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# Sets of morphisms between free modules#

The class FreeModuleHomset implements sets of homomorphisms between two free modules of finite rank over the same commutative ring.

AUTHORS:

• Eric Gourgoulhon, Michal Bejger (2014-2015): initial version

REFERENCES:

class sage.tensor.modules.free_module_homset.FreeModuleHomset(fmodule1, fmodule2, name=None, latex_name=None)#

Set of homomorphisms between free modules of finite rank over a commutative ring.

Given two free modules $$M$$ and $$N$$ of respective ranks $$m$$ and $$n$$ over a commutative ring $$R$$, the class FreeModuleHomset implements the set $$\mathrm{Hom}(M,N)$$ of homomorphisms $$M\rightarrow N$$. The set $$\mathrm{Hom}(M,N)$$ is actually a free module of rank $$mn$$ over $$R$$, but this aspect is not taken into account here.

This is a Sage parent class, whose element class is FiniteRankFreeModuleMorphism.

INPUT:

• fmodule1 – free module $$M$$ (domain of the homomorphisms), as an instance of FiniteRankFreeModule

• fmodule2 – free module $$N$$ (codomain of the homomorphisms), as an instance of FiniteRankFreeModule

• name – (default: None) string; name given to the hom-set; if none is provided, Hom(M,N) will be used

• latex_name – (default: None) string; LaTeX symbol to denote the hom-set; if none is provided, $$\mathrm{Hom}(M,N)$$ will be used

EXAMPLES:

Set of homomorphisms between two free modules over $$\ZZ$$:

sage: M = FiniteRankFreeModule(ZZ, 3, name='M')
sage: N = FiniteRankFreeModule(ZZ, 2, name='N')
sage: H = Hom(M,N) ; H
Set of Morphisms from Rank-3 free module M over the Integer Ring
to Rank-2 free module N over the Integer Ring
in Category of finite dimensional modules over Integer Ring
sage: type(H)
<class 'sage.tensor.modules.free_module_homset.FreeModuleHomset_with_category_with_equality_by_id'>
sage: H.category()
Category of homsets of modules over Integer Ring

Hom-sets are cached:

sage: H is Hom(M,N)
True

The LaTeX formatting is:

sage: latex(H)
\mathrm{Hom}\left(M,N\right)

As usual, the construction of an element is performed by the __call__ method; the argument can be the matrix representing the morphism in the default bases of the two modules:

sage: e = M.basis('e')
sage: f = N.basis('f')
sage: phi = H([[-1,2,0], [5,1,2]]) ; phi
Generic morphism:
From: Rank-3 free module M over the Integer Ring
To:   Rank-2 free module N over the Integer Ring
sage: phi.parent() is H
True

An example of construction from a matrix w.r.t. bases that are not the default ones:

sage: ep = M.basis('ep', latex_symbol=r"e'")
sage: fp = N.basis('fp', latex_symbol=r"f'")
sage: phi2 = H([[3,2,1], [1,2,3]], bases=(ep,fp)) ; phi2
Generic morphism:
From: Rank-3 free module M over the Integer Ring
To:   Rank-2 free module N over the Integer Ring

The zero element:

sage: z = H.zero() ; z
Generic morphism:
From: Rank-3 free module M over the Integer Ring
To:   Rank-2 free module N over the Integer Ring
sage: z.matrix(e,f)
[0 0 0]
[0 0 0]

The test suite for H is passed:

sage: TestSuite(H).run()

The set of homomorphisms $$M\rightarrow M$$, i.e. endomorphisms, is obtained by the function End:

sage: End(M)
Set of Morphisms from Rank-3 free module M over the Integer Ring
to Rank-3 free module M over the Integer Ring
in Category of finite dimensional modules over Integer Ring

End(M) is actually identical to Hom(M,M):

sage: End(M) is Hom(M,M)
True

The unit of the endomorphism ring is the identity map:

sage: End(M).one()
Identity endomorphism of Rank-3 free module M over the Integer Ring

whose matrix in any basis is of course the identity matrix:

sage: End(M).one().matrix(e)
[1 0 0]
[0 1 0]
[0 0 1]

