text:
# American Institute of Mathematical Sciences

eISSN:
2688-1594

All Issues

## Electronic Research Archive

March 2021 , Volume 29 , Issue 1

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2021, 29(1): 1625-1639 doi: 10.3934/era.2020083 +[Abstract](932) +[HTML](384) +[PDF](357.62KB)
Abstract:

In this paper, we show that a weak solution \begin{document}$(\mathbf{u},\mathbf{w},\mathbf{b})(\cdot,t)$\end{document} of the magneto-micropolar equations, defined in \begin{document}$[0,T)$\end{document}, which satisfies \begin{document}$\nabla u_3, \nabla_{h} \mathbf{w}, \nabla_{h} \mathbf{b}$\end{document} \begin{document}$\in L^{\frac{32}{7}}(0,T;$\end{document} \begin{document}$L^2(\mathbb{R}^3))$\end{document} or \begin{document}$\partial_3 u_3, \partial_3 \mathbf{w}, \partial_3 \mathbf{b} \in L^{\infty}(0,T;L^2(\mathbb{R}^3))$\end{document}, is regular in \begin{document}$\mathbb{R}^3\times(0,T)$\end{document} and can be extended as a \begin{document}$C^\infty$\end{document} solution beyond \begin{document}$T$\end{document}.

2021, 29(1): 1641-1660 doi: 10.3934/era.2020084 +[Abstract](1146) +[HTML](406) +[PDF](819.15KB)
Abstract:

The objective of the current paper is to investigate the dynamics of a new bioeconomic predator prey system with only predator's harvesting and Holling type Ⅲ response function. The system is equipped with an algebraic equation because of the economic revenue. We offer a detailed mathematical analysis of the proposed model to illustrate some of the significant results. The boundedness and positivity of solutions for the model are examined. Coexistence equilibria of the bioeconomic system have been thoroughly investigated and the behaviours of the model around them are described by means of qualitative theory of dynamical systems (such as local stability and Hopf bifurcation). The obtained results provide a useful platform to understand the role of the economic revenue \begin{document}$v$\end{document}. We show that a positive equilibrium point is locally asymptotically stable when the profit \begin{document}$v$\end{document} is less than a certain critical value \begin{document}$v^{*}_1$\end{document}, while a loss of stability by Hopf bifurcation can occur as the profit increases. It is evident from our study that the economic revenue has the capability of making the system stable (survival of all species). Finally, some numerical simulations have been carried out to substantiate the analytical findings.

2021, 29(1): 1661-1679 doi: 10.3934/era.2020085 +[Abstract](1034) +[HTML](406) +[PDF](387.44KB)
Abstract:

In this paper, we provide an effective method for estimating the thresholds of the stochastic models with time delays by using of the nonnegative semimartingale convergence theorem. Firstly, we establish the stochastic delay differential equation models for two diseases, and obtain two thresholds of two diseases and the sufficient conditions for the persistence and extinction of two diseases. Then, numerical simulations for co-infection of HIV/AIDS and Gonorrhea in Yunnan Province, China, are carried out. Finally, we discuss some biological implications and focus on the impact of some key model parameters. One of the most interesting findings is that the stochastic fluctuation and time delays introduced into the deterministic models can suppress the outbreak of the diseases, which can provide some useful control strategies to regulate the dynamics of the diseases, and the numerical simulations verify this phenomenon.

2021, 29(1): 1681-1689 doi: 10.3934/era.2020086 +[Abstract](921) +[HTML](414) +[PDF](297.34KB)
Abstract:

In this paper, we will investigate some properties for almost cellular algebras. We compare the almost cellular algebras with quasi-hereditary algebras, which are known to carry any homological and categorical structures. We prove that any almost cellular algebra is the iterated inflation and obtain some sufficient and necessary conditions for an almost cellular algebra \begin{document}$\mathrm{A}$\end{document} to be quasi-hereditary.

2021, 29(1): 1691-1708 doi: 10.3934/era.2020087 +[Abstract](880) +[HTML](358) +[PDF](415.27KB)
Abstract:

In present paper, we deal with the behavior of a solution beyond the occurrence of wave breaking for a modified periodic Coupled Camassa-Holm system. By introducing a new set of independent and dependent variables, which resolve all singularities due to possible wave breaking, this evolution system is rewritten as a closed semilinear system. The local existence of the semilinear system is obtained as fixed points of a contractive transformation. Moreover, this formulation allows us to continue the solution after wave breaking, and gives a global conservative solution where the energy is conserved for almost all times. Returning to the original variables. We finally obtain a semigroup of global conservative solutions, which depend continuously on the initial data. Additionally, our results repair some gaps in the pervious work.

2021, 29(1): 1709-1734 doi: 10.3934/era.2020088 +[Abstract](1686) +[HTML](491) +[PDF](484.76KB)
Abstract:

In this paper, we investigate a final boundary value problem for a class of fractional with parameter \begin{document}$\beta$\end{document} pseudo-parabolic partial differential equations with nonlinear reaction term. For \begin{document}$0<\beta < 1,$\end{document} the solution is regularity-loss, we establish the well-posedness of solutions. In the case that \begin{document}$\beta >1$\end{document}, it has a feature of regularity-gain. Then, the instability of a mild solution is proved. We introduce two methods to regularize the problem. With the help of the modified Lavrentiev regularization method and Fourier truncated regularization method, we propose the regularized solutions in the cases of globally or locally Lipschitzian source term. Moreover, the error estimates is established.

2021, 29(1): 1735-1752 doi: 10.3934/era.2020089 +[Abstract](1134) +[HTML](421) +[PDF](883.1KB)
Abstract:

In this paper, the second-order scalar auxiliary variable approach in time and linear finite element method in space are employed for solving the Cahn-Hilliard type equation of the phase field crystal model. The energy stability of the fully discrete scheme and the boundedness of numerical solution are studied. The rigorous error estimates of order \begin{document}$O(\tau^2+h^2)$\end{document} in the sense of \begin{document}$L^2$\end{document}-norm is derived. Finally, some numerical results are given to demonstrate the theoretical analysis.

2021, 29(1): 1753-1782 doi: 10.3934/era.2020090 +[Abstract](1075) +[HTML](563) +[PDF](492.02KB)
Abstract:

Unique identifiability by finitely many far-field measurements in the inverse scattering theory is a highly challenging fundamental mathematical topic. In this paper, we survey some recent progress on the inverse obstacle scattering problems and the inverse medium scattering problems associated with time-harmonic waves within a certain polyhedral geometry, where one can establish the unique identifiability results by finitely many measurements. Some unique identifiability issues on the inverse diffraction grating problems are also considered. Furthermore, the geometrical structures of Laplacian and transmission eigenfunctions are reviewed, which have important applications in the unique determination for inverse obstacle and medium scattering problems with finitely many measurements. We discuss the mathematical techniques and methods developed in the literature. Finally, we raise some intriguing open problems for the future investigation.

2021, 29(1): 1783-1801 doi: 10.3934/era.2020091 +[Abstract](1025) +[HTML](344) +[PDF](424.01KB)
Abstract:

We prove some results on the stability of slow stationary solutions of the MHD equations in two- and three-dimensional bounded domains for external force fields that are asymptotically autonomous. Our results show that weak solutions are asymptotically stable in time in the \begin{document}$L^2$\end{document}-norm. Further, assuming certain regularity hypotheses on the problem data, strong solutions are asymptotically stable in the \begin{document}$H^1$\end{document} and \begin{document}$H^2$\end{document}-norms.

2021, 29(1): 1803-1818 doi: 10.3934/era.2020092 +[Abstract](944) +[HTML](366) +[PDF](346.04KB)
Abstract:

The goal of this paper is to provide a new combinatorial meaning to two fifth order and four sixth order mock theta functions. Lattice paths of Agarwal and Bressoud with certain modifications are used as a tool to study these functions.

