of the resonant Jaynes-Cummings model describing the interaction of not ${\displaystyle N}$ but only one atom with one mode of the electromagnetic field. Starting from the exact Hamiltonian of the Jaynes-Cummings model at resonance

${\displaystyle {\hat {H}}_{\text{JC}}=\hbar \omega {\hat {a}}^{\dagger }{\hat {a}}+\hbar \omega {\frac {{\hat {\sigma }}_{z}}{2}}+{\frac {\hbar \Omega }{2}}\left({\hat {a}}{\hat {\sigma }}_{+}+{\hat {a}}^{\dagger }{\hat {\sigma }}_{-}+{\hat {a}}{\hat {\sigma }}_{-}+{\hat {a}}^{\dagger }{\hat {\sigma }}_{+}\right),}$

Applying the Holstein-Primakoff transformation for two spin levels and replacing the spin raising and lowering operators by those for the harmonic oscillators

${\displaystyle {\hat {\sigma }}_{-}\approx {\hat {b}}}$
${\displaystyle {\hat {\sigma }}_{+}\approx {\hat {b}}^{\dagger }}$
${\displaystyle {\hat {\sigma }}_{z}\approx 2{\hat {b}}^{\dagger }{\hat {b}}}$

one gets the Hamiltonian of two coupled harmonic oscillators:

${\displaystyle {\hat {H}}_{\text{JC}}=\hbar \omega {\hat {a}}^{\dagger }{\hat {a}}+\hbar \omega {\hat {b}}^{\dagger }{\hat {b}}+{\frac {\hbar \Omega }{2}}\left({\hat {a}}{\hat {b}}^{\dagger }+{\hat {a}}^{\dagger }{\hat {b}}+{\hat {a}}{\hat {a}}+{\hat {a}}^{\dagger }{\hat {b}}^{\dagger }\right),}$

which readily can be diagonalized. Postulating its normal form

${\displaystyle {\hat {H}}_{\text{JC}}=\Omega _{+}{\hat {A_{+}}}^{\dagger }{\hat {A_{+}}}+\Omega _{-}{\hat {A_{-}}}^{\dagger }{\hat {A_{-}}}+C}$

where

${\displaystyle {\hat {A_{\pm }}}=c_{\pm 1}{\hat {a}}+c_{\pm 2}{\hat {a}}^{\dagger }+c_{\pm 3}{\hat {b}}+c_{\pm 4}{\hat {b}}^{\dagger }}$

one gets the eigenvalue equation

${\displaystyle [{\hat {A_{\pm }}},{\hat {H}}_{\text{JC}}]=\Omega _{\pm }A}$

with the solutions

${\displaystyle \Omega _{\pm }=\omega {\sqrt {1\pm {\frac {\Omega }{\omega }}}}}$

The system collapses when one of the frequencies becomes imaginary i.e. when

${\displaystyle \Omega >\omega }$

or when the atom-field coupling is stronger than the frequency of the mode and atom oscillators. While there are physically higher terms in the true system the system in this regime will therefore undergo the phase transition.

### Criticality of James-Cummings model

The simplified Hamiltonian of the Jaynes-Cummings model neglecting the counter-rotating terms is

${\displaystyle {\hat {H}}_{\text{JC}}=\hbar \omega {\hat {a}}^{\dagger }{\hat {a}}+\hbar \omega {\frac {{\hat {\sigma }}_{z}}{2}}+{\frac {\hbar \Omega }{2}}\left({\hat {a}}{\hat {\sigma }}_{+}+{\hat {a}}^{\dagger }{\hat {\sigma }}_{-}\right),}$

and the energies for the case of zero detuning are

${\displaystyle E_{\pm }(n)=\hbar \omega \left(n+{\frac {1}{2}}\right)\pm {\frac {1}{2}}\hbar \Omega (n),}$
${\displaystyle \Omega (n)=\Omega {\sqrt {n+1}}}$

where ${\displaystyle \Omega _{n}}$ is the Rabi frequency. One can approximately calculate the canonical partition function

${\displaystyle Z=\sum _{\pm ,n}{\mathrm {e} }^{-\beta E_{\pm }(n)}\approx \sum _{\pm }\int {\mathrm {e} }^{-\beta E_{\pm }(n)}dn=\int {\mathrm {e} }^{\Phi (n)}dn}$,

where the discrete sum was replaced by the integral.