There is a canonical identification between endomorphisms of $$M$$ and tensors of type $$(1,1)$$ on $$M$$. Accordingly, coercion maps have been implemented between $$\mathrm{End}(M)$$ and $$T^{(1,1)}(M)$$ (the module of all type-$$(1,1)$$ tensors on $$M$$, see TensorFreeModule):

sage: T11 = M.tensor_module(1,1) ; T11
Free module of type-(1,1) tensors on the Rank-3 free module M over
the Integer Ring
sage: End(M).has_coerce_map_from(T11)
True
sage: T11.has_coerce_map_from(End(M))
True

See TensorFreeModule for examples of the above coercions.

There is a coercion $$\mathrm{GL}(M) \rightarrow \mathrm{End}(M)$$, since every automorphism is an endomorphism:

sage: GL = M.general_linear_group() ; GL
General linear group of the Rank-3 free module M over the Integer Ring
sage: End(M).has_coerce_map_from(GL)
True

Of course, there is no coercion in the reverse direction, since only bijective endomorphisms are automorphisms:

sage: GL.has_coerce_map_from(End(M))
False

The coercion $$\mathrm{GL}(M) \rightarrow \mathrm{End}(M)$$ in action:

sage: a = GL.an_element() ; a
Automorphism of the Rank-3 free module M over the Integer Ring
sage: a.matrix(e)
[ 1  0  0]
[ 0 -1  0]
[ 0  0  1]
sage: ea = End(M)(a) ; ea
Generic endomorphism of Rank-3 free module M over the Integer Ring
sage: ea.matrix(e)
[ 1  0  0]
[ 0 -1  0]
[ 0  0  1]

Element#
one()#

Return the identity element of self considered as a monoid (case of an endomorphism set).

This applies only when the codomain of self is equal to its domain, i.e. when self is of the type $$\mathrm{Hom}(M,M)$$.

OUTPUT:

EXAMPLES:

Identity element of the set of endomorphisms of a free module over $$\ZZ$$:

sage: M = FiniteRankFreeModule(ZZ, 3, name='M')
sage: e = M.basis('e')
sage: H = End(M)
sage: H.one()
Identity endomorphism of Rank-3 free module M over the Integer Ring
sage: H.one().matrix(e)
[1 0 0]
[0 1 0]
[0 0 1]
sage: H.one().is_identity()
True

NB: mathematically, H.one() coincides with the identity map of the free module $$M$$. However the latter is considered here as an element of $$\mathrm{GL}(M)$$, the general linear group of $$M$$. Accordingly, one has to use the coercion map $$\mathrm{GL}(M) \rightarrow \mathrm{End}(M)$$ to recover H.one() from M.identity_map():

sage: M.identity_map()
Identity map of the Rank-3 free module M over the Integer Ring
sage: M.identity_map().parent()
General linear group of the Rank-3 free module M over the Integer Ring
sage: H.one().parent()
Set of Morphisms from Rank-3 free module M over the Integer Ring
to Rank-3 free module M over the Integer Ring
in Category of finite dimensional modules over Integer Ring
sage: H.one() == H(M.identity_map())
True

Conversely, one can recover M.identity_map() from H.one() by means of a conversion $$\mathrm{End}(M)\rightarrow \mathrm{GL}(M)$$:

sage: GL = M.general_linear_group()
sage: M.identity_map() == GL(H.one())
True


text:
1. Math
2. Mathematics
3. 9 flour is sold in four different size packets which...

# Question: 9 flour is sold in four different size packets which...

###### Question details

9. Flour is sold in four different size packets. Which is the best buy?

A 450 g for $1.98 B 600 g for$2.60

C 800 kg for $3.42 D 1.5 kg for$6.80

10. The Coordinated Universal Time (UTC) of Darwin is + 9.5 hours and the UTC of Los Angeles is −8 hours. When the time in Darwin is 5 pm Friday, what is the time in Los Angeles?

