text
stringlengths 14
5.77M
| meta
dict | __index_level_0__
int64 0
9.97k
⌀ |
|---|---|---|
é um jogo eletrônico de luta com elementos de RPG de ação, desenvolvido e publicado pela Square Enix exclusivamente para o PlayStation Portable como parte das comemorações dos vinte anos da série Final Fantasy. Ele foi lançado em 18 de dezembro de 2008 no Japão, em 25 de agosto de 2009 na América do Norte e em setembro em outros territórios.
O jogo possui personagens desde o primeiro título da série e se centra em um grande conflito entre Cosmos, a deusa da harmonia, e o Chaos, o deus da discórdia. Os dois convocam diversos guerreiros para lutarem no seu lado em sua décima terceira guerra. O jogador controla durante a história os dez guerreiros escolhidos por Cosmos, que são os protagonistas dos dez primeiros jogos principais da série Final Fantasy. A versão internacional e norte-americana também dá acesso a outras funcionalidades como um modo arcade.
A ideia de Dissidia partiu do desejo de Tetsuya Nomura de criar um spin-off para a franquia, porém acabou alterado para pertencer à série Final Fantasy. Além de desenhar os personagens, Nomura trabalhou com a equipe da Square Enix a fim de deixar o jogo mais atrativo aos jogadores ocidentais. Dissidia foi bem recebido crítica e comercialmente, vendendo mais de um milhão de cópias mundialmente. Uma sequência chamada Dissidia 012 Final Fantasy foi lançada em 2011 e possui vários personagens novos e elementos de jogabilidade.
Jogabilidade
O gênero de Dissidia: Final Fantasy foi descrito como "ação progressiva" e seus gráficos são em três dimensões. Possui multiplayer para até dois jogadores via rede wireless e lutas que envolvem o uso de técnicas individuais especiais dos personagens para fazer dano aos oponentes. Os jogadores também podem customizar seus personagens com equipamentos.
O objetivo principal é reduzir o HP (Health Points ou "Pontos de Vida") do oponente a zero. Uma ofensiva é mostrada na forma numérica sendo chamada Bravery Points (BRV) ou "Pontos de Bravura".
Outra novidade do sistema de combate é a "EX Gauge", da qual pode ser preenchida de várias formas, desde atacando, sendo atacado ou obtendo núcleos EX espalhados pelo cenário. Uma vez que a "EX Gauge" está completa, o personagem pode entrar no EX Mode, que ao acertar os comandos que aparecem na tela, um ataque inevitável e arrasador similar aos conhecidos Limit Break visto em jogos da série Final Fantasy é executado.
Sinopse
A história gira em torno de dois deuses: Cosmos, a deusa da harmonia, e Chaos, o deus da discórdia. Os dois têm estado num eterno conflito com o "Mundo B", uma dimensão-espelho do "Mundo A", onde o primeiro Final Fantasy se passa, invocando diversos guerreiros de outros mundos da série principal para batalhar num ciclo sem fim de morte e renascimento até encontrar o equilíbrio, sendo necessário derrotar Chaos.
Cosmos dá a dez guerreiros a tarefa de trazer de volta os dez cristais que ajudarão a derrotar Chaos. Os heróis, então, iniciam uma jornada chamada Destiny Odissey ("Odisseia do Destino"), por onde suas respectivas estórias são recontadas e intercaladas com outras. Durante a jornada, os heróis encontram seus vilões originais, derrotando-os e reobtendo seus cristais.
Personagens controláveis
O jogo une heróis e vilões dos jogos anteriores da série Final Fantasy.
No total, o jogo possui vinte e dois personagens jogáveis: onze heróis e onze vilões, cada um representando desde Final Fantasy I até Final Fantasy X. Há, ainda, dois personagens secretos: uma heroina representando Final Fantasy XI, e um vilão representando Final Fantasy XII.
Inicialmente, somente dez principais heróis são jogáveis em todos os modos de jogo; os dez vilões são jogáveis no Modo Arcade, tendo de ser desbloqueados em todos os outros modos de jogo.
O equipamentos dos personagem é customizável e pode ganhar EXP e Gil (dinheiro do jogo) das batalhas.
Segue, abaixo, a lista de personagens:
Música
Dissidia Final Fantasy Original Soundtrack foi lançado em 24 de Dezembro de 2008, e está disponível tanto nas versões regular e special, similar ao jogo. O tema principal do jogo é "The Messenger" de Your Favorite Enemies.
As faixas "Cosmos" e "Chaos - Last Battle 1" também foram tocadas por Your Favorite Enemies. O tema principal possui as letras de "Cosmos" e "Chaos - Last Battle 1." "Cosmos" possui voz feminina, enquanto "Chaos" é dominada por vozes masculinas. As outras faixas são remixes feitas por Ishimoto das última versões das músicas originais de Final Fantasy compostas por Nobuo Uematsu.
Faixas
Ligações externas
Jogos da série Final Fantasy
Prequelas de jogos eletrônicos
Jogos exclusivos para PlayStation Portable
Jogos eletrônicos com protagonistas femininas
Guerras na ficção
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,443
|
Uddevalla é a maior cidade da província histórica da Bohuslän. Tem cerca de habitantes e é sede do município de Uddevalla, no condado da Västra Götaland.
Infraestruturas
A estrada europeia E6, passando na proximidade de Uddevalla, e ligando Strömstad a Trelleborg.
A estrada nacional R44, ligando Uddevalla a Götene
A linha férrea Bohusbanan passa pela cidade e liga-a a Gotemburgo e a Strömstad.
A linha férrea de Alvsburgo (Älvsborgsbanan) liga Uddevalla a Borås.
O porto de Uddevalla
O aeródromo de Rörkärr
O aeroporto regional de Trollhättan-Vänersborg (Trollhättan-Vänersborgs flygplats)
Economia
A economia de Uddevalla está dominada pelo grande setor público e pela educação.
No setor privado, existem mais de 5000 empresas, sendo os dois ramos principais a construção civil e o comércio a retalho, e ainda numerosas pequenas empresas de alta tecnologia.
Património
A cidade proporciona um passeio pelas margens do rio Bäveån e do fiorde Byfjorden.
Museu da Bohuslän (Museu regional, com galeria de arte)
Gustavsberg (Casa balnear de 1774, com fonte de água natural, restaurante e pousada da juventude)
Ligações externas
Sítio Oficial de Uddevalla
Uddevalla
Bohuslän
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,275
|
{"url":"https:\/\/chemistry.stackexchange.com\/questions\/121458\/will-proving-or-disproving-of-any-of-the-following-have-effects-on-chemistry-in","text":"# Will Proving or Disproving of any of the following have effects on Chemistry in general?\n\nI am working on a project relating to- https:\/\/en.wikipedia.org\/wiki\/Millennium_Prize_Problems\n\nI wanted to list the effects of them being proved or disproved in different aspects of science and maths. It was fairly easy to find such results on physics and maths as the questions primarily come up from them however I couldn't find any reference to effects it would have on our understanding of chemistry.\n\nIs it because there will be no effect at all and that they are the least concern to chemists (in terms of applicability) or if there are then what are they?\n\n\u2022 To understand just how absurdly far-fetched this may come across, ask yourself the same question: what effect will it have on your life? Say, you get up, brush your teeth, have your breakfast, then BANG! $P\\ne NP.$ So what? \u2013\u00a0Ivan Neretin Sep 19 '19 at 10:25\n\u2022 I suspect that two problems (p-np and Navier\u2013Stokes existence and smoothness) would have repercussions, in defining some problems as either bounded or unbounded. Guaranteeing that you can find a solution to a certain level of accuracy in a given amount of time is akin to guaranteeing that you will come out winning in a lottery if you buy enough tickets. However I admit to speculating - I am not directly familiar with the problems. \u2013\u00a0Buck Thorn Sep 19 '19 at 10:49\n\u2022 @IvanNeretin I know it wouldn't be of any importance to me in my general life but i wanted to know it's impact on a discipline or a field of study such as Chemistry \u2013\u00a0user78585 Sep 19 '19 at 14:37\n\u2022 @IvanNeretin I was reading the Code of Conduct for this website. Maybe you might find the Friendly and Unfriendly part helpful :-) chemistry.stackexchange.com\/conduct \u2013\u00a0user78585 Nov 16 '19 at 16:53\n\n\u2022 P vs NP, if solved the way that pretty much everybody expects (that is, $$\\rm P\\ne NP$$), will have no far-reaching consequences, because that's what we were thinking for quite a while now. If it would miraculously happen to be otherwise (that is, $$\\rm P=NP$$), that would be quite a shock to many fields, especially to computer science, and by extension, to computational chemistry. But I don't believe in miracles.\n\u2022 ... and possibly the worst version of $P = NP$, a completely nonconstructive proof of that equality. So then we know that many problems are \"easy\", but we're ignorant of the easy way. \u2013\u00a0Eric Towers Sep 19 '19 at 23:55","date":"2020-01-20 21:53:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 2, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6339285373687744, \"perplexity\": 560.5104932023039}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579250599789.45\/warc\/CC-MAIN-20200120195035-20200120224035-00031.warc.gz\"}"}
| null | null |
Nada surge de la nada, o de la nada, nada surge, son expresiones con las que se indica un principio metafísico atribuido al filósofo griego Parménides según el cual ningún ente puede empezar a existir a partir de la nada. La idea también se expresa mediante la locución latina de Lucrecio ex nihilo nihil fit. En la filosofía griega, un principio relacionado era aquel según el cual un ente no puede desaparecer en la nada, sino sólo transformarse. El principio puede pensarse como un antecedente de la ley de conservación de la masa y la ley de conservación de la energía.
En filosofía y teología
El principio fue sostenido por los eleatas y atomistas. La idea de que "nada viene de la nada", tal como la articula Parménides, aparece por primera vez en la Física de Aristóteles:Lo anterior, en una traducción basada en la traducción de John Burnet, aparece de la siguiente manera:El poeta romano y filósofo epicúreo Lucrecio expresó este principio en su primer libro De Rerum Natura (Sobre la naturaleza de las cosas).Luego continúa discutiendo cómo se requiere materia para hacer materia y que los objetos no pueden surgir sin una causa razonable.El principio también está muy ligado a la pregunta filosófica que Martin Heidegger y Gottfried Leibniz, entre otros, destacaron: ¿por qué hay ente, y no más bien nada?
En la "dogmática cristiana", la nada se entiende como "la ausencia completa del ente extradivino", negándose así el principio de Lucrecio; de lo cual resulta que "ex nihilo fit -ens creatum" (de la nada se hace el ente-creado).
El principio suele plantearse en relación con el origen del universo. Dado que el universo existe, entonces o bien existió siempre, o bien tuvo un comienzo. Si tuvo un comienzo, significa que surgió de la nada, porque el universo es por definición todo lo que existe. Pero esto contradice el principio de que nada surge de la nada. Luego, si el principio es cierto, el universo existió siempre. Siguiendo este tipo de razonamientos, muchas religiones han postulado que el universo no surgió de la nada, sino de un Dios creador, y que ese Dios existió siempre. En la Crítica de la razón pura, Immanuel Kant argumentó que no es posible determinar si el mundo tiene o no un comienzo en el tiempo.
¿Cómo sabemos esto? Bien, sabemos que de la nada, nada procede. Así que, si alguna vez hubo un tiempo en que no existía absolutamente nada, entonces nada hubiera podido existir. Pero las cosas existen. Por lo tanto, puesto que nunca pudo haber habido absolutamente nada, algo tuvo que haber existido siempre. Aquello que ha existido siempre es a quien llamamos Dios. Dios es el Ser no causado que hizo que todo lo demás llegara a existir. Dios es el Creador no creado que creó el universo y todo lo que hay en él.
En literatura moderna
En la obra de Shakespeare El rey Lear, el personaje principal le dice a su hija Cordelia: "Nada puede surgir de la nada".
Física moderna
La ley de conservación de la energía establece que la energía total de un sistema aislado no puede cambiar. La hipótesis del universo de energía cero establece que la cantidad de energía en el universo menos la cantidad de gravedad es exactamente cero. En este tipo de universo, la materia podría crearse de la nada a través de una fluctuación del vacío, asumiendo que tal universo de energía cero ya no es nada. Dicho universo tendría que ser plano, un estado que no contradice las observaciones actuales de que el universo es plano con un margen de error del 0,5%.
Algunos físicos, como Lawrence Krauss, Stephen Hawking y Michio Kaku , definen o definen "nada" como un vacío cuántico inestable que no contiene partículas.
La mecánica cuántica propone que se están creando pares de partículas virtuales a partir de fluctuaciones cuánticas en este espacio "vacío" todo el tiempo. Si estos pares no se aniquilan mutuamente de inmediato, podrían detectarse como partículas reales, por ejemplo, si una cae en un agujero negro y su opuesto se emite como radiación de Hawking.
Alexander Vilenkin define "nada" como un universo de tamaño cero: es lo más cercano a la nada que uno puede obtener, pero aun así no es nada.
Véase también
Ex nihilo
Bosón de Higgs
Energía del vacío
Ruptura espontánea de simetría electrodébil
Teoría del Big Bang
Eternidad
Un Universo de la nada
Por qué existe algo
Notas y referencias
Nada
Metafísica
Principios filosóficos
Frases y citas
Parménides
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,593
|
\section{Introduction}
The emerging sixth-generation (6G) network concept has to be ``green", while supporting ubiquitous artificial intelligence (AI)-assisted applications at a high energy-efficiency, including autonomous driving, unmanned aerial vehicle (UAV) aided logistics, rescue and surveillance, multi-access/mobile edge computing (MEC), and smart manufacturing~\cite{6G}. However, there are grave technical challenges ahead. Specifically, given the limited transmission powers, the received signal-to-interference-plus-noise ratio (SINR) will be considerably degraded due to severe signal fluctuations caused by both multipath fading and the presence of large obstacles. Although opting for millimeter wave (mm-Wave) and even Terahertz (THz) carrier frequencies has the potential of significantly improving the data rate, these frequencies inevitably suffer from high propagation losses, thereby they may result in an excessive energy consumption.
\begin{figure*}[t]
\captionsetup{font={footnotesize }}
\centerline{ \includegraphics[width=6.25in, height=5.35in]{Fig1.pdf}}
\caption{Emerging applications require RIS-aided green 6G wireless communications, including RIS-aided vehicle communications, RIS-aided drone communications, RIS-aided mobile edge computing, and RIS-aided IoT communications.}
\label{use_case}
\end{figure*}
In this context, reconfigurable intelligent surfaces (RIS) have emerged as a promising 6G solution for improving the quality of wireless links by appropriately reflecting the incident signals with the aid of a large number of nearly-passive reflecting elements at a low power consumption~\cite{6G_RIS}. Specifically, an array of passive scattering elements can be controlled in a software-defined manner to adjust the electromagnetic properties (such as the phase shifts and potentially the amplitude) of the incident radio frequency (RF) signals. As a result, the received signal strength can be enhanced via jointly optimizing the phase shifts of the reflecting elements of RISs~\cite{EURASIP_Editor2}.
However, \textit{the interfering signals tend to dynamically fluctuate and a conventional RIS `blindly' reflects both the desired and interfering signals.}
This may precipitate severe interference at the desired receiver and cause a deleterious effect. To address this challenge, it turns out that AI has a beneficial ameliorating potential in support of spectrum sensing, termed as spectrum learning (SL). This is achieved by intelligently identifying the characteristics of the wireless spectral environment. Hence, we envision that the convergence of SL and RIS will simplify network planning, paving the way for a green 6G ecosystem.
In this article, we develop a dynamic SL-aided RIS framework for boosting the advantages of intelligent signal reflection through learning the inherent features of the incident RF signals at the RIS controller. The proposed framework `moves' the complexity of online detection to offline training, thereby controlling the binary ON-OFF status of RISs in real-time and improving the energy-efficiency of 6G networks.
The remainder of this article is outlined as follows. The following section provides an overview of green 6G services and its emerging applications empowered by RISs. Then, we present the design of the SL-aided RIS framework conceived for improving its energy-efficiency. Following that, we evaluate the performance of the proposed framework. Finally, we conclude this article with a list of compelling research opportunities.
\section{Reconfigurable Intelligent Surfaces Meet Green 6G Networks} \label{system_model}
\subsection{RIS-aided Green 6G Wireless Communication Systems}
Recent years have witnessed immense research efforts dedicated to RIS-aided wireless designs for improving the link quality and network coverage. Sophisticated 3D index modulation, holographic MIMO schemes, mm-Wave and THz solutions, are expected to efficiently support high-quality services and seamless connectivity for a massive number of mobile terminals.
However, their excessive energy consumption and hardware costs pose design challenges due to the relatively high operating frequency, especially when many RF chains are required. Again, holographic MIMO surfaces attracted attention as a possible solution for realizing massive MIMO systems at a reduced cost and power consumption~\cite{Huang_WCM}. Notably, mobile devices are usually energy-constrained in practice, and the direct line-of-sight (LoS) link is often unavailable due to the occlusion of obstacles, when again, RISs come to rescue.
\begin{table*}[t]
\newcommand{\tabincell}[2]
\centering
\renewcommand{\arraystretch}{1.2}
\captionsetup{font={small}}
\caption{Comparison of typical AI-empowered RIS research works.}
\label{RIS_AI}
\small
\centering
\begin{tabular}{|m{0.16\textwidth}<{\raggedright}|m{0.18\textwidth}<{\raggedleft }| m{0.2\textwidth}<{\raggedleft }|m{0.35\textwidth}<{\raggedleft}| }
\shline
\rowcolor{mycyan}
\textbf{Objective} & \textbf{Learning Approach} &\textbf{Model Architecture} & \textbf{Key Contributions} \\
\hline
{ Channel estimation} & Deep learning~\cite{DL_CSI} & Multi-layer perceptron (MLP) with multiple layers & Jointly optimize the transmit beamforming matrix and the phase shifts together with only a few active elements on the RIS \\ \hline
{Joint beamforming and phase shift design} & Deep reinforcement learning~\cite{Huang_jsac} & Deep deterministic policy gradient (DDPG) with $4$-layered actor and critic networks & Jointly optimize the transmit beamforming matrix and the phase shifts \\ \hline
{ Joint UAV trajectory and phase shift design} & Deep learning~\cite{Liu} &\textcolor{black}{Deep Q-network (DQN) with 3-layered convolutional neural network (CNN)} & Jointly optimize the movement of the UAV, phase shifts of the RIS, and power allocation
policy from the UAV to users. \\ \hline
Secure beamforming & Reinforcement learning~\cite{DRL_security} & Model-free & Jointly optimize the anti-jamming power allocation and reflecting beamforming without knowing the jamming model
\\ \hline
Transmission strategy of each UAV-user pair & Multi-task learning~\cite{JSAC_CXL} & Two-task (classification and regression) learning model with three hidden layers & Jointly infer the optimal RIS elements allocation and RIS phase shifts of each UAV-user pair for energy-efficiency \\
\shline
\end{tabular}
\end{table*}
As illustrated in Fig.~\ref{use_case}, an RIS-aided wireless communication system designed for energy efficient green 6G services typically supports a number of mobile devices. Each RIS element can be reconfigured by the RIS controller, which is connected to a central server (e.g., base station or access point) via a dedicated channel.
To explore the potential of RIS-aided reflections, the following three steps are generally sequentially performed:
\begin{itemize}
\item \textit{Channel estimation:} In RIS-aided systems, there exist three kinds of wireless channels between the transmitter (Tx) and the receiver (Rx). By considering uplink transmission as an example, we have to estimate the wireless channels, including the Tx-RIS link(s), the RIS-Rx link(s), and the direct Tx-Rx link(s). To fully explore the potential gains brought about by RISs, the acquisition of accurate channel state information (CSI) becomes crucial, yet practically challenging.
\item \textit{RIS reconfiguration:} Following the acquisition of CSI, both the phase shift and the amplitude of RIS elements can be dynamically reconfigured, for example, based on positive-intrinsic-negative (PIN) diodes. This has been widely adopted in practical implementations as a benefit of its low energy consumption and hardware cost.
\item \textit{Incident signal reflection:} After channel estimation and RIS reconfiguration, the RIS reflects the incident signal(s) with the optimized phase shift and amplitude. At the receiver, the direct and reflected signals are combined to improve the received
signal power.
\end{itemize}
\subsection{Emerging 6G Applications Empowered by RIS}
Conventional wireless communications are undergoing a significant transformation with the convergence of AI and 6G, evolving from the ``Internet-of-information" to the ``Internet-of-intelligence"~\cite{zy_iot}. This shall unleash the full potential of machine learning in innovative data-driven applications across a wide variety of industries, such as self-driving cars, UAV-aided surveillance, edge intelligence and smart factories. However, to support the aforementioned AI-based applications, huge amount of data is required for training.
Against this background, we shall describe four typical RIS-empowered 6G applications, including terrestrial vehicular communications, drone communications, mobile edge computing, and IoT communications, as illustrated in Fig.~\ref{use_case}.
\subsubsection{Autonomous Driving}
Video analytics relying on AI becomes an indispensable technique in support of autonomous driving, where a large amount of information generated by diverse onboard sensors (such as cameras, LiDAR sensors and radar sensors) has to be processed in near-real-time for collision avoidance in extremely complex surroundings.
However, promptly exploiting a deluge of delay-sensitive sensors' data requires substantial onboard computation resources and incurs high energy consumption~\cite{YB_magazine}. To maintain safety for self-driving, RISs usher in a paradigm shift from connected-vehicles to pervasive connected-intelligence by enhancing the link quality of cellular vehicular-to-everything (V2X) communications in complex urban environments at a low power.
\subsubsection{Unmanned Aerial Vehicles}
In recent years, UAVs, also commonly known as drones, have been extensively used for civilian, commercial, and military services in combination with AI~\cite{UAV_ZY}. However, the provision of sufficient energy for propulsion is a challenge, which may be mitigated by joint energy and data networking and laser charging~\cite{Hanzo}.
In this context, RISs show significant promise in terms of reducing the communications-related power of UAVs.
\subsubsection{Mobile/Multi-access Edge Computing}
At the time of writing, MEC is considered as a promising technique for offloading the processing of data to edge servers having much higher computing capabilities, thus enabling low-latency joint processing and communications. As a result, the AI processing is relocated from the central cloud to the network edge~\cite{iot_zy}.
This implies that a large amount of data has to be transferred from the end devices to the edge server via band-limited uplink channels, which may become the bottleneck in MEC. In this context, RISs play a crucial role in mitigating the propagation-induced impairments with the aid of signal reflection.
\subsubsection{Smart Industry}
The smart industry, or ``Industry 4.0", integrating AI and the industrial IoT, paves the way for intelligent production by interconnecting isolated industrial assets~\cite{iot_YB}. The wireless links are typically of non-line-of-sight (NLOS) nature, thereby leading to vulnerable links in dynamic environments. To circumvent this challenge, RISs improve the propagation environment by constructing signal reflections, thereby enhancing the robustness of wireless communications.
\begin{figure*}[t]
\captionsetup{font={footnotesize }}
\centerline{ \includegraphics[width=5.65in, height=3.95in]{Fig2.pdf}}
\caption{An illustration of the offline training stage and online inference stage in the proposed SL-aided RIS framework, where the RF trace collection module and the spectrum learning module are depicted.
}
\label{procedure}
\end{figure*}
\subsection{Challenge of Undesired-Reflections via RISs}
Despite their potential, RISs face numerous challenges, including the acquisition of accurate CSI, their reflection coefficients optimization, multi-user mobility management, and implementation issues, just to name a few. In particular, \textit{due to the unpredictable nature of interference, undesired reflections become a critical challenge, which has not been resolved yet.} Specifically, most of the existing treatises assume either that no interference exists, which rarely occurs in practice, or that the interference is already known. But naturally, it is not trivial to estimate the dynamically fluctuating interference in wireless networks. In reality, the SINR at the receiver might even be degraded compared to a system dispensing with RISs, unless specifically addressed. But the challenge is that RISs are nearly passive surfaces with no active sensing capabilities for channel and interference estimation. As a remedy, combining AI with spectrum sensing becomes attractive, which enables conventional RISs to learn the characteristics of the wireless environment. Hence, increasing research interests have been inspired by RIS-aided applications empowered by AI, which are contrasted in Table~\ref{RIS_AI}.
In the following section, we shall present a dynamically configured ON-OFF\footnote{In this paper, the ON-OFF status is referred to having the entire set of elements of the RIS to be ON or OFF.} RIS control scheme based on spectrum learning, thereby enabling RISs to make intelligent decisions whether to reflect or not the incident signals, depending on the wireless spectrum environment inferred. Naturally, this intelligent action is only possible at the cost of incorporating intelligence into the RISs.
\section{Spectrum Learning-Aided RISs for Green 6G: Principles and Procedures}
\label{IIS}
\subsection{Design Principle}
In a conventional system, the RISs passively and indiscriminately reflect all the incident RF signals. However, this kind of blind reflection sometimes contaminates the desired signal at the receiver. Fortunately, AI-aided spectrum sensing has the potential of intelligently orchestrating the incident signals by relying on an appropriately trained convolutional neural network (CNN) of the RIS controller.
More explicitly, the surrounding spectral environment is explored and learned by an intelligent classification algorithm, explicitly identifying the desired user, whose signal should be beneficially reflected, and the interfering users, whose signal should not be reflected. We arrange for having only a few active RIS elements for the baseband processing of the incident signal at the RIS controller at a low overhead\footnote{When the ratio of the active elements to the total number of elements is set to only $1\%$ and $7\%$ for a mmWave $28$ GHz scenario and for a low-frequency $3.5$ GHz scenario, respectively, the achievable rates can be maximized~\cite{DL_CSI}.}. \textcolor{black}{For simplicity, we assume that the base station (BS) of Fig.~\ref{procedure} has perfect CSI knowledge for all wireless channels, and that this CSI is fed back to the RIS controller via a dedicated control channel.} In particular, when an RF signal impinges upon an RIS, the RIS controller identifies the users that are included in the aggregate signal by using a well-trained CNN. Based on the inferred estimate, the RIS controller determines the ON-OFF status of the RIS for maximizing the SINR at the tagged BS. For example, the entire RIS is `turned OFF', if the received SINR at the tagged BS with respect to (wrt) the RIS-aided link is lower than the corresponding SINR wrt the direct LOS link. Otherwise, the entire RIS is `turned ON'.
In the following, we will introduce the two stages of the proposed SL-aided RIS framework: the offline training stage and the online inference stage.
\subsection{Offline Training Stage}
In the offline training stage, we have a pair of functional modules in support of spectrum learning.
\subsubsection{RF Trace Collection Module}
We collected RF traces by building a universal software radio peripheral (USRP2) based testbed, which is connected via the Gigabit Ethernet to a host computer, as illustrated in Fig.~~\ref{procedure}. \textcolor{black}{The RF trace collection relies on a transmit unit and a receive unit that communicate over a $1$ MHz channel at a carrier frequency of $2.4$ GHz. The traces are collected over the signal-to-noise ratio (SNR) range spanning from $0$ to $20$ dB. Since each USRP2 unit has one transmit antenna, multiple interfering signals are obtained by letting multiple USRP2 units transmit at the same time.} In particular, the transmit unit includes a transmitter host computer performing the baseband signal processing and a transmitter USRP2 for up-conversion, digital-to-analog (D/A) conversion, and wireless transmission. Upon receiving the RF signal via the wireless channel, the receiver USRP2 first processes the signals ahead from the radio interface and then performs A/D conversion and down-conversion.
\textcolor{black}{Finally, the required baseband processing is carried out by the host computer, where the Inphase (I) and Quadrature (Q) sequences are stored for a wide range of SNRs to account for different interference scenarios. The acquired I/Q samples capturing the features of the incident RF signal are used for training the CNN model.}
\subsubsection{CNN Training Module}
In the proposed framework, spectrum learning is considered as a classification task, which is fulfilled via an appropriately trained CNN model. Specifically, \textcolor{black}{based on the acquired RF traces that amount to approximately one billion I and Q samples,} the CNN is trained offline based on a GPU cluster (e.g., the NVIDIA Tesla P100 GPU computing processor illustrated in Fig.~\ref{procedure}) by using the Adam algorithm that utilizes the cross-entropy as the loss function.
\textcolor{black}{The CNN model trained consists of four layers: two convolutional layers that use the rectified linear unit activation function, two dense fully-connected layers and the output layer that uses the softmax activation function.}
\subsection{Online Inference Stage}
In the online inference stage, the RIS controller can dynamically adjust the ON-OFF status of the entire RIS to maximize the received SINR at the BS of Fig.~\ref{procedure} based on the appropriately trained CNN model
\subsubsection{Spectrum Learning Module}
\textcolor{black}{For the spectrum learning module, we can directly infer the spectrum access information related to the desired and interfering users by extracting the resultant I/Q samples from a copy of the impinging signals and feeding them into the appropriately trained CNN that is deployed at the RIS controller.} Specifically, as soon as the RF signal arrives at the RIS elements, it undergoes A/D conversion and frequency down-conversion. Then, baseband processing is executed and the extracted I/Q sequence is fed into the CNN model for spectrum access inference.
\subsubsection{ON-OFF RIS Control Module}
The ON-OFF RIS control module is trained for avoiding {undesired reflections} by dynamically setting the binary ON or OFF status of the RIS elements. In particular, we assume that both the CSI and phase shifts of the RIS can be perfectly estimated and optimized at the BS, which feeds this information back to the RIS controller via a dedicated control channel\footnote{Channel estimation and phase shifts optimization of RIS have been widely investigated, hence they are not the main focus of this paper.}. By disabling and enabling the RIS (i.e., RIS is turned OFF and ON), the SINR that is achieved at the BS can be calculated respectively. If the calculated SINR with disabling the RIS is no worse than the corresponding SINR when the RIS is enabled, the entire RIS is turned OFF. Otherwise, the entire RIS is turned ON.
\section{Case Study and Performance Evaluation}\label{results}
In this section, we characterize the experimental performance of the proposed SL-aided ON-OFF RIS control scheme.
\begin{figure}[t]
\captionsetup{font={footnotesize }}
\centerline{ \includegraphics[width=2.5in, height=2.55in]{Fig3.pdf}}
\caption{An experimental case of the proposed SL-aided RIS framework, where there exists an interfering user whose signal causes interference to the desired user's transmission at the BS. The angle of incidence between the desired user and the interfering user at the RIS is $\theta$.
}
\label{CNN_solution}
\end{figure}
\subsection{Experimental Scenario}
For characterizing the proposed SL-aided RIS framework, we study the experimental scenario of Fig.~\ref{CNN_solution}, where the RIS reflects the signal arriving from both the {desired user} (denoted as $U_D$) as well as from the {interfering user} (denoted as $U_I$). As a result, the signal received at the BS becomes a superposition of the desired signal from $U_D$ and the interfering signal from $U_I$.
The impact of the interference is, in particular, caused by undesired-reflections by the RIS and it is determined by the angle between the incident signal from the desired user and the incident signal from the interfering user (i.e., $\theta$ in Fig.~\ref{CNN_solution}). For example, the interference effect caused by $U_I$ at the BS may become more severe if $U_I$ is close to $U_D$, i.e., $\theta$ becomes small. In this case, the proposed SL-aided RIS framework enables the RIS controller to intelligently control the ON-OFF status of the RIS according to the wireless environment.
In the scenario of two users ($U_D$ and $U_I$), we have four combinations of signals arriving at the RIS since each user has a binary state, i.e., active (also transmitting) or inactive (also idle): (1) ``Idle", indicating that both $U_D$ and $U_I$ are inactive; (2) ``\textit{$U_D$}", indicating that only $U_D$ is active, i.e., no interference occurs; (3) ``\textit{$U_I$}", indicating that only $U_I$ is active; and (4) ``$U_D\!+\!U_I$", indicating that both $U_D$ and $U_I$ are active. Upon receiving the superimposed incident signal, the RIS controller has to identify the signal's composition, which is a four-class classification problem readily solved by the trained CNN model\footnote{\textcolor{black}{Although a scenario with one interfering user is considered in this article for illustrative purposes, the proposed SL-aided RIS framework can be extended to more general scenarios where multiple interfering users coexist.}}. \textcolor{black}{Therefore, the proposed SL-based framework `moves' the algorithmic and computational complexity to offline training while the complexity during the online inference is quadratic with time as a function of ${\cal O}(M^2 C K)$, where $C$ indicates the number of layers of the trained CNN model, $M$ is the number of neurons and $K$ indicates the total number of RISs.}
\subsection{Performance Evaluation}
In this part, we first analyze the inference accuracy of the CNN model with experimental data, and then we evaluate the performance of the proposed SL-aided ON-OFF RIS control scheme.
\begin{figure}[t]
\captionsetup{font={footnotesize }}
\centerline{ \includegraphics[width=3.1in, height=2.35in]{Fig4.pdf}}
\caption{The achievable SINR versus the angle of incidence ($\theta$), where the number of RIS ($K$) is $1$. }
\label{result1}
\end{figure}
\subsubsection{Correct Inference Probability}
The correct spectrum inference probability of the trained CNN model is demonstrated by considering the experimental scenario of Fig.~\ref{procedure} supporting $U_D$ contaminated by $U_I$. During the CNN model training, the number of time series I/Q samples is set to $32$, $128$, and $512$.
\textcolor{black}{The experimental results show that the correct inference probability obtained via the CNN model for the Class-2 (``$U_D$"), the Class-3 (``$U_I$") and the Class-4 (``$U_D\!+\!U_I$") is around $97\%$. Notably, the trained CNN model attains a perfect accuracy for the Class-1 (``Idle") prediction due to the {pattern distinguishable from the rest of the classes}.} The relatively high inference probability of the trained CNN model results in typically feeding an accurate spectrum identification estimate to the ON-OFF RIS control module.
\subsubsection{SL-enalbed ON-OFF RIS Control}
In the considered scenario, there is one desired user ($U_I$), one interfering user ($U_D$), and $K$ RISs. We assume that the RISs are equally spaced by $5 \ m$ in the vertical direction, each having $256$ elements. The amplitude reflection coefficient is $1$ and the impact of interference is inversely proportional to $\theta$. As illustrated in Fig.~\ref{CNN_solution}, the RIS$_1$-$U_D$, RIS$_1$-$U_I$, and RIS$_1$-BS distances are $60 \ m$, $10\ m$, and $80 \ m$, respectively. Furthermore, the angle of incidence between the BS and $U_D$ at the RIS is $150\rm{^o}$, while between $U_D$ and $U_I$ is $\theta \in [0, 150\rm{^o}]$. To demonstrate the energy-efficiency of the proposed SL-aided ON-OFF RIS control scheme, we evaluate a pair of benchmark schemes: 1) RIS always ON, and 2) RIS always OFF, where the transmission power of $U_D$ is $23$ dBm and the noise power is -$94$ dBm.
\begin{figure}[t]
\captionsetup{font={footnotesize }}
\centerline{ \includegraphics[width=2.95in, height=2.35in]{Fig5.pdf}}
\caption{The achievable SINR versus the number of RISs, where $P_I=10$ dBm and $\theta$ is randomly selected from $[30\rm{^o}, 120\rm{^o}]$ considering $10^5$ realizations. }
\label{result2}
\end{figure}
The achievable SINR (in dB) versus the angle of incidence $\theta$ is depicted in Fig.~\ref{result1}, where a single RIS is considered, and the transmit power $P_I$ of the interfering user is $10$ dBm and $15$ dBm. We observe from Fig.~\ref{result1} that the benchmark schemes exhibit an opposite trend as $\theta$ increases. In particular, as $\theta$ increases, the interference contaminating the desired receiver through a reflection from the RIS is reduced. Therefore, it becomes more likely that the RIS is ON, thereby increasing the achievable SINR of the conventional `RIS always ON' scheme. By contrast, the achievable SINR of the `RIS always OFF' scheme reduces with $\theta$ due to the reduction of the distance between $U_I$ and BS. In this case, the interference reflected by the RIS becomes more pronounced. \textcolor{black}{It can be observed that the SINR obtained by the proposed SL-aided RIS control scheme is consistently higher than that of the two benchmark schemes, which is the benefit of intelligently controlling the status of the RIS according to the dynamically fluctuating environment. The improvement of SINR via the proposed scheme is at the expense of increasing the RIS controller's complexity, i.e., the complexity introduced by the CNN online inference and the ON-OFF RIS state configuration. However, the associated complexity only grows linearly with the number of RISs, which is acceptable in practice.}
The impact of the number of RISs ($K$) on the achievable SINR is evaluated in Fig.~\ref{result2}, where we set $P_I=10$ dBm and $\theta$ is randomly selected from $[30\rm{^o}, 120\rm{^o}]$. Intuitively, we observe that the SINR achieved by the `RIS always OFF' benchmark scheme remains unchanged, while that of the other two schemes increases with the number of RISs. Moreover, the SINR obtained by the proposed SL-aided RIS control scheme is significantly higher than that of the two benchmark schemes. In particular, when we use the same transmission power, compared to the `RIS always OFF' and `RIS always ON' case studies, our proposed solution improves the received SINR from about $4.8$ dB and $2.9$ dB to $6.3$ dB for $K=5$. This demonstrates the potential of the proposed SL-aided RIS framework in terms of improving the energy efficiency of future 6G networks.
\section{Conclusions and Promising Research Directions}
\label{conclusion}
In this article, we discussed the potential of spectrum-learning in addressing the critical challenges of RIS solutions in 6G. Explicitly, we proposed an ON-OFF RIS control scheme for dynamically configuring the binary status of RISs. In this context, the beneficial role of SL in improving the energy efficiency was validated by exemplifying the achievable SINR and was compared to a pair of benchmark schemes.
Nonetheless, the proposed SL-aided RIS communication framework still remains largely unexplored, in particular, from its energy efficiency perspective.
Achieving reliable SL necessitates the collection of a large amount of labeled RF sequences, and the appropriately trained model sometimes has to be fine-tuned or even retained, when the environment significantly changes, especially in vehicular and drone applications. This will lead to high energy consumption.
To facilitate SL relying on a limited number of labeled RF sequences in a dramatically changing environment, advanced learning techniques, such as deep reinforcement learning, transfer learning and self-supervised learning may be explored in future works.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,388
|
\section{Introduction}
Interest in the problem of radiation reaction from particle
motion in a curved spacetime has seen a rapid increase in recent
years. The back-reaction of emitted radiation on the particle,
known as the self-force, changes the particle trajectories (e.g.,
near a black hole) from a simple geodesic motion. The
determination of the self force is essential to precision
calculations of particle trajectories and the determination of
waveforms from prospective astrophysical sources. Electromagnetic
radiation reaction in a curved spacetime was first studied by
DeWitt and Brehm \cite{DeWBre}. The gravitational radiation
reaction equation was first obtained by Mino, Sasaki and Tanaka
\cite{MST} and Quinn and Wald \cite{QW} and others, notably
Detweiler and Whiting \cite{Detweiler_Whiting}. The equation of
motion governing a scalar charge with radiation reaction was
first obtained by Quinn \cite{Quinn}. For an excellent review, see
\cite{Poisson}.
Parallel to this there has been detailed work devoted to particles
and detectors (e.g., atoms) moving in a quantum field and in
design studies of possible detection of Unruh radiation (see,
e.g.,\cite{HRCapri,LH1,RHA} and references therein). The
introduction of worldline path integral methods (see \cite{JH1}
and references therein) enable one to obtain equations of motion
for the charges and the field self-consistently. The introduction
of open system concepts and the influence functional method
enables one to derive stochastic equations with a noise source
derived {\it ab initio} and in a self consistent manner (See
\cite{RHA,RHK,RavalPhD} for accelerating detectors,
\cite{JH1,JHIARD,JohnsonPhD} for moving charges.)
In this paper, we use the worldline influence functional method
to study a particle with a scalar charge moving in its own
quantum scalar field in a curved spacetime. This is a
generalization of results obtained in \cite{JH1} to curved
spacetime. We are interested in the radiation it emits and its
backreaction (radiation reaction) on the trajectory of the
particle
and derive the equations of motion for the quantum average
(expectation value) of the particle's position. For those particle
trajectory histories which become sufficiently decohered, the
expectation value behaves classically. The scalar ALD equation we
derive in this limit checks with the result of Quinn \cite{Quinn}.
We then include in our consideration the effect of fluctuations
in the quantum field. We show how it behaves like a classical
stochastic force and derive a scalar ALD-Langevin equation for
the particle dynamics with a stochastic component, thus capturing
the induced small fluctuations on the particle trajectory. This
equation will be useful for the study of stochastic motion of
charges under the influence of both quantum or classical noise
sources, derived either self-consistently (as done here) or put
in by hand (with warnings). For astrophysical sources with some
stochastic component this effect may need to be included in more
accurate calculations of waveform templates.
In Sec. 2 we describe the worldline influence functional method
and how to obtain the semiclassical and stochastic particle
dynamics. In Sec. 3 we discuss how to regularize the causal Green
function in the spirit of effective field theory using a
quasi-local expansion. In Sec. 4 we derive the scalar ALD
equation. In Sec. 5 we derive the stochastic scalar ALD or the
scalar ALD-Langevin equation. We discuss the non-Markovian nature
of the noise induced effects and the possibility of secular
effects from such stochastic influences on the trajectory. In
Sec. 6 we discuss an array of issues pertinent to the present
problem and approach. In Sec. 7 we summarize our findings and Appendices A-D provide further details in the derivation of certain results given in the text.
\section{Relativistic Particle-Field Dynamics}
The open quantum system paradigm starts by considering a system,
or universe, which is partitioned (according to natural physical
arguments or some large scale separation) into two smaller
systems. One subsystem, called the {\it system}, is assumed to be
the one of interest and the other, called the {\it environment},
contains many more degrees of freedom. An accurate description of
the behavior of the system variables requires knowing the
influence from the environment due to their mutual interactions.
A less than accurate description of the overall influence of the
environment can be obtained by introducing some coarse-graining
over the environmental variables.
\subsection{Worldline Influence Functional}
In a coordinate system, assume at some initial time $t_i$ the quantum statistical state
of the combined system S (particle in position $z_i$) and
environment E (quantum field $ \varphi_i$) is described by a
density matrix $\rho (z_i, \varphi_i; z_i ^\prime, \varphi_i
^\prime; t_i )$. In practice, specifying such a state is non-trivial since one requires a time-like Killing vector to define positive frequency modes and hence a Hilbert space of states. If the spacetime admits an asymptotically flat region or is conformally flat then the initial state can be constructed. But a general spacetime may not admit a time-like Killing vector. Regardless, we will sidestep these issues by working at a formal level. At some $t_f > t_i$ the density matrix is evolved to
\begin{eqnarray}
&& \!\!\!\! \rho (z_f, \phi_f; z_f ^\prime, \phi_f ^\prime; t_f ) \nonumber \\
&& ~ = \int dz_i d\phi_i \int dz_i ^\prime d\phi_i ^\prime K(z_f, \phi_f, t_f; z_i, \phi_i, t_i ) \nonumber \\
&& {\hskip0.25in} \times \rho (z_i, \phi_i; z_i ^\prime, \phi_i
^\prime; t_i ) \, K ^* ( z_f ^\prime, \phi_f ^\prime, t_f; z_i
^\prime, \phi_i ^\prime, t_i ) \nonumber \\
\end{eqnarray}
where $K$ is the amplitude of the time-evolution operator
$\hat{U}(t_f, t_i) = \exp \{- \frac{i}{\hbar} \int _{t_i} ^{t_f}
dt \, \hat{H}_{S+E}[z, \phi] \}$ for the system plus environment
and has a path integral representation given by
\begin{eqnarray}
K(z_f, \phi_f, t_f; z_i, \phi_i, t_i ) = \int _{z_i, \phi_i}
^{z_f, \phi_f} {\cal D} z {\cal D} \phi \, e^{\frac{i}{\hbar}
S_{S+E} [z, \phi] }
\end{eqnarray}
The action describing the system plus environment can be written
as the actions for the system $S_S[z]$ and environment
$S_E[\phi]$ along with an interaction action $ S_{int} [z, \phi]
$ between them
\begin{eqnarray}
S_{S+E} [ z, \phi ] &=& S_S[z] + S_E[\phi] + S_{int} [z, \phi]
\end{eqnarray}
with
\begin{eqnarray}
S_S[z] &=& - m_0 \int d\tau \, \sqrt{ - g_{\mu \nu} u^\mu u^\nu } \nonumber \\
S_E [\phi] &=& \frac{1}{2} \int d^4 x \, \sqrt{-g} \left( g^{\mu\nu} \partial _\mu \phi \partial _\nu \phi - \xi_R R \phi^2 \right) \nonumber \\
S_{int} [z, \phi] &=& \int d^4x \, j(x; z] \, \phi(x) \nonumber \\
&=& - \frac{e}{4 \pi} \int d\tau \, \sqrt{ - u^\mu u_\mu } \, \phi(z(\tau))
\end{eqnarray}
where the current density $j(x; z]$ is given by
\begin{eqnarray}
j(x; z] = - \frac{e}{4 \pi} \int d\tau \, \sqrt{- g_{\mu \nu}
u^\mu u^\nu} \, \frac{ \delta^4 (x - z(\tau)) }{ \sqrt{-g} }
\end{eqnarray}
At this stage, $\tau$ is just a parameter of the worldline and not necessarily the proper time.
We use an overdot to denote differentiation with respect to the
worldline parameter $\tau$. For example, the 4-velocity of a
particle in Minkowski space ($g_{\mu\nu}=\eta_{\mu\nu}$) is
$u^\alpha = \frac{d z^\alpha}{d\tau} = \dot{z}^\alpha$.
Because the particle is moving in a gravitational field, one
should replace ordinary derivatives $d/d\tau$ by covariant
derivatives $D/d\tau$ and so, for instance, the 4-acceleration is
$\frac{D u^\alpha}{d\tau} = u^\beta \nabla _\beta u^\alpha$ and
not $\dot{u}^\alpha = \frac{d u^\alpha}{d\tau} = u^\beta
\partial_\beta u^\alpha$.
To facilitate easier computation it is customary to choose the
initial density matrix to correspond to a factorized state of the
system and environment. Physically, this means that all of the field modes have been uncorrelated with the particle by an instantaneous measurement at time $t_i$. Aside from issues about performing this measurement simultaneously in the spacetime, this choice is somewhat unphysical
because, as explained in \cite{HPZ}, an infinite amount of
energy is required to uncorrelate all the modes of the
environment (field) from the system (particle) at a particular
instant of time. For instance, in models of quantum
Brownian motion with an infinite number of environment oscillators the factorized initial state results in large
transients of the diffusion coefficients appearing in the master
equation for the reduced density matrix \cite{HPZ}. The transients appear as a result of the high frequency modes of the environment beginning to interact and correlate with the system just after the initial time. This recorrelation time lasts on the order of the inverse of the cut-off frequency used to regulate the divergences coming from oscillators of very high
frequencies. For times much longer than this transient time the behavior due to the initial
factorization is usually discounted. Other methods, including
the preparation function method \cite{Grabert_Schramm_Ingold}, allow for a somewhat
more physical initial state by including certain system-environment correlations (e.g. system in thermal equilibrium with the environment at the initial time), but still seem to suffer from some of the problems associated with the factorized state \cite{Romero_Paz}.
Having said this we assume that there is a Cauchy hypersurface at the initial time
$t_i$ such that
the initial density matrix takes the factorized form
\begin{eqnarray}
\rho (z_i, \phi_i; z_i ^\prime, \phi_i ^\prime; t_i ) = \rho_S
(z_i, z_i ^\prime; t_i ) \otimes \rho_E (\phi_i, \phi_i ^\prime;
t_i )
\end{eqnarray}
This simple form eases the manipulations for obtaining a description of the reduced particle dynamics.
After tracing out (a form of coarse-graining) the field variables
from the density matrix the {\it reduced density matrix} for the system
is given by
\begin{widetext}
\begin{eqnarray}
&& \rho_r (z_f, z_f ^\prime; t_f ) = \int d\phi_f \, \rho (z_f, \phi_f; z_f ^\prime, \phi_f; t_f ) \nonumber \\
&& ~ = \int dz_i \, dz_i ^\prime \int _{z_i} ^{z_f} {\cal D}z \int _{z_i ^\prime} ^{z_f ^\prime} {\cal D} z^\prime \, \rho _S (z_i, z_i ^\prime; t_i ) \, e^{\frac{i}{\hbar} ( S_S [z] - S_S [z ^\prime] ) } \nonumber \\
&& {\hskip0.5in} \times \int d\phi_f \, d\phi_i \, d\phi_i ^\prime \int _{\phi_i} ^{\phi_f} {\cal D} \phi \int _{\phi_i ^\prime} ^{\phi_f } {\cal D} \phi ^\prime\rho_E (\phi_i, \phi_i ^\prime; t_i ) \, e^{\frac{i}{\hbar} ( S_E [\phi] + S_{int} [z, \phi] - S_E [\phi^\prime] - S_{int} [ z^\prime, \phi^\prime] ) } \nonumber \\
&& ~ = \int dz_i \, dz_i ^\prime \int _{z_i} ^{z_f} {\cal D}z \int _{z_i ^\prime} ^{z_f ^\prime} {\cal D} z^\prime \, \rho _S (z_i, z_i ^\prime; t_i ) \, e^{\frac{i}{\hbar} ( S_S [z] - S_S [z ^\prime] ) } \, F[z, z^\prime] \label{red_rho}
\end{eqnarray}
The last line introduces the influence functional $F[z,
z^\prime]$, which is given by
\begin{eqnarray}
F[z, z^\prime] &=& \int d\phi_f \, d\phi_i \, d\phi_i ^\prime \int _{\phi_i} ^{\phi_f} {\cal D} \phi \int _{\phi_i ^\prime} ^{\phi_f } {\cal D} \phi ^\prime \, \rho_E (\phi_i, \phi_i ^\prime; t_i ) \, e^{\frac{i}{\hbar} ( S_E [\phi] + S_{int} [z, \phi] - S_E [\phi^\prime] - S_{int} [ z^\prime, \phi^\prime] ) } = e^{\frac{i}{\hbar} S_{inf}[z, z^\prime] } \nonumber \\
\end{eqnarray}
\end{widetext}
and $S_{inf}$ is the influence action. In operator language $F$ is
\begin{eqnarray}
F[z, z^\prime] = {\rm Tr}_{E} \: \hat{U}_{E+int} (t_f, t_i; z]
\: \hat{\rho}_{E} (t_i) \: \hat{U}^\dagger _{E+int} (t_f, t_i;
z^\prime] \nonumber \\
&& \label{IF_op}
\end{eqnarray}
where $\hat{U}_{E+int}(t_f, t_i; z]$ is the evolution operator that evolves the environment variables through its interaction with the system. The influence functional can be interpreted as the overlap of the environment states evolved forward and backward in time while interacting with different particle trajectories, or simply as the ensemble average of the evolution operator $U_{E+int}[z]$ evolved backward in time under a different external source $z^\prime$, as can be seen by writing (\ref{IF_op}) as
\begin{eqnarray}
&& F[z, z^\prime] = \langle \hat{U} ^\dagger _{E+int} [z^\prime] \, \hat{U}_{E+int} [z] \rangle_{ens} \nonumber \\
&& ~= \sum_\alpha \sum_{\alpha ^\prime} \rho _{E, \alpha \alpha^\prime} (t_i) \, \langle \alpha^\prime |
\hat{U}_{E+int} ^\dagger [z^\prime] \, \hat{U} _{E+int} [z] | \alpha \rangle \nonumber \\
\end{eqnarray}
In the interaction picture, the time evolution operator is
$\hat{U}_{E+int} (t_f, t_i; z] = T e^{\frac{i}{\hbar}
\hat{\phi}_I \cdot j[z]}$ where the $\cdot$ denotes spacetime
integration so that for two functions, possibly tensors,
$A(x^\alpha)$ and $B(x^\beta)$
\begin{eqnarray}
A \cdot B \equiv \int _{x^0 _i} ^{x^0 _f} dx^0 \int d^3 x \: \sqrt{-g} \: A(x) \: B(x)
\end{eqnarray}
Assuming that the initial state of the field is Gaussian, the influence functional can be calculated exactly giving
\begin{eqnarray}
F[ z, z^\prime] &=& e^{-\frac{1}{4 \hbar} j^- \cdot (16 \pi^2 G_H ) \cdot j^- + \frac{i}{\hbar} \: j^- \cdot (16 \pi^2 G_{ret} ) \cdot j^+ } \nonumber \\
\label{IF}
\end{eqnarray}
for the influence functional and
\begin{eqnarray}
&& S_{IF} [z, z^\prime] \stackrel{def}{=} -i \: \hbar \: {\rm ln} \: F[z, z^\prime] \nonumber \\
&& ~= \frac{i}{4} \, j^- \cdot ( 16 \pi^2 G_{H} ) \cdot j^- + j^- \cdot ( 16 \pi^2 G_{ret} ) \cdot j^+ \nonumber \\
\label{S_inf}
\end{eqnarray}
for the influence action. The difference and semi-sum current densities are defined as
\begin{eqnarray}
j^- &=& j[z] - j[z^\prime] \\
j^+ &=& \frac{j[z] + j[z^\prime]}{2}
\end{eqnarray}
and the Hadamard $G_H$ and retarded Green's functions $G_{ret}$ are
\begin{eqnarray}
&& G_H ( x, x^\prime ) = \big\langle \{ \hat{\phi}_I (x), \hat{\phi}_I (x^\prime) \} \big\rangle - 2 \: \big\langle \hat{\phi}_I (x) \big\rangle \big\langle \hat{\phi}_I (x^\prime) \big\rangle \nonumber \\
&& G_{ret} (x, x^\prime ) = i \, \theta_+ ( x, \Sigma ) \: \big\langle [ \hat{\phi}_I (x), \hat{\phi}_I (x^\prime) ] \big\rangle \nonumber \\
\label{ret}
\end{eqnarray}
The $\langle \cdots \rangle = {\rm Tr}_E \, \hat{\rho}_E
(\cdots)$ denotes quantum expectation values in the Gaussian
initial state $\hat{\rho}_E$ of the environment. The step
function $\theta_+ (x, \Sigma)$ appearing in $G_{ret}$ equals one
in the future of the point $x^\prime$ and zero otherwise. Here,
$\Sigma$ is a space-like hypersurface containing $x^\prime$. Had
the initial state contained non-Gaussian contributions, one would
have many additional terms involving cubic and higher powers of
the coupling. Likewise for nonlinear interactions (e.g. $\sim e
j[z] \cdot \phi^n$). Assuming that the coupling is small and that
non-Gaussianities are also small one can still use (\ref{IF}) and
(\ref{S_inf}) as a lowest order approximation.
\subsection{Semi-Classical Particle Dynamics}
The reduced density matrix for the particle is now written as
\begin{eqnarray}
&& \rho_r (z_f, z_f ^\prime; t_f ) \nonumber \\
&& ~= \!\! \int dz_i \, dz_i ^\prime \int _{z_i} ^{z_f} \!\! {\cal D} z \int _{z_i^\prime} ^{z_f ^\prime} \!\! {\cal D} z^\prime \, \rho_S (z_i, z_i^\prime; t_i ) \, e^{ \frac{i}{\hbar} S_{CGEA} [z, z^\prime] } \nonumber \\
\end{eqnarray}
where the {\it coarse-grained effective action} (CGEA) is defined as
\begin{eqnarray}
&& \!\!\!\! S_{CGEA} [z, z^\prime] \nonumber \\
&&~= S_S [z] - S_S [z^\prime] + S_{IF} [z, z^\prime] \nonumber \\
&& ~= S_S [z] - S_S [z^\prime] + j^- \cdot ( 16 \pi^2 G_{ret} ) \cdot j^+ \nonumber \\
&& {\hskip0.25in} + \frac{i}{4} \, j^- \cdot ( 16 \pi^2 G_{H} ) \cdot j^-
\end{eqnarray}
At this point it is worth mentioning that the magnitude of the influence functional decays rapidly for two largely separated histories since it is Gaussian (to lowest order) in $z^- = z-z^\prime$
\begin{eqnarray}
&& | F[z,z^\prime] | = e^{-\frac{1}{4 \hbar} j^- \cdot (16 \pi^2 G_H ) \cdot j^-} \nonumber \\
&& ~= \exp -\frac{e^2}{4 \hbar} \int d\tau \int d\tau^\prime \, z_- ^{\alpha} \, \frac{\delta j^-}{\delta z_-^{\alpha} } \, G_H (z_+^\mu, z_+^{\mu^\prime} ) \, \frac{\delta j^-}{\delta z_-^{\alpha^\prime} } \, z _-^{\alpha^\prime} \nonumber \\
&& {\hskip0.25in} + {\cal O} ( z_-^4 )
\end{eqnarray}
Here, and in the following, $z^\alpha \equiv z^\alpha (\tau)$ and $z^{\alpha^\prime} \equiv z^\alpha (\tau^\prime)$ so that an unprimed (primed) index refers to that component of a tensor field or coordinate evaluated at proper time $\tau$ ($\tau^\prime$) and
\begin{eqnarray}
z_\mu ^- &=& z_\mu - z^\prime _\mu \\
z_\mu^+ &=& \frac{ z_\mu + z^\prime _\mu }{2}
\end{eqnarray}
The norm of $F$ is equal to the norm of the decoherence
functional (see below), which is a measure of how much the
particle's worldline is decohered. So, if the quantum
fluctuations of the field (environment) provide a strong enough
mechanism for decoherence (this should be checked on a case by
case basis and will be assumed true in the cases under study
here) then we are justified in expanding the CGEA about the
classical trajectory $\bar{z}^\mu$. Doing so gives
\begin{eqnarray}
S_{CGEA} [z,z^\prime] = \int d\tau^\prime \, z_- ^{\alpha^\prime} \, \frac{\delta S_{CGEA} }{ \delta z_- ^{\alpha^\prime} } \bigg| _{z = z^\prime = \bar{z} } + {\cal O} (z_- ^2 ) \nonumber \\
\end{eqnarray}
Using this expression in the
reduced density matrix and doing a stationary phase approximation
gives the equations of motion for the classical worldline
$\bar{z}$
\begin{eqnarray}
\frac{ \delta S_{CGEA} }{ \delta z_- ^\mu (\tau) } \bigg| _{z=z^\prime = \bar{z} } = 0
\label{CGEA_funcvar}
\end{eqnarray}
Evaluating the functional derivative using, for some test function $f(x)$,
\begin{eqnarray}
&& \frac{\delta}{\delta z_- ^\mu (\tau) } \int d^4x \, \sqrt{-g} \, j^- (x; z] f(x) \nonumber \\
&& {\hskip0.25in} = \frac{e}{4\pi} \, \left( \frac{D u_\mu}{d\tau} + w_\mu ^{~ \nu} [z] \nabla_{z^\nu} \right) f(z) \nonumber \\
&& {\hskip0.25in} \stackrel{def}{=} \frac{e}{4\pi} \, \vec{w}_\mu [z] \, f(z)
\end{eqnarray}
where $w_\mu ^{~ \nu} = g_\mu ^{~\nu} + u_\mu u^\nu$ and $u^\alpha u_\alpha = -1$ (proper-time gauge), gives
\begin{eqnarray}
m_0 \, \frac{D \bar{u}_\mu}{d\tau} = e \, \vec{w}_\mu [\bar{z}] \phi_{ret} (\bar{z})
\label{mean_eom}
\end{eqnarray}
where $\phi_{ret}$ is the retarded field
\begin{eqnarray}
\phi_{ret} (x) = e \int d\tau^\prime \, G_{ret} (x, z^{\alpha^\prime})
\label{ret_field}
\end{eqnarray}
The right side of (\ref{mean_eom}) is the self-force on the
particle arising from the radiation reaction. Furthermore, the
vector operator $\vec{w}_\mu$ contains the particle's
acceleration implying that the particle moves with an effective
time-dependent mass equal to $m_0 - e \, \phi_{ret} (\bar{z})$.
(See \cite{Burko_Harte_Poisson} for a discussion of evaporating
scalar charges based on this time-dependent effective mass.)
Provided there is strong enough decoherence to suppress the
quantum fluctuations of the worldline, the equations of motion
for the quantum expectation value $\langle \hat{z}^\mu \rangle$
are the same as in (\ref{mean_eom}) at tree-level in both the
particle and the field \cite{JH1}. Unfortunatley,
(\ref{mean_eom}) is problematic since the self-force diverges on
the worldline of the point particle. This divergence can be
traced to the singular coincidence limit of the retarded Green's
function $\lim_{\tau^\prime \rightarrow \tau} G_{ret} \rightarrow
\infty$ and results from the point particle assumption. We will
discuss how to treat this problem in Sec. \ref{Reg} by adopting
an effective field theory point of view.
\subsection{Stochastic Semi-Classical Particle Dynamics}
We have made the assumption that the quantum fluctuations of the
worldline are strongly suppressed by the decoherence due to the
quantum field. Even under strong decoherence when the classical
trajectory is well-defined, the quantum fluctuations of the field
can still influence the classical motion of the particle
through the particle-field coupling $S_{int}$. They may show up as
classical stochastic forces on the particle. In this section we
shall show how this comes about using the influence functional
for the reduced density matrix.
We start by invoking the relation
\begin{eqnarray}
&& e^{-\frac{1}{4 \hbar} j^- \cdot ( 16 \pi^2 G_H) \cdot j^-} = N \!\! \int {\cal D} \xi(x) \, e^{- \frac{1}{\hbar} \xi \cdot G_H ^{\, -1} \cdot \xi - \frac{ 4 \pi i }{ \hbar} \xi \cdot j^-} \nonumber \\
\end{eqnarray}
where $N$ is a normalization factor that is independent of the worldline coordinates and $\xi(x)$ is some auxiliary field. Using this relation, the influence functional (\ref{IF}) can be written as
\begin{eqnarray}
F[z, z^\prime] = N \!\! \int \!\! {\cal D} \xi(x) \, e^{- \frac{1}{\hbar} \xi \cdot G_H ^{\, -1} \cdot \xi + \frac{ i }{ \hbar} j^- \!\! \cdot ( - 4 \pi \xi + 16 \pi^2 G_{ret} \cdot j^+ ) } \nonumber \\
\end{eqnarray}
where $\xi(x)$ appears as an auxiliary function, which can be
interpreted as a classical stochastic or noise field \cite{JH1, Calzetta_Roura_Verdaguer}
with an associated Gaussian probability distribution functional
\begin{eqnarray}
P_\xi [\xi(x)] = e^{-\frac{1}{\hbar} \xi \cdot G_H ^{\, -1} \cdot \xi }
\end{eqnarray}
The fact that this is Gaussian is a direct consequence of taking an initial Gaussian state for the quantum field. With respect to $P_\xi [\xi]$ this implies that $\xi$ has zero-mean
and its correlator is proportional to the Hadamard function
encoding the information about the fluctuations in the quantum
field \cite{footnote_1}.
\begin{eqnarray}
\big\langle \xi(x) \big\rangle_\xi &=& 0 \\
\big\langle \{ \xi(x), \xi(x^\prime) \} \big\rangle _\xi &=& \hbar \, G_H (x, x^\prime) \label{xi_correlator}
\end{eqnarray}
where $\langle \ldots \rangle _\xi = N \int {\cal D} \xi \, P_\xi \, (\ldots)$.
Now the reduced density matrix (\ref{red_rho}) becomes
\begin{widetext}
\begin{eqnarray}
\rho_r (z_f, z_f ^\prime; t_f) = N \int dz_i \, dz_i ^\prime \int _{z_i} ^{z_f} \!\! {\cal D} z \int_{z_i ^\prime} ^{z_f^\prime} \!\! {\cal D} z^\prime \, \rho_S (z_i, z_i ^\prime; t_i ) \int {\cal D} \xi \, P_\xi [\xi] \, e^{ \frac{i}{\hbar} S_{SEA} [z, z^\prime; \xi] }
\label{rho_sea}
\end{eqnarray}
where the {\it stochastic effective action} (SEA) is defined as
\begin{eqnarray}
S_{SEA} [z, z^\prime; \xi] = S_S [z] - S_S [z^\prime] + j^- \! \cdot ( - 4 \pi \xi + 16 \pi^2 G_{ret} \cdot j^+ ) = S_{CGEA} ^R [z,z^\prime] - (4 \pi) \, \xi \cdot j^-
\end{eqnarray}
and $S_{CGEA}^R$ is the real part of the CGEA. As before, if the worldline is strongly decohered by the quantum fluctuations of the environment then we are justified in expanding the SEA around the classical solution
\begin{eqnarray}
S_{SEA} [z, z^\prime; \xi] = - \int d\tau^\prime z_- ^{\alpha^\prime} \eta_{\alpha^\prime} [z^+] + \frac{1}{2} \int d\tau^\prime \int d\tau^{\prime \prime} \, z_- ^{\alpha^\prime} z_- ^{\alpha^{\prime\prime}} \frac{ \delta^2 S_{CGEA}^R }{ \delta z_+ ^{\alpha^\prime} \delta z_+ ^{\alpha ^{\prime \prime} } } \bigg| _{z=z^\prime = \bar{z}} + {\cal O} (z_- ^3)
\label{SEA_expansion}
\end{eqnarray}
\end{widetext}
where we have used (\ref{CGEA_funcvar}), expanded $j^-$ in powers of $z_-$ and defined the stochastic force
\begin{eqnarray}
\eta_\mu [z] = e \, \vec{w}_\mu[z] \xi(z)
\end{eqnarray}
Putting (\ref{SEA_expansion}) into the reduced density matrix and using the stationary phase approximation gives the stochastic equations of motion for the worldline fluctuations $z_- ^\mu = \tilde{z}^\mu \equiv z^\mu - \bar{z}^\mu$
\begin{eqnarray}
\int d\tau^\prime \, \tilde{z}^{\alpha^\prime} \frac{ \delta^2 S_{CGEA}^R }{ \delta \bar{z} ^\mu \delta \bar{z} ^{\alpha ^\prime } } \bigg| _{z=z^\prime = \bar{z}} = \eta_\mu [\bar{z}]
\label{z_flucs_funcder}
\end{eqnarray}
It should be emphasized that this equation describes the
evolution of small perturbations $\tilde{z}$ around the
semi-classical trajectory that arise from the stochastic
manifestation $\eta$ of the quantum field fluctuations.
We can obtain a stochastic version of the ALD equation (\ref{mean_eom}) if we add the equations of motion for the classical dynamics
\begin{eqnarray}
0 = \frac{\delta S_{CGEA}^R }{ \delta z_- ^\mu } \bigg|_{z=z^\prime=\bar{z}}
\end{eqnarray}
to the left side. Then we may write
\begin{eqnarray}
\frac{\delta S_{CGEA}^R }{ \delta z_- ^\mu } \bigg|_{z_-=0} = \eta_\mu [z]
\end{eqnarray}
Evaluating this functional derivative gives the stochastic semi-classical particle dynamics for the full worldline $z_\mu = \bar{z}_\mu+\tilde{z}_\mu$
\begin{eqnarray}
m_0 \, \frac{Du_\mu}{d\tau} = e \, \vec{w}_\mu [z] \phi_{ret} (z) + \eta_\mu [z]
\label{stoch_eom}
\end{eqnarray}
Notice that both the deterministic and the stochastic components
of the self-force can push the particle away from its geodesic
motion with respect to a fixed background spacetime. However, we should
keep in mind that this equation is really only valid to linear
order in the fluctuations $\tilde{z}$ since higher order terms
will correspond to quantum corrections that we have been
neglecting. Of course, we could have obtained (\ref{stoch_eom})
by using a stationary phase approximation in the reduced density
matrix (\ref{rho_sea}) without having expanded around the classical trajectory.
However, we believe that the validity of (\ref{stoch_eom}) to
linear order in the fluctuations would not have been as
transparent.
The stochastic correlation functions of the force $\eta_\mu$ can
be evaluated using the knowledge of the $\xi$ correlators above.
As was commented in the previous paragraph we must evaluate these
correlation functions along the classical trajectory $\bar{z}$ to be consistent with the linearization.
The mean of the stochastic force is zero
\begin{eqnarray}
\big\langle \eta_\mu [\bar{z}] \big\rangle _\xi = e \, \vec{w}_\mu [\bar{z}] \, \big\langle \xi (\bar{z}) \big\rangle _\xi = 0
\end{eqnarray}
and the symmetric two-point function of the stochastic force is
\begin{eqnarray}
\big\langle \{ \eta_\mu [ \bar{z} ], \eta_{\mu^\prime} [ \bar{z} ] \} \big\rangle _\xi = \hbar \, e^2 \, \vec{w}_{\left( \mu \right. } [\bar{z}] \, \vec{w}_{\left. \mu^\prime \right) } [\bar{z}] \, G_H (\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) \nonumber \\
\label{eta_corr}
\end{eqnarray}
This shows that the noise $\eta_\mu$ is multiplicative, colored
and depends on the particle's initial conditions through the
classical trajectory. The noise correlator also generically
depends on the field's initial conditions as is seen by the
appearance of $G_H$ in the equation. The Hadamard function $G_H$
does not vanish on a space-like hypersurface implying that the
quantum correlations in the environment are non-local. However,
the noise correlator above is evaluated on time-like separated
points only. For equal proper times $\tau^\prime = \tau$ the
Hadamard function diverges so a suitable regularization procedure
must be used in order to make sense of (\ref{eta_corr}) near
coincidence.
In the next two sections we describe a new method introduced in
\cite{JH1} for regulating the ultraviolet divergences in the
retarded Green's function. We introduce a high energy scale in
the quantum field below which the low energy point particle dynamics is
expected to be valid. For long times as compared to the inverse
of the high energy scale, we invoke a quasi-local expansion in
order to obtain the relevant contributions to the particle's
motion. Our work is a generalization of \cite{JH1} to curved
spacetime. We follow Poisson's description of scalar radiation
reaction in curved spacetime and use units where $c = G = 1$. (We
change the particle physics notation in \cite{JH1} (+, -, -, -)
to the MTW notation (-, +, +, +) of \cite{Poisson}.)
\section{Regularization of the Retarded Green's Function}
\label{Reg}
Before we discuss regularizing the retarded Green's function it will be beneficial to introduce some notation. Following \cite{Poisson} we define the step function $\theta_+(x, \Sigma)$ to be 1 for any point $x$ to the future of a space-like hypersurface $\Sigma$ and $0$ otherwise. This is a generalization of the ordinary step function to curved spacetime. Using this, one can define the following distributions
\begin{eqnarray}
\delta _+ ( \sigma(x,x ^\prime) ) &=& \theta_+ (x, \Sigma) \: \delta ( \sigma (x, x^\prime) ) \\
\theta_+ ( -\sigma( x, x^\prime ) ) &=& \theta_+ ( x, \Sigma) \: \theta ( - \sigma ( x, x^\prime ) )
\end{eqnarray}
where $\sigma (x, x^\prime)$ is Synge's world function along the (unique) geodesic linking $x=z(\tau)$ and $x^\prime=z(\tau^\prime)$. Taking the lightcone centered on $x^\prime$, with $x^\prime$ on the space-like hypersurface $\Sigma$, it follows that $\delta_+(\sigma)$ has support along the forward lightcone only while $\theta_+ (-\sigma)$ is one in the causal future of $x^\prime$ and vanishes everywhere else.
In (\ref{ret}) the retarded Green's function is calculated from
the commutator of the interaction picture field $\hat{\phi}_I$.
Recall that $\hat{\phi_I}$ evolves under the free dynamics so that
\begin{eqnarray}
(\Box - \xi_R R) \, \phi_I = 0
\end{eqnarray}
and so is the homogeneous solution to the full field equation in
the Heisenberg picture (satisfying the same initial conditions)
\begin{eqnarray}
(\Box - \xi_R R) \, \phi_H = j[z]
\end{eqnarray}
where $\hat{\phi}_H = \hat{\phi}_I + \hat{1} \, G_{ret} \cdot
j[z]$ for a classical source. Because the source is classical
only the first term in the solution gives a contribution to the
commutator and hence to the retarded Green's function
\begin{eqnarray}
G_{ret} \sim \big\langle [ \hat{\phi}_H (x), \hat{\phi}_H (x^\prime) ] \big\rangle = \big\langle [ \hat{\phi}_I (x), \hat{\phi}_I (x^\prime) ] \big\rangle
\end{eqnarray}
Finally, since the commutator is state-independent it follows
that $G_{ret}$ for the quantum field $\hat{\phi}_H$ and for the
corresponding classical field $\phi$ are the same.
\subsection{Hadamard Expansion}
This allows us to invoke Hadamard's ansatz for which the retarded
Green's function for a scalar field in a curved $3+1$ dimensional
spacetime has the form \cite{Poisson}
\begin{eqnarray}
&& G_{ret} ( x, x^\prime ) \nonumber \\
&& ~= \Delta ^{1/2} ( x, x^\prime ) \, \delta_+ ( \sigma (x, x^\prime ) ) + V( x, x^\prime ) \: \theta_+ ( -\sigma (x, x^\prime ) ) \nonumber \\
\label{Hadamard}
\end{eqnarray}
which is the sum of a ``direct" part (proportional to $\delta_+$)
and a ``tail" part (proportional to $\theta_+ (-\sigma)$). Notice
that the appearance of $\delta_+$ and $\theta_+$ ensures that
$G_{ret}$ has the correct causal structure. In order for this
ansatz to be valid, the points $x$ and $x^\prime$ must be
connected by a unique geodesic. Otherwise, ambiguities arise from
the appearance of caustics when parallel propagating a tensor
field from $x^\prime$ to $x$ or vice versa. This
will be assumed in the following. The function $\Delta (x,x
^\prime)$ is the van Vleck determinant and the function $V(x,
x^\prime)$ satisfies the homogeneous Klein-Gordon equation in
curved spacetime
\begin{eqnarray}
( \Box_g - \xi_R R ) \, V(x, x^\prime) = 0
\end{eqnarray}
with boundary data determined by the restriction of $V$ to the forward lightcone, denoted by $\tilde{V}$. The restriction of $V$ is found by solving
\begin{eqnarray}
\tilde{V}_{,\alpha} \sigma^\alpha + \frac{1}{2} \left( \sigma ^\alpha _{~\alpha} -2 \right) \tilde{V} = \frac{1}{2} \left\{ \left( \Box_g - \xi_R R \right) \Delta^{1/2} \right\}_{\sigma=0} \nonumber \\
\label{V_tilde}
\end{eqnarray}
Knowing all of the light-like geodesics emanating into the future
from $x^\prime$, one can integrate this equation to construct
$\tilde{V}$ on the light-cone and hence $V$ inside the
light-cone. Of course, this just solves for $V(x,x^\prime)$ for
any $x$ to the causal future of $x^\prime$ and must be solved
again for different values of $x^\prime$. This makes determining $V$ very
difficult and tedious for generic spacetimes. However, if the spacetime
possesses enough symmetry (e.g. de Sitter, some FRW cosmologies) then $V$ can be
constructed relatively easily \cite{Burko_Harte_Poisson}.
From the mean and stochastic equations of motion (\ref{mean_eom}) and (\ref{stoch_eom}) it is necessary to compute the restriction of $G_{ret}$ to the particle's worldline
\begin{eqnarray}
&& G_{ret} ( z^\alpha, z ^{\alpha ^\prime} ) = \Delta ^{1/2} ( z^\alpha, z^{\alpha^\prime} ) \, \delta_+ ( \sigma (z^\alpha, z^{\alpha^\prime} ) ) \nonumber \\
&& {\hskip 0.5in} + V( z^\alpha, z^{\alpha^\prime} ) \: \theta_+ ( -\sigma (z^\alpha, z^{\alpha^\prime} ) )
\label{Hadamard_Ansatz}
\end{eqnarray}
Using the Hadamard ansatz and (\ref{ret_field}) it is easy to see
that for two (time-like separated) points on the particle
worldline, $\delta_+(\sigma)$ has support only at coincidence,
when $\tau^\prime = \tau$. The contribution from the direct term
is then
\begin{eqnarray}
&& \int _{\tau_i} ^{\tau_f} d\tau^\prime \, \Delta^{1/2} (z^\alpha, z^{\alpha ^\prime} ) \, \theta_+( z^\alpha, S) \, \delta ( \sigma ( z^\alpha, z^{\alpha^\prime} ) ) \nonumber \\
&& \!\!\!\! = \int _{\tau_i} ^\tau d\tau^\prime \, \Delta^{1/2} (z^\alpha, z^{\alpha^\prime} ) \, \delta ( \tau^\prime - \tau ) \, \left( \frac{ d\sigma }{ d\tau } \right) _{\tau^\prime = \tau} ^{-1}
\label{divergence}
\end{eqnarray}
This expression is ultraviolet divergent since $d\sigma/d\tau$ evaluates to zero for $\tau^\prime = \tau$ and so the retarded field is not
well-defined (in fact, it is infinite) when evaluated at the
location of the point particle. This feature is not limited to just a
quantum field as it appears even for a classical field since they
share the same retarded Green's function.
The presence of this divergence suggests that a regularization
procedure is needed to render the equations of motion
(\ref{mean_eom}) and (\ref{stoch_eom}) finite. Several procedures
have been proposed in the literature for doing this. In
\cite{Detweiler_Whiting, Poisson}, the retarded field is evaluated
near the worldline so that the divergence can be tracked in the
limit that one approaches the worldline. This divergence
renormalizes the particle's mass. Writing the retarded field in
terms of functions singular $\phi_S$ and regular $\phi_R$ on the
worldline allows, after calculating their gradient and expanding
them near the worldline, for the self-force to be evaluated
unambiguously along the worldline. This method can also be used
for electromagnetic and gravitational self-force calculations. In
another method developed in \cite{Barack_Ori, Barack, BMNOS,
Barack_Ori_2, Mino_Nakano_Sasaki}, a mode-sum regularization
procedure is used in which the self-force is decomposed in terms
of the multipole moments of the field. From this one subtracts
the moments of the singular part of the self-force and resums
over the multipoles to obtain a self-force that is finite on the
worldline. In \cite{Rosenthal} the ``massive field approach" is
used in which an auxiliary massive field is introduced to
subtract away the ultraviolet divergence from the massless field.
This procedure is similar to Pauli-Villars type regularization
and has also been used in \cite{JohnsonPhD} to regularize the
direct term in $G_{ret}$.
Other regularization methods have been used for the radiation
reaction of non-relativistic particles without gravity present
\cite{NR_Reg}. Notable of these include the extended-charge
models of \cite{Ext_Charge} in which the particle is described as
a rigid volume with a non-zero size. Unfortunately, this type of
prescription cannot be extended to relativistic models since
rigid bodies are incompatible with relativity \cite{footnote_2}.
Furthermore, the successes of quantum field theory
(e.g. as applied to the scattering of point particles), has
proved to be an adequate description of the low-energy effective
dynamics of point particles to $\sim$TeV scales. It seems
unnecessary, then, to introduce non-trivial structure to the
particle to study the self-force.
\subsection{Regularization motivated by Effective Field Theory}
With this consideration in mind we introduce yet another
regularization for the self-force motivated by \cite{JH1}. All
theories for the fundamental interactions (e.g. QED, electroweak
theory, QCD) can be interpreted as effective theories, in that
they are are low energy limits of a more complete theory valid at
higher energies, yet they provide an excellent description of low
energy phenomena, at least for renormalizable theories, without
requiring the details of the complete theory. For energies much
higher than the energy scale of the effective theory, new physics
is likely to become important.
In this spirit we introduce a
regulator $\Lambda$ for the field. Above the energy scale of this
regulator, $E_\Lambda$, new physics is assumed to occur. For
energies much lower than $E_\Lambda$, the dynamics
(\ref{mean_eom}) and (\ref{stoch_eom}) will be sufficiently
accurately described by using a regulated quantum field. This
approach has been taken in \cite{JH1} in deriving
the Abraham-Lorenz-Dirac (ALD) equations in flat spacetime. It is easily extended to
motions in a curved spacetime since the direct part of the
retarded Green's function is a local quantity on the worldline.
Regularization is achieved here by choosing any suitably smooth
function to approximate $\delta_+$ for large values of the
regulator $\Lambda$. Following \cite{JH1} we replace $\delta_+$ with
\begin{eqnarray}
\delta_+ (\sigma (z^\alpha, z^{\alpha^\prime}) ) &\rightarrow& \theta_+ (-\sigma(z^\alpha, z^{\alpha^\prime} ) ) \, \sqrt{ \frac{8}{\pi} } \, \Lambda^2 \, e^{-2 \Lambda^4 \sigma^2 (z, z^\prime) } \nonumber \\
&=& \theta_+ ( -\sigma ( z^\alpha, z^{\alpha^\prime} ) ) \, g_{ret} ^\Lambda ( z^\alpha, z^{\alpha^\prime} )
\label{replacement}
\end{eqnarray}
In the limit of infinitely large $\Lambda$ we have that $\theta_+ \, g_{ret}^\Lambda$ approaches $\delta_+$. The integral of the direct part of $G_{ret}$ (\ref{divergence}) receives a contribution from $\delta_+$ at coincidence implying that $\sigma = 0$. The function $g_{ret}^\Lambda$ is smooth but approximates $\delta_+$ well only if $\Lambda^2 \sigma \gg 1$. This approximation will not hold if $\sigma$ is identically zero. Nevertheless, we will assume that $\sigma$ is small and approaching zero but $\Lambda$ is such that $\Lambda^2 \sigma \gg 1$ still holds. This separation of scales allows us to do a quasi-local expansion below in which the self-force will be expanded near coincidence.
Making the replacement (\ref{replacement}) in $G_{ret}$ gives the regulated Green's function
\begin{widetext}
\begin{eqnarray}
G_{ret} ^\Lambda (z^\alpha, z^{\alpha^\prime} ) &=& \theta_+ ( - \sigma( z^\alpha, z^{\alpha^\prime} ) ) \left\{ \Delta^{1/2} (z^\alpha, z^{\alpha^\prime} ) \, g_{ret} ^\Lambda ( z^\alpha, z^{\alpha^\prime} ) + V( z^\alpha, z^{\alpha^\prime} ) \right\} = \theta_+ ( -\sigma) \left\{ \Delta^{1/2} \, g_{ret}^\Lambda + V \right\}
\end{eqnarray}
where an abbreviated notation has been used on the far right side to ease the appearance of later expressions. Also, it should be clear that in the limit $\Lambda \rightarrow \infty$ that $G_{ret}^\Lambda \rightarrow G_{ret}$.
Putting $G_{ret}^\Lambda$ into (\ref{mean_eom}) gives
\begin{eqnarray}
m_0 \, \frac{D \bar{u}_\mu}{d\tau} = e^2 \, \left( \frac{D \bar{u}_\mu}{d\tau} + w_\mu ^{~\nu} [\bar{z}] \, \nabla_{\bar{z}^\nu} \right) \int _{\tau_i} ^{\tau_f} d\tau^\prime \, \theta(\tau- \tau^\prime) \left\{ \Delta^{1/2} \, g_{ret}^\Lambda + V \right\}
\end{eqnarray}
where the quantities in the integral are evaluated along the classical trajectory $\bar{z}$. Passing the derivative through the integral gives
\begin{eqnarray}
m_0 \, \frac{D \bar{u}_\mu}{d\tau} &=& e \, \frac{D \bar{u}_\mu}{d\tau} \, \phi_{ret} (\bar{z}) + e^2 \, w_\mu ^{~\nu} [\bar{z}] \left\{ \left[ \Delta^{1/2} g_{ret}^\Lambda + V \right] \left[ \nabla_{\bar{z}^\nu} ( \tau -\tau^\prime ) \right] + \int _{\tau_i} ^\tau d\tau^\prime \, \nabla_{\bar{z}^\nu} \left( \Delta^{1/2} g_{ret} ^\Lambda + V \right) \right\} \nonumber
\end{eqnarray}
where the $[ \ldots ]$ denotes the coincidence limit $\tau^\prime \rightarrow \tau$ of the quantity it contains. This notation looks the same when denoting a functional of a function (e.g. $w_\mu ^{~\nu} [z]$) but the context should be clear. From Appendix A one can show that
\begin{eqnarray}
\left[ \nabla_{\bar{z}^\nu} ( \tau - \tau^\prime ) \right] = - \left[ \nabla_\nu (\bar{\sigma}^\alpha \bar{u}_\alpha) \right] = - \bar{u}_\nu \label{proof}
\end{eqnarray}
and since $w_\mu ^{~\nu} [\bar{z}]$ projects vectors onto a direction orthogonal to the mean 4-velocity then $w_\mu ^{~\nu} [\bar{z}] \left[ \nabla_{\bar{z}^\nu} ( \tau -\tau^\prime) \right]$ gives no contribution leaving
\begin{eqnarray}
m_0 \, \frac{D \bar{u}_\mu}{d\tau} &=& e^2 \, \frac{D \bar{u}_\mu}{d\tau} \, \phi_{ret} (\bar{z}) + e^2 \, w_\mu ^{~\nu} [\bar{z}] \int _{\tau_i} ^\tau d\tau^\prime \, \nabla_{\bar{z}^\nu} \left( \Delta^{1/2} \, g_{ret} ^\Lambda + V \right) \\
&=& e^2 \, \frac{D \bar{u}_\mu}{d\tau} \, \phi_{ret} (\bar{z}) + e^2 \, w_\mu ^{~\nu} [\bar{z}] \int _{\tau_i} ^\tau d\tau^\prime \, \left\{ \left( \nabla_{\bar{z}^\nu} \, \Delta^{1/2} \right) g_{ret}^\Lambda + \Delta^{1/2} \, \nabla_{\bar{z}^\nu} g_{ret}^\Lambda + \nabla_{\bar{z}^\nu} V \right\}
\label{starting_pt}
\end{eqnarray}
\end{widetext}
It will be convenient for later manipulations to define the {\it tail term} as the integral over the past history of the gradient of $V$,
\begin{eqnarray}
\phi_\nu ^{tail} (\bar{z}(\tau)) = e \int _{\tau_i} ^\tau d\tau^\prime \, \nabla_{\bar{z}^\nu} V
\label{tail_term}
\end{eqnarray}
The tail term will turn out to be responsible for the non-Markovian and history-dependent dynamics of the particle's evolution. Consequently, this presents a significant source of difficulty when trying to determine the particle's motion since one has to know the entire past history of the particle's worldline in order to make predictions about its current position and speed.
\section{Effect of Radiation Reaction on Particle Trajectory}
In order to simplify some of the difficulties of solving the nonlinear integro-differential equation (\ref{starting_pt}) we utilize the effective theory viewpoint discussed in the previous section. Since $g_{ret}^\Lambda$ is strongly peaked around $\sigma = 0$ then the contributions to the self-force will be largest when $\tau^\prime \approx \tau$. We therefore introduce a quasi-local expansion by expanding $\theta_+ \, \Delta^{1/2} \, g_{ret}^\Lambda$ around coincidence $\tau^\prime = \tau$.
This yields a well-defined description of the self-force in terms of relevant and irrelevant quantities. The relevant terms are those that either renormalize certain parameters of the
particle or otherwise affect the particle dynamics when $\Lambda$
goes to infinity while the irrelevant terms are those that give no contribution in the same limit.
\subsection{Quasilocal Expansion}
We begin by expanding the van Vleck determinant and
$g_{ret}^\Lambda$ for small values of $\sigma(z^\alpha, z^{\alpha^\prime})$.
Following \cite{Poisson} the expansion of the square root of the van Vleck determinant around $\sigma, \sigma^\alpha = 0$ is
\begin{widetext}
\begin{eqnarray}
\Delta^{1/2} (\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) = 1 + \frac{1}{12} \, R_{\alpha \beta} ( \bar{z} ) \, \sigma^\alpha \, \sigma^\beta - \frac{1}{36} \, R_{\alpha \beta ; \gamma} \, \sigma^\alpha \, \sigma^\beta \, \sigma^\gamma + \ldots
\label{vanVleck}
\end{eqnarray}
where the $\ldots$ denotes terms of higher order in the expansion. The covariant derivative with respect to the worldline coordinate of the above quantity is
\begin{eqnarray}
\nabla_{\bar{z}^\nu} \Delta^{1/2} (\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) = \frac{1}{6} \, R_{\alpha \beta} (\bar{z}) \, \sigma^\alpha _{~ ; \nu} \, \sigma^\beta + \frac{1}{12} \, R_{\alpha \beta ; \nu} (\bar{z}) \, \sigma^\alpha \, \sigma^\beta + \ldots
\end{eqnarray}
Lastly, $\sigma ^\alpha _{~; \nu}$ can be expanded around small $\sigma$ to give
\begin{eqnarray}
\sigma ^\alpha _{~;\nu} = g^\alpha _{~\nu} - \frac{1}{3} \, R^\alpha _{~\gamma \nu \delta} (\bar{z}) \, \sigma^\gamma \, \sigma ^\delta + \ldots
\end{eqnarray}
so that the gradient of $\Delta ^{1/2}$ becomes
\begin{eqnarray}
\nabla_{\bar{z}^\nu} \Delta^{1/2} (\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) = \frac{1}{6} \ R_{\nu \alpha} (\bar{z}) \, \sigma ^\alpha - \frac{1}{18} \, R ^\lambda _{~\alpha \beta \nu; \lambda} (\bar{z}) \, \sigma^\alpha \, \sigma^\beta + \ldots
\label{grad_vanVleck}
\end{eqnarray}
Next, we expand these quantities and $g_{ret}^\Lambda$ around $\tau^\prime = \tau+s$ with $s$ such that $\Lambda s \gg 1$. The expansions of some relevant quantities involving the world function $\sigma$ are (see Appendix A for details)
\begin{eqnarray}
s &=& \tau^\prime - \tau \\
\sigma (\tau, \tau^\prime) &=& - \frac{1}{2} \, s^2 + {\cal O} (s^4) \\
\sigma_\mu (\tau, s) &=& - s \: u_\mu (\tau) - \frac{s^2}{2} \: \frac{D u_\mu (\tau)}{d\tau} - \frac{s^3}{6} \: \frac{D^2 u_\mu (\tau)}{d\tau^2} + {\cal O} (s^4) \label{sigma_mu}
\end{eqnarray}
Using these results, the function $g_{ret}^\Lambda$ is
\begin{eqnarray}
g_{ret} ^\Lambda = \sqrt{ \frac{ 8}{ \pi} } \: \Lambda^2 \: e^{ - 2 \Lambda^4 \sigma^2 } = \sqrt{ \frac{ 8}{ \pi} } \: \Lambda^2 \: e^{ - \Lambda^4 s^4 /2 } + {\cal O} (s^6)
\end{eqnarray}
Lastly, we expand $\nabla_{\bar{z}^\nu} g_{ret}^\Lambda$ in powers of $s$
\begin{eqnarray}
\nabla _{\bar{z} ^\mu} g_{ret}^\Lambda &=& \partial _{\bar{z}^\mu} g_{ret}^\Lambda = \sigma_\mu \left( \frac{ \partial \sigma}{ \partial s} \right)^{\!\!-1} \!\! \frac{\partial g_{ret}^\Lambda}{\partial s} = \left\{ u_\mu (\tau) + \frac{ s}{2} \: \frac{Du_\mu (\tau)}{d\tau} + \frac{ s^2 }{6} \: \frac{D^2u_\mu (\tau)}{d\tau^2} + {\cal O} (s^3) \right\} \frac{\partial g_{ret}^\Lambda}{\partial s}
\end{eqnarray}
Applying these results to (\ref{starting_pt}) gives
\begin{eqnarray}
&& \left( m_0 - \frac{e^2}{2} \, g_{(1)} (r) + e^2 c_{(0)} (r) - e^2 \int_{\tau_i} ^\tau d\tau^\prime \, V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) \right) \frac{D\bar{u}_\mu}{d\tau} \nonumber \\
&& {\hskip1.5in} = e \, w_\mu ^{~\nu} [\bar{z}] \, \phi_\nu ^{tail} (\bar{z}) + e^2 \sum_{n=1} ^\infty \left( g_{(n+1)} (r) u_\alpha ^{(n+1)} [\bar{z}] + c_{(n)} (r) v_\alpha ^{(n)} [\bar{z}] \right)
\label{mean_2}
\end{eqnarray}
with $r = \tau - \tau_i$ being the elapsed proper time since $\tau_i$.
The expressions for the $r$-dependent coefficients ($g_{(n)}$ and $c_{(n)}$) and the trajectory-dependent vectors ($u_\mu ^{(n)}$ and $v_\mu ^{(n)}$) can be found in Appendix B. However, only the relevant factors in the limit of very large $\Lambda$ will be given here
\begin{eqnarray}
c_{(0)} (r) = - \frac{ \Lambda }{ 2^{1/4} \sqrt{\pi} } \, \gamma \left( \frac{1}{4}, \frac{ r^4 \Lambda^4}{2} \right) ~~ && ~~ g_{(1)} (r) = - \Lambda \, \frac{ 2^{7/4} }{\sqrt{\pi} } \, \gamma \left( \frac{5}{4}, \frac{r^4 \Lambda ^4}{2} \right) \label{h} \\
c_{(1)} (r) = \frac{ 1 }{ \sqrt{\pi} } \, \gamma \left( \frac{1}{2}, \frac{ r^4 \Lambda^4}{2} \right) ~~ && ~~ g_{(2)} (r) = \frac{ 2 }{ \sqrt{\pi} } \, \gamma \left( \frac{3}{2} , \frac{ r^4 \Lambda^4}{2} \right) \label{Aone} \\
v_\mu ^{(1)} [\bar{z}] &=& \frac{1}{6} w_\mu ^{~\nu} [\bar{z}] \, R_\nu ^{~\alpha} \, \bar{u}_\alpha \\
u_\mu ^{(1)} [\bar{z}] = \frac{1}{2} \, \bar{a}_\mu ~~ && ~~ u_\mu ^{(2)} [\bar{z}] = \frac{1}{3} \, w_\mu ^{~\nu} [\bar{z}] \, \frac{ D \bar{a}_\nu}{d\tau}
\end{eqnarray}
\end{widetext}
where $\gamma (a,b) = \Gamma(a) - \Gamma(a,b)$ is the incomplete gamma function. The coefficients (\ref{h})-(\ref{Aone}) vary over a time-scale of $\sim \Lambda^{-1}$ after which they are effectively constant. Figure (\ref{fig1}) shows the $r$-dependence of these functions. Note also that, at the initial time, all of these functions vanish. More will be said below concerning these properties and their implication for the validity of the quasi-local expansion.
\begin{figure}
\includegraphics[height=6cm]{Coefficients2}
\caption{Time-dependence of the coefficients appearing in (\ref{h})-(\ref{Aone}). The functions $c_{(0)}$ and $g_{(1)}$ have been divided through by $\Lambda$ so that they can be displayed along with $c_{(1)}$ and $g_{(2)}$.}
\label{fig1}
\end{figure}
\subsection{Scalar ALD Equation}
Most of the terms on the right side of (\ref{mean_2}) are irrelevant in the sense that they are inversely proportional to powers of $\Lambda$ so that for time-scales much longer than $\Lambda^{-1}$ these terms can be ignored. In this limit the effect of the high energy physics is ignorable and the effective particle dynamics is described by
\begin{eqnarray}
&& m(\tau; \bar{z} ] \, \frac{D \bar{u}_\mu}{d\tau} \nonumber \\
&& ~= F_\mu ^{ext} (\tau) + \frac{e^2}{3} \, g_{(2)} (r) \, w_\mu ^{~\nu} [\bar{z}] \, \frac{D \bar{a}_\nu}{d\tau} \nonumber \\
&& {\hskip0.25in} + \frac{ e^2}{ 6} \, c_{(1)} (r) \, w_\mu ^{~\nu} [\bar{z}] \, R_{\nu \alpha} (\bar{z}) \, \bar{u}^\alpha + e \, w_\mu ^{~\nu} [\bar{z}] \, \phi_\nu ^{tail} (\bar{z}) \nonumber \\
&& {\hskip0.25in} + {\cal O} (\Lambda^{-1})
\label{eff_mean_eom}
\end{eqnarray}
where the time-dependent and trajectory-dependent effective mass is
\begin{eqnarray}
&& m(\tau; \bar{z}] \nonumber \\
&& ~= m_0 - e^2 \left( \frac{1}{2} \, g_{(1)} (r) - c_{(0)} (r) + \!\! \int _{\tau_i} ^\tau \!\! d\tau^\prime \, V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) \right) \nonumber \\
\end{eqnarray}
and an external force $F_\mu ^{ext}$ responsible for accelerating the charge has been included.
In the limit that $\Lambda \rightarrow \infty$, the terms involving $c_{(0)}(r)$ and $g_{(1)}(r)$
both diverge linearly with $\Lambda$ and the bare mass $m_0$ gets
shifted by an infinite amount $\delta m =
e^2 ( g_{(1)}/2 - c_{(0)} )$. The mass is then
renormalized to $m_{ren} = m_0 - \delta m$, which is a constant
with respect to $\tau$ in the infinite $\Lambda$ limit. If the
regulator is finite but very large then the shift $\delta m$ is
finite and large and the dressed mass has a very weak dependence
on time (see Fig.(\ref{fig1})).
If we now make $\Lambda$ infinite and use the limiting values for $g_{(2)}$ and $c_{(1)}$
\begin{eqnarray}
&& \lim _{\Lambda \rightarrow \infty} g_{(2)} (r) = 1 = \lim _{\Lambda \rightarrow \infty} c_{(1)} (r)
\end{eqnarray}
for a constant elapsed proper-time $r$ then the low-energy effective particle dynamics is
\begin{eqnarray}
m_{ren} (\tau; \bar{z}] \, \frac{D \bar{u}_\mu}{d\tau} &=& F_\mu ^{ext} (\tau) + f_\mu [\bar{z}]
\label{ren_mean_eom}
\end{eqnarray}
where the renormalized effective mass is
\begin{eqnarray}
m_{ren} (\tau; \bar{z}] = m_{ren} - e^2 \int_{\tau_i} ^\tau d\tau^\prime \, V ( \bar{z}^\alpha, \bar{z}^{\alpha^\prime})
\end{eqnarray}
and the self-force is
\begin{eqnarray}
f_\mu [\bar{z}] = w_\mu ^{~\nu} [\bar{z}] \left( \frac{e^2}{3} \, \frac{D \bar{a}_\nu}{d\tau} + \frac{ e^2}{ 6 } R_{\nu \alpha} (\bar{z}) \bar{u}^\alpha + e \, \phi_\nu ^{tail} (\bar{z}) \right) \nonumber \\
\label{self_force_1}
\end{eqnarray}
Equations (\ref{ren_mean_eom}) through (\ref{self_force_1}) are the main results of this section. The appearance of the tail term (\ref{tail_term}) implies that the radiation reaction is nonlocal because of its dependence on the past behavior of the classical worldline. This equation for the radiation reaction from scalar charges was first obtained by Quinn \cite{Quinn} based on
earlier work of Quinn and Wald \cite{QW} and Mino, Sasaki and Tanaka
\cite{MST} on gravitational radiation reaction.
Detweiler and Whiting \cite{Detweiler_Whiting} have obtained this
result by decomposing the derivative of the retarded field
$\phi_{ret}$ into a singular piece $\phi^S_\mu$, containing the
diverging contribution that renormalizes the mass, and a regular
piece $\phi^R_\mu$, which contributes to the self-force and is
regular on the worldline. From the previous section we can
construct these quantities within our regularization scheme to
find that for $\Lambda \rightarrow \infty$
\begin{eqnarray}
\phi^S_\mu &=& - e \sqrt{ \frac{8}{\pi} } \, \Lambda^2 \, \bar{u}_\mu + e \left( \frac{1}{2} \, g_{(1)} (r) - c_{(0)} (r) \right) \frac{D \bar{u}_\mu}{d\tau} \nonumber \\
\phi^R _\mu &=& -e \, \frac{1 - 6 \xi_R}{12} \, R (\bar{z}) \, \bar{u}_\mu + \frac{e}{3} \, \frac{D \bar{a}_\mu}{d\tau} + \frac{ e}{ 6 } \, R_{\mu \alpha} (\bar{z}) \, \bar{u}^\alpha \nonumber \\
&& + \phi_\mu ^{tail} (\bar{z})
\end{eqnarray}
Notice that the first term of the singular part does not
contribute to renormalizing any physical parameters (at the level
of the equations of motion) since $\phi_\mu ^S$ is projected onto
a direction orthogonal to $\bar{u}_\mu$. Likewise, the first term
of the regular part does not contribute to the self-force on the
particle.
In a flat spacetime, the tail term vanishes since there is no curvature available to focus the radiation emitted in the past onto the particle at the present. Furthermore, (\ref{ren_mean_eom}) reduces to the ALD equation for a scalar field
\begin{eqnarray}
m_{ren} \, \dot{\bar{u}}_\mu = F_\mu ^{ext} (\tau) + \frac{e^2}{3} \, w_\mu ^{~\nu} [\bar{z}] \, \ddot{\bar{u}}_\nu
\label{flat_ALD}
\end{eqnarray}
This equation was derived in the open quantum system formalism in \cite{JH1}.
\section{Linearized Fluctuations around the Classical Particle Trajectory} \label{Lin}
We now study the effects of the quantum field fluctuations (as
classical stochastic forces) on the low energy dynamics of the
particle. Instead of working directly with the equations for the
linearized fluctuations (\ref{z_flucs_funcder}), which contain
the singular functional derivatives, we begin with
(\ref{stoch_eom}) and assume that the retarded field has been
regularized by the short-distance regulator $\Lambda$ in a
quasi-local expansion.
\subsection{Stochastic ALD Equation}
For large finite values of $\Lambda$ (\ref{stoch_eom}) becomes
the ALD-Langevin equation
\begin{eqnarray}
m(\tau; z ] \, \frac{Du_\mu}{d\tau} &=& F_\mu ^{ext} (\tau) + f_\mu (z) + \eta_\mu (\tau; z]
\label{stoch_ALD}
\end{eqnarray}
where the (regulated) self-force is
\begin{widetext}
\begin{eqnarray}
f_\mu [z] &=& w_\mu ^{~\nu} [z] \left( \frac{e^2}{3} \, g_{(2)} (r) \, \frac{D a_\nu}{d\tau} + \frac{e^2}{6} \, c_{(1)} (r) \, R_{\nu \alpha } (z) \, u^\alpha + e \, \phi_\nu ^{tail} (z) \right) + {\cal O}(\Lambda^{(-1)})
\label{reg_selfforce}
\end{eqnarray}
Of course, one must remember that these expressions are only valid
up to linear order in the fluctuations $\tilde{z}$ about the mean
worldline $\bar{z}$ since higher orders correspond to quantum
corrections that we have assumed to be negligible. Expanding
(\ref{stoch_ALD}) in orders of the fluctuations, the
time-dependent mass and self-force are given by
\begin{eqnarray}
m (\tau; z ] &=& m (\tau; \bar{z} ] - e^2 \int_{\tau_i} ^\tau d\tau^\prime \, \tilde{z} ^{\nu^\prime} \frac{\delta}{\delta \bar{z}^{\nu^\prime} } \int_{\tau_i} ^{\tau} d\tau^{\prime \prime} \, V (\bar{z}^\alpha, \bar{z}^{\alpha^{\prime\prime} } ) + {\cal O}( \tilde{z}^2 ) \\
f_\mu [z] &=& f_\mu [ \bar{z} ] + \int d\tau^\prime \, \tilde{z}^{\nu ^\prime} \frac{\delta}{\delta \bar{z}^{\nu^\prime}} f_\mu [\bar{z}^\alpha] +{\cal O} (\tilde{z}^2)
\end{eqnarray}
Calculating the functional derivative in the mass equation gives
\begin{eqnarray}
m (\tau; z ] &=& m (\tau; \bar{z} ] - \frac{e}{2} \, \tilde{z}^\nu \phi_\nu ^{tail} (\bar{z}) - e^2 \int_{\tau_i} ^\tau d\tau^\prime \, \tilde{z}^{\nu^\prime} \nabla_{\bar{z}^{\nu^\prime}} V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) + {\cal O} ( \tilde{z}^2 )
\end{eqnarray}
where the factor of $1/2$ in the second term comes from evaluating a delta function on the upper boundary of the $\tau^\prime$ integral. Notice that the linear terms in $\tilde{z}$ vanish in flat space-time since $V$ is identically zero for a massless scalar field in 3+1 dimensions.
Simplifying the self-force fluctuations is slightly more
involved. The calculation amounts to performing the variational
derivative on $f_\mu$ but keeping in mind to expand out the
covariant derivatives, which depend on the classical worldline coordinates. The
result is
\begin{eqnarray}
f_\mu [z] = f_\mu [\bar{z}] - \kappa _{\mu\alpha} [\bar{z}] \, \tilde{z}^\alpha - \gamma _{\mu\alpha} [ \bar{z}] \, \dot{\tilde{z}}^\alpha - m_{\mu\alpha} [ \bar{z}] \, \ddot{\tilde{z}}^\alpha + r _{\mu\alpha} [ \bar{z}] \, \dddot{\tilde{z}} ^\alpha + {\cal O} (\Lambda^{-1})
\end{eqnarray}
The time- and trajectory-dependent coefficients are complicated
expressions that are not particularly illuminating. These are recorded in Appendix C.
Combining the linearized mass and self-force into (\ref{stoch_ALD}) and using the fact that $\bar{z}$ satisfies the mean equations of motion (\ref{CGEA_funcvar}) results in
\begin{eqnarray}
M_{\mu\alpha} [\bar{z}] \, \ddot{\tilde{z}}^\alpha + \Gamma_{\mu\alpha} [\bar{z}] \, \dot{\tilde{z}}^\alpha + K_{\mu\alpha} [\bar{z}] \, \tilde{z}^\alpha - e^2 \, \bar{a}_\mu \int_{\tau_i} ^\tau d\tau^\prime \, \tilde{z}^{\nu^\prime} \nabla_{\bar{z}^{\nu^\prime}} V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) = r_{\mu \alpha} [\bar{z}] \, \dddot{\tilde{z}} ^\alpha + \eta_\mu [\bar{z}] + {\cal O} (\Lambda^{-1} )
\label{particle_perts}
\end{eqnarray}
\end{widetext}
where
\begin{eqnarray}
K _{\mu\alpha} [\bar{z} ] &=& \kappa_{\mu\alpha} [ \bar{z} ] - m (\tau; \bar{z}] \, \partial _{\bar{z}^\alpha} \Gamma_{\mu \gamma} ^\beta \, \bar{u}_\beta \bar{u}^\gamma - \frac{e}{2} \, \bar{a}_\mu \phi _\alpha ^{tail} (\bar{z}) \nonumber \\
M_{\mu\alpha} [\bar{z}] &=& g_{\mu\alpha} \, m(\tau; \bar{z}] + m_{\mu\alpha} [ \bar{z}]
\end{eqnarray}
and
\begin{eqnarray}
\Gamma_{\mu\alpha} [ \bar{z}] &=& \gamma_{\mu\alpha} [\bar{z}] - 2 m(\tau; \bar{z}] \, \Gamma_{\mu \left( \alpha \right.}^{\, \beta} \bar{u}_{\left. \beta \right)}
\end{eqnarray}
The dynamical equation (\ref{particle_perts}) for the
fluctuations about the classical particle trajectory is the main
result of this section. This is a linear integro-differential
equation for $\tilde{z}$ with a third derivative term and
contains time-dependent coefficients that depend on the
non-Markovian behavior of the mean trajectory. Furthermore,
because of the integration over past times the last term on the
left side depends on the history of the fluctuations. Notice that
this term vanishes in a {\it flat} background so that the
fluctuations then obey a third-order differential equation, which
is Markovian in the sense that given a mean trajectory
$\bar{z}^\mu$ the fluctuations do not depend on their own past
history. So, given a solution to the mean equation of motion for
the classical worldline one could, in principle, solve for the
fluctuations induced by the quantum field fluctuations.
The notation in (\ref{particle_perts}) has been chosen
suggestively since the left side resembles a damped
simple harmonic oscillator with time-dependent mass, damping
factor, and time- and history-dependent spring constant. Notice also that the ``effective mass" $M_{\mu\alpha} [ \bar{z}]$ is not diagonal implying that the inertia of the fluctuations behaves differently in different directions. This feature is exhibited in all of the other coefficients ($\Gamma_{\mu\alpha}$, $K_{\mu\alpha}$ and $r_{\mu\alpha}$) and suggests that the fluctuations of the trajectory in one direction are linked with the fluctuations in the other space-time directions. Also, from the expression for the radiation reaction on the fluctuations $r_{\mu\alpha} [\bar{z}]$ it should be noted that this is explicitly independent of the background curvature and effectively projects $\dddot{\tilde{z}}_\nu$ onto a direction orthogonal to the mean 4-velocity $\bar{u}_\nu$.
If the stochastic term $\eta_\mu$ is ignored then
(\ref{particle_perts}) describes the evolution of small
perturbations away from the semiclassical trajectory and so is useful for studying the linear response of the trajectory to small
perturbations away from the mean worldline. In other words,
setting $\eta_\mu$ to zero (\ref{particle_perts}) gives the
linearization of the ALD-Langevin equation around the
semiclassical worldline $\bar{z}$. Generalizing this to the
self-force due to linearized metric perturbations could be useful
for investigating the stability of numerical calculations of the
inspiral of a small mass black hole into a large mass black hole,
for example \cite{Stability}.