The normal approach is that the later integral is calculated by the Gaussian approximation around the maximum of the exponent:

${\displaystyle {\frac {\partial \Phi (n)}{\partial n}}=0}$
${\displaystyle \Phi (n)=-\beta \hbar \omega \left(n+{\frac {1}{2}}\right)+\log 2\cosh {\frac {\hbar \Omega (n)\beta }{2}}}$

This leads to the critical equation

${\displaystyle \tanh {\frac {\hbar \Omega (n)\beta }{2}}=4{\frac {\omega }{\Omega }}{\sqrt {n+1}}}$

This has the solution only if

${\displaystyle \Omega >4\omega }$

which means that the normal and the superradiant phase exist only if the field-atom coupling is significantly stronger than the energy difference between the atom levels. When the condition is fulfilled the equation gives the solution for the order parameter ${\displaystyle n}$ depending on the inverse of the temperature ${\displaystyle 1/\beta }$ which means non-vanishing ordered field mode. Similar considerations can be done in true thermodynamic limit of the infinite number of atoms.

## References

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text:
# Calculating the number of winners and losers [closed]

A contestant is said to be a winner if his number is strictly greater than the numbers to his left and his right and a contestant is said to be a loser if his number is strictly less than the numbers to his left and right so the contestant are on a circle and I should calculate the number of winners and losers.

            int winner=0;
int loser=0;
for (int i = 0; i < contestant.length; i++) {
if (i == 0) {
if (contestant[0] < contestant[contestant.length - 1]
&& contestant[0] < contestant[1])
loser++;
else if (contestant[0] > contestant[contestant.length - 1]
&& contestant[0] > contestant[1])
winner++;
} else if (i == contestant.length - 1) {
if (contestant[0] > contestant[contestant.length - 1]
&& contestant[contestant.length - 2] > contestant[contestant.length - 1])
loser++;
else if (contestant[0] < contestant[contestant.length - 1]
&& contestant[contestant.length - 2] < contestant[contestant.length - 1])
winner++;
} else {
if (contestant[i] < contestant[i + 1] &&contestant[i] < contestant[i - 1])
loser++;
else if (contestant[i] > contestant[i + 1]
&& contestant[i] > contestant[i - 1])
winner++;
}
}

## closed as unclear what you're asking by IEatBagels, 200_successApr 15 '15 at 5:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• CodeReview is about reviewing the complete code. – Shirishkumar Bari Apr 10 '15 at 12:35
• @Shree has a point - your code is not complete. There is enough to review, but it would make for a much better question if you could, for example, make it easy to run, give an example, or something. – rolfl Apr 10 '15 at 12:38
• It looks like nothing but embedding in a method is missing. But you should also add a main method calling it so we can try it out. It's not much work but it helps a lot. – maaartinus Apr 10 '15 at 12:42

One problem with your method is that you didn't write any. Maybe because you'd need to return both winner and loser from it. Returning two things is impossible, but you can always write a class encapsulating them. I did it below and let its constructor do the whole computation.

Another problem is that the code is far too long. Maybe you're handling the border cases correctly, maybe you're not, it's too much (similarly looking) code to read.

A simple trick for handling circles is to compute the indexes in a circular manner instead of repeating the lengthy expression again and again. Another trick would be to define a method isWinner(current, next, previous).

The code is also long because of the lengthy identifier contestant. Normally, I'd define a local variable and use it instead in order to keep the lines short.

You've forgot a space in &&contestant and your algorithm for breaking lines is a bit inconsistent. But these are very minor issues.

class Contestanter {
private int winner;
private int loser;

Contestanter(int contestant) {
int length = contestant.length;
for (int i = 0; i < length; i++) {
int next = i == length - 1 ? 0 : i + 1;
int prev = i == 0 ? length - 1 : i - 1;
if (contestant[i] < contestant[next] && contestant[i] < contestant[prev]) {
loser++;
} else if (contestant[i] > contestant[next] && contestant[i] > contestant[prev]) {
winner++;
}

int getWinner() {
return winner;
}

int getLoser() {
return loser;
}
}

There are two tricks you should be aware of when challenged with problems like this.

• The first is that you should be identifying changes in direction: inflection points.
• The second is that there are three states you are trying to identify: an entry is larger, the same, or smaller than the previous entry.

When scanning the data, you can calculate the 'direction' of a point, and then reuse that direction for the next test as well.

Java has a 'compareTo(...)' concept which returns a positive value if the first comared value is larger than the second. It returns a negative value if the first value is less than the second. It returns 0 if they are the same. For integer values, this is easy to do with: Integer.compare(int, int).

Using that, we can simply do:

current = Integer.compare(contestant[i+1], contestant[i]);

The value in current will be positive, 0, or negative if the i+1 value is larger, the same as, or less than the i value.