A 11:30 pm Thursday

B 11:30 am Friday

C 10:30 am Saturday

D 10:30 pm Saturday

11. Megan is 23 years old and has been driving for 3 years. She needs to make a claim on her comprehensive motor vehicle insurance as the driver at-fault in a car accident. The following excesses apply to claims for at-fault motor vehicle accidents:

• Basic excess of $600 for each claim • An additional age excess of$1100 for drivers under 25 years of age

• An additional age excess of $450 for drivers 25 years of age or over with no more than 2 years driving experience How much would Stefan be required to pay as excess? A$600

B $600 +$450

C $600 +$1100

D $600 +$1100 + $450 12. What is the interest earned if$900 is invested for 8 months at a simple interest rate of 2.5% p.a?

A $15 B$150

C $180 D$1800

format: plain


text:
# How to write superscripts and subscripts in math mode?

I just started using latex and the closest I've come is

$F\overset{fr}{\textsubscript{s}max}$

or

$F{fr \above 0pt \ \textsubscript{s}max}$

## 1 Answer

Try this one:

$F^{\mathrm{fr}}_{s\mathrm{MAX}}$.

In math mode ^ means superscript and _ means subscript.


text:
Article

# Secret Correlation in Repeated Games with Imperfect Monitoring

Authors:
To read the full-text of this research, you can request a copy directly from the authors.

## Abstract

We characterize the maximum payoff that a team can guarantee against another in a class of repeated games with im- perfect monitoring. Our result relies on the optimal trade-off for the team between optimization of stage-payoffs and generation of signals for future correlation.