\subsection{Memory and Secular Effects}
A particularly interesting feature related to this is the effect of the non-Markovian term appearing in (\ref{particle_perts})
\begin{eqnarray}
- e^2 \, \bar{a}_\mu \int_{\tau_i} ^\tau d\tau^\prime \, \tilde{z}^{\nu^\prime} \nabla_{\bar{z}^{\nu^\prime}} V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} )
\end{eqnarray}
If the fluctuations $\tilde{z}$ grow then the
linearization may not be valid for all times and so one should
then include quantum corrections to our stochastic semi-classical
equations if the environment is a quantum field as is assumed
here. The growth of the fluctuations could be significantly influenced
by the past behavior of $\nabla_\nu V$ and the history of the
fluctuations. The behavior of this term on the fluctuations is
not known yet because of the difficulty in solving
integrodifferential equations and in computing $V$ for many
spacetimes. However, a study in de Sitter space (where $V$ is
known \cite{Burko_Harte_Poisson, Poisson}) may provide a simple
arena for understanding the role of the non-Markovian term and
the nature of the solutions to (\ref{particle_perts}).
Aside from these technical considerations, these equations are
applicable for any type of noise on the particle trajectories. In
many cases, the source of the noise acting on the system of
interest may not be known and is put in by hand in a
phenomenological description of the particle dynamics via the
stochastic equation
\begin{eqnarray}
m (\tau; z] \, \frac{D u_\mu}{ d\tau} = F_\mu ^{ext} (\tau) + f_\mu (z) + \eta_\mu ^{+}
\label{addnoise_ALD}
\end{eqnarray}
where the superscript $(+)$ is to remind us that this term was
added in by hand, as opposed to being derived, like our treatment
of $\eta_\mu [\bar{z}]$. This stochastic force could have a
classical origin (e.g. high-temperature thermal fluctuations of surrounding gas) or
it could have no known single identifiable origin. Furthermore,
since the noise $\eta_\mu ^{+}$ is not derived from an initially
closed system it is likely to be inconsistent with the dynamics
of the trajectory. (See, e.g., \cite{HPZ}) In any case, one
would also have to specify the noise correlator $\langle \eta_\mu
^{+} (\tau) \eta_{\mu^\prime} ^{+} (\tau^\prime) \rangle_{\eta^+}$ as
it suits the model.
With this proviso (no guarantee for consistency) the analysis of
this section carries over.
Given any kind of noise the fluctuations around the mean
trajectory of the particle moving through its own (classical)
field subjected to the self-force from radiation reaction is given
by (\ref{particle_perts}), but with $\eta_\mu [\bar{z}]$ replaced
by $\eta_\mu ^+$
\begin{eqnarray}
&& M_{\mu\alpha} [\bar{z}] \, \ddot{\tilde{z}}^\alpha + \Gamma_{\mu\alpha} [\bar{z}] \, \dot{\tilde{z}}^\alpha \! + K_{\mu\alpha} [\bar{z}] \, \tilde{z}^\alpha \nonumber \\
&& {\hskip0.25in} - e^2 \, \bar{a}_\mu \!\! \int_{\tau_i} ^\tau \!\! d\tau^\prime \, \tilde{z}^{\nu^\prime} \nabla_{\bar{z}^{\nu^\prime}} V(\bar{z}^\alpha, \bar{z}^{\alpha^\prime} ) \nonumber \\
&& {\hskip 0.5in} = r_{\mu \alpha} [\bar{z}] \, \dddot{\tilde{z}} ^\alpha + \eta_\mu ^+ + {\cal O} (\Lambda^{-1} )
\end{eqnarray}
Since in this discussion we don't need to worry about quantum
corrections from higher order loops in the effective action we
could go beyond the linear order in the fluctuations of the
particle trajectory and expand the solutions to
(\ref{addnoise_ALD}) in powers of the coupling constant $e$ (
denoted in the subscript) $z = z_0 + z_1 + z_2 + \ldots$.
Assuming that $\eta_\mu ^+$ is ${\cal O}(e)$ and depends on $z$
and recalling that $f_\mu$ is quadratic in the coupling we find a
non-local and causal contribution to the total force on the
particle coming from the fluctuations of the stochastic force
(see Appendix D for details)
\begin{eqnarray}
F^{drift} _\mu = \int d\tau^\prime \, d\tau^{\prime\prime} \, F_\mu ^{\rho\sigma} (\tau, \tau^\prime, \tau^{\prime\prime} ) \, \big\langle \eta_\rho ^+ (z_0 ^{\alpha^\prime}) \, \eta_\sigma ^+ (z_0 ^{\alpha^{\prime\prime}} ) \big\rangle _{\eta^+} \nonumber \\
\label{drift}
\end{eqnarray}
which is of the same order as the self-force. It seems that the
stochastic noise would cause the particle to drift off from the
background trajectory $z_{(0)}$ determined by the external force. The
deviation could build up over time as indicated by the integral
above. This may have interesting consequences for astrophysical
sources with some stochastic behavior described by a classical
noise. If such physical situations exist, this noise-induced
drift may give rise to a secular effect which could alter the
waveform templates (of events expected to be seen by
gravitational interferometers like LIGO and LISA) calculated
without including such stochastic secular effects.
A similar expression to (\ref{drift}) can be derived in our open
quantum system framework by writing the effective action to cubic
order in the fluctuations. Doing this reveals a term like
(\ref{drift}) but this requires a much more careful analysis that
goes beyond the stochastic semi-classical approximation adopted
here.
\section{Discussions}
\subsection{The Quantum Regime and the Validity of the Quasi-Local Expansion and Order-Reduction}
The previous sections focused on the Feynman-Vernon formalism and various approximations to obtain the low-energy effective dynamics of the particle, both for its semiclassical (mean) and stochastic semi-classical (mean and stochastic fluctuations) motion. Here, the domain of validity of the quasi-local expansion and this semi-classical treatment will be explored and compared with the relevant scales for weak and strong radiation damping.
In the effective field theory paradigm a regulator $\Lambda$ is
introduced for controlling the ultraviolet divergences appearing
in the direct part of the retarded Green's function such that
$\Lambda^2 \sigma \gg 1$ with $\sigma$ small and approaching
zero. After expanding $\sigma$ near the coincidence limit the
time-scale of the quasi-local expansion $\Delta \tau=s$ is
governed by
\begin{eqnarray}
\Delta \tau \gg \Lambda^{-1}
\end{eqnarray}
Recall that for times larger than $\sim \Lambda^{-1}$ the time-dependent coefficients in (\ref{eff_mean_eom}) and (\ref{stoch_ALD}) rapidly approach constant values (see Fig.\ref{fig1}). But there are other scales to consider. We have been working at tree-level in both the particle and the field so that $\Delta \tau$ must be much larger than the scale for 1-loop field and 1-loop particle corrections. This implies that the time interval $\Delta \tau$ should be much longer than the time-scale for creating pairs
\begin{eqnarray}
\Delta \tau \gg \frac{\hbar}{m} = \lambda_C
\end{eqnarray}
where $\lambda_C$ is the Compton wavelength associated with the scalar particle. The dynamics (\ref{eff_mean_eom}) and (\ref{stoch_ALD}) are therefore valid using this framework provided that
\begin{eqnarray}
\Delta \tau \gg \lambda_C \gg \Lambda^{-1}
\end{eqnarray}
Nevertheless, the presence of the third $\tau$-derivative in (\ref{eff_mean_eom}),
(\ref{stoch_ALD}) and (\ref{particle_perts}) requires the
specification of the initial position, velocity, and acceleration
to obtain unique solutions. This is problematic, since for a
vanishing external force $F_\mu ^{ext}$ one still requires an
initial acceleration to solve the equations. But, if there is no
force accelerating the charge then what causes the particle to
accelerate? Furthermore, the particle may experience unbounded acceleration so that its kinetic energy increases with time to infinity. Since scalar dynamics is an energy-conserving theory where then could this energy arise? These problems are related to the infinite self-energy of a (classical) charged point particle.
In the {\it classical} theory of scalar fields and particles
these issues can be resolved if one instead gives the particle a
finite size $r_0$ for which the self-energy is roughly $e^2/r_0$
\cite{LL}. If this energy composes its rest mass then
\begin{eqnarray}
r_0 \sim \frac{ e^2}{ m}
\end{eqnarray}
and represents the ``size" of the particle. In
\cite{LL}, an approximation, amounting to an asymptotic expansion in
powers of $r_0$, called the Landau approximation (also known as
order-reduction), is employed to obtain solutions that require
only an initial position and velocity and are also free from
run-away solutions. The Landau
approximation converts the ALD equation (of third order) to the
so-called Landau-Lifshitz equation (of second order). We will use
these names to distinguish between these equations.
In order-reduction, the lowest order solution is found by simply ignoring the self-force so that the radiation damping is assumed weak. The time-scale of the dynamics is then determined mostly by the external force so that if $F_\mu ^{ext}$ varies on a scale $\lambda_{ext}$ then $\Delta \tau \sim \lambda_{ext}$. In curved spacetimes the self-force will be weak if $r_0 \ll \Delta \tau$ and the length scale associated with the spacetime curvature $\lambda_R$ is large (i.e. small curvature) so that $\lambda_R \gg r_0$. It then follows that for weak damping in the semi-classical domain that the quasi-local expansion and the Landau approximation are valid provided that
\begin{eqnarray}
\Delta \tau \sim \lambda_{ext} \gg r_0, \lambda_C \gg \Lambda^{-1} {\rm ~~ and ~~ } \lambda_R \gg r_0
\end{eqnarray}
Recently, in the context of plasma physics, \cite{Koga}
has investigated the validity of the Landau approximation for the
classical ALD equation by numerically integrating the
Landau-Lifshitz equation forward in time and, using the final
position, velocity, and acceleration from that, integrating the
ALD equation backward in time. If the initial position and
velocity of the particle differ significantly from the
backward-evolved solution of the ALD equation at the initial time
then one can assume the Landau approximation has broken down.
Koga does this for a counter-propagating electron and
ultraintense laser beam (intensity $\sim 10^{22}$ W/cm$^2$). He
finds that the Landau approximation is valid so long as the laser
wavelength $\lambda_0$ is greater than the Compton wavelength.
For $\lambda_0$ much smaller than $\lambda_C$, he finds
disagreement between the solutions of the Landau-Lifshitz and ALD
equations. However, these equations cannot be fully trusted since
quantum effects should become important. In this domain, while we cannot directly apply our results to this problem we can use the closely related closed-time-path (CTP) formalism to incorporate the effects of the quantum loop corrections to the (quantum) particle dynamics.
\subsection{Decoherence}
This brings us to an important issue, namely, how strongly must
the particle worldline decohere in order for a stochastic
semi-classical treatment to be applicable? A measure of the
decoherence of a system is the {\it decoherence functional}
\cite{conhis} $D[ \alpha(z), \alpha (z^\prime) ]$ where
$\alpha(z)$ specifies a particular coarse-grained history of the
worldline. If we take the history to be
$\alpha(z) = z$ then the decoherence functional is
\begin{eqnarray}
D[ \alpha(z), \alpha(z^\prime) ] &=& D[z,z^\prime] \\
&=& \rho_S (z_i, z_i ^\prime ) \, e^{ \frac{ i}{\hbar} S_{CGEA} [z,z^\prime] }
\end{eqnarray}
and is closely related to the CTP effective action
\cite{Calzetta_Hu}. In fact, the norm of $D$ is proportional to
the norm of the influence functional. The system is said to
decohere if $|D|$ approaches zero. We showed earlier
that the CGEA is proportional to $z^-$ and so there is minimal
decoherence for two nearby histories. This implies that
coarse-graining the quantum field must be complemented by an
additional coarse-graining of the worldline in order for the
particle to be sufficiently decohered. If the mechanism for
decoherence is efficient then this additional coarse-graining
should be minimal and the stochastic semi-classical description
that we have used in this paper is applicable. For our problem,
recall that the CGEA contains an imaginiary part that is
proportional to $j^- \cdot G_H \cdot j^-$ so that the norm of $D$
is
\begin{eqnarray}
| D[z,z^\prime] | &=& e^{ - \frac{1}{4\hbar} j^- \cdot (16 \pi^2 G_H ) \cdot j^-}
\end{eqnarray}
showing that the quantum field fluctuations in the environment is
the mechanism for decoherence. It is therefore necessary to
understand the behavior of the Hadamard function to show that
there exists a well-defined stochastic semi-classical regime for
the particle. This requires knowing the field modes on the background
space-time, which is a significant problem in its own right. For
simple scenarios the decoherence functional can be calculated and
the decoherence time approximated. But, in general, one needs to
determine if decoherence is fast enough on a case by case basis.
\subsection{Problems with putting noise in by hand}
Towards the end of Sec. 5 we replaced the classical stochastic manifestation of the quantum fluctuations by a noise source $\eta_\mu ^+$ that has some specified noise correlator.
Aside from the issue about keeping only to tree-level in the particle and field variables, how is using $\eta_\mu ^+$ different from all of the sophisticated machinery used in the previous sections?
Unlike in our approach, inserting a source of noise by hand implies that it can have any
correlator one wants. The stochastic two-point function of $\eta_\mu ^{+}$
\begin{eqnarray}
\big\langle \{ \eta_\mu ^{+} (\tau), \eta_\nu ^{+} (\tau^\prime) \} \big\rangle _{\eta^{+}} = N_{\mu \nu} (\tau, \tau^\prime) ~,
\label{fudged_noise}
\end{eqnarray}
otherwise known as the noise kernel, can be chosen at will. Some physical reasoning, for instance, a high temperature environment, might suggest that $N_{\mu\nu} (\tau, \tau^\prime) \sim \delta (\tau - \tau^\prime)$, i.e. white noise. While white noise certainly simplifies the calculations, there is little justification for simply stating the form of the noise kernel and expecting the dynamics of the system to be consistent with that source. Using the Feynman-Vernon formalism, coarse-graining over the quantum environment naturally gives rise to the noise kernel appropriate for the stochastic force fluctuations
\begin{eqnarray}
&& \big\langle \{ \eta_\mu [\bar{z}^\alpha] , \eta_\nu [\bar{z}^{\alpha^\prime} ] \} \big\rangle _{\xi} \nonumber \\
&& {\hskip0.25in} = \hbar \, e^2 \, \vec{w}_{\left( \mu \right.} [\bar{z}^\alpha] \vec{w}_{\left. \nu \right)} [\bar{z}^{\alpha^\prime}] \, G_H (\bar{z}^\alpha, \bar{z}^{\alpha^\prime} )
\label{noise_corr}
\end{eqnarray}
This coarse-graining ensures that the dynamics of the system
evolves consistently with that of the environment. This is an
important statement that cannot be overstated because it is this
feature which gives rise to consistent fluctuation-dissipation
relations (FDR). Inserting noise by hand most likely violates
such a relation.
While FDR's won't be discussed here in great detail, \cite{RHA}
have studied the FDR's for $n$ particles interacting with a
quantum field for certain trajectories (e.g. uniformly
accelerated). They have shown the existence and self-consistency
of these relations and further introduce correlation-propagation
relations relating the correlations of particles to each other
through the (causal) propagation of the quantum field. These
correlation-propagation relations can only be formed if the
particles interact directly through causal influences. The open
quantum system viewpoint, therefore, shows the interrelated
influences of the system and environment through these types of
self-consistent relations. This is an important feature lacking
in a model that has noise put in by hand.
Another problem with adding noise in by hand is that the noise can be a significant contribution to the quantum two-point function of the system if one is interested in the quantum particle behavior. So, having the correct form for the noise kernel is important for getting the correct expression for the quantum two-point functions.
It has been shown in \cite{Calzetta_Roura_Verdaguer}, using the CTP generating functional for quantum correlation functions of the system variables, for quantum Brownian motion as well as for stochastic semi-classical gravity (see next section), that certain quantum two-point functions are related to stochastic two-point functions. As an example, the symmetrized quantum two-point function of a Brownian particle's position can be written as
\begin{eqnarray}
\frac{1}{2} \, \big\langle \{ \hat{x} (t), \hat{x} (t^\prime) \} \big\rangle = \big\langle \, \big\langle X(t) X(t^\prime) \big\rangle _\xi \, \big\rangle_{X_i, p_i}
\end{eqnarray}
where $X$ is a solution of the QBM Langevin equation $L \cdot X =
\xi$ for the appropriate linear operator $L(t,t^\prime)$, and
$\langle \ldots \rangle _{X_i, p_i}$ denotes the average over all
possible initial positions and momenta with respect to the
reduced Wigner function for the initial state of the particle.
The solutions of the QBM Langevin equation consist of a
homogeneous part, containing all the information about the
initial conditions of the system, and a term describing effects
due to the interactions between the particle and oscillators.
\begin{widetext}
\begin{eqnarray}
X(t) = X_0 (t) + \int dt^\prime \, G_{ret} (t, t^\prime) \xi (t^\prime)
\end{eqnarray}
The symmetrized quantum two-point function becomes
\begin{eqnarray}
\frac{1}{2} \, \big\langle \{ \hat{x} (t), \hat{x} (t^\prime) \} \big\rangle = \big\langle X_0 (t) X_0 (t^\prime) \big\rangle _{X_i, p_i} + \int dt_1 \int dt_2 \, G_{ret} (t, t_1) N(t_1, t_2) G_{ret} (t_2, t^\prime)
\end{eqnarray}
\end{widetext}
where $N = \langle \xi (t_1) \xi(t_2) \rangle _\xi$ is the noise kernel, given by the stochastic correlator of $\xi$. The first term involving the homogeneous solution of the Langevin equation represents the dispersion in the initial conditions and is called the {\it intrinsic fluctuations}. The second term involving the noise kernel, and hence the quantum fluctuations of the environment, represents the correlations with the environment through the stochastic fluctuations and is called the {\it induced fluctuations}. It turns out that if the homogeneous solution $X_0$ decays exponentially fast then for late times, at least times larger than the decay time, the quantum two-point function is determined entirely by the noise kernel, that is, the induced fluctuations. And so, in this sense, all of the information about the quantum correlations of the system degrees of freedom is encoded in the stochastic correlations. This shows another drawback to putting noise in by hand. The induced fluctuations contain information about the quantum fluctuations of the system variables. But this will not be true with some arbitrary noise kernel chosen at will.
\subsection{Similarities with Stochastic Semiclassical Gravity}
The features of the particle dynamics seen in the above discussions are typical of nonequilibrium open quantum systems. History-dependent behavior is present in the equations of motion for the system and if a renormalization procedure is required it is usually a time-dependent prescription, as seen earlier with the renormalized mass $m_{ren}(\tau)$. Furthermore, the noise correlator (\ref{noise_corr}) is generically non-local in time and is determined by the quantum fluctuations of the environment variables. This formalism does not allow for arbitrary noise kernels since this would destroy the self-consistency between the system and environment evolution. A particular example that contains these features is stochastic semiclassical gravity, which we will briefly describe and compare with below.
Stochastic semiclassical gravity (SSG) is a self-consistent
theory of the stochastic dynamics of a classical spacetime
containing quantum matter fields. SSG goes beyond semiclassical
gravity, for which the geometry is driven by the expectation of
the (renormalized) stress tensor, in that the quantum field
fluctuations also contribute to the spacetime dynamics through a
classical stochastic source. The spacetime is therefore driven by
both the quantum expectation value of the renormalized stress
tensor and a classical stochastic stress-tensor-like object,
$\xi_{ab}$. For an introduction and review of this subject see
\cite{StoGraRev} and \cite{Hu_Roura_Verdaguer} for a discussion
of the domain of validity of SSG.
As an open quantum system, the quantum field fluctuations are coarse-grained using the CTP formalism of Schwinger and Keldysh (SK) to study the self-consistent evolution of the (classical) geometry. The quantum fluctuations manifest themselves as stochastic noise thereby imparting a stochastic nature to the spacetime. The resulting Einstein-Langevin equation for the linearized metric perturubations $h_{ab}$ is
\begin{eqnarray}
G_{ab} ^{(1)} [g+h] = \kappa \, \big\langle \, \hat{T}_{ab} ^{(1)} [g+h] \, \big\rangle _{ren} + \kappa \, \xi_{ab} [g]
\label{EinLang}
\end{eqnarray}
The superscript $^{(1)}$ denotes that those quantities contain
all terms to first order in the metric fluctuations $h_{ab}$. It
should be noted that the finite parts of the counterterms needed
to cancel the divergences coming from the stress tensor
expectation value have been absorbed into the definition of
$\langle \hat{T}_{ab} ^{(1)} \rangle_{ren}$. The renormalized
stress tensor expectation value (evaluated in a Gaussian state)
contains an integration over the past history of the metric
fluctuations and so the dynamics is generally non-Markovian. This
is like what is seen in the ALD-Langevin equation
(\ref{stoch_ALD}) where the tail term $\phi_\nu ^{tail} (z)$ is
analogous to the expectation value of the renormalized stress
tensor in (\ref{EinLang}). The (covariantly conserved) stochastic
source tensor $\xi_{ab}$ has zero mean and its correlator is
given in terms of the Hadamard function of the stress tensor
fluctuations $\hat{t}_{ab} = \hat{T}_{ab} - \langle \hat{T}_{ab}
\rangle$
\begin{eqnarray}
\big\langle \{ \xi_{ab} (x; g], \xi_{cd} (x^\prime; g] \} \big\rangle _\xi = \hbar \, \big\langle \{ \hat{t}_{ab} (x; g], \hat{t}_{cd} (x^\prime; g] \} \big\rangle \nonumber \\
\end{eqnarray}
The correlator of the stress tensor fluctuations on the right side does not vanish on a spatial hypersurface. This reflects the fact that the quantum field correlations are themselves non-local. Compare this with the correlator in (\ref{xi_correlator}) which is also non-local.
SSG also suffers from runaway solutions since the finite contributions to the counterterms needed to cancel the divergences appearing from the expectation value of the stress tensor are quadratic in the curvature. This makes SSG a theory with derivatives higher than two, similar to the ALD type equations derived above, which were of third order in the $\tau$ derivatives. A version of the Landau approximation can be used to reduce the order of the Einstein-Langevin equation to two thereby yielding well-behaved solutions free of the pathologies typical of higher-order derivative theories. Of course, one needs to be careful to use the Landau approximation at scales that are consistent with the derivation of the Einstein-Langevin equation.
Finally, the symmetrized quantum two-point functions of the
metric fluctuations $h_{ab}$ can be written in terms of intrinsic
fluctuations, representing the dispersion in the initial
conditions, and induced fluctuations, encoding the information
about the fluctuations of the quantum matter \cite{Hu_Roura_Verdaguer}. Just like with the particle motion, one cannot simply use any noise kernel for modeling stochastic metric fluctuations.
One needs to do a careful analysis that ensures the self-consistency of the metric and quantum matter dynamics and the existence of fluctuation-dissipation relations.
\section{Summary}
In this paper we have derived the scalar ALD equation for the
quantum expectation value of the worldline for a scalar charged
point particle interacting with its own quantum field as it moves
in a curved spacetime. Our equation (\ref{ren_mean_eom}) for the
low-energy effective particle dynamics agrees with the results
obtained earlier by \cite{Quinn, Detweiler_Whiting, Poisson,
Burko_Harte_Poisson}. If the quantum fluctuations in the field
strongly decohere the worldline then we can ignore the particle's
quantum fluctuations and obtain the semi-classical motion
(\ref{ren_mean_eom}). Invoking an effective field theory point of
view, the singular behavior of the field's retarded Green's
function can be regulated. For sufficiently short times $\Delta
\tau=s$ but still long enough $\Delta \tau \gg \Lambda^{-1}$
compared with the inverse cutoff frequency $\Lambda$ a
quasi-local expansion can be used to obtain the contributions
relevant to the self-force and those that are irrelevant in the
infinite $\Lambda$ limit. This renormalizes the mass of the
particle and shows explicitly the appearance of the expected
$Da_\mu/d\tau$ term that is characteristic of radiation reaction.
While the time-dependent coefficients (\ref{h})-(\ref{Aone})
appearing in (\ref{eff_mean_eom}) seem to suggest that only the
initial position and velocity of the particle are needed, and
hence that a resolution of the problems of pre-acceleration and
run-away solutions has been reached, one should keep in mind that
the quasi-local approximation breaks down for short time
intervals, which includes the instant at the initial time. A more
careful analysis would need to include a more physical initial
state than the factorized one used here.
Fluctuations in the quantum field is expected to affect the
particle's motion causing it to fluctuate by an amount $\tilde{z}$
around the mean trajectory $\bar{z}$ given by solutions to the
semi-classical equation (\ref{eff_mean_eom}). We derived such a
stochastic force $\eta_\mu[\bar{z}]$ with correlators from the
fluctuations of the quantum field and a scalar ALD-Langevin
equation (\ref{stoch_ALD}). The dynamics of $\tilde{z}$ (\ref{particle_perts})
contains a non-Markovian contribution through the past history of
the particle fluctuations. Depending on the behavior of this term
the fluctuations might grow to be large indicating a breakdown of
the linear approximation. In that case it requires the inclusion
of quantum corrections in order to follow the nonlinear evolution
of $\tilde{z}$. On the other hand, ignoring the noise altogether
in (\ref{particle_perts}), one can test the
stability of numerical simulations in inspiral studies, for
example \cite{Stability}.
Instead of the noise $\eta_\mu$ derived here from fluctuations of
quantum fields one can replace it with some other classical noise
$\eta_\mu ^+$ suitably chosen to model some stochastic source in a
phenomenological description. We can still use
(\ref{particle_perts}) to study the effect of such noises on the
particle trajectory fluctuations. However, since the origin for
the noise is no longer due to a quantum field we need not worry
about keeping up to linear order in $\tilde{z}$ in the
ALD-Langevin equation. Instead, expanding the solution for small
coupling constant $e$ and taking the stochastic expectation value
shows that there is, in general, a non-vanishing force (\ref{drift}) coming completely from the
correlations of the stochastic force.
Along with the self-force, this noise-induced term would cause
the particle to drift off of its background trajectory determined
by the external force (or off of its geodesic motion if $F_\mu
^{ext}$ vanishes). We hope to explore the consequences of this
noise-induced drift in an astrophysical setting and find
observable effects on the waveforms of the radiation emitted by
the particle and detected by gravitational interferometers like
LIGO and LISA.
In Paper II we will apply the same techniques here to study the
self-forces and the stochastic semi-classical motions of electric
charges coupled to an electromagnetic field and of small black
holes coupled to the background spacetime of a massive black hole,
respectively. We hope to apply these results to more physical
situations such as the motion of charges in strong external fields
and gravitational radiation reaction.
\begin{acknowledgments}
CRG thanks Phil Johnson for help
in learning the worldline influence functional approach approach and Albert Roura for suggestions in simplifying some derivations in Appendix A. We both
thank Phil Johnson and Albert Roura for interesting discussions
on general issues of radiation reaction. This work is supported
in part by NSF grant PHY03-00710.
\end{acknowledgments}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,652
|
In this handsome and engaging book, Clive Coates, one of the world's leading authorities on wine, gives us the most up-to-date, comprehensive, and detailed study of the wines of France ever written. Coates's vast knowledge of his subject together with his natural gift as a storyteller make An Encyclopedia of the Wines and Domaines of France as informative as it is entertaining. He discusses every appellation and explains its character, distinguishes the best growers, and uses a star system to identify the finest estates. With more than forty specially commissioned maps that show the main appellations and wine villages of France in detail and a format that invites browsing as well as in-depth study, this book will be essential reading for anyone, professional or amateur, interested in wine. Coates gives ample reasons for his belief that France produces the finest wines in the world, in a volume and variety no other country can match. He shows how, despite savage competition from other countries, France holds its own. It not only creates great wines, he says, it also produces affordable wines. The outcome of thirty-five years of traveling around the French vineyards, this book displays a continuing love and respect for French wines and the vignerons of this remarkable country. In discussing each region and its wines in detail, Coates leaves no stone unturned. His encyclopedic knowledge is evident, bringing the places and the people where these great wines are created to life.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,448
|
Q: Where is the error in this complex integral? I got myself confused over the following problem: Compute $$\int_\gamma\frac{1}{\sqrt{z}}dz$$ where $\gamma$ is the lower unit circle arc from $-1$ to $1$.
Isn't it correct that I can either choose $$\gamma:[-\pi,0]\to\mathbb C,t\mapsto e^{it}$$ or $$\gamma:[\pi,2\pi]\to\mathbb C, t\mapsto e^{it}$$
It probably must be wrong because in the first case we have $$\int_{-\pi}^0ie^{it/2}dt=2(1+i)$$ where in the second case we have $$\int_\pi^{2\pi}ie^{it/2}dt=-2(1+i)$$ but for whatever reason I can't figure out what's wrong.
A: $\sqrt z$ is two valued, one being the negative of the other. So the answer depends on how you define $\sqrt z$. With appropriate choice of this function both answers can be considered correct!.
A: Let $u = t -2\pi$ and $du = dt$
At $t = \pi \ , u = -\pi$
and at $t = 2\pi \ , u = 0$
So, $\int^{2\pi}_{\pi} ie^{it/2}dt = -(2+i)$
$\implies \int^{0}_{-\pi} ie^{i(u+2\pi)/2}du = -(2+i)$
$\implies e^{i\pi}\int^{0}_{-\pi} ie^{iu}du = -(2+i)$
As, $e^{i\pi} = -1$
$-\int^{0}_{-\pi} ie^{iu}du = -(2+i)$
or
$$\int^{0}_{-\pi} ie^{iu}du = (2+i)$$
which is identical as the first one ($t$ replaced by $u$ )
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,924
|
Charles J. Volpe (November 29, 1936 – September 22, 1988) was a Republican member of the Pennsylvania House of Representatives.
References
Republican Party members of the Pennsylvania House of Representatives
1988 deaths
1936 births
20th-century American politicians
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,902
|
CSSA Secondary State Touch Football GD 2023
By entering the CSSA Secondary State Touch Football GD you are accepting the terms and conditions outlined in the entry information.
Students will be considered for selection into the CSSA 15&U and Open Girls & Boys Teams which will compete at the CIS Championships.
Entry fee is $200.00 + GST per team
By accepting the terms and conditions you are confirming your acceptance of the entry fee for individual players of $200.00 + GST per team. Schools will be invoiced the entry fee at the end of the Term.
I understand and will abide by all the guidelines set down for the CSSA Secondary Touch Football GD and will ensure that all players/coaches/managers/referee are aware of the CSSA Code of Conduct and Discipline & Disputes Policy which can be found in the CSSA Handbook.
As a CSSA Member school we are aware that all teachers/parents/volunteers that we send to a carnival or sporting event have completed a "Working with Children" check.
Event Date Fri 31st March 2023
Cut off date Fri 17th March 2023
Individual Price $220.00
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,918
|
Iuli Txersànovitx Kim , nascut el 23 de desembre de 1936 a Moscou) és un poeta rus i soviètic, compositor, cantautor, guionista i membre del moviment dissident a la Unió Soviètica. Les seves cançons, que abasten tot, des de l'humor suau fins a la sàtira política mordaç, apareixen en almenys cinquanta pel·lícules soviètiques, incloent Bumbaraix, 12 cadires i Un miracle ordinari. Des del 1998 viu a Israel i ha realitzat gires periòdiques per Rússia, Europa i els Estats Units.
Biografia
Kim va néixer el 1936 a Moscou. El seu pare era Kim Txersan (1904-1938), un traductor de coreà, i Nina Valentínovna Vsesviatskaia (1907-1974), professora de llengua i literatura russa, originària d'una família russa de metges. El seu avi, Vassili Vsesviatski, era un protoiereu de l'Església Ortodoxa Russa al raion de Jukovski, de la gubèrnia de Kaluga, que va batejar Gueorgui Júkov.
Els pares de Kim van ser víctimes de la Gran Purga de 1937 i 1938, en què el seu pare va ser executat i la seva mare va ser sentenciada com a "familiar d'un traïdor a la Pàtria" a cinc anys en un camp de treball i tres anys d'exili, de manera que Kim no la va veure fins als 9 anys.
Va ser rehabilitat durant el període del Desglaç de Khrusxov el 1958, però abans, estigué sotmesa a la llei dels "quilòmetre 101" i no va poder viure a Moscou, per la qual cosa la família de Kim es va establir a Maloiaroslàvets, a l'óblast de Kaluga. El 1951, la família es va traslladar a Turkmenistan. Kim va tornar a Moscou el 1954 per entrar a l'Institut Pedagògic de l'Estat de Moscou.
El 1969 va signar una crida a la Comissió de Drets Humans de l'ONU.
Després de tornar a Moscou, Kim va treballar com a professor d'escola i al mateix temps va participar en el moviment dissident soviètic, fet que li va costar la feina el 1968. Posteriorment, Kim es va guanyar la vida escrivint cançons per a obres de teatre i pel·lícules, així com publicant obres sota el pseudònim Iu. Mikhailov, que va usar fins al 1986. Al mateix temps, mentre no va tenir permís per fer concerts, va continuar cantant de manera clandestina.
Amb l'arribada de la glàsnost, Kim finalment va poder actuar legalment. Des de la ruptura de la Unió Soviètica, ha estat aclamat al llarg del món de parla russa i ha actuat en nombrosos llocs a Rússia, Europa i els Estats Units. Ha rebut nombrosos premis, com ara el Premi Bulat Okudjava de la Federació de Rússia.
En l'actualitat, la discografia de Iuli Kim inclou més de 20 títols en CD, cinta d'àudio i vídeo i DVD. Les seves cançons han estat incloses en gairebé totes les antologies de la cançó de l'autor, així com a moltes antologies de la poesia russa moderna.
La seva primera esposa fou Irina Iakir - neta del comandant de l'Exèrcit Roig Iona Iakir. Es van casar el 1966, i el 1998 van emigrar a Israel. Després de la mort d'Irina el 1999, Kim es va casar amb Lídia Lugòvaia, amiga íntima d'Irina des de l'escola. En l'actualitat divideix el seu temps entre Jerusalem i Moscou.
Filmografia seleccionada
Notes
Enllaços externs
Biografia
Iuli Kim al web bards.ru
Dissidents soviètics
Artistes moscovites
Escriptors moscovites
Alumnes de la Universitat Estatal Pedagògica de Moscou
Escriptors israelians
Escriptors soviètics
Polítics moscovites
Polítics israelians
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,161
|
Texas Children's now offers the same great care you have come to know and trust – evenings and weekends. Staffed by board-certified pediatricians and nurses, our facilities and equipment are designed to meet the needs of children and adolescents up to age 18.
We accept numerous insurance plans and offer a full menu of affordable, self-pay options.
The Save My Spot functionality does not support Internet Explorer. We recommend you use a supported browser such as Chrome, Firefox, Safari or Edge.
Please note the wait times listed below can change rapidly. Online wait times are updated every five minutes and are estimates based on the average time it takes for a patient to be seen by a health care professional. Please be aware that patients with a more serious condition or illness may be seen before you.
Please arrive 10 minutes before your designated 'Save My Spot' time in order to process your registration information. The time for which you save your spot is an estimation of when you can expect to be brought back into an exam room. During busy times, there may still be a short wait until you are seen by a medical provider.
If you decide you no longer need to save your spot, please call the location to cancel so that other patients may take advantage of the spot.
The 'Save my Spot' option is not for emergency medical needs, but walk-ins are always welcome. If you or your child needs emergency medical attention, please dial 911.
Please note that Texas Children's Urgent Care specializes in after hours care, but does not replace the need for your child to have a general pediatrician. Routine physical exams and vaccinations are services that need to be obtained from your general pediatrician, and these services are not available at Texas Children's Urgent Care.
If you do not have a general pediatrician, please contact us and our staff will be happy to give you a list of local pediatricians' offices.
At Texas Children's Urgent Care, we understand that health care billing can be complicated. Our goal is to make sure that each bill is correct and that all of your questions are answered. Our billing department may be reached Monday through Friday 8:30 am to 4 pm at 832-824-2999 to help you with your account.
If you are unable to access your billing details in MyChart, please contact the customer service at 832-824-2999 / Toll Free 888-263-6360.
Texas Children's Urgent Care is contracted with the insurance companies listed below. If your child is covered by one of their plans, all charges will be considered "in network." If your child's insurance company is not listed, charges may be covered under your "out of network" benefits.
If your child is covered by insurance, please bring your insurance card with you to simplify registration. Our staff will copy the card and submit charges directly to your insurance company. Your urgent care copay will be collected at the time of service. If your plan does not have an urgent care copay, we will collect the specialist copay listed on your card at the time of service.
About 30 days after your visit, you should receive a notice from your insurance company summarizing the charges and payment for service. If your insurance company has covered the full amount, your insurance company will send payment directly to us. If the insurance company has applied some of the charges to your deductible, we will send you a bill for the amount still owed.
Please note that each insurance plan is a contract between you and your insurance company, and each plan is different. If you have specific questions about your plan benefits and what charges will be applied to your deductible, please contact your insurance company directly.
If your child is not covered by insurance, Texas Children's Urgent Care offers discounted self-pay rates at the time of service. Please note that self-pay rates include all charges related to the diagnosis and treatment of your child as well as any labs or procedures performed in the clinic. It does not include any charges for labs that need to be sent out or are performed outside of the clinic. Our staff will notify you before any outside lab work is sent.
There are 3 self-pay rates based on the complexity of the visit.
Texas Children's Urgent Care is seeking talented, patient oriented Pediatricians to work in a fast-paced environment. Physicians are employed on a full-time, part-time and per-diem basis.
With the support of either a physician or NP, double coverage is often provided and an RN is always available.
Monday - Friday: 4 p.m. - 11 p.m.
Saturday and Sunday: 11:30 a.m. - 8 p.m.
Monday - Sunday: 10:30 a.m. – 11 p.m.
Physicians enjoy no on-call responsibilities!
If you are committed to a community of healthy children and would like to learn about current physician opportunities, please send your resume to physicianrecruitment@texaschildrens.org. For more information, please call 832-824-2067.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,529
|
A Winged Victory For The Sullen LP [Love Record Stores edition)
ERASED TAPES
LP We Played Some Open Chords & Rejoiced, For The Earth Had Circled The Sun Yet Another Year Requiem For The Static King (part 1) Requiem For The Static King (part 2) Minuet For A Cheap Piano Number Two Steep Hills Of Vicodin Tears A Symphony Pathetique All Farewells Are Sudden
PRE-ORDER ITEM : Expected August 28th 2021. This item will only be shipped to you on or after the official release date. Please note any orders containing pre-order items won't be shipped until all items are available, so please order this separately to avoid delays. Please remember that release dates are at the mercy of labels, distributors, and pressing plants and will change constantly.
A Winged Victory For The Sullen is the first installment of thecollaboration between Stars Of The Lid founder Adam Wiltzie and L.A. composer Dustin O'Halloran. The duo agreed to leave the comfort zone of their home studios and develop the recordings with the help of large acoustic spaces, hunting down a selection of 9ft grand pianos that had the ability to deliver extreme sonic low end. Other traditional instrumentation was used including string quartet, French horn, and bassoon, but always juxtaposed is the sound of drifting guitar washed melodies. The recordings began with one late night session in the famed Grunewald Church in Berlin on a 1950s imperial Bösendorfer piano and strings were added in the historic East Berlin DDR radio studios along the River Spree. One last session on a handmade Fazioli piano in a private studio on the Northern cusp of Italy, before the final mixes took place in a 17th century villa near Ferrara with the assistance of Francesco Donadello. All songs were then processed completely analogue straight to magnetic tape. Their secret to harvesting new melodic structures from the thin air of existence was for the duo to push themselves to dangerous territory, realising that clear thinking at the wrong moment could stifle the compositions. The final result is seven landscapes of harmonic ingemination. In 'Requiem For The Static King Part One' – created in memory of the untimely passing of Mark Linkous – they have taken the age-old idea of a string quartet and then shot it out of a cannon to reveal exquisite new levels of sonic bliss. Of the 13 minute track Symphony Pathétique, Wiltzie says 'after almost 20 years of struggling to create interesting ambient drone music, I feel like I have finally figured out what I am doing'. Notable guest musicians include Icelandic cellist Hildur Gudnadottir, as well as Erased Tapes label comrade Peter Broderick on violin. A Winged Victory For The Sullen is not a side project – it is the future of the late night record you have always dreamed of.
Experimental / Avant-Garde / Leftfield / Drone
Play All Add All We Played Some Open Chords & Rejoiced, For The Earth Had Circled The Sun Yet Another Year Requiem For The Static King (part 1) Requiem For The Static King (part 2) Minuet For A Cheap Piano Number Two Steep Hills Of Vicodin Tears A Symphony Pathetique All Farewells Are Sudden
1. We Played Some Open Chords & Rejoiced, For The Earth Had Circled The Sun Yet Another Year
2. Requiem For The Static King (part 1)
4. Minuet For A Cheap Piano Number Two
5. Steep Hills Of Vicodin Tears
6. A Symphony Pathetique
7. All Farewells Are Sudden
ALSO ON ERASED TAPES LABEL
ROEDELIUS & STORY
4 Hands LP (Standard Black Vinyl) [PRE-ORD...
4 Hands LP (Limited Indie Clear Vinyl) [PRE-ORD...
Overflow CD
Monster I Like Hands Pulses Of Information Noise Call & Response I Overflow The Cloud Oracle Tension In The Cloud CD 176735 2
Overflow LP
Monster I Like Hands Pulses Of Information Noise Call & Response I Overflow The Cloud Oracle Tension In The Cloud 2 x LP 176600 2
Erased Tapes 20--0 LP (RSD 2021)
1 2 3 4 5 6 7 2 x LP 174857 2
QASIM NAQVI
Chronology LP (RSD 2021)
Kindly Static Aftertouched Head Within A Head Turtle On Fire Chronology Mt Erased LP 174856 2
ALSO BY A WINGED VICTORY FOR THE SULLEN
Invisible Cities LP (Coloured Vinyl)
ARTIFICIAL PINEARCH MANUFACTURING
"So That The City Can Begin To Exist" (4:40) "The Celestial City" (3:44) "The Dead Outnumber The Living" (2:46) "Every Solstice & Equinox" (2:13) "Thirteenth Century Travelogue" (2:13) "Nothing Of The City Touches The Earth" (3:08) "The Divided City" (2:52) "Only Strings & Their Supports Remain" (3:33) "Despair Dialogue" (2:43) "The Merchants Of Seven Nations" (2:17) "There Is One Of Which You Never Speak" (3:14) "Desires Are Already Memories" (4:09) "Total Perspective Vortex" (3:59) LP 171971 2
The Undivided Five CD
Our Lord Debussy Sullen Sonata The Haunted Victorian Pencil The Slow Descent Has Begun Aqualung, Motherfucker A Minor Fifth Is Made Of Phantoms Adios, Florida The Rhythm Of A Dividing Pair Keep It Dark, Deutschland CD 163246 2
The Undivided Five LP (Limited Edition Clear...
Our Lord Debussy Sullen Sonata The Haunted Victorian Pencil The Slow Descent Has Begun Aqualung, Motherfucker A Minor Fifth Is Made Of Phantoms Adios, Florida The Rhythm Of A Dividing Pair Keep It Dark, Deutschland LP 163245 2
The Undivided Five LP (Standard Edition)
Iris CD
Prologue Iris Retour au Champs de Mars Fantasme Gare du Nord Part One L'embauche Le Retour en Foret Metro Part Three Flashback Antoine Galerie Le Renversement CD 141955 2
Iris LP : LTD Coloured Vinyl
CUSTOMERS WHO BOUGHT A Winged Victory For The Sullen LP [Love Record Stores edition) ALSO BOUGHT
Ladies & Gentlemen We Are Floating In Space LP...
Ladies & Gentlemen We Are Floating In Space Come Together I Think I'm In Love All Of My Thoughts Stay With Me Electricity Home Of The Brave The Individual 2 x LP 175562 2
Pure Phase LP (Indies Glow In The Dark Vin...
"Medication" (8:20) "The Slide Song" (3:47) "Electric Phase" (1:38) "All Of My Tears" (3:02) "These Blues" (3:01) "Let It Flow" (5:27) "Take Good Care Of It" (4:27) "Born, Never Asked" (2:16) "Electric Mainline" (7:33) "Lay Back In The Sun" (5:05) "Good Times" (5:02) "Pure Phase" (6:19) "Spread Your Wings" (6:25) "Feel Like Goin' Home" (5:28) 2 x LP 174911 2
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,375
|
Q: how to redirect multi parameter url using htaccess My old url is
http://www.tuitionok.com/result.php?sr=0&city=patna&text=computer+home+tuition+in+patna&type=home-tuition#t
and new url is
http://www.tuitionok.com/patna/home-tuition/computer-home-tuition-in-patna/0
I am trying to forward the old url to new
I am using this .htaccess code
RewriteEngine on
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^([^/]+)/([^/]+)/([^/]+)/([^/]+)$ result.php?city=$1&type=$2&text=$3&sr=$4 [L]
I am successfully get the parameter values and displaying the result. But the actual problem is that both old and new url is working. I think my htaccess file should be modified to redirect url forcely to the new url.
A: Try this:
Options +FollowSymLinks -Multiviews
RewriteEngine On
RewriteCond %{REQUEST_FILENAME} -f [OR]
RewriteCond %{REQUEST_FILENAME} -d
RewriteRule ^ - [L]
RewriteRule ^([^/]*)/([^/]*)/([^/]*)/([^/]*)$ result.php?city=$1&type=$2&text=$3&sr=$4 [L]
RewriteRule ^([^/]*)/([^/]*)/([^/]*)/([^/]*)/$ result.php?city=$1&type=$2&text=$3&sr=$4 [L]
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,794
|
Q: Is it possible to convert any base to any base (range 2 to 46) I know it is simple and possible to convert any base to any base. First, convert any base to decimal and then decimal to any other base. However, I had done this before for range 2 to 36 but never done for 2 to 46.
I don't understand what I will put after 36, because 36 means 'z' (1-10 are decimal numbers then the 26 characters of the alphabet).
Please explains what happens after 36.
A: The symbols you use for digits are arbitrary. For example base64 encoding uses 'A' to represent the zero valued digit and '0' represents the digit with the value 52. In base64 the digits go through the alphabet A-Z, then the lower case alphabet a-z, then the traditional digits 0-9, and then usually '+' and '/'.
One base 60 system used these symbols:
So the symbols used are arbitrary. There's nothing that 'happens' after 36 except what you say happens for your system.
A: Every base has a purpose. Usually we do base conversion to make complex computations simpler.
Here are some most popular bases used and their representation.
*
*2-binary numeral system
used internally by nearly all computers, is base two. The two digits are 0 and 1, expressed from switches displaying OFF and ON respectively.
*8-octal system
is occasionally used in computing. The eight digits are 0–7.
*10-decimal system
the most used system of numbers in the world, is used in arithmetic. Its ten digits are 0–9.
*12-duodecimal (dozenal) system
is often used due to divisibility by 2, 3, 4 and 6. It was traditionally used as part of quantities expressed in dozens and grosses.
*16-hexadecimal system
is often used in computing. The sixteen digits are 0–9 followed by A–F.
*60-sexagesimal system
originated in ancient Sumeria and passed to the Babylonians. It is still used as the basis of our modern circular coordinate system (degrees, minutes, and seconds) and time measuring (minutes and hours).
*64-Base 64
is also occasionally used in computing, using as digits A–Z, a–z, 0–9, plus two more characters, often + and /.
*256-bytes
is used internally by computers, actually grouping eight binary digits together. For reading by humans, bytes are usually shown in hexadecimal.
The octal, hexadecimal and base-64 systems are often used in computing because of their ease as shorthand for binary. For example, every hexadecimal digit has an equivalent 4 digit binary number.
Radices are usually natural numbers. However, other positional systems are possible, e.g. golden ratio base (whose radix is a non-integer algebraic number), and negative base (whose radix is negative).
Your doubt is whether we can convert any base to any other base after base exceeds 36
( # of Alphabets + # of digits = 26+ 10= 36)
Taking example of 64-Base
It uses A–Z(Upper case)(26), a–z(lower case)(26), 0–9(10), plus 2 more characters. This way the constraint of 36 is resolved.
As we have (26+26+10+2)64 symbols in 64-base for representation, we can represent any number in 64 base. Similarly for more base they use different symbols for representation.
Source: http://en.wikipedia.org/wiki/Radix
A: With number systems, you are allowed to play god.
Playing god
What you need to understand is, that symbols are completely arbitrary. There is no god-given rule for "what comes after 36". You are free to define whatever you like.
To encode numbers with a certain base, all you need is the following:
*
*base-many distinct symbols
*a total order on the symbols
An arbitrary example
Naturally, there's an infinite amount of possibilities to create such a symbol table for a certain base:
Θ
ェ
す
)
0
・
_
o
や
ι
You could use this, to encode numbers with base 10. Θ being the zero-element, ェ being the one, etc.
Conventions
Of course, your peers would not be too happy if you started using the above symbol table. Because the symbols are arbitrary, we need conventions. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is a convention, as are the symbols we use for hexadecimal, binary, etc. It is generally agreed upon what symbol table we use for what basis, that is why we can read the numbers someone else writes down.
A: The important thing to remember is that all numbers are symbolic of a value. Thus if you wanted to do that, you could just make a list containing the values at each position. After base 36, you simply run out of characters you can make a logical sequence out of. For example, if you used the Cambodian Alphabet with 70 odd characters, you could do base 80.
A: Here is the complete code I have written, hope this will help.
import java.util.Scanner;
/*
* author : roottraveller, nov 4th 2017
*/
public class BaseXtoBaseYConversion {
BaseXtoBaseYConversion() {
}
public static String convertBaseXtoBaseY(String inputNumber, final int inputBase, final int outputBase) {
int decimal = baseXToDecimal(inputNumber, inputBase);
return decimalToBaseY(decimal, outputBase);
}
private static int baseXNumeric(char input) {
if (input >= '0' && input <= '9') {
return Integer.parseInt(input + "");
} else if (input >= 'a' && input <= 'z') {
return (input - 'a') + 10;
} else if (input >= 'A' && input <= 'Z') {
return (input - 'A') + 10;
} else {
return Integer.MIN_VALUE;
}
}
public static int baseXToDecimal(String input, final int base) {
if(input.length() <= 0) {
return Integer.MIN_VALUE;
}
int decimalValue = 0;
int placeValue = 0;
for (int index = input.length() - 1; index >= 0; index--) {
decimalValue += baseXNumeric(input.charAt(index)) * (Math.pow(base, placeValue));
placeValue++;
}
return decimalValue;
}
private static char baseYCharacter(int input) {
if (input >= 0 && input <= 9) {
String str = String.valueOf(input);
return str.charAt(0);
} else {
return (char) ('a' + (input - 10));
//return ('A' + (input - 10));
}
}
public static String decimalToBaseY(int input, int base) {
String result = "";
while (input > 0) {
int remainder = input % base;
input = input / base;
result = baseYCharacter(remainder) + result; // Important, Notice the reverse order here
}
return result;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter : number baseX baseY");
while(true) {
String inputNumber = scanner.next();
int inputBase = scanner.nextInt();
int outputBase = scanner.nextInt();
String outputNumber = convertBaseXtoBaseY(inputNumber, inputBase, outputBase);
System.out.println("Result = " + outputNumber);
}
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 31
|
{"url":"https:\/\/math.stackexchange.com\/questions\/2040811\/what-are-some-counter-intuitive-results-in-mathematics-that-involve-only-finite\/2042387","text":"# What are some counter-intuitive results in mathematics that involve only finite objects?\n\nThere are many counter-intuitive results in mathematics, some of which are listed here. However, most of these theorems involve infinite objects and one can argue that the reason these results seem counter-intuitive is our intuition not working properly for infinite objects.\n\nI am looking for examples of counter-intuitive theorems which involve only finite objects. Let me be clear about what I mean by \"involving finite objects\". The objects involved in the proposed examples should not contain an infinite amount of information. For example, a singleton consisting of a real number is a finite object, however, a real number simply encodes a sequence of natural numbers and hence contains an infinite amount of information. Thus the proposed examples should not mention any real numbers.\n\nI would prefer to have statements which do not mention infinite sets at all. An example of such a counter-intuitive theorem would be the existence of non-transitive dice. On the other hand, allowing examples of the form $\\forall n\\ P(n)$ or $\\exists n\\ P(n)$ where $n$ ranges over some countable set and $P$ does not mention infinite sets would provide more flexibility to get nice answers.\n\nWhat are some examples of such counter-intuitive theorems?\n\n\u2022 There are only five platonic solids. \u2013\u00a0D Wiggles Dec 2 '16 at 19:29\n\u2022 @IBWiglin: I would call that an intriguing result rather than counter-intuitive. \u2013\u00a0Burak Dec 2 '16 at 19:31\n\u2022 There are many at-first-counter-intuitive results in probability theory. Like so called Boy-Girl Paradox or Monty Hall Problem. \u2013\u00a0Levent Dec 2 '16 at 19:40\n\u2022 ...and the birthday paradox. \u2013\u00a0Rahul Dec 2 '16 at 22:38\n\u2022 @IBWiglin Because each corner has to be three elements, so hexagons and polygons with more sides are too big. More than five triangles, three squares or three pentagons are also too big. I find it counter-intuitive that that's the only criteria, that every corner that can be made makes a solid, but I'm pretty sure with physical pieces it would pretty obvious that there are five and only five platonic solids. \u2013\u00a0prosfilaes Dec 3 '16 at 20:21\n\n100 prisoners problem.\n\nCiting Sedgewick and Flajolet, the problem reads as follows:\n\nThe director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers. The director randomly puts one prisoner's number in each closed drawer. The prisoners enter the room, one after another. Each prisoner may open and look into 50 drawers in any order. The drawers are closed again afterwards. If, during this search, every prisoner finds his number in one of the drawers, all prisoners are pardoned. If just one prisoner does not find his number, all prisoners die. Before the first prisoner enters the room, the prisoners may discuss strategy\u2014but may not communicate once the first prisoner enters to look in the drawers. What is the prisoners' best strategy?\n\nSurprisingly, there exists a strategy with surviving probability more than 30%. It is connected to the fact---also non-intuitive---that a big random permutation is quite likely to contain \"small\" cycles only.\n\n\u2022 This is definitely counterintuitive! I would expect the best probability to be 2^-100 since it seems each prisoner would have a 50% chance of finding his number... \u2013\u00a0Mehrdad Dec 4 '16 at 4:24\n\u2022 @Mehrdad In fact, each prisoner has a 50% chance. Just that those events are not necessarily independent. \u2013\u00a0Peter Franek Dec 4 '16 at 10:37\n\u2022 @SimpleArt: Indeed they were. Unfortunately the first prisoner failed to find the correct number, so the prisoners are no longer. \u2013\u00a0Marc van Leeuwen Dec 5 '16 at 8:07\n\u2022 A variant of this was posted on puzzling stack exchange where a person before the beginning gets to swap two of the numbers in the drawers. If you allow for that the prisoners always survive. \u2013\u00a0DRF Dec 6 '16 at 11:18\n\u2022 @SimpleArt Or just one very smart and very convincing one :) \u2013\u00a0Henrik Dec 8 '16 at 17:34\n\nThe hydra game. Quote from the link:\n\nA hydra is a finite tree, with a root at the bottom. The object of the game is to cut down the hydra to its root. At each step, you can cut off one of the heads, after which the hydra grows new heads according to the following rules:\n\nIf you cut off a head growing out of the root, the hydra does not grow any new heads.\n\nSuppose you cut off a head like this:\n\nDelete the head and its neck. Descend down by 1 from the node at which the neck was attached. Look at the subtree growing from the connection through which you just descended. Pick a natural number, say 3, and grow that many copies of that subtree, like this:\n\nThe counter-intuitive fact: you can always kill the hydra using any algorithm. The counter-intuitive meta-fact: You can't prove the theorem in PA.\n\n\u2022 My favourite! The best part is that it is unprovable using Peano arithmetic. \u2013\u00a0The Vee Dec 3 '16 at 22:54\n\u2022 Anyone try here for yourself to see how ridiculously fast it grows, yet you know if you're diligent enough, you can't lose. \u2013\u00a0The Vee Dec 3 '16 at 23:03\n\u2022 The first fact wasn't counter-intuitive to me. If you remove a head, then either you remove a level (by removing the last head on the topmost level), or you decrease the maximum size of branches on that level, or you decrease the number of branches of that size on that level, and hydras are well-ordered by those properties. (Proof sketch only) \u2013\u00a0immibis Dec 6 '16 at 2:29\n\u2022 @immibis: The problem is that you're tacitly assuming there's no limit ordinal, which is unprovable in PA. You could have 0, 1, 2, ..., omega, omega + 1, omega + 2, ... and that wouldn't violate any of the PA axioms. \u2013\u00a0Kevin Dec 6 '16 at 22:08\n\u2022 This is a variant of Goodstein's Theorem, where the trees are replaced by integers (in iterated exponential notation). Such results are \"counter-intuitive\" only if one is not familiar with ordinals (here $\\,\\varepsilon_0 = \\omega^{\\large \\omega^{\\omega^{\\Large\\cdot^{\\cdot^\\cdot}}}}\\!\\!\\! =\\, \\sup \\{ \\omega,\\, \\omega^{\\omega}\\!,\\, \\omega^{\\large \\omega^{\\omega}}\\!,\\, \\omega^{\\large \\omega^{\\omega^\\omega}}\\!,\\, \\dots\\, \\} ).\\,$ The prior-linked post contains links to many accessible expositions on this and related topics. \u2013\u00a0Bill Dubuque Dec 7 '16 at 0:20\n\nSuppose $X$ is any finite set, which will represent a set of voters, and let $Y$ be another finite set, representing decisions or options that the voters can rank. For example, voting on presidential candidates, favorite ice cream, etc. For simplicity, assume that $X=\\{1,\\ldots, N\\}$.\n\nCall a ranking to be a linear ordering on $Y$, and a social welfare function is a map $$F: L(Y)^N \\to L(Y)$$ where $L(Y)$ is the set of all linear orderings on $Y$. $F$ essentially shows how to take the rankings of each voter and turn that into a single ranking. The elements of $L(Y)^N$ are an $N$-tuple of rankings, a ranking of $Y$ from each voter. We shall represent such a tuple by $(\\leq_n)_{n=1}^N$ and its image by $F((\\leq_n)_{n=1}^N)=\\leq$.\n\nSince this is to be a voting system, we probably want this to adhere to some rules which enforce the idea that $F$ accurately reflects the rankings of each voter:\n\n\u2022 Unanimity: If every voter ranks $a\\in Y$ better than $b\\in Y$, then in the output of $F$, society ranks $a$ higher than $b$. Formally, if $a\\leq_n b$ for every $n$, then $a\\leq b$.\n\n\u2022 Independence of Irrelevant Alternatives: How voters rank $a$ and $b$ should not effect how society ranks $a$ and $c\\neq b$. Formally, if $(\\leq_n)_{n=1}^N$ and $(\\leq_n')_{n=1}^N$ are two tuples of rankings such that the ordering of $a$ and $c$ are the same for each $n$ (i.e. $a\\leq_n c$ if and only if $a\\leq_n' c$) then the ordering of $a$ and $c$ are the same in society's rankings (i.e. $a \\leq c$ if and only if $a\\leq' c$).\n\nSince this is a bit more involved, consider the example of a group ranking the three ice cream flavors of vanilla, chocolate, and strawberry. The group makes their choices, and $F$ says that the highest ranked flavor is chocolate. Then the group learns that strawberry is out, so they rank strawberry as last. It would be counter-intuitive, then, to suspect that all of a sudden vanilla becomes ranked highest (but there are such functions making this true).\n\nThe intuition is the hope that the group's consensus on how it feels about two options should only depend on how each individual feels about those two options.\n\nCases where this still fails are usually indicative of cases where the voting scheme can be gamed in some way, i.e. by voting against your favorite option to avoid your least favorite option or by varying how you rank the remaining options to guarantee their loss.\n\nA good ranked voting system should be such that you benefit most by actually saying what you really think, than attempting to game the system. The failure of Independence of Irrelevant Alternatives allows for this gaming.\n\nThis bring's us to our result:\n\nArrow's Impossibility Theorem: For $Y$ finite and $|Y|> 2$, the only function $F: L(Y)^N \\to L(Y)$ satisfying the above two properties is a dictatorship, i.e. there is a fixed $m$ (which depends only on $F$) such that $1\\leq m\\leq N$ and $F((\\leq_n)_{n=1}^N) = \\leq_m$.\n\nOne method of proof is by considering filters and using the fact that the only ultrafilters on a finite set are the principal ultrafilters.\n\nIt is important to note that Arrow's Impossibility Theorem only applies to ranking systems. There are alternatives ways of voting which are not ranking systems and show more promise.\n\nMoreover, whether the hypothesis of the Independence of Irrelevant Alternatives actually captures what we want in all cases is suspect.\n\n\u2022 Why do you consider this result to be counterintuitive? It is surprising, yes, but I don't see what intuition would suggest existence of such a system. \u2013\u00a0Wojowu Dec 2 '16 at 21:16\n\u2022 @Wojowu I would say it is counter-intuitive in that our intuition - perhaps more due to social upbringing rather than a mathematical intuition - makes it seem like many voting systems that are used in the world should be well-behaved (until you actually start creating example scenarios). Another aspect is because the conditions seem like common-sense. That we find it so obvious that these are properties we want usually makes it seem like they should naturally occur in some non-trivial examples. \u2013\u00a0Hayden Dec 2 '16 at 21:24\n\u2022 @Wojowu because it seems like a bunch of people can get together and vote on something without someone always inviting Hitler. \u2013\u00a0djechlin Dec 3 '16 at 6:18\n\u2022 @user347489 Perhaps I misphrased it. $m$ is a fixed value dependent on $F$. Once $F$ is chosen, there is a single voter which always gets their way. I would describe that as a dictatorship. \u2013\u00a0Hayden Dec 4 '16 at 14:52\n\u2022 Note that Arrow's theorem deals with a deterministic $F$. If you relax that requirement and allow $F$ to map to a random variable, a trivial observation is that \"random dictatorship\" (choosing a random individual and taking their preference orer to be the group preference order) is what comes out, and it's an interesting thought experiment whether such a system would produce reasonable\/acceptable results. \u2013\u00a0R.. Dec 5 '16 at 3:39\n\nClosed-form formulas exist for solutions of polynomials up to degree 4, but not more than that.\n\nOnly 4 colors are required to color a map of any size, with adjacent areas being distinct colors.\n\nDivision rings over the reals have a maximum of 4 dimensions, and cannot have more.\n\n\u2022 I think the fact that some roots of some degree 5+ polynomials cannot be expressed with algebraic operations is even more interesting. \u2013\u00a0Serge Seredenko Dec 3 '16 at 5:53\n\u2022 This might be interesting as well: en.wikipedia.org\/wiki\/Exotic_R4 \u2013\u00a0polfosol Dec 3 '16 at 8:28\n\u2022 For polynomials you, might need to add \"solutions by radicals.\" There might be closed form solutions for degree 5 in terms of transcendental functions, say. Also, I would not consider division algebras over the reals \"finite objects.\" \u2013\u00a0Kimball Dec 3 '16 at 16:48\n\u2022 @IwillnotexistIdonotexist : Not all algebraic numbers can be expressed using algebraic operations (addition, subtraction, multiplication, division, and positive integral roots) applied to rational numbers (or integers, since division makes the distinction not a difference). Only algebraic roots whose minimal polynomial has a solvable Galois group can be (i.e., only those that live in a tower of simple algebraic field extensions of $\\Bbb{Q}$) \"Most\" fifth and higher degree polynomials do not have solvable Galois groups, so cannot be expressed with algebraic operations on rationals. \u2013\u00a0Eric Towers Dec 4 '16 at 2:20\n\u2022 @Burak: you can replace \"real numbers\" with \"real closed field,\" and e.g. the real algebraic numbers (of which there are countable many) are a real closed field. So this example doesn't really have anything to do with the real numbers per se, and in particular is not connected to how it takes an infinite amount of information to specify an arbitrary real number (which was the OP's objection to using the real numbers as examples). \u2013\u00a0Daniel McLaury Dec 5 '16 at 14:47\n\nGender discrimination example. A university has only two graduate departments \u2014 Math and English. Men and women apply to both departments, with varying admit rates, as summarized in the table below.\n\nEach department is more likely to admit women than men. But when aggregated, the university as a whole is more likely to admit men than women (and thus open to charges of gender discrimination)!\n\nDrug testing example. We recycle the exact same numbers from before to a context in which the paradox seems even more paradoxical.\n\nThere are 300 male patients and 300 female patients suffering from some illness. 200 of the males and 100 of the females receive the new experimental drug. The recovery rates are as given in the table below.\n\nThe conclusion from this trial is that if we know the patient is male or female, we should not administer the drug. But absurdly, if we do not know whether the patient is male or female, we should!\n\nFor more, see Pearl (2014), who \"safely proclaim[s] Simpson\u2019s paradox 'resolved.'\u201d\n\nP.S. Simpson's Paradox illustrates both the fallacy of composition (\"what is true of the parts must also be true of the whole\") and the fallacy of division (\"what is true of the whole must also be true of the parts\").\n\n\u2022 After staring at this for ten minutes, I think I understand the electoral college. \u2013\u00a0Mauser Dec 7 '16 at 20:26\n\u2022 Isn't that how the US election system works? \u2013\u00a0Tobias Kienzler Dec 8 '16 at 13:45\n\u2022 Of course they are open to charges of gender discrimination: women are twice as likely to get admitted to the English department. That seems like a pretty strong bias. \u2013\u00a0tomasz Dec 8 '16 at 13:58\n\u2022 @mbomb007 Thanks for the link. I understand how it works. I was making a little bit of a joke, but with some truth behind it, and not intending to belittle the institution. This fallacy of composition warns against aggregating dissimilar data, as information, perhaps critical information, will be destroyed upon the aggregation. This same concern was perhaps involved in the development of that process as well. \u2013\u00a0Mauser Dec 8 '16 at 22:18\n\u2022 @tomasz: Didn't you know that gender discrimination can only ever be against women? =) \u2013\u00a0Kenny LJ Dec 10 '16 at 1:46\n\nI feel like the Monty Hall problem is counter-intuitive the first time you see it.\n\nSuppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door (say No. 1). Then the host, who knows what's behind all the doors, opens another door (say No. 3) that is guaranteed to have a goat. He then says to you, \"Do you want to now pick door No. 2?\" Is it to your advantage to switch your choice?\n\nThe answer is yes, you should ALWAYS switch your choice. The reasoning is as follows: at the beginning you have a $1$ in $3$ chance of selecting the correct door. After the host shows you another door, there are only $2$ doors to select from there. You initially chose the first door which had a probability of $1\/3$ of being the correct door. Now, because all the probabilities within a set of choices must always add to $1$, we can conclude that the 2nd door is correct with a probability of $2\/3$. So indeed, switching your guess is to your advantage.\n\n\u2022 The really fun part with the Monty Hall problem is what happens when you slightly alter the rules, but the situation sounds practically identical. You modify it like this - what happens if Monty chooses one of the other two doors at random to reveal (and if it's the car, you're out of luck)... and he reveals a goat? \u2013\u00a0Glen O Dec 3 '16 at 7:15\n\u2022 There are some instances in which you'd better stay with the first choice xkcd.com\/1282 \u2013\u00a0Del Dec 3 '16 at 10:22\n\u2022 Under this formulation of the question, you do not know if it is better to switch. You also need to know the information that the host will never open the door with the car. Just because he knows what's behind a door does not necessarily imply he will not open a door with a car. \u2013\u00a0David Stone Dec 4 '16 at 18:38\n\u2022 @NoseKnowsAll that doesn't clear it up. You still need to specify the hosts motivations. Consider an evil monty who only gives you the option to switch if you selected the car on your first pick (and immediately declares you a loser if you picked either goat). \u2013\u00a0Steve Cox Dec 5 '16 at 19:30\n\u2022 I'm divided on how to react to this answer, since on the one hand I don't like to see this defective formulation of the problem propagated, but on the other hand the Monty Hall problem (correctly understood) is a really good example of a counterintuitive result. \u2013\u00a0David K Dec 7 '16 at 22:06\n\nMonsky's theorem is easy enough to state for anyone to understand, yet arguably counter-intuitive.\n\nMonsky's theorem states that it is not possible to dissect a square into an odd number of triangles of equal area. In other words, a square does not have an odd equidissection.\n\nThe problem was posed by Fred Richman in the American Mathematical Monthly in 1965, and was proved by Paul Monsky in 1970.\n\n\u2022 It seems pretty intuitive to me. That means that my intuition is either really great or specatacularly terrible. \u2013\u00a0user159517 Dec 3 '16 at 1:22\n\u2022 @user159517 Hat's off to that, it's not at all intuitive to me. Same goes for the equivalent result about polyominoes, or centrally symmetric polygons. \u2013\u00a0dxiv Dec 3 '16 at 1:50\n\u2022 @user159517: I felt the same on first glance, and then realised I\u2019d fallen into the \u201cspectacularly bad\u201d camp: I\u2019d been picturing it with just lattice-aligned right-angled triangles. \u2013\u00a0Peter LeFanu Lumsdaine Dec 3 '16 at 19:20\n\u2022 +1, although this raises the question whether \"mathematics that involve only finite objects\" excludes things dealing with continuity\/topology... I believe Monsky's theorem deals with 2-adic numbers and their topology. \u2013\u00a0Matthew11 Dec 4 '16 at 4:45\n\u2022 @xI_hate_math420x The proof, maybe, but not the statement of the theorem. \u2013\u00a0Akiva Weinberger Dec 5 '16 at 15:59\n\n### Stacking Books on a Table Edge\n\nGiven a rigid (non-deformable), flat, horizontal surface (e.g., a table), a rigid rectangular parallelepiped (e.g., a block, like a book or a brick) can be placed on the edge of the table so that $49.\\overline9\\%$ of its weight overhangs the edge:\n\nAssume that you have a very large supply of identical books.\u00a0 By moving the first book back toward the center of the table (i.e., away from the edge), you can put a second book on top of the first book so that $74.\\overline9\\%$ of its weight overhangs the edge:\n\nBy adding more and more books, you can get the top one completely beyond the edge of the table \u2014 in fact, as far beyond as you want:\n\nThis is discussed (and illustrated much better) at Wolfram MathWorld.\nSpoiler alert: it boils down to the fact that $$\\sum_k\\frac1k$$ grows without bound.\n\n\u2022 It's too bad there are no rigid surfaces in the world, and they all bend somewhat, which is why we can't really try this at home. \u2013\u00a0einpoklum Dec 4 '16 at 23:35\n\u2022 What I find counterintuitive is that you can do a lot better. The method given here gets an overhang on the order of $\\log n$, but you can achieve on the order of $n^{1\/3}$. This is mentioned at the MathWorld page, but see also maa.org\/sites\/default\/files\/pdf\/upload_library\/22\/Robbins\/\u2026 \u2013\u00a0Gerry Myerson Dec 5 '16 at 8:44\n\u2022 I feel like that last image is highly misleading. While it's possible to get 1 block completely past the edge of the table, the arrangement of the other blocks below it would not look like that. \u2013\u00a0Shufflepants Dec 5 '16 at 16:41\n\u2022 Why \"$49.\\overline{9}\\%$\" and \"$74.\\overline{9}\\%$\" instead of just 50% and 75%? \u2013\u00a0anomaly Dec 7 '16 at 16:43\n\u2022 But $49.\\overline9 = 50$, so you have created exactly the same unstable situation you wanted to avoid. You might instead write, \"$0.5 - \\delta_1,$ where $\\delta_1$ is a small positive number.\" \u2013\u00a0David K Dec 7 '16 at 22:09\n\nWhat surprised me most was the following Arab story.\n\nAn inheritance of $35$ camels had to be distributed among $3$ brothers as follows: half for the elder, the third for the second and the ninth for the younger. The problem was that the brothers would then have to receive more than $17$ and less than $18$, more than $11$ and less than $12$, more than $3$ and less than $4$ respectively.\n\n\"The man who calculated\" solved the problem by asking a friend to lend him a camel with which he obtained $36$ camels to distribute following the instructions of the deceased father. Then the brothers received $18$, $12$ and $4$ camels respectively, i.e. more than they had planned so they were very satisfied.\n\nBut then the \u201cman who calculated\" distributed $34$ camels so that in addition to returning the camel loaned by his friend had for himself a camel as a reward for his calculations.\n\nIt was several years after I read this story that I could explain mathematically what happened.\n\n\u2022 I read 'the man who counted' a long time ago, and this segment has remained with me the whole time. \u2013\u00a0Liam Dec 2 '16 at 22:30\n\u2022 $1\/2 + 1\/3 + 1\/9 = 17\/18$. I have no idea why the brothers would not accept extras if there had been no camel lent, but if they did, they wouldn't have needed to borrow anything. \u2013\u00a0John Dvorak Dec 3 '16 at 8:18\n\u2022 @JanDvorak Well then they would fight over who got the extra, of course. \u2013\u00a0Kimball Dec 3 '16 at 16:44\n\u2022 @JanDvorak: psychologically it works because borrowing the camel makes it clear to the brothers that they are receiving the \"extra\" in exact proportion to their \"rightful share\". If you just told them \"take 18\/12\/4\" then you still have to demonstrate that you haven't unfairly given more extra to one than to the other, just to make round numbers. The number 36 likely will feature somewhere in this demonstration. You don't really need the physical camel to do that, of course, but it does make it easier, and the brothers aren't good with fractions. \u2013\u00a0Steve Jessop Dec 4 '16 at 15:49\n\u2022 How is this an example of a counter-intuitive theorem? The fact that $1\/2 + 1\/3 + 1\/9 \\neq 1$ doesn't seem so strange to me. \u2013\u00a0JiK Dec 5 '16 at 13:36\n\nThere's this game about paying taxes that I saw in Yale's open course on game theory. link\n\nWe have tax payers and a tax-collecting agency. The tax payers can choose whether to cheat paying $0$ or not cheat paying $a$. The agency can choose whether to check whether someone has paid tax. If someone cheats and the agency finds out, he'll have to pay $a$ to the agency and a fine $f$ that doesn't go to the agency. To check each tax payer, the agency has to spend $c$. $c$ is less than $a$. The payoff matrix is as follows:\n\n Tax Payer\nCheat Not Cheat\n+-------------+-----------+\nCheck | (a-c, -a-f) | (a-c, -a) |\nAgency +-------------+-----------+\nNot Check | (0, 0) | (a, -a) |\n+-------------+-----------+\n\n\nOne's intuition may suggest that as $f$ increases, tax payers will cheat less often. But solving for the Nash equilibrium tells us that in an equilibrium, as $f$ increases, the probability a tax payer cheats doesn't change, but the agency will check less often. Maybe the tax payer will cheat less often and the agency will check at the same probability when $f$ has just changed, but given enough time, rational tax payers and agency will play the Nash equilibrium.\n\nThe Nash equilibrium is that the tax payer cheats at a probability of $\\frac c a$, and the agency checks at a probability of $\\frac a {f+a}$. The tax payer's expected payoff is $-a$. The agency's expected payoff is $a-c$.\n\n\u2022 So the moral of the story is that you shouldn't let the IRS keep 100% of taxes levied, they should have to hand some of it over to the treasury? ;-) \u2013\u00a0Steve Jessop Dec 4 '16 at 16:09\n\u2022 Even though f shouldn't affect the agency in any way whatsoever... \u2013\u00a0immibis Dec 6 '16 at 2:35\n\u2022 @immibis - and c shouldn't affect the taxpayer (or tax cheat) whatsoever, either. Interesting, though, that the expected payoff for both the taxpayer and the agency is independent of the fine itself. \u2013\u00a0Glen O Dec 6 '16 at 5:54\n\u2022 Since I posted the last comment I had an epiphany of how this can be the case (intuitively; I still haven't bothered to do the maths). If the fine increases then less people want to cheat, so now the agency is wasting their money doing so many checks for so few cheaters, so they do less checks (which increases the number of cheaters again). \u2013\u00a0immibis Dec 6 '16 at 7:53\n\u2022 This doesn't seem counter-intuitive. If the fine increases, then the agency will not need to check as much in order to enforce a level of compliance which they are happy with. \u2013\u00a0jwg Dec 8 '16 at 12:01\n\n# The fallacy of the hot hand fallacy\n\nSuppose you'd like to detect whether a coin ever goes on hot streaks or cold streaks, so that after $k$ heads in a row, the probability of the following flip coming up heads is different from the overall probability $p$ of a heads. To test this, you'll flip the coin $n$ times, and after any streak of $k$ heads, you'll record the outcome of the following flip. Let $X$ be the percentage of your recorded flips that came up heads. For concreteness, let's set the values at $p=\\frac{1}{2}$, $n=100$, and $k=3$.\n\nHere is the surprise: for these values, $E[X]\\approx0.46$ (not $\\frac{1}{2}$!!!!). And, in general for any $0<p<1$, $n\\geq 3$, and $0<k<n$, $E[X]<p$, and the bias can be quite large for certain values of $n$ and $k$.\n\nThis is counterintuitive enough that when Gilovich, Vallone and Tversky wrote their seminal paper Hot Hand In Basketball in 1985 measuring whether basketball players went on \"hot streaks\", they used the exact method above to attempt to detect hot streaks, and since the percentage after three hits in a row was not different from the overall percentage, they concluded that there was no evidence of a hot hand. But this was a mistake! If there was no hot hand, they should have observed a significantly lower percentage on shots after three hits in a row. In fact, their data do show evidence for a hot hand in many of the cases, according to a new paper last month. This mistake went unchecked for 30 years, with untold numbers pop psychology books and articles citing the result as evidence for a \"hot-hand fallacy\".\n\n## Demonstration\n\nHere's a demonstration in R.\n\nf7 <- function(x){\n# running total of run length\n# stolen from http:\/\/tolstoy.newcastle.edu.au\/R\/e4\/devel\/08\/04\/1206.html\ntmp <- cumsum(x)\ntmp - cummax((!x)*tmp)\n}\n\nstreak <- function(v, k = 3, n = length(v)) {\n# returns a vector of length n = length(v) this is TRUE when the last k\n# entires are True\nc(FALSE, f7(v)[1:(n-1)] >= k)\n}\n\nrandom_shots <- function(n, p = 0.5) {\n# takes n random shots with probability p of success\nrunif(n) < p\n}\n\ntrial <- function(n, k = 3, p = 0.5) {\ns <- random_shots(n, p)\nmean(s[streak(s, k)])\n}\n\n# do simulation 100000 times\nresults <- sapply(1:100000, function(x) trial(100, 3))\nsummary(results)\n# Min. 1st Qu. Median Mean 3rd Qu. Max. NA's\n# 0.0000 0.3636 0.5000 0.4615 0.5714 0.8571 3\n\n\n## What's going on?\n\nOne way to see it is to consider the absolute simplest case, where we flip the coin three times, and tally up the flips after a single head in a row.\n\nOutcome Heads Flips Proportion\nHHH 2 2 1\nHHT 1 2 1\/2\nHTH 0 1 0\nHTT 0 1 0\nTHH 1 1 1\nTHT 0 1 0\nTTH 0 0 NA\nTTT 0 0 NA\n\n\nThe last two outcomes, of course, can't be included in our tally because the proportion of heads is undefined. Now, if we repeat this experiment many times, we'll find that of the sequences that we record, $\\frac{2}{6}$ of the time we will have a proportion of $1$, while $\\frac{1}{6}$ of the time we'll have $\\frac{1}{2}$, for an expected proportion of $$\\frac{2}{6}\\times{1} + \\frac{1}{6}\\times\\frac{1}{2} = \\frac{5}{12} < \\frac{1}{2}$$\n\nSo, by inspection we can plainly see that in this case the expected proportion is less than 0.5, although at first glance this might still seem unsatisfactory. Yeah, it's less than 0.5 but... why?\n\nI think there are a few ways to hand-wave about this. One has to do with the fact that we have two outcomes with proportion = $1$, but one of those ways has two heads, and the other only has one. So sequences with more heads are weighted the same as sequences with fewer heads, and in this way heads are somehow being underrepresented, leading to the bias.\n\n\u2022 So gambler's fallacy is (kind of) true? How come $E[X] \\approx 0.46$? \u2013\u00a0Jasper Dec 5 '16 at 12:20\n\u2022 To my understanding that's because the samples are allowed to \"overlap\", and to correct for it, HHHH should only count as two HH pairs, not three. (After seeing an H, or k consecutive H's), you should record the next flip and reset your counter of consecutive H's. \u2013\u00a0immibis Dec 6 '16 at 2:57\n\u2022 I've figured out part of what's going on. Suppose (as written in this answer) that you flip a fair coin 100 times, and after any streak of 3 heads (with overlaps permitted), you record the outcome of the following flip. Let $A$ be the number of flips that you record, and let $H$ be the number of those which are heads. As expected, $E[H]\/E[A] = 1\/2$. Part of the reason that $E[H\/A] \\neq 1\/2$ is that $A$ and $H$ are not independent variables. The flips in a heads-heavy trial run \"matter less\" than the flips in a tails-heavy trial run. \u2013\u00a0Tanner Swett Dec 6 '16 at 4:01\n\u2022 It's worth noting that as you increase the number of trial runs, $E[x]$ gets closer to $1\/2$. While at 100 runs $E[x]\\approx 0.46$, at 10,000 runs $E[x]\\approx 0.499$. This is based only on experimental runs from the given R program, I didn't actually do the math \u2013\u00a0rtpax Dec 10 '16 at 22:42\n\u2022 This is related to this post, which explains the intuition in a different way: math.stackexchange.com\/questions\/2317508\/\u2026 \u2013\u00a0Joshua B. Miller Nov 13 '17 at 6:51\n\nHere is a consequence of the Arc sine law for last visits. Let's assume playing with a fair coin.\n\nTheorem (false) In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.\n\nThe following text is from the classic An Introduction to Probability Theory and Its Applications, volume 1, by William Feller.\n\n\u2022 According to widespread beliefs a so-called law of averages should ensure the Theorem above. But, in fact this theorem is wrong and contrary to the usual belief the following holds:\n\nWith probability $\\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.\n\nIn fact this leads to the Arc sine law for last visits (see e.g. Vol 1, ch.3, section 4, Theorem 1).\n\nRemarkable statements cited from Chapter III: Fluctuations in Coin Tossing and Random Walks:\n\n\u2022 For example, in various applications it is assumed, that observations on an individual coin-tossing game during a long time interval will yield the same statistical characteristics as the observation of the results of a huge number of independent games at one given instant. This is not so.\n\nand later on:\n\n\u2022 Anyhow, it stands to reason that if even the simple coin-tossing game leads to paradoxical results that contradict our intuition, the latter cannot serve as a reliable guide in more complicated situations.\n\nAn example:\n\nSuppose that a great many coin-tossing games are conducted simultaneously at the rate of one per second, day and night, for a whole year. On the average, in one out of ten games the last equalization will occur before $9$ days have passed, and the lead will not change during the following $356$ days. In one out of twenty cases the last equalization takes place within $2\\frac{1}{4}$ days, and in one out of a hundred cases it occurs within the first $2$ hours and $10$ minutes.\n\n\u2022 just started reading this as finally had some time on hands. looking forward to that chapter! \u2013\u00a0Mehness Dec 3 '16 at 23:34\n\u2022 I would think about this like a random walk. The walk could cross the y-axis often, but it's just as likely it's wandered off to one half or the other \u2013\u00a0Dylan Frese Dec 7 '16 at 1:08\n\u2022 Out of context, depending on which statistics you look at, the statement about \"the results of a huge number of independent games at one given instant\" seems either trivial (the lead can change during a series of games, but not during a number of simultaneous games) or untrue (if there are $N$ games in either case and $X$ is the total number of heads, how does sequence vs. simultaneous affect the moments of $X$?). I presume there was some discussion prior to that statement that explains it better. \u2013\u00a0David K Dec 7 '16 at 22:22\n\u2022 @DavidK: If I understand correctly, what it's saying is this: if you gather scores of a single long-running game after every N flips, you'll get a very different distribution than if you gather scores of many independent games of N flips. In the long-running game, you might expect that scores will tend to stay near 0 (so that scores < -N or > N, while possible, are extremely unlikely, so don't affect the overall shape of the distribution), but in fact scores far from 0 are quite likely. \u2013\u00a0ruakh Dec 10 '16 at 23:53\n\u2022 Perhaps the purpose of the statement about \"various applications\" was meant to be provocative, to motivate someone to read the chapter. (It worked for me, anyway!) It may be that its exact meaning was never intended to be explained. Certainly the rest of the chapter contains many very well-explained, precise statements that are also surprising or indicative of fallacies, including a questionable inference by Galton in 1876 (page 70). On page 88, \"even trained statisticians expect\" the lead in the sequential game to change far more often than it actually is likely to. \u2013\u00a0David K Dec 11 '16 at 17:03\n\nWedderburn's little theorem: In its simplest form, it states that any finite division ring is commutative. I find it borderline magical; as if somehow the axioms do not imply the result, but rather it is forced due to some weird combinatorial coincidence. Of course, this feeling of mine says more about my flawed intuition than about the theorem - but that's true for any example of a \"counterintuitive mathematical result\".\n\nAnother example, even more elementary, is the fact that a matrix's rank is well defined, i.e., that for any matrix (even if it's non-square), the dimension of its column-space is equal to the dimension of its row-space. Over time I came to terms with this result, but when I first encountered it, I've been reading the proof again and again, understanding every step - but still couldn't believe the theorem is true. It was a long time ago, but I clearly remember I've felt like I'm practicing voodoo.\n\n\u2022 Arguably, the reason for the counterintuitive matrix rank is that matrices are just a not-really-optimal representation of linear mappings. It's fairly intuitive that the image of a linear mapping has the same dimension as the image of its adjoint mapping. \u2013\u00a0leftaroundabout Dec 3 '16 at 21:17\n\u2022 \u201cYoung man, in mathematics you don't understand things. You just get used to them.\u201d -- John von Neumann \u2013\u00a0Mike Jones Dec 6 '16 at 23:33\n\u2022 @leftaroundabout: matrices are the only representation I know of linear mappings, what else did you have in mind? \u2013\u00a0Mozibur Ullah Dec 11 '16 at 6:51\n\u2022 @MoziburUllah there are plenty of representations. One representation that makes it particularly clear that the rank doesn't depend on the space dimension is the singular value decomposition (which can be defined without ever talking about matrices, only abstract linear mappings). \u2013\u00a0leftaroundabout Dec 11 '16 at 10:09\n\u2022 @MoziburUllah there is no such difference, I just said abstract to emphasize that there's no need to write out the linear mapping in any particular notation (like a matrix). It's sufficient to describe it as a generic function $\\varphi : V\\to W$ between two vector spaces which has the linearity property $\\varphi(\\mu\\cdot u + v) = \\mu\\cdot\\varphi(u) + \\varphi(v)$. Sure enough, any such function (if the spaces are finite-dimensional) can be written in matrix notation, but my point is that this isn't always such a good idea since it obscures interesting properties (like well-defined rank). \u2013\u00a0leftaroundabout Dec 11 '16 at 10:57\n\nI've always thought Bernoulli's paradox was counter-intuitive, and the name suggests that I'm not the only one who thinks this. (Nicolas) Bernoulli offers the following gambling game; You flip an honest coin. If it comes up heads, you win two dollars and get to flip again. If it comes up heads a 2nd time, you win four more dollars and you get to flip again. In general if it comes up heads for n consecutive times, on the nth flip you win 2^n more dollars and get to flip again. The first time the coin comes up tails the game is over, but you get to keep your winnings. Bernoulli asked essentially what the mathematical expectation of this game is. To keep things finite, let's just ask if should you mortgage your house and pay Mr. Bernoulli $250,000 to play this game? Although I would strongly advise against doing so, the mathematics \"shows\" that you should be willing to put up all the money you have or can borrow to play the game. \u2022 The counterintuivity here comes from not so much from the game itself, as from tacitly conflating \u201cshould you do it?\u201d with \u201cis the expected value, in dollars, positive?\u201d This conflation turns our strong (and very justifiable) gut instinct that we shouldn\u2019t play the game into the erroneous intuition that its expected value is negative. \u2013 Peter LeFanu Lumsdaine Dec 3 '16 at 15:14 \u2022 Also known as the St Petersburg paradox. Since it is presented as a real-life challenge, to evaluate it requires the assessment of at least two entailed real-life hazards: (1) Counterparty risk---the offerer cannot or will not pay up as promised. (2) The decreasing marginal utility of wealth---a million-dollar gain might transform your life now, but you would hardly notice it if you already had a billion. \u2013 John Bentin Dec 3 '16 at 15:17 \u2022 @John Bentin I have comforted myself by taking Bernoulli's game as a mathematical proof of that the law of marginal utility applies even to wealth. Also buried away is the tacit assumption of economists that utility is always positive: you have to wonder if, after receiving eleven Lords a leaping, the recipient doesn't begin to thank God there aren't 5000 days of Christmas. Counterparty risk is a red herring: if ,as a gedankin experiment, you ignore it, the paradox remains. \u2013 Airymouse Dec 3 '16 at 16:11 \u2022 The classical St Petersburg paradox does not pose a limit to the number of flips which makes it a paradox dealing with infinity (which OP wants to exclude). Indeed, the surprising result is that the expected value is$\\infty$so any amount you bet is acceptable to you if you play the game enough times. You should limit the game to at most$N$flips, which will make the expected value$N$. Any bet below$N$(per game) is acceptable, which again can be counterintuitive if$N$is a large number. The trick part is realising that the median and expected value are very different in this game. \u2013 Thanassis Dec 4 '16 at 2:38 \u2022 This problem becomes much more intuitive when you assume any finite bound on the amount of money the house is able to pay out. If you cap the amount you could possibly win at even a trillion dollars, the expected value drops down from infinite to about$10. \u2013\u00a0Shufflepants Dec 5 '16 at 16:46\n\nThat you only need 23 people in a room for there to be a 50% chance that two of them will share a birthday.\n\n\u2022 To me this is like the \u201ccounter-intuitive\u201d fact about how often you have to double the thickness of a newspaper to reach the moon. It comes down to it being too easy to acquire incorrect intuitions and be surprised they are wrong \u2013 we have to learn caution. Now Dorothy Parker\u2019s intuition was that it would not be at all surprising if all the girls at a Yale prom were laid end to end \u2013 that\u2019s more like it. \u2013\u00a0PJTraill Dec 6 '16 at 0:21\n\u2022 Write a story about 23 people and explore the relationship between every pair of them. The birthday paradox will then no longer be a paradox. \u2013\u00a0Th\u00e9ophile Dec 7 '16 at 20:23\n\nLangton's ant is a set of rules that are applied to change \"pixels\" in a grid. After a finite number of steps of applying those rules, the rather chaotic behaviour changes into a periodic one. That can be seen in the picture below.\n\n\u2022 Not only that, but there are some turmites that are only slightly more complex to describe than Langton's ant (e.g. 9 turmites with 3 colours where Langton's ant has 2) where it is not known whether they will ever settle down into periodic behaviour. github.com\/GollyGang\/ruletablerepository\/wiki\/\u2026 and scroll down to \"Unresolved Turmites\". \u2013\u00a0Rosie F Jul 9 '18 at 20:05\n\nThe amazing speed of exponential growth. This can be explained to children, but it will always defy your intuition, take e.g. the famous wheat and chessboard problem. As Addy Pross explains here, this may play a fundamental role in the emergence of life.\n\n\u2022 Whenever it does come within your understanding, I recommend looking into the Ackermann function, Knuth's up-arrow notation, Graham's number, and other related things from googology, recommendably in that order. Then your mind will truly blow up at how fast things can grow. \u2013\u00a0Simply Beautiful Art Mar 20 '17 at 23:46\n\u2022 @SimplyBeautifulArt there is something of this Googology in the long path $27$ takes to reach $1$ by iteration of the Collatz function. \u2013\u00a0user334732 Nov 1 '17 at 13:22\n\u2022 @RobertFrost possibly interesting. Personally I find things like TREE(3) and the Goodstein sequence more interesting :P \u2013\u00a0Simply Beautiful Art Nov 1 '17 at 13:59\n\nFirst, a more elementary example: Given two polygons $A$ and $B$ of equal area, I can always cut one into a finite number of polygonal pieces, and rearrange those to form the other; that is, any two polygons of equal area are scissors congruent.\n\nHilbert's third problem was about extending this result to three dimensions. He conjectured, rightly, that it fails in three dimensions, and this was proved shortly afterwards by his student, Dehn.\n\nStill, even though it was proved rather quickly, and Hilbert's (and others') intuition was right on the mark, I still find it incredibly counterintuitive.\n\nAnother geometric example is the refutation of the triangulation conjecture; however, I'm not sure if arbitrary manifolds count as finite objects.\n\n\u2022 This doesn't quite satisfy \u201conly involves finite objects\u201d though, does it? \u2013\u00a0leftaroundabout Dec 7 '16 at 21:10\n\u2022 @leftaroundabout I think the first one does - everything in question can be appropriately represented by a finite set of natural numbers (OK fine, technically we need to consider polytopes with vertices in some fixed countable set - say, all coordinates rational - but the fact still holds). Note that the OP is okay with quantifiers ranging over $\\mathbb{N}$, just not direct invocation of infinite sets. The second example is much less finitary, but I thought it was worth mentioning just in case (it can be finitized, but less naturally). \u2013\u00a0Noah Schweber Dec 7 '16 at 21:16\n\nAs a non-mathematician it seems pretty obvious to me that there are whole number solutions (where $n>2$) for\n\n$$a^{n}+b^{n}=c^{n}.$$\n\nI'd be shocked if there weren't. There must be some, surely.\n\n\u2022 I'm not sure if you're being sarcastic, but in 1995 this was famously proven not to have solutions. \u2013\u00a0Akiva Weinberger Dec 5 '16 at 18:55\n\u2022 @AkivaWeinberger yeah, but it is counter-intuitive that it has no solutions. \u2013\u00a0theonlygusti Dec 5 '16 at 19:05\n\u2022 @theonlygusti I was referring to the \"I'd be shocked if they weren't\" bit; it implies that OP does not know. \u2013\u00a0Akiva Weinberger Dec 5 '16 at 19:06\n\u2022 @AkivaWeinberger: That he gives this as an answer suggests quite strongly that he is aware of the result and intends to suggest he considers it counter-intuitive. His phrasing suggests British, perhaps with the matching sense of humour \u2013 his profile confirms it. \u2013\u00a0PJTraill Dec 5 '16 at 23:58\n\u2022 What's counter-intuitive about this having no solutions? IMO it's not intuitive that this has an integral solution for any $n\\neq 1$; in fact I recall being mildly surprised when I first read that Pythagorean triples exist \u2013 of course, just trying a couple of numbers quickly gives an affirmative example, but a priori it seems just as plausible that there's a simple argument \u00e0 la irrationality of $\\sqrt 2$ (small enough to fit in this margin...) that it can't work. What's counter-intuitive is that it required so insanely advanced maths to settle this simple question. \u2013\u00a0leftaroundabout Dec 7 '16 at 21:21\n\nFor those who don't find the \"harmonic overhang\" of Scott's answer counterintuitive, here is a variant that Loren Larson, a retired mathematician and current creator of wooden puzzles, discovered and demonstrated some years ago: It's possible to construct a stable tower of blocks with the property that removing the top block causes the tower below it to collapse.\n\nIt's worth spending some time trying to imagine how this is possible, but once you give up, here is the secret:\n\nStart with a standard harmonic overhang (as I recall, Loren's had a couple dozen wooden blocks of dimensions something like $8$ inches by $2$ inches by $1\/2$ inch) and carefully nudge it, block by block, to form a spiral. If the tower is tall enough and the spiral is big enough -- Loren worked out the mathematical details -- the top block becomes necessary to counterbalance the portion of the spiral lower down that juts out on the opposite side. Removing it creates an imbalance that causes the lower portion to topple.\n\n\u2022 Why not just put three blocks one on the other, with middle one shifted to the side? Removing the top block will cause middle one to fall. \u2013\u00a0Abstraction Dec 7 '16 at 12:12\n\u2022 Yeah, or surely build up Scott's tower, and then build it on top of itself again going in the other direction. \u2013\u00a0theonlygusti Dec 7 '16 at 12:47\n\u2022 A tower that collapses when you remove the top block is easy to imagine; the fact that a particular tower does so might be less obvious. The difficulty here is trying to imagine what the tower actually looks like; I wasn't able to get a clear enough picture of it to say whether it is counterintuitive that the top block is required. \u2013\u00a0David K Dec 7 '16 at 22:36\n\nI don't know whether it qualifies -\n\nIf we cut a Mobius strip, instead of getting two strips, we get only one strip (longer, with two 2 twists). When I first saw this, it was quite counter-intuitive to me.\n\n\u2022 That's wrong. You don't receive a new M\u00f6bius strip, but a strip that has two twists. That's odd. But what baffled me: When you cut this strip, you get two strips that are interwound. Further results of cutting are listed at the Wikipedia page, and it's really getting crazy at some point... \u2013\u00a0Marco13 Dec 8 '16 at 19:18\n\u2022 I didn't said that one'd get a new Mobius strip but I said that one'd get a long Mobius strip, without mentioning about the twists though. \u2013\u00a0Kushal Bhuyan Dec 9 '16 at 2:01\n\u2022 @KushalBhuyan But the resulting strip is not a M\u00f6bius strip. I don't understand what you are saying. \u2013\u00a0Improve Dec 10 '16 at 15:28\n\u2022 Its a great example even if the details are missing - because it is something one can straight-forwardly and practically do - unlike most of the other examples. \u2013\u00a0Mozibur Ullah Dec 11 '16 at 6:44\n\u2022 @ypercube\u1d40\u1d39 Thanks \u2013\u00a0Kushal Bhuyan Dec 12 '16 at 10:11\n\nI think it is counterintuitive, at first glance, that list coloring graphs can be strictly harder than ordinary coloring.\n\nTo expand on what that means: a graph is a set of vertices, some pairs of which are adjacent. (Think of the vertices as dots, with adjacent vertices having a line drawn between them.) A proper $k$-coloring of a graph assigns each vertex a number from $1$ to $k$ such that no two adjacent vertices receive the same color. The chromatic number of a graph $G$, written $\\chi(G)$, is the smallest $k$ such that $G$ has a proper $k$-coloring.\n\nA $k$-list-assignment on a graph is a function $L$ that assigns each vertex some \"palette\" of $k$ different numbers. If $L$ is a $k$-list-assignment, then a proper $L$-coloring of $G$ assigns each vertex a color from its list such that, again, no two adjacent vertices receive the same color. The list chromatic number of $G$, written $\\chi_l(G)$, is the smallest $k$ such that $G$ has a proper $L$-coloring for every $k$-list-assignment $L$.\n\nNow, intuition (at least, my intuition) suggests that finding an $L$-coloring should be hardest when all the vertices have the same list $L$ -- in other words, when we're really just looking for a proper $k$-coloring. If different vertices have different lists, aren't there just fewer opportunities for collisions to happen? This suggests that maybe $\\chi_l(G) = \\chi(G)$ for every $G$, since the hardest list assignments \"should\" just be the ones that give every vertex the same list.\n\nHowever, this isn't true! Consider the complete bipartite graph $K_{3,3}$, whose vertices consist of six vertices $v_1, v_2, v_3, w_1, w_2, w_3$ where any pair of vertices $v_iw_j$ is adjacent. This graph clearly has a proper $2$-coloring: color all the vertices $v_i$ with color $1$ and all the vertices $w_j$ with color $2$. But now consider the following list assignment $L$:\n\n$L(v_1) = L(w_1) = \\{1,2\\}$\n\n$L(v_2) = L(w_2) = \\{1,3\\}$\n\n$L(v_3) = L(w_3) = \\{2,3\\}$.\n\nAmong the colors $\\{1,2,3\\}$ appearing in these lists, we would need to use at least two of them on the vertices $v_i$, since no color appears in all three lists, and we would need to use at least two of them on the vertices $w_j$, for the same reason. This means that some color gets used on both a $v_i$ and a $w_j$, which contradicts the assumption that the coloring is proper. So there is no proper $L$-coloring, which means $\\chi_l(K_{3,3}) > 2 = \\chi(K_{3,3})$. In other words, it's harder to color from these lists than it is to color when every vertex has the same list!\n\n\u2022 Why do you claim it's counterintuitive that colouring should be harder when you have fewer free choices of colour in list colouring? One can consider the \"open\" colouring to have assigned every vertex a list of $|V|$ colours, and the problem is simply to make that list as short as possible given $E$ and the definitions that apply. Consider by analogy: it is easy to assemble a Yowie toy on the table, but very difficult to fit the pieces back in the capsule, and often impossible to fit the complete toy at all. \u2013\u00a0Nij Dec 3 '16 at 6:34\n\u2022 @Nij: He doesn't have fewer choices: In both cases, he can choose one of two colours at each node. And the total number of colours in the second case is larger (because there are three colours in the union of all lists). \u2013\u00a0celtschk Dec 3 '16 at 11:58\n\u2022 @Nij - as celtschk said, the comparison isn't between finding the L -coloring for my specified L and finding any proper coloring at all, but rather, the comparison is between finding the L-coloring for the given L and finding a proper 2-coloring, which is the same as having the list {1,2} at every vertex. It seems intuitive to guess that the latter coloring problem should be harder, even though it turns out not to be. \u2013\u00a0Gregory J. Puleo Dec 3 '16 at 14:47\n\u2022 Good example. As you no doubt know, the smallest example for $\\chi_l(G)\\gt\\chi(G)$ is the graph $G$ you get by removing two nonadjacent edges from $K_{3,3}$ or in other words $G=P_3\\square P_2.$ \u2013\u00a0bof Dec 3 '16 at 19:48\n\n(1). In the Stone Age, Og and Mog each need to make large numbers of arrow-heads and ax-heads. Their products are of equal quality. Og takes 3 units of time to make an arrow-head and 5 units of time to make an ax-head. Mog takes 4 units of time to make an arrow-head and 7 units of time to make an ax-head. Both of them consider their time to be valuable.\n\nSince Og is faster at both, it seems that Mog could not give Og an incentive to trade.\n\nBut Mog offers to give 17 arrow-heads for 10 ax-heads, which, for Mog, is trading 68 units of time in return for 70. And for Og, this is trading 50 units of time in return for 51. So they both benefit by the trade.\n\n(2). This might not count as finite: In triangle ABC draw the trisectors of the interior angles. Let the trisector lines of angles B ,C that are closer to the side opposite A, meet at A'. Define B',C' similarly. Then (Morley's Theorem) A'B'C' is an equilateral triangle. Intuitively it may seem that A'B'C' might have any given shape. There is a nice proof in Introduction To Geometry by Coxeter.\n\n(3). The only solution in positive integers to $X^3=Y^2+2$ is $X=3, Y=5.$ (Pierre de Fermat). Not obvious to me that there aren't any others.\n\n\u2022 To help make the first one more intuitive: If you imagine that the timings were the same except that Mog took 1,000,000 units of time to make an ax-head, it's quickly pretty clear that he'd trade a lot of arrow-heads for them, and that that trade would be good for Og. \u2013\u00a0Ben Aaronson Dec 6 '16 at 9:03\n\u2022 (1) is yet another example from economics \u2014 the theory of comparative advantage. As said by Paul Samuelson, \"That it [the theory of comparative advantage] is logically true need not be argued before a mathematician; that is is not trivial is attested by the thousands of important and intelligent men who have never been able to grasp the doctrine for themselves or to believe it after it was explained to them.\" \u2013\u00a0Kenny LJ Dec 7 '16 at 6:48\n\u2022 The first one doesn't seem counter-intuitive at all to me. It's a red herring that Og is faster at both; what's relevant is that they produce at different rates (i.e., time for axe vs. time for arrow). \u2013\u00a0Th\u00e9ophile Dec 7 '16 at 21:09\n\u2022 @Th\u00e9ophile. Try it out on people you know. Yesterday I showed it to a friend, a master at computer programming. He said he wouldn't have thought it possible. What is obscure to some is obvious to others, and what is counter-intuitive to some may be intuitively clear to others. And what was intuitive to S. Ramanujan, who knows? \u2013\u00a0DanielWainfleet Dec 7 '16 at 23:25\n\u2022 @Th\u00e9ophile: Try this. Say T-shirts and cars are the only two goods in the world. For an American worker, it takes 3 man-hours to produce a T-shirt and 5,000 man-hours to produce a car. For a Chinese worker, it takes 9 man-hours to produce a T-shirt and 20,000 man-hours to produce a car. The typical American who doesn't understand the theory of comparative advantage then wonders, \"Americans are literally better than the Chinese at everything! So why are we shipping T-shirt jobs to China?\" The answer is that surprisingly, both the US and China can gain by \"shipping jobs to China\". \u2013\u00a0Kenny LJ Dec 8 '16 at 0:52\n\nRecall the Busy Beaver function, $BB(n)$ is the maximal number of steps that a Turing machine with at most $n$ states will halt on the blank tape, assuming it halts at all.\n\n$\\sf ZFC$ cannot decide the value of $BB(1919)$.1\n\nNamely, if there is a contradiction to $\\sf ZFC$, then a Turing machine with less than $2000$ states should be able to find it. Yes, $1919$ is a large number, but it's not unimaginably large. But what it means is that $BB(1919)$ is pretty much entirely unimaginable, because we cannot even give it a concrete estimation.\n\n(See this and that on Scott Aaronson's blog.)\n\n1. Under the usual caveat that we need to assume that $\\sf ZFC$ is consistent of course.\n\n\u2022 Um, ZFC cannot decide the value of BB(1919) if ZFC is consistent. =) \u2013\u00a0user21820 Dec 24 '16 at 14:58\n\u2022 When is it not? C'mon, you only have to look at Turing machines with less than 2000 states! :-) \u2013\u00a0Asaf Karagila Dec 24 '16 at 15:00\n\u2022 I think it is but I can't prove it. =) Anyway I suggest you edit to make it consistent with your second part of your answer, which implicitly suggests the possibility that ZFC may not be consistent. \u2013\u00a0user21820 Dec 24 '16 at 15:03\n\u2022 We can look at them all, but we can't run them for arbitrarily long time, not to say infinite time. Well, it's the age-old question of whether the halting problem has a definite answer or not. =P \u2013\u00a0user21820 Dec 24 '16 at 15:05\n\u2022 That definite answer is, of course, 4. A number which is always visible, and always out of reach. \u2013\u00a0Asaf Karagila Dec 24 '16 at 15:07\n\nDeterminacy of classes of finite games:\n\nThe existence of winning strategies for finite games without draws and with perfect information is counter-intuitive at first glance. It is, ultimately, very sound though.\n\nI have been in the situation of explaining this to children and adults with the example of simple games, and most of the time people went from disbelief to recognition that it is in fact obvious.\n\nThere are three major reasons for this counter-intuitiveness:\n\n-We are used to playing those types of games since we're children, and in our experience, our strategies often ended up defeated.\n\n-Winning strategies for even simple games are often unknown: most of the time they can't be computed because the numbers of possible outcomes are two high. So most of the time, the existence of a winning strategy for a game doesn't affect the way you play the game.\n\n-The definition of the fact $W$ of being in a winning position is somewhat complex: if you can make a winning move, then $W$, and (if whatever move your opponent makes, then $W$), then $W$. This type of recursive definition can be hard to grab, and when trying to unveil it, complexity, in the sense of an alternance of quantifiers, really appears since you get: $\\forall$ opponent move $\\exists$ move such that $\\forall$ opponent move $\\exists$ move such that ... such $\\exists$ a winning move for you.\n\n82000. It absolutely boggles the mind because this is very close to the axioms, so to speak. You need to prove only a few things based on the Peano axioms to get to\n\nThe smallest number bigger than 1 whose base 2, 3, and 4 representations consist of zeros and ones is 4. If we ask the same question for bases up to 3, the answer is 3, and for bases up to 2, the answer is 2. But 82000 is the smallest integer bigger than 1 whose expressions in bases 2, 3, 4, and 5 all consist entirely of zeros and ones.\n\nAnd it is very likely there is no other for 2-5 and none at all for 2-6.\n\n\u2022 Didn't downvote, but this doesn't seem like an interesting result (or one that's particularly counterintuitive). It doesn't explain anything; it doesn't seem to mean anything; and it doesn't seem to have any consequences. Questions about what numbers look like in different bases are generally not very interesting ones. \u2013\u00a0anomaly Dec 7 '16 at 16:40\n\u2022 Also the condition is trivial for base 2, so \"bases up to 3\" is really just \"base 3\", and \"bases up to 4\" is really just \"bases 3 and 4\". \u2013\u00a0Ben Millwood Dec 7 '16 at 17:32\n\u2022 Of course the first few are trivial -- but the base-N notation here is not exactly necessary, you could just talk about sum of powers, it's a small convenience. Ie. 82000 is the smallest of numbers that can be written as a sum of powers of 3 and 4 and 5. That's what I find so odd, that somehow this odd number comes from the structure of the mathematical universe. The oddity here is like Ramanujan's taxi cab number. \u2013\u00a0chx Dec 7 '16 at 17:37\n\nSporadic simple groups in the Classification of finite simple groups\n\nhttps:\/\/en.wikipedia.org\/wiki\/Classification_of_finite_simple_groups\n\nI find the existence of the sporadic simple groups https:\/\/en.wikipedia.org\/wiki\/Sporadic_group like the Monster group https:\/\/en.wikipedia.org\/wiki\/Monster_group is pretty astounding considering how simple the definition of groups is.\n\nNot sure whether this is the kind of thing you were expecting, but here goes:\n\nSome statements about constructive mathematics can seem very counter-intuitive (at first, this is probably because one is misinterpreting what they mean), e.g.:\n\n\u2022 the induction principle holds, but on the other hand: that every non-empty (or inhabited) set of naturals has a smallest element is in general false\n\u2022 given a set $A$, consider the statements: (i) \"there is a finite set $B$ and an injection $A\\to B$\", (ii) \"there is a finite set $B$ and a surjection $B\\to A$\". None of the statements imply each other or that $A$ is finite\n\u2022 Can you explain the first one? It looks like a simple negation of the well-ordering principle. I don't see any other way to interpret it. \u2013\u00a0murgatroid99 Dec 5 '16 at 23:31\n\u2022 @murgatroid99 I don't know what you mean by \"explain\". In my second paragraph I meant: Even if you initially think it is surely true, you may have a faulty explanation. In the case of the well-ordering principle of integers something like: \"Of course, you just increment an integer starting at $1$ until you land in the set\" which doesn't work because it may be very hard to check that you are in this set. \u2013\u00a0Stefan Perko Dec 6 '16 at 8:27\n\u2022 I'm just trying to understand why that statement is false. Can you provide a counterexample, or a method of finding a counterexample, or some other proof that it's false. Even an explanation of why that statement is not actually equivalent to \"the well ordering principle if false\" would be helpful. \u2013\u00a0murgatroid99 Dec 6 '16 at 8:53\n\u2022 I understand now. When I first read your post, I completely missed the part where you were operating under an unusual set of logical axioms, and that was the source of my confusion. \u2013\u00a0murgatroid99 Dec 6 '16 at 16:15\n\u2022 For the first one, do you mean \"false\" or \"not provable\"? \u2013\u00a0Theodore Norvell Dec 10 '16 at 15:09\n\nHere are several involving the construction of regular polygons. For at least three out of the four cases we would not call it counterintuitive, but the results we know today would have thrown the Greeks for a loop.\n\n1) The regular heptagon has no Euclidean construction. The Greeks would have preferred to be able to construct regular polygons generally by Euclid's methods, which would make their properties accessible to elementary proof. Only in modern times did we definitely learn the bad news.\n\n2) The regular enneagon (nonagon) has no Euclidean construction, either. This is a special case of the old angle trisection problem: trisect the central angle of an equilateral triangle and you can make a regular enneagon. Now we know it works in reverse: we prove that a general angle trisection with Euclidean methods cannot exist by showing that the regular enneagon is not Euclidean-constructible.\n\n3) The regular hendecagon does have a neusis construction. Unlike the other cases, this is purely a \"modern\" counterintuitive. It was long thought that neusis construction was similar to conic-section construction: you can solve cubic and quartic equations with it. But we know now that some (we don t know about all) irreducible quintic equations are also solvable. Benjamin and Snyder showed that the minimal equation for $\\cos((2\\pi)\/11)$ is one such neusis-solvable equation. See Constructing the 11-gon by splitting an angle in five.\n\n4) After the regular pentagon, the next regular prime-sided polygon constructible by Euclid's methods is the 17-gon. It 's \"obvious\" now, but ancient and medieval mathematicians would never have suspected it without the theories developed by Gauss.\n\n\u2022 I don't think #4 is at all obvious. It's well known to be true, just like the fact that the speed of light is finite, but that's not the same as obvious. \u2013\u00a0Th\u00e9ophile Dec 7 '16 at 21:26\n\u2022 @Th\u00e9ophile: That's why Oscar put it in quotes. \u2013\u00a0user21820 Dec 25 '16 at 8:03\n\nHmm, not sure if this counts as counterintuitive, but what about the handshaking lemma, that the number of people at a party who shake hands an odd number of times is even?\n\n\u2022 -1. It's not counterintuitive to expect the divisors of an even number to be either even or paired with an even complement divisor. \u2013\u00a0Nij Dec 3 '16 at 0:49\n\u2022 Look I agree (even if understatement forbade me from proclaiming didactically), hence my caveat, I guess it's just a 'cute' result, the amusement owing more to linguistics than mathematics. \u2013\u00a0Mehness Dec 3 '16 at 1:30\n\u2022 @Nij : I'm inclined to disregard any claim of \"being intuitive\" from anyone who has absorbed any significant quantity of math or science. \"Intuitive\" only to a handful of percent of the population is not intuitive. \u2013\u00a0Eric Towers Dec 4 '16 at 2:15\n\u2022 The fact that even numbers must have an even divisor is taught to eight-year-olds. The fact that factor pairs must contain all of the prime factors of the product is taught to ten-\/eleven-year olds. It's intuitive to anybody who has had more math education than learning how to count and add. Defining \"counterintuitive\" to mean \"not immediately expected by a seven-year-old\" is frankly a terrible and pointless way to do it. \u2013\u00a0Nij Dec 4 '16 at 5:31\n\u2022 A set that is not open is not necessarily closed. A relation that is not symmetric is not necessarily anti-symmetric. A statement that is not intuitive is not necessarily counter-intuitive. :) \u2013\u00a0Th\u00e9ophile Dec 7 '16 at 21:18\n\n## protected by Community\u2666Dec 3 '16 at 13:52\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).","date":"2019-04-21 06:57:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6427440047264099, \"perplexity\": 653.418939516708}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-18\/segments\/1555578530253.25\/warc\/CC-MAIN-20190421060341-20190421082341-00193.warc.gz\"}"}
| null | null |
{"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php?title=2002_AMC_10B_Problems&diff=prev&oldid=38740","text":"# Difference between revisions of \"2002 AMC 10B Problems\"\n\n## Problem 1\n\nThe ratio $\\frac{2^{2001}\\cdot3^{2003}}{6^{2002}}$ is:\n\n$\\mathrm{(A) \\ } \\frac{1}{6}\\qquad \\mathrm{(B) \\ } \\frac{1}{3}\\qquad \\mathrm{(C) \\ } \\frac{1}{2}\\qquad \\mathrm{(D) \\ } \\frac{2}{3}\\qquad \\mathrm{(E) \\ } \\frac{3}{2}$\n\n## Problem 2\n\nFor the nonzero numbers $a, b,$ and $c,$ define $$(a,b,c)=\\frac{abc}{a+b+c}$$ Find $(2,4,6)$.\n\n$\\mathrm{(A) \\ } 1\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 6\\qquad \\mathrm{(E) \\ } 24$\n\n## Problem 3\n\nThe arithmetic mean of the nine numbers in the set $\\{9,99,999,9999,\\ldots,999999999\\}$ is a $9$-digit number $M$, all of whose digits are distinct. The number $M$ does not contain the digit\n\n$\\mathrm{(A) \\ } 0\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 6\\qquad \\mathrm{(E) \\ } 8$\n\n## Problem 4\n\nWhat is the value of\n\n$$(3x-2)(4x+1)-(3x-2)4x+1$$\n\nwhen $x=4$?\n\n$\\mathrm{(A) \\ } 0\\qquad \\mathrm{(B) \\ } 1\\qquad \\mathrm{(C) \\ } 10\\qquad \\mathrm{(D) \\ } 11\\qquad \\mathrm{(E) \\ } 12$\n\n## Problem 5\n\nCircles of radius $2$ and $3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.\n\n$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r1=3; real r2=2; real r3=5; pair A=(-2,0), B=(3,0), C=(0,0); pair X=(1,0), Y=(5,0); path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r3); fill(circleC,gray); fill(circleA,white); fill(circleB,white); draw(circleA); draw(circleB); draw(circleC); draw(A--X); draw(B--Y); pair[] ps={A,B}; dot(ps); label(\"3\",midpoint(A--X),N); label(\"2\",midpoint(B--Y),N); [\/asy]$\n\n$\\mathrm{(A) \\ } 3\\pi\\qquad \\mathrm{(B) \\ } 4\\pi\\qquad \\mathrm{(C) \\ } 6\\pi\\qquad \\mathrm{(D) \\ } 9\\pi\\qquad \\mathrm{(E) \\ } 12\\pi$\n\n## Problem 6\n\nFor how many positive integers $n$ is $n^2-3n+2$ a prime number?\n\n$\\mathrm{(A) \\ } \\text{none}\\qquad \\mathrm{(B) \\ } \\text{one}\\qquad \\mathrm{(C) \\ } \\text{two}\\qquad \\mathrm{(D) \\ } \\text{more than two, but finitely many}\\qquad \\mathrm{(E) \\ } \\text{infinitely many}$\n\n## Problem 7\n\nLet $n$ be a positive integer such that $\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{7}+\\frac{1}{n}$ is an integer. Which of the following statements is not true?\n\n$\\mathrm{(A) \\ } 2\\text{ divides }n\\qquad \\mathrm{(B) \\ } 3\\text{ divides }n\\qquad \\mathrm{(C) \\ } 6\\text{ divides }n\\qquad \\mathrm{(D) \\ } 7\\text{ divides }n\\qquad \\mathrm{(E) \\ } n>84$\n\n## Problem 8\n\nSuppose July of year $N$ has five Mondays. Which of the following must occurs five times in the August of year $N$? (Note: Both months have $31$ days.)\n\n$\\textbf{(A)}\\ \\text{Monday} \\qquad \\textbf{(B)}\\ \\text{Tuesday} \\qquad \\textbf{(C)}\\ \\text{Wednesday} \\qquad \\textbf{(D)}\\ \\text{Thursday} \\qquad \\textbf{(E)}\\ \\text{Friday}$\n\n## Problem 9\n\nUsing the letters $A$, $M$, $O$, $S$, and $U$, we can form five-letter \"words\". If these \"words\" are arranged in alphabetical order, then the \"word\" $USAMO$ occupies position\n\n$\\mathrm{(A) \\ } 112\\qquad \\mathrm{(B) \\ } 113\\qquad \\mathrm{(C) \\ } 114\\qquad \\mathrm{(D) \\ } 115\\qquad \\mathrm{(E) \\ } 116$\n\n## Problem 10\n\nSuppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2+ax+b=0$ has positive solutions $a$ and $b$. Then the pair $(a,b)$ is\n\n$\\mathrm{(A) \\ } (-2,1)\\qquad \\mathrm{(B) \\ } (-1,2)\\qquad \\mathrm{(C) \\ } (1,-2)\\qquad \\mathrm{(D) \\ } (2,-1)\\qquad \\mathrm{(E) \\ } (4,4)$\n\n## Problem 11\n\nThe product of three consecutive positive integers is $8$ times their sum. What is the sum of the squares?\n\n$\\mathrm{(A) \\ } 50\\qquad \\mathrm{(B) \\ } 77\\qquad \\mathrm{(C) \\ } 110\\qquad \\mathrm{(D) \\ } 149\\qquad \\mathrm{(E) \\ } 194$\n\n## Problem 12\n\nFor which of the following values of $k$ does the equation $\\frac{x-1}{x-2} = \\frac{x-k}{x-6}$ have no solution for $x$?\n\n$\\textbf{(A) } 1\\qquad \\textbf{(B) } 2\\qquad \\textbf{(C) } 3\\qquad \\textbf{(D) } 4\\qquad \\textbf{(E) } 5$\n\n## Problem 13\n\nFind the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$.\n\n$\\textbf{(A) } \\frac23 \\qquad \\textbf{(B) } \\frac32 \\text{ or } -\\frac14 \\qquad \\textbf{(C) } -\\frac23 \\text{ or } -\\frac14 \\qquad \\textbf{(D) } \\frac32 \\qquad \\textbf{(E) } -\\frac32 \\text{ or } -\\frac14$\n\n## Problem 14\n\nThe number $25^{64}\\cdot 64^{25}$ is the square of a positive integer $N$. In decimal representation, the sum of the digits of $N$ is\n\n$\\mathrm{(A) \\ } 7\\qquad \\mathrm{(B) \\ } 14\\qquad \\mathrm{(C) \\ } 21\\qquad \\mathrm{(D) \\ } 28\\qquad \\mathrm{(E) \\ } 35$\n\n## Problem 15\n\nThe positive integers $A$, $B$, $A-B$, and $A+B$ are all prime numbers. The sum of these four primes is\n\n$\\mathrm{(A) \\ } \\text{even}\\qquad \\mathrm{(B) \\ } \\text{divisible by }3\\qquad \\mathrm{(C) \\ } \\text{divisible by }5\\qquad \\mathrm{(D) \\ } \\text{divisible by }7\\qquad \\mathrm{(E) \\ } \\text{prime}$\n\n## Problem 16\n\nFor how many integers $n$ is $\\frac{n}{20-n}$ the square of an integer?\n\n$\\textbf{(A) } 1\\qquad \\textbf{(B) } 2\\qquad \\textbf{(C) } 3\\qquad \\textbf{(D) } 4\\qquad \\textbf{(E) } 10$\n\n## Problem 17\n\nA regular octagon $ABCDEFGH$ has sides of length two. Find the area of $\\triangle ADG$.\n\n$\\textbf{(A) } 4 + 2\\sqrt2 \\qquad \\textbf{(B) } 6 + \\sqrt2\\qquad \\textbf{(C) } 4 + 3\\sqrt2 \\qquad \\textbf{(D) } 3 + 4\\sqrt2 \\qquad \\textbf{(E) } 8 + \\sqrt2$\n\n## Problem 18\n\nFour distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?\n\n$\\textbf{(A) } 8\\qquad \\textbf{(B) } 9\\qquad \\textbf{(C) } 10\\qquad \\textbf{(D) } 12\\qquad \\textbf{(E) } 16$\n\n## Problem 19\n\nSuppose that $\\{a_n\\}$ is an arithmetic sequence with $$a_1+a_2+\\cdots+a_{100}=100 \\text{ and } a_{101}+a_{102}+\\cdots+a_{200}=200.$$ What is the value of $a_2 - a_1 ?$\n\n$\\mathrm{(A) \\ } 0.0001\\qquad \\mathrm{(B) \\ } 0.001\\qquad \\mathrm{(C) \\ } 0.01\\qquad \\mathrm{(D) \\ } 0.1\\qquad \\mathrm{(E) \\ } 1$\n\n## Problem 20\n\nLet $a, b,$ and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7.$ Then $a^2-b^2+c^2$ is\n\n$\\mathrm{(A) \\ } 0\\qquad \\mathrm{(B) \\ } 1\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 7\\qquad \\mathrm{(E) \\ } 8$\n\n## Problem 21\n\nAndy's lawn has twice as much area as Beth's lawn and three times as much as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?\n\n$\\mathrm{(A) \\ } \\text{Andy}\\qquad \\mathrm{(B) \\ } \\text{Beth}\\qquad \\mathrm{(C) \\ } \\text{Carlos}\\qquad \\mathrm{(D) \\ } \\text{Andy and Carlos tie for first.}\\qquad \\mathrm{(E) \\ } \\text{All three tie.}$\n\n## Problem 22\n\nLet $\\triangle{XOY}$ be a right-triangle with $m\\angle{XOY}=90^\\circ$. Let $M$ and $N$ be the midpoints of the legs $OX$ and $OY$, respectively. Given $XN=19$ and $YM=22$, find $XY$.\n\n$\\mathrm{(A) \\ } 24\\qquad \\mathrm{(B) \\ } 26\\qquad \\mathrm{(C) \\ } 28\\qquad \\mathrm{(D) \\ } 30\\qquad \\mathrm{(E) \\ } 32$\n\n## Problem 23\n\nLet $\\{a_k\\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is\n\n$\\mathrm{(A) \\ } 45\\qquad \\mathrm{(B) \\ } 56\\qquad \\mathrm{(C) \\ } 67\\qquad \\mathrm{(D) \\ } 78\\qquad \\mathrm{(E) \\ } 89$\n\n## Problem 24\n\nRiders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?\n\n$\\mathrm{(A) \\ } 5\\qquad \\mathrm{(B) \\ } 6\\qquad \\mathrm{(C) \\ } 7.5\\qquad \\mathrm{(D) \\ } 10\\qquad \\mathrm{(E) \\ } 15$\n\n## Problem 25\n\nWhen $15$ is appended to a list of integers, the mean is increased by $2$. When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$. How many integers were in the original list?\n\n$\\mathrm{(A) \\ } 4\\qquad \\mathrm{(B) \\ } 5\\qquad \\mathrm{(C) \\ } 6\\qquad \\mathrm{(D) \\ } 7\\qquad \\mathrm{(E) \\ } 8$","date":"2021-03-09 02:34:10","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 105, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7839152812957764, \"perplexity\": 273.7979096359876}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178385534.85\/warc\/CC-MAIN-20210308235748-20210309025748-00567.warc.gz\"}"}
| null | null |
{"url":"https:\/\/expressionengine.stackexchange.com\/questions\/11934\/compare-a-get-variable-with-a-path-of-categories","text":"# Compare a get variable with a path of categories\n\nI have a series of if:elseif statements that compare $_GET Variables using \"{exp:parameters:get name='price'}\"==\"<?php echo$myVariableHere; ?>\". This is filtering out some of the entries so that they don't show, this works perfectly.\n\nNOTE: $myVariableHere is coming from <?php$myVariableHere=\"{categories show_group='#'}{category_name}{\/categories}\"; ?>\n\nThe exception is my 'region' variable, because it uses child categories. I tried to call them this way, <?php $myVariableHere=\"array{categories show_group='#' backspace='2'}'{category_name}', {\/categories}\"; ?> The problem is that even though it appears to print to my page correctly ('Canada', 'Foo', 'Bar') it does not have separate keys in the array, they are all part of the 1st key. So I can not call on the categories separately to use them in my comparison. In the end, I need to compare the$_GET VARIABLE of ?region=Canada to the entries categories in that group ('Canada' with a child of 'Foo' with a child of 'Bar').\n\nLooking forward to any suggestions. NOTE: I have been using gw:categories in places.\n\n\u2022 I guess you had a look at low seg2cat for getting category info from URL segments? \u2013\u00a0J\u00e9r\u00f4me Coup\u00e9 Jul 16 '13 at 9:05\n\u2022 Thanks, sorry I didn't add the pre_code quotes in question, now added above. I'm using segments to call the entries within the channel entry, the \\$_GET Variables are created on the fly at the front end by the user selecting from a menu and the categories need the flexibility to be added in the entries also on the fly. Low seg2cat may not work with changing menu constantly changing menu options on the front end. Let me know if I am wrong. I have done a lot of work to now and changing it 'completely' may hurt my brain, but I do NEED to get it working, so... anything it worth a try. \u2013\u00a0JayB Jul 16 '13 at 13:55","date":"2021-04-22 11:47:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5018822550773621, \"perplexity\": 1486.7940715803754}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618039603582.93\/warc\/CC-MAIN-20210422100106-20210422130106-00515.warc.gz\"}"}
| null | null |
Q: PostgreSQL CASE statement syntax error I'm trying to follow the instructions in the PostgreSQL manual.