Now, if we have done the same with the i and i - 1 values (stored in previous), we can use some math-tricks to help:

• anything multiplied by 0 is 0.
• two positives multiplied is a positive.
• two negatives multiplied is a positive.
• a positive and negative multiplied is a negative.

i.e. if we multiply the previous and current, and the result is negative, then we have a peak, or trough (an inflection).

If current is positive, then the inflection is a low-point. If current is negative, then the inflection is a high point.

Using them all together, the code:

    int wincount = 0;
int losecount = 0;
if (contestants.length > 1) {
int current = Integer.compare(contestants[0], contestants[contestants.length - 1]);
for (int i = 0; i < contestants.length; i++) {
int previous = current;
current = Integer.compare(contestants[(i + 1) % contestants.length], contestants[i]);
if (current * previous < 0) {
if (current > 0) {
losecount++;
} else {
wincount++;
}
}
}
}

I have put this in to an ideone here too;

• You were 17 seconds faster... nice trick with the inflection points. For length==2 there should be IMHO wincount = losecount = contestants[0] != contestants[1] ? 1 : 0`, shouldn't it? – maaartinus Apr 10 '15 at 13:06
• I was working on the wrap-around immediately after I posted, @maaartinus - was the 'circular' part of the original question, or was it an in-the-grace-period edit? - the ideone is fixed too now. – rolfl Apr 10 '15 at 13:09
• Not sure, if I understand: I guess, the OP has never edited it. And the circularity is the only reason for such a complicated code. Otherwise they could simply forget the first and last elements. – maaartinus Apr 10 '15 at 13:14
• @maaartinus - edits made in the first 5 minutes of posting something do not show up in the edit history... I was just wondering whether the 'circular' nature of the problem was added in that 5 minutes, or whether I just did not see it until after I posted my answer... that's all. – rolfl Apr 10 '15 at 13:16


text:
Corpus ID: 221083370

# A descent view on Mitchell's theorem.

@article{Elmanto2020ADV,
title={A descent view on Mitchell's theorem.},
author={E. Elmanto and Denis Nardin and Lucy Yang},
journal={arXiv: K-Theory and Homology},
year={2020}
}
• Published 2020
• Mathematics
• arXiv: K-Theory and Homology
In this short note, we given a new proof of Mitchell's theorem that $L_{T\left(n\right)} K(Z) \cong 0$ for $n \geq 2$. Instead of reducing the problem to delicate representation theory, we use recently established hyperdescent technology for chromatically-localized algebraic K-theory.

#### References

SHOWING 1-10 OF 29 REFERENCES
Remarks on K(1)-local K-theory
• Mathematics
• 2020
We prove two basic structural properties of the algebraic $K$-theory of rings after $K(1)$-localization at an implicit prime $p$. Our first result (also recently obtained by Land--Meier--Tamme byExpand
Vanishing results for chromatic localizations of algebraic $K$-theory
• Mathematics
• 2020
We show that algebraic $K$-theory preserves $n$-connective $L_{n}^{f}$-equivalences between connective ring spectra, generalizing a result of Waldhausen for rational algebraic $K$-theory to higherExpand
Hyperdescent and étale K-theory
• Mathematics
• 2019
We study the etale sheafification of algebraic K-theory, called etale K-theory. Our main results show that etale K-theory is very close to a noncommutative invariant called Selmer K-theory, which isExpand
Topological cyclic homology, Handbook of homotopy theory
• 2019
Topological cyclic homology, Handbook of homotopy theory (Haynes Miller, ed.), CRC Press/Chapman
• 2019
Descent in algebraic $K$-theory and a conjecture of Ausoni-Rognes
• Mathematics
• 2016
Let $A \to B$ be a $G$-Galois extension of rings, or more generally of $\mathbb{E}_\infty$-ring spectra in the sense of Rognes. A basic question in algebraic $K$-theory asks how close the map \$K(A)Expand
Chromatic redshift
Notes for the author’s MSRI lecture in January 2014.
On the algebraic K-theory of the complex K-theory spectrum
Let p≥5 be a prime, let ku be the connective complex K-theory spectrum, and let K(ku) be the algebraic K-theory spectrum of ku. In this paper we study the p-primary homotopy type of the spectrumExpand
On motivic cohomology with Z/l -coefficients
In this paper we prove the conjecture of Bloch and Kato which relates Milnor’s K-theory of a field with its Galois cohomology as well as the related comparisons results for motivic cohomology withExpand
The homotopy limit problem for two-primary algebraic K-theory
• Mathematics
• 2005
Abstract We solve the homotopy limit problem for two-primary algebraic K-theory of fields, that is, the Quillen–Lichtenbaum conjecture at the prime 2.


text:
# Real analysis question: Suprema and Infima

Let $S$ be an ordered set with the $L.U.B$ Property, $S \supset B \neq \varnothing$, $B$ is bounded below. Write $L = \{ l : l \; \text{is a lower bound of } \; B \}$. Then, it follows that $\alpha = \sup L$ exists in $S$ and $\alpha = \inf B$.