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Chapter
Glossary Definition of the Subject Introduction Correlated Equilibrium: Definition and Basic Properties Correlated Equilibrium and Communication Correlated Equilibrium in Bayesian Games Related Topics and Future Directions Acknowledgment Bibliography
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This paper studies the interaction of automata of size m. We characterise statistical properties satisfied by random plays generated by a correlated pair of automata with m states each. We show that in some respect the pair of automata can be identified with a more complex automaton of size comparable to mlogm\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m\log m$$\end{document}. We investigate implications of these results on the correlated min–max value of repeated games played by automata.
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Résumé. Le but du cours est de présenter certains outils et résultats fondamentaux de la théorie des jeux. On étudiera principalement l'approche stratégique en considérant les problèmes liés à l'information et à la dynamique. Les auteurs remercient Miquel Oliu-Barton, Tristan Tomala, Cheng Wan ainsi que Vianney Perchet, Guillaume Vigeral et Yannick Viossat pour leurs lectures attentives et leurs remarques judicieuses.
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One of the main goals of bounded rationality models is to understand the limitations of agent's abilities in building representations of strategic situations as maximization problems and in solving these problems. Modern cryptography relies on the assumption that agents's computations should be implementable by polynominal Turing machines and on the exstence of a trapdoor function. Uder those assumption, we prove that very correlated equilibrium of the original infinitely repreated game can be implemented through public communication only.
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This paper examines repeated games in which each player observes a private and imperfect signal on the actions played and in which players are allowed to communicate using public messages. Providing incentives for players to reveal their observations generate revelation constraints which, combined with signal imperfections, may be a source of inefficiencies. However, the author shows that, by delaying the revelation of their observations, players may reduce the cost of deterring deviations. With at least three players, he obtains a Nash threat version of the folk theorem. With two players, the author shows that an efficient outcome can (almost) always be approximated.
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The Nash equilibrium concept may be extended gradually when the rules of the game are interpreted in a wider and wider sense, so as to allow preplay or even intraplay communication. A well-known extension of the Nash equilibrium is Aumann's correlated equilibrium, which depends only on the normal form of the game. Two other solution concepts for multistage games are proposed here: the extensive form correlated equilibrium, where the players can observe private extraneous signals at every stage and the communication equilibrium where the players are furthermore allowed to transmit inputs to an appropriate device at every stage. We show that the set of payoffs associated with each solution concept has a canonical representation (in the spirit of the revelation principle) and is a convex polyhedron. We also provide for each concept a "super canonical" game such that the set of payoffs associated with the solution concept is precisely the set of Nash equilibrium payoffs of this game.
Article
We prove the folk theorem for discounted repeated games under private, almost-perfect monitoring. Our result covers all finite, n-player games that satisfy the usual full-dimensionality condition. Mixed strategies are allowed in determining the individually rational payoffs. We assume no cheap-talk communication between players and no public randomization device. Copyright The Econometric Society 2006.
Article
The authors study repeated games in which players observe a public outcome that imperfectly signals the actions played. They provide conditions guaranteeing that any feasible, individually rational payoff vector of the stage game can arise as a perfect equilibrium of the repeated game with sufficiently little discounting. The central condition requires that there exist action profiles with the property that, for any two players, no two deviations--one by either player--give rise to the same probability distribution over public outcomes. The results apply to principal-agent, partnership, oligopoly, and mechanism-design models, and to one-shot games with transferable utilities. Copyright 1994 by The Econometric Society.
Article
Introduction Many interactions involve teams of participants that have coinciding interests but that have to act individually. For example, the individuals in a company work towards a common goal, but often have to make their decisions independently. Coordinating their actions may be impossible or too expensive. Similarly, the performance of a distributed computer system in a given situation depends on individual actions taken by the processors. In many situations, it may be useful to have the processors randomly choose their actions. However, randomization is only efficient if done locally by each processor. In the game of Bridge, the two players on each team are forbidden by the rules to coordinate their actions by secret communication. These situations can be modeled as noncooperative games. In such a game, we define a team as a set of n players with identical payoffs. 1 We are concerned with teams facing a single
On the optimal use of coordination
• O Gossner
• R Laraki
• T Tomala
Gossner, O., R. Laraki, T. Tomala. 2006. On the optimal use of coordination. Math. Programming B. Forthcoming.
Team max min equilibria. Games and Econom
• B Von Stengel
• D Koller
von Stengel, B., D. Koller. 1997. Team max min equilibria. Games and Econom. Behav. 21 309–321.


text:
# Inexact accelerated augmented Lagrangian methods

## Author Info

Listed author(s):
• Myeongmin Kang

()

• Myungjoo Kang

()

• Miyoun Jung

()

Registered author(s):

## Abstract

The augmented Lagrangian method (ALM) is a popular method for solving linearly constrained convex minimization problems, and it has been used in many applications such as compressive sensing or image processing. Recently, accelerated versions of the augmented Lagrangian method (AALM) have been developed, and they assume that the subproblem can be exactly solved. However, the subproblem of the augmented Lagrangian method in general does not have a closed-form solution. In this paper, we introduce an inexact version of an accelerated augmented Lagrangian method (I-AALM), with an implementable inexact stopping condition for the subproblem. It is also proved that the convergence rate of our method remains the same as the accelerated ALM, which is $${{\mathcal {O}}}(\frac{1}{k^2})$$ O ( 1 k 2 ) with an iteration number $$k$$ k . In a similar manner, we propose an inexact accelerated alternating direction method of multiplier (I-AADMM), which is an inexact version of an accelerated ADMM. Numerical applications to compressive sensing or image inpainting are also presented to validate the effectiveness of the proposed iterative algorithms. Copyright Springer Science+Business Media New York 2015

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## Bibliographic Info

Article provided by Springer in its journal Computational Optimization and Applications.