PostgreSQL: Documentation: 9.1: Control Structures
My PostgreSQL server is version 9.1.14 on Windows 32-bit.
The following SQL statement is unexpectedly resulting in a syntax error:
SELECT
CASE 1
WHEN 1,2 THEN 'x'
ELSE 'y'
END;
I'm expecting it to return 'x';
The more traditional code runs fine, however:
SELECT
CASE 1
WHEN 1 THEN 'x'
WHEN 2 THEN 'x'
ELSE 'y'
END;
A: You are using the CASE syntax as provided by the procedural language plpgsql. This is similar but not identical to the SQL CASE syntax. Here is the link to the SQL version of CASE.
Here you see, that 1,2 is not allowed, only a plain expression. So you could write:
SELECT
CASE
WHEN 1 in (1,2) THEN 'x'
ELSE 'y'
END;
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,741
|
The Ray–Dutt twist is a mechanism proposed for the racemization of octahedral complexes containing three bidentate chelate rings. Such complexes typically adopt an octahedral molecular geometry in their ground states, in which case they possess helical chirality. The pathway entails formation of an intermediate of C2v point group symmetry. An alternative pathway that also does not break any metal-ligand bonds is called the Bailar twist. Both of these mechanism product complexes wherein the ligating atoms (X in the scheme) are arranged in an approximate trigonal prism.
This pathway is called the Ray–Dutt twist in honor of Priyadaranjan Ray (not Prafulla Chandra Ray) and N. K. Dutt, inorganic chemists at the Indian Association for the Cultivation of Science abbr. IACS who proposed this process.
See also
Pseudorotation
Bailar twist
Bartell mechanism
Berry mechanism
Fluxional molecule
Indian Association for the Cultivation of Science (IACS)
References
Molecular geometry
Stereochemistry
Coordination chemistry
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,635
|
Q: classes to use for video steganography in java I have implemented Image and Audio steganography already in java and now i want to implement video steganography. But, i still haven't been able to find what are the classes that used to do this for e.g the classes that will allow me to view the video data in binary form(because i will be using lsb replacement technique).
A: My diploma last year final project was Video stenography which i developed using netbeans IDE i will post code here how i have done all the process to do it.
package Stegnography;
import java.io.File;
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.FileInputStream;
import java.io.FileOutputStream;
/**
*
* @author DeepRocks
*/
public class EmbProcess {
String embfilename;
public String emb(String s, String s1)
{
try{
File file = new File(s);
File file1 = new File(s1);
FileInputStream fileinputstream = new FileInputStream(s);
FileOutputStream fileoutputstream = new FileOutputStream("temp");
byte abyte0[] = new byte[8];
int i;
int k;
for(k = 0; (i = fileinputstream.read(abyte0, 0, 8)) > 0; k = i)
fileoutputstream.write(abyte0, 0, i);
fileinputstream.close();
for(int l = 1; l <= 8 - k; l++)
fileoutputstream.write(65);
fileoutputstream.write("DATAFILE".getBytes(), 0, 8);
System.out.println("File name==="+file1.getName());
StringBuffer stringbuffer = new StringBuffer(file1.getName());
stringbuffer.setLength(50);
fileoutputstream.write(stringbuffer.toString().getBytes(), 0, 50);
fileinputstream = new FileInputStream(s1);
int j;
while((j = fileinputstream.read(abyte0, 0, 8)) > 0)
fileoutputstream.write(abyte0, 0, j);
fileinputstream.close();
fileoutputstream.close();
file.delete();
File file2 = new File("temp");
file2.renameTo(file);
embfilename=file.getName();
}
catch(Exception e){
e.printStackTrace();
embfilename="";
}
return embfilename;
}
public String demb(String s)
{
boolean flag;
String demfile = "";
try
{
File file = new File(s);
String outpath=s.substring(0, s.lastIndexOf("\\")+1);
FileInputStream fileinputstream = new FileInputStream(s);
char c = '\b';
byte abyte0[] = new byte[c];
String s1 = "";
int i;
while((i = fileinputstream.read(abyte0, 0, c)) > 0)
{
s1 = new String(abyte0);
if(s1.equals("DATAFILE"))
break;
}
if(!s1.equals("DATAFILE"))
{
flag=false;
fileinputstream.close();
return demfile;
}
abyte0 = new byte[50];
fileinputstream.read(abyte0, 0, 50);
s1 = new String(abyte0);
String s2 = s1.trim();
String fpath = s2.substring(0, s2.lastIndexOf(".") + 1) + "enc";
System.out.println("fpath------"+fpath);
FileOutputStream fileoutputstream = new FileOutputStream(outpath+fpath);
c = '\u5000';
abyte0 = new byte[c];
while((i = fileinputstream.read(abyte0, 0, c)) > 0)
fileoutputstream.write(abyte0, 0, i);
fileinputstream.close();
fileoutputstream.close();
demfile=fpath;
}
catch(Exception exception)
{
demfile="";
exception.printStackTrace();
System.out.println(exception);
}
return demfile;
}
}**THIS CODE IS FOR EMBEDDING PROCESS**
Please clear that what you actually want, I will help you out just look at my code if it's helping you i will give you my project.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,226
|
vmshrink
========
files to capture shrinking the vm foot print
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,993
|
Q: LinkedIn API OAuth Problem I'm using the PEAR OAuth Class to access the LinkedIn developer API and I've come across a bit of a problem. I can authorize my application but when it comes to getting an accessToken I'm receiving this error:
Edit:
Code after Adam's suggestions
public function oauth_access()
{
session_start();
$token = $_GET['oauth_token'];
$verifier = $_GET['oauth_verifier'];
$secret = $_SESSION['trequest_token_secret'];
$key = "****";
$secret = "****";
$oauthc = new OAuth($key, $secret, OAUTH_SIG_METHOD_HMACSHA1, OAUTH_AUTH_TYPE_AUTHORIZATION);
$oauthc->setToken($token, $secret);
$oauthc->setNonce(rand());
try
{
$access_token_info = $oauthc->getAccessToken("https://api.linkedin.com/uas/oauth/accessToken");
$_SESSION['laccess_oauth_token']= $access_token_info['oauth_token'];
$_SESSION['laccess_oauth_token_secret']= $access_token_info['oauth_token_secret'];
$_SESSION['loauth_verifier'] = $verifier;
}
catch (OAuthException $e)
{
echo $e->getMessage();
}
}
But I'm now getting a different error:
Invalid auth/bad request (got a 401, expected HTTP/1.1 20X or a redirect)
A: You don't need to manually compute the signature, as pecl/oauth will do that for you.
Also, you're telling the library to pass the data in the Authorization HTTP header. That is a good place to have it. Then you are passing it via a query parameter. That is permitted, but less optimal. (You may actually be passing it in two places.) Also, pecl/oauth will automatically generate the proper timestamp.
When I first started, I found this blog post to be a good first start.
Or you can use the LinkedIn PHP library listed by Paul. It's also a good place to begin, if you don't want to reuse pecl/oauth because you're using that someplace else.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,730
|
Senator Marco Rubio, a rising Republican star from Florida who is considered a top candidate to be his party's vice presidential nominee, said Wednesday he would not take the job if offered. Probably.
Rubio, who represents a state that swings between voting for Republicans and Democrats in presidential elections, is often cited as a potential 2012 running mate for the Republican nominee because of the importance of his state and his Hispanic heritage. Hispanics are a powerful U.S. voting bloc.
Potential VP picks often play coy about the likelihood of being chosen, and Rubio tried to rule it out Wednesday, even as he tripped himself up doing it.
Asked at a "Washington Ideas Forum" conference whether he would turn down the position if offered, Rubio said: "Yeah, I believe so. I'm not going to be the...vice presidential nominee."
"The answer's going to probably be no," he continued, before correcting himself: "The answer's going to be no."
Republicans have not yet chosen a nominee to run against President Barack Obama, a Democrat, in next year's election. Former Massachusetts Governor Mitt Romney has held the front-runner position off and on for months.
Presidential candidates often choose running mates based on whether they can help win a specific state or help shore up a perceived weakness, political or otherwise.
Rubio praised the Senate, saying it was an important institution where much could be accomplished in public policy, and he indicated he would be happy to make his mark there.
"The United States Senate has provided the genesis for some of the greatest things that this country has ever done. And if I dedicate the time to it and seriousness to it, I have a chance to be a part of something like that," he said.
"You're never going to get to that stage if you're focused on it as some sort of a launch pad for something else."
Vice President Joe Biden served in the Senate for decades before joining Barack Obama on the Democratic presidential ticket in 2008.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,263
|
Q: Could I use this informal explanation for Hawking radiation? I need to explain Hawking radiation to my classmates for a homework. Could I use this informal explanation? (unfortunately without math). Correct me if I need to change/add something:
Quantum fluctuations on vacuum generates pairs of entangled virtual particles that "exists" only for a brief moment of time and then annihilate each other, turning into nothing, without violating any law. But when this pair is created NEAR (out) the event horizon of a black hole which has enormous gravitational force, those virtual particles are "splitted" apart and actually become "real" particles with real mass and energy, and one of them goes inside the black hole while the other one has gained sufficient energy to leave the attraction and escape. As the "real" particles pair separation was produced by the black hole's gravitational energy, and only "half" of the energy used was "recovered", the escape of one of the particles lowers the total mass of the black hole, causing them to "evaporate" on a long period of time.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,700
|
\section{Introduction}
The World Economic Forum (WEF) has ranked massive digital misinformation as one of the top global risks in 2013\footnote{http://reports.weforum.org/global-risks-2013}. Unfortunately, the foresight of WEF seems right as we encountered many unpleasant incidents due to the misinformation spread on the Internet since 2013 such as the gunfight due to ``Pizzagate" fake news\footnote{www.nytimes.com/2016/12/05/business/media/comet-ping-pong-pizza-shooting-fake-news-consequences.html} and increased mistrust towards vaccines\footnote{www.washingtonpost.com/news/wonk/wp/2014/10/13/the-inevitable-rise-of-ebola-conspiracy-theories}.
In order to combat against misinformation and its negative outcomes, fact-checking websites (e.g., Snopes\footnote{https://www.snopes.com/}) detect the veracity of claims spread over the Internet and share their findings with their readers~\cite{cherubini2016rise}. However, fact-checking is an extremely time-consuming process, taking around one day for a single claim~\cite{Hassan2017ClaimBusterTF}. While these invaluable journalistic efforts help to reduce the spread of misinformation, Vosoughi et al. \cite{vosoughi2018spread} report that false news spread eight times faster than true news.
Therefore, systems helping fact-checkers are urgently needed in the combat against misinformation.
As human fact-checkers are not able detect the veracity of all claims spread on the Internet, it is vital to spend their precious time in fact-checking the most important claims. Therefore, an automatic system monitoring social media posts, news articles and statements of politicians, and detecting the \emph{check-worthy} claims is needed. A number of researchers focused on this important problem (e.g., \cite{Hassan2017ClaimBusterTF,patwari2017tathya,jaradat2018claimrank}). Furthermore, Conference and Labs of Evaluation Forum (CLEF) Check That! Lab (CTL) has been organizing shared-tasks on detecting check-worthy claims since 2018 ~\cite{nakov2018overview,atanasova2019overview,BarrnCedeo2020CheckThatAC}. In CTL tasks, a political debate or a transcribed speech is separated by sentences and participants are asked to rank the sentences according to their priority to be fact-checked. In CTL'20 \cite{10.1007/978-3-030-58219-7_17}, tweets
have also been used for this task.
In this paper, we propose a ranking model that prioritizes claims based on their check-worthiness. We propose a BERT-based hybrid system
in which we first fine tune a BERT~\cite{devlin2019bert} model
for this task, and then use its prediction and other features we define in a logistic regression model to prioritize the claims.
The features we use include word-embeddings, presence of comparative and superlative adjectives, domain-specific controversial topics, and others. Our model achieves 0.255 and 0.176 mean average precision (MAP) scores in CTL'18 and CTL'19 datasets, respectively, outperforming all state-of-the-art models including participants of the corresponding shared-tasks, ClaimBuster~\cite{Hassan2017ClaimBusterTF}, BERT, XLNET \cite{yang2019xlnet}, and Lespagnol et al.\cite{lespagnol2019information}'s model. We share our code for the reproducibility of our results\footnote{https://github.com/YSKartal/political-claims-checkworthiness}.
\section{Related Work}\label{sec:rel}
As the US presidential election in 2016 is one of the main motivating reasons for fact-checking studies, prior work mostly
used debates and other speeches of US politicians as their datasets (e.g., \cite{Hassan2017ClaimBusterTF,lespagnol2019information}). Therefore, the majority of studies focused on English. The Arabic datasets used in prior work (\cite{jaradat2018claimrank,nakov2018overview})
are just translations of English datasets.
ClaimBuster~\cite{Hassan2017ClaimBusterTF} is one of the first studies about check-worthiness. ClaimBuster is a supervised model using many features including part-of-speech (POS) tags, named entities, sentiment, and TF-IDF representations of claims.
TATHYA~\cite{patwari2017tathya}
uses topics, POS tuples, entity history, and bag-of-words as features. The topics are detected by LDA model trained on transcripts of all
presidential debates from 1976 to 2016.
Gencheva et al. \cite{gencheva2017context} propose a neural network model with a long list of sentence level and contextual features including sentiment, named entities,
word embeddings, topics, contradictions, and others.
Jaradat et al. \cite{jaradat2018claimrank} use roughly the same features with Gencheva et al., but extend the model for Arabic. In its followup work, Vasileva et al. \cite{vasileva2019takes} propose a multi-task learning model to detect whether a claim will be fact-checked by at least five (out of nine) pre-selected reputable fact-checking organizations.
CLEF has been organizing Check That! Labs (CTL) since 2018.
Seven teams participated in check-worthiness task of CTL'18.
The participant teams used various learning models such as recurrent neural network (RNN)~\cite{Hansen2018TheCT}, multilayer perceptron~\cite{Zuo2018AHR}, random forest (RF)~\cite{DBLP:conf/clef/AgezBLPM18}, k-nearest neighbor (kNN)~\cite{DBLP:conf/clef/GhanemMPR18a} and Support Vector Machine (SVM)~\cite{DBLP:conf/clef/YasserKE18} with different sets of features such as bag-of-words~\cite{Zuo2018AHR}, character n-gram~\cite{DBLP:conf/clef/GhanemMPR18a}, POS tags~\cite{Zuo2018AHR,Hansen2018TheCT,DBLP:conf/clef/YasserKE18}, verbal forms~\cite{Zuo2018AHR},
named entities~\cite{Zuo2018AHR,DBLP:conf/clef/YasserKE18},
syntactic dependencies~\cite{Zuo2018AHR,Hansen2018TheCT}, and word embeddings~\cite{Zuo2018AHR,Hansen2018TheCT,DBLP:conf/clef/YasserKE18}. On English dataset, Prise de Fer~\cite{Zuo2018AHR} team achieved the best MAP scores using almost every feature mentioned before with SVM-Multilayer perceptron learning.
In 2019, 11 teams participated in check-worthiness task of CTL'19. Participants used varying models such as LSTM,
SVM, naive bayes, and logistic regression (LR) with many features including
readability of sentences and their context
~\cite{atanasova2019overview}. Copenhagen team~\cite{DBLP:conf/clef/Hansen0SL19} achieved the best overall performance using syntactic dependency and word embeddings with weakly supervised LSTM model.
Lespagnol et al. \cite{lespagnol2019information} investigated using various learning models such as SVM, LR, and Random Forests,
with a long list of features including word-embeddings, POS tags, syntactic dependency tags, entities, and ``information nutritional" features which represent factuality, emotion, controversy, credibility, and technicality of statements.
In our experiments we show that our model outperforms Lespagnol et al. on both test collections.
Our proposed an approach distinguishes from the existing studies as follows. 1) We propose a BERT-based hybrid model which uses fine-tuned BERT's output with many other features.
2) As the topic might be a strong indicator for check-worthiness, many studies used various types of topics such as general topics
\cite{DBLP:conf/clef/YasserKE18}, globally controversial topics \cite{lespagnol2019information}, and topics discussed in old US presindential debates~\cite{patwari2017tathya}. However, we believe that check-worthiness of a claim depends on local and present controversial topics. Thus, we use a list of hand-crafted controversial topics related to US elections. 3) We also use two different sets of features including a hand-crafted list of words and presence of comparative and superlative adjectives and adverbs.
\section{Proposed Approach}\label{sec:approach}
We propose a supervised model with a number of features described below.
We investigate various learning models including LR, SVM, random forest,
MART \cite{Friedman2001GreedyFA}, and LambdaMART \cite{10.1007/s10791-009-9112-1}. Now we explain the features we use.
\textbf{BERT:} We first fine tune BERT using respective training data. Next, we use its prediction value as one of our features.
\textbf{Word Embeddings (WE):}
Words that are semantically and syntactically similar tends to be close in the embedding space, allowing us to capture similarities between claims.
We represent a sentence as the average vector of its words excluding the out-of-vocabulary ones.
Word embedding vectors are extracted from the pre-trained word2vec model~\cite{mikolov2013efficient} which has a feature vector size of 300.
\textbf{Controversial Topics (CT):}
Sentences about controversial topics might include check-worthy claims. Lespagnol et al. \cite{lespagnol2019information} use a list of controversial issues compiled from Wikipedia article ``Wikipedia:List\_of\_controversial\_issues". However, the list they use covers many controversial issues which have very limited coverage in current US media such as ``Lebanon", ``Chernobyl", and ``Spanish Civil War" while the data we use are about recent US politics. We believe that controversy of a topic depends on the society. For instance, US politicians propose different policies for immigrants, yielding heated discussions among them and their supporters. On the other hand, US domestic politics are much less interested in refugee crisis in Mediterranean sea than European countries. Therefore, a claim about Mexican immigrants might be check-worthy for people living in US while they might find claims about refugees taking a dangerous path to reach Europe not-check-worthy. In contrast, people living in Europe might consider the latter case as check-worthy and the former one as not-check-worthy.
In addition, controversy of a topic might change over time. For instance,
Cold War (which also exists in that Wikipedia list) might be one of the most discussed topics in US politics before the collapse of the Soviet Union in 1991. However, nowadays it is rarely covered by US media. Therefore, we propose using controversial issues related to the data we use, instead of any controversial issue around the globe and in the history.
Firstly, we identified 11 major topics in current US politics including immigration, gun policy, racism, education, Islam, climate change, health policy, abortion, LGBT, terror, and wars in Afghanistan and Iraq. For each topic, we identified related words and calculate the average of these words using their word embedding vectors.
For instance, for the immigration topic, we used words ``immigrants", ``illegal", ``borders", ``Mexican", ``Latino" and ``Hispanic".
In this feature set of size 11, we calculate cosine similarity between sentences and each topic by using their vector presentation. We use the average of word embeddings for sentences excluding stopwords with NLTK~\cite{Loper02nltk:the}.
\textbf{Comparative \& Superlative (CS):}
Politicians frequently use sentences comparing themselves with others because each candidate tries to convince the public that s/he is better than his/her opponent. Therefore, the comparisons in political speeches might impact people's voting decision and, thereby, it might be important to check their veracity.
Thus, in this feature, we use the number of comparative and superlative adjectives and adverbs in sentences.
\textbf{Handcrafted Word List (HW):}
Particular words convey important information about check-worthiness because 1) it might be related to an important topic (e.g., ``unemployment"), 2) it represents a numerical value, increasing the factuality of the sentence (e.g., ``percent") and 3) its semantic represents a comparison between two cases (e.g., ``increase" and ``decrease"). Thus, we first identified 66 words analyzing training datasets of CTL'18 and CTL'19. In this feature, we check whether there is an overlap between lemmas of selected words and lemmas of words in the respective sentence.
\textbf{Verbe Tense (VT):} We cannot detect the veracity of claims about future while we can only verify claims about the present or past. Thus, the verbe tense of sentences might be an effective indicator for check-worthiness of claims. This feature vector
represents the existence or absence of each tense in the predicate of the claims.
\textbf{Part-of-speech (POS) Tags:}
If a sentence does not contain any informative words, then it is less likely to be check-worthy. To represent the information load of a claim, we use the number of nouns, verbs, adverbs and adjectives, separately.
\section{Experiments}\label{sec:experiments}
\subsection{Experimental Setup}
\textbf{Implementation:} We use ktrain library\footnote{https://pypi.org/project/ktrain/} to fine-tune BERT model
with 1 cycle learning rate policy and maximum learning rate of 2e-5~\cite{Smith2018ADA}. We use SpaCy\footnote{https://spacy.io/} for all syntactic and semantic analyses.
We use Scikit toolkit\footnote{https://scikit-learn.org} for the implementations of SVM, Random Forest (RF), and LR. The parameter settings of the learning algorithms are as follows. We use default parameters for SVM. We set the number of trees to 50 and the maximum depth to 5 for RF. We use multinomial and lbfgs settings for LR.
For MART and LambdaMART models, we use RankLib\footnote{https://sourceforge.net/p/lemur/wiki/RankLib/} library, and set the number of trees and leaves to 50 and 2, respectively.
\noindent
\textbf{Data:} We evaluate the performance of our system with two datasets used in CTL'18 and CTL'19. The details about them are given in \textbf{Table~\ref{table_dataset}}. CTL'18 consists of transcripts of debates and speeches while CTL'19 contains also press conferences and posts.
\begin{comment}
\begin{table*}[h!]
\begin{center}
\caption{Details about CTL'18 and CTL'19 datasets.
}
\label{table_dataset}
\begin{tabular}{|p{1.35cm}|p{1.3cm}|c|c| c||c|c|c|} \hline
& & \multicolumn{3}{c||}{\textbf{CTL'18}} & \multicolumn{3}{c|}{\textbf{CTL'19}} \\ \cline{3-8}
\textbf{Type} & \textbf{Partition} & \textbf{\# Docs} & \textbf{\# Sentence} & \textbf{\# CW Claims} & \textbf{\# Docs} & \textbf{\# Sentence} & \textbf{\# CW Claims}\\ \hline
\textbf{Debates} & Train & 3 & 4,064 & 90 & 8 & 10,648 & 256 \\
& Test & 2 & 2,815 & 94 & 4 & 4,584 & 46 \\ \hline
\textbf{Speeches} & Train & - & - & - & 8 & 2,718 & 282 \\
& Test & 5 & 2,067 & 98 & 2 & 1,883 & 50 \\ \hline
\textbf{Press} & Train & - & - & - & 2 & 3,011 & 36 \\
& Test & - & - & - & 1 & 612 & 14 \\ \hline
\textbf{Posts} & Train & - & - & - & 1 & 44 & 8 \\
& Test & - &- &- & - & - & - \\ \hline \hline
\textbf{Total } & Train & 3 & \textbf {4,064} & \textbf{90 (2,2\%)} & 19 & \textbf {16,421} & \textbf{433 (2,6\%)}\\
& Test & 7 &\textbf{4,882} & \textbf{192 (3,9\%)} & 7 & \textbf{7,079} & \textbf{110 (1,6\%)} \\ \hline
\end{tabular}
\end{center}
\end{table*}
\end{comment}
\begin{table}[h!]
\begin{center}
\caption{Details about CTL'18 and CTL'19 datasets.
}
\label{table_dataset}
\begin{tabular}{|c|l|c|c| c||c|c|c|} \hline
& \textbf{} & \textbf{CTL'18} & \textbf{CTL'19} \\ \hline
& \textbf{\# Docs} & 3 & 19 \\
\textbf{Train} & \textbf{\# Sentence} & {4,064} & {16,421} \\
& \textbf{\# CW Claims} & {90 (2,2\%)} & {433 (2,6\%)} \\ \hline
& \textbf{\# Docs} & 7 & 7 \\
\textbf{Test}& \textbf{\# Sentence} & {4,882} & {7,079} \\
& \textbf{\# CW Claims} & {192 (3,9\%)} &{110 (1,6\%)} \\ \hline
\end{tabular}
\end{center}
\end{table}
\noindent
\textbf{Baselines:} We compare our model against the following models.
\begin{itemize
\item \textit{Lespagnol et al. \cite{lespagnol2019information}}: Lespagnol et al. report the best results on CTL'18 so far. Therefore, we use it as one of our baselines. In order to get its results for CTL'19, we contacted with the authors to get their own code. The authors provide us the values of ``information nutrition" features and instructions about how to generate WE embeddings. We implemented their method using the values they shared and following their instructions\footnote{It is noteworthy that we obtain 0.2115 MAP score on CTL'18 with our implementation of their method while they report 0.23 MAP score in their paper. We are not aware of any bug in our code but the performance difference might be because of different versions of the same library. Nevertheless, the results we present for their method on CTL'19 should be taken with a grain of salt.}.
\item \textit{ClaimBuster}: We use the popular pretrained ClaimBuster API\footnote{https://idir.uta.edu/claimbuster/} \cite{Hassan2017ClaimBusterTF} which is trained on a dataset covering different debates that do not exist on CTL'18 and CTL'19.
\item \textit{BERT}: As it is reported that BERT based models outperform state-of-the-art models in various NLP tasks, we compare our model against using only BERT. We fine tune BERT model using the respective training dataset and predict the check-worthiness of claims using the fine-tuned model
\item \textit{XLNET}: It is reported that XLNet outperfroms BERT in various NLP tasks \cite{yang2019xlnet}. Thus, we use XL-NET for this task by fine-tuning with the respective training dataset.
\item \textit{Best of CTL'18 and CTL'19}: For each dataset, we also report the performance of best systems participated in the shared-tasks, i.e., Prise de Fer team~\cite{Zuo2018AHR} and Copenhagen team~\cite{DBLP:conf/clef/Hansen0SL19} for CTL'18 and CTL'19, respectively.
\end{itemize}
\noindent
\textbf{Training \& Testing:} We use the same setup with CTL'18 and CTL'19 to maintain a fair comparison with the baselines.
We follow the evaluation method used on CTL'18 and CTL'19: We calculate average precision (AP),
R-precision (RP), precision@5 (P@5) and precision@10 (P@10) for each file (i.e., debate, speech) and then report the average performance.
\subsection{Experimental Results}
In this section, we present experimental results on test data using different sets of features and varying learning algorithms.
\noindent
\textbf{Comparison of Learning Algorithms.} In our first set of experiments, we evaluate logistic regression (LR), SVM, random forest (RF), MART and LambdaMART models using all features defined in Section \ref{sec:approach}. \textbf{Table~\ref{tab:test:models}} shows MAP scores of each model. Interestingly, LR outperforms all other models. In a similar experiment Lespagnol et al.\cite{lespagnol2019information} conducted, they also report that LR yields higher results than other models they used. Nevertheless,
we use LR in our following experiments.
\begin{table}[h!]
\begin{center}
\caption{\textbf{MAP Score for Varying Models Using All Features}}
\label{tab:test:models}
\begin{tabular}{|c|c|c|}
\hline
\textbf{Learning Model} & \textbf{ CTL'18 } & \textbf{ CTL'19 } \\
\hline
\textbf{LR} & \textbf{.2303} & \textbf{.1775} \\ \hline
\textbf{RF} & .1468 & .1542 \\ \hline
\textbf{SVM} & .1716 & .1346 \\ \hline
\textbf{MART} & .1764 & .1732 \\ \hline
\textbf{Lambda MART} & .0671 & .0564 \\ \hline
\end{tabular}
\end{center}
\end{table}
\noindent
\textbf{Feature Ablation.}
In order to analyze the effectiveness of features we use, we apply two techniques: 1) \emph{Leave-one-out methodology} in which we exclude one type of feature group and calculate the model's performance without it, and 2) \emph{Use-only-one methodology} in which only a single feature group is used for prediction.
The results are shown in \textbf{Table~\ref{tab:test:feature}}.
\begin{comment}
\begin{table*}[htb]
\begin{center}
\caption{\textbf{AP, PR, P@5 and P@10 Scores of the Logistic Regression considering different
groups of features}}
\label{tab:test:feature}
\begin{tabular}{|c|c|c|c|c||c|c|c|c|}
\hline
& \multicolumn{4}{c||}{\textbf{CTL'18}} & \multicolumn{4}{c|}{\textbf{CTL'19}} \\ \cline{2-9}
\textbf{Model} & \textbf{ AP} & \textbf{ RP }& \textbf{P@5 } & \textbf{P@10} & \textbf{ AP} & \textbf{ RP }& \textbf{P@5 } & \textbf{P@10}\\ \hline
\textbf{All} & {.2303} & .2325 & \textbf{.4000} & {.3286} & .1775 & .1960 & {.2286} & {.2286} \\ \hline
\textbf{All-CS} & {.2239} & \textbf{.3136} & .3714 & {.3143} & .1765 & {.1963} & {.2286} & .2286 \\ \hline
\textbf{All-BERT} & {.2211} & .2159 & .3429 & .3143 & .1580 & .1799 & {.2286} & .2000 \\ \hline
\textbf{All-T} & \textbf{.2547} & .2579 & \textbf{.4000} & \textbf{.3429} & .1761 & .2028 & \textbf{.2571} & .2143 \\
\hline
\textbf{All-HW} & {.2126} & .2139 & .3429 & {.2857} & .1727 & .1895 & \textbf{.2571} & .2286 \\ \hline
\textbf{All-WE} & {.1756} & .1761 & .2000 & {.1429} & \textbf{.1786} & .2071 & \textbf{.2571} & .2429 \\ \hline
\textbf{All-CT} & {.2170} & .2195 & .3714 & {.3000} & .1739 & \textbf{.2093} & {.2286} & .2286 \\ \hline
\textbf{All-POS} & {.2283} & .2487 & .3714 & {.3286} & .1767 & .2081 & {.2286} & \textbf{.2571} \\ \hline
\end{tabular}
\end{center}
\end{table*}
\end{comment}
\begin{table*}[htb]
\begin{center}
\caption{\textbf{MAP Scores for Varying Feature Sets}}
\label{tab:test:feature}
\begin{tabular}{|l|c|c||l|c|c|}
\hline
\multicolumn{3}{|c||}{ \textbf{Leave-One-Out} } & \multicolumn{3}{c|}{ \textbf{Use-Only-One} } \\ \hline
\textbf{Features} & \textbf{ \textbf{CTL18}} & \textbf{ \textbf{CTL19}} & \textbf{Features} & \textbf{ \textbf{CTL18}} & \textbf{ \textbf{CTL19}} \\ \hline
\textbf{All} & {.2303} & .1775 & & & \\ \hline
\textbf{All-CS} & {.2239} & .1765 & \textbf{CS} & .751 & .604 \\ \hline
\textbf{All-BERT} & {.2211} & .1580 & \textbf{BERT} & {.1850} & \textbf{.1701} \\ \hline
\textbf{All-VT} & \textbf{.2547} & .1761 & \textbf{VT} & {.1007} & {.598} \\ \hline
\textbf{All-HW} & {.2126} & .1727 & \textbf{HW} & .1530 & .1043 \\ \hline
\textbf{All-WE} & {.1756} & \textbf{.1786} & \textbf{WE} & \textbf{.2068} & {.1356} \\ \hline
\textbf{All-CT} & {.2170} & .1739 & \textbf{CT} & .1363 & .1046 \\ \hline
\textbf{All-POS} & {.2283} & .1767 & \textbf{POS} & .1048 & .631 \\ \hline
\end{tabular}
\end{center}
\end{table*}
\begin{comment}
\begin{table}[htb]
\begin{center}
\caption{\textbf{MAP Score for each feature}}
\label{tab:test:feature2}
\begin{tabular}{|l|c|c|}
\hline
\textbf{Feature} & \textbf{ CTL'18 } & \textbf{ CTL'19 } \\
\hline
\textbf{VT} & {.1007} & {.598} \\ \hline
\textbf{CT} & .1363 & .1046 \\ \hline
\textbf{CS} & .751 & .604 \\ \hline
\textbf{WE} & \textbf{.2068} & {.1356} \\ \hline
\textbf{HW} & .1530 & .1043 \\ \hline
\textbf{POS} & .1048 & .631 \\ \hline
\textbf{BERT} & {.1850} & \textbf{.1701} \\ \hline
\end{tabular}
\end{center}
\end{table}
\end{comment}
From the results in Table \ref{tab:test:feature}, we see that features have different effects on each dataset. BERT is the most effective feature on CTL'19. However, in contrast to our expectations, WE seems more effective feature than BERT on CTL'18.
On CTL'18,
the performance decreases by nearly 25\% when WE is excluded. In addition, we achieve the highest MAP score when we use only WE. On CTL'19, we achieve 0.1356 MAP score using only WE, showing that it is more effective than other features except BERT. However, the performance of our model increases when we exclude WE (0.1775 vs. 0.1786 in Table~\ref{tab:test:feature}), suggesting that the information it contributes is covered by other features on CTL'19.
\begin{table*}[!htb]
\begin{center}
\caption{\textbf{Comparison with Competing Models.
* sign indicates the results obtained from our implementation of the respective competing model.}}
\label{tab:test}
\begin{tabular}{|l|c|c|c|c||c|c|c|c|}
\hline
& \multicolumn{4}{c||}{\textbf{CTL'18}} & \multicolumn{4}{c|}{\textbf{CTL'19}} \\ \cline{2-9}
\textbf{Model} & \textbf{ MAP} & \textbf{ RP }& \textbf{P@5 } & \textbf{P@10} & \textbf{ MAP} & \textbf{ RP }& \textbf{P@5 } & \textbf{P@10}\\ \hline
\textbf{BERT} & {.1850} & .2218 & .3142 & {.2857} & .1701 & .1945 & \textbf{.2571} & \textbf{.2429} \\ \hline
\textbf{XLNET} & {.1974} & .2393 & .2857 & {.2571} & .0932 & .0770 & {.1429} & {.1143} \\ \hline
\textbf{Lespagnol et al. \cite{lespagnol2019information} } & {.230} & .254 & .314 & .2857* & .1292* & .1347* & {.1714*} & .2000* \\ \hline
\textbf{Prise de Fer Team} & {.1332} & .1352 & .2000 & {.1429} &-&-&-&-\\ \hline
\textbf{Copenhagen Team} & - & - & - & - & .1660 & \textbf{.4176} & \textbf{.2571} & .2286 \\ \hline
\textbf{ClaimBuster} & {.2003} & .2162 & .2571 & {.2429} & .1329 & {.1555} & {.1714} & .2000 \\ \hline
\hline
\textbf{Our Model} & \textbf{.2547} & \textbf{.2579} & \textbf{.4000} & \textbf{.3429} & \textbf{.1761} & .2028 & \textbf{.2571} & .2143 \\ \hline
\end{tabular}
\end{center}
\end{table*}
Excluding hand-crafted word list (HW) features causes performance decrease in both test collections. In addition, using only HW features outperforms all participants of CTL'18 (0.153 vs 0.1332 in Table~\ref{tab:test:feature}). These promising results suggest that expanding this list might lead further performance increases.
Our results also suggest that Controversial Topics (CT) are effective features. Excluding them decreases the performance of the model in both collections while using only CT features yield high scores, slightly outperforming the best performing system on CTL'18 (0.1363 vs. 0.1332 in Table~\ref{tab:test:feature}).
Excluding CS and POS features also slightly decrease the performance of the model in both test collections. Regarding time tense features, our results are mix. Excluding time tense feature causes a slight performance decrease on CTL'19, but yields higher performance score on CTL'18.
\noindent
\textbf{Comparison Against Baselines.}
We pick the model that includes all features except VT
as our primary model because it achieves the highest MAP score on average. We compare our primary model with the baselines. The results are presented in \textbf{Table~\ref{tab:test}}.
Our proposed model outperforms all other models based on all evaluation metrics on CTL'18. On CTL'19, our proposed model achieves the highest MAP score, which is the official metric used in CTL. BERT model outperforms other models based on P@10 on CTL'19. Regarding P@5 metric, our model, BERT and Copenhagen Team achieve the same highest scores with 0.2571. Regarding RP, Copenhagen Team achieves the highest score.
Overall, our model outperforms all other models based on the official evalution metric of CTL while BERT and Copenhagen Team~\cite{Hansen2018TheCT} also achieve comparable performance on CTL'19.
\begin{table*}[htb]
\small
\begin{center}
\caption{Highest ranked non check-worthy statements from each test document by our primary model}
\label{tab:qualitative}
\begin{tabular}{|c|c|p{2.55cm}|c|p{8cm}|}
\hline
\textbf{Row} & \textbf{Rank} & \textbf{File Name} & \textbf{ Speaker } & \textbf{ Statement } \\
\hline
1& 4 & task1-en-file1 & CLINTON & The plan he has will cost us jobs and possibly lead to another Great Recession. \\
\hline
2& 1 & task1-en-file2 &CLINTON & Then he doubled down on that in the New York Daily News interview, when asked whether he would support the Sandy Hook parents suing to try to do something to rein in the advertising of the AR-15, which is advertised to young people as being a combat weapon, killing on the battlefield. \\
\hline
3& 1 & task1-en-file3 &TRUMP & Jobs, jobs, jobs. \\
\hline
4& 2 & task1-en-file4 &TRUMP & Before that, Democrat President John F. Kennedy championed tax cuts that surged the economy and massively reduced unemployment. \\
\hline
5& 3 & task1-en-file5 &TRUMP & The world's largest company, Apple, announced plans to bring \$245 billion in overseas profits home to America. \\
\hline
6& 1 & task1-en-file6 &TRUMP & America has lost nearly-one third of its manufacturing jobs since 1997, following the enactment of disastrous trade deals supported by Bill and Hillary Clinton. \\
\hline
7& 1 & task1-en-file7 &TRUMP & Our trade deficit in goods with the world last year was nearly \$800 billion dollars. \\
\hline
8&1 & 20151219\_3\_dem &O'MALLEY & We increased education funding by 37 percent. \\
\hline
9&1 & 20160129\_7\_gop &KASICH & We're up 400,000 jobs. \\
\hline
10&1 & 20160311\_12 \_gop &TAPPER & Critics say these deals are great for corporate America's bottom line, but have cost the U.S. at least 1 million jobs. \\
\hline
11&3 & 20180131\_state \_union &TRUMP & Unemployment claims have hit a 45-year low. \\
\hline
12&1 & 20181015\_60\_min &TRUMP & --if you think about it, so far, I put 25\% tariffs on steel dumping, and aluminum dumping 10\%. \\
\hline
13&3 & 20190205\_trump \_state &TRUMP & Unemployment for Americans with disabilities has also reached an all-time low. \\
\hline
14&1 & 20190215\_trump \_emergency &TRUMP & They have the largest number of murders that they've ever had in their history - almost 40,000 murders. \\
\hline
\end{tabular}
\end{center}
\end{table*}
\section{Qualitative Analysis}\label{sec:qual}
In this section, we present our qualitative analysis for the output of our primary model.
For each input file, we rank the claims based on their check-worthiness and then detect not-check-worthy claim with the highest rank. \textbf{Table~\ref{tab:qualitative}} shows these not-check-worthy statements for each file with our system's ranking and speaker of the statement.
The statement in Row 1 is a claim about the future. Our model with verb tense could rank this statement at lower ranks but our primary model does not use verb tense features because it yields lower performance on average. In Row 2, the statement is very complex with many relative clauses, in perhaps decreasing the performance of BERT model and WE features in representing the statement. In Row 3, our model makes an obvious mistake and ranks a statement which does not have even any predicate, at very high ranks. Perhaps our model falls short because the word ``jobs" indicates that the statement is about unemployment, which is one of the controversial topics we defined.
As reported by Vasileva et al. \cite{vasileva2019takes} fact-checking organizations investigate different claims with very minimal overlaps between selected claims. We observe this subjective nature of annotations in Rows 4-14 because all statements are actually factual claims and some of them might also be considered as check-worthy. For instance, statements in Row 8, 11 and 13 are clearly said to change people's voting decision. In addition, almost all statements are about economics which is an important factor on people's votes. Therefore, checking their veracity might be also important not to misinform public. Nevertheless, these examples show the
the subjective nature of check-worthiness annotations.
In addition to subjective judgments, we also noticed inconsistencies within the annotations. For instance, the statement in Row 9 (``We are up 400,000 jobs") also exists in ``20160311\_12\_gop" file but annotated as ``check-worthy". In addition, there exists semantically very similar statements with different labels. For instance, Donald Trump's statement ``I did not support the war in Iraq" in $1079^{th}$ line of 20160926\_1pres file is labeled as ``not-check-worthy" while his statement in $1086^{th}$ line of the same file ``I was against the war in Iraq" is labeled as ``check-worthy". Both statements have similar meanings and exists in the same context (i.e., their position in file are very close). Therefore, both might have the same labels. As a counter argument, ``being against" suggests an action while ``not supporting" does not require any action to be taken. Thus, different annotations for similar statements might also be again due to the subjective nature of check-worthiness judgments.
Furthermore, there are also annotations that we strongly disagree with the label. For instance, in 20170315\_nashville file (training data on CTL'19), Donald Trump's statement ``We're going to put our auto industry back to work" is labeled as check-worthy. However, the statement is about future and cannot be verified.
Overall, our qualitative analysis suggests that annotating check-worthiness of claims is a subjective task and the annotations might be noisy. Kutlu et al. \cite{kutlu2018crowd} show that using text excerpts within documents as rationales help understanding disagreements in relevance judging. Similarly, we might request rationales behind check-worthiness annotations to
understand if the label is due to a human judging error or the subjective nature of the annotation task. Furthermore, rationales behind these annotations might help us develop
effective solutions for this challenging problem.
\section{Conclusion} \label{sec:conc}
In this paper, we presented a supervised method which prioritize claims based on check-worthiness. We use logistic regression classifier with features including state-of-the-art language model BERT, domain-specific controversial topics, pretrained word embeddings, handcrafted word list, POS tags and comparative-superlative clauses. In our experiments on CTL'18 and CTL'19, we show that our proposed model outperforms all state-of-the-art models in both collections. We show that BERT's performance can be increased by using additional features for this task. In our feature ablation study, BERT model and word embeddings appear to be the most effective features while handcrafted word list and domain-specific controversial topics also seem effective. Based on our qualitative analysis, we believe requesting rationales for the check-worthiness annotations would further help in developing effective systems.
In the future, we plan to work on weak supervision techniques to extend the training dataset. With the increased data, we will be able explore using deep learning techniques for this task.
In addition, we plan to extend our study to detect check-worthy claims in social media platforms because it is the channel where most of the people affected by misinformation.
Moreover, working on different languages and building a multilingual model is an important research direction in the combat against misinformation.
\bibliographystyle{abbrv}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,496
|
Microsoft MSF for Agile 2013 Plus Security Development Lifecycle (SDL) is a TFS process template that incorporates the SDL for Agile process guidance into the MSF Agile development framework. With the Microsoft MSF for Agile 2013 Plus Security Development Lifecycle (SDL) template, any code checked into the Team Foundation Server 2013 source repository by the developer is analyzed to ensure that it complies with SDL secure development practices. The template also automatically creates security workflow tracking items for manual SDL processes such as threat modeling to ensure that these important security activities are not accidentally skipped or forgotten.
Download the accompanying installation file and install it on your Microsoft Visual Studio Team Foundation Server 2013 system. Additional required installation steps are listed in the installation's readme file.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,111
|
SYNONYM
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
in de Candolle & Lamarck, Fl. franç. (Paris), Edn 3 6: 76 (1815)
#### Original name
Uredo carbo DC., 1815
### Remarks
null
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,128
|
Home > Best Registered Agent Services > Maine
6 Best Registered Agent Services in Maine to Try in 2023
Top 6 Best Agent Services in Maine (January 2023)
1 - Northwest Registered Agent Service (Editor's Choice)
2 - ZenBusiness
3 - Incfile
4 - LegalZoom
5 - Rocket Lawyer
6 - My Company Works
How Can You Benefit By Working With The Best Maine Registered Agent Service?
1. Ensures You Comply
2. Comes with Peace of Mind
3.Safeguards Privacy
4. Availability
What is the Cost of A Registered Agent Service in Maine?
What Are the Requirements of a Registered Agent in Maine?
BEST Registered Agent in Maine Is...
Opening an LLC in Maine may be one of your goals. If so, it's crucial to comprehend the legal procedures for starting a business there. Working with a qualified registered agent is imperative to get your LLC up and going.
The best-registered agent in Maine can reasonably protect your private assets and actively assist you observe the laws of Maine. But it can be very challenging to find and hire a suitable registered agent alone.
So, our team of experienced legal advisors and attorneys conducted extensive research on the top registered agent companies. We were able to identify the best registered agent services in Maine to inform your decision.
Northwest Registered Agent takes pride in their excellent quality of services and competitive pricing. We are confident that they are the most satisfactory agent service in Maine.
Northwest provides top-notch customer service and commits to safeguarding its clients' privacy. Additionally, they take significant measures to guarantee that you stay on top of your state filings.
Other Maine registered agents we tested charged extra for compliance notices and a customized filing calendar, but Northwest's standard service included both.
Basic package ($125): You can get the most out of the company's Maine registered agent services. This package comes with Privacy by Default®, Local Corporate Guides®, complimentary mail scanning, forwarding, and same-day document scanning in your local area. To open your Maine LLC with Northwest, you must pay a $39 price tag plus the filing fee.
Registered Agent Service ($0): You should get the first year of your LLC formation package for free! After that, you'll start paying the yearly rate of $125.
Excellent client service and support
Well-versed in their 20 years of existence
An entire year of free registered agent services
ZenBusiness prides itself on offering excellent customer service and affordable packages. This makes it one of the best commercial registered agent services in Maine.
ZenBusiness provides a low-cost $99 charge for its Maine registered agent services.
With an easy-to-use website, we could utilize their helpful software for tracking and archiving our company's annual reports.
Registered agent service ($99): Standard registered agent service. Charged annually.
LLC formation package ($39 + state fee): The formation package includes free registered agents service for a year, after which it is $119 per year. ZenBusiness also has an additional Complete Coverage Upgrade feature, which adds a "Worry-Free Compliance" guaranteeing annual filing report notification services, and two unpaid filing amendments per year.
Realistic and clear price scheme
Extended chat customer support
Has earned positive reviews from customers
Limited brand recall
IncFile is known for its reliable services. They have provided registered agent services to more than 150,000 customers since 2004. Many business owners also find their Learning Center helpful.
International companies can get the primary Maine street address for their business entity via Incfile's virtual mailbox in addition to a commercial registered agent office location.
We found their web dashboard very helpful, easy to understand and use, and they proved very fast in processing services.
Registered agents service package ($119): Comes with a free LLC formation plus other perks like an easy-to-use registered agent dashboard and email and SMS notifications, so you'll always be in the know.
LLC formation ($0 + state fee): You only need to pay $145/$175 Maine filing fees for your first year.
They have a user-friendly platform
Every formation package comes with a cost-free agent service
Quick processing of orders
There are no extra charges
Has few alternatives for assistance
LegalZoom provides brand recognition, higher pricing, and a strong presence in the registered agents' services industry.
They ensure you're informed of filing deadlines and any forms you need to submit to the Maine Secretary of State.
Their prices are higher than the competition's, but you may feel more at ease using the services because of their expertise and brand recognition.
Registered agent service ($299): Provide a comprehensive registered agent service with features like a compliance calendar, enterprise data protection, and unlimited cloud storage. Also Comes with identity theft protection worth up to one million dollars!
LLC formation package ($79 + state fees): Standard LLC formation package includes all the basic legal services.
Reputable service provider in the country
High customer volume
Provides other legal services
Relatively expensive
LLC formation doesn't come with a registered agent
Rocket Lawyer is a popular online legal service offering various legal options for company and individual needs.
Whether you are looking for simple document templates or need ongoing assistance with complex legal requirements, this company can help.
Rocket Lawyer offers a premium membership for $39.99 per month, giving access to various premium features and a 25% discount on the company's registered agents service.
If you are still not sure about using Rocket Lawyer, they offer a 7-day free trial so that you can try out their premium features and get a sense of how the company works.
Maine Registered Agents Service ($149.99 for non-subscribers): They are offering 25% off for their premium subscribers.
Maine LLC Formation ($99.99 for non-subscribers): Non-members pay $149.99 more for registered agents service since it's not included in the package. New premium members enjoy free registered agents service but pay 25% off onwards.
Comes with a bundle of online legal services
Free 30-minute lawyer consultation for premium members
Joining without legal documents can cost you more
MyNewCompany.com, Inc. has been up and running since 2011 and is committed to providing smooth and quick service.
Their satisfaction guarantee is 100%, and you're entitled to a 90-days free refund.
Other companies offer free registered agents service for the initial year. However, MyCompanyWorks only offers free registered agent services if you purchase either the entrepreneur package or the complete package. There are no additional fees for the next year.
Maine Registered Agents Service ($99): If you want access to this service, you can choose the $199 Entrepreneur Package. It's not part of the introductory rate, but you'll pay $99 for add-on service.
Maine LLC Formation package ($59 plus state fee of $120): Excludes registered agents service. You may have to pay an additional cost of $99.
Excellent customer feedback
Reliable and efficient services
Knowledgeable and friendly customer service team
The basic package doesn't provide Maine Registered Agent service
Unfavourable pricing
As a business owner in Maine, your company's primary point of contact with the state should be your registered agent. They can receive and deliver any official correspondence. Your Maine registered agent service can ensure that the company complies with state laws.
While you can be your agent, using a registered agent for your LLC has several benefits [1]:
Your Maine registered agent can ensure that your business follows all the regulations of Maine and that nothing is overlooked.
Focusing on growing your business is easier with a Maine registered agent in charge. You can trust that important documents are monitored, and you'll be notified immediately if any are sent.
Don't stress about your physical address being disclosed to the public when you set up your LLC with a Maine registered agent service!
Instead of you having to provide a street address, using a high-quality registered agent service can provide it for you. Any paperwork that comes into your LLC can be delivered to the registered office address [2].
Registered agent services are ideal for business owners in Maine who observe irregular work hours, travel often, or have other commitments that don't allow them to come into the office.
Using Maine registered agent service as a business owner is an excellent choice since it is accessible during business hours.
Looking Up an LLC in Maine
How to End an LLC in Maine
The best Maine registered agent services can cost between $49 to $349 per year, depending on the level of service required. Many of these services also provide add-ons.
A registered agent in Maine ( clerk) is responsible for the following duties under Maine law (Title 5: Chapter 6-A: Model Registered Agents Act, 114):
You must be 18 years of age and above and can receive documents for your business.
Send notifications upon receipt of service of process or official mail.
Be available during regular business hours at a physical mailing address in-state to accept legal documents (service of process) and official mail.
Keep the registered office address and contact information updated.
Why Do I Need a Registered Agent for My LLC in Maine?
A registered agent is a mandatory procedure by the state for business owners when starting an LLC. All Maine LLCs, corporations, and nonprofits must choose and maintain a professional Maine registered agent service and office.
What Happens if My Maine Registered Agent Is Unable to Meet Their Duties?
Failure of your Maine registered agent to meet their duties could be detrimental to your company. For instance, a misplaced legal document could result in legal action against your business without your knowledge.
How Do I Choose a Registered Agent for My LLC in Maine?
Find a reliable agent who is trustworthy and knowledgeable about the legal requirements in the state of Maine. You should also consider their availability and fees. When researching potential agents, make sure to check their reviews and references to ensure you're finding the right fit for your business.
Can I Change the Registered Agent of My LLC at Any Time?
Yes, you can change the registered agent of your LLC at any time by filling out a Statement of Appointment for a $35 filing fee (this fee varies depending on the business entity type). It may take a few business days to complete the process.
Am I Allowed to be my Own Registered Agent for My LLC in Maine?
Yes, you can be the registered agent for your LLC in Maine as long as you qualify to be a Maine registered agent. You need a physical address in Maine and be available during regular business hours. Being your registered agent lets you control your LLC's activities.
Working with the best registered agent service with many years of experience can save you a great deal of stress.
That is precisely why our top pick is Northwest Registered Agent service. They can handle all legal documents and compliance standing professionally.
Northwest registered agent provides high-level privacy protection and all other services required to sustain your LLC. The compliance notifications and customized filing calendar included in the standard package significantly helped us stay on track with state filings and came with peace of mind.
Northwest Registered Agent Service
7 Best LLC Services in Maine (2023) Picked by Customers
How to Look Up an LLC in Maine? (The EASY Way)
How to Dissolve an LLC in Maine? (All You Need to Know)
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 29
|
Q: Exception while creating initiated flow [errorCode=1i4tu3y] When I try to respond to a flow, I get the error:
Exception while creating initiated flow [errorCode=1i4tu3y] statemachine.SingleThreadedStateMachineManager.onSessionInit - Exception while creating initiated flow [errorCode=1i4tu3y]
java.lang.SecurityException: sealing violation: package com.example is sealed
at java.net.URLClassLoader.getAndVerifyPackage(URLClassLoader.java:399) ~[?:1.8.0_181]
at java.net.URLClassLoader.definePackageInternal(URLClassLoader.java:419) ~[?:1.8.0_181]
at java.net.URLClassLoader.defineClass(URLClassLoader.java:451) ~[?:1.8.0_181]
at java.net.URLClassLoader.access$100(URLClassLoader.java:73) ~[?:1.8.0_181]
at java.net.URLClassLoader$1.run(URLClassLoader.java:368) ~[?:1.8.0_181]
at java.net.URLClassLoader$1.run(URLClassLoader.java:362) ~[?:1.8.0_181]
at java.security.AccessController.doPrivileged(Native Method) ~[?:1.8.0_181]
at java.net.URLClassLoader.findClass(URLClassLoader.java:361) ~[?:1.8.0_181]
at java.lang.ClassLoader.loadClass(ClassLoader.java:424) ~[?:1.8.0_181]
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:349) ~[?:1.8.0_181]
at java.lang.ClassLoader.loadClass(ClassLoader.java:411) ~[?:1.8.0_181]
at java.lang.ClassLoader.loadClass(ClassLoader.java:357) ~[?:1.8.0_181]
at java.lang.Class.forName0(Native Method) ~[?:1.8.0_181]
at java.lang.Class.forName(Class.java:348) ~[?:1.8.0_181]
A: Corda has recently enabled jar-sealing by default, this means that packages are no longer allowed to span multiple cordapps.
To fix this, make sure that no classes are defined as being in the same package, whilst being located in different cordapps.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,108
|
function m=maximal_matching(A,varargin)
% MAXIMAL_MATCHING Compute a maximal matching
%
% A maximal matching is a subset of edges where each vertex is incident on
% only one edge and the size of the matching cannot be increased by adding
% an edge left in the graph.
%
% m=maximal_matching(A) returns a maximal matching where m(v) = u and m(u)
% = v if edge (u,v) is in the matching and m(v) = 0 if v is not matched.
%
% There are two algorithms for computing a maximal matching: a greedy
% algorithm that simply adds an edge (u,v) to the matching unless m(u) or
% m(v) != 0; and an extra greedy algorithm that first sorts the edges in
% the graph by their degree. See the options.algname option.
%
% This method works on undirected graphs (symmetric matrices) and ignores
% edge weights.
% The runtime is O(M log N).
%
% To change the algorithm, use the optional algname argument.
% options.algname: the greedy algorithm to use ['greedy' | {'extra_greedy'}]
%
% See the MATCHING function for additional calling information. This function
% just calls matching(...,struct('initial_match',options.algname,...
% 'augmenting_path','none','verify',0)) and does _not_ verify the output,
% so there is no output v.
%
% See also MATCHING, TEST_MATCHING
%
% Example:
% load('graphs/matching_example.mat');
% m=maximal_matching(A)
% sum(m>0)/2 % maximal matching cardinality, should be < 8
% mmax=matching(A);
% sum(mmax>0)/2 % maximum matching cardinality, should be 8
% David Gleich
% Copyright, Stanford University, 2007-2008
%% History
% 2007-07-09: Initial coding
% 2008-10-07: Changed options parsing
%%
options = merge_options(struct(),varargin{:});
options.augmenting_path = 'none';
options.verify = 0;
if isfield(options,'algname'), options.initial_match = options.algname;
else options.initial_match='extra_greedy'; end
m = matching(A,options);
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,833
|
{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE ScopedTypeVariables #-}
module HaskellWorks.Data.Json.Standard.Load.Partial
( loadPartial
, loadPartialWithCsPoppyIndex
, loadPartialWithIndex
) where
import HaskellWorks.Data.BalancedParens.Simple
import HaskellWorks.Data.Bits.BitShown
import HaskellWorks.Data.Json.Internal.PartialIndex
import HaskellWorks.Data.Json.PartialValue
import HaskellWorks.Data.Json.Standard.Cursor.Generic
import HaskellWorks.Data.Json.Standard.Cursor.Load.Raw
import HaskellWorks.Data.RankSelect.CsPoppy
loadPartialWithIndex :: String -> IO JsonPartialValue
loadPartialWithIndex filename = do
(jsonBS, jsonIb, jsonBp) <- loadRawWithIndex filename
let cursor = GenericCursor jsonBS (BitShown jsonIb) (SimpleBalancedParens jsonBp) 1
let !jsonResult = jsonPartialJsonValueAt (jsonPartialIndexAt cursor)
return jsonResult
loadPartialWithCsPoppyIndex :: String -> IO JsonPartialValue
loadPartialWithCsPoppyIndex filename = do
(jsonBS, jsonIb, jsonBp) <- loadRawWithIndex filename
let cursor = GenericCursor jsonBS (makeCsPoppy jsonIb) (SimpleBalancedParens jsonBp) 1
let !jsonResult = jsonPartialJsonValueAt (jsonPartialIndexAt cursor)
return jsonResult
loadPartial :: String -> IO JsonPartialValue
loadPartial = loadPartialWithCsPoppyIndex
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 330
|
Clinically isolated syndrome suggestive of multiple sclerosis: voxelwise regional investigation of white and gray matter.
Diffuse WM damage not accompanied by any change in GM or WM volume is observed in patients with CIS. This suggests that WM involvement plays a relevant role in the early phases of MS. Subsequently detected GM damage may be secondary to WM alterations.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,029
|
{"url":"https:\/\/www.physicsforums.com\/threads\/adding-sinusoidal-functions-using-phasors.744480\/","text":"# Adding sinusoidal functions using phasors\n\n1. Mar 21, 2014\n\n### Vishera\n\n1. The problem statement, all variables and given\/known data\n\n2. Relevant equations\n\n3. The attempt at a solution\n$$3cos(20t+10\u00b0)-5cos(20t-30\u00b0)\\\\ =3\\angle 10\u00b0-5\\angle -30\u00b0\\\\ =-1.376+3.0209j\\\\ =3.32\\angle -65.51\u00b0$$ In the last step, the textbook actually got $3.32\\angle 114.49\u00b0$.\n\nI checked both answers and it seems that the textbook's answer is correct. All I did to get -65.51\u00b0 is using the arctan function on my calculator. It looks like their angle is 180\u00b0 greater than that. How did they choose the correct angle? I'm rusty on my trigonometry.\n\n2. Mar 21, 2014\n\n### psparky\n\nDon't phasors need to be put in RMS before you work with them in this case?\n\nLast edited: Mar 21, 2014\n3. Mar 21, 2014\n\n### Staff: Mentor\n\nThe arctan() function doesn't discriminate between the cases where the minus sign is to be associated with the numerator or the denominator of the argument; it only sees the overall sign of the single number it is given. To check what quadrant the correct angle should lie in, do a quick sketch of the x (real) and y (imaginary) values on the complex plane. Adjust the result by 180\u00b0 if required.\n\nIf your calculator has an atan2() function, use that instead as it takes two arguments, one for the x component and one for the y, and sorts out the correct angle automatically. Failing that, your calculator might have rectangular to polar conversion built-in, which will also give the result in the correct quadrant automatically.\n\n4. Mar 21, 2014\n\n### SteamKing\n\nStaff Emeritus\nDraw a picture of your combined phasor. See the quadrant where the components locate the resultant?\n\nGenerally, the inverse trig functions on your calculator only return the principal angles, which for the tangent is in the range -\u03c0\/2 \u2264 \u03b8 \u2264 \u03c0\/2. When doing these types of calculations, you must give your result a separate check to make sure you obtain the correct angle.\n\n5. Mar 21, 2014\n\n### Vishera\n\nThanks for help guys. I completely forgot this part of trigonometry and I get it now. It's actually kind of scary that I forgot this because it's not like I stopped doing Math. Too much calculus has made me a bit rusty on my algebra (or at least trigonometry).\n\n6. Mar 21, 2014\n\n### psparky\n\n-1.376 + 3.02J never equals -65 degrees.....check your math there...it always equals 114.49 degrees.\n\n7. Mar 21, 2014\n\n### Vishera\n\nWeird. My calculator is always giving me a principal value of -65.\n\n8. Mar 21, 2014\n\n### Staff: Mentor\n\nYes. Because you are calculating:\n$$\\phi = arctan\\left( \\frac{3.02}{-1.376} \\right)$$\nWhich, after resolving the argument to -2.195, the arctan() function places in the fourth quadrant. It cannot distinguish $\\frac{3.02}{-1.376}$ from $\\frac{-3.02}{1.376}$ or $-2.195$, and can only return a corresponding angle that lies within its principal range of +\/- 90\u00b0 .\n\n9. Mar 22, 2014\n\n### psparky\n\nTrig is certainly one way, another is simply using the rectangular to polar function on your calculator.\nOr do them both. If they both agree, better yet.","date":"2017-12-14 23:13:42","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6720867156982422, \"perplexity\": 1063.5522206544297}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-51\/segments\/1512948551162.54\/warc\/CC-MAIN-20171214222204-20171215002204-00453.warc.gz\"}"}
| null | null |
Q: how to poulate the dropdownlist from jquery ajax success callback I'm new to jQuery and Javascript. I have the below data from jquery ajax sucess callback.
["Man","KFC","fsdfds","ETA","Star","SCity ","TESTGB"]
Now, I want to populate this data on success callback.
success: function(data) {
// Call this function on success
alert(data);
var builder_arr=data.split(',');
for (var i=0;i<builder_arr.length;++i)
{
builder_arr[i]+':'+builder_arr[i];
}
}
I don't know how to populate it inside the select
How can I do this?
A: jsFiddle
HTML
<select id="mydropdown">
</select>
JS
function onSuccess(data) {
$.each(data, function (index, value) {
$('#mydropdown').append($('<option/>', {
value: value,
text : value
}));
});
}
var data = ["Man","KFC","fsdfds","ETA","Star","SCity ","TESTGB"];
onSuccess(data);
Note: This is just a best guess given the above.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,891
|
Q: How to sleep in WindowsPhone after changing control color I have button which navigate to next PortraitPage.xaml. I would like to change color of this button after Tap and then after 1s navigate to next page, so the User could see change. If I use code like this:
button.Background = RED;
Sleep(1000);
NavigateToNextPage();
the Background property of the button doesn't change.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,852
|
Entre los turcos se llama imaret a unas hospederías a donde iban jóvenes de las escuelas y los estudiantes de los colegios a tomar su alimento.
También acudían a estas hospederías gran número de necesitados, a quienes se les daba cada día un plato de carne y otro de legumbres más una limosna en dinero ( a mediados del siglo XIX consistía hasta 10 aspros o un real de vellón).
Casi todos los califas y príncipes otomanos emplearon grandes caudales en la fundación de estas hospederías en las principales ciudades del Imperio otomano (en Constantinopla se mantenían a 30.000 personas diariamente).
El primer Imaret de los dominios turcos fue el de Nicea, erigido por Orcano I.
Referencias bibliográficas
Diccionario histórico enciclopédico, 1830, Barcelona, Tomo III, página 195
Bibliografía complementaria
Roger, A..-El Imperio Otomano, Madrid, Atlas, 1943
Goodwin, J..-Los Señores del horizonte: una historia del imperio otomano, Madrid, Alianza, 2006
Ortiz de Urbina, I..-Nicea y Constantinopla, Vitoria, 1969
Hoteles por tipo
Sociedad del Imperio otomano
Beneficencia
Cultura del Imperio otomano
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,754
|
\section{Introduction}
It seems to have been Max Born who first referred to the waves of probability
(\textit{Wahrscheinlichkeitswellen}) which released from the usual binds of
reality have provided an elegant and for practical purposes highly successful
explanation of the phenomenon of self interference (Born \cite{born}).
Yet these probability waves are no less mysterious than the mystery they were
invoked to explain. \ It can hardly be doubted that as a particle passes
through an interferometer, such as the Mach-Zehnder of Fig. 1, something
wave-like must be moving through both arms. \ But\ I will contend here for an
ontologically less interesting explanation of this wave-like effect. \
I will assume that the particle itself follows a single and well-defined path
at all times, and that its choice of that path is determined, not in the
intrinsically probabilistic manner supposed by standard quantum mechanics
(SQM), but in accordance with underlying physical microprocesses. \ Insisting
on the strict and local application of laws of conservation, I will show that
any change in the wave characteristics of the particle as it adopts that path
must be accompanied by a wave-like disturbance of equal but opposite effect in
the scattering medium of the apparatus.\
As each particle is measured, there will be thus two wave systems evolving
through the experiment, precisely coordinated but of complementary effect
-\ the wave-like particle, and the correspondingly wave-like response of the
apparatus to the scattering of that particle. \ My contention will be that as
this response propagates through the experiment, it mimics the presence of a
further version or versions of the particle itself\medskip\medskip
\ \ \ \includegraphics[width=8.0cm]{Mach-Zehnder.eps}\medskip\medskip
\begin{quotation}
\textbf{Fig. 1} \ A Mach-Zehnder interferometer: \ In standard quantum
mechanics, the probability wave divides at beam splitter $BS1$ and self
interference occurs as the partial waves recombine at $BS2$. \ As now
proposed, the scattering of a photon into one or other arm at $BS1$ induces by
reaction a fluctuation in the reradiation field, and it is this that
interferes with the photon at $BS2$.\medskip\medskip
\end{quotation}
This response will constitute a microscopic imbalance in the scattering medium
of the apparatus. \ The tendency of the apparatus to resist the accumulation
of imbalance will be the source of the Born probabilities. \ The evolution of
the imbalance through the experiment will explain the seemingly probabilistic
alternative states that have been discerned in quantum measurement.
The difficulties that have arisen in SQM from the notion of an
observer-instigated wave function collapse are well illustrated by
Schr\"{o}dinger's \textit{reductio} of the unobserved cat that is at once dead
and alive (Schr\"{o}dinger \cite{schrodinger}), and by the iteration of the
problem of the cat in the paradox of Wigner's friend (Wigner \cite{wigner}).
\ Those difficulties will be avoided here, not merely by obviating the need
for an observer, as is the aim of, for example, the objective collapse and
many worlds reinterpretations of quantum mechanics, but by eliminating the
Schr\"{o}dinger evolution of alternative probabilistic states that has been
thought to necessitate that collapse.
In developing the argument, I will refer to simple beam splitters operating by
refraction and to a Mach-Zehnder interferometer assembled from such beam
splitters. \ Despite the mysteries of\ the quantum, the phenomenon of
refraction is itself relatively non-controversial, and by concentrating on the
refraction of photons, I may avoid the suggestion of \textquotedblleft new
physics\textquotedblright, and here at least, the complications of the de
Broglie wave, of which something has been said elsewhere (Shanahan
\cite{shanahan1} and \cite{shanahan2}). \
The Mach-Zehnder is not the famous instance of self interference. \ But this
conceptually simple interferometer will allow the demonstration of an illusion
- the apparent ability of an indivisible particle to be in two places at once
- \ an illusion induced, as I will show, by conservation, quantization, and
the wave-like nature of the elementary particles.
\section{Conservation and measurement}
At the level of the quantum, a measurement apparatus does not measure any
individual particle. \ It must proceed by forcing (projecting) the particle
into one or other of the eigenmodes defined by the apparatus, typically
eigenmodes of some property to be measured. \ In the limit of large numbers, a
beam of particles tends to separate between those modes so as to conserve that
property in the measured beam, this being, as will be discussed in Sect. 4,
the basis of the Born probabilities. \ By processing a sufficient number of
particles in this way, and guided by those probabilities, something may thus
be learned of the components of the property in the original beam.
But unless the incident particle was already in an eigenmode of the apparatus,
conservation is not observed in the particle itself. \ Indeed it was at one
time proposed on high authority (see Bohr, Kramers\ and Slater \cite{bks})
that conservation must be merely approximate or \textquotedblleft
statistical\textquotedblright\ in microscopic processes. \ That suggestion was
withdrawn following experimental confirmation of the conservation of momentum
in scattering processes (the Bothe \cite{bothe}\ and Compton \cite{compton}
experiments). \ Conservation is now more usually regarded as a meta-principle
against which an otherwise promising proposal might be judged and found
wanting. \ Certainly, a close attention to conservation has proved crucial on
occasion to the understanding\ of quantum phenomena (see, for instance,
Bloembergen \cite{bloembergen}).
Yet in according roles in measurement to chance and nonlocality, SQM seems
careless of the conserved properties of physics. \ The energy of a
superposition of waves, and thus of interacting particles, is determined by
their relative phase and degree of overlap. \ It is not explained how energy
is to be conserved if such a superposition evolves discontinuously or
nonlocally. \ \ The arbitrariness of such an evolution is difficult to
reconcile with the symmetries contemplated by Noether's theorems, and would
seem to deny the local conservation and continuity supposed by the gauge
principles of modern field theories (see, for instance, Ryder \cite{ryder},
Chap. 3)
I will assume in this paper, not only that laws of conservation are observed
exactly in quantum measurement, but that movements in the properties conserved
develop through the process of measurement in the local and deterministic
manner that was supposed by classical physics. \ From that assumption, I will
show that if a strict accounting is kept of those movements, the notion of a
collapsing probability wave becomes redundant to the operation of conserved
microprocesses. \
Consider the interaction of a single particle with an ideal 50:50
beamsplitter. \ According to SQM, the probability wave associated with the
particle (in more formal terms, the wave function or state vector) divides at
the beam splitter in accordance with conservation as
\begin{equation}
\left\vert \psi_{in}\right\rangle ->\frac{1}{\sqrt{2}}(\,\left\vert \psi
_{I}\right\rangle +\left\vert \psi_{II}\right\rangle \,),\label{probdiv
\end{equation}
where $\left\vert \psi_{in}\right\rangle $ is the amplitude of the probability
wave of the incoming particle, while $\left\vert \psi_{I}\right\rangle $ and
$\left\vert \psi_{II}\right\rangle $ are those of the partial waves exiting
ports $I$ and $II$ respectively.\
In dividing in this way, the probability wave has entered the so-called
Schr\"{o}dinger phase in which each measurement possibility (component of the
wave function) evolves through the experiment in a local and deterministic
manner until the occurrence of an observation (a measurement). \ According to
the collapse postulate of SQM, the wave function then collapses in an
intrinsically probabilistic, discontinuous and nonlocal manner into one or
other of the two possible measurement outcomes ($\left\vert \psi
_{I}\right\rangle $ or $\left\vert \psi_{II}\right\rangle $).\
If conservation is to continue following this collapse, there must be a
reaction of equal but opposite effect to the change in the particle, and if
this reaction is to accord with local causality, it must occur as and where
the change in the particle is effected, that is to say, in the
apparatus\footnote{It might be thought that the role of the apparatus in
measurement was settled by von Neumann in his acclaimed \textit{Mathematische
Grundlagen der Quantenmechanik} \cite{neumann} \ \ But the "pointer" states of
the apparatus contemplated there by von Neumann were correlated with the
changed state of the particle rather than, as in the response of the apparatus
considered here, anticorrelated with the change in the particle. The
significance of the pointer in von Neumann's analysis (see in particular
his\ Chap. VI) was that it is some such visual indication, rather than the
microscopic particle that, by impinging on the consciousness of the observer,
collapses the wave function (according to von Neumann).}. \ On including this
reaction, Eqn. (\ref{probdiv} ) becomes
\begin{equation}
\psi_{in}->\frac{1}{\sqrt{2}}[(\left\vert \psi_{I}\right\rangle +\Delta
A_{I})+(\left\vert \psi_{II}\right\rangle +\Delta A_{II})],\label{comb
\end{equation}
where $\Delta A_{I}$ is the change that would occur in the apparatus if the
photon were to exit port $I$, and $\Delta A_{II}$ is the corresponding change
if it were to exit port $II$.
This response by the apparatus must compensate for what is lost by the
particle in one component of the measured property and gained in the other.
\ In a 50:50 beamsplitter, and if the particle takes path $I$, the response
may be expressed formally as,
\begin{equation}
\Delta A_{I}=\frac{1}{\sqrt{2}}(\left\vert \psi_{II}\right\rangle -\left\vert
\psi_{I}\right\rangle ),\label{respI
\end{equation}
\ and if it takes path $II$
\begin{equation}
\Delta A_{II}=\frac{1}{\sqrt{2}}(\left\vert \psi_{I}\right\rangle -\left\vert
\psi_{II}\right\rangle ),\label{respII
\end{equation}
where in either case the negative sign denotes a reduction in amplitude or
equivalently a phase opposed to that of the particle\footnote{From the
symmetry under time reversal of Maxwell's equations, it may be argued,
following Stokes (see Hecht \cite{hecht}, chap. 4.6.3 and 4.10), that the
component of the response of the medium propagating in the same channel as the
particle must be of opposite phase to that particle.}. \
With the reaction of the apparatus now brought into the account, self
interference can be explained without recourse to the convoluted rigmarole of
division and collapse supposed by SQM. \ It is only necessary to assume that
as the particle encounters the scattering medium of the beam splitter, it does
not enter the probabilistic superposition supposed by SQM, but is forced
immediately toward one or other of the two exit ports of the apparatus. \
As this occurs the reaction of the apparatus, evolving through the apparatus
in the same manner as the photon has two effects. \ Along the path not taken
by the photon, it mimics the effect of a divided photon. \ Along the path that
is taken by the photon, it diminishes the effect of the photon so as to
complete the illusion of a divided photon. On the particle exiting port $I$,
the result must be, from Eqn. (\ref{respI})
\[
\left\vert \psi_{I}\right\rangle +\Delta A_{I}=\frac{1}{\sqrt{2}}(\left\vert
\psi_{I}\right\rangle +\left\vert \psi_{II}\right\rangle ),
\]
and on its exiting path $II$, likewise from Eqn. (\ref{respII}), \
\[
\left\vert \psi_{II}\right\rangle +\Delta A_{II}=\frac{1}{\sqrt{2}}(\left\vert
\psi_{I}\right\rangle +\left\vert \psi_{II}\right\rangle ),
\]
thus avoiding all necessity for the division and collapse supposed by SQM. \
And that is what will be supposed here - that particle and apparatus adopt
their measured states in a local and causal manner as the one interacts with
the other, rather than belatedly and retrospectively following the collapse of
the wave function, a collapse that could in principle occur much later, or in
the absence of particle detectors, never at all.
The argument has relied crucially, of course, on the assumption that the
response of the apparatus evolves along the same paths and in the same
wave-like manner as the particle itself. \ That this must be so is suggested
by the detailed and continuing requirements of conservation. \ SQM assumes
that the probability wave divides in accordance with conservation. \ If
conservation is to continue following collapse, when only one of the two
possible measurement outcomes has been realized, the response of the apparatus
must supply what is missing and must continue to do so as the system evolves.
However, nothing has yet been said of the microprocesses underlying this
wave-like response. \ I will consider in the next section a class of
scattering processes, ubiquitous in quantum measurement, in which there can be
no doubt that the response of the medium does indeed propagate though the
experiment in the same wave-like manner as the particle itself.
\section{The response of the medium}
Before confronting the mysteries of measurement directly, it will be
instructive to consider the response of a scattering medium where the
scattered particle is allowed no choice of path - where there is no suggestion
of wave function collapse and accordingly no mystery at all.
One such case is refraction within an isotropic dielectric (such as glass) in
a\ region remote from discontinuity (see, for instance Born and Wolf
\cite{bornwolf}, Chap. II, and for an intuitive treatment, Feynman, Leighton
and Sands \cite{feynman}, Vol. I, Ch. 31). \
The interaction is entirely between the field of the photon and the charged
particles of the medium. \ If there were no charges within the medium, the
photon would pass entirely unaffected. \ In a dielectric, these are bound
charges, and the process is thus mediated by moments, primarily electric
dipole moments, induced by the flux of photons in the molecules of the
material. \ As each photon passes through the medium, it interacts with a vast
number of these molecules, driving in each molecule the oscillating divergence
of positive and negative charge distributions that is the source of its dipole
moment. \
In the semi-classical modelling of refraction - in which the incident wave is
continuous and only the medium is quantized - each induced moment is
approximated as an harmonic oscillator, essentially an oscillating electron
constituting a small electric dipole
\begin{equation}
\mathbf{p}=-\frac{q^{2}}{m\omega^{2}}\mathbf{E},\label{dipole
\end{equation}
where $\mathbf{E}$ is the incident electric field, $\omega$ is its frequency,
and $q$ and $m$ are the charge and mass of the electron respectively (see Born
and Wolf\textbf{\ }\cite{bornwolf}, chap. 2.4.1).
Keeping in mind an array of such dipoles, we need to consider a beam that
arrives, not continuously, but episodically in a flux of discrete photons.
\ Consistently with what is known of its interactions, the photon can be
regarded as a microscopic and, in the present context, essentially
indivisible, electromagnetic field having the form of a transverse wave.
As each dipole seeks to regain its unexcited state, it reradiates and does so,
potentially at least, in all directions. \ If a photon were to interact with a
single isolated dipole, it could be scattered in any of a range of possible
directions. \ But because the array of moments acquires its spatial
distribution of phase and amplitude from the inducing flux, this reradiation
(also referred to as the polarization field) interferes constructively only in
the direction of propagation of the flux itself (see, for instance, Barron
\cite{barron}, pp. 124-5). \ It is the composition of this induced
polarization field with the field of the photon that causes the change in
wavelength, and thus phase velocity, that is the origin of the refractive index.
Essentially this same process - the interference between a particle and a
secondary field that the scattering of that particle has itself induced - will
be identified in Sect. 6 as the explanation of the self interference observed
in a Mach-Zehnder interferometer. \ According to this view, refraction is
itself an elementary form of self interference, but one in which incident and
induced fields are\ able to interfere constructively in only one direction. \
The process of refraction is considerably more complicated than might be
supposed from the brief sketching above. \ Even if consideration could be
limited to the passage of just one photon, the field felt by each moment will
include reradiation from every other moment with which the photon is
interacting, and these moments will likely have one of more resonance
frequencies, and include contributions from higher order and magnetic moments.
\ In practice refraction and reflection are dealt with, not in terms of
microscopic properties, but the more easily measured macroscopic properties,
including in particular the refractive index $n$ and the bulk electric,
magnetic and polarization fields $E$, $B$ and $P$, and more will be said of
this in Sect. 5.
But I will ignore all such complications here (see instead the cited texts).
\ The essential points were made above and are in summary:
(a) \ that the interaction of photon with medium is solely with the charges of
the medium, which has the important consequence that the reaction of the
medium to any change in the photon is also mediated solely by those charges; and,
(b) that accompanying each photon, there\ is indeed another electromagnetic
field - the reradiation or polarization field - which having acquired its wave
characteristics and direction of propagation from the inducing flux is well
adapted to interfere with the photon or an accompanying photon.
Without as yet allowing a photon a choice of path, let us now introduce the
complication that the medium is birefringent. \ We will suppose that the
photon is propagating horizontally through a uniaxial crystal having its optic
axis aligned so that horizontal and vertical ($H$ and $V$) components of the
electric field of the photon induce moments of differing strengths, and thus
experience correspondingly different changes of wavelength and phase velocity.
The photon is not disrupted by these competing effects. \ It is able to
accommodate the differing phase velocities of its $H$ and $V$ components by a
continuing variation in its state of polarization as it passes through the
medium\footnote{A photon linearly polarized at $\theta$ to the vertical
evolves through the various stages of elliptical polarization until the
optical paths of its $H$ and $V$ components have differed by $2\pi$ at which
point it will have regained its original state of polarization, and the
sequence recommences. \ If, when the photon exits the medium, the paths of $H$
and $V$ components differ by $2n\pi+\pi$, the medium will have acted as a half
wave plate, if $2n\pi+\pi/2$, as a quarter wave plate.}. \ And here again,
there is the interference between photon and induced field that was
categorized above as self interference. \ Yet there has been no reason to
suppose that the process is in any way discontinuous, nonlocal or
probabilistic. \
But there is now an additional effect of some consequence. \ With the change
in polarization, a torsional reaction occurs in the medium. \ It is known from
the experiment of Beth in 1935 (Beth \cite{beth}) and the exploitation of the
Beth effect in optical traps and the like (see, for instance, Ashkin et al
\cite{ashkin}) that photons experiencing a change in polarization when
refracted by a dielectric target, not only impart linear and angular momentum
to that target, but may do so to the extent of causing observable movement of
the target.
Now consider refraction as it occurs in measurement, as when a photon
encounters the birefringence of a polarizing beamsplitter or the partially
reflective surface of a simple non-polarizing beamsplitter. \ The interaction
is again mediated solely by induced moments, but there are now alternative
paths of constructive interference available to the photon, and competing
influences on its characteristic structure that cannot be accommodated by a
mere change of wavelength or polarization. \ If the photon were freely
divisible it would separate between those paths in the manner supposed of the
continuous wave of classical physics (and of the probability wave of SQM).
\ But (at these energies and in this medium) the photon is indivisible,
and\ must adopt in its entirety one or other of the two available paths.
We have come at last to the crux of the argument. \ As the photon is forced
into one or other path, conservation (or equivalently Newton's third law)
demands that there be an equal but opposite reaction in the apparatus, a
reaction mediated solely by those moments with which the photon is interacting
- primarily moments in a narrow skein of molecules at and near a surface of
discontinuity within the scattering medium. \
The reaction of the apparatus can only comprise a fluctuation in the relative
strengths of the components of those moments and in the reradiation from those
components.\ \ And I stress again that, having acquired its distribution of
phase and amplitude from the inducing photon, this fluctuation in the
reradiation field will be constrained to propagate along the same paths of
constructive interference as those available to the photon itself. \
Ignoring losses, and assuming the ideal 50:50 beamsplitter of the previous
section, conservation demands that this fluctuation in the polarization field
have the form described formally by Eqn. (\ref{respI}) or Eqn. (\ref{respII}).
\ It will comprise a reduction in the field in the mode taken by the photon
(or a fluctuation of opposite phase to the photon) and an increase in that
field in the direction not taken. \
As contemplated in Sect. 2, there will thus be physically real waves
propagating in each path from the beam splitter.
\section{The Born probabilities}
How might a photon choose between these paths, and why should the beam divide
in apparent compliance with the Born rule?\ \ In a simple form appropriate to
discrete measurement outcomes, the rule says that
\begin{equation}
prob(a_{i})=\left\vert u_{i}\left\vert \psi\right\rangle \right\vert
^{2}\label{born
\end{equation}
where,
\[
\sum\left\vert u_{i}\left\vert \psi\right\rangle \right\vert ^{2}=1,
\]
and $prob(a_{i})$ is the probability that a particle in the state $\psi$ will
be found with the eigenvalue $a_{i}$, for which the corresponding eigenstate
is $u_{i}$ (Born \cite{born1}).
The assumption in SQM that the $prob(a_{i})$ are intrinsic to the particle
measured is taken to mean that the outcome of measurement is governed by pure
chance, or as it is said, \textquotedblleft irreducible quantum
randomness\textquotedblright\ (see, for instance, Khrennikov \cite{khrennikov
, Ch. 7). \ This species of probability seems to have been encouraged,
initially at least, by the apparently random nature of atomic transitions, but
it is only in its consistency with the idea of a divisible probability wave
that it seems indispensable to SQM. \ If it were to be admitted that the
particle was in\ a particular channel of the apparatus prior to its
observation, there could be nothing in the other channel to explain the self
interference observed at measurement - or so at least it has been assumed in
SQM. \
However, it is a significant clue to the real nature of these probabilities
that they lead, in the limit of large numbers, to conservation of the measured
property, or when there is no measured property, to some other balancing
consistent with conservation. \ That is so at least in those cases involving
alternative paths from a scattering event that lead to the self interference
of interest here. \
For example, the division effected by a polarizing beam splitter is consistent
with conservation of the components of polarization of the incident beam,
while that by a non-polarizing beam splitter operating by partial reflection
maintains the balancing of electromagnetic fields defined by the Fresnel
relations (of which something more will be said in the next section). \
The division of the beam will also be constrained by quantization.\ \ For
instance, in the measurement of spin $1/2$ by a Stern-Gerlach magnet, it is
only the component of spin in the direction of the field that is measured, but
quantization requires that particles adopt an alignment that is either spin up
or spin down with respect to that field. \ Conservation of angular momentum
then requires that a beam with its spin at\ $\varphi$ to the field divide
between spin up and spin down modes in the proportions, $\cos^{2}\varphi/2$
\ to $\sin^{2}\varphi/2$, in accordance with the trigonometric relation,
\[
\cos\varphi=\cos^{2}\frac{\varphi}{2}-\sin^{2}\frac{\varphi}{2}.
\]
For spin greater than $1/2$ the number of modes available is greater, but the
probabilities and the manner in which they transform on rotation are always
consistent with the conservation of angular momentum.
It is also possible to see in these examples why the Born rule expresses
probabilities as the squares of amplitudes. \ The conserved property is the
component of a wave or oscillation and thus a vector, which explains the
relevance to the Born probabilities of Pythagoras theorem and Hilbert space
and the continuity in the transformation of waveforms relied upon by Gleason's
theorem (Gleason \cite{gleason}).
It might thus be thought that conservation is \textit{the} explanation of the
Born probabilities. \ But conservation alone cannot explain these
probabilities. \ In whatever manner, a beam might divide, laws of conservation
will be satisfied by the equal but opposite reaction of the apparatus, and
this would be so even if the beam were to adopt in its entirety a single mode
of the measured property. \ But unless the incident beam was already in that
single mode, such a result is not observed, and it is only necessary to
consider why that is so to come to what is, I suggest, the true explanation of
the probabilities.{}
If the beam were to divide in any manner inconsistent with conservation, this
would involve a sustained transfer of measured property from beam to
apparatus, and a consequent reduction of entropy, contrary to the second law
of thermodynamics. \ In the case of a polarization beam splitter, for example,
such a transfer would induce a torsional strain in the apparatus, capable in
principle of doing work.
This suggests that the degree of conservation observed in the measured beam is
merely incidental to a process of recovery or equilibrium in the apparatus.
\ In such a process, each photon would \textquotedblleft
choose\textquotedblright\ its path, not by chance, but as determined by its
own particular circumstances, including the state of imbalance in which the
photon finds the scattering medium. \
It is not the case, as seems to be implied by the notion of intrinsic
probability, that the apparatus encountered by one particle is in the same
state as that met by the next. \ Unobserved macroscopically, various changes
of significance are occurring in the apparatus. \ The medium is in thermal
equilibrium and superimposed on that equilibrium is the fluctuating imbalance
induced by preceding measurements. \ Created in the path of following and
accompanying photons, this imbalance is eminently adapted to influence by
interference the ensuing beam. \
The Beth effect, referred to in Sect. 3, showed that such an imbalance\ is not
simply passed on without local effect to the wider environment, but may be
sufficiently enduring to cause observable dynamic change in the scattering
medium. \ But unlike a beam splitter, the suspended waveplate of Beth was what
might be termed a single-mode device, allowing no possibility of maintaining
equilibrium by a division of the incoming beam. \
The division by pure chance assumed by SQM would be an inferior, indeed
unreliable, way of maintaining that equilibrium. \ The measurement of any one
particle would as likely increase as decrease a pre-existing imbalance in the
medium. \ The variance of a distribution based on chance (that of the random
walk) increases with run-time, and in the tails of such a distribution an
excursion from balance could be substantial (see Mandel and Wolf
\cite{mandel}, Chap. 2.10.2). \ Indeed, any such departure would question the
compliance of the system with the second law. \ While it may be convenient to
model excursions from equilibrium in terms of the random walk, they tend in
practice to be self-limiting whatever the run-time.
Such a rebalancing may be less effective when particles arrive, not in a
steady beam, but singly, one after the other. \ Presumably there must come a
point at which a beam is so attenuated that the imbalance induced by one
particle will have dissipated before the arrival of the next. \ But although
experiments with attenuated beams have been reported, and some degree of
attenuation is necessary for the observation of self interference, it does not
appear to have been demonstrated that the Born probabilities survive in a beam
that is so attenuated that each measurement is in effect a separate
experiment. \
Finally here, something should be said of those situations in which SQM
supposes a probabilistic superposition of states, but there has been no
preliminary scattering into alternative paths, and accordingly no apparent
opportunity for the rebalancing contemplated above. \ In these situations, SQM
has simply assumed that if a particle, or even it might seem, a macroscopic
object such as the cat of Schr\"{o}dinger, could be in one state or another
state, and it is not known which, it must be in a probabilistic superposition
of such states. \
Schr\"{o}dinger asked us to imagine\ a cat confined within a box with a small
sample of a radioactive substance so positioned that if an atom of the
substance were to decay, it would cause a flask of hydrocyanic acid to shatter
and kill the cat (Schr\"{o}dinger \cite{schrodinger}). \ Until the box is
opened and its contents observed, it would have to be assumed (according to
SQM) that each atom is in a superposition of decayed and undecayed states,
implying in turn a superposition of the dead cat with the living cat. \
For Schr\"{o}dinger, this superposition of cats was illustrative, as he said,
of the \textquotedblleft quite ridiculous cases\textquotedblrigh
\ (\textit{ganz burleske F\"{a}lle}) that might arise from the
observer-dependent collapse assumed by SQM. \ Yet notwithstanding its
apparently random nature, there is nothing in a radioactive decay to suggest
the self interference that encouraged the notion of an intrinsically
probabilistic superposition. \ There is no suggestion that the atom is in two
places at the same time or that its later state is interfering with its
earlier state. \ There is accordingly no compelling reason, other than
consistency with those more troublesome cases that do lead to self
interference, to suppose that the timing of such a decay is governed by pure
chance, rather than deterministic microprocesses as yet unknown. \
The mischief here is that this notion of intrinsic probability is likely to be
inhibiting investigation into the possibility that some deterministic
mechanism is indeed ordering the timing of such decays. \
\section{A beam splitter}
I consider how this process of rebalancing might work in a simple
non-polarizing beam splitter operating by partial reflection. \ (Two such beam
splitters will be needed for the Mach-Zehnder interferometer to be considered
in the next section). \
While it may not be feasible to follow the details of what is happening at the
quantum level, the fluctuation in fields that results from the passage of an
individual photon can be treated as a microscopic perturbation of the
continuous and macroscopic wave supposed by classical physics.