# My Try:

(1) $L \neq \varnothing$: We are given that $B$ is bounded below which implies $L \neq \varnothing$.

(2) $L$ is bounded above: Let $l \in L$ be arbitrary, hence $l \leq b$ for all $b$. In particular, every $b$ is an upper bound for $l$. It follows that $L$ is bounded above

(3) $L$ has a L.U.B: follows from $S$ possess the Least Upper Bound property, $\alpha = \sup L$ exists.

(4) $\alpha$ is a lower bound for $B$: If there is $y \in B$ with $y < \alpha$, then $y$ is not an upper bound of $L$, a clear contradiction since we have established above that every element of $B$ is an upper bound of $L$. Hence, $\alpha \leq x$ for all $x \in B$.

(5) This implies $\alpha \in L$

(6) $\alpha$ is the G.L.B (infimum) of $B$. Suppose $\beta$ is another arbitrary lower bound of $B$ with $\alpha < \beta$. I need to show that $\beta$ is not a lower bound of $B$: $\beta \notin L$. This is true since any number greater than the supremum of $L$ is not in $L$ and we have $\alpha < \beta$.

• Looks OK to me. I took the liberty of setting out the steps a little more clearly. – Tom Collinge Nov 27 '14 at 10:45
• Thank you very much. Just one question to see if I have understood the concept. This problem in fact shows that $S$ also has the greatest lower bound property. Correct? – user195835 Nov 27 '14 at 13:42
• Yes, the two are equivalent. See math.stackexchange.com/questions/470895/… – Tom Collinge Nov 27 '14 at 13:57


text:
# 10. Reduction Formulae

by M. Bourne

You may have noticed in the Table of Integrals that some integrals are given in terms of a simpler integral. These require a few steps to find the final answer.

Reduction formulae are integrals involving some variable n, as well as the usual x. They are normally obtained from using integration by parts.

We use the notation I_n when writing reduction formulae.

### Example 1

Given the reduction formula

I_n=intsin^nx\ dx =-1/ncos\ x\ sin^(n-1)x+(n-1)/nI_(n-2),

find intsin^4x\ dx.

### Example 2

We are given that if

I_n=inttan^nx\ dx,

then I_n=1/(n-1)tan^(n-1)x-I_(n-2).

Find inttan^3x\ dx.

See the Scientific Notebook version of the same Examples. (info: ).

Didn't find what you are looking for on this page? Try search:

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text:
2021 AIME II Problems/Problem 14

Problem

Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Diagram

~MRENTHUSIASM (by Geometry Expressions)

Solution 1

Let $M$ be the midpoint of $BC$. Because $\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o$, $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$, and $\angle{XOY}=\angle{AOM}$. Because $\angle{AOM}=2\angle{BCA}+\angle{BAC}$, it's easy to get $\frac{585}{7} \implies \boxed{592}$ from here.

~Lcz

Solution 2

In this solution, all angle measures are in degrees.

Let $M$ be the midpoint of $\overline{BC}$ so that $\overline{OM}\perp\overline{BC}$ and $A,G,M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$

Note that:

1. Since $\angle OGX = \angle OAX = 90,$ quadrilateral $OGAX$ is cyclic by the Converse of the Inscribed Angle Theorem.

It follows that $\angle OAG = \angle OXG,$ as they share the same intercepted arc $OG.$

2. Since $\angle OGY = \angle OMY = 90,$ quadrilateral $OGYM$ is cyclic by the supplementary opposite angles.

It follows that $\angle OMG = \angle OYG,$ as they share the same intercepted arc $OG.$

Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \angle XOY = 17k.$

Next, we express $\angle BAC$ in terms of $k.$ By angle addition, we have \begin{align*} \angle AOM &= \angle AOB + \angle BOM \\ &= \underbrace{2\angle BCA}_{\substack{\text{Inscribed} \\ \text{Angle} \\ \text{Theorem}}} + \underbrace{\frac12\angle BOC}_{\substack{\text{Perpendicular} \\ \text{Bisector} \\ \text{Property}}} \\ &= 2\angle BCA + \underbrace{\angle BAC}_{\substack{\text{Inscribed} \\ \text{Angle} \\ \text{Theorem}}}. \end{align*} Substituting back gives $17k=2(2k)+\angle BAC,$ from which $\angle BAC=13k.$