2021, 29(1): 1819-1839 doi: 10.3934/era.2020093 +[Abstract](1198) +[HTML](568) +[PDF](522.1KB)
Abstract:

In this paper, we propose a conservative semi-Lagrangian finite difference (SLFD) weighted essentially non-oscillatory (WENO) scheme, based on Runge-Kutta exponential integrator (RKEI) method, to solve one-dimensional scalar nonlinear hyperbolic equations. Conservative semi-Lagrangian schemes, under the finite difference framework, usually are designed only for linear or quasilinear conservative hyperbolic equations. Here we combine a conservative SLFD scheme developed in [21], with a high order RKEI method [7], to design conservative SLFD schemes, which can be applied to nonlinear hyperbolic equations. Our new approach will enjoy several good properties as the scheme for the linear or quasilinear case, such as, conservation, high order and large time steps. The new ingredient is that it can be applied to nonlinear hyperbolic equations, e.g., the Burgers' equation. Numerical tests will be performed to illustrate the effectiveness of our proposed schemes.

2021, 29(1): 1841-1857 doi: 10.3934/era.2020094 +[Abstract](861) +[HTML](321) +[PDF](368.99KB)
Abstract:

In this paper, we establish a new summation formula for Schur processes, called the complete summation formula. As an application, we obtain the generating function and the asymptotic formula for the number of doubled shifted plane partitions, which can be viewed as plane partitions "shifted at the two sides". We prove that the order of the asymptotic formula depends only on the diagonal width of the doubled shifted plane partition, not on the profile (the skew zone) itself. By using similar methods, the generating function and the asymptotic formula for the number of symmetric cylindric partitions are also derived.

2021, 29(1): 1859-1880 doi: 10.3934/era.2020095 +[Abstract](892) +[HTML](348) +[PDF](2296.84KB)
Abstract:

The mathematical model of a semiconductor device is described by a coupled system of three quasilinear partial differential equations. The mixed finite element method is presented for the approximation of the electrostatic potential equation, and the characteristics finite element method is used for the concentration equations. First, we estimate the mixed finite element and the characteristics finite element method solution in the sense of the \begin{document}$L^q$\end{document} norm. To linearize the full discrete scheme of the problem, we present an efficient two-grid method based on the idea of Newton iteration. The two-grid algorithm is to solve the nonlinear coupled equations on the coarse grid and then solve the linear equations on the fine grid. Moreover, we obtain the \begin{document}$L^{q}$\end{document} error estimates for this algorithm. It is shown that a mesh size satisfies \begin{document}$H = O(h^{1/2})$\end{document} and the two-grid method still achieves asymptotically optimal approximations. Finally, the numerical experiment is given to illustrate the theoretical results.

Bin Wang and
2021, 29(1): 1881-1895 doi: 10.3934/era.2020096 +[Abstract](914) +[HTML](364) +[PDF](1246.21KB)
Abstract:

In this paper, we develop a viscosity robust weak Galerkin finite element scheme for Stokes equations. The major idea for achieving pressure-independent energy-error estimate is to use a divergence preserving velocity reconstruction operator in the discretization of the right hand side body force. The optimal convergence results for velocity and pressure have been established in this paper. Finally, numerical examples are presented for validating the theoretical conclusions.

2021, 29(1): 1897-1923 doi: 10.3934/era.2020097 +[Abstract](905) +[HTML](350) +[PDF](430.32KB)
Abstract:

A weak Galerkin (WG) finite element method is presented for nonlinear conservation laws. There are two built-in parameters in this WG framework. Different choices of the parameters will lead to different approaches for solving hyperbolic conservation laws. The convergence analysis is obtained for the forward Euler time discrete and the third order explicit TVDRK time discrete WG schemes respectively. The theoretical results are verified by numerical experiments.

2020 Impact Factor: 1.833
5 Year Impact Factor: 1.833


text:
# What is the Cartesian form of ( 6,-pi/3 )?

$\left(3 , - 3 \setminus \sqrt{3}\right)$

#### Explanation:

The Cartesian coordinates $\left(x , y\right)$ of the point $\left(6 , - \frac{\setminus \pi}{3}\right) \setminus \equiv \left(r , \setminus \theta\right)$ are given as follows

$x = r \setminus \cos \setminus \theta$

$= 6 \setminus \cos \left(- \frac{\setminus \pi}{3}\right)$

$= 6 \setminus \cos \left(\setminus \frac{\pi}{3}\right)$

$= 6 \setminus \cdot \frac{1}{2}$

$= 3$

$y = r \setminus \sin \setminus \theta$

$= 6 \setminus \sin \left(- \frac{\setminus \pi}{3}\right)$

$= - 6 \setminus \sin \left(\setminus \frac{\pi}{3}\right)$

$= - 6 \setminus \cdot \setminus \frac{\sqrt{3}}{2}$

$= - 3 \setminus \sqrt{3}$

hence, the Cartesian coordinates are $\left(3 , - 3 \setminus \sqrt{3}\right)$

format: plain


text:
# zbMATH — the first resource for mathematics

Phase-type distributions and risk processes with state-dependent premiums. (English) Zbl 0876.62089
Summary: Consider a risk reserve process with initial reserve $$u$$, Poisson arrivals, premium rule $$p(r)$$ depending on the current reserve $$r$$ and claim size distribution which is phase-type in the sense of M. F. Neuts [see “Matrix-geometric solutions in stochastic models. An algorithmic approach” (1981; Zbl 0469.60002)]. It is shown that the ruin probabilities $$\psi(u)$$ can be expressed as the solution of a finite set of differential equations, and similar results are obtained for the case where the process evolves in a Markovian environment (e.g., a numerical example of a stochastic interest rate is presented). Further, an explicit formula for $$\psi(u)$$ is presented for the case where $$p(r)$$ is a two-step function. By duality, the results apply also to the stationary distribution of storage processes with the same input and release rate $$p(r)$$ at content $$r$$.

##### MSC:
 62P05 Applications of statistics to actuarial sciences and financial mathematics 91B30 Risk theory, insurance (MSC2010) 62M99 Inference from stochastic processes
Full Text:
##### References:
 [1] Sparre Andersen E., Transactions XVth International Congress of Actuaries, New York, II pp 219– (1957) [2] Asmussen S., Applied Probability and Queues (1987) [3] Asmussen S., Ruin Probabilities (1995) [4] Asmussen S., Does Markov-modulation increase the risk? (1994) [5] Asmussen S., Ruin Probabilities via local adjustment coefficients (1995) · Zbl 0834.60099 [6] DOI: 10.2307/1427367 · Zbl 0657.60111 [7] DOI: 10.1016/0167-6687(92)90058-J · Zbl 0748.62058 [8] Baker C. T. H., The Numerical Solution of Integral Equations (1977) [9] Bladt M., Ph.D. thesis (1993) [10] Cramer H., On the Mathematical Theory of Risk (1930) [11] Emanuel D. C., Scand. Actuarial J. pp 37– (1975) [12] Feller W., An Introduction to Probability Theory and its Applications. Vol. II, 2. ed. (1971) · Zbl 0219.60003 [13] Gerber H., Mitt. Verein Schweitz. Versich. Math. 71 pp 63– (1971) [14] Gerber H., An Introduction to the Mathematical Risk Theory (1979) · Zbl 0431.62066 [15] Graham A., Kronecker Products and Matrix Calculus with Applications (1981) · Zbl 0497.26005 [16] DOI: 10.1007/978-1-4613-9058-9 [17] DOI: 10.1016/0304-4149(77)90051-5 · Zbl 0361.60053 [18] Neuts M. F., Matrix-Geometric Solutions in Stochastic Models (1981) · Zbl 0469.60002 [19] Schock Petersen S., Scand. Actuarial J. pp 147– (1990) [20] Taylor G. C., Scand. Act. J. pp 57– (1980) · Zbl 0426.62069
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Fixing a messed table

I am trying to make a table in Latex. Now I have combined most of my Latex experience in making this (I am still new) and I thought I was making good progress until I saw the end result.