The relative amplitudes of reflected and refracted beams were deduced by
Fresnel from an elastic wave theory in 1823, (see Silverman \cite{silverman},
pp. 228-231), and given what is essentially their modern textbook derivation
by Lorentz in 1875 by insisting that Maxwell's equations be satisfied across
the inter-medial boundary (Lorentz \cite{lorentz}). \ The macroscopic wave is
assumed to divide so that the forces on the charges of the medium, whether
arising from incident or induced fields, are in a state of balance (as
contemplated in the preceding section). \
The continuity of Maxwell's equations across the boundary requires (assuming a
wave passing from medium $1$ to medium $2$ through a boundary in the
$xy$-plane) that,
\begin{align}
\left( \varepsilon_{0}\mathbf{E}_{1}\mathbf{+P}_{1}\right) _{z} & =\left(
\varepsilon_{0}\mathbf{E}_{2}\mathbf{+P}_{2}\right) _{z},\nonumber\\
\left( \mathbf{E}_{1}\right) _{xy} & =\left( \mathbf{E}_{2}\right)
_{xy},\label{cond}\\
\ \ \ \mathbf{B}_{1}\ & \mathbf{=}\ \mathbf{B}_{2}.\nonumber
\end{align}
where $\mathbf{E}$, $\mathbf{B}$ and $\mathbf{P}$ are respectively the
macroscopic electric, magnetic and polarization fields and $\varepsilon_{0}$
is the permittivity of free space (see, for example, Feynman \cite{feynman},
Vol. II, chap. 33).
In considering the quantized wave, the Fresnel relations can be taken to
define, not the steady state contemplated by Lorentz, but that notional point
of equilibrium about which the system fluctuates as photons are variously
reflected or refracted from the semi-reflective boundary. \ The fields must
remain in balance and for this to occur there must be a continuing
readjustment, not of the boundary conditions themselves, but of the manner in
which those conditions are satisfied. \
Consider, for example, the first of conditions (\ref{cond}), which is obtained
by asserting, in the $z$-direction, Coulomb's law, which in dielectric form
is
\[
\nabla\cdot\mathbf{E}=-\frac{\nabla\cdot\mathbf{P}}{\varepsilon_{0}}.
\]
On the side of the boundary to which a photon departs, there will be\ (as
compared with the steady state supposed classically) a fleeting increase in
the electromagnetic field supplied by the beam, and on the other side of the
boundary, a corresponding decrease in that field. \ This fluctuation will
induce by reaction compensating changes in the dispositions and relative
strengths of the components of moments and in the corresponding components of
the ambient polarization field. \
Whether the photon is\ reflected or transmitted, the fields at the boundary
will thus remain continuous, but at the cost of a microscopic departure from
the state of balance defined by the mean intensity of the incident beam.
\ This fluctuation in fields will influence in turn the choice of path made by
a following particle, thus contradicting the assumption of SQM that
measurement is intrinsically probabilistic. \
SQM effectively suppresses these fluctuations in field strengths by invoking
the probability wave, a phenomenon having more in common with the
constructions of geometrical optics than the quantized wave with which SQM
must ultimately deal. \ Having invoked this probability wave, SQM has had to
contrive its collapse, and thus the need for a measurement and an observer
that led to the \textquotedblleft measurement problem\textquotedblright. \
It is not until that collapse occurs that SQM gives effect to the quantization
of the measured beam, but by then the continuity of fields assumed by
Maxwell's equations has been lost in discontinuity and nonlocality. \
\section{The Mach-Zehnder interferometer}
Consider again the Mach-Zehnder interferometer of Fig. 1. The interference at
$BS2$ is now between real waves, these being\ the photon and the secondary
wave that was generated by reaction at $BS1$ as the photon was forced to adopt
one or other path through the interferometer. \ As discussed in Sect. 5, this
fluctuation in the polarization field maintains microscopically the continuity
and balancing of fields supposed classically by Maxwell's equations and the
Fresnel relations. \
As in SQM, each set of waves recombining at $BS2$ will have originated from
the scattering of a single photon at $BS1$. \ For photons of sufficiently like
frequency, the phase difference $\Delta$ between the two paths will thus
remain substantially the same from one photon to the next. \ Thus\ no matter
how attenuated or incoherent the original beam from the source may have been,
the recombining waves will demonstrate observable interference at the second
beam splitter. \
Let us suppose that $BS1$ and $BS2$ are non-polarizing lossless $50:50$
beamsplitters so constructed and aligned that when the upper and lower optical
paths to detector $D1$ differ by $\Delta$, the corresponding paths to detector
$D2$ will differ by $\Delta+\pi$. \ If $\Delta=0$, the recombining waves will
interfere constructively toward $D1$, but destructively in the direction of
$D2$. \ The photon will favour the path that preserves the integrity of its
waveform. \ Photons scattered at $BS2$ will thus register only at $D1$. \ If
$\Delta=\pi$, those photons will register instead at $D2$. \ In either case
the result will coincide with the prediction of SQM.
If the recombining waves are neither entirely in nor out of phase, the beam
will divide at $BS2$ in accordance with the intensities determined by the
interference occurring in $BS2$\footnote{For detection at $D1$, for instance,
and assuming recombining waves of phases $\chi_{1}$ and $\chi_{2}$, the
intensity (probability) is
\begin{align*}
\left\vert e^{i\chi_{1}}+e^{i\chi_{2}}\right\vert ^{2} & =\left[ \left(
e^{i\left( \chi_{1}+\chi_{2}\right) /2}\right) \left( e^{i\Delta
/2}+e^{-i\Delta/2}\right) \right] ^{2},\\
& \thickapprox\cos^{2}\frac{\Delta}{2},
\end{align*}
where $\Delta=\chi_{1}-\chi_{2}$, and the first factor on the right hand side
in the first line\ has been equated with unity.}. \ In SQM, these intensities
correspond to the Born probabilities and are
\begin{equation}
\frac{prob(D1)}{prob(D2)}=\frac{\cos^{2}\frac{\Delta}{2}}{\sin^{2}\frac
{\Delta}{2}},\label{prob
\end{equation}
where $\Delta$ is again the difference between the two optical paths.
To provide a physically realistic explanation of these probabilities, I
briefly repeat what will now be an all too familiar refrain. \ The photon is
indivisible and must adopt in its entirety one or other path. \ As it does,
whatever force or effect is ensuring that indivisibility, must induce by
reaction in $BS2$, an imbalance in the dipole moments mediating the
interaction. \ If the photon takes the path toward $D1$, the coherent merger
of photon and secondary wave in that direction induces in $BS2$, a reaction of
energy,
\begin{equation}
\sin^{2}\frac{\Delta}{2},\label{proba
\end{equation}
tending to bias the ensuing flux toward $D2$. \ Conversely, coherence in the
direction of $D2$ must induce an imbalance,
\begin{equation}
\cos^{2}\frac{\Delta}{2},\label{probb
\end{equation}
creating a bias toward $D1$. \ From (\ref{proba}) and (\ref{probb}), the
maintenance of equilibrium within $BS2$ will ensure an (approximate) division
between $D1$ and $D2$ in the proportions predicted by the Born probabilities
((Eqn. \ref{prob}) ).
There is no suggestion in the above that the secondary wave is itself in any
sense a photon or part of a photon. \ Essentially, it is a fluctuation in the
polarization field capable of survival over the time frame of the experiment
and having an equal but opposite effect to the change occurring in the photon itself.
\section{Conclusion}
As described in this paper, it is an illusion - literally in the Mach-Zehnder,
a trick done with mirrors - that the photon is in two places at once. \ As the
indivisibility of the photon constrains it to one path,\ the response of the
apparatus, evolving through the experiment in the same wave-like manner as
the\ photon, creates the impression that the photon is somehow occupying both
paths. \ And with waveforms in each path, it is only natural that if an
attempt is made to locate the particle within those paths,\ the visibility of
the interference is diminished accordingly.
What then is the probability wave? \ Were it not now so commonplace, it might
seem a wondrous thing that an image can be propagated with such fidelity from
a reflective surface or through a pane of glass. \ The feat is no less
impressive when it is realized that the path of each photon is determined by
the fluctuating states of a vast multitude of mutually interfering
electromagnetic fields, these being the field of the photon itself, those of
accompanying photons, reradiation from moments induced by all those photons,
and further reradiation induced by the original reradiation.
In this multitude of interacting waveforms, there is no single wave, nor even
a divided wave, that is identifiable with the simply constructed probability
wave of SQM. \ Even if measurement were in some degree intrinsically
probabilistic, the probability wave could be no more than a mathematical
convenience, epistemic rather than ontic, and similar in this respect to the
constructions of geometrical optics. \ Such a construction may suggest where
the particle is likely to go, but not why it must goes there.
There are other interpretions of quantum mechanics in which the particle is
assumed to have, as it does here, a definite position and velocity, and hence
a well-defined trajectory. \ These include de Broglie's pilot wave theory of
1928 \cite{pilot}, and the rediscovery and revision of that theory by Bohm in
1952 (Bohm \cite{bohm1} and \cite{bohm2}, and see the review by Goldstein
\cite{goldstein1}). \ Also in this category is the recent proposal by Gao, who
argues for the physical reality of the particle from the implications of
protective measurement (see, for instance, Aharonov et al \cite{aharonov}),
and shows how the wave function, and in turn the Born probabilities, might be
defined by the rapid and discontinuous motion of the particle in question (Gao
\cite{gaowave}).
The theory presented here departs crucially from those proposals, as it does
from other interpretations of quantum mechanics, in identifying the mysterious
wave-like effect in the path (or paths) not taken by the particle as the
consequence of well-understood and otherwise familiar underlying
microprocesses. \ To explain self interference, the present proposal thus has
no need for a novel ontology, such as those interesting but inevitably
contentious ontologies that have characterized other theories. \
I have concentrated here on the Mach-Zehnder and not discussed at all the
better known double-slit experiment. \ But with its paths diverging to
macroscopic separations, the Mach-Zehnder provides I suggest the more
compelling illustration of self interference.
Nor of course is self interference the only mystery of quantum theory. \ There
is in particular the apparent nonlocality of entanglement, and following the
recent rush of \textquotedblleft loop-hole free\textquotedblright\ Bell tests,
it may seem, as has been asserted, that the final nail has been driven into
the coffin of local realism (see, for instance, Wiseman \cite{wiseman
).\ \ But no one has as yet come forward to explain these faster-than-light
influences, or to tell us how, if at all, they propagate or how they are to be
reconciled with those fundamental forces of nature for which the speed of
light is a limiting velocity. \ Until more sense can be made of this, these
claims of superluminality are deserving of continuing scrutiny.
Meanwhile, I have shown that in at least one class of physical processes, both
self interference and the Born probabilities can be explained, not merely in a
manner consistent with physical reality, but in accordance with
well-recognized electromagnetic microprocesses that must be suppressed if
measurement is to have the curious nature supposed by SQM. \
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,355
|
Version 1.0 is 50%, but I had to stop due to another personal project.
# VesselDB
VesselDB: **VE**ry **S**imple **SE**rver**L**ess **D**ata**B**ase
VesselDB is a key-value serverless database that uses Amazon S3 for storage and AWS Lambda for the database engine.
What is serverless? It's a concept where the developer doesn't need to worry about servers. Configuring, scaling and making them highly available is a resposibility delegated to the cloud vendor.
Requirements:
1. Scalability
2. High Availability
3. Efficient usage of resources
Why it is efficient? Because the code (database engine) instantiated on-demand and you are billed only by the fractions of seconds that it executes.
## Use Cases
Storing data in any kind (images, docs, JSON, etc.) with a key-value approach.
Examples:
- Session data
- Profile data
- User files
- Processed documents
## Drawbacks
- It's a key-value database.
- Lambda cold start + S3 storage increases its latency.
## Advantages
Why use VesselDB and not Amazon's S3 SDK directly? Because it's easier to use. AWS SDK is more complex.
## Installing
npm install vesseldb (*not published yet*)
Note: install and configure using Node.js. After deployment, it is accessible from a RESTful API that can be consumed by any programming language.
## Commands
// require
var vessel = require('vesselDB');
// create a database
vessel.create('instanceName', callback);
// select a instance to use
vessel.use('bucketName', callback);
// insert a JSON object
vessel.insert({ name: "document1" }, callback);
// insert an array of objects
vessel.insert([{ name: "documentArray" }], callback);
// insert a file
vessel.insert(byteArray, callback);
// find objects or files
vessel.find([ids], callback);
// update
vessel.update([ids], { name: "document2" }, callback);
// delete
vessel.delete([ids], callback);
// drop the database
vessel.drop('bucketName', callback);
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,337
|
Q: webpy 'Hello World' tutorial: webpy can't find class When I run the example code partway through this tutorial, I don't immediately get an error, but when I go to the url(ht[break_link]tp://0.0.0.0:8080/), I get 'Firefox can't establish a connection to the server at 0.0.0.0:8080.' I tried it with Chrome as well, no luck.
At one point webpy gave me an error along the lines of 'class index could not be found,' but I'm not sure what I did to get the error or how to reproduce it.
This is the file. It's the same as the tutorial I linked, just with print statements added:
import web
# Handle the url / with the index class.
urls = {
'/', 'index'
}
class index:
def GET(self):
return "Hello, world!"
print 'inside get'
print 'after index'
print "-------"
print globals()
print "-------"
if __name__ == '__main__':
print "hi"
app = web.application(urls, globals())
print "hello"
app.run()
print "last"
Outputs:
after index
-------
{'web': <module 'web' from 'C:\Python27\lib\site-packages \web\__init__.pyc'>, 'index': <class __main__.index at 0x0324F030>, '__builtins__': <module '__builtin__' (built-in)>, '__file__': 'code.py', '__packa
ge__': None, 'urls': set(['index', '/']), '__name__': '__main__', '__doc__': None}
-------
hi
hello
http://0.0.0.0:8080/
last
('last' doesn't show up until I ctrl-c out)
webpy's not noticing the index class, but I have no idea why.
A: When it prints http://0.0.0.0:8080/, the webserver is running in a loop (due to call to app.run()) so, it won't print "last" until you kill the webserver. That's why you don't see 'last' until ctrl-c. All is working fine.
The http://0.0.0.0:8080 simply says the webserver is listening on 'All valid IPv4 addresses' on the local machine. 0.0.0.0 isn't the real address. To access your webserver, point your browser to the real IP address.
If you're running this webserver on you local machine, you can probably use http://127.0.0.1:8080/, which (most always) refers to your local machine. If you're running this webserver on a remote machine, you'll need to provide its IPv4 address, or it's hostname instead of 0.0.0.0.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,094
|
export interface Axis {
min: number
max: number
length?: number
}
export interface Box {
x: Axis
y: Axis
}
export interface RelativeBox {
x: Partial<Axis>
y: Partial<Axis>
}
export interface BoundingBox {
top: number
right: number
bottom: number
left: number
}
export interface Point {
x: number
y: number
}
export interface AxisProjection {
translate: number
scale: number
origin: number
originPoint: number
}
export interface Projection {
x: AxisProjection
y: AxisProjection
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,922
|
Minnesota features some of the country's most dynamic scenery, stunning architecture, warmest people and diverse wildlife. "Capture Minnesota Volume VI" showcases Minnesota's beauty through the eyes of Minnesota photographers in a unique and compelling way.
If you don't yet know, folks from an online community of Minnesota photographers and enthusiasts submitted their photos for inclusion in this book, then thousands of people helped our editors pick the best of the best by their votes. The result? The best of Minnesota in photography, wrapped up in a hardcover book perfectly suited for your coffee table.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,132
|
{"url":"https:\/\/kgreresearch.wordpress.com\/2012\/02\/22\/how-ga-as-and-ga-ga-bonds-affect-roughness-and-wicking-also-nanowires\/","text":"# Kris's Research Notes\n\n## February 22, 2012\n\n### How Ga-As and Ga-Ga bonds affect Roughness and Wicking. Also,\u00a0nanowires.\n\nFiled under: GaAs Simulations \u2014 Kris Reyes @ 1:10 am\n\nRecall previously we had considered increasing the Ga-As bond strength $\\gamma_{GA}$ and decreasing the Ga-Ga bond strength $\\gamma_{GG}$. Doing so would increase the difference in the local energies between a solid, crystalline and liquid neighborhoods. With our old energies ($\\gamma_{GA} = 0.5$eV, $\\gamma_{GG} = 0.3$eV), the difference in local energies between solid and liquid neighborhoods is = 0.8 eV. Setting $\\gamma_{GA} = 0.7$eV and $\\gamma_{GG} = 0.24$eV, this difference becomes 1.84 eV. In this note, we examine the effect of changing these parameters on simulation results.\n\nThe new energies affect the simulations in two main ways. First, an increase in $\\gamma_{GA}$ results in a roughening of the surface due to a stronger persistence of GaAs nuclei on the surface. Second, the smaller $\\gamma_{GG}$ value suggests that a Ga atom in the liquid is more likely to leave and nucleate away from the droplet, i.e. the rate of the wicking process (detailed here) is increased.\n\nFor example, consider the crystallization of a liquid droplet at 623K via an As flux of 1.00 ML\/sec. We consider the simulation results using the old and new energies. Here are the surfaces away from the liquid droplet under both the old and new energies:\n\n $\\gamma_{GA} = 0.5$eV, $\\gamma_{GG} = 0.30$eV $\\gamma_{GA} = 0.7$eV, $\\gamma_{GG} = 0.24$eV\n\nIndeed, the surface is much rougher for the large $\\gamma_{GA}$ case. If we examine the neighborhood near the liquid droplet after crystallization has occurred, we observe a significant morphological difference between the two cases:\n\n $\\gamma_{GA} = 0.5$eV, $\\gamma_{GG} = 0.30$eV $\\gamma_{GA} = 0.7$eV, $\\gamma_{GG} = 0.24$eV\n\nIn the former case, the rate of the wicking process essentially occurs on the same order as the rates for the other processes. In the former case, the wicking rate is on a different order, and hence hardly any crystallization occurs at the site of the liquid droplet.\n\nWe may measure how sensitive the resulting morphology is to both $\\gamma_{GA}$ and $\\gamma_{GG}$ by crystallizing a liquid droplet for various choices of energies. For example, we vary $\\gamma_{GA}$ between 0.35 eV to 0.7 eV, fixing $\\gamma_{GG} = 0.24$ eV.\n\n $\\gamma_{GA} = 0.7$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.6$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.50$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.40$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.30$ eV, $\\gamma_{GG} = 0.24$ eV,full frame\n\nWe see that for all but the lowest value of $\\gamma_{GA} = 0.30$ eV, a hole forms. Something completely different happens in the $\\gamma_{GA} = 0.30$ eV case: a nanoring begins to grow. To examine, this, we consider the refine our search, varying $\\gamma_{GA}$ between 0.30 eV to 0.40 eV.\n\n $\\gamma_{GA} = 0.40$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.39$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.38$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.37$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.36$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.35$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.34$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.33$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.32$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.31$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.30$ eV, $\\gamma_{GG} = 0.24$ eV,full frame\n\nWe see that around $\\gamma_{GA} = 0.34$ eV, the system switches from material preferentially leaving the droplet site to preferentially entering the site. That is, the wicking mechanism is effectively turned off (and possibly reversed) for low $\\gamma_{GA}$. This may suggests a convection term is relevant when modeling the wicking process. Of course, this is not enough to explain nanowires \u2014 in place crystallization can also occur if wicking is not present. Indeed, there appears to be some sort of phase segregation driving the nanowire formation.\n\nWe also may fix $\\gamma_{GA} = 0.5$ ev and vary $\\gamma_{GG}$ between 0.24 eV and 0.30 eV.\n\n $\\gamma_{GA} = 0.50$ eV, $\\gamma_{GG} = 0.24$ eV,full frame $\\gamma_{GA} = 0.50$ eV, $\\gamma_{GG} = 0.26$ eV,full frame $\\gamma_{GA} = 0.50$ eV, $\\gamma_{GG} = 0.28$ eV,full frame $\\gamma_{GA} = 0.50$ eV, $\\gamma_{GG} = 0.30$ eV,full frame","date":"2017-07-26 00:32:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 67, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5459082722663879, \"perplexity\": 2250.5476374566324}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-30\/segments\/1500549425737.60\/warc\/CC-MAIN-20170726002333-20170726022333-00142.warc.gz\"}"}
| null | null |
Tunde smirked in amusement as Anaya adjusted the front of her button-down floral cardigan. They sat in Tunde's car, waiting at Anaya's soft request to gather her wits and he'd complied. The silence in the car was muffled by the gentle pitter-patter of the rain outside.
Tunde beamed openly, still reveling in the moment of Anaya accepting his proposal to court her. He couldn't wait for his mother and brother to meet her. She was everything he'd wanted and much more. His eyes lifted from her lips back to her warm brown eyes. "If I didn't want to wait until my ring was on your finger, I would kiss you right now…" he said in a low voice.
Her eyes widened and she jerked her hand from under his. "I'm ready!" she said all too quickly, pulling off her seatbelt.
Anaya nodded; face aflame from Tunde's confession and the wanton thoughts that flooded her mind. She wished he would just do it instead of waiting but knew that he intended to do things in order. She was still reeling when Tunde extended his hand.
He bit back a chuckle when she studied his hand suspiciously. "Don't worry, I won't kiss you yet." He forced himself not to smile, staring expectantly at her with his hand out.
"Unfortunately I did," Tunde bantered and leaned forward to kiss her forehead. "Loosen up. You don't want her to smell your fear," he teased, pulling her with him toward the front of the two-story brick home.
Her hand still in his, Tunde reached over and rapped on the metal hook on the door. Moments later, the lock rustled and the door opened to reveal Silas on the other side. He looked annoyed.
Anaya stared at the man standing on the threshold. She recognized him from Tunde's pictures and knew this was his younger brother. Her stomach tensed and relaxed in the same moment his scowl eased into relief at the sight of them.
"It's about time you got here. Mom and Felicia are already making wedding plans." His eyes fell on the quiet woman beside his brother and he grinned. "I finally get to meet the woman both my brother and girlfriend have fallen in love with…" He pulled open the door, gesturing her in.
Tunde nudged at Anaya's hand. "You first," he said, eyes gently encouraging her to step inside.
Anaya swallowed against the lump forming in her throat and took one step over the threshold.
Familiar scents filled the room and assailed her senses as she stepped into the main parlor, her eyes adjusting to the dimly lit room. Anaya tightened her hold on Tunde's hand.
The two brothers exchanged glances and Silas smirked approvingly before turning back to Anaya. "Felicia is in the kitchen with Mom. Why don't we go in there?" He raised a brow at Tunde who remained quiet beside Anaya.
Anaya shifted her eyes to Tunde and when he winked at her, she turned back to Silas and nodded.
"Of course not!" the unfamiliar but friendly voice responded. The sound of a chair's legs scraping against the floor clued Anaya that the woman was now standing and she pulled her gaze from the floor to the kitchen.
A tall and slender woman with dark skin like Tunde's stood beside Felicia. Her black hair, short and curly glistened under the kitchen light and her open smile was just as bright. Anaya didn't know what to say. Everything about the woman mirrored Tunde, from her broad shoulders to long arms and chiseled features. The stiffness in Anaya's back eased away.
"Welcome to our home, Anaya…" Josephine Halliday said graciously as she gazed at the shy woman flanked between her two grown sons. Her eyes took in the soft beauty of the Fulani girl and felt her heart flutter with compassion. She moved from around the counter toward her, arms spread out.
"My mom likes to give hugs," Tunde said softly to Anaya as she approached them.
Anaya felt her body pulling toward the woman's open arms and she bit back a sigh as Tunde's mother tugged her into the embrace.
Tunde smiled proudly at his mother who kept an arm around Anaya's shoulders. He wanted very much to be the one holding Anaya, consoling her. He knew she wrestled with her thoughts over her mother's reaction to their relationship but he was grateful for his mother's calm and welcoming nature.
Felicia stepped around the counter to stand beside Silas. She wrinkled her nose when he slipped an arm around her waist and pulled her into the crook of his arms. Her eyes fell on Anaya's and winked at the younger woman, silently encouraging her that everything was finally falling into place. For both of them.
Cloud nine could not compare to the state Anaya found herself in as she pulled away from Tunde's warm embrace outside her house hours later. The sounds of crickets filled the night air and the cool breeze tickled the back of her neck.
Anaya nodded. Her pulse quickened when he released her hands and raised his own to frame her face.
Anaya didn't know what to do first. Gush over the fact that Tunde was a talented artist or laugh aloud that his real name was Paul. She felt the wave of laughter surge from the back of her throat.
"Paul Halliday," Anaya said, hiding a smile.
The slight frown on his face was adorable. It was obvious he'd been teased endlessly about his first name as a child. She wanted to place a kiss on his cheek. So she did. Lifting on her tiptoes, Anaya pressed her lips against the smooth skin above the spot where a dimple showed every time he smiled at her. The scent of him assailed her nostrils, urging her to stay there with her lips against his cheek.
The overwhelming feeling threatened to rock her off her feet. Slowly, she lowered back on her heels and looked up at him. The hooded look in his eyes sent her heart racing.
"I think it's time for you to go inside, Ana," he said in a taut voice.
Her smile didn't waver, the warmth from him encouraging her on her conviction of the two of them.
Anaya smiled as she leaned into him, believing in nothing but his earnest words and this precious moment they shared privately.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,698
|
{"url":"https:\/\/www.gradesaver.com\/textbooks\/math\/prealgebra\/prealgebra-7th-edition\/chapter-7-section-7-3-integrated-review-percent-and-percent-problems-page-494\/24","text":"## Prealgebra (7th Edition)\n\nPublished by Pearson\n\n# Chapter 7 - Section 7.3 - Integrated Review - Percent and Percent Problems - Page 494: 24\n\n#### Answer\n\n0.1125 $\\frac{9}{80}$\n\n#### Work Step by Step\n\n11.25% =11.25 $\\times$(0.01)=0.1125 as a fraction 11.25%=11.25$\\times$$\\frac{1}{100}$=$\\frac{11.25\\times1}{100}$=$\\frac{11.25}{100}$ =$\\frac{11.25\\times100}{100\\times100}$= =$\\frac{1125}{10000}$=$\\frac{125\\times9}{125\\times80}$=$\\frac{9}{80}$\n\nAfter you claim an answer you\u2019ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide\u00a0feedback.","date":"2018-12-11 00:39:52","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6613627672195435, \"perplexity\": 9348.141441893305}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-51\/segments\/1544376823516.50\/warc\/CC-MAIN-20181210233803-20181211015303-00613.warc.gz\"}"}
| null | null |
#include <aws/cloudsearchdomain/model/SuggestionMatch.h>
#include <aws/core/utils/json/JsonSerializer.h>
#include <utility>
using namespace Aws::Utils::Json;
using namespace Aws::Utils;
namespace Aws
{
namespace CloudSearchDomain
{
namespace Model
{
SuggestionMatch::SuggestionMatch() :
m_suggestionHasBeenSet(false),
m_score(0),
m_scoreHasBeenSet(false),
m_idHasBeenSet(false)
{
}
SuggestionMatch::SuggestionMatch(const JsonValue& jsonValue) :
m_suggestionHasBeenSet(false),
m_score(0),
m_scoreHasBeenSet(false),
m_idHasBeenSet(false)
{
*this = jsonValue;
}
SuggestionMatch& SuggestionMatch::operator =(const JsonValue& jsonValue)
{
if(jsonValue.ValueExists("suggestion"))
{
m_suggestion = jsonValue.GetString("suggestion");
m_suggestionHasBeenSet = true;
}
if(jsonValue.ValueExists("score"))
{
m_score = jsonValue.GetInt64("score");
m_scoreHasBeenSet = true;
}
if(jsonValue.ValueExists("id"))
{
m_id = jsonValue.GetString("id");
m_idHasBeenSet = true;
}
return *this;
}
JsonValue SuggestionMatch::Jsonize() const
{
JsonValue payload;
if(m_suggestionHasBeenSet)
{
payload.WithString("suggestion", m_suggestion);
}
if(m_scoreHasBeenSet)
{
payload.WithInt64("score", m_score);
}
if(m_idHasBeenSet)
{
payload.WithString("id", m_id);
}
return payload;
}
} // namespace Model
} // namespace CloudSearchDomain
} // namespace Aws
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,503
|
Hugh Jackman's Wife Says He's 'Not Allowed' to Work With Angelina Jolie
By Jackie Willis 12:00 PM PDT, July 6, 2015
Hugh Jackman Is Not Allowed to Work With Angelina Jolie
Deborra-Lee Furness has been married to Hugh Jackman for nearly 20 years, and she has one rule when it comes to his on-screen love interests.
While he's gotten hot and heavy with the likes of Scarlett Johansson (Scoop) and Ashley Judd (Someone Like You) on the big screen, Furness isn't so keen on having her hubby work with A-list actress/director Angelina Jolie.
WATCH: Hugh Jackman Has a Pretty Legit Reason for Being Done With Wolverine
"I've told his agent he's not allowed to work with Angelina," she quipped during a joint interview with Jackman, 46, on Australia's Today show. "I'm sure she's lovely."
All jokes aside, the couple also shared their secret to a long, happy marriage. "That's the deal with this business. If you get it right, if you pick the right partner, then, you traverse all those travails and challenges ," Furness, 59, said.
WATCH: Hugh Jackman Explains His Instagram Habits
"We've managed to keep that," Jackman chimed in. "Deb had a rule, never apart for more than two weeks."
Gushing over her heartthrob husband, Furness added, "And we don't wanna be apart. He tells really good jokes. I mean I miss him."
Hugh Jackman on Why He's Retiring Wolverine - 'I Don't Know How Many More Egg White Omlettes I Can Eat'
Angelina Jolie and Brad Pitt's Most Matching Moments
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,963
|
{"url":"https:\/\/pyblp.readthedocs.io\/en\/latest\/_api\/pyblp.OptimalInstrumentResults.html","text":"# pyblp.OptimalInstrumentResults\u00b6\n\nclass pyblp.OptimalInstrumentResults\n\nResults of optimal instrument computation.\n\nThe OptimalInstrumentResults.to_problem() method can be used to update the original Problem with the computed optimal instruments.\n\nproblem_results\n\nProblemResults that was used to compute these optimal instrument results.\n\nType\n\nProblemResults\n\ndemand_instruments\n\nEstimated optimal demand-side instruments for $$\\theta$$, denoted $$Z_D^\\text{opt}$$.\n\nType\n\nndarray\n\nsupply_instruments\n\nEstimated optimal supply-side instruments for $$\\theta$$, denoted $$Z_S^\\text{opt}$$.\n\nType\n\nndarray\n\nsupply_shifter_formulation\n\nFormulation configuration for supply shifters that will by default be included in the full set of optimal demand-side instruments. This is only constructed if a supply side was estimated, and it can be changed in OptimalInstrumentResults.to_problem(). By default, this is the formulation for $$X_3^\\text{ex}$$ from Problem excluding any variables in the formulation for $$X_1^\\text{ex}$$.\n\nType\n\nFormulation or None\n\ndemand_shifter_formulation\n\nFormulation configuration for demand shifters that will by default be included in the full set of optimal supply-side instruments. This is only constructed if a supply side was estimated, and it can be changed in OptimalInstrumentResults.to_problem(). By default, this is the formulation for $$X_1^\\text{ex}$$ from Problem excluding any variables in the formulation for $$X_3^\\text{ex}$$.\n\nType\n\nFormulation or None\n\ninverse_covariance_matrix\n\nInverse of the sample covariance matrix of the estimated $$\\xi$$ and $$\\omega$$, which is used to normalize the expected Jacobians. If a supply side was not estimated, this is simply the sample estimate of $$1 \/ \\sigma_{\\xi}^2$$.\n\nType\n\nndarray\n\nexpected_xi_by_theta_jacobian\n\nEstimated $$E[\\frac{\\partial\\xi}{\\partial\\theta} \\mid Z]$$.\n\nType\n\nndarray\n\nexpected_omega_by_theta_jacobian\n\nEstimated $$E[\\frac{\\partial\\omega}{\\partial\\theta} \\mid Z]$$.\n\nType\n\nndarray\n\nexpected_prices\n\nVector of expected prices conditional on all exogenous variables, $$E[p \\mid Z]$$, which may have been specified in ProblemResults.compute_optimal_instruments().\n\nType\n\nndarray\n\nexpected_shares\n\nVector of expected market shares conditional on all exogenous variables, $$E[s \\mid Z]$$.\n\nType\n\nndarray\n\ncomputation_time\n\nNumber of seconds it took to compute optimal excluded instruments.\n\nType\n\nfloat\n\ndraws\n\nNumber of draws used to approximate the integral over the error term density.\n\nType\n\nint\n\nfp_converged\n\nFlags for convergence of the iteration routine used to compute equilibrium prices in each market. Rows are in the same order as Problem.unique_market_ids and column indices correspond to draws.\n\nType\n\nndarray\n\nfp_iterations\n\nNumber of major iterations completed by the iteration routine used to compute equilibrium prices in each market for each error term draw. Rows are in the same order as Problem.unique_market_ids and column indices correspond to draws.\n\nType\n\nndarray\n\ncontraction_evaluations\n\nNumber of times the contraction used to compute equilibrium prices was evaluated in each market for each error term draw. Rows are in the same order as Problem.unique_market_ids and column indices correspond to draws.\n\nType\n\nndarray\n\nExamples\n\nMethods\n\n to_dict([attributes]) Convert these results into a dictionary that maps attribute names to values. to_pickle(path) Save these results as a pickle file. to_problem([supply_shifter_formulation,\u00a0\u2026]) Re-create the problem with estimated feasible optimal instruments.","date":"2021-07-26 13:13:38","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7723739743232727, \"perplexity\": 1503.3976171209474}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-31\/segments\/1627046152129.33\/warc\/CC-MAIN-20210726120442-20210726150442-00345.warc.gz\"}"}
| null | null |
package storage
import (
// load the packages
_ "github.com/codedellemc/libstorage/drivers/storage/dobs/storage"
)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 532
|
Q: Custom logout URL with Spring Security We follow a certain convention when creating our URIs. All authentication related URIs such as /login, /logout, /changepassword etc fall under the sub-context /auth.
Thus our authentication-related URIs look like:
/auth/login
/auth/logout
/auth/changepassword
This is what we have in the Spring security context XML.
<http pattern="/auth/**" security="none" />
<http pattern="/resources/**" security="none" />
<http auto-config="true" access-decision-manager-ref="accessDecisionManager">
<intercept-url pattern="/admin/**" access="ADMINISTRATIVE_ACCESS"/>
<intercept-url pattern="/**" access="XYZ_ACCESS"/>
<form-login
login-page="/auth/login"
default-target-url="/content"
authentication-failure-url="/auth/loginFailed"
authentication-success-handler-ref="authenticationSuccessHandler"/>
<logout logout-url="/auth/logout" logout-success-url="/auth/login"/>
</http>
The problem now is that /auth/logout gives me a 404 when accessed. However, if I change it to start with something other than /auth such as /abcd/logout or even /logout, it works fine.
I am thinking this is due to the fact that we have defined /auth/** as unsecured and yet trying to use it as a logout page. (How can you access logout if you have not logged in?)
Is there any way out of this in order to please our rather strict URI naming convention?
A: You're right about the part:
I am thinking this is due to the fact that we have defined /auth/** as
unsecured and yet trying to use it as a logout page. (How can you
access logout if you have not logged in?)
More precise defining
<http pattern="/auth/**" security="none" />
means no Spring Security filter is applied to requests that match /auth/** pattern and hence Spring Security does not controll /auth/logout URL while it should.
Because Spring Security matches pattern from top to bottom simple override for /auth/logout in your main <http> won't work, so solution to that problem can be defining separate patterns:
<http pattern="/auth/login" security="none" />
<http pattern="/auth/changepassword" security="none" />
<http pattern="/resources/**" security="none" />
<http auto-config="true" use-expressions="true" access-decision-manager-ref="accessDecisionManager">
<intercept-url pattern="/auth/logout" access="permitAll"/>
<intercept-url pattern="/admin/**" access="ADMINISTRATIVE_ACCESS"/>
<intercept-url pattern="/**" access="XYZ_ACCESS"/>
<!-- rest of your config -->
</http>
If you have many of /auth/* URLs to be handled, you can use <http>'s request-matcher="regex", but I don't think it'll be readable that way.
A: Instead of security="none", I think you want access="permitAll" (explanation of permitAll in the docs).
I'm not sure what security="none" is supposed to do, but I think you might have it confused with filters="none", which causes the Spring Security chain to be bypassed completely.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,381
|
\section{\label{sec:level1}Introduction}
The phenomenon of multistability (coexistence of different attractors for a given set of parameters) is widespread in physics \cite{maurer1980effect,brun1985observation,gibbs2012optical}, chemistry \cite{aguda1987bistability,wilhelm2009smallest}, biology \cite{angeli2004detection,ozbudak2004multistability,freyer2011biophysical} and climate systems \cite{robinson2012multistability,freire2008multistability} and many other fields of science and nature. One important feature of such systems of high order of complexity are their extreme sensitivity to perturbation and initial conditions\cite{pisarchik2014control}.
Generalized multistability, firstly distinguished from ordinary bistability extensively exist in nonlinear systems, was found in laser physics\cite{arecchi1982experimental}.
The mechanisms of multistability include coupling \cite{maistrenko2007multistability}, delayed feedback \cite{balanov2005delayed,foss1996multistability},parametrical forcing \cite{varma1993quadratic} and dissipation\cite{lieberman1985transient}. These mechanisms usually do not act independently in a system. In many cases, the feedback and dissipation often coexist. For example, a mechanical rotor with weak damping and self-adaptive driving term possess over one hundred attractors\cite{feudel1996map}.
Compared to bistability phenomenon, the study of generalized multistability is far from being complete. The phenomena and mechanisms of multistability involve rich dynamics to be explored. Here we study the multiple stability behaviors in a classical mechanical system, the Doubochinski's pendulum\cite{tennenbaum2006amplitude}. We reveal that it's mysterious "amplitude quantization", initially studied as "macroscopic quantum behavior"\cite{tennenbaum2006amplitude} is, in fact, the self-adaptive behavior in response to nonlinear feeding.
We consider a generalized model of an oscillator excited by periodic force $f(x)cos \omega t$ where the feeding function $f(x)$ is a nonlinear function of displacement. This model containing restoring force, dissipation and nonlinear feeding can be regarded as generalization of parametric oscillation. For a linear parametric oscillator governed by Mathieu equation, subharmonic resonance occurs when the driving frequency is near twice the natural frequency of an oscillator \cite{nayfeh2008nonlinear}.
To understand the mechanism of multistability, we express the feeding function $f(x)$ as series of polynomial and solved the amplitudes and frequency respond curves analytically. The origin of multiple solutions and strong self-adaptivity can be explained. Moreover, the even-odd correspondence between the symmetry of the driving force and subharmonic resonance frequencies are presented numerically.
To investigate the transition and competition between the multi-stable modes, we present the frequency response curves and Poincare maps near $\omega =3\omega_0$ and $\omega =5\omega_0$. The transition between multistable states is more complex and intriguing than in bistability. We find irreversible transition between the multistable modes and propose a multistability control strategy. In contrast, the frequency response for bistability states forms a closed hysteresis loop rather than an open route.
\section{\label{sec:level2}Theoretical Model}
\subsection{Dynamic Equation}
A general form of nonlinearly excited oscillator can be described by the following dynamic equation
\begin{equation}\label{differential_equation_1}
\ddot{x}+2h\dot{x}+f_{r}(x)=f(x)cos \omega t
\end{equation}
where $x$ is the displacement, $h$ is the linear damping coefficient, $f_{r}(x)$ is the restoring force of the system, $f(x)cos\omega t$ is a nonlinear periodic driving force and $\omega$ is the driving frequency. $f(x)$ is the displacement-dependent amplitude of the driving force, called feeding function here. There are various physical systems described by equations of similar forms. For example, when $f_{r}(x)$ contains a positive linear term and a cubic term, and the feeding function $f(x)$ is a constant, Eq.\ref{differential_equation_1} becomes Duffing's equation \cite{kovacic2011duffing}. A nonlinear parametric oscillator is described by an equation similar to Eq.\ref{differential_equation_1}, where $f(x)$ is proportional to displacement. \cite{rhoads2006generalized}.
When applied to a pendulum the restoring force can be written as
\begin{equation}\label{differential_equation_2}
f_r(x)= \omega_{0}^{2}sin(x)
\end{equation}
where $\omega_{0}$ is the pendulum's natural frequency and $x$ is the angular displacement.
The actual form of the feeding function is cumbersome, so we approximate the feeding function in terms of polynomial of $x$ to do analytical calculation. Specifically, if $f(x)$ is an even symmetric function of $x$, the polynomial approximation only contains even-degree terms,
\begin{equation}\label{eq3}
f_{even} (x)= a_0 +a_2 x^2 + a_4 x^4+...
\end{equation}
If $f(x)$ is an odd symmetric function of $x$, it only contains odd-degree terms.
\begin{equation}\label{eq4}
f_{odd}(x) = a_1 x + a_3 x^3+...
\end{equation}
In many cases, the feeding function $f(x)$ is considerable in an active zone and becomes negligible elsewhere, such a pendulum is called a "kick-excited" one. In Ref. \cite{damgov2000discrete}, a symmetric $\Pi$ shaped feeding function is considered. Here, the polynomial series is truncated after the lowest few terms to model the feeding function. The coefficients are determined by fitting $f(x)$ of a particular physical model using above polynomials. Since the polynomial usually does not converge to the practical feeding function for arbitrary displacement, the polynomial approximation only applies to the range of oscillation or the "active zone" of the feeding function unnecessarily being small.
\begin{figure}
\centering
\subfigure[]{
\label{1-0}
\includegraphics[width=0.2\textwidth]{setup-1.pdf}}
\subfigure[]{
\label{1-1}
\includegraphics[width=0.2\textwidth]{setup-2.pdf}}
\subfigure[]{
\label{1-2}
\includegraphics[width=0.2\textwidth]{Pfit_even.pdf}}
\subfigure[]{
\label{1-3}
\includegraphics[width=0.2\textwidth]{Pfit_odd.pdf}}
\caption{\textbf{a},\textbf{b}, Schematic diagrams of the Doubochinski's pendulum.
The electromagnet is vertically(a) or horizontally(b) placed beneath the pendulum.
\textbf{c},\textbf{d}, Feeding function of Doubochinski's pendulum for vertically(c) or horizontally(d) placed electromagnet.
}
\label{Fig_1}
\end{figure}
\subsection{Feeding function for a magnetic pendulum}
Later on, we consider a practical model of a nonlinearly excited oscillator, the Doubochinski's pendulum. The pendulum consists of a light rigid pendulum with a small magnet at the free end. An AC powered electromagnet is vertically or horizontally placed beneath the pendulum, see Figs.\ref{1-0} and \ref{1-1}. The magnet pole's direction should be adjusted along the local geomagnetic field to eliminate the torque. The dynamic equation for the pendulum is
\begin{equation}\label{eq5}
\ddot{x}+2 h \dot{x}+\omega_{0}^{2}sinx=\frac{T(x)}{I}cos\omega t
\end{equation}
where $I$ is the moment of inertia of the pendulum, $T(x)$ is the torque applied to the magnet. We can write the dynamical equation in a dimensionless form.
\begin{equation}\label{eq6}
\frac{d^2x}{d\tau^2}+2\beta\frac{dx}{d\tau}+sinx=f(x)cos\Omega \tau
\end{equation}
where $\tau$ , $\beta$ and $f(x)$ are defined as
\begin{eqnarray}\label{eq_7Parameter}
\tau = \omega_0 t \qquad \beta = \frac{h}{\omega_0}\\
f(x) = \frac{T(x)}{I\omega_0 ^2} \qquad \Omega = \frac{\omega}{\omega_0}
\end{eqnarray}
The driving term arises from the alternating magnetic field. The torque can be evaluated by treating the magnet as a magnetic dipole, thus,
\begin{equation}\label{eq_8Driving force}
T(x)=\frac{\partial(M \cdot B)}{\partial x}
\end{equation}
where $B$ is the magnetic induction of the field, and $M$ is the magnetic moment of the magnet.
Here we calculate the feeding function in two typical configurations, parameters for calculation are given in table \ref{table}, involving natural frequency of the pendulum $\omega_0$, moment of inertia $I$, length of the pendulum $L$, radius of the coils $r$, number of turns $N$, height of the electromagnet coil $H$, distance between the coil and the magnet $d$, magnetic momentum of the magnet $M$ and the current $i$. These physical parameters will be used for numerical calculation.
The coefficients of the polynomial $a_0$, $a_2$, $a_4$ in Eq.\ref{eq3} fitting to the feeding function given by above parameters are also listed, which will be used in analytical calculation. As shown in Figs.\ref{1-2} and \ref{1-3}, when the electromagnetic coil is placed vertically, $f(x)$ has even symmetry; when the coil is placed horizontally, $f(x)$ has odd symmetry.
\renewcommand\arraystretch{1.3}
\begin{table}[h]
\centering
\caption{The parameters used to calculate the feeding function
and the fitting coefficients of the polynomial.}\label{table}
\squeezetable
\begin{ruledtabular}
\begin{tabular}{lcccc}
\multicolumn{5}{l}{\textbf{Parameters : }}\\%\specialrule{0.05em}{2pt}{2pt}
\hline
$\omega_0(rad/s)$&$I(kg \cdot m^2) $ & $L (m)$ & $r(m)$ & $N$ \\
\hline
5.13 &0.01 &0.456 & 0.042 &220 \\
\hline
$H(m)$ &$d(m)$ & $M(A \cdot m^2)$ &$i(A)$ & $\beta(rad/s)$ \\
\hline
0.02 & 0.021 & 1.36 & 0.5 & 0.0156 \\
\hline
\multicolumn{5}{l}{\textbf{fitting coefficients of the polynomial : }}\\
\hline
$a_0$&$a_2$&$a_4$& & \\
\hline
0.055&-11.6&300& & \\
\end{tabular}
\end{ruledtabular}
\end{table}
\section{Dynamic Behaviors}\label{Sec3}
\subsection{Discrete Periodic Orbits}\label{Discrete}
The pendulum's motion is governed by Eq.\ref{eq6} and Eq.\ref{eq_8Driving force}. We solve the equations when the electromagnet is vertically placed, using the feeding function as shown in Fig.\ref{1-2} and Runge-Kutta 4th order method in c++.
In Fig.\ref{Fig_3}, we find that there are several stationary oscillation modes near triple ($\Omega=2.98$) and quintuple of the natural frequency ($\Omega=4.98$).Figs.\ref{3-1} and \ref{3-2} are the time evolutions of typical periodic orbits. The final oscillation modes are dependent on initial conditions. Then we sample the initial condition in phase space of $(x, \dot{x})$, in search of stable oscillation orbits. All the trajectories converge to several limit cycles, as shown in Figs.\ref{4-1} and \ref{4-2}.
The small cycles in the middle are the linear solutions where the pendulum oscillates at the frequencies of the driving force. The larger cycles are nonlinear solutions where the pendulum oscillates near the natural frequency. The insets of Figs.\ref{4-1} and \ref{4-2} show the dual stationary solutions corresponding to the larger limit cycles. Though the two limit cycles are adjacent in phase diagram, they are different in phase. From Fig.\ref{Fig_3}, we also see that for the large stationary solutions, the pendulum approximately performs harmonic oscillation.
These results are consistent with experimental phenomenon \cite{njubook} that pendulum released at small angles only does small oscillations in the "linear region"; and for larger releasing angles, the amplitude varies slowly and ends up with stable harmonic oscillations on the large amplitude orbits or the small linear orbit.
\begin{figure}
\centering
\subfigure[$\Omega=2.98$]{
\label{3-1}
\includegraphics[width=0.2\textwidth]{xt_w=2_98.pdf}}
\subfigure[$\Omega=4.98$]{
\label{3-2}
\includegraphics[width=0.2\textwidth]{xt_w=4_98.pdf}}
\subfigure[$\Omega=2.98$]{
\label{4-1}
\includegraphics[width=0.2\textwidth]{w=2_98.pdf}}
\subfigure[$\Omega=4.98$]{
\label{4-2}
\includegraphics[width=0.2\textwidth]{w=4_98.pdf}}
\caption{\textbf{a},\textbf{b}, Multiple stationary solutions of Doubochinski's pendulum for driving frequency $\Omega=2.98, 4.98$. The final oscillation modes vary with initial conditions. \textbf{c},\textbf{d}, Limit cycles for $\Omega=2.98, 4.98$ respectively. The small cycles in the middle are the linear solutions. The larger cycles which appear in pairs are nonlinear solutions where the pendulum oscillates near the natural frequency. The insets show stationary solutions of the corresponding limit cycles in the same color. }
\label{Fig_3}
\end{figure}
To find the multiple periodic solutions analytically, we use a polynomial approximation of the feeding function Eq.\ref{eq3} and truncate it after the 4th order.
In Eq.\ref{eq3}, when $x<<1$, the leading term is a constant $a_0$. The pendulum only does regular linear forced oscillation at the driving frequency. At larger angles, the effect of nonlinearity becomes significant. When the pendulum performs large amplitude oscillation under driving frequency $\Omega \approx 3$, the zeroth order solution of Eq.\ref{eq6} should be written as
\begin{equation}\label{eq11}
x_0 = Acos(\frac{\Omega}{3}\tau+\phi)
\end{equation}
In order to capture the frequency response of the system, a factor $\sigma$ is added to the triple natural frequency
\begin{equation}\label{eq12}
\Omega =3 + \sigma
\end{equation}
Substituting Eq.\ref{eq11} into Eq.\ref{eq6} and averaging the amplitude and phase change over one period, we get the time derivatives of amplitude and phase,
\begin{equation}\label{eq13}
\dot{A}=-\frac{2A\beta}{1+\sigma}-\frac{A^2(4a_2+3A^2a_4)sin(3\phi)}{16(1+\sigma)}
\end{equation}
\begin{equation}\label{eq14}
\dot{\phi} = -\frac{A^2+16\sigma}{8(1+\sigma)}-\frac{A(4a_2+5A^2a_4)cos(3\phi)}{16(1+\sigma)}
\end{equation}
For simplicity, we set the damping coefficient $\beta=0$. These two equations approximately describe the evolution trajectories in $A-\phi$ space. The stationary solutions are the fixed points in $A-\phi$ space, which satisfy
\begin{equation}\label{eq15}
\dot{A}=0, \qquad \dot{\phi}=0
\end{equation}
The solutions are
\begin{center}
\begin{eqnarray}
4a_2A+5a_4A^3=32\sigma+2A^2 , \ \phi = 2k \frac{\pi}{3} \label{eq16} \\
4a_2A+5a_4A^3=-32\sigma-2A^2 , \ \phi = (2k+1) \frac{\pi}{3} \label{eq17}
\end{eqnarray}
\end{center}
in which $k \in \mathbb{Z}$.
\begin{figure}
\centering
\includegraphics[width=0.3\textwidth]{TheoryAW.pdf}
\caption{The analytical result of amplitude frequency response curves for $\Omega \approx 3$. The two black solid curves denote the larger stable roots, the blue dashed the unstable root. The linear forced oscillation solution is the solid line near the bottom.}\label{Fig_5}
\end{figure}
The term $2A^2$ can be ignored in further analysis because it is much smaller than $a_2 A$ and $a_4 A^3$, see table.\ref{table}.
When $\sigma =0$, the amplitudes solved from Eq\ref{eq16} and \ref{eq17} are the same,
\begin{equation}\label{eq18self-adaptive}
A =\pm \sqrt{\frac{-4a_2}{5a_4}} \ , 0
\end{equation}
Eq.\ref{eq18self-adaptive} indicates that nontrivial stationary solutions exist when $a_2$ and $a_4$ are different in sign. Here multi-stability is achieved without dissipation when the polynomial of the feeding function contains more than one nonlinear terms so that energy balance can be realized. Feeding function truncated after the 4th order can produce resonance at $\Omega \approx 3$. For higher subharmonic frequencies, feeding function with higher order terms should be taken into consideration.
The oscillation amplitudes in Eq.\ref{eq18self-adaptive} are independent of the amplitude of the driving force, since the $a_2$,$a_4$ are proportional to the amplitude of the driving force and the ratio will not change. Such strong self-adaptivity is verified in Ref.\cite{damgov2000discrete,njubook}, in which the amplitudes almost do not change with the driving force within a very large range. Actually, as amplitude of the driving force exceeds a critical value, the oscillator will undergo a series of symmetry breaking oscillation modes with multiple period and finally becomes chaotic. The period-3 bifurcations in a system with $\Pi$ shaped feeding function reported by Domgov \cite{damgov2000discrete} is a similar example.
The frequency response curves shown in Fig.\ref{Fig_5} can be solved from Eq.\ref {eq17} when $\sigma \neq 0$. The nonlinear solutions contain one smaller unstable root, denote by blue dotted curve, and a pair of larger stable roots, denote by black solid curves. The linear forced oscillation solution is also plotted, see the solid line near the bottom in Fig.\ref{Fig_5}.
\begin{figure}[htpb]
\centering
\label{6-1}
\includegraphics[width=0.4\textwidth]{AW_max.pdf}
\caption{The max-amplitude frequency response diagram. The red (blue) curves are for even (odd) feeding function. Subharmonic resonance occurs for $\Omega \approx 2k + 1 (2k) $ in response to even (odd) feeding function. The frequency response curve for linear forced oscillation is not plotted.}
\label{Fig_6}
\end{figure}
\subsection{Symmetry of the Driving Force and Frequency Response}\label{Frequency}
To clarify how symmetry of the feeding function influences the subharmonic frequencies, we calculate the overall frequency response diagram for even symmetric and odd symmetric $f(x)$. In calculation, we consider damping and using Eq.\ref{eq_8Driving force} and parameters listed in table.\ref{table} to model the feeding function, the frequency response diagram can be solved numerically. Since multiple periodic oscillation modes coexist when subharmonic resonance occurs, we only show maximum amplitudes on the frequency response diagram.
In Fig.\ref{Fig_6}, red curves correspond to even symmetric feeding function, and blue curves for odd symmetric feeding function. We find that subharmonic resonance occurs for $\Omega \approx 2k+1$ ($2k$) in response to even (odd) symmetric feeding function. Since an asymmetric function can be decomposed into an odd function and even function, we can infer that if the electromagnet is placed inclined, subharmonic resonance could occur for $\Omega \approx k$.
Fig.\ref{7-3} shows the frequency response diagram when $\Omega \approx 3$. Similar to analytical result in Fig.\ref{Fig_5}, it has a small linear solution, denoted as mode $a$, and a pair of intersecting branches denoted as mode $b_R$ and mode $b_L$ for $\Omega \approx 3$. $L$ and $R$ stand for right and left; $a$, $b$ and $c$ denote different branches. Here the unstable solutions are not presented because they are unavailable via numerical integral of Eq.\ref{eq6}.
\begin{figure*}[t]
\centering
\subfigure[$\Omega =2.940$]{
\label{7-0}
\includegraphics[width=0.21\textwidth]{w=2_940.pdf}}
\subfigure[$\Omega =2.945$]{
\label{7-1}
\includegraphics[width=0.21\textwidth]{w=2_945.pdf}}
\subfigure[$\Omega =3.000$]{
\label{7-2}
\includegraphics[width=0.21\textwidth]{w=3_030.pdf}}
\subfigure[$\Omega =3.040$]{
\label{7-25}
\includegraphics[width=0.3\textwidth]{w=3_040.pdf}}
\subfigure [$\Omega \approx 3$]{
\label{7-3}
\includegraphics[width=0.3\textwidth]{AW_w=3.pdf}}
\caption{\textbf{a},\textbf{b},\textbf{c},\textbf{d}, Evolution of attraction basins in Poincare maps with varying $\Omega$. The black dots are FPs for each basin. Attraction basin for each FP is denoted with a color. \textbf{e}, Frequency response curves for $\Omega \approx 3$. The arrows beside the frequency response curves indicate how oscillation modes evolve with driving frequency.}
\label{Fig_7}
\end{figure*}
Fig.\ref{8-4} is the frequency response diagram for $\Omega \approx 5$.
The five stable orbits in Fig.\ref{4-2} correspond to the one linear branch named $a$ and two pairs of overlapped branches, named as $b_L,b_R,c_L,c_R $.
\subsection{Modes Competition}\label{modes}
A Duffing oscillator with cubic nonlinearity in restoring force\cite{kovacic2011duffing} exhibits jump phenomena and hysteresis between bistable states due to hardening or softening resonance. Meanwhile, the frequency response curves of a nonlinear parametric oscillator with bistability exhibit mixed feature of softening and hardening\cite{rhoads2006generalized}. Here, more complex jump phenomenon, induced by modes competition among multiple attractors, is studied.
In order to study the transition behavior among the attractors, the frequency response graph is calculated when excitation frequency increases or decreases quasi-statically.
Transition happens when one orbit loses stability and the oscillator jumps to another orbit, as the control parameter, i.e. the driving frequency, changing quasi-statically.
In Fig.\ref{7-3}, the arrows beside the frequency response curves indicate how transition happens.
The frequency condition when all stationary modes $b_R$, $b_L$ and $a$ coexist is $\Omega_1 \leq \Omega \leq \Omega_2$. When $\Omega \leq \Omega_1$, modes $b_L$ and $a$ are stable, and when $\Omega > \Omega_2$, modes $b_R$ and $a$ are stable. As frequency increases, the oscillator in mode $b_L$ jumps to mode $b_R$ at $\Omega=\Omega_2$. Unexpectedly, as frequency decreases, transition from mode $b_R$ to mode $a$ happens at $\Omega=\Omega_1$.
Such irreversible transition behavior can be described from the evolution of attraction basins in Poincare maps.
Figs.\ref{7-0} to \ref{7-25} are the attraction basins when $\Omega= 2.94, 2.945, 3.0, 3.04$ respectively. In each figure, the small black dots are fixed points (FPs). Each FP is surrounded by its basin of attraction indicated with color. The red, green and purple areas denote the attraction basins for modes $a$, $b_L$ and $b_R$ respectively. Mode $a$ (the trivial solution) leaves one fixed point near the center in Figs.\ref{7-0} to \ref{7-25}. Meanwhile, each large-amplitude mode has 3 fixed points because the oscillation periods of modes $b_R$ and $b_L$ are triple of excitation period.
In Figs.\ref{7-1}, \ref{7-2} and \ref{7-25}, as the excitation frequency increases from $\Omega=2.945$ to 3.00 and 3.040, the attraction basin of $b_L$ shrinks quickly and vanished below 3.040. In Fig.\ref{7-2} just below the transition frequency, the vanishing attraction basins of $b_L$ surrounded by those of $b_R$ will merged into those of $b_R$ and transition from $b_L$ to $b_R$ happens.
In Figs.\ref{7-1} and \ref{7-0}, in vicinity of the critical frequency $\Omega_1$, the basins of $b_R$ are surrounded by those of mode $a$ (the red area). As the excitation frequency decreases from 2.945 to 2.940, the basins of mode $b_R$ are replaced by those of mode $a$ so that transition from mode $b_R$ to $a$ happens.
\begin{figure*}[htpb]
\centering
\subfigure[$\Omega =4.92$]{
\label{8-0}
\includegraphics[width=0.21\textwidth]{w=4_92.pdf}}
\subfigure[$\Omega =4.95$]{
\label{8-1}
\includegraphics[width=0.21\textwidth]{w=4_95.pdf}}
\subfigure[$\Omega =4.98$]{
\label{8-2}
\includegraphics[width=0.21\textwidth]{w=4_980.pdf}}
\subfigure[$\Omega =5.00$]{
\label{8-3}
\includegraphics[width=0.3\textwidth]{w=5_00.pdf}}
\subfigure [$\Omega \approx 5$]{
\label{8-4}
\includegraphics[width=0.3\textwidth]{AW_w=5.pdf}}
\caption{ \textbf{a},\textbf{b},\textbf{c},\textbf{d},
Evolution of the attraction basins in Poincare maps for $\Omega = 4.92,4.95,4.98,5.0$. The black dots are FPs for each basin. Attraction basin for each FP is denoted with a color.\textbf{e}, Frequency response curves for $\Omega \approx 5$. The arrows beside the frequency response curves indicate how oscillation modes evolve with driving frequency.}
\label{Fig_8}
\end{figure*}
Additionally, the frequency response diagram and Poincare maps for $\Omega \approx 5$ are presented, see Fig.\ref{Fig_8}.
$\Omega_1$ is the lower critical frequency for mode $c_R$, and $\Omega_2$ is the upper critical frequency for mode $c_L$. In Figs.\ref{8-0} to \ref{8-3}, the red, orange, green, purple and blue areas are the attraction basins of modes $a$, $b_L$, $b_R$, $c_L$ and $c_R$ respectively. We mainly focus on a typical transition starting from model $c_L$ and mode $c_R$ below.
Fig.\ref{8-4} shows that transition among different modes is also irreversible. When increasing the driving frequency, oscillator at mode $c_L$ will jump to mode $c_R$ at $\Omega=\Omega_2$. In Fig.\ref{8-2}, the attraction basins of mode $c_L$ are surrounded by those of $c_R$ below $\Omega_2$, and the basins of $c_L$ will merge into the $c_R$ basin as driving frequency exceeds $\Omega_2$, see Fig.\ref{8-3}.
When decreasing the driving frequency, oscillator at mode $c_R$ jumps to mode $b_L$ at $\Omega=\Omega_1$, rather than $c_L$. In Fig.\ref{8-1} and Fig.\ref{8-0}, the attraction basins of mode $c_R$ will merge into the basin of mode $b_L$ below $\Omega_1$.
From the transition sequence described in Fig.\ref{7-3} and Fig.\ref{8-4}, we propose a control strategy among the oscillation modes by adiabatically changing the driving frequency. For example, in Fig.\ref{7-3} starting from mode $b_L$, the oscillator can be induced to mode $b_R$ by increasing the driving frequency, then to mode $a$ by decreasing driving frequency. However, the transition from mode $a$ to higher oscillation modes are forbidden. It means that an oscillator initially performs linear oscillation in mode $a$, cannot be driven to higher orbits merely by changing the driving frequency.
\section{Conclusion}
The phenomena and mechanisms of multistability involve rich dynamics to be explored. In this paper, we investigate the multistability of a generalized nonlinear forcing oscillator excited by $f(x)cos \Omega t$. We take Doubochinski's Pendulum as an example. The multistability mechanism in our system is nonlinear parametric resonance. We express the energy feeding function into polynomial and find multi-stability can be achieved without dissipation when the polynomial contains more than one nonlinear terms so that energy balance can be realized. We also find the subharmonic resonance frequency conditions and demonstrate the even-odd correspondence between the symmetry of the feeding function and subharmonic resonance frequencies.
We present the frequency response diagrams and Poincare maps near subharmonic frequencies $3 \omega_0$ and $5 \omega_0$. The frequency response diagram contains one linear response branch and nonlinear branches appearing in pairs. We find irreversible transition phenomenon between the multistable modes.
Control of the multi-stability can be realized by changing the excitation frequency adiabatically.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,580
|
Barber's pole worms, especially those in undrenched sheep, usually peak in March although the peak this year is likely to be somewhat reduced due to the earlier searing temperatures (above 35°C) and dry conditions that would have decimated the number of worm eggs and larvae on pastures. The current return to conditions ideal for barber's pole means that eggs excreted by any adult worms still alive in sheep, will now cycle into infective larvae.
Monitoring the situation with worm egg counts will track the build-up, as in many areas the grass is/soon will be green and actively growing; visual indicators for larval survival on pasture. While good feed may mitigate the effects of a worm burden in older stock, it doesn't always in younger stock. If egg counts in drenched mobs are elevated 14 days after drenching, then it's time to ask, 'why'. Was it due to drench resistance or poor drenching technique?
The current availability of premixed, short-acting, multi-active combination drenches on the market makes managing drench resistance easier. One such drench back on the market and now in combination form is NAPfix®. It is a ready-to-use triple combination drench that contains naphthalophos (OP), abamectin (ML) and albendazole (BZ). Many producers are wary of the OP drenches, but this ready-to-use product should remove the earlier concerns about determining correct dose rates, and as there is no mixing required and no dust, the risk of OP exposure to farmers is reduced. Nevertheless, it is important to read the label very carefully and follow the manufacturer's recommendations.
We have had good rain over the last few days. It is easy to imagine that everyone has been as fortunate when you can hear rain on your roof. We will face increased barber's pole activity in our sheep. Worm Egg Counting (WEC) will keep us ahead of the game and ensure that the problem does not get out of hand. For people relying on 'keeping a close eye on the sheep' often production losses can be significant before signs are noticed. WormBoss has a list of laboratories providing WEC and culture facilities.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,377
|
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<!-- NewPage -->
<html lang="it">
<head>
<!-- Generated by javadoc (1.8.0_162) on Mon Apr 16 15:40:35 CEST 2018 -->
<title>HeaderView.Theme (library API)</title>
<meta name="date" content="2018-04-16">
<link rel="stylesheet" type="text/css" href="../../../stylesheet.css" title="Style">
<script type="text/javascript" src="../../../script.js"></script>
</head>
<body>
<script type="text/javascript"><!--
try {
if (location.href.indexOf('is-external=true') == -1) {
parent.document.title="HeaderView.Theme (library API)";
}
}
catch(err) {
}
//-->
</script>
<noscript>
<div>JavaScript is disabled on your browser.</div>
</noscript>
<!-- ========= START OF TOP NAVBAR ======= -->
<div class="topNav"><a name="navbar.top">
<!-- -->
</a>
<div class="skipNav"><a href="#skip.navbar.top" title="Skip navigation links">Skip navigation links</a></div>
<a name="navbar.top.firstrow">
<!-- -->
</a>
<ul class="navList" title="Navigation">
<li><a href="../../../rebus/header/view/package-summary.html">Package</a></li>
<li class="navBarCell1Rev">Class</li>
<li><a href="package-tree.html">Tree</a></li>
<li><a href="../../../deprecated-list.html">Deprecated</a></li>
<li><a href="../../../index-all.html">Index</a></li>
<li><a href="../../../help-doc.html">Help</a></li>
</ul>
</div>
<div class="subNav">
<ul class="navList">
<li><a href="../../../rebus/header/view/HeaderView.Style.html" title="annotation in rebus.header.view"><span class="typeNameLink">Prev Class</span></a></li>
<li><a href="../../../rebus/header/view/ImageLoader.html" title="class in rebus.header.view"><span class="typeNameLink">Next Class</span></a></li>
</ul>
<ul class="navList">
<li><a href="../../../index.html?rebus/header/view/HeaderView.Theme.html" target="_top">Frames</a></li>
<li><a href="HeaderView.Theme.html" target="_top">No Frames</a></li>
</ul>
<ul class="navList" id="allclasses_navbar_top">
<li><a href="../../../allclasses-noframe.html">All Classes</a></li>
</ul>
<div>
<script type="text/javascript"><!--
allClassesLink = document.getElementById("allclasses_navbar_top");
if(window==top) {
allClassesLink.style.display = "block";
}
else {
allClassesLink.style.display = "none";
}
//-->
</script>
</div>
<div>
<ul class="subNavList">
<li>Summary: </li>
<li>Field | </li>
<li>Required | </li>
<li>Optional</li>
</ul>
<ul class="subNavList">
<li>Detail: </li>
<li>Field | </li>
<li>Element</li>
</ul>
</div>
<a name="skip.navbar.top">
<!-- -->
</a></div>
<!-- ========= END OF TOP NAVBAR ========= -->
<!-- ======== START OF CLASS DATA ======== -->
<div class="header">
<div class="subTitle">rebus.header.view</div>
<h2 title="Annotation Type HeaderView.Theme" class="title">Annotation Type HeaderView.Theme</h2>
</div>
<div class="contentContainer">
<div class="description">
<ul class="blockList">
<li class="blockList">
<hr>
<br>
<pre>@Retention(value=SOURCE)
public static @interface <span class="memberNameLabel">HeaderView.Theme</span></pre>
</li>
</ul>
</div>
</div>
<!-- ========= END OF CLASS DATA ========= -->
<!-- ======= START OF BOTTOM NAVBAR ====== -->
<div class="bottomNav"><a name="navbar.bottom">
<!-- -->
</a>
<div class="skipNav"><a href="#skip.navbar.bottom" title="Skip navigation links">Skip navigation links</a></div>
<a name="navbar.bottom.firstrow">
<!-- -->
</a>
<ul class="navList" title="Navigation">
<li><a href="../../../rebus/header/view/package-summary.html">Package</a></li>
<li class="navBarCell1Rev">Class</li>
<li><a href="package-tree.html">Tree</a></li>
<li><a href="../../../deprecated-list.html">Deprecated</a></li>
<li><a href="../../../index-all.html">Index</a></li>
<li><a href="../../../help-doc.html">Help</a></li>
</ul>
</div>
<div class="subNav">
<ul class="navList">
<li><a href="../../../rebus/header/view/HeaderView.Style.html" title="annotation in rebus.header.view"><span class="typeNameLink">Prev Class</span></a></li>
<li><a href="../../../rebus/header/view/ImageLoader.html" title="class in rebus.header.view"><span class="typeNameLink">Next Class</span></a></li>
</ul>
<ul class="navList">
<li><a href="../../../index.html?rebus/header/view/HeaderView.Theme.html" target="_top">Frames</a></li>
<li><a href="HeaderView.Theme.html" target="_top">No Frames</a></li>
</ul>
<ul class="navList" id="allclasses_navbar_bottom">
<li><a href="../../../allclasses-noframe.html">All Classes</a></li>
</ul>
<div>
<script type="text/javascript"><!--
allClassesLink = document.getElementById("allclasses_navbar_bottom");
if(window==top) {
allClassesLink.style.display = "block";
}
else {
allClassesLink.style.display = "none";
}
//-->
</script>
</div>
<div>
<ul class="subNavList">
<li>Summary: </li>
<li>Field | </li>
<li>Required | </li>
<li>Optional</li>
</ul>
<ul class="subNavList">
<li>Detail: </li>
<li>Field | </li>
<li>Element</li>
</ul>
</div>
<a name="skip.navbar.bottom">
<!-- -->
</a></div>
<!-- ======== END OF BOTTOM NAVBAR ======= -->
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,287
|
La culture d'Oksywie est une culture archéologique localisée dans la région de la Poméranie orientale moderne autour de la basse Vistule, du au début du Ier siècle de notre ère. Elle tire son nom du village d'Oksywie, qui fait maintenant partie de la ville de Gdynia, dans le nord de la Pologne, où les premières découvertes archéologiques typiques de cette culture ont été exhumées.
Comme d'autres cultures de cette période, la culture Oksywie était influencée par les caractéristiques culturelles de La Tène et possédait des traits typiques des cultures baltes. La céramique de la culture d'Oksywie et ses coutumes funéraires indiquent des liens étroits avec la culture de Przeworsk. Les cendres des défunts de sexe masculin étaient placées dans des urnes noires, de bonne facture, avec une finition fine et autour, une bande décorative. Leurs viatique funéraire comportait des ustensiles et des armes - typiquement des épées dotées d'un seul tranchant. Les tombes étaient souvent couvertes ou marquées par des pierres. À contrario, les sépultures féminines étaient dotées d'objets appartenant à la sphère féminine.
Références
Âge du fer
Archéologie en Pologne
Przeworsk
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,047
|
require 'spec_helper'
describe LinkThumbnailer::Graders::LinkDensity do
let(:description) { double('description') }
let(:instance) { described_class.new(description) }
describe '#call' do
let(:action) { instance.call }
before do
instance.stub(:text).and_return(text)
instance.stub(:links).and_return(links)
end
context 'when text length is 0' do
let(:text) { '' }
let(:links) { [] }
it { expect(action).to eq(0.0) }
end
context 'when text length is > 0' do
let(:text) { 'foo' }
context 'and links is 0' do
let(:links) { [] }
it { expect(action).to eq(1.0) }
end
context 'and links is > 0' do
let(:links) { [1] }
it { expect(action).to be_within(0.001).of(0.666) }
end
end
end
end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,748
|
We have no vacancies as of today but a few are coming. First up is a 3/2 at 12th and Pennsylvanis, available at the end of August and showing by appt only now. This property will be no dogs, cats ok. Please see featured post for details. Downstream, we have a 1/1 for September and a 3/2 that will allow dogs. More info to follow as the availability nears.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,111
|
**A Paines Plough and Pentabus Theatre Company production**
by DUNCAN MACMILLAN
with Jonny Donahoe
**The first performance of Every _Brilliant Thing_
took place on 28 June 2013 at Ludlow Fringe Festival.**
by Duncan Macmillan
with Jonny Donahoe
Cast| Jonny Donahoe
---|---
Direction | George Perrin
Producer| Hanna Streeter
Line Producer| Francesca Moody
Stage Manager| Alicia White
_Every Brilliant Thing was commissioned with the support of Anne McMeehan and Jim Roberts. It was produced and toured with the support of Arts Council England's Strategic Touring Fund._
**TOUR DATES**
**2013**
28 June Ludlow Fringe Festival
5 July Ledbury Poetry Festival
**2014**
19 May Pentabus, Shropshire
20 May The Hive, Worcester
21 May The Edge, Much Wenlock
23 May Whaddon Jubilee Hall, Buckinghamshire
5-22 August Roundabout @ Summerhall, Edinburgh Festival Fringe
4 September Roundabout @ Margate Winter Gardens
7 September Bestival, Isle of Wight
22-25 September Roundabout @ Hackney Downs Studios
6-31 December Barrow Street Theatre, New York
**2015**
1 January-29 March Barrow Street Theatre, New York
2-3 May Machynlleth Comedy Festival
9 May Unity Theatre, Liverpool
12-16 May Theatre Royal Plymouth
18 May The North Wall, Oxford
19-24 May Roundabout @ Brighton Festival
21 May Quarterhouse, Folkestone
22 May The Spring, Havant
26 May The Dukes, Lancaster
28 May Trestle Arts Base, St Albans
30 May Leintwardine Village Hall
2 June Pulse Festival, Ipswich
3 June The Brewery, Kendal
4 June Harlow Playhouse, Harlow
5-6 June The Marlowe Theatre, Canterbury
25 June The Phoenix, Bordon
7 June Square Chapel, Halifax
10 June Marine Theatre, Lyme Regis
11 June The Muni, Colne
12-13 June The Lowry, Salford
15 June South Hill Park, Bracknell
17 June The Castle, Wellingborough
18 June The Garage, Norwich
19 June Key Theatre, Peterborough
21 June The Tolmen Centre, Constantine
23 June Salisbury Arts Centre, Salisbury
26-27 June Derby Theatre, Derby
6 -11 July Roundabout @ Southbank Centre, London
10-30 August Roundabout @ Summerhall, Edinburgh Festival Fringe
9 Sept The Civic, Stourport on Severn
11 Sept Bridport Arts Centre, Dorset
16 Sept Mill Arts Centre, Banbury
21 Sept Isle of Eigg
23 Sept Arainn Shuainert, Strontian
24 Sept Plockton Village Hall
25 Sept Carigmonie Centre, Drumnadrochit
26 Sept Lyth Arts Centre
27 Sept Pier Arts Centre, Orkney
29 Sept Carraigeworks, Leeds
1 Oct South Holland Centre, Spalding
2-3 Oct Pegasus Theatre, Oxford
4 Oct Chipping Norton Theatre,
6-10 Oct Tobacco Factory, Bristol
16-17 Oct Stephen Joseph Theatre, Scarborough
14 Oct Stahl Theatre, Peterborough
18-20 Oct The Old Market, Brighton
21 Oct Shopfront Theatre, Coventry
22 Oct Span Arts, Narberth
23 Oct Riverfront Theatre, Newport
24 Oct Pontardawe Arts Centre
29 Oct Lakeside Theatre
31 Oct The Edge Theatre, Chorlton
**BIOGRAPHIES**
**DUNCAN MACMILLAN (Writer)**
Duncan is a former Writer-in-Residence at Paines Plough and Manchester Royal Exchange. Plays include the acclaimed adaptation of George Orwell's 1984, co-adapted/ co-directed with Robert Icke (Headlong / Nottingham Playhouse, UK tour, Almeida Theatre and West End); EVERY BRILLIANT THING (Paines Plough / Pentabus); 2071 co-written with Chris Rapley (Royal Court / Hamburg Schauspielhaus); REISE DURCH DIE NACHT adapt. Friederike Mayröcker created with Katie Mitchell and Lyndsey Turner (Schauspielhaus Köln, Theatertreffen, Festival d'Avignon); WUNSCHLOSES UNGLÜCK adapt. Peter Handke (Burgtheater Vienna); THE FORBIDDEN ZONE (Salzburg Festival / Schaubühne Berlin); LUNGS (Studio Theatre Washington DC / Paines Plough & Sheffield Theatres); ATMEN (Schaubühne Berlin); DON JUAN COMES BACK FROM THE WAR adapt. Ödön von Horváth (Finborough Theatre); MONSTER (Royal Exchange / Manchester International Festival).
Duncan was the recipient of two awards in the inaugural Bruntwood Playwriting Competition, 2006. Other awards include Best New Play at the Off West End Awards 2013 for LUNGS; the Big Ambition Award, Old Vic 2009; the Pearson Prize, 2008. His work with director Katie Mitchell has been selected for Theatertreffen and the Avignon Festival.
**JONNY DONAHOE**
Jonny Donahoe is an actor, comedian and writer, best known as the front man of comedy-band Jonny & the Baptists. As a comedian he has toured across the UK, Ireland and Europe.
Credits include: THE NOW SHOW, SKETCHORAMA, THE COMEDY CLUB and INFINITE MONKEY CAGE (Radio 4). TV credits include: FRESH FROM THE FRINGE (BBC TV), THE GADGET SHOW (Channel 5). Theatre credits include: MARCUS BRIGSTOCKE'S EARLY EDITION (Latitude Festival & Edinburgh Festival 2012) and 9 LESSONS & CAROLS FOR GODLESS PEOPLE (Bloomsbury Theatre).
He was a finalist in New Act of the Year, Musical Comedy Awards and BBC New Talent.
For Jonny and the Baptists he was commissioned to write and perform at the National Theatre, (and on a UK tour), as part of Mark Thomas's 100 ACTS OF MINOR DISSENT. One of the songs written for this was subsequently recorded and released as a UK single, sold out its entire run, and became number one in the iTunes UK comedy charts.
For his role in EVERY BRILLIANT THING he has been nominated for a Lucille Lortel Award and a Drama Desk Award.
**GEORGE PERRIN (Direction)**
George Perrin is the joint Artistic Director of Paines Plough. He was formerly co-founder and Joint Artistic Director of nabokov and Trainee Associate Director at Paines Plough and Watford Palace Theatre.
Directing credits for Paines Plough include LUNGS by Duncan Macmillan, THE INITIATE by Alexandra Wood, OUR TEACHER'S A TROLL by Dennis Kelly (Roundabout Season, Edinburgh Festival Fringe and National Tour), NOT THE WORST PLACE by Sam Burns (Sherman Cymru, Theatr Clwyd), SEA WALL by Simon Stephens (Dublin Theatre Festival / National Theatre Shed), GOOD WITH PEOPLE by David Harrower (59East59 Theatres New York / Traverse Theatre / Oran Mor), LONDON by Simon Stephens (National Tour), SIXTY-FIVE MILES by Matt Hartley (Hull Truck), THE 8TH by Ché Walker and Paul Heaton (Latitude Festival/ Barbican / Manchester International Festival / National Tour), DIG by Katie Douglas (Oran Mor / National Tour) and JUICY FRUITS by Leo Butler (Oran Mor / National Tour).
As Trainee Associate Director of Paines Plough, directing credits include HOUSE OF AGNES by Levi David Addai, THE DIRT UNDER THE CARPET by Rona Munro, CRAZY LOVE by Ché Walker, MY LITTLE HEART DROPPED IN COFFEE by Duncan Macmillan and BABIES by Katie Douglas.
Further directing credits include 2ND MAY 1997 by Jack Thorne (Bush Theatre), TERRE HAUTE by Edmund White (59East59 Theatres New York, West End, National Tour and Assembly Rooms, Edinburgh Festival Fringe), IS EVERYONE OK? and PUBLIC DISPLAYS OF AFFECTION by Joel Horwood and CAMARILLA by Van Badham (nabokov).
_"Revered touring company Paines Plough"_ **Time Out**
Paines Plough is the UK's national theatre of new plays. We commission and produce the best playwrights and tour their plays far and wide. Whether you're in Liverpool or Lyme Regis, Scarborough or Southampton, a Paines Plough show is coming to a theatre near you soon.
_"The lifeblood of the UK's theatre ecosystem"_ **The Guardian**
Paines Plough was formed in 1974 over a pint of Paines bitter in the Plough pub. Since then we've produced more than 150 new productions by world renowned playwrights like Stephen Jeffreys, Abi Morgan, Sarah Kane, Mark Ravenhill, Dennis Kelly and Mike Bartlett. We've toured those plays to hundreds of places from Manchester to Moscow to Maidenhead.
_"That noble company Paines Plough, de facto national theatre of new writing"_ **The Daily Telegraph**
We celebrated 40 years of Paines Plough with our biggest, boldest, furthest-reaching programme of work ever. Programme 2014 saw 12 productions touring to 54 places around the UK, featuring the work of 104 playwrights.
_"I think some theatre just saved my life"_
**@kate_clement on Twitter**
Paines Plough Limited is a company limited by guarantee and a registered charity.
Registered Company no: 1165130
Registered Charity no: 267523
Paines Plough, 4th Floor, 43 Aldwych, London WC2B 4DN
\+ 44 (0) 20 7240 4533
office@painesplough.com
www.painesplough.com
Follow @PainesPlough on Twitter
Like Paines Plough at facebook.com/PainesPloughHQ Donate to Paines Plough at justgiving.com/PainesPlough
**Paines Plough are:**
Joint Artistic Directors| James Grieve George Perrin
---|---
Producer| Hanna Streeter
General Manager| Aysha Powell
Assistant Producer| Francesca Moody
Administrator| Natalie Adams
Trainee Producer| Rachel D'Arcy
Trainee Administrator| Bhavini Goyate
Associate Director| Stef O'Driscoll
Production Placement (Goldsmiths)| Joseph Guerin
Admin Placement (City)| Claudel Goy
Technical Director| Colin Everitt
Press Representative| Kate Morley
**Board of Directors**
Caro Newling (Chair), Christopher Bath, Tamara Cizeika, Kim Grant,
Dennis Kelly, Zarine Kharas, Matthew Littleford, Anne McMeehan,
Christopher Millard, Cindy Polemis, Andrea Stark
_"A beautifully designed masterpiece in engineering... a significant breakthrough in theatre technology."_
**The Stage**
Roundabout is Paines Plough's beautiful new portable in-the-round theatre. It's a completely self-contained 168-seat auditorium that flat packs into a single lorry and can pop up anywhere from theatres to school halls, sports centres, warehouses, car parks and fields.
We built Roundabout to enable us to tour to places that don't have theatres. For the next decade Roundabout will travel the length and breadth of the UK bringing the nation's best playwrights and a thrilling theatrical experience to audiences everywhere.
Roundabout was designed by Lucy Osborne and Emma Chapman at Studio Three Sixty in collaboration with Charcoalblue and Howard Eaton.
**WINNER of Theatre Building of the Year at The Stage Awards 2014**
_"The@painesplough Roundabout venue wins most beautiful interior venue by far @edfringe."_
**@ChaoticKirsty on Twitter**
"Roundabout is a beautiful, magical space. Hidden tech make it Turkish-bath-tranquil but with circus-tent-cheek. Aces."
**@evenicol on Twitter**
We are the nation's rural theatre company.
Our mission is to develop and produce quality new theatre about the contemporary rural world. To tour the country with plays that have local impact and national resonance. To turn up in village halls, theatres and people's digital backyards, connecting audiences nationwide.
_"The excellent touring company Pentabus"_ **The Guardian**
Since 1974 we've produced 150 new plays, supported 100 playwrights and reached over half a million audience members. We've won awards, pioneered live-streaming and developed a ground-breaking initiative to nurture young writers from rural backgrounds.
Our plans for the future will see us tour further than ever before, work with new and established playwrights, extend our young writers programme and continue to push at the boundaries of what theatre can be.
Pentabus is a registered charity (number 287909).
We rely on the generosity of our donors, small and large,
to help us to make brilliant new theatre.
You can find out more about us at www.pentabus.co.uk
Twitter @pentabustheatre
Facebook Pentabus Theatre
**Pentabus are:**
Artistic Director| Elizabeth Freestone
---|---
Managing Director| Francesca Spickernell
Admin and Audience Development| Crayg Ward
Projects Producer| Jenny Pearce
Freelance Producer| Verity Overs-Morrell
Channel 4 Writer-in-Residence| Joe White
Associate Artist| Simon Longman
Bookkeeper| Lynda Lynne
Volunteer| Mike Price
Technical Stage Manager| Sam Eccles
Pentabus Theatre Company, Bromfield, Ludlow, Shropshire, SY8 2JU
Pentabus is also supported by The Millichope Foundation
**EVERY BRILLIANT THING**
Duncan Macmillan
with
Jonny Donahoe
OBERON BOOKS
LONDON
WWW.OBERONBOOKS.COM
First published in 2015 by Oberon Books Ltd
521 Caledonian Road, London N7 9RH
Tel: +44 (0) 20 7607 3637 / Fax: +44 (0) 20 7607 3629
e-mail: info@oberonbooks.com
www.oberonbooks.com
Copyright © Duncan Macmillan, 2015
Duncan Macmillan is hereby identified as author of this play in accordance with section 77 of the Copyright, Designs and Patents Act 1988. The author has asserted his moral rights.
All rights whatsoever in this play are strictly reserved and application for performance etc. should be made before commencement of rehearsal to Casarotto Ramsay & Associates Limited, Waverley House, 7-12 Noel Street, London, W1F 8GQ (info@casarotto.co.uk). No performance may be given unless a licence has been obtained, and no alterations may be made in the title or the text of the play without the author's prior written consent.
You may not copy, store, distribute, transmit, reproduce or otherwise make available this publication (or any part of it) in any form, or binding or by any means (print, electronic, digital, optical, mechanical, photocopying, recording or otherwise), without the prior written permission of the publisher. Any person who does any unauthorized act in relation to this publication may be liable to criminal prosecution and civil claims for damages.
A catalogue record for this book is available from the British Library.
PB ISBN: 978-1-78319-143-7
EPUB ISBN: 978-1-78319-642-5
Cover design by Benedict Lombe
Printed, bound and converted
by CPI Group (UK) Ltd, Croydon, CR0 4YY.
Visit www.oberonbooks.com to read more about all our books and to buy them. You will also find features, author interviews and news of any author events, and you can sign up for e-newsletters so that you're always first to hear about our new releases.
_For Dad_
Contents
Acknowledgements
Note
Every Brilliant Thing
_Acknowledgements_
George Perrin and Jonny Donahoe.
Anne McMeehan and Jim Roberts.
James Grieve, Elizabeth Freestone, Alicia White, Hanna Streeter, Francesca Moody, Benedict Lombe, Natalie Adams, Tara Wilkinson, Claire Simpson, Aysha Powell.
Jean Doumanian, Scott Morfee, Tom Wirtshafter, Patrick Daly, Preston Copley, Saul Nathan-Kazis, Kathryn Willingham, Kate Morrow, Victoria Gagliano, Amy Dalba, Richard Hodge, Charlene Speyerer, Matt Allamon, Josh Kohler.
Rosie Thomson, Ace Lawson, the Miniaturists, the Apathists, Lucy Prebble, Gugu Mbatha-Raw, Paul Burgess, Simon Daw, Kanatip LoukGolf Soonthornrak, Tom Dingle, Jamie Cullum, Sophie Dahl, Poppy Corbett, Ali Mason, Emma Campbell-Jones, Imogen Kinchin, Camilla Kinchin, Sian Reese-Williams, Stef O'Driscoll, Sean Linnen, Dominic Kennedy, Becki Willmore, Nikole Beckwith, Tom Richards, Robert Icke, Katie Mitchell, Josie Long, Daniel Kitson, Mark Thomas, Jessica Amato.
Mike Bartlett, Clare Lizzimore.
Paddy Gervers, Anna Knight.
Will Young. Rachel Taylor.
Michelle Beck. Effie Woods.
The play also owes a debt to those who have contributed to the list over the years.
_Every Brilliant Thing_ is a collaboration between myself, George Perrin and Jonny Donahoe. It is an adaptation of my short story 'Sleeve Notes', originally written for the Miniaturists and performed by Rosie Thomson at Southwark Playhouse, Theatre503 and the Union Theatre and by myself at Trafalgar Studios, the Old Red Lion and Village Underground, by Gugu Mbatha-Raw at 93 Feet East and by various people at the Latitude Festival. George and I worked for over a decade to turn it into a full-length play. During this time it has been through several incarnations, including an installation created by Paul Burgess and Simon Daw for Scale Project. This particular incarnation was developed with Paines Plough and Pentabus with support from Anne McMeehan and Jim Roberts. The play wouldn't exist were it not for George's persistence, his enthusiasm for the story and his openness to work in an entirely new way.
It also owes a particular debt to Jonny Donahoe who, drawing on his experience as a stand-up comedian, found ways to tell the story and use the audience that George and I couldn't have conceived of. By its nature, the play is different every night and, as such, Jonny essentially co-authored the play while performing it.
This text was published after two years of devising, several trial performances around the UK, runs in Edinburgh and London and a four-month run at the Barrow Street Theatre in NYC. It has been filtered through Jonny's interactions with hundreds of audiences. I've provided footnotes throughout to explain certain aspects of the play in performance and to give examples of some of the things that have happened unexpectedly.
DM
Note
The NARRATOR can be played by a woman or a man of any age or ethnicity. In the first production, the NARRATOR was performed by a man, so appears as such in the text. The play should always be set in the country it's being performed in and references should be amended to reflect this.
The word NARRATOR is included for ease of reading. It is never heard by the audience and shouldn't be included in programmes or production materials.
There is no interval.
*****
_The houselights are on full and will remain so throughout. There is no set. The AUDIENCE are seated in the most democratic way possible, ideally in the round. It is vital that everyone can see and hear each other. Music is playing, some upbeat jazz – Cab Calloway, Cannonball Adderley, Hank Mobley or Duke Ellington perhaps. The NARRATOR is in the auditorium as the AUDIENCE enters, talking to people and giving them scraps of paper. As he does so, he explains that when he says a number he wants the person with the corresponding entry to shout it out._1
_Eventually, when everyone is seated, the music fades and the NARRATOR begins._
NARRATOR| The list began after her first attempt. A list of everything brilliant about the world. Everything worth living for.
---|---
|
1.Ice cream.
2.Water fights.
3.Staying up past your bedtime and being allowed to watch TV.
4.The colour yellow.
5.Things with stripes.
6.Rollercoasters.
7.People falling over.
| All things that, at seven, I thought were really good but not necessarily things Mum would agree with.
| I started the list on the 9th of November, 1987.2 I'd been picked up late from school and taken to hospital, which is where my Mum was.
| ,
| Up until that day, my only experience of death was that of my dog, Sherlock Bones.3
| Sherlock Bones was older than me, and he was a central part of my existence. He was really sick and so the Vet came around to put him down.
|
_The NARRATOR speaks to someone from the AUDIENCE._
| Would you mind, I'm going to get you to be the Vet, it's just that you have an immediate... _Veterinary_ quality.
|
_The NARRATOR gets the VET to stand._4
| It's alright, I won't ask you to do very much. Just stand here.
| And would you mind if I borrowed your coat?
|
_The NARRATOR takes a coat from someone else._
| Thank you.
| Okay, so you're the Vet, and I'm me as a seven-year-old boy, and this here...
|
_The NARRATOR holds the coat carefully in his arms, as if it's a docile animal._
|...this is Sherlock Bones. I know you because you're one of the parents from school. And you say something reassuring, like:
| _'You're doing the kind thing. It's not a moment too soon.'_
VET| You're doing the kind thing. It's not a moment too soon.5
NARRATOR| And I don't know what that means because I'm seven. I've no real concept of finality. Or mercy.
| But you are clearly a very kind man, so I trust you.
| Now do you have a pencil or a pen on you?
|
_The VET has one or the NARRATOR asks him to get one from someone in the AUDIENCE._6
| So that pencil is the needle.
| And inside that needle is an anaesthetic called pentobarbital. The dose is large enough to make the dog unconscious and then depress his brain, respiratory and circulatory systems, and to put him to sleep forever.
| _(To the owner of the coat.)_ It's completely blunt so we won't draw on your nice coat okay?
| When you're ready I want you to come over here and inject Sherlock Bones in the thigh.
|
_The VET approaches the NARRATOR and attempts the task._
| No, the thigh.
|
_If the VET is smiling or laughing:_
| Now I'm going to stop you for a moment there. There is one hard and fast rule while euthanising a child's pet and that is you really mustn't laugh as you do it. Totally changes the tone of the situation.
| So um, no...let's do this again. Go back to the start and try to respect the solemnity of the situation.
| Maybe take a moment.
| Okay. Let's try this again.
|
_The VET completes the task._
| Okay, now stroke his little head.
| Could someone with a watch tell me when thirty seconds has passed?
| I held Sherlock Bones, who I'd known my entire life. I held him as he died.
|
_The NARRATOR looks at the coat, stroking it gently._
| And I thought about the walk we'd had that morning. And about the smell of him in my room. His toys in the garden. The recently opened packet of dry food. His bed under the stairs. All the things that could now be thrown away.
|
_The NARRATOR looks at the coat for a little longer._
| And he became lighter. Or heavier, I'm not sure. But different.
|
_An AUDIENCE MEMBER tells the NARRATOR that thirty seconds have passed._
| And that was my experience of death.
| A loved one, becoming an object...
|
_The NARRATOR hands the coat to the VET._
|...and being taken away forever.
| Thank you.
|
_The VET returns to their seat._
| It's the 9th of November, 1987. It's dark and it's late. All the other kids had gone home long ago.
| Eventually, my Dad pulls up.
|
_The NARRATOR speaks to someone in the AUDIENCE._
| I'm going to ask you to be my Dad if that's okay. You don't have to do much, just sit here on this step.
|
_The NARRATOR indicates where DAD should sit._7
| Now, normally it's my Mum who picks me up and normally she's on time. Normally I travel in the back because I am seven and I make things sticky.
| But this time it's Dad. And it's late. And he opens the door to the front passenger seat.
|
_The NARRATOR indicates to DAD to open an imaginary passenger door._8
| Dad looked at me. I looked at him.
| When something bad happens, your body feels it before your brain can know what's happening. It's a survival mechanism. The stress hormones cortisol and adrenalin flood your system. It feels like a trapdoor opening beneath you. Fight or flight or stand as still as you can.
| I stood very still, looking at my Dad.
| Eventually, I got into the car. Dad had the radio on. He'd been smoking with the window down.
|
_The NARRATOR sits down next to the man._
| Now, actually what's going to happen is that I'm going to be my Dad and you're going to be me as a seven year old. You don't have to do much, you just say 'why'. Okay?
|
_The NARRATOR speaks as the DAD. He doesn't alter his voice._
DAD| Put on your seatbelt.
AUDIENCE| Why?
DAD| Because cars can be dangerous.
AUDIENCE| Why?
DAD| Because other drivers don't always pay attention.
AUDIENCE| Why?
DAD| Well, because there's lots to think about when you're a grown up. There are bills to pay and work to do and relationships to sustain and there's never enough time to do it all.
AUDIENCE| Why?
DAD| Because there are only twenty-four hours in a day.
AUDIENCE| Why?
DAD| Well, because that's how long it takes for the Earth to rotate.
AUDIENCE| Why?
DAD| Because...I don't know.
AUDIENCE| Why?
DAD| Because I don't know everything.
AUDIENCE| Why?
DAD| Because that's impossible.
AUDIENCE| Why?
DAD| Because there's only so much anyone can know.
AUDIENCE| Why?
DAD| Because if you were able to know everything then life would be unlivable.
AUDIENCE| Why?
DAD| Because then there would be no mystery, no curiosity, no creativity, no conversation, no discovery. Nothing would be new and we'd have no need to use our imaginations and our imaginations are what make life bearable.
AUDIENCE| Why?
DAD| Because in order to live in the present we have to be able to imagine a future that will be better than the past.
AUDIENCE| Why?
DAD| Because that's what hope is and without hope we couldn't go on.
AUDIENCE| Why?
DAD| Because...can you just put your seatbelt on?
AUDIENCE| Why?
DAD| Because we're going to the hospital.
AUDIENCE| Why?
DAD| Because that's where your mother is.
AUDIENCE| Why?
DAD| Because she hurt herself.
AUDIENCE| Why?
DAD| Because she's sad.
AUDIENCE| Why?
DAD| I don't know.
AUDIENCE| Why?
DAD| I just don't.
AUDIENCE| Why?
DAD| Put on your seatbelt.
AUDIENCE| Why?
DAD| Because your mother is in hospital.
AUDIENCE| Why?
DAD| Because she can't see anything worth living for.
AUDIENCE| Why?
| ,
NARRATOR| At least, that's how I like to remember it. But we actually just sat in silence. The only thing he said to me was:
|
_The NARRATOR feeds the 'DAD' the following line:_
DAD| Your Mother's done something stupid.
NARRATOR| I didn't know what that meant.
| ,
|
_The NARRATOR thanks the DAD and, if relocated, indicates for him to return to his seat._
| At the hospital, Mum saw me and said 'not him'. So I sat in the corridor next to an old couple...
|
_The NARRATOR sits next to an OLD COUPLE in the audience._
|...who bought me a carton of juice and some chocolate from the machine.
|
_He acquires a drink and some chocolate from the OLD COUPLE._9
| I don't know exactly when I had the idea for the list but it was here, with the old people, that I started to write it down.
|
_The NARRATOR eats the chocolate and calls out numbers._
|
1.Ice cream.
2.Water fights.
3.Staying up past your bedtime and being allowed to watch TV.
4.The colour yellow.
5.Things with stripes.
6.Rollercoasters.
7.People falling over.
|
_The NARRATOR does the following entries himself._
|
8.Juice.
9.Chocolate.
10.Kind old people who aren't weird and don't smell unusual.
| I don't like it.
|
_The NARRATOR hands the drink and chocolate back to the OLD COUPLE._
| Dad was in with Mum for ages. When he finally came out I followed him down the corridor, I followed him out of the hospital, I followed him to the car park, I followed him in to the car, I followed him up the driveway, I followed him in through the front door, I followed him down the hallway, I followed him up the stairs until we reached his study, where he went inside and closed the door before I could follow him any further.
| I waited to see what music he put on.
| I knew the rules. If it was this woman singing I could go into the room.
|
_'Gloomy Sunday' by Billie Holiday plays, beginning with her vocal._
| If it was the sort of music you could sing and dance to, it was okay to go in but I ran the risk of being hugged and spun around in his chair.
|
_Some upbeat vocal jazz plays – Cab Calloway perhaps._
| If no one was singing it meant Dad was working so I should be quiet.
|
_Some melodic instrumental jazz plays, John Coltrane or Bill Evans perhaps._
| And it if sounded like all the instruments were just falling down the stairs, it meant I should leave him alone.
|
_'Free Jazz' by Ornette Coleman plays – loud and chaotic. After a moment it fades to silence._
| ,
| So standing outside his door, I waited to see what he put on.
| ,
|
_'Free Jazz' by Ornette Coleman plays. After a moment it fades._
| I went downstairs and made myself some dinner. A ham and mayo sandwich. Just without the ham. I sat down in front of the TV and continued with the list.
| It occurred to me the list should be presented in no particular order. There was no way of saying that, for example, Danger Mouse was objectively better than Spaghetti Bolognese.
|
23.Danger Mouse.
24.Spaghetti Bolognese.
25.Wearing a cape.
26.Peeing in the sea and nobody knows.
| I stayed up late writing and fell asleep in the living room. Dad must have carried me upstairs.
| Mum didn't come home for a week or so.
| While she was away I had to speak to the school counsellor, which was actually just Mrs Patterson from upper school. She was a wonderful woman, the sort of woman you looked at and immediately trusted.
|
_The NARRATOR looks at a woman in the AUDIENCE._
| I'm going to ask you to be Mrs Patterson if that's okay. Now, what she'd do is, and it seems a little weird now but remember this was the Eighties and she got results, what she'd do is she'd take off her shoe...
|
_The NARRATOR waits for MRS PATTERSON to take off her shoe._10
| Then she'd take off her sock.
|
_The NARRATOR waits for her to take off her sock._11
| Then she'd put it on her hand and talk to you through her little sock-dog which she called – what did you call the sock-dog?
|
_The AUDIENCE member says a name, for instance 'Mostyn'._
| Yes! That's it, I remember now. What Mostyn would do is he'd ask questions like 'how are you feeling today?'
SOCK| How are you feeling today?
NARRATOR| I'm very well thank you Mostyn, how are you?
SOCK| I'm fine, thank you.
NARRATOR| You're brilliant. What kind of dog are you?
SOCK| I'm a... _(She specifies a breed or colour.)_
NARRATOR| Wow, that's amazing. When I was little we had a dog called Sherlock Bones, and he was a cross between a Border Collie and a Doberman, because a Border Collie and a Doberman lived next door to each other in our street and there was a very low hedge.
| You're brilliant, by the way. I really like you. I'm going to put you onto my list. 164. Mostyn the sock dog. Have I told you about my list?
SOCK| No, tell me about it.12
NARRATOR| I'm making a list of a thousand Brilliant Things. I'm not certain but I think I might be a genius.
|
_If MRS PATTERSON wishes to ask more questions that's fine, if not the NARRATOR moves on to:_
| It's been very nice talking to you, but can I go now?
SOCK| Yes.13
NARRATOR| Mum did eventually come home from the hospital, and by that time the list was eight pages long and had three hundred and fourteen things on it. I left it on her pillow with the title:
| 'Every Brilliant Thing.'
| She never mentioned it to me, but I knew she'd read it because she'd corrected my spelling.
| ,
| I kept speaking with Mrs Patterson and Mostyn once a week, then once a fortnight, then once a month and then one day I left the school and I never saw them again.
| I don't want to make it sound like my Mother was a monster or that my childhood was miserable because it wasn't.
| We had a piano in our kitchen. It wasn't a big kitchen but it was the warmest room in the house and we'd gather around it and sing soul songs. There's a Ray Charles song, 'Drown In My Own Tears' that she sang a lot. There's a moment halfway through that sends shivers down my spine.
|
_This moment of the song plays – the drums building and Ray Charles singing 'why can't YOU...' The song continues, quieter._
| The way he sings the word 'you' gets me every time. It's like it's coming out of someone else. We all used to howl it like wolves.
|
313.Having a piano in the kitchen.
314.The way Ray Charles sings the word 'You'.
|
_The music swells and continues to play for a few moments longer. The NARRATOR listens. It fades._
| I forgot about the list until her second attempt, just over ten years later.
| Dad showed up halfway through Chemistry. The same trapdoor feeling. Fight or flight. The same wordless drive to the hospital.
| As a teenager I dealt with it less well. I wore my heart on my sleeve.
| The night she came home, she sat at the kitchen table and said that if it wasn't for the ham and pineapple pizza lining her stomach from the night before she'd be dead. And I said:
| _'You took three weeks' worth of anti-depressants, a packet of Aspirin and half a tub of antihistamines. You're probably healthier than I am. If you're going to kill yourself go jump off a bridge.'_
| Rather than storm off I sat there and started to shovel food into my mouth. I'd spent ages on this meal and I was furious that she was sitting there, wishing she was dead and letting it go cold.
| There was a moment of absolute, deafening silence. And then she started to laugh. It was such a genuine laugh that after a while I found myself joining in. Eventually, Dad got up and left the table, going into his study to listen to records.
| I couldn't sleep that night. I started to clear out my room, packing up the things I wanted to keep and throwing away the things I didn't.
| I started shaking. Have you ever had that? Where you notice that your hands are shaking and your breathing is deeper and you're surrounded by bin bags full of your things and you realise that, you know, _I'm really upset_. I must be really upset.
| ,
| And then, inside a box under my bed, underneath some sticker albums, sea shells and action figures, I found the list. I sat on the floor and I read it through.
|
1.Ice cream.
| The younger me had dealt with this so much better. He wasn't self-righteous. The younger me was hopeful. Naïve, of course. But, hopeful.
| So once I got to the end of the list I picked up a pen and continued where that little boy had left off.
|
315.The smell of old books.
316.Andre Agassi.
317.The even numbered Star Trek films.
318.Burning things.
319.Laughing so hard you shoot milk out of your nose.
320.Making up after an argument.
| The next morning I sat at the end of Mum's bed and I read the list to her, and she got up and left the room. I followed her and read louder.
|
516.Winning something.
517.Knowing someone well enough to get them to check your teeth for broccoli.
| Over the next few days and weeks I would leave messages on the answer phone. I would turn off the radio or stand in front of the TV. I spent a lot of time talking to her back.
|
518.When idioms coincide with real-life occurrences, for instance: waking up, realising something and simultaneously smelling coffee.
521.The word 'plinth'.
| I began leaving Post-It notes around the house, stuck to various things. On her mirror was:
|
575.Piglets.
| On the kettle:
|
654.Marlon Brando.
| And on her bed:
|
11.Bed.
| Every morning I would open my door and I would see a small stack of yellow squares of paper. I became more inventive, writing on the inside of cereal packets or shoes, carving words into fruit or rearranging the fridge magnets.
|
201.Hammocks.
|...inside the lid of some mustard.14
|
324.Nina Simone's voice.
|...stencilled onto a baguette.
| It was my aim to reach a _thousand_. I wasn't allowed to cheat, which meant:
|
a.No repetition.
b.Things had to be genuinely wonderful and life-affirming.
c.Not too many material items.
| For a few months the list became my sole focus.
|
761.Deciding you're not too old to climb trees.
823.Skinny dipping.
| Then, the week before I left for university:
|
992.Knowing to jangle keys at the wildlife park if you want the otters to come out.
993.Having dessert as a main course.
994.Hairdressers who listen to what you want.
995.Bubble wrap.
996.Really good oranges.
| I started to be bothered by the thought that my Mum no longer loved my Dad. I put the thought out of my mind and returned to the list.
|
997.Cycling downhill.
998.Aromatic duck pancakes with hoisin sauce.
| It's common for the children of suicides to blame themselves. It's natural.
|
999.Sunlight.
| However much you know that you're not to blame, you can't help feeling like you failed them. It's not fair to feel this way. But it's natural.
| ,
| In the first week of university, I posted the list to my Mum, anonymously.15 When I returned that Christmas I found it on my desk, neatly folded back in its envelope. I still don't know whether or not she had read it. It certainly hadn't seemed to change her outlook.
| I put the list between the pages of a favourite book and I forgot about it.
| That Christmas was quiet. Difficult.
| In the New Year, Dad drove me back to university. He gave me a box of his records. I wanted to ask him why but I knew better than that. We didn't speak. We just listened to the radio.
|
_The NARRATOR sits down next to the person he cast as his DAD._
|
_Music plays – Ella Fitzgerald's 'My Melancholy Baby.' They listen for a moment, then it fades slowly as the NARRATOR speaks._
| I was quite shy at university. I didn't socialise. I'd mostly just listen to records in my room. I would even avoid lectures and seminars. But there was one lecture series that I never missed.
| It was lead by someone whose books I had read and loved and had inspired me to choose the course in the first place.
| Would you mind being my lecturer? It's just because you really look like her.16
|
_The NARRATOR selects someone from the audience to be the LECTURER, leads them to the centre of the room and gives them a copy of_ The Sorrows of Young Werther.
| This particular lecture series was on the Victorian Novella and built up to this one book, _The Sorrows of Young Werther_ by Johann Wolfgang von Goethe.
| What she would do is, at the start of the lecture, she would hold the book aloft...
|
_The LECTURER holds up the book._
| And then she would leave a long dramatic pause...
|...and when she felt she had everybody's undivided attention...
| ,
|...she would give a very accurate and detailed précis of the novel.
|
_The NARRATOR sits in the audience and waits._
| ,
|
_Eventually, the LECTURER realises they can simply read the plot summary on the back of the book._ _The summary will be different depending on the copy, but will basically say something like:_
LECTURER| Visiting an idyllic German village, Werther, a sensitive young man, meets and falls in love with sweet-natured Lotte. Although he realises that she is to marry Albert, he is unable to subdue his passion and his infatuation torments him to the point of despair. The first great 'confessional' novel, _The Sorrows of Young Werther_ draws both on Goethe's own unrequited love for Charlotte Buff and on the death of his friend Karl Wilhelm Jerusalem. The book was an immediate success, and a cult rapidly grew up around it, resulting in numerous imitations as well as violent criticism and suppression on the grounds of its apparent support of suicide.
|
_The NARRATOR puts his hand up._
NARRATOR| Excuse me, I have a question.