For the sum of the interior angles of $\triangle ABC,$ we get \begin{align*} \angle ABC + \angle BCA + \angle BAC &= 180 \\ 13k+2k+13k&=180 \\ 28k&=180 \\ k&=\frac{45}{7}. \end{align*} Finally, we obtain $\angle BAC=13k=\frac{585}{7},$ from which the answer is $585+7=\boxed{592}.$

~Constance-variance (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 3 (Guessing in the Last 3 Minutes, Unreliable)

Notice that $\triangle ABC$ looks isosceles, so we assume it's isosceles. Then, let $\angle BAC = \angle ABC = 13x$ and $\angle BCA = 2x.$ Taking the sum of the angles in the triangle gives $28x=180,$ so $13x = \frac{13}{28} \cdot 180 = \frac{585}{7}$ so the answer is $\boxed{592}.$

Solution 4 (Easy and Simple)

Firstly, let $M$ be the midpoint of $BC$. Then, $\angle OMB = 90^o$. Now, note that since $\angle OGX = \angle XAO = 90^o$, quadrilateral $AGOX$ is cyclic. Also, because $\angle OMY + \angle OGY = 180^o$, $OMYG$ is also cyclic. Now, we define some variables: let $\alpha$ be the constant such that $\angle ABC = 13\alpha, \angle ACB = 2\alpha,$ and $\angle XOY = 17\alpha$. Also, let $\beta = \angle OMG = \angle OYG$ and $\theta = \angle OXG = \angle OAG$ (due to the fact that $AGOX$ and $OMYG$ are cyclic). Then, $$\angle XOY = 180 - \beta - \theta = 17\alpha \implies \beta + \theta = 180 - 17\alpha.$$ Now, because $AX$ is tangent to the circumcircle at $A$, $\angle XAC = \angle CBA = 13\alpha$, and $\angle CAO = \angle OAX - \angle CAX = 90 - 13\alpha$. Finally, notice that $\angle AMB = \angle OMB - \angle OMG = 90 - \beta$. Then, $$\angle BAM = 180 - \angle ABC - \angle AMB = 180 - 13\alpha - (90 - \beta) = 90 + \beta - 13\alpha.$$ Thus, $$\angle BAC = \angle BAM + \angle MAO + \angle OAC = 90 + \beta - 13\alpha + \theta + 90 - 13\alpha = 180 - 26\alpha + (\beta + \theta),$$ and $$180 = \angle BAC + 13\alpha + 2\alpha = 180 - 11\alpha + \beta + \theta \implies \beta + \theta = 11\alpha.$$ However, from before, $\beta+\theta = 180 - 17 \alpha$, so $11 \alpha = 180 - 17 \alpha \implies 180 = 28 \alpha \implies \alpha = \frac{180}{28}$. To finish the problem, we simply compute $$\angle BAC = 180 - 15 \alpha = 180 \cdot \left(1 - \frac{15}{28}\right) = 180 \cdot \frac{13}{28} = \frac{585}{7},$$ so our final answer is $585+7=\boxed{592}$.

~Osman Nal


text:
Q

# Tell me? - Complex numbers and quadratic equations - JEE Main-12

The complex number $z$ satisfying $\left |z \right |= 2$  will be

• Option 1)

$z= 1+i$

• Option 2)

$z= \sqrt{3}-i$

• Option 3)

$\sqrt{2}+i$

• Option 4)

$z= 1-i$

119 Views

if $z=1+i$, then $\left |z \right |=\sqrt{1^{2}+1^{2}}=\sqrt{2}$

if $z=\sqrt{3}-i$, then  $\left |z \right |=\sqrt{\left (\sqrt{3} \right )^{2}+\left (-1 \right )^{2}}=\sqrt{3+1}=2$

if $z=\sqrt{2}+i$, then  $\left |z \right |=\sqrt{\left (\sqrt{2} \right )^{2}+1^{2}}=\sqrt{2+1}=\sqrt{3}$

if $z=1+i$ then  $\left |z \right |=\sqrt{1^{2}+\left (-1 \right )^{2}}=\sqrt{1+1}=\sqrt{2}$

$\therefore$ Option (B)

Definition of Modulus of z(Complex Number) -

$\left | z \right |=\sqrt{a^{2}+b^{2}}$ is the distance of z from origin in Argand plane

- wherein

Real part of z = Re (z) = a & Imaginary part of z = Im (z) = b

Option 1)

$z= 1+i$

This is incorrect

Option 2)

$z= \sqrt{3}-i$

This is correct

Option 3)

$\sqrt{2}+i$

This is incorrect

Option 4)

$z= 1-i$

This is incorrect

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