The problem is that the text is all messed, and it makes the table too long. I honestly have no idea on how this can be fixed and I would appreciate any possible help.

Here is the MWE:

\documentclass[a4paper,10pt]{article}

\usepackage[table]{xcolor}
\usepackage[utf8]{inputenc}
\usepackage{tabularx}
\usepackage{array}

%define some nice colors
\definecolor{middleware_green}{RGB}{82, 119, 17}
\definecolor{middleware_white}{RGB}{255, 255, 255}
\definecolor{middleware_light_gray}{RGB}{239, 239, 239}
\definecolor{middleware_dark_gray}{RGB}{153, 153, 153}

%renewing commands for tabularx
\renewcommand{\tabularxcolumn}[1]{>{\small}m{#1}}

%new column type for test tables
\newcolumntype{L}{>{\leavevmode\ignorespaces\raggedleft\arraybackslash\sffamily}X}%
\newcolumntype{C}{>{\leavevmode\ignorespaces\centering\arraybackslash\sffamily}X}%

\newcolumntype{R}{|>{\leavevmode\ignorespaces\color{magenta!70!black}\raggedright\sffamily\bfseries\footnotesize}p{2.5cm}}%
\newcolumntype{H}{|>{\leavevmode\ignorespaces\color{orange!30!black}\centering\arraybackslash\sffamily\footnotesize}X}%
\newcolumntype{J}{|>{\leavevmode\ignorespaces\color{green!30!black}\centering\sffamily\footnotesize}X}%
\newcolumntype{W}{|>{\leavevmode\ignorespaces\color{blue!30!black}\centering\arraybackslash\sffamily\footnotesize}X|}%

\begin{document}

\begin{table}[ph!]
\footnotesize\centering
\rowcolors{2}{middleware_white}{middleware_light_gray}

\begin{tabularx}{\textwidth}{|L|L|L|L|L|L|}
\hline\rowcolor{middleware_green}
\multicolumn{ 6}{|c|}{\small\textcolor{white}{\textbf{Cloud Types}}} \\ \hline
\rowcolor{middleware_dark_gray}
&   \multicolumn{1}{|C}{\textcolor{white}{\textbf{Pu\-blic cloud}}}             &
\multicolumn{1}{|C}{\centering\textcolor{white}{\textbf{Pri\-va\-te cloud}}}    &
\multicolumn{1}{|C}{\centering\textcolor{white}{\textbf{Hy\-brid cloud}}}       &
\multicolumn{1}{|C|}{\textcolor{white}{\textbf{Vir\-tu\-al pri\-va\-te cloud}}}     &
\multicolumn{1}{|C|}{\textcolor{white}{\textbf{Com\-mu\-ni\-ty cloud}}}         \\ \hline
Definition  &   Focuses on sharing or renting services and infrastructures from external entities with several clients. In this type of cloud the resources are usually allocated dynamically during an undetermined amount of time.    &   Also known as internal clouds, these are designed for the exclusive use of a company. Depending on the contract, it may be maintained by the company itself, or by an external company. These clouds usually exist when the client has specific needs or when it wants a long term contract.    &   This is a combination of the public cloud with the private cloud, trying to overcome the limitations of both types while gaining the benefits from them. As the name suggest, in this type of cloud, part of the system is external to the client, while the other part is internally managed by him.   &   This is an alternative to attack the problems present in both private and public clouds. In this approach, the supplier of a public cloud has a has system of virtual private networks (VPNs) that allow him to define management and security protocols, thus effectively simulate a network of virtual servers and their connections. This is sometimes mistaken as a hybrid cloud, however the main difference is that in such a cloud, both public and private clouds are real, while in the one the private cloud is merely virtualized on the top of the public cloud.    &   This type of cloud is a shared cloud between several entities that share a common interest, such as politics, security or other.    \\ \hline
\item The client does not need initialization money to build the cloud because this money has already been invested by the supplier.
\item Migrates the responsibility and management of risks to the supplier.
\item Are usually a lot bigger than the private clouds.
\end{itemize}
&   \begin{itemize}
\item The client has more control over the infrastructures, and can therefore have its own security and failure protocols.
\item Easier and cheaper than buying a database.
\end{itemize}
&   \begin{itemize}
\item More flexible than a private cloud or a public cloud.
\item Provide more control than public clouds have easier scalability when compared to private clouds.
\item Helps isolate and delegate the computational resources used, thus increasing the availability of the company's own resources.
\end{itemize}
&   \begin{itemize}
\item Allows the client to use the provider's infrastructures as part of his own cloud.
\item Allows the dynamic manipulation of the protocols defined.
\item Easier to scale because the created servers are virtual.
\end{itemize}
&   \begin{itemize}
\item Because it shares the costs, this option is the cheapest one.
\end{itemize}
\\  \hline
\item The client has no control over the data, security and network.
\end{itemize}
&   \begin{itemize}
\item Criticized by being similar to private server farms, thus sharing all their problems including maintenance.
\item Requires an initial investment to buy the necessary infrastructures.
\end{itemize}
&   \begin{itemize}
\item Requires a careful analysis of what should go to the private cloud and what should go to the public one.
\item Moving big amounts of data between the two clouds results in additional costs and stateful applications are harder to maintain.
\end{itemize}
&   \begin{itemize}
\item Adds the complexity of migrating workload from the client to the supplier's cloud.
\end{itemize}
&   \begin{itemize}
\item Because each company has specific security restrictions, this type of cloud has to manage them all at he same time, consequently not being very effective at doing it.
\item There is little control over the resources because the environment in which they are is shared.
\end{itemize}
\\  \hline
\end{tabularx}
\caption{Cloud Types\label{tab:cloud_types}}
\end{table}

\end{document}

-
Please ask single questions like "how can I get a wider table" or "how do I use itemize in tables" or "how to align the first lines of text in table cells" or alike. It is not clear, what you are asking for. Until know, you have too much text in the cells. Easy to see this. Do you really need this? You will need to split your table then onto more pages or set the font to really small. You will find all this information on this homepage. I guess, all this would be much easier to solve in text with description or itemize. No need for green colours then... –  LaRiFaRi May 16 '14 at 14:05
The problem is that I do not know what I need in order to fix this table ! I think I did everything alright, but the output clearly disagrees with me :S –  Flame_Phoenix May 16 '14 at 15:01

An alternative of table version. The main issue, in my view, is the use of m column specifier, which is for text vertically aligned in the middle. Use p for paragraph text vertically aligned at the top instead is what the proposed method did.

\renewcommand{\tabularxcolumn}[1]{>{\small}m{#1}}  % use p{#1} in the proposal

Further to gain some spaces for itemized item, the solutions uses

\setlength{\leftmargini}{1em}

to move the item to the left a bit.