Volume (Year): 62 (2015)
Issue (Month): 2 (November)
Pages: 373-404

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 Handle: RePEc:spr:coopap:v:62:y:2015:i:2:p:373-404 DOI: 10.1007/s10589-015-9742-8 Contact details of provider: Web page: http://www.springer.com Order Information: Web: http://www.springer.com/math/journal/10589

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American Institute of Mathematical Sciences

December  2018, 11(6): 1333-1358. doi: 10.3934/krm.2018052

A pedestrian flow model with stochastic velocities: Microscopic and macroscopic approaches

 Department of Mathematics, University of Mannheim, 68131 Mannheim, Germany

* Corresponding author: S. Göttlich

Received  March 2017 Revised  December 2017 Published  June 2018

We investigate a stochastic model hierarchy for pedestrian flow. Starting from a microscopic social force model, where the pedestrians switch randomly between the two states stop-or-go, we derive an associated macroscopic model of conservation law type. Therefore we use a kinetic mean-field equation and introduce a new problem-oriented closure function. Numerical experiments are presented to compare the above models and to show their similarities.

Citation: Simone Göttlich, Stephan Knapp, Peter Schillen. A pedestrian flow model with stochastic velocities: Microscopic and macroscopic approaches. Kinetic & Related Models, 2018, 11 (6) : 1333-1358. doi: 10.3934/krm.2018052
References:
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Schultz, Stochastic transition model for pedestrian dynamics, in Pedestrian and Evacuation Dynamics 2012, Springer International Publishing, (2013), 971-985, arXiv: 1210.5554. doi: 10.1007/978-3-319-02447-9_81. Google Scholar [26] A. Tordeux and A. Schadschneider, A stochastic optimal velocity model for pedestrian flow, in Parallel Processing and Applied Mathematics, Springer International Publishing, 9574 (2016), 528-538. doi: 10.1007/978-3-319-32152-3_49. Google Scholar [27] A. Tordeux and A. Schadschneider, White and relaxed noises in optimal velocity models for pedestrian flow with stop-and-go waves, Journal of Physics A: Mathematical and Theoretical, 49 (2016), 185101, 16pp. doi: 10.1088/1751-8113/49/18/185101. Google Scholar