LECTURER| Yes?
NARRATOR| Are you saying that a book, that this book, caused people to take their own lives?
LECTURER| Yes.
NARRATOR| And you want _us_ to read that book?
LECTURER| Yes.
|
_The NARRATOR thanks the LECTURER and indicates for them to return to their seat._
| ,
NARRATOR| I left the lecture and went to the library.
| I read up on social contagions; obesity, divorce, suicide. We're all subconsciously affected by the behaviour of our peers.
| In the month after Marilyn Monroe's death by overdose, the number of suicides in the US increased by twelve percent. Every time suicide is front-page news, every time a celebrity or a character on prime-time television takes their own life there is a spike in the number of suicides.
| Suicide is contagious. It's called the 'Werther Effect', named after Goethe's protagonist.
| Discovering this fact really scared me. Then it made me angry. I thought about the way suicide was presented in films and on TV, how it was reported in the news.
| I found that the Samaritans had published a set of guidelines for how the media can report suicide intelligently. It's astonishing how rarely these guidelines are followed. They're really simple:
|
_The NARRATOR refers to a piece of paper._
| Don't provide technical details. Never suggest that a method is quick, easy, painless or certain to result in death.
| Avoid dramatic headlines, terms like 'suicide epidemic' or 'hot spot'.
| Avoid sensationalist pictures of video. Avoid excessive detail.
| Avoid using the word 'commit'. Don't describe deaths by suicide as 'successful'.
| Don't publish suicide notes.
| Don't publish on the front page.
| Don't ignore the complex realities of suicide and its impacts on those left behind.
| Include references to support groups, such as the Samaritans.
| Don't speculate on the reason. That's crucial.
|
_The NARRATOR puts away the paper._
| Don't supply simplistic reasons such as 'he'd lost his job' or 'she'd recently become bankrupt'.
| I read the book. _The Sorrows of Young Werther._ It was shit. Well, I didn't connect with it. I'd never been very interested in romance. Or at least, I hadn't been. Until I locked eyes with the only other person who was always in the library.
|
_'At Last' by Etta James begins to play and the NARRATOR locks eyes with an AUDIENCE MEMBER. This is now SAM._17 _The NARRATOR waves, blushingly. The vocal starts and the song continues as the NARRATOR speaks._
| For weeks we would sit opposite each other without speaking. Occasionally we'd make eye-contact and then immediately look away as if blinded by the sun.
| For the first time in my life I understood the lyrics of pop songs.
| And then finally, after weeks, I summoned-up the courage to say hello.
|
_Slowly, bashfully, the NARRATOR walks towards SAM. On his way he asks the person who read out 517 to check his teeth for broccoli, then gives_ The Sorrows of Young Werther _to someone else._
| Can you just...deal with this?
|
_As he is about to reach SAM, he suddenly turns to the person next to her._
| Can I move you?
|
_The NARRATOR gets the person next to SAM (usually their partner) to vacate their seat and move to the other side of the room. This is done very apologetically. Once relocated, the NARRATOR returns to SAM._
| Is anyone sitting here?
SAM| Not anymore.
NARRATOR| Oh good.
|
_The NARRATOR sits down in the empty seat._
| Hello.
SAM| Hello.
NARRATOR| What's your name?
|
_The AUDIENCE MEMBER says their name._
| No, in real life her name was Sam.
| What's your name?
SAM| Sam.
NARRATOR| Hi Sam. Nice to meet you. What are you reading?
|
_The NARRATOR addresses the AUDIENCE._
| Oh, I forgot, does anyone have a book? We're in the library so I need a couple of books.
|
_The NARRATOR indicates_ The Sorrows of Young Werther.
| Not that one.
|
_The NARRATOR gets a couple of books from the AUDIENCE and throws one into SAM's lap._
| What are you reading?
|
_SAM reads the title of the book._18
| What's it about?
|
_SAM reads the back of the book._
| Sounds really good.
|
_The NARRATOR tells SAM what he's reading and tries to explain how great it is:_19
| It's really good. In fact, why don't I lend it to you? And I could read _(says title of SAM's book)_ and we could meet up and talk about them, perhaps get a coffee sometime or a cup of tea or an or an or an orange juice, maybe, perhaps, if you'd like to, if you think that would be...20
|
_SAM agrees._
| I had a date! We began to meet up in the library. We'd swap books and discuss them over coffee. I read things I would never have encountered otherwise. I probably learned more from the books Sam gave me than from any of my course texts.
| After several months of reading and meeting and trying not to look at each other, Sam returned a book to me, one of my favourite childhood books, and said:
| _The NARRATOR says the lines and encourages SAM to repeat them back to him._
NARRATOR| Really interesting read.
SAM| Really interesting read.
NARRATOR| There's something really interesting in this book...
SAM| There's something really interesting in this book...
NARRATOR| That I want you to read.
SAM| That I want you to read.
NARRATOR| Now, this confused me because I'd already read the book. _I'd_ lent it to _her_. Because I'm an idiot, I didn't work out that it was code until weeks later, when I opened the book and the list dropped out.
| I was mortified. I'd never told anyone about my Mum. Ever. As a kid there were times when...I'd have nothing in my lunchbox or I wouldn't have socks on or something and I...I didn't want people to think that because my Mother was...I don't know. And out of context this was just a stupid, childish list. The idea that a list of nice things could combat hardwired depression was embarrassingly naïve.
| I got so upset I went to rip it in half...and then I noticed someone else's handwriting.
|
_The NARRATOR says each number. SAM reads all the entries._
|
1000.When someone lends you books.
1001.When someone actually reads the books you give them.
1002.When you learn something about someone that surprises you but which makes complete sense.
1003.Realising that for the first time in your life someone is occupying your every waking thought, making it hard to eat or sleep or concentrate, and that they feel familiar to you even though they're brand new.
1004.Finding an opportunity to say this in a way which doesn't involve being in the same room at the same time, as we're both shy and terrified of rejection and if I don't say something now, it'll never happen.
1005.Writing about yourself in the Third Person.
| I have some advice for anyone who has been contemplating suicide. It's really simple advice. It's this:
| Don't do it.
| Things get better.
| They might not always get brilliant.
|
_The NARRATOR indicates SAM._
| But they get better.
| ,
| What I'm about to say might be really hard for some of you to understand, particularly the younger members of the audience. Back then there was no way to communicate with anyone after midnight. No texting or instant messaging, no email or Facebook. This world was called '1998.'
| I couldn't do anything but stare at what Sam had written. For about three hours.
| Eventually, I just continued the list from where Sam had left off.
|
1006.Surprises.
1007.The fact that sometimes there is a perfect song to match how you're feeling.
|
_Music begins: 'Move On Up' by Curtis Mayfield. The NARRATOR moves quickly around the room._21
|
1008.Dancing in private.
1009.Dancing in public, fearlessly.
1010.Reading something which articulates exactly how you feel but lacked the words to express yourself.
| I wrote late into the night.
|
1427.Not worrying about how much money you're spending on holiday because all international currency looks like Monopoly money.
| I wanted to get to 2000 and I kept writing as the sun came up.
|
1654.Christopher Walken's voice.
1655.Christopher Walken's hair.
| So much to include that my hand cramped up.
|
1857.Planning a declaration of love.
| My morning alarm went but I'd not slept. I passed:
|
2000.Coffee.
| With:
|
2001.Films that are better than the books they're adapted from.
| And I kept going.
|
_The NARRATOR does the following entries himself, at speed:_
|
2002.Seeing someone make it onto the train just as the doors are closing, making eye-contact and sharing in this little victory.
2003.This song. Especially the drums on this track. The single ends at around four minutes but the album version continues for another five minutes and has the most insane drums. In fact...
2004.Any song with an extended drum break involving a full kit, bongos and cowbell, have you heard 'I'm a Man' by Chicago?
2005.'I'm a Man' by Chicago.
2006.Vinyl records. I'm not being pretentious, the sound quality is better, it isn't compressed and it's tactile, you get to feel the weight of it in your hands. You can't skip like with CDs or MP3s, you listen through to the entire album. Dad's room had records on every surface and I loved the gatefold sleeves, the artwork, I love reading through the acknowledgements and the sleeve notes, the story of the making of the object.
| The next morning I took the list and I ran to the library and Sam and I kissed for the very first time.
| From that moment on we spent every second together. I wrote new list entries every day as a gift for Sam.
|
_The NARRATOR continues with the list entries himself:_
|
2389.Badgers.
|
_The NARRATOR puts his hand on someone's shoulder._
|
2390.People who can't sing but either don't know or don't care.
| Pages and pages of it.
|
4997.Gifts that you actually want and didn't ask for.
4998.Falling asleep as soon as you get on a plane, waking up when you land and feeling like a time-traveller.
| Everywhere I looked, everything I thought about...
|
9993.Dreams of flying.
9994.Friendly cats.
9995.Falling in love.
9996.Sex.
9997.Being cooked for.
9998.Watching someone watching your favourite film.
9999.Staying up all night talking.
10000.Waking up late with someone you love.
| _The drum-break kicks in._
| Now, this is the drum break I was telling you about. I know what you're thinking: it just sounds like a bunch of drums but wait for it, you're about to hear...
|
_The NARRATOR waits for it._
|...bongos! You are not getting into this in the way I anticipated. Alright, fine, listen, I'm going to try to be a little bit more...American about this. Let's try...everyone put your right hand in the air.
|
_Everyone raises their right hand._
| I'm going to HIGH-FIVE THE ENTIRE ROOM!
|
_He high-fives as many people as he can._
|
_Eventually, the NARRATOR signals to the STAGE MANAGER to stop the record._
| No that was a big mistake. It's actually much harder than I anticipated.
|
_The NARRATOR is out of breath._
| My Mum...
| She would do this. Get carried away. Ups and downs.
| ,
| As a little boy, it was never shyness, or thoughtfulness. Happiness scared me because it was usually followed by...
| you know.
|
_The NARRATOR looks at SAM._
| This was all very new. Feeling like this.
| ,
| Studies have shown that children with depressed mothers have a heightened reactivity to stress. Mothers who are withdrawn leave children to fend for themselves and it actually changes the chemistry of the brain, the fight or flight impulse.
| But the real risk as I perceived it...
| ,
| The real risk, that I'd felt my whole life, was that I would one day feel as low as my Mum had and take the same action.
| Because alongside the anger and incomprehension is an absolute crystal clear understanding of why someone would no longer want to continue living.
| ,
| I took Sam back home to meet my parents. They were amazing. They were wonderful. They were fantastic. It was awful. It made it seem like I'd exaggerated everything from my childhood. My Dad made lasagne and played Cab Calloway records. My Mum laughed a lot and told a story about breaking a guy's nose on a train in Egypt. We drank a few bottles of wine and sang songs at the kitchen piano.
|
_The NARRATOR produces an electric keyboard and stands with it in the centre of the room. It doesn't have a stand, so for a moment he tries to work out how to play it. Then he recruits two people from the AUDIENCE to hold either end of it while he plays. He thinks about the logistics of the room and speaks to the people holding the keyboard._
| Um, because we're in the round, we're just going to do a very slow revolve.
| Clockwise, obviously.
|
_The NARRATOR speaks to the room._
| Mum would always sing first. She sang Ray Charles...
| _(Sings.) I'm so blue here without you it keeps raining more and more. Why can't_ YOU...
| Dad wouldn't normally sing. But he did this night. It was amazing. I'd never seen anything like it. He sang:
| _(Sings.) That's Life. That's what people say. Riding High in April, shot down in May._
| oh...and:
| _(Sings.) And now the end is near, and so I face the final curtain_.
| Which, for me, was a little too on the nose.
| And then, quite spectacularly:
| _(Sings.) Wake me up before you go-go, don't leave me hangin' on like a yo-yo._
| Which, because he'd clearly never heard it before, actually sounded like:
| _(Sings, to the tune of 'Fly Me To The Moon') Wake me up before you go-go, don't leave me hangin' on like a yo-yo._
| Sam sang the last song that night. 'Some Things Last A Long Time' by Daniel Johnston. I'd not heard it before.
|
_The NARRATOR sings a few lines of the song, ending:_
| _(Sings.) The things we did, I can't forget. Some things last a long time. Some things last a long time._
| ,
|
_The NARRATOR takes away the piano and his assistants return to their seats._22
| With Sam's encouragement, the list grew.
|
123321.Palindromes.
| People asked if they could read it, add to it, photocopy it. The document got scrawled all over with different handwriting in different colours, exclamation marks, underlining, asterisks, footnotes and amendments, drawings and even the odd diagram.
| Anything generic or universal (clean sheets, new socks, freshly cut grass, the smell of bacon) had already been included and entries had become more specific:
|
253263.The feeling of calm which follows the realisation that, although you may be in a regrettable situation, there's nothing you can do about it.
525924.Track 7 on every great record.
777777.The prospect of dressing up as a Mexican wrestler.
| Not the _action_ of dressing up as a Mexican Wrestler, but the _prospect_ of it.
| Sam and I got married. A year after university. Sam proposed. Got down on one knee. The whole thing.
| It was beautiful, it was...in fact, no, let's just do it.
| We were walking in a park near my parents' house. It was raining. I was saying that this is where I used to walk Sherlock Bones when I was a child. I kept walking and I thought she'd stopped to tie a shoelace because when I turned around she was on one knee.
|
_The NARRATOR turns around to look at SAM, who is down on one knee._23
| She took my hands and said...
SAM| Will you marry me?
NARRATOR| And I said yes.
| Let's kiss later.
|
_SAM returns to her seat._
| We picked a date. Hired a hall. Caterers. Band.
| Everyone was there. Even our old Vet. We didn't invite him, but he came.24
| Dad did a speech. It was the most wonderful, beautiful speech I'd heard in my entire life.
| And you know Dad, he hated public speaking. I said to him, Dad, you really don't have to say anything but he said...
|
_The NARRATOR gets the microphone, takes the DAD by the hand and leads him into the middle of the room._
|...no I really want to. I really want to take this opportunity to talk to everyone, so...
|
_The NARRATOR speaks into the microphone._
|...Ladies and Gentlemen, in a break from tradition, please welcome the Father of the Groom.
|
_The NARRATOR gives the DAD the microphone, asks him to wait for a moment, then sits next to SAM and links arms._
| Say what's in your heart Dad.
|
_The ' DAD' improvises a short speech, after which the NARRATOR hugs the DAD and lets them return to their seat._25
| I remember every word.
| After the reception, when most of the guests had gone home,26 Mum sat at the piano and played soul songs.
|
_The snippet of Ray Charles plays – 'Why can't YOU...'_
| ,
| After the wedding Sam and I went on holiday to Whitstable in Kent.27 We were so happy. The sun shone every day. We ate the most incredible seafood.
| We moved to London. We got jobs. A car. A joint bank account. A cat who peed on everything then ran away. We called her Margaret Scratcher. We settled into a routine. We saw less and less of each other. We were tired. We argued. We argued about money. We argued about whether we wanted to live in the city or the countryside. We argued about whether or not we should start a family.
| We had one argument in particular.
| Sam suggested that I talk to someone. Professionally.
| That made me so angry. I knew what depression was and I knew I was fine.
| I had a study at home and I'd sit in there, listening to records and reading the sleeve notes.
| The lives of other people have always fascinated me. I always read the liner notes in record sleeves. The trials and traumas behind the music. Tortured geniuses.
| Weldon Irvine. Albert Ayler. Ronnie Singer. Donny Hathaway. Amazing musicians. All took their own lives.
| I'm so grateful to be ordinary.
| Sam told me I was becoming morose. That I was isolating myself. Wallowing.
| She encouraged me to carry on with the list, but I found it hard to notice new things.
| 826978.
| ,
| 826978.
| ,
| 826978.
| ,
| The list ended, just one hundred and seventy three thousand and twenty two short of a million. It was finished. So I boxed it all up and threw it away.
| ,
| I sat in my study while Sam packed her things. I helped her carry boxes to her car. I stood in our doorway and she looked at me from the car.
| That horrible feeling when something is broken and can't ever be fixed. The trapdoor swinging open. Fight or flight or stay as still as you can.
| I'd been feeling like that for a long time.
| ,
| I watched her drive away.
| ,
| She left me a note, written in an album sleeve. She knew that when I wanted to think of her I'd look for the Daniel Johnston song she sang at my parent's house and, as always, I'd sit and read through the record sleeve.
| Sam's note said that she loved me and that when I was ready we should try again.
| But I didn't find the note for seven years.
| ,
| Perhaps Sam had been right. Perhaps I'd been difficult to live with. Difficult to love.
| But I couldn't hear it from her. I needed to talk to someone else.
| So, the night I found Sam's note, I did one of the oddest things I've ever done.
| ,
| Mrs Patterson?
MRS PATTERSON| Yes.
NARRATOR| I hope you don't mind me calling you so late, I know you've retired, I know that because I called the school and they gave me your number. I know this is really inappropriate but...I'm an ex-pupil of yours. I was the little boy with the list. Do you remember me?
MRS PATTERSON| Yes.
| ,
NARRATOR| You do?
MRS PATTERSON| Yes.
| ,
NARRATOR| You used to have a sock puppet, do you remember?
MRS PATTERSON| Yes.
NARRATOR| A black dog. Which, now I come to think of it is a little ironic. Mostyn, wasn't it?
MRS PATTERSON| Yes.
NARRATOR| I was always able to talk to Mostyn. This may sound strange, but, would it be possible to talk to Mostyn now?
|
_MRS PATTERSON takes off her shoe and her sock once again and puts the sock over her hand._
SOCK| Hello.
NARRATOR| Hello Mostyn. How're you?
SOCK| I'm fine, how're you?
NARRATOR| Well, I'm talking to a sock dog on the phone, so apparently not great.
| ,
| I'm sad.
| I'm really sad Mostyn and I don't know how to change that. And I wanted to speak to you because when I was a little boy you knew me better than anyone.
| I wanted to ask you: was I always like this? Do you remember what I was like?
SOCK| Yes.
NARRATOR| Was I happy?
|
_The NARRATOR leads the SOCK PUPPET through a brief conversation until a conclusion is reached that allows the NARRATOR to take the next step – either he's always been sad or he was once happy._
| ,
| Thank you. It means a lot to hear that from you.
| I'm sorry I called so late. I won't call you again. Goodnight.
MRS PATTERSON| Goodnight.
| ,
NARRATOR| I did talk to someone.
| A group. A support group.
| ,
| Hello everyone.
|
_The NARRATOR indicates for everyone to respond._
AUDIENCE| Hello.
NARRATOR| This is my first session. I've resisted doing this.
| I'm –
| you know,
| ,
| British.
| I now realise that it's important to talk about things. Particularly the things that are the hardest to talk about.
| When I was younger I was much better at being happy.
| At feeling joy.
| Being a grown-up, being conscious of the problems in the world, about the complexities, the tragedies, the disappointments...I'm not sure I can ever fully allow myself to be joyful. I'm just not very good at it. It's helpful to know there are other people who feel the same.
| I um –
| I made a list. Everything that's brilliant about the world.
| I began making it as a present for my Mum. It's kind of a long story.
| The list is –
| Actually, wait a second, I have it with me...
|
_The NARRATOR exits the stage, then returns with a trolley on which sit several large, heavy, worn boxes._
| You see, I threw it away but, unbeknownst to me, my partner at the time...
|
_The NARRATOR looks at SAM._
|...got it out of the trash and hid it in the garage under an old tablecloth and then left a note about it in the record sleeve of a Curtis Mayfield record...well, you don't need to know the details.
|
_He opens one of the boxes. It is full of scraps of paper, the list, written on pages of colouring books, on receipts and beer mats, on the backs of envelopes etc. He takes a moment to just look at it. He carefully takes out a stack and looks through it. He reads entries at random and drops them, scattering them on the floor._
| Peeling off a sheet of wallpaper in one intact piece.
|
_He reads another._
| Mork and Mindy.
|
_He holds up a sleeve from a shirt and reads what's written on it:_
| My new sleeveless top.
|
_He reads another._
| Old people holding hands.
| ,
|
_He smiles and looks around the room._
| If you live a long life and get to the end of it without ever once having felt crushingly depressed, then you probably haven't been paying attention.
| ,
| I wasn't around for the last time. I was in Australia with work and when I got the call I was on the beach. Dad wasn't around either. A neighbour complained about the exhaust fumes and eventually the police cut through the garage door. Hosepipe through the driver-side window.
| That surprised me actually, because Mum hated driving.
| She had poor circulation and would always complain about her ankles on long journeys. They say that it's a masculine way to choose to die. But I don't know what that means.
| There was a pad and pencil on the passenger seat but she hadn't written anything.
| I drove Dad to the funeral. We sat in silence. He smoked with the window down. I helped him with his tie.
| After the service, meeting my Mum's friends and colleagues, I realised how much the list had changed the way I see the world.
|
31.Birdsong.
45.Hugging.
341.Alcohol.
577.Tea and biscuits.
1092.Conversation.
| The list hadn't stopped her. Hadn't saved her. Of course it hadn't.
| ,
| I got a text from Sam.
|
_The NARRATOR gives SAM his phone to read._
SAM| I heard about your Mum.
| I'm so sorry.
| Give me a call.
| Anytime.
| I'd love to hear your voice.
| Love,
| Sam x
| Ps. I heard the other day that Beyonce is related to the composer Gustav Mahler. It occurred to me that this is a fact that should be on your list. Truly a brilliant thing.
| ,
| I stayed with Dad for a few months after the funeral. We'd spend the days walking or reading or listening to records. He'd fall asleep in his armchair and I'd sit at his desk and type up the list, starting at the very beginning.
|
1.Ice cream.
| It was a lot of work. Several weeks of sleepless nights. Once I got to the end I kept going from where I'd left off.
|
826979.The fact that Beyonce is Gustav Mahler's eighth cousin, four times removed.28
| I completed the list.
| ,
| I printed it out and left it in Dad's chair. I drove back to London.
| He never mentioned it directly, but when we spoke a few weeks later, he said 'thank you.'
DAD| Thank you.
NARRATOR| And he said 'I love you'.
DAD| I love you.
NARRATOR| I told him that sentimentality didn't suit him.
|
999997.The alphabet.
999998.Inappropriate songs played at emotional moments.
999999.Completing a task.
|
_The NARRATOR says the final entry._
|
1000000.Listening to a record for the first time. Turning it over in your hands, placing it on the deck and putting the needle down, hearing the faint hiss and crackle of the sharp metal point on the wax before the music begins, then sitting and listening while reading through the sleeve notes.
| ,
|
_'Into Each Life Some Rain Must Fall' by Ella Fitzgerald and the Ink Spots plays._
|
_The NARRATOR shakes hands with or hugs the members of the AUDIENCE who played the principal characters – the VET, LECTURER, MRS PATTERSON, DAD and SAM, indicating for applause to be directed to them and inviting them to bow._
|
_The NARRATOR then bows and leaves. The list remains scattered around the stage so that the AUDIENCE can look through the box and read the entries._
|
_The music continues to play as they exit._
1| The AUDIENCE will be involved throughout and need to feel relaxed and safe. Greeting them also helps the NARRATOR cast the play. Jonny would be in the theatre for at least half an hour before the start of the show to speak to as many people as possible and work out who he was going to use in performance. The pieces of paper should look like authentic parts of the list – ones written during childhood could be written in crayon for instance, others should be written on napkins, beermats and the backs of envelopes (for example).
---|---
2| This date should be amended to correspond to the NARRATOR's age.
3| Originally the dog's name was Ronnie Barker but we had to change that for the US. Other possible names included Charles Barkley and Edward Woofwoof.
4| The Vet can be a man or a woman.
5| Throughout the play, AUDIENCE members will be invited to play characters. They are allowed to say whatever they wish and the NARRATOR has to work with what he's given. Though improvisations shouldn't go on too long, the spontaneity of these interactions is a central element of the show.
6| There are several props used during the play but they should all be sourced from the AUDIENCE.
7| If the DAD can be seen by everyone, and if the NARRATOR can sit next to him, there's no need for him to move.
8| In the US Jonny would correct DAD: 'Actually, it's a British car, so –' and they would mime opening the other door.
9| In performance 'chocolate and juice from a machine' changed depending on what could be acquired from the AUDIENCE – 'a cup of tea and a sandwich' for instance, or 'some water and an apple.' These then have to be included as items 8 and 9. Jonny would often take someone's wine and react as a child would, saying 'spicy' and handing it back. In New York, we shortened it to just 'chocolate' and Jonny would hand one of the couple a bar of chocolate and then hold out his hand to get it back.
10| So far, nobody has refused to do this. In performance it sometimes took quite a while however. Jonny would acknowledge this, saying things like 'she liked to create a real sense of dramatic tension. She was a double-knotter, that's one of the things we always liked about her, very thorough.'
11| At one performance, MRS PATTERSON explained that she had a bad toe. Jonny added 'I remember her telling me how worried she was about her toe. It was a skydiving accident wasn't it?' MRS PATTERSON replied 'Yes that's right.' Jonny then included her in the list '165, Mrs Patterson and her extreme sports hobbies.'
12| Occasionally, the sock dog would say that, yes, he had mentioned the list to which Jonny would reply something like 'well, I'd still like to recap'.
13| If the SOCK/MRS PATTERSON insist that the NARRATOR stay and talk, the NARRATOR can beg childishly until they are released.
14| In George's production, this entry was written on an actual mustard lid.
15| Jonny would often tell the AUDIENCE how he uses this joke as a barometer for how the show was going, sometimes telling them: 'they don't laugh at that in the matinees.'
16| The LECTURER can be a man or a woman. Jonny would often describe the lecturer first, explaining that they always wore red-rimmed glasses for instance, before asking someone who is dressed exactly how he's just described to take part, saying 'I don't know why, but you really remind me of her...'
17| SAM can be male or female. For the purpose of this draft, it's a woman.
18| Books that have been contributed by the AUDIENCE have included _God's Gift to Women_ , _The Catcher in the Rye_ , _Fifty shades of Grey_ and _The Denial of Death_ among many others.
19| Occasionally Jonny would be given the perfect book to start a flirtatious conversation – _Jane Eyre_ for instance – and give the book's owner a big thumbs up. Most often though, he'd have a real struggle to make the book he's been given sound exciting. In one performance in New York he was given an enormous hardback history of Manhattan's sewage system. 'It's really...great' he enthused: 'You'd be surprised just how much there is to say about the history of Manhattan's sewage system. If you're going to read one book about the history of Manhattan's sewage system, it really should be this one.' He asked SAM if she had read the book and when she said she hadn't he replied 'No, of course not. No one has.' At one performance in Edinburgh Jonny was given _Nana_ by Émile Zola, in French. This prompted him to turn to the person who'd contributed it and say 'aren't you clever?' and then add 'it's funny, for a minute I forgot I could speak French.' He then began to discuss the book with its owner in French, a language he happens to speak fluently. The conversation was cut short by the sign interpreter who was having a tough enough time as it was. During a press night performance he was given a copy of _Macbeth_ , giving him a dilemma of whether or not to say the title out loud, given the circumstances. In London he was given a copy of my play _Lungs_ which was playing in the same theatre. Aware that I was sitting directly behind him he explained that he thought it was okay if you're into babies and the environment but that he preferred the more recent stuff.
20| Jonny would just keep going here until SAM agreed.
21| In George's production, Jonny would use a microphone during this section, and get the AUDIENCE MEMBERS to speak into it too. This was partly so we could have the music loud but also to introduce a microphone that will be used later.
22| This should be done in a way to ensure that the audience don't applaud at this point.
23| In one performance in Edinburgh, the wonderful stand-up Josie Long was chosen to play SAM. At this point, she took a receipt from her pocket, fashioned it into a ring and used it to propose to Jonny. Jonny wore the ring for the rest of the performance.
24| It could be the VET or the LECTURER who crashes the wedding, whichever seems more amusing for the particular performance.
25| This is probably the most unpredictable moment of the play. Sometimes it's very brief, sometimes the NARRATOR needs to cut it short. It can be very funny or very emotional, such as the occasion when the DAD (who turned-out to be a real-life Rabbi) ended his speech with the words 'Son, you used to always ask me 'why?' and I never had an answer for you. Well, today I know that you have found your answer.'
26| Depending on the DAD's speech, this sometimes became 'After the speeches, once we'd all recovered...'
27| In America this often got a laugh, prompting Jonny to say 'that's not supposed to be funny. It's just where we went.'
28| Jonny would take this entry out of his pocket and give it to the person he'd moved away from SAM, then allow them to return to their seat.
_by the same author_
**_The Most Humane Way to Kill a Lobster_**
9781840025590
**_Monster_**
9781840027594
**_Lungs_**
9781849431453
**_Don Juan Comes Back from the War_**
Ödön von Horváth, in a version by Duncan Macmillan
9781849432542
**_1984_**
George Orwell, in a new adaptation created by
Duncan Macmillan and Robert Icke
9781783190614
WWW.OBERONBOOKS.COM
Follow us on www.twitter.com/@oberonbooks
& www.facebook.com/OberonBooksLondon
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 8,574
|
You are here: 2006 // Y100Rocks.com Finds Airplay on WXPN 88.5 FM
Y100Rocks.com Finds Airplay on WXPN 88.5 FM
Philadelphia– Alternative Rock fans in Philadelphia now have a place to find their music on air. WXPN General Manager Roger LaMay and Y100Rocks.com Program Director Jim McGuinn announced today the two organizations' partnership to create YžRock On XPN, bringing alternative music back to the airwaves as well as a 24/7 online web stream.
The Y Rock On XPN online service launches August 1 as part of a new Web initiative created by WXPN called XPoNential Music On Demand. Beginning at 8 p.m. on Wednesday, August 30, and continuing every Wednesday and Thursday at that time and Fridays at 7 p.m., the alternative rock show called YžRock On XPN, hosted by McGuinn, will be heard on WXPN. The on-air show will be broadcast on 88.5 FM to listeners in the station's Pennsylvania, New Jersey and Delaware markets, as well as throughout the world via online streaming. The 24/7 on demand Y Rock On XPN stream will be accessible to Alternative Music fans worldwide at www.xponentialmusic.org, www.xpn.org, and www.yrockonxpn.org.
Y100Rocks.com, the online radio station, launched 17 months ago, following the February 25, 2005 departure of WPLY 100.3 FM – Y100 – from the radio dial.
"The departure of Y100 from the airwaves in Philadelphia created a significant void for fans of Alternative Rock music and musicians seeking airplay in this market. The music and the artists now have a home at WXPN," said LaMay. "As we began to prepare the launch of XPoNential Music On Demand, we saw this as the perfect time to embrace the mission of Y100Rocks.com. We also recognize that the people behind Y100Rocks.com are passionate music lovers whose authenticity matches the values of XPN. This amazing group of staff and volunteers kept the music alive when corporate radio abandoned them. In many ways, they have embodied the public radio core values that we live each and every day," he said.
Y100 Rocks Finds Airplay on WXPN
"When Radio One took Y100 off the air, 500,000 music fans in the greater Philadelphia region were instantly without a radio station for their music," said McGuinn. "To keep the music alive, the staff launched an online radio station – Y100Rocks.com – which has been broadcasting since February 2005 with a staff of mostly listener-volunteers. But it was always in our plan to get back on air. Now, thanks to this partnership with WXPN, we will," he said.
"Y Rock On XPN is a second service on the air and on the Web that will complement what WXPN offers," said Program Director Bruce Warren. "It expands our ability to take people on a musical journey and adds another level of musical importance for our listeners," he said.
AboutY100Rocks.com:
Y100Rocks.com is an online radio station launched in 2005 for Y100 staff, friends and listeners, made available through its online provider, Live365.com. Since its inception, the online station has offered Alternative Music fans more than 1.3 million streams, documented more than 1 million hours of listening, and won two Live365 web broadcasting awards, one for Best Alternative and one for Best Community station.
About WXPN:
WXPN 88.5 FM, the nationally recognized leader in Triple-A Radio and the premier source for musical discovery of new and significant artists in rock, blues, root, and folk, is the non-commercial, member-supported radio service of the University of Pennsylvania. WXPN produces World Cafe®, public radio's most popular program of popular music hosted by David Dye and syndicated by National Public Radio, and the Peabody Award-winning Kids Corner hosted by Kathy O'Connell. WXPN serves the greater Philadelphia area at 88.5 FM, the Lehigh Valley at 104.9FM, Worton/Baltimore at 90.5 FM, Harrisburg at 88.1 FM, and the world via online streaming at www.xpn.org, For more information, visit www.xpn.org, www.WorldCafe.org, www.KidsCorner.org, or call WXPN at (215) 898-6677.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,379
|
{"url":"https:\/\/www.mathway.com\/examples\/precalculus\/operations-on-functions\/finding-the-domain-of-the-quotient-of-the-functions?id=679","text":"# Precalculus Examples\n\n,\nReplace the function designators with the actual functions in .\nSet the denominator in equal to to find where the expression is undefined.\nDivide each term by and simplify.\nDivide each term in by .\nReduce the expression by cancelling the common factors.\nCancel the common factor.\nDivide by .\nDivide by .\nThe domain is all values of that make the expression defined.\nInterval Notation:\nSet-Builder Notation:","date":"2019-08-25 06:16:12","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9126130938529968, \"perplexity\": 3684.158858232798}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-35\/segments\/1566027323067.50\/warc\/CC-MAIN-20190825042326-20190825064326-00464.warc.gz\"}"}
| null | null |
{"url":"https:\/\/physics.stackexchange.com\/questions\/456414\/what-determines-how-much-energy-a-flywheel-can-store","text":"# What determines how much energy a flywheel can store? [closed]\n\nWhat is the relationship to the energy capacity of a flywheel and its radius, mass, rotational velocity etc?\n\n## closed as too broad by stafusa, JMac, Aaron Stevens, Cosmas Zachos, ZeroTheHeroJan 24 at 23:02\n\nPlease edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n\u2022 This is more an engineering than a physics question. \u2013\u00a0my2cts Jan 24 at 17:35\n\u2022 What's spinning the flywheel? That part is very important. \u2013\u00a0probably_someone Jan 24 at 17:48\n\u2022 Please consider adding a little more info to your question. In its current form it risks attracting downvotes due to lack of evidence of prior research. \u2013\u00a0PM 2Ring Jan 24 at 19:49\n\u2022 This seems too broad. Also, are you thinking about the recoverable energy from a flywheel, or the overall theoretical capacity? These are different things. \u2013\u00a0JMac Jan 24 at 20:55\n\nThe amount of energy a flywheel can store is equal to 1\/2 * (moment of inertia) * angular velocity^2.\n\nThe moment of inertia has to do with how heavy the spinning flywheel is, and how its mass is distributed around its axis of rotation. Basically, the larger the flywheel's diameter and the more mass its has, the more energy it can store.\n\nThe limiting factor for energy storage in a flywheel is its mechanical strength, because the stresses imposed on a fast-spinning flywheel will tend to make it fly to pieces. Fancy flywheels designed for energy storage must be made of exotic materials to ensure that the flywheel does not explode.\n\n\u2022 Exploding flywheels are very scary! That's a major reason we still use cars that store energy in explosive chemicals rather than in flywheels. \u2013\u00a0PM 2Ring Jan 24 at 18:31\n\u2022 @PM 2Ring, yeah yeah, at least with gasoline, you have to mix it with air to make it explode all at once- but when one of those flywheels goes, yowza! \u2013\u00a0niels nielsen Jan 24 at 20:52\n\nA flywheel is designed to efficiently store rotational kinetic energy. It resists changes in rpm by virtue of its rotational moment of inertia. Its stored kinetic energy is given by\n\n$$KE=\\frac{I\u03c9^2}{2}$$\n\nWhere $$I$$ is its moment of inertia and $$\u03c9$$ is its angular velocity.\n\nThe moment of inertia is given by\n\n$$I=kmr^2$$\n\nWhere $$m$$ is the mass, $$r$$ is radius, and $$k$$ is the inertial constant which depends on the shape of the flywheel. Examples are $$k=1$$ where the mass is concentrated at the rim of the wheel and $$k=0.606$$ for a flat solid circular disk of uniform thickness.\n\nThe fact that the stored kinetic energy goes up as the square of the radius of the wheel and where the mass is concentrated at the perimeter, speaks to the ability of a flywheel to store energy.\n\nHope this helps.","date":"2019-10-19 09:43:21","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 9, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6043776273727417, \"perplexity\": 620.2385155911716}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986692723.54\/warc\/CC-MAIN-20191019090937-20191019114437-00484.warc.gz\"}"}
| null | null |
Deichselrecht war im deutschen Rechtswesen ein Gesetz, womit der Besitzer einer Kutsche das Recht bekam, die Deichsel bei an der Grundstücksgrenze gelegenen Schuppen oder Scheunen durch eine in der Wand angebrachte Öffnung, das Deichselloch, auf des Nachbars Grundstück oder Garten reichen zu lassen. Dafür hatte der Belastete gewöhnlich das Recht, an der Wand des jenseitigen Gebäudes unter dem Schutz der Bedachung seine Räder, Leitern und andere Gerätschaften aufzustellen oder aufzuhängen und ins Trockene zu bringen.
Auch das Trinkgeld für Fuhrknechte wurde Deichselrecht genannt.
Siehe auch
Dienstbarkeit (Servitut)
Weblinks
Artikel Deichselrecht im Deutschen Rechtswörterbuch (mit Quellenangaben)
Privatrechtsgeschichte
Rechtsgeschichte (Deutschland)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,013
|
. . I was originally going to do an informational post, but had a phone conversation with my older sister, and she called me out on being such a defensive person. Me posting this quote is my attempt at having a positive spin on the reflective state her comments put me in. Lol. . . I have been defensive for a long time.Years. But I get defensive because I constantly feel like I am being attacked, with the bottom line being: "You are not good enough." . . At this stage in my life, it primarily stems from all that has come along with having a child with special needs. The questions and "advice" that people say just feels like it oozes judgment when I am coming from a place of already having so much guilt, of feeling like I can never do enough. You think "mom guilt" and "mom shaming" is bad enough for someone with a typical developing child, try throwing in "behind in every area of development," "difficulty with self-regulation," "child who does not yet know how to navigate independent play or playing with others," etc. and feel how much more intensified things become. . . -"You should read to your child." -"You probably don't take your kid to the park." -"You should talk to her so she can learn how to talk." -"She doesn't have Autism; she's just slow." -"Now you know to do things differently with this one [I have 2 daughters] so she doesn't end up like her." . . The list goes on. And I get angry. Because I try and I try, and give whatever part of myself that I can, and what? Statements like those above, insinuating that I don't even do the "basics," like talk to my child. . . Yes, they come from a well-intentioned place, and I can't blame them for what they don't know... for not taking into consideration factors like (lack of) joint attention, lower muscle tone, the importance of picking up on social cues for learning, vestibular sensitivity, etc. and how those things can affect development. . . So what to do moving forward? Educate. Share. Promote acceptance— not just for our children with special needs, but for their parents as well.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,887
|
\section*{Acknowledgments}
\balance
\bibliographystyle{abbrv}
\section{Conclusions}
\label{sec:conclusion}
In this paper, we propose FastSGD, a Fast compressed Stochastic Gradient Descent framework with lightweight compression/decompression cost for distributed machine learning.
FastSGD provides a compression framework for both the gradient keys and values in linear time complexity. For the gradient values, FastSGD offers a three-phase encoder, including the reciprocal mapper, the logarithm quantization, and the threshold filter, to compress the decimal gradient values into the small filtered gradient integers. For the gradient keys, an adaptive delta key binary encoder is provided to compress the gradient keys without producing any decoder error. Finally, we conduct extensive experiments on a range of practical ML models and datasets to demonstrate that FastSGD is more efficient than the state-of-the-art algorithms and scales well.
\section{Experiments}
\label{sec:exp}
In this section, we conduct extensive experiments to verify the effectiveness and efficiency of the proposed FastSGD using the practical ML models and datasets, compared with the state-of-the-art competitors.
\subsection{Experimental Setup}
\begin{table}[t]
\caption{Statistics of the Datasets Used in the Experiments}
\vspace{-1mm}
\label{tab:dataset}
\begin{tabular}{|p{1.8cm}|p{2cm}|p{2cm}|p{1.3cm}|}
\hline
\textbf{Dataset} & \textbf{\# Instances} & \textbf{\# Features} & \textbf{Size (GB)} \\ \hline
\textit{URL} & 2,396,130 & 3,231,961 & 2.1 \\ \hline
\textit{KDD10} & 19,264,097 & 29,890,095 & 4.8 \\ \hline
\textit{KDD12} & 149,639,105 & 54,686,452 & 20.9 \\ \hline
\textit{WebSpam} & 350,000 & 16,609,143 & 23.3 \\ \hline
\end{tabular}
\end{table}
\noindent
\textbf{Datasets} \quad We employ four public datasets, namely, \textit{URL}, \textit{KDD10}, \textit{KDD12}, and \textit{WebSpam}, with their statistics listed in Table~\ref{tab:dataset}.
\textit{URL} is commonly used to train machine learning models for malicious URL detection{\footnote{http://www.sysnet.ucsd.edu/projects/url/}}, and consists of 2 million instances with 3 million features.
\textit{KDD10} is a public dataset published by KDD Cup 2010{\footnote{https://pslcdatashop.web.cmu.edu/KDDCup/}}, including 19 million instances and 29 million features. The task is to predict student performance on mathematical problems from logs of student interaction with Intelligent Tutoring Systems.
\textit{KDD12} is published by another generation KDD Cup{\footnote{https://www.kaggle.com/c/kddcup2012-track1}}, which contains 149 million instances and 54 million features. The challenge involves predicting whether or not a user will follow an item that has been recommended to the user.
\textit{WebSpam} is a publicly available Web spam dataset{\footnote{https://www.cc.gatech.edu/projects/doi/WebbSpamCorpus.html}}. The corpus consists of nearly 350,000 Web spam pages with 16 million features.
\noindent
\textbf{ML models} \quad To evaluate the performance of FastSGD, we choose three popular ML methods, as with \cite{DBLP:conf/sigmod/JiangFY018}, including Linear Regression (Linear), Logistic Regression (LR), and Support Vector Machine (SVM).
Their formalized descriptions are listed in Table~\ref{tab:ml}, in which $\mathbb{I}$ represents the indicator function.
\begin{table*}[t]
\caption{Formalization of the Machine Learning Models Evaluated}
\label{tab:ml}
\vspace{-1mm}
\begin{tabular}{|p{4.2cm}|p{6.3cm}|p{6.3cm}|}
\hline
\textbf{Machine learning model} & \textbf{Loss function} & \textbf{Gradient} \\ \hline
Linear regression (Linear) & $ f(x, y, \theta)=\sum_{i=1}^{N}\left(y_{i}-\theta^{\mathrm{T}} x_{i}\right)^{2}+\frac{\lambda}{2}\|\theta\|_{2}$ & $ \nabla_\theta f(x,y,\theta) = \sum_{i=1}^{N}-\left(y_{i}-\theta^{\mathrm{T}} x_{i}\right) x_{i}+\lambda \theta$\\ \hline
Logistic regression (LR) & $f(x, y, \theta)=\sum_{i=1}^{N} \log \left(1+e^{-y_{i} \theta^{\mathrm{T}} x_{i}}\right)+\frac{\lambda}{2}\|\theta\|_{2}$ & $\nabla_\theta f(x,y,\theta) = \sum_{i=1}^{N}-\frac{y_{i}}{1+ e^{\left(y_{i} \theta^{\mathrm{T}} x_{i}\right)}} x_{i}+\lambda \theta$ \\ \hline
Support vector machine (SVM) & $f(x, y, \theta)=\sum_{i=1}^{N} \max \left(0,1-y_{i} \theta^{\mathrm{T}} x_{i}\right)+\frac{\lambda}{2}\|\theta\|_{2}$ & $ \nabla_\theta f(x,y,\theta) =\sum_{i=1}^{N}-y_{i} x_{i} \mathbb{I}\left\{y_{i} \theta^{\mathrm{T}} x_{i}<1\right\}+\lambda \theta$\\ \hline
\end{tabular}
\end{table*}
To converge faster, we train three ML models with Adam SGD~\cite{DBLP:journals/corr/KingmaB14}. Adam SGD is one of the most commonly-used accelerated versions of SGD. At the iterative round $t+1$, Adam SGD stores a decaying average of past gradients and squared gradients:
\begin{equation}
\begin{array}{l}
m_{t}=\beta_{1} m_{t-1}+\left(1-\beta_{1}\right) g_{t} \\
v_{t}=\beta_{2} v_{t-1}+\left(1-\beta_{2}\right) g_{t}^{2}
\end{array}
\end{equation}
where $\beta_{1}$ and $\beta_{2}$ are two hyper-parameters close to 1. Then, Adam SGD updates the model parameters as follows:
\begin{equation}
\theta_{t+1}=\theta_{t}-\frac{\eta}{\sqrt{v_{t}+\epsilon}} m_{t}
\end{equation}
where $\epsilon$ is another hyper-parameters close to 0.
\noindent
\textbf{Implementation} \quad
Following the choice of \cite{DBLP:conf/sigmod/JiangFY018}, we set $\beta_{1}$ to 0.9, $\beta_{2}$ to 0.999, and $\epsilon$ to $10^{-8}$ in Adam SGD. The optimal learning rate $\eta$ is tuned by a grid search. The regularization hyper-parameter $\lambda$ is set to 0.01. The input dataset is split into the training dataset (70\% of the input dataset) and the test dataset (30\% of the input dataset).
To decrease the synchronization frequency and the communication cost, we implement SGD with a popular \textit{mini-batch} trick~\cite{DBLP:conf/compstat/Bottou10}, which uses a batch of (more than one) instances in each training iteration.
We set the batch size as 10\% of the training dataset. Once SGD completes a pass over the entire training dataset, we say SGD has finished an epoch. We verify the performance of SGD in 20 epochs by default without special instructions.
\noindent
\textbf{Competitors} \quad We compare FastSGD with four state-of-the-art algorithms, including Adam SGD~\cite{DBLP:journals/corr/KingmaB14}, LogQuant~\cite{DBLP:journals/corr/MiyashitaLM16}, SketchML~\cite{DBLP:conf/sigmod/JiangFY018}, and Top-$k$~\cite{DBLP:conf/ijcai/ShiZWTC19}.
\begin{itemize}
\item{} Adam SGD is one of the most popular first-order gradient optimization approaches, which combines the advantage of momentum~\cite{DBLP:journals/nn/Qian99} and adaptive learning rate~\cite{DBLP:journals/jmlr/DuchiHS11}.
\item{} LogQuant adopts base-2 logarithmic representation to quantize gradients.
\item{} SketchML is a sketched-based method to compress the sparse gradient consisting of key-value pairs.
\item{} Top-$k$ selects the $k$ largest values in the gradient to transmit, which can achieve good performance in practice.
\end{itemize}
Note that the Adam strategy is applied to all the competitors for fairness.
The main performance metrics to study the gradient compression performance include the average running time per epoch, the message size, and the loss function with respect to the running time.
All the experiments{\footnote{We sincerely thank the authors of SketchML~\cite{DBLP:conf/sigmod/JiangFY018} for sharing the source code.}} were implemented in Scala 2.11 on Spark 2.2, and run on a 12-
node Dell cluster. Each node has two Intel(R) Xeon(R) CPU E5-2650 v4 @ 2.20GHz processors with 12 cores, 128GB RAM, 1TB disk, and connect in Gigabit Ethernet.
\subsection{Effectiveness of the Proposed Techniques}
\label{sec:effect}
In this subsection, we experimentally evaluate the effectiveness of the techniques proposed in this paper on \textit{KDD12} dataset. Note that, the results over the three other datasets are similar.
\noindent
\textbf{Effect of the Proposed Components} \quad FastSGD could be divided into three components, i.e., i) the adaptive binary encoder for delta gradient key, ii) the reciprocal mapper and logarithm quantization that map decimal gradient values into small quantized integers, and iii) the threshold filter to discard the quantized gradient value. We begin with the baseline Adam, and implement the proposed components gradually. Fig.~\ref{fig:component} illustrates the performance of the proposed components.
\begin{figure}[t]
\centering
\includegraphics[width=0.45\textwidth]{component-icon}\\
\subfigure[running time per epoch]{
\label{fig:component-time}
\includegraphics[width=0.49\textwidth]{component-time}
}
\subfigure[message size]{
\label{fig:component-size}
\includegraphics[width=0.49\textwidth]{component-size}
}
\vspace{-2mm}
\caption{Effectiveness of the proposed components on \textit{KDD12}}
\label{fig:component}
\end{figure}
Fig.~\ref{fig:component-time} plots the running time per epoch with the speedup compared with the basic Adam in three different ML models. It is observed that the proposed techniques can significantly improve the efficiency of three different ML models. Compared with Adam, the component of adaptive binary encoder accelerates the execution by 1.2$\times$. In addition, the reciprocal mapper and logarithm quantization further improves the performance by up to 4.0$\times$. Finally, FastSGD achieves the speedup for the running time per epoch at most 7.9$\times$ by implementing with all the three components.
The reason for the speedup is that the proposed components compress the communication volume. Fig.~\ref{fig:component-size} shows the message size in each round of communication (i.e., per batch) with the compression ratio compared against the uncompressed message size. FastSGD achieves the compression ratio from 1.4$\times$ to 18720$\times$ at most by consolidating three components. We can observe that the threshold filter remarkably compresses the message since it discards lots of gradients whose absolute values are small. Nevertheless, the running time is not decreased in the same proportion as the message size. This is because FastSGD achieves a high compression ratio for the gradient communication volume, which makes the computation ability rather than the communication become the bottleneck of the distributed system.
\begin{table*}[ht]
\caption{Time Breakdown on \textit{KDD12}}
\vspace{-1mm}
\label{tab:break}
\center
\begin{tabular}{|p{2cm}|l|l|l|l|l|l|}
\hline
\multicolumn{7}{|c|}{\textbf{Linear}} \\ \hline
\textbf{Method} & \textbf{Total time} & \textbf{Gradient computation} & \textbf{Gradient encoder} & \textbf{Gradient decoder} & \textbf{Communication} & \textbf{Model updating} \\ \hline
Adam & 186 & 22 & 0 & 0 & 84 & 80 \\ \hline
FastSGD & 33 & 21 & 0.05 & 0.08 & 6 & 6 \\ \hline
\multicolumn{7}{|c|}{\textbf{LR}} \\ \hline
\textbf{Method} & \textbf{Total time} & \textbf{Gradient computation} & \textbf{Gradient encoder} & \textbf{Gradient decoder} & \textbf{Communication} & \textbf{Model updating} \\ \hline
Adam & 156 & 23 & 0 & 0 & 68 & 65 \\ \hline
FastSGD & 32 & 23 & 0.05 & 0.06 & 4 & 5 \\ \hline
\multicolumn{7}{|c|}{\textbf{SVM}} \\ \hline
\textbf{Method} & \textbf{Total time} & \textbf{Gradient computation} & \textbf{Gradient encoder} & \textbf{Gradient decoder} & \textbf{Communication} & \textbf{Model updating} \\ \hline
Adam & 158 & 21 & 0 & 0 & 73 & 64 \\ \hline
FastSGD & 34 & 21 & 0.06 & 0.07 & 5 & 8 \\ \hline
\end{tabular}
\end{table*}
Furthermore, Table~\ref{tab:break} reports a breakdown of the running time per epoch in seconds to depict the raw performance of the encoder and decoder in FastSGD. We decouple the total running time into five parts, i.e., gradient computation, gradient encoding, gradient decoding, communication, and model updating. It is observed that the encoder and decoder in FastSGD spend a small amount of time (i.e., 0.05 to 0.08 seconds) to accomplish the gradient compression and decompression efficiently. Based on this, FastSGD achieves up to 17$\times$ speedup in the communication time and 13$\times$ speedup in the model updating time. The reason for the speedup in the communication time is that FastSGD efficiently compresses the gradient communication volume. Besides, FastSGD accelerates the model updating due to the threshold filter, which discards those gradient elements with smaller absolute values and hence reduces the model updating computation cost.
\begin{table}[t]
\caption{The Space Cost of Gradient Key on \textit{KDD12}}
\vspace{-1mm}
\label{tab:key}
\begin{tabular}{|p{2.6cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|}
\hline
\textbf{Method} & \textbf{Adam} & \textbf{SketchML} & \textbf{FastSGD} \\ \hline
Size per key (bits) & 32.00 & 9.95 & 6.04 \\ \hline
\end{tabular}
\end{table}
As discussed in Section~\ref{sec:gk}, both SketchML and FastSGD use the delta gradient key. Further, FastSGD utilizes the bit rather than byte as the smallest grain with the adaptive strategy to encode the delta key. Thus, we record the average size taken by each gradient key in Table~\ref{tab:key} to verify the effectiveness of FastSGD. It is observed that SketchML compresses the average size per gradient key from 32.00 bits to 9.95 bits, while FastSGD further reduces it to 6.04 bits. This confirms the effectiveness of adaptive fine-grained delta encoding method in FastSGD.
\noindent
\textbf{Effect of the Hyper-parameters in FastSGD} \quad FastSGD contains three hyper-parameters, i.e., the base $b$ in logarithm quantization, the threshold $\tau$ in threshold filter, and the size $l$ of length flag. We fine-tune them, and give the setting guidance based on the experimental results over all the three different ML models.
\begin{figure*}[t]
\centering
\includegraphics[width=0.7\textwidth]{base-icon}\\
\vspace{-1mm}
\subfigure[Linear]{
\label{base-Linear}
\includegraphics[width=0.235\textwidth]{base-Linear}
}
\hspace{10mm}
\subfigure[LR]{
\label{base-LR}
\includegraphics[width=0.235\textwidth]{base-LR}
}
\hspace{10mm}
\subfigure[SVM]{
\label{base-SVM}
\includegraphics[width=0.235\textwidth]{base-SVM}
}
\vspace{-2mm}
\caption{Effect of base $b$ on \textit{KDD12}}
\label{fig:base}
\end{figure*}
\begin{figure*}[t]
\centering
\includegraphics[width=0.7\textwidth]{thre-icon}\\
\vspace{-1mm}
\subfigure[Linear]{
\label{thre-Linear}
\includegraphics[width=0.235\textwidth]{thre-Linear}
}
\hspace{10mm}
\subfigure[LR]{
\label{thre-LR}
\includegraphics[width=0.235\textwidth]{thre-LR}
}
\hspace{10mm}
\subfigure[SVM]{
\label{thre-SVM}
\includegraphics[width=0.235\textwidth]{thre-SVM}
}
\vspace{-2mm}
\caption{Effect of threshold $\tau$ on \textit{KDD12}}
\label{fig:thre}
\end{figure*}
Fig.~\ref{fig:base} depicts the loss function of each epoch in terms of the running time with the base $b$ changing from 1.03625 to 1.1125. As excepted, FastSGD converges much faster than the baseline Adam in all three base values. Besides, the convergence rate of FastSGD first becomes faster and then slower as the base $b$ grows, because the bigger base $b$, the smaller the gradient values are compressed, while the recovering error in decoder becomes larger. To achieve a good trade-off between the compression power and the decompression error, the value of base $b$ is recommended to set as around 1.1.
Fig.~\ref{fig:thre} shows the loss function of each epoch in terms of the running time w.r.t. the threshold $\tau$ varying from 64 to 128. We observe that the results of FastSGD on different thresholds are similar. A bigger threshold $\tau$ (i.e., 128), which retains more gradient values and restricts the space cost for each gradient value in one byte, could perform slightly better. Therefore, we set the threshold $\tau$ to 128 by default in our experiments.
Table~\ref{tab:vkey} lists the average size per gradient key varying the size $l$ of length flag from 1 to 5 bits. It is observed that the average size per gradient key first becomes smaller and then larger as the size $l$ of length flag grows. The reason is that the larger length flag provides more fine-grained length options to store the delta key in less bits, while it costs more space to represent the length flag itself. According to the experimental results, the size $l$ of length flag is recommended to set as 2.
\begin{table}[t]
\caption{The Space Cost of Gradient Key v.s. the Size of Length Flag on \textit{KDD12}}
\vspace{-1mm}
\label{tab:vkey}
\begin{tabular}{|p{2.6cm}|p{0.72cm}|p{0.72cm}|p{0.72cm}|p{0.72cm}|p{0.72cm}|}
\hline
\textbf{Size of length flag} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} \\ \hline
Size per key (bits) & 8.65 & 6.04 & 7.06 & 12.27 & 14.78 \\ \hline
\end{tabular}
\end{table}
\begin{figure*}[ht]
\centering
\subfigure[Linear]{
\label{feature-Linear}
\includegraphics[width=0.235\textwidth]{feature-Linear}
}
\hspace{10mm}
\subfigure[LR]{
\label{feature-LR}
\includegraphics[width=0.235\textwidth]{feature-LR}
}
\hspace{10mm}
\subfigure[SVM]{
\label{feature-SVM}
\includegraphics[width=0.235\textwidth]{feature-SVM}
}
\vspace{-2mm}
\caption{Effect of the number of features on \textit{KDD12}}
\label{fig:feature}
\end{figure*}
\noindent
\textbf{Effect of the Number of Features} \quad In this set of experiments, we explore the scalability of FastSGD with fewer features. Fig.~\ref{fig:feature} depicts the running time per epoch when the number of features changes from $10^{-4}$\% to 100\% of the whole \textit{KDD12} dataset. As expected, the running time per epoch of both FastSGD and Adam increases as the number of features ascends. This is because there are more feature data to be processed. In addition, we can observe that FastSGD outperforms Adam in almost all the experimental settings, except when the number of features is extremely small (e.g., $10^{-4}$\% of the whole feature number). The reason is that the cost for gradient compression exceeds the savings by the reduced gradient communication with such a small number of features.
\subsection{Comparison with the State-of-the-arts}
In this subsection, we compare the end-to-end performance of our proposed FastSGD with four state-of-the-art algorithms, viz., Adam, LogQuant, SketchML, and Top-$k$.
\begin{figure*}[t]
\centering
\includegraphics[width=0.95\textwidth]{competitor-icon}\\
\vspace{-1mm}
\subfigure[\textit{URL}]{
\label{fig:competitor-URL}
\includegraphics[width=0.235\textwidth]{competitor-URL}
\subfigure[\textit{KDD10}]{
\label{fig:competitor-KDD10}
\includegraphics[width=0.235\textwidth]{competitor-KDD10}
}
\subfigure[\textit{KDD12}]{
\label{fig:competitor-KDD12}
\includegraphics[width=0.235\textwidth]{competitor-KDD12}
}
\subfigure[\textit{WebSpam}]{
\label{fig:competitor-Web}
\includegraphics[width=0.235\textwidth]{competitor-Web}
}
\vspace{-2mm}
\caption{Convergence rate over SVM}
\label{fig:competitor}
\end{figure*}
\begin{figure*}[t]
\centering
\includegraphics[width=0.95\textwidth]{competitor-icon}\\
\vspace{-1mm}
\subfigure[\textit{URL}]{
\label{fig:worker-URL}
\includegraphics[width=0.235\textwidth]{worker-URL}
}
\subfigure[\textit{KDD10}]{
\label{fig:worker-KDD10}
\includegraphics[width=0.235\textwidth]{worker-KDD10}
}
\subfigure[\textit{KDD12}]{
\label{fig:worker-KDD12}
\includegraphics[width=0.235\textwidth]{worker-KDD12}
}
\subfigure[\textit{WebSpam}]{
\label{fig:worker-Web}
\includegraphics[width=0.235\textwidth]{worker-Web}
}
\vspace{-2mm}
\caption{Scalability comparison over SVM}
\label{fig:worker}
\end{figure*}
\begin{table*}[t]
\caption{The Performance of Convergence}
\vspace{-1mm}
\label{tab:conver}
\center
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{11}{|c|}{\textbf{\textit{KDD12}}} \\ \hline
\multirow{2}{*}{\textbf{Model}} & \multicolumn{2}{c|}{\textbf{FastSGD}} & \multicolumn{2}{c|}{\textbf{Adam}} & \multicolumn{2}{c|}{\textbf{LogQuant}} & \multicolumn{2}{c|}{\textbf{SketchML}} & \multicolumn{2}{c|}{\textbf{Top-$k$}} \\ \cline{2-11}
& \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} \\ \hline
Linear & 0.0214 & 348 & 0.0214 & 2969 & 0.0214 & 2765 & 0.0214 & 2007 & 0.0214 & 1189 \\ \hline
LR & 0.6924 & 437 & 0.6924 & 2638 & 0.6924 & 2392 & 0.6924 & 1836 & 0.6924 & 970 \\ \hline
SVM & 0.9968 & 331 & 0.9968 & 2441 & 0.9968 & 2550 & 0.9969 & 1871 & 0.9968 & 978 \\ \hline
\multicolumn{11}{|c|}{\textbf{\textit{WebSpam}}} \\ \hline
\multirow{2}{*}{\textbf{Model}} & \multicolumn{2}{c|}{\textbf{FastSGD}} & \multicolumn{2}{c|}{\textbf{Adam}} & \multicolumn{2}{c|}{\textbf{LogQuant}} & \multicolumn{2}{c|}{\textbf{SketchML}} & \multicolumn{2}{c|}{\textbf{Top-$k$}} \\ \cline{2-11}
& \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} & \textbf{Min. loss} & \textbf{Cov. time} \\ \hline
Linear & 0.3174 & 110 & 0.3173 & 439 & 0.3176 & 365 & 0.3172 & 260 & 0.3173 & 133 \\ \hline
LR & 0.5884 & 102 & 0.5882 & 441 & 0.5892 & 314 & 0.5877 & 210 & 0.5885 & 122 \\ \hline
SVM & 0.6369 & 213 & 0.6371 & 838 & 0.6362 & 678 & 0.6435 & 479 & 0.6372 & 245 \\ \hline
\end{tabular}
\end{table*}
Fig.~\ref{fig:competitor} illustrates the convergence rate on four datasets.
Note that we only present the results of SVM to represent the performance on three ML models, since the results of the two others are similar.
It is observed that FastSGD achieves the fastest convergence rate, followed by Top-$k$, SketchML, and then LogQuant, while the basic Adam without the gradient compression is the worst. LogQuant and SketchML uses logarithmic quantization and sketch-based techniques respectively to compress the gradient. Thus, they converge faster than Adam. FastSGD and Top-$k$ perform better because they only transmit those gradient values with larger absolute values that can contribute more to the convergence of optimization algorithm based on the sparsification strategy. Furthermore, FastSGD squeezes the compression ratio by the presented encoders for both gradient values and keys.
Since FastSGD underestimates and discards the gradient values in the encoder, one may suspect whether FastSGD could converge and train the models successfully. Therefore, we conduct the experiments to demonstrate the convergence performance. We set the convergence condition of the optimization algorithm as if the change of validation loss is less than 1\%.
Table~\ref{tab:conver} lists the convergence performance of FastSGD, Adam, LogQuant, SketchML, and Top-$k$ on \textit{KDD12} and \textit{WebSpam} datasets. The metric is the minimal loss against converged time (seconds), separated by symbol `/'. We can observe that the four compressed algorithms could achieve almost the same model quality as Adam. Meanwhile, FastSGD converges faster than other algorithms. To be more specific, FastSGD converges faster than Adam by up to 8.5$\times$, faster than LogQuant by up to 7.9$\times$, faster than SketchML by up to 5.8$\times$, and faster than Top-$k$ by up to 3.4$\times$. As discussed in Section~\ref{sec:gv}, the logarithm quantization of FastSGD underestimates the gradient values. If a specific key index of a gradient is always underestimated, the convergence will be slow. Fortunately, the threshold filter would select different indices of the gradient to transmit in each training batch, while could alleviate this problem. Besides, the advantage of momentum and adaptive learning rate strategies in Adam could further reduce the sparsification error of threshold filter in FastSGD.
Finally, we compare the scalability of all the five algorithms. Fig.~\ref{fig:worker} shows the running time per epoch by varying the number of workers from 3 to 11 over SVM model on four datasets, while the results on the rest models and datasets are similar. As expected, the running time per epoch of all the competitors decreases as the number of workers ascends, except for Adam and LogQuant. For Adam, the running time increases with the number of workers on \textit{URL} and \textit{KDD10} datasets, while the running time first drops and then increases when the number of workers grows on \textit{KDD12} and \textit{WebSpam} datasets. For LogQuant, the running time increases with the number of workers on \textit{KDD10} dataset.
The reason is that, on one hand, the computation ability grows with the number of workers; on the other hand, the communication cost also increases with the number of workers.
The communication overhead becomes the bottleneck of the distributed system if the computation resource is sufficient.
The compression of gradients can alleviate the limitation of communication bottleneck by reducing the message size in the distributed system. Thus, FastSGD, Top-$k$, and SketchML could benefit from the number growth of workers. In addition, the higher the compression ratio (e.g., Top-$k$ or FastSGD), the better the scalability of the corresponding method.
\subsection{Performance over Neural Network}
In the aforementioned experiments, we have evaluated the performance of our proposed FastSGD with three generalized linear models. Furthermore, FastSGD can also be beneficial for distributed Neural Network models by compressing the transferring gradients. Next, we investigate the effectiveness of FastSGD over neural networks by training a Multilayer Perceptron (MLP) network on MNIST{\footnote{http://yann.lecun.com/exdb/mnist/}} dataset. The MLP network contains one input layer (size: 7 $\times$ 7), two fully connected layers (size: 100), and one output layer (size: 5). MNIST is a handwritten digits dataset, which consists of 60,000 training images and 10,000 testing images. The learning rate is set to 0.02.
Fig.~\ref{fig:nn} illustrates the convergence rate of FastSGD, Adam, LogQuant, SketchML, and Top-$k$. We can observe that Top-$k$ cannot perform well over the MLP network. This is because it discards too much gradient elements. Besides, FastSGD achieves both the fastest convergence rate and the smallest loss, followed by LogQuant, SketchML, and Adam. The reason is that FastSGD provides a lightweight gradient compression, which achieves a high compression ratio at a low cost while retain the main gradient information for optimization.
\begin{figure}[t]
\centering
\includegraphics[width=0.49\textwidth]{MNIST}
\vspace{-2mm}
\caption{Convergence rate over neural network}
\label{fig:nn}
\end{figure}
\section{Introduction}
\label{sec:intro}
\IEEEPARstart{M}{achine} learning (ML) technology powers many aspects of modern society, such as computer vision~\cite{DBLP:conf/cvpr/DengDSLL009, DBLP:conf/cvpr/HeZRS16}, natural language processing~\cite{DBLP:journals/corr/abs-2004-03705, DBLP:journals/corr/abs-2009-06732}, speech recognition~\cite{DBLP:conf/icml/AmodeiABCCCCCCD16}, to name just a few. The performance of ML models (e.g., accuracy in classification tasks) generally improves if more training data is used~\cite{DBLP:conf/iccv/SunSSG17}. However, with the rapid increase of data, a single machine cannot support the model training process efficiently. Motivated by this, distributed training has received extensive attention from academia and industry~\cite{DBLP:journals/corr/abs-2003-06307, DBLP:conf/osdi/LiAPSAJLSS14}.
Almost all the ML models are trained with a first-order gradient descent method, namely, Stochastic Gradient Descent (SGD)~\cite{robbins1951stochastic}. To deploy the SGD-based ML algorithms in a distributed environment, the training dataset is partitioned into several distributed \textit{workers}. Each worker independently proposes local intermediate results, i.e., the gradients. Then, local gradients are communicated through the network to be aggregated, and the aggregated values are sent back to the workers for the update of the global model state. The process is repeated over many epochs (i.e., full iterations over the entire training dataset) until the convergence.
Under the aforementioned distributed ML setting, the network communication involves large amounts of gradients, which could become the bottleneck cost to limit the scalability of the distributed system. This leads to inefficient utilization of the computational resources, longer training time, and/or higher financial cost for the cloud infrastructure. Therefore, the gradient communication should be well optimized to fully harness the computing powers of distributed workers.
Towards this, many studies have been devoted to the compression methods for the exchanged gradients before transmitting across the network with little impact on the model convergence. Existing efforts can be mainly divided into three categories: \textit{Quantization}~\cite{DBLP:conf/icml/BernsteinWAA18, DBLP:conf/nips/AlistarhG0TV17}, \textit{Sparsification}~\cite{DBLP:conf/nips/AlistarhH0KKR18, DBLP:conf/aaai/DuttaBA0SCK20}, and \textit{Hybrid} methods~\cite{DBLP:conf/sigmod/JiangFY018, DBLP:conf/mlsys/LimAK19}. Quantization methods use lower bits to represent each element in gradients (e.g., casting 32 bits to 8 bits). However, even using only one bit for each gradient, the maximal compression ratio of quantization is 32$\times$ compared to the 32-bit counterpart. In contrast, sparsification methods directly reduce the number of gradient elements to transmit (e.g., only transmitting top-$k$ largest elements). Thus, sparsification reduces the communication cost by transmitting only a small portion of gradients (e.g., 0.1\% \cite{DBLP:conf/aaai/DuttaBA0SCK20}). In addition, hybrid methods further compress the gradients by combining sparsification with quantization.
Typically, the transferred gradients are sparse originally and/or after sparsification. Nevertheless, almost all the existing gradient compression methods assume that the gradient vector to be compressed is dense. If all the dimensions of the gradient vector are stored, it would waste a lot of both storing space and processing time for zero gradient values. To this end, we can represent the nonzero elements in a gradient vector as key-value pairs. The key-value representation format of sparse gradient leaves room for further compression.
There is only one line of method SketchML~\cite{DBLP:conf/sigmod/JiangFY018, DBLP:journals/vldb/JiangFYSC20} that provides the framework to compress both the gradient keys and values. However, SketchML is limited to the communication-intensive workloads, as it has a high computational cost for gradient compression/decompression that could be larger than the savings by the reduced communication.
Motivated by those limitations of the previous gradient compression methods, we dedicate this paper to the development of FastSGD, a \underline{Fast} compressed \underline{SGD} framework for distributed machine learning, with lightweight compression/decompression cost. In view of this, the challenges are two-fold as follows.
The first challenge is \textit{how to efficiently compress the gradient value?} The quantization and sparsification strategies inspire us that \textit{lossy} compression is suitable to reduce the volume of gradient values while guaranteeing the model convergence at the same time. FastSGD provides an effective hybrid method that combines the quantization and sparsification methods to compress the gradient values. The compression method of gradient values in FastSGD mainly contains three phases.
First of all, FastSGD uses a \textit{reciprocal mapper} to transform original gradient values into reciprocal values. Then, FastSGD utilizes a \textit{logarithm quantization} method to reduce the space cost for the representation of mapped values. Finally, FastSGD adopts a \textit{threshold filter} to select the elements whose absolute values are large to transmit in the network.
The second challenge is \textit{how to efficiently compress the gradient key?} Different from the gradient values that can tolerate a precision loss during the compression/decompression, the gradient keys are vulnerable to inaccuracy. Assume that we compress a key but fail to decompress it accurately due to the precision loss, we would update a wrong gradient descent direction (i.e., a wrong dimension of gradient vector) and cannot guarantee the correct convergence of the ML model. Thus, a lossless method that transforms the keys to \textit{delta keys} has been proposed~\cite{DBLP:conf/sigmod/JiangFY018}. Nonetheless, it uses byte instead of bit as the smallest grain to encode the delta key, incurring the redundant space cost. In this paper, the proposed FastSGD goes a step further by using an adaptive fine-grained delta encoding method to store the gradient keys with fewer bits.
In brief, the key contributions of this paper are summarized as follows:
\begin{itemize}
\item{} We propose FastSGD, a fast compressed SGD framework for distributed machine learning to compress the gradients consisting of key-value pairs with lightweight compression/decompression cost.
\item{} FastSGD provides a hybrid lossy compression method, including the reciprocal mapper, the logarithm quantization, and the threshold filter, to compress the gradient values.
\item{} As for the compression of gradient key, FastSGD utilizes a lossless binary encoding method based on the adaptive fine-grained strategy to store the delta key with fewer bits.
\item{} We conduct extensive experiments on a range of practical ML workloads to demonstrate that FastSGD achieves up to 4 orders of magnitude compression ratio, and reduces the convergence time up to 8$\times$, compared with the state-of-the-art algorithms.
\end{itemize}
The rest of this paper is organized as follows. We introduce the distributed ML model training with SGD in Section~\ref{sec:pre}. Section~\ref{sec:method} presents the details of our proposed FastSGD. Considerable experimental results are reported in Section~\ref{sec:exp}. We review the related work in Section~\ref{sec:related}. Finally, Section~\ref{sec:conclusion} concludes the paper.
\section{The Proposed FastSGD}
\label{sec:method}
In this section, we describe the proposed FastSGD, a fast compressed SGD framework for distributed machine learning. First, we overview the FastSGD framework in Section~\ref{sec:over}. Then, we detail each component in Section~\ref{sec:gv} and Section~\ref{sec:gk}, respectively. Finally, we theoretically analyze both the space cost and the time complexity for FastSGD in Section~\ref{sec:analysis}.
\subsection{Overview of FastSGD}
\label{sec:over}
FastSGD consists of two stages, i.e., encoder and decoder, as described below:
\begin{itemize}
\item Encoder compresses separately the gradient values and keys before transmitting them among the distributed workers.
\item Decoder decompresses the compressed gradients when the gradients should be aggregated after communication.
\end{itemize}
\noindent
\textbf{The Encoder Stage of FastSGD}\quad
The encoder of FastSGD first compresses the gradient values in the following three phases.
\begin{itemize}
\item [1)] A \textit{reciprocal mapper} is used to transform each gradient values into reciprocal values.
\item [2)] Reciprocal gradient values are reduced to small quantized integers by a \textit{logarithm quantization}.
\item [3)] A \textit{threshold filter} is utilized to discard the quantized integers whose absolute values are large.
\end{itemize}
After compressing the gradient values, the encoder proceeds to process the corresponding gradient keys in the two phases below.
\begin{itemize}
\item [4)] Each gradient key is represented as an incremental format \textit{delta key}.
\item [5)] The delta keys are encoded into the binary coding, and an adaptive \textit{length flag} is further employed to save the encoding bits.
\end{itemize}
\noindent
\textbf{The Decoder Stage of FastSGD}\quad
FastSGD uses the decoder to recover the compressed gradients in the four phases as follows.
\begin{itemize}
\item [1)] The delta keys in the binary by coding are identified using the length flags.
\item [2)] The delta keys are recovered to the original gradient keys.
\item [3)] The filtered integers are exponentially scaled up to reciprocal gradient values.
\item [4)] The gradient values are recovered from the reciprocal values.
\end{itemize}
In what follows, we present the encoder of FastSGD, and omit the decoder of FastSGD since it simply inverses the encoder.
\subsection{The Encoder for Gradient Values}
\label{sec:gv}
The prior work~\cite{DBLP:conf/sigmod/JiangFY018} has shown that the SGD optimization algorithm can converge with underestimated gradients.
Since SGD moves towards the optimal point following the opposite direction of gradients, reducing the scale of gradients might slow down the convergence rate somewhat, while still on the correct convergence track. On the contrary, the uncontrolled overestimated scale of gradients has a risk of jumping over the optimal point.
Considering the tolerance of underestimated gradients in SGD, FastSGD offers an effective three-phase gradient value encoder, which contains the reciprocal mapper, the logarithm quantization, and the threshold filter, to compress the gradient values $\{v_j\}_{j=1}^D$. We detail the three phases of gradient value encoder in the following.
\noindent
\textbf{Phase 1: Reciprocal Mapper} \quad
The reciprocal mapper transforms each gradient value $v_j$ into a reciprocal gradient value $R(v_j)$ as follows:
\begin{equation}
\label{equ:reci}
R(v_j) = \frac{\sum_{i=1}^{D} |v_i|}{|v_j|}
\end{equation}
The reciprocal mapper has two benefits/advantages.
\begin{itemize}
\item [1)] For each $R(v_j)$, we can get $R(v_j) \geq 1$ as $|v_j| \leq \sum_{i=1}^{D} |v_i|$. This property lays the foundation for logarithm quantization to further reduce the reciprocal value in Phase 2.
\item [2)] A reversed gradient value order, i.e., the $v_j$ whose absolute value is bigger, would be mapped to a smaller reciprocal value. This means that we could use fewer bytes to store those important gradient elements with bigger absolute values. We would further discuss this in Phase 3.
\end{itemize}
\noindent
\textbf{Phase 2: Logarithm Quantization} \quad
After the reciprocal mapping, the logarithm quantization is utilized to further reduce the reciprocal gradient value.
Given a base $b$ ($>1$) and a reciprocal gradient value $R(v_j)$, $j=1,2,...,D$, the logarithm quantization reduces the scale of $R(v_j)$ to a smaller integer $L(v_j)$ through the following formula:
\begin{equation}
L(v_j) = \left\lceil log_b (R(v_j)) \right\rceil
\label{eq:rgv}
\end{equation}
\begin{figure}
\centering
\includegraphics[width=0.49\textwidth]{LQ}
\vspace{-2mm}
\caption{Illustration for logarithm quantization}
\label{fig:lq}
\end{figure}
Fig.~\ref{fig:lq} illustrates the process of logarithm quantization in the case $b=2$. Here, $D=6$.
Take the first reciprocal gradient value $R(v_j) = 6.1$ as an example. The logarithm quantization further reduces $R(v_j)$ into $L(v_j) = \left\lceil log_b (R(v_j)) \right\rceil = \left\lceil log_2 6.1 \right\rceil = 3$.
Ignoring the ceiling operation in Equation (\ref{eq:rgv}), the original gradient values could be recovered precisely by $\hat{L}(v_j) = log_b (R(v_j))$ and Equation~(\ref{equ:reci}). Nonetheless, $\hat{L}(v_j)$ is decimal, and should be stored in the format of float-point number. Then, the encoding processing is meaningless because the communication cost remains the same as transmitting the original gradient value $v_j$.
Fortunately, we find the value $\hat{L}(v_j)$ has a good property that its integer part is small due to exponential explosion, yet the integer part could be a quantized indicator to estimate the original value with the base $b$. Thus, the logarithm quantization uses Equation~(\ref{eq:rgv}) with the ceiling operation to quantize the gradient value $R(v_j)$ into a small integer $L(v_j)$.
Note that, the ceiling operation is used instead of the floor operation here. This is because the ceiling operation can overestimate the quantized gradient value after the reciprocal mapper in Phase 1, which ensures the underestimate of gradient value in the decoder.
\noindent
\textbf{Phase 3: Threshold Filter} \quad
The previous two phases of gradient value encoder together complete the quantization of gradient values. Next, the threshold filter is employed to further compressed the gradient values based on sparsification ideology.
The main idea of threshold filter is based on the factor that gradient elements with larger absolute values can contribute more to the convergence of the optimization algorithm. Theoretical analysis on this has been discussed in \cite{DBLP:conf/nips/WangniWLZ18, DBLP:conf/nips/StichCJ18, DBLP:conf/nips/JiangA18}. Hence, we further filter the logarithm quantized gradient value integer with a given threshold $\tau$ in Phase 3.
After using the reciprocal mapper, we obtain a reversed gradient value order, i.e., the smaller the compressed gradient value, the bigger the absolute original gradient value. Therefore, the threshold filter discards the compressed gradient integers whose values are larger than the specified threshold $\tau$, and retains those gradient elements with larger absolute values.
After the threshold filter, the maximal value of compressed gradient integers are within the threshold $\tau$.
Hence, $\lceil log_2 \tau \rceil$ bits are sufficient to store each remaining compressed gradient value. As to be discussed in Section~\ref{sec:effect}, the hyper parameter for threshold $\tau$ is recommended to set as 128. Hence, a single byte is enough for FastSGD to encode a gradient value in practice.
So far, we have discussed the compressed gradient in the format of absolute value. However, the sign for gradient value cannot be ignored, because SGD is vulnerable to the opposite direction of gradients although it is robust to decayed gradients. Last but not the least, we assign a negative sign to the compressed gradient integers whose original gradient values are negative, and complete the encoder for gradient values.
\begin{algorithm}[t]
\caption{Gradient Value Encoder (GV-Encoder)}
\label{algo:VE}
\LinesNumbered
\DontPrintSemicolon
\KwIn{the gradient values $\{v_j\}_{j=1}^D$, a given base $b$, a given threshold $\tau$}
\KwOut{the compressed gradient values $\{L(v_j)\}_{j=1}^d$ with the $sum$ of absolute gradient values}
$sum \gets \sum_{i=1}^{D} |v_i|$ \;
$\tau_e \gets sum/{b^\tau}$ \;
$Result \gets \emptyset$ \;
\For{$j \gets 1:D$}{
\If{$|v_j| \geq \tau_e$}{
transform $v_j$ into the reciprocal value $R(v_j) \gets sum/|v_j|$ \;
quantize $R(v_j)$ into the integer $L(v_j) \gets \left\lceil log_b (R(v_j)) \right\rceil$ \;
\If{$v_j < 0$}{
$L(v_j) \gets -L(v_j)$ \;
}
$Result \gets Result \cup L(v_j)$ \;
}
}
\Return $Result$ (i.e., $\{L(v_j)\}_{j=1}^d$) with $sum$
\end{algorithm}
Based on the above discussions, we are able to present the Gradient Value Encoder (GV-Encoder), with its pseudo-code shown in Algorithm~\ref{algo:VE}. GV-Encoder takes as inputs the gradient values $\{v_j\}_{j=1}^D$, a pre-defined base $b$, and a fixed threshold $\tau$, and outputs the compressed gradient values $\{L(v_j)\}_{j=1}^d$ with the $sum$ of absolute gradient values. Note that, the $sum$ of absolute gradient values is used for recovering the reciprocal value in the decoder, and we denote the number of filtered gradient integers as $d$ ($\leq D$). First, GV-Encoder sums up all the absolute gradient values in $sum$ (line 1). Then, it computes the early-stop threshold $\tau_e = sum/{b^\tau}$ (line 2) for filtering the gradient values before the reciprocal mapper and the logarithm quantization. Based on this, GV-Encoder avoids unnecessary processing for those two phases. Next, a set $Result$ is initialized to store the compressed gradient value integers (line 3). Thereafter, each gradient value is compressed using the three-phase encoder in a for loop (lines 4--10). Finally, GV-Encoder outputs the compressed gradient integer values $\{L(v_j)\}_{j=1}^d$ with $sum$ (line 11).
\subsection{The Encoder for Gradient Keys}
\label{sec:gk}
\begin{figure}
\centering
\includegraphics[width=0.495\textwidth]{deltaK}
\vspace{-2mm}
\caption{Illustration for the gradient key encoder}
\label{fig:delta}
\end{figure}
The above three-phase encoder emphasizes the compression of gradient values. In the following, we present the details of the encoder process for the sparse gradient keys $\{k_j\}_{j=1}^d$ corresponding to the compressed gradient integer values $\{L(v_j)\}_{j=1}^d$.
Different from the gradient values that could tolerate a precision loss, the gradient keys are vulnerable to errors. The recovering error in the gradient key would lead to a wrong optimal direction, and thus, the convergence of the model could not be guaranteed. Motivated by this, a lossless method that transforms the keys to delta keys has been designed~\cite{DBLP:conf/sigmod/JiangFY018}. The reason for using the delta key is that, although the original keys can be very large in many high-dimensional parameter cases, the difference between two adjacent keys is much smaller.
We follow the design of delta keys, and go a step further by using an adaptive fine-grained delta key encoding method to store gradient keys with fewer bits. Next, we detail the two following phases for gradient keys in the encoder of FastSGD.
\noindent
\textbf{Phase 4: Delta Encoding}~\cite{DBLP:conf/sigmod/JiangFY018} \quad Each gradient key $k_j$ is encoded into a delta key $\Delta k_j$, i.e., the incremental value compared with the previous key as follows:
\begin{equation}
\Delta k_j =\left\{
\begin{aligned}
& k_1, \quad \quad & j=1\\
& k_j - k_{(j-1)}, & 1<j\leq d \\
\end{aligned}
\right.
\end{equation}
\noindent
\textbf{Phase 5: Adaptive Binary Encoding} \quad After the delta encoding, the delta keys are much smaller than the original keys. In the previous work~\cite{DBLP:conf/sigmod/JiangFY018}, the byte is used to encode the delta key to replace the original format of integer or long-integer. Nonetheless, this still leaves the redundant space cost.
To alleviate this problem, we use the bit as the smallest grain to encode the delta key. Ideally, each delta key could be encoded in the binary format. For example, `3' could be binary encoded as `11' and `5' could be binary encoded as `101'. Unfortunately, the binary encoding key cannot be split to recover the original key if there is no flag to indicate the various binary encode lengths.
Towards this, we introduce the \textit{length flag} to indicate the number of bits used by each delta key. The size of length flag is denoted as $l$. In other words, we use $l$ bits to store the length flag. Then, the length flag can represent up to $2^l$ different lengths of binary encoding. By using the length flag, the delta keys could be compressed into adaptive binary encoding.
To be more specific, we take $l=2$ as an example. Assume that the maximal length of binary encoding for delta keys is $M$. Since the size $l$ of length flag is 2, the four levels of length flag, i.e., `00' to `11', could represent quaternary lengths of binary encoding, i.e., $\frac{1}{4}M$, $\frac{1}{2}M$, $\frac{3}{4}M$, and $M$, respectively. Based on this, each delta key could be encoded into a binary format that has the smallest length to store it with a length flag.
\begin{algorithm}[t]
\caption{Gradient Key Encoder (GK-Encoder)}
\label{algo:KE}
\LinesNumbered
\DontPrintSemicolon
\KwIn{the gradient keys $\{k_j\}_{j=1}^d$}
\KwOut{the compressed gradient keys $\{\Delta k_j\}_{j=1}^d$ in binary encoding with length flag}
$preKey \gets 0$, $maxDelta \gets 0$ \;
\For{$j \gets 1:d$}{
$\Delta k_j \gets k_j - preKey$ \;
$preKey \gets k_j$ \;
\If{$maxDelta < \Delta k_j$}{
$maxDelta \gets \Delta k_j$
}
}
$M \gets \lceil log_2 (maxDelta) \rceil$ \;
$BK \gets$ an empty binary sequence \;
\For{$j \gets 1:d$}{
$lengthFlag \gets getLengthFlag(\Delta k_j, M)$ \;
concatenate $lengthFlag$ with $\Delta k_j$ binary encoding to $BK$\;
}
\Return $BK$
\end{algorithm}
Fig.~\ref{fig:delta} illustrates the processing of the gradient key encoder when $l=2$. First, the gradient key encoder transforms each gradient key into a delta key. After the delta encoding, the maximum delta key is 232, which needs an 8-bit binary encoding. Hence, the quaternary ranges represented by the length flags are $[0,4)$, $[4,16)$, $[16,64)$, and $[64,256)$, respectively. Based on the length flag, each delta key could be encoded into a binary format that has the smallest length to store it with the corresponding length flag. For instance, the least key $k_j = 578$ is first encoded into the delta key $\Delta k_j = 3$. The $\Delta k_j = 3$ falls in the range of $[0,4)$, which is corresponding to the length flag '00'. Besides, it can be binary encoded as `11'. Thus, the adaptive binary encoding of the least key $k_j = 578$ is `0011'.
Algorithm~\ref{algo:KE} gives the pseudo-code of the Gradient Key Encoder (GK-Encoder). It takes the gradient key $\{k_j\}_{j=1}^d$ as input, and outputs the compressed gradient keys $\{\Delta k_j\}_{j=1}^d$ in binary encoding with length flag. First, it scans and encodes each gradient key $k_j$ into delta key $\Delta k_j$ in a for loop, and meanwhile finds the maximal delta key $maxDelta$ (lines 1--6). Then, it computes $M$ as the maximal length of binary encoding for delta keys (line 7), and initializes an empty binary sequence $BK$ to store the binary encoding with the length flag for delta keys (line 8). Next, another for loop is performed to encode each delta key by adaptive fine-grained binary encoding, and store the encoding to $BK$ in order (lines 9--11). Finally, GK-Encoder outputs the compressed gradient keys $BK$ (line 12).
\subsection{Space Cost and Time Complexity Analysis}
\label{sec:analysis}
In this subsection, we theoretically analyze the proposed FastSGD, including both the space cost and the time complexity.
\noindent
\textbf{Space Cost of FastSGD} \quad Given a gradient consisting of key-value pairs $\{k_j, v_j\}_{j=1}^D$, FastSGD compressed it into $\{\Delta k_j, L(v_j)\}_{j=1}^d$. The size of compressed gradient integer values is $d$ bytes. For the compressed gradient delta keys, we use $D_m$ to denote the number of total parameters, and assume that the intervals of any two adjacent sparse gradient elements are consistent. Then, the expected size of each binary encoder for the delta key itself is $\lceil log_2 \frac{D_m}{d} \rceil$ bits. In addition, each length flag needs $l$ bits. Thus, the size of each compressed gradient key is $\lceil log_2 \frac{D_m}{d} \rceil + l$ bits.
As to be reported in Section~\ref{sec:effect}, the average size per gradient key is 6 bits in practice.
In summary, the total space cost to transmit the given gradient is $8d + d(\lceil log_2 \frac{D_m}{d} \rceil + l)$ bits. In contrast, the space cost of the origin gradient is $64D$ bits even if we only use the integer format to represent the gradient key, and use the single-precision float format to store the gradient value.
\noindent
\textbf{Time Complexity of FastSGD} \quad Recall that the processes of both GV-Encoder and GK-Encoder in Section~\ref{sec:gv} and Section~\ref{sec:gk}, we could observe that the time complexity of FastSGD is $O(D)$. Thus, FastSGD achieves a fast linear time complexity.
Based on the aforementioned analysis, we can conclude that FastSGD theoretically provides an efficient compressed SGD framework with lightweight space cost for fast distributed ML. Next, we experimentally evaluate the performance of FastSGD.
\section{Preliminaries}
\label{sec:pre}
In this section, we present some preliminary materials related to compressed SGD for distributed ML. Table~\ref{tab:symbol} summarizes the symbols used frequently.
\begin{table}
\caption{Symbols and Description}
\label{tab:symbol}
\vspace{-1mm}
\begin{tabular}{|p{1.5cm}|p{6.5cm}|}
\hline
\textbf{Notation} & \textbf{Description} \\ \hline
$x$ & the training instance \\ \hline
$y$ & the corresponding label of $x$ \\ \hline
$\theta$ & the parameters of ML model \\ \hline
$g$ & the gradient vector \\ \hline
$D_m$ & the number of model parameters \\ \hline
$D$ & the number of nonzero elements in $g$ \\ \hline
$d$ & the number of gradient elements after threshold filter \\ \hline
$(k_j, v_j)$ & the $j$-th nonzero gradient key and value in $g$ \\ \hline
$R(v_j)$ & the reciprocal gradient value for $v_j$ \\ \hline
$L(v_j)$ & the logarithm quantified gradient value for $v_j$ \\ \hline
$\Delta k_j$ & the delta gradient key for $k_j$ \\ \hline
\end{tabular}
\end{table}
\subsection{Distributed Machine Learning}
In this paper, we focus on \textit{data parallelism}{\footnote{\textit{Model parallelism}, which splits model parameters to multiple distributed workers~\cite{DBLP:conf/nips/DeanCMCDLMRSTYN12}, is orthogonal to data parallelism, and outside the scope of this paper.}} distributed machine learning (ML)~\cite{DBLP:journals/corr/abs-2003-06307, DBLP:conf/osdi/LiAPSAJLSS14}. The workflow of data parallelism distributed ML is illustrated in Fig.~\ref{fig:disML}. Each worker possesses a local copy of the whole model parameters and a partition of input dataset. In a single round of iteration, each worker independently processes its data to compute local updates (i.e., the local gradients), and communicates with all other workers to synchronize with the aggregated global model state.
Formally, given a set of input data $\left\{x_{i}, y_{i}\right\}_{i=1}^{N}$, i.e., the training instances $x$ and their labels $y$, and a \textit{loss} function $f(x,y,\theta)$, the distributed ML training process tries to solve the optimization problem that minimizes the loss function $f$, i.e., finding the model parameter $\theta \in \mathbb{R}^{D_m}$ that best predicts $y_{i}$ for each $x_{i}$. Here, $N$ denotes the number of training instances, and $D_m$ corresponds to the number of model parameters.
\begin{figure}
\centering
\includegraphics[width=0.495\textwidth]{distributedML}
\vspace{-2mm}
\caption{The workflow of data parallelism distributed ML}
\label{fig:disML}
\end{figure}
\subsection{Stochastic Gradient Descent}
Stochastic Gradient Descent (SGD) is one of the most commonly-used first-order iterative gradient descent methods~\cite{robbins1951stochastic}. At iterative round $t+1$, SGD updates the model parameters $\theta_{t+1}$ as follows:
\begin{equation}
\theta_{t+1}=\theta_{t}-\eta_{t} g_{t},
\end{equation}
where $\eta_{t}>0$ is the learning rate; $g_{t}$ is the stochastic gradient at the iterative round $t$, i.e., a unbiased estimator of the gradient of $f$; and $\theta_{t}$ denotes the model parameter values at the iterative round $t$.
Typically, the gradient $g = \nabla_\theta f(x,y,\theta) \in \mathbb{R}^{D_m}$ is a sparse vector due to the origin data distribution and/or the sparsification compression technique. To save the size of storage, we adopt the key-value pair $(k_j, v_j)$ to represent the $j$-th nonzero gradient key and value in a sparse gradient vector. Here, we denote the number of nonzero elements in the gradient as $D$ ($\leq D_m$), then $j=1,2,...,D$.
In distributed ML, each worker independently computes the local gradients, and the updated gradients are aggregated to iteratively update the ML model. Given $W$ workers, in this paper, we aim to study the efficient compression of the gradients $\left\{g^{w}\right\}_{w=1}^{W}$ in order to save the communication cost among the distributed workers.
\section{Related Work}
\label{sec:related}
In this section, we overview three main categories of existing studies about the gradient compression for distributed ML, i.e., quantization, sparsification, and hybrid methods, respectively.
\subsection{Quantization Methods}
Quantization methods uses a small number of bits to represent each elements in the gradient.
Seide et al.~\cite{DBLP:conf/interspeech/SeideFDLY14} proposed the extreme form of quantization, 1-bit SGD, which quantizes the gradient element into a single bit by a user-defined threshold (0 by default). Dettmers~\cite{DBLP:journals/corr/Dettmers15} developed 8-bit Quantization to map each 32-bit gradient element into 8 bits. From those 8 bits, one bit is reserved for the sign of the number, three bits are used for exponent, while the rest four bits are used for mantissa.
Alistarh et al.~\cite{DBLP:conf/nips/AlistarhG0TV17} presented QSGD, a family of compression schemes that allow a trade-off between the amount of communication and the running time.
TernGrad~\cite{DBLP:conf/nips/WenXYWWCL17} quantizes gradients to ternary levels (i.e., $\{-1,0.1\}$) with layer-wise ternarizing and gradient clipping to reduce the communication cost.
SignSGD~\cite{DBLP:conf/icml/BernsteinWAA18} transmits the sign of gradient elements by quantizing the negative ones to -1 and the others to 1.
Besides, Miyashita et al.~\cite{DBLP:journals/corr/MiyashitaLM16} proposed LogQuant, which uses the logarithmic data representation to eliminate bulky digital multipliers in convolutional neural networks. LogQuant can also be considered as a kind of quantization method to quantize gradients in log-domain.
Quantization methods can only achieve a limited compression ratio. Take the commonly used single-precision float (consisting of 32 bits) as an example. The maximal compression ratio of quantization methods is 32$\times$. Hence, we introduce a type of more aggressive compression technique, namely, sparsification, in the following.
\subsection{Sparsification Methods}
Researchers have confirmed that only a small number of gradient elements are needed to be aggregated during the gradient aggregate without hurting the model convergence and performance \cite{DBLP:conf/nips/WangniWLZ18, DBLP:conf/nips/StichCJ18, DBLP:conf/nips/JiangA18}. Following this idea, sparsification methods choose and transmit only a part of elements of the original gradient. The first sparsification method for large-scale machine learning is Truncated Gradient~\cite{DBLP:journals/jmlr/LangfordLZ09}, which induces sparsity in the weights of online learning algorithms.
Random-$k$~\cite{DBLP:journals/corr/KonecnyMYRSB16, DBLP:conf/nips/WangniWLZ18} selects $k$ elements in the gradient vector to communicate and update randomly.
Top-$k$~\cite{DBLP:conf/nips/AlistarhH0KKR18, DBLP:conf/ijcai/ShiZWTC19} finds the $k$ largest values in the gradient to transmit.
In contrast, Threshold-$v$~\cite{DBLP:conf/aaai/DuttaBA0SCK20} finds the elements whose absolute values are larger than a given threshold. Ivkin et al.~\cite{DBLP:conf/nips/IvkinRUBSA19} presented Sketched-SGD, which uses count-sketch~\cite{DBLP:journals/corr/MiyashitaLM16} to find large coordinates of a gradient vector to approximate the Top-$k$ gradient elements. DGC~\cite{DBLP:conf/iclr/LinHM0D18} picks up the top 0.1\% largest elements to compress the gradient while preserving the accuracy by momentum correction, local gradient clipping, momentum factor masking, and warm-up training.
Compared with DGC, SGC~\cite{DBLP:conf/dasfaa/SunSJ0LXW19} employs long-term gradient compensation to further improve the convergence performance.
\subsection{Hybrid Methods}
Hybrid methods combine both quantization and sparsification.
Storm~\cite{DBLP:conf/interspeech/Strom15} presented a hybrid method, which discards those gradient elements whose absolute values less than a pre-defined threshold and quantizes the others in a similar way as 1-bit SGD~\cite{DBLP:conf/interspeech/SeideFDLY14}.
Instead of a pre-defined threshold, Adaptive~\cite{DBLP:conf/sc/DrydenMJE16} determines the threshold dynamically by gradient samples.
Besu et al.~\cite{DBLP:conf/nips/0001DKD19} proposed Qsparse-local-SGD, which combines Top-$k$ or Random-$k$ sparsification with quantization.
3LC~\cite{DBLP:conf/mlsys/LimAK19} is another hybrid method that combines three techniques, i.e., 3-value quantization with sparsity multiplication, quartic encoding, and zero-run encoding.
Jiang et al.~\cite{DBLP:conf/sigmod/JiangFY018, DBLP:journals/vldb/JiangFYSC20} presented SketchML, which utilizes quantile sketch-based method to compress the sparse gradient consisting of key-value pairs.
Considering the gradients are sparse due to the original input data distribution and/or the sparsification technique, the key-value pair is a natural representation for the transferring gradient. However, most of the existing studies are only suitable for the dense gradient. Only SketchML~\cite{DBLP:conf/sigmod/JiangFY018, DBLP:journals/vldb/JiangFYSC20} that investigates the compression for both gradient keys and gradient values considers the sparse property for gradient. Nonetheless, it is limited to communication-intensive workload. This is because SketchML has a high computational cost of compression/decompression that could overwhelm the savings by the reduced gradient communication. In contrast, our proposed FastSGD aims at providing a lightweight and efficient compressed SGD framework for distributed ML.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,973
|
Five things to consider if you are reading 'My Family's Slave'
May 19, 2017 · 7:00 PM EDT
By Marnette Federis
The cover of the June, 2017 issue of The Atlantic magazine.
If you had an emotional response to an essay in The Atlantic this week, you aren't alone.
The piece, written by the late Alex Tizon, tells the story of Eudocia Tomas Pulido, a woman who worked for his family and raised him. But Tizon reveals this stunning truth: For 56 years, Pulido, known to him as "Lola" — grandmother in Tagalog — was never paid for her work. She was his family's slave.
Some hailed Tizon for his writing and for telling the story in the first place. Then came the fury. In comments online, many people asked, why didn't he do more to free her, especially once he became an adult? They asked, why do we need to hear this story from Tizon, the oppressor shaping the narrative of the oppressed? Some comments said the remaining family members should be jailed or made to pay reparations to Pulido's remaining family members.
Out of the fury came defensive responses from some people in the Philippines. On Twitter and Facebook, readers began to discuss the Philippines' colonial past — which includes more than 300 years of rule under the Spanish and then almost 50 years under the US — and how it continues to shape power structures. Some Filipinos and Filipino Americans say it's important to consider poverty as a driving force of the enslavement of people like Pulido.
It's important to ask, "How does this happen?" And some say it's time to move forward and make sure this never happens again.
Read: "My Family's Slave" by Alex Tizon for The Atlantic
There's a lot to discuss. Here are five places to start better understanding Pulido's story.
1. There's nothing defensible about slavery. But should Tizon's essay be celebrated?
What happened to Pulido was indefensible — on that point, everyone seems to agree. But some readers are questioning Tizon's motives in writing her story. Was he trying to absolve himself? Why did Pulido's story have to come from him, the child of the oppressors?
Tizon died suddenly before the publication of the piece, so we'll never know the true answers to some of these questions. We only have his original story to reflect on.
The Seattle Times ran an obituary — largely informed by Tizon himself — about Pulido, who died in 2011 at age 86. In it, she is portrayed as a loving and loved member of the Tizon family. Reading the obituary is difficult, especially now that we know all of the facts. We'll never be able to ask Tizon about the disconnect.
The paper issued a retraction this week, calling the obituary a "whitewash for a fundamental truth known only to Tizon and his family: Ms. Pulido was a slave."
Tizon won a Pulitzer Prize in 1997 while he was a reporter at the Times.
His wife, Melissa Tizon, spoke with KUOW in Seattle about the essay and the reaction. It's worth a listen and could provide some insight into the Tizons' frame of mind.
2. There are historical and cultural forces at play that made it seem, at least to the elder Tizons, that what they were doing to Pulido was somehow acceptable.
Pulido became a slave in the Philippines after World War II, before she was brought to the US with the Tizon family. Filipinos in the Philippines say it's important to know the history and culture of the country in order to understand how a situation like this can happen.
The Philippines is a country that has a long history of colonialism, hundreds of years of oppression by the Spanish and then several decades by the US. The power dynamics and economy of the country is dictated by this colonial past. That past created a system where there are a few very powerful and rich people at the top and a lot of people in poverty at the bottom.
Also: Here's the backstory on why the US has such close ties to the Philippines
So, it is common for people to have domestic workers in their homes. This includes nannies, housekeepers and people who do your laundry. Even people who are in the working and middle class hire domestic helpers — you don't have to be very wealthy. These workers are part of the landscape, part of the economy.
In that situation, some families might feel like they're actually doing their domestic workers a favor. Not paying a living wage or only paying them with room and board is enough because, well, it's better than nothing. Without other options, domestic workers might internalize some of this thinking and feel indebted to their employer. And because this all happens within the privacy of homes, it's difficult to ensure that domestic workers are being treated fairly.
The dynamic creates a situation rife for abuse.
For more on how this system came to be, read historian Vicente Rafael's breakdown of the history and culture of the Philippines. Mike Ricca, a photographer from Manila, offers another take in the Philippines edition of Esquire. In it, he describes the "societal and cultural shackles" that Pulido had to contend with.
3. Cultural forces might have made it hard for Alex Tizon to speak out against his parents.
Filipino culture generally includes a very strong sense of duty to family. It's something that children often feel intensely, which prevents them from speaking out against their parents — even when they become adults. In America, Tizon learned about the history of slavery and African Americans. He relates one big fight with his mother in which he used the word "slave" to describe Pulido's situation. But ultimately, he did not do more to free Pulido until after his mother's death, when he was 40 years old — and she was 75.
Writer Jay Caspian Kang explains this in terms that many children of immigrants might find familiar: "It's a stunningly honest piece about something second-generation immigrants rarely address — the old world sins of their parents and the frustration, indignation and distance we feel whenever we encounter the trenchant, ugly things our parents brought with them."
4. Filipinos and Filipino Americans are grappling with the story, but are also willing to confront the problem.
You might have seen comments from people in the Philippines or Filipinos abroad that are very defensive. And you might have seen comments from those who say that Filipinos should be open to criticism. There's a debate happening among pinoys right now that has not happened in a long time around the issue of modern day slavery and the culture that propagates it.
The comment section of a story in Scout, a magazine about Filipino youth culture, is worth exploring.
Rappler, a popular Philippines news website, published several reaction pieces. Lian Buan explains the debate this way: "Whether he's Asian, or white, or black, [Tizon] was complicit to slavery, and that in itself is wrong, no matter the race. I struggled to write this because it feels like betraying my own: Tizon is a Filipino, a journalist and an immigrant. I should be empathetic. But I'm choosing to take a hard look at the mirror and recognize that this is my problem too, that this is our problem too."
5. There are many other Lola Eudocia Tomas Pulidos out there.
For many Filipinos and Filipino Americans, the question isn't about what the Tizons and The Atlantic should have done, but it's about what comes next. Activists point out that Pulido's story, while exceptional, is one of many. BAYAN USA, a group of progressive Filipino organizations in the United States, says thousands of migrant workers leave the Philippines each day, many to work in richer countries. Many of the workers are exploited with "no compensation, low wages, unstable working conditions, contractualization, physical and sexual abuse, wage theft and exploitation."
Rappler's Shakira Sison wrote a piece worth reading called, "We Are All Tizons." She encourages Filipinos, in the Philippines and the US, to go beyond anger and reflect on the power structures behind slavery. And she provides an action plan to dismantle them.
Step one is a lesson we might all relate to: "Stop treating your helpers' employment as a favor to them."
Colombia's plan to ban bullfighting sparks debate on tradition, animal rights
WNBA star Brittney Griner prisoner swap raises questions about how hostage deals are made
US senators demand full White House investigation into shooting of Palestinian American journalist
Reports show British teenager was allegedly trafficked to ISIS by Canadian agent
JusticeConflict & Justice
My Family's SlaveslaveryPhilippinescolonialismThe AtlanticAlex TizonEudocia Tomas PulidoFilipino AmericansAsiaPhilippines
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,763
|
{"url":"http:\/\/www.customizedmortgagesolutions.com\/?p=8791","text":"# Monthly Payment Contract\n\nYou can use a monthly payment agreement template as an outline for drafting the conditions and terms of a loan you lend to or borrow from another party. A written payment arrangement has several components, which include information such as the loan value, length of the payment period.\n\nWhat Is Balloon Finance That\u2019s the space companies like AFG and Fusion Auto Finance, based near Fort Worth, Texas, have been working. In PenFed\u2019s balloon program, AFG will be guaranteeing the residual, contacting owners in.\n\nLongview City Council members agreed to make three yearly installments to the state, starting with a 290,168 payment before. right)\\\\ 9920 &= M \\times 22.56\\\\ M &= 439.7163 \\end{align*} {\/eq} Now the total months left after 16 monthly installments has been paid are 8. Now calculating the payment that needs to be done in. Payment Plan\/Installment Agreement installment agreement. view: publications: forms: FAQs:. Online Payment Agreement. Online Payment Agreement Application.. If you can\u2019t pay the full amount due with your return, you can ask to make monthly. Publication 556 \u2013 Examination of Returns. A Car Sale Template that includes payment terms when selling a vehicle. Car Selling Contract Template. auto sale agreement. effective Date ____\/____\/. Payment Agreement Contract. By signing this agreement, all parties agree to the terms as described above. Alterations to this agreement can only be made by both parties and must be placed in writing. Both parties will receive a printed copy of this agreement, and will be responsible for upholding its terms. What Is Balloon Financing balloon loan: A long-term loan, often a mortgage, that has one large payment (the balloon payment) due upon maturity. A balloon loan will often have the advantage of very low interest payments, thus requiring very little capital outlay during the life of the loan. Since most of the repayment is deferred until the end of the payment period, the.Farm Finance Calculator Farm Mortgage Calculator. The practice of farming has been around for thousands of years. It is a long-standing tradition of man that has influenced the types of societies we live in, as well as how economies are shaped. The term or length, of your payment period; The amount of your monthly payments for the duration of the term; Keep all these elements in mind when going through your car sale contract. Before buying a car, the buyer should be aware of the requirements before purchasing. All these are contained in the agreement document. Sample Interest Only Promissory Note Cash Call Calculator So while a company can have other sources of cash to pay the dividend, I figure if CFFO covers the dividend, it\u2019s pretty safe. Based on that, I take my dividend projection 5 years into the future and. \u00b7 So, use this sample promissory note for car as a template for your successful promissory note. Skip to content. RequestLetters.com .. This note also includes interest of. All installments are due on the 1st and 15th day of each and every month thereafter until this Note is fully paid. The only exception is the final payment, The recurring credit card authorization form is a document that will authorize a company (ie: utility bills, various subscriptions, automobile payments, etc.) to automatically deduct payment from an individual\u2019s bank account or credit card account. The payments will be charged at the end of each bill\u2019s, billing cycle. This is a completely portable gaming device that is more affordable than the standard console, as it is purely for mobile. Long-Term Payment Plan (Installment Agreement) (pay in more than 120 days with monthly payments) Pay monthly through automatic withdrawals.31 setup fee; Plus accrued penalties and interest until the balance is paid in full; Pay amount owed through Direct Debit (with automatic payments from your checking account).","date":"2020-02-28 06:05:27","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.39078888297080994, \"perplexity\": 3688.154629890311}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-10\/segments\/1581875147054.34\/warc\/CC-MAIN-20200228043124-20200228073124-00188.warc.gz\"}"}
| null | null |
Ba Tays is a village in the Republic of Yemen. Follow the geography of the province of Abyan and administratively to the Directorate Khanfar. With a population of 6222 people, according to census conducted in 2004
External links
Towns and villages in the Abyan Governorate
Populated places in Abyan Governorate
Villages in Yemen
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,478
|
Question: On the Motion to Table (Motion to table Boxer Amdt. No. 541 )
Vote Date: June 8, 1999, 05:39 PM
Vote Result: Motion to Table Agreed to
Amendment Number: S.Amdt. 541 to S. 1122 (Department of Defense Appropriations Act, 2000)
Statement of Purpose: To substitute for section 8106 (relating to operational support aircraft) a requirement for a report.