Code

\documentclass[a4paper,10pt]{article}
\usepackage[margin=0.5cm]{geometry}
\usepackage[table]{xcolor}
\usepackage[utf8]{inputenc}
\usepackage{tabularx,multirow}
\usepackage{array}

%define some nice colors
\definecolor{middleware_green}{RGB}{82, 119, 17}
\definecolor{middleware_white}{RGB}{255, 255, 255}
\definecolor{middleware_light_gray}{RGB}{239, 239, 239}
\definecolor{middleware_dark_gray}{RGB}{153, 153, 153}

%renewing commands for tabularx
\renewcommand{\tabularxcolumn}[1]{>{\small} p{#1}}
%new column type for test tables
\newcolumntype{L}{>{\raggedright\arraybackslash\sffamily}X}%
\newcolumntype{C}{>{\centering\arraybackslash\sffamily}X}%

\begin{document}

\begin{table}[ph!]
\footnotesize\centering
\rowcolors{2}{middleware_white}{middleware_light_gray}

\begin{tabularx}{\textwidth}{|L|L|L|L|L|L|}
\hline\rowcolor{middleware_green}
\multicolumn{6}{|c|}{\small\textcolor{white}{\textbf{Cloud Types}}} \\ \hline
\rowcolor{middleware_dark_gray}
&   \multicolumn{1}{C|}{\textcolor{white}{\textbf{Pu\-blic cloud}}}             &
\multicolumn{1}{C|}{\centering\textcolor{white}{\textbf{Pri\-va\-te cloud}}}    &
\multicolumn{1}{C|}{\centering\textcolor{white}{\textbf{Hy\-brid cloud}}}       &
\multicolumn{1}{C|}{\textcolor{white}{\textbf{Vir\-tu\-al pri\-va\-te cloud}}}     &
\multicolumn{1}{C|}{\textcolor{white}{\textbf{Com\-mu\-ni\-ty cloud}}}  \\ \hline
Definition
&
Focuses on sharing or renting services and infrastructures from external entities with several clients. In this type of cloud the resources are usually allocated dynamically during an undetermined amount of time.    &
Also known as internal clouds, these are designed for the exclusive use of a company. Depending on the contract, it may be maintained by the company itself, or by an external company. These clouds usually exist when the client has specific needs or when it wants a long term contract.
&
This is a combination of the public cloud with the private cloud, trying to overcome the limitations of both types while gaining the benefits from them. As the name suggest, in this type of cloud, part of the system is external to the client, while the other part is internally managed by him.
&
This is an alternative to attack the problems present in both private and public clouds. In this approach, the supplier of a public cloud has a has system of virtual private networks (VPNs) that allow him to define management and security protocols, thus effectively simulate a network of virtual servers and their connections. This is sometimes mistaken as a hybrid cloud, however the main difference is that in such a cloud, both public and private clouds are real, while in the one the private cloud is merely virtualized on the top of the public cloud.
&
This type of cloud is a shared cloud between several entities that share a common interest, such as politics, security or other.    \\ \hline  % 2nd column
\setlength{\leftmargini}{1em}
\begin{itemize}
\item The client does not need initialization money to build the cloud because this money has already been invested by the supplier.
\item Migrates the responsibility and management of risks to the supplier.
\item Are usually a lot bigger than the private clouds.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item The client has more control over the infrastructures, and can therefore have its own security and failure protocols.
\item Easier and cheaper than buying a database.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item More flexible than a private cloud or a public cloud.
\item Provide more control than public clouds have easier scalability when compared to private clouds.
\item Helps isolate and delegate the computational resources used, thus increasing the availability of the company's own resources.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Allows the client to use the provider's infrastructures as part of his own cloud.
\item Allows the dynamic manipulation of the protocols defined.
\item Easier to scale because the created servers are virtual.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Because it shares the costs, this option is the cheapest one.
\end{itemize}        \\  \hline
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item The client has no control over the data, security and network.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Criticized by being similar to private server farms, thus sharing all their problems including maintenance.
\item Requires an initial investment to buy the necessary infrastructures.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Requires a careful analysis of what should go to the private cloud and what should go to the public one.
\item Moving big amounts of data between the two clouds results in additional costs and stateful applications are harder to maintain.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Adds the complexity of migrating workload from the client to the supplier's cloud.
\end{itemize}
&
\setlength{\leftmargini}{1em}
\begin{itemize}
\item Because each company has specific security restrictions, this type of cloud has to manage them all at he same time, consequently not being very effective at doing it.
\item There is little control over the resources because the environment in which they are is shared.
\end{itemize}  \\  \hline
\end{tabularx}
\caption{Cloud Types\label{tab:cloud_types}}
\end{table}
\end{document}

-
Wow, this is exactly what I was looking for. Now I can play with the text a little more, but at least I have a basic structure to see ! Thanks! –  Flame_Phoenix May 16 '14 at 15:22
You are welcome. –  Jesse May 16 '14 at 15:25
\documentclass[a4paper,10pt]{article}

\usepackage[table]{xcolor}
\usepackage[utf8]{inputenc}
\usepackage{tabularx}
\usepackage{array}

%define some nice colors
\definecolor{middleware_green}{RGB}{82, 119, 17}
\definecolor{middleware_white}{RGB}{255, 255, 255}
\definecolor{middleware_light_gray}{RGB}{239, 239, 239}
\definecolor{middleware_dark_gray}{RGB}{153, 153, 153}

%renewing commands for tabularx
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\begin{document}

\subsection{Definitions}
\begin{description}
\item[Public Cloud]
Focuses on sharing or renting services and infrastructures from external entities with several clients. In this type of cloud the resources are usually allocated dynamically during an undetermined amount of time.
\item[Private Cloud]
Also known as internal clouds, these are designed for the exclusive use of a company. Depending on the contract, it may be maintained by the company itself, or by an external company. These clouds usually exist when the client has specific needs or when it wants a long term contract.
\item[Hybrid Cloud]
This is a combination of the public cloud with the private cloud, trying to overcome the limitations of both types while gaining the benefits from them. As the name suggest, in this type of cloud, part of the system is external to the client, while the other part is internally managed by him.
\item[Virtual Private Cloud]
This is an alternative to attack the problems present in both private and public clouds. In this approach, the supplier of a public cloud has a has system of virtual private networks (VPNs) that allow him to define management and security protocols, thus effectively simulate a network of virtual servers and their connections. This is sometimes mistaken as a hybrid cloud, however the main difference is that in such a cloud, both public and private clouds are real, while in the one the private cloud is merely virtualized on the top of the public cloud.
\item[Community by Cloud]
This type of cloud is a shared cloud between several entities that share a common interest, such as politics, security or other.
\end{description}

\begin{description}

\item[Public Cloud]
\begin{itemize}
\item The client does not need initialization money to build the cloud because this money has already been invested by the supplier.
\item Migrates the responsibility and management of risks to the supplier.
\item Are usually a lot bigger than the private clouds.
\end{itemize}
\item[Private cloud]
\begin{itemize}
\item The client has more control over the infrastructures, and can therefore have its own security and failure protocols.
\item Easier and cheaper than buying a database.
\end{itemize}
\item[Hybrid cloud]
\begin{itemize}
\item More flexible than a private cloud or a public cloud.
\item Provide more control than public clouds have easier scalability when compared to private clouds.
\item Helps isolate and delegate the computational resources used, thus increasing the availability of the company's own resources.
\end{itemize}
\item[Virtual private cloud]
\begin{itemize}
\item Allows the client to use the provider's infrastructures as part of his own cloud.
\item Allows the dynamic manipulation of the protocols defined.
\item Easier to scale because the created servers are virtual.
\end{itemize}
\item[Community by cloud]
\begin{itemize}
\item Because it shares the costs, this option is the cheapest one.
\end{itemize}
\end{description}

\subsection{Disdvantages}
\begin{description}

\item[Public Cloud]
\begin{itemize}
\item The client has no control over the data, security and network.
\end{itemize}
\item[Private Cloud]
\begin{itemize}
\item Criticized by being similar to private server farms, thus sharing all their problems including maintenance.
\item Requires an initial investment to buy the necessary infrastructures.
\end{itemize}
\item[Hybrid cloud]
\begin{itemize}
\item Requires a careful analysis of what should go to the private cloud and what should go to the public one.
\item Moving big amounts of data between the two clouds results in additional costs and stateful applications are harder to maintain.
\end{itemize}
\item[Virtual private cloud]
\begin{itemize}
\item Adds the complexity of migrating workload from the client to the supplier's cloud.
\end{itemize}
\item[Community by cloud]
\begin{itemize}
\item Because each company has specific security restrictions, this type of cloud has to manage them all at he same time, consequently not being very effective at doing it.
\item There is little control over the resources because the environment in which they are is shared.
\end{itemize}
\end{description}

\end{document}

-
So, in your opinion, a Table is a terrible solution to output this information right? –  Flame_Phoenix May 16 '14 at 15:04
@Flame_Phoenix Not quite what I said, but I don't think I'd use a table for this much text. That much footnotesize text on an actual paper would be unreadable (I probably ought to wear my glasses more often:-) –  David Carlisle May 16 '14 at 16:02


text:
# Why is it considered acid rain with pH <5.6?