show all references

References:
 [1] D. Armbruster, S. Martin and A. Thatcher, Elastic and inelastic collisions of swarms, Physica D: Nonlinear Phenomena, 344 (2017), 45-57. doi: 10.1016/j.physd.2016.11.008. Google Scholar [2] D. Armbruster, S. Motsch and A. Thatcher, Swarming in bounded domains, Physica D: Nonlinear Phenomena, 344 (2017), 58-67. doi: 10.1016/j.physd.2016.11.009. Google Scholar [3] H. Bauer, Probability Theory, vol. 23 of De Gruyter Studies in Mathematics, Walter de Gruyter & Co., Berlin, 1996, Translated from the fourth (1991) German edition by Robert B. Burckel and revised by the author. doi: 10.1515/9783110814668. Google Scholar [4] N. Bellomo, C. Bianca and V. Coscia, On the modeling of crowd dynamics: An overview and research perspectives, S$\vec{\rm e}$MA J., 54 (2011), 25-46. Google Scholar [5] C. Cercignani, R. Illner and M. Pulvirenti, The Mathematical Theory of Dilute Gases, vol. 106 of Applied Mathematical Sciences, Springer-Verlag, New York, 1994. doi: 10.1007/978-1-4419-8524-8. Google Scholar [6] L. Chen, S. Göttlich and Q. Yin, Mean field limit and propagation of chaos for a pedestrian flow model, Journal of Statistical Physics, 166 (2017), 211-229. doi: 10.1007/s10955-016-1679-5. Google Scholar [7] A. Chertock, A. Kurganov, A. Polizzi and I. Timofeyev, Pedestrian flow models with slowdown interactions, Math. Models Methods Appl. Sci., 24 (2014), 249-275. doi: 10.1142/S0218202513400083. Google Scholar [8] E. Cristiani, B. Piccoli and A. Tosin, Multiscale Modeling of Pedestrian Dynamics, vol. 12 of MS&A. Modeling, Simulation and Applications, Springer, Cham, 2014. doi: 10.1007/978-3-319-06620-2. Google Scholar [9] P. Degond, C. Appert-Rolland, M. Moussaïd, J. Pettré and G. Theraulaz, A hierarchy of heuristic-based models of crowd dynamics, J. Stat. Phys., 152 (2013), 1033-1068. doi: 10.1007/s10955-013-0805-x. Google Scholar [10] P. Degond and C. Ringhofer, Stochastic dynamics of long supply chains with random breakdowns, SIAM J. Appl. 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Schultz, Stochastic transition model for pedestrian dynamics, in Pedestrian and Evacuation Dynamics 2012, Springer International Publishing, (2013), 971-985, arXiv: 1210.5554. doi: 10.1007/978-3-319-02447-9_81. Google Scholar [26] A. Tordeux and A. Schadschneider, A stochastic optimal velocity model for pedestrian flow, in Parallel Processing and Applied Mathematics, Springer International Publishing, 9574 (2016), 528-538. doi: 10.1007/978-3-319-32152-3_49. Google Scholar [27] A. Tordeux and A. Schadschneider, White and relaxed noises in optimal velocity models for pedestrian flow with stop-and-go waves, Journal of Physics A: Mathematical and Theoretical, 49 (2016), 185101, 16pp. doi: 10.1088/1751-8113/49/18/185101. Google Scholar
Velocity vector at the boundary
Overview of the deterministic and stochastic model hierarchy equations
Densities at different times: $u_{ij}^{\text{Mic}, n}$ for the microscopic and $u_{ij}^{\text{Mac}, n}$ for the macroscopic model
Mass balances at $x = -1$ and $x = 0$
$L^1$ and $L^2$ error
Densities for $\lambda_1$ at different times: $u_{ij}^{\text{Mic}, n}$ for the microscopic and $u_{ij}^{\text{Mac}, n}$ for the macroscopic model
Densities for $\lambda_2$ at different times: $u_{ij}^{\text{Mic}, n}$ for the microscopic and $u_{ij}^{\text{Mac}, n}$ for the macroscopic model
Mass balances at $x = 1$
$L^1$ and $L^2$ errors
Numerical error and EOOC for the first example
 $\mathsf{err}$ EOOC $\Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.3251 - $\Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.1755 0.8897 $\Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.0717 1.2919
 $\mathsf{err}$ EOOC $\Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.3251 - $\Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.1755 0.8897 $\Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.0717 1.2919
Numerical error and EOOC for the second example with rate function $\lambda_1$
 $\mathsf{err}$ EOOC $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.4457 - $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.2215 1.0085 $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.0889 1.3170
 $\mathsf{err}$ EOOC $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.4457 - $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.2215 1.0085 $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.0889 1.3170
Numerical error and EOOC for the second example with rate function $\lambda_2$
 $\mathsf{err}$ EOOC $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.5203 - $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.2873 0.8567 $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.1153 1.3176
 $\mathsf{err}$ EOOC $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{5}\;$ 0.5203 - $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{10}\;$ 0.2873 0.8567 $\displaystyle \Delta x = {}^{1}\!\!\diagup\!\!{}_{20}\;$ 0.1153 1.3176
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text:
### AST_ELLIPSE

Create a Ellipse

#### Description:

This function creates a new Ellipse and optionally initialises its attributes.

A Ellipse is a Region which represents a elliptical area within the supplied 2-dimensional Frame.

#### Invocation

RESULT = AST_ELLIPSE( FRAME, FORM, CENTRE, POINT1, POINT2, UNC, OPTIONS, STATUS )