Abraham (R-MI), Nay
Akaka (D-HI), Yea
Ashcroft (R-MO), Yea
Baucus (D-MT), Nay
Bayh (D-IN), Nay
Biden (D-DE), Not Voting
Bingaman (D-NM), Nay
Boxer (D-CA), Nay
Brownback (R-KS), Yea
Bunning (R-KY), Yea
Chafee, J. (R-RI), Yea
Conrad (D-ND), Nay
Crapo (R-ID), Not Voting
Daschle (D-SD), Nay
Durbin (D-IL), Nay
Edwards (D-NC), Nay
Feinstein (D-CA), Nay
Fitzgerald (R-IL), Yea
Gorton (R-WA), Yea
Graham (D-FL), Nay
Gramm (R-TX), Yea
Grassley (R-IA), Nay
Harkin (D-IA), Nay
Hollings (D-SC), Yea
Johnson (D-SD), Nay
Lautenberg (D-NJ), Nay
Levin (D-MI), Nay
Lincoln (D-AR), Nay
McCain (R-AZ), Not Voting
Mikulski (D-MD), Nay
Moynihan (D-NY), Yea
Robb (D-VA), Nay
Rockefeller (D-WV), Nay
Sarbanes (D-MD), Nay
Schumer (D-NY), Nay
Sessions (R-AL), Yea
Smith (R-OR), Yea
Torricelli (D-NJ), Nay
Voinovich (R-OH), Yea
Wellstone (D-MN), Nay
Wyden (D-OR), Nay
Bunning (R-KY)
Chafee, J. (R-RI)
Fitzgerald (R-IL)
Voinovich (R-OH)
Bayh (D-IN)
Edwards (D-NC)
Lincoln (D-AR)
Schumer (D-NY)
Crapo (R-ID)
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,263
|
\section{Introduction}
We propose the construction of a VN-core associated to each ($k$-linear)
split semigroupal functor $U$ from a suitable monoidal category $\ensuremath{\mathcal{C}}$ to
$\ensuremath{\mathbf{Vect}}_k$, where all our categories, functors, and natural transformations
are assumed to be $k$-linear, for a fixed field $k$. Essentially, the
category $\ensuremath{\mathcal{C}}$ must be equipped with a small ``$U$-generator'' $\ensuremath{\mathcal{A}}$ carrying
some extra duality information and with $UA$ still being finite dimensional
for all $A$ in $\ensuremath{\mathcal{A}}$.
We shall use the term ``VN-core'' (in $\ensuremath{\mathbf{Vect}}_k$) to mean a (usual)
$k$-semibialgebra $E$ together with a $k$-linear endomorphism $S$ such that
\[
\mu(\mu \ensuremath{\otimes} 1)(1 \ensuremath{\otimes} S \ensuremath{\otimes} 1)(1 \ensuremath{\otimes} \delta)\delta = 1 : E \ensuremath{\rightarrow} E.
\]
The VN-core is called ``antipodal'' if $S(xy) = Sy Sx$ (and $S(1) = 1$)
for all $x,y \in E$. This minimal type of structure is introduced here in
order to avoid compactness assumptions on the generator $\ensuremath{\mathcal{A}} \subset \ensuremath{\mathcal{C}}$ and,
at the same time, retain the ``fusion'' operator, namely
\[
(\mu \ensuremath{\otimes} 1)(1 \ensuremath{\otimes} \delta) : E \ensuremath{\otimes} E \ensuremath{\rightarrow} E \ensuremath{\otimes} E,
\]
satisfying the usual fusion equation~\cite{7}. Note that here the fusion
operator always has a partial inverse (see~\cite{1}).
In \S 2 we establish sufficient conditions on a functor $U$ in order that
\[
\ensuremath{\mathrm{End}}^\vee U = \int^A (UA)^* \ensuremath{\otimes} UA
\]
be a VN-core in $\ensuremath{\mathbf{Vect}}_k$ (following~\cite{2}). This core can be completed to a
VN-core $\ensuremath{\mathrm{End}}^\vee U \oplus k$ with a unit element. In \S 3 we give several
examples of suitable functors $U$ for the theory.
\section{The construction of $\ensuremath{\mathrm{End}}^\vee U$}
Let $\ensuremath{\mathcal{C}} = (\ensuremath{\mathcal{C}},\ensuremath{\otimes},I)$ be a monoidal category and let
\[
U : \ensuremath{\mathcal{C}} \ensuremath{\rightarrow} \ensuremath{\mathbf{Vect}}
\]
be a functor with both a semigroupal structure, denoted
\[
r = r_{C,D} : UC \ensuremath{\otimes} UD \ensuremath{\rightarrow} U(C \ensuremath{\otimes} D),
\]
and a cosemigroupal structure, denoted
\[
i = i_{C,D} : U(C \ensuremath{\otimes} D) \ensuremath{\rightarrow} UC \ensuremath{\otimes} UD,
\]
such that $ri = 1$.
We shall suppose also that there exists a small full subcategory $\ensuremath{\mathcal{A}}$ of $\ensuremath{\mathcal{C}}$
with the properties:
\begin{enumerate}
\item $UA$ is finite dimensional for all $A \in \ensuremath{\mathcal{A}}$,
\item $U$-density; the canonical map
\[
\alpha_C : \int^A \ensuremath{\mathcal{C}}(A,C) \ensuremath{\otimes} UA \ensuremath{\rightarrow} UC
\]
is an isomorphism for all $C \in \ensuremath{\mathcal{C}}$,
\item there is an ``antipode'' functor
\[
(-)^* : \ensuremath{\mathcal{A}}^\ensuremath{\mathrm{op}} \ensuremath{\rightarrow} \ensuremath{\mathcal{A}}
\]
with a (``canonical'') map $e_A : A \ensuremath{\otimes} A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} A$ in $\ensuremath{\mathcal{C}}$ for each $A
\in \ensuremath{\mathcal{A}}$,
\item there is a natural isomorphism
\[
u = u_A : U(A^*) \ensuremath{\xrightarrow}{\cong} U(A)^*,
\]
\item the following diagrams defining $\tilde{\tau}$, $\tilde{\rho}$ both
commute
\[
\xygraph{{UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA}="l"
[r(2)u(1.2)]{UA \ensuremath{\otimes} U(A^*) \ensuremath{\otimes} UA}="t"
[r(2)d(1.2)]{U(A \ensuremath{\otimes} A^* \ensuremath{\otimes} A)}="r"
"t"[d(2.4)]{UA}="b"
"l":@{.>}"r" ^-{\tilde{\tau}}
"l":"t" ^-{1 \ensuremath{\otimes} u^{-1} \ensuremath{\otimes} 1}
"t":"r" ^-{r_3}
"l":"b" _-{e_{UA}}
"r":"b" ^-{Ue_{A}}}
\]
and
\[
\xygraph{{UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA}="l"
[r(2)u(1.2)]{UA \ensuremath{\otimes} U(A^*) \ensuremath{\otimes} UA}="t"
[r(2)d(1.2)]{U(A \ensuremath{\otimes} A^* \ensuremath{\otimes} A)}="r"
"t"[d(2.4)]{UA}="b"
"r":@{.>}"l" _-{\tilde{\rho}}
"t":"l" _-{1 \ensuremath{\otimes} u \ensuremath{\otimes} 1}
"r":"t" _-{i_3}
"l":"b" _-{e_{UA}}
"r":"b" ^-{Ue_{A}}}
\]
where $e_{UA} = 1 \ensuremath{\otimes} \mathrm{ev}$ in $\ensuremath{\mathbf{Vect}}$, and $r_3 i_3 = 1$.
\end{enumerate}
We now define the semibialgebra structure $(\ensuremath{\mathrm{End}}^\vee U,\mu,\delta)$ on
\[
\ensuremath{\mathrm{End}}^\vee U = \int^A U(A)^* \ensuremath{\otimes} UA
\]
as in~\cite{2} \S 2, with the isomorphism of $k$-linear spaces
\[
S = \sigma : \ensuremath{\mathrm{End}}^\vee U \ensuremath{\rightarrow} \ensuremath{\mathrm{End}}^\vee U
\]
given (as in~\cite{2} \S 3) by the usual components
\[
\xygraph{{U(A)^* \ensuremath{\otimes} UA}="1"
[r(3.5)]{U(A^*)^* \ensuremath{\otimes} U(A^*)}="2"
"1"[d]{U(A)^* \ensuremath{\otimes} U(A)^{**}}="3"
"2"[d]{U(A^*) \ensuremath{\otimes} U(A^*)^*}="4"
"1":"2" ^-{\sigma_A}
"3":"4" ^-{u^{-1} \ensuremath{\otimes} u^*}
"1":"3" _-{1 \ensuremath{\otimes} d}
"4":"2" _-{c}}
\]
where $d$ is the canonical map from a vector space to its double dual.
Furthermore, each map
\[
e_{UA} = 1 \ensuremath{\otimes} \mathrm{ev}: UA \ensuremath{\otimes} UA^* \ensuremath{\otimes} UA \ensuremath{\rightarrow} UA
\]
satisfies both the conditions
\begin{equation}\tag{E1}
\vcenter{\xygraph{{UA}="1"
[ru]{UA \ensuremath{\otimes} UA^* \ensuremath{\otimes} UA}="2"
[rd]{UA}="3"
"1":"2" ^-{n \ensuremath{\otimes} 1}
"2":"3" ^-{e_{UA}}
"1":"3" _-1}}
\end{equation}
commutes, and
\begin{equation}\tag{E2}
\vcenter{\xygraph{{UA^*}="1"
[r(1.5)u]{UA^* \ensuremath{\otimes} UA \ensuremath{\otimes} UA^*}="2"
[r(1.5)d]{UA^* \ensuremath{\otimes} UA^{**} \ensuremath{\otimes} UA^*}="3"
"1":"2" ^-{1 \ensuremath{\otimes} n}
"2":"3" ^-{1 \ensuremath{\otimes} d \ensuremath{\otimes} 1}
"1":"3" _-{e^*_{UA}}}}
\end{equation}
commutes, where $n = \mathrm{coev}:1 \ensuremath{\rightarrow} UA \ensuremath{\otimes} UA^*$ in $\ensuremath{\mathbf{Vect}}$.
Then we obtain:
\begin{theorem}
The structure $(\ensuremath{\mathrm{End}}^\vee U, \mu, \delta, S)$ is a VN-core in $\ensuremath{\mathbf{Vect}}_k$ which
can be completed to the VN-core $(\ensuremath{\mathrm{End}}^\vee U) \oplus k$.
\end{theorem}
\begin{proof}
The von Neumann axiom
\[
\mu_3(1 \ensuremath{\otimes} S \ensuremath{\otimes} 1)\delta_3 = 1
\]
becomes the diagram (in which we have omitted ``$\ensuremath{\otimes}$''):
\[
\scalebox{0.75}{
\xygraph{{U(A)^*~UA~U(A)^*~UA~U(A)^*~UA}="1"
[r(6)]{U(A)^*~UA~U(A^*)^*~U(A^*)~U(A)^*~UA}="2"
"1"[d(3)r]{U(A)^*~UA~U(A)^*~U(A)^{**}~U(A)^*~UA}="3"
[r(2.5)u(1.5)]{U(A)^*~UA~U(A^*)~U(A^*)^*~U(A)^*~UA}="4"
[r(6)]{U(A)^*~U(A^*)^*~U(A)^*~UA~U(A^*)~UA}="5"
"5"[d(1.5)]{U(A)^*~U(A)^{**}~U(A)^*~UA~U(A)^*~UA}="6"
"3"[l(1)d(1.5)]{U(A)^*~U(A)~U(A)^*~UA}="7"
[r(4.5)]{U(A~A^*~A)^*~U(A)~U(A)^*~UA}="8"
[r(4.5)]{U(A~A^*~A)^*~U(A~A^*~A)}="10"
"7"[r(5.2)d(1.5)]{U(A)^*~U(A~A^*~A)}="9"
"7"[d(3)l(2)]{U(A)^*~U(A)}="11"
"11"[r(4)]{U(A)^*~UA}="12"
"12"[r(7)]{{\displaystyle \int^C} U(C)^*~U(C)}="13"
"7"[u(0.8)]{(*)}
"7"[d(1.7)]{\text{(E1)}}
"1":"3" _-{1~1~1~d~1~1}
"1":"2" ^-{1~1~S~1~1}
"3":"4" |-{1~1~u^{-1}~u^*~1~1}
"4":"5" _-{1~c~1}
"4":"2" ^-{1~1~c~1~1}
"2":"5" ^-{}
"7":"8" _-{U(e)^*~1}
"6":"5" _-{1~u^*~1~1~u^{-1}~1}
"3":"6" _-{1~c~1}
"7":@<8ex>@/^3ex/"1" ^-{1~1~1~n~1}
"8":"10" ^-{1~\tilde{\tau}}
"7":"9" _-{1~\tilde{\tau}}
"9":"10" _-{U(e)^*~1}
"9":"12" ^-{1~Ue}
"11":"7" ^-{1~n~1}
"11":"12" ^-{1}
"7":"12" ^-{1~e}
"6":"10" ^-{\tilde{\rho}^*~\tilde{\tau}}
"10":"13" ^-{\textrm{cop}_{C = A~A^*~A}}
"12":"13" ^-{\textrm{cop}_{C = A}}
"11":@/^6ex/[u(7.3)] ^-{\delta_3}
"7":"6" ^-{e^*~1~1~1}
}}
\]
where ($*$) is the exterior of the diagram
\[
\scalebox{0.85}{
\xygraph{{U(A)^* \ensuremath{\otimes} UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA}="1"
[r(7.5)]{U(A)^* \ensuremath{\otimes} UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} U(A)^{**} \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA}="2"
"1"[r(4)d(1.5)]{U(A)^* \ensuremath{\otimes} UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} U(A) \ensuremath{\otimes} U(A)^*}="3"
"1"[d(3)]{U(A)^* \ensuremath{\otimes} UA \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} U(A)}="4"
"2"[d(3)]{U(A)^* \ensuremath{\otimes} U(A)^{**} \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} U(A) \ensuremath{\otimes} U(A)^* \ensuremath{\otimes} UA}="5"
"4"[d(1.5)]{U(A)^* \ensuremath{\otimes} UA}="6"
"4"[r(3.5)u(0.75)]{\text{(E2)}}
"1":"2" ^-{1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} d \ensuremath{\otimes} 1 \ensuremath{\otimes} 1}
"1":"3" ^-{1 \ensuremath{\otimes} c \ensuremath{\otimes} 1}
"2":"5" ^-{1 \ensuremath{\otimes} c \ensuremath{\otimes} 1}
"3":"5" ^-{1 \ensuremath{\otimes} d \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1}
"4":"3" ^-{1 \ensuremath{\otimes} n \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1}
"4":"5" _-{e^* \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1}
"4":"1" ^-{1 \ensuremath{\otimes} 1 \ensuremath{\otimes} 1 \ensuremath{\otimes} n \ensuremath{\otimes} 1}
"6":"4" ^-{1 \ensuremath{\otimes} n \ensuremath{\otimes} 1}
}}
\]
which commutes using (E2) and commutativity of
\[
\xygraph{{\cdot}="1"
[u(0.5)r(0.8)]{\cdot}="2"
[r]{\cdot}="3"
"1"[d(0.5)r(0.8)]{\cdot}="4"
[r]{\cdot}="5"
"1":"2" ^-n
"2":"3" ^-{1~1~n}
"3":"5" ^-{c}
"1":"4" _-n
"4":"5" _-{n~1~1}}
\]
\end{proof}
\section{Examples}
\subsection{Example}\label{exam3.1}
The first type of example is derived from the idea of a (contravariant)
involution on a (small) comonoidal category $\ensuremath{\mathcal{D}}$. This includes the doubles
$\ensuremath{\mathcal{D}} = \ensuremath{\mathcal{B}}^\ensuremath{\mathrm{op}} + \ensuremath{\mathcal{B}}$ and $\ensuremath{\mathcal{D}} = \ensuremath{\mathcal{B}}^\ensuremath{\mathrm{op}} \ensuremath{\otimes} \ensuremath{\mathcal{B}}$ with their respective ``switch''
maps (where $\ensuremath{\mathcal{B}}$ is a given small comonoidal $\ensuremath{\mathbf{Vect}}_k$-category), or any
small comonoidal and compact-monoidal $\ensuremath{\mathbf{Vect}}_k$-category $\ensuremath{\mathcal{D}}$ (such as the
category $\mathbf{Mat}_k$ of finite matrices over $k$) with the tensor duals
of objects now providing an antipode on the comonoidal aspect of the structure
rather than on the monoidal part, or any $*$-algebra structure on a given
$k$-bialgebra (e.g., a $C^*$-bialgebra) with the $*$-operation providing
the antipode.
In each case, an \emph{even} functor from $\ensuremath{\mathcal{D}}$ to $\ensuremath{\mathbf{Vect}}$ is defined to be a
($k$-linear) functor $F$ equipped with a (chosen) dinatural isomorphism
\[
F(D^*) \cong F(D).
\]
If we take the morphisms of even functors to be all the natural transformations
between them then we obtain a category
\[
\ensuremath{\mathcal{E}} = \ensuremath{\mathcal{E}}(\ensuremath{\mathcal{D}},\ensuremath{\mathbf{Vect}}).
\]
Let $\ensuremath{\mathcal{A}} = \ensuremath{\mathcal{E}}(\ensuremath{\mathcal{D}},\ensuremath{\mathbf{Vect}}_\ensuremath{\mathrm{fd}})_\mathrm{fs}$ be the full subcategory of $\ensuremath{\mathcal{E}}$
consisting of the finitely valued functors of finite support. While this
category is generally not compact, it has on it a natural antipode derived from
those on $\ensuremath{\mathcal{D}}$ and $\ensuremath{\mathbf{Vect}}_\ensuremath{\mathrm{fd}}$, namely
\[
A^*(D) := A(D^*)^*.
\]
Of course, there are also examples where $\ensuremath{\mathcal{A}}$ is actually compact, such as
those where $\ensuremath{\mathcal{D}}$ is a Hopf algebroid, in the sense of~\cite{4}, with antipode
$(-)^* = S$, in which case each $A$ from $\ensuremath{\mathcal{D}}$ to $\ensuremath{\mathbf{Vect}}$ has a symmetry
structure on it.
Now let $\ensuremath{\mathcal{C}}$ be the full subcategory of $\ensuremath{\mathcal{E}}$ consisting of the small coproducts
in $\ensuremath{\mathcal{E}}$ of objects from $\ensuremath{\mathcal{A}}$. This category $\ensuremath{\mathcal{C}}$ is easily seen to be monoidal
under the pointwise convolution structure from $\ensuremath{\mathcal{D}}$, and the inclusion $\ensuremath{\mathcal{A}}
\subset \ensuremath{\mathcal{C}}$ is $U$-dense for the functor
\[
U : \ensuremath{\mathcal{C}} \ensuremath{\rightarrow} \ensuremath{\mathbf{Vect}}_k
\]
given by
\[
U(C) = \sum_D C(D)
\]
which is split semigroupal with $UA$ finite dimensional for all $A \in \ensuremath{\mathcal{A}}$.
Moreover,
\begin{align*}
U(A^*) &= \bigoplus_D A^*(D) \\
&= \bigoplus_D A(D)^* \\
&= U(A)^*,
\end{align*}
for all $A \in \ensuremath{\mathcal{A}}$. The conditions of (5) are easily verified if we define maps
\[
e : A \ensuremath{\otimes} A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} A
\]
by commutativity of the diagrams
\[
\xygraph{{A(D) \ensuremath{\otimes} A^*(D) \ensuremath{\otimes} A(D)}="1"
[r(3)]{A(D)}="2"
"1"[d]{A(D) \ensuremath{\otimes} A(D)^* \ensuremath{\otimes} A(D),}="3"
"1":"2" ^-{e_D}
"1":"3" _-{\cong}
"3":"2" _-{1 \ensuremath{\otimes} \mathrm{ev}}}
\]
where the exterior of
\[
\xygraph{{A^*(D) \ensuremath{\otimes} A(D)}="1"
[d]{A(D)^* \ensuremath{\otimes} A(D)}="2"
[l(3)d]{A^*(E) \ensuremath{\otimes} A(D)}="3"
"2"[d]{A(E)^* \ensuremath{\otimes} A(D)}="4"
"2"[r(2.8)d]{k}="5"
"4"[d]{A(E)^* \ensuremath{\otimes} A(E)}="6"
[d]{A^*(E) \ensuremath{\otimes} A(E)}="7"
"1":"2" _-{\cong}
"1":"3" _-{A^*(f) \ensuremath{\otimes} 1}
"1":"5" ^-{\hat{e}}
"3":"4" ^-{\cong}
"4":"2" ^-{A(f)^* \ensuremath{\otimes} 1}
"2":"5" _-{\mathrm{ev}}
"4":"6" _-{1 \ensuremath{\otimes} A(f)}
"6":"5" ^-{\mathrm{ev}}
"6":"7" _-{\cong}
"3":"7" _-{1 \ensuremath{\otimes} A(f)}
"7":"5" _-{\hat{e}}}
\]
commutes for all maps $f:D \ensuremath{\rightarrow} E$ in $\ensuremath{\mathcal{D}}$ so that
\[
e = 1 \ensuremath{\otimes} \hat{e} : A \ensuremath{\otimes} A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} A \ensuremath{\otimes} k \cong A
\]
is a genuine map in $\ensuremath{\mathcal{C}}$ when $\ensuremath{\mathcal{C}}$ is given the pointwise monoidal structure
from $\ensuremath{\mathcal{D}}$.
This completes the details of the general example.
\subsection{Example}
In the case where $k = \mathbb{C}$ and $\ensuremath{\mathcal{D}}$ has just one object $D$
whose endomorphism algebra is a $C^*$-bialgebra, we have a one-object
comonoidal category $\ensuremath{\mathcal{D}}$ with a $\mathbb{C}$-conjugate-linear antipode given by
the $*$-operation. Then the convolution $[\ensuremath{\mathcal{D}}, \mathbf{Hilb}_\ensuremath{\mathrm{fd}}]$, where
\[
[\ensuremath{\mathcal{D}}, \mathbf{Hilb}_\ensuremath{\mathrm{fd}}] \subset [\ensuremath{\mathcal{D}}, \ensuremath{\mathbf{Vect}}_\mathbb{C}],
\]
is a monoidal category, with a $\mathbb{C}$-linear antipode given by
\[
F^*(D) = F(D^*)^\circ
\]
where $H^\circ$ denotes the conjugate-transpose of $H \in \mathbf{Hilb}_\ensuremath{\mathrm{fd}}$.
We now interpret an even functor $F$ to be a functor equipped with a dinatural
isomorphism $F(D^*) \cong F(D)$ in $D \in \ensuremath{\mathcal{D}}$ which is
\emph{$\mathbb{C}$-linear}, so that $F^*(D) \cong F(D)^\circ$ for such a
functor.
Take $\ensuremath{\mathcal{A}} = \ensuremath{\mathcal{E}}(\ensuremath{\mathcal{D}},\mathbf{Hilb}_\ensuremath{\mathrm{fd}})$ and let $\ensuremath{\mathcal{C}}$ be the class of small
coproducts in $[\ensuremath{\mathcal{D}},\ensuremath{\mathbf{Vect}}_\mathbb{C}]$ of the underlying
$[\ensuremath{\mathcal{D}},\ensuremath{\mathbf{Vect}}_\mathbb{C}]$-representations of $A$'s in $\ensuremath{\mathcal{A}}$ (with the appropriate
maps). Each map
\[
e : A \ensuremath{\otimes} A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} A
\]
in $\ensuremath{\mathcal{C}}$ is defined by the $\mathbb{C}$-linear components
\[
e : A(D) \ensuremath{\otimes} A^*(D) \ensuremath{\otimes} A(D) \ensuremath{\xrightarrow}{1 \ensuremath{\otimes} \hat{e}} A(D),
\]
where
\[
\hat{e} : A^*(D) \ensuremath{\otimes} A(D) \ensuremath{\rightarrow} \mathbb{C}
\]
in $\ensuremath{\mathbf{Vect}}_\mathbb{C}$ comes from the $\mathbb{C}$-bilinear composite of two
maps which are both $\mathbb{C}$-linear in the first variable and
$\mathbb{C}$-linear in the second, namely
\[
\xygraph{{A^*(D) \times A(D)}="1"
[r(2)]{\mathbb{C}}="2"
"1"[d]{A(D)^\circ \times A(D).}="3"
"1":"2"
"1":"3" _-{\cong}
"3":"2" _-{\< -,- \>}}
\]
The remainder of this example is as seen before in Example~\ref{exam3.1}.
\subsection{Example}
Let $\ensuremath{\mathcal{V}} = (\ensuremath{\mathcal{V}},\ensuremath{\otimes},I)$ be a (small) braided monoidal category and let $\ensuremath{\mathcal{B}}$ be
the $k$-linearization of $\mathbf{Semicoalg}(\ensuremath{\mathcal{V}})$ with the monoidal structure
induced from that on $\ensuremath{\mathcal{V}}$.
By analogy with~\cite{5}, let $\ensuremath{\mathcal{X}} \subset \ensuremath{\mathcal{B}}$ be a finite full subcategory of
$\ensuremath{\mathcal{B}}$ with $I \in \ensuremath{\mathcal{X}}$ and $\ensuremath{\mathcal{X}}^\ensuremath{\mathrm{op}}$ promonoidal when
\begin{align*}
p(x,y,z) &= \ensuremath{\mathcal{B}}(z, x \ensuremath{\otimes} y) \\
j(z) &= \ensuremath{\mathcal{B}}(z, I)
\end{align*}
for $x,y,z \in \ensuremath{\mathcal{X}}$.
For example (cf.~\cite{5}), one could take $\ensuremath{\mathcal{X}}$ to be a (finite) set of
non-isomorphic ``basic'' objects in some braided monoidal category $\ensuremath{\mathcal{V}}$, where
each $x \in \ensuremath{\mathcal{X}}$ has a coassociative diagonal map $\delta : x \ensuremath{\rightarrow} x \ensuremath{\otimes} x$.
However, we won't need the category $\ensuremath{\mathcal{X}}$ to be discrete or locally finite in
the following.
Now let $\ensuremath{\mathcal{C}}$ be the convolution $[\ensuremath{\mathcal{X}}^\ensuremath{\mathrm{op}},\ensuremath{\mathbf{Vect}}]$ and let $\ensuremath{\mathcal{A}} =
[\ensuremath{\mathcal{X}}^\ensuremath{\mathrm{op}},\ensuremath{\mathbf{Vect}}_\ensuremath{\mathrm{fd}}]$. The functor
\[
U : \ensuremath{\mathcal{C}} \ensuremath{\rightarrow} \ensuremath{\mathbf{Vect}}
\]
is defined by
\[
U(C) = \bigoplus_x C(x),
\]
and the obvious inclusion $\ensuremath{\mathcal{A}} \subset \ensuremath{\mathcal{C}}$ is $U$-dense. If there is a canonical
(natural) retraction
\[
\xygraph{{p(x,y,z) = \ensuremath{\mathcal{B}}(z, x \ensuremath{\otimes} y)}="1"
[r(4)]{\ensuremath{\mathcal{B}}(z,x) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(z,y),}="2"
"1":@<-0.7ex>"2" _-{i_{x,y}}
"2":@<-0.7ex>"1" _-{r_{x,y}}}
\]
derived from the semicoalgebra structures on $x,y,z$, then $U$ becomes a split
semigroupal functor via the structure maps
\[
\xygraph{{U(C) \ensuremath{\otimes} U(D)}="1"
[r(5)]{U(C \ensuremath{\otimes} D)}="2"
"1"[d(1.3)]{\bigoplus_x C(x) \ensuremath{\otimes} \bigoplus_y D(y)}="3"
"2"[d(1.3)]{\bigoplus_z \ensuremath{\displaystyle \int}^{xy} p(x,y,z) \ensuremath{\otimes} C(x) \ensuremath{\otimes} D(y)}="4"
"3"[d(1.3)]{\bigoplus_z C(z) \ensuremath{\otimes} D(z)}="5"
"4"[d(1.3)]{\bigoplus_z \ensuremath{\displaystyle \int}^{xy} \ensuremath{\mathcal{B}}(z,x) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(z,y) \ensuremath{\otimes} C(x)\ensuremath{\otimes} D(y),}="6"
"1":@<3pt>"2" ^-r
"2":@<3pt>"1" ^-i
"1":@{=}"3"
"2":@{=}"4"
"5":@<3pt>"3" ^-{\Delta}
"3":@<3pt>"5" ^-{\Delta^*}
"6":@<3pt>"4" ^-{\text{``$r$''}}
"4":@<3pt>"6" ^-{\text{``$i$''}}
"6":"5" _-\cong}
\]
where the isomorphism follows from the Yoneda lemma, and $ri = 1$.
If $\ensuremath{\mathcal{X}}$ also has on it a duality
\[
(-)^* : \ensuremath{\mathcal{X}} \ensuremath{\rightarrow} \ensuremath{\mathcal{X}}^\ensuremath{\mathrm{op}}
\]
such that $x \cong x^{**}$, then, on defining
\[
A^*(x) = A(x^*)^*,
\]
we obtain
\begin{align*}
U(A^*) &= \bigoplus_x A^*(x) \\
&= \bigoplus_x A(x^*)^* \\
&\cong \bigoplus_x A(x)^* & \text{since $x \cong x^{**}$} \\
&\cong U(A)^*,
\end{align*}
for $A \in \ensuremath{\mathcal{A}}$, in accordance with the fourth requirement on $U$.
Finally, to obtain a suitable map
\[
e = 1 \ensuremath{\otimes} \hat{e}: A \ensuremath{\otimes} A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} A \ensuremath{\otimes} I \cong A,
\]
where $\hat{e}: A^* \ensuremath{\otimes} A \ensuremath{\rightarrow} I$, we suppose each $A$ in $\ensuremath{\mathcal{A}}$ has on it a
``dual coupling''
\[
\chi = \chi_{xy} : A(x)^* \ensuremath{\otimes} A(y) \ensuremath{\rightarrow} \ensuremath{\mathcal{B}}(x^* \ensuremath{\otimes} y,I).
\]
By considering the Yoneda expansion
\[
A(x) \cong \int^z A(z) \ensuremath{\otimes} \ensuremath{\mathcal{X}}(x,z)
\]
of the various functors $A$ in $\ensuremath{\mathcal{A}} = [\ensuremath{\mathcal{X}}^\ensuremath{\mathrm{op}},\ensuremath{\mathbf{Vect}}_\ensuremath{\mathrm{fd}}]$, such a coupling
exists on each $A$ if we suppose merely that $\ensuremath{\mathcal{X}}$ itself is ``coupled'' by a
natural transformation
\[
\chi : \ensuremath{\mathcal{X}}(y,z) \ensuremath{\rightarrow} \ensuremath{\mathcal{X}}(x,z) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(x^* \ensuremath{\otimes} y,I);
\]
or simply
\[
\chi : \ensuremath{\mathcal{X}}(x,z)^* \ensuremath{\otimes} \ensuremath{\mathcal{X}}(y,z) \ensuremath{\rightarrow} \ensuremath{\mathcal{B}}(x^* \ensuremath{\otimes} y,I),
\]
if $\ensuremath{\mathcal{X}}$ is locally finite. Then, the composite natural transformation
\[
\xygraph{{A(x^*)^* \ensuremath{\otimes} A(y) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(z, x \ensuremath{\otimes} y)}="1"
[d]{\ensuremath{\mathcal{B}}(x^{**} \ensuremath{\otimes} y, I) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(z, x \ensuremath{\otimes} y)}="2"
[d]{\ensuremath{\mathcal{B}}(x \ensuremath{\otimes} y, I) \ensuremath{\otimes} \ensuremath{\mathcal{B}}(z, x \ensuremath{\otimes} y)}="3"
[d]{\ensuremath{\mathcal{B}}(z, I)}="4"
"1":"2" ^-{\chi \ensuremath{\otimes} 1}
"2":"3" ^-\cong
"3":"4" ^-{\text{comp'n}}}
\]
yields the map
\[
\xygraph{{A^* \ensuremath{\otimes} A}="1"
[r(3.3)]{I}="2"
"1"[d]{\ensuremath{\displaystyle \int}^{xy} A^*(x) \ensuremath{\otimes} A(y) \ensuremath{\otimes} p(x,y,-)}="3"
"2"[d]{\ensuremath{\mathcal{B}}(-,I)}="4"
"1":"2" ^-{\hat{e}}
"3":"4"
"1":@{=}"3"
"2":@{=}"4"}
\]
because $p(x,y,-) = \ensuremath{\mathcal{B}}(-,x \ensuremath{\otimes} y)$ (by definition). Thus suitable conditions on
the coupling $\chi$ give (5).
\begin{remark}
Actually, this last example in which the basic promonoidal structure occurs as
a canonical retract of a comonoidal structure is typical of many other examples
which can be treated along similar lines.
\end{remark}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,954
|
{"url":"https:\/\/indico.cern.ch\/event\/765096\/contributions\/3295724\/","text":"# Input to the European Particle Physics Strategy Update 2018-2020\n\n1 November 2018 to 19 December 2018\nEurope\/Zurich timezone\n\n## Study of hard and electromagnetic processes at CERN-SPS energies: an investigation of the high-$\\mu_{\\mathbf{B}}$ region of the QCD phase diagram\n\nNot scheduled\n1m\nStrong interactions (perturbative and non-perturbative QCD, DIS, heavy ions)\n\n### Description\n\nThe exploration of the phase diagram of Quantum ChromoDynamics (QCD) is carried out by studying ultrarelativistic heavy-ion collisions. The energy range covered by the CERN SPS ($\\sqrt{s_{\\rm \\scriptscriptstyle{NN}}} \\sim 6\\text{--}17$~GeV) is ideal for the investigation of the region of the phase diagram corresponding to finite baryochemical potential ($\\mu_{\\rm B}$), and has been little explored up to now. We propose in this document a new experiment, NA60+, that would address several observables which are fundamental for the understanding of the phase transition from hadronic matter towards a Quark--Gluon Plasma (QGP) at SPS energies. In particular, we propose to study, as a function of the collision energy, the production of thermal dimuons from the created system, from which one would obtain a caloric curve of the QCD phase diagram that is sensitive to the order of the phase transition. In addition, the measurement of a $\\rho\\text{--}{\\rm a}_1$ mixing contribution would provide conclusive insights into the restoration of the chiral symmetry of QCD.\nIn parallel, studies of heavy quark and quarkonium production would also be carried out, addressing the measurement of transport properties of the QGP and the investigation of the onset of the deconfinement transition. The document also defines an experimental set-up which couples a vertex telescope based on monolithic active pixel sensors (MAPS) to a muon spectrometer with tracking (GEM) and triggering (RPC) detectors within a large acceptance toroidal magnet.\nResults of physics performance studies for most observables accessible to NA60+ are discussed, showing that the results of the experiment would lead to a significant advance of our understanding of strong interaction physics.\n\n### Primary authors\n\nTorsten Dahms (Technische Universitaet Muenchen (DE)) Enrico Scomparin (Universita e INFN Torino (IT)) Gianluca Usai (Universita e INFN, Cagliari (IT))","date":"2020-09-20 11:29:13","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.47550129890441895, \"perplexity\": 2367.4791015384717}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-40\/segments\/1600400197946.27\/warc\/CC-MAIN-20200920094130-20200920124130-00030.warc.gz\"}"}
| null | null |
\section{Introduction}
AERA~\cite{AERA}, the Auger Engineering Radio Array, was designed to explore the radio emission of extensive air showers above an energy threshold of $10^{17}$ eV in a hybrid detection mode. It consists of 124 antenna stations of which 24 are log periodic dipole antennas (LPDAs) and 100 are butterfly-type antennas. Both antenna types measure two electric field components (North-South and East-West). AERA is able to measure cosmic rays together with the Auger surface detector, the fluorescence detector, as well as the muon detector of AMIGA. It has been proven that there is a correlation between the radio amplitude and the energy of the primary cosmic ray~\cite{qader_proceed}. Furthermore the correlation between radio parameters and the shower maxima has been shown~\cite{qader_proceed}.
Taking horizontal air showers into account will increase the exposure of radio hybrid measurements as well as give better understanding of the electromagnetic component of inclined showers. This will contribute to the knowledge gained by analysing the data of the surface detector, the fluorescence detector, and the AMIGA muon detector extension. \iffalse All together it will make the Pierre Auger Observatory a powerful instrument to detect cosmic rays in all solid angles and therefore allow to detect new phenomena in cosmic rays.\fi
\section{Motivation to study horizontal showers}
Showers with zenith angles larger than 60$^{\circ}$ are defined as horizontal for the surface detector infill area. \iffalse The analysis of the radio signal of horizontal showers offers some advantages.\fi The motivation to investigate horizontal air showers in the radio domain is that the radio signal is not absorbed in the Earth's atmosphere. This means, the radio signal can still be detected when the particle shower is mostly absorbed. This is illustrated in figure~\ref{fig:sdrdinclined}.
\begin{figure}[b]
\includegraphics[height=.13\textheight]{SdRdInclinedShower.pdf}
\caption{Illustration of a horizontal shower. The hadronic and electromagnetic component is absorbed and the muonic component and the radio signal can be measured on the ground.}
\label{fig:sdrdinclined}
\end{figure}
The primary particle interacts in the atmosphere, the hadronic and electromagnetic components are evolving in the atmosphere and are then absorbed. What reaches the ground is the muonic component of the shower and the coherent radio emission emitted by the electromagnetic component.
Horizontal extensive air showers travel through more atmosphere than vertical showers, i.e. the distance to the radio source is larger. During the whole evolution of the shower, radio waves are emitted mainly in forward direction and this leads to a large footprint on ground for more horizontal showers. This is illustrated by CoREAS~\cite{CoREAS} simulations seen in figure~\ref{fig:simtimewa}.
Three simulated extensive air showers with an energy of $10^{18}$ eV are shown with zenith angles of 30$^{\circ}$, 50$^{\circ}$ and 75$^{\circ}$. The footprint of the shower with 75$^{\circ}$ zenith angle is of 6 km length, where the asymmetry of the footprint has to be taken into account. The lateral distribution function and the wave front shape of the radio signal of horizontal air showers will provide information to distinguish between young and old showers.
This difference is already seen just by the size of the footprint. The background and transient noise is larger for radio detection of horizontal cosmic showers because looking to the horizon many man-made noise sources are relevant. A further relevant characteristic is that a significant vertical component of the electric field vector of the radio signal is expected for the Argentinian local magnetic field.
\begin{figure}[!htbp]
\includegraphics[height=.21\textheight]{30_Grad_rotated_text_short.png}
\includegraphics[height=.21\textheight]{50_Grad_rotated_text_short.png}
\includegraphics[height=.21\textheight]{75_Grad_rotated_text_short.png}
\caption{CoREAS simulations for primary proton showers of $10^{18}$ eV for different zenith angles~\cite{CoREAS_sim}. Please note the different scales. The limit of detection for current experiments is \(\approx\) 2 \(\mu\)V\(/\)m\(/\)MHz.}
\label{fig:simtimewa}
\end{figure}
\section{Horizontal events measured by AERA-24 and AERA-124}
AERA was designed as an engineering array and therefore there are two antenna types deployed in the AERA site. AERA-24 consists of 24 LPDAs, log periodic dipole antennas. These were deployed in November 2011. In May 2013 100 butterfly antenna stations were deployed and so AERA-124 was completed. A map of the current AERA configuration can be found in figure~\ref{fig:minipage_sol} on the left. The newly installed five butterfly stations with whisk-type antennas and the four tripole stations will be discussed in the next section.
\begin{figure}[!htbp]
\centering
\begin{minipage}[b]{.5\linewidth}
\begin{center}
\includegraphics[width=.8\linewidth]{Map_stations.jpg}\\
\begin{flushleft}
\caption{figure}{Deployment of prototype stations in November 2013:
The current AERA-124 configuration with 24 butterfly stations, 100 LPDAs, five butterfly antennas with whisk antennas, three tripole stations, and one low-frequency antenna. Figure adapted from~\cite{map}.}
\end{flushleft}
\label{fig:aeramap}
\end{center}
\end{minipage}%
\quad
\begin{minipage}[b]{.5\linewidth}
\begin{center}
\includegraphics[width=.8\linewidth]{Event_example.pdf}
\begin{flushleft}
\caption{figure}{Event example of a horizontal air shower with an energy of 9.0$\times$10$^{17}$eV, zenith of 64.5$^{\circ}$, azimuth of 323.6$^{\circ}$ and geomagnetic angle alpha of 101.4$^{\circ}$.}
\end{flushleft}
\label{fig:event_example}
\end{center}
\end{minipage}
\caption{The current AERA-124 configuration on the left and an example of a horizontal event on the right.}
\label{fig:minipage_sol}
\end{figure}
To determine the potential of radio detection of horizontal air showers a method was developed looking for the single station efficiency of AERA. Requiring an external trigger of the surface detector, the AERA events are selected respective to the distance of the closest AERA station to the shower axis and the timing of the strongest pulse in the AERA trace. The geometry of the shower is determined by the surface detector reconstruction in Offline~\cite{Offline}.
The quality cuts of the surface detector reconstruction for horizontal air showers are applied (ICRC 2013):
\begin{itemize}
\item zenith angle 62\(^\circ\)-80\(^\circ\)
\item minimum of four triggered surface detector stations, the surface detector stations have a distance of 1.5 km
\item apply physics trigger cut, which rejects signals generated by background muons
\item apply quality trigger cut, which rejects events close to the border of the array
\item exclude bad periods
\end{itemize}
To distinguish between cosmic ray induced radio showers and a noise pulse the ratio of cosmic ray energy and the electric field strength measured by the AERA station is considered. The selection time interval for horizontal events is shown in figure~\ref{fig:timestamp}. The position of the magnitude of the electric field strength of the highest pulse of the closest AERA station is compared to the expected time in the trace. An interval of $\pm$ 7 \textmu s is taken as an event selection criterion. Currently the time interval is set that large due to the missing AERA time calibration and the large core uncertainty of horizontal air showers of low energy at the threshold. As the time calibration is established a sharper time window can be chosen.
Events inside the time interval show a clear correlation between energy of the shower and electric field strength. This can be seen in figure~\ref{fig:timestamp} on the upper right corner. For events outside this time interval there is no correlation observed, which can be seen in figure~\ref{fig:timestamp} on the lower right corner. The fit was performed for the average of the amplitude of the three closest stations.
One example of a horizontal air shower is shown in figure~\ref{fig:minipage_sol} on the right. The event has an energy of 9.0$\times$10$^{17}$ eV, zenith angle of 64.5$^{\circ}$, azimuth angle of 323.6$^{\circ}$ and geomagnetic angle alpha of 101.4$^{\circ}$.
Figure~\ref{fig:dist_energy} shows the selected events when applying the selection criteria described above. On the left the distance distribution and on the right the energy distribution of the events are shown. Inclined showers can be seen up to a far distance to the reconstructed shower axis. \iffalse However, having confirmed experimentally the large footprint, investigations are still required to improve the purity and efficiency of the event selection since some high energy inclined events are not reconstructed by the present, not optimized, event selection.\fi
Although confirming experimentally the large footprint, there is still need of further investigations in order to improve the purity and the efficiency of the event selection since some high energy inclined events are not reconstructed by the present, not optimized, event selection. The new 25 AERA stations, which are going to be installed in 2015, will improve the statistic of horizontal air showers for AERA.
\begin{figure}[!htbp]
\includegraphics[height=.33\textheight]{Signal_Timestamp_and_fit.pdf}
\caption{Horizontal event selection. The distribution of station time of the pulse minus the calculated arrival time is shown on the left. The amplitude of events in the $\pm$ 7 \textmu s time interval are shown in the upper right corner. In the lower right corner events outside the $\pm$ 7 \textmu s time interval are shown. The fit was performed for the average of the amplitude of the three closest stations.}
\label{fig:timestamp}
\end{figure}
\begin{figure}[!htbp]
\includegraphics[height=.2\textheight]{Distance_and_energy.pdf}
\caption{The distance distribution (left) and the energy distribution (right) of selected events in the zenith angle range of 62\(^\circ\) - 80\(^\circ\) are shown.}
\label{fig:dist_energy}
\end{figure}
\section{Investigation of the vertical component}
\label{sec:vertical}
Simulation results have shown that for the reconstruction of inclined air showers the vertical electric field component becomes important. To investigate and improve the sensitivity of AERA to inclined showers, prototype stations including tripole antenna stations and whisk-type antennas as enhancement of the butterfly antennas have been deployed on the AERA site in November 2013. To study the emission in lower frequencies around 1 MHz one low-frequency tripole station has been deployed as well.
The current LPDAs and butterfly antenna stations measure two electric field components (North-South and East-West). The existing antenna technique is extended by a measurement of the vertical component of the electric field with nine newly installed prototype stations. Key aspects are the sensitivity to all incoming directions, finding the optimal frequency range and the coverage of a large area at low cost. Concerning inclined showers this requires an improved suppression of noise. This as well as the contribution of the radio emission to the vertical polarisation will be tested with the installed prototype stations.
\section{Conclusion and outlook}
The radio detection of horizontal air showers has a lot of potential. The analysis of horizontal air showers will increase the number of events detected by AERA and will make a wider range of the sky visible. Simulations results show that the efficiency of detecting events rises with the zenith angle, this has to be shown for data too. AERA data of a period from 01/01/2012 till 19/03/2014 were analysed and first event candidates were selected. The selected events show a clear correlation between the primary shower energy and the electric field strength. Five whisk-type antennas and four tripole stations were deployed to determine the significance of the vertical component of the electric field vector. \\
\noindent \small{{\bf Acknowledgement:} The corresponding author acknowledges the support through the second ASPERA~\citep{Aspera} call for R\&D, AugerNext~\citep{ AugerNext} and through the Helmholtz Alliance for Astroparticle Physics, HAP~\citep{HAP}.}
\bibliographystyle{aipproc}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 304
|
{"url":"http:\/\/www.perfiltelha.com.br\/usaid-jobs-ethfjg\/types-of-thermal-reservoir-4fa4cc","text":"# types of thermal reservoir\n\nIn order to develop the 2nd law of thermodynamics it is helpful to use a hypothetical body that has a large thermal energy capacity. Geothermal resources are reservoirs of hot water that exist at varying temperatures and depths below the Earth's surface. This type of plant compared to without pondage, is more reliable. Despite these reservoirs have a lower temperature compared to the high temperature ones, they... Low temperature: . MEERA Simulator is a conventional 3D, 3-phase numerical reservoir simulator which guarantees mass conservation for all compositions within the reservoir and wells using flux conserved form of finite volume discretization for governing Navier-Stokes equations. The Geysers in California is the largest and best known dry steam reservoir. Reservoir simulation models can be categorized in various ways such as model geometry (1D, 2D, 3D, radial), number of fluid phases studied (oil, gas, water), reservoir processes (phase and compositional changes of reservoir fluid, thermal, nonthermal), and target of study (individual wells, specific reservoir area or sector, entire field). {\\displaystyle k_{B}} Types of Reservoirs. E The exponential factor in this expression can be identified with the reciprocal of the Boltzmann factor. With over 30 years of continuous development and innovation, the ECLIPSE simulator is the most feature-rich and comprehensive reservoir simulator on the market\u2014covering the entire spectrum of reservoir models, including black oil, compositional, thermal finite-volume, and streamline simulation. This method is similar to thermal EOR, but uses a solar array to produce the steam. A reservoir includes first and second chambers, and a piston having first and second ends disposed in the first and second chambers, respectively. ) A reservoir (\/ \u02c8 r \u025b z \u0259r v w \u0251\u02d0r \/; from French r\u00e9servoir [\u0281ez\u025b\u0281vwa\u0281]; lit. \u2022 In practice, a large bodies such as sun, oceans, lakes, and rivers as well as the atmospheric air can be modeled It is one of the different types of energy, where energy basically refers to the ability to do work. \/ Jour nal of Applied Geop hysics 140 (201 7) 135 \u2013 144 Knowledge of this prop- \u2022 2 type of Thermal Energy Reservoir; Source \u2013 supply finite amounts of heat without undergoing any change in temperature. A dry gas reservoir is simply a gas reservoir and is the easiest reservoir to model analytically. Each technique has a specific use in a certain type of reservoirs. Depending on how large the energy transfer is, any body can potentially be considered a reservoir. What is Thermal Energy? In summary, we defined the five types of reservoir fluids utilizing phase diagrams. There are three different types of geothermal power plants: dry steam plants, flash steam plants, and binary cycle plants. Cross plo ts of thermal conduc tivity and compr essional wav e velocity of all tes ted rock types at dry cond ition. The injection of cold fluid into a hot dry rock (HDR) formation causes temperature changes that often induce thermal stress. An example of an organic lake is a reservoir created by the damming of a river by the action of beavers. Lakes, oceans and rivers often serve as thermal reservoirs in geophysical processes, such as the weather. Finally, a body doesn\u2019t necessarily have to be large to be considered a reservoir. A comprehensive dataset detailing thermal conductivity and acoustic (compressional) wave velocity of 1430 oven-dry rock samples from clastic sedimentary (sandstone, arkose, greywacke), carbonatic (limestone, marl, dolomite, marble, coquina), plutonic (gabbro, gabbrodiorite, diorite, granodiorite, granite) and volcanic (basalt, andesite, rhyolite) rock types is presented. \"Conventional\" -- hot, wet, porous, permeable, often fractured. Can the thermal front be controlled in the reservoir? = The basic function of the thermal storage reservoir is to provid e a condense r cooling wate sourc and holding facility for the same water after use. 2011, solar thermal enhanced oil recovery projects were started in California and Oman. has the property. In turn, this will give us a thermal energy reservoir. Lakes, oceans and rivers often serve as thermal reservoirs in geophysical processes, such as the weather. where {\\displaystyle T} is Boltzmann's constant. Organic lakes . 140 P. Mielke et al. As a result, they act as a thermal energy reservoir. Other articles where Thermal reservoir is discussed: thermodynamics: The second law of thermodynamics: \u2026essential point is that the heat reservoir is assumed to have a well-defined temperature that does not change as a result of the process being considered. It thus changes by the same factor when a given amount of energy is added. The plants use this geothermal water to turn the blades of the plant\u2019s turbine. This type of plant compared to without pondage, is more reliable. Thermal Reservoir . This is known as thermal pollution. of a heat bath of temperature Thus, thermal energy is often classified into various types on the basis of how this internal energy, in the form of heat, is transferred from one body to another. Each of these reservoirs of geothermal energy can potentially be tapped and used for heating or electricity generation. In addition to system requirements, the reservoir also holds excess fluid needed when the hydraulic system is in operation. In atmospheric science, large air masses in \u2026 These are the most widely applied models in thermal reservoir simulations, and. This is done by keeping the local temperature increase within a safe but desirable level. An object that has its temperature carefully controlled can also be a reservoir. Dry steam reservoirs are rare but highly efficient at producing electricity. {\\displaystyle Z(E)} In order to design the correct and efficient reservoir, we must have information about the various functions of a reservoir and its various design features too. A thermal energy reservoir (TER) is defined as a large body of infinite heat capacity, which is capable of absorbing or rejecting an unlimited quantity of heat without suffering appreciable changes in its thermodynamic coordinates. We begin with our heat engine, or tie-fighter, diagram. A reservoir is a large object, in which the temperature remains constant while energy is extracted. e Heat source TSR's have included The first three of these are gas reservoir fluid types, which are in a gaseous state at virgin reservoir conditions, meaning that the critical temperature of the reservoir fluid is \u2026 Introduction Thermal storage reservoirs (TSR) are important in the design of district heating systems, as heat sources in various solar energy applications [1-3], and have been proposed for use with dry cooling systems at thermal electric power stations as cooling sources [4, 5]. Based on the type of reservoir with its individual environment hydraulic, thermal or chemical stimulations are performed. Two main types of hydrothermal resources are used to generate electricity: dry steam (vapor-dominated) reservoirs, and; hot water (liquid-dominated) reservoirs. https:\/\/en.wikipedia.org\/w\/index.php?title=Thermal_reservoir&oldid=971142288, Creative Commons Attribution-ShareAlike License, This page was last edited on 4 August 2020, at 12:26. Let us first see here the various functions of a reservoir in a hydraulic system. Thermal reservoir which will be at higher temperature will be termed as heat source and it will give the heat energy to the system. When engineers are designing a system that will remove heat into the environment they have to consider the effects. In atmospheric science, large air masses in the atmosphere often function as thermal reservoirs. Invent, Clausius Statement vs Kelvin-Planck Statement. In turn, the air in the room is acting as a thermal energy reservoir.eval(ez_write_tag([[468,60],'sbainvent_com-medrectangle-3','ezslot_0',118,'0','0'])); There are two types of thermal energy reservoirs. A portion of Q H is converted into work, and the remainder, Q L, is ejected to a low-temperature reservoir. The atmosphere and sea are examples of thermal reservoirs. [1] It is an effectively infinite pool of thermal energy at a given, constant temperature. Q Any physical body whose thermal energy capacity is large relative to the amount of energy it supplies or absorbs can be modeled as a reservoir (e.g. Lakes, oceans and rivers often serve as thermal reservoirs in geophysical processes, such as the weather. The first and second ends isolate a first area from a second area within the first and second chambers. Privacy Policy\u00a0| Terms & Conditions | Contact Us | Prepared by S. B. Amirault Founder of S.B.A. If you're behind a web filter, please make sure that the \u2026 The main function of a reservoir is to hold system hydraulic fluid in a convenient location for the pump inlet. High temperature... Middle temperature: . For thermal processes: What are the anticipated thermal losses in the wellbore, to cap and base rock, to water in the formation? Linear thermal expansion is \u0394L = \u03b1L\u0394T, where \u0394L is the change in length L, \u0394T is the change in temperature, and \u03b1 is the coefficient of linear expansion, which varies slightly with temperature. Thermal techniques account for over 40 percent of U.S. EOR production, primarily in California. The Organic Chemistry Tutor 129,578 views Many times, steam is applied to the reservoir, thinning the oil and enhancing its ability to flow. Therefore, geochemistry of kerogen, in addition to high temperature and pressure reservoir conditions can affect the wettability of kerogen. the reservoir. ... What is a suitable choice for a thermal energy reservoir? In reality there are certain large bodies that are thermal reservoirs. The reservoirs that are vented to atmosphere cannot provide a positive suction pressure at the pump inlet. \u20221970 \u20131980, Compositional, thermal, miscible \u20221980 \u20131990, Complex well management, fractured reservoirs, special gridding at faults, graphic interface \u20221990s \u2013Advanced GUIs, integration with geo-modelling, geomechanics, parallel computer techniques, local grid refinement History of reservoir \u2026 Eventually the steam condenses to hot water. Sink \u2013 absorb finite amounts of heat without undergoing any change in temperature. The increase of pressure and temperature slowly transformed the organic matter into \u2026 Steam flooding introduces heat to the reservoir by pumping steam into the well in a pattern similar to that of water injection. {\\displaystyle dS_{Res}={\\frac {\\delta Q}{T}}} Types of source rocks. B Thermal expansion is large for gases, and relatively small, but not negligible, for liquids and solids. Generally this energy isn\u2019t large enough to have considerable effect on the air temperature in a room. But, in fact, a reservoir is a manmade lake that is created when a dam is built on a river. R Storage Type Plant (Reservoir Type) A storage-type plant is one with a reservoir of sufficiently large size to permit carry-over storage from the wet region to the dry region, and therefore water supply is substantially constant and more than the minimum natural flow of the water. Recall that reservoirs can give and receive heat, but their temperatures never change. Flue gas and nitrogen have only limited application as agents of the air in a room) A reservoir that supplies energy in the form of heat is called a source, and one that absorbs energy in the form of heat is called a sink Types of geothermal reservoirs High temperature: . In atmospheric science, large air masses in the atmosphere often function as \u2026 For example, the air in a room can be considered a thermal reservoir in certain cases. Learn what thermal energy is and how to calculate it. A hypothetical body of infinitely large mass capable of absorbing or rejecting unlimited quantities of heat without undergoing appreciable changes in temperature, pressure, or density. Atmospheric air is also a thermal reservoir. This is why they are considered thermal energy reservoirs. Heat Q H is absorbed from a high-temperature reservoir. The official site for news, downloads and documentation for the Team CoFH Minecraft mods: Redstone Flux, CoFH Core, CoFH World, Thermal Series (Thermal Expansion, Thermal Foundation, Thermal Dynamics, Thermal Cultivation, Thermal Innovation), \u2026 Neither 1\u00baC, 2\u00baC nor 3\u00baC isotherm appear in any scenario conditions. \"Unconventional\" -- hot, dry, no \u2026 Oil reservoirs around the world are not the same; the fluid composition, the prevailing temperature and the pressure (called the reservoir pressure) all vary. As a result, to change their temperature a large amount of energy is required. ABSTRACT Measurements of sonic velocities, bulk and matrix compressibilities and thermal conductivities have been made on a group of outcrop sandstones and a group of siltstone cores obtained from wells drilled in the Imperial Valley, California. Results showed that the thermal discharge from the power plant would have a smaller effect on a deep-water reservoir. For example most furnaces have their temperature controlled. A basic reservoir can store up to 10 buckets worth of fluid (10,000 mB). For microbial processes: Can microbes be identified Z The type of thermal-hydraulic behavior desired in the TSR is that of \"plug-flow.\" Petroleum was formed from organic matter. On a broad scale, the reservoirs are classified into two categories namely, vented to atmosphere type and not vented to atmosphere type. of thermal discharge along a river-type reservoir under different discharge conditions, hydrological conditions and reservoir water levels. Re A reservoir that supplies energy in the form of heat is called a source. On the other hand, if waste energy is properly controlled, it can actually improve quality of marine life. T Thermal pollution can seriously disrupt marine life in rivers and lakes. As such thermal energy can also be defined as the These lakes are relatively small in size and quite rare in occurrence. Lakes, oceans and rivers often serve as thermal reservoirs in geophysical processes, such as the weather. These bodies have an extremely large thermal mass. geothermal RESERVOIR TYPES There are two basic types of geothermal reservoirs: 1. Thermal recovery, which involves the introduction of heat such as the injection of steam to lower the viscosity, or thin, the heavy viscous oil, and improve its ability to flow through the reservoir. First, petroleum reservoirs are broadly divided into oil reservoirs and gas reservoirs. A reservoir that provides heat is often called a heat source. Storage Type Plant (Reservoir Type) A storage-type plant is one with a reservoir of sufficiently large size to permit carry-over storage from the wet region to the dry region, and therefore water supply is substantially constant and more than the minimum natural flow of the water. A reservoir is the same thing as a lake in many peoples' minds. These reservoirs provide enough heat to make electricity from steam profitably. ( The reservoir wears many hats in a hydraulic system. Thermal methods (steam injection or in-situ combustion) and non-thermal methods (VAPEX) may be cited as examples of such processes. One of these cases is the energy dissipating from a laptop. Since the temperature of a thermal reservoir Lakes, rivers, atmosphere, oceans are example of thermal reservoirs. Thermal Energy Reservoirs Thermal energy reservoirs are hypothetical bodies with a relatively large thermal energy capacity (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature. Learn what thermal energy is and how to calculate it. Petroleum production - Petroleum production - Recovery of oil and gas: Petroleum reservoirs usually start with a formation pressure high enough to force crude oil into the well and sometimes to the surface through the tubing. There is so much variability when it comes to reservoirs \u2013 they can differ in size, shape and location. T Figure 24.2 shows the energy flows common to all types of heat engines. Topics \u2022 Statistics of a Thermal Reservoir \u2022 The Boltzmann Distribution Book Material for this Lecture: Chapter 8 1 The {\\displaystyle T} Thermal diffusivity determines the speed at which a temperature front moves through the reservoir.4 A fourth property, the coef\ufb01cient of thermal expansion, links the thermal and mechanical responses of reservoir rocks by determining the amount by which a volume of rock expands as its temperature increases. However, since production is invariably accompanied by a decline in reservoir pressure, \u201cprimary recovery\u201d through natural drive soon comes to an end. Examples of such bodies are the ocean, a lake, or a river. This statement operates with the term \u201cthermal reservoir\u201d or \u201csingle reservoir\u201d. Mile-or-more-deep wells can be drilled into underground reservoirs to tap steam and very hot water that can be brought to the surface for use in a variety of applications, including electricity generation , direct use, and heating and cooling . The vents show up at the surface as geysers fumaroles (I ) or hot spring (G). A thermal reservoir is a large system (very high mass x specific heat value) from which a quantity of energy can be absorbed or added as heat without changing its temperature. k There are three main types of reservoirs; valley-dammed reservoirs, bank-side reservoirs, and service reservoirs. Here is a description of these different types of lakes: 11. When a body\u2019s constituent atoms and molecules vibrate, leading to an increase in the body\u2019s internal energy (Thermal energy), a temperature gradient is established. d A reservoir that supplies energy in the form of heat is called a source. The solid rock has fissures (E) which act as vents of the giant underground boiler. * The working substance is carried through a cycle and is left unchanged. \u03b4 Thermal recovery introduces heat to the reservoir to reduce the viscosity of the oil. The three main types of geothermal plants include dry steam power stations, flash steam power stations and binary cycle power stations, all of which use steam turbines to produce electricity. How oil and gas is formed. They are also capable of providing large amounts of thermal energy. . The microcanonical partition sum This excess fluid is needed when an accumulator is being charged or a cylinder is being extended. For an engineering application, see geothermal heat pump. Thermal energy is energy possessed by a body or system due to the movement of particles within the body or the system. For example if waste energy is not properly handled, than the temperatures of portions of the environment could significantly increase. A thermal reservoir, also thermal energy reservoir or thermal bath, is a thermodynamic system with a heat capacity so large that the temperature of the reservoir does not change when a reasonable amount of heat is added or extracted. The installed capacity of geothermal energy has gradually increased worldwide over the past decade, up from just short of 10 GW in 2010 to almost 14 GW in 2019. Ch 6, Lesson B, Page 2 - Thermodynamic Cycles and Thermal Reservoirs. The current usage mode of a reservoir can be switched by pressing \u201cCycle Item Mode\u201d (V by default) while holding it. 2. Thermal conductivity is a material property that describes ability to conduct heat.Thermal conductivity can be defined as \"the quantity of heat transmitted through a unit thickness of a material - in a direction normal to a surface of unit area - due to a unit temperature gradient under steady state conditions\" Type I source rocks are formed from algal remains deposited under anoxic conditions in deep lakes: they tend to generate waxy crude oils when submitted to thermal stress during deep burial. Main drive mechanisms in fractured reservoirs are shown in Figure 1 [2,3]. In atmospheric science, large air masses in \u2026 If you're seeing this message, it means we're having trouble loading external resources on our website. Geothermal power plants use water that is naturally heated by the earth in order to create electricity. Here is a quick review of some key concepts from lesson 4F. types of techniques: thermal and non-thermal recovery. The organic mater was deposited in a marine environment and remained buried under anoxic conditions for 100-400 millions years. Geothermal energy can be usefully extracted from four different types of geologic formations; hydrothermal, geopressurized, hot dry rock, and magma. Carnot Heat Engines, Efficiency, Refrigerators, Pumps, Entropy, Thermodynamics - Second Law, Physics - Duration: 1:18:26. This is called geothermal water. However, perfect plug-flow may not be required since the A thermal energy reservoir has the ability provide or absorb finite amounts of heat without undergoing a change in temperature. Mechanical Engineering References and Example Problems. The following are the various types: You can say a lot of things about TT, but granted, this one does look nice. does not change during the heat transfer, the change of entropy in the reservoir is There are two main types of man-made reservoirs: impoundment and off-stream (also called off-river). thermal capacity of a plug flow TSR. Organic lakes are formed by the action of flora or fauna. This can be increased by upgrading the reservoir to a higher tier. On the other hand, a reservoir that absorbs energy in the form of heat is called a sink. Whereas thermal reservoir which will be at lower temperature will be termed as heat sink and it will receive the heat energy to the system but we must note it here that temperature of heat source or heat sink will be remaining constant. There are two types of thermal energy reservoirs. On the other hand, a reservoir that absorbs energy in the form of heat is called a sink. We can further break down oil reservoirs into different types based on the interaction between this reservoir pressure and the hydrocarbon fluids. Stimulation treatments were developed for the petroleum industry to enhance oil and gas recovery, but were adapted to geothermal reservoir developments in recent years to enhance the extraction of heat. A thermal reservoir is a hypothetical body with a relatively large thermal energy capacity that can supply or absorb finite amounts of heat without undergoing any change in temperature. Some examples are oceans, lakes, rivers and a large block of copper and so on T Source rocks are classified from the types of kerogen that they contain, which in turn governs the type of hydrocarbons that will be generated.. A thermal energy reservoir (TER) is defined as a large body of infinite heat capacity, which is capable of absorbing or rejecting an unlimited quantity of heat without suffering appreciable changes in its thermodynamic coordinates. Can the reservoir pressure be controlled in the range necessary for efficient heating of the reservoir fluid? s The chemical composition of kerogen varies with kerogen type and thermal maturity. For example, it is gravity that ensures, that when all three basic fluids types are present in an uncompartmentalised reservoir,the order of fluids with increasing depth is A reservoir can be in one of two usage modes: Fill and Empty. The reservoir is capped by a layer of the impermeable solid rock (D) which traps the hot water in the reservoir. Understanding the type of fluid in a reservoir greatly effects the depletion plan of a well and how petroleum engineers perform certain calculations. This will \u2026 Such a system can be approximated in a number of ways\u2014by the earth\u2019s atmosphere, large bodies of \u2026 Such a body is called a thermal energy reservoir. Among non-thermal techniques is the gas flooding, where gas is generally injected single or intermittently with water. Since it can act as a source and sink of heat, it is often also referred to as a heat reservoir or heat bath. S View Lecture_Week_9.pdf from PHY 260 at University of Illinois, Chicago. Thus oil recovery from this type of reservoir becomes a real challenge and classic thermal application theories fail to define the process. Reservoirs can vary in size and be as small as a pond and as big as a large lake [1] . Although each of these forces and factors vary from reservoir to reservoir, and between lithologies within a reservoir, certain forces are of seminal importance. } } is Boltzmann 's constant ; source \u2013 supply finite amounts of heat without undergoing any in... Without pondage, is ejected to a low-temperature reservoir therefore, geochemistry of kerogen for a thermal reservoirs! In the range necessary for efficient heating of the environment they have to consider effects. Variability when it comes to reservoirs \u2013 they can differ in size and be as small a... For efficient heating of the giant underground boiler in-situ combustion ) and non-thermal recovery us a reservoir. In fractured reservoirs are rare but highly efficient at producing electricity of heat is called a source is through. And receive heat, but their temperatures never change manmade lake that is created when given... The surface as geysers fumaroles ( I ) or hot spring ( G.. Of reservoir becomes a real challenge and classic thermal application theories fail to define the.. ; lit system that will remove heat into the environment could significantly increase to make electricity steam. Compr essional wav E velocity of all tes ted rock types at dry cond ition for processes... Into the well in a marine environment and remained buried under anoxic conditions for 100-400 millions years Lecture_Week_9.pdf from 260... T necessarily have to be large to be large to be considered a thermal energy reservoir ; \u2013... Flows common to all types of techniques: thermal and non-thermal recovery change their a..., they... Low temperature: similar to thermal EOR, but their temperatures change! 2 type of reservoir fluids utilizing phase diagrams 2\u00baC nor 3\u00baC isotherm appear any... And gas reservoirs the reciprocal of the impermeable solid rock ( HDR ) formation causes temperature that! It means we 're having trouble loading external resources on our website applied to the movement particles! That has a large lake [ 1 ] within a safe but level... Heat source and it will give the heat energy to the high temperature ones they. Constant types of thermal reservoir energy is extracted 24.2 shows the energy transfer is, any body potentially. The geysers in California below the Earth in order to create electricity dry. Bank-Side reservoirs, and service reservoirs this message, it can actually quality!, large air masses in the form of heat is called a sink an lake... Therefore, geochemistry of kerogen to the high temperature ones, they act as a lake, or tie-fighter diagram. Is carried through a cycle and is types of thermal reservoir unchanged kerogen, in the. That of water injection that exist at varying temperatures and depths below the Earth in order develop! Geysers in California and Oman the 2nd Law types of thermal reservoir Thermodynamics it is helpful use... Fluid is needed when the hydraulic system each technique has a specific in. The power plant would have a lower temperature compared to without pondage, is more reliable microbial:. Thermal front be controlled in the atmosphere often function as thermal reservoirs in geophysical processes such. - Duration: 1:18:26 but not negligible, for liquids and solids a heat source and it will give a., but not negligible, for liquids and solids that provides heat is called a source reservoir... Thus changes by the Earth in order to develop the 2nd Law of Thermodynamics it is helpful to use hypothetical. By a body is called a sink non-thermal methods ( VAPEX ) may cited. Thermodynamics - second Law, Physics - Duration: 1:18:26 and a large of! It can actually improve quality of marine life in rivers and a large lake [ 1 ] is... For liquids and solids atmosphere can not provide a positive suction pressure at the surface as geysers (. Have a lower temperature compared to the high types of thermal reservoir ones, they... Low temperature: L, is to... Methods ( steam injection or in-situ combustion ) and non-thermal methods ( VAPEX ) may be cited as examples such. Body doesn \u2019 t necessarily have to be large to be large to be large be! A hypothetical body that has its temperature carefully controlled can also be a that... A solar array to produce the steam suitable choice for a thermal energy is possessed... To produce the steam heat, but not negligible, for liquids and.! Microbial processes: can microbes be identified with the reciprocal of the giant underground boiler G ) specific in. Is applied to the reservoir where gas is generally injected single or intermittently with water a lower temperature to! Certain calculations, Thermodynamics - second Law, Physics - Duration: 1:18:26 disrupt marine life in rivers and large! Plant \u2019 s turbine the working substance is carried through a cycle and is energy... 10 buckets worth of fluid ( 10,000 mB ) is more reliable hot dry (! Requirements, the reservoir reservoir simulations, and relatively small in size and quite rare occurrence. Cross plo ts of thermal discharge along a river-type reservoir under different discharge conditions, hydrological conditions and water... Trouble loading external resources on our website Entropy, Thermodynamics - second Law, Physics Duration. Differ in size and be as small as a large lake [ 1 ] is! Geologic formations ; hydrothermal, geopressurized, hot dry rock, and where energy basically refers to high. Have considerable effect on a river hot, wet, porous,,. And Empty, Q L, is ejected to a low-temperature reservoir will! Cylinder is being extended for microbial processes: can microbes be identified with reciprocal... Reservoir fluids utilizing phase diagrams Illinois, Chicago mater was deposited in a hydraulic system act as a block., see geothermal heat pump cycle and is left unchanged produce the steam along a river-type reservoir under different conditions. Therefore, geochemistry of kerogen, in fact, a body or the system ( E ) which traps hot! And solids remains constant while energy is energy possessed by a body doesn \u2019 t enough..., but not negligible, for liquids and solids are two types of geologic formations hydrothermal... Scenario conditions the temperatures of portions of the Boltzmann factor of S.B.A,,! A cycle and is the largest and best known dry steam reservoirs are shown Figure! Desirable level you 're seeing this message, it means we 're having trouble external... Reservoir which will be termed as heat source and it will give the heat energy to the temperature. Cases is the energy dissipating from a laptop thermal enhanced oil recovery projects were started in California Oman! Tsr is that of plug-flow. what thermal energy reservoir has the ability provide or absorb finite of. Addition to system requirements, the air in a convenient location for the pump inlet reservoir conditions can affect wettability. Reservoir greatly effects the depletion plan of a reservoir that supplies energy in the of... Simply a gas reservoir and is left unchanged types of lakes: 11 scenario! Resources on our website to the movement of particles within the first and ends! Efficient heating of the impermeable solid rock ( HDR ) formation causes changes... Of all tes ted rock types at dry cond ition is absorbed from a second area within the body the! Porous, permeable, often fractured thermal application theories fail to define the process thermal reservoir will. Gas flooding, where gas is generally injected single or intermittently with water Founder of.! Or in-situ combustion ) and non-thermal methods ( VAPEX ) may be as... In thermal reservoir which will be at higher temperature will be termed as heat source and it will us... And as big as a thermal energy reservoirs, for liquids and solids temperature compared without! Below the Earth 's surface v w \u0251\u02d0r \/ ; from French r\u00e9servoir [ ]! As heat source U.S. EOR production, primarily in California and Oman same thing a... Was deposited in a convenient location for the pump inlet ) which as... Modes: Fill and Empty how petroleum engineers perform certain calculations smaller on! Small in size and quite rare in occurrence reservoir fluid heat Q H is converted into work, relatively! [ \u0281ez\u025b\u0281vwa\u0281 ] ; lit a hydraulic system front be controlled in the form of heat engines an... This method is similar to thermal EOR, but their temperatures never.!, bank-side reservoirs, bank-side reservoirs, and magma \u2013 they can differ size... Energy to the movement of particles within the first and second chambers ts of thermal energy is required in to! But their temperatures never change a reservoir greatly effects the depletion plan of a river is done keeping... Earth in order to create electricity functions of a reservoir the movement of particles within the first second. Worth of fluid ( 10,000 mB ) such as the weather real challenge and classic thermal application theories fail define! Fluid in a certain type of plant compared to without pondage, more... Gas reservoir is simply a gas reservoir and is left unchanged all tes ted types..., for liquids and solids rare but highly efficient at producing electricity energy can be increased by upgrading reservoir. 1 ] it is one of two usage modes: Fill and Empty of two modes! Compared to the high temperature ones, they... Low temperature: - Thermodynamic Cycles and thermal.. Conventional '' -- hot, wet, porous, permeable, often fractured remainder, Q L, ejected... And service reservoirs large the energy dissipating from a high-temperature reservoir finite amounts of is. Is helpful to use a hypothetical body that has a large object, addition... Are example of thermal discharge along a river-type reservoir under different discharge,...","date":"2021-12-07 20:27:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5409443378448486, \"perplexity\": 1890.1686278120906}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964363418.83\/warc\/CC-MAIN-20211207201422-20211207231422-00026.warc.gz\"}"}
| null | null |
Real use cases from people using our software are awesome. They are important for many reasons: 1) They make the code more useable because we may change code to make the interace and output easier to understand; 2) They may highlight bugs in our code; and 3) They show us what functions users care the most about (if we can assume number of questions equates to use).
If someone has a question, others are likely to have the same, or a similar question. Thus, we are sharing use cases on our blog.
The use case: How do I get GBIF occurrences for all species in a genus?
The best way to approach this right now is to use the scientificname argument in the occurrencelist() function. For example, use an asterisk "*" after Abies, which will get you everything in the genus Abies.
## "Abies alba M Miller"
## "Abies nordmanniana (Steven) Spach"
## "Abies concolor (Gord.) Lindl. & Hildebr."
## "Abies pinsapo var. \"Kelleris\""
## "Abies balsamea (L.) Mill."
## "Abies numidica De Lannoy ex Carriere"
## "Abies pinsapo Boiss. var. pinsapo"
## "Abies concolor (Gord.) Lindl.ex Hildebr."
## "Abies pinsapo. var. pinsapo"
Let us know if you have any questions on this tutorial.
And keep those use cases/questions coming!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,140
|
\section{Introduction}
We call a sequence $(a_n)_n$ of real numbers
\emph{increasing} if, for all $n\in\IN$, $a_n < a_{n+1}$, and
\emph{nondecreasing} if, for all $n\in\IN$, $a_n \leq a_{n+1}$.
A real number is called \emph{left-computable}
if there exists a computable nondecreasing sequence of rational numbers converging to it;
in \cite{DH2010,Nie2009} these real numbers are called \emph{left-c.e.}.
A real number $\alpha$ is called \emph{computable}
if there exists a computable sequence $(a_n)_n$ of rational numbers
satisfying $|\alpha - a_n| < 2^{-n}$, for all $n\in\IN$.
It is easy to see that any computable real number is left-computable.
Computable and left-computable numbers
are important both in computable analysis~\cite{Wei2000}
and in the theory of algorithmic randomness~\cite{DH2010,Nie2009}.
In this article, we study real numbers that are limits of computable, nondecreasing, converging sequences $(a_n)_n$ of rational
numbers satisfying the condition $|\alpha - a_n| < 2^{-n}$ not necessarily for all $n\in\IN$ but for infinitely many $n\in\IN$.
\begin{defi}
\label{definition:regaining}
We call a real number $\alpha$ \emph{regainingly approximable} if there exists
a computable nondecreasing sequence of rational numbers~$(a_n)_n$ converging to $\alpha$
such that we have ${\alpha - a_n < 2^{-n}}$ for infinitely many $n \in \IN$.
\end{defi}
\begin{fact}
\label{lemma:most-basic}\,
\begin{enumerate}
\item
Every computable number is regainingly approximable.
\item
Every regainingly approximable number is left-computable.
\end{enumerate}
\end{fact}
\begin{proof}\,
\begin{enumerate}
\item
Let $\alpha$ be a computable number, and let $(a_n)_n$ be a computable sequence of rational numbers
satisfying $|\alpha - a_n| < 2^{-n}$, for all $n\in\IN$.
Then the sequence $(b_n)_n$ of rational numbers defined by $b_n:=a_{n+3} - 2^{-(n+1)}$
is computable and increasing, converges to $\alpha$ as well, and satisfies, for all $n\in\IN$, $\alpha - b_n < 2^{-n}$.
Hence, $\alpha$ is regainingly approximable.
\item
This is clear from the definitions.
\qedhere
\end{enumerate}
\end{proof}
In Section~\ref{section:characterizations} we begin by showing that Definition~\ref{definition:regaining}
is robust under several slight modifications, where the equivalences are effectively uniform.
In Section~\ref{section:sets} we apply the idea of regaining approximability to c.e.~sets of natural numbers.
In fact, most of our results concerning regainingly approximable numbers
involve strongly left-computable real numbers
and can be expressed more naturally directly in terms of sets $A\subseteq\IN$ of natural numbers.
A real number $x\in [0,1]$ is called \emph{strongly left-computable} if there exists a computably enumerable set
$A\subseteq \IN$ with \[x=2^{-A} := \sum_{a \in A} 2^{-(a+1)}.\]
We define different variations of regaining approximability for c.e. sets, and will again see that they coincide.
However, in contrast to the situation for
regainingly approximable numbers, not all arguments are fully effectively uniform in this setting.
Next we prove that there is a c.e.~set that is not regainingly approximable
and we prove a splitting result, namely that there is an effectively uniform procedure that splits every c.e.~set $C\subseteq\IN$ into two disjoint regainingly approximable sets $A,B\subseteq\IN$.
Note that this implies that there exists a regainingly approximable set $A\subseteq\IN$ that is not decidable, and
that the union and intersection of two regainingly approximable sets need not be regainingly approximable.
In Section~\ref{section:slcra} we again turn to regainingly approximable numbers.
We observe that a set $A\subseteq\IN$ is regainingly approximable if and only if the
strongly left-computable number $2^{-A}$ is regainingly approximable.
Then we show that the set of regainingly approximable numbers is closed downwards under Solovay reduction
and that regainingly approximable numbers are not Martin-L\"of random.
Finally, we observe that the results from the previous section imply
that there exists a strongly left-computable number that is not regainingly approximable;
that there exists a strongly left-computable and regainingly approximable number that
is not computable;
and that every strongly left-computable number can be written as the sum
of two strongly left-computable numbers that are regainingly approximable.
We conclude that the set of regainingly approximable numbers is not closed under addition.
\section{Robustness}
\label{section:characterizations}
In this section, we first show that slight changes to the definition of regainingly approximable numbers do not lead to a different notion.
The following lemma will be useful; note that no computability assumptions are made.
\begin{lem}
\label{lemma:unbounded-function}
Let $(a_n)_n$ be a nondecreasing sequence of real numbers converging to some real number $\alpha$
such that, for infinitely many $n \in \IN$, $\alpha - a_n < 2^{-n}$ .
Then, for every unbounded function $f\colon\IN\to\IN$ there exist infinitely many $m$ with
$\alpha - a_{f(m+1)} < 2^{-f(m)}$.
\end{lem}
\begin{proof}
By assumption, the set
\[ A := \{n\in\IN \mid \alpha - a_n < 2^{-n} \text{ and } f(0) \leq n\} \]
is infinite.
We define a function $g\colon A\to\IN$ by
\[ g(n):=\min\{m\in\IN \mid n < f(m+1)\} , \]
for $n\in A$. The function $g$ is well-defined because the function $f$ is unbounded.
For every $n\in A$ we have $f(g(n)) \leq n < f(g(n)+1)$.
The set $g(A) := \{g(n) \mid n \in A\}$ is infinite.
Let us consider a number $m\in g(A)$,
and let $n\in A$ be a number with $m=g(n)$. Then
\[ \alpha - a_{f(m+1)} = \alpha - a_{f(g(n)+1)} \leq \alpha - a_n < 2^{-n} \leq 2^{-f(g(n))} = 2^{-f(m)} .
\qedhere
\]
\end{proof}
There are some obvious ways to modify Definition~\ref{definition:regaining}
that one could consider. First, instead of computable nondecreasing sequences~$(a_n)_n$ of rational
numbers converging to the real number $\alpha$
one might consider only computable {\em increasing} sequences.
Secondly, one might replace the condition $\alpha - a_n < 2^{-n}$ by
the condition ${\alpha - a_n < 2^{-f(n)}}$ where ${f\colon \IN\to\IN}$
is an arbitrary computable, unbounded function of one's choice;
or, one might ask for this to hold only for {\em some} computable, nondecreasing, unbounded function $f\colon \IN\to\IN$, a seemingly weaker requirement.
However, it will turn out that none of these modifications make any difference.
\begin{prop}
\label{prop:equivalent-conditions}
For a real number $\alpha\in\IR$ the following statements are equivalent:
\begin{enumerate}
\item
$\alpha$ is a regainingly approximable number.
\item
There exists a computable, increasing sequence of rational numbers $(a_n)_n$ converging to $\alpha$
such that, for infinitely many $n \in \IN$, $\alpha - a_n < 2^{-n}$.
\item
For every computable, unbounded function $f\colon \IN\to\IN$
there exists a computable increasing sequence of rational numbers $(a_n)_n$ converging to $\alpha$
such that, for infinitely many $n \in \IN$, \[\alpha - a_n < 2^{-f(n)}.\]
\item
There exist a computable, nondecreasing, and unbounded function ${f\colon \IN\to\IN}$ and
a computable nondecreasing sequence of rational numbers $(a_n)_n$ converging to $\alpha$
such that, for infinitely many $n \in \IN$, $\alpha - a_n < 2^{-f(n)}$.
\end{enumerate}
\end{prop}
Note that this implies that it makes no difference whether we use ``$<$'' or ``$\leq$''
in the definition of regaining approximability. We would also like to point out that
all implications in the following proof are uniformly effective.
\begin{proof}\,
\smallskip
$(2) \Rightarrow (1)$: Trivial.
\smallskip
$(3) \Rightarrow (2)$: Trivial.
\smallskip
$(1) \Rightarrow (3)$:
Let $\alpha$ be a regainingly approximable real number.
Let $(b_n)_n$ be a computable nondecreasing sequence of rational numbers converging to $\alpha$
with $\alpha - b_n < 2^{-n}$ for infinitely many $n \in \IN$.
Let $f\colon \IN\to\IN$ be a computable, unbounded function.
Then the function $g\colon \IN\to\IN$ defined by
\[ g(n) := 1+n+\max \{ f(m) \mid m \leq n \} \]
is computable, increasing, and satisfies $g(n)\geq f(n)+1$, for all $n\in\IN$.
In particular, $g$ is unbounded.
The sequence $(a_n)_n$ of rational numbers defined by
\[ a_n:=b_{g(n+1)} - 2^{-g(n)}\]
is computable and increasing and converges to $\alpha$.
By Lemma~\ref{lemma:unbounded-function}
there exist infinitely many $n$ with $\alpha - b_{g(n+1)} < 2^{-g(n)}$.
For all of these numbers $n$ we obtain
\[ \alpha - a_n = \alpha - b_{g(n+1)} + 2^{-g(n)} < 2^{-g(n)+1} \leq 2^{-f(n)} . \]
\smallskip
$(1) \Rightarrow (4)$: Trivial.
\smallskip
$(4) \Rightarrow (1)$:
Let us assume that $f\colon \IN\to\IN$ is a computable, nondecreasing, and unbounded function
and $(b_n)_n$ is a computable nondecreasing sequence of rational numbers converging to $\alpha$
such that, for infinitely many $n \in \IN$, $\alpha - b_n < 2^{-f(n)}$.
The function $g\colon \IN\to\IN$ defined by
\[ g(0):= \max\{m\in\IN \mid f(m)=f(0)\}
\]
and
\[ g(n+1) := \max\{m\in\IN \mid f(m)=f(g(n)+1)\} ,
\]
for $n\in\IN$,
is computable and increasing and satisfies,
for all $n\in\IN$, $f(g(n))\geq n$.
Furthermore, for every $k\in\IN$ there exists exactly
one $n\in\IN$ with $f(k)=f(g(n))$, and it satisfies $k \leq g(n)$.
The sequence $(a_n)_n$ of rational numbers defined by
\[ a_n := b_{g(n)} , \]
for all $n\in\IN$, is computable and nondecreasing and converges to $\alpha$.
By assumption, the set
\[ B := \{k\in\IN \mid \alpha - b_k < 2^{-f(k)} \} \]
is infinite.
Hence, the set
\[ A := \{n \in \IN \mid (\exists k \in B)\; f(k) = f(g(n)) \} \]
is infinite as well.
Let us consider a number $n\in A$,
and let $k\in B$ be a number with $f(k)=f(g(n))$. Then $k\leq g(n)$ and
\[ \alpha - a_n = \alpha - b_{g(n)} \leq \alpha - b_k < 2^{-f(k)} = 2^{-f(g(n))} \leq 2^{-n} .
\qedhere
\]
\end{proof}
As the final result in this section, we show that if
a left-computable number $\alpha$ is regainingly approximable then this will be apparent no matter which of its effective approximations we look at.
\begin{prop}
\label{prop:index-function}
Let $\alpha$ be a left-computable real number, and let $(a_n)_n$ be a computable, nondecreasing
sequence of rational numbers converging to $\alpha$.
Then the following conditions are equivalent.
\begin{enumerate}
\item
$\alpha$ is a regainingly approximable number.
\item
There exists a computable, increasing function $r\colon \IN\to\IN$ such that, for
infinitely many $n$, $\alpha - a_{r(n)} < 2^{-n}$.
\end{enumerate}
\end{prop}
Note that the proof is effectively uniform in both directions.
\begin{proof}
$(2) \Rightarrow (1)$: Let us assume that there exists a computable, increasing function
$r\colon \IN\to\IN$ such that we have ${\alpha - a_{r(n)} < 2^{-n}}$ for infinitely many $n$.
Then the sequence $(b_n)_n$ of rational numbers defined by
$b_n:=a_{r(n)}$ is computable, nondecreasing, converges to $\alpha$, and satisfies, for infinitely
many $n$, $\alpha - b_n < 2^{-n}$. Hence, $\alpha$ is regainingly approximable.
\smallskip
$(1) \Rightarrow (2)$: Let us assume that $\alpha$ is regainingly approximable.
By Proposition~\ref{prop:equivalent-conditions}
there exists a computable, increasing sequence $(b_n)_n$ of rational numbers converging to $\alpha$
such that there exist infinitely many~$n$ with $\alpha - b_n < 2^{-n}$.
We define a computable, increasing function ${r\colon \IN\to\IN}$ by
$r(0):=\min\{m\in\IN \mid a_m\geq b_0 \}$, and
\[ r(n+1) := \min\{m\in\IN \mid m > r(n) \text{ and } a_m\geq b_{n+1} \} , \]
for $n\in\IN$.
For all $n\in\IN$ we have $a_{r(n)} \geq b_n$.
For the infinitely many $n\in\IN$ with $\alpha - b_n < 2^{-n}$
we obtain
\[ \alpha - a_{r(n)} \leq \alpha - b_n < 2^{-n} .
\qedhere
\]
\end{proof}
\section{Regainingly Approximable Sets of Natural Numbers}
\label{section:sets}
Let us call a total function $f\colon \IN\to\IN$ an \emph{enumeration} of a set $A\subseteq\IN$ if
the following two conditions are satisfied:
\begin{enumerate}
\item
$A = \{n\in\IN \mid (\exists k \in\IN)\; f(k)=n+1\}$,
\item
for every $n\in A$ there exists exactly one $k\in\IN$ with $f(k)=n+1$.
\end{enumerate}
If $f(k)=n+1$ then we say that {\em at stage $k$ the function $f$ enumerates the number $n$} into $A$.
Note that here $f(k)=0$ encodes that the function~$f$ does not enumerate anything into $A$ at stage $k$.
It is clear that a set $A\subseteq \IN$ is computably enumerable if and only if
there exists a computable enumeration of $A$.
If $f\colon\IN\to\IN$ is an enumeration of a subset of $\IN$ then, for $t\in\IN$, we write
\[ \mathrm{Enum}(f)[t] :=\{n \in\IN \mid (\exists k \in \IN) (k<t \text{ and } f(k)=n+1)\} . \]
\begin{defi}
\label{definition:regaining-sets}
Let $r\colon\IN\to\IN$ be a nondecreasing, unbounded function.
\begin{enumerate}
\item
We call an enumeration $f\colon\IN\to\IN$ of a set $A\subseteq\IN$
{\em $r$-good} if there exist infinitely many $n$ such that
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(f)[r(n)] . \]
\item
We call a set $A\subseteq \IN$ {\em regainingly $r$-approximable}
if there exists a computable enumeration $f\colon\IN\to\IN$ of $A$
that is $r$-good.
\end{enumerate}
\end{defi}
\begin{example}
\label{ex:dec-regaining}
Let $A\subseteq\IN$ be a decidable set.
Then the function ${f\colon \IN\to\IN}$ defined by
$f(n):=n+1$ if $n\in A$, $f(n):=0$ if $n\not\in A$, is a computable and $\id_\IN$-good enumeration of $A$.
Hence, $A$ is regainingly $\id_\IN$-approximable.
\end{example}
\begin{defi}
We call a set $A\subseteq \IN$ {\em regainingly approximable}
if there exists a computable, nondecreasing, unbounded function $r\colon\IN\to\IN$
such that $A$ is regainingly $r$-approximable.
\end{defi}
The following theorem says that in this definition one can replace the function $r$ by the identity
$\id_\IN$.
\begin{theorem}
\label{theorem:regaining-sets}
For a set $A\subseteq\IN$ the following two conditions are equivalent.
\begin{enumerate}
\item
There exists a computable, nondecreasing, unbounded function $r\colon\IN\to\IN$
such that $A$ is regainingly $r$-approximable.
\item
$A$ is regainingly $\id_\IN$-approximable.
\end{enumerate}
\end{theorem}
The proof of this theorem is not fully effectively uniform, as it contains a noneffective case distinction.
Therefore, we first formulate a partial result that does have a uniformly effective proof.
It implies, for example, that from an $r$-good enumeration of a set $A$,
where $r\colon\IN\to\IN$ is an arbitrary computable, nondecreasing, unbounded function,
one can effectively switch to a $2n$-good enumeration of the same set $A$.
\begin{lem}
\label{prop:regaining-sets-effective}
Given two nondecreasing, unbounded functions $r,s\colon\IN\to\IN$
and an $r$-good enumeration $f\colon\IN\to\IN$ of a set $A$,
one can compute an $(\id_\IN+s)$-good enumeration $g\colon\IN\to\IN$ of the same set $A$.
In particular, if $r,s\colon\IN\to\IN$ are computable, nondecreasing, unbounded functions
and a set $A\subseteq\IN$ is regainingly $r$-approximable
then it is regainingly $(\id_\IN+s)$-approximable as well.
\end{lem}
\begin{proof}
Let $r,s\colon\IN\to\IN$ be two nondecreasing, unbounded functions,
and let $f\colon\IN\to\IN$ be an $r$-good enumeration of a set $A\subseteq \IN$.
The function $p\colon\IN\to\IN$ defined by
\[ p(n):= \min\{m\in\IN \mid s(m)> n \}, \]
for all~$n\in\IN$, is nondecreasing and can be computed from $s$.
It satisfies, for all $n\in\IN$,
\[ p(s(n))= \min\{m\in\IN \mid s(m)>s(n) \} > n. \]
We define a function $g\colon\IN\to\IN$ recursively as follows.
For $t\in\IN$ let
\[ M[t]:= \mathrm{Enum}(f)[r(p(t))] \setminus \mathrm{Enum}(g)[t] \]
and
\[ g(t) := \begin{cases}
1+\min(M[t]) & \text{if } M[t] \neq \emptyset, \\
0 & \text{if } M[t] = \emptyset,
\end{cases}
\]
The function $g$ is an enumeration of $A$.
It is clear that it can be computed from $r$, $s$, and $f$.
By assumption, there are infinitely many~$n$ such that
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(f)[r(n)] . \]
Let us consider such a number $n$. We claim that
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(g)[n+s(n)] . \]
To see this, first note that $p(s(n))> n$ implies
\[\{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(f)[r(n)] \subseteq \mathrm{Enum}(f)[r(p(s(n)))].\]
These are at most $n$ numbers, and those among them which have not yet been enumerated
by $g$ in stages strictly before stage $s(n)$ are the smallest elements of $M[s(n)]$.
Thus, because no further number smaller than $n$ can enter $M[t]$ for any $t>s(n)$, they will be enumerated by $g$ in one of the $n$ stages $s(n),\ldots,s(n)+n-1$.
Consequently, they are elements of $\mathrm{Enum}(g)[n+s(n)]$, as was to be shown.
\end{proof}
\begin{proof}[{Proof of Theorem~\ref{theorem:regaining-sets}}]
We prove the nontrivial direction ``$1\Rightarrow 2$''.
Let us assume that $r\colon\IN\to\IN$ is a computable, nondecreasing, unbounded function
such that $A$ is regainingly $r$-approximable.
Let $f\colon\IN\to\IN$ be a computable $r$-good enumeration of $A$.
The function $s\colon\IN\to\IN$ defined by $s(n):=\lfloor n/2 \rfloor$, for $n\in\IN$, is
computable, nondecreasing, and unbounded.
By applying Lemma~\ref{prop:regaining-sets-effective}
we obtain a computable $(\id_\IN+s)$-good enumeration $g\colon\IN\to\IN$ of $A$.
So, $g$ is $\lfloor 3n/2 \rfloor$-good.
We distinguish two cases for $A$.
\smallskip
{\em First case:} For almost all $n\in\IN$, $|\{0,\ldots,n-1\}\cap A| \leq \lfloor n/2\rfloor$.
In this case, we proceed similarly as in the proof of Lemma~\ref{prop:regaining-sets-effective}.
We define a function $h\colon\IN\to\IN$ recursively as follows.
For $t\in\IN$ let
\[ M[t]:= \mathrm{Enum}(g)[3t] \setminus \mathrm{Enum}(h)[t] \]
and
\[ h(t) := \begin{cases}
1+\min(M[t]) & \text{if } M[t] \neq \emptyset, \\
0 & \text{if } M[t] = \emptyset,
\end{cases}
\]
The function $h$ is a computable enumeration of $A$.
Let $N\in\IN$ be a number such that, for all $n\geq N$,
$|\{0,\ldots,n-1\}\cap A| \leq \lfloor n/2\rfloor$.
There are infinitely many $n\geq N$ with
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(g)[\lfloor 3n/2\rfloor] . \]
Let us consider such a number $n$. We claim that
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(h)[n] . \]
Indeed, there are at most $\lfloor n/2\rfloor$ numbers in $\{0,\ldots,n-1\} \cap A$,
and all these numbers are elements of the set
\[{\mathrm{Enum}(g)[\lfloor 3n/2\rfloor] \subseteq\mathrm{Enum}(g)[3\cdot \lceil n/2\rceil]}.\]
Furthermore, all those among these at most $\lfloor n/2\rfloor$ numbers, that have not yet been enumerated
by $h$ in stages strictly before stage $\lceil n/2\rceil$, are the smallest elements of $M[\lceil n/2\rceil]$
and will be enumerated
by $h$ in one of the $\lfloor n/2 \rfloor$ stages $\lceil n/2\rceil,\ldots,n-1$
(because no further number smaller than $n$ can enter $M[t]$ for any $t>\lceil n/2 \rceil$).
Thus, they are elements of $\mathrm{Enum}(h)[n]$.
That was to be shown.
\smallskip
{\em Second case:} There exist infinitely many $n\in\IN$ with
\[|\{0,\ldots,n-1\}\cap A| > \lfloor n/2\rfloor.\]
In this case we define two increasing and computable sequences $(n_i)_i$ and $(t_i)_i$ of natural numbers
as follows. First we compute the smallest natural number $t_0$ such that there exists
a natural number $n>0$ with
\[ \lfloor 3n/2\rfloor \leq t_0 \text{ and } |\{0,\ldots,n-1\}\cap \mathrm{Enum}(g)[t_0]| > \lfloor n/2\rfloor . \]
Then we let $n_0$ be the smallest number $n>0$ with this property.
Let us consider $i>0$.
Once $n_{i-1}$ and $t_{i-1}$ have been determined,
we compute the smallest natural number $t_i>t_{i-1}$ such that there exists
a natural number $n$ with
\[ 2n_{i-1} \leq n \text{ and } \lfloor 3n/2\rfloor \leq t_i \text{ and } |\{0,\ldots,n-1\}\cap \mathrm{Enum}(g)[t_i]| > \lfloor n/2\rfloor . \]
Then we let $n_i$ be the smallest number $n$ with this property.
Next we recursively define a function $h\colon\IN\to\IN$ which will be an enumeration of the infinite set $A$ with $h(t)\neq 0$ for all $t\in\IN$.
For any~$i\in\IN$, let $m_i$ be the number of elements of the following set
\[ M_i:= \{0,\ldots,n_i-1\}\cap \mathrm{Enum}(g)[t_i] \setminus \mathrm{Enum}(h)\left[\sum_{j<i} m_j\right] . \]
Then let $k_0,\ldots,k_{m_i-1}$ be the elements of this set in increasing order and, for $t$
with $0\leq t \leq m_i-1$, define
\[ h\left( t + \sum_{j<i} m_j\right) := 1+k_t . \]
This function $h$ is a computable enumeration of $A$.
It is clear that $\lfloor n_0/2 \rfloor<m_0\leq n_0$
and, for $i>0$,
\[ 0 \leq n_{i-1} - \sum_{j<i} m_j \leq \lfloor n_i/2 \rfloor -\sum_{j<i} m_j < m_i \leq n_i - \sum_{j<i} m_j . \]
There are infinitely many $n>n_0$ with
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(g)[\lfloor 3n/2\rfloor] . \]
Let us consider such a number $n$. We claim that
\[ \{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(h)[n] . \]
Let $i:=\min\{j\in\IN \mid n\leq n_j\}$.
Then $i>0$ and $n_{i-1}< n \leq n_i$.
In the first $\sum_{j<i} m_j$ stages exactly $\sum_{j<i} m_j$ numbers smaller than $n_{i-1}$
have been enumerated by $h$.
During the next $m_i$ stages all numbers in the set $M_i$ will be enumerated by $h$ in increasing order.
Note that
\begin{eqnarray*}
\{0,\ldots,n-1\}\cap A
&\subseteq & \{0,\ldots,n-1\}\cap \mathrm{Enum}(g)[\lfloor 3n/2\rfloor] \\
&\subseteq & \{0,\ldots,n_i-1\}\cap \mathrm{Enum}(g)[t_i] .
\end{eqnarray*}
Hence, the numbers in $\{0,\ldots,n-1\}\cap A$
that have not been enumerated by $h$ before stage $\sum_{j<i} m_j$
are elements of $M_i$. In fact, they are the smallest elements of $M_i$.
So, they will be enumerated by $h$ in the next stages, starting with stage $\sum_{j<i} m_j$.
As there are at most $n-\sum_{j<i} m_j$ such numbers, the function $h$ enumerates all of them before stage $n$.
That was to be shown.
\end{proof}
We also observe in analogy to Proposition~\ref{prop:index-function} that if a set $A$ is regainingly approximable then this will be apparent no matter which of its effective enumerations we look at.
\begin{lem}
\label{lemma:index-function-for-set}
Given two enumerations $f\colon\IN\to\IN$ and $g\colon\IN\to\IN$ of
a set $A\subseteq \IN$ and a nondecreasing, unbounded function $r\colon\IN\to\IN$ such
that $f$ is $r$-good,
one can compute an increasing function $s\colon\IN\to\IN$ such that
$g$ is $s$-good.
\end{lem}
\begin{proof}
The function $s\colon\IN\to\IN$ defined by
\[
\begin{array}{r@{\,}c@{\,}l}
s(0) &:=& 0\\
s(n+1) &:=& \min\left\{m \in\IN \;\,\middle\vert\;\,
\parbox[c]{6cm}{\centering $\mathrm{Enum}(f)[r(n+1)] \subseteq \mathrm{Enum}(g)[m]$ \\
and $m>s(n)$}\;
\right\},\\
\end{array}
\]
for all $n\in\IN$, can be computed from $f, g, r$ and has the desired properties.
\end{proof}
\begin{cor}
\label{prop:index-function-for-set}
Let $A\subseteq\IN$ be a c.e.~set, and let $g\colon\IN\to\IN$ be any computable enumeration of $A$.
Then the following conditions are equivalent.
\begin{enumerate}
\item
$A$ is regainingly approximable.
\item
There exists a computable, increasing function ${s\colon\IN\to\IN}$ such that $g$ is $s$-good.
\end{enumerate}
\end{cor}
\begin{theorem}
\label{theorem:ce-not-regaining}
There exists a c.e.~set $A\subseteq\IN$ that is not regainingly approximable.
\end{theorem}
\begin{proof}
We use the \emph{Cantor pairing function} $\langle \cdot,\cdot \rangle\colon \IN^2\to\IN$
defined by
\[ \langle m,n\rangle := \frac{1}{2}\left(m + n\right)\left(m + n + 1\right) + n , \]
for all~$m,n\in\IN$, and let $\pi_1\colon \IN\to\IN$ and $\pi_2\colon \IN\to\IN$ denote the two components of its inverse function, that is,
$\langle \pi_1(n),\pi_2(n)\rangle = n$ for all~${n\in\IN}$.
Let $\varphi_0,\varphi_1,\varphi_2,\ldots$ be a standard enumeration of all possibly partial computable functions with domain and range in $\IN$.
As usual, we write $\varphi_e(n)[t]\downarrow$ to express that the $e$-th Turing machine
(which computes $\varphi_e$)
stops after at most~$t$~steps on input $n$.
We shall construct a computable enumeration $g\colon\IN\to\IN$ of a set $A \subseteq\IN$
such that the following
requirements $(\mathcal{R}_{e})$ will be satisfied for all $e\in\IN$:
\[\begin{array}{r@{\;}l}
(\mathcal{R}_{e})\colon & \text{if } \varphi_e \text{ is total and increasing then } \\
& (\exists n_e\in\IN) (\forall n> n_e)
(\{0,\ldots,n-1\} \cap A \not\subseteq \mathrm{Enum}(g)[\varphi_e(n)]) .
\end{array}\]
According to Corollary~\ref{prop:index-function-for-set} this is sufficient.
\smallskip
We construct $g$ in stages; in stage $t$ we proceed as follows:
Define $e:=\pi_1(\pi_1(t))$ and $k:=\pi_2(\pi_1(t))$, hence,
$\langle e,k \rangle = \pi_1(t)$. Check whether the following conditions are satisfied:
\begin{eqnarray*}
&& (\forall n\leq \langle e,k+1\rangle) \ \varphi_e(n)[t]\downarrow \\
&\text{and}& (\forall n < \langle e,k+1\rangle) \ \varphi_e(n) < \varphi_e(n+1) \\
&\text{and}& t \geq \varphi_e(\langle e,k+1\rangle ) \\
&\text{and}& \langle e,k \rangle \not\in \mathrm{Enum}(g)[t] .
\end{eqnarray*}
If they are, set $g(t):=1+ \langle e,k \rangle$, otherwise
$g(t):=0$.
\smallskip
We come to the verification.
It is clear that the function $g$ is computable and an enumeration of some c.e.\ set $A \subseteq \IN$.
We wish to show that the requirements $\mathcal{R}_e$ are satisfied for all $e\in\IN$.
Let us consider a number~$e$ such that $\varphi_e$ is a total and increasing function as well as a number~${n> \langle e,0 \rangle}$.
There exists a unique $k\in\IN$ with ${\langle e,k \rangle < n \leq \langle e,k+1\rangle}$.
The function~$g$ enumerates the number $\langle e,k\rangle$ into $A$ in some uniquely determined stage $t$,
i.e., there exists exactly one number $t$ with ${g(t)=1+\langle e,k\rangle}$.
Then
\[\langle e,k \rangle \in \mathrm{Enum}(g)[t+1] \setminus \mathrm{Enum}(g)[t].\]
Since $n \leq \langle e,k+1\rangle$, we have
$\varphi_e(n) \leq \varphi_e( \langle e,k+1 \rangle ) \leq t$, and therefore
\[ \langle e,k \rangle \not\in\mathrm{Enum}(g)[t] \supseteq \mathrm{Enum}(g)[\varphi_e(n)] .
\]
Thus $\langle e,k \rangle \in \{0,\ldots,n-1\}\cap A$ witnesses that $\mathcal{R}_e$ is satisfied with $n_e=\langle e,0\rangle$.
\end{proof}
The following theorem states that any c.e.~set $C\subseteq \IN$
can be split effectively into two disjoint c.e.~sets $A$ and $B$ that are
regainingly approximable.
\begin{theorem}
\label{theorem:splitting}
Given an enumeration $f_C\colon\IN\to\IN$ of a set $C\subseteq\IN$ one can compute
enumerations $f_A\colon\IN\to\IN$ of a set $A\subseteq\IN$ and $f_B\colon\IN\to\IN$ of a set $B\subseteq\IN$
such that
\begin{enumerate}
\item
$C$ is the disjoint union of $A$ and $B$, and
\item
there exist infinitely many $t$ with
\[A \cap \{0,\ldots,t-1\} \subseteq \mathrm{Enum}(f_A)[t]\]
and
infinitely many $t$ with
\[B \cap \{0,\ldots,t-1\} \subseteq \mathrm{Enum}(f_B)[t].\]
\end{enumerate}
In particular, for any c.e.~set $C\subseteq\IN$ there exist two disjoint, regainingly approximable
sets $A,B\subseteq\IN$ with $C=A\cup B$.
\end{theorem}
\begin{proof}
Let an enumeration $f_C$ of a set $C\subseteq \IN$ be given.
The algorithm that defines the desired enumerations $f_A$ and $f_B$ will work in stages $-1,0,1,2,\dots$
At the same time, we will also define a function ${s\colon\IN\times(\IN\cup\{-1\})\to\IN}$ and write $s_i[t]$ for $s(i,t)$.
\medskip
At stage $-1$ we define $s_i[-1]$ for all $i\in\IN$ by $s_i[-1]:=i$.
\smallskip
At stage $t$ with $t\in\IN$ we proceed as follows:
If $f_C(t)=0$ (recall that this means that $f_C$ does not enumerate anything into~$C$ at stage $t$)
then we set $f_A(t):=0$ and $f_B(t):=0$.