I recently read in a book that rain is considered acid rain if the pH falls below 5.6. However a substance is acidic when the pH is below 7; so why is the boundary for acid rain 5.6?

I was thinking pH between 5.6-7 would be too diluted to have an effect on limestone and other materials, and that's why the pH has to be under 5.6. But I wasn't sure if this is true so I wanted to find out.

• Because "normal rain" still is a bit acidic. – Mithoron Jul 27 at 22:11
• Actually, just the carbon dioxide the rain picks up on the way down is enough to have an effect on limestone - that's how limestone is dissolved with no human activity, and that's how all those wonderful caves with stalagmites and stalactites are formed. – Luaan Jul 29 at 10:13
• Would you consider something at a pH of 6,9 acidic? How about 6,99? What's the point of that if the substance is relatively neutral? Keep in mind pH is logarithmic, a couple of points is a major difference. A line is drawn, it had to be drawn somewhere, and this is where they put it. – Mast Jul 30 at 6:17
• @Luaan This, of course, begs the question of how those big limestone masses had managed to form in the first place. :-) – oakad Jul 31 at 1:42
• @oakad It's quite interesting, actually - rain water is acidic enough to dissolve limestone, yes. But the deposits form out of seawater, which is basic rather than acidic. And then you have life, of course. Those deposits then either get uplifted, or release their carbon dioxide as they get subducted. If the deposit gets enough rain, it will slowly be dissolved away back into the oceans (building up caves and sinkholes in the meantime). – Luaan Jul 31 at 7:22

You are forgetting an important component of the air: carbon dioxide. When it dissolves in pure water (=rain water), it makes it acidic. It is not considered that harmful.

Acid rain has a negative connotation; it is mainly caused by anthropogenic activities. The low pH of acid rain is due to sulfur oxides and nitrogen oxides and it is indeed below 5.7.

IUPACs definition "Rain with pH values < about 5; commonly results from acids formed from pollutants. 'Pure' rain water equilibrated with atmospheric CO2 and naturally occurring acids in relatively clean air usually has a pH>5."

• but why is it considered acid rain only below pH 5.7? Why not below 7? – muhammad haider Jul 27 at 19:38
• Because natural rain, which is free from all types of anthropogenic pollution has a pH < 7. You need to "distinguish" natural acidity vs. excessive acidity due man's activities. Otherwise what is the point of calling each an every rain as acid rain. Also remember there is little bit of nitric acid formed from lightning which is the plant food. That will not be termed as acid rain either. As I said, acid rain has a negative meaning. – M. Farooq Jul 27 at 21:37
• "Why not below 7?" I think this has to do with how you parse the term "acid rain". You seem to consider it to mean "an acid (anything with a pH < 7) which is raining". As the other answers have explained, acid rain means rain which is more acid than normal rain. – Philip Ngai Jul 28 at 20:48
• @M.Farooq So why choose 5.7 as a boundary value? Acidic soils having pH < 5 or even 4.5 are not uncommon, so such a rain would not harm them. Of course, such pH values are harmless for humans. – trolley813 Jul 29 at 8:10
• @trolley813, see Loong's answer. A pH of 5.67 is the border pH of CO2 in water. Soil chemistry is a complete different story. Acid rain was not named on the basis of its effect on soil by Angus Smith in 1872. He saw deleterious effects on plants and chlorophyll from "acid gases". His book is available online. – M. Farooq Jul 29 at 13:04

The $$\mathrm{pH}$$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($$p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$$) can be calculated as follows.

$$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$

where $$[\ce{H2CO3^*}]$$ is the total analytical concentration of dissolved $$\ce{CO2}$$, i.e. $$[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$$, and
$$K_\mathrm H= 3.39\times10^{-2}\ \mathrm{mol\ l^{-1}\ atm^{-1}}$$ is Henry's law constant for $$\ce{CO2}$$.

\begin{align} \log[\ce{H2CO3^*}]&=\log K_\mathrm H+\log p_{\ce{CO2}}\\ &=-1.5-3.5\\ &=-5.0 \end{align}

The commonly used first acid dissociation constant of carbonic acid $$\mathrm pK_{\mathrm a1}=6.3$$ (at $$25\ \mathrm{^\circ C}$$) actually is a composite constant that includes both the hydration reaction $$\ce{H2O + CO2(aq) <=> H2CO3}$$ and the protolysis of true $$\ce{H2CO3}$$ $$\ce{H2CO3 <=> H+ + HCO3-}$$ For a weak acid \begin{align} \log[\ce{H+}]&\approx\frac12\left(\log K_\mathrm a+\log[\ce{H2CO3^*}]\right)\\ &=\frac12\left(-6.3-5.0\right)\\ &=-5.65\\ \mathrm{pH}&=5.65 \end{align}

Thus, pure rain in equilibrium with the atmosphere has about $$\mathrm{pH}=5.65$$. Any acid rain with lower $$\mathrm{pH}$$ would be caused by additional acids.

Under atmospheric pressure, dissolved carbon dioxide can reach an equilibrium state in water that yields a pH of as low as 5.7


text:
henselization and completion

This might not be a question appropriate for this forum, I apologize in this case... Is it true that any element of the completion of a valued ring $R$ that is algebraic over the field of fractions of $R$ is in the henselization of $R$ ?

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@Sell: What do you mean by the Henselization of a valued field $R$? Do you mean to ask whether every element in the completion of some valuation ring $R$ that is algebraic over $R$ is in the Henselization of $R$? –  Jason Starr Aug 24 '12 at 12:26
Yes, that's exactly what I meant. Thank you for having reformulated it in a better way. Is it possible for me to edit my question so I can reformulate it ? –  seli Aug 24 '12 at 12:59
This is true if $R$ is a discrete valuation ring (I don't know whether it is true in general). See Ex. 8.3.34 (p360 first ed.) of Liu's book "Algebraic geometry and arithmetic curves". In fact in that case, the henselization is actually equal to the set of elements of $\widehat{R}$, which are algebraic over $R$. –  Damian Rössler Aug 24 '12 at 14:22
It is false: if $R$ is a non-excellent dvr (so of equichar. $p > 0$) there's $a \in R$ not a $p$th power that is a $p$th power in $\widehat{R}$, so $a^{1/p} \in \widehat{R} - R^h$. By Artin-Popescu approximation, for any excellent normal local noetherian ring, the "algebraic closure" in the completion is the henselization. (The henselization is also excellent.) This includes any excellent dvr. The omission of excellence is an error in Liu's book, at least in the 1st edition; I don't know if this was corrected in later editions. –  user22479 Aug 24 '12 at 17:43
@quasi-coherent: you are quite right, one has to suppose that $R$ is excellent. Otherwise, the algebraic closure in the completion might not be separable (in fact this is what you point out). Sorry for that. –  Damian Rössler Aug 24 '12 at 17:51

Edit Add noetherian hypothesis because $R$ might no be a subring of $\hat{R}$ otherwise.

The answer is no as pointed out by quasi-coherent in the comments. But suppose $R$ is a discrete valuation ring and denote $K=\mathrm{Frac}(R)$, then

$R^h=\hat{R}\cap K^{sep}$.