#### Arguments

##### FRAME = INTEGER (Given)
A pointer to the Frame in which the region is defined. It must have exactly 2 axes. A deep copy is taken of the supplied Frame. This means that any subsequent changes made to the Frame using the supplied pointer will have no effect the Region.
##### FORM = INTEGER (Given)
Indicates how the ellipse is described by the remaining parameters. A value of zero indicates that the ellipse is specified by a centre position and two positions on the circumference. A value of one indicates that the ellipse is specified by its centre position, the half-lengths of its two axes, and the orientation of its first axis.
##### CENTRE( 2 ) = DOUBLE PRECISION (Given)
An array containing the coordinates at the centre of the ellipse.
##### POINT1( 2 ) = DOUBLE PRECISION (Given)
If FORM is zero, this array should contain the coordinates of one of the four points where an axis of the ellipse crosses the circumference of the ellipse. If FORM is one, it should contain the lengths of semi-major and semi-minor axes of the ellipse, given as geodesic distances within the Frame.
##### POINT2( 2 ) = DOUBLE PRECISION (Given)
If FORM is zero, this array should containing the coordinates at some other point on the circumference of the ellipse, distinct from POINT1. If FORM is one, the first element of this array should hold the angle between the second axis of the Frame and the first ellipse axis (i.e. the ellipse axis which is specified first in the POINT1 array), and the second element will be ignored. The angle should be given in radians, measured positive in the same sense as rotation from the positive direction of the second Frame axis to the positive direction of the first Frame axis.
##### UNC = INTEGER (Given)
An optional pointer to an existing Region which specifies the uncertainties associated with the boundary of the Ellipse being created. The uncertainty in any point on the boundary of the Ellipse is found by shifting the supplied " uncertainty" Region so that it is centred at the boundary point being considered. The area covered by the shifted uncertainty Region then represents the uncertainty in the boundary position. The uncertainty is assumed to be the same for all points.

If supplied, the uncertainty Region must be of a class for which all instances are centro-symetric (e.g. Box, Circle, Ellipse, etc.) or be a Prism containing centro-symetric component Regions. A deep copy of the supplied Region will be taken, so subsequent changes to the uncertainty Region using the supplied pointer will have no effect on the created Ellipse. Alternatively, a null Object pointer (AST__NULL) may be supplied, in which case a default uncertainty is used equivalent to a box 1.0E-6 of the size of the Ellipse being created.

The uncertainty Region has two uses: 1) when the AST_OVERLAP function compares two Regions for equality the uncertainty Region is used to determine the tolerance on the comparison, and 2) when a Region is mapped into a different coordinate system and subsequently simplified (using AST_SIMPLIFY), the uncertainties are used to determine if the transformed boundary can be accurately represented by a specific shape of Region.

##### OPTIONS = CHARACTER $\ast$ ( $\ast$ ) (Given)
A character string containing an optional comma-separated list of attribute assignments to be used for initialising the new Ellipse. The syntax used is identical to that for the AST_SET routine.
##### STATUS = INTEGER (Given and Returned)
The global status.

#### Returned Value

##### AST_ELLIPSE = INTEGER
A pointer to the new Ellipse.

#### Notes:

• A null Object pointer (AST__NULL) will be returned if this function is invoked with STATUS set to an error value, or if it should fail for any reason.


text:
# Revision history [back]

I am sorry that i can not access to the cocalc file. The difference is basically that a lambda function is anonymous: you do not need to give it a name to use it in a function.

For example, if a function named foo takes a function as input, you can do:

def f(x):
return blah
foo(f)

or

foo(lambda x : blah)

In the first construction, you have to give the name f to define the function, in the second case, you can pass the function to foo without having to give it a name first.


text:
# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective

As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.

Is the above correct?

The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.

If my solution was correct, which choice is more "canonical"?

Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra..

• I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– ℋolo
Sep 30 '19 at 8:58
• I was a bit puzzled because I did not find my solution anywhere in the web..
– Tom
Sep 30 '19 at 9:02
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine"..
– Tom
Sep 30 '19 at 9:03
• I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋolo
Sep 30 '19 at 12:47

Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear).

One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same.

This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$.

By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition.

So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$.

Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required.

• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12

Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$.

Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$.

Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).