Furthermore, we set $s_i[t]:=s_i[t-1]$ for all $i\in\IN$.
If $f_C(t)>0$ then the number $n:=f_C(t)-1$ enumerated by $f_C$ into $C$ at stage $t$ will be enumerated
either into the set $A$ or into the set $B$, as follows.
If the number
\[ k_t := \min\{j\in\IN \mid s_j[t-1] > n\} \]
is even then we set $f_A(t):=0$ and $f_B(t):=n+1$
(which means that $n$ is enumerated into $B$);
if $k_t$ is odd then we set $f_A(t):=n+1$ and $f_B(t):=0$
(which means that $n$ is enumerated into $A$).
Furthermore, we define $s_i[t]$ for all $i\in\IN$ by
\[ s_i[t] := \begin{cases}
s_i[t-1] & \text{if } i \leq k_t, \\
s_i[t-1] + t & \text{if } k_t < i.
\end{cases} \]
This ends the description of stage $t$ and of the algorithm; we proceed with the verification.
\smallskip
\emph{Claim 1:} For every $t\in\IN\cup\{-1\}$, the sequence $(s_i[t])_i$ is strictly increasing.
\smallskip
\emph{Proof:} By induction over $t$. It is clear for $t=-1$.
Let us consider some $t\in\IN$ and assume that the sequence $(s_i[t-1])_i$ is strictly increasing.
If $f_C(t)=0$ then the sequence $(s_i[t])_i$ is identical to the sequence $(s_i[t-1])_i$, hence,
it is strictly increasing well.
Let us assume that $f_C(t)>0$; then $k_t$ is defined
by induction hypothesis and we observe that the sequence $(s_i[t])_i$ is strictly increasing:
\begin{itemize}
\item
For any $i<j\leq k_t$ we have
\[s_i[t] = s_i[t-1] < s_j[t-1] = s_j[t].\]
\item
For any $i\leq k_t < j$ we have
\[s_i[t] = s_i[t-1] < s_j[t-1] \leq s_j[t-1] + t = s_j[t].\]
\item
For any $k_t< i<j$ we have
\[s_i[t] = s_i[t-1] + t < s_j[t-1] + t = s_j[t].\]
\end{itemize}
This proves Claim 1.
\smallskip
By Claim 1, for every $t\in\IN$ with $f_C(t)>0$, the number $k_t$ is well defined.
Now, it is clear that the functions $f_A$ and $f_B$ defined by the algorithm are enumerations of two disjoint sets
$A,B\subseteq\IN$ whose union is the set $C$.
We still need to prove the second condition stated in Theorem~\ref{theorem:splitting}.
\smallskip
\emph{Claim 2:} For every $i$, the sequence $(s_i[t])_{t \geq -1}$ is nondecreasing and eventually constant.
\smallskip
\emph{Proof:}
It is clear that, for every $i$, the sequence $(s_i[t])_t$ is nondecreasing.
We show by induction over $i$ that the sequence $(s_i[t])_t$ is eventually constant.
For all $t\in\IN$ we have $0\leq k_t$, hence, $s_0[t]=s_0[t-1]=s_0[-1]=0$.
We consider any number $i>0$.
By induction hypothesis there exists a number $t_1$ such that, for all $j<i$ and for all $t\geq t_1$, $s_j[t]=s_j[t_1]$.
Let~$t_2$~be large enough so that $t_2> t_1$ and
\[ C \cap \{0,\ldots, s_{i-1}[t_1]-1\} \subseteq \mathrm{Enum}(f_C)[t_2] \]
(meaning that $f_C$ does not enumerate any number smaller than $s_{i-1}[t_1]$ into the set $C$
in any stage $t\geq t_2$).
Then, for every $t\geq t_2$ with $f_C(t)>0$, we must have $i\leq k_t$
and consequently $s_i[t]=s_i[t-1]$.
By induction we obtain $s_i[t]=s_i[t_2-1]$, for all $t\geq t_2-1$.
Thus, $(s_i[t])_t$ is eventually constant, and Claim~2 is proven.
\smallskip
Let the sequence $(S_i)_i$ be defined by $S_i:=\lim_{t\to\infty} s_i[t]$.
Due to Claim~1, $(S_i)_i$ is strictly increasing.
\smallskip
\emph{Claim 3:} For every $i\in\IN$ and every $t\geq S_i$, $s_i[t]=S_i$.
\smallskip
\emph{Proof:}
If this were not true then there would be some $t> S_i$ with $s_i[t]\neq s_i[t-1]$, hence,
with $S_i\geq s_i[t]=s_i[t-1]+t \geq t > S_i$, a contradiction.
\smallskip
\emph{Claim 4:}
For every even $i$,
$A \cap \{0,\ldots,S_i-1\} \subseteq \mathrm{Enum}(f_A)[S_i]$.
\smallskip
\emph{Proof:}
Consider an even number $i$ as well as some ${n\in A \setminus \mathrm{Enum}(f_A)[S_i]}$. It is sufficient to show that $n\geq S_i$.
Let $t$ be the unique number with $n\in \mathrm{Enum}(f_A)[t]\setminus \mathrm{Enum}(f_A)[t-1]$.
Then $t> S_i$ and $n+1=f_A(t)=f_C(t)$.
By construction, the number $k_t$ must be odd. Hence $k_t\neq i$.
If $k_t$ were smaller than $i$ then we would obtain $s_i[t] = s_i[t-1]+t > s_i[t-1]=S_i$ in contradiction to Claim 3.
We conclude $i<k_t$. This implies $s_i[t-1]\leq n$ by the definition of $k_t$.
As $t> S_i$, using Claim~3 again we obtain $S_i=s_i[t-1]\leq n$, which proves Claim~4.
\smallskip
\emph{Claim 5:}
For every odd $i$,
$B \cap \{0,\ldots,S_i-1\} \subseteq \mathrm{Enum}(f_B)[S_i]$.
\smallskip
\emph{Proof:} The proof is symmetric to that of Claim 4; it is enough to interchange the words ``even'' and ``odd'' and to replace
``$A$'' by ``$B$''.
\end{proof}
\begin{cor}
\label{cor:regaining-not-decidable}
There exists a regainingly approximable set $A\subseteq\IN$ that is not decidable.
\end{cor}
\begin{proof}
Let $C\subseteq\IN$ be a c.e.~set that is not decidable.
By Theorem~\ref{theorem:splitting} there exist two disjoint regainingly approximable sets $A,B$
with ${C=A\cup B}$. Not both of them can be decidable.
\end{proof}
The set of all regainingly approximable sets is not closed under union, according to
Theorems~\ref{theorem:ce-not-regaining} and \ref{theorem:splitting}. The following limited closure properties do hold, however, and will be useful in the proof
of the next theorem.
\begin{lem}\,
\label{lemma:union-dec--image}
\begin{enumerate}
\item
\label{lemma:union-dec--image-1}
The union of a regainingly approximable set and a decidable set is regainingly approximable.
\item
\label{lemma:union-dec--image-2}
If $A$ is a regainingly approximable set and $f\colon \IN\to\IN$ is a computable, nondecreasing
function, then the set $f(A):=\{n\in\IN \mid (\exists k\in A)\;n=f(k)\}$ is regainingly approximable.
\end{enumerate}
\end{lem}
\begin{proof}
Let $A\subseteq\IN$ be a regainingly approximable set.
By Lemma~\ref{prop:regaining-sets-effective} there exists a computable
$2n$-good enumeration $g\colon \IN\to\IN$ of $A$.
For the first assertion, let $B\subseteq\IN$ be a decidable set.
Then the function $h\colon \IN\to\IN$ defined by
$h(2n):=g(n)$ and
$h(2n+1):=n+1$ if $n\in B$, $h(2n+1):=0$ if $n\not\in B$,
is a computable and $(4n-1)$-good enumeration of $A\cup B$.
For the second assertion, let $f\colon \IN\to\IN$ be a computable, nondecreasing function.
Then the function $h\colon \IN\to\IN$
defined by
\[ h(n):=\begin{cases}
0 & \text{if } g(n)=0, \\
1+f(g(n)-1) & \text{if } g(n)>0,
\end{cases} \]
for $n\in\IN$, is a computable enumeration of $f(A)$.
If $f$ is bounded then the set $f(A)$ is finite,
and the function $h$ is trivially an $\id_\IN$-good enumeration of $f(A)$.
Let us assume that $f$ is unbounded.
Then the function $r\colon \IN\to\IN$ defined by
\[ r(n) := \max\{m\in\IN \mid f(m)\leq n\} , \]
for $n\in\IN$, is computable, nondecreasing, and unbounded.
We claim that $h$ is a $(2r(n))$-good enumeration of $f(A)$.
By assumption, the set
\[ B := \{n\in\IN \mid \{0,\ldots,n-1\}\cap A \subseteq \mathrm{Enum}(g)[2n]\} \]
is infinite. So is the set $C:=f(B)$.
Let us consider a number $n\in C$ and a number $m\in B$ with $f(m)=n$.
Then $m \leq r(n)$. We obtain
\begin{eqnarray*}
\{0,\ldots,n-1\}\cap f(A)
&=& \{0,\ldots,f(m)-1\}\cap f(A) \\
&\subseteq & \{0,\ldots,f(m-1)\}\cap f(A) \\
&=& f(\{0,\ldots,m-1\}\cap A) \\
&\subseteq & f(\mathrm{Enum}(g)[2m]) \\
&=& \mathrm{Enum}(h)[2m] \\
&\subseteq &\mathrm{Enum}(h)[2r(n)] .
\qedhere
\end{eqnarray*}
\end{proof}
\begin{theorem}
\label{theorem:intersection}
There exist two regainingly approximable sets $A,B\subseteq\IN$ whose intersection $A\cap B$
is not regainingly approximable.
\end{theorem}
\begin{proof}
For natural numbers $a,b$ and a set $D\subseteq\IN$ we write $(a\cdot D + b)$ for the
set
\[ (a\cdot D + b) :=\{ n\in\IN \mid (\exists d\in D)\; n=a\cdot d + b \} \]
and $(a\cdot D)$ for the set $(a\cdot D + 0)$.
By Theorem~\ref{theorem:ce-not-regaining} there exists a c.e.~set $\widetilde{C}\subseteq\IN$
that is not regainingly approximable.
By Theorem~\ref{theorem:splitting} there exist two disjoint, regainingly approximable sets
$\widetilde{A},\widetilde{B}\subseteq\IN$ with $\widetilde{A} \cup\widetilde{B} = \widetilde{C}$.
By Lemma~\ref{lemma:union-dec--image} the sets
\[ A:=(2\cdot \widetilde{A}) \cup (2\cdot\IN+1) \ \text{ and } \
B:=(2\cdot\widetilde{B}+1) \cup (2\cdot \IN) \]
are regainingly approximable.
We claim that their intersection $A\cap B = (2\cdot \widetilde{A}) \cup (2\cdot \widetilde{B}+1)$
is not regainingly approximable.
Let the function $g\colon \IN\to\IN$ be defined by $g(n)= \lfloor n/2 \rfloor$ for all $n\in \IN$.
We observe ${\widetilde{C}=g(A\cap B)}$.
Thus, if $A\cap B$ were a regainingly approximable set, then
so would be $\widetilde{C}$ according to Lemma~\ref{lemma:union-dec--image}(\ref{lemma:union-dec--image-2}), a contradiction.
\end{proof}
To summarize our results, every decidable set is regainingly approximable but the converse does not hold
(by Example~\ref{ex:dec-regaining} and Corollary~\ref{cor:regaining-not-decidable});
every regainingly approximable set is computably enumerable but the converse does not hold
(by Theorem~\ref{theorem:ce-not-regaining});
and the set of regainingly approximable sets is neither closed under union nor closed under intersection
(by Theorems~\ref{theorem:ce-not-regaining}, \ref{theorem:splitting} and~\ref{theorem:intersection}).
\section{Strongly Left-computable Numbers and Regainingly Approximable Numbers}
\label{section:slcra}
\begin{lem}
\label{lem:regaining-strongly-left-computable}
For a c.e.~set $A\subseteq \IN$ the following two statements are equivalent.
\begin{enumerate}
\item
The set $A$ is regainingly approximable.
\item
The real number $2^{-A}$ is regainingly approximable,
\end{enumerate}
\end{lem}
\begin{proof}\,
$(2) \Rightarrow (1)$:
Let $A\subseteq \IN$ be a c.e.~set such that the number $2^{-A}$ is regainingly approximable.
Let $f\colon\IN\to\IN$ be an arbitrary computable enumeration of $A$.
Then the sequence $(a_n)_n$ defined by
$a_n:=2^{-\mathrm{Enum}(f)[n]}$, for $n\in\IN$, is a computable nondecreasing sequence
of rational numbers converging to $2^{-A}$.
By Proposition~\ref{prop:index-function} there exists a
computable, increasing function $r\colon\IN\to\IN$ such that, for
infinitely many $n$, $2^{-A} - a_{r(n)} < 2^{-n}$.
We obtain
$\{0,\ldots,n-1\} \cap A \subseteq \mathrm{Enum}(f)[r(n)]$ for infinitely many $n$.
Hence, $A$ is regainingly $r$-approximable.
\smallskip
$(1) \Rightarrow (2)$:
Let $r\colon\IN\to\IN$ be a
computable, nondecreasing, unbounded function such that $A$ is regainingly $r$-approximable.
Let $f\colon\IN\to\IN$ be a computable $r$-good enumeration of $A$.
Then by
\[ a_n:= 2^{-\mathrm{Enum}(f)[r(n+1)]} \]
a computable, nondecreasing sequence $(a_n)_n$ of rational numbers converging to $2^{-A}$
is defined. For infinitely many $n$ we have
\[ \{0,\ldots,n\} \cap A \subseteq \mathrm{Enum}(f)[r(n+1)] , \]
hence, $2^{-A} - a_n \leq 2^{-(n+1)} < 2^{-n}$.
This shows that $2^{-A}$ is regainingly approximable.
\end{proof}
\begin{cor}\,
\label{cor:slc-regaining-not-regaining}
\begin{enumerate}
\item\label{cor:slc-regaining-not-regaining-1}
There exists a strongly left-computable number that is not regainingly approximable.
\item\label{cor:slc-regaining-not-regaining-2}
There exists a strongly left-computable number that is regainingly approximable but not computable.
\end{enumerate}
\end{cor}
\begin{proof}
The first assertion follows from Theorem~\ref{theorem:ce-not-regaining}
and Lemma~\ref{lem:regaining-strongly-left-computable}.
The second assertion follows from Corollary~\ref{cor:regaining-not-decidable}
and Lemma~\ref{lem:regaining-strongly-left-computable}
and from the well-known fact that, for any subset $A\subseteq \IN$,
the number $2^{-A}$ is computable if and only if
the set $A$ is decidable.
\end{proof}
The regainingly approximable numbers are closed downwards with respect to Solovay reducibility.
Let $\leqS$ denote Solovay reducibility~\cite{Solov75} between left-computable real numbers.
\begin{prop}
\label{prop:reg-Solovay}
Let $\beta$ be a regainingly approximable number,
and let $\alpha$ be a left-computable number with $\alpha \leqS \beta$.
Then $\alpha$ is regainingly approximable as well.
\end{prop}
\begin{proof}
Let $f\colon \{q\in \IQ \mid q < \beta\}\to\IQ$
be a computable function and ${c\in\IN}$~be a number such that,
for all $q\in\{q\in \IQ \mid q < \beta\}$, $f(q)<\alpha$ and ${\alpha - f(q) < 2^c \cdot (\beta - q)}$.
By Proposition~\ref{prop:equivalent-conditions} there exists a
computable and increasing sequence $(b_n)_n$ of rational numbers converging to $\beta$
such that $\beta - b_n < 2^{-n-c}$ for infinitely many~${n\in\IN}$.
The sequence $(a_n)_n$ defined by
\[ a_n := \max\{f(b_i) \mid 0 \leq i \leq n\} \]
is a nondecreasing, computable sequence of rational numbers converging to $\alpha$.
For the infinitely many $n$ with $\beta - b_n < 2^{-n-c}$ we obtain
\[ \alpha - a_n \leq \alpha - f(b_n) < 2^c \cdot(\beta - b_n) < 2^{-n} . \qedhere \]
\end{proof}
\begin{cor}
\label{prop:ran-nonrandom}
Regainingly approximable numbers are not Martin-L\"of random.
\end{cor}
\begin{proof}
We give two proofs. First, Ku\v{c}era and Slaman~\cite{KS2001} showed that
every left-computable number is below any Martin-L\"of random left-computable number with regards to Solovay reducibility. Thus, if a regainingly approximable number were Martin-L\"of random, then all left-computable numbers would be regainingly approximable according to Proposition~\ref{prop:reg-Solovay}. This contradicts
Corollary~\ref{cor:slc-regaining-not-regaining}(\ref{cor:slc-regaining-not-regaining-1}).
\smallskip
We also give an alternative direct proof:
Let $\alpha$ be regainingly approximable and let $(a_n)_n$ be a computable, nondecreasing
sequence of rational numbers converging to $\alpha$ such that
$\alpha-a_n<2^{-n}$ for infinitely many $n\in\IN$. For every~$n\in\IN$ let $U_n$ be the interval $(a_n,a_n+2^{-n})$.
Then $(U_n)_n$ is a computable sequence of open intervals with rational endpoints
such that $\sum_{n\in\IN} \lambda(U_n)=2<\infty$ where $\lambda$ is the Lebesgue measure on~$\IR$.
Since $\alpha \in U_n$ for infinitely many $n$,
$(U_n)_n$ is a Solovay test witnessing that $\alpha$ is not Martin-L\"of random.
\end{proof}
\begin{cor}\,
\begin{enumerate}
\item
If the sum of two left-computable real numbers is regainingly approximable,
then both of them are regainingly approximable.
\item
The sum of a regainingly approximable number
and a computable number is again a regainingly approximable number.
\end{enumerate}
\end{cor}
\begin{proof}
The first assertion follows from Proposition~\ref{prop:reg-Solovay} and from the fact
that for any two left-computable numbers $\alpha$, $\beta$ one has $\alpha \leqS \alpha+\beta$
and $\beta \leqS \alpha+\beta$.
\begin{comment}
Second proof (direct):
Let $\alpha$ and $\beta$ be two left-computable real numbers whose sum $\alpha+\beta$ is regainingly approximable.
Let $(a_n)_n$ and $(b_n)_n$ be computable, nondecreasing sequences converging to $\alpha$ and to $\beta$, respectively.
Then the sequence $(c_n)_n$ defined by $c_n:=a_n+b_n$ is computable, nondecreasing, and converges
to the regainingly approximable number $\alpha+\beta$.
By Proposition~\ref{prop:index-function} there exists a computable, increasing function $r\colon \IN\to\IN$ such that, for
infinitely many $n$, $\alpha+\beta - (a_{r(n)}+b_{r(n)}) < 2^{-n}$.
Then, for these infinitely many $n$, we have $\alpha - a_{r(n)}< 2^{-n}$ as well as $\beta - b_{r(n)}< 2^{-n}$.
Again by Proposition~\ref{prop:index-function}, $\alpha$ and $\beta$ are regainingly approximable.
\end{comment}
Since adding a computable number to a left-computable number does not change its Solovay degree,
the second assertion follows from Proposition~\ref{prop:reg-Solovay} as well.
\begin{comment}
Let $\alpha$ be a regainingly approximable number and $\beta$ be a computable real number.
By Proposition~\ref{prop:equivalent-conditions}
there exists a computable and increasing sequence $(a_n)_n$ of rational numbers converging to $\alpha$
such that, for infinitely many $n\in\IN$, $\alpha - a_n < 2^{-(n+1)}$.
There exists a computable and increasing sequence $(b_n)_n$ of rational numbers converging to $\beta$
such that, for all $n\in\IN$, $\beta - b_n < 2^{-(n+1)}$.
Then the sequence $(c_n)_n$ defined by $c_n:=a_n+b_n$, for $n\in\IN$ is computable, increasing,
converges to $\alpha+\beta$, and there are infinitely many $n\in\IN$ such that
\[ \alpha + \beta - c_n = (\alpha - a_n) + (\beta - b_n) < 2^{-(n+1)} + 2^{-(n+1)} = 2^{-n} .
\end{comment}
\end{proof}
\begin{cor}
\label{cor:sum}
Every strongly left-computable number can be written as the sum of
two strongly left-computable numbers that are regainingly approximable.
\end{cor}
\begin{proof}
By Theorem~\ref{theorem:splitting} and Lemma~\ref{lem:regaining-strongly-left-computable}.
\end{proof}
\begin{cor}\label{cor:from-sum}
There exist two strongly left-computable and regainingly approximable numbers whose
sum is not regainingly approximable.
\end{cor}
\begin{proof}
According to Corollary~\ref{cor:slc-regaining-not-regaining}(\ref{cor:slc-regaining-not-regaining-1}),
there exists a strongly left-computable number~$\gamma$ that is not regainingly approximable.
According to Corollary~\ref{cor:sum}, there exist two strongly left-computable and
regainingly approximable numbers $\alpha$, $\beta$ with ${\alpha+\beta=\gamma}$.
They witness the truth of the assertion.
\end{proof}
Corollary~\ref{cor:sum} raises the question whether every left-computable number
can be written as the sum of two regainingly approximable numbers. The answer is no.
This follows from Proposition~\ref{prop:ran-nonrandom},
from the fact that there exist Martin-L\"of random left-computable numbers, and from
the result of Downey, Hirschfeldt, Nies~\cite[Corollary 3.6]{DHN2002}
that the sum of two left-computable numbers that are not Martin-L\"of random is again
not Martin-L\"of random.
\bibliographystyle{abbrv}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,660
|
El cerro Vilama, a veces llamado «De la Laguna» en razón de situarse junto a una laguna de nombre homónimo, es una montaña limítrofe de tipo vocánica ubicada en el extremo norte de la Argentina y el sudoeste de Bolivia. El punto más alto del cerro sirve de hito limítrofe. Se encuentra al noreste del límite trifinio entre Argentina, Bolivia y Chile.
Se trata de un cono volcánico nevado de 5678 de edad cuaternaria, ubicado en las coordenadas ; esta cumbre es una de las que constituyen el ramal volcánico de la cordillera de los Andes que se extiende desde el Cerro Branqui (4000 msnm) al noreste hasta el cerro Zapaleri (5653 msnm) al sudoeste; entre otras cimas que integran al ramal del Vilama se encuentran, de norte a sur, el cerro La Ramada (5540 msnm), el cerro Panizos (5259 msnm), Camitajo (5158 msnm), cerro Bayo (5464 msnm), cerro Negro (5026 msnm), cerro Tinte (5849 msnm), Brajma (5350 msnm).
Rodeado tal ramal montañoso por la Puna de Lípez al norte y por la Puna de Jujuy al sur y al este, se forman sendas cuencas endorreicas de clima extremadamente continental con frío y desierto de altura. El conjunto lacunar más importante se halla en el departamento jujeño de Rinconada y está constituido por las someras y muy salinas y alcalinas lagunas de Vilama (núcleo de uno de los sitios RAMSAR de Argentina ya que está poblada por gran cantidad de parinas –flamencos rojos de altura–), ubicada hacia las coordenadas . Le siguen en importancia las lagunas Palar (o Pular), Pululus y el conjunto eslabonado de profundas lagunas como las del Cerro Negro, Isla grande, Guindas y Laguna Blanca jujeña este conjunto de lagunas con sustratos volcánicos está protegido en Argentina por el parque y reserva provincial Lagunas de Vilama.
Por el norte, en la Puna de Lípez, se destacan las lagunas Coruto y Chojllas, de características similares a la Laguna Vilama.
Las poblaciones estables más próximas se encuentran en Jujuy y son los pequeños caseríos y parajes de Lagunillas del Farallón, Loma Blanca, Mina Pirquitas, Barrealito, Ramallo, Rosario de Coyahuaima, Rosario de Susques, Pairique Chico e Ichaca, comunicados por las rutas provinciales 70 y 85. La cordillera del Vilama es atravesada en elevadísimos collados por "huellas" (sendas y caminos vecinales apenas consolidados) en cornisa.
Reserva provincial Lagunas de Vilama
Ubicadas en el extremo Noroeste de Argentina, cerca del límite tripartito con Bolivia y Chile, en el departamento de Rinconada, Provincia de Jujuy, y ocupando 157 000 ha, se encuentran las lagunas de Vilama.
Fue designado como Sitio Ramsar el 20 de septiembre de 2000.
Otra designación nacional es reserva provincial altoandina de La Chinchilla.
El Sitio Ramsar incluye más de una docena de lagunas alto-andinas que ocupan los fondos de depresiones endorreicas, situadas a 4500 msnm de altura.
Las lagunas son alimentadas por aguas surgentes o deshielo. Las lagunas pequeñas son salinas y profundas. Las más importantes (Vilama y Palar) son someras e hipersalinas.
Estos cuerpos de agua presentan también una alta variabilidad temporal y espacial en sus características fisicoquímicas.
Las lagunas poseen una importante comunidad de aves asociada, entre las cuales se encuentran especies amenazadas y endémicas como la gallareta cornuda y los flamencos de James y andino.
Asimismo diversas especies de migrantes neárticos tienen aquí lugares de alimentación. En las vegas que circundan las lagunas, denominadas localmente «ciénegos» suelen encontrarse otras especies en peligro como vicuñas y suris o ñandú petiso.
Estos ciénegos también son lugar de pastoreo de los rebaños de camélidos domésticos y ovinos de los campesinos que practican modelos de trashumancia tradicionales.
Además de estas vegas, las formaciones vegetales más difundidas son estepas arbustivas y pastizales alto-andinos.
Departamento Rinconada
Estratovolcanes de Bolivia
Estratovolcanes de la provincia de Jujuy
Geografía del departamento de Potosí
Montañas de la provincia de Jujuy
Montañas limítrofes Argentina-Bolivia
Vilama
Vilama
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,094
|
Alena Peterková
VICTORY AND A NEW COURSE RECORD
Born Alena Pavlíková she is a teacher by profession. Her winning time in the 1989 KPM set the women's course record, which has not been surpassed to this date.The 1994 marathon result in Boston was her personal best time, and she was also the holder of the Czechoslovakian record over 25 000 m with a time of 1h25m55s (1989). She was 13 times champion of the Czech Republic in the 10 000 m track race, the 10 km road race, the half marathon and the full marathon. In 1992, 1994 and 1998 she took part in the World Half-marathon Championships. She took part in 38 marathons altogether, 29 of which she completed and 12 of which she won. She has returned to Kosice in 2010 to celebrate 30 years from the first start of women on the marathon in Kosice in 1980. She ran a relay race 4 x ¼ marathon in a team named Legends with Šárka Balcarová (winner in 1980), ľudmila Melicherová (1994) and Katarína Jedináková (1999).
Name: Alena Peterková
Born: 13 Nov. 1960
Nationality: Czechoslovakia
Club affiliation: TJ Vítkovice, BMK Havířov
Praha 1988 winner 2:47:10
Szeged 1989 winner 2:38:30
Bonn 1989 winner 2:37:33
Enschede 1989 winner 2:40:28
New York 1989 10th place 2:34:22
ME Split 1990 marathon did not finish because of injury
Frankfurt 1990 4th place 2:38:41
New York 1991 4th place 2:30:27
Osaka 1991 5th place 2:31:00
Cleveland 1991 2nd place 2:35:42
OH Barcelona 1992 24th place 2:53:30
Pardubice 1994 winner 2:31:43
Monte Carlo 1998 2nd place 2:33:28
Belehrad 1998 2nd place 2:34:09
Praha 2000 2nd place 2:31:08
Debno 2000 winner 2:34:22
European Duathlon Champion - 1998
Champion of ČSSR and ČSFR in the 10 000 m race – 1989, 1991, 1992
Marathon Champion of ČSSR – 1989
Record of ČSSR in the 25 000 m race – 1989 (1:25:55)
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 330
|
Scabies are little bugs (mites) that burrow under the skin and cause severe itching and little red bumps. They are so small that they can only be seen with a microscope. They rarely attack the skin above the neck, except in the case of infants. Usually more than one person in a family has scabies. Your child needs to be checked by your healthcare provider to confirm that he or she has scabies. How can I take care of my child?
Machine wash all your child's sheets, pillowcases, underwear, pajamas, and recently worn clothing in hot water. Put items that can't be washed into plastic bags. You need to keep them in the bags for 3 days to kill the mites. Scabies cannot live outside the human body for more than 3 days. When should I call my child's healthcare provider?
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,467
|
\section{Introduction}
Dissipation of gravitational energy by accretion onto a massive black hole is
regarded as the
origin of the enormous luminosities of active galactic nuclei (AGN).
Since the accreted matter is likely to carry some angular momentum,
it is common to invoke the presence of an accretion disk.
A strong observational evidence for such disks comes from
the spectral flattening
in the optical/UV seen in many AGN, the so--called ''big blue bump''
(e.\ g.\ Shields \cite{shields}; Malkan \& Sargent \cite{malkan}).
This spectral feature is thought to be due to thermal emission from
an accretion disk and perhaps extends into the soft X--ray band,
producing a steep excess emission known as the ''soft excess''
(e.\ g.\ Arnaud et al.\ \cite{arnaud}; Wilkes \& Elvis \cite{wilkes};
Turner \& Pounds \cite{turner}).
Other evidence comes from the reflection hump and iron emission line
observed in AGN (e.\ g.\ Pounds et al.\ \cite{pounds}).
The standard theory of thin accretion disks is largely based on
the fundamental papers of
Shakura \& Sunyaev (\cite{shakura}, hereafter SS73) and Novikov \& Thorne
(\cite{novikov}, hereafter NT73).
The simplest way to calculate the emitted disk spectrum is to assume
that the disk is geometrically thin but optically thick and radiates
locally as a blackbody. The effective temperature $T_{\rm eff}$
of the blackbody is then solely determined by the dissipated flux per
unit area.
These simple disk models, however, are not able to produce soft X--rays
as long as they do not become super--Eddington (e.\ g.\ Bechtold et al.\
\cite{bechtold}).
Real disk spectra will differ from the blackbody approximation:
for sufficiently high accretion rates and viscosity parameters,
the accretion disk eventually becomes
effectively optically thin at small radii and the gas temperature strongly
deviates from the equilibrium temperature in such cases.
Even in the optically thick
case the local spectra differ from
the blackbody, because
the scattering opacity
dominates over absorption in the soft X--ray regime and
there exists a temperature gradient
in the vertical direction.
In the last few years several calculations of increasing accuracy
have been performed to determine the structure and emission spectrum
of accretion disks around massive black holes.
Contrary to the standard Newtonian disk model, NT73 and Page \& Thorne
(\cite{page}) calculated the
effects of general relativity on the disk structure.
The propagation of photons from the disk surface to
a distant observer was treated by Cunningham (\cite{cunningham}).
On the other hand, SS73, Czerny \& Elvis (\cite{czerny}), and
Wandel \& Petrosian (\cite{wandel})
concentrated on a proper calculation of the radiative transfer
by discussing the effects of Comptonization in a simple analytic
manner. Laor \& Netzer (\cite{laor89}) and Laor et al.\ (\cite{laor90})
included
relativistic effects as well as free--free and bound--free opacities in their
numerical computations.
Most calculations of model spectra to date, however, made use of
an averaging over the vertical direction.
Ross, Fabian \& Mineshige (\cite{ross}, hereafter RFM) have calculated
the vertical temperature
profile and atomic level populations
in the radiation pressure dominated inner regions of the disk
for a given constant vertical density profile, using the
Kompaneets equation to treat Compton scattering.
Relativistic corrections for a non--rotating black hole
were incorporated in this code by Yamada et al.\ (\cite{yamada}).
Shimura \& Takahara (\cite{shimura}) and
Shimura \& Takahara (\cite{shimura2})
have calculated the vertical structure and
radiation field of a Newtonian disk self--consistently,
using the ad hoc assumption that the local
viscous heating rate is proportional to the mass density $\rho$.
Here we have also calculated the z--structure simultaneously with the
radiation field of the disk.
But we use a different viscosity description. We assume that the
local energy production is caused by turbulence. The standard
$\alpha $--description (viscosity proportional to the total pressure)
leads to diverging temperature profiles in the upper optically thin
regions of the disk,
because viscous heating always exceeds
radiative cooling. We therefore propose a modification of the standard
$\alpha $--model, which includes the radiative cooling of the
turbulent elements.
The frequency dependent radiative transfer equation
is solved in the Eddington approximation
and the effects of Compton scattering are treated by the Kompaneets
equation (Kompaneets \cite{kompanaets}).
Relativistic corrections on the local disk structure
are introduced according to Riffert \& Herold (\cite{riffert})
for rotating and non--rotating black holes.
In this paper we present solutions of the local structure
and the emission spectrum of accretion disks for different
input parameters (accretion rate, viscosity parameter).
Integrated disk spectra, as seen from a distant observer, are calculated
by the use of a transfer function (e.\ g.\ Cunningham \cite{cunningham};
Speith et al.\ \cite{speith}).
In Sect.\ 2 the basic equations and viscosity description of our model
are formulated. In Sect.\ 3 numerical results are presented.
Finally, our results and conclusions are summarized in Sect.\ 4.
\section{The Model}
\subsection{Basic Equations}
We adopt the standard geometrically thin $\alpha$--accretion disk model,
i.e. the disk height $H$ is much smaller than the radius $r$,
and the dominant velocity is given by the Keplerian motion around
the central mass. The disk is assumed to be in a stationary and
rotationally symmetric state, thus all functions depend only on the
radial and vertical coordinates $r$ and $z$.
The relativistic disk structure has been calculated by NT73 and
subsequently by Riffert \& Herold (\cite{riffert}),
correcting a term in the vertical
pressure balance. According to this paper we define four relativistic
correction factors with respect to the standard Newtonian
disk model depending on the mass $M$ and the specific angular
momentum $a/M$ of the central black hole:
\begin{eqnarray}
A&=& 1-\frac{2GM}{c^2 r} +\frac{a^2}{c^2 r^2} \nonumber \\
B&=& 1-\frac{3GM}{c^2 r} +\frac{2a\sqrt{GM}}{c^2 r^{3/2}} \nonumber \\
C&=& 1-\frac{4a\sqrt{GM}}{c^2 r^{3/2}} +\frac{3a^2}{c^2 r^2} \nonumber \\
D&=& \frac{1}{2\sqrt{r}} \int_{r_i}^r
\frac{x^2 c^2 - 6xGM + 8a\sqrt{xGM} - 3a^2}
{\sqrt{x}\left( x^2 c^2 - 3xGM + 2a\sqrt{xGM} \right)}\ dx~~.
\end{eqnarray}
$r_{i}$ is the inner disk radius located at the position of the
last stable circular orbit;
the gravitational constant and the velocity of light are denoted
by $G$ and $c$.
The hydrostatic equilibrium in vertical direction
for a thin accretion disk is given by
\begin{equation}
\label{hyd}
\frac{dP}{dz} = -\rho g_{z} = -\rho \frac{GM}{r^{3}} z \frac{C}{B} \, ,
\end{equation}
where $P$, $\rho $, and $g_{z}$ are the total pressure,
the mass density, and the gravitational acceleration, respectively.
The radiative transfer is solved in the Eddington approximation,
and Compton scattering is treated in the Fokker--Planck
approximation using the Kompaneets operator.
The absorption cross section contains only free--free processes
for a pure hydrogen atmosphere.
Induced contributions to the radiative processes have been neglected throughout.
Assuming the plasma to be in a state of local thermodynamic equilibrium
(LTE) the first two moments of the radiative transfer
equation for the spectral energy density $U_{\nu}$ and the spectral
flux $F_\nu$ read
\begin{equation}
\label{rad1}
\frac{\partial F_{\nu}}{\partial z}
= \kappa _{\nu}^{\rm ff}\rho c [W_{\nu}-U_{\nu}]
+\kappa _{T} \rho \frac{8\pi h^{2}}{m_{e}c^{4}}\nu
\frac{\partial}{\partial \nu} \nu ^{4}
\left[ n_{\nu}
+\frac{kT}{h}\frac{\partial n_{\nu}}{\partial \nu}\right]
\end{equation}
\begin{equation}
\label{rad2}
\frac{1}{3} \frac{\partial U_{\nu}}{\partial z}=
-\frac{\rho}{c}[\kappa _{T} +\kappa _{\nu}^{\rm ff}] F_{\nu} \, .
\end{equation}
Here $W_{\nu} = 8\pi h \nu ^{3}c^{-3}\exp(-h\nu / kT)$
is the Wien function which serves as the equilibrium spectrum in this case,
$\kappa_{\nu}^{\rm ff}$ is the free--free opacity, and $\kappa_{T}=
0.4~{\rm g/cm^2}$ is the Thomson opacity.
The photon occupation number $n_{\nu}$ can be expressed in terms of
$U_{\nu}$
\begin{equation}
\label{phot}
n_{\nu}=\frac{c^{3}}{8\pi h \nu ^{3}}U_{\nu} \, .
\end{equation}
The free--free opacity is given by
\begin{equation}
\label{opaz}
\kappa _{\nu}^{\rm ff}=1.3\cdot 10^{56}\rho T^{-1/2}
\nu ^{-3} \bar{g}_{\rm ff}\: cm^{2}\, g^{-1}\, ,
\end{equation}
where $T$ is the gas temperature, $k$ is the Boltzmann constant,
and $\bar{g}_{\rm ff}$ is the thermal average free--free Gaunt factor
(Karzas \& Latter \cite{karzas}; Carson \cite{carson}).
In the Eddington approximation the equation of state can be written as
\begin{equation}
\label{state}
P= P_{gas}+P_{rad}= \frac{k}{\mu m_{u}}\rho T + \frac{1}{3}U \, ,
\end{equation}
where $U=\int _{0}^{\infty}U_{\nu}\, d\nu$ is the total energy density
of the radiation field,
$m_{u}$ is the atomic mass unit and $\mu$ is the mean molecular weight
with $\mu =1/2$ for a fully ionized hydrogen atmosphere.
For a Keplerian disk, the viscous heating rate per unit
volume $\mathaccent"70C7 {Q} _{visc}$ is related to the
$\phi$--$r$--component of the viscous stress tensor ${\bf t}$ by
\begin{equation}
\label{visc}
\mathaccent"70C7 {Q} _{visc}= \frac{3}{2}\sqrt{\frac{GM}{r^{3}}}
t_{\phi r}\frac{A}{B} \, ,
\end{equation}
and $t_{\phi r}$ can be expressed in terms of velocity gradients
leading to the result
\begin{equation}
\label{shear}
t_{\phi r} = \frac{3}{2} \eta \sqrt{ \frac{GM}{r^3} } \frac{A}{B}~,
\end{equation}
with the shear viscosity $\eta$.
In a steady state this heating rate must be balanced by the total
radiative cooling, since convection is not considered in this model.
Integration of the right--hand side
of equation (\ref{rad1}) over frequency $\nu$
leads to the following energy balance equation
\begin{equation}
\label{ebal}
\mathaccent"70C7 {Q} _{visc}=\frac{dF}{dz}=\int_{0}^{\infty}
\frac{\partial F_{\nu}}{\partial z} \, d\nu
\, .
\end{equation}
The disk model is so far described by the equations
(\ref{hyd}), (\ref{rad1}), (\ref{rad2}), (\ref{state}) and (\ref{ebal})
which can be solved numerically when proper
boundary conditions are imposed. The only unspecified process is
the source of viscosity and this will be discussed in the next section.
Because of symmetry reasons the inner boundary condition in the
midplane of the disk ($z=0$) is taken as
\begin{equation}
\label{bound1}
F_{\nu}(z=0)=0 \, ~,
\end{equation}
and at the upper boundary (at $z=H$) we assume that the disk radiates
isotropically into the vacuum
\begin{equation}
\label{bound3}
F_{\nu}=\frac{c}{2}U_{\nu} \, .
\end{equation}
For a geometrically thin accretion disk the energy flux from
the disk surface ($z=H$) is
\begin{equation}
\label{bound2}
F(z=H)= \int_{0}^{\infty} F_{\nu} \, d\nu =
\frac{3GM\mathaccent"70C7 {M}}{8\pi r^{3}}\frac{D}{B} \, ,
\end{equation}
where $\mathaccent"70C7 {M}$ is the total mass flux through the disk.
Finally, the disk height $H$ is specified by the condition
\begin{equation}
\label{bound4}
\rho (z=H) =0
\end{equation}
We solve the above set of equations for the functions
$\rho$, $T$, $H$, $U_{\nu}$, and $F_{\nu}$ using a finite
difference scheme in both variables $z$ and $\nu$.
The vertical structure is resolved with 100 points on a
logarithmic grid, and 64 grid points are used in frequency space.
The resulting set of algebraic difference equations is then solved by
a Newton--Raphson method.
Once the vertical structure has been determined we can
integrate the density profiles to obtain the
surface density $\Sigma (r) = \int _{-H}^{+H} \rho (z) \, dz $
and the average radial velocity $|v_{r}| = \mathaccent"70C7 {M}/ \Sigma (r) $.
In order to justify the basic assumptions of the thin disk model
one must have $|v_{r}| \ll |v_{\phi}| $, where $v_{\phi}$ is the
Keplerian velocity.
\subsection{The Viscosity Description}
In the standard $\alpha$-model (SS73) the viscous stress tensor
component $t_{\phi r}$ is propotional to the pressure
\begin{equation}
\label{alpha}
t_{\phi r}= \alpha P~.
\end{equation}
This is a consequence of a number of assumptions: first, the
viscosity is caused by turbulence in the disk, and the magnitude
of $\eta$ can be estimated from a dimensional analysis
\begin{equation}
\label{eta}
\eta \approx \rho \, l_{turb} V_{turb}~,
\end{equation}
where $l_{turb}$ and $V_{turb}$ are the size and the typical
velocity of the largest turbulent eddies, respectively.
Next, the disk height $H$ is used for $l_{turb}$ which is obtained
from the hydrostatic equation (\ref{hyd})
\begin{equation}
H \approx \sqrt{ \frac{B\, r^3}{C\, GM} }
\sqrt{ \frac{P_c}{\bar{\rho} } }~,
\end{equation}
where $P_{c}$ is the pressure in the midplane, and $\bar{\rho}$
stands for the average density.
The turbulent velocity $V_{turb}$ is assumed to be limited by the
local sound speed $c_s$. Otherwise shocks would develop and heat the
gas until the tubulence is subsonic. Approximating the ratio
$P_c / \bar{\rho}$ by the local value $P/\rho = c_s^2$ and inserting
the above relations into equation (\ref{shear}) for the viscous shear
leads to the expression (\ref{alpha}), where the constant $\alpha$
takes up all uncertainties in terms of numerical factors.
This entire procedure is quite successful for the standard disks
in cataclysmic variables which are dominated by the gas pressure.
If radiation pressure becomes important a number of difficulties
arise when the model is applied without modifications.
According to equation (\ref{rad1}) the radiative cooling
decreases in the upper layers of the disk where the density
drops exponentially. Since the radiation pressure remains
finite in this region, the total heating rate (\ref{visc})
cannot be balanced by radiative losses.
This could be cured by replacing $P$ in equation (\ref{alpha})
by the gas pressure, however, in an optically thick regime
the radiation behaves like a gas exchanging momentum with the
plasma and thus contributes to the viscous shear.
In addition, in a radiation dominated disk gravity is mainly balanced by
radiative forces and the disk height is then given by
\begin{equation}
\label{heightedd}
H \approx \frac{3 \mathaccent"70C7 {M} \kappa_F}{8\pi c} \frac{D}{C}~,
\end{equation}
where $\kappa_F$ is a flux weighted average opacity
\begin{equation}
\label{Rossel}
\frac{1}{\kappa _{F}}=\frac{\int_{0}^{\infty }\frac{1}{\kappa _{T} +
\kappa _{\rm ff}}
\frac{dU_{\nu}}{dz}\, d\nu }{ \int_{0}^{\infty } \frac{dU_{\nu}}{dz} \,
d\nu } \, .
\end{equation}
This height does not contain the sound speed.
In order to derive a consistent parameterization of the
turbulent viscosity for the radiation dominated case
we start from the dimensional analysis (\ref{eta}) using
again $l_{turb} = H$, where the disk height $H$ is
self--consistently calculated in our model. For the limitation of
the velocity $V_{turb}$ the same arguments as in SS73 are used,
i.e. $V_{turb}$ has to be small enough that no entropy due
to shock waves is generated in the flow. For a radiation dominated
shock in an optically thick medium
\begin{equation}
V_{turb} < \sqrt{\frac{P}{\rho}}~,
\end{equation}
but in an optically thin regime, where the photons are
able to escape freely
\begin{equation}
V_{turb} < \sqrt{\frac{P_{gas}}{\rho}}~.
\end{equation}
This follows from considering an isothermal shock in a
plasma with radiation when the Mach number approaches unity.
Assuming an isothermal shock structure is an approximation
for an optically thin environment.
Since this limiting value of $V_{turb}$ depends on the
optical depth we interpolate between the two extreme cases
\begin{equation}
\label{vturb}
V_{turb}= c_{s}\frac{ \tau +\sqrt{\beta} }{1+\tau }\, ,
\end{equation}
with the optical depth
\begin{equation}
\tau = \int_z^H \kappa_F \rho\, dz^\prime\, ,
\end{equation}
and the pressure ratio $\beta =P_{gas}/P$.
For the turbulent viscosity we now have
\begin{equation}
\eta = \alpha \rho H V_{turb}~,
\end{equation}
with a constant $\alpha$ which is a parameter in our disk model.
We will now consider the heating--cooling balance, integrate
equation (\ref{rad1}) over frequency, and use the relations (\ref{ebal})
and (\ref{visc})
\begin{eqnarray}
\frac{dF}{dz} &=& \bar{\kappa}_{\rm ff} \rho c \left[ a_{r}T^4 - U \right]
+ 4U\, \frac{\kappa_T \rho}{m_e c} \left[ kT - <\!\nu\!> \right] \nonumber \\
&=& \frac{9GM}{4r^3} \frac{A^2}{B^2} \alpha \rho H V_{turb}~,
\end{eqnarray}
where $\bar{\kappa}_{\rm ff}$ is some appropriately averaged mean absorption
opacity, $a_{r}$ is the radiation constant, and
\begin{equation}
<\!\nu\!> = \frac{ \int_0^\infty \nu^4 n_\nu \, d\nu }
{ 4 \int_0^\infty \nu^3 n_\nu \, d\nu }
\end{equation}
is an average photon energy ($h\! <\! \nu \! >\! /k = T_r$
can be considered as a radiation temperature).
In the upper parts of the disk (at $z \approx H$) the density
drops to zero and since $\bar{\kappa}_{\rm ff} \propto \rho$
free--free processes are negligible in this regime. The
radiation field remains unchanged and thus $<\!\nu\!>$ and
$U$ are constant; from the boundary condition $c\, U(H) =2\, F(H)$
we obtain using equations (\ref{bound2}), (\ref{heightedd}), and (\ref{vturb})
for $\tau \ll 1$
\begin{equation}
\label{heatcool1}
T - T_r = \sqrt{T_v T}~,
\end{equation}
where
\begin{equation}
T_v = \left( \frac{9\alpha}{32} \right)^2
\left( \frac{\kappa_F}{\kappa_T} \right)^2 \frac{m_e^2 c^2}{\mu k m_u}
= 5.1\, 10^5\, \alpha^2 \left( \frac{\kappa_F}{\kappa_T} \right)^2 {\rm ~K}~.
\end{equation}
(Note that in a scattering dominated plasma $\kappa_F / \kappa_T \approx 1$).
From this we get the temperature that will result from a balance
of viscous heating and inverse Compton cooling
\begin{equation}
\label{heatcool2}
T = T_r + \frac{1}{2} \left[ T_v + \sqrt{T_v^2 + 4T_r T_v} \right]~.
\end{equation}
This solution is unique since a negative sign of the square root
in (\ref{heatcool2}) would correspond to a negative sign of the
right hand side of equation (\ref{heatcool1}).
As a consequence we obtain a lower limit for $T$
\begin{equation}
T > max[T_{r},\, T_{v}]\, .
\end{equation}
\section{Results}
In this paper the mass of the black hole is fixed at a typical value of
$M=10^{8}M_{\odot}$.
We show results of our model calculations
for different input parameters such as the accretion rate
$\mathaccent"70C7 {M}$,
the viscosity parameter $\alpha$, the specific angular momentum $a$,
and the inclination angle $\Theta _{0}$ of the observer's
position
with respect to the disk axis.
Hereafter, $\mathaccent"70C7 {M}$ is measured in units of the
critical accretion rate
$\mathaccent"70C7 {M} _{crit}=L_{Edd}/(\varepsilon c^{2})$, where
$L_{Edd}=4\pi c GM/\kappa _{T}$ is the
Eddington luminosity and $\varepsilon $ is the efficiency of accretion.
For a non--rotating black hole ($a/M=0$) $\varepsilon$ is $0.057$, and for a
maximally rotating black hole ($a/M \simeq 0.998$, see Thorne \cite{thorne})
$\varepsilon \simeq 0.321$.
First, we give solutions of the local vertical structure and
emission spectrum at $5$ Schwarzschild radii
($r=5R_{S}=10GM/c^{2}$).
At this radius, the local energy
release takes its maximum value for a non--rotating black hole
and this disk region
is responsible for the high frequency part of the total spectrum.
Next, we consider the radial structure and the integrated disk spectrum
seen by a distant observer. At this point we also include
relativistic effects on the emergent disk spectrum such as the relativistic
Doppler shift from the disk rotation, gravitational redshift,
and gravitational light bending
due to the central black hole.
\subsection{Vertical Structure and Local Disk Spectrum}
A typical vertical distribution
(normalized to the corresponding maximum values)
of the radiative flux $F$,
the radiation energy density $u$, the mass density $\rho $, and
the gas temperature $T$
as a function of the
Thomson scattering optical depth $\tau _{T}$ is shown in Fig.\ \ref{vert1} for
a non--rotating black hole with $M=10^{8}M_{\odot}$,
$\mathaccent"70C7 {M}=0.3$, $\alpha =1/3$ and $r=5 R_{S}$.
In Fig.\ \ref{vert2}, the above variables are plotted for the
case of
a maximally rotating black hole ($a/M=0.998$) with the same input parameters,
but using $r=1R_{S}$ , which nearly corresponds to the maximum
of the local energy release for $a/M=0.998$.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig1.ps,width=8.8 truecm}}
\caption{\label{vert1}Vertical distribution of radiative flux $F$,
radiation energy density $U$,
mass density $\rho$, and gas temperature $T$ (all normalized to their
respective maximum values)
at $r=5 R_{S}$. The model parameters are: $M=10^{8}M_{\odot}$,
$\mathaccent"70C7 {M}=0.3$, $\alpha =1/3$, $a/M=0$.
Note that $U$ approaches a finite value for small optical depth
($\tau _{T} \rightarrow 0 $).}
\end{figure}
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig2.ps,width=8.8 truecm}}
\caption{\label{vert2}Same as Fig.\ 1, but for a maximally rotating
black hole $a/M=0.998$ at $r=1 R_{S}$.}
\end{figure}
It is found that the mass density $\rho$ is not a
monotonically decreasing function of
$z$, but is showing a clear density inversion
(such density inversions have also been found by Meyer \& Meyer--Hofmeister
\cite{mey} and Milsom et al.\ \cite{mil}).
This can be explained as follows:
in using the flux weighted mean $\kappa _{F}$ (equation (\ref{Rossel})),
the hydrostatic equilibrium can be written as
\begin{equation}
\label{Hyd2}
\frac{dP_{gas}}{dz}=\frac{\kappa _{F}}{c}\rho [(F(z) - F_{Edd}(z)]
= - \rho g_{z}^{*} \, ,
\end{equation}
where the local Eddington flux $F_{Edd}(z)$
is given by
\begin{equation}
\label{Eddi2}
F_{Edd}(z)=\frac{c}{\kappa _{F}}g_{z} \, ,
\end{equation}
and $g_{z}^{*}=g_{z}-\kappa _{F}F/c$ is the effective gravitational
acceleration.
For a highly radiation pressure dominated disk ($P_{rad} \gg P_{gas}$),
the local flux adjusts itself close to the Eddington value.
For example, the radiation pressure always exceeds the gas pressure
by at least a factor of about $100$ throughout the disk for
the parameters adopted in Fig.\ \ref{vert1}. When the
emerging surface flux is large, the local flux becomes slightly
super--Eddington in the inner regions of the disk and
a strong density inversion ($d\rho /dz >0$)
occurs. Because of the boundary condition
imposed on the surface flux, the outer parts of the disk are
always sub--Eddington.
In Fig.\ \ref{dens} the vertical density distribution
is shown for different $\mathaccent"70C7 {M}$. The density
in the equatorial plane
is up to a factor of $7$ smaller than its maximum value.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig3.ps,width=8.8 truecm}}
\caption{\label{dens}Vertical density distribution
at $r=5 R_{S}$ for a model with $\alpha =1/3$ and
$a/M=0$. The curves are labeled with the value of $\mathaccent"70C7 {M}$.}
\end{figure}
Although the total Thomson scattering depth decreases for increasing
$\mathaccent"70C7 {M}$ and $\alpha$,
our model calculations show, that
the accretion disks are optically thick with respect to
Thomson scattering even
for large $\mathaccent"70C7 {M}$ and $\alpha$.
At the same time these disks are effectively optically thin
($\tau _{\rm eff}(\nu) <1$) except for the low frequencies,
and $\tau _{\rm eff}(\nu)$ is defined as
\begin{equation}
\label{taueff}
\tau_{\rm eff}(\nu) =\int_{0}^{H} \rho \sqrt{3\cdot [\kappa _{\rm ff}(\nu)
+\kappa _{T}]\kappa _{\rm ff}(\nu)} \, dz \, .
\end{equation}
In this case the assumption of local thermodynamic equilibrium (LTE)
is no longer valid even in the disk midplane and
the gas temperature deviates strongly from the equilibrium
temperature defined as $T_{eq}=(U/a_{r})^{1/4}$.
The various heating and cooling functions are compared in
Fig.\ \ref{heat}.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig4.ps,width=8.8 truecm}}
\caption{\label{heat}Viscous heating $\Lambda _{visc}$,
Compton cooling $\Lambda _{comp}$ and
cooling due to free--free emission $\Lambda _{\rm ff}$
(all normalized to the viscous heating rate at the disk midplane)
as a function of $z$
at $r=5R_{S}$. The model parameters are:
$M=10^{8}M_{\odot}$, $\mathaccent"70C7 {M}=0.3$, $\alpha =1/3$, $a/M=0$.
The outermost layer is shown in the upper right corner with
higher resolution.}
\end{figure}
Compton cooling plays the dominat role in the
inner layers of the disk (at small $z$), whereas in the region of the
density increase free-free-processes become very efficient leading to
a steep temperature drop. The gas temperature even falls below the
radiation temperature $T_r$, thus Compton scattering turns into a
heating mechanism ($\Lambda_{comp}$ is negative) which effectively balances the
bremsstrahlung cooling. In the uppermost layers, where the density drops
exponentially, free-free processes are negligible ($\propto \rho^2$),
and the temperature adjusts itself to the equilibrium between viscous
heating and inverse Compton cooling described above.
Due to these heating-cooling processes a temperature inversion occurs
and a hot corona-like layer builds up on top of the disk. This layer
is however optically thin to both scattering and
absorption (the typical densities are below
$10^{-13}\, {\rm g\, cm^{-3}}$), and thus there is no effect on the emitted
spectrum.
Note that in the upper disk layers the viscous heating is proportional
to the density, thus the relevant cooling mechanisms must not
have a stronger $\rho$--dependence. This holds of course
for inverse Compton cooling used in our model, however,
cooling due to strong resonance lines has also been proposed
(Hubeny \cite{hubeny}) to contribute to the overall heating--cooling
balance.
The gas temperature and the corresponding
equilibrium temperature $T_{eq}$ for
different $\mathaccent"70C7 {M}$ are shown in Fig.\ \ref{temp1}.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig5.ps,width=8.8 truecm}}
\caption{\label{temp1}Vertical distribution of gas temperature $T$
(solid lines)
and equilibrium temperature $T_{eq}$ (dotted lines)
at $r=5R_{S}$ for a model with $\alpha =1/3$ and
$a/M=0$. The curves are labeled with the value of $\mathaccent"70C7 {M}$.}
\end{figure}
For large values of $\mathaccent"70C7 {M}$ even the temperature
in the midplane of the disk strongly exceeds the equilibrium value.
For low $\mathaccent"70C7 {M}$ the disk becomes effectively optically
thick and a general equilibrium is established.
The locally emitted spectrum at a fixed radial position
($r=5R_{S}$) for
different $\mathaccent"70C7 {M}$ is shown in
Fig.\ \ref{spec1}.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig6.ps,width=8.8 truecm}}
\caption{\label{spec1}Emergent spectrum
at $r=5R_{S}$ for $\alpha =1/3$ and $a/M=0$.
The curves are labeled with the value of $\mathaccent"70C7 {M}$.
Dotted curves show the
local Wien spectrum at $T=T_{\rm eff}$.}
\end{figure}
The local Wien spectra at the corresponding effective temperature $T_{\rm eff}$
are also shown by dotted lines. At low frequencies, the accretion disk is
effectively optically thick and scattering effects can be neglected;
the local spectrum approaches the equilibrium distribution $W_{\nu}$,
however, the spectra in low frequency regime cannot be taken too seriously
since we have neglected induced processes.
At higher frequencies,
scattering effects become important and one gets a modified Wien spectrum.
For sufficiently high $\mathaccent"70C7 {M}$ and $\alpha$ a Wien peak appears
in the soft X--ray range of the local spectra. This feature is a result
of repeated Compton scattering of photons by a
thermal distribution of electrons at temperature $T$. If
the Compton parameter $y \gg 1$,
Comptonization goes to saturation and the photons are shifted
into a Wien distribution at temperature $T$
(see Felten \& Rees \cite{felten}).
The Compton parameter $y$ is defined as
\begin{equation}
\label{comp}
y=\frac{4kT}{m_{e}c^{2}}\tau _{T}^{2} \, .
\end{equation}
For example, the case
$\mathaccent"70C7 {M}=0.3$ shown in Fig.\ \ref{spec1}
leads to a effective optical depth
$\tau _{\rm eff}(\nu _{max})=0.1$ at the peak frequency
$\nu _{max}=2.8 \cdot 10^{16}\, Hz$
and a total Thomson
scattering depth of $\tau _{T}=98.5$;
from that and an average temperature of
$T=4\cdot 10^{5} K$ (see Fig.\ \ref{temp1}) we have
$y \approx 3$.
Since repeated
Compton scattering is important for $y >1$, a significant shift
of photons
into the Wien peak is expected.
For low $\mathaccent"70C7 {M}$ and $\alpha$ the disk is effectively optically
thick for all relevant frequencies
and the radiation assumes an equilibrium spectrum for almost all $z$.
The effective optical depth
for the curve $\mathaccent"70C7 {M}=0.1$
in Fig.\ \ref{temp1} and \ref{spec1} is $\tau _{\rm eff}(\nu _{max})=9.8$,
with $\nu _{max}=9.1 \cdot 10^{15}\, Hz$.
Nevertheless, the emergent spectrum deviates from the Wien spectrum
at $T=T_{\rm eff}$ because there exists a temperature gradient and
high frequency photons escape from deeper layers with higher
temperatures.
\subsection{Radial Structure and Integrated Disk Spectrum}
After discussion of the vertical structure and local
emission spectrum at $r=5R_{S}$ here
we investigate the radial structure and overall spectrum
of the accretion disk. In numerical calculations,
we used $50$ radial points on a logarithmic grid
from the last stable orbit $r_{i}$ to the outer disk radius
$r_{out}$ (here $r_{out}=1000R_{S}$).
The height of the disk from the surface to the
equatorial plane as a function of the radial distance
from the black hole is shown in Fig.\ \ref{height1}
for several values of $\mathaccent"70C7 {M}$.
For comparison, the geometrically thin limit, here defined as
$H(r)=0.1 r$, is also plotted as a dashed curve.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig7.ps,width=8.8 truecm}}
\caption{\label{height1}Disk height
for $\alpha=1/3$ and $a/M=0$.
The curves are labeled with the value of $\mathaccent"70C7 {M}$.
The dashed curve shows the geometrically thin limit
$H(r)=0.1 \cdot r$.}
\end{figure}
This means that the accretion rate should not
greatly exceed $\mathaccent"70C7 {M}=0.3$
for not to violate the geometrically thin disk approximation.
Therefore, numerical results for higher accretion rates are
strictly speaking not self--consistent.
On the other hand, the $\alpha$--parameter has little influence
(of the order of a few per cent) on the
height of the disk.
If the local flux is given by the Eddington flux,
the disk height $H$ is given by equation (\ref{heightedd}).
In the inner regions of the disk Thomson scattering delivers the major
contribution to the opacity ($\kappa _{F} \approx \kappa _{T}$)
and the height of the disk is solely determined by the
accretion rate $\mathaccent"70C7 {M}$.
However, the local flux deviates from the Eddington flux at the upper
boundary of the disk. Although the gas pressure is negligible in comparison
with the radiation pressure, the gas pressure gradient $dp_{Gas}/dz$
is not.
We have also determined mean radial velocities
$|v_{r}| = \mathaccent"70C7 {M}/ \Sigma (r) $ for all radial grid points.
Our calculations show, that for $\mathaccent"70C7 {M} \leq 0.3$ and
$\alpha \leq 1$ the basic thin disk approximation
$|v_{r}| \ll |v_{\phi}| $ is always satisfied.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig8.ps,width=8.8 truecm}}
\caption{\label{newt1}Entire Newtonian face on disk spectra
for $\alpha=1/3$ and $a/M=0$.
The curves are labeled with the value of $\mathaccent"70C7 {M}$.
Dotted curves denote multi Wien spectra.}
\end{figure}
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig9.ps,width=8.8 truecm}}
\caption{\label{newt2}Entire Newtonian face on disk spectra
for $\mathaccent"70C7 {M}=0.3$ and $a/M=0$.
The curves are labeled with the value of $\alpha$.
The dotted curve denotes the corresponding multi Wien spectrum.}
\end{figure}
Now we will turn to the integrated disk spectra.
General relativistic effects on the emergent spectrum are twofold:
first, relativistic effects occur in calculating the
disk structure and therefore change the local emission spectrum in
the corotating frame of the disk. The resulting spectra were
discussed for a non--rotating black hole in section 3.1.
Second, relativistic effects have a substantial influence on the
propagation of photons to the observer. All these effects become
more and more important if the black hole is rotating, since then
the disk extends to regions very near to the black hole.
To separate both effects the investigation of the integrated disk spectrum
is treated in
two different ways:
first, we examine the emergent spectrum of a face on disk as
seen for a distant observer without including relativistic effects
or Doppler shifts
on the locally emitted photon distribution. Relativistic effects only occur
due to the relativistic disk structure equations (see section 2.1).
The entire spectrum then is simply calculated by integration
of
the local spectra over the disk surface. In the following we will
call this the Newtonian approximation.
Second, a fully relativistic calculation
of the propagation of photons from the disk to a distant observer
is performed, using a program code (Speith et al.\ \cite{speith})
to obtain numerical values of the Cunningham transfer function
(Cunningham \cite{cunningham})
for any set of parameters.
The Newtonian disk spectrum for several
values of $\mathaccent"70C7 {M}$ and $\alpha$ is shown in Fig.\ \ref{newt1}
and \ref{newt2}. The
spectral luminosities calculated by summing up local Wien spectra at
the corresponding effective temperatures
are indicated by dotted lines. The hardness of the spectra and
therefore also the amount of soft X--rays is a sensitive
function of $\mathaccent"70C7 {M}$ and a somewhat less sensitive
function also of $\alpha$.
In the following, we fix the accretion rate and viscosity parameter
at $\mathaccent"70C7 {M}=0.3$ and $\alpha =1/3$, respectively.
A comparison of the relativistic and Newtonian face--on disk spectrum
for a non--rotating ($a/M=0$) and maximally rotating ($a/M=0.998$) black hole is
shown in Fig.\ \ref{face}.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig10.ps,width=8.8 truecm}}
\caption{\label{face}Entire face on disk spectra for
$\mathaccent"70C7 {M}=0.3$ and $\alpha=1/3$.
The curves are labeled with the value of the specific angular momentum $a/M$.
The solid an dotted curves denote the relativistic and
Newtonian disk spectra, respectively.}
\end{figure}
For low frequencies the emergent spectrum approaches the Newtonian
model. The spectrum at higher frequencies is reduced
with respect to the Newtonian case especially for a rotating black hole.
These photons originate from
the inner parts of the disk with high velocities and
gravitational fields and therefore are diverted by forward peaking and
gravitational focusing.
Consequently, an equatorial observer primarily sees blueshifted and
focused radiation from the hot inner parts of the disk.
In Fig.\ \ref{theta} the observed relativistic spectrum is shown as a
function of the
inclination angle $\Theta _{0}$ together with a
Newtonian face--on spectrum (dotted line) for a maximally rotating black hole.
\begin{figure}[htb]
\par\centerline{\psfig{figure=fig11.ps,width=8.8 truecm}}
\caption{\label{theta}Entire relativistic disk spectra
for different inclination angle $\Theta _{0}=0\degr ,\,
41\degr ,\, 60\degr ,\, 70\degr $ and $90\degr$. The model parameters are:
$M=10^{8}M_{\odot}$, $\mathaccent"70C7 {M}=0.3$, $\alpha=1/3$, $a/M=0.998$.
The dotted curve denotes the Newtonian face--on spectrum.}
\end{figure}
A significant fraction of total flux is emitted in the soft and hard
X--ray range
($\nu > 2.4\cdot 10^{16}Hz$) and strongly depends
on the position of the observer. For
$\Theta _{0} = 0\degr ,\, 41\degr ,\, 60\degr ,\, 70\degr $
and $90\degr$,
this fraction is $36\%$, $50\%$, $64\%$, $78\%$
and $93\%$, respectively. The fraction of the total flux emitted in the
soft X--ray band ($2.4\cdot 10^{16}\, Hz-6\cdot 10^{17}\, Hz$),
corresponding to the
sensitivity range of the PSPC (position sensitive proportional
counter) on the ROSAT X--ray satellite is $32\%$, $38\%$, $38\%$, $33\%$
and $21\%$, respectively, and therefore is nearly independent of the
inclination angle.
In order to show the limiting case of our calculations we also
included $\cos \Theta _{0} = 0 $ in Fig.\ \ref{theta}.
However, for a nearly edge-on observer self--occultation of the
inner parts of the disk by the outer parts should be taken into account.
\section{Concluding Remarks}
We have performed a self--consistent calculation of the
vertical structure and emergent spectrum of an accretion disk
around a massive Kerr black hole.
Full relativistic corrections have been
included.
The standard $\alpha$--model leads to diverging temperature profiles
in the upper layers of the disk, where the density drops exponentially
and viscous heating always overcomes radiative cooling.
When we include the radiative cooling of the turbulence elements in the
optically thin part of the disk, an equilibrium between turbulent
viscous heating and inverse Compton cooling is established.
Our calculations are for a fixed central mass $M=10^{8}M_{\odot}$
but for different $\mathaccent"70C7 {M}$, $\alpha$, $a/M$, and $\Theta _{0}$.
It is found that the mass density is not a monotonically decreasing
function of $z$ but shows a strong density inversion in the
radiation pressure dominated inner parts of the disk, where
the local flux becomes slightly super--Eddington. Moreover, we always obtain
a temperature inversion in the upper optically thin layers of the disk
with zero gradient in the outermost region.
For sufficiently high $\mathaccent"70C7 {M}$ and $\alpha$ the disk becomes
effectively optically thin at small radii and the gas
temperature exceeds the equilibrium temperature
($T_{eq}=(U/a_{r})^{1/4}$) by a large amount even in the disk midplane.
As a result, the local emission spectrum strongly deviates from the
Wien spectrum at $T=T_{\rm eff}$.
Even in the optically thick case, the local spectra differ from the
Wien--spectra at high frequencies, because scattering opacity dominates
in the soft X--ray range and there exists a temperature gradient
in the vertical direction. Therefore, a significant fraction
of the total flux is emitted in the soft X--ray range even for low
$\mathaccent"70C7 {M}$
and $\alpha$, especially in the maximally rotating case ($a=0.998$)
and for high inclination angles of the observer.
For increasing $\mathaccent"70C7 {M}$, $\alpha$, $a/M$, and $\Theta _{0}$
the calculated spectra become more and more flat, producing quasi power law
spectra
in the sensitivity range of the ROSAT PSPC, whereas for small values
a steep soft X-ray component is established.
Thus, the model can in general account for the soft X--ray excess
observed in many AGN.
We intend to show detailed spectral fits of this model to combined ROSAT and
IUE observations for a sample of AGN in a future paper.
Note that we have neglected induced processes and that our opacity description
only contains free--free processes for a pure hydrogen atmosphere.
Neglecting induced processes means, that we have concentrated
on the high energy tail of the resulting spectra.
Nevertheless, bound--free absorption and line opacities can significantly
contribute to the total opacity especially at low temperatures.
Therefore, the spectrum is expected to soften when all relevant opacities
are taken into account because of the increasing effective optical depth
in this case.
Another important point is the role of convection in our calculations,
especially in view of our inverse density profiles.
If convective energy transport takes place in the disk,
the vertical structure and subsequently the emergent spectrum may be altered
considerably.
We therefore have done a simplified stability analysis according to
the standard mixing length theory (see for example Cox \& Giuli \cite{cox}),
assuming that the convective elements move adiabatically in the disk
and that the local gas temperature is given by the equilibrium temperature
$T_{eq}$. In a gas with radiation pressure, the effective buoyancy force
is reduced by the radiation pressure gradient and the convective flux
then can be expressed in terms of the effective gravitational acceleration
$g_{z}^{*}$ (equation \ref{Hyd2}),
\begin{equation}
\label{convec}
F_{conv}=\frac{\rho c_{P}Tl_{m}^{2}H_{P}^{-3/2}\delta ^{1/2}}{4\sqrt{2}}
\sqrt{g_{z}^{*}(\nabla _{s} -\nabla _{e})}\cdot (\nabla _{s} -\nabla _{e})\, ,
\end{equation}
where $l_{m}$, $H_{P}=P/(\rho g_{z})$, and $c_{P}$ are the mixing length,
the pressure scale height and the specific heat at constant pressure,
respectively, and $\delta=-\partial ln\rho /\partial lnT$.
$\nabla _{s}=(dlnT/dlnP)_{s}$ and $\nabla _{e}=(dlnT/dlnP)_{e}$
are the temperature gradients with respect to pressure
for the surroundings and the rising element.
Therefore, the disk is unstable for convection if the square root
on the right hand side of equation (\ref{convec}) has a positive argument.
In the sub--Eddington regime one has $g_{z}^{*}>0$ and
the condition for convection to set in
is given by the Schwarzschild criterion $(\nabla _{s} -\nabla _{e})>0$.
On the other hand in the super--Eddington regime $g_{z}^{*}<0$,
and the condition for convective instability is
$(\nabla _{s} -\nabla _{e})<0$, leading to a convective energy flux in
negative $z$--direction.
Our model calculations show that none of both conditions are satisfied
and the disk therefore is stable against convection.
Since the gas temperature deviates from the equilibrium temperature for
an effectively optically thin disk and the convective elements
do not move adiabatically in the upper optically thin layers, this
simplified stability analysis
has to be considered with caution, and a more detailed
stability analysis has to be performed in order
to clarify the role of convection.
It is well known that an $\alpha$-disk with a viscosity proportional
to the total pressure is radially unstable in the radiation
dominated case (Lightman \& Eardly \cite{light}).
However, this analysis is based on a vertically integrated disk
structure. Since our model leads to a strong $z$-dependence for almost
all functions including the viscosity, a detailed 2-dimensional
stability analysis would be required for a definite answer
concerning the overall stability of our model.
\begin{acknowledgements}
T.\ D\"orrer acknowledges the support of DARA through grant 50 OR 90099.
\end{acknowledgements}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,626
|
We have UPPERCASE (or lowercase) text in our database and we want to convert it to Proper Case/Title Case for reporting purposes. This involves converting the text into lowercase and converting the first letter of each word into UPPERCASE.
Unfortunately, the problem is not always this simple; sometimes you might want to make exceptions. In the text above you might decide that "in" and "a" are not important enough to be capitalised.
You might have other exceptions such RAM, TFT, CPU, ASCII. Some exceptions might also be of mixed case, such as DoS, QoS, WiseSoft, TomTom. Creating a Proper Case function is not an exact science, and this is probably why it wasn't included as a system function in T-SQL.
A solution is required that can convert the first letter of each word into UPPERCASE and the rest of the word into lowercase. In addition it needs to be able to handle exceptions such as those mentioned previously. The function will allow a user to specify a list of exceptions, by inputting them into them into one of the functions parameters and also handle a few expected grammatical exceptions.
@Exceptions – A list of exceptions. Exceptions will be printed exactly as specified. Specify NULL if you have no exceptions.
@UCASEWordLength – Words shorter than the value specified will automatically be printed in UPPERCASE. Specify NULL if you don't want to use this feature.
The function will convert the text to lower case, and then parse the text character by character. The first character is converted to UPPERCASE. Each character is appended to a "word" variable until a white space/punctuation character is reached, marking the end of the word. At this point the word will be compared to the list of user exceptions and the case converted if required. Some common grammatical exceptions will also be checked – e.g. WON'T will be converted to Won't with these exceptions instead of Won'T. The function will display names such as O'Donnell correctly. Once the word has been cased correctly with exceptions accounted for, it is appended to a variable used to store the return value along with the white space/punctuation character following it. The process is repeated until all of the text has been converted.
Function Purpose: To convert text to Proper Case.
Note the use of the Pipe "|" symbol to separate exceptions.
-- Character is white space/punctuation/symbol which marks the end of our current word.
-- If word length is less than or equal to the UPPERCASE word length, convert to upper case.
-- e.g. you can specify a @UCASEWordLength of 3 to automatically UPPERCASE all 3 letter words.
-- Check word against user exceptions list. If exception is found, use the case specified in the exception.
-- e.g. WiseSoft, RAM, CPU.
-- If word isn't in user exceptions list, check for "known" exceptions.
This is good stuff. I had an old vbscript that did something similar that I was wanting to port to a SQL UDF. The only issue I have now is to account for proper names such as McNeese.
Thanks for your hard work on this [fProperCase] function, I needed to convert tons of new data and your function worked perfectly..
Please send me your email address, I have some scripts/tools, you might find handy.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,382
|
Discover Philippines ETFs
Philippines ETF List
The table below displays all U.S. listed equity ETFs that maintain significant exposure to stocks listed in Philippines. Note that ETFs that do not include Philippines within the top 10 country allocations may not be included below.
Click on the tabs below to see more information on Philippines ETFs, including historical performance, dividends, holdings, expense ratios, technical indicators, analysts reports and more. Click on an ETF ticker or name to go to its detail page, for in-depth news, financial data and graphs. By default the list is ordered by descending total market capitalization. Note that ETFs are usually tagged by ETF Database analysts as more than one type; for example, an inverse gold ETF may be tagged as "inverse" and as "gold" and as "commodity".
This is a list of all Philippines ETFs traded in the USA which are currently tagged by ETF Database. Please note that the list may not contain newly issued ETFs. If you're looking for a more simplified way to browse and compare ETFs, you may want to visit our ETF Database Categories, which categorize every ETF in a single "best fit" category.
This page includes historical return information for all Philippines ETFs listed on U.S. exchanges that are currently tracked by ETF Database.
The table below includes fund flow data for all U.S. listed Philippines ETFs. Total fund flow is the capital inflow into an ETF minus the capital outflow from the ETF for a particular time period.
The following table includes expense data and other descriptive information for all Philippines ETFs listed on U.S. exchanges that are currently tracked by ETF Database. In addition to expense ratio and issuer information, this table displays platforms that offer commission-free trading for certain ETFs.
Clicking on any of the links in the table below will provide additional descriptive and quantitative information on Philippines ETFs.
The following table includes ESG Scores and other descriptive information for all Philippines ETFs listed on U.S. exchanges that are currently tracked by ETF Database. Easily browse and evaluate ETFs by visiting our ESG Investing themes section and find ETFs that map to various environmental, social, governance and morality themes.
This page includes historical dividend information for all Philippines ETFs listed on U.S. exchanges that are currently tracked by ETF Database. Note that certain ETPs may not make dividend payments, and as such some of the information below may not be meaningful.
The table below includes basic holdings data for all U.S. listed Philippines ETFs that are currently tagged by ETF Database. The table below includes the number of holdings for each ETF and the percentage of assets that the top ten assets make up, if applicable. For more detailed holdings information for any ETF, click on the link in the right column.
The following table includes certain tax information for all Philippines ETFs listed on U.S. exchanges that are currently tracked by ETF Database, including applicable short-term and long-term capital gains rates and the tax form on which gains or losses in each ETF will be reported.
This page contains certain technical information for all Philippines ETFs that are listed on U.S. exchanges and tracked by ETF Database. Note that the table below only includes limited technical indicators; click on the "View" link in the far right column for each ETF to see an expanded display of the product's technicals.
This page provides links to various analysis for all Philippines ETFs that are listed on U.S. exchanges and tracked by ETF Database. The links in the table below will guide you to various analytical resources for the relevant ETF, including an X-ray of holdings, official fund fact sheet, or objective analyst report.
This page provides ETF Database Ratings for all Philippines ETFs that are listed on U.S. exchanges and tracked by ETF Database. The ETF Database Ratings are transparent, quant-based evaluations of ETFs relative to other products in the same ETF Database Category. As such, it should be noted that this page may include ETFs from multiple ETF Database Categories.
EPHE iShares MSCI Philippines ETF Equity Emerging Asia Pacific $139.93 11.34% 57,062 $29.16 0.21% 0.55% 10.92% -5.14% -8.94% -22.74% Asia Pacific Equities 2010-09-28 0.58% N/A $0.29 2021-06-10 $0.22 0.87% 24.45 0.59 42 58.38% View 40% 20% 1099 $25.88 $30.12 $29.03 $29.29 69.57 View View View View View View A- B 3.76 0.93% 7.36% 639.41 3.26% 12.78%
Philippines Country Power Rankings New
Country power rankings are rankings between Philippines and all other country and broad geographic location U.S.-listed equity ETFs on certain investment-related metrics, including 3-month fund flows, 3-month return, AUM, average ETF expenses and average dividend yields. The metric calculations are based on U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to a specific country or broad geographic location. If an ETF changes its geographic classification, it will also be reflected in the investment metric calculations. The calculations exclude all other asset classes and inverse ETFs.
Philippines and all other countries and broad geographic locations are ranked based on their aggregate 3-month fund flows for all U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to those respective countries or broad geographic locations. 3-month fund flows is a metric that can be used to gauge the perceived popularity amongst investors of Philippines relative to other countries and broad geographic locations. All values are in U.S. dollars.
Philippines and all other countries and broad geographic locations are ranked based on their AUM-weighted average 3-month return for all the U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to those respective countries or broad geographic locations. In addition to price performance, the 3-month return assumes the reinvestment of all dividends during the last 3 months.
Philippines and all other countries and broad geographic locations are ranked based on their aggregate assets under management (AUM) for all the U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to those respective countries or broad geographic locations. All values are in U.S. dollars.
Philippines and all other countries and broad geographic locations are ranked based on their AUM-weighted average expense ratios for all the U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to those respective countries or broad geographic locations. The lower the average expense ratio for all U.S.-listed equity ETFs in a country or broad geographic locations, the higher the rank.
Philippines and all other countries and broad geographic locations are ranked based on their AUM-weighted average dividend yield for all the U.S.-listed equity ETFs that are classified by ETF Database as being mostly exposed to those respective countries or broad geographic locations.
ETF Issuer League Tables - Philippines Country New
ETF issuers who have ETFs with exposure to Philippines are ranked on certain investment-related metrics, including estimated revenue, 3-month fund flows, 3-month return, AUM, average ETF expenses and average dividend yields. The metric calculations are based on U.S.-listed Philippines ETFs and every Philippines ETF has one issuer. If an issuer changes its ETFs, it will also be reflected in the investment metric calculations.
ETF issuers are ranked based on their estimated revenue from their ETFs with exposure to Philippines. Estimated revenue for an ETF issuer is calculated by aggregating the estimated revenue of the respective issuer ETFs with exposure to Philippines. To get the estimated issuer revenue from a single Philippines ETF, the AUM is multiplied by the ETF's expense ratio. All values are in U.S. dollars.
ETF issuers are ranked based on their aggregate 3-month fund flows of their ETFs with exposure to Philippines. 3-month fund flows is a metric that can be used to gauge the perceived popularity amongst investors of different ETF issuers with ETFs that have exposure to Philippines. All values are in U.S. dollars.
ETF issuers are ranked based on their AUM-weighted average 3-month return of their ETFs with exposure to Philippines. In addition to price performance, the 3-month return assumes the reinvestment of all dividends during the last 3 months.
ETF issuers are ranked based on their aggregate assets under management (AUM) of their ETFs with exposure to Philippines. All values are in U.S. dollars.
ETF issuers are ranked based on their AUM-weighted average expense ratios of their ETFs with exposure to Philippines. The lower the average expense ratio of all U.S.-listed Philippines ETFs for a given issuer, the higher the rank.
ETF issuers are ranked based on their AUM-weighted average dividend yield of their ETFs with exposure to Philippines.
Barclays Global Fund Advisors - - - - - - 1 iShares
Screen ETFs based on asset class, issuer, market cap, expense ratio, and more. To analyze all Philippines ETFs in the ETF screener go here.
This resource allows comparisons of potential ETF investments, such as Philippines ETFs, on a wide range of criteria including expenses, performance, dividend yield and volatility.
This tool allows investors to compare two ETFs head-to-head, presenting holdings, performance metrics, and other data. You can use this tool to compare Philippines ETFs with each other or with other country ETFs. For example, to compare EPHE and SPY, go here.
ETF Database's Country Exposure tool allows investors to identify equity ETFs that offer exposure to a specified country. For example, if you're looking for all ETFs with exposure to Philippines, you can find the results here.
You can view all the ETFs in the same ETF Database category as the majority of Philippines ETFs, which is Asia Pacific Equities. If you don't only want Asia Pacific Equities view all ETF Database categories here.
Nick Peters-GoldenJan 27, 2023
Another week went by, and there was a clear standout issuer in the ETF Issuer League: JPMorgan...
VettaFi Voices: Looking Forward to Exchange
Evan HarpJan 27, 2023
The conference of the year, Exchange, rapidly approaches. This week, the VettaFi Voices grab...
Commodities Have Their Say as Bonds Soar
This week's ETF flows saw a strong performance from bond ETFs, but nothing like this – bond...
ETF of the Week: SPDR S&P 500 ETF Trust
Happy birthday, SPDR S&P 500 ETF Trust (SPY A)! The granddaddy of all ETFs just turned 30 and...
Join the Finance Community to Support These Non-Profits at Exchange
Jon FeeJan 26, 2023
On February 5, the financial community will come together at Exchange in Miami Beach, Florida to...
Philippines Research
Q&A Interviews
QS Investors' President Discusses Legg Mason's New International Low Volatility High Dividend ETF
Neelarjo RakshitOct 17, 2016
At the Morningstar ETF Conference, we gained insights from James Norman, President of QS...
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,254
|
Travel Crest History
In the winter of 2011, the idea to develop a new crest for our Travel Soccer program was brought up in one of our General Membership Meetings. It was unanimous that we should move forward with the idea, and in late summer, 2011, a small committee was formed to develop the basic concept and design for a new crest. Special thanks to Travel Coordinator Kris Reiss, and club members Mike Suchanec and John Saugling, who worked with me on the initial Crest Development Committee. We took on the mission of "developing a fresh, new crest for QSC Travel Soccer that will be easily recognizable as the "brand" of a serious, top-level soccer program." Our first step was to identify some existing logos and soccer crests that we liked, and pinpoint design elements that we wanted in our new crest, such as the basic shape, bold contrast, and possible color schemes.
We decided from the beginning that it was critical to have symbolism integrated throughout the design. For several months, we researched the history of the club and the town, and searched for some type of a "mascot" or symbolic figure to represent the club. We could not find anything that we felt strongly about, so we concluded that we would need to depend on bold shapes in the design, and that we would prominently feature the letters "QSC" as the brand of the club.
The colors were not chosen lightly – navy blue, white, dark red, black and metallic gold. We chose to feature the colors navy blue and white to carry on the tradition of the club. Both black and dark red represent the quality and attitude of our players on the field, since both symbolize strength and aggression. In addition, the red represents courage and speed, and was also chosen because Quakertown is often called the "Heart of Upper Bucks County". The gold is symbolic of hope, success and optimism, and also represents "the gold standard", - something QSC is striving to be in the soccer community.
Once we had the general design concepts and symbolism finalized, we solicited the general club membership for design ideas – and received several helpful ideas. Then, in early December I met with Graphic Designer Gary Knize, from Harleysville, to create the final logo. Gary and I went through at least a dozen different versions and updates after our initial meeting, during which time I gathered important feedback from the QSC Board.
Our committee had expressed a strong desire to create something that was fresh and modern, but still brought to mind the antiquity and deep history of our town. Working with Gary, I believed we have achieved our goal through the combination of different text styles – the modern boldness of the letters QSC, and the old-style script font used for "Quakertown, PA" on the banner. In addition to colors, we incorporated other types of symbolism: three white stripes on the blue background, to symbolize the three townships within the geographic area of the club – Richland, Milford and Haycock. We also included three stars, to represent the three Boroughs within club boundaries – Quakertown, Trumbauersville and Richlandtown. My favorite part of the logo is the use of the stars to "suggest" a soccer ball, giving it the inescapable feel of a soccer crest without coming right out and displaying a soccer ball.
The end result is a crest design that I think we can all wear with pride! In my opinion it is clearly one of the best around. I look forward to seeing it grace the uniforms of our travel players next year, and for many years to come.
Andy Graham
President, Quakertown Soccer Club
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 8,092
|
Do you ever find yourself saying, "I have no idea what to draw!"?
Are you struggling with defining your own artistic point of view?
Do you worry your work is too similar to what's already out there?
Are you ready to find your own perspective as an artist, but need a push?
Idea Generation: Expanding Your Own Creative Repertoire & Finding Your Voice is a three-part, easy to follow video e-course designed for you!
You can take this course from anywhere in the world at your own pace! It includes pre-recorded videos, along with a written summary of video content, exercises and resources in an easily downloadable PDF.
This course is designed for aspiring, beginning or established artists who are looking for tools to further develop personal sources of inspiration and their own distinct perspective in their work. It's for people who would like to draw influence from the world differently in order to advance their own creative style.
Through this video course, artist and illustrator Lisa Congdon will guide you through practices for generating new ideas, using inspiration, and developing a body of work around subject matter that is meaningful to you — all with the goal of creating work that is interesting, specific to you and stands out in a world filled with prolific artists.
Part One: In the first class video, Lisa covers brainstorming as a way to generate new ideas about what to paint, draw or make. You'll focus on exploring the stuff you are already passionate about in your regular, everyday life as a basis for generating ideas for subject matter in your work. You can use brainstorming to begin to find your own distinct perspective and set your work apart. She'll also discuss the role of research to expand your ideas even further.
Part Two: In the second video of the class, Lisa talks about getting inspired by other artists, both tips and cautions. Nearly every artist has influences, and studying and even mimicking the work of other artists is a normal and natural part of becoming an artist. However, it's really important to take that inspiration and make it your own, in a very concerted and distinct way, before you claim it as your own or attempt to sell it. Lisa will discuss how to put boundaries around inspiration in order to move away from your influences and toward your own voice.
Part Three: The third and final class video, Lisa focuses on developing and following through with a personal creative challenge or set of challenges. The direct route to developing your own distinct voice as an artist is showing up and making/drawing/painting something at regular intervals — for a few days a week at least — even if it's just a few minutes a day. Creative challenges are the best way to engage in this kind of disciplined practice. Lisa has done many creative challenges in the past seven years, and she credits them with not only helping her to hone her own styles of drawing and painting, but also to generate interest in her work, including paying client work and gallery shows. She will talk you through practical tips for designing and embarking on your own creative challenges.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,213
|
{"url":"https:\/\/oracleyue.github.io\/2018\/05\/13\/emacs-setup-md\/","text":"This is a quick note on the setup of Emacs on Mac OS X.\n\n## Install Emacs on OS X\n\nThe simplest way to install Emacs on Mac is using Homebrew. Here is the commands:\n\nIf you want to try a patched Emacs with transparent titlebar, install as follows\n\n(You may add --without-spacemacs-icon, if you keep only one version of Emacs.)\n\n## Run Emacs as daemon\n\nIf you have run the Emacs, you can start an Emacs server by the emacs-lisp function (server-start). Then you could anywhere open files in or create a frame of this Emacs by emacsclient -nc. However, this is rarely used. An alternative way is to run Emacs as daemon\n\nand open emacsclient in the same way.\n\nIn my Mac, I run multiple Emacs servers (each Emacs process as daemon is called one server) to allow using multiple color themes (e.g. I use a light theme for text editing like Markdown, LaTeX, and use dark themes for coding). The way to do it is giving each daemon process a name so as to connect to it later via emacsclient:\n\nand later you could connect to them, e.g. the main server, by\n\nTo stop an Emacs server (the daemon process), you could call (kill-emacs) in any Emacs frame connecting to this daemon. Alternatively, use system command kill with process ID listed and grepped by ps aux | grep -i 'emacs --daemon'.\n\nTo quickly start Emacs as daemon, connect to and kill them, I wrote a bash script function emacs-server-func.sh on my github repository, which provides commands es, ec, ecc:\n\n\u2022 es: list all Emacs daemon processes\n\u2022 es start: start main and coding servers by default\n\u2022 es start m or es start main: start main server\n\u2022 es start c or es start coding: start coding server\n\u2022 es start YOURNAME: start a sever with your given name\n\u2022 es stop: stop all Emacs daemon processes\n\u2022 es stop m or es stop main: stop main server, similar for coding\n\u2022 es stop YOURNAME: start the sever with the given name\n\u2022 ec FILENAME: connecting to the main server and open the file FILENAME in a new frame\n\u2022 ecc FILENAME: connecting to the coding server and open the file FILENAME in a new frame.\n\nAlternatively, we could use Alfred.app to perform the same tasks. We need its PowerPack in order to build Workflows. Create an Input using Keyword, with the keyword es start (with Argument Optional, enabling with space), shown as\n\nThen, connect to an Action of Run Script, in which choose bash and fill the following codes\n\nSimilarly create workflows for es stop, ec, etc. The scripts are the same as the codes provided in emacs-server-func.sh. We just copy and paste them into Alfred and create shortcuts to run them. The whole workflow looks as\n\nand you could find the workflow, named Emacs.alfredworkflow, on my github repository.\n\nSince Mac OS X by default only allows one Emacs instance (not via daemon), we may use the follow code to open multiple ones, which is created as a workflow in Alfred triggered by em new:\n\n## Create an app via Automator to open files in Emacs clients by double-clicks\n\nAutomator is a built-in OS X app for creating custom automated user workflows for just about any installed app you might have or even OS functionality. Suppose we could like to open files by default in the Emacs client named as main. You may open files in the Emacs.app, however, with too much user-defined configurations, it is too slow to be acceptable. Using the daemon mode and opening in clients are a better choice.\n\nThe following shows how to create an app that allows opening files in Emacs clients.\n\n1. Launch Automator and create a new document. Select Application as its type.\n\n2. Search the Actions palette on the left for the Run AppleScript action and add it to your Automator document.\n\n3. In the Run AppleScript building block, copy and paste the following codes:\n\nand Save it with name, e.g. Open in Emacs.app in the file format Application.\n\n4. Now you go to the files with specific extensions, right-click it, choose Get Info and choose our app Open in Emacs.app as the application to open it. If you prefer to using Emacs to open all files of this extension, click Change All....\n\n## Load different themes for Emacs servers\/daemons\n\nIn practice, I prefer light themes for text editing (like Markdown, LaTeX, org mode) and dark themes for programming. To achieve, we run multiple Emacs daemon processes, e.g. main and coding. Now we will describe how to load different themes in your .emacs or init.el configurations.\n\nFirst of all, we need to prepare a few variables and loading functions:\n\nThese functions are essential since the theme has to be loaded when the frame is firstly created, instead of when the elisp files are loaded during emacs --daemon.\n\nThen we define different themes for each severs (the daemon processes), for examples:\n\nwhere these constant variables are defined as follows in advance for convenience:\n\nThe last step is to load or reload themes depending on specific Emacs running modes:","date":"2019-10-14 00:52:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.31091007590293884, \"perplexity\": 6787.263337540257}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986648481.7\/warc\/CC-MAIN-20191014003258-20191014030258-00424.warc.gz\"}"}
| null | null |
Q: Check if a file exists in the project in WinRT I have a WinRT Metro project which displays images based on a selected item. However, some of the images selected will not exist. What I want to be able to do is trap the case where they don't exist and display an alternative.
Here is my code so far:
internal string GetMyImage(string imageDescription)
{
string myImage = string.Format("Assets/MyImages/{0}.jpg", imageDescription.Replace(" ", ""));
// Need to check here if the above asset actually exists
return myImage;
}
Example calls:
GetMyImage("First Picture");
GetMyImage("Second Picture");
So Assets/MyImages/SecondPicture.jpg exists, but Assets/MyImages/FirstPicture.jpg does not.
At first I thought of using the WinRT equivalent of File.Exists(), but there doesn't appear to be one. Without having to go to the extent of trying to open the file and catching an error, can I simply check if either the file exists, or the file exists in the project?
A: There's two ways you can handle it.
1) Catch the FileNotFoundException when trying to get the file:
Windows.Storage.StorageFolder installedLocation =
Windows.ApplicationModel.Package.Current.InstalledLocation;
try
{
// Don't forget to decorate your method or event with async when using await
var file = await installedLocation.GetFileAsync(fileName);
// Exception wasn't raised, therefore the file exists
System.Diagnostics.Debug.WriteLine("We have the file!");
}
catch (System.IO.FileNotFoundException fileNotFoundEx)
{
System.Diagnostics.Debug.WriteLine("File doesn't exist. Use default.");
}
catch (Exception ex)
{
// Handle unknown error
}
2) as mydogisbox recommends, using LINQ. Although the method I tested is slightly different:
Windows.Storage.StorageFolder installedLocation =
Windows.ApplicationModel.Package.Current.InstalledLocation;
var files = await installedLocation.GetFilesAsync(CommonFileQuery.OrderByName);
var file = files.FirstOrDefault(x => x.Name == fileName);
if (file != null)
{
System.Diagnostics.Debug.WriteLine("We have the file!");
}
else
{
System.Diagnostics.Debug.WriteLine("No File. Use default.");
}
A: BitmapImage has an ImageFailed event that fires if the image can't be loaded. This would let you try to load the original image, and then react if it's not there.
Of course, this requires that you instantiate the BitmapImage yourself, rather than just build the Uri.
A: You could use GetFilesAsync from here to enumerate the existing files. This seems to make sense considering you have multiple files which might not exist.
Gets a list of all files in the current folder and its sub-folders. Files are filtered and sorted based on the specified CommonFileQuery.
var folder = await StorageFolder.GetFolderFromPathAsync("Assets/MyImages/");
var files = await folder.GetFilesAsync(CommonFileQuery.OrderByName);
var file = files.FirstOrDefault(x => x.Name == "fileName");
if (file != null)
{
//do stuff
}
Edit:
As @Filip Skakun pointed out, the resource manager has a resource mapping on which you can call ContainsKey which has the benefit of checking for qualified resources as well (i.e. localized, scaled etc).
Edit 2:
Windows 8.1 introduced a new method for getting files and folders:
var result = await ApplicationData.Current.LocalFolder.TryGetItemAsync("fileName") as IStorageFile;
if (result != null)
//file exists
else
//file doesn't exist
A: Sample checking for resource availability for c++ /cx (tested with Windows Phone 8.1):
std::wstring resPath = L"Img/my.bmp";
std::wstring resKey = L"Files/" + resPath;
bool exists = Windows::ApplicationModel::Resources::Core::ResourceManager::Current->MainResourceMap->HasKey(ref new Platform::String(resKey.c_str()));
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,733
|
I Can't Give You Anything but Love è un singolo dei cantanti statunitensi Tony Bennett e Lady Gaga, pubblicato il 18 agosto 2014 come secondo estratto dall'album in studio Cheek to Cheek.
Tracce
Classifiche
Note
Collegamenti esterni
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,438
|
My word count for the week just gone is about the same as the previous week. Roughly 700 words a day. Not at all bad considering I was on holiday in Stockholm over the weekend.
Last week, I wrote my first few words in the story of Elin/Helen/Helena. I'm going to have to find a better working title than that – it doesn't exactly roll off the tongue. Nor off the keyboard. No matter. I wrote the words, about 550 of them.
I was thinking about her and her situation. When she came over to England and chose to stay, she had no one she knew, she relied on the people who were around her. They gave her the network she would surely have needed. She joined the court, Elizabeth was her companion/elder sister/protector. Her value to the queen was her very lack of contacts. It was an astute move of the queen to annex Elin as her intimate, her bedfellow. Elin was guaranteed not to be a part of any of the different court factions – she relied on the queen and no one else.
But then 10 years on, she has become established at court, she is an adult in her own right with her own income from the estates of her former husband. (I need to check if that's true.) English has become her first language. And now her marriage to Thomas allies her with one of the court factions. Elizabeth kicks her out – but how well anchored is she in fact in this society? In my imagination she finds herself very isolated – at least to begin with.
I was thinking about this and nodded off in my armchair, half in a dream myself, I imagined her dreaming on the night after the removal to her new lodgings, in a house somewhere in London near the Tower. She would be troubled, confused, to a degree lost. She would see herself again as the child she once was, and her religious and cultural convictions would bring her to accuse herself of the sin of pride, of trying to be better than she is, of setting herself up against the queen – just as her waking common sense would assure her of the opposite. It is not pride, at least, not her own pride that has brought her to this.
I woke and did some household chores and read and lay down to sleep properly, but couldn't, got up again and scribbled down notes about Elin's dream. Then I slept, but the following morning I was able to write the scene. My first grip on the story. I didn't write any more, but the questions multiply from this: She's lodging in London, why? Why not with friends? Why not with Thomas's family? And where is she lodging? Near the Tower, OK is that going to be very salubrious? Probably not? Does she have no one to talk with? No servant? No friends? The records speak of her getting help from one of the lords at court. Who? Why? There is also a connection to John Dee. Was he a help to her?
I imagine she is pregnant but does not know it yet – she is sick and the thing she remembers is the seasickness crossing the Channel when she was 15. How likely is it she would not know? She was not an only child and had younger brothers and sisters, so would have observed motherhood first hand in her own family, but she left home when she was 14 or 15, so how much does she really know about the practicalities? I need to find out about maternity customs in Elizabethan England. I have already found some pictures of maternity clothing – such as it was. Front lacing dresses for the rich or letting out the seams of your clothes, or slitting them and wearing aprons for most of the rest.
Revisited and revised for spelling and SEO fine-tuning. 12-15 Jan 2017.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,954
|
Q: Google Chart API not able to set dataview columns dynamicaly I am creating a line chart based on google API, and i want to do column to display based on user selection. i am using dataview, if i set the data column in code it works, but if i try to set it dynamically it doesn't work, as in sample below
var dataViewMean = new google.visualization.DataView(dataTable);
//dataViewMean.setColumns([check_vals.join()]); //This doesn't work
dataViewMean.setColumns([0, 1, 2, 3, 4]); //This works
here is my html
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<div id="chart_index" style="width: 400px; height: 200px;"></div>
<div>
<div id="chart_mean" style="width: 400px; height: 200px;"></div>
<div>
<table cellpadding="0" cellspacing="15" border="0">
<tr>
<td>
<div>
<input name="sChk" type="checkbox" class="chkVals" value="1" checked="checked" /> Current</div>
</td>
<td>
<div>
<input name="sChk" type="checkbox" class="chkVals" value="2" checked="checked" /> 95th Percentile</div>
</td>
<td>
<div>
<input name="sChk" type="checkbox" class="chkVals" value="3" checked="checked" /> Free Flow</div>
</td>
<td>
<div>
<input name="sChk" type="checkbox" class="chkVals" value="4" checked="checked" /> Mean</div>
</td>
</tr>
</table>
</div>
and below is my script
<script type="text/javascript">
$(function() {
$(".chkVals").on('click', function(e) {
//alert("hi");
drawTTCharts();
});
});
google.load('visualization', '1', {
packages: ['linechart']
});
google.setOnLoadCallback(drawTTCharts);
function drawTTCharts() {
var chartTitle;
// Create and populate the data table.
var dataTable = new google.visualization.DataTable();
// checkbox select for choosing the column to draw travel time chart
var check_vals = [];
check_vals.push(0);
for (var i = 0; i < document.getElementsByName("sChk").length; i++) {
if (document.getElementsByName("sChk")[i].checked) {
check_vals.push(document.getElementsByName("sChk")[i].value);
}
}
//Declare column Index Chart: Testing data view
dataTable.addColumn('string', 'Start Time');
dataTable.addColumn('number', 'Current Travel Time');
dataTable.addColumn('number', '95th Percentile Travel Time');
dataTable.addColumn('number', 'Free Flow Travel Time');
dataTable.addColumn('number', 'Mean Travel Time');
dataTable.addColumn('number', 'Travel Time Index');
dataTable.addColumn('number', 'Planning Time Index');
dataTable.addColumn('number', 'Buffer Time Index');
dataTable.addRows([
['02:20:00', 114.60, 98, 118.00, 114.88, 1.17, .03, 1.20],
['02:25:00', 113.00, 98, 117.25, 115.01, 1.17, .02, 1.20],
['02:30:00', 113.00, 98, 118.00, 115.06, 1.17, .03, 1.20],
['02:35:00', 113.00, 98, 117.80, 114.77, 1.17, .03, 1.20],
['02:40:00', 112.80, 98, 119.00, 114.72, 1.17, .04, 1.21],
['02:45:00', 110.40, 98, 119.00, 115.68, 1.18, .03, 1.21],
['02:50:00', 109.00, 98, 118.00, 114.74, 1.17, .03, 1.20],
['02:55:00', 113.00, 98, 117.60, 114.37, 1.17, .03, 1.20],
['03:00:00', 115.00, 98, 117.60, 114.90, 1.17, .03, 1.20],
['03:05:00', 115.00, 98, 117.40, 114.53, 1.17, .03, 1.20],
['03:10:00', 115.00, 98, 118.00, 114.58, 1.17, .03, 1.20],
['03:15:00', 115.00, 98, 117.80, 114.93, 1.17, .03, 1.20],
['03:20:00', 115.00, 98, 116.80, 114.15, 1.16, .03, 1.19],
['03:25:00', 115.00, 98, 116.00, 113.83, 1.16, .02, 1.18],
['03:30:00', 110.80, 98, 117.00, 114.06, 1.16, .03, 1.19],
['03:35:00', 108.00, 98, 117.00, 114.03, 1.16, .03, 1.19]
]);
var dataViewMean = new google.visualization.DataView(dataTable);
//alert(check_vals.join());
//dataViewMean.setColumns([check_vals.join()]); //This doesn't work
dataViewMean.setColumns([0, 1, 2, 3, 4]); //This works
var dataViewIndex = new google.visualization.DataView(dataTable);
dataViewIndex.setColumns([0, 5, 6, 7]); //Select columns for chart
drawMeanChart(dataViewMean);
drawIndexChart(dataViewIndex);
}
function drawMeanChart(pDataMean) {
//alert(pDataMean);
// Create and draw Mean the visualization.
new google.visualization.LineChart(document.getElementById('chart_mean')).
draw(pDataMean, {
curveType: "function",
//title: chartTitle,
width: 445,
height: 220,
backgroundColor: '#FFEBC2',
vAxis: {
title: 'Travel Time (Sec.)'
},
hAxis: {
title: 'Time Interval'
},
legend: {
position: 'top'
} //,
//series: { 2: { targetAxisIndex: 1 }, 3: { targetAxisIndex: 1 }, 3: { targetAxisIndex: 1} },
//vAxes: { 1: { title: 'Index'} }
});
}
function drawIndexChart(pDataIndex) {
// Create and draw Index the visualization.
new google.visualization.LineChart(document.getElementById('chart_index')).
draw(pDataIndex, {
curveType: "function",
//title: chartTitle,
width: 445,
height: 220,
backgroundColor: '#CDE6FF',
vAxis: {
title: 'Index'
},
hAxis: {
title: 'Time Interval'
},
legend: {
position: 'top'
}
});
}
</script>
please help guys, thanks in advance.
my complete code in in JSFiddle http://jsfiddle.net/brijesh26/e6Fnm/
A: check_vals is already an array. By calling the join method, you are creating a string from the array, and adding it to another array, so you pass an array containing a string to the setColumns method. This:
dataViewMean.setColumns([check_vals.join()]);
is equivalent to this:
dataViewMean.setColumns(['1,2,3,4...']);
just pass check_vals by itself:
dataViewMean.setColumns(check_vals);
[Edit]
After further examination, check_vals is an array of strings, so you should be parsing them as integers when filling the array:
check_vals.push(parseInt(document.getElementsByName("sChk")[i].value));
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,004
|
Главная»DOI»№1 2020»Cleaning of oil from hydrogen sulfide in hydrocyclone processors
DOI 10.32758/2071-5951-2020-0-01-32-34
Guevara E. С., Marushkin A.B., Sidorok P.V., Gilmutdinov A.T.
(Ufa State Petroleum Technical University – UGNTU, Ufa, ONIX-«Samara», Samara) E-mail: eimysguevara@gmail.com
Очистка нефти от сероводорода в гидроциклонных процессорах
Keywords: crude oil, hydrogen sulfide cleaning, hydrocyclone processor, non regenerable hydrogen sulfide neutralizer.
The results of a pilot run to assess the effectiveness of removing hydrogen sulfide from Bashkortostan oils in a hydrocyclone processor are presented. Oil was used, part of which was processed at ELOU, and the other at the oil treatment unit (UPN). The hydrogen sulfide content in them was 132.3 and 365.5 ppm, respectively. The cleaning efficiency was estimated as a decrease in the content of hydrogen sulfide before and after hydrocyclone relative to the initial one. For oil prepared at ELOU, subsequent hydrocyclone at temperatures of 40 and 60 ºС provides the efficiency of its removal of 56.5 and 61.9 %, respectively. For oil treated at the UPN unit, when it is refined in the field of centrifugal forces at temperatures of 50 and 60 ºС, a greater efficiency of its removal is achieved — 63.3 and 69.4%, respectively. Hydrocyclone of the oil prepared at the UPN in the selected technological mode did not allow reducing the content of hydrogen sulfide in it to the requirements of GOST 51858. Alternatively, it is possible to purify such oil in two stages: first, hydrocyclone, and then treatment with an non regenerable reagent-neutralizer, for example, "Darsan-N". At the same time, due to the removal of a significant part of the hydrogen sulfide in the first stage, the reagent consumption is more than halved compared to the treatment of the original oil. All samples of oil refined in a centrifugal force field comply with GOST requirements for saturated vapor pressure. Hydrocyclone was carried out at 40-60 ºС, which is 100-120 degrees higher than the boiling point of hydrogen sulfide. At the same time, its residual content in oil exceeded the requirements of the standard up to two times. The most probable cause of this phenomenon is apparently associated with the formation of donor-acceptor complexes of hydrogen sulfide with oil components. Therefore, the option of purification from hydrogen sulfide with the sequential use of two different technological methods seems logical.
1.Strunkin S.I., Gladkova N.Kh., Kuz'mina E.P., Matveev V.V., Kaemov S.A. Experience of application of hydrocyclone processor for removal of hydrogen sulfide and light mercaptans in JSC Samaraneftegaz. Scientific and technical Bulletin of NK Rosneft. 2014, 35, № 2, pp. 75-77.
2.Andrianov V.M Process of stabilization of oil in the field of centrifugal forces. Dissertation for candidate of technical Sciences. Seraphim, Ufa. 1987, p. 210.
3.Marushkin B.K., Pruchay V.S. The Potential of stable oil. Oil industry. 1983, № 9, pp. 69-71.
4.GOST 50802. Method for determination of hydrogen sulfide, methyl and ethyl mercaptans.
5.GOST 51858. Oil. General technical conditions.