There are several equivalent properties defining henselien local rings. I will use two of them: let $(A, m)$ be a local ring and $k=A/m$. Then $A$ is henselian iff

(a) for any monic polynomial $f(X)\in A[X]$, any simple root of $\bar{f}(X)\in k[X]$ lifts to root of $f(X)$ in $A$.

or equivalently

(b) If $A\to A'$ is a local homomorphism, étale and with trivial residue extension, then $A'=A$.

One can find a proof in Raynaud: Anneaux locaux henséliens, IV, §3. For discrete valuation rings, one can certainly find easier references.

Now let us prove the claim above $R^h$. As $\hat{R}$ is henselian (Hensel Lemma), we have $R\subseteq \hat{R}$.

(1) For any algebraic separable extension $F/K$ contained in $\hat{K}$, $B_F=\hat{R}\cap F$ is a DVR with ramification index $1$ over $R$ and trivial residue extension. This is easy.

(2) Let $B=B_L$ where $L$ is the separable closure of $K$ in $\hat{K}$. Let us prove that $B$ is henselian using Property (a) above. Let $f(X)\in B[X]$ as in (a). Then any simple root of $\bar{f}(X)\in k[X]$ lifts to a root $\lambda\in \hat{K}$. It is automatically a simple root, so $\lambda$ is separable over $L$, hence $\lambda\in L\cap \hat{R}=B$.

(3) Let $R\to R'$ be an extension to a henselian DVR. Let's us prove it factorizes throught $R\to B$, which will show that $B$ is a henselization of $R$. It is enough to prove this factorization for $B_F$ for any finite separable exension $F/K$ contained in $\hat{K}$. As $B_F$ is an étale extension of $R$, $B_F\otimes_R R'$ is étale over $R'$, with a unique maximal ideal above the maximal ideal of $R'$ and trivial residue extension at this maximal ideal (because the quotient $B_F\otimes_R R'/(m')\simeq R'/m'$). By Property (b), this implies that $F\otimes_K K'$ has a direct factor equal to $K'$. Hence $F\subseteq K'$ (the minimal polynomial of a primitive element of $F$ has a root in $K'$) and $B_F\subseteq R'$.

Remark Parts (1)-(2) work for any local ring $R$ such that $\hat{R}$ is a domain and such that $R\to \hat{R}$ is injective. In particular $\hat{R}\cap K^{sep}$ is henselian. (3) works for any valuation ring But in general, I don't think (3) holds because $B_F$ has no reason to be étale over $R$.

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Warning: this is OK only if by "valuation" you mean "valuation of rank one", i.e. the value group is a subgroup of $\mathbb{R}$. There are valuation rings (of rank $>1$) which are complete but not henselian! –  Laurent Moret-Bailly Aug 30 '12 at 6:23
Laurent, yes thanks ! I had in mind the $m$-adic completion, but a non-noetherian valuation ring is not separated in general and so doesn't embedd in its $m$-adic completion (can be equal to its residue field). And you remark makes the situation even worse. I will edit the answer. –  Qing Liu Aug 30 '12 at 8:54

I found that henselization R^h is algebraically closed in generally completion \widehat{R} for an arbitrary noetherian normal local ring (R, m) in Nagata's Book Local rings".

Please inform me of counter-example if any.

pmatsumi@gmail.com

-


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Register now and save up to $200 Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code ## Leila tagged by: Brent@GMATPrepNow This topic has 3 expert replies and 4 member replies j_shreyans Legendary Member Joined 07 Aug 2014 Posted: 510 messages Followed by: 5 members Thanked: 3 times #### Leila Wed Oct 29, 2014 10:03 am Elapsed Time: 00:00 • Lap #[LAPCOUNT] ([LAPTIME]) Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A)1/5^4 B)1/5^3 C)6/5^4 D)13/5^4 E)17/5^4 OAE Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 10862 messages Followed by: 1215 members Thanked: 5200 times GMAT Score: 770 Wed Oct 29, 2014 11:00 am j_shreyans wrote: Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A)1/5⁴ B)1/5³ C)6/5⁴ D)13/5⁴ E)17/5⁴ OAE Give: P(succeeds on 1 throw) = 1/5 P(succeeds at least 3 times) = P(succeeds 4 times OR succeeds 3 times) = P(succeeds 4 times) + P(succeeds 3 times) P(succeeds 4 times) P(succeeds 4 times) = P(succeeds 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time) = P(succeeds 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time) = 1/5 x 1/5 x 1/5 x 1/5 = 1/5⁴ P(succeeds 3 times) Let's examine one possible scenario in which Leila succeeds exactly 3 times: P(FAILS the 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time) = P(FAILS the 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time) = 4/5 x 1/5 x 1/5 x 1/5 = 4/5⁴ Keep in mind that this is only ONE possible scenario in which Leila succeeds exactly 3 times (Leila fails the 1st time). Leila can also FAIL the 2nd time, or the 3rd time or the 4th time. Each of these probabilities will also equal 4/5³ So, P(succeeds 3 times) = 4/5⁴ + 4/5⁴ + 4/5⁴ + 4/5⁴ = 16/5⁴ So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times) + P(succeeds 3 times) = 1/5⁴ + 16/5⁴ = 17/5⁴ = E Cheers, Brent _________________ Brent Hanneson – Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources Thanked by: j_shreyans GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! Mathsbuddy Master | Next Rank: 500 Posts Joined 08 Nov 2013 Posted: 447 messages Followed by: 1 members Thanked: 25 times Tue Nov 11, 2014 5:43 am Possible throw combinations (using 0 = lose, 1 = win): 1111 -> p = (1/5)^4 1110 -> p = (1/5)^3 x 4/5 = 4(1/5)^4 1101 -> p = (1/5)^3 x 4/5 = 4(1/5)^4 1011 -> p = (1/5)^3 x 4/5 = 4(1/5)^4 0111 -> p = (1/5)^3 x 4/5 = 4(1/5)^4 Add them up: TOTAL P = 17 x (1/5)^4 ANSWER E GMATinsight Legendary Member Joined 10 May 2014 Posted: 1001 messages Followed by: 21 members Thanked: 205 times Fri Nov 28, 2014 7:11 am j_shreyans wrote: Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A)1/5^4 B)1/5^3 C)6/5^4 D)13/5^4 E)17/5^4 OAE Chances of Success = 1/5 i.e.e Chances of Failure = 1-(1/5) = 4/5 Probability of Atleast 3 Successful includes 1) Probability of exactly 3 Successful attempt = 4C3 (1/5)^3 x (4/5) 2) Probability of exactly 4 Successful attempt = (1/5)^4 Total Probability = [4C3 (1/5)^3 x (4/5)] + [(1/5)^4] = (4x4 + 1) / (5^4) = 17/5^4 Answer: Option E _________________ Prosper!!! Bhoopendra Singh & Sushma Jha "GMATinsight" Contact Us Testimonials To register for One-on-One FREE ONLINE DEMO Class Call/e-mail e-mail: info@GMATinsight.com Mobile: +91-9999687183 / +91-9891333772 Get in touch for SKYPE-Based Interactive Private Tutoring One-On-One Classes fee - US$40 per hour &
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P(wins atleast 3 times)= P(wins 3 times)+P(wins 4 times)
= No. of ways of selecting 3 of 4*P(wins)*P(wins)*P(wins)P(loses) + P(wins)P(wins)P(wins)P(wins)

= 4C3 * 1/5 */15 * 1/5 * 4/5 + 1/5 * 1/5 */15 * 1/5
= 16/5^4 + 1/5^4
= 17 / 5^4

~Binit.

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Wed Jun 07, 2017 11:53 pm
j_shreyans wrote:
Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?