• in the second line "gives rise to the injective mapping".. you meant surjective?
– Tom
Sep 30 '19 at 8:56
• Try \sim which gives $\sim$ Sep 30 '19 at 8:56


text:
# Bash script to ftp only files of type gz in current directory

I have a bash manuscript that cd to the appropriate directory site after that makes use of the mput command to ftp all the files in the directory site to a remote server, as listed below:

for from_dir in $DATE_LIST do cd${FILE_LANDING_DIR}/deltas/${from_dir} ftp_files done ftp_files () { ftp -inv$HOST <<-EOF > $LOG_DIR/tempftplog user$USER $PASS cd${ID_LANDING_DIR}
mput *
bye
EOF
}

I attempted modifying the ftp_files function to ftp just the .gz files because directory site by transforming the line from mput * to mput *.gz yet that did not function.

Can a person inform me just how I can do this as soon as in the ftp command?

1
2022-06-07 14:32:17
Source Share

Does mput *.gz not function? Otherwise, attempt the ftp glob command first.

0
2022-06-07 14:51:05
Source

What concerning executing a directory site search and also piping it to the mput?

ls *.gz | mput

I can not examine this now due to the fact that Im not at a linux box so it could require to be fine-tuned to get it to function 100% appropriately.

0
2022-06-07 14:50:23
Source


text:
# Using Laplace Transform (LT) and Inverse Laplace Transform (LT) solve the following system of equations: 1....

###### Question:

Using Laplace Transform (LT) and Inverse Laplace Transform (LT) solve the following system of equations: 1. X'=- 2x + 5oy y' = x - y With x(0) = 25, and y(0) = 0 2. x' + 4x - y = 7t x' + y' - 2y = 3t With x(0) = 1, and y(0) = 0

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##### C++ Program What is the output of the following code fragment?(beta is of type int.) beta...
C++ Program What is the output of the following code fragment?(beta is of type int.) beta = 5; do { switch (beta) { case 1: cout <<'R'; break; case 2: cout case 4: cout << 'O'; break; case 5: cout << 'L'; } beta--; }while (beta>1); cout <<'X'...
##### 3000 exercise books are arranged in 3 piles
3000 exercise books are arranged in 3 piles.the first pile has 10 more books than the second pile.the number of books in the second pile is twice the number of books in the third pile.how many books are in the third pile?...
##### Problem 2115 Points) Solve the following initial-boundary-value prob- lem: u(0,t) = 1 uz(1,1) + ßu(1,1) = 1 BCs: 0 < t < oo 0 IC: u(z,0)= sin(nx)+x, 1 x by transforming it into homogeneous...
Problem 2115 Points) Solve the following initial-boundary-value prob- lem: u(0,t) = 1 uz(1,1) + ßu(1,1) = 1 BCs: 0 < t < oo 0 IC: u(z,0)= sin(nx)+x, 1 x by transforming it into homogeneous BCs and then solving the transformed problem Problem #2115 Points) Solve the following initial-bo...
##### Homework: 6-2 MyStatLab: Module Six Problem Set Score: 0 of 5 pts 5 of 14 (4...
Homework: 6-2 MyStatLab: Module Six Problem Set Score: 0 of 5 pts 5 of 14 (4 complete) 4.1.40-T Lyme disease is an inflammatory disease that results in a skin rash and flulike symptoms. It is transmitted through the bite of an infected deer tick. The following data represent the number of reported c...
##### An object is located 15.0 cm to the left of a diverging lens whose focal points...
An object is located 15.0 cm to the left of a diverging lens whose focal points are 35.2 cm from the lens. (a) Determine the distance (unsigned) of the image from the lens. distance     cm location     ---Select--- to the right of the lens to the left of ...
##### QUESTION 2 3 p Compute the multifactor productivity measure for an 8 hour day in which...
QUESTION 2 3 p Compute the multifactor productivity measure for an 8 hour day in which the usable output was 630 units, produced by 5 workers who used 905 pounds of materials. Workers have an hourly wage of 17 and material cost is 2.2 per pound. Overhead is 1.5 times labor cost The multifactor produ...