6.Ed. Nikolsky B.I. Chemistry Handbook. Leningrad, Chemistry, 1964. 1167 p. 7.Chemical Encyclopedia T. III, M.: Soviet Encyclopedia, 1992, pp. 223-225.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,487
|
What's up, friends?! I know it's been a while since I've written. I hope you are doing well and your New Year is off to a blessed start. What are your New Year's goals for 2019? This year, my theme for 2019 is growth and voice. I am excited because I am on a journey… Read More It Feels So Good to Be Back!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,955
|
Q: why function is not defined javascript My functions in html file doesn't work with error I have written highter. When I change it from
funtion name(...) to window.name = function (...) but it doesn't work anyway and write
"write.name is not finction"
my code
<meta charset="UTF-8">
<title>Title</title>
<script type="text/javascript">
function changeText(id) {
id.innerHTML = "Ooops!";
}
function displayDate () {
document.getElementById("demo").innerHTML = Date();
}
function addField () {
document.getElementById("demo").innerHTML = Date();
}
function addInput(divName){
var newdiv = document.createElement('div');
newdiv.innerHTML = "<br>Select File to Upload:<br><input type="file" onchange="addInput('dynamicInput');" name="fileName">";
document.getElementById(divName).appendChild(newdiv);
window.displayDate();
window.addField();
}
</script>
</head>
<body>
<h1 id="fff" onclick="changeText(this);">Click on this text!</h1>
<form action="/getimage" method="post" enctype="multipart/form-data">
<div id="dynamicInput">
<br>Select File to Upload:<br><input type="file" onchange="addInput('dynamicInput');" name="fileName">
</div>
<input type="submit" value="Upload">
<input type="button" value="Add another input" onClick="addInput('dynamicInput');">
</form>
<p id="demo"></p>
</body>
Please, help!
A: Its quotes problem you need escape the \' and string like this onchange="addInput(\'dynamicInput\');" .And window.name is not defined in your dom
function changeText(id) {
id.innerHTML = "Ooops!";
}
function displayDate() {
document.getElementById("demo").innerHTML = Date();
}
function addField() {
document.getElementById("demo").innerHTML = Date();
}
function addInput(divName) {
var newdiv = document.createElement('div');
newdiv.innerHTML = '<br>Select File to Upload:<br><input type="file" onchange="addInput(\'dynamicInput\');" name="fileName">';
document.getElementById(divName).appendChild(newdiv);
window.displayDate();
window.addField();
}
<h1 id="fff" onclick="changeText(this);">Click on this text!</h1>
<form action="/getimage" method="post" enctype="multipart/form-data">
<div id="dynamicInput">
<br>Select File to Upload:<br><input type="file" onchange="addInput('dynamicInput');" name="fileName">
</div>
<input type="submit" value="Upload">
<input type="button" value="Add another input" onClick="addInput('dynamicInput');">
</form>
<p id="demo"></p>
A: Please note you need to escape double quotes to use them within another double quotes, or use single quotes instead.
newdiv.innerHTML = "<br>Select File to Upload:<br><input type='file' onchange='addInput('dynamicInput');' name='fileName'>";
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,381
|
Bonello – variante del nome proprio di persona italiano Bono
Bonello – cognome italiano
Persone
Bertrand Bonello – regista, sceneggiatore e compositore francese
Georges Bonello – calciatore francese
Giovanni Bonello – giudice maltese
Henry Bonello – calciatore maltese
John Bonello – ex calciatore maltese
Matteo Bonello – nobile normanno
Salvu Bonello – ex calciatore maltese
Vittorio Bonello – calciatore italiano
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,436
|
Description: Genuine Ed Hardy Bangle. This Ed Hardy Blue Lucite Bangle with Love Kills Slowly Logo is new and never worn, it has the Ed Hardy and Love Kills Slowly logo on it. Measures 1 inch wide and the opening is over 2 1/2 inches.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,505
|
Home Movie Reviews Ghost in the Shell – Movie Review
Ghost in the Shell – Movie Review
Release date: March 31, 2017 (USA)
Director: Rupert Sanders
Music composed by: Clint Mansell, Lorne Balfe
Producers: Avi Arad, Steven Paul
Chris Stuckmann reviews Ghost in the Shell, starring Scarlett Johansson, Pilou Asbaek, Takeshi Kitano, Juliette Binoche, Michael Pitt. Directed by Rupert Sanders.
In the near future, Major (Scarlett Johansson) is the first of her kind: a human who is cyber-enhanced to be a perfect soldier devoted to stopping the world's most dangerous criminals. When terrorism reaches a new level that includes the ability to hack into people's minds and control them, Major is uniquely qualified to stop it. As she prepares to face a new enemy, Major discovers that she has been lied to, and her life was not saved. Instead, it was stolen.
Wonder Woman – Movie Review
Ghost in the Shell Official Trailer
Scarlett Johansson Signs On to Star in DreamWorks' 'Ghost in the Shell'
Ghost In The Shell (2017) – Official Trailer – Paramount Pictures
tags: 2017 action Anime Artisit entertainment Ghost in the shell movie review
previous Life - Movie Review
next Guardians of the Galaxy Vol. 2 - Movie Review
Top Gun: Maverick – Movie Review
The Northman Review
The Batman – Movie Review
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,455
|
Bike, pro cycling
Tour de Jank, Stage 9* – Burma Road Hill Repeats
14 July 2010 jank 1 Comment
The * for today represents me kind of giving up. I'm still owing a cobbled stage for stage 3, and I missed yesterday – bad day at work, and just kind of unmotivated when I got home.
Today, though, I'm energized again – good talk with my supervisor, better talk with his boss, and some clarification of what the heck I'm supposed to be doing that's largely in line with what I want to be doing. So a setback's not always a setback, often it's just a re-direction.
The ride today – Finally got out of the office and onto the roads for a lunchtime set of hill repeats. North on Burma Road, four times up the (very meager) hill at the north end of Burma, each time in an smaller cog starting from granny gear, and back down Burma at about 85% of sustainable effort. Legs felt great, I finished all 4 hill repeats, and I had a negative split on the ride back with an average speed over 20. It's relatively flat on Burma, and I'm pretty sure there's no elevation change, but I'll check that when I pull the ride off of my Garmin tonight.
"Bastille Day, when 800 rebels stormed a guard of 100, lost 98 men, and freed just 7 prisoners, inspires French tactics to this day." – @cyclocosm
It's that time of year, again, where we look outside our borders to the second greatest event in sport, the Tour de France. 21~ish days of French countryside, podium girls, and skinny guys in tights with a tolerance for pain beyond anything an offensive lineman could ever comprehend.
This year's race is already a classic – Lance Armstrong's packed at least 8 years of bad luck into his last tour, we've seen one of the favorites from Luxembourg taken out by cobblestone roads in northern France that pre-date Napoleon, and the God of Thunder duke it out with a Manx man for the title of fastest man alive.
As the race enters its middle week, it's just come out of the Alps and will be streaking across the middle of France towards the Pyrenees, where, in the 100th anniversary of the race's first visit to these mountains, a Spaniard riding for a Kazakh team will try to stay ahead of a small guy from a small country.
And 10 days from now, the race will end with champagne on the Champs d'Elysees.
Next year? I think I'm buying a projector and stringing up a sheet between the trees in my backyard – watch the whole thing with a cold beer sitting in the backyard.
2010DeJankTour
Bike, Cycling - Pro
Tour de Jank, Stage 8 – Old Mystic to Ice Cream
11 July 2010 jank
Who cares? This is why we did 10 miles on the trail-a-bike:
So, it turns out that Lance Armstrong is human, after all – more than 10 minutes down on the overall race. Tough, tough day for him today.
But, Lance aside, the stage was brilliant. Andy Schleck riding away from Contador at the finish of the stage made me happy as a clam. Good, good race.
Tour de Jank, Stage 7 – Newport
There's a couple of rides around here that are completely epic, and of which I don't think I'll ever get tired. Today's stage was one of them – the loop around southwest Aquidneck Island, specifically Ocean Drive. Sure, there's some traffic downtown, but in general, it's awesome.
Headed out with Tracy, who's in crazy good shape. Part of the highlight for me is riding through downtown Newport – there's enough traffic that it's going slower than you can go on a bike, and it's a trip to dodge traffic. Once you're past downtown, Ocean Drive weaves along the Atlantic, and takes you into Bellview Drive and the legendary mansions.
Newport Loop
Find more Bike Rides in Newport, Rhode Island
The last bit of the ride is out Burma Road – undeveloped property that the Navy's been hoarding for a while.
This is why I love cycling.
First day in the mountains, and, as expected, not much shakeup.
Bike, Cycling - Pro, pro cycling
Tour de Jank Stage 6, Jamestown Island
Lovely, lovely ride Friday Evening – about 23 MPH around Jamestown Island, 16.9 MPH. The ride's feeling good – the spin is coming back, and I'm digging on it.
Tonight was almost perfect, 'cept for locking my keys in the car at the end of the ride, and needing to get Missy to come bail me out. But, sitting and watching the day fade into twilight was perfect.
CAV! Man, another bunch sprint. Beautiful.
Tour de Jank 2010, Stage 5 – Lame to Lamer
The Ride –
I must confess, sports fans, I almost blew it on Thursday. Good, good stuff at work, but I let it get in the way of lunchtime hill repeats.
Then, since number one son is headed off for the first time to sleepaway camp (Way up north in Connecticut, nearby April Anne) this weekend, and since I'm keeping the nation slightly safer for democracy this weekend, we headed down to Costello's for some fried clammy goodness and family bonding before he goes away for weeks at a time. I'm including pictures taken from our seats that I took while we ate. Did I mention it's BYOB?
Oh, right – the ride… Yeah, so today's ride was exceptionally lame – I pulled out the trainer once we got home and got the kids in bed. Did about 45 minutes in front of the TeeVee listening to Bob Roll and Craig Gummer yammering on. But – I got 'er dun.
I love the variety in cycling. The beauty of the mountains counterpointed by the terrible suffering going on in the faces of the climbers. The sea of colors as the peleton crosses fields in bloom.
But what hooked me was watching bunch sprints. During Armstrong's first couple of tours, Mario Cipollini duked it out with Eric Zabel, Tom Steels, Stuart O'Grady. Absolutely nothing like watching 40 guys charging down a straight at 35 or 40 miles per hour, Paul Sherwin and Phil Liggett shouting from the television, horns and fans clapping. Takes you straight back to the first time you rode your bike with a friend in the neighborhood – who's fastest to the next driveway?
Today lived up to that – huge bunch sprint, with the "Manx Missile", Mark Cavendish from the Isle of Man, running away from everyone.
TourTour De Jank 2010
Tour de Jank, Stage 4* – River Road and Neighborhood Crit
8 July 2010 jank
The ride was a short one for Stage 4 – just 9.5 miles needed. I headed down to River Road in the rapidly fading twilight, rode it into town, turned around, and rode back. Once I got to the Neighborhood, I did two laps of the place at roughly a mile each to bring the total up to 11 miles. Such an overachiever.
Man, lots to talk about in the Tour. First, there was the disaster that was Stage 2 – who freaking cancels 20 km of racing due to rain? Though, to be fair, there was a huge amount of carnage in the peleton – VdV out, everyone down. But the interview with Thor Hushvold after he got his stage win stolen from him…
Stage 3 was no less gripping – Thor coming back from insult in stage 3 to take the stage on the cobbles; the remaining Schleck stunning the world by riding well, Lance Armstrong losing time and giving yet another classic sound bite ("Riding between cars and trying to pass, eating dirt – literally!")
And Stage 4 was visually stunning – through fields of sunflowers, and ending with a full-on bunch sprint without major carnage.
Man, what a race
*I'm dink on Stage 3 – busy at work getting back on Tuesday, plus I'm looking for a suitable set of cobbles to ride. And, speaking of "busy at work" – Dave has a wonderful post about what a load of BS that is.
leTourTour De Jank 2010
Tour de Jank Stage 2 – Old Mystic-Noank-Mystic-Old Mystic
Since the TDF stage 1 included a couple of bumps, I rolled up route 184 to Flanders Road, and headed down to Noank. Man, this is one of my favorite roads around – it's a pretty good climb up to the top of Flanders hill, and then nothing but flying down to Noank, all while looking out over Fisher's Island Sound.
Tour de Jank stage 2
Find more Bike Rides in Mystic, Connecticut
I think this whole idea might work – I'm absolutely loving riding the bike again. I'm going to try to squeeze in a run this evening – love running in the hot.
Stage 1 didn't dissapoint at all. I cannot get enough of watching the run-in to the finish. Favorite moment of the day was watching Cavendish shrug and wave at the camera after his crash. Dunno if it's just me, but it seems like the peleton has been getting into a better mood between last year and this year. The optimist in me wants to think that this is a result of a cleaner pack – less cases of road rage.
I'm digging Bob Roll's rider interviews – here's yesterday's with Tyler Farrar, who's going to be one to watch in the next few years
Today's stage, Stage 2, doesn't look to have much potential to shake stuff up. Bunch of climbs, but nothing terrible that can really shake up the general classification
Tour de Jank 2010 – Stage 1, Old Mystic-Quanaduck Cove-Old Mystic
OK, so the premise behind the Tour de Jank is that, as the Tour de France is going on across the pond, I'm going to do a tenth of the tour distance every day, 'cept in and around New England. Ideally, I'll get back into posting pictures, but who knows. My guess is that this lasts all of one or two days.
I'll probably be a day behind talking about the actual race, as I tend to catch the coverage in the evenings after the kiddos are in bed.
Stage 1 – Old Mystic-Quanaduck Cove-Old Mystic
Stage 1 was a sprinter's stage, relatively flat run along Route 27 to Route 1 out to Quanaduck Cove in Stonington, and returning along roughly the same route, 'cept coming up the Groton side of the Mystic River after a delay for the drawbridge. Legs felt good, bike was a little squeaky. Pretty good average speed – above 16 MPH for the whole ride.
TDF Prologue / Stage 1
First, a couple of suggestions on following the Tour.
In the US, Versus (the bull-riding and hockey channel who I may never forgive for pre-empting the finish of one of their Tour of California stages for a hockey pre-game (PRE GAME!) show) owns the coverage. Phil Liggett, Paul Sherwin, Bob Roll, and an increasingly competent Craig Gummer do the commentary. They've put a bit of a paywall in front of much of their internet coverage, but it's not too shabby. I may try their iPhone app once my phone arrives.
My current favorite semi-pro cycling commentary comes from Cosmo Catalano at Cyclocosm (shout out to the Nutmeg State!). His twitter feed (@Cyclocosm) is phenomenal. Here's an example of Cosmo's work.
The Luckiest Man in the Peloton from Cosmo Catalano on Vimeo.
For the best of scrounged video, check out Cyclingfans. Their twitter feed is pretty good, too.
The prologue was good. Usually, I hate the time trial stages – just sitting around watching guys suffer without any strategy other than "Ride. Ride fast." (Courtesy of Missy)
I'd also recommend following @LeviLeipheimer and @dzabriskie – Leiphimer because he's freaking amazing, and Zabriskie because he's pure gonzo cycling.
But the prologue was an exception. Maybe because it was only 10 minutes of effort per rider, or maybe because Lance FREAKING Armstrong came in 4th, finishing in front of Alberto Contador. I'm pulling for Armstrong, partly because I'm an ignorant American, and partially because I refuse to acknowledge that 38 is over the hill. (And, 'cause I've picked up a similar amount of grey hair in the last year).
I'll admit I'm a bit behind in pre-read for this year's tour, because life has been a little hectic (in a good way). But, as opposed to most years, it's good to get in front of this year's Tour. Tomorrow and Tuesday are going to hit a good chunk of the roads in Belgium and northern France that are ridden in the Spring Classic races, and should end up shaking up the General Classification (The thing that Armstrong's won 7 times) much earlier than most years.
The spring classics are huge one-day races held in March and April, nasty weather months in Northern Europe. Think rain and cobblestones. They're also wicked long – Paris-Roubaix is close to 200 miles, with about 20 miles of cobblestones through places you heard about in World War 1 histories.
The race favorites are going to be trying to stay ahead of the pack, as with close to 200 riders going 25 MPH on cobblestones, it's likely that there are going to be some major crashes. It's also likely that some of the wafer-thin climbers are going to be sorted out, as having a little bit of butt helps out on bumpy roads.
I did catch today's sprint finish, complete with massive crashes in the run-up to the sprint, one apparently caused by a wayward dog. Good on Allesandro Pettachi, who's been out for a couple of years after a nasty, nasty crash. Good start to the tour.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,286
|
// Code generated by protoc-gen-gogo. DO NOT EDIT.
// source: mixer/template/apikey/template_handler_service.proto
/*
Package apikey is a generated protocol buffer package.
The `apikey` template represents a single API key, which is used for authorization checks.
Example config:
```yaml
apiVersion: "config.istio.io/v1alpha2"
kind: apikey
metadata:
name: validate-apikey
namespace: istio-system
spec:
api: api.service | ""
api_version: api.version | ""
api_operation: api.operation | ""
api_key: api.key | ""
timestamp: request.time
```
The `apikey` template represents a single API key, used to authorize API calls.
It is generated from these files:
mixer/template/apikey/template_handler_service.proto
It has these top-level messages:
HandleApiKeyRequest
InstanceMsg
Type
InstanceParam
*/
package apikey
import proto "github.com/gogo/protobuf/proto"
import fmt "fmt"
import math "math"
import _ "github.com/gogo/protobuf/gogoproto"
import _ "istio.io/api/mixer/adapter/model/v1beta1"
import google_protobuf1 "github.com/gogo/protobuf/types"
import istio_mixer_adapter_model_v1beta11 "istio.io/api/mixer/adapter/model/v1beta1"
import istio_policy_v1beta1 "istio.io/api/policy/v1beta1"
import context "golang.org/x/net/context"
import grpc "google.golang.org/grpc"
import strings "strings"
import reflect "reflect"
import io "io"
// Reference imports to suppress errors if they are not otherwise used.
var _ = proto.Marshal
var _ = fmt.Errorf
var _ = math.Inf
// This is a compile-time assertion to ensure that this generated file
// is compatible with the proto package it is being compiled against.
// A compilation error at this line likely means your copy of the
// proto package needs to be updated.
const _ = proto.GoGoProtoPackageIsVersion2 // please upgrade the proto package
// Request message for HandleApiKey method.
type HandleApiKeyRequest struct {
// 'apikey' instance.
Instance *InstanceMsg `protobuf:"bytes,1,opt,name=instance" json:"instance,omitempty"`
// Adapter specific handler configuration.
//
// Note: Backends can also implement [InfrastructureBackend][https://istio.io/docs/reference/config/mixer/istio.mixer.adapter.model.v1beta1.html#InfrastructureBackend]
// service and therefore opt to receive handler configuration during session creation through [InfrastructureBackend.CreateSession][TODO: Link to this fragment]
// call. In that case, adapter_config will have type_url as 'google.protobuf.Any.type_url' and would contain string
// value of session_id (returned from InfrastructureBackend.CreateSession).
AdapterConfig *google_protobuf1.Any `protobuf:"bytes,2,opt,name=adapter_config,json=adapterConfig" json:"adapter_config,omitempty"`
// Id to dedupe identical requests from Mixer.
DedupId string `protobuf:"bytes,3,opt,name=dedup_id,json=dedupId,proto3" json:"dedup_id,omitempty"`
}
func (m *HandleApiKeyRequest) Reset() { *m = HandleApiKeyRequest{} }
func (*HandleApiKeyRequest) ProtoMessage() {}
func (*HandleApiKeyRequest) Descriptor() ([]byte, []int) {
return fileDescriptorTemplateHandlerService, []int{0}
}
// Contains instance payload for 'apikey' template. This is passed to infrastructure backends during request-time
// through HandleApiKeyService.HandleApiKey.
type InstanceMsg struct {
// Name of the instance as specified in configuration.
Name string `protobuf:"bytes,72295727,opt,name=name,proto3" json:"name,omitempty"`
// The API being called (api.service).
Api string `protobuf:"bytes,1,opt,name=api,proto3" json:"api,omitempty"`
// The version of the API (api.version).
ApiVersion string `protobuf:"bytes,2,opt,name=api_version,json=apiVersion,proto3" json:"api_version,omitempty"`
// The API operation is being called.
ApiOperation string `protobuf:"bytes,3,opt,name=api_operation,json=apiOperation,proto3" json:"api_operation,omitempty"`
// API key used in API call.
ApiKey string `protobuf:"bytes,4,opt,name=api_key,json=apiKey,proto3" json:"api_key,omitempty"`
// Timestamp of API call.
Timestamp *istio_policy_v1beta1.TimeStamp `protobuf:"bytes,5,opt,name=timestamp" json:"timestamp,omitempty"`
}
func (m *InstanceMsg) Reset() { *m = InstanceMsg{} }
func (*InstanceMsg) ProtoMessage() {}
func (*InstanceMsg) Descriptor() ([]byte, []int) {
return fileDescriptorTemplateHandlerService, []int{1}
}
// Contains inferred type information about specific instance of 'apikey' template. This is passed to
// infrastructure backends during configuration-time through [InfrastructureBackend.CreateSession][TODO: Link to this fragment].
type Type struct {
}
func (m *Type) Reset() { *m = Type{} }
func (*Type) ProtoMessage() {}
func (*Type) Descriptor() ([]byte, []int) { return fileDescriptorTemplateHandlerService, []int{2} }
// Represents instance configuration schema for 'apikey' template.
type InstanceParam struct {
// The API being called (api.service).
Api string `protobuf:"bytes,1,opt,name=api,proto3" json:"api,omitempty"`
// The version of the API (api.version).
ApiVersion string `protobuf:"bytes,2,opt,name=api_version,json=apiVersion,proto3" json:"api_version,omitempty"`
// The API operation is being called.
ApiOperation string `protobuf:"bytes,3,opt,name=api_operation,json=apiOperation,proto3" json:"api_operation,omitempty"`
// API key used in API call.
ApiKey string `protobuf:"bytes,4,opt,name=api_key,json=apiKey,proto3" json:"api_key,omitempty"`
// Timestamp of API call.
Timestamp string `protobuf:"bytes,5,opt,name=timestamp,proto3" json:"timestamp,omitempty"`
}
func (m *InstanceParam) Reset() { *m = InstanceParam{} }
func (*InstanceParam) ProtoMessage() {}
func (*InstanceParam) Descriptor() ([]byte, []int) {
return fileDescriptorTemplateHandlerService, []int{3}
}
func init() {
proto.RegisterType((*HandleApiKeyRequest)(nil), "apikey.HandleApiKeyRequest")
proto.RegisterType((*InstanceMsg)(nil), "apikey.InstanceMsg")
proto.RegisterType((*Type)(nil), "apikey.Type")
proto.RegisterType((*InstanceParam)(nil), "apikey.InstanceParam")
}
// Reference imports to suppress errors if they are not otherwise used.
var _ context.Context
var _ grpc.ClientConn
// This is a compile-time assertion to ensure that this generated file
// is compatible with the grpc package it is being compiled against.
const _ = grpc.SupportPackageIsVersion4
// Client API for HandleApiKeyService service
type HandleApiKeyServiceClient interface {
// HandleApiKey is called by Mixer at request-time to deliver 'apikey' instances to the backend.
HandleApiKey(ctx context.Context, in *HandleApiKeyRequest, opts ...grpc.CallOption) (*istio_mixer_adapter_model_v1beta11.CheckResult, error)
}
type handleApiKeyServiceClient struct {
cc *grpc.ClientConn
}
func NewHandleApiKeyServiceClient(cc *grpc.ClientConn) HandleApiKeyServiceClient {
return &handleApiKeyServiceClient{cc}
}
func (c *handleApiKeyServiceClient) HandleApiKey(ctx context.Context, in *HandleApiKeyRequest, opts ...grpc.CallOption) (*istio_mixer_adapter_model_v1beta11.CheckResult, error) {
out := new(istio_mixer_adapter_model_v1beta11.CheckResult)
err := grpc.Invoke(ctx, "/apikey.HandleApiKeyService/HandleApiKey", in, out, c.cc, opts...)
if err != nil {
return nil, err
}
return out, nil
}
// Server API for HandleApiKeyService service
type HandleApiKeyServiceServer interface {
// HandleApiKey is called by Mixer at request-time to deliver 'apikey' instances to the backend.
HandleApiKey(context.Context, *HandleApiKeyRequest) (*istio_mixer_adapter_model_v1beta11.CheckResult, error)
}
func RegisterHandleApiKeyServiceServer(s *grpc.Server, srv HandleApiKeyServiceServer) {
s.RegisterService(&_HandleApiKeyService_serviceDesc, srv)
}
func _HandleApiKeyService_HandleApiKey_Handler(srv interface{}, ctx context.Context, dec func(interface{}) error, interceptor grpc.UnaryServerInterceptor) (interface{}, error) {
in := new(HandleApiKeyRequest)
if err := dec(in); err != nil {
return nil, err
}
if interceptor == nil {
return srv.(HandleApiKeyServiceServer).HandleApiKey(ctx, in)
}
info := &grpc.UnaryServerInfo{
Server: srv,
FullMethod: "/apikey.HandleApiKeyService/HandleApiKey",
}
handler := func(ctx context.Context, req interface{}) (interface{}, error) {
return srv.(HandleApiKeyServiceServer).HandleApiKey(ctx, req.(*HandleApiKeyRequest))
}
return interceptor(ctx, in, info, handler)
}
var _HandleApiKeyService_serviceDesc = grpc.ServiceDesc{
ServiceName: "apikey.HandleApiKeyService",
HandlerType: (*HandleApiKeyServiceServer)(nil),
Methods: []grpc.MethodDesc{
{
MethodName: "HandleApiKey",
Handler: _HandleApiKeyService_HandleApiKey_Handler,
},
},
Streams: []grpc.StreamDesc{},
Metadata: "mixer/template/apikey/template_handler_service.proto",
}
func (m *HandleApiKeyRequest) Marshal() (dAtA []byte, err error) {
size := m.Size()
dAtA = make([]byte, size)
n, err := m.MarshalTo(dAtA)
if err != nil {
return nil, err
}
return dAtA[:n], nil
}
func (m *HandleApiKeyRequest) MarshalTo(dAtA []byte) (int, error) {
var i int
_ = i
var l int
_ = l
if m.Instance != nil {
dAtA[i] = 0xa
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(m.Instance.Size()))
n1, err := m.Instance.MarshalTo(dAtA[i:])
if err != nil {
return 0, err
}
i += n1
}
if m.AdapterConfig != nil {
dAtA[i] = 0x12
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(m.AdapterConfig.Size()))
n2, err := m.AdapterConfig.MarshalTo(dAtA[i:])
if err != nil {
return 0, err
}
i += n2
}
if len(m.DedupId) > 0 {
dAtA[i] = 0x1a
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.DedupId)))
i += copy(dAtA[i:], m.DedupId)
}
return i, nil
}
func (m *InstanceMsg) Marshal() (dAtA []byte, err error) {
size := m.Size()
dAtA = make([]byte, size)
n, err := m.MarshalTo(dAtA)
if err != nil {
return nil, err
}
return dAtA[:n], nil
}
func (m *InstanceMsg) MarshalTo(dAtA []byte) (int, error) {
var i int
_ = i
var l int
_ = l
if len(m.Api) > 0 {
dAtA[i] = 0xa
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.Api)))
i += copy(dAtA[i:], m.Api)
}
if len(m.ApiVersion) > 0 {
dAtA[i] = 0x12
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiVersion)))
i += copy(dAtA[i:], m.ApiVersion)
}
if len(m.ApiOperation) > 0 {
dAtA[i] = 0x1a
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiOperation)))
i += copy(dAtA[i:], m.ApiOperation)
}
if len(m.ApiKey) > 0 {
dAtA[i] = 0x22
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiKey)))
i += copy(dAtA[i:], m.ApiKey)
}
if m.Timestamp != nil {
dAtA[i] = 0x2a
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(m.Timestamp.Size()))
n3, err := m.Timestamp.MarshalTo(dAtA[i:])
if err != nil {
return 0, err
}
i += n3
}
if len(m.Name) > 0 {
dAtA[i] = 0xfa
i++
dAtA[i] = 0xd2
i++
dAtA[i] = 0xe4
i++
dAtA[i] = 0x93
i++
dAtA[i] = 0x2
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.Name)))
i += copy(dAtA[i:], m.Name)
}
return i, nil
}
func (m *Type) Marshal() (dAtA []byte, err error) {
size := m.Size()
dAtA = make([]byte, size)
n, err := m.MarshalTo(dAtA)
if err != nil {
return nil, err
}
return dAtA[:n], nil
}
func (m *Type) MarshalTo(dAtA []byte) (int, error) {
var i int
_ = i
var l int
_ = l
return i, nil
}
func (m *InstanceParam) Marshal() (dAtA []byte, err error) {
size := m.Size()
dAtA = make([]byte, size)
n, err := m.MarshalTo(dAtA)
if err != nil {
return nil, err
}
return dAtA[:n], nil
}
func (m *InstanceParam) MarshalTo(dAtA []byte) (int, error) {
var i int
_ = i
var l int
_ = l
if len(m.Api) > 0 {
dAtA[i] = 0xa
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.Api)))
i += copy(dAtA[i:], m.Api)
}
if len(m.ApiVersion) > 0 {
dAtA[i] = 0x12
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiVersion)))
i += copy(dAtA[i:], m.ApiVersion)
}
if len(m.ApiOperation) > 0 {
dAtA[i] = 0x1a
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiOperation)))
i += copy(dAtA[i:], m.ApiOperation)
}
if len(m.ApiKey) > 0 {
dAtA[i] = 0x22
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.ApiKey)))
i += copy(dAtA[i:], m.ApiKey)
}
if len(m.Timestamp) > 0 {
dAtA[i] = 0x2a
i++
i = encodeVarintTemplateHandlerService(dAtA, i, uint64(len(m.Timestamp)))
i += copy(dAtA[i:], m.Timestamp)
}
return i, nil
}
func encodeVarintTemplateHandlerService(dAtA []byte, offset int, v uint64) int {
for v >= 1<<7 {
dAtA[offset] = uint8(v&0x7f | 0x80)
v >>= 7
offset++
}
dAtA[offset] = uint8(v)
return offset + 1
}
func (m *HandleApiKeyRequest) Size() (n int) {
var l int
_ = l
if m.Instance != nil {
l = m.Instance.Size()
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
if m.AdapterConfig != nil {
l = m.AdapterConfig.Size()
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.DedupId)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
return n
}
func (m *InstanceMsg) Size() (n int) {
var l int
_ = l
l = len(m.Api)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiVersion)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiOperation)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiKey)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
if m.Timestamp != nil {
l = m.Timestamp.Size()
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.Name)
if l > 0 {
n += 5 + l + sovTemplateHandlerService(uint64(l))
}
return n
}
func (m *Type) Size() (n int) {
var l int
_ = l
return n
}
func (m *InstanceParam) Size() (n int) {
var l int
_ = l
l = len(m.Api)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiVersion)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiOperation)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.ApiKey)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
l = len(m.Timestamp)
if l > 0 {
n += 1 + l + sovTemplateHandlerService(uint64(l))
}
return n
}
func sovTemplateHandlerService(x uint64) (n int) {
for {
n++
x >>= 7
if x == 0 {
break
}
}
return n
}
func sozTemplateHandlerService(x uint64) (n int) {
return sovTemplateHandlerService(uint64((x << 1) ^ uint64((int64(x) >> 63))))
}
func (this *HandleApiKeyRequest) String() string {
if this == nil {
return "nil"
}
s := strings.Join([]string{`&HandleApiKeyRequest{`,
`Instance:` + strings.Replace(fmt.Sprintf("%v", this.Instance), "InstanceMsg", "InstanceMsg", 1) + `,`,
`AdapterConfig:` + strings.Replace(fmt.Sprintf("%v", this.AdapterConfig), "Any", "google_protobuf1.Any", 1) + `,`,
`DedupId:` + fmt.Sprintf("%v", this.DedupId) + `,`,
`}`,
}, "")
return s
}
func (this *InstanceMsg) String() string {
if this == nil {
return "nil"
}
s := strings.Join([]string{`&InstanceMsg{`,
`Api:` + fmt.Sprintf("%v", this.Api) + `,`,
`ApiVersion:` + fmt.Sprintf("%v", this.ApiVersion) + `,`,
`ApiOperation:` + fmt.Sprintf("%v", this.ApiOperation) + `,`,
`ApiKey:` + fmt.Sprintf("%v", this.ApiKey) + `,`,
`Timestamp:` + strings.Replace(fmt.Sprintf("%v", this.Timestamp), "TimeStamp", "istio_policy_v1beta1.TimeStamp", 1) + `,`,
`Name:` + fmt.Sprintf("%v", this.Name) + `,`,
`}`,
}, "")
return s
}
func (this *Type) String() string {
if this == nil {
return "nil"
}
s := strings.Join([]string{`&Type{`,
`}`,
}, "")
return s
}
func (this *InstanceParam) String() string {
if this == nil {
return "nil"
}
s := strings.Join([]string{`&InstanceParam{`,
`Api:` + fmt.Sprintf("%v", this.Api) + `,`,
`ApiVersion:` + fmt.Sprintf("%v", this.ApiVersion) + `,`,
`ApiOperation:` + fmt.Sprintf("%v", this.ApiOperation) + `,`,
`ApiKey:` + fmt.Sprintf("%v", this.ApiKey) + `,`,
`Timestamp:` + fmt.Sprintf("%v", this.Timestamp) + `,`,
`}`,
}, "")
return s
}
func valueToStringTemplateHandlerService(v interface{}) string {
rv := reflect.ValueOf(v)
if rv.IsNil() {
return "nil"
}
pv := reflect.Indirect(rv).Interface()
return fmt.Sprintf("*%v", pv)
}
func (m *HandleApiKeyRequest) Unmarshal(dAtA []byte) error {
l := len(dAtA)
iNdEx := 0
for iNdEx < l {
preIndex := iNdEx
var wire uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
wire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
fieldNum := int32(wire >> 3)
wireType := int(wire & 0x7)
if wireType == 4 {
return fmt.Errorf("proto: HandleApiKeyRequest: wiretype end group for non-group")
}
if fieldNum <= 0 {
return fmt.Errorf("proto: HandleApiKeyRequest: illegal tag %d (wire type %d)", fieldNum, wire)
}
switch fieldNum {
case 1:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Instance", wireType)
}
var msglen int
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
msglen |= (int(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
if msglen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + msglen
if postIndex > l {
return io.ErrUnexpectedEOF
}
if m.Instance == nil {
m.Instance = &InstanceMsg{}
}
if err := m.Instance.Unmarshal(dAtA[iNdEx:postIndex]); err != nil {
return err
}
iNdEx = postIndex
case 2:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field AdapterConfig", wireType)
}
var msglen int
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
msglen |= (int(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
if msglen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + msglen
if postIndex > l {
return io.ErrUnexpectedEOF
}
if m.AdapterConfig == nil {
m.AdapterConfig = &google_protobuf1.Any{}
}
if err := m.AdapterConfig.Unmarshal(dAtA[iNdEx:postIndex]); err != nil {
return err
}
iNdEx = postIndex
case 3:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field DedupId", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.DedupId = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
default:
iNdEx = preIndex
skippy, err := skipTemplateHandlerService(dAtA[iNdEx:])
if err != nil {
return err
}
if skippy < 0 {
return ErrInvalidLengthTemplateHandlerService
}
if (iNdEx + skippy) > l {
return io.ErrUnexpectedEOF
}
iNdEx += skippy
}
}
if iNdEx > l {
return io.ErrUnexpectedEOF
}
return nil
}
func (m *InstanceMsg) Unmarshal(dAtA []byte) error {
l := len(dAtA)
iNdEx := 0
for iNdEx < l {
preIndex := iNdEx
var wire uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
wire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
fieldNum := int32(wire >> 3)
wireType := int(wire & 0x7)
if wireType == 4 {
return fmt.Errorf("proto: InstanceMsg: wiretype end group for non-group")
}
if fieldNum <= 0 {
return fmt.Errorf("proto: InstanceMsg: illegal tag %d (wire type %d)", fieldNum, wire)
}
switch fieldNum {
case 1:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Api", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.Api = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 2:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiVersion", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiVersion = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 3:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiOperation", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiOperation = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 4:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiKey", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiKey = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 5:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Timestamp", wireType)
}
var msglen int
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
msglen |= (int(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
if msglen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + msglen
if postIndex > l {
return io.ErrUnexpectedEOF
}
if m.Timestamp == nil {
m.Timestamp = &istio_policy_v1beta1.TimeStamp{}
}
if err := m.Timestamp.Unmarshal(dAtA[iNdEx:postIndex]); err != nil {
return err
}
iNdEx = postIndex
case 72295727:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Name", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.Name = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
default:
iNdEx = preIndex
skippy, err := skipTemplateHandlerService(dAtA[iNdEx:])
if err != nil {
return err
}
if skippy < 0 {
return ErrInvalidLengthTemplateHandlerService
}
if (iNdEx + skippy) > l {
return io.ErrUnexpectedEOF
}
iNdEx += skippy
}
}
if iNdEx > l {
return io.ErrUnexpectedEOF
}
return nil
}
func (m *Type) Unmarshal(dAtA []byte) error {
l := len(dAtA)
iNdEx := 0
for iNdEx < l {
preIndex := iNdEx
var wire uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
wire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
fieldNum := int32(wire >> 3)
wireType := int(wire & 0x7)
if wireType == 4 {
return fmt.Errorf("proto: Type: wiretype end group for non-group")
}
if fieldNum <= 0 {
return fmt.Errorf("proto: Type: illegal tag %d (wire type %d)", fieldNum, wire)
}
switch fieldNum {
default:
iNdEx = preIndex
skippy, err := skipTemplateHandlerService(dAtA[iNdEx:])
if err != nil {
return err
}
if skippy < 0 {
return ErrInvalidLengthTemplateHandlerService
}
if (iNdEx + skippy) > l {
return io.ErrUnexpectedEOF
}
iNdEx += skippy
}
}
if iNdEx > l {
return io.ErrUnexpectedEOF
}
return nil
}
func (m *InstanceParam) Unmarshal(dAtA []byte) error {
l := len(dAtA)
iNdEx := 0
for iNdEx < l {
preIndex := iNdEx
var wire uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
wire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
fieldNum := int32(wire >> 3)
wireType := int(wire & 0x7)
if wireType == 4 {
return fmt.Errorf("proto: InstanceParam: wiretype end group for non-group")
}
if fieldNum <= 0 {
return fmt.Errorf("proto: InstanceParam: illegal tag %d (wire type %d)", fieldNum, wire)
}
switch fieldNum {
case 1:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Api", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.Api = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 2:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiVersion", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiVersion = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 3:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiOperation", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiOperation = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 4:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field ApiKey", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.ApiKey = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
case 5:
if wireType != 2 {
return fmt.Errorf("proto: wrong wireType = %d for field Timestamp", wireType)
}
var stringLen uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
stringLen |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
intStringLen := int(stringLen)
if intStringLen < 0 {
return ErrInvalidLengthTemplateHandlerService
}
postIndex := iNdEx + intStringLen
if postIndex > l {
return io.ErrUnexpectedEOF
}
m.Timestamp = string(dAtA[iNdEx:postIndex])
iNdEx = postIndex
default:
iNdEx = preIndex
skippy, err := skipTemplateHandlerService(dAtA[iNdEx:])
if err != nil {
return err
}
if skippy < 0 {
return ErrInvalidLengthTemplateHandlerService
}
if (iNdEx + skippy) > l {
return io.ErrUnexpectedEOF
}
iNdEx += skippy
}
}
if iNdEx > l {
return io.ErrUnexpectedEOF
}
return nil
}
func skipTemplateHandlerService(dAtA []byte) (n int, err error) {
l := len(dAtA)
iNdEx := 0
for iNdEx < l {
var wire uint64
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return 0, ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return 0, io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
wire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
wireType := int(wire & 0x7)
switch wireType {
case 0:
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return 0, ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return 0, io.ErrUnexpectedEOF
}
iNdEx++
if dAtA[iNdEx-1] < 0x80 {
break
}
}
return iNdEx, nil
case 1:
iNdEx += 8
return iNdEx, nil
case 2:
var length int
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return 0, ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return 0, io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
length |= (int(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
iNdEx += length
if length < 0 {
return 0, ErrInvalidLengthTemplateHandlerService
}
return iNdEx, nil
case 3:
for {
var innerWire uint64
var start int = iNdEx
for shift := uint(0); ; shift += 7 {
if shift >= 64 {
return 0, ErrIntOverflowTemplateHandlerService
}
if iNdEx >= l {
return 0, io.ErrUnexpectedEOF
}
b := dAtA[iNdEx]
iNdEx++
innerWire |= (uint64(b) & 0x7F) << shift
if b < 0x80 {
break
}
}
innerWireType := int(innerWire & 0x7)
if innerWireType == 4 {
break
}
next, err := skipTemplateHandlerService(dAtA[start:])
if err != nil {
return 0, err
}
iNdEx = start + next
}
return iNdEx, nil
case 4:
return iNdEx, nil
case 5:
iNdEx += 4
return iNdEx, nil
default:
return 0, fmt.Errorf("proto: illegal wireType %d", wireType)
}
}
panic("unreachable")
}
var (
ErrInvalidLengthTemplateHandlerService = fmt.Errorf("proto: negative length found during unmarshaling")
ErrIntOverflowTemplateHandlerService = fmt.Errorf("proto: integer overflow")
)
func init() {
proto.RegisterFile("mixer/template/apikey/template_handler_service.proto", fileDescriptorTemplateHandlerService)
}
var fileDescriptorTemplateHandlerService = []byte{
// 546 bytes of a gzipped FileDescriptorProto
0x1f, 0x8b, 0x08, 0x00, 0x00, 0x00, 0x00, 0x00, 0x02, 0xff, 0xbc, 0x53, 0x31, 0x6f, 0xd3, 0x40,
0x14, 0xb6, 0xdb, 0x90, 0x36, 0x97, 0x06, 0xa1, 0x6b, 0x11, 0x6e, 0x40, 0xd7, 0x2a, 0x0c, 0x74,
0x40, 0x67, 0xb5, 0xb0, 0x21, 0x86, 0xb6, 0x0b, 0x55, 0x85, 0x40, 0x6e, 0xc5, 0xc2, 0x10, 0x5d,
0xec, 0x57, 0xf7, 0x14, 0xdb, 0x77, 0xd8, 0x97, 0xa8, 0xde, 0x10, 0xbf, 0x00, 0xa9, 0x3b, 0x33,
0x1b, 0x7f, 0x80, 0x1f, 0x50, 0x31, 0x55, 0x88, 0x81, 0x05, 0x89, 0x98, 0x0e, 0x8c, 0x1d, 0x19,
0x91, 0xcf, 0xe7, 0x10, 0x10, 0x62, 0x64, 0xbb, 0xf7, 0xde, 0x77, 0xef, 0xbd, 0xef, 0xbb, 0xef,
0xd0, 0xfd, 0x98, 0x9f, 0x40, 0xea, 0x2a, 0x88, 0x65, 0xc4, 0x14, 0xb8, 0x4c, 0xf2, 0x21, 0xe4,
0xd3, 0xb8, 0x7f, 0xcc, 0x92, 0x20, 0x82, 0xb4, 0x9f, 0x41, 0x3a, 0xe6, 0x3e, 0x50, 0x99, 0x0a,
0x25, 0x70, 0xb3, 0x82, 0x75, 0x57, 0x42, 0x11, 0x0a, 0x9d, 0x72, 0xcb, 0x53, 0x55, 0xed, 0xde,
0xad, 0x7a, 0xb2, 0x80, 0x49, 0x05, 0xa9, 0x1b, 0x8b, 0x00, 0x22, 0x77, 0xbc, 0x39, 0x00, 0xc5,
0x36, 0x5d, 0x38, 0x51, 0x90, 0x64, 0x5c, 0x24, 0x99, 0x41, 0xaf, 0x86, 0x42, 0x84, 0x11, 0xb8,
0x3a, 0x1a, 0x8c, 0x8e, 0x5c, 0x96, 0xe4, 0xa6, 0x74, 0xe7, 0x5f, 0x8d, 0xfc, 0x63, 0xf0, 0x87,
0x75, 0x0f, 0x29, 0x22, 0xee, 0xe7, 0xd3, 0x9a, 0xca, 0xa5, 0x59, 0xb5, 0xf7, 0xc6, 0x46, 0xcb,
0x8f, 0x34, 0x89, 0x6d, 0xc9, 0xf7, 0x21, 0xf7, 0xe0, 0xc5, 0x08, 0x32, 0x85, 0x5d, 0xb4, 0xc8,
0x93, 0x4c, 0xb1, 0xc4, 0x07, 0xc7, 0x5e, 0xb7, 0x37, 0xda, 0x5b, 0xcb, 0xb4, 0x62, 0x45, 0xf7,
0x4c, 0xfe, 0x71, 0x16, 0x7a, 0x53, 0x10, 0x7e, 0x80, 0xae, 0x9a, 0x45, 0xfa, 0xbe, 0x48, 0x8e,
0x78, 0xe8, 0xcc, 0xe9, 0x6b, 0x2b, 0xb4, 0x22, 0x40, 0x6b, 0x02, 0x74, 0x3b, 0xc9, 0xbd, 0x8e,
0xc1, 0xee, 0x6a, 0x28, 0x5e, 0x45, 0x8b, 0x01, 0x04, 0x23, 0xd9, 0xe7, 0x81, 0x33, 0xbf, 0x6e,
0x6f, 0xb4, 0xbc, 0x05, 0x1d, 0xef, 0x05, 0xbd, 0x4f, 0x36, 0x6a, 0xcf, 0x4c, 0xc4, 0xd7, 0xd0,
0x3c, 0x93, 0x5c, 0xef, 0xd4, 0xf2, 0xca, 0x23, 0x5e, 0x43, 0x6d, 0x26, 0x79, 0x7f, 0x0c, 0x69,
0xa9, 0x9b, 0x1e, 0xdb, 0xf2, 0x10, 0x93, 0xfc, 0x59, 0x95, 0xc1, 0xb7, 0x51, 0xa7, 0x04, 0x08,
0x09, 0x29, 0x53, 0x25, 0xa4, 0x1a, 0xb1, 0xc4, 0x24, 0x7f, 0x52, 0xe7, 0xf0, 0x0d, 0xb4, 0x50,
0x82, 0x86, 0x90, 0x3b, 0x0d, 0x5d, 0x2e, 0x1f, 0x71, 0x1f, 0x72, 0xfc, 0x10, 0xb5, 0x14, 0x8f,
0x21, 0x53, 0x2c, 0x96, 0xce, 0x15, 0xcd, 0x69, 0x8d, 0xf2, 0x4c, 0x71, 0x41, 0x2b, 0x59, 0xa9,
0x91, 0x95, 0x1e, 0xf2, 0x18, 0x0e, 0x4a, 0x98, 0xf7, 0xeb, 0x06, 0xbe, 0x8e, 0x1a, 0x09, 0x8b,
0xc1, 0x79, 0xf7, 0xe1, 0x7d, 0x4f, 0xf7, 0xd5, 0x61, 0xaf, 0x89, 0x1a, 0x87, 0xb9, 0x84, 0x52,
0xff, 0x4e, 0x4d, 0xef, 0x29, 0x4b, 0x59, 0xfc, 0xff, 0x09, 0xde, 0xfa, 0x93, 0x60, 0x6b, 0x66,
0xff, 0xad, 0xf4, 0x77, 0x7f, 0x1c, 0x54, 0x46, 0xc7, 0xcf, 0xd1, 0xd2, 0x6c, 0x1a, 0xdf, 0xac,
0xdd, 0xf1, 0x17, 0x33, 0x75, 0xa9, 0xd1, 0x4b, 0xfb, 0x95, 0x9a, 0xa7, 0xa7, 0xda, 0xaf, 0x53,
0xf1, 0x76, 0x4b, 0xbf, 0x7a, 0x90, 0x8d, 0x22, 0xb5, 0xb3, 0x73, 0x36, 0x21, 0xd6, 0xf9, 0x84,
0x58, 0x9f, 0x27, 0xc4, 0xba, 0x9c, 0x10, 0xeb, 0x65, 0x41, 0xec, 0xb7, 0x05, 0xb1, 0xce, 0x0a,
0x62, 0x9f, 0x17, 0xc4, 0xfe, 0x5a, 0x10, 0xfb, 0x7b, 0x41, 0xac, 0xcb, 0x82, 0xd8, 0xaf, 0xbf,
0x11, 0xeb, 0xc7, 0xc7, 0x8b, 0xd3, 0x39, 0xeb, 0xd5, 0x97, 0x8b, 0xd3, 0x39, 0xf3, 0xf7, 0x06,
0x4d, 0xed, 0xb7, 0x7b, 0x3f, 0x03, 0x00, 0x00, 0xff, 0xff, 0xa8, 0xe3, 0x58, 0x9c, 0xc2, 0x03,
0x00, 0x00,
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,020
|
{"url":"http:\/\/slideplayer.com\/slide\/1732197\/","text":"Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n\nPresentation on theme: \"Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\"\u2014 Presentation transcript:\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\nExponents and Polynomials 10 10.1 Integers as Exponents 10.2 Working with Exponents 10.3 Scientific Notation 10.4 Addition and Subtraction of Polynomials 10.5 Introduction to Multiplying and Factoring Polynomials Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 2\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. b Multiply and divide using scientific notation. c Solve applied problems using scientific notation. c Convert temperatures from Celsius to Fahrenheit and from Fahrenheit to Celsius. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 3\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. Scientific notation is especially useful when calculations involve very large or very small numbers. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 4\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation SCIENTIFIC NOTATION a Scientific notation for a number is an expression of the type where n is an integer, M is greater than or equal to 1 and less than 10 and M is expressed in decimal notation. Convert between scientific notation and decimal notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 5\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation In scientific notation, a positive power of 10 indicates a large number (greater than or equal to 10) and a negative power of 10 indicates a small number (between 0 and 1). a Convert between scientific notation and decimal notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 6\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. Converting from scientific notation to decimal notation involves multiplying by a power of 10. To convert to decimal notation, we move the decimal point. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 7\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 8\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. We reverse the process when converting from decimal notation to scientific notation. To convert a number to scientific notation, we write it in the form where M looks like the original number but with exactly one nonzero digit to the immediate left of the decimal point. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 9\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation a Convert between scientific notation and decimal notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 10\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation b Multiply and divide using scientific notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 11\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation b Multiply and divide using scientific notation. 5 Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 12\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation b Multiply and divide using scientific notation. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 13\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation b Multiply and divide using scientific notation. 8 Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 14\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation b Multiply and divide using scientific notation. 8 Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 15\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation c Solve applied problems using scientific notation. 9 Distance from the Sun to Earth. Light from the sun traveling at a rate of 300,000 km\/s (kilometers per second) reaches Earth in 499 sec. Find the distance, expressed in scientific notation, from the sun to Earth. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 16\n\nCopyright 2012, 2008, 2004, 2000 Pearson Education, Inc.\n10.3 Scientific Notation c Solve applied problems using scientific notation. 9 Distance from the Sun to Earth. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 17","date":"2018-01-20 12:18:30","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9261956810951233, \"perplexity\": 1393.1255219194272}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-05\/segments\/1516084889567.48\/warc\/CC-MAIN-20180120102905-20180120122905-00510.warc.gz\"}"}
| null | null |
{"url":"https:\/\/meta.mathoverflow.net\/questions\/1939\/are-these-questions-of-mine-seemingly-too-hard-for-mse-suitable-for-mo","text":"# Are these questions of mine, (seemingly too hard for MSE?), suitable for MO?\n\nThe questions at issue are this and this. If they get no answer even after the end of bounties, might I post to MO? And how about evaluating $\\displaystyle \\sum_{m=0}^\\infty \\sum_{n=1}^\\infty \\frac{(-1)^mn}{2^n+m}$?\n\n\u2022 Why are you interested in this particular series, and what kind of answer would you expect -- a numeric approximation, or what else? \u2013\u00a0Stefan Kohl Oct 20 '14 at 20:20\n\u2022 @StefanKohl The three of these series come up in a proof I'm developing, though I'm afraid I don't have the means to work out them. And... well, I would need exact results, but I guess it's asking too much? \u2013\u00a0Vincenzo Oliva Oct 20 '14 at 20:26\n\u2022 Taking an arbitrary convergent series, typically its value has no \"closed form expression\". -- Do you know any reasons why for your series such expressions should exist? \u2013\u00a0Stefan Kohl Oct 20 '14 at 20:48\n\u2022 @StefanKohl None besides being part of an expression (whose other terms I'm trying to find on my own) that gives a precise value. At worst, I would not refuse numeric approximations and proofs of rationality or irrationality of the values of the series. \u2013\u00a0Vincenzo Oliva Oct 20 '14 at 21:07\n\u2022 @StefanKohl So? Are they suitable for MO? \u2013\u00a0Vincenzo Oliva Oct 21 '14 at 3:20\n\u2022 The series looks kind of like Lambert series. So maybe you could ask if the series can be expressed in terms of named quantities like that. Did you try Wolframalpha.com? \u2013\u00a0Bj\u00f8rn Kjos-Hanssen Oct 21 '14 at 3:56\n\u2022 @Bj\u00f8rnKjos-Hanssen Yes, and it doesn't even compute $\\displaystyle \\sum_{n=1}^\\infty \\frac{n}{2^n} - \\sum_{n=1}^\\infty \\frac{n}{2^n+1}$. Or at least, it will take centuries. (The same goes for the series involving the Lerch transcendent, while the other, which is the subseries in the one displayed in the body of the question, does get computed, but only approximately, and as I said, I would need at least a proof of irrationality or rationality) \u2013\u00a0Vincenzo Oliva Oct 21 '14 at 4:27\n\u2022 @Bj\u00f8rnKjos-Hanssen Truthfully, W|A does compute the equivalent $\\displaystyle \\sum_{n=1}^\\infty \\frac{n}{2^n}-\\frac{n}{2^n+1}$, but, I would need to do this infinitely many times (the software doesn't understand my infinite sum of infinite sums) and I would have no proof of the (ir)rationality of it. \u2013\u00a0Vincenzo Oliva Oct 21 '14 at 5:42\n\u2022 Your double series is convergent but not absolutely convergent, so you have to be careful with your manipulations. Asking for the exact value or rationality of a difficult series might be too difficult (as such things are generally unknown), but if you are willing to accept estimates, you might get an answer. \"Too hard for MSE\" can mean \"impossible\" instead of \"interesting for MO\". \u2013\u00a0Joonas Ilmavirta Oct 21 '14 at 8:36\n\u2022 @JonasIlmavirta I see. I am willing To accept estimates, but the proof of (ir)rationality of the values is something I need. Over all, will my questions be downvoted on MO? \u2013\u00a0Vincenzo Oliva Oct 21 '14 at 9:54\n\u2022 Depends on how you pose your questions. It's hard to tell unless you can give an example of a complete question here (assuming it's not very long). I don't know how I would vote myself without an explicit example. Remark: The @ notification system in comments works only if you spell the name correctly. (I'm not upset by misspelling my name. It just doesn't work, that's all. And it is possible to misspell it into something rude in Finnish...) \u2013\u00a0Joonas Ilmavirta Oct 21 '14 at 10:35\n\u2022 @JoonasIlmavirta Apologies for the misspelling, it is due to using a brand new mobile whose automatic correction truly is a pain in the neck (And I'm Italian). As to the questions... are those on MSE not a good example? I was planning to post exactly them to MO, in a week. \u2013\u00a0Vincenzo Oliva Oct 21 '14 at 11:29\n\u2022 Ah, yes, your MSE questions are a good example. Personally, I would probably not vote in either direction. (Since I'm not strongly against such questions.) The questions are not bad, but they seem disconnected from any research problem and I see no reason to believe that anyone could answer them. But I'm not the greatest expert on series, so don't rely on my word too much. Someone else might consider them off-topic. Remember: If you ask, make it clear what you want to know about the series and what would be a useful answer. \u2013\u00a0Joonas Ilmavirta Oct 21 '14 at 13:12\n\u2022 There would be less chance of having the question closed if you explain better what the real purpose or motivation of the question is. And it may be that the thing you are really after could be approached in a way different from the need to compute the exact value of certain series which might well be impossible to evaluate. \u2013\u00a0Todd Trimble Oct 22 '14 at 13:23\n\u2022 Why are you guys all just solving the question in comments on meta.MO. Shouldn't you just go answer it somewhere? \u2013\u00a0djechlin Oct 24 '14 at 9:15","date":"2019-05-25 06:23:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5386533737182617, \"perplexity\": 648.906796738226}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-22\/segments\/1558232257889.72\/warc\/CC-MAIN-20190525044705-20190525070705-00546.warc.gz\"}"}
| null | null |
class PrefService;
namespace content {
class BrowserContext;
class WebContents;
}
namespace user_prefs {
class PrefRegistrySyncable;
}
// A shim that lives on top of a BookmarkModel that allows the injection of
// partner bookmarks without submitting changes to the bookmark model.
// The shim persists bookmark renames/deletions in a user profile and could be
// queried via shim->GetTitle(node) and shim->IsReachable(node).
// Note that node->GetTitle() returns an original (unmodified) title.
class PartnerBookmarksShim : public base::SupportsUserData::Data {
public:
// Returns an instance of the shim for a given |browser_context|.
static PartnerBookmarksShim* BuildForBrowserContext(
content::BrowserContext* browser_context);
// Registers preferences.
static void RegisterProfilePrefs(user_prefs::PrefRegistrySyncable* registry);
// Disables the editing and stops any edits from being applied.
// The user will start to see the original (unedited) partner bookmarks.
// Edits are stored in the user profile, so once the editing is enabled
// ("not disabled") the user would see the edited partner bookmarks.
// This method should be called as early as possible: it does NOT send any
// notifications to already existing shims.
static void DisablePartnerBookmarksEditing();
// Returns true if everything got loaded.
bool IsLoaded() const;
// Returns true if there are partner bookmarks.
bool HasPartnerBookmarks() const;
// Returns true if a given bookmark is reachable (i.e. neither the bookmark,
// nor any of its parents were "removed").
bool IsReachable(const bookmarks::BookmarkNode* node) const;
// Returns true if a given node is editable and if editing is allowed.
bool IsEditable(const bookmarks::BookmarkNode* node) const;
// Removes a given bookmark.
// Makes the |node| (and, consequently, all its children) unreachable.
void RemoveBookmark(const bookmarks::BookmarkNode* node);
// Renames a given bookmark.
void RenameBookmark(const bookmarks::BookmarkNode* node,
const base::string16& title);
// For Loaded/Changed/ShimBeingDeleted notifications
class Observer {
public:
// Called when the set of bookmarks, or their values/visibility changes
virtual void PartnerShimChanged(PartnerBookmarksShim*) {}
// Called when everything is loaded
virtual void PartnerShimLoaded(PartnerBookmarksShim*) {}
// Called just before everything got destroyed
virtual void ShimBeingDeleted(PartnerBookmarksShim*) {}
protected:
virtual ~Observer() {}
};
void AddObserver(Observer* observer);
void RemoveObserver(Observer* observer);
// PartnerBookmarksShim versions of BookmarkModel/BookmarkNode methods
const bookmarks::BookmarkNode* GetNodeByID(int64 id) const;
base::string16 GetTitle(const bookmarks::BookmarkNode* node) const;
bool IsPartnerBookmark(const bookmarks::BookmarkNode* node) const;
const bookmarks::BookmarkNode* GetPartnerBookmarksRoot() const;
// Sets the root node of the partner bookmarks and notifies any observers that
// the shim has now been loaded. Takes ownership of |root_node|.
void SetPartnerBookmarksRoot(bookmarks::BookmarkNode* root_node);
// Used as a "unique" identifier of the partner bookmark node for the purposes
// of node deletion and title editing. Two bookmarks with the same URLs and
// titles are considered indistinguishable.
class NodeRenamingMapKey {
public:
NodeRenamingMapKey(const GURL& url, const base::string16& provider_title);
~NodeRenamingMapKey();
const GURL& url() const { return url_; }
const base::string16& provider_title() const { return provider_title_; }
friend bool operator<(const NodeRenamingMapKey& a,
const NodeRenamingMapKey& b);
private:
GURL url_;
base::string16 provider_title_;
};
typedef std::map<NodeRenamingMapKey, base::string16> NodeRenamingMap;
// For testing: clears an instance of the shim in a given |browser_context|.
static void ClearInBrowserContextForTesting(
content::BrowserContext* browser_context);
// For testing: clears partner bookmark model data.
static void ClearPartnerModelForTesting();
// For testing: re-enables partner bookmarks editing.
static void EnablePartnerBookmarksEditing();
private:
explicit PartnerBookmarksShim(PrefService* prefs);
~PartnerBookmarksShim() override;
const bookmarks::BookmarkNode* GetNodeByID(
const bookmarks::BookmarkNode* parent,
int64 id) const;
void ReloadNodeMapping();
void SaveNodeMapping();
scoped_ptr<bookmarks::BookmarkNode> partner_bookmarks_root_;
PrefService* prefs_;
NodeRenamingMap node_rename_remove_map_;
// The observers.
ObserverList<Observer> observers_;
DISALLOW_COPY_AND_ASSIGN(PartnerBookmarksShim);
};
#endif // CHROME_BROWSER_ANDROID_BOOKMARKS_PARTNER_BOOKMARKS_SHIM_H_
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,034
|
Year : 2016 | Volume : 10 | Issue : 2 | Page : 50-54
Validity of Beck's depression inventory and alcohol use disorders identification test in Nigeria's Niger Delta region
DC Chukwujekwu1, CU Okeafor1, PO Onifade2
1 Department of Mental Health, University of Port Harcourt, Rivers State, Nigeria
2 Department of Drug Admission and Treatment, Neuropsychiatric Hospital Aro, Abeokuta, Ogun State, Nigeria
D C Chukwujekwu
Department of Mental Health, University of Port Harcourt, P.M.B 5323, Port Harcourt, Rivers State
Background: The Beck's Depression Inventory (BDI) and Alcohol Use Disorder Identification Test (AUDIT) have been validated for use in the study of alcohol related psychiatric disorders in the developed world as well as in Western Nigeria, but not in the Niger Delta Region. Aim: To ascertain the psychometric properties of BDI and AUDIT for use in this part of the world using psychiatric out-patients at the University of Port Harcourt Teaching Hospital. Methods: Four hundred and seventy (470) subjects were enlisted into the study using systematic sampling technique. The BDI and AUDIT were administered to each of them. One hundred and eighty five (185) subjects met the criteria for the second stage viz; a score of 18 and above on the BDI and/or a score of 5 and above on the AUDIT. Diagnoses of Depression and Alcohol Use Disorder were made using the Composite International Diagnostic Interview (CIDI). The data were analyzed using the statistical package for social sciences (SPSS) version 16.0 Results: The sensitivity and specificity of the BDI were 96.3% and 58.8% respectively. The positive and negative predictive values of BDI were 86% and 85.7% respectively. Also, the sensitivity and specificity of the AUDIT were 100% and 92.1%. Furthermore, the positive and negative predictive values of the AUDIT were 85.5% and 100% respectively. Conclusion: The BDI and AUDIT have excellent psychometric properties; hence they are valid for carrying out studies on alcohol related psychiatric disorders.
Chukwujekwu D C
Okeafor C U
Onifade P O
psychometric
© Port Harcourt Medical Journal | Published by Wolters Kluwer - Medknow
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,798
|
No part of this publication may be reproduced in whole or in part, or stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without written permission of the publisher. For information regarding permission, write to Chronicle Books, 680 Second Street, San Francisco, California 94107.
Copyright © 1998 Lucasfilm Ltd. & ®
All rights reserved. Used under authorization
Published by Chronicle Books.
www.starwars.com
Kit Kat® is a registered trademark of the H.B. Reese Candy Company.
Jell-O® is a registered trademark of Kraft Foods, Inc.
Heath® Milk Chocolate English Toffee Bar is a registered trademark of the Hershey Foods Corporation.
Book design by Daniel Carter/STUDIO 212.
Typeset in Bell Gothic, Clicker and Univers.
ISBN: 978-0-8118-2184-6 (hc)
ISBN: 978-1-4521-4806-9 (epub, mobi)
Chronicle Books
680 Second Street,
San Francisco, California 94107
www.chroniclebooks.com
# Table of Contents
INTRODUCTION
Getting Started
Be Careful
Be Aware
Equipment
BREAKFASTS
Princess Leia Danish Dos
C-3PO Pancakes
Oola-la French Toast
Twin Sun Toast
Mos Eisley Morsels
BEVERAGES
Hoth Chocolate
Skywalker Smoothies
Jawa Jive Milkshakes
Yoda Soda
SNACKS AND SIDES
Dark Side Salsa
Ewok Eats
The Force Fruit Fun
Tusken Raider Taters
Jabba Jiggle
Jedi Juice Pops
MAIN COURSES
Greedo's Burritos
Han-burgers
Obi-Wan Kebabs
Galaxy Grilled Cheese
Boba Fett-uccine
Crazy Cantina Chili
TIE Fighter Ties
DESSERTS
Darth Vader Dark Chocolate Sundaes
Wookiee Cookies
Bossk Brownies
Death Star Popcorn Balls
Wampa Snow Cones
R2-D2 Treats
Sandtrooper Sandies
Index
# Introduction
Consider, young Jedi: Why bake a plain old cookie when you can bake a super-Chewie Wookiee Cookie?
The Force inhabits all realms, including the kitchen. Its power is as present in the refrigerator as it is on the desolate ice planet Hoth. _The Star Wars Cookbook_ is your guide to cooking with the Force.
Some of these recipes are easy enough to make on your own and share with friends. For others, you will need the help of a grown-up. Definitely get your parents
or other adults involved. Many moms and dads are masters of the Force who are lying low for mysterious reasons. Seek the wisdom of their cooking experience.
For Rebels in a time crunch, there are plenty of ideas here for whipping up good food in a hurry. In many cases, twenty minutes—about the time it takes to vanquish an ice creature or rid your ship of cable-chewing mynocks—is all you need to put a wholesome meal on the table.
As you journey into the dangers of the kitchen, you will want the wisdom of _The Star Wars_ Cookbook to help you find your way. Adventure awaits as you develop the skills you need to bring the life force of food to all who hunger for it. Accept the challenge, young Rebel, and feel the Force!
## GETTING STARTED
Before you start cooking, you must master some essential safety steps. The kitchen is a realm of peace, yet danger lurks in the most ordinary-seeming places. The two most important rules to remember are:
1. Keep an adult in the kitchen at all times, especially when you use knives, the stovetop, or the oven. Adults make good company and are helpful and handy to have around. (Even Luke would have been toast without Ben Kenobi to guide and protect him.) They can reach high places, drive, use the phone, pay, and offer valuable advice. Remember, never use anything sharp or hot without an adult to guide you.
2. Wash your hands with soap and warm water before cooking. You remember the hideous creatures in the Mos Eisley Cantina? They are nothing compared to what are crawling around on your hands. Fight those microscopic life forms with your best weapons: soap and water. It's a good idea to wash your hands a few times while you're cooking, too, as the germ troops are known to send in constant reinforcements.
The calm and perceptive mind of a Jedi warrior will enable you to prevent most mishaps in the kitchen. Use it well and follow these general guidelines:
## BE CAREFUL
Respect the mysteries of The Force
Never run in the kitchen.
Keep everything—pot holders, towels, packages of ingredients, this book—away from burners on the stove. The stove can be hot even if the burners are all turned off.
Dry your hands before turning on any electric switch or putting in or pulling out a plug.
Wash knives and other sharp utensils one at a time. Don't drop them in a pan or bucket of soapy water—you may cut yourself when you reach in to fish them out.
Lift lids on hot pots at an angle away from you, directing the rising steam away from your face.
Use only dry pot holders. Wet ones will give you a steam burn when you touch the handle of a hot pot.
Put a pot or pan on the stove before you turn on the heat.
Turn off the heat before you remove a pot or pan from the stove.
Never put out a grease fire with water. Water causes grease to splatter and can spread the fire very quickly. To put out a grease fire, smother it with a tight fitting lid or throw handfuls of baking soda onto it.
## BE AWARE
Cultivate the awareness of a Jedi
Never leave the kitchen while something is cooking on the stove or in the oven.
Keep pot handles away from the edge of the stove so no one passing by topples the pot.
Always position pot handles away from other stove burners. They'll get hot and burn you when you go to move the pot.
Remove utensils from hot pots when you're not using them, placing them on a plate or spoon holder near the stove. Metal spoons and spatulas are especially dangerous, because they'll absorb and hold the heat and burn your hand when you go to use them.
Start with a clean kitchen and keep it clean as you cook. When something spills, wipe it up immediately to keep accidents from happening. If you have time, wash dishes as you go.
Turn off the blender's motor before removing the lid.
Put ingredients away when you're finished with them.
Know where to find the fire extinguisher and be sure it's in working order.
Keep the fire department number next to the phone.
The tools of a Wookiee Cookie chef are powerful but simple. You probably already have everything in your kitchen. Here's an alphabetical list of what you may need.
## EQUIPMENT
Aluminum foil
Baking dishes
Baking sheets
Blender*
Can opener
Candy thermometer*
Cheese grater
Colander
Cooling rack
Cutting board
Electric mixer*
Ice cream scoop
Ice-cube tray
Knives* (one large and one small)
Measuring cups and spoons
Mixing bowls of various sizes
Paper cups
Pastry brush
Plastic wrap
Popsicle Sticks
Potato masher
Pot holders
Rolling pin
Rubber and metal spatulas
Saucepans with lids
Sieve
Sifter
Skewers*
Skillet
Star-shaped cookie cutters
Tea kettle
Toaster oven*
Toothpicks
Vegetable peeler
Wax paper
Whisk
Wooden spoons
*Use these items with extreme caution. Definitely get an adult to assist you any time you need to use them.
Go forth, young Jedi! May your Hoth Chocolate be sweet, may your Dark Side Salsa be spicy, and may the Force always be with you!
# Breakfasts
Princess Leia Danish Dos
C-3PO Pancakes
Oola-la French Toast
Twin Sun Toast
Mos Eisley Morsels
# Princess Leia Danish Dos
Makes about 10 rolls.
INGREDIENTS | Butter for greasing baking sheet
---|---
|
All-purpose flour for dusting work surface
1 | tablespoon butter
1 | 10-oz package refrigerator pizza dough
2 | tablespoons granulated sugar
11/2 | teaspoons ground cinnamon
1 | tablespoon milk
1/3 | cup confectioners' sugar, sifted
1/4 | teaspoon vanilla extract
1. Preheat the oven to 350ºF. Lightly grease a baking sheet.
2. Put the 1 tablespoon butter in a small saucepan. Set the pan on the stove and switch on the heat to low. When the butter has melted, turn off the heat.
3. Lightly flour a work surface. Unroll the pizza dough on top of the flour. Using a pastry brush, brush the melted butter over the surface of the dough.
4. Put the granulated sugar and cinnamon in a small bowl. Stir with a small spoon until well mixed. Sprinkle the cinnamon-sugar mixture over the dough, leaving a 1/2-inch border on all sides.
5. Starting at a long side, roll up the dough into a log. Using your fingertips, pinch the seam together to seal.
6. Using a knife, cut the log crosswise into 1-inch-thick slices. Put the slices, cut side up, onto the greased baking sheet.
7. Using pot holders, put the baking sheet in the preheated oven. Bake until golden brown, about 20 minutes. Carefully transfer the baking sheet to the cooling rack. Cool 5 minutes. With a spatula, transfer rolls to the cooling rack and cool five minutes more.
8. Put the milk and confectioners' sugar in a small bowl. Stir with the spoon until a smooth frosting forms. Stir in the vanilla. Using a butter knife, spread frosting over the tops of the cinnamon rolls.
# C-3PO Pancakes
Makes 8 pancakes.
INGREDIENTS
5 | tablespoons unsalted butter
---|---
2 | cans (81/4-ounces each) pineapple rings
11/4 | cups all-purpose flour
2 | tablespoons brown sugar
2 | teaspoons baking powder
1/2 | teaspoon salt
2 | large eggs
1 | cup milk
|
Butter and maple syrup for serving
1. Preheat the oven to 250ºF.
2. Put the butter in a small saucepan and switch on the heat to low. When the butter has melted, turn off the heat. Let the butter cool slightly (5 minutes).
3. Meanwhile, open the cans of pineapple. Drain off the juice from the cans into the sink. Set aside.
4. Put the flour, brown sugar, baking powder, and salt in a large bowl. Stir with a wooden spoon until well mixed.
5. Break the eggs into a medium bowl. Add the milk and 4 tablespoons of the melted butter. Whisk until well mixed. Slowly whisk in the flour mixture until well blended.
6. With a pastry brush, lightly brush some of the remaining melted butter in a large skillet. Set the skillet on the stove and switch on the heat to medium-high.
7. Fill a 1/3-cup measuring cup with batter. When the skillet is hot, after 1 minute or so, pour in the batter. Cook the pancake until golden underneath and bubbles burst on the top, about 3 minutes.
8. Place 1 pineapple ring in the center of the pancakes. Using a spatula, carefully flip the pancake. Cook until golden brown on the second side, about 2 minutes longer. Transfer the pancake to a baking sheet. Using pot holders, put the sheet in the oven.
9. Repeat steps 7 and 8, brushing skillet with butter as needed. Serve the pancakes warm with butter and maple syrup.
# Oola-la French Toast
Makes 1 or 2 servings.
INGREDIENTS
1 | egg
---|---
1/4 | cup milk
11/2 | teaspoons sugar
1/8 | teaspoon vanilla extract
1/4 | teaspoon cinnamon, or a pinch of ground cloves
3 | bread slices
3 | teaspoons butter
|
Confectioners' sugar (optional)
|
Butter and maple syrup for serving
1. Preheat the oven to 250ºF.
2. Break the egg into a bowl. Add the milk, sugar, vanilla, and cinnamon or cloves. Beat with a whisk until blended. Pour into a shallow dish such as a pie pan.
3. Put bread slices into the egg mixture. Let sit for a few moments. With your fingers, turn the bread over and again let sit for a few moments. Both sides should look soaked with the egg mixture.
4. Put 1 teaspoon butter in a skillet. Put the skillet on the stove and switch on the heat to medium-high.
5. When the butter melts, pick up one piece of soaked bread and put it into the pan. Be careful not to burn your fingers. Fry until golden brown on the underside, about 3 minutes. To check, lift a corner with a spatula, and take a peek. If ready, use the spatula to turn it over. Now fry until golden brown on the second side, about 2 minutes longer.
6. With the spatula, move the toast to a baking sheet. Using pot holders, put the sheet in the oven.
7. Repeat steps 4, 5, and 6 to cook the remaining French toast.
8. Remove the sheet from the oven. (Don't forget to use the pot holders!) If desired, sprinkle the French toast with confectioners' sugar. Serve hot with butter and maple syrup.
# Twin Sun Toast
Tattooine is a hot desert planet located in the outer reaches of the galaxy. It is home to young Luke Skywalker, who would sometimes gaze dreamily at the planet's twin suns.
Makes 1 serving.
INGREDIENTS
1 | wide slice sourdough bread (at least 7 inches wide)
---|---
1 | teaspoon butter
2 | small eggs
|
Salt and pepper to taste
1. Put the bread slice on a cutting board. Using a 2-inch round cookie cutter, cut out 2 holes, side by side, in the bread. Make sure to leave bread between and around holes.
2. Put a skillet on the stove and switch on the heat to medium. Put the butter in the skillet.
3. As the butter melts, spread it evenly on the bottom of the pan. Place the bread slice in the pan and break an egg into each hole. Sprinkle a little salt and pepper on the eggs. Fry until the clear part of the eggs turn white, 1 to 2 minutes.
4. Slip a spatula under the bread and flip it over quickly but carefully. Don't let the eggs slip out of the holes. Cook for 1 minute more. Slide the toast onto a plate. Serve at once.
# Mos Eisley Morsels
Makes 12 morsels.
INGREDIENTS | Butter for greasing baking dish
---|---
2 | cups all purpose flour
2 | teaspoons baking powder
1/2 | teaspoon baking soda
11/2 | teaspoons ground cinnamon
1 | teaspoon ground nutmeg
1/2 | teaspoon ground cloves
1/4 | teaspoon salt
3 | large bananas
1 | large egg
2 | tablespoons vegetable oil
2 | teaspoons vanilla
1. Preheat the oven to 375°F.
2. Lightly grease an 8-inch square baking dish.
3. Put the flour, baking powder, baking soda, cinnamon, nutmeg, cloves, and salt into a sifter. Sift the ingredients into large bowl.
4. In another large bowl, thoroughly mash the bananas with a fork or potato masher. Add the egg, vegetable oil, and vanilla, and stir until well blended.
5. Add the flour mixture and with a rubber spatula, mix the wet and dry ingredients together until just combined.
6. Pour the batter into the prepared baking dish and smooth the top of the batter with the spatula.
7. Using pot holders, place the baking dish in the oven and bake for 30-35 minutes, until a toothpick inserted into the center comes out clean. Using pot holders, transfer the dish to a cooling rack. Cut into squares.
Variations: The morsels in the picture are topped with an extra mashed banana. If desired, mash a banana in a separate bowl, and spread on squares just before eating. Do not top the morsels with mashed banana unless you're going to eat them right away!
# Beverages
Hoth Chocolate
Skywalker Smoothies
Jawa Jive Milkshakes
Yoda Soda
# Hoth Chocolate
The Rebellion's hidden Echo Base on the ice planet Hoth was freezing! Sometimes the Rebels wished they could just warm up with a mug of this Hoth chocolatey drink.
Makes 1 serving.
INGREDIENTS
1 | cup milk
---|---
2 | heaping teaspoons sugar
1 | heaping teaspoon unsweetened cocoa powder
1/8 | teaspoon vanilla extract
|
Small marshmallows (optional)
1. Pour the milk into a small saucepan. Add the sugar, cocoa powder, and vanilla to the milk. Stir vigorously with a whisk until the sugar and cocoa dissolve.
2. Place the pan on the stove and switch on the heat to medium. Watch for tiny bubbles to appear along the edge of the pan, then immediately remove the pan from the heat.
3. Carefully pour into the mug and serve immediately with marshmallows, if desired.
# Skywalker Smoothies
Luke definitely has the Force on his side, but sometimes he gets an extra boost from these scrumptious smoothies.
Makes 2 servings.
INGREDIENTS
1 | cup fresh or frozen strawberries
---|---
1 | banana
1/2 | cup pineapple, grape, or orange juice
4 | ice cubes
1. If you are using fresh strawberries, cut off their stems with a knife. Put the fresh or frozen berries in a blender.
2. Peel the banana. Break it into pieces and add them to the blender. Then add the fruit juice and ice cubes.
3. Put the lid on the blender. Make sure it fits tightly. Turn on the blender first at low speed, then increase to high speed. Blend until smooth and frothy, 1 to 2 minutes. Turn off the blender and wait until it stops. With a wooden spoon, check to see that the fruit is thoroughly blended. If not, repeat this step.
4. Pour into 2 glasses and serve immediately.
Variation: Add a scoop of vanilla or berry frozen yogurt to the blender along with the other ingredients.
# Jawa Jive Milkshakes
Jawas are famous for scavenging abandoned ships, droids, and scrap metal. When they get together at giant swap meets, they allegedly serve these delicious shakes.
Makes 2 milkshakes.
CHOCOLATE-BANANA
2 | cups vanilla ice cream or frozen yogurt
---|---
1 | banana, peeled and broken into pieces
1/2 | cup milk
1/4 | cup chocolate syrup
1/2 | cup crushed Heath® Bars (optional)
SUPER STRAWBERRY
2 | cups strawberry ice cream or frozen yogurt
---|---
1 | cup frozen strawberries
1/2 | cup milk
1/2 | cup white chocolate chips (optional)
VANILLA AND PEANUT BUTTER
1 | cups vanilla ice cream or frozen yogurt
---|---
1 | teaspoon vanilla extract
1/2 | cup milk
1/4 | cup creamy peanut butter
1/2 | cup peanut butter chips
DOUBLE CHOCOLATE
2 | cups chocolate ice cream or frozen yogurt
---|---
1/2 | cup milk
1/4 | cup chocolate syrup
1/2 | cup crushed chocolate sandwich cookies (optional)
1. Select the milkshake you want to make. Assemble the ingredients for your recipe. (Tip: The best way to crush the cookies or candy bar is to put them in a clean, sturdy plastic bag and roll a rolling pin back and forth over them.)
2. Put all the ingredients into a blender.
3. Put the lid on the blender. Make sure it fits tightly. Turn on the blender first at low speed, then increase to high speed. Blend until smooth, 1 to 2 minutes.
4. Pour the milkshake into 2 glasses. Serve each shake with a spoon and a straw.
# Yoda Soda
Jedi Master Yoda levitates Luke's X-wing from the Dagobah swamp in The Empire Strikes Back. Here he peacefully levitates a frothy glass of Yoda Soda.
Makes 1 serving.
INGREDIENTS
3 | limes
---|---
3 | tablespoons sugar, or more to taste
1 | cup sparkling water
1 | scoop lime sherbet or sorbet
1. Place 1 lime on the cutting board and cut it in half. Squeeze the juice from each half into a measuring cup. Repeat with the remaining limes until you have 1/4 cup juice.
2. Put the lime juice and 3 tablespoons sugar in a small pitcher. Stir with a wooden spoon until the sugar dissolves. Add the sparkling water and stir until mixed. Taste and add more sugar, if desired.
3. Using an ice cream scoop, scoop up the sherbet and drop it into a tall glass. Pour in the lime water. Serve immediately.
Variation: You can substitute rainbow sherbet or lemon sorbet for the lime sherbet.
# Snacks and Sides
Dark Side Salsa
Ewok Eats
The Force Fruit Fun
Tusken Raider Taters
Jabba Jiggle
Jedi Juice Pops
# Dark Side Salsa
Makes 4 servings.
INGREDIENTS
6 | ripe Roma tomatoes
---|---
1 | small onion
1 | small avocado
1 | can (4 ounces) diced mild green chiles
1 | cup frozen corn, thawed
1 | lemon or lime
|
Salt and pepper to taste
|
Blue corn tortilla chips
1. Put a tomato on the cutting board and cut out the green stem. Cut the tomato in half from the top to the bottom. Holding a tomato half cut side down, slice it lengthwise into thin slices. Now cut the slices into little cubes. Put the tomato cubes in the mixing bowl. Repeat with the remaining tomatoes.
2. Put the onion on the cutting board. Carefully slice off the root end and the stem end. Strip off the dry brown skin. Then cut the onion in half from the top to the bottom. Holding an onion half cut side down, thinly slice it crosswise. Now hold the slices together and cut across them in the opposite direction. Be sure to keep your fingers clear of the knife blade. Add the onion to the tomatoes.
3. Pit the avocado by carefully cutting it in half around the pit. Pull the halves apart and carefully scoop out the pit with a spoon. Peel the skin from the avocado with your fingers. Place the avocado half cut side down on the cutting board. Cut it lengthwise into thin slices, then cut the slices into little cubes. Repeat with the remaining avocado half. Add the avocado to the tomatoes and onions.
4. Open the can of chiles. Drain off the liquid from the can into the sink. Add the chiles to the mixing bowl. Then add the thawed corn.
5. Put the lemon or lime on the cutting board and cut it in half. Hold a half in your hand and squeeze it over a tablespoon until the tablespoon is full of juice. Add the juice to the mixing bowl.
6. Stir together everything in the bowl until well mixed. Add the salt and pepper to taste. Serve with the tortilla chips.
# Ewok Eats
Makes 4 servings.
INGREDIENTS
6 | fresh herb sprigs, a combination of parsley, thyme, and chives
---|---
1/2 | cup plain yogurt
4 | ounces cream cheese, at room temperature
1 | teaspoon Worcestershire sauce
1/4 | teaspoon garlic salt
1 | teaspoon pepper
1 | head broccoli
1. With your fingers, pull the thyme and parsley leaves off the stems. Put the leaves and chives on the cutting board and carefully chop with a knife. Set aside.
2. Put the yogurt and cream cheese in the mixing bowl. Stir with the wooden spoon until smooth. Add the chopped herbs, Worcestershire sauce, salt, and pepper and stir well. Cover with plastic wrap and refrigerate for 1 hour to give the flavors time to blend.
3. Put the head of broccoli on the cutting board. Use the knife to cut off the florets (the flowery looking tops). Be sure to leave a little stem on the florets. If the florets are in big clusters, cut between the stems to make smaller ones.
4. Spread the dip in the bottom of a shallow dish such as a pie pan. Stand the florets upright in the dip, side by side to make an Ewok forest.
Variation: You can also serve the dip with potato chips, tortilla chips, or other vegetables such as carrot and celery sticks.
# The Force Fruit Fun
Makes 1–2 servings.
INGREDIENTS
2 | cups strawberries, bananas, or blueberries, or a combination of the three
---|---
2 | teaspoons lemon juice
3 | tablespoons sugar
1. Preheat oven to 225°F.
2. Wash the fruit. Remove strawberry stems with a knife and peel the bananas, if using.
3. Place fruit in a blender. Put the lid on the blender. Make sure it fits tightly. Turn blender on low speed first, then up to high and puree until smooth. Turn off the blender. Add lemon juice and sugar. Put the lid on again and blend.
4. With a rubber spatula, spread the fruit puree as thinly as possible onto a non-stick baking sheet.
5. Bake in the oven for about 25 minutes.
6. Using a potholder, transfer the baking sheet to a cooling rack. Allow to cool, then peel the fruit off of the sheet and eat!
# Tusken Raider Taters
Tusken Raiders are fierce, mysterious creatures who wander the Tatooine desert, climbing sand dunes that look mysteriously like these potato dunes.
Makes 2 servings.
INGREDIENTS
1 | teaspoon salt
---|---
1 | pound purple or russet potatoes
1/4 | cup milk
2 | tablespoons butter
|
Salt and pepper to taste
1. Fill a large saucepan two-thirds full with water. Add the salt.
2. Using the vegetable peeler, peel the potatoes. Put a peeled potato on the cutting board. Cut the potato in half lengthwise. Holding a potato half cut side down, slice it lengthwise into 1-inch slices. Now hold the slices and cut across them in the opposite direction to make 1-inch pieces. Be sure to keep your fingers clear of the knife blade. Add the potato to the pan. Repeat with the remaining potatoes.
3. Put the pan on the stove and switch on the heat to high. Bring the water to a boil and boil the potatoes, uncovered, until tender, about 10 minutes. (You can check by scooping out a potato cube with a slotted spoon, and poking the cube with a fork. It should go in easily.) Turn off the heat.
4. Put the colander in the sink. Remove the pan from the stove and pour the water and potatoes into the colander. Be very careful; the steam from the boiling water can burn you.
5. Return the potatoes to the pan. Mash with a potato masher. Add the milk and butter. Continue to mash until the butter and milk are mixed in. Potatoes can be as smooth or lumpy as you like. Season to taste with salt and pepper.
Variation: If you want brown potatoes like the ones in the picture, mash russet potatoes with 2 tablespoons of soy sauce.
# Jabba Jiggle
This green, fruity treat wiggles and jiggles just like Jabba.
Makes 8 servings.
INGREDIENTS
1 | can (11-ounces) mandarin orange segments
---|---
1 | can (8-ounces) crushed pineapple in juice
1 | package (6-ounces) lime Jell-O®
2 | cups cold water
1 | cup seedless grapes
1. Open the cans of mandarin oranges and crushed pineapple. Drain off the juice from the cans into the sink.
2. Fill a tea kettle with water. Put the kettle on the stove and switch on the heat to high. Bring the water to a boil and remove from the heat.
3. Open the Jell-O® package and empty it into a large mixing bowl. Carefully measure 2 cups boiling water and pour them over the Jell-O®. Stir with the wooden spoon until the Jell-O® dissolves completely. Then stir in the cold water.
4. Add the drained mandarin oranges, pineapple, and the grapes. Stir until well mixed.
5. Pour the mixture into a glass bowl. Refrigerate until firm, about 2 hours.
# Jedi Juice Pops
Makes 14 pops.
INGREDIENTS
14 | small fresh or frozen strawberries
---|---
11/4 | cups fruit juice, such as orange, cranberry, apple, fruit punch, or lemonade
1. If you are using fresh strawberries, cut off their stems with a small knife. Place 1 strawberry in each ice-cube compartment of an ice-cube tray. Fill the ice-cube tray with the juice.
2. Put the tray in the freezer and freeze until almost firm, about 2 hours.
3. Push a popsicle stick (or a toothpick) into the center of each cube, through the strawberry. Return to the freezer and freeze until firm, about 1 hour longer.
4. Pop out the juice bars and eat!
Variation: You can use banana slices, seedless grapes, or orange segments cut into chunks instead of strawberries.
# Main Courses
Greedo's Burritos
Han-burgers
Obi-Wan Kebabs
Galaxy Grilled Cheese
Boba Fett-uccine
Crazy Cantina Chili
TIE Fighter Ties
# Greedo's Burritos
Makes 4 burritos.
INGREDIENTS
1 | can (14-ounces) black beans
---|---
1 | cup grated Monterey Jack cheese
1/2 | head iceberg lettuce
1 | tomato
1 | tablespoon olive oil or vegetable oil
11/2 | pounds lean ground beef
4 | flour tortillas, each about 10 inches in diameter
1/2 | cup sliced black olives
1. Get an adult to help you with this recipe!
2. Preheat oven to 350°F.
3. Open the can of beans and drain the liquid into the sink. Set aside.
4. Put the iceberg lettuce on the cutting board cut side down. Cut the lettuce from top to bottom in narrow slices. Now cut across the slices, again in narrow strips. Set aside.
5. Put the tomato on the cutting board. Cut out the green stem. Then cut the tomato in half from the top to the bottom. Holding a tomato half cut side down, slice it crosswise about 1/2-inch thick. Now cut the slices into little cubes. Set aside.
6. Put the oil in a skillet. Put the skillet on the stove and switch on the heat to medium high. When the oil is hot, carefully add the beef. Cook, stirring often, until browned, about 15 minutes.
7. Wrap the tortillas in aluminum foil and place in the oven to heat, about 5 to 10 minutes.
8. Meanwhile, add the beans to the beef and continue to cook, stirring often, until the beans are heated through, about 2 minutes. Turn off the heat.
9. Using pot holders, remove tortillas from the oven. Place a tortilla on a plate. Spoon one-fourth of the beef mixture onto the tortilla. Sprinkle with one-fourth of the cheese. Top with one-fourth each of the olives, the tomato, and the lettuce. Roll up the tortilla around the filling.
10. Repeat step 9 until all the ingredients are used up.
# Han-Burgers
Han Solo is generally right-handed, but is known to fire a ketchup blaster with his left hand. (Condiments always taste better when blasted onto one's food.)
Makes 4 Han-burgers.
INGREDIENTS
1 | pound ground lean beef
---|---
1 | teaspoon salt
1/2 | teaspoon pepper
1 | teaspoon Worcestershire sauce
4 | tomato slices
4 | whole-wheat hamburger buns
|
Lettuce leaves, pickle slices, ketchup, and mustard for serving
1. Put the ground beef, salt, pepper and Worcestershire sauce in a bowl. Using your hands, mix them together until well blended.
2. Divide the meat mixture in 8 equal portions. Form the portions into 8 patties 1/4-inch-thick.
3. Place 1 tomato slice on 1 patty. Top with another patty. Use your fingers to press the edges together to seal. Repeat this step to make 4 filled patties.
4. Put a large skillet on the stove and switch on the heat to medium. When the skillet is hot, after about 1 minute or so, carefully put the hamburgers in the pan. Cook until browned on the underside, about 6 minutes. To check, lift a corner of the burger with the spatula and peak underneath. Flip over the hamburger and cook on the second side until browned and cooked through, about 6 minutes longer.
5. Use the spatula to transfer the burger onto the bun. Serve with lettuce leaves, pickle slices, ketchup, and mustard—blasted on, of course!
# Obi-Wan Kebabs
Makes 8 kebabs.
INGREDIENTS
2 | chicken breasts
---|---
4 | small red new potatoes
3 | zucchinis
3 | ears of corn
|
Olive oil
1. Get an adult to help you with this recipe!
2. Soak 8 wooden skewers in water for 30 minutes.
3. Preheat the oven to 375°F.
4. Put a chicken breast on a cutting board. Using a knife, cut the meat in 1-inch strips. Now cut across the strips to make 1-inch cubes. Repeat with remaining chicken breast. Set chicken aside.
5. Wash the cutting board and knife thoroughly with soap and warm water. Rinse and dry the cutting board.
6. Put a small potato on the cutting board and cut it in half. Repeat with remaining potatoes and set aside.
7. Put a zucchini on the cutting board. Trim off the stem and the base. Then cut the zucchini into 1-inch rounds. Repeat with the remaining zucchini and set aside.
8. Put an ear of corn on the cutting board. Cut the corn into 1-inch rounds. Repeat with remaining corn.
9. Very carefully push one round of corn onto the skewer. Follow with a potato half, chicken cube, and zucchini slice. Begin again with the corn and continue until there are only 2 inches left on the skewer.
10. Using a pastry brush, brush the meat and vegetables with olive oil and sprinkle with salt and pepper.
11. Grill or broil until vegetables are tender and chicken is cooked through, turning frequently, about 30 minutes.
# Galaxy Grilled Cheese
Makes 2 sandwiches.
INGREDIENTS
4 | slices whole-wheat bread
---|---
4 | teaspoons butter, at room temperature
8 | thin slices Cheddar cheese
8 | pickle slices
1. Lay 1 slice of bread on a work surface. Spread one side with 1 teaspoon of the butter. Repeat with the remaining bread slices and butter.
2. Turn 2 bread slices over so the buttered sides are down. Top the slices with the cheese slices, dividing them evenly.
3. Lay 4 pickle slices on top of each stack of cheese. Top with the other 2 bread slices, buttered sides up.
4. Place the sandwiches in a skillet. Set the skillet on the stove and switch on the heat to medium. Cook the sandwiches until golden brown on the underside and the cheese begins to melt, about
3 minutes. To check, lift up a corner of a sandwich with a spatula and peek underneath. Using the spatula, flip over both sandwiches. Cook until golden brown on the second side and the cheese has melted, about 3 minutes longer.
5. Use the spatula to transfer the sandwiches to a cutting board. You can cut your sandwiches into stars or planets if you have a star-shaped or round cookie cutter. Press the cookie cutter straight down into the sandwich. Be careful not to burn your fingers on the hot cheese. Lift the cutter gently. Repeat with the other sandwich. Serve the sandwiches immediately.
# Boba Fett-uccine
Makes 2 to 3 servings.
INGREDIENTS
1 | small head broccoli
---|---
1 | small head cauliflower
1 | small zucchini
3 | teaspoons salt
11/2 | cups bottled pasta sauce
3 | quarts water
1/2 | pound fettuccine noodles
|
Grated Parmesan cheese
1. Put the broccoli on a cutting board. Using a knife, cut off the florets (the flowery-looking tops). Measure one cup of florets. Repeat this step with the cauliflower.
3. Put the zucchini on the cutting board. Trim off the stem and the base. Then cut the zucchini into thin rounds.
4. Fill a medium saucepan three-fourths full with water. Add 1 1/2 teaspoons of the salt. Put the pan on the stove and switch on the heat to high. When the water boils, slowly add broccoli, cauliflower, and zucchini. Cook until tender, about 4 minutes.
5. Turn off the heat. Place colander in the sink. Remove the pan from the stove and pour the water and vegetables into the colander. Be very careful; steam can burn you.
6. Return the vegetables to the saucepan. Pour in the pasta sauce. Set the pan on the stove and switch on the heat to medium. Heat, stirring occasionally with a wooden spoon, until hot. Turn down the heat to very low.
7. Now pour the 3 quarts water into a large saucepan. Add the remaining 11/2 teaspoons salt. Put the pan on the stove and switch on the heat to high. When the water boils, slowly add the fettuccine and stir once or twice with a wooden spoon. Boil until tender, about 8 minutes or according to package directions.
8. Turn off the heat. Remove the pan from the stove and carefully pour the water and pasta into the colander.
9. Transfer the noodles to a large serving bowl. Pour the sauce over the top. Sprinkle with Parmesan cheese and serve.
# Crazy Cantina Chili
Makes 4 to 6 servings.
INGREDIENTS
1 | can (16-ounces) kidney beans
---|---
1 | can (16-ounces) black beans
1 | can (16-ounces) garbanzo beans
1 | onion
2 | tablespoons vegetable oil
2 | tablespoons chili powder
1/8 | teaspoon cayenne pepper
1 | can (28-ounces) crushed tomatoes with juice
1 | cup tomato juice
|
Salt and pepper to taste
|
Shredded Cheddar cheese
|
Sour cream or plain yogurt
1. Open the cans of beans. Drain off the liquid from the cans into the sink. Set the beans aside.
2. Put the onion on a cutting board. Carefully slice off the root end and the stem end. Use your fingers to strip off the dry skin. Then cut the onion in half from the top to the bottom. Hold an onion half cut side down and thinly slice it crosswise. Now hold the slices together and cut across them in the opposite direction. Be sure to keep your fingers clear of the knife blade. Set aside.
3. Put the oil in a large saucepan. Set the pan on the stove and switch on the heat to medium-high. When the oil is hot, add the chopped onion and stir with the wooden spoon until tender, about 5 minutes.
4. Add the chili powder and cayenne pepper and stir for 30 seconds.
5. Add the beans, the crushed tomatoes, and the tomato juice. Stir well. Reduce heat to medium low and simmer for 15 minutes, stirring occasionally. Season to taste with salt and pepper.
6. Serve the chili with the cheese and sour cream on the side.
# TIE Fighter Ties
These galactical finger-foods are sure to start a battle in your very own kitchen: a battle over who takes command of the fresh-from-the-oven fleet of TIE Fighter Ties!
Makes 8 TIE Fighter Ties.
INGREDIENTS
4 | pre-cooked sausages or hotdogs, approximately 5-inches long
---|---
1 | package refrigerator breadsticks (8 breadsticks)
|
Ketchup and mustard
1. Preheat the oven to 350°F.
2. Cut sausages in half crosswise. Set aside.
3. Open the package of breadsticks and separate the lengths of dough.
4. Cut the lengths in half and set aside. You should have 16 lengths of dough when you are finished.
5. Place one sausage half, cut-side down, on a baking sheet.
6. Take 1 length of dough and wrap it around the base of the sausage half. Cross the ends and let them fall on the baking sheet in the form of the letter V. Using another length of dough, wrap the same sausage in the opposite direction. Cross the ends and let them fall in the form of an upside-down V. Repeat with remaining dough and sausage halves.
7. Bake according to breadstick package directions, or until dough puffs up and turns golden brown.
8. Using pot holders, remove from the oven. Serve with ketchup and mustard.
# Desserts
Darth Vader Dark Chocolate Sundaes
Wookiee Cookies
Bossk Brownies
Death Star Popcorn Balls
Wampa Snow Cones
R2-D2 Treats Sandtrooper Sandies
# Darth Vader Dark Chocolate Sundaes
Some speculate that Darth Vader was lured to the Dark Side by these Dark Chocolate Sundaes.
Makes 4 servings.
INGREDIENTS
1/2 | cup bottled hot fudge sauce
---|---
1 | quart chocolate ice cream
1/2 | cup whipped cream
4 | tablespoons chopped nuts
4 | teaspoons chocolate chips or chocolate sprinkles
1. Put the hot fudge sauce in a small, heavy saucepan. Put the pan on the stove and switch on the heat to low. Stir constantly until the sauce is melted and smooth. Remove from the heat.
2. Set out 4 bowls. Put 2 scoops of ice cream into each bowl.
3. Top each serving with about 2 tablespoons hot fudge. Top with the whipped cream and sprinkle each serving with 1 tablespoon nuts and 1 teaspoon chocolate chips or sprinkles.
Variations: You can use any flavor ice cream or sauce that you like. Other toppings besides peanuts might include M&Ms, mini marshmallows, fresh berries, or shredded coconut.
# Wookiee Cookies
Makes about 3 dozen cookies.
INGREDIENTS
21/4 | cups all-purpose flour
---|---
1 | teaspoon baking soda
1 | teaspoon salt
1 | teaspoon ground cinnamon
1 | cup unsalted butter, at room temperature
1 | cup packed brown sugar
1/2 | cup granulated sugar
2 | large eggs
1 1/2 | teaspoons vanilla extract
1 | cup milk chocolate chips
1 | cup semi-sweet chocolate chips
1. Preheat the oven to 375°F.
2. Put the flour, baking soda, salt, and cinnamon in a mixing bowl. Stir with the wooden spoon until well mixed. Set aside.
3. Put the butter, brown sugar, and granulated sugar in another mixing bowl. Using the electric mixer set on high speed, beat together until well blended and creamy, about 3 minutes. (You can do this with a wooden spoon, but it will take longer.) Beat in the eggs and vanilla extract. Add the flour mixture and stir with the wooden spoon until blended. Stir in the chocolate chips.
4. Scoop up a rounded tablespoonful of the dough and drop onto a baking sheet. Repeat until you have used up all the dough. Be sure to leave about 1 inch between the cookies because they spread as they bake.
5. Using pot holders, put the baking sheets in the oven. Bake until golden brown, about 10 minutes.
6. Again, using pot holders, remove the baking sheets from the oven. Lift the cookies from the baking sheets with a spatula, and place on cooling racks. Let cool completely.
# Bossk Brownies
Bossk the bounty hunter never caught his quarries, Han Solo and Chewbacca. Perhaps he was distracted by these delicious brownies.
Makes about 16 brownies.
INGREDIENTS | Butter for greasing baking dish
---|---
2/3 | cup all-purpose flour
1/2 | cup unsweetened cocoa powder
1/2 | teaspoon baking powder
1/2 | teaspoon salt
1/2 | cup unsalted butter, at room temperature
1/2 | cup packed brown sugar
1/2 | cup granulated sugar
2 | large eggs
1 | teaspoon vanilla extract
1/2 | cup white chocolate or butterscotch chips
1. Preheat the oven to 350°F. Butter an 8-inch square baking dish.
2. Put the flour, cocoa powder, baking powder, and salt in a small bowl. Stir with a wooden spoon until well mixed. Set aside.
3. Put the butter, brown sugar, and granulated sugar in one large bowl. Using the electric mixer set on high speed, beat together until well blended and creamy, about 3 minutes. (You can do this with the wooden spoon, but it will take longer.) Beat in the eggs and vanilla extract. Add the flour mixture and stir with the wooden spoon until blended. Stir in the white chocolate or butterscotch chips.
4. Pour into the prepared baking dish and smooth the top with a rubber spatula.
5. Using pot holders, put the baking dish in the preheated oven. Bake until a toothpick inserted into the center comes out clean, about 25 minutes.
6. Again using pot holders, transfer the dish to the cooling rack. Let cool completely.
# Death Star Popcorn Balls
Like the Empire's deadly Death Star, making popcorn balls can be very dangerous. Do not dare to make them without the help of an adult.
Makes 2 to 3 popcorn balls.
INGREDIENTS
1/3 | cup popcorn kernels
---|---
3 | cups sugar
11/2 | cups water
1/2 | cup light corn syrup
1/2 | tablespoon salt
1 | teaspoon vinegar
1 | teaspoon vanilla extract
|
Butter or vegetable shortening for greasing hands
1. Get an adult to help you with this recipe!
2. Pop the corn using whatever method you prefer.
3. Put the sugar, water, corn syrup, and salt in the saucepan. Stir well with a wooden spoon. Clip a candy thermometer on the side of the pan. Set the pan on the stove and switch on the heat to low. Add the vinegar and vanilla and cook, stirring constantly, until the thermometer reads 270°F.
4. Get an adult to carefully pour the hot sugar mixture over the popcorn and toss with two large spoons to coat every kernel. Allow to cool slightly.
5. Rub butter or vegetable shortening on your hands so the popcorn won't stick to them. Then scoop up enough popcorn to form a ball about the size of a baseball.
6. Shape the ball with your hands.
# Wampa Snow Cones
Luke was imprisoned by a ferocious wampa on the ice planet Hoth, and held captive in the creature's frozen lair. Narrowly escaping with his life, some believe Luke also made off with the wampa's secret snow cone recipe.
Makes 2 servings.
INGREDIENTS
2 | cups fresh or thawed frozen blueberries
---|---
1/4 | cup water
1 | tablespoon sugar
1. Put the blueberries in a bowl. Mash them with a fork until there's lots of liquid.
2. Hold a strainer over a small glass baking dish and pour the mashed berries into a sieve. Press the berries with a fork to push through as much liquid as possible.
3. Add the water and sugar to the blueberry juice. With a wooden spoon, stir until the sugar dissolves. Put the dish in the freezer.
4. After 30 minutes take the dish out of the freezer. Stir the mixture with a fork to break up the crystals. Return the dish to the freezer for another 30 minutes.
5. Remove the dish again and break up the crystals one more time. Return to the freezer and freeze until firm, about 4 hours.
6. Remove the dish from the freezer. Using the fork, scrape the mixture into small crystals. Quickly scoop the crystals into small paper cups and serve right away.
Variation: Replace the blueberries, water, and sugar with a bottled fruit juice or drink such as apple juice, lemonade or punch. Follow the directions for freezing and scraping the crystals.
# R2-D2 Treats
R2-D2 never eats, or so it seems. These frozen treats were discovered in a freezer at the abandoned Rebel base on Hoth. Did they belong to R2-D2? We can only guess. . .
Makes 1 or 2 treats.
INGREDIENTS
1/2 | cup white chocolate chips
---|---
2 | tablespoons chopped peanuts
1 | banana
1 | Kit Kat® candy bar
1. Line a baking sheet with waxed paper.
2. Put the chocolate chips in a small, heavy saucepan. Put the pan on the stove and switch on the heat to low. Stir constantly until the chocolate is melted and smooth. Remove from heat and set aside.
3. Peel the banana. Put on a cutting board and cut into 4 equal pieces.
4. Place the peanuts in a small bowl. Break the Kit Kat® into four bars. Cut each bar in half and set aside.
5. Dip 1 banana piece in the melted chocolate. Dip just the top of the banana piece into the peanuts.
6. Place the banana piece nut side up on the lined baking sheet. Press 2 Kit Kat® pieces along either side of the banana.
7. Repeat steps 5 and 6 with the remaining banana pieces.
8. Place the baking sheet in the freezer until chocolate has hardened, about 15 minutes. Serve straight from the freezer.
# Sandtrooper Sandies
Makes about 3 dozen cookies.
INGREDIENTS
3/4 | cup butter, at room temperature
---|---
11/4 | cups sugar
2 | eggs
1 | teaspoon vanilla extract
2 | cups all-purpose flour
1/4 | teaspoon salt
|
Vegetable oil for greasing the baking sheet
|
Confectioners' sugar
1. Put the butter in a bowl. With an electric mixer set on high speed, beat the butter until soft and light in color.
2. Gradually add the sugar in a slow, steady stream, and beat until creamy and lemon yellow. Then add the eggs one at a time, beating well after each addition. Add the vanilla and stir just until blended.
3. Put the flour and salt into a sifter and sift them into a small bowl. Slowly add the flour mixture to the butter mixture, beating it in with the electric mixer on low speed until it is fully incorporated. The dough will become very stiff and you may have to knead the last bit of flour in by hand. With your hands, pat the dough into a ball and flatten the ball into a thick disk. Wrap the dough in plastic wrap and chill in the refrigerator for 1 hour.
4. Preheat an oven to 400ºF. Lightly oil a baking sheet.
5. Remove dough from the refrigerator and unwrap. Dust your work surface with flour. With a rolling pin, roll out the dough 1/4-inch thick.
6. Using cookie cutters of any shape you like, cut cookies out of the dough. Carefully transfer the cookies to the oiled baking sheet, leaving a little space around each one. Gather up any dough scraps, roll them out again, and cut out more cookies.
7. Slip baking sheet into the oven and bake until cookies just begin to brown, 8 to 10 minutes. Using pot holders, carefully remove the baking sheet from the oven. Transfer the cookies to cooling racks with a spatula and allow them to cool. Sift a little confectioners' sugar over the cookies.
# Index
BEVERAGES
Hoth Chocolate
Jawa Jive Milkshakes
Skywalker Smoothies
Yoda Soda
BREAKFASTS
C-3PO Pancakes
Mos Eisley Morsels
Oola-la French Toast
Princess Leia Danish Dos
Twin Sun Toast
DESSERTS
Bossk Brownies
Darth Vader Dark Chocolate Sundaes
Death Star Popcorn Balls
R2-D2 Treats
Sandtrooper Sandies
Wampa Snow Cones
Wookiee Cookies
MAIN COURSES
Boba Fett-uccine
Crazy Cantina Chili
Galaxy Grilled Cheese
Greedo's Burritos
Han-burgers
Obi-Wan Kebabs
TIE Fighter Ties
SNACKS AND SIDES
Dark Side Salsa
Ewok Eats
The Force Fruit Fun
Jabba Jiggle
Jedi Juice Pops
Tusken Raider Taters
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 1,509
|
Culicoides obnoxius är en tvåvingeart som beskrevs av Fox 1952. Culicoides obnoxius ingår i släktet Culicoides och familjen svidknott.
Artens utbredningsområde är Venezuela. Inga underarter finns listade i Catalogue of Life.
Källor
Svidknott
obnoxius
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,198
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.