A)1/5^4

B)1/5^3

C)6/5^4

D)13/5^4

E)17/5^4

OAE

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Thu Jun 08, 2017 5:04 pm
P(at least 3) = P(exactly 3) + P(exactly 4)

P(exactly 4) is easy => 1/5 * 1/5 * 1/5 * 1/5

P(exactly 3) is trickier, since we have four different arrangements to consider. (The miss could come on any of the four throws.) That means we've got four orders:

Hit, Hit, Hit, Miss
Hit, Hit, Miss, Hit
Hit, Miss, Hit, Hit
Miss, Hit, Hit, Hit

Each of these has the same probability (1/5 * 1/5 * 1/5 * 4/5), and since we've got four of them, we multiply that by 4:

P(exactly 3) = 1/5 * 1/5 * 1/5 * 4/5 * 4

Adding the two up, we're done!

(1/5)⁴ + (1/5)⁴ * 4² =>

(1/5)⁴ * (1 + 16) =>

17/625

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j_shreyans wrote:
Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?

A)1/5^4

B)1/5^3

C)6/5^4

D)13/5^4

E)17/5^4
We need to determine the probability that Leila succeeds on exactly 3 throws or on all 4 throws.

Scenario 1: succeeds on exactly 3 throws

We can let Y denote a successful throw and N denote a non-successful throw:

P(Y-Y-Y-N) = 1/5 x 1/5 x 1/5 x 4/5 = 4/(5^4)

However, we must account for the order of Y-Y-Y-N. Using our formula for indistinguishable items, Y, Y, Y, and N can be arranged in 4!/3! = 4 ways.

Thus, the probability of succeeding on exactly 3 throws (out of 4 attempts) is 4/(5^4) x 4 = 16/(5^4).

Now we can determine scenario 2:

Scenario 2: succeeds on all 4 attempts

P(Y-Y-Y-Y) = 1/5 x 1/5 x 1/5 x 1/5 = 1/(5^4)

Thus, the probability of succeeding on 3 throws out of 4 or 4 throws out of 4 is 16/(5^4) + 1/(5^4) = 17/5^4.

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text:
Triangle Medians

A median of a triangle is a line segment from a vertex of a triangle to the midpoint of the opposite side of the triangle. The medians to the legs of a certain right triangle have lengths 13 and 19. What is the length of the hypotenuse of the triangle?

I don't know how to find the length of the hypotenuse with such little information given. Any ideas?

Draw a picture. Let our triangle be $ABC$, with right angle at $C$. Let the legs be $a$ and $b$. Let the median to side $a$ have length $13$. By the Pythagorean Theorem, we have $\frac{a^2}{4}+b^2=13^2$. Similarly, $a^2+\frac{b^2}{4}=19^2$. Solve.
Or better, don't solve. Add. We get $\frac{5}{4}(a^2+b^2)=13^2+19^2$.
In the above diagram, The relevant information is that $ADC$ and $ABE$ are right triangles too, and their hypotenuses are known and their legs are each one of the main triangle legs and half the other main triangle leg. Therefore, you get two equations using the Pythagorean Theorem, where $a=AB, b=AC, c=BC, d=DC, e=BE$ $$d^2=\frac{a^2}{4}+b^2$$ $$e^2=a^2+\frac{b^2}{4}$$ Solving this yields $$a=2\sqrt{\frac{(4e^2-d^2)}{15}}$$ $$b=2\sqrt{\frac{(4d^2-e^2)}{15}}$$ Using the Pythagorean Theorem again, this means the hypotenuse is: $$c=2\sqrt{\frac{(d^2+e^2)}{5}}$$


text:
# CGMO-2012 (China Girls Math Olympiad 2012) Problem 4

There is a stone at each vertex of a given regular $13$-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the $13$-gon

I am a young student who loves math . I like number theory and inequalities part the most , and preparing for Math Olympiads :)

Posted on Saturday,August 18, 2012, in Problems and tagged , , , , . Bookmark the permalink. 2 Comments.


text:
# Notes on an Emacs workflow for academic documents

I've put together a workflow for authoring in raw text using Emacs. These are my notes:

## 1. Why author in emacs?

The sociologist Kieran Healy argues that Emacs is a very useful tool if you want to present research results in a professional, clear, and well-organized manner. Frankly, to get to the point where this is true, one must climb a steep learning curve. I think the productivity gains make it worth the effort.

I do a lot of document preparation in Markdown. I can move documents from Markdown to LaTeX, a sophisticated document preparation system favored by many academics, using Pandoc. There is a "Pandoc mode" available for Emacs. Importantly, I can add LaTeX preamble commands to a YAML snippet at the top of each document, like so:

---
title: 'An Example Title.'
author: "Nick Judd"
abstract: "Paper abstract."
date: "February 15, 2016"
classoption: titlepage
- \usepackage{txfonts}
and so on ...

---

I can also add math which will be immediately typeset the "right way." This is one of the biggest perks to using something that is built on top of TeX. For instance, I can write $a^2 + b^2 = c^2$ and it will be converted from Markdown into LaTeX the same way it is converted using Mathjax on this site: $$a^2 + b^2 = c^2$$.

### Bibliographies and citations

Academic work involves fastidious recordkeeping about external sources. I prefer Zotero because it has two key features:

1. Firefox integration that downloads a journal article citation, the PDF (where available), and relevant metadata with one click, meaning it is very easy to get data in;

2. It has many options for generating citations in different formats and for exporting entire bibliographies -- it is easy to get data out, for research purposes and to make sure no software can hold my work hostage.

Several packages integrate Zotero with Pandoc and Emacs. The one that works for me is zotxt. I use Zotxt to satisfy another set of criteria:

1. On-the-fly bibliography management;

2. On-the-fly citations --- I don't want to do any cut-pasting or hunting around for the right citation key to use;

3. Relative ease of use.

### Professional results

Tools like ggplot2 and stargazer in R or pandas, seaborn, and statsmodels in Python generate beautiful tables and charts. They can also be used to generate raw LaTeX output suitable for integrating directly into a document.

With many tables and figures in each document, this sort of thing is easy to do and quick to replicate in Markdown and LaTeX, but would start to get tedious in Word.

## 2. How-to

### Installation

I'll assume you have figured out how to download and install the following:

• LaTeX/Texlive/BibLaTeX
• Python (should be on your computer already) and pip (instructions here).
• Zotero Standalone (I use the Firefox extension to acquire PDFs, but got the citation workflow going using Standalone.)
• Pandoc and pandoc-citeproc
• Emacs
• pandoc-mode
• pandoc-zotxt

For some reason my Emacs and Pandoc did not come with citeproc. In Fedora, I installed it using the dnf package manager like this: sudo dnf install pandoc-citeproc.x86_64. You might consider Kieran Healy's Emacs Starter Kit if acquiring everything seems mystifying. (Zotero should be pretty straightforward to get separately.)

Now, the part that is especially non-obvious:

1. Go to zotxt's Firefox Add-ons page; right-click on "Add to Firefox;" save the file.
2. With Zotero Standalone open, go to Tools -> Add-ons and select "Install Add-on From File." I found this option by clicking on the crossed hammer and screwdriver at upper right; perhaps your interface is a little different. In the dialog box, select the file you just downloaded.
3. Again in Zotero Standalone, go to Preferences -> Export and change the "Default Output Format" for QuickCopy to "Easy Citekey." If that option doesn't exist, restart Zotero Standalone and try again; it might need to be restarted for it to pick up on the fact that zotxt is there now.

The Easy Citekey format is Pandoc-friendly --- you'll see how it works a little later.

Now you should have all of the tools you need.

### The Workflow

1. Try out the Emacs interactive tutorial if you haven't used Emacs before. The tutorial is comprehensive, assumes no prior knowledge, and is very user-friendly.
2. Author your work in an Emacs file in Markdown. With Emacs open, you can open a new file by holding down Control and pressing x, then holding down Control and pressing f. Emacs users call these "chords" and would abbreviate that combo like this: C-x C-f. Notice an open "minibuffer" at the bottom of the Emacs screen; you can use this to specify out The link again to the basic syntax is here. How to use YAML metadata blocks for Pandoc is here.
3. Adding a citation: With Zotero Standalone open and while working in Emacs, place your cursor where you want the citation to go and use this keychord: C-c " k (Control-c, then shift-single-quote for double quotes, then the letter k).
4. Notice the minibuffer at the bottom of the screen is now active; it wants you to hit Enter, so do that.
5. In the next prompt in the minibuffer, type the author's last name and the date of the citation you want and hit Enter.
6. Use the up and down arrow keys to navigate to the correct citation and hit Enter when you've found it. You should now see a citekey in your document beginning with the "@" symbol. For Pandoc to make use of this, the ref must be in brackets ([]). You can additionally add multiple citations by separating them with semicolons (;) so the whole package looks like this: [@mayhew:1974;@fenno:1978].
7. Generating a document: After saving the Markdown file, open up a terminal, console, or shell window. Navigate to the appropriate directory and run Pandoc from the command line, invoking both zotxt and citeproc like this: pandoc -F pandoc-zotxt -F pandoc-citeproc -s inputfile.md -o outputfile.pdf.
8. Notice that the bibliography has been inserted at the end of the document by default. You could also output LaTeX and edit that file to customize where your bibliography goes.

### Troubleshooting and Customization

Sometimes citeproc won't pick up on a reference because it is a duplicate. You can fix this by specifying a citekey of your own within Zotero. If a reference isn't showing up, open up Zotero and add a citekey as a tag. See the section "Creating a custom citekey" here.

The syntax for invoking zotxt from Pandoc is adapted from Warren Knight's recipe.

## Improve This Workflow

I'd love to hear about other productivity hacks that beat this procedure for speed or ease of use.


text:
Explaining Tensorflow Code for a Convolutional Neural Network

In this post, we will go through the code for a convolutional neural network. We will use Aymeric Damien’s implementation. I recommend you have a skim before you read this post. I have included the key portions of the code below.

If you’re not familiar with TensorFlow or neural networks, you may find it useful to read my post on multilayer perceptrons (a simpler neural network) first.

Feature image credits: Aphex34 (Wikimedia Commons)

1. Code

Here are the relevant network parameters and graph input for context (skim this, I’ll explain it below). This network is applied to MNIST data – scans of handwritten digits from 0 to 9 we want to identify.

Here is the model (I will explain this below):

2. Translating the code

Let’s draw the model the function conv_net represents. The batch size given is 128. That means that each time, at most 128 images are fed into our model.

The big picture:

In more detail:

We can see that there are fives types of layers here:

• convolution layers,
• max pooling layers,
• layers for reshaping input,
• fully-connected layers and
• dropout layers.

2.1 What is conv2d (convolution layer)?

A convolution layer tries to extract higher-level features by replacing data for each (one) pixel with a value computed from the pixels covered by the e.g. 5×5 filter centered on that pixel(all the pixels in that region).

Credits: Fletcher Bach

We slide the filter across the width and height of the input and compute the dot products between the entries of the filter and input at each position. I explain this further when discussing tf.nn.conv2d() below.

Stanford’s CS231n course provides an excellent explanation of how convolution layers work (complete with diagrams). Here we will focus on the code.

This function comprises three parts:

1. Conv2D layer from Tensorflow
1. tf.nn.conv2d()
2. This is analogous to xW (multiplying input by weights) in a fully connected layer.
3. ReLU activation
1. This transforms the output like so:$ReLU(x) = max(0,x)$. (See previous post for more details).
2. X axis: input, y axis: output. Credits: ml4a on GitHub

You can see it is structurally the same as a fully connected layer, except we multiply the input with weights in a different way.

Conv2D layer

The key part here is tf.nn.conv2d(). Let’s look at each of its arguments.

• x is the input.
• W are the weights.
• The weights have four dimensions:  [filter_height, filter_width, input_depth, output_depth].
• What this means is that we have  output_depth filters in this layer.
• Each filter considers information with dimensions [filter_height, filter_width, input_depth] at a time. Yes, each filter goes through ALL the input depth layers.
• This is like how, in a fully connected layer, we may have ten neurons, each of which interacts with all the neurons in the previous layer.
• stride is the number of units the filter shifts each time.
• Why are there four dimensions? This is because the input tensor has four dimensions: [number_of_samples, height, width, colour_channels].
• strides = [1, strides, strides, 1] thus applies the filter to every image and every colour channel and to every  strides image patch in the height and width dimensions.
• You don’t usually skip entire images or entire colour channels, so those positions are hardcoded as 1 here.
• E.g. strides=[1, 2, 2, 1] would apply the filter to every other image patch in each dimension. (Image below has width stride 1.)
• "SAME" padding: the output size is the same as the input size. This requires the filter window to shift out of the input map. The portions where the filter window is outside of the input map is the padding.
• “SAME” padding: output (green) is the same size as the input (blue). Stride = 1.

• The alternative is "VALID" padding, where there is no padding. The filter window stays inside the input map the whole time (in valid positions), so the output size is smaller than the input.

2.2 What is maxpool2d (max_pool)?

Pooling layers reduce the spatial size of the output by replacing values in the kernel by a function of those values. E.g. in max pooling, you take the maximum out of every pool (kernel) as the new value for that pool.

Here the kernel is square and the kernel size is set to be the same as the stride. It resizes the input as shown in the diagram below:

Max Pooling with k = 2. Credits: CS231n

2.3 Layers for reshaping input

We reshape input twice in this model. The first time is at the beginning:

Recall the input was

That is, each sample inputted to the model was a one-dimensional array: an image flattened into a list of pixels. That is, the person who was preprocessing the MNIST dataset did this with each image:

And now we’re reversing the process:

The second time we reshape input is right after the convolutional layer before the first fully connected layer:

We’re doing this again to prepare 1D input for the fully connected layer:

2.4 What is a fully connected layer?

Do a Wx + b: For each neuron (number of neurons = number of outputs), multiply each input by a weight, sum all those products up and then add a bias to get your output for that neuron.

2.5 What is Dropout?

Just before the output layer, we apply Dropout:

Dropout sets a proportion 1-dropout of activations (neuron outputs) passed on to the next layer to zero. The zeroed-out outputs are chosen randomly.

• What happens if we set the dropout parameter to 0?

This reduces overfitting by checking that the network can provide the right output even if some activations are dropped out.

And that’s a wrap – hope you found this useful! If you enjoyed this or have any suggestions, do leave a comment.

4 Comments on “Explaining Tensorflow Code for a Convolutional Neural Network”

1. l3v0

great! I love you post. Could you explain: why we use “-1” in the shape parameter of
x = tf.reshape(x, shape=[-1, 28, 28, 1]) thanks.

1. Jessica Yung

Hi there, putting `-1` means that the length can take any value there. Specifically, the first position refers to the number of examples you feed in. So the -1 in the first position means you can feed in as many or as few examples as you like in one go.

You want this because you may change your batch size (or your final batch may have a different number of examples if the total number of training examples may not be divisible by your batch size). Hope this helps!

2. The Anh

Thank you for your good explanation. If you don’t mind, please explain conv2d in the formulas. For example: We have an input I of shape [batch_size, w, h, i_channels] and weights (filters) W of shape [fw, fh, i_channels, o_channels]. So how does conv2d compute to output O = f(I, W) of shape [batch_size, w, h, o_channels] (in case of padding “SAME”).
The convolution demo in Stanford’s course on CNNs (CS231n) explains this well. Basically you take the dot product of the filter and the input once for every entry of your output volume, shifting the filter (width and height-wise) by `stride` units around the input to fill up the output.
For TensorFlow’s implementation of conv2d, you can read the code for the function `convolution` here.